Codeforces Round #502-C-The Phone Number（贪心）

C. The Phone Number

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

Mrs. Smith is trying to contact her husband, John Smith, but she forgot the secret phone number!

The only thing Mrs. Smith remembered was that any permutation of nn can be a secret phone number. Only those permutations that minimize secret value might be the phone of her husband.

The sequence of nn integers is called a permutation if it contains all integers from 11 to nn exactly once.

The secret value of a phone number is defined as the sum of the length of the longest increasing subsequence (LIS) and length of the longest decreasing subsequence (LDS).

A subsequence ai1,ai2,…,aikai1,ai2,…,aik where 1≤i1<i2<…<ik≤n1≤i1<i2<…<ik≤n is called increasing if ai1<ai2<ai3<…<aikai1<ai2<ai3<…<aik. If ai1>ai2>ai3>…>aikai1>ai2>ai3>…>aik, a subsequence is called decreasing. An increasing/decreasing subsequence is called longest if it has maximum length among all increasing/decreasing subsequences.

For example, if there is a permutation [6,4,1,7,2,3,5][6,4,1,7,2,3,5], LIS of this permutation will be [1,2,3,5][1,2,3,5], so the length of LIS is equal to 44. LDScan be [6,4,1][6,4,1], [6,4,2][6,4,2], or [6,4,3][6,4,3], so the length of LDS is 33.

Note, the lengths of LIS and LDS can be different.

So please help Mrs. Smith to find a permutation that gives a minimum sum of lengths of LIS and LDS.

Input

The only line contains one integer nn (1≤n≤1051≤n≤105) — the length of permutation that you need to build.

Output

Print a permutation that gives a minimum sum of lengths of LIS and LDS.

If there are multiple answers, print any.

Examples

input

Copy

output

Copy

3 4 1 2

input

Copy

output

Copy

2 1

Note

In the first sample, you can build a permutation [3,4,1,2][3,4,1,2]. LIS is [3,4][3,4] (or [1,2][1,2]), so the length of LIS is equal to 22. LDS can be ony of [3,1][3,1], [4,2][4,2], [3,2][3,2], or [4,1][4,1]. The length of LDS is also equal to 22. The sum is equal to 44. Note that [3,4,1,2][3,4,1,2] is not the only permutation that is valid.

In the second sample, you can build a permutation [2,1][2,1]. LIS is [1][1] (or [2][2]), so the length of LIS is equal to 11. LDS is [2,1][2,1], so the length of LDS is equal to 22. The sum is equal to 33. Note that permutation [1,2][1,2] is also valid.

题意：给你一个数字n，问你能否构造一个排列，使得这个排列的最长递增子序列+最长递减子序列最小。

题解：我们可以考虑找到一个x和一个y使得x*y>=n并且x+y尽可能的小，然后我们就将n个数分成x组，并且每组有y个数，这y个数是按顺序排列的，但是要满足前一组的权值大于后一组。

例如：n=6，此时x=2，y=3，生成的排列就是

5 6 3 4 1 2

可以证明这样得出的排列是满足题意的。

#include<stdio.h>
#include<algorithm>
using namespace std;
#define ll long long
int p,q,n,sm=1000000000,pp,qq;
int main(void)
{
	scanf("%d",&n);
	for(int i=1;i<=n;i++)
	{
		p=i;q=n/p+(n%p!=0);
		if(p+q<sm)
			pp=p,qq=q,sm=p+q;
	}
	p=pp;q=qq;
	int t=q*(p-1);
	for(int i=t+1;i<=n;i++)
		printf("%d ",i);
	n=n-n%q;
	for(int i=p-1;i>0;i--)
	{
		for(int j=q*i-q+1;j<=q*i;j++)
			printf("%d ",j);
	}
	printf("\n");
	return 0;
}

Codeforces Round #502-C-The Phone Number（贪心）

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