The Phone Number CodeForces - 1017C（构造+思维）-优快云博客

The Phone Number CodeForces - 1017C

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 n

can be a secret phone number. Only those permutations that minimize secret value might be the phone of her husband.

The sequence of n
integers is called a permutation if it contains all integers from 1 to n

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,…,aik
where $1≤i1<i2<…<ik≤n$ is called increasing if $ai1<ai2<ai3<…<aik. If ai1>ai2>ai3>…>aik$

, 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 4. LDS can be [6,4,1], [6,4,2], or [6,4,3], so the length of LDS is 3

.

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 n

(1≤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

4

Output

3 4 1 2

Input

2

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 2. 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.

题意：

给定一个数n，构造一个1-n的序列使得其中的最长上升子序列和最长下降子序列的和最短

分析：

构造。

说实话这种东西真的就是灵光乍现就想出来了，要不然真的想不出来

其实构造就直接分段就行了，怎么个分段法呢

比如123456789

如果我们让每段4个数就变成下面样子

6789 2345 1

这样最长的上升子序列就是每段的长度，这里是4，最长的下降子序列的长度就是段数，这里是3

这样就可以求出和

我们可以暴力枚举1-n看每段几个数时最优，记录下来，然后按照上面规则输出就行了

code：

#include <bits/stdc++.h>
using namespace std;
int main(){
    int n;
    scanf("%d",&n);
    int x;
    int ans = 100010;
    for(int i = 1; i <= n; i++){
        int tmp = i + n / i + (n % i == 0 ? 0 : 1);
        if(ans > tmp){
            x = i;
            ans = tmp;
        }
    }
    for(int i = n; i >= 1; i -= x){
        int j,flag;
        if(i - x < 0) j = 1,flag = 1;
        else j = i - x + 1,flag = 0;
        for(; j <= i; j++){
            printf("%d",j);
            if(j == i){
                if(flag) puts("");
                else putchar(' ');
            }
            else putchar(' ');
        }
    }
    return 0;
}