我正在尝试编写一个程序,通过随机数生成器根据蒙特卡洛方法估算 Pi。我试图在 1、2、3、4、5 和 6 位数字的精度内估算 Pi,并让程序在屏幕上打印出需要多少点才能达到 Pi 的 .1 位数字,然后是 Pi 的 .01 位数字依此类推,直到 Pi 的 .000001 位。我允许用户输入他们想要运行的试验数量,因此它将打印“试验 1、2、3、4”等,其中包含我上面列出的所有信息。我被困在最后一点,那就是让它通过计算循环回来(它不会打印超过试验 1)。虽然我没有收到程序已终止的消息,但我无法判断是我的 while 循环失败还是我的嵌套 for 循环。请帮忙! :)
我尝试过切换 for 循环以及尝试不同的可变 if 语句。这是我最接近的运行方式,除了允许用户运行多个试验之外。
#include "pch.h"
#include <iostream> //need this by default for cin
#include <math.h> //includes math functions
#include <cmath> //includes basic math
#include <cfloat> //includes floating point numbers
#include <iomanip> //includes setprecision for decimal places
#include <cstdlib> //needed for rand and srand functions
#include <ctime> //needed for time function used to seed generator
using namespace std;
int main()
{
cout << "The purpose of this program is to estimate pi using the monte
carlo method and a random number generator" << endl << endl;
unsigned seed = time(0);
srand(seed);
float radius;
int trialcount = 0;
int trials;
float accuracy;
const float pi = 3.14159265;
float randpi = 0;
int squarecount = 0;
int circlecount = 0;
float x;
float y;
int n;
cout << "The value of PI can be found as the ratio of areas of a circle of radius r located within a square of side 2r" << endl;
cout << "This program runs a MonteCarlo Simulation that generates numbers located randomly within a square" << endl;
cout << "The count of values within the square and the count of numbers within the circle approximate their areas" << endl;
cout << "An input value of radius determines the size of the circle and square" << endl;
cout << "The user specifies how many trials or test runs are desired" << endl << endl;
cout << "The true value of PI to 8 decimal places is 3.14159265" << endl << endl;
cout << "Input a value for radius: ";
cin >> radius;
cout << endl;
cout << "How many trials would you like? ";
cin >> trials;
cout << endl << endl;
cout << "Square count gives the Total number of random samples (they are within the square)" << endl;
cout << "Circle count gives the number of random samples that also fall within the circle" << endl << endl;
while (trialcount != trials)
{
accuracy = .1;
cout << "Trial " << trialcount + 1 << endl;
cout << "Accuracy \t\t" << "Square Count \t\t" << "Circle Count \t\t" << "Pi" << endl << endl;
for (int j = 0; randpi >= pi - accuracy || randpi <= pi + accuracy; j++)
{
cout << setprecision(6) << fixed << accuracy << " \t\t" << squarecount << " \t\t" << circlecount << " \t\t" << randpi << endl << endl;
accuracy = accuracy / 10;
for (int i = 0; randpi >= pi + accuracy || randpi <= pi - accuracy; i++)
{
x = (float)(rand());
x = (x / 32767) * radius;
y = (float)(rand());
y = (y / 32767) * radius;
squarecount++;
if ((x * x) + (y * y) <= (radius * radius))
{
circlecount++;
}
randpi = float(4 * circlecount) / squarecount;
}
}
trialcount++;
}
}
最佳答案
我看到的问题:
问题1
第一个 for
循环没有任何意义。如果您想确保使用 0.1、0.01、0.001 等精度,您只需要一个简单的 for
循环。应执行以下操作:
for ( int j = 0; j < 6; ++j )
{
...
