hdu 4335 What is N?

此题用到的公式：a^b%c=a^(b%phi(c)+phi(c))%c (b>=phi(c)).

1.当n!<phi(p)时，直接暴力掉；

2.当n!>=phi(p) && n!%phi(p)!=0,用上面公式求；

3.当n!>=phi(p) && n!%phi(p)==0,变为n^(phi(p))%p,找循环节，就可以了

 #include<iostream>
 #include<stdio.h>
 #include<algorithm>
 #include<iomanip>
 #include<cmath>
 #include<string>
 #include<cstdlib>
 #include<vector>
 #define ll unsigned __int64
 using namespace std;
 ll an[];
 ll euler(ll n)
 {
     ll ans=;
     for(int i=;i*i<=n;i++)
     {
         if(n%i==)
         {
             ans*=i-;
             n/=i;
             while(n%i==)
             {
                 ans*=i;
                 n/=i;
             }
         }
     }
     if(n>) ans*=n-;
     return ans;
 }
 ll pows(ll a,ll b, ll mod)
 {
     ll ans=;
     while(b)
     {
         if(b&) ans=(ans*a)%mod;
         b>>=;
         a=(a*a)%mod;
     }
     return ans%mod;
 }
 int main()
 {
     int t,k=;
     ll m,ans,fac,c,i,j,phi,b,p;
     cin>>t;
     while(t--)
     {
         scanf("%I64u%I64u%I64u",&b,&p,&m);
         printf("Case #%d: ",++k);
         if(p==)
         {
             if(m==18446744073709551615U)
                printf("18446744073709551616\n");
             else printf("%I64u\n",m+);
             continue;
         }
         ans=;fac=;
         phi=euler(p);
         //n!<phi(p)
         for(i=;i<=m&&fac<=phi;i++)
         {
             if(pows(i,fac,p)==b)
                 ans++;
             fac*=(i+);
         }
         fac%=phi;
         //n!>=phi(p) && n!%phi(p)!=0
         for(;i<=m&&fac;i++)
         {
             if(pows(i,fac+phi,p)==b)
                 ans++;
             fac=(fac*(i+))%phi;
         }
         //n!>=phi(p) && n!%phi(p)==0
         if(i<=m)
         {
             ll cnt=;
         //    memset(an,0,sizeof(an));
             for(j=;j<p;j++)
             {
                 an[j]=pows(i+j,phi,p);
                 if(an[j]==b)
                     cnt++;
             }
             c=(m-i+)/p;
             ans+=c*cnt;
             ll remind=(m-i+)%p;
             for(j=;j<remind;j++)
                 if(an[j]==b)
                     ans++;
         }
         printf("%I64u\n",ans);
     }
     return ;
 }