For two matrices to be equal, the dimensions of both matrices have to be equal and the corresponding values should also be equal. I would follow niek_e's method to compare two matrices.

I dont know if it can be classified as anime or not, but I like Avatar.

Insert line 20 inside the while loop and add the statement prev=prev->link; at the end of the loop.

Take a good book on computer architecture and start reading, once you have finished a chapter start reading the refrences given at the end of the chapter. Most of them will be papers from IEEE or ACM...

An alternative could be using link list if you want to do it without STL. That way your first issue would be solved, but still the best way would be using vectors.

Spend the day adventure sporting with my friends.


40C here and I am burning :(

You are not using your head and instead simply copy pasting the code. If you had read any of the posts above the mistake in my code would have been clear to you.

number1 = number % 10;
number=number/10;
number2 = number % 10;
number=number/10;
number3 = number % 10;

No you are not. The base of the numbers is 10 not 100 or 1000. And you have used cout the wrong way


cout << endl << number<< number2<<number3;

A hint, divide a number by 10. The quotient would be the number except for the least significant digit and the remainder would be the least significant digit. Use these two in combination to get your...

I am not sure, it has worked for me and I have never bothered to search. I will leave it for more experienced programmers to answer your question.

There is no need to use floor() as int truncates any decimal place as it is.

for any 4 digit number this should do the trick to extract all 4 digits

i=0;
while(number>0)
{
a[i]=number%10;...

Its not

cin >> number,first,second,third,fourth;


but

cin >> number>>first>>second>>third>>fourth;

I dont mind being called a geek because that is more or less what I am :)

Note that the '==' operator does not apply to float. You would get the output as C++ if you tried

if(f==.7)

The formula can be generalized as

index= ((row_num-1)*3) + (col_num-1)

where index is the index in the array
row_num is the row number of the element you want to refer and
col_num is the...

Hi James

Right click on My Computer and select Manage. On the left select disk management and select your drive. On the unallocated partition, right click and select format. You will have options...

Format the partitions from disk management in either FAT32 or NTFS.

for(int i=countDown;i>=0;i-- )

Number of colums in the first matrix have to be equal to the number of rows in the second one. If you have only a row or a column matrix i.e. two arrays, their lengths have to be equal in order for...

Also note how m and n get interchanged if n>m, consider the case of m=6 and n=15

r1=6 MOD 15 = 6

m<-n implies m=15
n<-r implies n=6

the process is repeated to get gcd as 3

LCM is the tricky one. One needs to know that every number greater than 1 can be represented as products of prime numbers. Once factors each number m,n into its corresponding prime factors and then...

Okay here is the pseudo code for hcf, credit to Euclid.

gcd(m,n)

r<-(m MOD n);
if r=0 then
{ return n }
else
{
m<-n

Are you using a dhcp to assign ip's to your ethernet card? What is your isp? Also if you are using pppoe make sure that your username and password are correct :P

Or maybe


arr1 = 3 0 0 5 size=4
arr2= 3 0 5 size=3

// array sol of size 5 here


then we need the array solution as size 4+1. It will still be simulation...

For this you can do what Ancient Dragon said and find out the maximum/minimum value and then traverse the array once again and compare each element with the maximim/minimum value, if it is equal then...

In the above loop you would have to check for every i if it is odd or not using

if(i%2!=0)


alternatively you could use i=i-2 instead of i-- and simply display the numbers.

There is a strrev() function in string.h for char array. I dont know enough to help you with your code but I suggest you to follow a different method for checking palindrome in a string.

Instead...

I dont know what you are trying to do but in line 12 during the last iteration of the loop i=4, i+1 will be 5 that will go out of bounds for the array pass. Run the loop till i<4 because there will...

At line 7 when the loop ends, f will be 10. You are simply multiplying 1*10 and storing it in result. You can make result an array of size 10 then use

result[f-1]=k*f;

If you want to do it by using regular expressions you would need a lexical analyzer. For C/C++ lex is usually used. Here is a tutorial page
http://dinosaur.compilertools.net/lex/index.html

as for...

Tennis, sleeping and movies.

Factorial of any number greater than 5 ends with a zero. For a larger number, the number of zeroes at the end constitute a large part of the number. If you can eliminate those zeroes it will help.
...

Oops correction
f(209)=f(f(20)+9)=f(f(f(2)+0)+9))=f(f(2)+9)=f(11)=f(1)+1=2

So many f's . But again I might be wrong again :) . I havnt tested it for values where sum of the digits is a number of...

Use else if statement. According to your code, if a=0 and b=0 then both the first and the second if statements will be executed. And the output will be
"There are no solutions"
"There is one...

From the test cases I used, the code was working fine (till the max value of int) . I usually write down a mathematical expression before coding it. The function for sum of the digits of a whole...

Okay here is another shorter alternative.

#include<iostream.h>

int eig(int num)
{

if(num<10)
{
return num;

Okay I made this code on Turbo C

#include<iostream.h>

int eig(int num)
{

if(num<10)
{
return num;

From what I remember of my coordinate geometry, testing for a point inside a triangle wasnt easy. Try this
http://mathworld.wolfram.com/TrianglePointPicking.html

you might have to translate one...

I am presently working on emulation of human movements through robotics as my college project.

You can clear the entire screen using the cleardevice() function and redraw the square in a loop with the new positions.