BestCoder Round #65 (ZYB's Game)
ZYB's Game
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].
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
For each testcase,print the ans.
1
3
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
- BestCoder Round #65 (ZYB's Premutation)
ZYB's Premutation Accepts: 220 Submissions: 983 Time Limit: 2000/1000 MS (Java/Others) Memory Limit: ...
- BestCoder Round #65 (ZYB's Biology)
ZYB's Biology Accepts: 848 Submissions: 1199 Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 13 ...
- 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 ...
- 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 ...
- 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 ...
- BestCoder Round #65 HDOJ5592 ZYB'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 ...
- 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 ...
- 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 ...
- 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
- Item 4 ----通过私有构造器强化不可实例化的能力
场景: 在创建工具类的时候,大部分是无需实例化的,实例化对它们没有意义.在这种情况下,创建的类,要确保它是不可以实例化的. 存在问题: 在创建不可实例化的类时,虽然没有定义构造器.但是,客户端在使 ...
- 「6月雅礼集训 2017 Day4」暴力大神hxx
[题目大意] 给出一个n重循环,每重循环有范围$[l, r]$,其中$l$,$r$可能是之前的变量,也可能是常数.求循环最底层被执行了多少次. 其中,保证每个循环的$l$,$r$最多有一个是之前的变量 ...
- Python与RPC -- (转)
XML-RPC xmlrpc是使用http协议做为传输协议的rpc机制,使用xml文本的方式传输命令和数据. 一个rpc系统,必然包括2个部分: 1)rpc client,用来向rpc server调 ...
- I题 hdu 1234 开门人和关门人
题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=1234 开门人和关门人 Time Limit: 2000/1000 MS (Java/Others) ...
- E题hdu 1425 sort
题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=1425 sort Time Limit: 6000/1000 MS (Java/Others) M ...
- Low-overhead enhancement of reliability of journaled file system using solid state storage and de-duplication
A mechanism is provided in a data processing system for reliable asynchronous solid-state device bas ...
- zip函数的应用
#!/usr/bin/env python # encoding: utf-8 from itertools import zip_longest # ➍ # zip并行从输入的各个可迭代对象中获取元 ...
- python爬虫模块之URL管理器模块
URL管理器模块 一般是用来维护爬取的url和未爬取的url已经新添加的url的,如果队列中已经存在了当前爬取的url了就不需要再重复爬取了,另外防止造成一个死循环.举个例子 我爬www.baidu. ...
- pycharm模板参数
# -*- coding: utf-8 -*-# @Time : ${DATE} ${TIME}# @Author : cxa# @File : ${NAME}.py# @Software: ${PR ...
- 【LabVIEW技巧】策略模式
前言 在之前的文章提到了如何学习OOP以及对应的简单工厂模式,由于时间比较长,我们先回顾一下原有内容,然后继续了解新的模式. 为什么学习OOP 在测控系统的软件开发过程中,LabVIEW工程师一直认为 ...