CF 327B. Hungry Sequence

B. Hungry Sequence

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

Iahub and Iahubina went to a date at a luxury restaurant. Everything went fine until paying for the food. Instead of money, the waiter wants Iahub to write a Hungry sequence consisting of n integers.

A sequence a₁, a₂, ..., a_n, consisting of n integers, is Hungry if and only if:

 

• Its elements are in increasing order. That is an inequality a_i < a_j holds for any two indices i, j (i < j).
• For any two indices i and j (i < j), a_j must not be divisible by a_i.

 

Iahub is in trouble, so he asks you for help. Find a Hungry sequence with n elements.

Input

The input contains a single integer: n (1 ≤ n ≤ 10⁵).

Output

Output a line that contains n space-separated integers a₁ a₂, ..., a_n (1 ≤ a_i ≤ 10⁷), representing a possible Hungry sequence. Note, that each a_i must not be greater than 10000000 (10⁷) and less than 1.

If there are multiple solutions you can output any one.

Sample test(s)

input

3

output

2 9 15

input

5

output

11 14 20 27 31

//我火星了，其实从100000,开始输出就可以了，但我。。。唉
#include<iostream>
#include<cstring>
#include<cstdio>
using namespace std;
#define N 1000000
int a[N+1]={0},p[N]={2,3};
int main(){
	int n,m=2,i,j,k=4;
	int x,y;
	for(i=5;i<N;i+=k^=6){
		if(a[i]==0){
			p[m++]=i;
			for(j=i;j<=N/i;j++)a[j*i]=-1;
		}
	}
	//for(i=0;i<100;i++)cout<<p[i]<<" ";
	//cout<<m<<endl;
	while(~scanf("%d",&n)){
		if(n<=m){
			printf("%d",p[0]);
			for(i=1;i<n;i++)printf(" %d",p[i]);printf("
");
		}else{
			printf("10");//5,7,11,....
			for(i=j=4,x=14,y=15;--n;){
				if(x>y){
					printf(" %d",y);
					y=p[j++]*3;
				}else{
					printf(" %d",x);
					x=p[i++]*2;
				}
			}
			printf("
");
		}
	}
return 0;
}