一个求数字字根的题：原题：
The digital root of a positive integer is found by summing the digits of the integer. If the resulting value is a single digit then that digit is the digital root. If the resulting value contains two or more digits, those digits are summed and the process is repeated. This is continued as long as necessary to obtain a single digit.

For example, consider the positive integer 24. Adding the 2 and the 4 yields a value of 6. Since 6 is a single digit, 6 is the digital root of 24. Now consider the positive integer 39. Adding the 3 and the 9 yields 12. Since 12 is not a single digit, the process must be repeated. Adding the 1 and the 2 yeilds 3, a single digit and also the digital root of 39.

输入提示：
24
39
0
对应的输出提示：
6
3
我的程序有两个，，但是在杭电上面都是出现Wrong Answer..
1:
#include"stdio.h"
int main()
{
    int n;
    while(scanf("%d",&n)!=EOF&&n!=0)
    {
        if(n%9==0)
        {
            printf("%d\n",9);
        }
        else
        {
            printf("%d\n",n%9);
        }
    }
}
----------------------------------------
#include"stdio.h"
int main()
{
    int n,a,c,i;
    int b[100];
    while(scanf("%d",&n)!=EOF&&n!=0)
    {
st:
        c=n;
        for(i=1;;i++)
        {
            c=c/10;
            if(c==0)
                break;
        }
        a=i;
        for(i=0;i<a;i++)
        {
            b[i]=n%10;
            n=n/10;
        }
        for(i=0;i<a;i++)
        {
            c=c+b[i];
        }
        if(c<10)
            printf("%d\n",c);
        else if(c==10)
            printf("%d\n",1);
        else if(c>10)
        {
            n=c;
            goto st;
        }
    }
}   

有没有中文

题目都是英文的...

题目的意思就是。。24   字根就是 2+4=6，，如果得出来的是9以下的，，那么就结束了输出。以上的就继续做，，比如39  字根是  3+9=12  不行。在做  1+2=3，，所以3就是字根

中文啊、、、
题目的意思就是。。24   字根就是 2+4=6，，如果得出来的是9以下的，，那么就结束了输出。以上的就继续做，，比如39  字根是  3+9=12  不行。在做  1+2=3，，所以3就是字根

改成 unsigned int 试试？

[ 本帖最后由 voidx 于 2011-8-3 19:21 编辑 ]

貌似没区别吧、、我试了下。。没有用的

#include<stdio.h>
#include<string.h>
int root(int n){
    int sum=0;
    while(n){
        sum+=n%10;
        n/=10;
    }
    return sum;
}
int main(){
    int s,i;
    char n[1001];
    while(scanf("%s",n)!=EOF){
        if(!strcmp(n,"0")) break;
        for(i=0,s=0;n[i]!='\0';i++)
            s+=(n[i]-'0');
        while(s>9)
            s=root(s);
        printf("%d\n",s);
    }
    return 0;
}

很巧妙

A了。。好厉害。。！谢谢！谢谢。。我要研究研究！