Bob is attending a chess competition. Now the competition is in the knockout phase. There are 2^r2r players now, and they will play over rr rounds.

In each knockout round, the remaining players would be divided into pairs, and the winner of each pair would advance to the next knockout round. Finally only one player would remain and be declared the champion.

Bob has already assigned all players in an order while he assigned himself to the k-th site. A better player is located at a site with a smaller number indicating a higher order. The probability that a player with higher order wins a player with lower order is p (0 \le p \le 1)p(0≤p≤1).

Bob notices that arrangement of matches is crucial for the final result.

For example, if there are 44 players and Bob is the second best player (he is the second site), and p = 0.9p=0.9. In the first round, if Bob meets the best player, he will have only 0.1 \times 0.9 = 0.090.1×0.9=0.09 probability to become the champion. However if he does not meet the best player in the first round, he will have 0.9 \times (0.9 \times 0.1 + 0.1 \times 0.9) = 0.1620.9×(0.9×0.1+0.1×0.9)=0.162 probability to become the champion. Now Bob wants to know, what is the winning probability for him in the best arrangement of matches.

Input

The first line in the input contains an integer t (1 \le t \le 63000)t(1≤t≤63000) which is the number of test cases.

For each case, there is only one line containing two integers rr and kk (1 \le r < 64,1 \le k \le 2^r)(1≤r<64,1≤k≤2r) and a float-point number p (0 \le p \le 1)p(0≤p≤1) as described above.

Output

For each case, calculate the winning probability for Bob in the best arrangement. Output the probability with the precision of 66 digits.

样例输入

2
1 1 0.8
2 2 0.9

样例输出

0.800000
0.162000 题意:2^r个人打比赛,一共比r伦决出冠军,主角的实力排在第k位,并且对于所有人,打败比他弱的人概率是p,打败比他强的人概率是(1-p);主角要尽可能的提高获胜的概率,求这个概率即可
思路:如果当前这一轮还有比主角弱的人,主角选择和弱的人对决,若只剩比主角强的人,就只能和强的人比赛。dfs记得记忆化搜索。。
AC代码:
#define _CRT_SECURE_NO_DEPRECATE
#include<iostream>
#include<algorithm>
#include<cmath>
#include<vector>
#include<cstring>
#include<map>
using namespace std;
#define INF 0x3f3f3f3f
#define EPS 1e-7
typedef unsigned long long ll;
const int N_MAX = +;
const int MOD = 1e10+;
int r;
ll k, qiang, ruo, num;
map<pair<ll, ll>, double>mp;//记忆化搜素
double p; double dfs(ll qiang,ll ruo) {
if (qiang == && ruo == )return ;
if (mp[make_pair(qiang, ruo)] != )return mp[make_pair(qiang, ruo)];
ll nxt_qiang = qiang >> , nxt_ruo = ruo >> ;
if (ruo & ) {//弱的为奇数个,说明强的为偶数个,只要主角和一个弱的打就行了
return mp[make_pair(qiang, ruo)] =p*dfs(nxt_qiang, nxt_ruo);
}
else {//否则弱的是偶数个,强的奇数个
if (ruo != ) {//弱的个数不为0,此时主角还是选择和弱的打,但是总会有一个强的会多出来和弱的打,所以这两者谁赢谁输就要分两种情况
return mp[make_pair(qiang, ruo)] = p*(p*dfs(nxt_qiang + , nxt_ruo - ) + ( - p)*dfs(nxt_qiang, nxt_ruo));
}
else {//没有弱的选手了,主角只能和强的打
return mp[make_pair(qiang, )] = ( - p)*dfs(nxt_qiang, );
}
}
} int main() {
int t; scanf("%d",&t);
while (t--) {
scanf("%d%lld%lf",&r,&k,&p);
mp.clear();
num = 1LL << r;
qiang = k - , ruo = num - k;
double res=dfs(qiang, ruo);
printf("%.6f\n",res);
}
return ;
}

2017南宁现场赛E The Champion的更多相关文章

  1. 2017acm南宁现场赛 J题 Rearrangement

    题意: 给定一个2 * n的矩阵, 和 2 * n 个数, 问能不能通过重排列, 使得任意相邻两数不能被3整除 分析: 这题一直卡到最后, 赛后经对面大佬提醒后, 发现统计所有数模三的结果(0,1,2 ...

  2. 2017 青岛现场赛 Suffix

    Consider n given non-empty strings denoted by s1 , s2 , · · · , sn . Now for each of them, you need ...

