pi.cc

MPICH是MPI的重要研究,提供了一系列的接口函数,为并行计算的实现提供了编程环境.

// -*- c++ -*-
//
// Copyright 1997-2000, University of Notre Dame.
// Authors: Jeremy G. Siek, Jeffery M. Squyres, Michael P. McNally, and
//          Andrew Lumsdaine
// 
// This file is part of the Notre Dame C++ bindings for MPI.
// 
// You should have received a copy of the License Agreement for the Notre
// Dame C++ bindings for MPI along with the software; see the file
// LICENSE.  If not, contact Office of Research, University of Notre
// Dame, Notre Dame, IN 46556.
// 
// Permission to modify the code and to distribute modified code is
// granted, provided the text of this NOTICE is retained, a notice that
// the code was modified is included with the above COPYRIGHT NOTICE and
// with the COPYRIGHT NOTICE in the LICENSE file, and that the LICENSE
// file is distributed with the modified code.
// 
// LICENSOR MAKES NO REPRESENTATIONS OR WARRANTIES, EXPRESS OR IMPLIED.
// By way of example, but not limitation, Licensor MAKES NO
// REPRESENTATIONS OR WARRANTIES OF MERCHANTABILITY OR FITNESS FOR ANY
// PARTICULAR PURPOSE OR THAT THE USE OF THE LICENSED SOFTWARE COMPONENTS
// OR DOCUMENTATION WILL NOT INFRINGE ANY PATENTS, COPYRIGHTS, TRADEMARKS
// OR OTHER RIGHTS.
// 
// Additional copyrights may follow.
//

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

int
main(int argc, char *argv[])
{
  int n, rank, size, i;
  double PI25DT = 3.141592653589793238462643;
  double mypi, pi, h, sum, x;

  MPI::Init(argc, argv);
  size = MPI::COMM_WORLD.Get_size();
  rank = MPI::COMM_WORLD.Get_rank();

  // Prompt the user for how many iterations to use
  
  if (rank == 0) {
    //cout << "Enter the number of intervals: (0 quits)" << endl;
    //cin >> n;

    // Hardwired for simplicity.  Feel free to uncomment the previous
    // lines to get user input.

    n = 10000;
  }

  // Broadcast the number of intervals
  
  MPI::COMM_WORLD.Bcast(&n, 1, MPI::INT, 0);
  
  if (n > 0) {
    h = 1.0 / (double) n;
    sum = 0.0;
    for (i = rank + 1; i <= n; i += size) {
      x = h * ((double)i - 0.5);
      sum += (4.0 / (1.0 + x*x));
    }
    mypi = h * sum;
    
    // Combine all the partial results
    MPI::COMM_WORLD.Reduce(&mypi, &pi, 1, MPI::DOUBLE, MPI::SUM, 0);
    
    if (rank == 0)
      cout << "After " << n << " iterations, pi is approximately " 
	   << pi << ", Error is " << fabs(pi - PI25DT) << endl;
  }

  MPI::Finalize();
  return 0;
}