fortran习题参考答案
fortran编程习题答案
fortran编程习题答案Fortran编程习题答案Fortran是一种古老而强大的编程语言，广泛应用于科学计算和工程领域。
在学习Fortran编程的过程中，解决习题是一种非常有效的方法。
本文将为您提供一些Fortran编程习题的答案，帮助您更好地理解和掌握这门语言。
1. 习题一：编写一个Fortran程序，计算并输出1到100之间所有整数的平方。
程序代码如下：```fortranprogram squareimplicit noneinteger :: ido i = 1, 100print *, i, i**2end doend program square```2. 习题二：编写一个Fortran程序，计算并输出斐波那契数列的前20个数。
程序代码如下：```fortranprogram fibonacciimplicit noneinteger :: i, n, fib(20)fib(1) = 0fib(2) = 1do i = 3, 20fib(i) = fib(i-1) + fib(i-2)end dodo i = 1, 20print *, fib(i)end doend program fibonacci```3. 习题三：编写一个Fortran程序，计算并输出一个给定整数的阶乘。
程序代码如下：```fortranprogram factorialimplicit noneinteger :: i, n, resultprint *, "请输入一个整数："read *, nresult = 1do i = 1, nresult = result * iend doprint *, n, "的阶乘为：", resultend program factorial```4. 习题四：编写一个Fortran程序，计算并输出一个给定整数是否为素数。
程序代码如下：```fortranprogram primeimplicit noneinteger :: i, n, countprint *, "请输入一个整数："read *, ncount = 0do i = 2, n-1if (mod(n, i) == 0) thencount = count + 1end ifend doif (count == 0) thenprint *, n, "是素数"elseprint *, n, "不是素数"end ifend program prime```5. 习题五：编写一个Fortran程序，计算并输出一个给定整数的所有因子。
fortran95期末考试题及答案
fortran95期末考试题及答案FORTRAN95期末考试题及答案一、选择题（每题2分，共20分）1. 下列哪个是FORTRAN95中的合法变量名？A. 123abcB. _123abcC. 123D. variable-name答案：B2. 在FORTRAN95中，以下哪个语句用于定义数组？A. DIMENSIONB. DEFINEC. ARRAYD. DECLARE答案：A3. 下列哪个是FORTRAN95中的内建数据类型？A. INTEGERB. FLOATC. DOUBLED. STRING答案：A4. 在FORTRAN95中，以下哪个语句用于实现循环？A. IFB. DOC. THEND. ELSE答案：B5. 在FORTRAN95中，以下哪个语句用于条件判断？A. IFB. DOC. SELECTD. CASE答案：A6. 下列哪个是FORTRAN95中的文件打开语句？A. OPENB. CLOSEC. READD. WRITE答案：A7. 在FORTRAN95中，以下哪个语句用于实现模块化编程？A. MODULEB. FUNCTIONC. SUBROUTINED. PROGRAM答案：A8. 下列哪个是FORTRAN95中的参数传递方式？A. PASS BY VALUEB. PASS BY REFERENCEC. BOTH A AND BD. NEITHER A NOR B答案：C9. 在FORTRAN95中，以下哪个语句用于定义常量？A. DEFINEB. CONSTANTC. PARAMETERD. EQUIVALENCE答案：C10. 下列哪个是FORTRAN95中的文件关闭语句？A. OPENB. CLOSEC. READD. WRITE答案：B二、简答题（每题5分，共30分）1. 简述FORTRAN95中模块化编程的优点。
答案：模块化编程允许程序被分解成独立的模块，每个模块可以独立编译和测试，提高了代码的可读性和可维护性。
FORTRAN答案
第三章C 3.11分期付款的计算(利率不可以很大!!!)DOUBLE PRECISION MPRINT *,'请输入从银行贷款数，每月偿还数，以及月利率'READ(*,*) D,P,RM=(LOG(P)-LOG(P-D*R))/LOG(1+R)PRINT *,'你需要还款',M,'个月'ENDC 3.12 借款额计算（年利率要小一点）DOUBLE PRECISION D,IPRINT *, '请输入每年准备还银行多少钱，年利率，以及打算还多少年' READ (*,*) A,R,NI =1+RD=A*(I**N-1)/((I-1)*(I**N))PRINT *, '你可以借款',D,'元'END第四章C 4.6 判断一个数能不能被7，11，17整除INTEGER IPRINT *, '请输入一个整数'READ (*,*) IIF((MOD(I,7)) .EQ. 0) THENPRINT *,'这个数能被7整除'END IFIF( (MOD(I,11)) .EQ. 0) THENPRINT *,'这个数能被11整除'END IFIF( (MOD(I,17)) .EQ. 0) THENPRINT *,'这个数能被17整除'END IFIF( (MOD(I,7) .NE. 0 ) .AND. (MOD(I,11) .NE. 0 ) .AND.& (MOD(I,17) .NE. 0 ) ) THENPRINT *, '这个数不能被7，11，17 整除' END IFENDC 4.8 输入四个数后排序（我用的是数组冒泡法！！！！，大家可以参照87页4.4看着写一下）INTEGER A(1:4)PRINT *,'请输入四个数'READ (*,*) A(1),A(2),A(3),A(4)TEMP=A(1)DO 100 I=1,4DO 10 J=I+1,4IF(A(I).LT.A(J)) THENTEMP=A(I)A(I)=A(J)A(J)=TEMPEND IF10 CONTINUE100 CONTINUEPRINT *, A(1), A(2), A(3), A(4)END第五章C 5.4 求SIN X 的值（这里N如果大了就算不出来了！！！）DOUBLE PRECISION SIN,Q,SQ=1SIN=0PRINT *, '请输入X的值'READ (*,*) XDO 10,N=1,4DO 100,M=1,(2*N-1)Q=M*Q100 CONTINUES=((-1)**(N+1))*(X**M)/QSIN=S+SIN10 CONTINUEPRINT *, 'SIN X=' ,SINENDC 5.8 找出所有的水仙花数DO 10,B=1,9DO 100,S=0,9DO 1000,G=0,9IF((B*100+S*10+G) .EQ. (B**3+S**3+G**3)) THENPRINT *, B*100+S*10+G ,'是水仙花数'END IF1000 CONTINUE100 CONTINUE10 CONTINUEEND第六章C 6.4 求三角形面积和重心坐标COMPLEX A,B,C,DDOUBLE PRECISION SA=(1.5,2.0)B=(4.5,4.5)C=(18.0,10.5)D=(A+B+C)/3AB=ABS(A-B)CB=ABS(C-B)CA=ABS(C-A)COSB=(CB**2+AB**2-CA**2)/(2*CB*AB)SINB=SQRT(1-COSB**2)S=1/2.0*SINB*AB*CBPRINT *,'三角形面积为',S,' 三角形重心为', D ENDC 6.8 使单词逆序输出PROGRAM MAINCHARACTER *(100)LINEWRITE (*,*)'请输入一个字符串'READ (*,*)LINEL=LENGTH(LINE)WRITE(*,*)(LINE(I:I),I=L,1,-1)ENDCFUNCTION LENGTH(LINE)CHARACTER *(*)LINEI=LEN(LINE)DO 10 WHILE((I.GT.0).AND.(LINE(I:I).EQ.' '))I=I-110 CONTINUELENGTH=IENDC 6.9 输出星花菱形(中间为30列我没管，大家可以参照复印的材料上的)CHARACTER * 9 LINEDO 10 I=1,5DO 20 J=6-I,4+ILINE(J:J)='*'20 CONTINUEWRITE (*,*)LINE10 CONTINUEDO 30 I=6,9LINE=' 'DO 40 J=I-4,14-ILINE(J:J)='*'40 CONTINUEWRITE(*,*)LINE 30 CONTINUEEND。
