xtuoj 1235 CQRXLB(博弈论）

CQRXLB
Accepted : 19		Submit : 40
Time Limit : 1000 MS		Memory Limit : 65536 KB

CQRXLB

Problem Description:

CQR and XLB are best friends. One night, they are staring each other and feel boring, and XLB says let's play game!

They place n piles of stones and take turns remove arbitrary amount(at least one) of stones in at least one pile at most x piles they chose. The one who can not remove any stone lose out.

CQR is a girl so she always move first. Duoxida wants to know who will win if they are both smart enough.

Input

The first line contains a integer T(no more than 100) indicating the number of test cases.

In each test case, each test case includes two lines. the first line contains two integers n and x \((1\le n\le 100, 1\le x\le 9)\). The second line contains n positive integers indicates the amount of stones in each pile.

All inputs are no more than \(10^9\).

Output

For each test case, puts the name of winner if they both acts perfect.

Sample Input

2
3 1
1 3 2
2 2
1000 1000

Sample Output

XLB
CQR

Source

XTU OnlineJudge 

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题意：有N堆石子，两个人在玩游戏。游戏规则是可以取不超过x堆中任意石子数，至少取一个，不能取者败，问先手还是后手赢。

那我们就需要设计出该石子堆的平衡状态和非平衡状态。

显然发现，这道题类似于NIM+BASH博弈。

别问为什么，赛场上我肯定推不出来的，只能靠猜想+证明。

猜想：将每个石子堆$n_{k}$ 变为二进制数，对所有的$n_{k}$，把各位分别加起来，并%（x+1）,然后把各位求和sum。若sum==0则后手赢，否则先手赢。

公平组合博弈的平衡和非平衡态满足的条件：
• 1、平衡态时，不可能转移到另外的平衡态。
• 2、非平衡态时，一定可以转移到平衡态的状态。
• 3、最终的状态是平衡态。且有限步数内会结束。

显然上面的sum==0对应平衡态，sum！=0对应非平衡态，读者可以自己证明下。

#include<cstdio>
#include<iostream>
#include<cstring> 
#define clr(x) memset(x,0,sizeof(x))
using namespace std;
int a[100]; 
int max(int a,int b)
{
    return a>b?a:b;
}
int main()
{
    int n,m,k,t,l,maxt,sum;
    int T;
    scanf("%d",&T);
    while(T--)
    {
        scanf("%d%d",&n,&m);
        clr(a);
        maxt=0;
        for(int i=1;i<=n;i++)
        {
            scanf("%d",&l);
            t=0;
            while(l)
            {
                t++;
                a[t]=(a[t]+(l%2))%(m+1);
                l/=2;
            }

            maxt=max(maxt,t);
        }
        sum=0;
        for(int i=1;i<=maxt;i++)
        {
            sum+=a[i];
        }
        if(sum)
            printf("CQR
");
        else
            printf("XLB
");
    }
    return 0;
}