ACM Ignatius and the Princess II

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 4

11 8

 

 

Sample Output

1 2 3 5 6 4

1 2 3 4 5 6 7 9 8 11 10

题解：　　

　　　　给定数n，求1.....n-1,n的第m个全排列，最原始的数组算第一个，那么只要用next_permutation()函数循环m-1次，然后输出即可。

#include<bits/stdc++.h>
using namespace std;
int main()
{
    int n,m;
    int num[1005];
    while(cin>>n>>m)
    {
        for(int i = 0; i < n+1; i++)
            num[i] = i+1;
        for(int i = 0; i < m-1; i++)
            next_permutation(num,num+n);
        for(int i = 0; i < n-1; i++)
            cout<<num[i]<<" ";
        cout<<num[n-1]<<endl;
    }
    return 0;
}

网上摘抄补充：

　　　在STL中，除了next_permutation外，还有一个函数prev_permutation，两者都是用来计算排列组合的函数。前者是求出下一个排列组合，而后者是求出上一个排列组合。所谓“下一个”和“上一个”，书中举了一个简单的例子：对序列 {a, b, c}，每一个元素都比后面的小，按照字典序列，固定a之后，a比bc都小，c比b大，它的下一个序列即为{a, c, b}，而{a, c, b}的上一个序列即为{a, b, c}，同理可以推出所有的六个序列为：{a, b, c}、{a, c, b}、{b, a, c}、{b, c, a}、{c, a, b}、{c, b, a}，其中{a, b, c}没有上一个元素，{c, b, a}没有下一个元素。