hdu 2588 GCD （欧拉函数扩展）

原题

题目：

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.

题意：

当gcd（x, n） >= m时，求X的个数

题解：

gcd(x, n) >= m   求x 

设gcd(x,n) = p;    所以p是 n的一个因子（n % p == 0）；两边同时除p得     gcd(x/p, n/p) = 1;

所以 n/p   与  x/p  互质， x 的取值范围  为       x >= m  && x <= n ，用欧拉函数求质数个数

#include
using namespace std;
#define ll long long
long long euler(long long  n){ 
   long long  res = n, a = n;
   for(long long  i = 2; i * i <= a; i++){
      if(a % i == 0){
         res = res / i * (i-1);   // 先除再乘，防止溢出
         while(a % i==0){
            a /= i;
         }
      }
   }
   if(a > 1) res = res / a * (a-1);
   return res;
}
int main(){
   int T;cin >> T;
   while(T--){
   	ll n, m;
	cin >> n >> m;
	ll ans = 0;
	for(int i = 1; i * i <= n; i++){
            if(n % i != 0){
               continue;
            } 
            if(i >= m && i * i != n){
               ans = ans + euler(n / i);
            }
            if(n / i >= m){
              ans = ans + euler(i);
            }
         }
       if(m == 1){
         ans = n;
        }
         cout << ans << endl;
    }
	return 0;
}