codeforce--Vasya and Petya's Game

网址：http://codeforces.com/contest/576/problem/A

A. Vasya and Petya's Game

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

Vasya and Petya are playing a simple game. Vasya thought of number x between 1 and n, and Petya tries to guess the number.

Petya can ask questions like: "Is the unknown number divisible by number y?".

The game is played by the following rules: first Petya asks all the questions that interest him (also, he can ask no questions), and then Vasya responds to each question with a 'yes' or a 'no'. After receiving all the answers Petya should determine the number that Vasya thought of.

Unfortunately, Petya is not familiar with the number theory. Help him find the minimum number of questions he should ask to make a guaranteed guess of Vasya's number, and the numbers y_i, he should ask the questions about.

Input

A single line contains number n (1 ≤ n ≤ 10³).

Output

Print the length of the sequence of questions k (0 ≤ k ≤ n), followed by k numbers — the questions y_i (1 ≤ y_i ≤ n).

If there are several correct sequences of questions of the minimum length, you are allowed to print any of them.

Sample test(s)

Input

4

Output

3
2 4 3 

Input

6

Output

4
2 4 3 5 

Note

The sequence from the answer to the first sample test is actually correct.

If the unknown number is not divisible by one of the sequence numbers, it is equal to 1.

If the unknown number is divisible by 4, it is 4.

If the unknown number is divisible by 3, then the unknown number is 3.

Otherwise, it is equal to 2. Therefore, the sequence of questions allows you to guess the unknown number. It can be shown that there is no correct sequence of questions of length 2 or shorter.

题解：

Task A. Div1.

If Petya didn't ask p^k, where p is prime and k ≥ 1, he would not be able to distinguish p^k - 1 and p^k.

That means, he should ask all the numbers p^k. It's easy to prove that this sequence actually guesses all the numbers from 1 to n

The complexity is O(N^1.5) or O(NloglogN) depending on primality test.

其实没有必要用到vector ，并且开始的时候还因为用了直接WR了，因为开始的时候N取1005，这样的话就遍历越界了，应该增加vec.size()的限制，或者是下面的这样把数组开大点

#include <iostream>
#include <stdio.h>
#include <math.h>
#include <string>
#include <queue>
#include <string.h>
#include <map>
#include <set>
#include <vector>
#include <algorithm>
#include <stdlib.h>
using namespace std;
#define rd(x) scanf("%d",&x)
#define N 1005
bool isprm[N];
vector <int> vec;
void isprime()
{
    int i,j,k=0;
    int s,e=sqrt( double(N) )+1;    	//sqrt是对于double数开平方

    memset(isprm,1,sizeof(isprm));
                                                  //prm[k++]=2;
    isprm[0] = isprm[1] = 0;
    for(i=4 ;i < N; i=2+i)
		isprm[i]=0;

    for(i=3;i<e;i=2+i)
        if(isprm[i])
            for(s=i*2,j=i*i;j<N;j=j+s)
                isprm[j]=0;
    for(int i=0;i<1000;i++)
        if(isprm[i])
           vec.push_back(i);
}
int main()
{
    int t,b[1005],num=0;
    rd(t);
    isprime();
    for(int i=1; i <= t ; i++)
    {
        if(isprm[i]){
            int temp=i,j=1;
            while(temp<=t)
                b[num++]=temp,temp=temp*i;
        }
    }
    printf("%d
",num);
    for(int i=0; i<num; i++)
        printf("%d ",b[i]);
    return 0;
}