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=2;i*i<=x;i++)
    {
        if(x%i==0)
        {
            res=res/i*(i-1);
            while(x%i==0)
               x=x/i;
        }
    }
    if(x>1)
        res=res/x*(x-1);
    return res;
}

int main()
{

    ll t,n,m,sum;
    scanf("%lld",&t);
    while(t--)
    {
        sum=0;
        scanf("%lld%lld",&n,&m);
        int q=sqrt(n);
        ll i;
        for(i=1;i*i<=n;i++)
        {
            if(n%i==0)
            {
                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
",sum);
    }
    return 0;
}