首先统计对以任意一个位置结尾的子数组的长，在suma和sumb数组中记录下，即suma[i]为a数组中以该点结尾的子数组长度为i的点的数量，然后计算后缀和就是长度为该值的子数组的数。
19：32过了，呜呜呜

#include <iostream>
#include <cstring>
#include <algorithm>
using namespace std;
int const N = 40010;
int n,m,k;
int a[N],b[N];
int suma[N],sumb[N];
int main()
{
    scanf("%d%d%d", &n, &m,&k);
    for (int i = 0; i < n; i ++ ) scanf("%d", &a[i]);
    for (int i = 0; i < m; i ++ ) scanf("%d", &b[i]);

    int sa = a[0];
    memset(suma, 0, sizeof suma);
    for (int j = 1; j <= sa; j ++ ) suma[j]++;

    for (int i = 1; i < n; i ++ ){
        if(a[i]==0){
            sa = 0;
        }else{
            sa++;
            suma[sa]++;

        }
    }
    for (int j = n-1; j >=1; j -- ) suma[j] += suma[j+1];

    int sb = b[0];
    memset(sumb,0, sizeof sumb);
    for (int j = 1; j <= sb; j ++ ) sumb[j]++;

    for (int i = 1; i < m; i ++ ){
        if(b[i]==0){
            sb = 0;
        }else{
            sb++;
            sumb[sb]++;

        }
    }
    for (int j = m-1; j >=1; j -- ) sumb[j]+=sumb[j+1];

    // for (int i = 1; i <= n; i ++ ) cout<<suma[i]<<" ";
    // cout<<endl;
    // for (int i = 1; i <= m; i ++ ) cout<<sumb[i]<<" ";
    // cout<<endl;

    long long res=0;
    for (int i = 1; i <= k/i; i ++ ){
        int t = k/i;
        if(k%i==0){
            if(t<=n && i<=m)
                res += suma[t] * sumb[i];
            if(t<=m && i<=n && t!=i)
                res += suma[i] * sumb[t];
            //cout<<i<<endl;
        }
    }
    cout<<res<<endl;
    return 0;
}