}
问题2
x
和 y
值计算不正确。您要确保它们的值小于或等于 radius
。但是,当您使用:
x = (float)(rand());
x = (x / 32767) * radius;
y = (float)(rand());
y = (y / 32767) * radius;
不保证它们小于或等于radius
。它们将比radius
更频繁。你需要使用
x = (float)(rand() % 32768);
x = (x / 32767) * radius;
y = (float)(rand() % 32768);
y = (y / 32767) * radius;
问题3
您需要在内部 for
的每次迭代中重置 randpi
、squarecount
和 circlecount
的值> 循环。否则,您的计算将受到上一次迭代计算的影响。
外层 for
循环必须以:
for (int j = 0; j < 6; j++)
{
accuracy /= 10;
randpi = 0;
squarecount = 0;
circlecount = 0;
问题4
内部 for
循环必须被限制为只能运行一定次数。如果由于某种原因没有达到准确性,您要确保不会溢出 i
。例如:
int stopAt = (INT_MAX >> 8);
for (int i = 0; (randpi >= pi + accuracy || randpi <= pi - accuracy) && i < stopAt; i++)
对于使用 32 位 int
的机器,这是当今实践中最常见的,您不会运行循环超过 0x7FFFFF
( 8388607
十进制)次。
这是您代码中的核心问题。您的计算有时不会收敛,并且您无法确保在一定次数的循环迭代后退出。
进一步改进
您不需要将 radius
作为程序中的变量。您可以将 x
和 y
计算为:
x = (float)(rand() % 32768);
x = (x / 32767);
y = (float)(rand() % 32768);
y = (y / 32767);
并将检查这是否是圆内的点的逻辑更改为
if ((x * x) + (y * y) <= 1.0 )
您还应该尝试仅在需要它们的范围内定义变量。这将确保您最终不会使用上一次迭代运行中的陈旧值。
修改后的程序
以下修改后的程序对我有用。
#include <iostream> //need this by default for cin
#include <math.h> //includes math functions
#include <cmath> //includes basic math
#include <cfloat> //includes floating point numbers
#include <iomanip> //includes setprecision for decimal places
#include <cstdlib> //needed for rand and srand functions
#include <ctime> //needed for time function used to seed generator
#include <climits>
using namespace std;
int main()
{
cout << "The purpose of this program is to estimate pi using the monte "
"carlo method and a random number generator" << endl << endl;
unsigned seed = time(0);
srand(seed);
int trialcount = 0;
int trials;
float accuracy;
const float pi = 3.14159265;
cout << "The value of PI can be found as the ratio of areas of a circle of radius r located within a square of side 2r" << endl;
cout << "This program runs a MonteCarlo Simulation that generates numbers located randomly within a square" << endl;
cout << "The count of values within the square and the count of numbers within the circle approximate their areas" << endl;
cout << "An input value of radius determines the size of the circle and square" << endl;
cout << "The user specifies how many trials or test runs are desired" << endl << endl;
cout << "The true value of PI to 8 decimal places is 3.14159265" << endl << endl;
cout << endl;
cout << "How many trials would you like? ";
cin >> trials;
cout << endl << endl;
cout << "Square count gives the Total number of random samples (they are within the square)" << endl;
cout << "Circle count gives the number of random samples that also fall within the circle" << endl << endl;
while (trialcount != trials)
{
accuracy = 0.1;
cout << "Trial " << trialcount + 1 << endl;
cout << "Accuracy \t\t" << "Square Count \t\t" << "Circle Count \t\t" << "Pi" << endl << endl;
for (int j = 0; j < 6; j++)
{
accuracy /= 10;
float randpi = 0;
int squarecount = 0;
int circlecount = 0;
int stopAt = (INT_MAX >> 8);
for (int i = 0; (randpi >= pi + accuracy || randpi <= pi - accuracy) && i < stopAt; i++)
{
float x = ((float)(rand() % 32768) / 32767);
float y = ((float)(rand() % 32768) / 32767);
squarecount++;
if ((x * x) + (y * y) <= 1.0 )
{
circlecount++;
}
randpi = float(4 * circlecount) / squarecount;
}
cout << setprecision(8) << fixed << accuracy << " \t\t" << squarecount << " \t\t" << circlecount << " \t\t" << randpi << endl << endl;
}
trialcount++;
}
}
在 https://ideone.com/laF27X 查看它的工作情况.
关于c++ - 用蒙特卡洛方法估计 Pi,循环似乎提前终止,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/55527800/