GCD

http://acm.hdu.edu.cn/showproblem.php?pid=2588

GCD

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1311    Accepted Submission(s): 591

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

 

欧拉函数

#include <cstdio>
int a[40000];
int euler(int n){ //返回euler(n)
     int res=n,a=n;
     for(int i=2;i*i<=a;i++){
         if(a%i==0){
             res=res/i*(i-1);//先进行除法是为了防止中间数据的溢出
             while(a%i==0) a/=i;//除去a的因子
         }
     }
     if(a>1) res=res/a*(a-1);//最多仅有一个素因子比根号n大
     return res;
}
int main()
{
    int t,n,m;
    scanf("%d",&t);
    while(t--){
        int ans=0;
        scanf("%d%d",&n,&m);
        int cnt=0;
        for(int i=1;i*i<=n;i++){
            if(n%i==0){
            a[cnt++]=i;
            if(i*i<n)
                a[cnt++]=n/i;
            }
        }
        for(int i=0;i<cnt;i++){
            if(a[i]>=m)
                ans+=euler(n/a[i]);
        }
        printf("%d\n",ans);
    }
    return 0;
}