转载请注明出处: http://www.cnblogs.com/fraud/          ——by fraud

Cyclic antimonotonic permutations

Time Limit: 20000/5000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)

Problem Description
A permutation is a sequence of integers which contains each integer from 1 to n exactly once. In this problem we are looking for permutations with special properties:

1. Antimonotonic: for each consecutive 3 values pi-1, pi, pi+1 (1 < i < n), pi should either be the smallest or the biggest of the three values. 
2. Cyclic: The permutation should consist of only one cycle, that is, when we use pi as a pointer from i to pi, it should be possible to start at position 1 and follow the pointers and reach all n positions before returning to position 1. 

 



Input
The input file contains several test cases. Each test case consists of a line containing an integer n, (3 ≤ n ≤ 106), the number of integers in the permutation. Input is terminated by n=0.

 



Output
For each test case print a permutation of the integers 1 to n which is both antimonotonic and cyclic. In case there are multiple solutions, you may print any one. Separate all integers by whitespace characters.
 



Sample Input
3
5
10
0
 



Sample Output
3 1 2
4 5 2 3 1
6 10 2 9 3 5 4 7 1 8
 



Source

分奇数和偶数,随便搞一下

hdu由于没有special judge,无法AC

 //#####################
//Author:fraud
//Blog: http://www.cnblogs.com/fraud/
//#####################
#include <iostream>
#include <sstream>
#include <ios>
#include <iomanip>
#include <functional>
#include <algorithm>
#include <vector>
#include <string>
#include <list>
#include <queue>
#include <deque>
#include <stack>
#include <set>
#include <map>
#include <cstdio>
#include <cstdlib>
#include <cmath>
#include <cstring>
#include <climits>
#include <cctype>
using namespace std;
#define XINF INT_MAX
#define INF 0x3FFFFFFF
#define MP(X,Y) make_pair(X,Y)
#define PB(X) push_back(X)
#define REP(X,N) for(int X=0;X<N;X++)
#define REP2(X,L,R) for(int X=L;X<=R;X++)
#define DEP(X,R,L) for(int X=R;X>=L;X--)
#define CLR(A,X) memset(A,X,sizeof(A))
#define IT iterator
typedef long long ll;
typedef pair<int,int> PII;
typedef vector<PII> VII;
typedef vector<int> VI;
#define MAXN 1000010
int a[MAXN];
int main()
{
int n;
while(scanf("%d",&n)&&n){
a[]=;
a[]=;
for(int i=;i<=n;i++){
if(i&)a[i]=i+;
else a[i]=i-;
}
if(n&)a[n]=n-;
else{
a[n]=n-;
a[n-]=n;
}
printf("%d",a[]);
for(int i=;i<=n;i++){
printf(" %d",a[i]);
}
printf("\n");
}
return ;
}

代码君

uva11630 or hdu2987 Cyclic antimonotonic permutations(构造水题)的更多相关文章

  1. 构造水题 Codeforces Round #206 (Div. 2) A. Vasya and Digital Root

    题目传送门 /* 构造水题:对于0的多个位数的NO,对于位数太大的在后面补0,在9×k的范围内的平均的原则 */ #include <cstdio> #include <algori ...

  2. codeforces 22C System Administrator(构造水题)

    转载请注明出处: http://www.cnblogs.com/fraud/          ——by fraud System Administrator Bob got a job as a s ...

  3. Educational Codeforces Round 94 (Rated for Div. 2) A. String Similarity (构造水题)

    题意:给你一个长度为\(2*n-1\)的字符串\(s\),让你构造一个长度为\(n\)的字符串,使得构造的字符串中有相同位置的字符等于\(s[1..n],s[2..n+1],...,s[n,2n-1] ...

  4. hdu 2044:一只小蜜蜂...(水题,斐波那契数列)

    一只小蜜蜂... Time Limit: / MS (Java/Others) Memory Limit: / K (Java/Others) Total Submission(s): Accepte ...

  5. hdu 2041:超级楼梯(水题,递归)

    超级楼梯 Time Limit: / MS (Java/Others) Memory Limit: / K (Java/Others) Total Submission(s): Accepted Su ...

  6. codeforces Gym 100187L L. Ministry of Truth 水题

    L. Ministry of Truth Time Limit: 20 Sec Memory Limit: 256 MB 题目连接 http://codeforces.com/gym/100187/p ...

  7. POJ 水题(刷题)进阶

    转载请注明出处:優YoU http://blog.csdn.net/lyy289065406/article/details/6642573 部分解题报告添加新内容,除了原有的"大致题意&q ...

  8. Atcoder 水题选做

    为什么是水题选做呢?因为我只会水题啊 ( 为什么是$Atcoder$呢?因为暑假学长来讲课的时候讲了三件事:不要用洛谷,不要用dev-c++,不要用单步调试.$bzoj$太难了,$Topcoder$整 ...

  9. CF330 C. Purification 认真想后就成水题了

    C. Purification time limit per test 1 second memory limit per test 256 megabytes input standard inpu ...

随机推荐

  1. BZOJ 1003 物流运输 (动态规划 SPFA 最短路)

    1003: [ZJOI2006]物流运输 Time Limit: 10 Sec Memory Limit: 162 MB Submit: 5590 Solved: 2293 [Submit][Stat ...

  2. QT5-控件-QComboBox

    #ifndef MAINWINDOW_H #define MAINWINDOW_H #include <QMainWindow> #include <QComboBox> cl ...

  3. web api简单验证实现办法

    需要使用WEBAPI,但是有验证问题没解决.后来参考网上文章做了一下DEMO 思路: 就是根据用户的账号在服务端加密一个字符串,然后返回给用户端. 具体: 一个用户编号用于唯一身份识别,密码,一个密钥 ...

  4. [TYVJ] P1026 犁田机器人

    犁田机器人 背景 Background USACO OCT 09 2ND   描述 Description Farmer John為了让自己从无穷无尽的犁田工作中解放出来,於是买了个新机器人帮助他犁田 ...

  5. qt之窗口换肤(一个qss的坑:当类属性发现变化时需要重置qss,使用rcc资源文件)

    1.相关文章 Qt 资源系统qt的moc,uic,rcc命令的使用 2.概要    毕业两年了,一直使用的是qt界面库来开发程序,使用过vs08.10.13等开发工具,并安装了qt的插件,最近在做客户 ...

  6. poj 3321

    题目链接 题意:一开始1-n都有苹果,Q查询以x为根下存在多少. 树状数组+DFS+队列转换 这题纠结了2天,一开始一点思路都没有,看大神都是吧树状数组转换成队列来做 看了好久都不知道怎么转换的, 解 ...

  7. RFC 2327--SDP

    Network Working Group M. Handley Request for Comments: 2327 V. Jacobson Category: Standards Track IS ...

  8. spring mvc 安全

    1,使用 spring form 标签 防 csrf 攻击 2,标明请求方法:RequestMethod.GET,RequestMethod.POST, PATCH, POST, PUT, and D ...

  9. poj 3616 Milking Time(dp)

    Description Bessie ≤ N ≤ ,,) hours (conveniently labeled ..N-) so that she produces as much milk as ...

  10. NavMeshAgent 动态加载障碍物

    如果你想让游戏人物绕开一些物体, 这些物体动态生成出来的.只需要给物体添加NavMeshObstacle组件即可 1. 绿色方块添加NavMeshObstacle组件 2. 红色方块没有添加NavMe ...