苏州大学2005年双控考研真题
2005年普通高等学校招生全国统一考试数学及详细解析(江苏卷)
2005年普通高等学校招生全国统一考试数学（江苏卷）第一卷（选择题共60分）参考公式：三角函数的和差化积公式sin sin 2sin cos sin sin 2cos sin 2222cos cos 2cos cos cos cos 2sin sin 2222αβαβαβαβαβαβαβαβαβαβαβαβ+-+-+=-=+-+-+=-=-若事件A 在一次试验中发生的概率是p ，则它在n 次独立重复试验中恰好发生k 次的概率()(1)k k n k n n P k C p p -=-一组数据12,,,n x x x 的方差2222121()()()n S x x x x x x n ⎡⎤=-+-++-⎣⎦ 其中x 为这组数据的平均数值一、选择题：本大题共12小题，每小题5分，共60分。
在每小题给出的四个选项中，只有一项是符合题意要求的。
（1） 设集合A={1,2}，B={1,2,3}，C={2,3,4}，则()A B C ⋂⋃=（A ）{1,2,3} （B ）{1,2,4} （C ）{2,3,4} （D ）{1,2,3,4}（2） 函数123()x y x R -=+∈的反函数的解析表达式为（A ）22log 3y x =- （B ）23log 2x y -= （C ）23log 2x y -= （D ）22log 3y x =- （3） 在各项都为正数的等比数列{a n }中，首项a 1＝3，前三项和为21，则a 3＋a 4＋a 5＝（A ）33 （B ）72 （C ）84 （D ）189（4） 在正三棱柱ABC-A 1B 1C 1中，若AB=2，AA 1=1则点A 到平面A 1BC 的距离为（A）4 （B）2 （C）4（D（5） △ABC 中，,3,3A BC π==则△ABC 的周长为 （A）)33B π++ （B）)36B π++ （C ）6sin()33B π++ （D ）6sin()36B π++（6） 抛物线y=4x 2上的一点M 到焦点的距离为1，则点M 的纵坐标是（A ）1716 （B ）1516 （C ）78（D ）0 （7） 在一次歌手大奖赛上，七位评委为歌手打出的分数如下：9.4 8.4 9.4 9.9 9.6 9.4 9.7去掉一个最高分和一个最低分后，所剩数据的平均值和方差分别为（A ）9.4, 0.484 （B ）9.4, 0.016 （C ）9.5, 0.04 （D ）9.5, 0.016（8） 设,,αβγ为两两不重合的平面，l ,m ,n 为两两不重合的直线，给出下列四个命题：①若,,αγβγ⊥⊥则α∥β；②若,,m n m αα⊂⊂∥,n β∥,β则α∥β；③若α∥,,l βα⊂则l ∥β；④若,,,l m n l αββγγα⋂=⋂=⋂=∥,γ则m ∥n .其中真命题的个数是（A ）1 （B ）2 （C ）3 （D ）4（9） 设k=1,2,3,4,5,则（x +2）5的展开式中x k 的系数不可能是（A ）10 （B ）40 （C ）50 （D ）80（10） 若1sin(),63πα-=则2cos(2)3πα+= （A ）79- （B ）13- （C ）13 （D ）79 （11） 点P （-3,1）在椭圆22221(0)x y a b a b+=>>的左准线上.过点P 且方向为a =(2,-5)的光线，经直线y=-2反射后通过椭圆的左焦点，则这个椭圆的离心率为（A ）3 （B ）13 (C)2 （D ）12（12） 四棱锥的8条棱代表8种不同的化工产品，有公共点的两条棱代表的化工产品放在同一仓库是危险的，没有公共顶点的两条棱代表的化工产品放在同一仓库是安全的，现打算用编号为①、②、③、④的4个仓库存放这8种化工产品，那么安全存放的不同方法种数为（A ）96 （B ）48 （C ）24 （D ）0参考答案：DACBD CDBCA AB第二卷（非选择题共90分）二、填空题：本大题共6小题，每小题4分，共24分。
2005考研数一真题及解析
2005考研数一真题及解析2005年全国硕士研究生入学统一考试数学(一)试卷一、填空题(本题共6小题,每小题4分,满分24分.把答案填在题中横线上)(1)曲线122+=x x y 的斜渐近线方程为 _____________.(2)微分方程x x y y x ln 2=+'满足91)1(-=y 的解为____________.(3)设函数181261),,(222z y x z y x u +++=,单位向量}1,1,1{31=n ,则)3,2,1(nu∂∂=.________.(4)设Ω是由锥面22y x z +=与半球面222y x R z --=围成的空间区域,∑是Ω的整个边界的外侧,则⎰⎰∑=++zdxdy ydzdx xdydz ____________.(5)设123,,ααα均为3维列向量,记矩阵123(,,)=A ααα,123123123(,24,39)=++++++B ααααααααα,如果1=A ,那么=B .(6)从数1,2,3,4中任取一个数,记为X , 再从X ,,2,1 中任取一个数,记为Y , 则}2{=Y P =____________.二、选择题(本题共8小题,每小题4分,满分32分.每小题给出的四个选项中,只有一项符合题目要求,把所选项前的字母填在题后的括号内)(7)设函数n n n x x f 31lim )(+=∞→,则()f x 在),(+∞-∞内(A)处处可导 (B)恰有一个不可导点(C)恰有两个不可导点 (D)至少有三个不可导点(8)设()F x 是连续函数()f x 的一个原函数,""N M ⇔表示"M 的充分必要条件是",N 则必有(A)()F x 是偶函数()f x ⇔是奇函数 (B)()F x 是奇函数()f x ⇔是偶函数(C)()F x 是周期函数()f x ⇔是周期函数 (D)()F x 是单调函数()f x ⇔是单调函数(9)设函数⎰+-+-++=yx y x dt t y x y x y x u )()()(),(ψϕϕ, 其中函数ϕ具有二阶导数,ψ 具有一阶导数,则必有(A)2222y ux u ∂∂-=∂∂(B)2222yu x u ∂∂=∂∂(C)222yu y x u ∂∂=∂∂∂(D)222x uy x u ∂∂=∂∂∂ (10)设有三元方程ln e 1xz xy z y -+=,根据隐函数存在定理,存在点(0,1,1)的一个邻域,在此邻域内该方程(A)只能确定一个具有连续偏导数的隐函数(,)z z x y = (B)可确定两个具有连续偏导数的隐函数(,)x x y z =和(,)z z x y = (C)可确定两个具有连续偏导数的隐函数(,)y y x z =和(,)z z x y = (D)可确定两个具有连续偏导数的隐函数(,)x x y z =和(,)y y x z =(C))1(~)1(--n t SXn (D)2122(1)~(1,1)nii n X F n X=--∑三 、解答题(本题共9小题,满分94分.解答应写出文字说明、证明过程或演算步骤) (15)(本题满分11分)设}0,0,2),{(22≥≥≤+=y x y x y x D ,]1[22y x ++表示不超过221y x ++的最大整数. 计算二重积分⎰⎰++Ddxdy y x xy .]1[22(16)(本题满分12分) 求幂级数∑∞=--+-121))12(11()1(n n n x n n 的收敛区间与和函数()f x .(17)(本题满分11分)如图,曲线C 的方程为()y f x =,点(3,2)是它的一个拐点,直线1l 与2l 分别是曲线C 在点(0,0)与(3,2)处的切线,其交点为(2,4).设函数()f x 具有三阶连续导数,计算定积分⎰'''+32.)()(dx x f x x(18)(本题满分12分)已知函数()f x 在[0,1]上连续,在(0,1)内可导,且(0)0,(1)1f f ==. 证明:(1)存在),1,0(∈ξ 使得ξξ-=1)(f .(2)存在两个不同的点)1,0(,∈ζη,使得.1)()(=''ζηf f(19)(本题满分12分)设函数)(y ϕ具有连续导数,在围绕原点的任意分段光滑简单闭曲线L 上,曲线积分24()22Ly dx xydyx y φ++⎰的值恒为同一常数.(1)证明:对右半平面0x >内的任意分段光滑简单闭曲线,C 有24()202Cy dx xydyx yφ+=+⎰.(2)求函数)(y ϕ的表达式.(20)(本题满分9分)已知二次型21232221321)1(22)1()1(),,(x x a x x a x a x x x f +++-+-=的秩为2.(1)求a 的值；(2)求正交变换x y =Q ,把),,(321x x x f 化成标准形. (3)求方程),,(321x x x f =0的解.(21)(本题满分9分)已知3阶矩阵A 的第一行是c b a c b a ,,),,,(不全为零,矩阵12324636k ⎡⎤⎢⎥=⎢⎥⎢⎥⎣⎦B (k 为常数),且=AB O ,求线性方程组0x =A 的通解.(22)(本题满分9分)设二维随机变量(,)X Y 的概率密度为(,)f x y =1001,02x y x <<<<其它求:(1)(,)X Y 的边缘概率密度)(),(y f x f Y X . (2)Y X Z -=2的概率密度).(z f Z(23)(本题满分9分)设)2(,,,21>n X X X n 为来自总体(0,1)N 的简单随机样本,X 为样本均值,记.,,2,1,n i X X Y i i =-=求:(1)i Y 的方差n i DY i ,,2,1, =. (2)1Y 与n Y 的协方差1Cov(,).n Y Y2005年考研数学一真题解析一、填空题（本题共6小题，每小题4分，满分24分. 把答案填在题中横线上） （1）曲线122+=x x y 的斜渐近线方程为 .4121-=x y 【分析】 本题属基本题型，直接用斜渐近线方程公式进行计算即可.【详解】 因为a=212lim )(lim 22=+=∞→∞→x x x x x f x x ，[]41)12(2lim )(lim -=+-=-=∞→∞→x x ax x f b x x ， 于是所求斜渐近线方程为.4121-=x y （2）微分方程x x y y x ln 2=+'满足91)1(-=y 的解为.91ln 31x x x y -=.【分析】直接套用一阶线性微分方程)()(x Q y x P y =+'的通解公式：⎰+⎰⎰=-])([)()(C dx e x Q e y dxx P dxx P ，再由初始条件确定任意常数即可.【详解】 原方程等价为x y xy ln 2=+'，于是通解为 ⎰⎰+⋅=+⎰⋅⎰=-]ln [1]ln [2222C xdx x xC dx ex ey dxx dxx=2191ln 31xC x x x +-， 由91)1(-=y 得C=0，故所求解为.91ln 31x x x y -= （3）设函数181261),,(222z y x z y x u +++=，单位向量}1,1,1{31=n ，则)3,2,1(nu∂∂=33.【分析】 函数u(x,y,z)沿单位向量γβαcos ,cos ,{cos =n}的方向导数为：γβαcos cos cos zu y u x u n u ∂∂+∂∂+∂∂=∂∂因此，本题直接用上述公式即可.【详解】 因为 3x x u =∂∂，6y y u =∂∂，9zz u =∂∂，于是所求方向导数为)3,2,1(nu∂∂=.33313131313131=⋅+⋅+⋅（4）设Ω是由锥面22y x z +=与半球面222y x R z --=围成的空间区域，∑是Ω的整个边界的外侧，则⎰⎰∑=++zdxdy ydzdx xdydz 3)221(2R -π.【分析】本题∑是封闭曲面且取外侧，自然想到用高斯公式转化为三重积分，再用球面（或柱面）坐标进行计算即可.【详解】 ⎰⎰∑=++zdxdy ydzdx xdydz ⎰⎰⎰Ωdxdydz 3=.)221(2sin 3320402R d d d R⎰⎰⎰-=πππθϕϕρρ（5）设321,,ααα均为3维列向量，记矩阵),,(321ααα=A ，)93,42,(321321321ααααααααα++++++=B ，如果1=A ，那么=B 2 .【分析】 将B 写成用A 右乘另一矩阵的形式，再用方阵相乘的行列式性质进行计算即可.【详解】 由题设，有)93,42,(321321321ααααααααα++++++=B=⎥⎥⎥⎦⎤⎢⎢⎢⎣⎡941321111),,(321ααα，于是有 .221941321111=⨯=⋅=A B（6）从数1,2,3,4中任取一个数，记为X, 再从X ,,2,1 中任取一个数，记为Y, 则}2{=Y P = 4813 . 【分析】 本题涉及到两次随机试验，想到用全概率公式, 且第一次试验的各种两两互不相容的结果即为完备事件组或样本空间的划分.【详解】}2{=Y P =}12{}1{===X Y P X P +}22{}2{===X Y P X P+}32{}3{===X Y P X P +}42{}4{===X Y P X P=.4813)4131210(41=+++⨯ 二、选择题（本题共8小题，每小题4分，满分32分. 每小题给出的四个选项中，只有一项符合题目要求，把所选项前的字母填在题后的括号内）（7）设函数nnn xx f 31lim)(+=∞→，则f(x)在),(+∞-∞内(A)处处可导.(B) 恰有一个不可导点.(C) 恰有两个不可导点. (D) 至少有三个不可导点. [ C ]【分析】 先求出f(x)的表达式，再讨论其可导情形.【详解】 当1<x 时，11lim)(3=+=∞→nnn xx f ；当1=x 时，111lim)(=+=∞→nn x f ；当1>x 时，.)11(lim )(3133x xxx f nnn =+=∞→即.1,11,1,,1,)(33>≤≤--<⎪⎩⎪⎨⎧-=x x x x x x f 可见f(x)仅在x=1±时不可导，故应选(C).（8）设F(x)是连续函数f(x)的一个原函数，""N M ⇔表示“M 的充分必要条件是N ”，则必有(A) F(x)是偶函数⇔f(x)是奇函数. （B ） F(x)是奇函数⇔f(x)是偶函数. (C) F(x)是周期函数⇔f(x)是周期函数.(D) F(x)是单调函数⇔f(x)是单调函数. [ A ]【分析】 本题可直接推证，但最简便的方法还是通过反例用排除法找到答案.【详解】 方法一：任一原函数可表示为⎰+=x C dt t f x F 0)()(，且).()(x f x F ='当F(x)为偶函数时，有)()(x F x F =-，于是)()1()(x F x F '=-⋅-'，即 )()(x f x f =--，也即)()(x f x f -=-，可见f(x)为奇函数；反过来，若f(x)为奇函数，则⎰x dt t f 0)(为偶函数，从而⎰+=xC dt t f x F 0)()(为偶函数，可见(A)为正确选项.方法二：令f(x)=1, 则取F(x)=x+1, 排除(B)、(C); 令f(x)=x, 则取F(x)=221x , 排除(D); 故应选(A).（9）设函数⎰+-+-++=yx yx dt t y x y x y x u )()()(),(ψϕϕ, 其中函数ϕ具有二阶导数，ψ 具有一阶导数，则必有(A) 2222yux u ∂∂-=∂∂. （B ）2222yux u ∂∂=∂∂.(C)222y uy x u ∂∂=∂∂∂. (D)222x uy x u ∂∂=∂∂∂.[ B ]【分析】 先分别求出22x u∂∂、22y u ∂∂、yx u ∂∂∂2，再比较答案即可.【详解】 因为)()()()(y x y x y x y x xu--++-'++'=∂∂ψψϕϕ，)()()()(y x y x y x y x yu-+++-'-+'=∂∂ψψϕϕ，于是 )()()()(22y x y x y x y x xu-'-+'+-''++''=∂∂ψψϕϕ， )()()()(2y x y x y x y x yx u-'++'+-''-+''=∂∂∂ψψϕϕ，)()()()(22y x y x y x y x yu-'-+'+-''++''=∂∂ψψϕϕ，可见有2222y ux u ∂∂=∂∂，应选(B).（10）设有三元方程1ln =+-xzey z xy ，根据隐函数存在定理，存在点(0,1,1)的一个邻域，在此邻域内该方程(A) 只能确定一个具有连续偏导数的隐函数z=z(x,y).(B) 可确定两个具有连续偏导数的隐函数x=x(y,z)和z=z(x,y).(C) 可确定两个具有连续偏导数的隐函数y=y(x,z)和z=z(x,y).(D) 可确定两个具有连续偏导数的隐函数x=x(y,z)和y=y(x,z).[ D ]【分析】 本题考查隐函数存在定理，只需令F(x,y,z)=1ln -+-xzey z xy , 分别求出三个偏导数yx z F F F ,,，再考虑在点(0,1,1)处哪个偏导数不为0，则可确定相应的隐函数.【详解】 令F(x,y,z)=1ln -+-xzey z xy , 则z e y F xzx+='， yz x F y-='，x e y F xzz+-='ln ， 且 2)1,1,0(='xF ，1)1,1,0(-='yF ，0)1,1,0(='zF . 由此可确定相应的隐函数x=x(y,z)和y=y(x,z). 故应选(D).（11）设21,λλ是矩阵A 的两个不同的特征值，对应的特征向量分别为21,αα，则1α，)(21αα+A 线性无关的充分必要条件是(A) 01≠λ. (B) 02≠λ. (C) 01=λ.(D) 02=λ. [ B ]【分析】 讨论一组抽象向量的线性无关性，可用定义或转化为求其秩即可.【详解】 方法一：令 0)(21211=++αααA k k ，则 022211211=++αλαλαk k k ， 0)(2221121=++αλαλk k k.由于21,αα线性无关，于是有⎩⎨⎧==+.0,022121λλk k k当02≠λ时，显然有0,021==k k，此时1α，)(21αα+A 线性无关；反过来，若1α，)(21αα+A 线性无关，则必然有02≠λ(,否则，1α与)(21αα+A =11αλ线性相关)，故应选(B).方法二：由于⎥⎦⎤⎢⎣⎡=+=+21212211121101],[],[)](,[λλαααλαλααααA ，可见1α，)(21αα+A 线性无关的充要条件是.001221≠=λλλ故应选(B).（12）设A 为n （2≥n ）阶可逆矩阵，交换A 的第1行与第2行得矩阵B, **,B A 分别为A,B 的伴随矩阵，则(A) 交换*A 的第1列与第2列得*B . (B)交换*A 的第1行与第2行得*B .(C) 交换*A 的第1列与第2列得*B -. (D)交换*A 的第1行与第2行得*B -.[ C ]【分析】 本题考查初等变换的概念与初等矩阵的性质，只需利用初等变换与初等矩阵的关系以及伴随矩阵的性质进行分析即可.【详解】 由题设，存在初等矩阵12E （交换n阶单位矩阵的第1行与第2行所得），使得BA E =12，于是 12*11212*12***12*)(E A E E A E A A E B-=⋅===-，即*12*B EA -=,可见应选(C).（13）设二维随机变量(X,Y) 的概率分布为 X Y 0 1 0 0.4 a 1 b 0.1 已知随机事件}0{=X 与}1{=+Y X 相互独立，则(A) a=0.2, b=0.3 (B) a=0.4, b=0.1(C) a=0.3, b=0.2 (D) a=0.1, b=0.4 [ B ] 【分析】 首先所有概率求和为1，可得a+b=0.5, 其次，利用事件的独立性又可得一等式，由此可确定a,b 的取值.【详解】 由题设，知 a+b=0.5 又事件}0{=X 与}1{=+Y X 相互独立，于是有 }1{}0{}1,0{=+===+=Y X P X P Y X X P ，即 a=))(4.0(b a a ++, 由此可解得 a=0.4, b=0.1, 故应选(B).（14）设)2(,,,21≥n X XX n 为来自总体N(0,1)的简单随机样本，X 为样本均值，2S 为样本方差，则(A) )1,0(~N X n (B) ).(~22n nS χ(C))1(~)1(--n t SXn (D)).1,1(~)1(2221--∑=n F XX n ni i[ D ]【分析】 利用正态总体抽样分布的性质和2χ分布、t 分布及F 分布的定义进行讨论即可. 【详解】 由正态总体抽样分布的性质知，)1,0(~10N X n nX =-，可排除(A);又)1(~0-=-n t SXn nS X ，可排除(C); 而)1(~)1(1)1(2222--=-n S n S n χ，不能断定(B)是正确选项.因为∑=-ni i n X X 222221)1(~),1(~χχ，且∑=-n i i n X X 222221)1(~)1(~χχ与相互独立，于是).1,1(~)1(1122212221--=-∑∑==n F XX n n XX ni ini i故应选(D).三 、解答题（本题共9小题，满分94分.解答应写出文字说明、证明过程或演算步骤.） （15）（本题满分11分） 设}0,0,2),{(22≥≥≤+=y x y x y x D ，]1[22y x++表示不超过221y x ++的最大整数. 计算二重积分⎰⎰++Ddxdy y xxy .]1[22【分析】 首先应设法去掉取整函数符号，为此将积分区域分为两部分即可.【详解】 令 }0,0,10),{(221≥≥<+≤=y x y x y x D ， }0,0,21),{(222≥≥≤+≤=y x y x y x D .