UVA 202

 这个题一开始没什么好思路，借鉴了大神的思路，直接考虑商并不好想，要考虑余数！余数有相等的时候循环节就开始了

#include<iostream>
#include <cstdio>
#include <cstring>
using namespace std;
int main()
{
    int m,n;
    int flag[5000];
    int res[5000];
    int temp[5000];
    while(cin>>m>>n) // m/n
    {
        memset(flag,0,sizeof(flag));
        memset(res,0,sizeof(res));
        memset(temp,0,sizeof(temp));
        int interger = m/n;//整数部分
        int num=0;//记录小数点位数
        int t=m;
        m=m%n;   
        while(flag[m]==0 && m!=0)
        {
            flag[m]=1;
            temp[m]=num;
            res[num]=m*10/n;
            num++;
            m=m*10%n;   
        }
        if (m==0)//能除尽
        {
            printf("%d/%d = %d.",t,n,interger);
            for (int i = 0; i < num; ++i)
              cout<<res[i];
            printf("(0)\n   1 = number of digits in repeating cycle\n");
        }
        else
        {
            printf("%d/%d = %d.",t,n,interger);
            for(int i=0;i<temp[m];++i)
                cout<<res[i];
            printf("(");
            if(num-temp[m]>50)
            {
                for (int i = temp[m]; i < 50 + temp[m]  ; ++i)
                    cout<<res[i];
                cout<<"...";
            }
            else
                for (int i = temp[m]; i < num; ++i)
                    cout<<res[i];
            printf(")\n   %d = number of digits in repeating cycle\n",num-temp[m]);
        }
        printf("\n");
    }
    return 0;
}