Question

这段代码中有一部分我需要帮助。该代码计算板上的稳态温度分布。我得到的提示声明：

您应该继续迭代，直到阵列中没有任何单元格变化超过0.1度，计算每次迭代时所有内部单元格的温度。您的程序应该监视阵列中任何单元格的最大变化，以确定何时停止重复。

我被卡住了！我目前正在使用while循环来获得正确的答案，但我无法弄清楚如何让它完成我在提示中提出的问题。任何帮助将非常感激！

#include <iostream>
#include <string>
#include <iomanip>
#include <fstream>

using namespace std;

const int SIZE = 20;
const double HEAT = 100;
const double EDGES = 0;
const int SURROUNDING = 4;
const int STABLE = .1;

// Initializes the first array
void begining_plate ( double plate[][SIZE]) {}


// Calculates one ideration of the steady-state distribution
double average_temp_calc(double plate[][SIZE], int a, int b) {}

// Prints the array
void print_plate( double plate[][SIZE]) {

// Exports the array to a .csv file
bool send_plate_info(double plate[][SIZE])


int main() {


    // Part 1 - Initialize and Print 2D Array
    cout << "Here is the initial heat: " << endl;
    double plate[SIZE][SIZE];
    begining_plate(plate);
    print_plate(plate);


    // Part 2 - Update Elements once
    double plate_saved[SIZE][SIZE];

    cout << "\nHere is the first run of the averaged plate.\n";
    for (int a = 0; a < SIZE; a++) {
        for (int b = 0; b < SIZE; b++) {
            if (a != 0 && a != SIZE - 1 && b != 0 && b != SIZE - 1) {
                plate_saved[a][b] = plate[a][b];
                plate[a][b] = average_temp_calc(plate, a, b);
            }
            else {
                plate_saved[a][b] = plate[a][b];
                plate[a][b] = plate[a][b];
            }
        }
    }
    print_plate(plate);
    cout << endl << endl;

    // Part 3 - Repeat update until stalbe

   ******* HERE IS THE PART I NEED HELP WITH **********

    int count = 0;
    int stable = 150;
    while (count < stable) {
        for (int a = 0; a < SIZE; a++) {
            for (int b = 0; b < SIZE; b++) {
                if (a != 0 && a != SIZE - 1 && b != 0 && b != SIZE - 1) {
                    plate_saved[a][b] = plate[a][b];
                    plate[a][b] = average_temp_calc(plate, a, b);
                }
                else {
                    plate_saved[a][b] = plate[a][b];
                    plate[a][b] = plate[a][b];
                }
            }
        }
    count++;
}
// Part 4 - Using Excel to Display Results

        if (send_plate_info(plate))
        {
            cout << "File wrote correctly\n";
        }
        else
        {
            cout << "The file did not write!\n";
        }



    system("pause");
    return 0;
}

Answer 1

应该对不稳定点的数量而不是循环次数进行计数：

for (int a = 0; a < SIZE; a++) {
    for (int b = 0; b < SIZE; b++) 
    {
        if (a != 0 && a != SIZE - 1 && b != 0 && b != SIZE - 1) {
            plate_saved[a][b] = plate[a][b];
            plate[a][b] = average_temp_calc(plate, a, b);
        }
        else {
            plate_saved[a][b] = plate[a][b];
            plate[a][b] = plate[a][b];
        }
        if ( abs(plate_saved[a][b]-plate[a][b]) > STABLE )
            ++count;
    }

这样你只计算不稳定点，并在没有任何点时停止：

do
{...}
while (count>0);

修改

请注意，必须在每次迭代开始时重置不稳定点的计数器，以便解决方案看起来像这样：

do { count = 0; for (int a = 0; a < SIZE; a++) { for (int b = 0; b < SIZE; b++) { double plate_saved = plate[a][b]; // Compute new value of plate[a][b] if ( fabs(plate_saved-plate[a][b]) > STABLE ) ++count; } } while (count>0);

我需要帮助结束更新2D数组的循环

1 个答案: