simple C problem

Question

Hi, I have had to start to learning C as part of a project that I am doing. I have started doing the 'euler' problems in it and am having trouble with the first one. I have to find the sum of all multiples of 3 or 5 below 1000. Could someone please help me. Thanks.

#include<stdio.h>
int start;
int sum;

int main() {
    while (start < 1001) {
        if (start % 3 == 0) {
            sum = sum + start;
            start += 1;
        } else {
            start += 1;
        }

        if (start % 5 == 0) {
            sum = sum + start;
            start += 1;
        } else {
            start += 1;
        }
        printf("%d\n", sum);
    }
    return(0);
}

Answer 1

You haven't said what the program is supposed to do, or what your problem is. That makes it hard to offer help.

At a guess, you really ought to initialize start and sum to zero, and perhaps the printf should be outside the loop.

Answer 2

You would be much better served by a for loop, and combining your conditionals.

Not tested:

int main()
{
  int x;
  int sum = 0;

  for (x = 1; x <= 1000; x++)
    if (x % 3 == 0 || x % 5 == 0)
      sum += x;

  printf("%d\n", sum);
  return 0;
}

Answer 3

You could change your ifs:

 if  ((start % 3 == 0) || (start % 5 == 0)) 
     sum += start;
 start ++;

and don´t forget to initialize your sum with zero and start with one. Also, change the while condition to < 1000.

Answer 4

Eh right, well i can see roughly where you are going, I'm thinking the only thing wrong with it has been previously mentioned. I did this problem before on there, obviously you need to step through every multiple of 3 and 5 and sum them. I did it this way and it does work:

int accumulator = 0;
int i;

for (i = 0; i < 1000; i += 3)
 accumulator += i;

for (i = 0; i < 1000; i +=5) {
 if (!(i%3==0)) {
  accumulator += i;
 }
}
printf("%d", accumulator);

EDIT: Also note its not 0 to 1000 inclusive, < 1000 stops at 999 since it is the last number below 1000, you have countered that by < 1001 which means you go all the way to 1000 which is a multiple of 5 meaning your answer will be 1000 higher than it should be.

Answer 5

Really you need a debugger, and to single-step through the code so that you can see what it's actually doing. Your basic problem is that the flow of control isn't going where you think it is, and rather than provide correct code as others have done, I'll try to explain what your code does. Here's what happens, step-by-step (I've numbered the lines):

1:    while (start < 1001) {
2:        if  (start % 3 == 0) {
3:            sum = sum + start;
4:            start += 1;
5:        }
6:        else {
7:            start += 1;
8:        }
9:
10:       if (start % 5 == 0) {
11:           sum = sum + start;
12:           start += 1;
13:       }
14:       else {
15:           start += 1;
16:       }
17:       printf("%d\n", sum);
18:    }

• line 1. sum is 0, start is 0. Loop condition true.
• line 2. sum is 0, start is 0. If condition true.
• line 3. sum is 0, start is 0. sum <- 0.
• line 4. sum is 0, start is 0. start <- 1.
• line 5. sum is 0, start is 1. jump over "else" clause
• line 10. sum is 0, start is 1. If condition false, jump into "else" clause.
• line 15. sum is 0, start is 1. start <- 2.
• line 16 (skipped)
• line 17. sum is 0, start is 2. Print "0\n".
• line 18. sum is 0, start is 2. Jump to the top of the loop.
• line 1. sum is 0, start is 2. Loop condition true.
• line 2. sum is 0, start is 2. If condtion false, jump into "else" clause.
• line 7. sum is 0, start is 2. start <- 3.
• line 10. sum is 0, start is 3. If condition false, jump into "else" clause.
• line 15. sum is 0, start is 3. start <- 4.
• line 17. sum is 0, start is 4. Print "0\n".

You see how this is going? You seem to think that at line 4, after doing sum += 1, control goes back to the top of the loop. It doesn't, it goes to the next thing after the "if/else" construct.

Answer 6

The problem with your code is that your incrementing the 'start' variable twice. This is due to having two if..else statements. What you need is an if..else if..else statement as so:

           if  (start % 3 == 0) {
                    sum = sum + start;
                    start += 1;
            }
            else if (start % 5 == 0) {
                    sum = sum + start;
                    start += 1;
            }
            else {
                    start += 1;
            }

Or you could be more concise and write it as follows:

if(start % 3 == 0)
    sum += start;
else if(start % 5 == 0)
    sum += start;
start++;

Either of those two ways should work for you.

Good luck!

Answer 7

The answers are all good, but won't help you learn C.

What you really need to understand is how to find your own errors. A debugger could help you, and the most powerful debugger in C is called "printf". You want to know what your program is doing, and your program is not a "black box".

Your program already prints the sum, it's probably wrong, and you want to know why. For example:

printf("sum:%d start:%d\n", sum, start);

instead of

printf("%d\n", sum);

and save it into a text file, then try to understand what's going wrong.

• does the count start with 1 and end with 999?
• does it really go from 1 to 999 without skipping numbers?
• does it work on a smaller range?

Answer 8

Here's a general solution which works with an arbitrary number of factors:

#include <stdio.h>

#define sum_multiples(BOUND, ...) \
    _sum_multiples(BOUND, (unsigned []){ __VA_ARGS__, 0 })

static inline unsigned sum_single(unsigned bound, unsigned base)
{
    unsigned n = bound / base;
    return base * (n * (n + 1)) / 2;
}

unsigned _sum_multiples(unsigned bound, unsigned bases[])
{
    unsigned sum = 0;

    for(unsigned i = 0; bases[i]; ++i)
    {
     sum += sum_single(bound, bases[i]);

     for(unsigned j = i + 1; bases[j]; ++j)
      sum -= sum_single(bound, bases[i] * bases[j]);
    }

    return sum;
}

int main(void)
{
    printf("%u\n", sum_multiples(999, 3, 5));
    return 0;
}

Answer 9

You have forgotten to initialize your variables,

Answer 10

Starting new here, I see plenty of good advice above. Without reading it all (something I should do, but I want to see how the site works) I note your algorithm is broken.

You want to sum multiples of 3 and 5, but increment your way through all the possible integers without regard to the possibility a number could satisfy both. Such mysterious numbers as 15, 30, etc...

So, perform both modulus tests before presuming to increment to the next int?