  3. 2017 青岛现场赛 I The Squared Mosquito Coil

    Lusrica designs a mosquito coil in a board with n × n grids. The mosquito coil is a series of consec ...

  4. 2017南宁网络赛 Problem J Minimum Distance in a Star Graph ( 模拟 )

    题意 : 乱七八糟说了一大堆,实际上就是问你从一个序列到另个序列最少经过多少步的变化,每一次变化只能取序列的任意一个元素去和首元素互换 分析 : 由于只能和第一个元素去互换这种操作,所以没啥最优的特别 ...

  5. ACM-ICPC 2017 西安赛区现场赛 K. LOVER II && LibreOJ#6062. 「2017 山东一轮集训 Day2」Pair(线段树)

    题目链接:西安:https://nanti.jisuanke.com/t/20759   (计蒜客的数据应该有误,题目和 LOJ 的大同小异,题解以 LOJ 为准)     LOJ:https://l ...

  6. 2017 ICPC区域赛(西安站)--- J题 LOL(DP)

    题目链接 problem description 5 friends play LOL together . Every one should BAN one character and PICK o ...

  7. 2013ACM/ICPC亚洲区南京站现场赛---Poor Warehouse Keeper(贪心)

    题目链接 http://acm.hdu.edu.cn/showproblem.php?pid=4803 Problem Description Jenny is a warehouse keeper. ...

  8. HDU 5920 Ugly Problem 高精度减法大模拟 ---2016CCPC长春区域现场赛

    题目链接 题意:给定一个很大的数,把他们分为数个回文数的和,分的个数不超过50个,输出个数并输出每个数,special judge. 题解:现场赛的时候很快想出来了思路,把这个数从中间分为两部分,当位 ...

  9. 2013杭州现场赛B题-Rabbit Kingdom

    杭州现场赛的题.BFS+DFS #include <iostream> #include<cstdio> #include<cstring> #define inf ...

随机推荐

  1. 初步学习pg_control文件之六

    接前文:初步学习pg_control文件之五 ,DB_IN_ARCHIVE_RECOVERY何时出现? 看代码:如果recovery.conf文件存在,则返回 InArchiveRecovery = ...

  2. Hadoop:WordCount分析

    相关代码: package com.hadoop; import org.apache.hadoop.conf.Configuration; import org.apache.hadoop.fs.P ...

  3. 初识Continuation

    本文来自网易云社区 作者:陆艺 上学时看了SICP之后就对scheme这个看上去比较古怪的语言产生了兴趣. 虽然书里并没有涉及scheme太多语法以及语言上特性的一些东西, 作为一个喜欢折腾的人, 手 ...

  4. 玩转VIM之将Vim全副武装

    玩转VIM之将Vim全副武装 懒癌末期的我貌似很久没有写博客了,已经欠了多少篇在计划中的博客我已然不好意思说了.好了,言归正传,在前三篇介绍了Vim作为代码编辑器之后可能会有人说,要学习那么多指令真的 ...

  5. VSCode 前端必备插件

    VSCode 前端必备插件 Debugger for Chrome 让 vscode 映射 chrome 的 debug功能,静态页面都可以用 vscode 来打断点调试 { "versio ...

  6. java编程思想 内容总结

    Java编程思想重点笔记(Java开发必看) Java编程思想,Java学习必读经典,不管是初学者还是大牛都值得一读,这里总结书中的重点知识,这些知识不仅经常出现在各大知名公司的笔试面 试过程中,而且 ...

  7. 支持ie的时间控件 html

    连接:http://www.my97.net/demo/resource/2.4.asp#m248 下载测试:链接: https://pan.baidu.com/s/17AdRa2OTLPI7ndiA ...

  8. STL中的set容器的一点总结2

    http://blog.csdn.net/sunshinewave/article/details/8068326 1.关于set C++ STL 之所以得到广泛的赞誉,也被很多人使用,不只是提供了像 ...

  9. esayui combotree 只能选择子节点

    esayui combotree 只能选择子节点用onBeforeSelect:参数是node,节点被选中之前触发,返回false取消选择动作. 网上找了好多都没一个可用的,要想知道他是子节点还是根节 ...

  10. 算法(10)Subarray Sum Equals K

    题目:在数组中找到一个子数组,让子数组的和是k. 思路:先发发牢骚,这两天做题是卡到不行哇,前一个题折腾了三天,这个题上午又被卡住,一气之下,中午睡觉,下午去了趟公司,竟然把namespace和cgr ...