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

3

4

2 3 4 5

5

1 1 2 4 10

1

1

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=1;
    cin>>t;
    while(t--)
    {
        cin>>k;
        chongyu=ans=0;
        for(i=0;i<k;i++)
        {
            cin>>data;
            if(data>1)
            chongyu++;
            ans^=data;
        }
        if((ans&&chongyu)||(!ans&&!chongyu)) //s1,s2,t0
        printf("Case %d: Alice
",cas++);
        else
        printf("Case %d: Bob
",cas++);
    }
    return 0;
}