About Question enthuware.ocajp.i.v7.2.960 :

[disznoperzselo]

It is easy if you try to apply highschool mathematics.
As a first step understand what the innermost loop does:

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for(int k = 0; k < 3; k++){ 
   c++;  
   if(k>j)
     break; 
}

Clearly, it increments c with the following sum:

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sum{ 1: 0 <=  k < = j + 1}  =  j + 2
 

Now, you have to run the refactored nested loops in your head
- or sum it up using an equivalent math formula -

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for(int i = 0; i < 2; i++){
   for(int j = 0; j < 2; j++){ 
        c += j + 2;
   }
}

That is

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sum { j + 2 :  j = 0,1}  =  5 
c = sum { 5 : i = 0,1 } = 10.

[EelcoD]

I did it this way:
If you leave out the break-statement, the "c++" statement is executed 12 times (3x2x2).

Quickly write down all the loops:
i-j-k
1) 0-0-0
2) 0-0-1
3) 0-0-2
4) 0-1-0
5) 0-1-1
6) 0-1-2
7) 1-0-0
8) 1-0-1
9) 1-0-2
10) 1-1-0
11) 1-1-1
12) 1-1-2

As soon as (k>j) the next loop (if any) is not executed.

Above this means that the loop 3) is not executed, since loop 2) already satisfied the condition.
Loop 9) is also not executed, since loop 8) already satisfied the condition.

I hope this helps anyone.

[JuergGogo]

The tip of computerinfoscience works fine.
As a first step forget about the outer loop A, since it has no effect on the if-break statement. So the question is reduced to two nested loops which can be analyzed easily. --> c=5
At the end double c *= 2 since Loop A runs twice. --> c=10

[RobinDRG]

I added an statement in the inner loop to help me understand how it works. Hope help someone else:


i:0 j:0 k:0 C:1
i:0 j:0 k:1 C:2
i:0 j:1 k:0 C:3
i:0 j:1 k:1 C:4
i:0 j:1 k:2 C:5
i:1 j:0 k:0 C:6
i:1 j:0 k:1 C:7
i:1 j:1 k:0 C:8
i:1 j:1 k:1 C:9
i:1 j:1 k:2 C:10
final c= 10

[Dreamweaver]

I simplify on paper like this:

c = 0 | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10
i = 0 | 0 | 0 | 0 | 0 | 0 | 1 | 1 | 1 | 1 | 1
j = 0 | 0 | 0 | 1 | 1 | 1 | 0 | 0 | 1 | 1 | 1
k = 0 | 0 | 1 | 0 | 1 | 2 | 0 | 1 | 0 | 1 | 2