#1

hey UG, so i have to make a program that tells you if the number you type in is a positive pair number using the "if" instruction and i dont know where to begin...any ideas?

Thanks in advance and sorry for my bad english. Saludos de Tijuana

Thanks in advance and sorry for my bad english. Saludos de Tijuana

#2

if

{ x>0;

printf ("positive");

}

else;

{

printf ("negative");

}

my programming was a yr ago, sumthin similar to that, is how we did it.. if i get your question correctly

{ x>0;

printf ("positive");

}

else;

{

printf ("negative");

}

my programming was a yr ago, sumthin similar to that, is how we did it.. if i get your question correctly

*Last edited by JimmyBanks6 at Oct 14, 2009,*

#3

thank you Jimmy, how about getting c++ to tell me if that same number is a pair number or impair number...

Thank you in advance

Thank you in advance

#4

thank you Jimmy, how about getting c++ to tell me if that same number is a pair number or impair number...

Thank you in advance

I'm not sure what you mean by pair or impair. If you could explain a little further I might be able to help you. From what I've seen, C++ if statements are almost exactly like Java's, so I should be of some use.

#5

hmm i think its similar to you set the value to a long, then you divide the x value by 2, if it results in a whole number you set it to print pair, if it is a decimal value, then you set it to print impair, i cant remember exactly the programming but that should be enough info for you to play around and get it

that is if you mean by pair:even, impair: odd . that is what i thought you meant?

that is if you mean by pair:even, impair: odd . that is what i thought you meant?

*Last edited by JimmyBanks6 at Oct 14, 2009,*

#6

sorry my english is very crappy, by pair number i mean like 2,4,6,8,10,12 etc.. and impair numbers i mean like 1,3,5,7,9,11,13 etc... thanks again for your quick response

#7

You're all WRONG...

Use the modulo (%) operator. If the number modulo 2 returns 1, print impair and if it returns 0, print pair.

Use the modulo (%) operator. If the number modulo 2 returns 1, print impair and if it returns 0, print pair.

#8

if((x%2)==0)

cout << "Even";

else

cout << "Odd";

done.

^^ shakes fist at guy above me.

cout << "Even";

else

cout << "Odd";

done.

^^ shakes fist at guy above me.

#9

if((x%2)==0)

cout << "Even";

else

cout << "Odd";

done.

ohh wow as if i forgot the %.. meh mechanical eng dont need programing

#10

sorry my english is very crappy, by pair number i mean like 2,4,6,8,10,12 etc.. and impair numbers i mean like 1,3,5,7,9,11,13 etc... thanks again for your quick response

Ah okay, in that case it should go something like this (int x set to whatever number you're evaluating):

int x;

if(x % 2 == 0){

cout<<"pair"<<endl;

}else if(x % 2 == 1){

cout<<"pair"<<endl;

}

I believe that's correct syntactically. If not you might have to play around with it a little. The '%' is the modulus operator, which returns the remainder after dividing by the following number, so if it's one, the number is odd, and if it's zero, the number is even.

EDIT: Wow, I got beat. That'll teach me not to respond with a novel.

*Last edited by Ajax413 at Oct 14, 2009,*

#11

Use the if and the mod ( % ) function, the mod function returns the difference of a divison. So 5 % 2 = 1, because 5 = 2*2 + 1. Hope you understand...

A pair number is dividable by 2, so x % 2 == 0, and odd ( impair ) number is not so

x % 2 != 0.

if( (x % 2) == 0) cout << "Even";

if ( (x% 2) != 0) cout <<"Odd";

if you have questions just ask.

A pair number is dividable by 2, so x % 2 == 0, and odd ( impair ) number is not so

x % 2 != 0.

if( (x % 2) == 0) cout << "Even";

if ( (x% 2) != 0) cout <<"Odd";

if you have questions just ask.

#12

wow thank you for the quick responses everyone very helpful!!