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
is called increasing if
, 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;
}