快速幂

long long quick_pow(int a,int n)  
{  
    long long result;  
  
    if(n == 1) return a;  
    if(!n)    return 1;  
    long long mid = quick_pow(a,n/2);  
    if(n%2) result = mid*mid%modx*a%modx;  
    else    result = mid*mid%modx;  
    return result;  //zzzzz
}