C语言程序填空练习题(二)

将下面的程序补充完整，以完成指定的功能。

1．尼科彻斯定理This program is to verify Theorem of Nicoqish.That is the cube of any integer can be represented as the sum of some continue odd numbers.For example, 8^3=512=57+59+61+63+65+67+69+71.#include <iostream>using namespace std;int main(){int n,a,i;while(1){cout<<"Please input a integer to verify(0 to quit): ";cin>>n;if(n==0) __________; // （1）// 输出等差数列，首项为a*a-a+1，公差为2，项数为na=n*n-n+1;cout<<n<<"*"<<n<<"*"<<n<<"="<<n*n*n<<"="<<a;for (i=1; __________;i++) // （2）cout<<"+"<<__________; // （3）cout<<endl;}return 0;}2．角谷猜想This program is to verify Jiaogu Guess.That is given any natural number, if it is an even, divides 2, if it is an odd, multiple 3 and add 1, the result continues to be calculated analogously. After some times, the result is always 1.#include <iostream>using namespace std;int main(){int n,a,i,cnt;while(1){cout<<"Please input a integer to verify(0 to quit): ";cin>>n;if(n==0) __________break; // （1）cnt=0;cout<<" ------ Results of verification: ------------\n";do{if(__________)// （2）{n=n*3+1;cout<<"Step No."<<++cnt<<":"<<(n-1)/3<<"*3+1="<<n<<endl;}else{n/=2;cout<<"Step No."<<++cnt<<":"<<2*n<<"/2="<<n<<endl;}} while(__________); // （3）cout<<endl;}return 0;}3．四方定理This program is to verify Theorem of Four Squares.That is all natural numbers can be represented as sum of no more than 4 squares of the numbers.e.g., 123=7*7+7*7+4*4+3*3.#include <iostream>using namespace std;int main(){int i,j,k,l,number;while(1){cout<<"Please input a number to verify(0 to quit): ";cin>>number;if(number==0) __________; // （1）cout<<" ------ Results of verification: ------------\n";for(i=1;i<=number/2;i++)for(j=0;j<=i;j++)for(k=0;k<=j;k++)for(l=0;l<=k;l++)if(__________) // （2）{cout<<number<<"="<<i<<"*"<<i<<"+"<<j<<"*"<<j<<"+"<<k<<"*"<<k<<"+"<<l< <"*"<<l<<endl;goto exit;}exit: cout<<" ---------------------------------------------\n";}return 0;}4．亲密数This is a program to find friendly numbers pair.Which means the sum of integer A's all factors (except A) equals to the sum of integer B's all factors (except B)< e.g. sum of integer 220's all factors are:1+2+4+5+10+11+20+22+44+55+110=284,and sum of integer 284's all factors are:1+2+4+71+142=220>#include <iostream>using namespace std;int main(){int a,i,b,n,m;cout<<"Please input the scale you want to find n: ";cin>>n;cout<<"\n There are following friendly--numbers pair smaller than "<<n<<endl;for(a=1;a<n;a++){for(__________;i<=a/2;i++) // （1）if(!(a%i)) b+=i;for(__________;i<=b/2;i++) // （2）if(!(b%i)) m+=i;if(__________&&a<b) // （3）cout<<a<<".."<<b<<" ";}cout<<endl;return 0;}5．自守数This program will find the automorphic numbers.The defination of a automorphic number is: the mantissa of a natural number's square equals to itself. e.g., 5^2=25, 76^2=5776, 9376^2=87909376.#include <iostream>using namespace std;int main(){int mul,number,k,kk;for(number=0;number<10000;number++){for(mul=number,k=1; __________;k*=10);// （1）kk=k*10;mul=number*number%kk;if(__________) // （2）cout<<number<<" ";}cout<<endl;return 0;}6．特殊的四位数This program will find the four figures which have the characteristic as follows: abcd=(ab+cd)^2. e.g., 3025=(30+25)*(30+25).#include <iostream>using namespace std;int main() {int n,a,b;for(n=1000;n<10000;n++){________________; // （1）________________; // （2）if((a+b)*(a+b)==n)cout<<n<<" ";}cout<<endl;return 0;}7．求方程cos(x)-x=0的根This program is to find the real root of function cos(x)-x=0.#include <iostream>#include <cmath>using namespace std;int main() {float x0,x1=0.0;while(1){________________; // （1）________________; // （2）if(fabs(x0-x1)<1e-6)break;}cout<<"The real root is "<<x1<<endl;return 0;}8．特殊的3位数This program is to find the Perfect Square Numbers. which have 3 digits, and 2 of them are the same. e.g.100, 121, 144, 225, 400, 441, 484, 676, 900.#include <iostream>#include <cmath>using namespace std;int main(){int i,n,a,b,c;for (i=10;i<=sqrt(1000);i++){_______________;// （1）a=n/100;_______________;// （2）c=n%10;if (a==b || a==c || b==c)cout<<n<<" ";}cout<<endl;return 0;}9．三重回文数This program is to find the Palindrome Numbers. whose square and cubic are also Palindrome Numbers.#include <iostream>using namespace std;bool isPalindrome(int n){int x,y=0;x=n;while (x!