LightOJ - 1341 (Fundamental Theorems of Arithmetic)

B - Aladdin and the Flying Carpet  

It's said that Aladdin had to solve seven mysteries before getting the Magical Lamp which summons a powerful Genie. Here we are concerned about the first mystery.

Aladdin was about to enter to a magical cave, led by the evil sorcerer who disguised himself as Aladdin's uncle, found a strange magical flying carpet at the entrance. There were some strange creatures guarding the entrance of the cave. Aladdin could run, but he knew that there was a high chance of getting caught. So, he decided to use the magical flying carpet. The carpet was rectangular shaped, but not square shaped. Aladdin took the carpet and with the help of it he passed the entrance.

Now you are given the area of the carpet and the length of the minimum possible side of the carpet, your task is to find how many types of carpets are possible. For example, the area of the carpet 12, and the minimum possible side of the carpet is 2, then there can be two types of carpets and their sides are: {2, 6} and {3, 4}.

Input

Input starts with an integer T (≤ 4000), denoting the number of test cases.

Each case starts with a line containing two integers: a b (1 ≤ b ≤ a ≤ 10¹²) where a denotes the area of the carpet and b denotes the minimum possible side of the carpet.

Output

For each case, print the case number and the number of possible carpets.

Sample Input

2

10 2

12 2

Sample Output

Case 1: 1

Case 2: 2

I'm so rubbish, this question has been done for so long. The main point of this question is how many pairs of factors are there in [b, a]. At first glance, this question is known to be the Fundamental Theorem of Arithmetic, but I don't know how to find a factor less than b. Enumerates numbers less than b and subtracts them if the number is divisible by a. I first used a prime number to sieve the prime numbers within 1e6, because if this number is a non-prime number, even if it is greater than 1e6, there must be a factor less than 1e6, and we must not forget that this number is a prime number. Don't forget the constraints when pushing, I will forget the condition if this is a prime number greater than 1e6

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cstring>
#include <cmath>

using namespace std;
const int numsize = 1000005;
bool isprime[numsize];
int prime[numsize];
intmain()
{
    memset(isprime, 1, sizeof(isprime));
    isprime[1] = false;
    int cnt = 1;
    for(int i = 2; i <= numsize; i ++)
    {
        if(isprime[i])
        {
            prime[cnt ++] = i;
        }
        for(int j = 1; j < cnt&&prime[j]*i <= numsize; j ++)
        {
            isprime[i*prime[j]] = false;
            if(i % prime[j] == 0)
                break;
        }
    }
   int t;
   scanf("%d", &t);
   int time = 0;
   while(t --)
   {
       long long  a, b;
       scanf("%lld %lld",&a, &b);
        if(b*b >= a)
        {
            printf("Case %d: 0\n", ++tem);
            continue;
        }
       long long x = a;
       long long cnt1 = 1;
       for(int i = 1; i < cnt&&prime[i] <= x; i ++)
       {
           int c = 0;
                while(x % prime[i] == 0)
           {
               x /= prime[i];
               c ++;
           }
           cnt1 *= (c + 1);
       }
        if(x > 1)
            cnt1 *= 2;
        cnt1/= 2;
        for(int i = 1; i < b; i ++)
        {
            if(a % i == 0)
                cnt1 --;
        }
        printf("Case %d: %lld\n", ++tem, cnt1);
   }
    return 0;
}