算数基本定理：

基本形式：（Pn表示质数）

导出结论（ACM中会用到的）：

1. （正因数个数）

2. （正因数之和）

例题：

1341 - Aladdin and the Flying Carpet

PDF (English)

Statistics

Forum

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.

Sample Input

Output for Sample Input

10 2

12 2

Case 1: 1

Case 2: 2

题意：给出a,b找出[b,a]中乘积为a的因子对个数（例如a=10,b=2,则只存在因子对{2,5}符合条件）

利用算数基本定理1求得[1,a]中因子对（求出因子数除以2），在暴力求解[1,b]中因子对，相减即可。

#include <cstdio>
#include <iostream>
#include <cmath>
#include <cstring>
#define MAX 1000047
typedef long long ll;
using namespace std;
ll a,b,p[MAX],prime[MAX],m,ans,t,cnt=0;void init() //标记素数并保存到prime数组中
{m=0;memset(p,0,sizeof(p));for (ll i=2;i<=MAX;i++){if (!p[i]){prime[m++]=i; //保存到prime数组 for (ll j=i+i;j<=MAX;j+=i)p[j]=1;}}
}int main()
{init();scanf("%d",&t);while(t--){scanf("%lld%lld",&a,&b);if (b>=sqrt(a)) //类似剪枝（否则会超时） {printf("Case %d: 0\n",++cnt);continue;}ll i=0,suma=1,sumb=0,x=a;while(prime[i]<a&&i<m) //为{if (a%prime[i]==0) //是因子{ll an=0;while(a%prime[i]==0){a/=prime[i];an++;} //求出指数 suma*=(an+1);}i++;}if (a>1) suma*=2;for (ll i=1;i<=sqrt(b);i++)if (x%i==0) sumb++; //找出[1,b)中的因子 ans=suma/2-sumb; //求出的是因子个数，除以2就是因子对个数 printf("Case %d: %lld\n",++cnt,ans);}return 0;
}

核心代码：

1.求素数（详见https://blog.csdn.net/Radium_1209/article/details/80232925）

2.求各个指数

while(a%prime[i]==0)
{a/=prime[i];an++; //an即为所求指数
}

3.带入公式

suma*=(an+1);

转载于:https://www.cnblogs.com/Radium1209/p/10415371.html

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