1 条题解
-
0
C++ :
#include <bits/stdc++.h> using namespace std; int n, m, a[100001]; long long l = 0, r = 1e18, mid, ans = 1e18, cnt; int main() { cin >> n >> m; for (int i = 1; i <= n; i++) cin >> a[i]; while (l <= r) { mid = (l + r) / 2; cnt = 0; for (int i = 1; i <= n; i++) cnt += (mid / a[i]); if (cnt < m) l = mid + 1; else { ans = min(ans, mid); r = mid - 1; } } cout << ans; return 0; }
- 1
信息
- ID
- 535
- 时间
- 1000ms
- 内存
- 128MiB
- 难度
- (无)
- 标签
- (无)
- 递交数
- 0
- 已通过
- 0
- 上传者