二分快速幂 [FOJ ] 1752

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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

Source

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;
}