A. Football
time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

Petya loves football very much. One day, as he was watching a football match, he was writing the players' current positions on a piece of paper. To simplify the situation he depicted it as a string consisting of zeroes and ones. A zero corresponds to players of one team; a one corresponds to players of another team. If there are at least 7 players of some team standing one after another, then the situation is considered dangerous. For example, the situation 00100110111111101 is dangerous and 11110111011101 is not. You are given the current situation. Determine whether it is dangerous or not.

Input

The first input line contains a non-empty string consisting of characters "0" and "1", which represents players. The length of the string does not exceed 100 characters. There's at least one player from each team present on the field.

Output

Print "YES" if the situation is dangerous. Otherwise, print "NO".

Examples
input
001001
output
NO
input
1000000001
output
YES
#include<bits/stdc++.h>
using namespace std;
int main()
{
string s;
cin>>s;
if(s.find("")!=string::npos||s.find("")!=string::npos)cout<<"YES";
else cout<<"NO";
cout<<endl;
return ;
}

odeforces Beta Round #77 (Div. 2 Only)的更多相关文章

  1. Codeforces Beta Round #77 (Div. 2 Only)

    Codeforces Beta Round #77 (Div. 2 Only) http://codeforces.com/contest/96 A #include<bits/stdc++.h ...

  2. Codeforces Beta Round #77 (Div. 1 Only) C. Volleyball (最短路)

    题目链接:http://codeforces.com/contest/95/problem/C 思路:首先dijkstra预处理出每个顶点到其他顶点的最短距离,然后如果该出租车到某个顶点的距离小于等于 ...

  3. codeforces水题100道 第二十三题 Codeforces Beta Round #77 (Div. 2 Only) A. Football (strings)

    题目链接:http://www.codeforces.com/problemset/problem/96/A题意:判断一个0-1字符串中出现的最长的0字串或者1字串的长度是否大于等于7.C++代码: ...

  4. Codeforces Beta Round #77 (Div. 2 Only) A. Football【字符串/判断是否存在连续7个0或7个1】

    A. Football time limit per test 2 seconds memory limit per test 256 megabytes input standard input o ...

  5. Codeforces Beta Round #80 (Div. 2 Only)【ABCD】

    Codeforces Beta Round #80 (Div. 2 Only) A Blackjack1 题意 一共52张扑克,A代表1或者11,2-10表示自己的数字,其他都表示10 现在你已经有一 ...

  6. Codeforces Beta Round #83 (Div. 1 Only)题解【ABCD】

    Codeforces Beta Round #83 (Div. 1 Only) A. Dorm Water Supply 题意 给你一个n点m边的图,保证每个点的入度和出度最多为1 如果这个点入度为0 ...

  7. Codeforces Beta Round #79 (Div. 2 Only)

    Codeforces Beta Round #79 (Div. 2 Only) http://codeforces.com/contest/102 A #include<bits/stdc++. ...

  8. Codeforces Beta Round #76 (Div. 2 Only)

    Codeforces Beta Round #76 (Div. 2 Only) http://codeforces.com/contest/94 A #include<bits/stdc++.h ...

  9. Codeforces Beta Round #75 (Div. 2 Only)

    Codeforces Beta Round #75 (Div. 2 Only) http://codeforces.com/contest/92 A #include<iostream> ...

随机推荐

  1. python中random模块使用

  2. Faster-rnnlm代码分析1 - 词表构建,Nnet成员

    https://github.com/yandex/faster-rnnlm     Gdb ./rnnlm r -rnnlm model-good.faster -train thread.titl ...

  3. 微信支付开发(1) JS API支付V3版(转)

    http://www.cnblogs.com/txw1958/p/wxpayv3-jsapi.html 本文介绍微信支付下的jsapi实现流程 前言 微信支付现在分为v2版和v3版,2014年9月10 ...

  4. .net转的时间戳用java去解析的代码

    /// <summary> /// 转换成java解析一致的时间戳 /// </summary> /// <param name="time"> ...

  5. Analysis Services OLAP 概述

    1. 什么是OLAP •定义1 :OLAP(联机分析处理)是针对特定问题的联机数据访问和分析.通过对信息(维数据)的多种可能的观察形式进行快速.稳定一致和交互性的存取,允许管理决策人员对数据进行深入观 ...

  6. oracle删除用户下所有的表

    需要创建这些删除语句,通过oracle的数据字典找到该用户下的所有表.视图等对象,拼接成语句.如下select 'drop table '||table_name|| ' cascade constr ...

  7. jquery判断当前设备是手机还是电脑并跳转

    <script type="text/javascript"> var commonURL = 'http://xxxx.com/'; function mobile_ ...

  8. java学习笔记(2):获取文件名和自定义文件过滤器

    //自定义文件过滤器import java.io.File; import javax.swing.filechooser.*; public class JavaChooser extends Fi ...

  9. 【微信Java开发 --1---番外1】在windows下,使用JAVA执行多条DOS命令+文件夹/路径中有空格怎么解决【目的是实现内容穿透外网】

    内网穿透外网的那一篇,参正集1 但是每次都要Ctrl+R 启动DOS窗口,也就是CMD,一句一句的去粘,略显繁琐. 所以将这些任务写在JAVA程序中,启动一次程序就可以实现[内网穿透]的功能,多好啊! ...

  10. MySQL的多表查询(笛卡尔积原理)

    先确定数据要用到哪些表. 将多个表先通过笛卡尔积变成一个表. 然后去除不符合逻辑的数据(根据两个表的关系去掉). 最后当做是一个虚拟表一样来加上条件即可. 注意:列名最好使用表别名来区别. 笛卡尔积 ...