Mathematically the result of the above program can be written as follows
$$\sum_{i=0}^{98}\sum_{j=1}^{46}\sum_{k=1}^{3}1$$
Explanation:
The inner most loop controlled by the value of $k$ breaks out of loop when its value reaches 4. Thus, the increment in the inner most loop is performed for only $3$ time.
The middle level loop controlled by the value of $j$ will continue until it reaches $49$. Therefore, it can be said that the iteration will run for $48$ times (loop terminated when $j == 49$). However, out of those $48$ iterations, two of them are ineffective because of second if condition which simply continues with the loop for $j == 20$ and $j == 40$. Thus, we are adding the result of innermost iteration for $46$ time
Similar argument can be given for the outer most loop and hence the summation proposed above correctly represents the program.
Solution:
$$\sum_{i=0}^{98}\sum_{j=1}^{46} 3 = \sum_{i=0}^{98}46*3 = 99*46*3 = 13662$$
The given program can be rewritten as follows
The lines from original code are commented and replaced by lines just below them.
#include <stdio.h>
int main(void) {
int i, j, k;
int s = 0;
for(i = 0; i <= 100; i++)
{
if(i == 25 || i == 30)
continue;
// for(j = 1; j <= 100; j++)
for(j = 1; j <= 48 ; j++)
{
// if(j == 49)
// break;
if(j == 20 || j == 40)
continue;
// for(k = 1; k <= 10; k++)
for(k = 1; k <= 3; k++)
{
// if(k == 4)
// break;
s = s+1;
}
}
}
printf("%d", s);
return 0;
}
HTH