HDU 4994 拿石子

Description

Nim is a mathematical game of strategy in which two players take turns removing objects from distinct heaps. On each turn, a player must remove at least one object, and may remove any number of objects provided they all come from the same heap. 
---Wikipedia 

Today, Nim takes revenge on you. The rule of the game has changed a little: the player must remove the objects from the current head(first) heap. Only the current head heap is empty can the player start to remove from the new head heap. As usual, the player who takes the last object wins.

Input

The first line contains a single integer T, indicating the number of test cases. 

Each test case begins with an integer N, indicating the number of heaps. Then N integer Ai follows, indicating the number of each heap successively, and the player must take objects in this order, from the first to the last. 

[Technical Specification] 
1. 1 <= T <= 100 
2. 1 <= N <= 1 000 
3. 1 <= Ai <= 1 000 000 000

Output

For each test case, output “Yes” if the first player can always win, otherwise “No”.

Sample Input

2
1
2
2
1 1

Sample Output

Yes
No
题意：给n个堆，每个堆有a[i]个石子，每次必须拿一个石子，第一个人先拿然后第二个人拿，最后一个石子如果是第一个人拿的话输出No，否则输出Yes。

思路：在最初的一个不是1的堆的时候，保证A先拿，
则前n-1个数（都是1的时候）的和保证是偶数，则A可以胜利。

#include <cstring>
#include <cstdio>
#include <iostream>
#include <algorithm>
#include <cmath>
#include <stack>
#include <vector>
#include <queue>
#include <cmath>
using namespace std;
#define N 10500
int main()
{
    int n,t;
    int sum;
    int a[N];
    scanf("%d", &t);
    while(t--)
    {
        memset(a,0,sizeof(a));
        sum = 0;
        scanf("%d",&n);
        for(int i=0; i<n; i++)
            scanf("%d",&a[i]);

        for(int i=0; i<n-1; i++)
        {
            if(a[i]==1)
                sum ++;
            else
                break;
        }
        if(sum % 2 != 0)
            printf("No
");
        else
            printf("Yes
");
    }

    return 0;
}