Openmp并行计算pi π

Posted 2021-07-23 各可

Openmp并行计算pi π

#include"stdio.h"
#include"omp.h"
#include<time.h>
#include<Windows.h>
#include<ctime>
#include<sys/timeb.h>
#define NUM_THREADS 8
static long n = 10000000;
double h;
int main()
{
	int i;
	double pi, sum[NUM_THREADS];
	for (i = 0;i < NUM_THREADS;i++) {
		sum[i] = 0.0;
	}
	h = 1.0 / (double)n;
	omp_set_num_threads(NUM_THREADS);
#pragma omp parallel
	{
		double x;
		int tid;
#pragma omp for
		for (i = 0;i < n;i++)
		{
			tid = omp_get_thread_num();
			x = (i + 0.5) * h;
			sum[tid] += 4.0 / (1.0 + x * x);
		}

	}
	for (i = 0, pi = 0.0;i < NUM_THREADS;i++) {
		pi += sum[i] * h;
	}
	printf("Approximation of pi%15.13f\\n", pi);
}

编译运行结果如下：

使用的编译器是g++；工具借助vscode；大家也可以使用visual studio配置一下openmp就行。

可以添加时间戳对比并行与非并行效率

#include<iostream>
using namespace std;
#include<cstdio>
#include"omp.h"
#include"time.h"
#include<ctime>
#include<windows.h>
#include<sys/timeb.h>
#define NUM_THREADS 8
static long n = 10000000;
double h;
LARGE_INTEGER t1, t2, tc;
int  main()
{
    int i;
    double x,pi,sum = 0.0;
    h=1.0/(double)n;
    omp_set_num_threads(NUM_THREADS);
    QueryPerformanceFrequency(&tc);
    QueryPerformanceCounter(&t1);
    #pragma omp parallel for reduction(+:sum) private(x)
    for (i = 0; i < n;i++)
    {
        x = (i + 0.5) * h;
        sum += 4.0 / (1.0 + x * x);
    }
    QueryPerformanceCounter(&t2);
    long time = (double)(t2.QuadPart - t1.QuadPart) * 1000000 / (double)tc.QuadPart;
    cout << "time = " << time / 1000.0 << "ms" << endl;
    pi = sum * h;
    printf("Appromxation of pi:%15.13f\\n", pi);
    return 0;
}

编译运行结果如下：