A. Vasya and Petya's Game（cf第一题）

最新推荐文章于 2021-07-16 22:51:34 发布

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本文介绍了一个简单的猜数游戏，玩家需通过提问来确定对手所想的数字。文章详细解释了如何利用最少的问题次数来猜出1到n之间的任意数字，并提供了一段C++代码实现。该算法通过询问特定数值的整除性来缩小答案范围。

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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 yi, 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 yi (1 ≤ yi ≤ 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.

#include <iostream>
#include<cstring>
#include<cmath>
using namespace std;

const int maxn=1103;
int b[maxn];

int a[maxn];
int prim()
{
    memset(a,0,sizeof(a));
    for(int i=2; i<=maxn; i++)
    {
        if(!a[i])a[++a[0]]=i;
        for(int j=1; j<=a[0]&&a[j]<=maxn/i; j++)
        {
            a[a[j]*i]=1;
            if(i%a[j]==0)
                break;
        }
    }

}
int main()
{
    int num=0;
    int t;
    cin>>t;
    prim();
    for(int i=1; i<a[0]; i++)
    {
        int temp=a[i];
        int p=1;
        for(int j=1; j<maxn; j++)
        {

            p=p*a[i];

            if(p<=t)
            {
                b[num++]=p;
            }
            else break;
        }
        if(temp>t)break;
    }
    cout<<num<<endl;
    if(num>=1)
        cout<<b[0];
    for(int i=1; i<num; i++)
        cout<<" "<<b[i];
    cout<<endl;
    return 0;
}