FORTRAN习题答案
FORTRAN习题答案习题⼆⼀、问答题1. 给出下⾯变量名称,哪些是合法变量？哪些是⾮法变量？说明原因。
Count 、num_2、x&y 、4x+5y 、china-suzhou 、$us 、AbCdE 、Mr.bai 、t5、_another 、school_class_25、#125、2002Y 、π、β、A01/02、alpha 、date(1) 1. 判定下⾯整数,指出哪些是合法整数,哪些是⾮法整数？说明原因。
-0、+ 215、$125、3,245,895、5.3245、5#384、-524_3、#5DFE 、23-345、16#1A2B 、38#ABCD 、8#275_2、+327890、4 #3212. 判定下⾯实数,指出哪些是合法实数,哪些是⾮法实数？说明原因。
-0E2、45.2345E3.5、-5489E25_8、-.2345E-35、$185.45E 、+ 2.753425E24_3、 58D85、+0.E-0、-00000.001E5、5,443,223.44、-12 34E+2、+ 18.5E 18、2.5E42习题三⼀、选择题1．下⾯是V isual Fortran 中正确的表达式是。
（A ）A*COS(X)+∣B ∣（B ）2*EXP(2*X)/SQRT(16.0)（C ）B 2-4AC （D ）MOD （24.5,0.5）2．下⾯算术赋值语句中正确的语句是。
（A ）M*N=(X-Y)/Z （B ）+R=A+B/C（C ）X=Y=Z-1.0 （D ）Y=A*B/C/D3．算术表达式1/3+2/3的值为。
（A ）0 （B ） 1 （C ） 0.99999999 （D ）值不确定⼆、问答题1. 将下列代数式⽤Visual Fortran 表达式描述：①②③ 4sin 3A-3sinA+sin3A ④ 2．执⾏下列赋值语句后，变量中的值。
变量的类型遵循I —N 规则。
Fortran95第一章第六大题习题与答案
1. 从键盘输入a,b,c 的值，计算f=cos ｜a+b ｜/sin ｜b｜｜a｜++tan c 上机执行该程序，输入a=-4.6°,b=10°,c=21.85°,观察计算结果。
Program ex1_1implicit nonereal a,b,c,fprint*,'请输入a,b,c(角度值)'read*,a,b,ca=a*3.14159/180.0b=b*3.14159/180.0c=c*3.14159/180.0f=cos(abs(a+b))/sin(sqrt(abs(a)+abs(b)))+tan(c)write(*,*)'f=',fstopEnd2.设圆锥体底面半径r 为6，高h 为5，从键盘输入r 、h ，计算圆锥体体积。
计算公式为V=32h r π。
Program ex1_2implicit nonereal r,h,vprint*,'请输入r,h 的值'read*,r,hv=3.14159*r*r*h/3write(*,*)'v=',vstopEnd3.求一元二次方程02=++c bx ax 的两个根1x 和2x 。
方程的系数a 、b 、c 值从键盘输入并假定042>-ac b 。
Program ex1_3implicit nonereal a,b,c,x1,x2print*,'请输入a,b,c 的值'read*,a,b,cx1=(b+sqrt(b*b-4*a*c))/2*ax2=(b-sqrt(b*b-4*a*c))/2*awrite(*,*)'x1=',x1,'x2=',x2stopEnd4.从键盘输入一个三位十进制整数，分别输出其个位、十位、百位上的数字。
Program ex1_4implicit noneinteger xprint*,'请输入一个三位十进制整数'read*,xwrite(*,*)'个位数=',mod(x,10)write(*,*)'十位数=',mod(x/10,10)write(*,*)'百位数=',x/100stopEnd5.已知ysin(⋅)+=+，分别计算等号两边的算式并输出计算⋅sinyxcosxycosx sin结果（x=30°,y=45°从键盘输入）。
fortran考试题及答案分开
fortran考试题及答案分开1. 以下哪个选项是Fortran语言中正确的整型变量声明？A. INTEGER xB. REAL xC. COMPLEX xD. LOGICAL x答案：A2. Fortran程序中，哪个关键字用于定义数组？A. ARRAYB. LISTC. VECTORD. DIMENSION答案：D3. 在Fortran中，以下哪个选项是正确的条件语句？A. IF (x > 0) THENPRINT *, 'x is positive'B. IF x > 0 THENPRINT *, 'x is positive'C. IF (x > 0)PRINT *, 'x is positive'D. IF x > 0PRINT *, 'x is positive'答案：A4. Fortran中用于循环结构的关键字是什么？A. LOOPB. ITERATEC. DOD. FOR答案：C5. 如何在Fortran程序中包含另一个文件？A. 使用INCLUDE语句B. 使用IMPORT语句C. 使用INCLUDE关键字D. 使用IMPORT关键字答案：A6. Fortran中，哪个函数用于计算数组元素的总和？A. SUMB. TOTALC. AGGREGATED. ACCUMULATE答案：A7. 在Fortran中，如何声明一个具有默认值的变量？A. INTEGER :: x = 0B. INTEGER x = 0C. INTEGER x DEFAULT 0D. INTEGER x = DEFAULT 0答案：A8. Fortran程序中的主程序必须以哪个关键字开始？A. PROGRAMB. MAINC. PROCEDURED. FUNCTION答案：A9. 在Fortran中，如何声明一个二维数组？A. INTEGER :: matrix(10, 10)B. INTEGER :: matrix[10][10]C. INTEGER :: matrix(10)(10)D. INTEGER :: matrix(10,10)答案：A10. Fortran中用于计算数组元素平均值的函数是什么？A. AVGB. MEANC. AVERAGED. SUM答案：C。
Fortran95程序设计习题答案
