问题描述

Sky从小喜欢奇特的东西，而且天生对数字特别敏感，一次偶然的机会，他发现了一个有趣的四位数 2992，这个数，它的十进制数表示，其四位数字之和为2+9+9+2=22，它的十六进制数BB0，其四位数字之和也为22，同时它的十二进制数表示 1894，其四位数字之和也为22，啊哈，真是巧啊。Sky非常喜欢这种四位数，由于他的发现，所以这里我们命名其为Sky数。但是要判断这样的数还是有点麻烦啊，那么现在请你帮忙来判断任何一个十进制的四位数，是不是Sky数吧。
 

输入

输入含有一些四位正整数，如果为0，则输入结束。

输出

若n为Sky数，则输出“#n is a Sky Number.”，否则输出“#n is not a Sky Number.”。每个结果占一行。注意：#n表示所读入的n值。

输入样例

2992
1234
0

输出样例

2992 is a Sky Number.
1234 is not a Sky Number.

代码如下：
#include<stdio.h>
int main()
{
    int b[4],c[4];
    int d,e,f,g,i,n,h;
    while(scanf("%d",&n)!=EOF)
    {
        if(n==0)
            break;
        h=n;
        d=e=f=g=0;
        d=n/1000+(n%1000/100)+(n%1000%100/10)+(n%1000%100%10/1);
        for(i=0;n!=0 && i<4;i++)
        {
            b[i]=n%16;
            n=n/16;
        }
        n=h;
        for(i=0;n!=0 && i<4;i++)
        {
            c[i]=n%12;
            n=n/12;
        }
        for(i=0;i<4;i++)
        {
            f+=b[i];
            g+=c[i];
        }
        if(d==f==g)
        {
            printf("%d is a Sky Number.\n",h);
        }
        else
        {
            printf("%d is not a Sky Number.\n",h);
        }
    }
    return 0;
}

#include<stdio.h>
int main()
{
    int b[4],c[4];
    int d,e,f,g,i,n,h;
    while(scanf("%d",&n)!=EOF)
    {
        if(n==0)
            break;
        h=n;
        d=e=f=g=0;
        d=n/1000+(n%1000/100)+(n%1000%100/10)+(n%1000%100%10/1);
        for(i=0;n!=0 && i<4;i++)
        {
            b[i]=n%16;
            n=n/16;
        }/*2992%16=0，2992/16=178；178%16=2，178/16=11；11%16=11，11/16=0；0%16=0，0/16=0*/
        n=h;
        for(i=0;n!=0 && i<4;i++)
        {
            c[i]=n%12;
            n=n/12;
        }
        for(i=0;i<4;i++)
        {
            f+=b[i];
            g+=c[i];
        }
        if(d==f==g)/*d==f&&f==g*/
        {
            printf("%d is a Sky Number.\n",h);
        }
        else
        {
            printf("%d is not a Sky Number.\n",h);
        }
    }
    return 0;
}

#include<stdio.h>
 #include<conio.h>
int main()
{
    int b[4],c[4];
    int d,e,f,g,i,n,h;
    while(scanf("%d",&n)!=EOF)
    {
        if(n==0)
            break;
        h=n;
        d=e=f=g=0;
        d=n/1000+(n%1000/100)+(n%1000%100/10)+(n%1000%100%10/1);
        for(i=0;n!=0 && i<4;i++)
        {
            b[i]=n%16;
            n=n/16;
        }
        n=h;
        for(i=0;n!=0 && i<4;i++)
        {
            c[i]=n%12;
            n=n/12;
        }
        for(i=0;i<3;i++)              //2992转化成16进制后占3位，
        f+=b[i];
        for(i=0;i<4;i++)
        {

            g+=c[i];
        }  printf("%d ",f);getch();
        if(d==f&&d==g)
        {
            printf("%d is a Sky Number.\n",h);
        }
        else
        {
            printf("%d is not a Sky Number.\n",h);
        }
    }getch();
    return 0;
}

  for(i=0;n!=0 && i<4;i++)
        {
            b[i]=n%16;
            n=n/16;
        }
b[]中的最后一个元素不是没有数值，系统给定任意值；他的累加和自然也不正确

这是在VC里的，你的 if(d==f==g)也不正确

能不能这样判断啊 我刚开始学c 不知道这样写有什么问题，请各位高手帮忙指点一下

 d=n/1000+(n%1000/100)+(n%1000%100/10)+(n%1000%100%10/1);
    if（d!=n/4096+(n%4096/256)+(n%4096%256/16)+(n%4096%256%16)）
       {
            printf("%d is not a Sky Number.\n",h);
            break;
        }
      if（d==n/1728+(n%1728/144)+(n%1728%144/12)+(n%1728%144%12)）
       {
            printf("%d is a Sky Number.\n",h);

        }
       else
        {
            printf("%d is not a Sky Number.\n",h);
         }


     

#include "stdio.h"
int main(void)
{
    int b[10],c[10];
    int d,f,g,i,j,n,h,number;
    printf("input the number:");
    scanf("%d",&n);
    while(n<1000||n>9999)
    {
        printf("the data error,input again:");
        scanf("%d",&n);
    }
    while(n>=1000&&n<=9999)
    {
        number=h=n;
        f=g=0;
        d=n/1000+(n%1000/100)+(n%100/10)+(n%10);
        b[0]=n%16;
        i=1;
        f=b[0];
        while(b[i])
        {
            i++;
            n=n/16;
            b[i]=n%16;
            f+=b[i];
        }
        c[0]=h%12;
        j=1;
        g=c[0];
        while(c[j])
        {   
            j++;
            h=h/12;
            c[i]=h%12;
            g+=c[i];
        }
        if(d==f&&f==g)
            printf("%d is a Sky Number.\n",number);
        else
            printf("%d is not a Sky Number.\n",number);
        exit(0);
    }
    return 0;
}

#include<stdio.h>
int main()
{
    int b[4],c[4];
    int d,f,g,i,n,h;
    while(scanf("%d",&n)!=EOF)
    {
        if(n==0)
        {
            break;
        }
        h=n;
        d=f=g=0;
        d=n/1000+(n%1000/100)+(n%1000%100/10)+(n%1000%100%10/1);
        for(i=0;n!=0 && i<4;i++)
        {
            b[i]=n%16;
            n=n/16;
        }
        n=h;
        for(i=0;n!=0 && i<4;i++)
        {
            c[i]=n%12;
            n=n/12;
        }
        for(i=0;i<3;i++)
            f+=b[i];
        for(i=0;i<4;i++)
        {
            g+=c[i];
        }
        if(d==f&&d==g)
        {
            printf("%d is a Sky Number.\n",h);
        }
        else
        {
            printf("%d is not a Sky Number.\n",h);
        }
    }
    return 0;
}

我这样交上去，为什么答案错误？

不好意思我知道哪错了，数组长度短了。变成b[1000],c[1000]就对了

#include<stdio.h>
main()
{
    int b[4]={0},c[4]={0};
    int d,f=0,g=0,i,n,h;
   
    scanf("%d",&n);
    while(n!=0)
    {
        h=n;
        d=n/1000+n%1000/100+n%100/10+n%10;//十位数相加
        while(n>0)
        {
            f=f+n%16;
            n=n/16;
        }
        n=h;
        while(n>0)
        {
            g=g+n%16;
            n=n/16;
        }
        if((d==f)&&(f==g))
        {
            printf("%d is a Sky Number.\n",h);
        }
        else
        {
            printf("%d is not a Sky Number.\n",h);
        }   
        scanf("%d",&n);
    }
    return 0;
}