Integer division problem recursive version

Recursive partitioning is easier to understand but time out

To divide n by a number m not greater than n

Total of four cases

1. If N==m has only one but to continue recursion, it is equal to 1+q (n,m-1)

2. If N<m does not consider the inability to partition then continue to use Q (n,n)

3. If there are two cases of n>m 1. Dividing the remainder with M is n-m so equals Q (n-m,m)

2. Continue dividing by a number less than M is Q (n,m-1)

3. So together is Q (n-m,m) +q (n,m-1)

4. Finally, the bottom of the recursive exit is definitely 1.

1 if (n==1 | | m==1)2     return1;

The complete code is as follows

1#include"stdio.h"2 intQintNintm)3 {4     if(n==1|| m==1)5         return 1;6     if(n<m)7         returnQ (n,n);8     if(n==m)9         return 1+q (n,m-1);Ten     if(n>m) One         returnQ (n-m,m) +q (n,m-1); A }  - Main () - { the     intN; -      while(SCANF ("%d", &n)! =EOF) -     { -printf"%d\n", Q (n,n)); +     } -}

Integer division problem recursive version