D—GCD （HDU 2588）

传送门

GCD

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

Total Submission(s): 1438 Accepted Submission(s): 673



[align=left]Problem Description[/align]
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.

[align=left]Input[/align]
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.

[align=left]Output[/align]
For each test case,output the answer on a single line.

[align=left]Sample Input[/align]

3
1 1
10 2
10000 72

[align=left]Sample Output[/align]

1
6
260

题目大意：

给定T组数据，然后两个数 N和 M, 让你求1<=X<=N ,GCD (X,N)>=M,让你X的个数

解题思路：

假设 GCD(X,N) = d,那么 X = q * d , N = p * d,而且p 和 q 肯定是互素的，如果d >= M的话，我们要求的只是在 <=p的情况下与 p 互素的个数，然后累加 ，就是假设 N的约数 i >= M，就是求与 <= N/i中与 N/i的个数，也就是求一个欧拉函数，所以这个题就是相当于 欧拉函数的延伸，还是比较不错的，当知道这个之后就是从 1 - N 中找，但是这样还是会超时的，所以我们就从 sqrt(N)中找 ，N%i
== 0 和 N%(N/i) ==0,最后特判一下是不是完全平方数就ok了，剩下的就是敲代码了：

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;
}