Super A^B mod C 快速幂+欧拉函数降幂

uper A^B mod C 

Time Limit:1000MS     Memory Limit:32768KB     64bit IO Format:%I64d & %I64u

Submit Status

Description

Given A,B,C, You should quickly calculate the result of A^B mod C. (1<=A,C<=1000000000,1<=B<=10^1000000).

Input

There are multiply testcases. Each testcase, there is one line contains three integers A, B and C, separated by a single space.

Output

For each testcase, output an integer, denotes the result of A^B mod C.

Sample Input

3 2 4

2 10 1000

Sample Output

1

24

 

首先要降幂，由降幂公式：

 （然而不知道公式是怎么来的。。。求指教）

降幂后用快速幂计算

#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cstring>
#include<cmath>
#define ll __int64
#define N 1000100
using namespace std;
char b[N];
ll p[N];
ll a, c;
ll quick(ll a, ll b){ //快速幂
    ll k = 1;
    while(b){
        if(b%2==1){
            k = k*a;
            k %=c;
        }
        a = a*a%c;
        b /=2;
    }
    return k;
}
void priem(){
    memset(p, 0, sizeof(p));
    ll i, j;
    p[1] = 1;
    for(i=2; i<=sqrt(N); i++){
        for(j=2; j<=N/i; j++)
            p[i*j] = 1;
    }
}
ll ola(ll n){ //欧拉函数
    ll i, j, r, aa;
    r = n;
    aa = n;
    for(i=2; i<=sqrt(n); i++){
        if(!p[i]){
            if(aa%i==0){
                r = r/i*(i-1);
                while(aa%i==0)
                    aa /= i;
            }
        }
    }
    if(aa>1)
        r = r/aa*(aa-1);
    return r;
}
int main(){
    ll d, i, j;
    priem();
    while(~scanf("%I64d%s%I64d",&a,b,&c)){
        ll l = strlen(b);
        ll B=0;
        ll oc = ola(c);
      //  cout<<"oc = "<<oc<<endl;
        for(i=0; i<l; i++){
            B = B*10+b[i]-'0';
            if(B>oc)
                break;
        }
        //cout<<i<<endl;
        if(i==l)
            d = quick(a,B);
        else{
            B=0;
            for(i=0; i<l; i++){ //降幂
                B = (B*10+b[i]-'0')%oc;
            }
            d = quick(a,B+oc);
        }
      //  printf("B= %I64d
",B);
        printf("%I64d
",d);
    }
    return 0;
}