[C][Algorithm]The logic to get the divisors and the codes in C

- Here we go

Let's get the divisors of n!

1) Basic - Divide n in i ( i substitutes from 1 to n). If the remainder is 0, the i would be a divisor. 

1 2 3 4 5 6 7 8 9 10 11 12 #include <stdio.h>   int main(void) {     int i,n;     scanf("%d", &n);          for(i=1; i<=n; i++)     {         if((n%i)==0)              printf("%d\n",i);      } }

2) Avoid repetition - Suppose n is 20.

1    20

2    10

4    5

5    4     ☜ It's repeated from here.

10    2

20    1

√n is the boundary to be able to avoid repetition.

1 2 3 4 5 6 7 8 9 10 11 12 13 14 #include <stdio.h>   int main(void) {         int input,m,a,b;     scanf("%d", &input);       for(m=1;m*m<input;m++)     {         if(input % m == 0)             printf("%d, %d, ",m,input/m);      }     if(m*m == input)         printf("%d, ", m); }

You can get the number of the divisors with this logic above.

Happy coding~!!