blaze12364
March 24th, 2008, 09:36 PM
Below is a piece of my program, i have printed out all the expressions. As the number of expressions it prints is incredibly large (over 5,000) and takes time i want to be able to count the number of expressions it will print instead of actually printing all the expressions.
Can anyone help?
newCalcWaiting=true; // to deal with a later possible equality
i = iva_level(level);
for (j = iva_level(level+1) ; j < iva_level(level+2) ; j++)
{if (i == highV[level])
{// ignore the first of the contributors to calculation
i++;
}
if (i == lowV[level])
{// ignore the second of the contributors to calculation
i++;
}
if (newCalcWaiting && (sorted[i]<=thiscalc
|| iva_level(level+1)<=i ) )
{// insert the new calculation into the next level
sorted[j]=thiscalc;
expression[j]=newform;
newCalcWaiting=false;
System.out.println("expression = " + expression[j]);
System.out.print(sorted[j]);
System.out.println();
}
Can anyone help?
newCalcWaiting=true; // to deal with a later possible equality
i = iva_level(level);
for (j = iva_level(level+1) ; j < iva_level(level+2) ; j++)
{if (i == highV[level])
{// ignore the first of the contributors to calculation
i++;
}
if (i == lowV[level])
{// ignore the second of the contributors to calculation
i++;
}
if (newCalcWaiting && (sorted[i]<=thiscalc
|| iva_level(level+1)<=i ) )
{// insert the new calculation into the next level
sorted[j]=thiscalc;
expression[j]=newform;
newCalcWaiting=false;
System.out.println("expression = " + expression[j]);
System.out.print(sorted[j]);
System.out.println();
}