则 ⎰⎰++Ddxdyy xxy ]1[22=⎰⎰⎰⎰+122D D xydxdy xydxdydrr d dr r d ⎰⎰⎰⎰+=202131320cos sin 2cos sin ππθθθθθθ=.874381=+ （16）（本题满分12分） 求幂级数∑∞=--+-121))12(11()1(n nn x n n 的收敛区间与和函数f(x).【分析】 先求收敛半径，进而可确定收敛区间. 而和函数可利用逐项求导得到.【详解】 因为11)12()12()12)(1(1)12)(1(lim =+--⨯+++++∞→n n n n n nn n n ，所以当21x<时，原级数绝对收敛，当21x>时，原级数发散，因此原级数的收敛半径为1，收敛区间为（－1,1） 记 121(1)(),(1,1)2(21)n nn S x x x n n -∞=-=∈--∑，则1211(1)(),(1,1)21n n n S x x x n -∞-=-'=∈--∑，122211()(1),(1,1)1n n n S x x x x ∞--=''=-=∈-+∑.由于 (0)0,(0)0,S S '==所以201()()arctan ,1xxS x S t dt dt x t '''===+⎰⎰2001()()arctan arctan ln(1).2x x S x S t dt tdt x x x '===-+⎰⎰又 21221(1),(1,1),1n nn x xx x∞-=-=∈-+∑从而22()2()1x f x S x x =++2222arctan ln(1),(1,1).1x x x x x x =-++∈-+（17）（本题满分11分）如图，曲线C 的方程为y=f(x)，点(3,2)是它的一个拐点，直线1l 与2l 分别是曲线C 在点(0,0)与(3,2)处的切线，其交点为(2,4). 设函数f(x)具有三阶连续导数，计算定积分⎰'''+302.)()(dx x f x x【分析】 题设图形相当于已知f(x)在x=0的函数值与导数值，在x=3处的函数值及一阶、二阶导数值.【详解】 由题设图形知，f(0)=0, 2)0(='f ; f(3)=2, .0)3(,2)3(=''-='f f由分部积分，知⎰⎰⎰+''-''+=''+='''+3303022302)12)(()()()()()()(dxx x f x f x x x f d x x dx x f x x=dxx f x f x x f d x ⎰⎰'+'+-='+-33030)(2)()12()()12(=.20)]0()3([216=-+f f （18）（本题满分12分）已知函数f(x)在[0，1]上连续，在(0,1)内可导，且f(0)=0,f(1)=1. 证明： （I ）存在),1,0(∈ξ 使得ξξ-=1)(f ；（II ）存在两个不同的点)1,0(,∈ζη，使得.1)()(=''ζηf f【分析】 第一部分显然用闭区间上连续函数的介值定理；第二部分为双介值问题，可考虑用拉格朗日中值定理，但应注意利用第一部分已得结论.【详解】 （I ） 令x x f x F +-=1)()(，则F(x)在[0，1]上连续，且F(0)=-1<0, F(1)=1>0,于是由介值定理知，存在),1,0(∈ξ 使得0)(=ξF ，即ξξ-=1)(f . （II ） 在],0[ξ和]1,[ξ上对f(x)分别应用拉格朗日中值定理，知存在两个不同的点)1,(),,0(ξζξη∈∈，使得0)0()()(--='ξξηf f f ，ξξζ--='1)()1()(f f f于是 .1111)(1)()()(=-⋅-=--⋅=''ξξξξξξξξζηf f f f（19）（本题满分12分）设函数)(y ϕ具有连续导数，在围绕原点的任意分段光滑简单闭曲线L 上，曲线积分⎰++Ly x xydydx y 4222)(ϕ的值恒为同一常数.（I ）证明：对右半平面x>0内的任意分段光滑简单闭曲线C ，有022)(42=++⎰Cyxxydydx y ϕ； （II ）求函数)(y ϕ的表达式.【分析】 证明（I ）的关键是如何将封闭曲线C 与围绕原点的任意分段光滑简单闭曲线相联系，这可利用曲线积分的可加性将C 进行分解讨论；而（II ）中求)(y ϕ的表达式，显然应用积分与路径无关即可.【详解】 （I ）1ll 2Co Xl 3如图，将C 分解为：21l lC +=，另作一条曲线3l围绕原点且与C 相接，则=++⎰Cyxxydydx y 4222)(ϕ-++⎰+314222)(l l y x xydydx y ϕ022)(3242=++⎰+l l y x xydydx y ϕ.（II ） 设2424()2,22y xyP Q xyx y ϕ==++，,P Q 在单连通区域0x >内具有一阶连续偏导数，由（Ⅰ）知，曲线积分24()22Ly dx xydyx y ϕ++⎰在该区域内与路径无关，故当0x >时，总有Q P x y∂∂=∂∂. 24252422422(2)4242,(2)(2)Q y x y x xy x y y x x y x y ∂+--+==∂++①Y243243242242()(2)4()2()()4().(2)(2)P y x y y y x y y y y y y x y x y ϕϕϕϕϕ'''∂+-+-==∂++ ②比较①、②两式的右端，得435()2,()4()2.y y y y y y y ϕϕϕ'=-⎧⎨'-=⎩ 由③得2()y yc ϕ=-+，将()y ϕ代入④得535242,y cy y -=所以0c =，从而2().y y ϕ=-（20）（本题满分9分） 已知二次型21232221321)1(22)1()1(),,(x x a x x a xa x x x f +++-+-=的秩为2.（I ） 求a 的值；（II ） 求正交变换Qy x =，把),,(321x x x f 化成标准形；（III ） 求方程),,(321x x x f =0的解.【分析】 （I ）根据二次型的秩为2，可知对应矩阵的行列式为0，从而可求a 的值；（II ）是常规问题，先求出特征值、特征向量，再正交化、单位化即可找到所需正交变换； （III ）利用第二步的结果，通过标准形求解即可.【详解】 （I ） 二次型对应矩阵为⎥⎥⎥⎦⎤⎢⎢⎢⎣⎡-++-=200011011a a a a A ，③ ④由二次型的秩为2，知 02011011=-++-=a aa a A ，得a=0.（II ） 这里⎥⎥⎥⎦⎤⎢⎢⎢⎣⎡=200011011A ， 可求出其特征值为,2321===λλλ.解 0)2(=-x A E ，得特征向量为：⎪⎪⎪⎭⎫⎝⎛=⎪⎪⎪⎭⎫ ⎝⎛=100,01121αα，解 0)0(=-x A E ，得特征向量为：.0113⎪⎪⎪⎭⎫ ⎝⎛-=α由于21,αα已经正交，直接将21,αα，3α单位化，得：⎪⎪⎪⎭⎫⎝⎛-=⎪⎪⎪⎭⎫ ⎝⎛=⎪⎪⎪⎭⎫ ⎝⎛=01121,100,01121321ηηη令[]321ααα=Q ，即为所求的正交变换矩阵，由x=Qy ，可化原二次型为标准形：),,(321x x x f =.222221y y +（III ） 由),,(321x x x f ==+222122y y0，得ky y y===321,0,0（k 为任意常数）.从而所求解为：x=Qy=[]⎥⎥⎥⎦⎤⎢⎢⎢⎣⎡-==⎥⎥⎥⎦⎤⎢⎢⎢⎣⎡0003321c c k k ηηηη，其中c 为任意常数.（21）（本题满分9分）已知3阶矩阵A 的第一行是c b a c b a ,,),,,(不全为零，矩阵⎥⎥⎥⎦⎤⎢⎢⎢⎣⎡=k B 63642321（k 为常数），且AB=O, 求线性方程组Ax=0的通解.【分析】 AB=O, 相当于告之B 的每一列均为Ax=0的解，关键问题是Ax=0的基础解系所含解向量的个数为多少，而这又转化为确定系数矩阵A 的秩.【详解】 由AB=O 知，B 的每一列均为Ax=0的解，且.3)()(≤+B r A r（1）若k 9≠, 则r(B)=2, 于是r(A)1≤, 显然r(A)1≥, 故r(A)=1. 可见此时Ax=0的基础解系所含解向量的个数为3-r(A)=2, 矩阵B 的第一、第三列线性无关，可作为其基础解系，故Ax=0 的通解为：2121,,63321k k k k k x ⎪⎪⎪⎭⎫⎝⎛+⎪⎪⎪⎭⎫ ⎝⎛=为任意常数.(2) 若k=9，则r(B)=1, 从而.2)(1≤≤A r 1） 若r(A)=2, 则Ax=0的通解为：11,321k k x ⎪⎪⎪⎭⎫⎝⎛=为任意常数.2） 若r(A)=1,则Ax=0 的同解方程组为：321=++cx bx ax ,不妨设≠a ,则其通解为2121,,1001k k a c k a b k x ⎪⎪⎪⎪⎪⎭⎫ ⎝⎛-+⎪⎪⎪⎪⎪⎭⎫ ⎝⎛-=为任意常数.（22）（本题满分9分）设二维随机变量(X,Y)的概率密度为.,20,10,0,1),(其他x y x y x f <<<<⎩⎨⎧= 求：（I ） (X,Y)的边缘概率密度)(),(y f x f Y X；（II ）Y X Z -=2的概率密度).(z fZ【分析】 求边缘概率密度直接用公式即可；而求二维随机变量函数的概率密度，一般用分布函数法，即先用定义求出分布函数，再求导得到相应的概率密度.【详解】 （I ） 关于X 的边缘概率密度)(x f X =⎰+∞∞-dyy x f ),(=.,10,0,20其他<<⎪⎩⎪⎨⎧⎰x dy x=.,10,0,2其他<<⎩⎨⎧x x 关于Y 的边缘概率密度)(y f Y =⎰+∞∞-dxy x f ),(=.,20,0,12其他<<⎪⎩⎪⎨⎧⎰y dx y=.,20,0,21其他<<⎪⎩⎪⎨⎧-y y（II ） 令}2{}{)(z Y X P z Z P z F Z≤-=≤=，1） 当0<z 时，0}2{)(=≤-=z Y X P z F Z；2） 当20<≤z 时，}2{)(z Y X P z F Z≤-= =241z z -; 3) 当2≥z 时，.1}2{)(=≤-=z Y X P z F Z即分布函数为：.2,20,0,1,41,0)(2≥<≤<⎪⎩⎪⎨⎧-=z z z z z z F Z故所求的概率密度为：.,20,0,211)(其他<<⎪⎩⎪⎨⎧-=z z z f Z（23）（本题满分9分） 设)2(,,,21>n X XX n 为来自总体N(0,1)的简单随机样本，X 为样本均值，记.,,2,1,n i X X Yi i=-=求：（I ） iY 的方差n i DY i,,2,1, =； （II ）1Y 与nY 的协方差).,(1nY Y Cov【分析】 先将iY 表示为相互独立的随机变量求和，再用方差的性质进行计算即可；求1Y 与nY 的协方差),(1nY Y Cov ，本质上还是数学期望的计算，同样应注意利用数学期望的运算性质.【详解】 由题设，知)2(,,,21>n X XX n 相互独立，且),,2,1(1,0n i DX EX i i ===，.0=X E（I ）∑≠--=-=nij j i i i X n X n D X X D DY ]1)11[()(=∑≠+-nij ji DX nDX n 221)11(=.1)1(1)1(222n n n n n n -=-⋅+-（II ） )])([(),(111n nnEY YEY Y E Y Y Cov --==)])([()(11X X X XE Y Y E n n--==)(211X X X X X XX E n n+-- =211)(2)(X E X X E XX E n+-=22121)(][20X E X D X X X E n nj j +++-∑==.112nn n -=+-。
2005年硕士研究生入学试题(答案)
答案：一、选择题1. c,2. b,3. c,4. c,5. a,6. b,7. d,8. d,9. c, 10. c, 11. c, 12. d, 13. d 14.b 15.a 16.a 17.a 18.a 19.c 20.c二、三、1、根据相图，固相从液相中结晶时，成分于液相不同。
因此在结晶过程中固相不断排除溶质原子，导致在液－固液界面液相侧溶质原子积聚。
此时原子的运动路线是固相－界面－液相开始时固相中排除的原子比从界面层向液相排出的原子多，导致界面层上溶质原子浓度升高，此时称之为初瞬态。
当溶质原子在界面层上的浓度高达一定程度时，达到动态平衡，即从固相到界面与从界面到液相的原子流量相同，此时在固液相之间形成了一层溶质原子浓度高于液相的过渡层，称之为边界层。
2、平衡分配叙述是指在平衡凝固时固相与液相中溶质浓度之比，即k0=ρS/ρL;有效分配系数是指在非平衡凝固是，当边界层建立后，边界固相侧溶质浓度和边界层以外的液相区中溶质浓度之比，即上图中k e＝(ρS)i/(ρL)B;3、当k e=k0时出现正常凝固，此时没有边界层；液相内成分完全均匀。
当k e＝1时，铸锭内成分最均匀，当k e=k0时即正常凝固是成分最不均匀。
4、如图虚线所示，当边界层中温度梯度与边界层浓度分布曲线相切时，是成分过冷的临界条件。
若温度分布曲线斜率小于切线斜率时则有成分过冷，反之则没有成分过冷。
四、1、形核功∆G*是指形成一个临界晶胚所需要的能量，这个能量是靠系统能量起伏提供。
L L －γ L －γ+Fe 3C(L d ) γ－Fe 3C Ⅱ γ－P P+ Fe 3C Ⅱ＋L d ’2、 临界半径是指临界晶胚的半径。
临界晶核是指结晶开始时，晶胚的形成与长大需要靠系统能量起伏提供能量；当晶胚长大到一定大小时，能量达到极大值，继续长大系统能量下降，此时的晶胚半径称之为领结半径；3、 形核率是指单位时间和单位体积内形成的晶核数。
苏州大学考研复试历年上机题目及答案
if (is_prime(i) &&num-i>1) {printf("%d ", i);split(num-i);return;}}}int main(void){int i;while (scanf("%d", &i)){split(i);printf("\n");}getchar();}要求：统计篇文章中各英文字母的个数，并排序.程序：#include <stdio.h>#include <stdlib.h>typedef struct{char c;int n;} Letter;void swap(Letter*a, Letter*b){Letter temp=*a;*a=*b;*b=temp;}int compare(Letter*a, Letter*b){if (a->n<b->n)return-1;else if (a->n>b->n)return1;elsereturn0;void selection_sort(Letter*ptr, int count){for (int i=0; i<count; i++)for (int j=i+1; j<count; j++)if (compare(&ptr[i], &ptr[j]) >0)swap(&ptr[i], &ptr[j]);}bool is_upper(char c) { return c>='A'&&c<='Z'; } bool is_lower(char c) { return c>='a'&&c<='z'; }int main(void){FILE*fp=fopen("address.txt", "r");Letter letter[26];char c;if(!fp){printf("File opening failed");return EXIT_FAILURE;}for (int i=0; i<26; i++){letter[i].c=i+'a';letter[i].n=0;}while ((c=fgetc(fp)) !=EOF){if (is_upper(c))letter[c-'A'].n++;else if (is_lower(c))letter[c-'a'].n++;}selection_sort(letter, 26);for (int i=0; i<26; i++)printf("%c: %d\n", letter[i].c, letter[i].n);fclose(fp);}2006年复试上机题要求：找出100到1000内的不含9的素数，存到result文件中.程序：#include <stdio.h>bool is_prime(int num){if (num<2)return false;for (int i=2; i*i<=num; i++)if (num%i==0)return false;return true;}bool has9(int num){if (num<0)return false;while (num>0){if (num%10==9)return true;num/=10;}return false;}int main(){FILE*fp=fopen("result.txt", "w");for (int i=100; i<1000; i++)if (is_prime(i) &&!has9(i))fprintf(fp, "%d\n", i);fclose(fp);}2007年复试上机题要求：把10到1000之间满足以下两个条件的数，存到result.txt文件中.是素数.它的反数也是素数，如：123的反数是321.程序：#include <stdio.h>bool is_prime(int num){if (num<2)return false;for (int i=2; i*i<=num; i++)if (num%i==0)return true;}int reverse(int num){int rev=0;while (num>0){rev=rev*10+num%10;num/=10;}return rev;}int main(){FILE*fp=fopen("result.txt", "w");for (int i=100; i<1000; i++)if (is_prime(i) &&is_prime(reverse(i)))fprintf(fp, "%d %d\n", i, reverse(i));fclose(fp);}2008年复试上机题要求：用IE 从FTP 上下载org.dat ，并保存在D盘的根目录中.此文件中按文本方式存放了一段其他文章，其中有若干长度小于15的英文单词，单词之间用空格分开，无其他符号.顺序读取这段文章的不同的单词(大小写敏感)，同时在读取的过程中排除所有的单词THE以及变形，即这些单词不能出现在读取的结果中.将读取的所有单词的首字母转大写后，输出D 根目录下new.txt，每个单词一行.程序：#include <stdio.h>#include <stdlib.h>#include <string.h>bool is_the(char word[15]){char the[] ="the";if (strlen(word) !=strlen(the))return false;for (int i=0; i<strlen(word); i++)word[i] |=0x20;return strcmp(word, the) ==0;}int main(){FILE*fporg=fopen("org.dat", "r");FILE*fpnew=fopen("new.txt", "w");char word[15];if(!fporg||!fpnew){printf("File opening failed");return EXIT_FAILURE;}printf("org.dat:\n");while (fscanf(fporg, "%s", word) !=EOF){printf("%s ", word);if (!is_the(word)){word[0] = (word[0] |0x20) -0x20;fprintf(fpnew, "%s\n", word);}}printf("\n");fclose(fporg);fclose(fpnew);fpnew=fopen("new.txt", "r");printf("new.txt:\n");while (fscanf(fporg, "%s", word) !=EOF)printf("%s ", word);printf("\n");fclose(fpnew);return0;}输出：org.dat:The constructor is used to initialize the object The destructor is used to delete the Object the calling seqence of constructor is opposite to the calling sequence of destructornew.txt:Constructor Is Used To Initialize Object Destructor Is Used To Delete Object Calling Seqence Of Constructor Is Opposite To Calling Sequence Of Destructor2009年复试上机题要求：用IE 浏览器从FTP 上下载org.dat ，并保存在D盘的根目录下.此文件中按文本方式存放了一段其他文章，其中有若干长度小于15的十进制或八进制数字，数字之间用,分开，数字内部存在且仅存在空格.顺序读取这些数字将他们转变为十进制数后按从大到小的顺序排序后，输出到D盘根目录下new.txt，每个数字一行.eg：_235_,34__2,_043_1_,1_3，分别是：十进制235，十进制342，八进制431，十进制13，_代表空格.