LightOJ 1341 Aladdin and the Flying Carpet

http://acm.hust.edu.cn/vjudge/contest/view.action?cid=70017#problem/C

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	Output for Sample Input
2 10 2 12 2	Case 1: 1 Case 2: 2

题目大意：给两个数a，b，求满足c*d==a且c>=b且d>=b的c,d二元组对数，(c,d)和(d,c)属于同一种情况

题目分析：根据唯一分解定理(一个数约数的个数等于将这个数分解成质数的所有因子的质数+1后相乘， 例如，12 = 2^2*3^1, 则12的约数的个数为（2+1）*（1+1）)，先将a唯一分解，则a的所有正约数的个数为ans =
(1 + a1) * (1 + a2) *...(1 + ai)，这里的ai是素因子的指数，见唯一分解定理，因为题目说了不会存在c==d的情况，因此ans要除2，去掉重复情况，然后枚举小于b的a的约数，拿ans减掉就可以了

#include <iostream>
#include <cstring>
#include <cstdio>
#include <algorithm>
#include <cmath>
#include <cstdlib>
#include <limits>
#include <queue>
#include <stack>
#include <vector>
#include <map>

using namespace std;

#define N 1000005
#define INF 0x3f3f3f3f
#define PI acos (-1.0)
#define EPS 1e-8
#define met(a, b) memset (a, b, sizeof (a))
typedef long long LL;

LL vis
= {1, 1, 0}, prim
, k = 0, a, b, aa;

void isprim ();

int main ()
{
int t;
LL flag = 1;
scanf ("%d", &t);

isprim ();

while (t--)
{
scanf ("%lld %lld", &a, &b);

if (a / b < b)
{
printf ("Case %lld: 0\n", flag++);
continue;
}

LL ans = 1, aa = a;

for (LL i=0; i<k&&prim[i]*prim[i]<=a; i++)
{
LL cnt = 0;

while (a % prim[i] == 0)
{
cnt++;
a /= prim[i];
}
ans *= (cnt+1);
}

if (a > 1) ans <<= 1;
ans >>= 1;//ans/2种矩形，另若为正方形，ans必为奇数，ans/2也可以去掉这种情况

for (LL i=1; i<b; i++)
if (aa % i == 0) ans--;
printf ("Case %lld: %lld\n", flag++, ans);
}
return 0;
}

void isprim ()
{
for (LL i=2; i*i<=N; i++)
if (!vis[i])
for (LL j=i*i; j<=N; j+=i)
vis[j] = 1;

for (LL i=0; i<=N; i++)
if (!vis[i]) prim[k++] = i;
}