排列生成器，
next_permutation(num, num+n) num表示指针，指向数组开始的位置，num+n,
指向数组结束的下一位 [first, last ) ， 左闭右开。
意义：按字典序的下一个排列。
头文件：#include<algorithm> ,
Ignatius and the Princess II
Time Limit: 2000/1000 MS
(Java/Others) Memory Limit: 65536/32768 K
(Java/Others)
Total Submission(s):
4117 Accepted Submission(s):
2468
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
Sample Output
1 2 3 5 6 4
1 2 3 4 5 6 7 9 8 11 10
代码如下：
1 #include<iostream>
2 #include<stdio.h>
3 #include<string>
4 #include<string.h>
5 #include<map>
6 #include<math.h>
7 #include<algorithm>
8 #define N 1005
9 using namespace std;
10 int n,m;
11 int num[N];
12 int main()
13 {
14 while(scanf("%d%d",&n,&m)!=EOF)
15 {
16 for(int i=1;i<=n;i++)
17 num[i]=i;
18 for(int i=1;i<=m-1;i++)
19 next_permutation(num+1,num+n+1);
20 for(int i=1;i<=n-1;i++)
21 printf("%d ",num[i]);
22 printf("%d\n",num[n]);
23 }
24 return 0 ;
25 }
hdu 1027 排列生成器,布布扣,bubuko.com
hdu 1027 排列生成器
原文：http://www.cnblogs.com/zn505119020/p/3581915.html
