Hello everyone! I have QUITE the load of homework to do tonight, one portion of which being a computer program. I had the program nearly finished, and I emailed it to myself from school. Unfortunately, it didn't get through - I have no program, and it's due tomorrow. Mine was like 500 lines of code, though a friend of mine finished his in 100 .. If anyone has any creative ideas of how to do this program, I'd be much obliged:
- Has to be constructed in WinOOT
- Input a number up to seven digits long, e.g - 1523682
- Outputs the number converted to english, e.g - 1 million 5 hundred twenty three thousand six hundred eighty two
-> As you can see from the above examples, the english number is fairly basic; no punctuation, no 'and', etc!
Thanks in advance!
Help with computer program!
-
Eldaran Celuril
- Posts: 9
- Joined: Mon Feb 24, 2003 5:25 pm
- Location: Where the wind takes me ..
- Contact:
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Eldaran Celuril
- Posts: 9
- Joined: Mon Feb 24, 2003 5:25 pm
- Location: Where the wind takes me ..
- Contact:
-
Eldaran Celuril
- Posts: 9
- Joined: Mon Feb 24, 2003 5:25 pm
- Location: Where the wind takes me ..
- Contact:
Have no idea what WinOOT is - but haveing a hard time to back down from programming problems I've solved it in Java 
The code isn't very optimized and can be improved massive, also presently it requiers a 7 digit number as input, but that is easy to fix:
public class NumberInterp
{
private int input;
private int[] temp = new int[7];
private int temp2 = 1000000;
private int index = 6;
private String output;
public NumberInterp(int input)
{
this.input = input;
output = "";
}//end constructor
public String start()
{
while (input > 0)
{
temp[index] = input/temp2;
input = input%temp2;
index--;
temp2 = temp2/10;
}//end while
if (temp[6] > 0)
{ output += buildNumber1(temp[6]) + " million "; }
if (temp[5] > 0)
{ output += buildNumber1(temp[5]) + " hundred "; }
if (temp[4] > 0)
{ output += buildNumber2(temp[4]); }
if (temp[3] > 0)
{ output += buildNumber1(temp[3]); }
output += " thounsand ";
if (temp[2] > 0)
{ output += buildNumber1(temp[2]) + " hundred "; }
if (temp[1] > 0)
{ output += buildNumber2(temp[1]); }
if (temp[0] > 0)
{ output += buildNumber2(temp[0]); }
return output;
}
private String buildNumber1(int index)
{
switch (index)
{
case 1: return " one ";
case 2: return " two ";
case 3: return " three ";
case 4: return " four ";
case 5: return " five ";
case 6: return " six ";
case 7: return " seven ";
case 8: return " eight ";
case 9: return " nine ";
}//end switch
return null;
}//end method
private String buildNumber2(int index)
{
switch (index)
{
case 1: return " ten ";
case 2: return " twenty ";
case 3: return " thirty ";
case 4: return " forty ";
case 5: return " fifty ";
case 6: return " sixty ";
case 7: return " seventy ";
case 8: return " eighty ";
case 9: return " ninty ";
}//end switch
return null;
}//end method
}//end class
called with the parameter:
System.out.println(new NumberInterp(1234567).start());
it will return:
one million two hundred thirty four thounsand five hundred sixty seven
Basicall there are two "tricks" to this problem:
one you need to convert your input number so you know how many of "each" you have, eg 1 million and 2 hundred thousands.
This can easiestly (imo) be done with the modulus operation:
Interger divide in programming will return the number of times one integer is dividable by another eg:
10 / 3 = 3.
Modulus will return the remainder after a integer division:
10 % 3 = 1. (java and most other uses % as modulus, Pascal uses the keyword "mod")
This can be used to do the main calculation:
1234567 / 1000000 = 1
1234567 % 1000000 = 234567
Thus you know you have 1 millions and the remainder is 234567.
Running next modulus and integer dividing:
234567 / 100000 = 2
234567 % 100000 = 34567
and so on.
This is what happens in the first while loop.
Secondly you need to translate the 1, 2 etc into one, two, twenty etc.
