C ++ Pi近似程序 [英] C++ Pi approximation program

问题描述

我只是学习C ++，所以我已经开始做一个简单的程序来近似pi的价值，使用系列：Pi ^ 6/960 = 1 + 1/3 ^ 6 + 1/5 ^ 6。这是我的代码：

  / *  - ----------------------------------------- 
 *作者：* 
 *程序名称：Pi计算器* 
 *程序函数：使用迭代* 
 *过程计算pi到16 * 
 *小数位* 
 * --- --------------------------------------- * / 
 #include< iostream> ; 
 #include< iomanip> 
 #include< cmath> 
 using namespace std; 
 double pi_approximation = 0; // pi 
的近似值double iterative = 3; //用于增加准确性的数字
 double sum = 1; //累积临时总计
 
 int main（）
 {
 while（true）//无限循环
 {
 sum = sum + pow迭代，-6）; //在系列中的下一步
 iteration = iteration + 2; // increment iteration 
 pi_approximation = pow（（sum * 960），（1/6））; //解方程为pi 
 cout<< setprecision（20）<< pi_approximation<< \\\
; // print pi to maximum precision allowed with a double 
} 
} 
  

代码似乎工作正常（变量'sum'和'iteration'正确增加）到这里：

  pi_approximation = pow（（sum * 960），（1/6））; //解决pi 
  

的方程，因为某些原因'pi_approximation'保留其值为1，这样打印到'cout'的文本是1。

解决方案

问题是整数除法：

（1/6）将返回0.我相信你知道，0的幂的任何值为1。 >

对于浮点除法，更改为（（double）1 /（double）6）或 （1.0 / 6.0）。

I'm just learning C++ and so I have started to make a simple program to approximate the value of pi, using the series: Pi ^ 6 / 960 = 1 + 1 / 3 ^ 6 + 1 / 5 ^ 6... and so on continuing with denominators of odd numbers to the power of 6. Here is my code:

/*-------------------------------------------
 *  AUTHOR:                                 *
 *  PROGRAM NAME: Pi Calculator             *
 *  PROGRAM FUNCTION: Uses an iterative     *
 *      process to calculate pi to 16       *
 *      decimal places                      *
 *------------------------------------------*/
#include <iostream>
#include <iomanip>
#include <cmath>
using namespace std;
double pi_approximation = 0; // the approximated value of pi
double iteration = 3; // number used that increases to increase accuracy
double sum = 1; // the cumulative temporary total

int main ()
{
    while (true) // endlessly loops
    {
        sum = sum + pow(iteration,-6); // does the next step in the series
        iteration = iteration + 2; // increments iteration
        pi_approximation = pow((sum * 960),(1 / 6)); // solves the equation for pi
        cout << setprecision (20) << pi_approximation << "\n"; // prints pi to maximum precision permitted with a double
    }
}

The code seems to work fine (both the variables 'sum' and 'iteration' increase correctly) up to this line here:

pi_approximation = pow((sum * 960),(1 / 6)); // solves the equation for pi

as for some reason 'pi_approximation' retains its value of 1 and as such the text printed to 'cout' is "1".

解决方案

The problem is integer division:

(1 / 6) will return 0. As I am sure you know, anything to the power of 0 is 1.

For floating point division, change to ((double)1 / (double)6) or (1.0 / 6.0).

这篇关于C ++ Pi近似程序的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！