ZYB's Game

Accepts: 672
Submissions: 1207
Time Limit: 2000/1000 MS (Java/Others)
Memory Limit: 65536/65536 K (Java/Others)
Problem Description

ZYBZYBZYB played a game named NumberBombNumber BombNumberBomb with his classmates in hiking:a host keeps a number in [1,N][1,N][1,N] in mind,then players guess a number in turns,the player who exactly guesses XXX loses,or the host will tell all the players that the number now is bigger or smaller than XXX.After that,the range players can guess will decrease.The range is [1,N][1,N][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 XXX,if two persons all use the best strategy,and the first player guesses first.You are asked to find the number of XXX that the second player will win when XXX is in [1,N][1,N][1,N].

Input

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

For each teatcase:

the first line there is one number NNN.

1≤T≤1000001 \leq T \leq 1000001≤T≤100000,1≤N≤100000001 \leq N \leq 100000001≤N≤10000000

Output

For each testcase,print the ans.

Sample Input
1
3
Sample Output
1
心得:总共才两个人,而且都知道X是多少(出题人在想什么?)先选的人先赢啊;
#include <iostream>
using namespace std;
int t,n;
int main(){
cin>>t;
for(int i=;i<t;i++){
cin>>n;
     cout<<n%<<endl;
}
return ;
}

BestCoder Round #65 (ZYB's Game)的更多相关文章

  1. BestCoder Round #65 (ZYB's Premutation)

    ZYB's Premutation Accepts: 220 Submissions: 983 Time Limit: 2000/1000 MS (Java/Others) Memory Limit: ...

  2. BestCoder Round #65 (ZYB's Biology)

    ZYB's Biology Accepts: 848 Submissions: 1199 Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 13 ...

  3. hdu 5592 BestCoder Round #65(树状数组)

    题意: ZYB有一个排列PP,但他只记得PP中每个前缀区间的逆序对数,现在他要求你还原这个排列. (i,j)(i < j)(i,j)(i<j)被称为一对逆序对当且仅当A_i>A_jA ...

  4. hdu 5591 BestCoder Round #65(博弈)

    题意: 问题描述 ZYBZYB在远足中,和同学们玩了一个“数字炸弹”游戏:由主持人心里想一个在[1,N][1,N]中的数字XX,然后玩家们轮流猜一个数字,如果一个玩家恰好猜中XX则算负,否则主持人将告 ...

  5. BestCoder Round #85(ZOJ1569尚未验证)

    A题 子序列和啊,就要想到前缀和的差.这个转换一定要!记着!那么i到j的一段子序列和Sij%m ==  0就等价于(Sj-Si-1)%m == 0 了,那么什么意思呢?就是如果有两段前缀和%m的模是一 ...

  6. BestCoder Round #65 HDOJ5592 ZYB&#39;s Premutation(树状数组+二分)

    ZYB's Premutation Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Othe ...

  7. Bestcoder round #65 && hdu 5593 ZYB's Tree 树形dp

    Time Limit: 3000/1500 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)Total Submissio ...

  8. Bestcoder round #65 && hdu 5592 ZYB's Premutation 线段树

    Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)Total Submissio ...

  9. hdu 4956 Poor Hanamichi BestCoder Round #5(数学题)

    题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=4956 Poor Hanamichi Time Limit: 2000/1000 MS (Java/Ot ...

随机推荐

  1. Zabbix 通过 JMX 监控 java 进程

    参考: [ JMX monitoring ] [ Zabbix Java gateway ] [ JMX Monitoring (Java Gateway) not Working ] [ Monit ...

  2. 【洛谷 P4934】 礼物 (位运算+DP)

    题目链接 位运算+\(DP\)=状压\(DP\)?(雾 \(a\&b>=min(a,b)\)在集合的意义上就是\(a\subseteq b\) 所以对每个数的子集向子集连一条边,然后答案 ...

  3. setTimeout()和setInterval()方法的区别

    setTimeout(); //5秒后执行yourFunction(),只执行一次 setInterval(); //每隔5秒执行一次 1.setTimeout(funhander,time)的作用是 ...

  4. bind类成员函数

    首先描述一个情景: 先贴出代码: class Solution { public: bool compare(int a, int b) { return a > b; } int functi ...

  5. exit()与_exit()区别

    exit()与_exit()都是用来终止进程的函数,当程序执行到两者函数时,系统将会无条件停止剩下操作,清除进程结构体相应信息,并终止进程运行. 二者的主要区别在于:exit()函数在执行时,系统会检 ...

  6. 网络设备之pci_device_id

    标准PCI设备都有一个配置寄存器,用来存储各种参数: /* pci设备配置寄存器 */ struct pci_device_id { /* 厂商id,设备id */ __u32 vendor, dev ...

  7. 基础的语法知识(static关键字)

    1.C++中的局部变量.全局变量.局部静态变量.全局静态变量的区别 局部变量(Local variables)与 全局变量: 在子程序或代码块中定义的变量称为局部变量,在程序的一开始定义的变量称为全局 ...

  8. wifi驱动移植

    目标板:Hi3518 内核版本:linux3.0.8   1.修改makefile #PLATFORM = PC //注释掉 PLATFORM = HI3518 //支持平台 ifeq ($(PLAT ...

  9. vue 开始开发

    1,引入vue.js文件 2,在body里用标签 编辑一个入口 <div id="app">{{msg}}</div> <-- 用双大括号 取数据显示 ...

  10. django开发项目实例1--建立一个项目并初步运行

    1:进入目标目录新建一个项目 D:\>django-admin.py startproject qiweijie 新建完成后,进入项目文件夹查看目录 D:\>cd qiweijie D:\ ...