字符串基础 O(n) 
第一次用了substr()函数 复杂度变成O(n^2)TLE
代码：
#include<bits/stdc++.h>
using namespace std;
const int N = 5e5 + 10;
int n;
string ans;
string s;
bool isg[N];
int cnt;

int main()
{
    ios::sync_with_stdio(0);
    cin.tie(0);
    cout.tie(0);

    cin >> n >> ans;
//  for(int i = 1;i <= n;i++)
//  {
//      cin >> s[i];
//  }

    for(int i = 1;i <= n;i++)
    {
        cin >> s;
        string x, y;
        int ls = s.size(), lt = ans.size();
        if(abs(ls - lt) >= 2)continue;
        if(ls == lt)//#1
        {
            if(s == ans)
            {
                cnt++;
                isg[i] = true;
                continue;
            }

            int k;
            for(int j = 0;j < ls;j++)//#4
            {
                if(s[j] != ans[j])
                {
                    k = j;
                    break;
                }
            }

            string one = "", two = "";
            for(int j = 0;j < k;j++)
            {
                one += s[j];
                two += ans[j];
            }
            for(int j = k + 1;j < ls;j++)
            {
                one += s[j];
                two += ans[j];
            }
            if(one == two)
            {
                cnt++;
                isg[i] = true;
                continue;
            }
        }

        else{
            if(ls > lt)//#3
            {
                int k = ls - 1;
                for(int j = 0;j < lt;j++)
                {
                    if(s[j] != ans[j])
                    {
                        k = j;
                        break;
                    }
                }

                string end = "";
                for(int j = 0;j < k;j++)end += s[j];
                for(int j = k + 1;j < ls;j++)end += s[j];
                if(end == ans)
                {
                    cnt++;
                    isg[i] = true;
                    continue;
                }
            }

            else{//#2
                int k = lt - 1;
                for(int j = 0;j < ls;j++)
                {
                    if(ans[j] != s[j])
                    {
                        k = j;
                        break;
                    }
//                  cout << k << endl;
                }

                string end = "";
                for(int j = 0;j < k;j++)end += ans[j];
                for(int j = k + 1;j < lt;j++)end += ans[j];
                if(end == s)
                {
                    cnt++;
                    isg[i] = true;
                    continue;
                }
            }
        }
    }

    cout << cnt << endl;
    for(int i = 1;i <= n;i++)
    {
        if(isg[i])cout << i << " ";
    }
    return 0;
}