算法练习-4-开根号-递归/非递归方式

 1 #include <stdio.h>
 2 
 3 float fabs(float x)
 4 {
 5     return x>=0 ? x:(-1)*x;
 6 }
 7 
 8 float sqrt1(float a,float p,float e)//求根号a,p是根号a的大致近似(可以令p=a/2),e是结果允许误差(取0.00001)
 9 {
10     while(fabs(p*p-a)>=e)
11         p=(p+a/p)/2;
12     return p;
13 }
14 
15 float sqrt2(float a,float p,float e)//递归方式
16 {
17     if(fabs(p*p-a)<e)
18         return p;
19     else
20         return sqrt2(a,(p+a/p)/2,e);
21 }
22 
23 int main(int argc, char const *argv[])
24 {
25     printf("%f
", sqrt1(88,44,0.0001));
26     printf("%f
", sqrt2(44,22,0.0001));
27     return 0;
28 }

运行结果:

原文地址:https://www.cnblogs.com/gilgamesh-hjb/p/11924177.html