Repeating Decimals UVA - 202

觉得这道题主要还是理解停止条件是什么,用两个数组分别存储除数和余数,当出现一组除数和余数存在于数组当中时就出现了循环

#include <iostream>
#include <vector>

using namespace std;

int main() {
	int a, b, flag = 0, length = 0, start = 0;  // length不计入0号单元
	vector<int> v1, v2;
	int m, n;
	while (cin >> a >> b) {
		v1.push_back(a/b);
		v2.push_back(a%b);
		v1.push_back(v2[0]*10/b);
		v2.push_back(v2[0]*10%b);
		length = 1;
		while (1) {
			m = v2[length]*10/b;
			n = v2[length]*10%b;
			length++;
			for (int i = 1; i < length; i++)
				if (m == v1[i] && n == v2[i]) {
					flag = 1;
					start = i;
					break;
				}
			if (flag)
				break;
			v1.push_back(m);
			v2.push_back(n);
		}
		printf("%d/%d = %d.", a, b, v1[0]);
		for (int i = 1; i < length; i++) {
			if (start == i)
				putchar('(');
			if (i > 50) {
				printf("...");
				break;
			}
			cout << v1[i];
		}
		cout << ')' << endl;
		printf("   %d = number of digits in repeating cycle\n\n", length - start);
		flag = 0, length = 0, start = 0;
		v1.clear();
		v2.clear();
	}

	return 0;
}

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转载自blog.csdn.net/malazhuzai/article/details/83590961