[FOJ 1752] A^B mod C

Problem 1752 A^B mod C

Accept: 750    Submit: 3205
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,B,C<2^63).

 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

 

二分快速幂模板题

#include <iostream>

#include <cstdio>

using namespace std;

#define ll __int64



ll quickadd(ll a,ll b,ll c)    //运用快速幂的思想快速加，这样就不会溢出

{

    ll ret=0;

    while(b)

    {

        if(b&1)

        {

            ret+=a; 

            if(ret>=c) ret-=c; //这样比直接取模（ret%=c）更快

        }

        a<<=1;

        if(a>=c) a-=c;

        b>>=1;

    }

    return ret;

}

ll quickpow(ll a,ll b,ll c)

{

    ll ret=1;

    while(b)

    {

        if(b&1) ret=quickadd(a,ret,c);

        a=quickadd(a,a,c);

        b>>=1;

    }

    return ret;

}

int main()

{

    ll a,b,c;

    while(scanf("%I64d%I64d%I64d",&a,&b,&c)!=EOF)

    {

        printf("%I64d\n",quickpow(a%c,b,c));

    }

    return 0;

}