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 ofX that the second player
will win when X is in [1,N].

 

Input

In the first line there is the number of testcasesT.

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

13

 

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");elseprintf("0\n");}return 0;}