Find the smallest
possible integer x, that has exactly n divisors.
Input. One positive
integer n (1
≤ n ≤ 16).
Output. Print the smallest
possible integer x that has exactly n divisors.
|
Sample input 1 |
Sample output 1 |
|
2 |
2 |
|
|
|
|
Sample input 2 |
Sample output 2 |
|
4 |
6 |
number
theory -
divisors
Iterate through the numbers x = 1, 2, 3, ... and print
the first one, which contains exactly n divisors. This solution will pass time
limit, since n is not
big.
The function Divisors counts the number of divisors in
the number n using a simple full search.
int Divisors(int
n)
{
int res = 0, i;
for(i = 1; i <= n; i++)
if (n % i == 0) res++;
return res;
}
The main part of the program. Read the value of n.
scanf("%d",&n);
Iterate over the numbers i = 1, 2,
3, … . Stop the loop as soon as the
number i contains n divisors.
for(i = 1; ; i++)
{
d = Divisors(i);
if (d == n) break;
}
Print the answer.
printf("%d\n",i);