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+7;
const int mx=1e7;
bool v[mx+5];
int pr[mx+5],phi[mx+5],s[mx+5],a[100010],r,n,m,t,p;

void resolve(int n)
{
    r=0;
    for (int i=1; i<=t; i++)
        if (n%pr[i]==0) {a[++r]=pr[i]; n/=pr[i];}
        else if (n<pr[i]) break;
}

int quick(ll x,ll y,int p)
{
    ll s=1;
    while (y)
    {
        if (y&1) s=s*x%p;
        x=x*x%p;
        y>>=1;
    }
    return (s==0)?p:s;
}

ll f(int i,int n,int m)
{
    if (n==1) return s[m];
    if (m==0) return 0;
    return (1ll*(a[i]-1)*f(i+1,n/a[i],m)%mo+f(i,n,m/a[i]))%mo;
}

int ans(int k,int p)
{
    if (p==1) return 1;
    return quick(k,ans(k,phi[p]),p);
}

int main()
{
    phi[1]=1;
    for (int i=2; i<=mx; i++)
    {
        if (!v[i])
        {
            pr[++t]=i;
            v[i]=1;
            phi[i]=i-1;
        }
        for (int j=1; j<=t; j++)
        {
            if (i*pr[j]>mx) break;
            v[i*pr[j]]=1;
            if (i%pr[j]) phi[i*pr[j]]=phi[i]*(pr[j]-1);
            else {
                phi[i*pr[j]]=phi[i]*pr[j];
                break;
            }
        }
    }
    for (int i=1; i<=mx; i++) s[i]=(s[i-1]+phi[i])%mo;
    while (scanf("%d%d%d",&n,&m,&p)!=EOF)
    {
        if (p==1) {puts("0"); continue;}
        resolve(n);
        ll k=f(1,n,m);
        printf("%d
",ans(k,p)%p);
    }
}