hdu2588 GCD

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 题意：Given integers N and M, how many integer X satisfies 1<=X<=N and (X,N)>=M.
    0.d=gcd(x,n) >= m  d是x,n的公共的最大公约数、
    1.找到n的因数(p)>=m gcd(x,n)=p 
    2.=> gcd(x/p,n/p) = 1 
    3.欧拉函数:φ(n) = 小于 n 且和 n 互质的正整数(包括1)的个数 (n为正整数) 

#include <bits/stdc++.h>
#define X 10005
#define inF 0x3f3f3f3f
#define PI 3.141592653589793238462643383
#define IO  ios::sync_with_stdio(false),cin.tie(0), cout.tie(0);
#pragma comment(linker, "/STACK:1024000000,1024000000")
using namespace std;
typedef long long ll;
typedef unsigned long long Ull; //2^64
const int maxn = (int)1e6 + 10;
const int MOD = 9973;
const ll inf = 9223372036854775807;
ll primer[maxn];
ll a[maxn];
void ex_gcd(ll a, ll b, ll &d, ll &x, ll &y) { if (!b) { x = 1; y = 0; d = a; } else { ex_gcd(b, a%b, d, y, x); y -= x * (a / b); }; }
ll gcd(ll a, ll b) { return b ? gcd(b, a%b) : a; }
ll lcm(ll a, ll b) { return b / gcd(a, b)*a; }
ll inv_exgcd(ll a, ll m) { ll d, x, y;ex_gcd(a, m, d, x, y);return d == 1 ? (x + m) % m : -1; }
ll inv1(ll b) { return b == 1 ? 1 : (MOD - MOD / b)*inv1(MOD%b) % MOD; }  
ll crt(int n,int *c,int *m){ll M=1,ans=0;for(int i=0;i<n;++i) M*=m[i];
for(int i=0;i<n;++i) ans=(ans+M/m[i]*c[i] %M *inv_exgcd(M/m[i],m[i]))%M; return ans;}
ll N;
int cnt;
ll fac[maxn];
void factor(ll n)
{
    memset(fac,0,sizeof(fac)),cnt=0;
    for(int i=1;i*i<=n;++i)
    {
        if(n%i==0)
        {
            fac[cnt++]=i;
            if(i*i!=n)
                fac[cnt++]=n/i;
        }
    }
}
ll eular(ll n)
{
    ll ans=n;
    for(int i=2;i*i<=n;++i)
    {
        if(n%i==0)
        {
            ans-=ans/i;
            while(n%i==0)
                n/=i;
        }
    }
    if(n>1) ans-=ans/n;
    return ans;
}
int main()
{
    int t;
    cin>>t;
    ll n,m;
    while(t--)
    {
        cin>>n>>m;
        factor(n);
        sort(fac,fac+cnt);
        int pos=lower_bound(fac,fac+cnt,m)-fac;
        ll ans=0;
        for(int i=pos;i<cnt;++i)
        {
            ans+=eular(n/fac[i]);
        }
        cout<<ans<<endl;
    }
    return 0;
}