Arrays of different lengths and consist of numbers. It is necessary to find the same elements and display.
- Repetitive elements to print or not? - avp
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2 answers
f(int a[], int b[]) { print (intersect (a, b)) ; } What is the question, is the answer.
avp avp
37.4k 3 35 90
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Since the language is not known, the algorithm is:
Масив А \\Первый масив Масив Б \\Второй масив Масив В \\Массив в которм будут результат числа : i : счетчик для массива А j : счетчик для массива Б C : счетчик для массива В c := 1; for i= 1 to длина массива А do begin for j = 1 to длинна массива Б do begin if A[i] = Б[j] then // сравниваем значения и если совпало то выполняем begin B[c] := A[c]; c := c + 1; end; end; end; Well, here we derive B in the screen.
This is a mixture of Russian and Pascal ^)
gecube
13.4k 25 45
SoftR SoftR
963 four eight
- And if there are more than one identical elements? Say, two units in array A and five in array B? How many units will then be in array B? Also it is not indicated how to allocate memory for the array B =) - gecube
- one figure, I think enough - MAMN54
- By the way, thanks, I think it will come in handy on pascal) - MAMN54
- The bottom line is that when I wrote the answer, the question was not according to the language not according to the conditions. I just wrote one of the solutions to the problem, in a mixture of Russian and Pascal. - SoftR
- @gecube I did not write the working code, but showed only a solution to the problem. If in A 2 fields 1 and in B 5 fields 1 then in B there will be 10 fields with 1. Nobody, when setting the task, set the task unique. - SoftR
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