zoukankan      html  css  js  c++  java
  • 杭电 1027 Ignatius and the Princess II

    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
     
    Author
    Ignatius.L
     
         一开始就知道stl库中有一个next_permutation()函数,但是不想用那个,想用递归写出来,但是失败啦!所以又回到了使用next_permutation()函数上!
     
    View Code
     1 #include <cstdio>
     2 #include <algorithm>
     3 using namespace std;
     4 
     5 int main()
     6 {
     7     int p[1005], n, m, i;
     8     while( scanf( "%d%d", &n, &m ) != EOF )
     9     {
    10            memset( p, 0, 1005 * sizeof(int) );
    11            for( i = 0; i < n; i++ )
    12                 p[i] = i+1;
    13            m = m - 1;
    14            while( m-- )
    15                   next_permutation(p, p+n);
    16            for( i = 0; i < n; i++ )
    17            {
    18                 printf( "%d", p[i] );
    19                 if( i != (n-1) )
    20                     printf( " " );
    21                 else
    22                     printf( "\n" );
    23            }
    24            
    25     }
    26     return 0;
    27 }
  • 相关阅读:
    C#开发Unity游戏教程之游戏对象的属性变量
    C#开发Unity游戏教程之Scene视图与脚本的使用
    BeagleBone Black教程之BeagleBone Black使用到的Linux基础
    BeagleBone Black教程之BeagleBone Black设备的连接
    ARP协议详解之Gratuitous ARP(免费ARP)
    ARP协议详解之ARP动态与静态条目的生命周期
    ArduinoYun教程之Arduino环境与Linux环境的桥梁Bridge
    ArduinoYun教程之OpenWrt-Yun与CLI配置Arduino Yun
    ArduinoYun教程之配置Arduino Yun环境
    IIR数字滤波器
  • 原文地址:https://www.cnblogs.com/yizhanhaha/p/3073573.html
Copyright © 2011-2022 走看看