Light OJ 1253 Misere Nim （尼姆博弈（2））

LightOJ1253 :Misere Nim

时间限制:1000MS 内存限制:32768KByte 64位IO格式:%lld & %llu

描述

Alice and Bob are playing game of Misère Nim. Misère Nim is a game playing on k piles of stones, each pile containing one or more stones. The players alternate turns and in each turn a player can select one of the piles and can remove as many stones from that pile unless the pile is empty. In each turn a player must remove at least one stone from any pile. Alice starts first. The player who removes the last stone loses the game.

输入

Input starts with an integer T (≤ 200), denoting the number of test cases.

Each case starts with a line containing an integer k (1 ≤ k ≤ 100). The next line contains k space separated integers denoting the number of stones in each pile. The number of stones in a pile lies in the range [1, 10⁹].

输出

For each case, print the case number and 'Alice' if Alice wins otherwise print 'Bob'.

样例输入

2 3 4 5

1 1 2 4 10

样例输出

Case 1: Bob

Case 2: Alice

Case 3: Bob

题目来源

Problem Setter: Jane Alam Jan

题意

k堆石子，每堆个数不定，每次取一堆中的任意个，谁取到最后一个谁输。

题解

详见上篇博客尼姆博弈 s1，s2，t0为必胜态。

#include <iostream>

#include <stdio.h>

using namespace std;

int main()

{

    int i,n,t,k,data,ans,chongyu,cas=;

    cin>>t;

    while(t--)

    {

        cin>>k;

        chongyu=ans=;

        for(i=;i<k;i++)

        {

            cin>>data;

            if(data>)

            chongyu++;

            ans^=data;

        }

        if((ans&&chongyu)||(!ans&&!chongyu)) //s1,s2,t0

        printf("Case %d: Alice\n",cas++);

        else

        printf("Case %d: Bob\n",cas++);

    }

    return ;

}

巴特西

Light OJ 1253 Misere Nim （尼姆博弈（2））

LightOJ1253 :Misere Nim

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