程序：#include <stdio.h>#include <stdlib.h>#include <string.h>void swap(int*a, int*b){int temp=*a;*a=*b;*b=temp;}int compare(int*a, int*b){if (a<b)return-1;else if (a>b)return1;elsereturn0;}void selection_sort(int*ptr, int count){for (int i=0; i<count; i++)for (int count=i+1; count<count; count++)if (compare(&ptr[i], &ptr[count]) >0)swap(&ptr[i], &ptr[count]);}bool is_num(char c) { return c>='0'&&c<='9'; }bool valid(char c) { return is_num(c) ||c==' '; }int main(){FILE*fporg=fopen("org.dat", "r");FILE*fpnew=fopen("new.txt", "w");int i;int count=0;int arr[128];char c;char num[15];printf("org.dat:\n");{if (is_num(c)){i=0;num[i++] =c;while ((c=fgetc(fporg)) !=EOF&&valid(c))if (is_num(c)) num[i++] =c;num[i] ='\0';printf("%s\n", num);arr[count++] =num[0] =='0'?strtol(num, NULL, 8) :strtol(num, NULL, 10);}}selection_sort(arr, count);for (i=0; i<count; i++)fprintf(fpnew, "%d\n", arr[i]);fclose(fporg);fclose(fpnew);fpnew=fopen("new.txt", "r");printf("new.txt:\n");while (fgets(num, sizeof(num), fpnew))printf("%s", num);fclose(fpnew);}输出：org.dat:235342043113new.txt:235342281132010年复试上机题要求：从FTP 上下载make.exe 和org.dat ，运行make.exe 输入准考证后三位生成data.txt，文件为二进制编码.data.txt内存有2048个整数，其中前n个为非0数，后2048-n个数为0，将其读入数组，计算非零数的个数n.选出n个数中的最大数和最小数.选出n个数中最大素数.将n个数从大到小排序，并平均分成三段(若n非3的整数倍，则不考虑最后的1-2个数)，选出中间段的最大数和最小数.程序：#include <math.h>#include <stdio.h>#include <stdlib.h>#include <string.h>#include <time.h>bool is_prime(int num){if (num<2)return false;for (int i=2; i*i<=num; i++)if (num%i==0)return false;return true;}void swap(int*a, int*b){int temp=*a;*a=*b;*b=temp;}int compare(int*a, int*b){if (a<b)return-1;else if (a>b)return1;elsereturn0;}void selection_sort(int*ptr, int count){for (int i=0; i<count; i++)for (int count=i+1; count<count; count++)if (compare(&ptr[i], &ptr[count]) >0)swap(&ptr[i], &ptr[count]);}int main(void){FILE*fp;int num[2048];int min=RAND_MAX;int max_prime=0;int n, i;fp=fopen("data.txt", "wb");if(!fp){printf("File opening failed");return EXIT_FAILURE;}srand(time(NULL));for (i=0; i<1000; i++)num[i] =rand() %4096;for (; i<2048; i++)num[i] =0;fwrite(num, sizeof(int), 2048, fp);fclose(fp);fp=fopen("data.txt", "rb");if(!fp){printf("File opening failed");return EXIT_FAILURE;}fread(num, sizeof(int), 2048, fp);for (n=0; n<2048; n++)if (num[n] ==0)break;printf("n = %d\n", n);for (i=0; i<n; i++){if (min>num[i]) min=num[i];if (max<num[i]) max=num[i];if (is_prime(num[i]) &&num[i] >max_prime) max_prime=num[i];}printf("min = %d\nmax = %d\nmax_prime = %d\n", min, max, max_prime);selection_sort(num, n);printf("min in mid_seg = %d\nmax in mid_seg = %d\n", num[n/3*2-1], num[n/3]);fclose(fp);}输出：n = 1000min = 1max = 4095max_prime = 4093min in mid_seg = 3630max in mid_seg = 18632011年复试上机题要求：输出1000-9999中满足以下条件的所有数：该数是素数.十位数和个位数组成的数是素数，百位数和个位数组成的数是素数.个位数和百位数组成的数是素数，个位数和十位数组成的数是素数. 比如1991，个位和十位组成的数就是19.程序：#include <stdio.h>bool is_prime(int num){if (num<2)return false;for (int i=2; i*i<=num; i++)if (num%i==0)return false;return true;}int main(void){for (int i=1000; i<=9999; i++){int g=i%10;int s= (i/10) %10;int b= (i/100) %10;if (is_prime(i) &&is_prime(i%100) &&is_prime(g*10+b) &&is_prime(g*10+s) &&is_prime(b*10+g))printf("%d ", i);}}输出：1117 1171 1997 2111 2113 2131 2137 2311 2371 2711 2713 2731 2917 3137 3331 3371 3779 3917 4111 4337 4397 4937 5113 5171 5197 5711 5779 6113 6131 6173 6197 6311 6317 6337 6397 6779 6917 6997 7331 7937 8111 8117 8171 8311 8317 8713 8731 8779 9137 9173 9311 9337 9371 93972012年复试上机题要求：从服务器上下载数据文件org.dat文件以二进制方式存放一系列整数，每个整数占4个字节. 从第一个整数规定处于第一象限的坐标点为有效点，请问数据文件中所有点的个数n为多少？有效点的个数k为多少？每个有效点与坐标原点构成一个的矩形，请问k个有效点与坐标原点构成的k个矩形的最小公共区域面积为多少？寻找有效点钟符合下列条件的点：以该点为坐标原点，其它有效点仍然是有效点即处于第一象限(不包括坐标轴上的点). 输出这些点.对所有有效点进行分组，每个有效点有且只有属于一个分组，分组内的点符合下列规则：若对组内所有点的x 坐标进行排序，点p1(x1, y1)在点p2(x2, y2)后面，即x1>x2那么y1>y2，请输出所有的分组.程序：#include <math.h>#include <stdio.h>#include <stdlib.h>#include <string.h>#include <time.h>typedef struct{int x;int y;} Coordinate;void swap(Coordinate*a, Coordinate*b){Coordinate temp=*a;*a=*b;*b=temp;}int compare(const Coordinate*lhs, const Coordinate*rhs){if (lhs->x<rhs->x)return-1;if (lhs->x>rhs->x)return1;elsereturn0;}void selection_sort(Coordinate*ptr, int begin, int end){for (int i=begin; i<end; i++)for (int j=i+1; j<end; j++)if (compare(&ptr[i], &ptr[j]) >0)swap(&ptr[i], &ptr[j]);}void display(Coordinate*ptr, int count){for (int i=0; i<count; i++)printf("\n");}int main(void){FILE*fp;Coordinate*xy;int*num;int n, count;fp=fopen("org.dat", "wb");num= (int*) malloc(sizeof(int) *128);srand(time(NULL));for (int i=0; i<128; i++)num[i] =rand() %128-64;fwrite(num, sizeof(int), 128, fp);free(num);fclose(fp);fp=fopen("org.dat", "rb");fseek(fp, 0, SEEK_END);count=ftell(fp) /sizeof(int);n=count/2;num= (int*) malloc(sizeof(int) *count);xy= (Coordinate*) malloc(sizeof(Coordinate) *n);rewind(fp);fread(num, sizeof(int), count, fp);fclose(fp);int k=0;int minx=RAND_MAX;int miny=RAND_MAX;for (int i=0; i<count; i+=2){if (num[i] >0&&num[i+1] >0){xy[k].x=num[i];xy[k].y=num[i+1];if (minx>xy[k].x)minx=xy[k].x;if (miny>xy[k].y)miny=xy[k].y;k++;}}selection_sort(xy, 0, k);printf("valid points:\n");display(xy, k);printf("n = %d k = %d\n", n, k);printf("min area = %d\n", minx*miny);int sorted=0;{selection_sort(xy, sorted, k);printf("points grouped by (%2d, %2d): ", xy[sorted].x, xy[sorted].y);miny=xy[sorted++].y;for (int i=sorted; i<k; i++){if (xy[i].y>=miny){miny=xy[i].y;printf("(%d, %d) ", xy[i].x, xy[i].y);swap(&xy[i], &xy[sorted++]);}}printf("\n");}}输出：valid points:( 3, 8) (10, 62) (11, 40) (20, 6) (26, 7) (26, 20) (29, 31) (33, 10) (34, 37) (37, 33) (50,55) (52, 59) (53, 49) (53, 8) (57, 16) (58, 20)n = 64 k = 16min area = 18points grouped by ( 3, 8): (10, 62)points grouped by (11, 40): (50, 55) (52, 59)points grouped by (20, 6): (26, 20) (29, 31) (34, 37) (53, 49)points grouped by (26, 7): (33, 10) (37, 33)points grouped by (53, 8): (57, 16) (58, 20)2013年复试上机题要求：IntroductionThe project will read flight data from an input file and flight path requests from another input file and output the required information.Your TaskYour program should determine if a particular destination airport can be reached from a particular originating airport within a particular number of hops. A hop (leg of a flight) is a flight from oneairport to another on the path between an originating and destination airports. For example, the flight plan from PVG to PEK might be PVG -> CAN -> PEK. So PVG -> CAN would be a hop and CAN -> PEK would be a hop.Input Data FilesPath Input File(PathInput.txt)This input file will consist of a number of single origination/destination airport pairs (direct flights).The first line of the file will contain an integer representing the total number of pairs in the rest of the file.6[PVG, CAN][CAN, PEK][PVG, CTU][CTU, DLC][DLC, HAK][HAK, LXA]Path Request File(PathRequest.txt)This input file will contain a sequence of pairs of origination/destination airports and a max number of hops. The first line of the file will contain an integer representing the number of pairs in the file.2[PVG, DLC, 2][PVG, LXA, 2]Output File(Output.txt )For each pair in the Path Request File, your program should output the pair followed by YES or NO indicating that it is possible to get from the origination to destination airports within the max number of hops or it is not possible, respectively.[PVG, DLC, YES][PVG, LXA, NO]Assumptions you can make:You may make the following simplifying assumptions in your project:C/C++ is allowed to be used.All airport codes will be 3 letters and will be in all capsOrigination/destination pairs are unidirectional. To indicate that both directions of flight arepossible, two entries would appear in the file. For example, [PVG, PEK] and [PEK, PVG] wouldhave to be present in the file to indicate that one could fly from Shanghai to Beijing and fromBeijing to Shanghai.程序：#include <math.h>#include <stdio.h>#include <stdlib.h>#include <string.h>#include <time.h>typedef struct Edge{int index;char city[4];struct Edge*next;typedef struct{char city[4];Edge*first;} Vertex;typedef struct{int size;int capacity;Vertex*vertices;} Graph;int find(Vertex*vertices, int count, char src[4]){int i;for (i=0; i<count; i++)if (strcmp(src, vertices[i].city) ==0)break;return i;}void add_edge(Graph*g, char src[4], char dst[4]){int i, j;Edge*e;i=find(g->vertices, g->size, src);if (i==g->size){strcpy(g->vertices[g->size].city, src);g->vertices[g->size].first=NULL;g->size++;}j=find(g->vertices, g->size, dst);if (j==g->size){strcpy(g->vertices[g->size].city, dst);g->vertices[g->size].first=NULL;g->size++;}for (e=g->vertices[i].first; e!=NULL; e=e->next) if (strcmp(e->city, dst) ==0)break;if (e==NULL){e= (Edge*) malloc(sizeof(Edge));strcpy(e->city, dst);e->index=j;e->next=g->vertices[i].first;g->vertices[i].first=e;}bool is_possible(Graph*g, bool*visited, char src[], char dst[], int step){int current;Edge*e;current=find(g->vertices, g->size, src);if (step==0||visited[current])return false;else{visited[current] =true;for (e=g->vertices[current].first; e!=NULL; e=e->next){if (strcmp(e->city, dst) ==0&&step==1)return true;else if (is_possible(g, visited, e->city, dst, step-1))return true;}visited[current] =false;}return false;}void destruct(Graph*g){for (int i=0; i<g->size; i++){while (g->vertices[i].first!=NULL){Edge*e=g->vertices[i].first->next;free(g->vertices[i].first);g->vertices[i].first=e;}}free(g->vertices);}void display(Graph*g){printf("adjacency list:\n");for (int i=0; i<g->size; i++){printf("%s: ", g->vertices[i].city);for (Edge*edge=g->vertices[i].first; edge!