Educational Codeforces Round 7-D. Optimal Number Permutation_educational codeforces round 7 d. optimal number p-优快云博客

本文链接：https://blog.youkuaiyun.com/qq_26122039/article/details/52883670

本文介绍了一个算法问题，即如何在包含从1到n的整数两次的数组中通过任意排列来最小化每个数字i的位置距离之和。文章提供了一段C++代码实现，该实现在输入整数n后，输出一个排列方案以达到最小化目标。

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You have array a that contains all integers from 1 to n twice. You can arbitrary permute any numbers in a.

Let number i be in positions xi, yi (xi < yi) in the permuted array a. Let's define the value di = yi - xi — the distance between the positions of the number i. Permute the numbers in array a to minimize the value of the sum $\text{[math]}$ .

Input

The only line contains integer n (1 ≤ n ≤ 5·105).

Output

Print 2n integers — the permuted array a that minimizes the value of the sum s.

 
       Examples
      
         input
        
2
 
         output
        
1 1 2 2
 
         input
        
1
 
         output
        
1 1

#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
#include <map>
#include <vector>
#define maxn 1000005
#define MOD 1000000007
using namespace std;
typedef long long ll;

int num[maxn];
int main(){
	int n;
	scanf("%d", &n);
	int l = 1, r = n;
	for(int i = 1; i <= n; i += 2){
		num[l] = i;
		num[r] = i;
		l++;
		r--;
	}
	num[n+1] = n;
	l = n + 2;
	r = 2 * n;
	for(int i = 2; i <= n; i += 2){
		num[l] = i;
		num[r] = i;
		l++;
		r--; 
	}
	printf("%d", num[1]);
	for(int i = 2; i <= 2 * n; i++)
	  printf(" %d", num[i]);
	return 0; 
}