Educational Codeforces Round 7 D. Optimal Number Permutation 构造题

D. Optimal Number Permutation

题目连接:

http://www.codeforces.com/contest/622/problem/D

Description

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 .

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.

Sample Input

2

Sample Output

1 1 2 2

Hint

题意

构造题,让构造一个1-n中每个数出现两次的一个序列,使得这个序列的权值和最小

权值 = (n-i)(di+i-n),di是指第i个数出现两次的距离

题解:

我们猜想一下,我们能不能让di = n-i,这样权值和就是0了

于是就这样构造吧,我们分奇数偶数去摆放,这样就可以错开了,就不会重叠在一起

因为n这个数可以使得(n-i)=0,所以就用它来调整这个序列

唔,如果我题解不是很清楚,可以跑跑我的代码,可以发现出来的结果是两个类似回文串的东西~

喵。

代码

#include<bits/stdc++.h>
using namespace std;
const int maxn = 1e6+7;
int a[maxn];
int d[maxn];
int main()
{
    int n;
    scanf("%d",&n);
    int tot = 1;
    for(int i=1;i<=n;i+=2)
        a[tot++]=i;
    int p = n;
    if(p%2==0)p--;
    else p-=2;
    for(int i=p;i>=1;i-=2)
    {
        a[tot++]=i;
    }
    for(int i=2;i<=n;i+=2)
        a[tot++]=i;
    p=n;
    if(p%2==1)p--;
    else p-=2;
    for(int i=p;i>=1;i-=2)
        a[tot++]=i;
    a[tot++]=n;
    int Ans=0;
    /*
    for(int j=1;j<=2*n;j++)
    {
        if(d[a[j]])
            Ans+=(n-a[j])*abs(j-d[a[j]]+a[j]-n);
        else
            d[a[j]]=j;
    }
    cout<<Ans<<endl;
    */
    for(int i=1;i<=2*n;i++)
        cout<<a[i]<<" ";
    cout<<endl;
}
posted @ 2016-02-12 14:57  qscqesze  阅读(292)  评论(2编辑  收藏  举报