GCD（欧拉函数）

GCD

http://acm.hdu.edu.cn/showproblem.php?pid=2588

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 3782    Accepted Submission(s): 2053

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<bits/stdc++.h>
using namespace std;
#define lson l,mid,rt<<1
#define rson mid+1,r,rt<<1|1
#define IT set<ll>::iterator
#define pb push_back
#define eb emplace_back
#define maxn 200005
#define eps 1e-6
#define PI acos(-1.0)
#define rep(k,i,j) for(int k=i;k<j;k++)
typedef long long ll;
typedef pair<int,int> pii;
typedef pair<ll,ll>pll;
typedef pair<ll,int> pli;
typedef pair<pair<int,string>,pii> ppp;
typedef unsigned long long ull;
const long long MOD=1000000007;
const double oula=0.57721566490153286060651209;
const int INF=0x3f3f3f3f;
using namespace std;


ll Euler(ll n){
    ll ans=n;
    for(ll i=2;i*i<=n;i++){
        if(n%i==0) ans=ans/i*(i-1);
        while(n%i==0) n/=i;
    }
    if(n>1) ans=ans/n*(n-1);
    return ans;
}

int main(){
    std::ios::sync_with_stdio(false);
    int t;
    cin>>t;
    while(t--){
        ll n,m;
        cin>>n>>m;
        ll ans=0;
        for(int i=1;i*i<=n;i++){
            if(n%i==0){
                if(i>=m) ans+=Euler(n/i);
                if(i*i!=n&&n/i>=m) ans+=Euler(i);
            }
        }
        cout<<ans<<endl;
    }
}