John
Time Limit: 5000/1000 MS (Java/Others) Memory Limit: 65535/32768 K (Java/Others)
Total Submission(s): 1312 Accepted Submission(s): 683
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.
Constraints:
1 <= T <= 474,
1 <= N <= 47,
1 <= Ai <= 4747
//博弈?呵呵、感觉和2进制好有关系的东西、是门大学问
#include <iostream>
#include <stdio.h>
#include <string.h>
#include <algorithm>
using namespace std;
int main()
{
int T;
int n,m;
scanf("%d",&T);
int sum,num,t;
while(T--)
{
scanf("%d",&n);
sum=num=0;
m=n;
while(m--)
{
scanf("%d",&t);
if(t==1) num++;
sum^=t;
}
// printf("%d\n",num);
if(num==n)
{
if(n%2)
printf("Brother\n");
else
printf("John\n");
}
else
{
if(sum==0)
printf("Brother\n");
else
printf("John\n");
}
}
return 0;
}
#include <iostream>
#include <stdio.h>
#include <string.h>
#include <algorithm>
using namespace std;
int main()
{
int T;
int n,m;
scanf("%d",&T);
int sum,num,t;
while(T--)
{
scanf("%d",&n);
sum=num=0;
m=n;
while(m--)
{
scanf("%d",&t);
if(t==1) num++;
sum^=t;
}
// printf("%d\n",num);
if(num==n)
{
if(n%2)
printf("Brother\n");
else
printf("John\n");
}
else
{
if(sum==0)
printf("Brother\n");
else
printf("John\n");
}
}
return 0;
}