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; }