hdu 2588 GCD (欧拉函数)

GCD

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 890    Accepted Submission(s): 396

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

 

Source

ECJTU 2009 Spring Contest

 

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//15MS    208K    784 B    G++
/*
 
   题意：
       求[1,n]内与n公约数大于m的数的个数。
       
   思路：
       想了挺长时间，没什么思路。
       后来得知结果为：
           对于所有大于m的n的公约数k，求出所有n/k的欧拉数的和即为题解。
       一个数m的欧拉数即为小于m且与m互质的数的和。
       至于怎么得出的自己推一下吧。 

*/
#include<stdio.h>
int euler(int n)
{
    int ret=1;
    for(int i=2;i*i<=n;i++)
        if(n%i==0){
            n/=i;ret*=i-1;
            while(n%i==0){
                n/=i;ret*=i;
            }
        }
    if(n>1) ret*=(n-1);
    return ret;
}
int cul(int n,int m)
{
    int sum=0;
    for(int i=1;i*i<=n;i++){
        if(n%i==0){ 
            if(n/i>=m && i*i!=n)
                sum+=euler(i);
            if(i>=m)
                sum+=euler(n/i);
        } 
    }
    return sum;
}
int main(void)
{
    int t,n,m;
    scanf("%d",&t);
    while(t--)
    {
        scanf("%d%d",&n,&m);
        printf("%d
",cul(n,m));
    }
    return 0;
}