#include <iostream>
#include <iomanip>
#include <fstream>
#include <math.h>

using namespace std;

void possible(int,int); 
void viable(int,int);
void solutions(int,int);

int main()
{
  int x,s; 

  cout << endl;
  cout << "This program enumerates all possible restricted" << endl;
  cout << "and unrestricted Sarvate-Beam triple systems" << endl;
  cout << "with v=5. Note that f(12) is 1, 4, 7 or 10." << endl;

  for(x= 1; x<= 10; x+= 3)
    for(s= 3; s<= floor((55- x)/6); s++)
    {
      possible(x,s); 
      viable(x,s);
      solutions(x,s);
    }

  cout << endl;
  cout << "Solutions are stored in u53solnst.dat" << endl;
  cout << endl;

  return(0);
}

void possible(int x, int y)
{
  int i,j,k,m,z= (55-x)/3-y;
  ofstream output1, output2;

  output1.open("apossible.dat");
  for(i= 0; i<= y; i++)
    for(j= 0; j<= y; j++)
      for(k= 0; k<= y; k++)
        if(i+ j+ k== y)
          output1 << i << " " << j << " " << k << endl;
  output1.close();

  output2.open("bpossible.dat");
  for(i= 0; i<= z; i++)
    for(j= 0; j<= z; j++)
      for(k= 0; k<= z; k++)
        for(m= 0; m<= z; m++)
          if(i+ j+ k+ m== z)
            output2 << i << " " << j << " " << k << " " << m << endl;
  output2.close();
}

void viable(int x, int y)
{
  int inp[4],a[3],i,j,v;
  ifstream input1,input2;
  ofstream output1,output2;

  input1.open("apossible.dat");
  output1.open("aviable.dat");

  input1 >> inp[0] >> inp[1] >> inp[2];
 
  while(! input1.eof())
  {
    a[0]= inp[0]+ inp[1];
    a[1]= inp[0]+ inp[2];
    a[2]= inp[1]+ inp[2];
    v= 0;

    for(i= 0; i< 2; ++i)
      for(j= i+ 1; j< 3; ++j)
          if(a[i]==a[j] || a[i]<1 || a[i]>10 || a[j]<1 || a[j]>10)
            v+= 1;

    switch(v)
    {
      case 0:  output1 << inp[0] << " " << inp[1] << " " << inp[2] << endl;
               break;
      default: ;
    }

    input1 >> inp[0] >> inp[1] >> inp[2];
  }

  input1.close();
  output1.close();

  input2.open("bpossible.dat");
  output2.open("bviable.dat");

  input2 >> inp[0] >> inp[1] >> inp[2] >> inp[3];
 
  while(! input2.eof())
  {
    a[0]= inp[0]+ inp[1];
    a[1]= inp[0]+ inp[2];
    a[2]= inp[1]+ inp[2];
    v= 0;

    for(i= 0; i< 2; ++i)
      for(j= i+ 1; j< 3; ++j)
          if(a[i]==a[j] || a[i]<1 || a[i]>10 || a[j]<1 || a[j]>10)
            v+= 1;

    switch(v)
    {
      case 0:  output2 << inp[0] << " " << inp[1] << " " << inp[2] << " " << inp[3] << endl;
               break;
      default: ;
    }

    input2 >> inp[0] >> inp[1] >> inp[2] >> inp[3];
  }

  input2.close();
  output2.close();
}

void solutions(int x, int y)
{
  int inp1[3],inp2[4],c[9],i,j,v;
  ifstream input1,input2;
  ofstream output1;

  input1.open("aviable.dat");
  input2.open("bviable.dat");
  output1.open("u53solnst.dat", ios::app );

  input1 >> inp1[0] >> inp1[1] >> inp1[2];
  input2 >> inp2[0] >> inp2[1] >> inp2[2] >> inp2[3];

  do
  {
    do
    {
      c[0]= inp1[0]+ inp1[1];
      c[1]= inp1[0]+ inp1[2];
      c[2]= inp1[1]+ inp1[2];
      c[3]= inp2[0]+ inp2[1];
      c[4]= inp2[0]+ inp2[2];
      c[5]= inp2[1]+ inp2[2];
      c[6]= inp1[0]+ inp2[0]+ inp2[3];
      c[7]= inp1[1]+ inp2[1]+ inp2[3];
      c[8]= inp1[2]+ inp2[2]+ inp2[3];
      v= 0;

      for(i= 0; i< 8; ++i)
        for(j= i+ 1; j< 9; ++j)
          if(c[i]==c[j] || c[i]<1 || c[i]>10 || c[j]<1 || c[j]>10 || c[i]==x || c[j]==x)
            v+= 1;

      switch(v)
      {
        case 0:  output1 << x << "  " << inp1[0] << " " << inp1[1] << " " << inp1[2] << "  " << inp2[0] << " " << inp2[1] << " " << inp2[2] << "  " << inp2[3] << endl;
                 break;
        default: ;
      }

      input2 >> inp2[0] >> inp2[1] >> inp2[2] >> inp2[3];
    } while(! input2.eof());

    input2.close();
    input2.open("bviable.dat");

    input2 >> inp2[0] >> inp2[1] >> inp2[2] >> inp2[3];
    input1 >> inp1[0] >> inp1[1] >> inp1[2];
  } while(! input1.eof());

  input1.close();
  input2.close();
  output1.close();
}