C - Aladdin and the Flying Carpet （唯一分解定理 + 约数定理）

C - Aladdin and the Flying Carpet

 1341 - Aladdin and the Flying Carpet

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Time Limit: 3 second(s)

Memory Limit: 32 MB

[/align]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	Output for Sample Input
2 10 2 12 2	Case 1: 1 Case 2: 2

       思路：直接计算n的约数的个数+唯一分解定理！！！！学习一下公式（百度百科）    


这里约数的个数需要变为一般，因为两边会有重复的，这里题目指出不是正方形，所以排除x*x = a的情况；
如果b*b > a 则输出0个
否则从1 -> (b-1)判断是否有a的约数，找到则删去就行啦！！！
代码：#include<bits/stdc++.h>
#define ll long long
using namespace std;
const int maxn = 1e6+7;

int t, n, cnt = 0;

ll prim[maxn];
bool vis[maxn];

void init()
{
memset(vis, 0, sizeof(vis));
for(int i = 2; i*i <= maxn; i++)
if(!vis[i])
for(int j = i*i; j <= maxn; j+= i)
vis[j] = true;
for(int i = 2; i < maxn; i++)
if(!vis[i]) prim[cnt++] = i;
}

ll FIND(ll a, ll b) {
ll ans = 1, k = a;
for(int i = 0; i < cnt&& k > 1&& prim[i]*prim[i] <= k; i++)
{
if(k%prim[i] == 0){
int c = 0;
while(k%prim[i] == 0) {
k /= prim[i];
c++;
}
ans *= (c + 1);
}
}
if(k > 1) ans *= 2;
ans /= 2;
if(b * b > a) ans = 0;
else {
for(int i = 1; i < b; i++)
{
if(a%i == 0) ans -= 1;
}
}
return ans;
}

int main()
{
init();
scanf("%d",&t);
int cas = 0;
while(t--)
{
ll a, b;
scanf("%lld%lld", &a, &b);
printf("Case %d: %lld\n", ++cas, FIND(a, b));
}
return 0;
}