HDU 2588 GCD------欧拉函数变形

GCD

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 812    Accepted Submission(s): 363

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

 /*
 题意：求1<=X<=N 满足GCD(X,N)>=M.

 思路：if(n%p==0 && p>=m)  那么gcd(n,p)=p，
       求所有的满足与n的最大公约数是p的个数，
       就是n/p的欧拉值。
       因为与n/p互素的值x1,x2,x3....
       满足 gcd(n,x1*p)=gcd(n,x2*p)=gcd(n,x3*p)....

       枚举所有的即可。
 */

 #include<stdio.h>
 #include<stdlib.h>
 #include<string.h>

 int Euler(int n)
 {
     int i,temp=n;
     for(i=;i*i<=n;i++)
     {
         if(n%i==)
         {
             while(n%i==)
             n=n/i;
             temp=temp/i*(i-);
         }
     }
     if(n!=)
     temp=temp/n*(n-);
     return temp;
 }

 void make_ini(int n,int m)
 {
     int i,sum=;
     for(i=;i*i<=n;i++)//！！
     {
         if(n%i==)
         {
           if(i>=m)
           sum=sum+Euler(n/i);
           if((n/i)!=i && (n/i)>=m)//！！
           sum=sum+Euler(i);
         }
     }
     printf("%d\n",sum);
 }

 int main()
 {
     int T,n,m;
     while(scanf("%d",&T)>)
     {
         while(T--)
         {
             scanf("%d%d",&n,&m);
             make_ini(n,m);
         }
     }
     return ;
 }