CodeForces 482A Diverse Permutation

最新推荐文章于 2019-02-17 19:42:36 发布

Hytai

最新推荐文章于 2019-02-17 19:42:36 发布

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分类专栏：数学构造

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本文探讨如何生成长度为n的排列p，使得p中相邻元素之差的集合恰好包含k个不同的元素。通过输入n和k值，实现特定条件下的排列生成。

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Permutation p is an ordered set of integers p1, p2, ..., pn, consisting of n distinct positive integers not larger than n. We'll denote as n the length of permutation p1, p2, ..., pn.

Your task is to find such permutation p of length n, that the group of numbers |p1 - p2|, |p2 - p3|, ..., |pn - 1 - pn| has exactly kdistinct elements.

Input

The single line of the input contains two space-separated positive integers n, k (1 ≤ k < n ≤ 105).

Output

Print n integers forming the permutation. If there are multiple answers, print any of them.

   Sample test(s) 
 
     input 
   
3 2

     output 
   
1 3 2

     input 
   
3 1

     output 
   
1 2 3

     input 
   
5 2

     output 
   
1 3 2 4 5

给定n和k，即有一个1~n的序列，现在要求找到一个序列，使得所有的 | Pi - Pi+1 | 有 k 种不同

#include <iostream>  
#include <cstdio>  
#include <vector>  
#include <string>
#include <cstring>
#include <queue>
#include <algorithm>
using namespace std;  

int main()
{  
	int i,j,k,t,m,n; 
    int cnt ;
	while(scanf("%d%d",&n,&k)!=EOF){
		int l = 1;
		int r = n;

		if(k%2==0){
			if(k==2){
				cout<<1<<' ';
				for(i=r;i>=2;i--)
					cout<<i<<' ';
			}
			else{
				cnt = 0;
				int tp = (k%2==0)?k/2:(k-1)/2;
				for(;cnt<tp;l++,r--){
					cout<<l<<' ';
					cout<<r<<' ';
					cnt++;
				}
				for(;l<=r;r--)
					cout<<r<<' ';		
			}
		}
		else{
			cnt = 0;
			if(k==1){
				for(i=1;i<=n;i++){
					cout<<i<<' ';
				}
			}
			else{
				cnt =0;
				for(;cnt<(k-1)/2;l++,r--){
					cout<<l<<' ';
					cout<<r<<' ';
					cnt++;
				}
				for(;l<=r;l++){
					cout<<l<<' ';
				}
			}
		}
		cout<<endl;
	}	
    return 0;  
}