Output Of C Program

Question

#include<stdio.h>

int main(){

    int x,sum;

    sum=0;

    for(x=0;x<=500;x+=10){

        sum=sum+x;

    }

     printf("%d",sum);

   return 0;

}

 

What is output of above C-program?

 

Actually, Answer when compiled on computer is 12750.

But i want explanation about How for loop is running and what is behaviour of sum value ?

Comments

is there any shortcut method available to solve this problem becoz counting iteration takes time

---

x=0,sum=0

x=10, sum=10

x=20, sum=30

first remove the zeroes in the both x and sum then

x=1, sum=1

x=2, sum=3

x=3, sum=6 …..

x=50 then sum of first 50 numbers

sum = 50*(50+1)/2 => 1275

now add that 0 to ones place => 12750

Answer 1

Iteration 1: s$um=0$ and $x=0$ $\Rightarrow$ $sum = 0+0$ and $x$ increase by $10$

Iteration 2; $sum=0+10=10$ and $x$ increase by $10$ so it becomes $x=20$

Iteration 3: $sum=10+20=30$ and $x$ increase by $10$ so it becomes $x=30$
.
.
.
.

Iteration 51: $sum=12250+500=12750$ and $x$ increase by $10$ so it becomes $x=510$, now the condition $x\leq500$ will become false and it will come out of for loop

so we see that $sum$ is nothing but A.P  $0+10+20+30+…….$, with $n=51, a=0, d=10$

using $S_n=\frac{n}{2}(2a+(n-1)d)$ we will get $12750$

Comments

We can also do something like this, sum = 0 + 10 + 20 + .. (51 times)

We can now take the common 10 out of the series, 10 * [0 + 1 + 2 + 3 + .. + 50]

Now sum of n counting numbers is n(n+1)/2, so 10 * [0 + 50(51)/2]

50/2 = 25 and 25 * 51 = 1275

So the sum should be 10 * [0 + 1275] which is obviously 12750

Answer 2

Total iterations: ${500 \over 10} + 1 = 51$

Using ap sum formula: $S = {n \over 2}.(2.a + (n - 1).d)$

Here, $a = 0, d = 10, n = 51$

 

answer = ${51 \over 2}.(2.0 + (51 - 1).10) = 12750$