博客
关于我
Ignatius and the Princess II(全排列)
阅读量:613 次
发布时间:2019-03-13

本文共 2135 字,大约阅读时间需要 7 分钟。

Ignatius and the Princess II

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 6359    Accepted Submission(s): 3760

Problem Description
Now our hero finds the door to the BEelzebub feng5166. He opens the door and finds feng5166 is about to kill our pretty Princess. But now the BEelzebub has to beat our hero first. feng5166 says, "I have three question for you, if you can work them out, I will release the Princess, or you will be my dinner, too." Ignatius says confidently, "OK, at last, I will save the Princess."
"Now I will show you the first problem." feng5166 says, "Given a sequence of number 1 to N, we define that 1,2,3...N-1,N is the smallest sequence among all the sequence which can be composed with number 1 to N(each number can be and should be use only once in this problem). So it's easy to see the second smallest sequence is 1,2,3...N,N-1. Now I will give you two numbers, N and M. You should tell me the Mth smallest sequence which is composed with number 1 to N. It's easy, isn't is? Hahahahaha......" Can you help Ignatius to solve this problem?
 

 

Input
The input contains several test cases. Each test case consists of two numbers, N and M(1<=N<=1000, 1<=M<=10000). You may assume that there is always a sequence satisfied the BEelzebub's demand. The input is terminated by the end of file.
 

 

Output
For each test case, you only have to output the sequence satisfied the BEelzebub's demand. When output a sequence, you should print a space between two numbers, but do not output any spaces after the last number.
 

 

Sample Input
6 411 8
 

 

Sample Output
1 2 3 5 6 41 2 3 4 5 6 7 9 8 11 10
 

题解:水题,让找第M小的序列;它这里的小应该就是逆序数从小到大;

其实就是个全排列,递归下就好了;

代码:

#include
#include
#include
#include
#include
using namespace std;#define mem(x,y) memset(x,y,sizeof(x))#define SI(x) scanf("%d",&x)#define PI(x) printf("%d",x)#define P_ printf(" ")const int MAXN=1010;int ans[MAXN];int vis[MAXN];int N,M;int cnt;int flot;void dfs(int num){ if(flot)return; if(num==N){ cnt++; if(cnt==M){ for(int i=0;i

 

转载地址:http://zveaz.baihongyu.com/

你可能感兴趣的文章
wxPython和PyOpenGL视频
查看>>
精美的湿度和温度传感器
查看>>
在30分钟内学习PHP
查看>>
软考高项之风险管理-攻坚记忆
查看>>
Spark程序运行常见错误解决方法以及优化
查看>>
Python http.server 服务器
查看>>
Python svm 支持向量机
查看>>
Flask与Django的区别
查看>>
python使用kafka收发消息
查看>>
iptables开放22、80端口,以及允许本机访问本机所有端口协议
查看>>
Docker之常用命令
查看>>
SAS进阶《深入分析SAS》之数据汇总和展现
查看>>
OpenStack 最小化安装配置(一):物理机网桥配置
查看>>
centos下sftp连接配置
查看>>
shell中将字符中换行符'\n'替换为空格
查看>>
Shell变量while循环内改变无法传递到循环外
查看>>
PS快速美白照片
查看>>
ubuntu 16.04 镜像下载
查看>>
CUDA9.1、cuDNN7在Ubuntu16.04上的安装
查看>>
Elasticsearch 搜索数据基本用法
查看>>