Uva 202 Repeating Decimals

关键是判断循环的长度，若某一步的余数已出现过，则循环确定
#include <cstdio>
#include <cstring>

int s[10000], y[10000],sign[10000];

int main()
{
    int a, b, sz, yu, k , cnt;
    while(scanf("%d%d",&a,&b)==2)
    {
        memset(sign,0,sizeof(sign));
        getchar();
        k  = 0, cnt = 0;
        sz = a / b;
        yu = a % b;
        y[0] = yu;
        while(sign[yu] != 1)
        {
            s[cnt++] = yu * 10 / b;
             sign[yu] = 1;
             yu = yu * 10 % b;
             y[cnt] = yu;
        }
        printf("%d/%d = %d.",a,b,sz);
        for(int i = 0; i < cnt && i < 50; i++)
        {
            if(y[i] == yu)
            {
                k = i;
                printf("(");
            }
            printf("%d",s[i]);
        }
        if((cnt - k) > 50) printf("...");
        printf(")\n   %d = number of digits in repeating cycle\n\n",cnt - k);

    }
    return 0;
}