HDOJ 5591 ZYB's Game(简单博弈)



ZYB's Game

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 72    Accepted Submission(s): 64

Problem Description

ZYB played a game named NumberBomb with his classmates in hiking:a host keeps a number in [1,N] in mind,then
players guess a number in turns,the player who exactly guesses X loses,or the host will tell all the players that
the number now is bigger or smaller than X .After that,the range players can guess will decrease.The range is [1,N] at first,each player should guess in the legal range.

Now if only two players are play the game,and both of two players know the X ,if two persons all use the best strategy,and the first player guesses first.You are asked to find the number of X that the second player
will win when X is in [1,N] .

 

Input

In the first line there is the number of testcases T .

For each teatcase:

the first line there is one number N .

1≤T≤100000 , 1≤N≤10000000

 

Output

For each testcase,print the ans.

 

Sample Input

  
  

   
   1
3
  
  

 

Sample Output

  
  

   
   1
  
  

 

题意： 有1~N个数，由两个人猜数，谁猜到X这个数，谁就输。 若猜的不是X，则主持人就会说当前这个数比X大还是小，限定接下来猜的范围。 现在这两个人都知道X的具体数值，均采用最优策略猜数，X在1~N中，问能让后手赢的X的个数有几个。

题解：很明显的博弈问题，当N=1时，后手必赢，能让他赢的X的个数只有一个，当N=2时，先手是不会猜X的，那么剩下的数就是X，后手必输，能让他赢的X的个数为0。  当N=3时，只有当X=2时，先手猜1或3，后手再猜剩下的3或者1，那么后手就能赢，能让他赢的X=2，只有这一个。就这么推下去，发现当N为奇数时，能让后手赢的数只有一个，X=(1+N)/2。  当N为偶数时，不存在让后手赢的X。

代码如下：

#include<cstdio>
#include<cstring>
int main()
{
	int t,n;
	scanf("%d",&t);
	while(t--)
	{
		scanf("%d",&n);
		if(n&1)
			printf("1\n");
		else
			printf("0\n");
	}
	return 0;
}