Programming Forums > Application Development > Java

LAYAN-2008 · May 2nd, 2008, 2:23 AM

hi , I want to know if my answer is true??and I want your opinion in my answer..
• Write a program fragment to find out the average of odd numbers between two numbers given by the user using a do-while statement?

int sum=0;
int X=50;
int Y = 70;
int count=0;
do{
X++;
if(X%2!=0)
{
sum +=X;
count++;
}
}
while(X<Y);
int average = sum/count;
System.out.println("average=" + average );

Freaky Chris · May 2nd, 2008, 2:41 AM

Looks like it would give the correct results, without testing i will say that you may have to change < to <= and possible change the point at which you increment X.

P.S Use [code] tags

Chris

JONY SMITH · May 2nd, 2008, 3:40 AM

Quote:

Originally Posted by Freaky Chris

Looks like it would give the correct results, without testing i will say that you may have to change < to <= and possible change the point at which you increment X.

P.S Use [code] tags

Chris

yes it will give you right answer possibly. Try with <= and note out different hope one will be the right answer.

tomisinbamgbelu · May 2nd, 2008, 7:45 AM

your code is quite right but the question says given by the user. ending with an odd no is not provided for, so the while statement should read x<=y; so as 2 test d last number, in case its odd.

misho · May 6th, 2008, 3:24 PM

The code as you have written it should work; however, that's based on X and Y being even. If X were odd, the "X++;" line would be bad. If Y were odd, the "X<Y" condition would be bad.
For arbitrary X and Y, you'd either need to modify the rest of the code, or add something like the following near the beginning:

 c++ Syntax (Toggle Plain Text) 
if(X%2 != 0) X--;
if(Y%2 != 0) Y++;
if(X%2 != 0) X--;
if(Y%2 != 0) Y++;

I'd also add that since consecutive odd numbers are 2 apart, incrementing by 1 then checking if your number is odd or even is kind of wasteful. Something like the following would be better:

 c++ Syntax (Toggle Plain Text) 
//assuming X and Y have been declared and are even by this point
//and that Y>X
int count = 0, sum = 0;
X++;
do {
sum += X;
count++;
X += 2; //or X++;X++; , not really sure which is faster
} while (X < Y);
int average = sum / count;
//assuming X and Y have been declared and are even by this point
//and that Y>X
int count = 0, sum = 0;
X++;
do {
sum += X;
count++;
X += 2; //or X++;X++; , not really sure which is faster
} while (X < Y);
int average = sum / count;

I was going to point out that average may not be an int, but you can prove that the average is (X+Y)/2 (again, X and Y are both even, or both odd, I guess), so it has to be.

May 2nd, 2008, 2:23 AM	#1
LAYAN-2008 Newbie Join Date: Apr 2008 Posts: 3 Rep Power: 0	find out the average of odd numbers -do-while statement. hi , I want to know if my answer is true??and I want your opinion in my answer.. • Write a program fragment to find out the average of odd numbers between two numbers given by the user using a do-while statement? int sum=0; int X=50; int Y = 70; int count=0; do{ X++; if(X%2!=0) { sum +=X; count++; } } while(X<Y); int average = sum/count; System.out.println("average=" + average );

May 2nd, 2008, 2:41 AM	#2
Freaky Chris Hobbyist Programmer Join Date: Dec 2007 Location: England Posts: 169 Rep Power: 1	Re: find out the average of odd numbers -do-while statement. Looks like it would give the correct results, without testing i will say that you may have to change < to <= and possible change the point at which you increment X. P.S Use [code] tags Chris __________________ Who said i couldn't program `sarcasm = raw_input('Type in a sarcastic remark: ') print sarcasm`

May 2nd, 2008, 7:45 AM	#4
tomisinbamgbelu Programmer Join Date: May 2008 Posts: 5 Rep Power: 0	Re: find out the average of odd numbers -do-while statement. your code is quite right but the question says given by the user. ending with an odd no is not provided for, so the while statement should read x<=y; so as 2 test d last number, in case its odd.

May 6th, 2008, 3:24 PM	#5
misho Newbie Join Date: Apr 2008 Posts: 10 Rep Power: 0	Re: find out the average of odd numbers -do-while statement. The code as you have written it should work; however, that's based on X and Y being even. If X were odd, the "X++;" line would be bad. If Y were odd, the "X<Y" condition would be bad. For arbitrary X and Y, you'd either need to modify the rest of the code, or add something like the following near the beginning: c++ Syntax (Toggle Plain Text) if(X%2 != 0) X--; if(Y%2 != 0) Y++; if(X%2 != 0) X--; if(Y%2 != 0) Y++; I'd also add that since consecutive odd numbers are 2 apart, incrementing by 1 then checking if your number is odd or even is kind of wasteful. Something like the following would be better: c++ Syntax (Toggle Plain Text) //assuming X and Y have been declared and are even by this point //and that Y>X int count = 0, sum = 0; X++; do { sum += X; count++; X += 2; //or X++;X++; , not really sure which is faster } while (X < Y); int average = sum / count; //assuming X and Y have been declared and are even by this point //and that Y>X int count = 0, sum = 0; X++; do { sum += X; count++; X += 2; //or X++;X++; , not really sure which is faster } while (X < Y); int average = sum / count; I was going to point out that average may not be an int, but you can prove that the average is (X+Y)/2 (again, X and Y are both even, or both odd, I guess), so it has to be.

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Programming Forums > Application Development > Java

find out the average of odd numbers -do-while statement.