codeforces 622D D. Optimal Number Permutation(找规律)
2016-02-26 20:10
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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
output
input
output
题意:使给的上式的值最小,可以发现,最小值可以为0,把简单的几个列一下可以发现规律;
AC代码:
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
题意:使给的上式的值最小,可以发现,最小值可以为0,把简单的几个列一下可以发现规律;
AC代码:
#include <bits/stdc++.h> using namespace std; int main() { int n; scanf("%d",&n); if(n%2) { for(int i=1;i<=n;i+=2) printf("%d ",i); for(int i=n-2;i>0;i-=2) printf("%d ",i); for(int i=2;i<n;i+=2) printf("%d ",i); for(int i=n-1;i>0;i-=2) printf("%d ",i); printf("%d\n",n); } else { for(int i=1;i<=n;i+=2) printf("%d ",i); for(int i=n-1;i>0;i-=2) printf("%d ",i); for(int i=2;i<=n;i+=2) printf("%d ",i); for(int i=n-2;i>0;i-=2) printf("%d ",i); printf("%d\n",n); } return 0; }
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