codeforces 96A （字符串模拟水题）

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本文介绍了一种通过输入由0和1组成的字符串来判断足球比赛中是否存在至少七个连续相同队伍球员的算法。该算法用于确定比赛是否处于危险状态，并提供了一个具体的C++实现示例。

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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".

Sample test(s)

Input

Output

NO

Input

1000000001

Output

YES

字符串的简单处理，求是否有连续七个的0或者连续七个的1；
╮(╯▽╰)╭此题也wa几次，细节又忽略掉了，↖(^ω^)↗
代码：
#include <iostream>
#include<string.h>
using namespace std;
int main()
{
    char s[101];
    bool flag;
    int i,j,k,ans;
    while(cin>>s)
    {
        flag=true;
        k=strlen(s);
        for(i=0; i<k; i++)
        {
            if(s[i]=='0')
            {
                //cout<<i<<endl;
                ans=0;
                for(j=i; j<k; j++)
                {
                    ans++;
                    if(s[j]=='1')
                    {
                        i=j-1;
                        ans--;
                        break;
                    }
                    else if(ans==7)
                    {
                        break;
                    }

                }
                if(ans==7)
                {
                    cout<<"YES"<<endl;
                    flag=false;
                    break;
                }
            }
            else if(s[i]=='1')
            {
                ans=0;
                for(j=i; j<k; j++)
                {
                    ans++;
                    if(s[j]=='0')
                    {
                        i=j-1;
                        ans--;
                        break;
                    }
                    else if(ans==7)
                    {
                        break;
                    }

                }
                if(ans==7)
                {
                    cout<<"YES"<<endl;
                    flag=false;
                    break;
                }
            }
        }
        if(flag)
            cout<<"NO"<<endl;
    }
    return 0;
}