hdu5728

详细题解：

http://blog.csdn.net/wust_zzwh/article/details/51966450

……化简公式的能力还不够啊……

 #include<bits/stdc++.h>

 using namespace std;
 typedef long long ll;
 const int mo=1e9+;
 const int mx=1e7;
 bool v[mx+];
 int pr[mx+],phi[mx+],s[mx+],a[],r,n,m,t,p;

 void resolve(int n)
 {
     r=;
     for (int i=; i<=t; i++)
         if (n%pr[i]==) {a[++r]=pr[i]; n/=pr[i];}
         else if (n<pr[i]) break;
 }

 int quick(ll x,ll y,int p)
 {
     ll s=;
     while (y)
     {
         if (y&) s=s*x%p;
         x=x*x%p;
         y>>=;
     }
     return (s==)?p:s;
 }

 ll f(int i,int n,int m)
 {
     if (n==) return s[m];
     if (m==) return ;
     return (1ll*(a[i]-)*f(i+,n/a[i],m)%mo+f(i,n,m/a[i]))%mo;
 }

 int ans(int k,int p)
 {
     if (p==) return ;
     return quick(k,ans(k,phi[p]),p);
 }

 int main()
 {
     phi[]=;
     for (int i=; i<=mx; i++)
     {
         if (!v[i])
         {
             pr[++t]=i;
             v[i]=;
             phi[i]=i-;
         }
         for (int j=; j<=t; j++)
         {
             if (i*pr[j]>mx) break;
             v[i*pr[j]]=;
             if (i%pr[j]) phi[i*pr[j]]=phi[i]*(pr[j]-);
             else {
                 phi[i*pr[j]]=phi[i]*pr[j];
                 break;
             }
         }
     }
     for (int i=; i<=mx; i++) s[i]=(s[i-]+phi[i])%mo;
     while (scanf("%d%d%d",&n,&m,&p)!=EOF)
     {
         if (p==) {puts(""); continue;}
         resolve(n);
         ll k=f(,n,m);
         printf("%d\n",ans(k,p)%p);
     }
 }