F - A^B mod C （快速幂+优化）

F - A^B mod C
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

#include<iostream>

using namespace std;
//ACM高精度模板综合偶然发现的，收藏一下。https://blog.csdn.net/u013615904/article/details/43373601 

//思路参考博客https://blog.csdn.net/aaaaacmer/article/details/47281441

/*
乘  函数f的含义为计算a*b%c
*/
unsigned long long f(unsigned long long a,
unsigned long long b,
unsigned long long c)
{
    //快速幂中间的乘法会炸，需要优化，用加法取代乘，用减法取代mod。
/*  while(b>0)
    {
        if(b&1)
            ans = (ans%c)*(a%c)%c;
        a = (a%c)*(a%c)%c;
        b>>=1;
    }
    */
    unsigned long long ans = 0;

    while(b)
    {
        if(b&1)
            ans+=a;
        if(ans>=c)
            ans-=c;
        a=a+a;
        if(a>=c)
            a-=c; 
        b>>=1;
    }
    return ans;
}
int main()
{
    unsigned long long a,b,c;
    unsigned long long ans = 1;
    while(cin>>a>>b>>c)
    {
        ans = 1;
//      f();
        a = a%c;
        //快速幂 
        while(b)
        {
            if(b&1) 
                ans = f(ans,a,c);//计算ans*a%c 
            a=f(a,a,c);//计算a*a%c
            b>>=1;
        }
        cout<<ans<<endl;
    }
    return 0;
}