/*If we list all the natural numbers below 10 that are multiples of 3 or 5, we get 3, 5, 6 and 9.
The sum of these multiples is 23.
Find the sum of all the multiples of 3 or 5 below 1000. */
When I put 10 the answer is 23,when I try for 1000,output:266333,which is a wrong answer according to the eulerproject.netCode:#include<stdio.h> int main() { int i=1,m3=0,n; for(;i<10;i++) { if(((n=i*3)<10)) m3+=n; if((n=i*5)<10) m3+=n; } printf("res=%d\n",m3); return 0; }
What is wrong in the program?
I'd imagine that it's because you're summing a lot of numbers twice. E.g. 15 is a multiple of 3 and of 5, and your code will add it twice. Since there aren't any common multiples of 3 and 5 below 10, you wouldn't see the problem there.
Hi Shruthi,
Your code is adding numbers which are multiples of both 3 and 5 twice.
That's why you got wrong answer.
Try this
Code:if(((n=i*3)<10)) { m3+=n; } else { if((n=i*5)<10) m3+=n; }
Or to write less code
Code:if( ( (n=i*3)<10 ) || ((n=i*5)<10) m3+=n;
I've used the above suggestion and still haven't got the right solution.It's not even satisfying the first condition,23.it's giving 18 because it not considering 5*1=5,it's only considered 3*1=3 and given the result 18.
Your idea in the code you posted in post #1 looks like it should work, but your code is rather messy. Here's an improved version:
Notice that I got rid of the n variable, because you should not have been adding n to the sum (i.e., I fixed another bug for you, for free). However, I have left your double counting problem in the code for you to fix. It really is very easy to fix it nowCode:#include<stdio.h> int main(void) { int sum = 0; int i; for (i = 1; i < 1000; i++) { if (i % 3 == 0) { sum += i; } if (i % 5 == 0) { sum += i; } } printf("sum=%d\n", sum); return 0; }
Hint: notice that for both the if statement, the body is exactly the same. Can you combine them into a single if statement?
Look up a C++ Reference and learn How To Ask Questions The Smart WayOriginally Posted by Bjarne Stroustrup (2000-10-14)
Thank you laserlight,finally got the answer.
Code:#include<stdio.h> int main() { int i,sum=0; for(i=1;i<1000;i++) { if((i%3==0) && (i%5==0)) sum+=i; else { if(i%3==0) sum+=i; if(i%5==0) sum+=i; } } printf("sum=%d\n",sum); return 0; }
That's good, though what I had in mind was:
Code:#include<stdio.h> int main(void) { int sum = 0; int i; for (i = 1; i < 1000; i++) { if (i % 3 == 0 || i % 5 == 0) { sum += i; } } printf("sum=%d\n", sum); return 0; }
Look up a C++ Reference and learn How To Ask Questions The Smart Way
and also you can club this in a single statement
Code:if((i*3==0) || (i*5==0) sum+=i;
Ok,got ur point