=NULL; edge=edge->next) printf("%s ", edge->city);printf("\n");}printf("\n");}int main(void)FILE*fpi=fopen("PathInput.txt", "r");FILE*fpr=fopen("PathRequest.txt", "r");FILE*fpo=fopen("Output.txt", "w");char line[16];char src[4];char dst[4];bool*visited;int step;int num;Graph g;if (!fpi||!fpr||!fpo){printf("failed to open files.\n");exit(-1);}memset(line, 0, sizeof(line));if (fgets(line, sizeof(line), fpi) !=NULL){g.size=0;g.capacity=atoi(line) *2;g.vertices= (Vertex*) malloc(g.capacity*sizeof(Vertex));memset(g.vertices, 0, g.capacity);}while (fgets(line, sizeof(line), fpi) !=NULL){memcpy(src, line+1, 3);src[3] ='\0';memcpy(dst, line+6, 3);dst[3] ='\0';add_edge(&g, src, dst);}display(&g);printf("flight info:\n");visited= (bool*) malloc(g.size);if (fgets(line, sizeof(line), fpr) !=NULL)num=atoi(line);while (fgets(line, sizeof(line), fpr) !=NULL){memcpy(src, line+1, 3);src[3] ='\0';memcpy(dst, line+6, 3);dst[3] ='\0';step=line[11] -'0';memset(visited, false, g.size);if (is_possible(&g, visited, src, dst, step)){printf("[%s, %s, %d, YES]\n", src, dst, step);fprintf(fpo, "[%s, %s, YES]\n", src, dst);}elseprintf("[%s, %s, %d, NO]\n", src, dst, step);fprintf(fpo, "[%s, %s, NO]\n", src, dst);}}destruct(&g);free(visited);fclose(fpi);fclose(fpr);fclose(fpo);}输出：adjacency list:PVG: CTU CANCAN: PEKPEK:CTU: DLCDLC: HAKHAK: LXALXA:flight info:[PVG, DLC, 2, YES][PVG, LXA, 2, NO]2014年复试上机题要求：从网页上下载input.dat 文件，里面是用二进制编写的，里面放了一堆int 型的数，每个数占4个字节，每次读取两个，这两个数构成一个坐标.规定处于第一象限的数是有效点(即x>0, y>0的坐标)，问这么多点中有效点有多少个？现在用户从键盘输入一个坐标和一个数字k，设计算法输出k个离该坐标距离最近的点的坐标和每个坐标到该点的距离，写入到output.txt文件中.程序：#include <math.h>#include <stdio.h>#include <stdlib.h>#include <string.h>#include <time.h>typedef struct{int x;int y;} Coordinate;Coordinate point;double distance(const Coordinate*a, const Coordinate*b) {int val= (a->x-b->x) * (a->x-b->x) +(a->y-b->y) * (a->y-b->y);return sqrt((double) val);}void swap(Coordinate*a, Coordinate*b){Coordinate temp=*a;*a=*b;*b=temp;}int compare(const Coordinate*a, const Coordinate*b){double dista=distance(a, &point);double distb=distance(b, &point);if (dista<distb)return-1;if (dista>distb)return1;elsereturn0;}void selection_sort(Coordinate*ptr, int begin, int end) {for (int i=begin; i<end; i++)for (int j=i+1; j<end; j++)if (compare(&ptr[i], &ptr[j]) >0)swap(&ptr[i], &ptr[j]);}void display(Coordinate*ptr, int count){for (int i=0; i<count; i++)printf("(%2d, %2d) ", ptr[i].x, ptr[i].y);printf("\n");}int main(void){FILE*fp;Coordinate*xy;int*num;int count, coord_count;num= (int*) malloc(sizeof(int) *128);srand(time(NULL));for (int i=0; i<128; i++)num[i] =rand() %128-64;fwrite(num, sizeof(int), 128, fp);free(num);fclose(fp);fp=fopen("org.dat", "rb");fseek(fp, 0, SEEK_END);count=ftell(fp) /sizeof(int);coord_count=count/2;num= (int*) malloc(sizeof(int) *count);xy= (Coordinate*) malloc(sizeof(Coordinate) *coord_count);rewind(fp);fread(num, sizeof(int), count, fp);fclose(fp);int valid=0;fp=fopen("output.txt", "w");for (int i=0; i<count; i+=2){if (num[i] >0&&num[i+1] >0){xy[valid].x=num[i];xy[valid].y=num[i+1];valid++;}}printf("valid points:\n");display(xy, valid);printf("total: %d\n", valid);int k;printf("coordinate: ");scanf("%d %d", &point.x, &point.y);printf("k = ");scanf("%d", &k);selection_sort(xy, 0, valid);printf("the nearest %d points to (%d, %d) are:\n", k, point.x, point.y);for (int i=0; i<k; i++){printf("(%2d, %2d) %7.2f\n", xy[i].x, xy[i].y, distance(&xy[i], &point));fprintf(fp, "(%2d, %2d) %7.2f\n", xy[i].x, xy[i].y, distance(&xy[i], &point));}fclose(fp);}输出：valid points:(49, 38) (59, 45) (53, 19) (34, 55) (23, 62) ( 2, 11) (14, 14) (55, 52) (62, 37) (46, 29) (19,57) (12, 18) ( 4, 7)total: 13coordinate: 0 0k = 5the nearest 5 points to (0, 0) are:( 4, 7) 8.06( 2, 11) 11.18(14, 14) 19.80(12, 18) 21.63(46, 29) 54.382015年复试上机题要求：从网页上下载input.dat 文件，里面是用二进制编写的，里面放了一堆int 型的数，每个数占4个字节，每次读取两个，这两个数构成一个坐标.规定处于第一象限的数是有效点(即x>0, y>0的坐标)，问这么多点中有效点有多少个？从键盘上输入k 和n ，从第一问中的有效点中找出距离小于n ，距离小于n 的点的个数要大于k，将它们以文本格式输出到文件中.程序：#include <math.h>#include <stdio.h>#include <stdlib.h>#include <string.h>#include <time.h>typedef struct{int x;int y;} Coordinate;Coordinate point;double distance(const Coordinate*a, const Coordinate*b){int val= (a->x-b->x) * (a->x-b->x) +(a->y-b->y) * (a->y-b->y);return sqrt((double) val);}void swap(Coordinate*a, Coordinate*b){Coordinate temp=*a;*a=*b;*b=temp;}int compare(const Coordinate*a, const Coordinate*b){double dista=distance(a, &point);double distb=distance(b, &point);if (dista<distb)return-1;if (dista>distb)return1;elsereturn0;}void selection_sort(Coordinate*ptr, int begin, int end){for (int i=begin; i<end; i++)for (int j=i+1; j<end; j++)if (compare(&ptr[i], &ptr[j]) >0)swap(&ptr[i], &ptr[j]);}void display(Coordinate*ptr, int count){for (int i=0; i<count; i++)printf("(%2d, %2d) ", ptr[i].x, ptr[i].y);printf("\n");}int main(void){FILE*fp;Coordinate*xy;int*num;int count, coord_count;fp=fopen("org.dat", "wb");num= (int*) malloc(sizeof(int) *128);srand(time(NULL));for (int i=0; i<128; i++)num[i] =rand() %128-64;fwrite(num, sizeof(int), 128, fp);free(num);fclose(fp);fp=fopen("org.dat", "rb");fseek(fp, 0, SEEK_END);count=ftell(fp) /sizeof(int);coord_count=count/2;num= (int*) malloc(sizeof(int) *count);xy= (Coordinate*) malloc(sizeof(Coordinate) *coord_count);输出：rewind(fp);fread(num, sizeof(int), count, fp); fclose(fp);int valid=0;fp=fopen("output.txt", "w");for (int i=0; i<count; i+=2){if (num[i] >0&&num[i+1] >0){xy[valid].x=num[i];xy[valid].y=num[i+1];valid++;}}printf("valid points:\n");display(xy, valid);printf("total: %d\n", valid);int k, n, pivot;printf("k = ");scanf("%d", &k);printf("n = ");scanf("%d", &n);pivot=0;while (pivot<valid){point.x=xy[pivot].x;point.y=xy[pivot].y;pivot++;selection_sort(xy, pivot, valid);printf("the points with a distance of less than %d to (%2d, %2d) are:\n", n, point.x, point.y);if (distance(&xy[pivot+k], &point) < (double)n){while (pivot<valid&&distance(&xy[pivot], &point) < (double)n){printf("(%2d, %2d) %7.2f\n", xy[pivot].x, xy[pivot].y, distance(&xy[pivot],&point));fprintf(fp, "(%2d, %2d) %7.2f\n", xy[pivot].x, xy[pivot].y, distance(&xy[pivot], &point));pivot++;}}}fclose(fp);}valid points:2016年复试上机题要求：文本文件input.txt 由若干英文单词和分隔符(空格，回车，换行)构成. 根据如下说明编写程序统计不同单词出现的次数(频度). 将统计结果按出现频度从高到低排序，并将出现频度大于5的单词及其频度输出到文件output.txt 中. 文件格式如图所示多个连续的分隔符被视为一个分隔符.大小写敏感. 每个单词的长度不超过20个字符.单词的数量未知. 如使用定义静态大数组的方式来统计，将被扣除5分.(11, 2) (26, 6) (24, 19) ( 8, 50) ( 2, 18) (60, 32) ( 4, 7) (50, 21) (21, 6) (34, 37) (30,40) ( 5, 16) (31, 59) (60, 34) (18, 32)total: 15k = 3n = 30the points with a distance of less than 30 to (11, 2) are:( 4, 7) 8.60(21, 6) 10.77( 5, 16) 15.23(26, 6) 15.52( 2, 18) 18.36(24, 19) 21.40the points with a distance of less than 30 to (18, 32) are:(30, 40) 14.42(34, 37) 16.76( 8, 50) 20.59(31, 59) 29.97the points with a distance of less than 30 to (50, 21) are:the points with a distance of less than 30 to (60, 32) are:the points with a distance of less than 30 to (60, 34) are:程序：#include <math.h>#include <stdio.h>#include <stdlib.h>#include <string.h>typedef struct Word{int freq;char str[20];} Word;int compare(const Word*lhs, const Word*rhs){if (lhs->freq>rhs->freq) return-1;else if (lhs->freq<rhs->freq) return1;else return0;}void swap(Word*e1, Word*e2){Word tmp=*e1;*e1=*e2;*e2=tmp;}void insertion_sort(Word*ptr, int count){for (int i=1; i<count; i++){int j=i;Word ref=ptr[i];while (j>0&&compare(&ptr[j-1], &ref) >0) {ptr[j] =ptr[j-1];j--;}ptr[j] =ref;}}int find(Word*words, int count, char*str){int i;for (i=0; i<count; i++)if (strcmp(words[i].str, str) ==0)break;return i;}int main(void){FILE*fpr=fopen("input.txt", "r");FILE*fpw=fopen("output.txt", "w");Word*words;char buf[20];int count=0;while (fscanf(fpr, "%s", buf) !=EOF)count++;words= (Word*) malloc(sizeof(Word)*count);rewind(fpr);count=0;while (fscanf(fpr, "%s", buf) !=EOF){int i=find(words, count, buf);if (i==count){strcpy(words[count].str, buf);words[count].freq=0;count++;}words[i].freq++;}insertion_sort(words, count);for (int i=0; i<count&&words[i].freq>5; i++) {fprintf(fpw, "%s %d\n", words[i].str, words[i].freq);printf("%s %d\n", words[i].str, words[i].freq);}free(words);fclose(fpr);fclose(fpw);}输出：that 13The 11It 9to 8we 8here 8a 7and 62017年复试上机题要求：已知：二进制数据文件data.bin中存放了若干个整数，请编写程序完成如下功能：编写程序读取所有数据.以每相邻两个整数为一对按顺序构成二维平面上的坐标点. 例如：有数据12，34，53，25，61，28，78等，则构成六个坐标点如下：(12, 34)、(34, 53)，(53, 25), (25, 61), (61, 28), (28,78)；以每个坐标点为圆心，以该点与其后面第一个点的欧氏距离为半径r. 计算每个圆包含的坐标点数. 计算最后一个点时以其和第一个点的欧氏距离为半径.例如：坐标点(12, 34)的圆半径$r=\sqrt{(12-34)^2+(34-53)^2}$是坐标点(12, 34)与(34, 53)的欧式距离.坐标点(28, 78)的圆半径$r=\sqrt{(28-12)^2+(78-34)^2}$是坐标点(28, 78)与(12, 34)的欧式距离.计算所有圆的点密度值，然后输出点密度值最大的5个坐标点以及相应圆中包含的点数和点密度值. 输出格式要求：坐标点包含点数点密度(x坐标，y坐标)(占5列，右对齐)(占7列，右对齐，保留2位小数)上述文字部分不需要显示.其中：圆的点密度为圆包含的点数除以圆面积，如果点在圆上，则也算圆包含该点，在计算点密度时，圆心也算一个点. 计算圆面积时$\pi=3.14$. 例如：坐标点(2, 1)，则该坐标点也属该坐标点的圆内的一个点.程序：#include <math.h>#include <stdio.h>#include <stdlib.h>#include <string.h>#include <time.h>#define PI 3.14typedef struct{int x;int y;} Coordinate;typedef struct{double r;Coordinate point;int points_num;double density;} Circle;static int compare(const Circle*a, const Circle*b){if (a->density>b->density) return-1;else if (a->density<b->density) return1;else return0;。
苏州大学考研真题数学分析2005(含答案)
