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;}