计算PI的值

在fedora.unix-center.net上的实验结果，加速比基本上约等于2。

-bash-3.2$gcc –xopenmp –o 2 2.c

-bash-3.2$ ./2

the whole processing time of the program is 0.951687 seconds

pi=3.141593

-bash-3.2$gcc –o 1 1.c

-bash-3.2$ ./1

the whole processing time of the program is 1.791129 seconds

pi=3.141593

 

同样的程序运行在t10000.unix-center.net上，效果完全不一样，具体原因还在寻找中。

 

-bash-3.00$ CC -o 11 1.c

-bash-3.00$ ./11

the whole processing time of the program is 33.903705 seconds

pi=3.141593

-bash-3.00$ CC –xopenmp –O3 –o22  2.c

 

-bash-3.00$ ./22

the whole processing time of the program is 7.478114 seconds

pi=3.141593

 

串行代码1.c

 

#include

#include

#include

#include

static long num_steps = 100000000;

double step;

int   main(void)

{

 

 struct timeval starttime,endtime;

 double timepast;

 int i; 

 double x, pi, sum = 0.0;

 step = 1.0/(double) num_steps;

 gettimeofday(&starttime,NULL);

     for (i=0;i<= num_steps; i++){

                                  x = (i+0.5)*step; 

                                sum = sum + 4.0/(1.0+x*x);  

              }

gettimeofday(&endtime,NULL);

        timepast=((double)(endtime.tv_sec-starttime.tv_sec)*1000000+(double)(endtime.tv_usec-starttime.tv_usec))/1000000;

          printf("the whole processing time of the program is %lf seconds\n",timepast);

 

                 pi = step * sum;

    printf("pi=%lf\n",pi);

     return 0;

}

 

 

 

并行代码2.c

 

 

#include

#include

#include

#include

#include

static long num_steps = 100000000;

double step;

int   main(void)

{

 

 struct timeval starttime,endtime;

 double timepast;

 int i; 

 double x, pi, sum = 0.0;

 step = 1.0/(double) num_steps;

 omp_set_num_threads(4);

 gettimeofday(&starttime,NULL);

 #pragma omp parallel for private(x)  reduction(+:sum)

           for (i=0;i<= num_steps; i++){

                                  x = (i+0.5)*step; 

                                sum = sum + 4.0/(1.0+x*x);  

              }

gettimeofday(&endtime,NULL);

        timepast=((double)(endtime.tv_sec-starttime.tv_sec)*1000000+(double)(endtime.tv_usec-starttime.tv_usec))/1000000;

          printf("the whole processing time of the program is %lf seconds\n",timepast);

 

                 pi = step * sum;

         printf("pi=%lf\n",pi);

        return 0;

}