2017年第0届浙江工业大学之江学院程序设计竞赛决赛—F

最新推荐文章于 2018-08-03 10:53:49 发布

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#2017年第0届浙江工业大学之江学院程序

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本文探讨了一个关于博弈论的游戏问题，具体为两个玩家轮流从若干堆硬币中取走一部分，目标是移除最后一枚硬币以赢得比赛。文章提供了一段C++代码示例，用于计算首位玩家确保获胜的不同策略数量。

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Problem F: qwb has a lot of Coins

Time Limit: 1 Sec Memory Limit: 128 MB
Submit: 781 Solved: 264
[ Submit][ Status][ Web Board]

Description

qwb has a lot of coins. One day, he decides to play a game with his friend using these coins. He first puts some of his coins into M piles, each of which is composed of Ni (1<=i<=M) coins. Then, the two players play the coin game in turns. Every step, one can remove one or more coins from only one pile. The winner is the one who removes the last coin.
Then comes the question: How many different ways the first player can do that will ensure him win the game?

Input

Input contains multiple test cases till the end of file. Each test case starts with a number M (1 <= M<= 1000) meaning the number of piles. The next line contains M integers Ni (1 <= Ni <= 1e9, 1 <= i<= M) indicating the number of coins in pile i.

Output

For each case, put the method count in one line.
If the first player can win the game, the method count is the number of different ways that he can do to ensure him win the game, otherwise zero.

Sample Input

Sample Output

0
1

想法：

尼姆博弈

    #include<iostream>
    #include<cstdio>
    using namespace std;
    int main()
    {
       int a[102],m,i,sum,s,count;
       while(scanf("%d",&m)!=EOF&&m)
       {
           sum=count=0;
          for(i=0;i<m;i++)
          {
             scanf("%d",&a[i]);
             sum=sum^a[i];
          }
          for(i=0;i<m;i++)
          {
             s=sum^a[i];
             if(s<a[i])
                 count++;
          }
          printf("%d\n",count);
       }
       return 0;
    }