BestCoder Round #65 1002 ZYB's Game HDU5591

最新推荐文章于 2016-08-06 15:05:24 发布

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最新推荐文章于 2016-08-06 15:05:24 发布

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本文介绍了一个名为Number Bomb的游戏，该游戏涉及两名玩家之间的策略博弈。在该游戏中，一名主持人会在心中默想一个1到N之间的数字，玩家轮流猜测该数字。若猜中，则猜中者输；若未猜中，主持人会告知数字相对于猜测值的大小，从而缩小后续猜测的范围。文章探讨了当两名玩家都知晓该数字并采取最佳策略时，第二名玩家获胜的情况数量。

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ZYB's Game

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 412 Accepted Submission(s): 340

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

Source

BestCoder Round #65

出题人：

我们会发现这个模型其实是类比于左右两堆石子,每次可以在一堆里取任意多,

取完的人胜利.当左右两堆石子相同时,我们可以简单的构造后手胜利的方法:

即在另一堆石子中取走同样多的石子,否则,先手可以取一些石子使得两堆石子相同.

所以,当 $N$ 是奇数输出 $1$ ,否则输出 $0$ .

#include
 
  
#include
  
   
using namespace std;
int main()
{
    int t,n;
    scanf("%d",&t);
    while(t--)
    {
        scanf("%d",&n);
        printf("%d\n",n%2);
    }
    return 0;
}