test2.cpp

一个用c写的机遇MPI的并行程序

#include <iostream>
#include "mpi.h"
#include <stdio.h>
#include <math.h>

double f(double);
double f(double x)
/* ????f(x) */
{
return (4.0 / (1.0 + x*x));
}
void main(int argc,char *argv[])
{
int done = 0, n, myid, numprocs, i;
double PI25DT = 3.141592653589793238462643;
/* ???????????p?*/
double mypi, pi, h, sum, x;
double startwtime = 0.0, endwtime;
int namelen;
char processor_name[MPI_MAX_PROCESSOR_NAME];
MPI_Init(&argc,&argv);
MPI_Comm_size(MPI_COMM_WORLD,&numprocs);
MPI_Comm_rank(MPI_COMM_WORLD,&myid);
MPI_Get_processor_name(processor_name,&namelen);
fprintf(stdout,"Process %d of %d on %s\n",
myid, numprocs, processor_name);
n = 0;
if (myid == 0)
{ printf("Please give N=");
 scanf("&d",&n);
startwtime = MPI_Wtime();
}
MPI_Bcast(&n, 1, MPI_INT, 0, MPI_COMM_WORLD);/*?n?????*/
h = 1.0 / (double) n;/*?????????????????*/
sum = 0.0; /*????????*/
for (i = myid + 1; i <= n; i += numprocs)
/*????????????????????numprocs?4 ?0-1?????100?
??????????????
144
0?? 1 5 9 13 ... 97
1?? 2 6 10 14 ... 98
2?? 3 7 11 15 ... 99
3?? 4 8 12 16 ... 100
*/
{
x = h * ((double)i - 0.5);
sum += f(x);
}
mypi = h * sum;/*??????????????*/
MPI_Reduce(&mypi, &pi, 1, MPI_DOUBLE, MPI_SUM, 0,
MPI_COMM_WORLD);
/*???????????????????????p?*/
if (myid == 0)
/*?????0???????????*/
{
printf("pi is approximately %.16f, Error is %.16f\n",
pi, fabs(pi - PI25DT));
endwtime = MPI_Wtime();
printf("wall clock time = %f\n", endwtime-startwtime);
fflush( stdout );
}
MPI_Finalize();
}