N bulbs（规律）

N bulbs

 

 Accepts: 408

 

 Submissions: 1224

 Time Limit: 10000/5000 MS (Java/Others)

 

 Memory Limit: 65536/65536 K (Java/Others)

Problem Description

N bulbs are in a row from left to right,some are on, and some are off.The first bulb is the most left one. And the last one is the most right one.they are numbered from 1 to n,from left to right.

in order to save electricity, you should turn off all the lights, but you're lazy. coincidentally,a passing bear children paper(bear children paper means the naughty boy), who want to pass here from the first light bulb to the last one and leave.

he starts from the first light and just can get to the adjacent one at one step. But after all,the bear children paper is just a bear children paper. after leaving a light bulb to the next one, he must touch the switch, which will change the status of the light.

your task is answer whether it's possible or not to finishing turning off all the lights, and make bear children paper also reach the last light bulb and then leave at the same time.

Input

The first line of the input file contains an integer T, which indicates the number of test cases.

For each test case, there are 2 lines.

The first line of each test case contains 1 integers n.

In the following line contains a 01 sequence, 0 means off and 1 means on.

• 1 \leq T \leq 101≤T≤10
• 1 \leq N \leq 10000001≤N≤1000000

Output

There should be exactly T lines in the output file.

The i-th line should only contain "YES" or "NO" to answer if it's possible to finish.

Sample Input

1
5
1 0 0 0 0

Sample Output

YES

Hint

Child's path is: 123234545 all switchs are touched twice except the first one.

题解：找了好久的规律，如果是0是偶数就可以走过去再拐回来关了，如果是1走过去就好了，现在就考虑奇数个0时走过了再拐回来关时会有01的向右传递，所以说只要0的奇数段为偶数就好了；

代码：

#include<iostream>
#include<cstdio>
#include<cmath>
#include<cstring>
#include<vector>
#include<map>
#include<algorithm>
#include<queue>
using namespace std;
const int INF=0x3f3f3f3f;
#define mem(x,y) memset(x,y,sizeof(x))
#define PI(x) printf("%d",x)
#define SI(x) scanf("%d",&x)
#define T_T while(T--)
const int MAXN=1000010;
int a[MAXN];
int main(){
	int T,N;
	SI(T);
	T_T{
		SI(N);
		int temp=0,ans=0;
		for(int i=0;i<N;i++){
			SI(a[i]);
		}
		for(int i=0;i<N;i++){
			if(a[i]){
				if(temp&1)ans++;
				temp=0;
			}
			else temp++;
		}
		if(temp&1)ans++;
		//printf("%d\n",ans);
		if(ans%2==0)puts("YES");
		else puts("NO");
	}
	return 0;
}