What is wrong wirh this code .

Question

#include <stdio.h>

void main() {
int z , x , y , a = 0 , b = 0 ;
for ( x = 1 ; x <= 49 ; x++ ) {
for ( y = 0 ; y < x ; y++ ) {
a = a + y ;
}
for ( z = x + 1 ; z <= 49 ; z++) {
b = b + z ;
}
if ( a == b ) {
printf ( "%d" , x ) ;
break ;
}
}
return 0;
}

-------------------------------------------------

This code should calculate the sum of numbers before and after any number between 1 - 49 .. if the sum of the numbers before equal to the numbers after .. print that number .

Answer 1

Try this

#include <stdio.h>

void main()
{
int z , x , y , a = 0 , b = 0 ;
int n=49;
for ( x = 1 ; x <= n ; x++ ) {
a=0;
for ( y = 0 ; y < x ; y++ ) {
a = a + y ;
}
b=0;
for ( z = x + 1 ; z <=n ; z++) {
b = b + z ;
}
if ( a == b ) {
printf ( "%d" , x ) ;
break ;
}
}
}

Answer 2

Firstly you should change return type "void" to "int", because the function will return an integer, 0 in "return 0;".

Answer 3

Re-initialize value of <a> and <b> to zero on each iteration of the <x> for-loop:

for (x = 1; x <= 49; x++)
{
a = 0; b = 0; // start over
for (y = 0; y < x; y++)

You didn't get a match because value of <a> and <b> isn't starting by zero when you go and check the next number in sequence.

Answer 4

I did not get what you wanted to achieve with it. Can you explain better or give some examples?

Answer 5

Sorry my english is weak ..
I ment for example if the summation of the numbers 1,2,3,4 which is before "5" is equal to the summation of the numbers 6,7,8 which is after "5"
But here it's between 1 and 49

Answer 6

Do you mean if sum of 1, 2, 3 and 1, 2, 3, 4 are equal, print their sum?

Answer 7

Sai Ram it's working .. thanks ❤