描述

You are given two arrays aa and bb of positive integers, with length nn and mm respectively.

Let cc be an n×mn×m matrix, where ci,j=ai⋅bjci,j=ai⋅bj.

You need to find a subrectangle of the matrix cc such that the sum of its elements is at most xx, and its area (the total number of elements) is the largest possible.

Formally, you need to find the largest number ss such that it is possible to choose integers x1,x2,y1,y2x1,x2,y1,y2 subject to 1≤x1≤x2≤n1≤x1≤x2≤n, 1≤y1≤y2≤m1≤y1≤y2≤m,

$\sum_{i=x_1}^{x_2}{\sum_{j=y_1}^{y_2}{c_{i,j}}} \leq x.$

Input

The first line contains two integers nn and mm (1≤n,m≤20001≤n,m≤2000).

The second line contains nn integers a1,a2,…,ana1,a2,…,an (1≤ai≤20001≤ai≤2000).

The third line contains mm integers b1,b2,…,bmb1,b2,…,bm (1≤bi≤20001≤bi≤2000).

The fourth line contains a single integer xx (1≤x≤2⋅1091≤x≤2⋅109).

Output

If it is possible to choose four integers x1,x2,y1,y2x1,x2,y1,y2 such that 1≤x1≤x2≤n1≤x1≤x2≤n, 1≤y1≤y2≤m1≤y1≤y2≤m, and ∑x2i=x1∑y2j=y1ci,j≤x∑i=x1x2∑j=y1y2ci,j≤x, output the largest value of (x2−x1+1)×(y2−y1+1)(x2−x1+1)×(y2−y1+1) among all such quadruplets, otherwise output 00.

input

output

input

output

Note

Matrix from the first sample and the chosen subrectangle (of blue color):

Matrix from the second sample and the chosen subrectangle (of blue color):

思路

给你两个数组 $a[],b[]$ ,现在我们可以由这两个数组构造出一个矩阵 $c$ ,矩阵 $c$ 中的元素是 $c[i][j]=a[i]*b[j]$

然后给你一个数 $x$ ,让你在矩阵c中找出一个子矩阵，使得这个矩阵的面积最大，且满足这个子矩阵中所有元素之和小于x.

因为c中的每一个元素是由数组a和b推导出来的，所以我们要求矩阵c中的任意一个矩阵的和就可以通过a数组对应位置的和(就是c中的行)乘以b数组对应位置的和(就是c中的列).

我们枚举a数组和b数组的间隔，我们可以求出间隔为0~n-1和0~m-1的和的最小值(用前缀和实现).然后计算出h[i]*c[j]<=x的值的最大的i*j就是答案。

代码

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
#define mem(a, b) memset(a, b, sizeof(a))
#define inf 0x3f3f3f
const ll N = 2000 + 10;
ll a[N], b[N], n, m, x;
ll h[N], c[N];
int main()
{
    //freopen("in.txt", "r", stdin);
    scanf("%lld%lld", &n, &m);
    for (ll i = 1; i <= n; i++)
    {
        scanf("%lld", &a[i]);
        a[i] += a[i - 1];
    }
    for (ll i = 1; i <= m; i++)
    {
        scanf("%lld", &b[i]);
        b[i] += b[i - 1];
    }
    scanf("%lld", &x);
    mem(h, inf), mem(c, inf);
    for (ll l = 0; l <= n; l++)
    {
        for (ll i = 1; i + l <= n; i++)
        {
            ll j = i + l;
            h[l] = min(h[l], a[j] - a[i - 1]);
        }
    }
    for (ll l = 0; l <= m; l++)
    {
        for (ll i = 1; i + l <= m; i++)
        {
            ll j = i + l;
            c[l] = min(c[l], b[j] - b[i - 1]);
        }
    }
    ll ans = 0;
    for (ll i = 1; i <= n; i++)
        for (ll j = 1; j <= m; j++)
            if (h[i - 1] * c[j - 1] <= x)
                ans = max(ans, i * j);
    printf("%lld\n", ans);
    return 0;
}

Codeforces Round #513 by Barcelona Bootcamp (rated, Div. 1 + Div. 2) C. Maximum Subrectangle(思路)

描述

Input

Output

input

output

input

Note

思路

代码

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