• The 15th UESTC Programming Contest Preliminary J


    地址:http://acm.uestc.edu.cn/#/problem/show/1567

    题目:

    Jermutat1on

    Time Limit: 3000/1000MS (Java/Others)     Memory Limit: 65535/65535KB (Java/Others)
     

    You are given two numbers nn and kk.

    You are required to construct a permutation p1,p2,...,pnp1,p2,...,pn of numbers 1,2,...,n1,2,...,n such that there are exactly kk different numbers in |p1p2||p1−p2|, |p2p3||p2−p3|, ..., |pn1pn||pn−1−pn|.

    Input

    Only one line contains two integers nn and kk.

    1k<n1000001≤k<n≤100000.

    Output

    Print nn integers forming the permutation.

    If there are multiple answers, print any of them.

    If there are no such permutation, print -1.

    Sample input and output

    Sample InputSample Output
    3 1
    1 2 3

    Source

    The 15th UESTC Programming Contest Preliminary
     思路:水题,见代码。
     1 #include<iostream>
     2 using namespace std;
     3 
     4 int n,k;
     5 
     6 int gcd(int x,int y)
     7 {
     8     if(y==0)
     9         return x;
    10     else
    11         return gcd(y,x%y);
    12 }
    13 
    14 int main()
    15 {
    16     cin>>n>>k;
    17     if(k>=n)
    18         cout<<-1<<endl;
    19     else
    20     {
    21         if(k==1)
    22             for(int i=1;i<=n;i++)
    23                 cout<<i<<' ';
    24         else if(k%2)
    25         {
    26             int mid=(k+1)/2;
    27             cout<<mid<<' ';
    28             for(int i=1;i<=k;i++)
    29             {
    30                 if(i%2)
    31                     cout<<(mid+=i)<<' ';
    32                 else    
    33                     cout<<(mid-=i)<<' ';
    34             }
    35             for(int i=mid+1;i<=n;i++)
    36                 cout<<i<<' ';
    37         }
    38         else
    39         {
    40             int mid=k/2+1;
    41             cout<<mid<<' ';
    42             for(int i=1;i<=k;i++)
    43             {
    44                 if(i%2)
    45                     cout<<(mid-=i)<<' ';
    46                 else    
    47                     cout<<(mid+=i)<<' ';
    48             }
    49             for(int i=mid+1;i<=n;i++)
    50                 cout<<i<<' ';
    51         }
    52     }
    53     return 0;
    54 }
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  • 原文地址:https://www.cnblogs.com/weeping/p/6633299.html
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