Question

我正在制作一个解决积分的程序。到目前为止，我设法制作了一个简单的函数积分（a，b）f（x）dx。现在我需要使用MPI在并行编程中制作多线程程序。这就是我的看法。需要为a，b，n创建一个元素数组（我希望我的函数解析得多么精确），然后为一个线程获取每个函数元素的组合。问题：我是否正确理解了这个问题？这是我的函数代码，我的史诗失败试图将其更改为数组。请帮忙！

   #include "stdafx.h"
#include <iostream>
#include "conio.h"
#include <stdio.h>
using namespace std;

void WaitForEnter(void) 
{ 
printf("Press Enter to continue: "); 
fflush(stdout); 
while ( _getch() != '\n' ) ; 
}

double fun (double x){return x;}//changing f(x)dx

double sumIntegral(double lowBound, int n, double dx)
{
    double rectangleArea;
    double cumSum=0;

        for (int i=0; i<n; i++)
            {
                double xi = lowBound+i*dx;//x1,x2...
                double funValue = fun(xi);
                double rectangleArea =  dx*funValue;
                 cumSum += rectangleArea;
            }

    return cumSum;
}

int main()
{
    double lowBound = 4;//a
    double upBound = 7 ;//b
    int n = 5;//how precise is solution
    double dx;

        dx = (upBound-lowBound)/n;

    double result = sumIntegral(lowBound , n, dx);
    cout<<"rez je = "<<result;
    WaitForEnter();


    return 0;
}

//epic fail

#include "stdafx.h"
#include <iostream>
#include "conio.h"
#include <stdio.h>
using namespace std;

void WaitForEnter(void) 
{ 
printf("Press Enter to continue: "); 
fflush(stdout); 
while ( _getch() != '\n' ) ; 
}

double fun (double x){return x;}//menja funkciju

double sumIntegral(double lowBound[], int n, double dx[],int numBound)
{
    double rectangleArea;
    double cumSum;
    double sumIntegralAll[] = {NULL};

    for(int j=0;j<numBound;j++)
    {
        cumSum=0;
        for (int i=0; i<n; i++)
            {
                double xi = lowBound[j]+i*dx[j];//x1,x2...
                double funValue = fun(xi);
                rectangleArea = dx[j]*funValue;
                cumSum+=rectangleArea;
                cout<<cumSum;
            }
        //sumIntegralAll[j]=cumSum;
    }

    for(int i=0;i<numBound;i++){cout<<sumIntegralAll;}//pise
    return cumSum;
}

int main()
{
    int numBound = 2;//broj do 
    double lowBound[] = {4,4};//a
    double upBound[] = {7,7} ;//b
    int n = 5;//broj intervala, preciznost
    double dx[]={NULL};

    for(int i=0;i<=numBound;i++){
        dx[i] = (upBound[i]-lowBound[i])/n;
    }

    double result = sumIntegral(lowBound , n, dx,numBound);
    //cout<<"rez je = "<<result;
    WaitForEnter();


    return 0;
}

c ++ - 多线程（积分函数）

0 个答案: