[USACO11FEB] Cow Line
https://www.luogu.org/problem/show?pid=3014
题目描述
The N (1 <= N <= 20) cows conveniently numbered 1...N are playing yet another one of their crazy games with Farmer John. The cows will arrange themselves in a line and ask Farmer John what their line number is. In return, Farmer John can give them a line number and the cows must rearrange themselves into that line.
A line number is assigned by numbering all the permutations of the line in lexicographic order.
Consider this example:
Farmer John has 5 cows and gives them the line number of 3.
The permutations of the line in ascending lexicographic order: 1st: 1 2 3 4 5
2nd: 1 2 3 5 4
3rd: 1 2 4 3 5
Therefore, the cows will line themselves in the cow line 1 2 4 3 5.
The cows, in return, line themselves in the configuration '1 2 5 3 4' and ask Farmer John what their line number is.
Continuing with the list:
4th : 1 2 4 5 3
5th : 1 2 5 3 4
Farmer John can see the answer here is 5
Farmer John and the cows would like your help to play their game. They have K (1 <= K <= 10,000) queries that they need help with. Query i has two parts: C_i will be the command, which is either 'P' or 'Q'.
If C_i is 'P', then the second part of the query will be one integer A_i (1 <= A_i <= N!), which is a line number. This is Farmer John challenging the cows to line up in the correct cow line.
If C_i is 'Q', then the second part of the query will be N distinct integers B_ij (1 <= B_ij <= N). This will denote a cow line. These are the cows challenging Farmer John to find their line number.
输入输出格式
输入格式:
Line 1: Two space-separated integers: N and K
- Lines 2..2*K+1: Line 2*i and 2*i+1 will contain a single query.
Line 2*i will contain just one character: 'Q' if the cows are lining up and asking Farmer John for their line number or 'P' if Farmer John gives the cows a line number.
If the line 2*i is 'Q', then line 2*i+1 will contain N space-separated integers B_ij which represent the cow line. If the line 2*i is 'P', then line 2*i+1 will contain a single integer A_i which is the line number to solve for.
输出格式:
- Lines 1..K: Line i will contain the answer to query i.
If line 2*i of the input was 'Q', then this line will contain a single integer, which is the line number of the cow line in line 2*i+1.
If line 2*i of the input was 'P', then this line will contain N space separated integers giving the cow line of the number in line 2*i+1.
输入输出样例
5 2
P
3
Q
1 2 5 3 4
1 2 4 3 5
5 题意:
询问这是第几个全排列,询问第k个全排列是什么
#include<cstdio>
#include<cstring>
using namespace std;
long long jc[];
bool v[];
int p[];
int main()
{
int n,q,cnt,now,sum;
long long x,tot;
char c[];
scanf("%d%d",&n,&q);
jc[]=;
for(int i=;i<=n;i++) jc[i]=jc[i-]*i;
while(q--)
{
scanf("%s",c);
if(c[]=='P')
{
scanf("%lld",&x);
memset(v,,sizeof(v));
for(int i=;i<n;i++)
{
cnt=now=;
while(x>jc[n-i]) x-=jc[n-i],cnt++;
for(int i=;i<=;i++)
if(!v[i])
{
now++;
if(now==cnt+)
{
printf("%d ",i);
v[i]=true;
continue;
}
}
}
for(int i=;i<=;i++)
if(!v[i])
{
printf("%d\n",i);
break;
}
}
else
{
tot=,sum;
for(int i=;i<=n;i++) scanf("%d",&p[i]);
memset(v,,sizeof(v));
for(int i=;i<n;i++)
{
sum=;
for(now=;now<=n;now++)
if(!v[now] && now<p[i]) sum++;
v[p[i]]=true;
tot+=sum*jc[n-i];
}
printf("%lld\n",tot+);
}
}
}
[USACO11FEB] Cow Line的更多相关文章
- P3014 [USACO11FEB]牛线Cow Line && 康托展开
康托展开 康托展开为全排列到一个自然数的映射, 空间压缩效率很高. 简单来说, 康托展开就是一个全排列在所有此序列全排列字典序中的第 \(k\) 大, 这个 \(k\) 即是次全排列的康托展开. 康托 ...