Fortran95程序设计习题答案第四章 1.program main implicit none write(*,*) "Have a good time." write(*,*) "That's not bad." write(*,*) '"Mary" isn''t my name.' end program 2.program main real, parameter :: PI=3 implicit none.14159real radius write(*,*) "请输入半径长" read(*,*) radius write(*,"(' 面积='f8. 3)") radius*radius*PI end program 3.program main implicit none real grades write(*,*) "请输入成绩" read(*,*)grades write(*,"(' 调整后成绩为 'f8.3)") SQRT(grades)*10.0 end program 4.integer a,b real ra,rb a=2 b=3 ra=2.0 rb=3.0 write(*,*) b/a ! 输出1, 因为使用整数计算, 小数部分会无条件舍去 write(*,*) rb/ra ! 输出1.5 5.program main implicit none type distance real meter, inch, cm end type type(distance) :: d write(*,*) "请输入长度:" read(*,*) d%meter d%cm = d%meter*100 d%inch = d%cm/2.54 write(*,"(f8.3'米 ='f8.3'厘米='f8.3'英寸')") d%meter, d%cm, d%inch end program 第五章 1.program main implicit none integer money real tax write(*,*) "请输入月收入" read(*,*) money if ( money<1000 ) then tax = 0.03 else if ( money<5000) then tax = 0.1 else tax = 0.15 end if write(*,"(' 税金为 'I8)") nint(money*tax) end program 2.program main implicit none integer day character(len=20) :: tv write(*,*) "请输入星期几" read(*,*) day select case(day) case(1,4) tv = "新闻" case(2,5) tv = "电视剧" case(3,6) tv = "卡通" case(7) tv = "电影" case default write(*,*) "错误的输入" stop end select write(*,*) tv end program 3.program main implicit none integer age, money real tax write(*,*) "请输入年龄"write(*,*) "请输入月收入" read(*,*) money if ( age<50 ) thenread(*,*) ageif ( money<1000 ) then tax = 0.03 else if ( money<5000 )then tax = 0.10 else tax = 0.15 end if else if ( money<1000 ) then tax = 0.5 else if ( money<5000 )then tax = 0.7 else tax = 0.10 end if end ifwrite(*,"(' 税金为 'I8)") nint(money*tax) end program 4.program main implicit none integer year, days logical mod_4, mod_100, mod_400write(*,*) "请输入年份" read(*,*) year mod_4 = ( MOD(year,4) == 0 ) mod_100 = ( MOD(year,100) == 0 ) mod_400 = ( MOD(year,400) == 0 ) if ( (mod_4 .NEQV. mod_100) .or. mod_400 ) then days = 366 else days = 365 end if write(*,"('这一年有'I3'天')") days stop end program 第六章1.program main implicit none integer i do i=1,5 write(*,*) "Fortran" end do stop end program2.program main implicit none integer i,sum sum = 0 do i=1,99,2 sum = sum+i end do write(*,*) sum stop end program3.program main implicit none integer, parameter :: answer = 45 integer, parameter :: max = 5 integer weight, i do i=1,max write(*,*) "请输入体重" read(*,*) weight if ( weight==answer ) exit end do if ( i<=max ) then write(*,*) "猜对了" else write(*,*) "猜错了" end if stop end program4.program main implicit none integer, parameter :: max=10 integer i real item real ans ans = 1.0 item = 1.0 do i=2,max item = item/real(i) ans = ans+item end do write(*,*) ans stop end program5.program main implicit none integer, parameter :: length = 79 character(len=length) :: input, output integer i,j write(*,*) "请输入一个字串" read(*,"(A79)") input j=1 do i=1, len_trim(input) if( input(i:i) /= ' ' ) then output(j:j)=input(i:i) j=j+1 end if end do write(*,"(A79)") output stop end program 第七章 1.program mainimplicit none integer, parameter :: max = 10 integer i integer ::a(max) = (/ (2*i, i=1,10) /) integer :: t ! sum()是fortran库函数write(*,*) real(sum(a))/real(max) stop end program2.integer a(5,5) ! 5*5=25 integer b(2,3,4) ! 2*3*4=24 integerc(3,4,5,6) ! 3*4*5*6=360 integer d(-5:5) ! 11 integer e(-3:3, -3:3) ! 7*7=49 3.program main implicit none integer, parameter :: max=10integer f(max) integer i f(1)=0 f(2)=1 do i=3,max f(i)=f(i-1)+f(i-2) end do write(*,"(10I4)") f stop end program 4.program main implicit none integer, parameter :: size=10 integer :: a(size) = (/5,3,6,4,8,7,1,9,2,10 /) integer :: i,j integer :: t do i=1, size-1 do j=i+1, size if ( a(i) < a(j) ) then ! a(i)跟a(j)交换 t=a(i)a(i)=a(j) a(j)=t end if end do end do write(*,"(10I4)") a stop end5.a(2,2) ! 1+(2-1)+(2-1)*(5) = 7 a(3,3) ! 