1 条题解
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0
#include<bits/stdc++.h> using namespace std;
int main() { int n; cin >> n; int l; cin >> l; int ants[n]; for (int i = 0; i < n; i++) { cin >> ants[i]; } int minTime = 0; for (int i = 0; i < n; i++) { int timeToLeftEnd = ants[i]; int timeToRightEnd = l - ants[i]; int currentMinTime = min(timeToLeftEnd, timeToRightEnd); minTime = max(minTime, currentMinTime); } cout << minTime << endl; return 0; } (这都不会)
- 1
信息
- ID
- 3899
- 时间
- 1000ms
- 内存
- 128MiB
- 难度
- 2
- 标签
- (无)
- 递交数
- 42
- 已通过
- 27
- 上传者