1063.txt

北大ACM题目例程 详细的解答过程 程序实现 算法分析

#include <stdio.h>


int main()
{
	int cas, i, n, j, s1, s2, a[30];
	scanf( "%d", &cas );
	
	while(cas-- )
	{
		scanf( "%d", &n );
		
		for( i=0; i<n; i++ )
		{
			scanf( "%d", &a[i] );
		}

		for( i=0; i<n; i++ )
		if( a[i] == 1 )
		{
			s1 = s2 = 0;
			for( j=1; j<n; j++ )
			if( a[(j+i)%n] == 1 )
			{
				if( j%2 )s1++;
				else s2++;
			}

			if( s1 == s2 || s1 == s2+1 )
			{
				printf( "YES\n" );
				break;
			}
		}
		
		if( i >= n )
		printf( "NO\n" );
	}

	return 0;
}