John(博弈)
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\n",(cnt%2!=1)?"John":"Brother");
continue;
}
if(ans!=0)
printf("John\n");
else
printf("Brother\n");
}
}
return 0;
}
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