HDU 1027：Ignatius and the Princess II ← next_permutation()

【题目来源】
http://acm.hdu.edu.cn/showproblem.php?pid=1027

【题目描述】
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?

【题目大意】
给定 N 和 M。
若第一个序列为 1,2,3...N-1,N，第二个序列为 1,2,3...N,N-1，……。求第 M 个排列。

【输入格式】
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.

【输出格式】
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.

【输入样例】
6 4
11 8

【输出格式】
1 2 3 5 6 4
1 2 3 4 5 6 7 9 8 11 10

【算法分析】
在使用 next_permutation() 的时候，初始序列一般是一个字典序最小的序列。
如果不是，可以用 sort() 排序，得到最小序列，然后再使用 next_permutation()。

#include <bits/stdc++.h>
using namespace std;const int maxn=100;
int a[maxn];
int n;int main() {cin>>n;for(int i=0; i<n; i++) cin>>a[i];sort(a,a+n);do {for(int i=0; i<n; i++)cout<<a[i]<<" ";cout<<endl;} while(next_permutation(a,a+n));return 0;
}/*
in:
3
1 3 2
out:
1 2 3
1 3 2
2 1 3
2 3 1
3 1 2
3 2 1
*/

next_permutation() 用法详见官网：
https://cplusplus.com/reference/algorithm/next_permutation/


【算法代码】

#include <bits/stdc++.h>
using namespace std;const int maxn=1005;
int a[maxn];int main() {int n,m;while(cin>>n>>m) {for(int i=1; i<=n; i++) a[i]=i;int t=1;do {if(t==m) break;t++;} while(next_permutation(a+1,a+1+n));for(int i=1; i<n; i++) cout<<a[i]<<" ";cout<<a[n]<<endl;}return 0;
}/*
in:
6 4
11 8out:
1 2 3 5 6 4
1 2 3 4 5 6 7 9 8 11 10
*/

【参考文献】
https://blog.csdn.net/hnjzsyjyj/article/details/118616680
https://blog.csdn.net/hnjzsyjyj/article/details/125808608
https://cplusplus.com/reference/algorithm/next_permutation/