如何计算代码的并行和串行部分？

Question

我有一个使用OpenMP和C ++编写的代码，我想知道其中有多少是串行和并行的。 我读了一些加速比定律和公式，但是我不知道如何计算并行代码和串行代码的百分比
并且如果有人可以引导我使用一些免费软件和网站来帮助我得出1-8核的加速比

#include <iostream>
#include <vector>
#include <cstdlib>
#include <ctime>
#include <cmath>
#include <omp.h>

using namespace std;

int userInput(int num) {
    const int len = num;
    return len;
}


void main() {

    long double dataSetOneSum = 0, dataSetTwoSum = 0, multiplicationSum = 0, setOnePowSum = 0, setTwoPowSum = 0;
    long double correlationCoefficient;

    int i;
    int num;
    cout << "enter size of data sets: " << endl;
    cin >> num;
    int length = userInput(num);

    double wtime;
    long double* dataSetOne = new long double[length];
    long double* dataSetTwo = new long double[length];
    long double* dataSetMultiplication = new long double[length];
    long double* dataSetOnePow = new long double[length];
    long double* dataSetTwoPow = new long double[length];


    srand(time(0));

    omp_set_num_threads(3);
    wtime = omp_get_wtime();



#pragma omp parallel for
        for (i = 0; i < length; i++) {
            dataSetOne[i] = 1 + (rand() % 99);
            cout << "thread num: " << omp_get_thread_num() << endl;
        }
#pragma omp parallel for
        for (i = 0; i < length; i++) {
            dataSetOneSum = dataSetOneSum + dataSetOne[i];
            //cout << "thread num: " << omp_get_thread_num() << endl;
        }
        cout << "the sum of the first data set :" << fixed << dataSetOneSum << endl;

#pragma omp parallel for
        for (i = 0; i < length; i++) {
            dataSetTwo[i] = 1 + (rand() % 99);
            //cout << "thread num: " << omp_get_thread_num() << endl;
        }
#pragma omp parallel for
        for (i = 0; i < length; i++) {
            dataSetTwoSum = dataSetTwoSum + dataSetTwo[i];
            //cout << "thread num: " << omp_get_thread_num() << endl;
        }

        cout << "the sum of the second data set :" << dataSetTwoSum << endl;;
#pragma omp parallel for
        for (i = 0;i < length;i++) {
            dataSetMultiplication[i] = dataSetOne[i] * dataSetTwo[i];
            multiplicationSum = multiplicationSum + dataSetMultiplication[i];
            //cout << "thread num: " << omp_get_thread_num() << endl;
        }

        cout << "the sum of the two data sets multiplied  : " << multiplicationSum << endl;
#pragma omp parallel for
        for (i = 0;i < length;i++) {
            dataSetOnePow[i] = pow(dataSetOne[i], 2);
            setOnePowSum = setOnePowSum + dataSetOnePow[i];
            //cout << "thread num: " << omp_get_thread_num() << endl;
        }

        cout << "the sum of the first data set numbers raised to the power of 2  : " << setOnePowSum << endl;
#pragma omp parallel for
        for (i = 0;i < length;i++) {
            dataSetTwoPow[i] = pow(dataSetTwo[i], 2);
            setTwoPowSum = setTwoPowSum + dataSetTwoPow[i];
            //cout << "thread num: " << omp_get_thread_num() << endl;
        }

        cout << "the sum of the second data set numbers raised to the power of 2  : " << setTwoPowSum << endl;

        correlationCoefficient = ((length * multiplicationSum) - (dataSetOneSum * dataSetTwoSum)) / (sqrt((length * setOnePowSum) - (pow(dataSetOneSum, 2))) * sqrt((length * setTwoPowSum) - (pow(dataSetTwoSum, 2))));

        cout << "the correlation = " << correlationCoefficient << endl;

        if (correlationCoefficient > 0) {
            cout << "the correlation between data sets is: " << correlationCoefficient << ",  data sets is positively correlated " << endl;
        }
        else if (correlationCoefficient == 0) {
            cout << "the correlation between datasets is: " << correlationCoefficient << ",  data sets is independent" << endl;
        }
        else if (correlationCoefficient < 0) {
            cout << "the correlation between data sets is: " << correlationCoefficient << ",  data sets is negatively correlated " << endl;
        }
        wtime = omp_get_wtime() - wtime;
        cout << "  Elapsed wall clock time = " << wtime << "\n";
        system("pause");
    }