1+(3-1)+(3-1)*(5) = 13 第八章1.program main implicit none real radius, area write(*,*) "请输入半径长" read(*,*) radius call CircleArea(radius, area) write(*,"(' 面积 ='F8.3)") area stop end program subroutine CircleArea(radius, area) implicit none real, parameter :: PI=3.14159 real radius, area area = radius*radius*PI return end subroutine 2.program main implicit nonereal radius real, external :: CircleArea write(*,*) "请输入半径长" read(*,*) radius write(*,"(' 面积 = 'F8.3)") CircleArea(radius) stop end program real function CircleArea(radius) implicit none real, parameter :: PI=3.14159 real radius CircleArea = radius*radius*PI returnend function 3.program main implicit none call bar(3) call bar(10) stop end program subroutine bar(length) implicit none integer, intent(in) :: length integer i character(len=79) :: string string=" " do i=1,length string(i:i)='*' end do write(*,"(A79)") string return end subroutine 4.program main implicit none integer, external :: add write(*,*)add(100) end program recursive integer function add(n)integer, intent(in) :: n if ( n<0 ) then sum=0 return elseresult(sum) implicit noneif ( n<=1 ) then sum=n return end if sum = n + add(n-1) return end function 5.program main implicit none integer, external :: gcdwrite(*,*) gcd(18,12) end program integer function gcd(A,B) implicit none integer A,B,BIG,SMALL,TEMP BIG=max(A,B) SMALL=min(A,B) dowhile( SMALL /= 1 )TEMP=mod(BIG,SMALL) if ( TEMP==0 ) exit BIG=SMALL SMALL=TEMP enddo gcd=SMALL return end function 6.program main use TextGraphLib implicit none integer, parameter :: maxx=60, maxy=20 real, parameter :: StartX=0.0, EndX=3.14159*2.0 real, parameter :: xinc = (EndX-StartX)/(maxx-1) real x integer i,px,py call SetScreen(60,20) call SetCurrentChar('*') x=StartX do px=1,maxx py = (maxy/2)*sin(x)+maxy/2+1 call PutChar(px,py) x=x+xinc end docall UpdateScreen() stop end program 第九章 1.program main implicitnone character(len=79) :: filename character(len=79) :: buffer integer, parameter :: fileid = 10 integer count integer :: status = 0 logical alive write(*,*) "Filename:" read (*,"(A79)") filenameinquire( file=filename, exist=alive) if ( alive ) then open(unit=fileid, file=filename, & access="sequential", status="old") count = 0 dowhile(.true.) read(unit=fileid, fmt="(A79)", iostat=status ) bufferif ( status/=0 ) exit ! 没有资料就跳出循环 write(*,"(A79)") buffercount = count+1 if ( count==24 ) then pause count = 0 end if end do else write(*,*) TRIM(filename)," doesn't exist." end if stop end2.program main implicit none character(len=79) :: filenamecharacter(len=79) :: buffer integer, parameter :: fileid = 10 integer i integer :: status = 0 logical alive write(*,*) "Filename:" read (*,"(A79)") filename inquire( file=filename, exist=alive) if ( alive ) then open(unit=fileid, file=filename, & access="sequential",status="old") do while(.true.) read(unit=fileid, fmt="(A79)",iostat=status ) buffer if ( status/=0 )exit ! 没有资料就跳出循环 do i=1, len_trim(buffer) buffer(i:i) = char( ichar(buffer(i:i))-3 ) end do write(*,"(A70)") buffer enddo else write(*,*) TRIM(filename)," doesn't exist." end if stop end3.program main implicit none type student integer chinese, english, math, science, social, total end type type(student) :: s, total integer, parameter :: students=20, subjects=5 integer iopen(10,file="grades.bin",access="direct",recl=1) write(*,"(7A10)") "座号","中文","英文","数学","自然","社会","总分" total =student(0,0,0,0,0,0) do i=1, students read(10,rec=(i-1)*subjects+1)s%chinese read(10,rec=(i-1)*subjects+2) s%english read(10,rec=(i-1)*subjects+3) s%math read(10,rec=(i-1)*subjects+4) s%scienceread(10,rec=(i-1)*subjects+5) s%social s%total =s%chinese+s%english+s%math+s%science+s%social total%chinese =total%chinese+s%chinese total%english = total%english+s%englishtotal%math = total%math+s%math total%science = total%science+s%science total%social = total%social+s%social total%total = total%total+s%total write(*,"(7I10)") i, s end do