codeforces 96A （字符串模拟水题）

A. Football
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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 least7 players of some team standing one after another, then the situation is considered dangerous. For example, the situation00100110111111101 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 exceed100 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

001001

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;}