一、K - next_permutation

最新推荐文章于 2025-02-20 00:33:22 发布

Jenny-z

最新推荐文章于 2025-02-20 00:33:22 发布

阅读量226

点赞数

分类专栏： vjudge

本文链接：https://blog.youkuaiyun.com/shuyizhao/article/details/81478081

版权

vjudge 专栏收录该内容

43 篇文章

订阅专栏

本文介绍了一个使用STL库next_permutation解决特定排列组合问题的方法。通过实例详细展示了如何求解给定数字范围内的第M小序列，适用于竞赛编程及算法训练。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

K - next_permutation

HDU - 1027

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

题意：给定n和m n表示1……n的数字排列。求第m个排列。

用stl库给定的next_permutation，输出第m个排列就可以。

#include <iostream>
#include <algorithm>
#define max 1000

using namespace std;

int main()
{
    int m,n,i;
    while(cin>>n)
    {
        int a[max],time=1;
        for(i=0;i<n;i++)
        {
            a[i]=i+1;
        }
        cin>>m;
        do{
            if(time==m)
            {
                for(i=0;i<n-1;i++) cout<<a[i]<<' ';
                cout<<a[n-1]<<endl;
                break;
            }
            time++;
        }while(next_permutation(a,a+n));
    }
    return 0;
}