This I use a swich structure for - it is basically just if constructions. And to avoid writing it all several times I've placed the code in methods: buildNumber1 and buildNumber2.
buildNumber1 returns a String with "one", "two" etc depending if the number is a 1,2 etc where as buildnumber2 returns "ten, twenty" etc.s if the number is a 1, 2....
Hope this helps a bit, if not I can try to explain it better
The code isn't very optimized and can be improved massive, also presently it requiers a 7 digit number as input, but that is easy to fix:
public class NumberInterp
{
private int input;
private int[] temp = new int[7];
private int temp2 = 1000000;
private int index = 6;
private String output;
public NumberInterp(int input)
{
this.input = input;
output = "";
}//end constructor
public String start()
{
while (input > 0)
{
temp[index] = input/temp2;
input = input%temp2;
index--;
temp2 = temp2/10;
}//end while
if (temp[6] > 0)
{ output += buildNumber1(temp[6]) + " million "; }
if (temp[5] > 0)
{ output += buildNumber1(temp[5]) + " hundred "; }
if (temp[4] > 0)
{ output += buildNumber2(temp[4]); }
if (temp[3] > 0)
{ output += buildNumber1(temp[3]); }
output += " thounsand ";
if (temp[2] > 0)
{ output += buildNumber1(temp[2]) + " hundred "; }
if (temp[1] > 0)
{ output += buildNumber2(temp[1]); }
if (temp[0] > 0)
{ output += buildNumber2(temp[0]); }
return output;
}
private String buildNumber1(int index)
{
switch (index)
{
case 1: return " one ";
case 2: return " two ";
case 3: return " three ";
case 4: return " four ";
case 5: return " five ";
case 6: return " six ";
case 7: return " seven ";
case 8: return " eight ";
case 9: return " nine ";
}//end switch
return null;
}//end method
private String buildNumber2(int index)
{
switch (index)
{
case 1: return " ten ";
case 2: return " twenty ";
case 3: return " thirty ";
case 4: return " forty ";
case 5: return " fifty ";
case 6: return " sixty ";
case 7: return " seventy ";
case 8: return " eighty ";
case 9: return " ninty ";
}//end switch
return null;
}//end method
}//end class
called with the parameter:
System.out.println(new NumberInterp(1234567).start());
it will return:
one million two hundred thirty four thounsand five hundred sixty seven
Basicall there are two "tricks" to this problem:
one you need to convert your input number so you know how many of "each" you have, eg 1 million and 2 hundred thousands.
This can easiestly (imo) be done with the modulus operation:
Interger divide in programming will return the number of times one integer is dividable by another eg:
10 / 3 = 3.
Modulus will return the remainder after a integer division:
10 % 3 = 1. (java and most other uses % as modulus, Pascal uses the keyword "mod")
This can be used to do the main calculation:
1234567 / 1000000 = 1
1234567 % 1000000 = 234567
Thus you know you have 1 millions and the remainder is 234567.
Running next modulus and integer dividing:
234567 / 100000 = 2
234567 % 100000 = 34567
and so on.
This is what happens in the first while loop.
Secondly you need to translate the 1, 2 etc into one, two, twenty etc.
This I use a swich structure for - it is basically just if constructions. And to avoid writing it all several times I've placed the code in methods: buildNumber1 and buildNumber2.
buildNumber1 returns a String with "one", "two" etc depending if the number is a 1,2 etc where as buildnumber2 returns "ten, twenty" etc.s if the number is a 1, 2....
Hope this helps a bit, if not I can try to explain it better
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Originally posted by Tamerlane
@ Xandax
Just had that sick feeling going through me. Thanks for reminding me to do my homework.
Never thought I'd live to see the day when Java is shown in all its horrible glory on SYM.
no problem - anytime
BTW - just looked at the code and found out that both switch methods could be replaced by 2 array of strings wich would be much easier.
Making a string array holding "one"; two"; etc... and another "ten"; "twenty"..... and then using index-1 to get ahold of prober conversion - thus index of 1 subtracted 1 = 0, and then the arrayindex of 0 could be one or ten, depending on the array
Bah - hate finding easier ways just after I posted it
Actually I think I could shorten it down to about half the code with a few modifications
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