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- function zadanie3(tabA,tabB,ma,mb,n)
- tabC is array[max(ma,mb)][2]
- begin
- max=0;
- if(ma>mb)then
- begin
- max=ma
- end
- else
- max=mb
- end
- for wiersze:=0...max-1 step1
- begin
- for kolumna:=0...1 step1
- begin
- tabC[wiersze][kolumna]=0
- end
- end
- kolumna:=0
- for wiersze:=0...max-1 step1
- begin
- if (tabA[wiersze]>tabB[wiersze])
- begin
- tabC[wiersz]=tabA[wiersz]
- end
- else
- begin
- tabC[wiersze]=tabB[wiersze]
- end
- end
- return tabC
- end
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