/* PASSFAIL: print out pass (>=45)/fail(<45)/pass distinction(>=81) on a score
             Note: should also verify input 0..100 but not that
             the code does not check this - bug. 
*/

/* 
1. FLOWGRAPH:
      
        R1       enter       artificial 
                   1 . . . . . . .      
                                 . 
                   |a            .
                   v             .
                                 .
                   2             .
                   |             .
           +-------+-------+     .
        b  |   F      T    | c   .
           v               v     .
                   R2            .
           4               3     .
                                 .
        d  |               | e   .
           +-----> 5 <-----+     .
                   |         R3  . i
                   + ------+     .
                   |   T   | g   .
                   |F      v     .
                 f |             .
                   |   R4  6     .
                   v             .
                           | h   .
                   7 <-----+     .
                  exit . . . . . .

2. CYCLOMETRIC COMPLEXITY:
  Predicates (2) + 1 = Edges (9) - Nodes (7) + 1 = Regions (4) - 1 = 3

3. BASIS SET OF CYCLES:
  1. 1-2-4-5-6-7-1   PASS WITH DISTINCTION
  2. 1-2-3-5-7-1     FAIL
  3. 1-2-4-5-7-1     PASS

  Note: The following is NOT a basis cycle because it is not independent:
  4. 1-2-3-5-6-7-1

    Proof:
      Consider nine edges: a,b,c,d,e,f,g,h,i
                                
                                    a b c d e f g h i
      In Cycle 1: <a,b,d,g,h,i> :  (1 1 0 1 0 0 1 1 1)
      In Cycle 2: <a,c,e,f,i>   : +(1 0 1 0 1 1 0 0 1)
      In Cycle 3: <a,b,d,f,i>   : -(1 1 0 1 0 1 0 0 1)
                                    -----------------
      In Cycle 4: <a,c,e,g,h,i> : =(1 0 1 0 1 0 1 1 1)

      Cycle 4 = Cycle 1 + Cycle 2 - Cycle 3
  
  Note: This path is also INFEASIBLE: FAIL WITH DISTINCTION
  4. 1-2-3-5-6-7-1

4. LINEARLY INDEPENDENT PATHS   5. FEASABLE (YES/NO)   6. TEST CASES   7. RESULT
                                                            (score)       (printf)
  1. 1-2-4-5-6-7                     YES                      85       PASS WITH DISTINCTION
  2. 1-2-3-5-7                       YES                      40       FAIL
  3. 1-2-4-5-7                       YES                      60       PASS

  Note: Although this white box strategy did 100% path coverage, it
        did not meet the specification. It did not test for 0..100.
        An input of 120 would pass the test. A BLACK BOX test of
        4 different values (40, 60, 85, 110) would have discovered
        the problem. 

        Therefore, even with 100% path coverage, no white box testing
        strategy on its own can guarantee adequate software testing.

See passfail.bat
*/

#include <stdio.h>
int debug = 0;
#define PASS 45
#define DISTINCTION 81
#define MIN 0
#define MAX 100

/* instrumentation for the line #'s*/
int L(int i) { if (debug) printf("%d-",i); return 1; } 

void passfail() {
int score;
  
  L(1); scanf("%d ",&score);

  if (L(2)&&(score<PASS)) {
      
      L(3); printf(" FAIL ");

  } else {

      L(4); printf(" PASS ");
  
  }
  
  if (L(5)&&(score>=DISTINCTION)) {
    
    L(6); printf(" WITH DISTINCTION ");
  }

  L(7); printf("\n");
}
       
int main(int argc, char *argv[]) {
int a, b, c;
  if (argc > 1) {
    debug = 1;
    printf("Path   : ");
  }
  passfail();
}