[Quick power] 1

Introduction:

• Common exponentiation:
  \ (A ^ B \) and consecutive multiplying \ (B \) times \ (A \) .

  Obviously, efficiencyVery TMlow

• First, we need to know,
  \ (Y = X ^ X ^ {Y_1 + + ... + y_n Y_2} \)
  For example, when the \ (b = 11 \) , the
  position of the maximum temporary base unit, they must be spelled All number in the range, it is a binary split is the fastest and optimize multiple backpack is also this idea.
  The \ (B \) split into two 229, the \ (b = (1011) _2 = (1 × 2 ^ 3 + 0 × 2 ^ 2 + 1 × 2 ^ 1 + 1 × 2 ^ 0) _ {10 } \)
  so \ (a ^ b = a ^
  {2 ^ 3 + 2 + 1} \) Code:

#include<cstdio>
inline int quick_pow(int A , int B){
    int base = A, ans = 1;
    while(B){
        if(B & 1){//取B的末尾进行算位权
            ans *= base;
        }
        base *= base;//及base=base^2,因为每次次数要按位加一
        B >>= 1;//去掉一位
    }
    return  ans;
}  
int main(){
    int a , b;
    scanf("%d%d",&a , &b);
    printf("%d",quick_pow(a , b));
    return 0;
}