HDU 2588（欧拉函数的应用）

GCD

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)

Total Submission(s): 1788    Accepted Submission(s): 895

Problem Description

The greatest common divisor GCD(a,b) of two positive integers a and b,sometimes written (a,b),is the largest divisor common to a and b,For example,(1,2)=1,(12,18)=6.
(a,b) can be easily found by the Euclidean algorithm. Now Carp is considering a little more difficult problem:
Given integers N and M, how many integer X satisfies 1<=X<=N and (X,N)>=M.

 

Input

The first line of input is an integer T(T<=100) representing the number of test cases. The following T lines each contains two numbers N and M (2<=N<=1000000000, 1<=M<=N), representing a test case.

 

Output

For each test case,output the answer on a single line.

 

Sample Input

  
  

   
   3
1 1
10 2
10000 72
  
  

 

Sample Output

  
  

   
   1
6
260
  
  
  
  


  
  
  
  


  
  
  
  

   
   分析：
  
  
  
  

   
   
    
    

     
     若X与n存在大于m的最大公约数，设d=(x,n);
    
    
    
    

     
     则X=q*d,n=p*d; 并且 (p,q)=1;
    
    
    
    

     
     我们可以枚举公约数，由欧拉函数的定义可知 phi(p)即为所求
    
    
    
    

     
     代码如下：
    
    
   
   
#include <cstdio>
#include <cstdlib>
#include <algorithm>
#include <iostream>
#include <cmath>
using namespace std;

typedef long long int ll; 

int euler(int N){
	int ans = N;
	for(int i=2; i*i<=N; i++){
		if(N%i==0)
			ans = ans/i*(i-1);
		while(N%i==0)
			N /= i;
	}
	if(N>1)
		ans = ans/N*(N-1); 
	return ans;
}

int main(){
	
	int T;
	scanf("%d",&T);
	int N,M;
	int ans;
	while(T--){
		ans = 0;
		scanf("%d %d",&N,&M);
		for(int i=1; i*i<=N; i++){
			if(i*i==N && i>=M)
				ans += euler(i);
			else if(N%i==0){
				if(i>=M)
					ans += euler(N/i);
				if(N/i>=M)
					ans += euler(i);
			}
		}
		printf("%d\n",ans);
	}

	return 0;
}