View Modular Operation from the Perspective of Unique Decomposition Theorem

View Modular Operation from the Perspective of Unique Decomposition Theorem

• Problem background

Given two numbers p and q, if p% q == 0, find the maximum value of x that satisfies the condition (p%x==0 && x% q !=0 ).

• First look at a set of examples

6 % 4 != 0

6 % 2 == 0

6 The prime factor of decomposition is: 2 ¹ ×3 ¹

4 Decomposition prime factors are: 2 ²

2 Decomposition prime factors are: 2 ¹

According to the power of each term of the prime factorization formulas of 6, 4, and 2 combined with the modular operation, have you found that the reason for 6% 4! = 0 is 2 ¹ of 6

2 ² that is less than 4 , and the result of the modulo operation is 0 because the prime factor term of 2 is 2 ¹ <= 2 ¹

• get conclusion

If two numbers a and b are modulo 0, then each term in the prime factorization formula of a is completely contained or completely contained in the prime factorization formula of b (where the inclusion refers to each of a The power of the prime term is greater than or equal to the power of each prime term of b)

• Solve the problem

Since p% q == 0 &&p>q, each prime factor term of p completely includes each prime factor term of q. If you want to find a satisfying condition of x, you can make x the initial value of p, and then let some A prime factor term that is less than q, so x %p !=0, and since each prime factor term of p completely contains x, so p% x == 0

Title from: CodeForces 1445C

撸Code:

#include<stdio.h>
#include<string.h>
#include<algorithm>
#include<vector>
using namespace std;

int main() {
    
    
	int tt;
	scanf("%d",&tt);
	while(tt--){
    
    
		long long p,q;
		scanf("%lld%lld",&p,&q);
		long long Q=q;
		long long ans=0;
		for(long long i=2;i*i<=q;i++)//将 q 分解质因子  为 i 
			if(q%i==0){
    
    
				while(q%i==0) q/=i;
				long long ss=p;
				while(ss%Q==0) ss/=i;//若 p%q==0  说明 p 的质因子 i 的幂次大于等于 q 的质因子 i 的幂次 
				ans=max(ans,ss);
			}
		if(q>1){
    
    
				long long ss=p;
				while(ss%Q==0) ss/=q;
				ans=max(ans,ss);			
		}
		printf("%lld\n",ans);
	}
	
	return 0;
}