三个连续数字的C乘积

Question

I was trying to solve this tricky question, but for some reason my code is doing something wrong... I don't exactly know why, but I'll try to explain as much as I can.

Consecutive products: Write a program that reads a positive integer from standard input and verifies if it's equal to the product of three natural and consecutive numbers. For example, the number 120 is equal to 4x5x6, as for number 90 there aren't any three consecutive natural numbers whose product is 90. Your program should generate as output 'S' if there are 3 consecutive natural numbers whose product is the value read, or 'N' if none.

Input

120

Expected Output

"S"


Input

60

Expected Output

"S"


Input

80

Expected Output

"N"


Input

120

Expected Output

"S"

And this is my code:

 #include <stdio.h>
int main(){
    int int1,i,count=10,j,k,w=0;
    scanf("%i",&int1);
    for (i = 1; i <= count; ++i)
    {
        for (j = 1; j <= count+1; ++j)
        {
            for ( k = 1; k <= count+2; ++k)
            {
                if ((i==j+1 && i==k+2) && (i*j*k==int1)){
                    w=1;
                }
            }
        }
    }
    if (w==0)
    {
        printf("N");
    }
    else{
        printf("S");
    }
}

So basically what this does is I have 3 loops that will generate random numbers in a k*i*j form... and it checks if we are getting what we want(the product of three natural and consecutive numbers) . This is for an assignment.

Answer 1

You don't need 3 loops. One trival approach would be:

int test(int num)
{
    for (int i = 1; i < num; i++)
    {
        int product = i * (i + 1) * (i + 2);
        if ( product == num )
            return true;
        else if (product > num)
            break;
    }
    return false;
}

Answer 2

I modified your code. Please let me know if the problem still exists. The change made is exactly as WDS said.

#include <stdio.h>
int main(){
int int1,i,count=10,j,k,w=0,comp;
scanf("%i",&int1);
for (i = 1; i <= count; ++i)
{
    comp = i*(i+1)*(i+2);
    if(comp==int1)
    {
        w = 1;
    }
}
if (w==0)
{
    printf("N");
}
else
{
    printf("S");
}
return 0;
}

Answer 3

You might want to add the algorithm tag on this question. That said, my approach would be to consider what the product of 3 consecutive numbers is. You could write it as x * (x+1) * (x+2). But there is a better way.

Write it as (x-1) * x * (x+1). Then multiply and simplify. The result is x^3-x.

Now for any given number, start a single loop on x from x = 2 (because if x=1 then x-1=0 and this will never be a solution) and incrementing by 1 each loop. Check on each loop for a match with the input number. If it is a match, return true. If it is not a match and exceeds the input number return false. If it is not a match and is less than the input number, loop again.