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(NEW)苏州大学医学部《626预防综合》历年考研真题汇编
2015年苏州大学626预防综合考研真题（B 卷）
2014年苏州大学626预防综合考研真题（B 卷）
2013年苏州大学626预防综合考研真题（B 卷）
2012年苏州大学626预防综合考研真题（A 卷）（部分）
2012年苏州大学626预防综合考研真题（回 忆版）
1年苏州大学626预防综合考研真题（A 卷）
2010年苏州大学626预防综合考研真题（A 卷）
2009年苏州大学626预防综合考研真题（A 卷）
2008年苏州大学626预防综合考研真题（A 卷）
2007年苏州大学626预防综合考研真题（A 卷）
目 录
2015年苏州大学626预防综合考研真题（B卷） 2014年苏州大学626预防综合考研真题（B卷） 2013年苏州大学626预防综合考研真题（B卷） 2012年苏州大学626预防综合考研真题（A卷）（部分） 2012年苏州大学626预防综合考研真题（回忆版） 2011年苏州大学626预防综合考研真题（A卷） 2010年苏州大学626预防综合考研真题（A卷） 2009年苏州大学626预防综合考研真题（A卷） 2008年苏州大学626预防综合考研真题（A卷） 2007年苏州大学626预防综合考研真题（A卷） 2005年苏州大学359预防综合考研真题（A卷）
2005年苏州大学359预防综合考研真题（A 卷）
2005-考研真题
2005年全国硕士研究生入学统一考试英语试题Section I Use of EnglishDirections:Read the following text. Choose the best word(s) for each numbered blank and mark [A], [B], [C] or [D] on ANSWER SHEET 1(10 point s)The human nose is an underrated tool. Humans are often thought to be insensitive smellers compared with animals, （1）_____ this is l argely because, （2）______ animals, we stand upright. This means t hat our noses are （3）______ to perceiving those smells which float through the air, （4）______ the majority of smells which stick to surf aces. In fact, （5）______ , we are extremely sensitive to smells, （6 ______ we do not generally realize it. Our noses are capable of （7）_ _____ human smells even when these are （8）______ to far below o ne part in one million.Strangely, some people find that they can smell one type of flowe r but not another, （9）______ others are sensitive to the smells of b oth flowers. This may be because some people do not have the genes necessary to generate （10）______ smell receptors in the nose. The se receptors are the cells which sense smells and send （11）______ to the brain. However, it has been found that even people insensitive to a certain smell （12）______ can suddenly become sensitive to it when （13）______ to it often enough.The explanation for insensitivity to smell seems to be that the bra in finds it （14）______ to keep all smell receptors working all the ti me but can （15）______ new receptors if necessary. This may （16_ _____ explain why we are not usually sensitive to our own smells—w e simply do not need to be. We are not （17）______ of the usual s mell of our own house, but we （18）______ new smells when we visi t someone else’s. The brain finds it best to keep smell receptors （19）______ for unfamiliar and emergency signals （20）______ the smell of smoke, which might indicate the danger of fire.1. [A] although[B] as[C] but[D] while2. [A] above[B] unlike[C] excluding[D] besides3.[A] limited[B] committed[C] dedicated[D] confined4.[A] catching[B] ignoring[C] missing[D] tracking5.[A] anyway[B] though[C] instead[D] therefore6.[A] even if[B] if only[C] only if[D] as if7.[A] distinguishing[B] discovering[C] determining[D] detecting8.[A] diluted[B] dissolved[C] dispersed[D] diffused9. [A] when[B] since[C] for[D] whereas10. [A] unusual[B] particular[C] unique[D] typical11. [A] signs[B] stimuli[C] messages[D] impulses12. [A] at first[B] at all[C] at large[D] at times13. [A] subjected[B] left[C] drawn[D] exposed14. [A] ineffective[B] incompetent[C] inefficient[D] insufficient15. [A] introduce[B] summon[C] trigger[D] create16. [A] still[B] also[C] otherwise[D] nevertheless17. [A] sure[B] sick[C] aware[D] tired18. [A] tolerate[B] repel[C] neglect[D] notice19. [A] available[B] reliable[C] identifiable[D] suitable20. [A] similar to[B] such as[C] along with[D] aside fromSection II Reading ComprehensionPart ADirections:Read the following four texts. Answer the questions below each text b y choosing [A], [B], [C] or D. Mark your answers on ANSWER SHEE T 1(40 points)Text 1Everybody loves a fat pay rise. Yet pleasure at your own can vani sh if you learn that a colleague has been given a bigger one. Indeed,if he has a reputation for slacking, you might even be outraged. Suc h behaviour is regarded as “all too human,” with the underlying assu mption that other animals would not be capable of this finely develop ed sense of grievance. But a study by Sarah Brosnan and Frans de W aal of Emory University in Atlanta, Georgia, which has just been publi shed in Nature, suggests that it is all too monkey, as well.The researchers studied the behaviour of female brown capuchin monkeys. They look cute. They are good-natured, co-operative creatur es, and they share their food readily. Above all, like their female hum an counterparts, they tend to pay much closer attention to the value of “goods and services” than males.Such characteristics make them perfect candidates for Dr. Brosna n’s and Dr. de Waal’s study. The researchers spent two years teaching their monkeys to exchange tokens for food. Normally, the monkeys were happy enough to exchange pieces of rock for slices of cucumber. However, when two monkeys were placed in separate but adjoining c hambers, so that each could observe what the other was getting in re turn for its rock, their behaviour became markedly different.In the world of capuchins, grapes are luxury goods (and much pr eferable to cucumbers). So when one monkey was handed a grape in exchange for her token, the second was reluctant to hand hers over for a mere piece of cucumber. And if one received a grape without ha ving to provide her token in exchange at all, the other either tossed her own token at the researcher or out of the chamber, or refused to accept the slice of cucumber. Indeed, the mere presence of a grape i n the other chamber (without an actual monkey to eat it) was enoug h to induce resentment in a female capuchin.The researchers suggest that capuchin monkeys, like humans, are guided by social emotions. In the wild, they are a co-operative, grou p-living species. Such co-operation is likely to be stable only when ea ch animal feels it is not being cheated. Feelings of righteous indignati on, it seems, are not the preserve of people alone. Refusing a lesser reward completely makes these feelings abundantly clear to other me mbers of the group. However, whether such a sense of fairness evolv ed independently in capuchins and humans, or whether it stems from the common ancestor that the species had 35 million years ago, is, as yet, an unanswered question.21. In the opening paragraph, the author introduces his topic by ________.[A] posing a contrast[B] justifying an assumption[C] making a comparison[D] explaining a phenomenon22. Th e statement “it is all too monkey” (Last line, Paragraph l) implies that ________.[A] monkeys are also outraged by slack rivals[B] resenting unfairness is also monkeys’ nature[C] monkeys, like humans, tend to be jealous of each other[D] no animals other than monkeys can develop such emotions23. Female capuchin monkeys were chosen for the research most probably because they are ________.[A] more inclined to weigh what they get[B] attentive to researchers’ instructions[C] nice in both appearance and temperament[D] more generous than their male companions24. Dr. Brosnan and Dr. de Waal have eventually found in their studythat the monkeys ________.[A] prefer grapes to cucumbers[B] can be taught to exchange things[C] will not be co-operative if feeling cheated[D] are unhappy when separated from others25. What can we infer from the last paragraph?[A] Monkeys can be trained to develop social emotions.[B] Human indignation evolved from an uncertain source.[C] Animals usually show their feelings openly as humans do.[D] Cooperation among monkeys remains stable only in the wild. Text 2Do you remember all those years when scientists argued that smo king would kill us but the doubters insisted that we didn’t know for s ure? That the evidence was inconclusive, the science uncertain? That the antismoking lobby was out to destroy our way of life and the gov ernment should stay out of the way? Lots of Americans bought that n onsense, and over three decades, some 10 million smokers went to e arly graves.There are upsetting parallels today, as scientists in one wave after another try to awaken us to the growing threat of global warming. T he latest was a panel from the National Academy of Sciences, enliste d by the White House, to tell us that the Earth’s atmosphere is defini tely warming and that the problem is largely man-made. The clear m essage is that we should get moving to protect ourselves. The preside nt of the National Academy, Bruce Alberts, added this key point in th e preface to the panel’s report: “Science n ever has all the answers. B ut science does provide us with the best available guide to the future, and it is critical that our nation and the world base important policie s on the best judgments that science can provide concerning the futu re consequences of present actions.”Just as on smoking, voices now come from many quarters insistin g that the science about global warming is incomplete, that it’s OK to keep pouring fumes into the air until we know for sure. This is a da ngerous game: by the time 100 percent of the evidence is in, it may be too late. With the risks obvious and growing, a prudent people wo uld take out an insurance policy now.Fortunately, the White House is starting to pay attention. But it’s obvious that a majority of the president’s advisers still don’t take glob al warming seriously. Instead of a plan of action, they continue to pre ss for more research -- a classic case of “paralysis by analysis.”To serve as responsible stewards of the planet, we must press for ward on deeper atmospheric and oceanic research. But research alone is inadequate. If the Administration won’t take the legislative initiativ e, Congress should help to begin fashioning conservation measures. A bill by Democratic Senator Robert Byrd of West Virginia, which would offer financial incentives for private industry, is a promising start. Ma ny see that the country is getting ready to build lots of new power pl ants to meet our energy needs. If we are ever going to protect the a tmosphere, it is crucial that those new plants be environmentally sou nd.26. An argument made by supporters of smoking was that ________.