zoukankan      html  css  js  c++  java
  • 【HDU 1027】Ignatius and the Princess II

    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

    题解:求全排列哦
    #include<cstdio>
    #include<iostream>
    #include<cstring>
    #include<cstdlib>
    #include<algorithm>
    #include<cmath>
    typedef long long ll;
    using namespace std;
    int n,m,a[1005];
    int main(){
        while(scanf("%d %d",&n,&m)!=EOF){
            for(int i=1;i<1005;i++) 
                a[i]=i;
            for(int i=1;i<m;i++)
                next_permutation(a+1, a+1+n);
            printf("%d",a[1]);
            for(int i=2;i<=n;i++)
                printf(" %d",a[i]);
            printf("
    ");
        }
        return 0;
    }
  • 相关阅读:
    Myeclipse下使用Maven搭建spring boot项目
    Dubbo+Zookeeper视频教程
    dubbo项目实战代码展示
    流程开发Activiti 与SpringMVC整合实例
    交换两个变量的值,不使用第三个变量的四种法方
    数据库主从一致性架构优化4种方法
    数据库读写分离(aop方式完整实现)
    在本地模拟搭建zookeeper集群环境实例
    box-sizing布局
    盒子模型
  • 原文地址:https://www.cnblogs.com/wuhu-JJJ/p/11328283.html
Copyright © 2011-2022 走看看