将k分成两个数相乘的形式（假设为k = a * b）

在row[]中找出有多少个a个连续的1；
在col[]中找出有多少个b个连续的1；

将两者相乘再累加；

#include <iostream>
#include <algorithm>
#include <cstring>

using namespace std;

typedef long long LL;

const int N = 40010;

int n, m, k;
bool row[N], col[N];

LL count(int x, bool arr[], int len)
{
    int res = 0, cnt = 0;
    for(int i = 1;i <= len;i ++)
        if(arr[i]) cnt ++;
        else
        {
            if(cnt >= x) res += cnt - x + 1;
            cnt = 0;
        }
    if(cnt >= x) res += cnt - x + 1;

    return res;
}

int main()
{
    cin >> n >> m >> k;
    for(int i = 1;i <= n;i ++) cin >> row[i];
    for(int i = 1;i <= m;i ++) cin >> col[i];

    LL res = 0;
    for(int i = 1;i <= k && i <= 40010;i ++)
        if(k % i == 0)
        {
            int a = i, b = k / i;
            LL cnta = count(a, row, n), cntb = count(b, col, m);
            res += cnta * cntb;
        }

    cout << res << endl;

    return 0;
}