LightOJ 1341 - Aladdin and the Flying Carpet

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## LightOJ 1341 - Aladdin and the Flying Carpet

angel_imp 2016-03-20 16:51:00 浏览1787

1341 - Aladdin and the Flying Carpet
 Time Limit: 3 second(s) Memory Limit: 32 MB

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.

# Output for Sample Input

2

10 2

12 2

Case 1: 1

Case 2: 2

12 = 1 * 12（不符合条件 最小的边长<2）
12 = 2 * 6（符合）
12 = 3 * 4（符合）

s = p1^e1 * p2^e2 *……* pk^ek，我们要得到的是因子的个数，这里所说的因子个数默认为正的，得到因子个数的方法是 num = (e1+1) * (e2+1) * ... *(ek+1)，然后又因为没有正方形，而且我们要得到的是有多少对，所以将 num除以2，就得到了可以组成矩形面积为 s 的矩形个数，然后我们只需要在 [1,a)的区间内（注意区间开闭）将 s 的因子减掉就行了（num--），这样就可以了。

```#include <iostream>
#include <cmath>
#include <cstring>
#include <algorithm>
#include <cstdio>
using namespace std;
typedef long long LL;
const int MAXN = 1e6+5;
bool prime[MAXN];
LL p[MAXN],k;
void isprime()
{
memset(prime, true, sizeof(prime));
prime[1] = false;
k = 0;
for(LL i=2; i<MAXN; i++)
{
if(prime[i])
{
p[k++] = i;
for(LL j=i*i; j<MAXN; j+=i)
prime[j] = false;
}
}
}

LL Solve(LL m)
{
LL ret = 1;
for(LL i=0; p[i]*p[i]<=m&&i<k; i++)
{
LL cnt = 0;
if(m%p[i] == 0)
{
while(m%p[i] == 0)
{
cnt++;
m /= p[i];
}
ret *= (cnt+1);
}
}
if(m > 1)
ret *= 2;
return ret;
}
int main()
{
isprime();
int T;
cin>>T;
for(int cas=1; cas<=T; cas++)
{
LL s,a;
cin>>s>>a;
if(a*a >= s)
printf("Case %d: 0\n",cas);
else
{
LL ret = Solve(s);
ret /= 2;
for(LL i=1; i<a; i++)
if(s % i == 0)
ret--;
printf("Case %d: %lld\n",cas,ret);
}
}
return 0;
}
```

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