Codeforces Round #191 (Div. 2) B. Hungry Sequence（素数筛选法）

. 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

输出n个从小到大排列、每个数不超过范围的素数即可

AC Code：

#include <iostream>
#include <cstdio>
#include <cstring>

using namespace std;

const int maxn = 2000000;
bool prime[maxn];

int main()
{
    int n;
    memset(prime, false, sizeof(prime));
    for(int i = 2; i * i < maxn; i++)
    {
        for(int j = 2; i * j < maxn; j++)
            prime[i*j] = true;
    }
    while(scanf("%d", &n) != EOF)
    {
        printf("2");
        for(int i = 1, j = 3; i < n; j++)
        {
            if(prime[j] == false)
            {
                printf(" %d", j);
                i++;
            }
        }
        printf("
");
    }
    return 0;
}