hdu2588-GCD-(欧拉函数+分解因子)

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.

InputThe 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.OutputFor each test case,output the answer on a single line.Sample Input

3
1 1
10 2
10000 72

Sample Output

1
6
260
翻译：给出n和m，1<=x<=n,求x符合gcd(x,n)>=m有多少个。
解题过程：
令d=gcd(x,n),显然d是n的因子，并且是x和n的最大公因子,则gcd(x/d,n/d)=1
对于每个d,令y=n/d,找有多少个x/d满足gcd(x/d,y)=1。
欧拉函数登场，累加y的欧拉函数值。

 #include <iostream>
 #include<stdio.h>
 #include <algorithm>
 #include<string.h>
 #include<cstring>
 #include<math.h>
 #define inf 0x3f3f3f3f
 #define ll long long
 using namespace std;

 ll euler(ll x)
 {
     ll res=x;
     for(ll i=;i*i<=x;i++)
     {
         if(x%i==)
         {
             res=res/i*(i-);
             while(x%i==)
                x=x/i;
         }
     }
     if(x>)
         res=res/x*(x-);
     return res;
 }

 int main()
 {

     ll t,n,m,sum;
     scanf("%lld",&t);
     while(t--)
     {
         sum=;
         scanf("%lld%lld",&n,&m);
         int q=sqrt(n);
         ll i;
         for(i=;i*i<=n;i++)
         {
             if(n%i==)
             {
                 if(i>=m)
                     sum=sum+euler(n/i);
                 if((n/i)>=m)
                     sum=sum+euler(i);
             }
         }
         i--;
         if(i*i==n && i>=m)
             sum=sum-euler(i);
         printf("%lld\n",sum);
     }
     return ;
 }