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;
}