You are given an array a of length n and array b of length m both consisting of only integers 0 and 1. Consider a matrix c of size n×m formed by following rule: ci,j=ai?bj (i.e. ai multiplied by bj). It’s easy to see that c consists of only zeroes and ones too.

How many subrectangles of size (area) k consisting only of ones are there in c?

A subrectangle is an intersection of a consecutive (subsequent) segment of rows and a consecutive (subsequent) segment of columns. I.e. consider four integers x1,x2,y1,y2 (1≤x1≤x2≤n, 1≤y1≤y2≤m) a subrectangle c[x1…x2][y1…y2] is an intersection of the rows x1,x1+1,x1+2,…,x2 and the columns y1,y1+1,y1+2,…,y2.

The size (area) of a subrectangle is the total number of cells in it.

Input
The first line contains three integers n, m and k (1≤n,m≤40000,1≤k≤n?m), length of array a, length of array b and required size of subrectangles.

The second line contains n integers a1,a2,…,an (0≤ai≤1), elements of a.

The third line contains m integers b1,b2,…,bm (0≤bi≤1), elements of b.

Output
Output single integer — the number of subrectangles of c with size (area) k consisting only of ones.

Examples
Input
3 3 2
1 0 1
1 1 1
Output
4
Input
3 5 4
1 1 1
1 1 1 1 1
Output
14
Note
In first example matrix c is:

There are 4 subrectangles of size 2 consisting of only ones in it:

In second example matrix c is:

思路：思路感覺挺好想的，但是有可能容易超時。對于連續的1個數相同的情況，我們可以合并考慮，因為他們的情況都是一樣的。因此，我們可以分別計算出a序列，連續的1的個數有多少種情況，并記錄每一種情況的個數。同理求出b序列的。然后我們分別遍歷a，b序列的這些情況，求出符合條件的就可以了。
代碼如下：

#include<bits/stdc++.h> #define ll long long using namespace std;const int maxx=4e4+100; map<int,int> p1,p2; int a[maxx]; int b[maxx]; int n,m,k;inline void dfs(int c[],int &j,int limit) {if(j>limit) return ;if(c[j]==0) return ;dfs(c,j+=1,limit); } inline void fcs() {int j;for(int i=1;i<=n;){if(a[i]==1){j=i;dfs(a,j,n);p1[j-i]++;i=j;}else i++;}for(int i=1;i<=m;){if(b[i]==1){j=i;dfs(b,j,m);p2[j-i]++;i=j;}else i++;} } inline void solve() {int len1,len2,num1,num2,_min,_max;ll sum=0;for(map<int,int> ::iterator it1=p1.begin();it1!=p1.end();it1++){len1=it1->first;num1=it1->second;for(map<int,int> ::iterator it2=p2.begin();it2!=p2.end();it2++){len2=it2->first;num2=it2->second;_min=min(len1,len2);_max=max(len1,len2);for(int i=1;i<=min(_min,k);i++){if(k%i==0&&k/i<=_max) sum+=((_max-k/i+1)*(_min-i+1))*num1*num2;}}}cout<<sum<<endl; } int main() {scanf("%d%d%d",&n,&m,&k);for(int i=1;i<=n;i++) cin>>a[i];for(int i=1;i<=m;i++) cin>>b[i];p1.clear();p2.clear();fcs();solve();return 0; }

努力加油a啊，(^o)/~

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