/* C05RC_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;

extern "C"
{
  static void (NAG_CALL fcn)(void* &ad_handle, const Integer &n,
                             const double x[], double fvec[], double fjac[],
                             Integer iuser[], double ruser[], Integer &iflag);
}

int main (void)
{
  // Scalars
  int           exit_status = 0;
  const Integer maxfev = 1000, n = 7, nprint = 0;

  cout << "C05RC_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], xtol;
  Integer iuser[1], nfev, njev;

  xtol = sqrt(X02AJC);
  factor = 100.;
  for (int i=0; i<n; ++i) {
    diag[i] = 1.;
  }

  Integer ifail = 0, mode = 2;
  void    *ad_handle = 0;
  c05rc_p0w_f_(ad_handle,fcn,n,x,fvec,fjac,xtol,maxfev,mode,
               diag,factor,nprint,nfev,njev,r,qtf,iuser,ruser,ifail);

  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;
}

static void (NAG_CALL fcn)(void* &ad_handle, const Integer &n,
                           const double x[], double fvec[], double fjac[],
                           Integer iuser[], double ruser[], Integer &iflag)
{
  if (iflag!=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];
    }
  } else {
    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];
  }
  iflag = 0;
  return;
}