用递归函数求m的n次方

Answer 1

#include <stdio.h>
#include <assert.h>
//note: 只能处理n是正整数的情况
float f(float m, int n)
{
assert(n>=0);
if(n == 0) return 1.0;
if(n == 1) return m;
return m * f(m, n-1);
}

int main()
{
float m;
int n;
printf("input m: ");
scanf("%f", &m);
printf("input n: ");
scanf("%d", &n);

if(n<0) printf("n error. must > 0");

printf("m pow n is %f\n", f(m, n));
return 0;
}

Answer 2

#include <iostream>
using namespace std;

double fun(double m, int n)
{
if (n == 0)
return 1;
else
return m*fun(m, n-1);
}
int main()
{
cout << fun(1.1,2) << endl;
}

// fun函数就是用的递归，main里面做了示范，m可以用小数哦~ ，如果要看c版本的，就在下面
#include <stdio.h>

double fun(double m, int n)
{
if (n == 0)
return 1;
else
return m*fun(m, n-1);
}
int main()
{
printf("%f\n", fun(5, 3));
}