fzu 1759Super A^B mod C 指数循环节

Problem 1759 Super A^B mod C

Time Limit: 1000 mSec    Memory Limit : 32768 KB

Problem 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

Source

FZU 2009 Summer Training IV--Number Theory

欧拉定理的推广——有关的高次幂取模（指数循环节）

公式： a^x mod(c)=a^{(x mod phi(c) +phi(c))} mod(c)， (x>=phi(c))

#include <cstdio>
#include <iostream>
#include <cstring>
using namespace std;
typedef long long int ll;

const int MAXN = 1e6;
int prime[MAXN+10], cnt = 0;
int a[MAXN+10];

ll pow_mod(ll a, ll n, ll mod){
    ll ans = 1;
    while(n > 0){
        if(n&1) {
            n--;
            ans = ans*a%mod;
        }
        n >>= 1;
        a = a*a %mod;
    }
    return ans;
}

void init(){
    for(int i = 2; i <= MAXN; i++) a[i] = true;
    for(int i = 2; i <= MAXN; i++){
        if(a[i]){
            prime[++cnt] = i;
        }
        for(int j = 1; j <= cnt; j++){
            if(prime[j]*i > MAXN) break;
            a[prime[j]*i] = false;
            if(i%prime[j] == 0) break;
        }
    }
}
ll Euler(ll n){
    ll ans = n;
    for(int i = 1; i <= cnt && prime[i] <= n; i++){
        if(n%prime[i] == 0){
            while(n%prime[i] == 0){
                n /= prime[i];
            }
            ans = ans/prime[i]*(prime[i]-1);
        }
    }
    return ans;
}
ll slove(ll a, char s[], ll c, ll mod){
    ll sum = 0;
    int len = strlen(s);
    sum = (s[0] - '0') %mod;
    for(int i = 1; i < len; i++){
        sum *= 10;
        sum = (sum + s[i] - '0')%mod;
    }
   // printf("sum = %I64d
", sum);
    return pow_mod(a, mod+sum, c);
}
int main(){
    ll a, c;
    char b[MAXN+10];
    init();
    while(scanf("%I64d%s%I64d", &a, b, &c) != EOF){
        ll p = Euler(c);
       // printf("p = %I64d
", p);
        printf("%I64d
", slove(a, b, c, p));
    }

}