Algorithm Problem:Telephone Number

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博客围绕判断字符串能否通过删除字符操作变成电话号码展开。电话号码要求是 11 位且首位为 8。给定不同长度由数字组成的字符串，通过一系列操作判断能否转化为电话号码，最后给出示例及对应输出结果。

A. Telephone Number
time limit per test1 second
memory limit per test256 megabytes
inputstandard input
outputstandard output
A telephone number is a sequence of exactly 11 digits, where the first digit is 8. For example, the sequence 80011223388 is a telephone number, but the sequences 70011223388 and 80000011223388 are not.

You are given a string s of length n, consisting of digits.

In one operation you can delete any character from string s. For example, it is possible to obtain strings 112, 111 or 121 from string 1121.

You need to determine whether there is such a sequence of operations (possibly empty), after which the string s becomes a telephone number.

Input
The first line contains one integer t (1≤t≤100) — the number of test cases.

The first line of each test case contains one integer n (1≤n≤100) — the length of string s.

The second line of each test case contains the string s (|s|=n) consisting of digits.

Output
For each test print one line.

If there is a sequence of operations, after which s becomes a telephone number, print YES.

Otherwise, print NO.
Example
input
2
13
7818005553535
11
31415926535
output
YES
NO
My Answer Code:

#include<bits/stdc++.h>
using namespace std;
int main()
{
    ios::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);
    int t;
    cin>>t;
    vector<string>s(t);
    vector<int>a(t);
    vector<int>firstindex(t,-1);
    vector<bool>ans(t);
    for(int i=0;i<t;i++)
    {
        cin>>a[i]>>s[i];
        if(a[i]<11){ans[i]=false;continue;}
        for(int j=0;j<a[i];j++)
        {
            if(s[i][j]=='8'){firstindex[i]=j;break;}
        }
        if(a[i]-firstindex[i]>=11&&firstindex[i]!=-1)ans[i]=true;
        else ans[i]=false;
    }
    for(int i=0;i<t;i++){if(ans[i])cout<<"YES"<<'\n';else cout<<"NO"<<'\n';}
    return 0;
}