详细题解：
http://blog.csdn.net/wust_zzwh/article/details/51966450
……化简公式的能力还不够啊……
1 #include<bits/stdc++.h> 2 3 using namespace std; 4 typedef long long ll; 5 const int mo=1e9+7; 6 const int mx=1e7; 7 bool v[mx+5]; 8 int pr[mx+5],phi[mx+5],s[mx+5],a[100010],r,n,m,t,p; 9 10 void resolve(int n) 11 { 12 r=0; 13 for (int i=1; i<=t; i++) 14 if (n%pr[i]==0) {a[++r]=pr[i]; n/=pr[i];} 15 else if (n<pr[i]) break; 16 } 17 18 int quick(ll x,ll y,int p) 19 { 20 ll s=1; 21 while (y) 22 { 23 if (y&1) s=s*x%p; 24 x=x*x%p; 25 y>>=1; 26 } 27 return (s==0)?p:s; 28 } 29 30 ll f(int i,int n,int m) 31 { 32 if (n==1) return s[m]; 33 if (m==0) return 0; 34 return (1ll*(a[i]-1)*f(i+1,n/a[i],m)%mo+f(i,n,m/a[i]))%mo; 35 } 36 37 int ans(int k,int p) 38 { 39 if (p==1) return 1; 40 return quick(k,ans(k,phi[p]),p); 41 } 42 43 int main() 44 { 45 phi[1]=1; 46 for (int i=2; i<=mx; i++) 47 { 48 if (!v[i]) 49 { 50 pr[++t]=i; 51 v[i]=1; 52 phi[i]=i-1; 53 } 54 for (int j=1; j<=t; j++) 55 { 56 if (i*pr[j]>mx) break; 57 v[i*pr[j]]=1; 58 if (i%pr[j]) phi[i*pr[j]]=phi[i]*(pr[j]-1); 59 else { 60 phi[i*pr[j]]=phi[i]*pr[j]; 61 break; 62 } 63 } 64 } 65 for (int i=1; i<=mx; i++) s[i]=(s[i-1]+phi[i])%mo; 66 while (scanf("%d%d%d",&n,&m,&p)!=EOF) 67 { 68 if (p==1) {puts("0"); continue;} 69 resolve(n); 70 ll k=f(1,n,m); 71 printf("%d\n",ans(k,p)%p); 72 } 73 }
