OpenMP求完数

源代码：

#include "stdafx.h"  //必须写在首行，因为其前面的include都会被忽略
#include "omp.h"  
#include <Windows.h>  
#include "time.h" 
#include <iostream>
#include <set>
using namespace std;

　　//串行方式
set<int> FinishedNumber(int n)
{
    set<int> s;
    for (int i = 2; i <= n; i++)
    {
        int maxv = (int)(sqrt(double(i)) + 1);
        int sum = 1;
        for (int j = 2; j<maxv; j++)
        if (i%j == 0)
            sum += j + i / j;
        if (sum == i)
            s.insert(i);
    }
    return s;
}


　　//并行方式
set<int> FinishedNumber(int n, int p)
{
    set<int> s;
    omp_set_num_threads(p);
#pragma omp parallel  
#pragma omp for  
    for (int i = 2; i <= n; i++)
    {
        int maxv = (int)(sqrt(double(i)) + 1);
        int sum = 1;
        for (int j = 2; j<maxv; j++)
        if (i%j == 0)
            sum += j + i / j;
        if (sum == i)
            s.insert(i);
    }
    return s;
}

void Test(int step, int count)
{
    /*int step=100000;
    int count=5;*/
    for (int i = 1; i <= count; i++)
    {
        int n = i*step;
        clock_t start_Serial = clock();

        set<int> result = FinishedNumber(n);
        printf("n=%d以内的所有完数为：
", n);
        for (set<int>::iterator it = result.begin(); it != result.end(); it++)
        {
            /*cout<<*it<<endl;*/
            printf("%d
", *it);
        }

        clock_t end_Serial = clock();
        double timeCollapsedSerial = end_Serial - start_Serial;
        printf("n=%d, 串行方式运行耗时：%f
", n, timeCollapsedSerial);

        int p = 0;
        double timeCollapsedParallel = 0;
        for (int power = 0; power<6; power++)
        {
            p = pow(2, power);
            clock_t start_Parallel = clock();
            FinishedNumber(n, p);
            clock_t end_Parallel = clock();
            timeCollapsedParallel = end_Parallel - start_Parallel;
            float accelerationRate = timeCollapsedSerial / timeCollapsedParallel;
            printf("n=%d, p=%d, 并行方式运行耗时: %f,加速比: %.3f
", n, p, timeCollapsedParallel, accelerationRate);
        }

        printf("
");
    }
}
int _tmain(int argc, _TCHAR* argv[])
{
    int step = 1000 * 1000;
    int count = 5;
    Test(step, count);
    system("pause");
}

运行结果：