UVa202

https://odzkskevi.qnssl.com/249e6049be24dee57a1c3b28ebf8f12d?v=1494220347

#include<cstdio>
#include<cstring>
using namespace std;
const int N = 3005;
int a[N], v[N];
int main()
{
	//freopen("in.txt", "r", stdin);
	//freopen("out.txt", "w", stdout);
	int n, m, cnt;
	while (~scanf("%d%d", &n, &m))
	{
		cnt = 0;
		memset(v, 0, sizeof(v));
		printf("%d/%d = %d.", n, m, n / m);
		n = n%m; // n为 n/m的小数部分
		while (!v[n])
		{
			a[++cnt] = (n * 10) / m;
			v[n] = cnt;
			n = n * 10 % m;
		}
		for (int i = 1; i <= cnt&&i<51; ++i)
		{
			if (i == v[n]) printf("(");
			printf("%d", a[i]);
			if (i == 50) printf("...");
		}
		printf(")\n %d = number of digits in repeating cycle\n\n", cnt - v[n] + 1);
	}
	return 0;
}