HDU 1027 Ignatiusand the Princess II

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本文介绍了一个经典的编程竞赛题目“Ignatius与公主II”，任务是找到由1到N组成的第M小的序列。文章提供了解题思路及C++实现代码，利用STL中的next_permutation函数进行全排列。

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Ignatiusand the Princess II

Time Limit: 2000/1000 MS (Java/Others) MemoryLimit: 65536/32768 K (Java/Others)
Total Submission(s): 2675 Accepted Submission(s): 1622

Problem Description

Now ourhero finds the door to the BEelzebub feng5166. He opens the door and findsfeng5166 is about to kill our pretty Princess. But now the BEelzebub has tobeat our hero first. feng5166 says, "I have three question for you, if youcan work them out, I will release the Princess, or you will be my dinner,too." Ignatius says confidently, "OK, at last, I will save thePrincess."

"Now I will show you the first problem." feng5166 says, "Given asequence of number 1 to N, we define that 1,2,3...N-1,N is the smallestsequence among all the sequence which can be composed with number 1 to N(eachnumber can be and should be use only once in this problem). So it's easy to seethe 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 withnumber 1 to N. It's easy, isn't is? Hahahahaha......"
Can you help Ignatius to solve this problem?

Input

The inputcontains several test cases. Each test case consists of two numbers, N andM(1<=N<=1000, 1<=M<=10000). You may assume that there is always asequence satisfied the BEelzebub's demand. The input is terminated by the endof file.

Output

For eachtest case, you only have to output the sequence satisfied the BEelzebub'sdemand. 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

Author

Ignatius.L

题目大意：给定一个序列（递增有序的序列），输出这个序列第M小的序列；

解题思路：其实就是求序列的全排列，按字典序排列，可以使用stl中的next_permutation进行全排列操作；

#include <iostream>
#include <cstdio>
#include <algorithm>
#define nMax 1024
using namespace std;

int a[nMax];
int main(int argc, char *argv[])
{
    int n,m;
    while(cin>>n>>m)
    {
        for(int i=1; i<=n; i++)
        {
            a[i]=i;
        }
        for(int i=1; i<=m-1; i++)
        {
            next_permutation(a+1,a+n+1);
        }
        for(int i=1; i<=n-1; i++)
        {
            cout<<a[i]<<" ";
        }
        cout<<a[n]<<endl;
    }
    return 0;
}