hdu2588 GCD

GCD

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 3559    Accepted Submission(s): 1921

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：输入测试样例数 t,输入n ,m。判断满足gcd(x,n)>=m条件x的个数

因为gcd(x,n)>=m

所以可以推出gcd(x/m,n/m)==1

 

所以题目转化为满足gcd(x/m,n/m)==1中X的个数 <==> 求 不大于n/m且与其互质的 n/m的个数 即求ϕ(n/m)

 

#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<algorithm>
using namespace std;
typedef long long ll;

ll euler(ll x)
{
    ll res = x;
    for(int i=2 ;i*i<=x ;i++)
    {
        if(x%i == 0)
        {
            res = res/i*(i-1);
            while(x%i==0) 
              x/=i;
        }
    }
    if(x>1) 
      res = res/x*(x-1);
    return res;
}

int main(){
    int T;
    scanf("%d",&T);
    while(T--)
    {
        ll n,m;
        scanf("%I64d%I64d",&n,&m);
        ll ans = 0;
        for(ll i=1 ;i*i<=n ;i++)
        {
            if(n%i == 0)//i是n的因数
            {
                if(i >= m)
                {
                    ans += euler(n/i);
                }
                if((n/i)>=m && n/i != i)//i*(n/i)==n,判断i对应的另一个因数是否符合
                {
                    ans += euler(i);
                }
            }
        }
        printf("%I64d\n",ans);
    }
    return 0;
}