【CodeForces - 1060C】【Maximum Subrectangle 】

最新推荐文章于 2022-01-28 16:26:00 发布

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最新推荐文章于 2022-01-28 16:26:00 发布

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分类专栏： acm训练

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本文探讨了在给定两个正整数数组的情况下，通过构造矩阵并寻找子矩阵，使得子矩阵元素之和不超过特定值的同时，子矩阵的面积达到最大。文章详细介绍了算法思路，包括如何将矩阵求和转化为一维数组的求和与乘积，以及如何通过固定边长来优化子矩阵的大小。

题目：

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

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, (x2−x1+1)×(y2−y1+1)=s(x2−x1+1)×(y2−y1+1)=s, and

∑i=x1x2∑j=y1y2ci,j≤x.∑i=x1x2∑j=y1y2ci,j≤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.

Examples

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):

解题报告：给我们两个矩阵（行，列），二者做矩阵乘法，然后得到新的矩阵C，其中我们找到其子矩阵，和小于等于X 且范围尽可能的大。这里用到了一个矩阵求和的小装换，子阵的和等于乘出来他的行列和的乘积(很溜的规律）。这样咱们就可以将而为矩阵的求和变为最初两个一维矩阵的求和求积。然后矩阵的话，是要求范围尽可能的大，所以在固定一边的长度情况下，这边的值越小，对方的值就要越大，那么长度就会越大。然后面积就变大。

ac代码：

#include<cstdio>
#include<cstring>
#include<algorithm>
#include<iostream>
using namespace std;
typedef long long ll;
const int inf=0x3f3f3f3f;
const int maxn = 2050;
ll a[maxn],b[maxn];
ll suma[maxn],sumb[maxn];

int main()
{
	ll x,n,m;
	while(scanf("%lld%lld",&n,&m)!=EOF)
	{
		a[0]=b[0]=0;
		for(int i=1;i<=n;i++)
		{
			scanf("%lld",&a[i]);
			a[i]=a[i]+a[i-1];	
		}	
		for(int i=1;i<=m;i++)
		{
			scanf("%lld",&b[i]);
			b[i]=b[i]+b[i-1];
		}
		scanf("%lld",&x);
		memset(suma,inf,sizeof(suma));
		memset(sumb,inf,sizeof(sumb));
		for(int i=1;i<=n;i++)
			for(int j=1;j<=n;j++)
			{
				if(i+j-1>n) break;
				suma[i]=min(suma[i],a[i+j-1]-a[j-1]);
			}
		for(int i=1;i<=m;i++)
			for(int j=1;j<=m;j++)
			{
				if(i+j-1>m) break;
				sumb[i]=min(sumb[i],b[i+j-1]-b[j-1]);
			}
		ll ans=0;
		for(int i=1;i<=n;i++)
			for(int j=1;j<=m;j++)
			{
				if((ll)suma[i]*sumb[j]<=x)
					ans=max(ans,(ll)i*j*1);
			}
		printf("%lld\n",ans);
	}	
	return 0;
}