1 条题解

  • 0
    @ 2024-3-31 15:33:56
    #include <iostream>
    #include <cstdio>
    
    using namespace std;
    
    int arr[100005];
    double sum[100005];
    int n, F;
    
    bool fun(double aver) {
        for(int i = 1; i <= n; i++) {
            sum[i] = sum[i - 1] + arr[i] - aver;    
        }
        double minn = 1e9, ans = -1e9;
        for(int i = F; i <= n; i++) {
            minn = min(minn, sum[i - F]);       
            ans = max(ans, sum[i] - minn);
        }
        return ans >= 0;
    }
    
    int main() {
        scanf("%d %d", &n, &F);
        for(int i = 1; i <= n; i++) {
            scanf("%d", &arr[i]);
        }
        double l = -1e6, r = 1e6;
        while(r - l > 1e-5) {
            double mid = (l + r) / 2.0;
            if(fun(mid)) {
                l = mid;
            } else {
                r = mid;
            }
        }
        cout << (long long)(r * 1000) << endl;
        return 0;
    }
    
    • 1

    信息

    ID
    14
    时间
    1000ms
    内存
    128MiB
    难度
    5
    标签
    递交数
    303
    已通过
    115
    上传者