Educational Codeforces Round 65 (Rated for Div. 2) A. Telephone Number

链接:https://codeforces.com/contest/1167/problem/A

题意:

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 70011223388and 80000011223388 are not.

You are given a string ss of length nn, consisting of digits.

In one operation you can delete any character from string ss. 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 ss becomes a telephone number.

思路:

找到第一个8出现的位置。再把多余的减掉看是否符合。

代码:

#include <bits/stdc++.h>
using namespace std;

typedef long long LL;
const int MAXN = 1e5+10;

int main()
{
    int t;
    cin >> t;
    while (t--)
    {
        int n;
        string s;
        cin >> n >> s;
        int cnt = -1;
        for (int i = 0;i < s.length();i++)
        {
            if (s[i] == '8')
            {
                cnt = i;
                break;
            }
        }
        if (cnt == -1)
            cout << "NO" << endl;
        else if (s.length() - cnt < 11)
            cout << "NO" << endl;
        else
            cout << "YES" << endl;
    }

    return 0;
}

  

转载于:https://www.cnblogs.com/YDDDD/p/10900190.html

评论
添加红包

请填写红包祝福语或标题

红包个数最小为10个

红包金额最低5元

当前余额3.43前往充值 >
需支付:10.00
成就一亿技术人!
领取后你会自动成为博主和红包主的粉丝 规则
hope_wisdom
发出的红包
实付
使用余额支付
点击重新获取
扫码支付
钱包余额 0

抵扣说明:

1.余额是钱包充值的虚拟货币,按照1:1的比例进行支付金额的抵扣。
2.余额无法直接购买下载,可以购买VIP、付费专栏及课程。

余额充值