程序填空题(二)

风格不统一 提交于 2021-01-04 16:41:32

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n

               a=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 NINE

This 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;

}

#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*2

2.(1)break    (2)n%2==1    (3)n!=1

3.(1)break    (2)number==i*i+j*j+k*k+l*l

4.(1)b=0,i=1  (2)m=0,i=1  (3)m==a

5.(1)(mul/=10)>0   (2)number==mul

6.(1)a=n/100      (2)b=n%100

7.(1)x0=x1    (2)x1=cos(x0)

8.(1)n=i*i    (2)b=n/10%10

9.(1)y=y*10+x%10    (2)x=x/10 

10.(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*nine

11.(1)y=20-2*x     (2)z=30-x-y   (3)3*x+2*y+z==50

12.(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)<=6

14.(1)'A'+x-1   (2)'A'+y-1  (3)'A'+z-1

15.(1)s=i*3+j*5   (2)a[s]==0   (3)a[i]==1

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