BZOJ4802:欧拉函数(Pollard-Rho,欧拉函数)

Description

已知N，求phi(N)

Input

正整数N。N<=10^18

Output

输出phi(N)

Sample Input

Sample Output

Solution

一开始读错题了……以为是求约束个数和……

读对题之后然后发现我不会就问旁边的宽嫂……

宽嫂：这不是欧拉函数的定义式么？我初中就会了啊。

我：aaarticlea/jpeg;base64,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" alt="" />

然后去百度了一发，发现一个数的欧拉函数就是

$\prod (p_i-1)*p_i^{k_i-1}$，其中$p_i$是这个数的质因子，$k_i$是这个质因子的次数……

然后套板子就行了。

发现我之前抄了个假的快速乘然后还花了我一会儿改代码+改博客……QAQ

Code

 #include<iostream>

 #include<cstring>

 #include<cstdio>

 #include<map>

 #include<algorithm>

 #define LL long long

 using namespace std;

 LL n,ans=,Num[],cnt;

 LL prime[]={,,,,,,,,,,};

 map<LL,LL>Keg;

 LL Mul(LL a,LL b,LL MOD)

 {

     LL tmp=a*b-(LL)((long double)a*b/MOD+0.1)*MOD;

     return tmp<?tmp+MOD:tmp;

 }

 LL Qpow(LL a,LL b,LL MOD)

 {

     LL ans=;

     while (b)

     {

         if (b&) ans=Mul(ans,a,MOD);

         a=Mul(a,a,MOD); b>>=;

     }

     return ans;

 }

 LL gcd(LL a,LL b) {return b==?a:gcd(b,a%b);}

 bool Miller_Rabin(LL n)

 {

     if (n==) return ;

     if (n< || n%==) return ;

     LL m=n-,l=;

     while (m%==) m>>=, l++;

     for (int i=; i<; ++i)

     {

         LL p=prime[i],w=Qpow(p,m,n);

         if (w==n- || w== || p==n) continue;

         for (int j=; j<=l; ++j)

         {

             LL u=Mul(w,w,n);

             if (u== && w!=n- && w!=) return ;

             w=u;

         }

         if (w!=) return ;

     }

     return ;

 }

 LL Pollard_Rho(LL n,LL c)

 {

     LL x=(rand()+)%n,y=x,p=,k=;

     for (LL i=; p==; ++i)

     {

         x=(Mul(x,x,n)+c)%n;

         p=x>y?x-y:y-x;

         p=gcd(p,n);

         if (i==k) y=x, k=k+k;

     }

     return p;

 }

 void Solve(LL n)

 {

     if (n==) return;

     if (Miller_Rabin(n))

     {

         if (!Keg[n]) Num[++cnt]=n;

         ++Keg[n]; return;

     }

     LL t=n;

     while (t==n) t=Pollard_Rho(n,rand()%(n-)+);

     Solve(t); Solve(n/t);

 }

 int main()

 {

     scanf("%lld",&n);

     Solve(n);

     for (int i=; i<=cnt; ++i)

         ans=ans*(Num[i]-)*Qpow(Num[i],Keg[Num[i]]-,2e18);

     printf("%lld\n",ans);

 }