介绍算法：遍历s2，若s2[j]一直等于s1[j - i]*，则y++

* ：如： s1：'liu'，s2：'shizaishiliuliuliua'，则s2[9(j)] = 'l' = s1[9(j) - 9(i)]……s2[11(j)] = 'u' = s1[11(j) - 9(i)]，则y++

AC代码，请查收！

#include <bits/stdc++.h>
using namespace std;
int main()
{
    string s1,s2;
    cin >> s1 >> s2;
    int y = 0,l1 = s1.length(),l2 = s2.length();
    for (int i = 0;i < l2 - l1;i++)
    {
        bool fl = true;
        for (int j = i;j < i + l1;j++)
        {
            if (s1[j - i] != s2[j])
            {
                fl = false;
            }
        }
        if (fl)
        {
            y++;
        }
    }
    cout << y;
}