write(*,"(A10,6F10.3)") "平均", & real(total%chinese)/real(students),&real(total%english)/real(students),&real(total%math)/real(students),&real(total%science)/real(students),&real(total%social)/real(students),& real(total%total)/real(students) stop end 4.program main implicit none character(len=79) :: filename character(len=79) :: buffer integer, parameter :: fileid = 10 integer i integer :: status = 0 logical alive write(*,*) "Filename:" read (*,"(A79)") filename inquire( file=filename, exist=alive) pen(unit=fileid, file=filename, & access="sequential", if ( alive ) then ostatus="old") do while(.true.) read(unit=fileid, fmt="(A79)",iostat=status ) buffer if ( status/=0 ) exit ! 没有数据就跳出循环 doi=1,len_trim(buffer) buffer(i:i) = char( ichar(buffer(i:i))-(mod(i-1,3)+1) ) end do write(*,"(A70)") buffer end do else write(*,*)TRIM(filename)," doesn't exist." end if stop end 5.module typedef typestudent integer :: num integer :: Chinese, English, Math, Natural, Social integer :: total integer :: rank end type end module program main use typedef implicit none integer, parameter :: fileid=10 integer, parameter :: students=20 character(len=80) :: tempstrtype(student) :: s(students) ! 储存学生成绩 type(student) :: total ! 计算平均分数用 integer i, num, error open(fileid,file="grades.txt",status="old", iostat=error) if ( error/=0 ) then write(*,*) "Open grades.txt fail." stop end if read(fileid, "(A80)") tempstr ! 读入第一行文字 total=student(0,0,0,0,0,0,0,0) ! 用循环读入每位学生的成绩 do i=1,students read(fileid,*) s(i)%num, s(i)%Chinese,s(i)%English, & s(i)%Math, s(i)%Natural, s(i)%Social ! 计算总分s(i)%Total = s(i)%Chinese + s(i)%English + & s(i)%Math + s(i)%Natural + s(i)%Social ! 累加上各科的分数, 计算各科平均时使用 total%Chinese = total%Chinese +s(i)%Chinese total%English = total%English + s(i)%Englishtotal%Math = total%Math + s(i)%Math total%Natural = total%Natural +s(i)%Natural total%Social = total%Social + s(i)%Social total%Total = total%Total + s(i)%Total end do call sort(s,students) ! 重新输出每位学生成绩 write(*,"(8A7)") "座号","中文","英文","数学","自然","社会","总分","名次" do i=1,students write(*,"(8I7)") s(i) end do ! 计算并输出平圴分数 write(*,"(A7,6F7.1)") "平均", &real(total%Chinese)/real(students),&real(total%English)/real(students),&real(total%Math) /real(students),&real(total%Natural)/real(students),& real(total%Social)/real(students),& real(total%Total) /real(students) stop end program subroutine sort(s,n) use typedef implicit none integer ntype(student) :: s(n), t integer i,j do i=1,n-1 do j=i+1,n if( s(i)%total < s(j)%total ) then t = s(i) s(i)=s(j) s(j) = t end if end do end do forall(i=1:n) s(i)%rank = i end forall end subroutine 第十章 1.integer(kind=4) ::4 bytes real(kind=4) :: b ! 4 bytes real(kind=8) :: c ! 8 bytes character(len=10) :: a !str ! 10 bytes integer(kind=4), pointer :: pa ! 4 bytesreal(kind=4), pointer :: pb ! 4 bytes real(kind=8), pointer :: pc ! 4 bytes character(len=10), pointer :: pstr ! 4 bytes type studentinteger Chinese, English, Math end type type(student) :: s ! 12 bytes type(student), pointer :: ps ! 4 bytes 2.integer, target :: a = 1 integer, target :: b = 2 integer, target :: c = 3 integer, pointer :: p p=>a write(*,*) p ! 1 p=>b write(*,*) p ! 2 p=>c p=5 write(*,*) c ! 