ACM HDU 1907 John(博弈)
John
Time Limit: 5000/1000 MS (Java/Others) Memory Limit: 65535/32768 K (Java/Others)
Total Submission(s): 831 Accepted Submission(s): 437
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
2
3
3 5 1
1
1
Sample Output
John
Brother
Source
Recommend
lcy
代码:
#include<stdio.h>
int a[50];
int main()
{
int i,ans,T,n,temp;
scanf("%d",&T);
while(T--)
{
scanf("%d",&n);
temp=0;//孤单堆
for(i=0;i<n;i++)
{
scanf("%d",&a[i]);
if(i==0)
ans=a[i];
else
ans=ans^a[i];
if(a[i]>1) temp=1;
}
if(temp==0)
{
if(n%2==1)
printf("Brother\n");
else
printf("John\n");
continue;
}
if(ans==0)
printf("Brother\n");
else
printf("John\n");
}
return 0;
}
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