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 no te 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.
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.
Constrains:
1 <= T <= 474,
1 <= N <= 47,
1 <= Ai <= 4747
Sample Input
Sample Output
分析:博弈题,最后拿者输!
代码:
#include<iostream> #include<stdio.h> #include<cstring> using namespace std; int main() { //freopen("a.in","r",stdin); //freopen("aa.out","w",stdout); int t,n,i,in,cnt,ans; while(scanf("%d",&t)!=EOF) { while(t--) { scanf("%d",&n); cnt=0;ans=0; for(i=0;i<n;i++) { scanf("%d",&in); if(in==1) cnt++; ans^=in; } if(cnt==n) { printf("%s ",(cnt%2!=1)?"John":"Brother"); continue; } if(ans!=0) printf("John "); else printf("Brother "); } } return 0; }