- 洛谷 P3014 [USACO11FEB]牛线Cow Line
P3014 [USACO11FEB]牛线Cow Line 题目背景 征求翻译.如果你能提供翻译或者题意简述,请直接发讨论,感谢你的贡献. 题目描述 The N (1 <= N <= 20) ...
- [洛谷P3014][USACO11FEB]牛线Cow Line (康托展开)(数论)
如果在阅读本文之前对于康托展开没有了解的同学请戳一下这里: 简陋的博客 百度百科 题目描述 N(1<=N<=20)头牛,编号为1...N,正在与FJ玩一个疯狂的游戏.奶牛会排成一行 ...
- POJ 3617 Best Cow Line (贪心)
Best Cow Line Time Limit: 1000MS Memory Limit: 65536K Total Submissions: 16104 Accepted: 4 ...
- POJ 3617 Best Cow Line(最佳奶牛队伍)
POJ 3617 Best Cow Line Time Limit: 1000MS Memory Limit: 65536K [Description] [题目描述] FJ is about to t ...
- 【BZOJ】【3301】【USACO2011 Feb】Cow Line
康托展开 裸的康托展开&逆康托展开 康托展开就是一种特殊的hash,且是可逆的…… 康托展开计算的是有多少种排列的字典序比这个小,所以编号应该+1:逆运算同理(-1). 序列->序号:( ...
- BZOJ3403: [Usaco2009 Open]Cow Line 直线上的牛
3403: [Usaco2009 Open]Cow Line 直线上的牛 Time Limit: 3 Sec Memory Limit: 128 MBSubmit: 48 Solved: 41[S ...
- BZOJ3301: [USACO2011 Feb] Cow Line
3301: [USACO2011 Feb] Cow Line Time Limit: 10 Sec Memory Limit: 128 MBSubmit: 67 Solved: 39[Submit ...
- BZOJ1640: [Usaco2007 Nov]Best Cow Line 队列变换
1640: [Usaco2007 Nov]Best Cow Line 队列变换 Time Limit: 5 Sec Memory Limit: 64 MBSubmit: 493 Solved: 2 ...
随机推荐
- Thunder团队第二周 - Scrum会议4
Scrum会议4 小组名称:Thunder 项目名称:爱阅app Scrum Master:邹双黛 工作照片: 宋雨同学在拍照,所以不再照片中. 参会成员: 王航:http://www.cnblogs ...
- 针对XX系统的可用性和易用性构想
可用性是与系统故障有关的一个质量属性,是指系统正常运行的时间的比例,一般通过两次故障之间的时间长度或在系统崩溃情况下能恢复正常运行的速度来衡量,同时此概念涉及一个公式的计算,就是系统正常运行时间的百分 ...
- Python基础1 Hello World!
从今天开始和大家分享一下python最基础的知识,以便帮助初学者快速入门. 最最基础的当然是hello world 了,无论哪门语言都会从它开始... 简单的‘Hello World!’ 1. 直接运 ...
- j2ee—框架(1):Servlet+JSP实现基本的登录功能(v1.0)
主要分为四个部分:LoginController.web.xml.login.jsp和login_success.jsp(login_fail.jsp). 第一部分 LoginController p ...
- Ubuntu16.0.4 安装mysql
1. sudo apt-get install mysql-server 2. sudo apt-get install mysql-client 3. sudo apt-get install l ...
- xpath教程二 ---- 通过ID和Class检索
必备知识点 在html中,id是唯一的 在html中,class是可以多处引用的 工具 Python3版本 lxml库[优点是解析快] HTML代码块[从网络中获取或者自己杜撰一个] requests ...
- nginx的平滑升级,不间断服务
nginx的平滑升级,不间断服务 Nginx更新真的很快,最近nginx的1.0.5稳定版,nginx的0.8.55和nginx的0.7.69旧的稳定版本已经发布.我一项比较喜欢使用新版本的软件, ...
- Github Atom汉化方式
1.下载:Atom https://atom.io/ 2.安装 3.菜单栏 -- Setting --- Install --- 搜索Chinese --安装汉化包 4.重启 生效.
- ASP.NET MVC 多语言解决方案
1:打开VS,新建ASP.NET MVC4项目 2:创建一个放本地化资源的文件夹并命名为"Language",右键选择添加新项,选择资源文件并命名为"Com" ...
- Java多线程中的join方法
新建一个Thread,代码如下: package com.thread.test; public class MyThread extends Thread { private String name ...