1 条题解

  • 0
    @ 2025-11-6 20:37:19

    #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