2021牛客暑期多校训练营4 J. Average（最长连续子序列平均值）

链接：https://ac.nowcoder.com/acm/contest/11255/J
来源：牛客网

题目描述

Bob has an n×mn×m matrix WW.

This matrix is very special, It's calculated by two sequences a1...n,b1...ma1...n,b1...m, ∀i∈[1,n],∀j∈[1,m]∀i∈[1,n],∀j∈[1,m], Wi,j=ai+bjWi,j=ai+bj

Now Bob wants to find a submatrix of WW with the largest average value.

Bob doesn't want the size of submatrix to be too small, so the submatrix you find must satisfy that the width(the first dimension of matrix) of it is at least xx and the height(the second dimension of matrix) of it is at least yy.

Now you need to calculate the largest average value.

输入描述:

The first line has four integers n,m,x,yn,m,x,y.

The second line has nn integers a1...na1...n.

The third line has mm integers b1...mb1...m.

1≤n,m≤1051≤n,m≤105

1≤x≤n,1≤y≤m1≤x≤n,1≤y≤m

0≤ai,bi≤1050≤ai,bi≤105

输出描述:

Output the largest average value.

Your answer will be considered correct if the absolute or relative error is less than 10−610−6

示例1

输入

复制

3 4 2 2
3 1 2
4 1 3 2

输出

复制

4.6666666667

不妨设选择的那块矩形长为$x$，宽为$y$，对应的行列的起始位置分别为$x_l,x_r,y_l,y_r$，则平均值为$average = \frac{ySum[x_l...x_r]+xSum[y_l...y_r]}{xy}=\frac{Sum[x_l...x_r]}{x}+\frac{Sum[y_l...y_r]}{y}$，其中$Sum[x_l...x_r]$为a数组的$x_l$到$x_r$位置的和。最大化均值相当于分别求a、b两个数组的连续子区间的最大均值然后再求和。而这个问题可以参考poj2018https://blog.csdn.net/weixin_43191865/article/details/90612344

#include <bits/stdc++.h>
using namespace std;
int n, m, x, y;
double a[100005], b[100006];
double sum1[100005], sum2[100005];

int main() {
    cin >> n >> m >> x >> y;
    double ans1 = 0, ans2 = 0;
    for(int i = 1; i <= n; i++) {
        cin >> a[i];
    }
    for(int i = 1; i <= m; i++) {
        cin >> b[i];
    }   
    double l=-1e7;
    double r=1e7;
    while(r-l>1e-9)
    {
        double mid=(l+r)/2;
        for(int i=1;i<=n;i++)
        {
            sum1[i]=sum1[i-1]+a[i]-mid;
        }
        double minn=1e9;
        double maxx=-1e9;
        for(int i=x;i<=n;i++)//这里多想一下就明白了。
        {
            minn=min(sum1[i-x],minn);
            maxx=max(sum1[i]-minn,maxx);
        }
        if(maxx>0)
        {
            l=mid;
        }
        else
        {
            r=mid;
        }
    }
    ans1 = r;
    l=-1e7;
    r=1e7;
    while(r-l>1e-9)
    {
        double mid=(l+r)/2;
        for(int i=1;i<=m;i++)
        {
            sum2[i]=sum2[i-1]+b[i]-mid;
        }
        double minn=1e9;
        double maxx=-1e9;
        for(int i=y;i<=m;i++)
        {
            minn=min(sum2[i-y],minn);
            maxx=max(sum2[i]-minn,maxx);
        }
        if(maxx>0)
        {
            l=mid;
        }
        else
        {
            r=mid;
        }
    }
    ans2 = r;
    cout << fixed << setprecision(8) << ans1 + ans2;
}