[A] there was no scientific evidence of the correlation between smoking and death[B] the number of early deaths of smokers in the past decades was insignificant[C] people had the freedom to choose their own way of life[D] antismoking people were usually talking nonsense27. According to Bruce Alberts, science can serve as ________.[A] a protector[B] a judge[C] a critic[D] a guide28. What does the author mean by “paralysis by analysis” (Last line,Paragraph 4)?[A] Endless studies kill action.[B] Careful investigation reveals truth.[C] Prudent planning hinders progress.[D] Extensive research helps decision-making.29. According to the author, what should the Administration do aboutglobal warming?[A] Offer aid to build cleaner power plants.[B] Raise public awareness of conservation.[C] Press for further scientific research.[D] Take some legislative measures.30. The author associates the issue of global warming with that of smoking because ________.[A] they both suffered from the government’s negligence[B] a lesson from the latter is applicable to the former[C] the outcome of the latter aggravates the former[D] both of them have turned from bad to worseText 3Of all the components of a good night’s sleep, dreams seem to be least within our control. In dreams, a window opens into a world wh ere logic is suspended and dead people speak. A century ago, Freud f ormulated his revolutionary theory that dreams were the disguised sh adows of our unconscious desires and fears; by the late 1970s, neuro logists had switched to thinking of them as just “mental noise” -- the random byproducts of the neural-repair work that goes on during sle ep. Now researchers suspect that dreams are part of the mind’s emot ional thermostat, regulating moods while the brain is “off-line.” And o ne leading authority says that these intensely powerful mental events can be not only harnessed but actually brought under conscious control, to help us sleep and feel better, “It’s your dream,” says Rosalind Cartwright, chair of psychology at Chicago’s Medical Center. “If you do n’t like it, change it.”Evidence from brain imaging supports this view. The brain is as a ctive during REM (rapid eye movement) sleep -- when most vivid dre ams occur -- as it is when fully awake, says Dr, Eric Nofzinger at the University of Pittsburgh. But not all parts of the brain are equally inv olved; the limbic system (the “emotional brain”) is especially active, while the prefrontal cortex (the center of intellect and reasoning) is re latively quiet. “We wake up from dreams happy or depressed, and tho se feelings can stay with us all day.” says Stanford sleep researcher D r. William Dement.The link between dreams and emotions shows up among the patie nts in Cartwright’s clinic. Most people seem to have more bad dreams early in the night, progressing toward happier ones before awakening, suggesting that they are working through negative feelings generated during the day. Because our conscious mind is occupied with daily lif e we don’t always think about the emotional significance of the day’s events -- until, it appears, we begin to dream.And this process need not be left to the unconscious. Cartwright b elieves one can exercise conscious control over recurring bad dreams. As soon as you awaken, identify what is upsetting about the dream. Visualize how you would like it to end instead; the next time it occur s, try to wake up just enough to control its course. With much practic e people can learn to, literally, do it in their sleep.At the end of the day, there’s probably little reason to pay attenti on to our dreams at all unless they keep us from sleeping or “we wa ke up in a panic,” Cartwright says. Terrorism, economic unc ertainties and general feelings of insecurity have increased people’s anxiety. Tho se suffering from persistent nightmares should seek help from a thera pist. For the rest of us, the brain has its ways of working through ba d feelings. Sleep -- or rather dream -- on it and you’ll feel better in t he morning.31. Researchers have come to believe that dreams ________.[A] can be modified in their courses[B] are susceptible to emotional changes[C] reflect our innermost desires and fears[D] are a random outcome of neural repairs32. By referring to the limbic system, the author intends to show ________.[A] its function in our dreams[B] the mechanism of REM sleep[C] the relation of dreams to emotions[D] its difference from the prefrontal cortex33. The negative feelings generated during the day tend to ________.[A] aggravate in our unconscious mind[B] develop into happy dreams[C] persist till the time we fall asleep[D] show up in dreams early at night34. Cartwright seems to suggest that ________.[A] waking up in time is essential to the ridding of bad dreams[B] visualizing bad dreams helps bring them under control[C] dreams should be left to their natural progression[D] dreaming may not entirely belong to the unconscious35. What advice might Cartwright give to those who sometimes havebad dreams?[A] Lead your life as usual.[B] Seek professional help.[C] Exercise conscious control.[D] Avoid anxiety in the daytime.Text 4Americans no longer expect public figures, whether in speech or i n writing, to command the English language with skill and gift. Nor d o they aspire to such command themselves. In his latest book, Doing Our Own Thing: The Degradation of Language and Music and Why We Should, Like, Care, John McWhorter, a linguist and controversialist of mixed liberal and conservative views, sees the triumph of 1960s cou nter-culture as responsible for the decline of formal English.Blaming the permissive 1960s is nothing new, but this is not yet another criticism a gainst the decline in education. Mr. McWhorter’s aca demic speciality is language history and change, and he sees the grad ual disappearance of “whom,” for example, to be natural and no moreregrettable than the loss of the case-endings of Old English.But t he cult of the authentic and the personal, “doing our own thi ng,” has spelt the death of formal speech, writing, poetry and music. While even the modestly educated sought an elevated tone when they put pen to paper before the 1960s, even the most well regarded wri ting since then has sought to capture spoken English on the page. Eq ually, in poetry, the highly personal, performative genre is the only fo rm that could claim real liveliness. In both oral and written English, t alking is triumphing over speaking, spontaneity over craft.Illustrated with an entertaining array of examples from both high and low culture, the trend that Mr. McWhorter documents is unmistak able. But it is less clear, to take the question of his subtitle, why we should, like, care. As a linguist, he acknowledges that all varieties of human language, including non-standard ones like Black English, can be powerfully expressive -- there exists no language or dialect in the world that cannot convey complex ideas. He is not arguing, as many do, that we can no longer think straight because we do not talk prop er.Russians have a deep love for their own language and carry large chunks of memorized poetry in their heads, while Italian politicians t end to elaborate speech that would seem old-fashioned to most Englis h-speakers. Mr. McWhorter acknowledges that formal language is not strictly necessary, and proposes no radical education reforms -- he is really grieving over the loss of something beautiful more than useful. We now take our English “on paper plates instead of china.” A shame, perhaps, but probably an inevitable one.36. According to McWhorter, the decline of formal English ________.[A] is inevitable in radical education reforms[B] is but all too natural in language development[C] has caused the controversy over the counter-culture[D] brought about changes in public attitudes in the 1960s37. The word “talking” (Line 6, Paragraph 3) denotes ________.[A] modesty[B] personality[C] liveliness[D] informality38. To which of the following statements would McWhorter most likelyagree?[A] Logical thinking is not necessarily related to the way we talk.[B] Black English can be more expressive than standard English.[C] Non-standard varieties of human language are just as entertaining.[D] Of all the varieties, standard English can best convey complex ideas.39. The description of Russians’ love of memorizing poetry shows theauthor’s ________.