Question

I am trying to solve one problem from on-line judging system. I have a solution which works, but not efficient enough. Here is the problem:

Which the least number n can we imagine in product n = a∙b like k ways? Products a∙b and b∙a is one of the way, where all numbers is natural (1≤ k ≤50).

Input One number k.
Output One number n.

My code did not pass four tests. It is too slow for k=31, 37, 47. I have been thinking on this problem 2 days,but no improvement. Here is my code, please share, if you have any ideas.

    #include<stdio.h>
    #include<stdlib.h>
    #include<math.h>

    int prime[10000];
    long x,j,i,flag,k,length,p,checker,count,number;

    int main()
    {

    prime[0]=2;
    scanf("%ld",&k);
    //I find prime numbers between 1 and 1000. 1000 can be changed, just for testing

    for (i=3;i<=1000;i=i+2)
            {
            flag=0;
            for (j=2;j<=sqrt(i);j++)
                    {
                    if(i%j==0)
                            {
                            flag=1;
                            break;
                            }
                    }
            if(flag==0)
                    {
                    x++;
                    prime[x]=i;
                    }
            }

    length=x;
    //this loop is too big I know, again for testing. I suspect, there must be a way to make some changes to this for loop 

    for (i=1;i<10000000000;i++)
            {
            number=i;
            p=1;
            for(x=0;x<=length;x++)
                    {
                    if(prime[x]>sqrt(i))
                    break;
                    count=0;
                    while(number%prime[x]==0)
                            {
                            number=number/prime[x];
                            count++;       
                            }
                    p=p*(count+1); 
//I find prime factors of numbers and their powers, then calculate number of divisors

                    }
            //printf("%d\n",p);
            //number of ways is just number of divisors/2 or floor (divisors/2)+1
            if(p%2==0)
            checker=p/2;
            else
            checker=floor(p/2)+1;
            if(checker==k)
                    {
                    printf("%ld\n",i);
                    break;
                    }
            }

    return 0;
    }

Answer 1

If I understand the problem correctly it’s asking you which is the least number n with exactly 2k divisors (should I consider 1 and n?)

in fact if a number has a divisor a, then n / a = b is an integer and n = a* b (counting only one time a and b, so you should divide by two the number of divisors)

edit

Doing that is time consuming indeed. So this is the idea;

for a number n in the form n = p1^(a1)*p2^(a2)…pn^(an) (this is the prime factorization of the number) the number of divisor is (a1 + 1)(a2+1)…(an+1)

Hence, if you want to find a number that has k divisor, factorize k. then assign the biggest factor to the smallest prime; eg if k = 2*5*7, then n should be 2^7*3^5*5^2

I know it is not since i didnt take into account that (a, b) is equal to (b, a) but play around it a little and it should work

example

take k = 37. Then double the number – (to consider the symmetry). You get 74.
Now, if you can imagine n as n = n * 1, then you just need to factor 74 (that is 2 * 37);
then give 36 to 2 and 1 to 3, leading n = 2^(36)*3 = 206158430208

if you can’t, then you need to add 1 to the number you got previously (in this case, 74 + 1 = 75 = 25*3); this way you get n = 2^24 * 3^2 = 150994944

If it’s none of the above, then I am probably wrong…