D—GCD （HDU 2588）

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## D—GCD （HDU 2588）

angel_imp 2016-03-21 20:39:00 浏览892

# GCD

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1438    Accepted Submission(s): 673

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

My Code:

```#include <iostream>
#include <cstdio>
#include <cmath>
using namespace std;
int Eular(int m)
{
int ret = m;
for(int i=2; i*i<=m; i++)
{
if(m%i == 0)
{
ret -= ret/i;
while(m%i == 0)
m /= i;
}
}
if(m > 1)
ret -= ret/m;
return ret;
}
int main()
{
int T, m, n;
cin>>T;
while(T--)
{
cin>>n>>m;
int sum = 0;
for(int i=1; i*i<n; i++)///这样比较省时间
{
if(n%i == 0)///n的约数
{
if(i >= m)
{
sum += Eular(n/i);
}
if(n/i >= m)
sum += Eular(i);
}
}
if((int)sqrt(n)>=m)///特判
{
if((int)sqrt(n)*(int)sqrt(n) == n)
sum += Eular((int)sqrt(n));
}
cout<<sum<<endl;
}
return 0;
}
```

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