题目：计算分数的循环节。

分析：数论，组合。

            n除以m的余数只能是0~m-1，根据抽屉原则，当计算m+1次时至少存在一个余数相同，

            即为循环节；存储余数和除数，输出即可。

说明：(⊙_⊙)。

#include <iostream>
#include <cstdlib>
#include <cstring>
#include <cstdio>

using namespace std;

int r[3003],u[3003],s[3003];

int main()
{
	int n,m,t;
	while (cin >> n >> m) {
		t = n;
		memset(r, 0, sizeof(r));
		memset(u, 0, sizeof(u));
		int count = 0;
		r[count ++] = n/m;
		n = n%m;
		while (!u[n] && n) {
			u[n] = count;
			s[count] = n;
			r[count ++] = 10*n/m;
			n = 10*n%m;
		}
		printf("%d/%d = %d",t,m,r[0]);
		printf(".");
		for (int i = 1 ; i < count && i <= 50 ; ++ i) {
			if (n && s[i] == n) printf("(");
			printf("%d",r[i]);
		}
		if (!n) printf("(0");
		if (count > 50) printf("...");
		printf(")\n");
		printf("   %d = number of digits in repeating cycle\n\n",!n?1:count-u[n]);
	}
	return 0;
}

UVa 202 - Repeating Decimals