D. Notepad

• 题意：求(b-1)b^(n-1)%mod 2<=b<=10⁽¹⁰6),1<=n<=10⁽¹⁰6),1<=c<=10^9;

//(b-1)*b^(n-1)%mod
//2<=b<=10^(10^6),1<=n<=10^(10^6),1<=c<=10^9;
#include<bits/stdc++.h>
using namespace std;
typedef long long int ll;
const int N=1000000+10;
ll get_phi(ll x)
{
    ll res=x;
    for(int i=2;i*i<=x;i++)
    {
        if(x%i==0)
        {
            res=res-res/i;
            while(x%i==0)
                x/=i;
        }
    }
    if(x>1)res=res-res/x;
    return res;
}
char b[N],n[N];
ll c;
ll quick(ll m,ll n,ll mod)
{
    ll res=1;
    while(n)
    {
        if(n&1)res=res*m%mod;
        m=m*m%mod;
        n>>=1;
    }
    return res;
}
ll cal(char *s,ll k)
{
    ll res=0;
    for(int i=0;s[i];i++)
    {
        res=(res*10LL+s[i]-'0')%k;
    }
    return res;
}
int main()
{
    while(~scanf("%s%s%d",b,n,&c))
    {
        ll A=cal(b,c);
        ll d=get_phi(c);
        bool mark=false;
        ll ans=0;
        for(int i=0;n[i];i++)
        {
            ans=(ans*10LL+n[i]-'0');
            if(ans>d)
            mark=true,ans%=d;
        }
        ans--;
        ll res=0;
        if(mark)
        {
            res=quick(A,ans+d,c);
        }
        else
        {
            res=quick(A,ans,c);
        }
        ll sum=(A-1+c)%c;
        sum=sum*res%c;
        if(sum==0)sum=c;
        printf("%lld\n",sum);
    }
    return 0;
}