hdu 4994 Revenge of Nim（尼姆博奕变形）

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本文介绍了一种基于经典尼姆博弈的游戏变体——复仇的尼姆。在这个版本中，玩家必须按照顺序从每一堆中移除物品，并且只能从当前的第一堆开始移除。文章详细解释了游戏规则和获胜策略，提供了具体的示例输入输出以及实现这一策略的代码。

Revenge of Nim

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 557 Accepted Submission(s): 288

Problem 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

Sample Output

Yes
No

题意：有n 堆石子，从第一堆到最后一堆，每次可以取任意多个，但只能取同一堆的，最后取完的人赢，问先手能否胜利？

思路：尼姆博奕的变形，应该来说比尼姆博奕要好理解。这里当某一堆里石头个数为1时，无奈，此人只能把该堆取完。

当某一堆石头个数大于1时，此时此人可以根据后面所有堆是必胜态还是必败态来决定是把该堆取完让下一个人进行那个状态还是把该堆取完只剩一个，让自己进行那个状态。所以不管后面如何变化，我们只要考虑第一次出现的大于1的堆之前有多少个大小为1的堆即可。因为谁先进入这个第一次出现的大于1的堆即可获得胜利。

代码：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
#define N 1100
int a[N];
int main()
{
    int T,n;
    scanf("%d",&T);
    while(T--)
    {
        scanf("%d",&n);
        int flag=0,ans=0;
        for(int i=1; i<=n; i++)
        {
            scanf("%d",&a[i]);
            if(a[i]>1)
                flag=1;
            if(!flag) ans++;
        }
        if(!flag)
        {
            if(ans&1) printf("Yes\n");
            else printf("No\n");
        }
        else
        {
            if(ans&1) printf("No\n");
            else printf("Yes\n");
        }
    }
    return 0;
}