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

    Ignatius and the Princess II

    Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
    Total Submission(s): 9095    Accepted Submission(s): 5311


    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
     1 #include<iostream>
     2 #include<string>
     3 #include<vector>
     4 #include<fstream>
     5 using namespace std;
     6 struct data
     7 {
     8     int zhi;
     9     int state;
    10 };
    11 int main()
    12 {
    13     //ifstream in("data.txt");
    14     int n,m;
    15     while(cin>>n>>m)
    16     {
    17         vector<data> vec;
    18         int i;
    19         int sum=1;
    20         int inc=1;
    21         int count;
    22         while(sum<m)
    23             sum*=++inc;
    24         for(i=1;i<=n-inc;i++)
    25             cout<<i<<" ";
    26         while(i<=n)
    27         {
    28             data temp={i++,0};
    29             vec.push_back(temp);
    30         }
    31         count=n-inc;
    32         int fir=0;
    33         while(count<n)
    34         {
    35             int jw=(m-1)/(sum/inc)+1;
    36             int j=1;
    37             i=0;
    38             //cout<<"jw: "<<jw<<endl;
    39             while(j<=jw)
    40                 if(vec[i++].state==0)
    41                     j++;
    42             if(fir==0)
    43                 fir=1;
    44             else
    45                 cout<<" ";
    46             cout<<vec[i-1].zhi;
    47             vec[i-1].state=1;
    48             count++;
    49             m=m%(sum/inc);
    50             if(m==0)
    51             {
    52                 i=vec.size()-1;
    53                 while(count<n)
    54                 {
    55                     if(vec[i].state==0)
    56                     {
    57                         cout<<" "<<vec[i].zhi;
    58                         count++;
    59                     }
    60                     i--;
    61                 }
    62                 break;
    63             }
    64             else
    65             {
    66                 sum/=inc;
    67                 inc--;
    68             }
    69         }
    70         cout<<endl;
    71     }
    72     return 0;
    73 }
  • 相关阅读:
    [C++]2-5 分数化小数
    [C++]2-4 子序列的和
    [C++]2-3 倒三角形
    [C++]2-2 韩信点兵
    [C++]2-1 水仙花数
    [C++]竞赛模板·数据统计与IO(重定向版与非重定向版)
    数学建模·经验小结
    信息检索·论文写作
    PPT制作
    演讲与语言表达
  • 原文地址:https://www.cnblogs.com/zhaopeng938/p/7900776.html
Copyright © 2011-2022 走看看