2 条题解
-
1
#include <cstdio> #include <iostream> using namespace std; int a, b, l, ka = 1000001, kb = 1; double c; inline int gcd(int a, int b) { return !b ? a : gcd(b, a % b); } int main() { freopen("ratio.in", "r", stdin); freopen("ratio.out", "w", stdout); scanf("%d%d%d", &a, &b, &l); c = double(a) / double(b); for (int i = 1; i <= l; i++) for (int j = 1; j <= l; j++) if (gcd(i, j) == 1 && (double)i / (double)j >= c) if ((double)i / (double)j < (double)ka / (double)kb) ka = i, kb = j; printf("%d %d", ka, kb); return 0; }
信息
- ID
- 453
- 时间
- 1000ms
- 内存
- 256MiB
- 难度
- 6
- 标签
- (无)
- 递交数
- 59
- 已通过
- 20
- 上传者