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): 1369    Accepted Submission(s): 632

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：寻找小于等于n且gcd（i，n）>=m的数的数量 思路：n = a * b，i = a * d（b >= d），若gcd(i, n)=a，那么b和d一定互质 ， 而此时i的个数即b的欧拉函数。这样问题就变成在满足b>=m的情况下,对应n/b的欧拉函数之和。 注意：数据量太大，普通的肯定超时 

ac代码：

    
    
    
    
#include<stdio.h>
#include<string.h>
#include<math.h>
#include<iostream>
#include<algorithm>
#define MAXN 100100
#define MOD 1000000007
#define LL long long
#define INF 0xfffffff
#define fab(a)(a)>0?(a):(-a)
using namespace std;
LL eular(LL n)  
{    
    LL i,res=n;    
    for(i=2;i*i<=n;i++)    
        if(n%i==0)  
        {    
            res=res/i*(i-1);    
            while(n%i==0)  
            n/=i;    
        }    
    if(n>1)   
    res=res/n*(n-1);    
    return res;    
}  
int main()
{
    int n,i,m;
    int t;
    scanf("%d",&t);
    while(t--)
    {
    	scanf("%d%d",&n,&m);
    	LL ans=0;
    	for(i=1;i<=sqrt(n);i++)
    	{
    		if(n%i)
    		continue;
    		if(i>=m)
    		ans+=eular(n/i);
    		if(n/i>=m&&i*i!=n)
    		ans+=eular(i);
		}
		printf("%lld\n",ans);
	}
    return 0;
}