[A] interest in their language[B] appreciation of their efforts[C] admiration for their memory[D] contempt for their old-fashionedness40. According to the last paragraph, “paper plates” is to “china” as ________.[A] “temporary” is to “permanent”[B] “radical” is to “conservative”[C] “functional” is to “artistic”[D] “humble” is to “noble”Part BDirections:In the following text, some sentences have been removed. For Questi ons 41-45, choose the most suitable one from the list A-G to fit into each of the numbered blanks. There are two extra choices, which do not fit in any of the gaps. Mark your answers on ANSWER SHEET 1.(10 points)Canada’s premiers (the leaders of provincial governments), if they have any breath left after complaining about Ottawa at their late Jul y annual meeting, might spare a moment to do something, together, to reduce health-care costs.They’re all groaning about soaring health budgets, the fastest-gro wing component of which are pharmaceutical costs.41. ________What to do? Both the Romanow commission and the Kirby commit tee on health care -- to say nothing of reports from other experts -- recommended the creation of a national drug agency. Instead of each province having its own list of approved drugs, bureaucracy, procedu res and limited bargaining power, all would pool resources, work with Ottawa, and create a national institution.42. ________But “national” doesn’t have to mean that. “National” could mean i nterprovincial -- provinces combining efforts to create one body.Either way, one benefit of a “national” organization would be to n egotiate better prices, if possible, with drug manufacturers. Instead of having one province -- or a series of hospitals within a province -- n egotiate a price for a given drug on the provincial list, the national ag ency would negotiate on behalf of all provinces.Rather than, say, Quebec, negotiating on behalf of seven million p eople, the national agency would negotiate on behalf of 31 million pe ople. Basic economics suggests the greater the potential consumers, t he higher the likelihood of a better price.43. ________A small step has been taken in the direction of a national agency with the creation of the Canadian Co-ordinating Office for Health Tech nology Assessment, funded by Ottawa and the provinces. Under it, a Common Drug Review recommends to provincial lists which new drug s should be included. Predictably, and regrettably, Quebec refused to j oin.A few premiers are suspicious of any federal-provincial deal-makin g. They (particularly Quebec and Alberta) just want Ottawa to fork ov er additional billions with few, if any, strings attached. That’s one reas on why the idea of a national list hasn’t gone anywhere, while drug c osts keep rising fast.44. ________Premiers love to quote Mr. Romanow’s report selectively, especially the parts about more federal money. Perhaps they should read what he had to say about drugs: “A national drug agency would provide go vernments more influence on pharmaceutical companies in order to co nstrain the ever-increasing cost of drugs.”45. ________So when the premiers gather in Niagara Falls to assemble their us ual complaint list, they should also get cracking about something in their jurisdiction that would help their budgets and patients.[A] Quebec’s resistance to a national agency is provincialist ideology.One of the first advocates for a national list was a researcher at L aval University. Quebec’s Drug Insurance Fund has seen its costs skyrocket with annual increases from 14.3 per cent to 26.8 per ce nt![B] Or they could read Mr. Kirby’s report: “the substantial buying power of such an agency would strengthen the public prescription-drug insurance plans to negotiate the lowest possible purchase pricesfrom drug companies.”[C] What doe s “national” mean? Roy Romanow and Senator Michael Kirby recommended a federal-provincial body much like the recently created National Health Council.[D] The problem is simple and stark: health-care costs have been, are, and will continue to increase faster than government revenues.[E] According to the Canadian Institute for Health Information, prescription drug costs have risen since 1997 at twice the rate of overall health-care spending. Part of the increase comes from drugs being used to replace other kinds of treatments. Part of it arises fromnew drugs costing more than older kinds. Part of it is higher pric es.[F] So, if the provinces want to run the health-care show, they should prove they can run it, starting with an interprovincial health listthat would end duplication, save administrative costs, prevent one province from being played off against another, and bargain forbetter drug prices.[G] Of course, the pharmaceutical companies will scream. They like divided buyers; they can lobby better that way. They can use the t hreat of removing jobs from one province to another. They can ho pe that, if one province includes a drug on its list, the pressure w ill cause others to include it on theirs. They wouldn’t like a nation al agency, but self-interest would lead them to deal with it.Part CDirections:Read the following text carefully and then translate the underlined se gments into Chinese. Your translation should be written clearly on AN SWER SHEET 2. (10 points)It is not easy to talk about the role of the mass media in this overwhelmingly significant phase in European history. History and news become confused, and one’s impressions tend to be a mixture of skep ticism and optimism. 46) Television is one of the means by which the se feelings are created and conveyed -- and perhaps never before ha s it served so much to connect different peoples and nations as in th e recent events in Europe. The Europe that is now forming cannot be anything other than its peoples, their cultures and national identities. With this in mind we can begin to analyze the European television sc ene. 47) In Europe, as elsewhere, multi-media groups have been incr easingly successful: groups which bring together television, radio, new spapers, magazines and publishing houses that work in relation to on e another. One Italian example would be the Berlusconi group, while abroad Maxwell and Murdoch come to mind.Clearly, only the biggest and most flexible television companies ar e going to be able to compete in such a rich and hotly-contested mar ket. 48) This alone demonstrates that the television business is not a n easy world to survive in, a fact underlined by statistics that show t hat out of eighty European television networks, no less than 50% too k a loss in 1989.Moreover, the integration of the European community will oblige t elevision companies to cooperate more closely in terms of both produ ction and distribution.49) Creating a “European identity” that respects the different cult ures and traditions which go to make up the connecting fabric of the Old Continent is no easy task and demands a strategic choice -- that of producing programs in Europe for Europe. This entails reducing our dependence on the North American market, whose programs relate t o experiences and cultural traditions which are different from our ow n.In order to achieve these objectives, we must concentrate more o n co-productions, the exchange of news, documentary services and tr aining. This also involves the agreements between European countries for the creation of a European bank for Television Production which, on the model of the European Investments Bank, will handle the fina nces necessary for production costs. 50) In dealing with a challenge o n such a scale, it is no exaggeration to say “United we stand, divided we fall”-- and if I had to choose a slogan it would be “Unity in our diversity.” A unity of objectives that nonetheless respect the varied pe culiarities of each country.Section III WritingPart A51. Directions:Two months ago you got a job as an editor for the magazine Desi gns & Fashions. But now you find that the work is not what you expe cted. You decide to quit. Write a letter to your boss, Mr. Wang, telling him your decision, stating your reason (s), and making an apology.Write your letter with no less than 100 words. Write it neatly on ANSWER SHEET 2.Do not sign your own name at the end of the letter; use “Li Min g” instead.You do not need to write the address. (10 points)Part B52. Directions:Write an essay of 160-200 words based on the following drawing. In your essay, you should first describe the drawing, then interpret its m eaning, and give your comment on it.You should write neatly on ANSWER SHEET 2. (20 points)。
历年苏州大学考研真题试卷与答案汇总-苏州大学考研真题试卷-苏大考研真题答案-苏大考研真题哪里找？
历年苏州大学考研真题试卷与答案汇总-苏大考研真题哪里找？东吴苏大考研网（苏州大学考研在线咨询入口）汇集了苏州大学各专业历年考研真题试卷（原版），同时与苏州大学专业课成绩前三名的各专业硕士研究生合作编写了配套的真题答案解析，答案部分包括了（解题思路、答案详解）两方面内容。
