傳送門

解析：

對於 $n={\prod }_{i=1}^t$

所以我們只需要用Pollard-Rho把 $n$的質因子分解出來就行了。

程式碼：

#include<bits/stdc++.h>
using namespace std;
#define ll long long
#define re register
#define gc getchar
#define pc putchar
#define cs const

cs int P=25007;
bool mark[P];
int prime[P],pcnt;

inline void linear_sieves(int len=P-7){
	mark[1]=true;
	for(int re i=2;i<=len;++i){
		if(!mark[i])prime[++pcnt]=i;
		for(int re j=1;j<=pcnt&&i*prime[j]<=len;++j){
			mark[i*prime[j]]=true;
			if(i%prime[j]==0)break;
		}
	}

}

inline ll mul(ll a,ll b,ll mod){return (a*b-(ll)((long double)a/mod*b)*mod+mod)%mod;}

inline ll quickpow(ll a,ll b,ll mod){
	re ll ans=1;
	for(;b;b>>=1,a=mul(a,a,mod))if(b&1)ans=mul(ans,a,mod);
	return ans;
}

inline bool isprime(ll x){
	if(x<=P-7)return !mark[x];
	if(x==46856248255981ll)return false;
    if(!(x&1)||!(x%3)||!(x%7)||!(x%61)||!(x%24251))return false;
    ll t=x-1,s=0;
    while(!(t&1))t>>=1,++s;
    for(int re i=1;i<=5;++i){
    	re ll p=prime[rand()%pcnt+1]%x;
    	re ll num=quickpow(p,t,x),pre=num;
    	for(int re j=0;j<s;++j){
    		num=mul(num,num,x);
    		if(num==1&&pre!=1&&pre!=x-1)return false;
    		pre=num;
		}
		if(num!=1)return false;
	}
	return true;
}

inline ll gcd(ll a,ll b){
	if(!a||!b)return a+b;
	re int t=__builtin_ctzll(a|b);
	a>>=__builtin_ctzll(a);
	do{
		b>>=__builtin_ctzll(b);
		if(b<a)a^=b,b^=a,a^=b;
		b-=a;
	}while(b);
	return a;
}

vector<ll> fact;

inline ll Pollard_Rho(ll x){
    if(x%2==0)return 2;
    if(x%3==0)return 3;
    if(x%5==0)return 5;
    if(x%7==0)return 7;
    if(x%61==0)return 61;
    if(x%24251==0)return 24251;
    ll n=0,m=0,t=1,q=1,c=rand()%(x-1)+1;
    for(ll k=2;;k<<=1,m=n,q=1){
    	for(ll i=1;i<=k;++i){
    		n=(mul(n,n,x)+c)%x;
    		q=mul(q,abs(m-n),x);
        }
        t=gcd(x,q);if(t>1)return t;
    }
}

inline void sieves(ll x){
	if(x==1)return ;
	if(isprime(x))return fact.push_back(x);
	re ll p=x;
	while(p>=x)p=Pollard_Rho(p);
	sieves(p);
	while(x%p==0)x/=p;
	sieves(x);
}

ll n;
signed main(){
	srand(time(0));
	linear_sieves();
	scanf("%lld",&n);
	sieves(n);
	sort(fact.begin(),fact.end());
	fact.erase(unique(fact.begin(),fact.end()),fact.end());
	for(int re i=0;i<fact.size();++i){
		n=n/fact[i]*(fact[i]-1);
	}
	printf("%lld",n);
	return 0;
}