[noip2001 popularization group] the problem of maximum common divisor and minimum common multiple

Topic link [NOIP2001 Popularization group ] The problem of maximum common divisor and minimum common multiple - Luogu https://www.luogu.com.cn/problem/P1029

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Title Description

Enter two positive integers x0​,y0​, Find out... That satisfies the following conditions  P, Q The number of ：

1. P,Q  It's a positive integer. .

2. requirement  P, Q With x0​ For the greatest common divisor , With y0​ Is the minimum common multiple .

Try to ask for ： All possible that meet the conditions P,Q The number of .

Input
         Two positive integers in a row x0​,y0​.

Output

         Count one line at a time , It means to find out if the condition is satisfied P, Q The number of .

Sample group

 Input 
3 60

 Output  4

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Topic ideas

        This is a way Water problem A relatively simple topic . First type x0​,y0​. And then x0,y0 Multiply and store in a variable R in . And then determine R Is it a complete square ( Is the square of a number ), If yes, the answer variable ans--; Then open a double cycle from 1 Cycle to the root R, Judge R Whether it can be cycled by variables i Divisible and i And R/i The greatest common factor of is x0( Just find one ), If the conditions are met ans+=2;

        The final output ans that will do .( Note that these variables need to be turned on long long)

        As for the greatest common factor , You can directly use __gcd, Or write a special gcd function ：

#define ll long long
ll gcd(ll a,ll b)// And __gcd Equivalent  
{return (b==0?a:gcd(b,a%b));}// Ternary operator 

         The main program of this topic is as follows ：

int getans(int n,int m)
{
	int r=n*m,ans=0;
	if(n==m) ans--; 
	for(int i=1;i*i<=r;i++)// This is better than using sqrt(r) faster 
		if(n%i==0&&gcd(i,n/i)==m) ans+=2;// It can also be used directly here __gcd
	return ans;	 
}

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AC Code

#include<bits/stdc++.h>
#define ll long long
using namespace std;
long long m,n,k,ans;
ll gcd(ll a,ll b)// And __gcd Equivalent  
{return (b==0?a:gcd(b,a%b));}
int main()
{
	cin>>m>>n;k=n*m;
	if(n==m) ans--;
	for(ll i=1;i*i<=k;i++)
		if(k%i==0&&gcd(i,k/i)==m) ans+=2;// It can also be used directly here __gcd
	cout<<ans;
	return 0;
}

There are so many questions ( Don't drive O2 I can still live ).

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