HDOJ1027 Ignatius and the Princess II&&STL神器

Ignatius and the Princess II

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 8279 Accepted Submission(s): 4890

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

本题的意思是给定数字的个数，要你求出按顺序的第M个全排列。
通过这道题发现了一个STL神器:next_permutation（后一个）和prev_permutation（前一个）函数
依照STL文档的描写叙述，next_permutation函数将按字母表顺序生成给定序列的下一个较大的序列。直到整个序列为减序为止。prev_permutation函数与之相反。是生成给定序列的上一个较小的序列。

这样这题就很简单了。
代码如下

#include <iostream>#include <algorithm>#include <cstdio>#include <cstring>#include <cmath>#include <cstdlib>#include <set>#include <queue>#include <cctype>#define N 1010int num[N];using namespace std;int main(){    int n,m;    while(cin>>n>>m)    {        memset(num,0,N);        for(int i=0;i<n;i++)            num[i]=i+1;        for(int i=0;i<m-1;i++)            next_permutation(num,num+n);//求下一个全排列，求M-1次        for(int i=0;i<n-1;i++)            cout<<num[i]<<' ';        cout<<num[n-1]<<endl;    }    return 0;}