/* C05RD_P0W_F C++ Header Example Program.
 *
 * Copyright 2019 Numerical Algorithms Group.
 * Mark 27, 2019.
 */

#include <nag.h>
#include <nagad.h>
#include <stdio.h>
#include <math.h>
#include <nagx02.h>
#include <nagx04.h>
#include <iostream>
#include <string>
using namespace std;

int main (void)
{
  // Scalars
  int           exit_status = 0;
  const Integer n = 7;

  cout << "C05RD_P0W_F C++ Header Example Program Results\n\n";

  // problem parameters and starting value
  double        ruser[5], x[7];

  ruser[0] = -1.0;
  ruser[1] =  3.0;
  ruser[2] = -2.0;
  ruser[3] = -2.0;
  ruser[4] = -1.0;

  for (int i=0; i<n; ++i) {
    x[i] = -1.0;
  }


  // Call passive routine
  double  diag[n], fjac[n*n], factor, fvec[n],
          qtf[n], r[n*(n+1)/2], rwsav[4*n+10], xtol;
  Integer irevcm, iwsav[17];

  xtol = sqrt(X02AJC);
  factor = 100.0;
  for (int i=0; i<n; ++i) {
    diag[i] = 1.0;
  }

  Integer ifail = 0, mode = 2;
  void    *ad_handle = 0;
  irevcm = 0;
  do {
    c05rd_p0w_f_(ad_handle,irevcm,n,x,fvec,fjac,xtol,mode,
                 diag,factor,r,qtf,iwsav,rwsav,ifail);

    switch (irevcm) {
    case 1:
      // Monitoring exit
      continue;
    case 2:
      for (int i=0; i<n; ++i) {
        fvec[i] = (ruser[1] + ruser[2]*x[i])*x[i] - ruser[4];
      }
      for (int i=1; i<n; ++i) {
        fvec[i] = fvec[i] + ruser[0]*x[i-1];
      }
      for (int i=0; i<n-1; ++i) {
        fvec[i] = fvec[i] + ruser[3]*x[i+1];
      }
      break;
    case 3:
      for (int i=0; i<n*n; ++i) {
        fjac[i] = 0.0;
      }
      fjac[0] = ruser[1] + 2.0*ruser[2]*x[0];
      fjac[n] = ruser[3];
      for (int i=1; i < n-1; ++i) {
        int k = i*n + i;
        fjac[k-n] = ruser[0];
        fjac[k] = ruser[1] + 2.0*ruser[2]*x[i];
        fjac[k+n] = ruser[3];
      }
      fjac[n*n-n-1] = ruser[0];
      fjac[n*n-1] = ruser[1] + 2.0*ruser[2]*x[n-1];
      break;
    }

  } while (irevcm != 0);

  cout.setf(ios::scientific,ios::floatfield);
  cout.precision(3);
  cout << "           Solution:\n";
  for (int i=0; i<n; ++i) {
    cout.width(8);
    cout << i+1;
    cout.width(18);
    cout << x[i] << endl;
  }

  return exit_status;
}