Creating a finite state machine in C language

 Finite state machine implementation in C

This is an implementation of FSM in C as you can see there are 3 functions named
cheese_day(),grapes_day() & lettuce_day() which represents the state/nodes
fsm is an array of pointers to function, which is used to access the functions
i.e. void (*fsm[])()= { cheese_day , grapes_day, lettuce_day };Now we are using the index of the fsm to jump to the specific function
i.e. fsm[input_symbol]();

1. #include <cstdio>
2. #include <cstdlib>
3. #include <cstring>
4. #define IM_STARVING 0
5. #define LETS_GO_SHOPPING 0
6. #define ILL_HAVE_CHEESE_TOMMORROW 1
7. #define ILL_HAVE_LETTUCE_TOMMORROW 2
8. #define ILL_HAVE_GRAPES_TOMMORROW 3
9. int inp[5]= {ILL_HAVE_CHEESE_TOMMORROW ,ILL_HAVE_LETTUCE_TOMMORROW,ILL_HAVE_CHEESE_TOMMORROW,ILL_HAVE_GRAPES_TOMMORROW,IM_STARVING};  // automated Input!
10. void cheese_day()
11. {
12.   printf("state :- cheese day\n");
13. }
14. void grapes_day()
15. {
16.   printf("state :- grapes day\n");
17. }
18. void lettuce_day()
19. {
20.   printf("state :- lettuce day\n");
21. }
22. void (*fsm[])()= { cheese_day , grapes_day, lettuce_day };
23. int main()
24. { int j=0;
25.   while(j< 5)
26.   { int input_symbol=inp[j++];
27.    switch(input_symbol) {
28.     case 1:
29.     input_symbol=ILL_HAVE_CHEESE_TOMMORROW - 1;
30.     fsm[input_symbol]();
31.     break;
32.     case 2:
33.     input_symbol=ILL_HAVE_LETTUCE_TOMMORROW - 1;
34.     fsm[input_symbol]();
35.     break;
36.     case 3:
37.     input_symbol=ILL_HAVE_GRAPES_TOMMORROW - 1;
38.     fsm[input_symbol]();
39.     break;
40.     default:
41.     fsm[input_symbol]();
42.    }
43.   }
44.  return 0;
45. }

Die Hard 3 Hackerrank Solution

   The primary idea is to shift the water in the larger jug to smaller jug, hence the difference of  the capacities of the jugs will be used as a constraint to find out variable c volume. we will also terminate the process of finding c if the difference between both the jugs is zero.

     There are 4 conditions 
      1)    if  diff >= min then diff=diff-min_capacity
        2)    if  diff < min then diff=max_capacity - ( min_capacity - diff)
        3)    if  diff == c then return "YES"
          4)     if   diff  < 0 then return "NO"

#include <cmath>
#include <cstdio>
#include <cstdlib>
#include <vector>
#include <iostream>
#include <algorithm>
#define max(a,b) a>b?a:b
#define min(a,b) a<b?a:b
using namespace std;
int jug(int a,int b,int c)
{
       int mx=max(a,b);
        int mn=min(a,b);
         if(a==c || b==c)
             return 1;
        int diff=abs(b-a);
        while(diff > 0)
        {
            if( diff==c )
              return 1;
           
           
            if(diff < mn)
            {
                diff=mx-(mn-diff);
               
            } else if(diff >= mn)
                {
                    diff=diff-mn;
                }
        }
             
                
   
    return 0;
}

int main() {
     int a,b,c,mx,mn,diff,t;
      cin >> t;
      while(t--)
       {  
       cin >>a>>b>>c;
          if(jug(a,b,c)==1)
              printf("YES\n");
          else printf("NO\n");
     
      }
   
    return 0;
}