hdu 5600 BestCoder Round #67 (div.2)

N bulbs
Accepts: 275  Submissions: 1237 Time Limit: 10000/5000 MS (Java/Others)  Memory Limit: 65536/65536 K (Java/Others)Problem DescriptionN 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.InputThe 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
OutputThere 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 Input1
5
1 0 0 0 0

YES

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

#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <functional>
#include <cmath>
#include <cstring>
#include <algorithm>
using namespace std;
const int maxn = 1000000;
int a[maxn];

int main()
{
    int T;
    scanf("%d",&T);
    while(T--)
    {
        int n;
        scanf("%d",&n);
        for(int i= 0; i < n; i++)
        {
            scanf("%d",&a[i]);
        }
        int num = 0;
        int flag = 1;
        for(int i = 0; i < n; i++)
        {
            if(a[i] != 1)
                num++;
            if(a[i] == 1)
            {
                if(num % 2 )
                {
                    num = 1;
                }
                else
                    num = 0;
            }
            if(i == n-1)
            {
                if(num % 2)
                    flag = 0;
                else
                    flag = 1;
            }
        }
        if(flag == 0)
            printf("NO\n");
        else
            printf("YES\n");
    }
    return 0;
}