Kate has a set SS of nn integers {1,…,n}{1,…,n}.
She thinks that imperfection of a subset M⊆SM⊆S is equal to the maximum of gcd(a,b)gcd(a,b) over all pairs (a,b)(a,b) such that both aa and bb are in MM and a≠ba≠b.
Kate is a very neat girl and for each k∈{2,…,n}k∈{2,…,n} she wants to find a subset that has the smallest imperfection among all subsets in SS of size kk. There can be more than one subset with the smallest imperfection and the same size, but you don't need to worry about it. Kate wants to find all the subsets herself, but she needs your help to find the smallest possible imperfection for each size kk, will name it IkIk.
Please, help Kate to find I2I2, I3I3, ..., InIn.
The first and only line in the input consists of only one integer nn (2≤n≤5⋅1052≤n≤5⋅105) — the size of the given set SS.
Output contains only one line that includes n−1n−1 integers: I2I2, I3I3, ..., InIn.
2
1
3
1 1
First sample: answer is 1, because gcd(1,2)=1gcd(1,2)=1.
Second sample: there are subsets of SS with sizes 2,32,3 with imperfection equal to 1. For example, {2,3}{2,3} and {1,2,3}{1,2,3}.
题目的大致意思是这样的，输入一个数N然后输出N-1个数，其中每一项都是从1~当前那个数两两相比最小的公倍数；
（既然是gcd，那么可以通过筛法的思路来写，比如2那么所有2的倍数都标记一下，然后以此类推。。。。
1 #include <bits/stdc++.h>
2 using namespace std;
3 int a[500005];
4 int main(){
5 int n;
6 cin>>n;
7 for(int i=1;i<=n;i++){//从一开始循环n次
8 for(int j=2*i;j<=n;j+=i){//like线性筛一样标记每一个尽可能小的可能出现的共倍数
9 a[j]=i;
10 }
11 }
12 /* for(int i=1;i<n;i++){
13 cout<<a[i]<<" ";
14 }
15 cout<<'
';
16 */ sort(a+1,a+n+1);//排一下序
17 for(int i=2;i<=n;i++){//i==1时a[i]肯定是0 没有gcd所以从2开始到n结束
18 cout<<a[i]<<" ";
19 }
20 cout<<'
';
21
22 return 0;
23 }