【数据结构与算法分析】2.３ 求Ｘ的Ｎ次幂

问题描述：

这还描述啥

解决思路：

递归，递推，以及logn的解法，第三种解法中，若ｎ为偶数，ｘ*n = x*(n/2)*x*(n/2);若ｎ为奇数,x*n=x*(n/2)*x*(n/2)*x;

算法实现：

#include<stdio.h>double recursion(double x, int n);double recurrence(double x, int n);double logn(double x, int n);int main() {        double x;        int n;        while(scanf("%lf%d",&x,&n)){                printf("%lf\n",recursion(x,n));                printf("%lf\n",recurrence(x,n));                printf("%lf\n",logn(x,n));        }        return 0;}double recursion(double x,int n) {        if(n == 0)                return 1;        if(n == 1)                return x;        return recursion(x,n-1)*x;}double recurrence(double x,int n) {        if(n == 0)                return 1;        if(n == 1)                return x;        int i;        int result = x;        for(i = 2;i<=n;i++){                result = result * x;        }        return result;}double logn(double x,int n) {        if(n == 0)                return 1;        if(n == 1)                return x;        if(n % 2 == 0)                return logn(x*x,n/2);　　 else                return logn(x*x,n/2)*x;｝