hdoj 2588 GCD【欧拉函数】

GCD

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1320    Accepted Submission(s): 597

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：计算1-N区间里有多少数和N的GCD是大于M的。

解题思路：

这道题一看到  2<=N<=1000000000,应该想到普通的做法肯定要超时，于是要想方法避免超时，“二分思想”   在这得以利用，左右开弓同时查找可以节省大量时间，并且许多合数都不必考虑，代码也从原来的TLE变成0ms.

#include<stdio.h>
int eular(int n)
{
	int rt = 1,i;
	for( i =2; i*i<=n; i++)
	if(n%i==0)
	{
		n/=i;
		rt=rt*(i-1);
		while(n%i==0)
		{
			n/=i;
			rt*=i;
		}
	}
	if(n >1 ) rt =rt*(n-1);
	return rt;
}
int main()
{
	int t,n,m,x,i,j;
	scanf("%d",&t);
	while(t--)
	{
		scanf("%d%d",&n,&m);
		int ans = 0;
		for(i = 1; i*i <= n ;i++)
		{
			if(n%i==0)
			{
				if(i>=m) 
					ans+=eular(n/i);
					
				if(n/i>=m&&i*i!=n) 
					ans+=eular(i);
			}
		}
		printf("%d\n",ans);
	}
	return 0;
}