Interesting prime number problem hdu5750

A positive proper divisor is a positive divisor of a number n, excluding n itself. For example, 1, 2, and 3 are positive proper divisors of 6, but 6 itself is not.

Peter has two positive integers n and d. He would like to know the number of integers below n whose maximum positive proper divisor is d.
Input
There are multiple test cases. The first line of input contains an integer T (1≤T≤106), indicating the number of test cases. For each test case:

The first line contains two integers n and d (2≤n,d≤109).
Output
For each test case, output an integer denoting the answer.
Sample Input
9
10 2
10 3
10 4
10 5
10 6
10 7
10 8
10 9
100 13
Sample Output
1
2
1
0
0
0
0
0
4

Worth considering ,HDU5750
Especially where notes are added ！

#include<cstdio>
#include<cstring>
#include<iostream>
using namespace std;
bool isPrime[100000100];
int Prime[600010], cnt = 0;

void GetPrime(int n)// Sieve to n
{
    
	memset(isPrime, 1, sizeof(isPrime));
	isPrime[1] = 0;//1 Not primes 
	
	for(int i = 2; i <= n; i++)
	{
    
		if(isPrime[i])// Didn't sift out  
			Prime[++cnt] = i; //i Become the next prime 
			
		for(int j = 1; j <= cnt && i*Prime[j] <= n/* Don't exceed the upper limit */; j++) 
		{
    
			isPrime[i*Prime[j]] = 0;
            
			if(i % Prime[j] == 0)//i Also contains Prime[j] This factor 
				break; // major measure . See principle 
		}
	}
}
int main(){
    
	int n;
	int a = 0,b,x,ans,pri=0;
	GetPrime(300005);
	cin >> n;
	while(n--){
    
		ans=0;
		scanf("%d %d",&a,&b);
		for(int i = 1;Prime[i]*b < a&&Prime[i]<=b;i++){
    
			ans++;
			if(b%Prime[i]==0)break;
			// If d Is a multiple of the current prime , namely d = pri[i] * x',  So when it comes to the next prime number , because d' = pri[i + 1] * x' > d,d There is no guarantee that it will be the largest , So when d When it can be divided by a prime number ,  It's about to exit the loop 
			}
			printf("%d\n",ans);
		}
return 0;
	}