HDU 1907 John nim博弈变形

John



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
 

 

题意:
  今有若干堆火柴,两人依次从中拿取,规定每次只能从一堆中取若干根, 可将一堆全取走,但不可不取,最后取完者为负。
题解:
  T态:异或和为0+
  S态:异或和不为0+
  定义:若一堆中仅有1根火柴,则被称为孤单堆。若大于1根,则称为充裕堆。
  定义:T态中,若充裕堆的堆数大于等于2,则称为完全利他态,用T2表示;若充裕堆的堆数等于0,则称为部分利他态,用T0表示。
 
孤单堆的根数异或只会影响二进制的最后一位,但充裕堆会影响高位(非最后一位)。一个充裕堆,高位必有一位不为0,则所有根数异或不为0。故不会是T态。
[定理5]:S0态,即仅有奇数个孤单堆,必败。T0态必胜。 
证明:
S0态,其实就是每次只能取一根。每次第奇数根都由己取,第偶数根都由对 
方取,所以最后一根必己取。败。同理,  T0态必胜#
[定理6]:S1态,只要方法正确,必胜。 
证明:
若此时孤单堆堆数为奇数,把充裕堆取完;否则,取成一根。这样,就变成奇数个孤单堆,由对方取。由定理5,对方必输。己必胜。  # 
[定理7]:S2态不可转一次变为T0态。 
证明:
充裕堆数不可能一次由2变为0。得证。  # 

 

[定理8]:S2态可一次转变为T2态。 
证明:
由定理1,S态可转变为T态,态可一次转变为T态,又由定理6,S2态不可转一次变为T0态,所以转变的T态为T2态。  # 
[定理9]:T2态,只能转变为S2态或S1态。 
证明:
由定理2,T态必然变为S态。由于充裕堆数不可能一次由2变为0,所以此时的S态不可能为S0态。命题得证。 
[定理10]:S2态,只要方法正确,必胜. 
证明:
方法如下: 
      1)  S2态,就把它变为T2态。(由定理8) 
      2)  对方只能T2转变成S2态或S1态(定理9)
    若转变为S2,  转向1) 
    若转变为S1,  这己必胜。(定理5) 
[定理11]:T2态必输。 
证明:同10。 
综上所述,必输态有:  T2,S0 
          必胜态:    S2,S1,T0. 

#include <iostream>
#include <cstdio>
#include <cmath>
#include <cstring>
#include <algorithm>
using namespace std;
#pragma comment(linker, "/STACK:102400000,102400000")
#define ls i<<1
#define rs ls | 1
#define mid ((ll+rr)>>1)
#define pii pair<int,int>
#define MP make_pair
typedef long long LL;
const long long INF = 1e18+1LL;
const double Pi = acos(-1.0);
const int N = 5e5+10, M = 2e5+20, mod = 1e9+7, inf = 2e9;

int sg[N],n,x,ans,vis[N];
int main() {
    int T;
    scanf("%d",&T);
    while(T--) {
        scanf("%d",&n);
        int ans = 0, cnt = 0;
        for(int i = 1; i <= n; ++i) {
            scanf("%d",&x);
            ans = ans ^ x;
            if(x > 1) cnt++;
        }
        if((ans && cnt >= 2) || (cnt == 1) || (cnt == 0 && !ans)) puts("John");
        else puts("Brother");
    }
    return 0;
}

 

posted @ 2016-10-29 15:01  meekyan  阅读(318)  评论(0编辑  收藏  举报