（Nim博弈）湘潭市赛，CQRXLB_湘潭搏弈乒乓-优快云博客

本文链接：https://blog.youkuaiyun.com/better_life30/article/details/46687795

本文探讨了两人石子博弈的策略，通过分析游戏规则，揭示了如何在限制条件下获胜的关键，适合对游戏数学和博弈论感兴趣的读者。

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CQRXLB
Accepted : 5 Submit : 15
Time Limit : 1000 MS Memory Limit : 65536 KB

CQRXLBProblem 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.
InputThe 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 .
 The second line contains n positive integers indicates the amount of stones in each pile.
All inputs are no more than .
OutputFor each test case, puts the name of winner if they both acts perfect.
Sample Input2

3 1

1 3 2

2 2

1000 1000

Sample OutputXLB

CQR


Source
XTU OnlineJudge

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。

NIM博弈回顾：
• 1、平衡态时，不可能转移到另外的平衡态。
• 2、⾮非平衡态时，⼀一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。

猜想：
• n的石子的二进制位的每⼀一位的和都是(x+1)的倍数时为平衡态。

证明可行性：
• 1、平衡态时，不可能转移到另外的平衡态。
• 2、⾮非平衡态时，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。

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;

int n,x;
int a[33];

void add(int tmp) {
    for(int i = 0; i < 32; i ++) {
        a[i] = (a[i] + tmp % 2) % (x + 1);
        tmp /= 2;
    }
}

int main()
{
    int t;
    scanf("%d",&t);
    while(t --) {
        scanf("%d%d",&n,&x);
        memset(a,0,sizeof(a));
        while(n --) {
            int y;
            scanf("%d",&y);
            add(y);
        }
        int sum = 0;
        for(int i = 0; i < 32; i ++) sum += a[i];
        if(sum) printf("CQR\n");
        else printf("XLB\n");
    }
    return 0;
}