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=2; i*i<=n; i++)
    {
        if(n%i==0)
        {
            sum=sum/i*(i-1);
            while(n%i==0)n/=i;
        }
    }
    if(n!=1)
    {
        sum=sum/n*(n-1);
    }
    return sum;
}

int main()
{
    int t,i,n,m,ans,size;
    scanf("%d",&t);
    while(t--)
    {
        scanf("%d%d",&n,&m);
        ans=0;
        size=sqrt(n+0.5);
        for(i=1; i<=size; i++)
        {
            if(n%i==0)
            {
                if(i>=m)
                    ans+=fun(n/i);
                if(i!=n/i&&n/i>=m)ans+=fun(i);
            }
        }
        printf("%d
",ans);
    }
}

 

Source

ECJTU 2009 Spring Contest