CF 96A Football

这个题其实很简单；

大致题意：YES代表危险，NO代表不危险；

当一串0和1的数字中连续出现一种数字大于等于七次，则为危险，否则不危险；

则在输入时候单独判断，如果1或者0大于7次则YES；否则1和0都必须小于7次；

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

AC代码：

#include <bits/stdc++.h>
using namespace std ;
int main()
{
char a[5000];
cin>>a;
int sum0,sum1,i;
sum0=sum1=0;
int len = strlen(a);
for(i=0;i<len;i++)
{
if(a[i]=='1'&&a[i+1]=='1')
{
sum1++;
if(sum1==6)
break;
}
if(a[i]=='1'&&a[i+1]=='0')
{
sum1=0;
}
}

for(i=0;i<len;i++)
{

if(a[i]=='0'&&a[i+1]!='1')
{
sum0++;
if(sum0==6)
break;
}
if(a[i]=='0'&&a[i+1]=='1')
{
sum0=0;
}
}

sum1+=1; //*判断规则导致最后一位会被少加上；
sum0+=1;
if(sum1>=7||sum0>=7)
{
printf("YES\n");
}
else if(sum1<7&&sum0<7)
{
printf("NO\n");
}
return 0 ;
}