Super A^B mod C 快速幂+欧拉函数降幂

Super A^B mod C 同余方程

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 42 10 1000

Sample Output

124

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
#include
#include
#include
#include
#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)
                break;
        }
        //cout<