数论 欧拉函数

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

题目大意：求m到n之间有多少个x,x满足 gcd(x,n)>=m，
        用到欧拉函数。

思路：先求出 n 大于等于 m 的因子 i,再计算 n/i 的欧拉函数,最后相加即可 

 

 

具体代码如下：

 

#include <stdio.h>
#include <iostream>
using namespace std;
int ol(int n)
{
    int s=n,i,j;
    for(i=2;i*i<=n;i++)
    {
        if(n%i==0)
        {
            while (!(n%i))
            n/=i;
            s=s/i*(i-1);
        }
    }
    if(n!=1) s=s/n*(n-1);
    return s;
}

int main()
{
   int t,n,m,sum,i,j;
   scanf("%d",&t);
   while(t--)
   {
       sum=0;
       scanf("%d%d",&n,&m);
       for(i=1;i*i<=n;i++)
       if(n%i==0)
       {
           if(i>=m)
           sum+=ol(n/i);
           if((n/i)!=i&&(n/i)>=m)
           sum+=ol(i);
       }
       printf("%d
",sum);
   }
   return 0;
}