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具体请点击进入【历年苏州大学考研真题答案下载】[东吴苏大考研网] 2019苏大考研817国际法学考试解读与真题答案详解[东吴苏大考研网] 苏州大学347心理学专业综合考研真题试卷（2011-2018）[东吴苏大考研网] 苏州大学333教育综合历年考研真题试卷（2007-2018年）[东吴苏大考研网] 2015年苏州大学335出版综合素质与能力考研真题与答案[东吴苏大考研网] 苏州大学889英语教学论考研真题（2017-2018年，回忆版）[东吴苏大考研网] 苏州大学808管理学考研真题试卷（2000-2015）[东吴苏大考研网] 苏州大学884高分子材料化学考研真题与答案（2015-2018年）[东吴苏大考研网] 苏州大学823艺术史考研真题试卷(2004-2015年)[东吴苏大考研网] 苏州大学822新闻传播专业基础考研真题试卷（2000-2018）[东吴苏大考研网] 苏州大学633教育学专业基础综合真题答案（2007-2018）[东吴苏大考研网] 苏州大学832普通物理考研真题与答案（2003-2015年）[东吴苏大考研网] 苏州大学807微观与宏观经济学考研真题试卷（1999-2017，不含06）[东吴苏大考研网] 苏大637药学基础综合（1）考研真题试卷（09-17）[东吴苏大考研网] 苏大872数据结构与操作系统考研真题答案（真题99-17，答案05-17）[东吴苏大考研网] 2019苏州大学考研889英语教学论复习全析（含真题，共三册）[东吴苏大考研网] 2013-2015年苏州大学612绘画基础（素描命题画）考研真题试卷[东吴苏大考研网] 2019苏大考研398法硕联考专业基础课（非法学）强化冲刺题库（共两册）[东吴苏大考研网] 2019苏大考研498法硕联考综合课（非法学）强化冲刺题库（共三册）[东吴苏大考研网] 2019苏大878电磁学考研强化冲刺题库[东吴苏大考研网] 2019苏大842自动控制原理考研强化冲刺题库（共两册）[东吴苏大考研网] 2019苏大电子技术基础考研强化冲刺题库（共两册）【历年苏州大学考研真题答案下载】[东吴苏大考研网] 2019苏大理论力学考研强化冲刺题库[东吴苏大考研网] 2019苏大837信号系统与数字逻辑考研强化冲刺题库（共三册）[东吴苏大考研网] 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2019苏大考研613社会学原理强化冲刺题库[东吴苏大考研网] 2019苏大戏剧与影视学考研核心考点与名校真题详解（共两册）[东吴苏大考研网] 2019苏州大学出版专硕考研强化冲刺题库【共4册】[东吴苏大考研网] 2019苏大822新闻传播专业基础考研强化冲刺题库（共两册）[东吴苏大考研网] 2019苏大334新闻与传播专业综合能力考研强化冲刺题库[东吴苏大考研网] 2019苏大440新闻与传播专业基础考研强化冲刺题库（共两册）[东吴苏大考研网] 2019苏大820评论写作（1）考研强化冲刺题库[东吴苏大考研网] 2019苏大625汉语言文字基础考研冲刺题库【共5册】[东吴苏大考研网] 2019苏大445汉语国际教育基础考研强化冲刺题库[东吴苏大考研网] 2019苏大354汉语基础考研强化冲刺题库[东吴苏大考研网] 2019苏州大学心理学考研强化冲刺题库（共七册）【历年苏州大学考研真题答案下载】[东吴苏大考研网] 2019苏大802中西政治思想史考研强化冲刺题库[东吴苏大考研网] 2019苏大801中西哲学史考研强化冲刺题库（共两册）[东吴苏大考研网] 2019苏大654公共管理基础理论考研强化冲刺题库（共两册）[东吴苏大考研网] 2019苏大652政治学原理考研强化冲刺题库（共两册）[东吴苏大考研网] 2019苏大651哲学概论考研强化冲刺题库[东吴苏大考研网] 2019苏大633教育学专业基础综合考研冲刺题库【共六册】[东吴苏大考研网] 2019苏大665中外音乐史考研复习全析（含历年真题）[东吴苏大考研网] 2019苏州大学666生物化学（农）考研复习全析（含历年真题，共两册）[东吴苏大考研网] 2019苏大842自动控制原理考研复习全析（含历年真题）[东吴苏大考研网] 2019苏大841电子技术基础（机电）考研复习全析（含历年真题）【共两册】[东吴苏大考研网] 2019苏大839管理信息系统与数据结构考研复习全析（含历年真题，共两册）[东吴苏大考研网] 2019苏大850高等数学基础考研复习全析（含历年真题，共两册）[东吴苏大考研网] 2019苏大627生物化学考研复习全析（含历年真题，共两册）[东吴苏大考研网] 2019苏大862材料科学基础考研复习全析（含历年真题）[东吴苏大考研网] 2019苏大858材料学（F）考研复习全析（共两册，含历年真题）[东吴苏大考研网] 2019苏大857细胞生物学（F）考研复习全析（含历年真题）[东吴苏大考研网] 2019苏大631生物化学（F）考研复习全析（含历年真题，共两册）[东吴苏大考研网] 2019苏大856物理化学（F）考研复习全析（含历年真题）[东吴苏大考研网] 2019苏大629无机化学（F）考研复习全析（含历年真题，共三册）[东吴苏大考研网] 2019苏大628有机化学（F）考研复习全析（含历年真题，共两册）[东吴苏大考研网] 2019苏大602高等数学（F）考研复习全析（含历年真题，共两册）[东吴苏大考研网] 2019苏大838信号与线性系统考研复习全析（含历年真题）[东吴苏大考研网] 2019苏大601高等数学考研复习全析（含历年真题，共两册）[东吴苏大考研网] 2019苏大625汉语言文字基础考研复习全析（含历年真题，共六册）[东吴苏大考研网] 2019苏大847金融学概论考研复习全析（含历年真题，共两册）[东吴苏大考研网] 2019苏州大学432统计学考研复习全析（含历年真题）[东吴苏大考研网] 2019苏大831高等代数考研复习全析（含真题答案，共三册）[东吴苏大考研网] 2019苏大618数学分析考研复习全析（含真题答案，共四册）[东吴苏大考研网] 2019苏大616基础俄语考研复习全析（含真题答案，共五册）[东吴苏大考研网] 2019苏大考研243日语强化冲刺题库[东吴苏大考研网] 2019苏大考研241英语强化冲刺题库[东吴苏大考研网] 2019苏大考研245德语强化冲刺题库[东吴苏大考研网] 2019苏大241英语考研复习全析（含历年真题，共五册）[东吴苏大考研网] 2019苏大826社会保障学考研复习全析（含历年真题）[东吴苏大考研网] 2019苏大658公共管理学考研复习全析（含历年真题，共三册）[东吴苏大考研网] 2019苏大660世界史专业基础综合考研复习全析（含历年真题，共六册）【历年苏州大学考研真题答案下载】[东吴苏大考研网] 2019苏大824社会研究方法考研复习全析（含历年真题，共三册）[东吴苏大考研网] 2019苏大613社会学原理考研复习全析（含历年真题，共三册）[东吴苏大考研网] 2019苏大437社会工作实务考研复习全析（含历年真题，共四册）[东吴苏大考研网] 2019苏大331社会工作原理考研复习全析（含历年真题，共三册）[东吴苏大考研网] 2019苏大337设计基础考研复习全析（含历年真题）[东吴苏大考研网] 2019苏大875设计艺术史考研复习全析（含历年真题）[东吴苏大考研网] 2019苏大346体育综合考研强化冲刺题库（共三册）[东吴苏大考研网] 2019苏大考研《经济法》强化冲刺题库[东吴苏大考研网] 2019苏大817国际法学考研复习全析（含真题答案，共四册）[东吴苏大考研网] 2019苏大816环境与资源保护法学考研复习全析（含历年真题）[东吴苏大考研网] 2019苏大815经济法学考研复习全析（含历年真题，共三册）[东吴苏大考研网] 2019苏大813民商法学考研复习全析（含历年真题，共三册）[东吴苏大考研网] 苏大《行政法与行政诉讼法》考试重难点与历年考研真题[东吴苏大考研网] 2019苏大809法学理论考研复习全析（含历年真题）[东吴苏大考研网] 2019苏大852土地资源管理考研复习全析（含历年真题）[东吴苏大考研网] 2019苏大805行政管理学考研复习全析（含真题答案，共三册）[东吴苏大考研网] 2019苏大804管理学基础考研复习全析（含历年真题，共两册）[东吴苏大考研网] 2019苏大802中西政治思想史考研复习全析（含历年真题）[东吴苏大考研网] 2019苏大654公共管理基础理论考研复习全析（共两册，含真题答案）[东吴苏大考研网] 2019苏州大学652政治学原理考研复习全析（含真题，共两册）[东吴苏大考研网] 2019苏州大学333教育综合考研强化冲刺题库[东吴苏大考研网] 2019苏州大学333教育综合考研复习全析（含真题答案，共八册）[东吴苏大考研网] 2019苏大637药学基础综合（1）考研复习全析[东吴苏大考研网] 2019苏大623生物综合考研复习全析（含历年真题）[东吴苏大考研网] 2019苏大854电子技术基础（城市轨道）考研复习全析（含历年真题）[东吴苏大考研网] 2019苏大818理论力学（城市轨道）考研复习全析（含历年真题）[东吴苏大考研网] 2019苏大832普通物理考研复习全析（含历年真题答案，共三册）[东吴苏大考研网] 2019苏大659中国史专业基础综合考研复习全析（含历年真题答案）[东吴苏大考研网] 2019苏大872数据结构与操作系统考研复习全析（含历年真题答案，共三册）[东吴苏大考研网] 2019苏大837信号系统与数字逻辑考研复习全析（含历年真题，共三册）[东吴苏大考研网] 2019苏大848植物生理与生物化学考研复习全析（含历年真题）[东吴苏大考研网] 2019苏大840理论力学（机电）考研复习全析（含历年真题）[东吴苏大考研网] 2019苏大346体育综合考研复习全析（含历年真题答案，共五册）[东吴苏大考研网] 2019苏大656体育学专业基础综合考研复习全析（含历年真题，共三册）[东吴苏大考研网] 2019苏州大学867化工原理考研复习全析（含历年真题）【历年苏州大学考研真题答案下载】[东吴苏大考研网] 2019苏大835有机化学B考研复习全析（含历年真题）[东吴苏大考研网] 2019苏大884高分子材料化学考研复习全析（含真题答案，共三册）[东吴苏大考研网] 2019苏大843材料科学基础考研复习全析（含历年真题）[东吴苏大考研网] 2019苏大861高分子化学考研复习全析（含历年真题答案，共两册）[东吴苏大考研网] 2019苏大853物理化学考研复习全析（含历年真题）[东吴苏大考研网] 2019苏大834分析化学考研复习全析（含历年真题）[东吴苏大考研网] 2019苏大635有机化学A考研复习全析（含历年真题）[东吴苏大考研网] 2019苏大620无机化学考研复习全析（含历年真题，共三册）[东吴苏大考研网] 2019苏大307中医综合考研复习全析（共五册）[东吴苏大考研网] 2019苏大645西医医学综合考研复习全析（共五册）[东吴苏大考研网] 2019苏大653马克思主义基本原理概论考研复习全析（含历年真题）[东吴苏大考研网] 2019苏大803思想政治教育学考研复习全析（含历年真题，共三册）[东吴苏大考研网] 2019苏大634心理学专业基础综合考研复习全析（含真题，共10册）[东吴苏大考研网] 2019苏大347心理学专业综合考研复习全析（含真题答案）（共12册）[东吴苏大考研网] 2019苏大考研397法硕联考专业基础（法学）强化冲刺题库（共两册）[东吴苏大考研网] 2019苏大考研497法硕联考综合课（法学）强化冲刺题库（共三册）[东吴苏大考研网] 2019苏大中国法制史考研强化冲刺题库（张晋藩版）[东吴苏大考研网] 2019苏大考研《刑法》强化冲刺题库[东吴苏大考研网] 2019苏大考研《民法》强化冲刺题库[东吴苏大考研网] 2019苏大考研《行政法与行政诉讼法》强化冲刺题库[东吴苏大考研网] 2019苏大考研《宪法》强化冲刺题库[东吴苏大考研网] 2019苏大考研《法理学》强化冲刺题库[东吴苏大考研网] 2019苏大考研808管理学强化冲刺题库[东吴苏大考研网] 2019苏州大学微观经济学考研强化冲刺题库[东吴苏大考研网] 2019苏州大学宏观经济学考研强化冲刺题库[东吴苏大考研网] 2019苏大801中西哲学史考研复习全析（含历年真题，共三册）[东吴苏大考研网] 2019苏州大学651哲学概论考研复习全析（含历年真题，共三册）[东吴苏大考研网] 2019苏大433税务专业基础考研复习全析（含真题与答案）[东吴苏大考研网] 2019苏州大学431金融学综合考研复习全析（含历年真题答案，共六册）[东吴苏大考研网] 2019苏大398法硕联考基础课(非法学)考研复习全析（含真题答案，共三册）[东吴苏大考研网] 2019苏大397法硕联考专业基础课(法学)考研复习全析【含真题答案，共三册】东吴苏大考研网（苏州大学考研在线咨询入口）【历年苏州大学考研真题答案下载】[东吴苏大考研网] 2019苏大497法硕联考综合课(法学)考研复习全析（含真题答案，共四册）[东吴苏大考研网] 2019苏大498法硕联考综合课(非法学)考研复习全析（含真题答案，共四册）[东吴苏大考研网] 2019苏大考研《文体与翻译》复习全析（刘宓庆，含历年真题答案）[东吴苏大考研网] 2019苏大考研661教育经济学强化冲刺题库[东吴苏大考研网] 2019苏大考研819教育管理学强化冲刺题库（共两册）[东吴苏大考研网] 2019苏大考研819教育管理学复习全析（含历年真题）[东吴苏大考研网] 2019苏大考研661教育经济学复习全析（含历年真题）[东吴苏大考研网] 2019苏大考研845细胞生物学强化冲刺题库[东吴苏大考研网] 2019苏大考研845细胞生物学复习全析（含历年真题）[东吴苏大考研网] 2019苏大考研621生物化学复习全析（含历年真题，共三册）[东吴苏大考研网] 2019苏大857细胞生物学（F）考研冲刺题库[东吴苏大考研网] 2019苏大生物化学考研强化冲刺题库【共两册】[东吴苏大考研网] 2012-2015年苏州大学880和声写作与分析考研真题[东吴苏大考研网] 2012-2015年苏州大学665中外音乐史考研真题[东吴苏大考研网] 2019苏州大学考研308护理综合考试解读与真题答案[东吴苏大考研网] 2011-2013年苏州大学850普通物理（生物医学工程）考研真题[东吴苏大考研网] 2011-2015年苏州大学859普通动物学考研真题[东吴苏大考研网] 2005、2013-2015年苏州大学882遗传学考研真题[东吴苏大考研网] 2011-2015年苏州大学879普通动物学与普通生态学考研真题[东吴苏大考研网] 2009-2010、2014-2015年苏州大学878电磁学考研真题[东吴苏大考研网] 2011-2015年苏州大学874药剂学考研真题[东吴苏大考研网] 2011-2015年苏州大学868生物教学论考研真题[东吴苏大考研网] 2013-2015年苏州大学860水产养殖学考研真题[东吴苏大考研网] 2009-2015年苏州大学851社会医学考研真题[东吴苏大考研网] 2004-2015年苏州大学845细胞生物学考研真题[东吴苏大考研网] 2013-2015年苏州大学666生物化学（农）考研真题[东吴苏大考研网] 2013-2015年苏州大学664动物生物化学与鱼类生理考研真题[东吴苏大考研网] 2006/2009-2016年苏州大学632卫生事业管理学考研真题[东吴苏大考研网] 2005-2015年苏州大学626预防综合考研真题[东吴苏大考研网] 2001-2015年苏州大学621生物化学考研真题[东吴苏大考研网] 2010-2015年苏州大学353卫生综合考研真题[东吴苏大考研网] 2011-2015年苏州大学349药学综合考研真题[东吴苏大考研网] 2011-2015年苏州大学340农业知识综合二考研真题[东吴苏大考研网] 2011-2013年苏州大学627园林设计基础考研真题[东吴苏大考研网] 2011-2013年苏州大学846中外建筑史考研真题[东吴苏大考研网] 2011、2013年苏州大学622建筑设计基础考研真题[东吴苏大考研网] 2014-2015年苏州大学846设计理论考研真题【历年苏州大学考研真题答案下载】[东吴苏大考研网] 2014-2015年苏州大学622快题设计与表现考研真题[东吴苏大考研网] 2009-2015年苏州大学848植物生理与生物化学考研真题[东吴苏大考研网] 2009-2015年苏州大学662植物学考研真题[东吴苏大考研网] 2013年苏州大学663普通昆虫学考研真题[东吴苏大考研网] 2014-2015年苏州大学877土力学考研真题[东吴苏大考研网] 2012-2015年苏州大学873交通工程学考研真题[东吴苏大考研网] 2009-2015年854苏州大学电子技术基础(城市轨道)考研真题[东吴苏大考研网] 2011-2014年苏州大学818理论力学(城市轨道)考研真题[东吴苏大考研网] 2014-2015年苏州大学870生产计划与控制考研真题[东吴苏大考研网] 2004-2015年苏州大学842自动控制原理考研真题[东吴苏大考研网] 2008-2015年苏州大学841电子技术基础考研真题[东吴苏大考研网] 2000-2015年苏州大学840理论力学（机电）考研真题[东吴苏大考研网] 2014-2015年苏州大学833钢铁冶金学考研真题[东吴苏大考研网] 1999-2017年苏州大学872数据结构与操作系统考研真题[东吴苏大考研网] 2009-2015年苏州大学839管理信息系统与数据结构考研真题[东吴苏大考研网] 2010-2013年苏州大学871数字电路考研真题[东吴苏大考研网] 2014-2015年苏州大学850专业综合（2）考研真题【高等数学基础+生物信息学基础】[东吴苏大考研网] 2004-2015年苏州大学836半导体物理考研真题（换05、08）[东吴苏大考研网] 2014-2015年苏州大学627专业综合（1）考研真题（线性代数+生物化学）[东吴苏大考研网] 2010-2015年苏州大学858材料学（F）考研真题（不含11年）[东吴苏大考研网] 2011-2015年苏州大学857细胞生物学（F）考研真题[东吴苏大考研网] 苏州大学856物理化学（F）考研真题（2001-2006、2010-2015年）[东吴苏大考研网] 2010-2015年苏州大学855普通物理（F）考研真题[东吴苏大考研网] 2010-2015年苏州大学631生物化学（F）考研真题[东吴苏大考研网] 2010-2015年苏州大学629无机化学（F）考研真题[东吴苏大考研网] 2012-2015年苏州大学628有机化学（F）考研真题[东吴苏大考研网] 2010-2015年苏州大学602高等数学（F）考研真题[东吴苏大考研网] 2010-2011年苏州大学867化学教学论考研真题[东吴苏大考研网] 2009-2011年苏州大学834化学原理考研真题[东吴苏大考研网] 2001-2006、2010-2015年苏州大学853物理化学考研真题[东吴苏大考研网] 2013-2015年苏州大学867化工原理考研真题[东吴苏大考研网] 2001-2015年苏州大学835有机化学B考研真题[东吴苏大考研网] 2004-2006、2013-2015年苏州大学834分析化学考研真题[东吴苏大考研网] 2001-2015年苏州大学635有机化学A考研真题[东吴苏大考研网] 1999-2018年苏州大学861高分子化学考研真题[东吴苏大考研网] 2005-2015年苏州大学843材料结构基础考研真题[东吴苏大考研网] 2011-2015年苏州大学838普通物理（光学工程）考研真题[东吴苏大考研网] 2003-2015年苏州大学832普通物理考研真题[东吴苏大考研网] 2009-2015年苏州大学838信号与线性系统考研真题【历年苏州大学考研真题答案下载】[东吴苏大考研网] 2014-2015年苏州大学821材料物理化学考研真题[东吴苏大考研网] 2010-2015年苏州大学866中学物理教学法考研真题[东吴苏大考研网] 2014-2015年苏州大学化学综合考研真题[东吴苏大考研网] 2003-2015年苏州大学601高等数学考研真题[东吴苏大考研网] 2012年苏州大学632艺术理论考研真题[东吴苏大考研网] 2012-2013年苏州大学619戏剧戏曲艺术学考研真题[东吴苏大考研网] 2009-2015年苏州大学245德语考研真题（不含13）[东吴苏大考研网] 2001-2015年苏州大学244法语考研真题（不含03、13）[东吴苏大考研网] 2003-2015年苏州大学243日语考研真题（不含13）[东吴苏大考研网] 2005-2015年苏州大学242俄语考研真题与答案【不含07、08、13】[东吴苏大考研网] 2005-2015年苏州大学241英语考研真题（不含08、13）[东吴苏大考研网] 2010-2015年苏州大学211翻译硕士英语考研真题答案[东吴苏大考研网] 2002-2004、2011年苏州大学会计学考研真题[东吴苏大考研网] 2013-2015年苏州大学661教育学原理考研真题[东吴苏大考研网] 2002-2015年苏州大学634心理学专业基础综合（自命题）考研真题[东吴苏大考研网] 2012-2015年苏州大学881知识产权法学考研真题（无13）[东吴苏大考研网] 2011-2015年苏州大学434国际商务专业基础考研真题与答案[东吴苏大考研网] 1999、2007、2009-2015年苏州大学844纺织材料学考研真题[东吴苏大考研网] 2002-2017年苏州大学431金融学综合考研真题与答案解析[东吴苏大考研网] 2019苏州大学考研新闻与传播专业硕士复习全析（含真题与答案，共六册）[东吴苏大考研网] 2003-2015年苏州大学612绘画基础（色彩命题画）考研真题试卷[东吴苏大考研网] 2019苏大考研633教育学专业基础综合复习全析【含真题答案，共八册】[东吴苏大考研网] 2019苏大考研汉语国际教育硕士复习全析（共四册，含真题答案）[东吴苏大考研网] 2013-2015年苏州大学847金融学概论考研真题[东吴苏大考研网] 2007-2015年苏州大学617综合日语考研真题[东吴苏大考研网] 1998-2015年苏州大学658公共管理学考研真题（不含1999）[东吴苏大考研网] 2000、2004、2013-2016年苏州大学660世界史专业基础综合（自命题）考研真题[东吴苏大考研网] 2000-2017年苏州大学659中国史专业基础综合（自命题）考研真题与答案[东吴苏大考研网] 2002、2006-2015年苏州大学624药学基础综合（2）考研真题[东吴苏大考研网] 2006、2009-2017年苏州大学623生物综合考研真题[东吴苏大考研网] 2012-2015年苏州大学865影视评论写作考研真题[东吴苏大考研网] 2011-2015年苏州大学611影视艺术学考研真题[东吴苏大考研网] 2000-2015年苏州大学819教育管理学考研真题（不含03、04、07、08、09、11年）[东吴苏大考研网] 苏州大学311教育学专业基础综合真题与答案（2007-2017年）[东吴苏大考研网] 2000-2006，2010-2015年苏州大学856物理化学（F）考研真题。