John
Time Limit: 5000/1000 MS (Java/Others) Memory Limit: 65535/32768 K (Java/Others)
Total Submission(s): 2412 Accepted Submission(s): 1303
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
Sample Output
Source
当每一堆都为1时，即都是孤单堆时，需要特殊考虑。
否则对每一项抑或，假如等于0，则后手胜，否则先手胜。
代码：
/* ***********************************************
Author :xianxingwuguan
Created Time :2014-2-8 0:11:46
File Name :1.cpp
************************************************ */
#pragma comment(linker, "/STACK:102400000,102400000")
#include <stdio.h>
#include <iostream>
#include <algorithm>
#include <sstream>
#include <stdlib.h>
#include <string.h>
#include <limits.h>
#include <string>
#include <time.h>
#include <math.h>
#include <queue>
#include <stack>
#include <set>
#include <map>
using namespace std;
#define INF 0x3f3f3f3f
#define eps 1e-8
#define pi acos(-1.0)
typedef long long ll;
int a[10000];
int main(){
int T,i,j,k,m,n;
cin>>T;
while(T--){
cin>>n;
int sum=0;
for(i=1;i<=n;i++)
cin>>a[i],
sum+=a[i]==1;
if(sum==n){
if(n%2==0)puts("John");
else puts("Brother");
}
else {
int flag=0;
for(i=1;i<=n;i++)
flag^=a[i];
if(flag)puts("John");
else puts("Brother");
}
}
}
hdu 1907 Nim游戏
原文：http://blog.csdn.net/xianxingwuguan1/article/details/18976581
