GCD hdu2588

GCD

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 897    Accepted Submission(s): 400

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

 

 

 #include <iostream>
 #include <algorithm>
 #include <cstdio>
 #include <math.h>

 using namespace std;
 int fun(int n)//传入一数，返回此数的欧拉函数，用素数表会更快
 {
     int sum=n;
     int i;
     for(i=; i*i<=n; i++)
     {
         if(n%i==)
         {
             sum=sum/i*(i-);
             while(n%i==)n/=i;
         }
     }
     if(n!=)
     {
         sum=sum/n*(n-);
     }
     return sum;
 }

 int main()
 {
     int t,i,n,m,ans,size;
     scanf("%d",&t);
     while(t--)
     {
         scanf("%d%d",&n,&m);
         ans=;
         size=sqrt(n+0.5);
         for(i=; i<=size; i++)
         {
             if(n%i==)
             {
                 if(i>=m)
                     ans+=fun(n/i);
                 if(i!=n/i&&n/i>=m)ans+=fun(i);
             }
         }
         printf("%d\n",ans);
     }
 }

 

Source

ECJTU 2009 Spring Contest