快速幂（递归，循环）

/*
    calculate a num y which is base x to exp n
    2^n=2*2^n-1 (n is an odd)
    2^n=2^(n/2)*2^(n/2) (n is an even)
    judge method:
    if:b&1 <=> b%2==1 (b is an odd)
    else:!(b&1) <=> b%2==0 (b is an even)
*/
#include <stdio.h>
//recursive
long long pow1(int a,int b){
    if(b==0)
        return 1;
    if(b&1){
        return a*pow(a,b-1);
    }else{
        long long mul;
        mul=pow(a,b/2);
        return mul*mul;
    }
}
//loop
long long pow2(int base, int exp)
{
    long long result = 1;
    while(1){
        if (exp & 1)
            result *= base;
        exp >>= 1;
        if (!exp)
            break;
        base *= base;
    }
    return result;
}

int main(){
    int base,exp;
    long long y;
    
    printf("enter a base and its exp:\n");
    while(~scanf("%d %d",&base,&exp)){
        y=pow1(base,exp);
        printf("%d^%d=%ld\n",base,exp,y); 
    }
    return 0;
}