LightOJ-1341-Aladdin and the Flying Carpet-数论-唯一分解定理

LightOJ-1341-Aladdin and the Flying Carpet-数论-唯一分解定理

【Description】

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 ≤ 10^12) 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.

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【Examples】

Sample Input

2
10 2
12 2

Sample Output

Case 1: 1
Case 2: 2

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【Problem Description】

给你矩形的面积a，和一个b，b为矩形的边的最小长度。求xy的对数，满足x*y=a,且x>b,y>b.

【Solution】

(唯一分解定理）对于每一个数a，都可以分解为以下形式，且唯一：

$$> a=a_1^{p1}\times a_2^{p2}\dots a_m^{pm} >$$

其中，ai为素数，pi为正整数。
则a的正因数的个数为num =（1+p1)*(1+p2)*...*(1+pm)
所以满足x*y=a,x,y的对数为cnt=num/2;
但题目还要求x>b,y>b所以暴力枚举一下，将小于b的边长减去即可。

最后还要注意特判a<b*b的情况，

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【Code】

/*
 * @Author: Simon 
 * @Date: 2018-09-05 20:09:09 
 * @Last Modified by: Simon
 * @Last Modified time: 2018-09-05 21:41:11
 */
#include<bits/stdc++.h>
using namespace std;
typedef int Int;
#define int long long
#define INF 0x3f3f3f3f
const int maxn=1e6+10;
Int prime[maxn];
void get_prime()//线性筛素数
{
    for(Int i=2;i<=maxn;i++)
    {
        if(!prime[i])
        {
            prime[++prime[0]]=i;
        }
        for(Int j=1;j<=prime[0]&&i*prime[j]<=maxn;j++)
        {
            prime[i*prime[j]]=1;
            if(i%prime[j]==0) break;
        }
    }
}
int getfactor(int x)
{
    int ans=1;
    for(int i=1;prime[i]*prime[i]<=x&&i<=prime[0];i++)
    {
        if(x%prime[i]==0)
        {
            int sum = 0;
            while(x%prime[i]==0)
            {
                sum++;//pi
                x/=prime[i];
            }
            ans*=(1+sum);//正因数的个数
        }
    }
    if(x>1) ans*=2;
    return ans;
}
Int main()
{
    get_prime();
    Int t;
    int a, b;
    scanf("%d",&t);
    for(Int ca=1;ca<=t;ca++)
    {

        scanf("%lld%lld",&a,&b);
        if(a<b*b){ printf("Case %d: 0\n",ca); continue;}//特判
        int cnt=getfactor(a)/2;
        for(int i=1;i<b;i++)//减去小于b的边长
        {
            if(a%i==0)
            {
                cnt--;
            }
        }
        printf("Case %d: %lld\n",ca,cnt);
    }
    cin.get(),cin.get();
    return 0;
}