我寫了下面的代碼,用牛頓法通過逐次逼近找到平方根,但它沒有給我正確的答案。可以請某人解釋它嗎?牛頓法求平方根的邏輯有什麼問題?
#include<stdio.h>
#include<stdlib.h>
#define square(x) x*x
double rootByNewtonApprox(int n);
double improve(double n);
double average(double a,double b);
int goodEnough(double guess);
double guess(int n);
int number;
int main(void)
{
double root;
printf("\nEnter the number you want square root of: ");
scanf("%d",&number);
if(number<0)
number = -1* number;
root = rootByNewtonApprox(number);
printf("\nThe square root of %d is %lf\n",number,root);
return 0;
}
double guess(int n)
{
return n/2;
}
double rootByNewtonApprox(int n)
{
if(goodEnough(guess(n)))
return guess(n);
else
rootByNewtonApprox(improve(guess(n)));
}
double improve(double guess)
{
return average(guess,(number/guess));
}
double average(double a,double b)
{
return ((a+b)/2);
}
int goodEnough(double guess)
{
if(abs(square(guess) - number) <= 0.001)
return 1;
else
return 0;
}
現在,當我給n = 2
它給人的輸出中nan
,當我給n = 9
它告訴segmentation Fault
。
見@ J.S。泰勒的回答如下代碼。你遇到的是整數除法(不允許餘數),所以你的一些函數沒有返回正確的浮點值。 – 2011-04-30 06:19:14