=0){_______________;// （1）_______________;// （2）}return y==n;}int main(){int m;for(m=11;m<1000;m++){if(isPalindrome(m) && isPalindrome(m*m) && isPalindrome(m*m*m)){cout<<m<<" "<<m*m<<" "<<m*m*m<<endl;}}return 0;}10．SIX and NINEThis program is to find the numbers of SIX and NINE, which satisfy the formula SIX+SIX+SIX=NINE+NINE, where S,I,X,N,E stand for digits between 0 and 9.#include <iostream>using namespace std;int main(){int i,x,a,b,c,d;for(i=668;i<=999; _____________)// （1）{x=3*i/2;a=i/10%10; b=x/100%10; c=x/1000; d=x%100/10;if(____________________)// （2）{cout<<i<<" "<<x<<endl;}}return 0;}或#include <iostream>using namespace std;int main(){int s,i,x,n,e,six,nine;for(s=1;s<10;s++)for(i=0;i<10;i++)for(x=0;x<10;x++)for(n=1;n<10;n++)for(e=0;e<10;e++){six=_____________;// （3）nine=_____________;// （4）if(_____________)// （5）cout<<six<<" "<<nine<<endl;}return 0;}11．马克思手稿中的数学题This program is to solve an interesting math question in Karl Marx's manuscript.The Problem is as follows: 30 persons spent 50 yuan in a restaurant, amony them, each man spent 3 yuan, each woman spent 2 yuan, and each child spent 1 yuan. The question is how many men, women and children are there?#include <iostream>using namespace std;int main(){int x,y,z;cout<<"Men \tWomen \tChildren"<<endl;for(x=0;x<=10;x++){________________; // （1）________________; // （2）if(__________________________) // （3）cout<<x<<"\t"<<y<<"\t"<<z<<endl;}return 0;}12．百钱百鸡问题This program is to solve Problem of Hundred Yuan Hundred Fowls.Which is presented by Zhang Qiujiang, a Chinese ancient mathematician, in his work Bible of Calculation: 5 Yuan can buy 1 cock,3 Yuan can buy 1 hen, 1 Yuan buy 3 chickens, now one has 100 Yuan to buy 100 fowls, the question is how many cocks, hens, chickens to buy?#include <iostream>using namespace std;int main(){int x,y,z;for(x=0; _____________;x++) // 外层循环控制鸡翁数x （1）for(y=0; _____________;y++) // 内层循环控制鸡母数y （2）{z=100-x-y;if(_____________) // （3）cout<<"cock="<<x<<", hen="<<y<<", chicken="<<z<<endl;}return 0;}13．三色球问题This program is to solve Problem of Three Color Ball.The Problem is as follows:There are 12 balls in the pocket.Amony them, 3 balls are red,3 balls are white and 6 balls are black. Now take out any 8 balls from the pocket,how many color combinations are there?#include <iostream>using namespace std;int main(){int i,j,cnt=0;cout<<"RED \tWHITE \tBLACK"<<endl;for(i=0; _____________;i++) // （1）for(j=0; _____________;j++) // （2）if(_____________) // （3）cout<<++cnt<<" : "<<i<<"\t"<<j<<"\t"<<8-i-j<<endl return 0;}14．配对新郎和新娘This program is to solve Problem of Bridegroom and Bride.The Problem is as follows: Someone goes to 3 couples lovers'wedding. The bridegrooms are A,B,C and the brides are X,Y,Z. He wants to know who marries who and asks them. A says he will marry to X, X says her fiance is C, C says he will marry to Z. The man knows that they are all kidding. What they said is not true. So try to find who will marry to who? #include <iostream>using namespace std;int main(){int x,y,z;for(x=1;x<=3;x++) // 穷举x的全部可能配偶for(y=1;y<=3;y++) // 穷举y的全部可能配偶for(z=1;z<=3;z++) // 穷举z的全部可能配偶if(x!=1&&x!=3&&z!=3&&x!=y&&x!=z&&y!=z){cout<<" X will marry to "<<char(_____________)<<endl; // （1）cout<<" Y will marry to "<<char(_____________)<<endl; // （1）cout<<" Z will marry to "<<char(_____________)<<endl; // （1）}return 0;}15．邮票组合This program is to solve Problem of Stamp Combination.The Problem is as follows. John has 4 stamps with value of 3 cents and 3 stamps with value of 5 cents. Use one or more of these stamps, how many kinds of postages can John provide?#include <iostream>using namespace std;int main(){int i,j,s,n=0,a[28]={0};for(i=0;i<=4;i++)for(j=0;j<=3;j++){____________; // （1）if (____________) { a[s]=1; n++; } // （2）}cout<<"There are "<<n-1<< " kinds of postages:\n";for (i=1;i<=27;i++)if (____________) cout<<i<<" "; // （3）cout<<endl;return 0;}参考答案：1．（1）break （2）i<n （3）a+i*22．（1）break （2）n%2==1 （3）n!=13．（1）break （2）number==i*i+j*j+k*k+l*l4．（1）b=0,i=1 （2）m=0,i=1 （3）m==a5．（1）(mul/=10)>0 （2）number==mul6．（1）a=n/100 （2）b=n%1007．（1）x0=x1 （2）x1=cos(x0)8．（1）n=i*i （2）b=n/10%109．（1）y=y*10+x%10 （2）x=x/1010．（1）i=i+2 （2）a==b && c==d （3）s*100+i*10+x（4）n*1000+i*100+n*10+e （5）3*six==2*nine11．（1）y=20-2*x （2）z=30-x-y （3）3*x+2*y+z==5012．（1）x<=20 （2）y<=33 （3）z%3==0&&5*x+3*y+z/3==100 13．（1）i<=3 （2）j<=3 （3）(8-i-j)<=614．（1）'A'+x-1 （2）'A'+y-1 （3）'A'+z-115．（1）s=i*3+j*5 （2）a[s]==0 （3）a[i]==1。