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：思路感觉挺好想的，但是有可能容易超时。对于连续的1个数相同的情况，我们可以合并考虑，因为他们的情况都是一样的。因此，我们可以分别计算出a序列，连续的1的个数有多少种情况，并记录每一种情况的个数。同理求出b序列的。然后我们分别遍历a，b序列的这些情况，求出符合条件的就可以了。
代码如下：
#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)/~
