CF 502 C- The Phone Number(思维)

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本文介绍了一个算法问题的解决方案，旨在找到一个由1到n整数构成的排列，使得该排列的最长递增子序列（LIS）与最长递减子序列（LDS）长度之和达到最小。通过观察和数学思考，提出了一种有效的生成策略。

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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], LIS of this permutation will be [1,2,3,5], so the length of LIS is equal to 44. LDScan be [6,4,1], [6,4,2], or [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≤10^5 1≤n≤10^5) — 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

output

3 4 1 2

input

output

2 1

Note

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

In the second sample, you can build a permutation [2,1]. LIS is [1] (or [2]), so the length of LIS is equal to 1. LDS is [2,1], so the length of LDS is equal to 2. The sum is equal to 3. Note that permutation [1,2] is also valid.

题目大意：给出一个n，输出1-n的全排列中最长上升子序列+最长下降子序列和最小的方案

思路：完全没有往数学方面想，，，，看大佬们的代码才明白怎么回事，观察样例，发现它是以对半输出的，大胆试一发：过8个点(这也能过8个点......),这也算是向ac靠近了一点点。要求只输出一种满足的情况就行，那上升下降子序列都按顺序好了，然后问题转化为类似矩形求面积（例如16，对半8+8，开根号4+4)，在面积一定的时候，周长最小的是正方形。

代码如下：

#include <iostream>
#include<cstring>
#include<cstdio>
#include<cmath>
#include<algorithm>
int n,mid,flag,l;
using namespace std;
int t1;
int main() 
{
    while(~scanf("%d",&n))
    {
        t1=0 ;
        l=(int)sqrt(n);
        for(int i=n-l+1;i>=1;i-=l)
        {
            for(int j=0;j<=l-1;j++)
            {
                cout<<i+j;
                    t1++;
                    if(t1!=n) cout <<" ";
            }
        }
        if(n%l!=0)
        {
            for(int i=1;i<=n%l;i++)
            {
                cout<<i;
                if(i!=n%l) cout<<" ";
            }
        }
        cout<<endl;
    }
    return 0;
}