Informatik für Mathematiker und Physiker HS16 Solutions 3

Lecturer: Prof. B. Gärtner
Informatik für Mathematiker und Physiker
HS16
Solutions 3
Solution 1
Summarized:
(i)
(ii)
(iii)
(iv)
(v)
type
int
bool
bool
bool
bool
value
2
false
true
true
true
Now about how to obtain these: here we do it in more detail than you’re usually asked for,
if they appear in an exam. We first show you the fully-parenthesized expressions indicating in
which order subexpressions are evaluated. The parentheses are applied according to the rules of
precedence and associativity.
a) The parenthesized expressions are:
(i) (n / f) * d
(ii) u == (4 * (++d))
(iii) ((0 < d) < 2) || (2 > f)
(iv) ((0 < 1) && (1 != 1)) || ((++x) > 3)
(v) ((3 * d) > d) || (((1 / z) != 0) && ((3 + 4) >= 7))
b) Possible step-by-step evaluations:
(i) (n / f)
(n / f)
(9 / f)
(9 / 5)
1 * 2→
2
*
*
*
*
d
2
2
2
→
→
→
→
Type: int
1
(ii) (Note:
u ==
u ==
u ==
4 ==
false
this exercise was slightly changed compared to the one in the exam.)
(4 * (++d)) →
(4 * 3) →
12 →
12 →
Type: bool
(iii) ((0 < d) < 2) || (2 > f) →
((0 < 2) < 2) || (2 > f) →
(true < 2) || (2 > f) →
(1 < 2) || (2 > f) →
true || (2 > f) →
true
Type: bool
Note: Due to short circuit evaluation, the right part of the or statement will not
be executed (ever!). If you tried to do so, your solution is wrong (independent of the
result you might have found).
(iv) ((0 < 1) && (1 != 1)) || ((++x) > 3) →
(true && (1 != 1)) || ((++x) > 3) →
(true && false) || ((++x) > 3) →
false || ((++x) > 3) →
false || (4 > 3) →
false || true →
true
Type: bool
(v) ((3 * d) > d) || (((1 / x) !=
((3 * 2) > d) || (((1 / x) !=
(6 > d) || (((1 / x) != 0) &&
(6 > 2) || (((1 / x) != 0) &&
true || (((1 / x) != 0) && ((3
true
0) && ((3 + 4) >= 7)) →
0) && ((3 + 4) >= 7)) →
((3 + 4) >= 7)) →
((3 + 4) >= 7)) →
+ 4) >= 7)) →
Type: bool
Note: Due to short circuit evaluation, the right part of the or statement will not
be executed (ever!). If you tried to do so, your solution is wrong (independent of the
result you might have found).
Solution 2
1
2
// I n f o r m a t i k − S e r i e 3 − Aufgabe 2
// Programm : c r o s s s u m . cpp
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3
4
5
6
// A u t o r : . . . ( Gruppe . . . )
// Compute t h e c r o s s sum o f a n a t u r a l number .
//#i n c l u d e ” t e s t s . h” // remove s l a s h e s a t b e g i n n i n g o f l i n e t o t e s t ←or submit
7 #include <i o s t r e a m >
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25
int main ( )
{
// i n p u t
unsigned int n ;
std : : cin >> n ;
// c o m p u t a t i o n
unsigned int cross = 0 ;
for ( unsigned int m = n ; m > 0 ; m /= 1 0 )
cross += m % 1 0 ;
// o u t p u t
std : : cout << "Cross sum of " << n << " is: "
<< cross << "\n" ;
return 0 ;
}
Solution 3
1
2
3
4
5
6
//
//
//
//
I n f o r m a t i k − S e r i e 3 − Aufgabe 3
Programm : d e c 2 b i n . cpp
A u t o r : . . . ( Gruppe . . . )
Output r e v e r s e b i n a r y r e p r e s e n t a t i o n o f a number n .
//#i n c l u d e ” t e s t s . h” // remove s l a s h e s a t b e g i n n i n g o f l i n e t o t e s t ←or submit
7 #include <i o s t r e a m >
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17
int main ( )
{
// i n p u t
unsigned int n ;
std : : cin >> n ;
// c o m p u t a t i o n and o u t p u t
std : : cout << "Reverse binary representation is: " ;
if ( n == 0 )
// s p e c i a l c a s e !
3
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25
std : : cout << n ;
for ( ; n > 0 ; n /= 2 )
std : : cout << n % 2 ;
std : : cout << "\n" ;
// no o u t p u t f o r n==0
return 0 ;
}
4

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