fzoj1759（歐拉定理）

Problem 1759 Super A^B mod C

Accept: 572    Submit: 1919
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 42 10 1000

Sample Output

124

歐拉函數的應用

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

#include<set>
#include<queue>
#include<cmath>
#include<cstdio>
#include<vector>
#include<string>
#include<utility>
#include<cstring>
#include<iostream>
#include<algorithm>
#include<functional>
#define Inf (1<<30)
#define LL long long
#define MOD 1000000009
#pragma comment(linker, "/STACK:102400000,102400000")
using namespace std;
LL a,c;
char b[1000100];
LL phi(LL x)
{
    LL res=x;
    for(LL i=2;i*i<=x;i++)
    {
        if(x%i==0)
        {
            res=res/i*(i-1);
            while(x%i==0)x/=i;
        }
    }
    if(x>1)res=res/x*(x-1);
    return res;
}
LL qpow(LL x,LL y)
{
    LL res=1;
    while(y>0)
    {
        if(y&1)res=(res*x)%c;
        y>>=1;
        x=(x*x)%c;
    }
    return (res+c)%c;
}
int main()
{
    int T;
    while(~scanf("%lld%s%lld",&a,b,&c))
    {
        LL mod=phi(c),L=strlen(b),res=0;
        for(LL i=0;i<L;i++)
            res=(res*10+b[i]-'0')%mod;
        res+=mod;
        printf("%lld\n",qpow(a,res));
    }
    return 0;
}