由题意可知，anw =  (b-1)*b^(n-1)%c，则重点为求b^(n-1)。

弱渣推不出来只能上公示。

phi(c)为小于c且与c互质的个数。

当x >= phi(c)时：A^x = A(x%phi(c) + phi(c)) 。

当x < phi(c)时：直接求即可。

#include <algorithm>
#include <iostream>
#include <cstring>
#include <cstdlib>
#include <cstdio>
#include <queue>
#include <cmath>
#include <stack>
#include <map>
#include <ctime>
#include <iomanip>

#pragma comment(linker, "/STACK:1024000000");
#define EPS (1e-6)
#define _LL long long
#define ULL unsigned long long
#define LL __int64
#define INF 0x3f3f3f3f
#define Mod 1000000007

/** I/O Accelerator Interface .. **/
#define g (c=getchar())
#define d isdigit(g)
#define p x=x*10+c-'0'
#define n x=x*10+'0'-c
#define pp l/=10,p
#define nn l/=10,n
template<class T> inline T& RD(T &x)
{
    char c;
    while(!d);
    x=c-'0';
    while(d)p;
    return x;
}
template<class T> inline T& RDD(T &x)
{
    char c;
    while(g,c!='-'&&!isdigit(c));
    if (c=='-')
    {
        x='0'-g;
        while(d)n;
    }
    else
    {
        x=c-'0';
        while(d)p;
    }
    return x;
}
inline double& RF(double &x)      //scanf("%lf", &x);
{
    char c;
    while(g,c!='-'&&c!='.'&&!isdigit(c));
    if(c=='-')if(g=='.')
        {
            x=0;
            double l=1;
            while(d)nn;
            x*=l;
        }
        else
        {
            x='0'-c;
            while(d)n;
            if(c=='.')
            {
                double l=1;
                while(d)nn;
                x*=l;
            }
        }
    else if(c=='.')
    {
        x=0;
        double l=1;
        while(d)pp;
        x*=l;
    }
    else
    {
        x=c-'0';
        while(d)p;
        if(c=='.')
        {
            double l=1;
            while(d)pp;
            x*=l;
        }
    }
    return x;
}
#undef nn
#undef pp
#undef n
#undef p
#undef d
#undef g
using namespace std;

char s[1000010];
int B[1000010],N[1000010];
LL c,b,n;

bool vis[1000010];
LL prime[1000010];
int Top;

LL Euler(LL n)
{
    LL ret=1,i;
    for(i=2; i*i<=n; i++)
    {
        if(n%i==0)
        {
            n/=i,ret*=i-1;
            while(n%i==0)
                n/=i,ret*=i;
        }
    }
    if(n>1)
        ret*=n-1;
    return ret;
}

LL Cal(int *n,LL c,bool &mark)
{
    LL re = 0;

    for(int i = 0; n[i] != -1; ++i)
    {
        re *= 10,re += n[i];
        if(re >= c)
            mark = true,re %= c;
    }
    return re;
}

LL qm(LL a,LL b,LL n)
{
    LL ret=1;
    LL tmp=a;
    while(b)
    {
        if(b&1) ret=ret*tmp%n;
        tmp=tmp*tmp%n;
        b>>=1;
    }
    return ret;
}

int main()
{
    int i,j;

    scanf("%s",s);
    for(i = 0; s[i] != '\0'; ++i)
        B[i] = s[i]-'0';
    B[i] = -1;

    scanf("%s",s);
    for(i = 0; s[i] != '\0'; ++i)
        N[i] = s[i]-'0';
    N[i] = -1;

    scanf("%I64d",&c);

    LL MAXN = sqrt(c);
    memset(vis,false,sizeof(vis));
    Top = 0;

    for(i = 2; i <= MAXN; ++i)
        if(vis[i] == false)
            for(j = i+i,prime[Top++] = i; j <= MAXN; j += i)
                vis[j] = true;

    LL phi = Euler(c),b1;
    bool mark = false;
    b = Cal(B,c,mark);
    mark = false;
    n = Cal(N,phi,mark);
    b1 = (b-1+c)%c;

    if(mark == true)
        n = (n-1+phi)%phi + phi;
    else
        n--;

    LL anw = b1*qm(b,n,c)%c;

    if(anw == 0)
        anw = c;
    printf("%I64d\n",anw);

    return 0;
}

Codeforces 17D Notepad 简单数论