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Description: 单片机上的开方程序
比传统的牛顿迭代法要快-microcontroller on the prescribing process than the traditional Newton iteration to be fast
Platform: |
Size: 1819 |
Author: 王宝成 |
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Description: 单片机上的开方程序
比传统的牛顿迭代法要快-microcontroller on the prescribing process than the traditional Newton iteration to be fast
Platform: |
Size: 2048 |
Author: 王宝成 |
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Description: Its functions are: n order within (excluding n) At the same time, 3 and 7 can be integral to all natural and a few of the square root of s, and function as a value to return, the final result s output to file out.dat China.
Platform: |
Size: 1024 |
Author: 食肉鸟 |
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Description: 利用平方根法解线性方程组源程序及流程图。-Solving the use of the square root of linear equations and the flow chart of the source code.
Platform: |
Size: 5120 |
Author: kelly |
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Description: vhdl源代码,可以开16比特的平方根，算法简单，速度快-this is a vhdl code for square root
Platform: |
Size: 1024 |
Author: lei |
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Description: import java.io.* public class SquareRoot { public static void main ( String [ ] args )
{ System.out.print("Enter a positive value: ") double value = 0 try { BufferedReader in
= new BufferedReader(new InputStreamReader(System.in)) String inputLine = in.readLine()
value = Double.valueOf(inputLine).doubleValue() } catch (Exception exc)
{ System.out.println("Input error!") return } if (value < 0) System.out.println("Input
error!") else { double root = 1 double pre = 0 do { pre = root root = (value / root +
root) / 2 } while ((pre - root > 0.000001) || (root - pre > 0.000001)) System.out.println("The
root of " + value + " is " + root) } } }
你容易理解这个程序吗？在计算机上运行该程序，看看该程序完成什么功能？尝试利用
分行、缩进、空行、空格等为该程序重新排版，并加上合适的注释。（提示：所谓程序的排
版不同于WPS 2000 那种对文档的排版，后者主要是对字体、字形、字号、字间距、行间距等格式的处理。
-Import Java. IO.* Public class SquareRoot {public static void main (String [] args)
{System. Print out. (" Enter a positive value: ") Double value = 0 Try {BufferedReader in
= new BufferedReader (new InputStreamReader (System..)) String inputLine = in. ReadLine ()
Value = Double. ValueOf (inputLine). DoubleValue () } catch (Exception exc)
{System. Out. Println (" Input error! ") Return } if value (< 0) System. Println. (" Input out
Error!!!!!") else {double root = 1 double pre = 0 do {pre = root root = (value/root+
Root)/2 } while ((pre-root > 0.000001) | | (root-pre > 0.000001)) System. Out. Println (The "
Root of "value++" is "+ root) }}}
You easy to understand this program? On the computer to run the program, check to see if the program finish what function? Try to take advantage of
Branch, indentation, empty line for the program, such as space to typesetting, and add the right note. (hint: the so-called program row
Version is different from that of d
Platform: |
Size: 1024 |
Author: 何俊乐 |
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Description: 迭代法求平方根。根据输入数字，求其平方根-Solute square root with iterative method according to input
Platform: |
Size: 205824 |
Author: LIU Tche |
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