C微分方程

Question

我似乎想不出如何解决这个问题。程序使用一秒钟来表示无限循环中运行期间的一年。在循环中，我需要解这个方程。

微分方程: dP/dt = rP(1-P/K) - hP

其中P(t)是随时间t(以年为单位)变化的人口。

这些符号的含义是: dP/dt是瞬时人口增长率(以年增长率为单位)，r是自然生殖增长率(以每年部分增长)K是人口承载能力(环境所能支持的最大人口)h是种群被剔除的收获率(作为每年的部分损失)。

取决于运行时秒(每秒钟代表一年)，下面是正确的示例

Simulation run time minute = 0; second = 3; millisec = 156
Year of simulation = 3
Rate of population change = 5180.209961

正如您所看到的，在执行我的代码时，我无法计算出正确的人口变化。这取决于秒，所以它会变化。

#define _CRT_SECURE_NO_WARNINGS

#include <stdio.h>
#include <stdlib.h>
#include <time.h>

int main(void)
    {
    int option = 0;
    float initialPopulation = 0; //starting population
    float maximumPopulation = 0; //max population
    float harvestingRate = 0;
    float fractionGrowthPopulationRate = 0;
    int minutes = 0;
    int milliseconds = 0;
    char quit;
    int forever = 1;
    float total = 0.0;

    printf("Population Simulation\n");
    printf("1. run the simulation\n");
    printf("2. Quit\n");
    scanf("%d", &option);

    if (option == 1) {
        printf("ENTER SIMULATION PARAMETERS\n");
        printf("Initial population (typically 900000)?");
        scanf("%d", &initialPopulation);
        printf("Maximum population the environment can support (typically 1000000)? \n");
        scanf("%d", &maximumPopulation);
        printf("Initial harvesting rate (fraction per year - 0 for no\n harvesting)? ");
        scanf("%f", &harvestingRate);
        printf("Natural fractional growth population rate (typically 0.2 per year)?\n");
        scanf("%f", &fractionGrowthPopulationRate);
        printf("\n");
    }
    else if (option == 2) {
        return 0;
    }

    while (forever != 'q') {
        printf("starting simulation\n");

        clock_t t;
        t = clock();
        //confused on how to code the formula
        total = fractionGrowthPopulationRate * initialPopulation* (1 - initialPopulation / maximumPopulation) - harvestingRate*initialPopulation;

        t = clock() - t;
        int seconds = ((int)t) / CLOCKS_PER_SEC; // in seconds

        printf("Simulation run time minute = %d; second = %d; millisec = %d\n", minutes, seconds, milliseconds);

         printf("*****************************************************************\n");
        printf("Year of simulation = %d\n", seconds); //every seconds equals 1 year
        printf("Rate of population change = %f\n", total); //total should be calculated 
        printf("*****************************************************************\n");
        printf("Press w / e to increase / decrease harvesting rate.\n");
        printf("Current harvesting rate : %f\n", harvestingRate);
        printf("Press p / o to increase / decrease Max population supported.\n");
        printf("Current Max population : %d\n", maximumPopulation);
        printf("Press k / l to increase / decrease growth rate.\n");
        printf("Current growth rate : %f\n", fractionGrowthPopulationRate);
        printf("Press q to quit.\n");

        fflush(stdin);
        scanf("%c", &forever);
    }
    return 0;
  }

Answer 1

以下代码：

1. 干净地编译
2. 将功能分离为单独的可调用函数。
3. 包含事件的适当顺序。
4. 暗示模拟必须考虑到先前迭代的结果。
5. 正确检查用户输入错误
6. 您仍然需要完成‘processParameters()’函数
7. 在processParameters()中，确保包含达到最大人口时的代码。

以下是拟议的守则：

#define _CRT_SECURE_NO_WARNINGS
#include <stdio.h>   // printf(), scanf()
#include <stdlib.h>  // exit(), EXIT_FAILURE
//#include <time.h>
#include <string.h>  // strerror()
#include <errno.h>   // errno

