codeforces_622D. Optimal Number Permutation(构造)

D. Optimal Number Permutation
time limit per test
1 second
memory limit per test
256 megabytes
input
standard input
output
standard output

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.

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

构造题,看出构造出di-(n-i)=0的数列就行,空缺的部分再扫一遍补上。

#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#define Si(a) scanf("%d",&a)
#define Sl(a) scanf("%lld",&a)
#define Sd(a) scanf("%lf",&a)
#define Ss(a) scanf("%s",a)
#define Pi(a) printf("%d\n",(a))
#define Pl(a) printf("%lld\n",(a))
#define Pd(a) printf("%lf\n",(a))
#define Ps(a) printf("%s\n",(a))
#define W(a) while(a--)
#define mem(a,b) memset(a,(b),sizeof(a))
#define FOP freopen("data.txt","r",stdin)
#define inf 0x3f3f3f3f
#define maxn 500010*2
#define mod 1000000007
#define PI acos(-1.0)
#define LL long long
using namespace std;

int ans[maxn];
int main()
{
    int i,n,top=1;
    Si(n);
    mem(ans,0);
    for(i=1;i<=n;i++)
    {
        if(i%2)
        {
            ans[top]=i;
            ans[top+n-i]=i;
            top++;
        }
    }
    top=n+1;
    for(i=1;i<=n;i++)
    {
        if(i%2==0)
        {
            ans[top]=i;
            ans[top+n-i]=i;
            top++;
        }
    }
    for(i=1;i<=2*n;i++)
    {
        if(!ans[i])ans[i]=n;
    }
    for(i=1;i<=2*n;i++)
    {
        printf("%d",ans[i]);
        if(i!=2*n)printf(" ");
    }
    return 0;
}


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