Easy Math



图片说明 

#include<bits/stdc++.h>
#define me(a,x) memset(a,x,sizeof(a))
#define sc scanf
#define itn int
#define IN freopen("in.txt","r",stdin);
#define OUT freopen("out.txt","w",stdout);
#define STR clock_t startTime = clock();
#define END clock_t endTime = clock();cout << double(endTime - startTime) / CLOCKS_PER_SEC *1000<< "ms" << endl;
using namespace std;
const int N=1e7+5;
const long long mod=1e9+7;
const long long mod2=998244353;
const int oo=0x7fffffff;
const int sup=0x80000000;
typedef long long ll;
typedef unsigned long long ull;
template <typename it>void db(it *begin,it *end){while(begin!=end)cout<<(*begin++)<<" ";puts("");}
template <typename it>
string to_str(it n){string s="";while(n)s+=n%10+'0',n/=10;reverse(s.begin(),s.end());return s;}
template <typename it>int o(it a){cout<<a<<endl;return 0;}
inline ll mul_64(ll x,ll y,ll c){return (x*y-(ll)((long double)x/c*y)*c+c)%c;}
ll ksm(ll a,ll b,ll c){ll ans=1;for(;b;b>>=1,a=a*a%c)if(b&1)ans=ans*a%c;return ans;}
void exgcd(ll a,ll b,ll &x,ll &y){if(!b)x=1,y=0;else exgcd(b,a%b,y,x),y-=(a/b)*x;}
int prime[N],tot=0;
bool vis[N]={0};
short int mu[N];
int f[N];
void pre(){
    mu[1]=1;f[1]=1;
    for(int i=2;i<N;i++){
        if(!vis[i])prime[++tot]=i,mu[i]=-1;
        for(int j=1;j<N&&i*prime[j]<N;j++){
            vis[i*prime[j]]=1;
            if(i%prime[j]==0){
                mu[i*prime[j]]=0;
                break;
            }else mu[i*prime[j]]=-mu[i];
        }
        f[i]=f[i-1]+mu[i];
    }
}
int get_mu(ll n){
    if(n==1)return 1;
    int cnt=0;
    for(int i=1;i<=tot&&1LL*prime[i]*prime[i]<=n;i++){
        if(n%prime[i]==0){
            n/=prime[i];
            if(n%prime[i]==0)return 0;
            cnt++;
        }
    }
    if(n>1)cnt++;
    return cnt%2==1?-1:1;
}
map<ll,ll>p;
map<ll,bool>q;
ll cal(ll n){
    if(n<N)return f[n];
    if(q[n])return p[n];
    ll ans=1;
    for(ll d=2,last;d<=n;d=last+1){
        last=n/(n/d);
        ans-=1LL*(last-d+1)*cal(n/d);
    }
    q[n]=true;
    return p[n]=ans;
}
map<ll,short int>MU;
ll F(int m,ll n){
    if(m==0)return 0;
    if(n==1)return cal(m);
    ll ans=0;
    for(ll d=1;d*d<=n;d++){
        if(n%d==0){
            ans+=1LL*MU[d]*F(m/d,d);
            if(d!=n/d)ans+=1LL*MU[n/d]*F(m/(n/d),n/d);
        }
    }
    return ans*MU[n];
}
int main(){
    pre();
    int m;ll n;
    sc("%d%lld",&m,&n);
    for(int i=1;1LL*i*i<=n;i++){
        if(n%i==0){
            MU[i]=mu[i];
            if(n/i<N)MU[n/i]=mu[n/i];
            else MU[n/i]=get_mu(n/i);
        }
    }
    if(MU[n]==0)return o(0);
    printf("%lld\n",F(m,n));
}