思路：素数筛处理出所有素数，枚举每个素数p，可以知道如果p^k<n，则p^k一定需要选，根据这个原则求出所有要猜的数。
#include<bits\stdc++.h>
using namespace std;
const int maxn = 1105;
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 n;
cin >> n;
prim();
int pos = 0;
for(int i = 1;i<a[0];i++)
{
int temp = a[i];
int p = 1;
for(int j = 1;j<maxn;j++)
{
p*=a[i];
if(p<=n)
b[pos++]=p;
else
break;
}
if(temp > n)
break;
}
cout << pos << endl;
if(pos>=1)
cout << b[0];
for(int i = 1;i<pos;i++)
cout << " " << b[i];
cout << endl;
}
C. Vasya and Petya's Game
time limit per test
memory limit per test
input
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
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.
Examples
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.
