zoukankan      html  css  js  c++  java
  • 全排列

    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? 
    InputThe 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. 
    OutputFor 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

    利用STL里面的next_permutation()做这道题不要太简单。。

    #include<iostream>
    #include<algorithm>
    using namespace std;
    #define N 1005
    void permutation(int *a,int length,int end)
    {
        int count=0;
        sort(a,a+length);
        do
        {
            count++;
            if(count==end)
                {
                    for(int i=0;i<length;i++)
                        {
                            if(i==length-1)
                                cout<<a[i];
                            else
                                cout<<a[i]<<" ";
                        }
                    break;
                }
        }
        while(next_permutation(a,a+length));
    }
    int main()
    {
        int x,a[N],y;
        while(cin>>x>>y)
        {
            for(int i=0;i<x;i++)
                a[i]=i+1;
            permutation(a,x,y);
            cout<<endl;
        }
        return 0;
    }
    

      

  • 相关阅读:
    理解OO 思想 架构好一个程序之基石!~
    WP7 ZIP 压缩与解压缩
    Gis LBS 应用 剧本 (自己乱想的)
    PHP 入门 环境搭建
    Android 入门必须知道的 英文缩写
    Android 开发 数据结构理解 队列和栈 分析及实现
    c函数中形参为引用的情况;C++中*a和*&a的区别
    如何在.NET中使用尾递归
    [ProjectEuler.net] 25 找到第一个fib数,数位为1000位
    NFA_DFA
  • 原文地址:https://www.cnblogs.com/zzzying/p/7202226.html
Copyright © 2011-2022 走看看