hdoj 2588 GCD(欧拉函数)



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): 1111    Accepted Submission(s): 507

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
  
  

分析：假设x<=n，n=p*d,x=q*d.假设n与x的最大公约数为d，则能够推出p与q肯定是互质的，因为x<=n所以要求的就是p的欧拉函数值了，那么我们就转化成求满足：n=p*d,并且d>=m的p的欧拉函数值之和。

#include<stdio.h>
#include<string.h>
int Euler(int n)
{
	int ret=n,i;
	for(i=2;i*i<=n;i++)
	if(n%i==0)
	{
	     ret=ret-ret/i;
		 while(n%i==0) 
		 n/=i;
	}
	if(n>1) ret=ret-ret/n;
	return ret;
}
int main()
{
	 int T,N,M,X,i,j,k;
	 __int64 sum;
	 scanf("%d",&T);
	 while(T--)
	 {
	      scanf("%d %d",&N,&M);
//		  for(i=M,sum=0;i<=N;i++)//这样写超时 
//		  if(N%i==0)
//		  sum+=Euler(N/i);
          for(i=1,sum=0;i*i<=N;i++)//这个for循环是对上面的for循环的改进。 
          if(N%i==0)
          {
          	  if(i>=M) 
          	  sum+=Euler(N/i);
          	  if(i*i!=N/*如果满足i*i==N,上面的if已经加过了，所以这里要排除*/&&N/i>=M)
          	  sum+=Euler(i);
          }
		  
		  printf("%I64d\n",sum);
	}
	return 0;
}