53.module linklist type student integer :: num integer :: Chinese, English, Math, Science, Social end type type datalink type(student) :: item type(datalink), pointer :: next end type contains function SearchList(num, head) implicit none integer :: num type(datalink), pointer :: head, p type(datalink), pointer :: SearchList p=>headnullify(SearchList) do while( associated(p) ) if ( p%item%num==num ) then SearchList => p return end if p=>p%next end do return end function end module linklist program ex1016 use linklist implicit nonecharacter(len=20) :: filename character(len=80) :: tempstrtype(datalink), pointer :: head type(datalink), pointer :: ptype(student), allocatable :: s(:) integer i,error,size write(*,*) "filename:" read(*,*) filename open(10, file=filename, status="old", iostat=error) if ( error/=0 ) then write(*,*) "Open file fail!" stop end if allocate(head) nullify(head%next) p=>head size=0 read(10,"(A80)") tempstr ! 读入第一行字符串, 不需要处理它 ! 读入每一位学生的成绩do while(.true.) read(10,fmt=*, iostat=error) p%item if ( error/=0 )exit size=size+1 allocate(p%next, stat=error) ! 新增下一个数据 if( error/=0 ) then write(*,*) "Out of memory!" stop end if p=>p%next ! 移动到链表的下一个数据 nullify(p%next) end do write(*,"('总共有',I3,'位学生')") size allocate( s(size) ) p=>head do i=1,size s(i)=p%itemp=>p%next end do do while(.true.) write(*,*) "要查询几号同学的成绩?" read (*,*) i if ( i<1 .or. i>size ) exit ! 输入不合理的座号write(*,"(5(A6,I3))") "中文",s(i)%Chinese,& "英文",s(i)%English,& "数学",s(i)%Math,& "自然",s(i)%Science,& "社会",s(i)%Social end do write(*,"('座号',I3,'不存在, 程序结束.')") i stop end program 4.module typedef implicit none type :: datalink integer :: i type(datalink), pointer :: next end type datalink end module typedef program ex1012 use typedef implicit none type(datalink) , pointer :: p, head, nextinteger :: i,n,err write(*,*) 'Input N:' read(*,*) n allocate( head ) head%i=1 nullify(head%next) p=>head do i=2,n allocate( p%next,stat=err ) if ( err /= 0 ) then write(*,*) 'Out of memory!' stop endif p=>p%next p%i=i end do nullify(p%next) p=>head dowhile(associated(p)) write(*, "(i5)" ) p%i p=>p%next end do ! 释放链表的存储空间 p=>head do while(associated(p)) next => p%nextdeallocate(p) p=>next end do stop end program 第十一章 1.moduleutility implicit none interface area module procedure CircleArea module procedure RectArea end interface contains real function CircleArea(r) real, parameter :: PI=3.14159 real rCircleArea = r*r*PI return end function real function RectArea(a,b) real a,b RectArea = a*b return end function end module program main use UTILITY implicit none write(*,*) area(1.0) write(*,*) area(2.0,3.0)stop end program 2.module time_utility implicit none type :: timeinteger :: hour,minute,second end type time interface operator(+) module procedure add_time_time end interface contains functionadd_time_time( a, b ) implicit none type(time) :: add_time_timetype(time), intent(in) :: a,b integer :: seconds,minutes,carryseconds=a%second+b%second carry=seconds/60minutes=a%minute+b%minute+carry carry=minutes/60add_time_time%second=mod(seconds,60)add_time_time%minute=mod(minutes,60)add_time_time%hour=a%hour+b%hour+carry return end functionadd_time_time subroutine input( a ) implicit none type(time),intent(out) :: a write(*,*) " Input hours:" read (*,*) a%hourwrite(*,*) " Input minutes:" read (*,*) a%minute write(*,*) " Input seconds:" read (*,*) a%second return end subroutine input subroutine output( a ) implicit none type(time), intent(in) :: a write(*, "(I3,'hours',I3,' minutes',I3,' seconds')" ) a%hour,a%minute,a%second return end subroutine output end module time_utility program main usetime_utility implicit none type(time) :: a,b,c call input(a) callinput(b) c=a+b call output(c) stop end program main 3.modulerational_utility implicit none private public :: rational, &operator(+), operator(-), operator(*),& operator(/),assignment(=),operator(>),& operator(<), operator(==), operator(/=),& output, input type :: rational integer :: num, denom end type rational interface operator(+) module procedure rat__rat_plus_rat end interface interface operator(-)module procedure