hdu 1907
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John
Time Limit: 5000/1000 MS (Java/Others) Memory Limit: 65535/32768 K (Java/Others)Total Submission(s): 4885 Accepted Submission(s): 2824
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.
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
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
233 5 111
Sample Output
JohnBrother思路:这是一道Nim博弈的题目,不过这是最后一个拿到的算输。代码:#include <iostream>#include <stdio.h>#include <string.h>#include <algorithm>#include <queue>#include <cmath>#include <stdlib.h>using namespace std;int main(){ int T; cin>>T; while(T--) { int a[100]; int i; int sum=0; int n; cin>>n; for(i=0;i<n;i++) { scanf("%d",&a[i]); if(a[i]==1)sum++; } if(sum==n) { if(sum%2!=0)cout<<"Brother"<<endl; else cout<<"John"<<endl; } else { int e=a[0]^a[1]; for(i=2;i<n;i++) { e^=a[i]; } if(e==0)cout<<"Brother"<<endl; else cout<<"John"<<endl; } } return 0;}
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