FZU1759Super A^B mod C(快速冪取模) 公式

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

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

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

3 2 4
2 10 1000

1
24

對於A^B%C 有一個公式 即

A^x = A^(x % Phi(C) + Phi(C)) (mod C)

公式的具體證明：http://hi.baidu.com/aekdycoin/item/e493adc9a7c0870bad092fd9

// A^x = A^(x % Phi(C) + Phi(C)) (mod C)，其中x≥Phi(C)
#include <iostream>
#include <cstring>
#include <cstdio>
using namespace std;
typedef long long  LL;
char bb[1000001];
LL a,m;
LL phi(LL n)
{
    LL rea=n;
    for(int i=2;i*i<=n;i++){
        if(n%i==0){
            rea=rea-rea/i;
            while(n%i==0)
                n/=i;
        }
    }
    if(n>1)
        rea=rea-rea/n;
    return rea;
}
LL quick_mod(LL a,LL b,LL m)
{
    LL ans=1;
    a%=m;
    while(b){
        if(b&1){
            ans=ans*a%m;
            b--;
        }
        b>>=1;
        a=a*a%m;
    }
    return ans;
}
int main()
{
    while(~scanf("%lld",&a)){
        scanf("%s",bb);
        scanf("%lld",&m);
        LL t=phi(m);
        int l=strlen(bb);
        LL b=0;
        for(int i=0;i<l;i++){
            b=b*10+bb[i]-'0';
            while(b>=t)
                b-=t;
        }
        b+=t;
        printf("%lld\n",quick_mod(a,b,m));
    }
    return 0;
}