快速幂取余

#include <iostream>

using namespace std;
typedef long long LL;
LL quickmod(LL a, LL b, LL c)
{
    LL ans=1;
    a%=c;
    while(b)
    {
        if(b&1)
        ans=(ans*a)%c;
        b/=2;
        a=(a*a)%c;
    }
    return ans;
}
int main()
{
    LL a,b,c;
    while(cin>>a>>b>>c)
    {
        LL ans=quickmod(a, b, c);
        cout<<ans<<endl;
    }
    return 0;
}