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# Nim博弈

2019-03-23 · 6 min read · CC BY-NC-ND 2.5

# Nim博弈

### 问题分析

C. Bouton证明了以下定理

nim-sum即每堆的大小的异或和。

#### 证明：

x 1，...， x n是移动前堆的大小，y 1，...， y n是移动后的相应大小。让$s = x_1 \bigoplus ... \bigoplus x_n​$$t = y_1 \bigoplus ... \bigoplus y_n​$。假设一次移动在堆k中，则对于所有ik，我们有$x_i = y_i​$，并且$x_k > y_k​$。根据上面提到的异或的属性，我们有

t = 0 ⊕ t
= s ⊕ s ⊕ t
= s ⊕ (x1 ⊕ ... ⊕ xn) ⊕ (y1 ⊕ ... ⊕ yn)
= s ⊕ (x1 ⊕ y1) ⊕ ... ⊕ (xn ⊕ yn)
= s ⊕ 0 ⊕ ... ⊕ 0 ⊕ (xk ⊕ yk) ⊕ 0 ⊕ ... ⊕ 0
= s ⊕ xk ⊕ yk

(*) t = s ⊕ xk ⊕ yk


• lemma 1

• Lemma 2

t = s ⊕ xk ⊕ yk
= s ⊕ xk ⊕ (s ⊕ xk)
= 0


P.S. 此定理仅在有堆的大小为2或大于2时成立

### 范例问题

#### problem

HDU1907

John

Time Limit: 5000/1000 MS (Java/Others) Memory Limit: 65535/32768 K (Java/Others)Total Submission(s): 6413 Accepted Submission(s): 3698

Problem Description

Little John is playing very funny game with his younger brother. There is one big box filled with M&Ms of different colors. At first John has to eat several M&Ms of the same color. Then his opponent has to make a turn. And so on. Please note that each player has to eat at least one M&M during his turn. If John (or his brother) will eat the last M&M from the box he will be considered as a looser and he will have to buy a new candy box.

Both of players are using optimal game strategy. John starts first always. You will be given information about M&Ms and your task is to determine a winner of such a beautiful game.

Input

The first line of input will contain a single integer T – the number of test cases. Next T pairs of lines will describe tests in a following format. The first line of each test will contain an integer N – the amount of different M&M colors in a box. Next line will contain N integers Ai, separated by spaces – amount of M&Ms of i-th color.

Constraints:
1 <= T <= 474,
1 <= N <= 47,
1 <= Ai <= 4747

Output

Output T lines each of them containing information about game winner. Print “John” if John will win the game or “Brother” in other case.

Sample Input

2
3
3 5 1
1
1


Sample Output

John
Brother


#include<cstdio>
int main()
{
int t,n,a,sum,s;
scanf("%d",&t);
while(t--)
{
sum=0;
s=0;
scanf("%d",&n);
for(int i=0;i<n;i++)
{
scanf("%d",&a);
s^=a;
sum+=a;
}
if(sum==n)//所有堆大小都为1
{
if(sum%2==0) printf("John\n");
else printf("Brother\n");
}
else
{
if(s==0) printf("Brother\n");
else printf("John\n");
}
}
return 0;
}