LA 5916(GCD Guessing Game-质数分组)

现在有一个数x，1 ≤ x≤ n，告诉你n，每次你可以猜一个数y，如果x==y则结束，否则返回gcd(x,y)，问最少只要几次就可以保证猜出答案。

Input

The input file contains several test cases, each of them as described below.
The input contains one integer n, 2≤ n≤ 10000.

Output

For each test case, write to the output on a line by itself.
Output one integer -- the number of guesses Andrew will need to make in the worst case.

Sample Input

6

Sample Output

2

首先把所有n以内素分组，每次询问一组素数的积——根据Gcd的性质确定这个数

每次贪心拿一个大质数与一堆小质数配（最右X最左）

#include<cstdio>
#include<cstring>
#include<algorithm>
#include<functional>
#include<iostream>
#include<cstdlib>
#include<cmath>
using namespace std;
#define MAXN (10000)
#define eps (1e-9)
#define For(i,n) for(int i=1;i<=n;i++)
#define ForD(i,n) for(int i=n;i;i--)
#define Fork(i,k,n) for(int i=k;i<=n;i++)
int n,a[MAXN],size=0;
bool b[MAXN];
int main()
{
memset(b,0,sizeof(b));b[1]=1;
Fork(i,2,MAXN)
{
if (!b[i]) b[i]=1,a[++size]=i;
For(j,size)
{
if (i*a[j]>MAXN) break;
b[i*a[j]]=1;
if (!i%a[j]) break;
}
}
//	For(i,100) cout<<a[i]<<' ';
while (cin>>n)
{
int i=0,head=1,tail=size;
while (a[tail]>n) tail--;
while (head<=tail)
{
int p=a[tail];
while (p*a[head]<=n) p*=a[head++];
tail--;i++;
}
cout<<i<<endl;
}
return 0;
}