PP 2017 Zadanie 5

From Łukasz Graczykowski

Revision as of 18:20, 9 April 2017 (view source) Lgraczyk (Talk | contribs) ← Older edit		Revision as of 18:20, 9 April 2017 (view source) Lgraczyk (Talk | contribs) Newer edit →
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	'''K01:''' Dla <code>j = n - 1, n - 2, ..., 0</code>: wykonuj '''K02'''...'''K04'''<br>		'''K01:''' Dla <code>j = n - 1, n - 2, ..., 0</code>: wykonuj '''K02'''...'''K04'''<br>
	'''K02:''' <code>x ← tablica[j];  i ← j + 1</code><br>		'''K02:''' <code>x ← tablica[j];  i ← j + 1</code><br>
-	'''K03:''' Dopóki <code>( i n )  ∧  ( x > tablica[i] )</code>: wykonuj <code>tablica[i - 1] ← tablica[i];  i ← i + 1</code><br>	+	'''K03:''' Dopóki <code>( i <= n )  ∧  ( x > tablica[i] )</code>: wykonuj <code>tablica[i - 1] ← tablica[i];  i ← i + 1</code><br>
	'''K04:'''    <code>tablica[i - 1] ← x</code><br>		'''K04:'''    <code>tablica[i - 1] ← x</code><br>
	'''K05:'''    Zakończ		'''K05:'''    Zakończ

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Revision as of 18:20, 9 April 2017

Lista kroków (przykład dla n-elementowej tablicy):
K01: Dla j = n - 1, n - 2, ..., 0: wykonuj K02...K04
K02: x ← tablica[j]; i ← j + 1
K03: Dopóki ( i <= n ) ∧ ( x > tablica[i] ): wykonuj tablica[i - 1] ← tablica[i]; i ← i + 1
K04: tablica[i - 1] ← x
K05: Zakończ