AtCoder Beginner Contest 142D题判断素数的模板+求一个数的因子的模板

Posted 2020-11-15 pengge666

D - Disjoint Set of Common Divisors

Problem Statement

Given are positive integers $\mathrm{AA}$ and $\mathrm{BB}$.

Let us choose some number of positive common divisors of $\mathrm{AA}$ and $\mathrm{BB}$.

Here, any two of the chosen divisors must be coprime.

At most, how many divisors can we choose?

Definition of common divisorDefinition of being coprimeDefinition of dividing

Constraints

• All values in input are integers.
• $\mathrm{1\le A,B\le 10121\le A,B\le 1012}$

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Input

Input is given from Standard Input in the following format:

$\mathrm{AA}$ $\mathrm{BB}$

Output

Print the maximum number of divisors that can be chosen to satisfy the condition.

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Sample Input 1 Copy

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12 18

Sample Output 1 Copy

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3

$1212$ and $1818$ have the following positive common divisors: $11$, $22$, $33$, and $66$.

$11$ and $22$ are coprime, $22$ and $33$ are coprime, and $33$ and $11$ are coprime, so we can choose $11$, $22$, and $33$, which achieve the maximum result.

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Sample Input 2 Copy

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420 660

Sample Output 2 Copy

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4

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Sample Input 3 Copy

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1 2019

Sample Output 3 Copy

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1

$11$ and $20192019$ have no positive common divisors other than $1$

$\mathrm{思路：找出有多少个公因子，并且公因子必须是素数。然后再加一。【比赛时思路一模一样，代码写挫了QAQ】}$

$\mathrm{AC代码：}$

 1 #include<bits/stdc++.h>
 2 
 3 using namespace std;
 4 
 5 #define int long long
 6 bool isprime(int num){ // 判断是否是素数
 7     if(num==1){
 8         return false;
 9     }
10     if(num==2)
11         return true;
12     if(num%2==0)
13         return false;
14     double sqrtNum = sqrt(num);
15     for (int  i = 3; i <= sqrtNum; i += 2)
16     {
17         if (num % i == 0)
18         {
19             return false;
20         }
21     }
22     return true;
23 }
24 vector<int> v;
25 signed main(){
26     int n,m;
27     cin>>n>>m;
28     int temp=min(n,m);
29     for(int i=1;i*i<=temp;i++){ // 求一个数的因子的模板
30         if(temp%i==0){
31             v.push_back(i);
32             if(i*i!=temp){
33                 v.push_back(temp/i);
34             }
35         }
36     }
37     int ans=0;
38     int t=max(n,m);
39     for(int i=0;i<v.size();i++){
40         if(t%v[i]==0&&isprime(v[i]))
41             ans++;
42     }
43     cout<<ans+1;
44     return 0;
45 }