牛客多校4J二分答案连续子段最大平均值

Posted 2021-08-29 hesorchen

题目

Average

问题可以转化为 求长度不小于K的子段的最大平均值。POJ2018原题。

其中一种比较简单的方法是，二分出答案$mid$，对原数组每个数减去$mid$后做前缀和$sum$，如果存在一段区间$[l,r]$满足$su{m}_r-su{m}_{l-1}>=0$，那么最大平均值$mid$就存在。

代码

include <bits/stdc++.h>
using namespace std;
const int N = 2e5 + 10;
int n, m, x, y;
int a[N], b[N];
double c[N], pre[N];

double eps = 1e-6;
double sov(int a[], int n, int x)
{
    double l = 0, r = 100000;
    while (l + eps <= r)
    {
        double mid = (l + r) / 2;
        double minn = 100000;
        int flag = 0;
        for (int i = 1; i <= n; i++)
        {
            c[i] = 1.0 * a[i] - mid;
            pre[i] = pre[i - 1] + c[i];
            if (i >= x)
            {
                minn = min(minn, pre[i - x]);
                if (pre[i] >= minn)
                    flag = 1;
            }
        }
        flag ? l = mid + eps : r = mid - eps;
    }
    return l;
}

int main()
{
    scanf("%d %d %d %d", &n, &m, &x, &y);
    for (int i = 1; i <= n; i++)
        scanf("%d", &a[i]);
    for (int i = 1; i <= m; i++)
        scanf("%d", &b[i]);
    double ans = sov(a, n, x) + sov(b, m, y);
    printf("%.10f\\n", ans);
}