題干：

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

3 3 1 2 3 1 2 3 9

Output

4

Input

5 1 5 4 2 4 5 2 5

Output

1

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):

題目大意：

? ? ?給出兩個數組A和B，他們能組成一個矩陣，然后求子矩陣的和<=X的，最大的子矩陣的面積。

?

解題報告：

? ?聽說有人往滑窗問題上想了？

? ? 如果題目規定是正方形就好做了啊，二維前綴和+二分寬度，直接o(nmlog(nm))，如果矩形的話只能帶倆log了。。。沒嘗試過。但應該也不難寫。

? ?想想啊，為什么題目不直接給你一個矩陣，而是給你倆數組讓你自己造矩陣？肯定是因為矩陣只是個幌子，真正要用到的還是那兩個數組。所以我們推公式：

于是我們知道了和x比較的值只與連續子段和有關，且A和B這兩者之間沒關系（只是最后做一下乘法）所以算出每個長度的最小和，最后再o(nm)遍歷一遍維護答案就行了。

AC代碼：

#include<bits/stdc++.h> #define ll long long using namespace std; const ll INF = 0x3f3f3f3f; ll a[2005],b[2005],suma[2005],sumb[2005]; ll dpa[2005],dpb[2005]; ll x,ans; int main() {int n,m;cin>>n>>m;memset(dpa,INF,sizeof dpa);memset(dpb,INF,sizeof dpb);for(int i = 1; i<=n; i++) scanf("%lld",a+i),suma[i] = suma[i-1] + a[i];for(int i = 1; i<=m; i++) scanf("%lld",b+i),sumb[i] = sumb[i-1] + b[i]; scanf("%lld",&x);for(int len = 1; len<=n; len++) {for(int i = 1; i+len-1<=n; i++) {dpa[len] = min(dpa[len],suma[i+len-1] - suma[i-1]);}}for(int len = 1; len<=m; len++) {for(int i = 1; i+len-1<=m; i++) {dpb[len] = min(dpb[len],sumb[i+len-1] - sumb[i-1]);} }for(int i = 1; i<=n; i++) {for(int j = 1; j<=m; j++) {if(dpa[i] * dpb[j] <= x) ans = max(ans,(ll)i*j);}}printf("%lld\n",ans);return 0 ; }

?

網絡版AC代碼：（博客鏈接）

#include<cstdio> #include<cstring> #include<algorithm> using namespace std; typedef long long ll; const int N=2000; ll lA,lB,x; int A[N+5],B[N+5]; struct NODE{ll len,sum; }node[N*N+5]; int cnt_NODE; bool cmp_NODE(NODE A,NODE B){return A.sum<B.sum; } int Solve(ll sum){int l=0,r=cnt_NODE-1;if(node[r].sum<=sum) return r;while(l+1<r){int mid=(l+r)>>1;if(node[mid].sum<=sum) l=mid;else r=mid;}if(node[r].sum<=sum) return r;return l; } int main(){scanf("%d%d",&lA,&lB);for(int i=1;i<=lA;i++)scanf("%d",&A[i]);for(int i=1;i<=lB;i++)scanf("%d",&B[i]);scanf("%d",&x);for(int i=1;i<=lA;i++){ll sum=0;for(int j=i;j<=lA;j++){sum+=A[j];node[cnt_NODE].len=j-i+1;node[cnt_NODE].sum=sum;cnt_NODE++;}}sort(node,node+cnt_NODE,cmp_NODE);for(int i=1;i<cnt_NODE;i++)node[i].len=max(node[i].len,node[i-1].len);ll ans=0;for(int i=1;i<=lB;i++){ll sum=0;for(int j=i;j<=lB;j++){sum+=B[j];if(x/sum==0) continue;int res=Solve(x/sum);if(node[res].sum>x/sum) continue;ans=max(ans,node[res].len*(j-i+1));}}printf("%lld\n",ans);return 0; }

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