題目：LightOJ:1341 - Aladdin and the Flying Carpet(因子個數）

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

題意：

　　根據唯一分解定理，先將a唯一分解，得a的所有正約數的個數為sum ,
　　因為題目說了不會存在c==d的情況，因此sum要除2 去掉重復情況，
　　然后枚舉小于b的a的約數，拿sum減掉就可以了

數論原理：

　　正整數n有素因子分解n=p1^a1*p2^a2*p3^a3*···*pn^an 因子個數=(a1+1)(a2+2)(a3+3) ···(an+n)

代碼：

#include<bits/stdc++.h> using namespace std; const int maxn=1e6+10; int prime[maxn]={1,1,0}; int prime1[maxn]; int sum,num,cnt; long long a,b,temp; void db() //埃式篩 {cnt=0;for(int i=2;i<=sqrt(maxn);i++){if(!prime[i]){for(int j=i+i;j<=maxn;j+=i){prime[j]=1;}}}for(int i=2;i<=maxn;i++){if(!prime[i]){prime1[cnt++]=i;}} } void reslove()　　求所有素因子 {for(int j=0;j<cnt&&prime1[j]<=sqrt(temp);j++){int num=0;if(temp%prime1[j]==0){while(temp%prime1[j]==0){num++;temp/=prime1[j];}sum*=(num+1);}}if(temp>1) //如果temp不能被整分，說明還有一個素數是它的約數，此時num=1sum*=2; } int main() {db();int t;cin>>t;for(int i=1;i<=t;i++){cin>>a>>b;if(a<b*b)printf("Case %d: 0\n", i);else{sum=1;temp=a;reslove();sum/=2;for(int k=1;k<b;k++){if(a%k==0){sum--;}}printf("Case %d: %d\n",i,sum);} }return 0; }

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轉載于:https://www.cnblogs.com/BYBL/p/10969698.html

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