// prototypes
void inputParameters( void );
void processParameters( void );
void displayResults( void );
int  handleMenu( void );

// global data
float initialPopulation = 0.0f; //starting population
float maximumPopulation = 0.0f; //max population
float harvestingRate = 0.0f;
float fractionGrowthPopulationRate = 0.0f;
float total = 0.0f;

int minutes = 0;
int seconds = 0;
int milliseconds = 0;


int main(void)
{
    int quit = 0;

    while( !quit )
    {
        int option = handleMenu();

        switch( option )
        {
            case 1:
                inputParameters();
                processParameters();
                displayResults();
                break;

            case 2:
                quit = 1;
                break;

            default:
                printf( "invalid option: %d try again\n", option );
                break;
        } // end switch
    } // end while

    return 0;
} // end function: main


void inputParameters()
{
    printf("ENTER SIMULATION PARAMETERS\n");

    printf("Initial population (typically 900000)?");
    if( 1 != scanf("%f", &initialPopulation) )
    {
        fprintf( stderr, "input of initial population failed due to: %s\n", strerror( errno ) );
        exit( EXIT_FAILURE );
    }

    printf("Maximum population the environment can support (typically 1000000)? \n");
    if( 1 != scanf("%f", &maximumPopulation) )
    {
        fprintf( stderr, "input of maximum population failed due to: %s\n", strerror( errno ) );
        exit( EXIT_FAILURE );
    }

    printf("Initial harvesting rate (fraction per year - 0 for no harvesting)? ");
    if( 1 != scanf("%f", &harvestingRate) )
    {
        fprintf( stderr, "input of omotoa; jarvestomg rate failed due to: %s\n", strerror( errno ) );
        exit( EXIT_FAILURE );
    }

    printf("Natural fractional growth population rate (typically 0.2 per year)?\n");
    if( 1 != scanf("%f", &fractionGrowthPopulationRate) )
    {
        fprintf( stderr, "input of initial population failed due to: %s\n", strerror( errno ) );
        exit( EXIT_FAILURE );
    }

    printf("\n");
} // end function: inputParameters


void displayResults()
{
    printf("Simulation run time minute = %d; second = %d; millisec = %d\n",
            minutes, seconds, milliseconds);

    printf("*****************************************************************\n");
    printf("Year of simulation = %d\n", seconds); //every seconds equals 1 year
    printf("Rate of population change = %f\n", total); //total should be calculated
    printf("*****************************************************************\n");
    printf("Press w / e to increase / decrease harvesting rate.\n");
    printf("Current harvesting rate : %f\n", harvestingRate);
    printf("Press p / o to increase / decrease Max population supported.\n");
    printf("Current Max population : %f\n", maximumPopulation);
    printf("Press k / l to increase / decrease growth rate.\n");
    printf("Current growth rate : %f\n", fractionGrowthPopulationRate);
    printf("Press q to quit.\n");
} // end function: displayResults


int handleMenu()
{
    int option = 0;

    printf("\nPopulation Simulation\n");
    printf("1. run the simulation\n");
    printf("2. Quit\n");

    if( 1 != scanf("%d", &option) )
    {
        fprintf( stderr, "scanf failed due to: %s\n", strerror( errno ) );
        exit( EXIT_FAILURE );
    }

    return( option );
} // end function: handleMenu


void processParameters()
{
    int simulationDuration = 0;

    printf( "how long is simulation to run:");
    if( 1 != scanf( "%d", &simulationDuration ) )
    {
        fprintf( stderr, "input of simulation Duration failed due to: %s\n", strerror( errno ) );
        exit( EXIT_FAILURE );
    }

    printf("\nstarting simulation\n");

    total = initialPopulation;
    for( int i=0; i<simulationDuration; i++ )
    {
        //confused on how to code the formula
        total = fractionGrowthPopulationRate * total * (1 - total / maximumPopulation) - harvestingRate*initialPopulation;

        seconds = i; // in seconds
    }

    printf( "\nsimulation complete\n");
} // end function: processParameters