rat__rat_minus_rat end interface interfaceoperator(*) module procedure rat__rat_times_rat end interfaceinterface operator(/) module procedure rat__rat_div_rat end interface interface assignment(=) module procedure rat_eq_rat module procedureint_eq_rat module procedure real_eq_rat end interface interface operator(>) module procedure rat_gt_rat end interface interface operator(<) module procedure rat_lt_rat end interface interface operator(==) module procedure rat_compare_rat end interface interface operator(/=) module procedure rat_ne_rat end interface containsfunction rat_gt_rat(a,b) implicit none logical :: rat_gt_rattype(rational), intent(in) :: a,b real :: fa,fbfa=real(a%num)/real(a%denom)fb=real(b%num)/real(b%denom) if ( fa > fb ) then rat_gt_rat=.true. else rat_gt_rat=.false. end if return end function rat_gt_ratfunction rat_lt_rat(a,b) implicit none logical :: rat_lt_rattype(rational), intent(in) :: a,b real :: fa,fbfa=real(a%num)/real(a%denom) fb=real(b%num)/real(b%denom) if ( fb > fa ) then rat_lt_rat=.true. else rat_lt_rat=.false. end if return end function rat_lt_rat function rat_compare_rat(a,b) implicit nonelogical :: rat_compare_rat type(rational), intent(in) :: a,btype(rational) :: c c=a-b if ( c%num == 0 ) thenrat_compare_rat=.true. else rat_compare_rat=.false. end if returnend function rat_compare_rat function rat_ne_rat(a,b) implicit none logical :: rat_ne_rat type(rational), intent(in) :: a,btype(rational) :: c c=a-b if ( c%num==0 ) then rat_ne_rat=.false.else rat_ne_rat=.true. end if return end function rat_ne_ratsubroutine rat_eq_rat( rat1, rat2 ) implicitnone type(rational), intent(out):: rat1 type(rational),intent(in) :: rat2 rat1%num = rat2%num rat1%denom = rat2%denom return end subroutine rat_eq_rat subroutine int_eq_rat( int, rat ) implicit none integer, intent(out):: int type(rational), intent(in) :: rat int = rat%num / rat%denom return end subroutine int_eq_rat subroutinereal_eq_rat( float, rat ) implicit none real, intent(out) :: floattype(rational), intent(in) :: rat float = real(rat%num) /real(rat%denom) return end subroutine real_eq_rat function reduse( a ) implicit none type(rational), intent(in) :: a integer :: btype(rational) :: reduse b=gcv_interface(a%num,a%denom) reduse%num =a%num/b reduse%denom = a%denom/b return end function reduse functiongcv_interface(a,b) implicit none integer, intent(in) :: a,b integer :: gcv_interface if ( min(a,b) .eq. 0 ) then gcv_interface=1 return end if if (a==b) then gcv_interface=a return else if ( a>b ) thengcv_interface=gcv(a,b) else if ( a<b ) then gcv_interface=gcv(b,a)end if return end function gcv_interface recursive function gcv(a,b) result(ans) implicit none integer, intent(in) :: a,b integer :: m integer :: ans m=mod(a,b) select case(m) case(0) ans=b returncase(1) ans=1 return case default ans=gcv(b,m) end select return end function gcv function rat__rat_plus_rat( rat1, rat2 ) implicit none type(rational) :: rat__rat_plus_rat type(rational), intent(in) :: rat1,rat2 type(rational) :: act act%denom= rat1%denom * rat2%denom act%num = rat1%num*rat2%denom + rat2%num*rat1%denom rat__rat_plus_rat = reduse(act) return end function rat__rat_plus_rat functionrat__rat_minus_rat( rat1, rat2 ) implicit none type(rational) ::rat__rat_minus_rat type(rational), intent(in) :: rat1, rat2type(rational) :: temp temp%denom = rat1%denom*rat2%denom temp%num =rat1%num*rat2%denom - rat2%num*rat1%denom rat__rat_minus_rat = reduse( temp ) return end function rat__rat_minus_ratfunction rat__rat_times_rat( rat1, rat2 ) implicit nonetype(rational) :: rat__rat_times_rat type(rational), intent(in) :: rat1, rat2 type(rational) :: temp temp%denom = rat1%denom* rat2%denom temp%num = rat1%num * rat2%num rat__rat_times_rat = reduse(temp)return end function rat__rat_times_rat function rat__rat_div_rat( rat1, rat2 ) implicit none type(rational) :: rat__rat_div_rattype(rational), intent(in) :: rat1, rat2 type(rational) :: temptemp%denom = rat1%denom* rat2%num temp%num = rat1%num * rat2%denomrat__rat_div_rat = reduse(temp) return end function rat__rat_div_rat subroutine input(a) implicit none type(rational), intent(out) :: awrite(*,*) "分子:" read(*,*) a%num write(*,*) "分母:" read(*,*)a%denom return end subroutine input subroutine output(a) implicit none type(rational), intent(in) :: a if ( a%denom/=1 ) then write(*, "(' (',I3,'/',I3,')' )" ) a%num,a%denom else write(*, "(I3)" ) a%num end if return end subroutine output end module rational_utility program main use rational_utility implicit none type(rational) :: a,b,c call input(a) call input(b) c=a+b write(*,*) "a+b=" call output(c) c=a-bwrite(*,*) "a-b=" call output(c) c=a*b write(*,*) "a*b=" call output(c)c=a/b write(*,*) "a/b=" call output(c) if (a>b) write(*,*) "a>b" if(a<b) write(*,*) "a<b" if (a==b) write(*,*) "a==b" if (a/=b) write(*,*) "a/=b" stop end program main 4.module vector_utility implicit none type vector real x,y end type interface operator(+) module procedurevector_add_vector end interface interface operator(-) module procedurevector_sub_vector end interface interface operator(*) module procedure real_mul_vector module procedure vector_mul_real module procedure vector_dot_vector end interface interface operator(.dot.) module procedure vector_dot_vector end interface contains type(vector) functionvector_add_vector(a,b) type(vector), intent(in) :: a,bvector_add_vector = vector(a%x+b%x, a%y+b%y) end function type(vector) functionvector_sub_vector(a,b) type(vector), intent(in) :: a,bvector_sub_vector = vector(a%x-b%x, a%y-b%y) end function type(vector) function real_mul_vector(a,b) real, intent(in) :: a type(vector), intent(in) :: b real_mul_vector= vector( a*b%x, a*b%y ) end functiontype(vector) functionvector_mul_real(a,b) type(vector), intent(in) :: a real, intent(in) :: b vector_mul_real = real_mul_vector(b,a) end function real function vector_dot_vector(a,b) type(vector), intent(in) :: a,bvector_dot_vector = a%x*b%x + a%y*b%y end function subroutineoutput(vec) type(vector) :: vec write(*,"('('F6.2','F6.2')')") vec end subroutine end module program main use vector_utility implicit none type(vector) a,b,c a=vector(1.0, 2.0) b=vector(2.0, 1.0) c=a+b call output(c) c=a-b call output(c) write(*,*) a*b end program main。
fortran课后习题答案
fortran课后习题答案Fortran课后习题答案在学习Fortran编程语言时，课后习题是巩固知识、提高编程能力的重要途径。
通过解答课后习题，学生可以加深对Fortran语法和逻辑的理解，提高编程实践能力。
以下是一些Fortran课后习题答案，供大家参考。
1. 编写一个Fortran程序，计算并输出1到100的所有偶数的和。
程序代码如下：```fortranprogram sum_even_numbersimplicit noneinteger :: i, sumsum = 0do i = 2, 100, 2sum = sum + iend doprint *, 'The sum of even numbers from 1 to 100 is:', sumend program sum_even_numbers```2. 编写一个Fortran程序，找出一个整数数组中的最大值和最小值，并输出它们的位置。
程序代码如下：```fortranprogram find_max_minimplicit noneinteger :: i, n, max_val, min_val, max_pos, min_pos integer, dimension(10) :: arr! 初始化数组arr = (/3, 7, 2, 8, 5, 10, 1, 6, 4, 9/)! 初始化最大值和最小值max_val = arr(1)min_val = arr(1)max_pos = 1min_pos = 1! 找出最大值和最小值do i = 2, 10if (arr(i) > max_val) thenmax_val = arr(i)max_pos = iendifif (arr(i) < min_val) thenmin_val = arr(i)min_pos = iendifend doprint *, 'The maximum value is', max_val, 'at position', max_posprint *, 'The minimum value is', min_val, 'at position', min_posend program find_max_min```通过这些课后习题的答案，我们可以看到Fortran语言的一些基本特性和常用语法的运用。