Question

Var k, m: integer; Dat: array[1..10] of integer; Begin A [1] := 20; A [2] := 16; A [3] := 20; A [4] := 36; A [5] := 15; A [6] := 21; A [7] := 28; A [8] := 17; A [9] := 26; A [10] := 35; m := 1; for k := 1 to 10 do if A [k]< А[m] then begin m := m+1 end; writeln(m); End. С подробным пояснением пож.)))

Answer 1

Вы вводите 2 целочисленные переменные: k, m

И объявляете целочисленный массив: Dat , в котором выделяется память под 10 ячеек.

Далее вы присваиваете каждому элементу массива Dat, собственное значение A[1] := 20, A[4] := 36 и т.д

После этого вы присваиваете m значение 1

И создаете цикл от 1 до 10, в котором у вас происходит сравнение элемента A[k] и элемента A[m]. В случае, если А[k] < A[m]. То вы добавляете к переменной m единицу.

После этого цикла, вы выводите на экран это самое m, которое по сути показывает количество чисел, которые больше чем предыдущее

Answer 2

Sure! Let's go through the provided Pascal code step by step:

1. Variable Declaration:

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Var k, m: integer; 
Dat: array[1..10] of integer;

In this section, two integer variables k and m are declared, and an integer array Dat is also defined with 10 elements.

2. Array Initialization:

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Begin 
    A [1] := 20; 
    A [2] := 16; 
    A [3] := 20; 
    A [4] := 36; 
    A [5] := 15; 
    A [6] := 21; 
    A [7] := 28; 
    A [8] := 17; 
    A [9] := 26; 
    A [10] := 35; 

In this section, an array A is initialized with 10 values at different indices.

3. Loop and Comparison:

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    m := 1; 
    for k := 1 to 10 do 
        if A [k] < A[m] then 
        begin 
            m := m + 1 
        end; 

The code starts by setting m to 1. Then, it enters a loop where the variable k takes values from 1 to 10. During each iteration of the loop, it checks if the element at index k in array A is less than the element at index m in array A. If this condition is true, it increases the value of m by 1. This process effectively finds the index of the smallest element in array A.

4. Output:

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    writeln(m); 
End.

Finally, the code prints the value of m, which is the index of the smallest element in the array A.

Let's evaluate the code step by step using the provided array values:

A = [20, 16, 20, 36, 15, 21, 28, 17, 26, 35]

1. Initially, m is set to 1.

2. Loop iterations:

   • For k = 1, A[k] is 20, and since A[k] is not less than A[m] (which is also 20 at this point), m remains 1.
   • For k = 2, A[k] is 16, and since A[k] is less than A[m] (which is 20 at this point), m is updated to 2.
   • For k = 3, A[k] is 20, and since A[k] is not less than A[m] (which is 16 at this point), m remains 2.
   • The loop continues, and for k = 5, A[k] is 15, which is less than A[m] (which is 16 at this point), so m is updated to 5.
   • The loop continues until k = 10.

3. After the loop completes, m will be the index of the smallest element, which is 5 (corresponding to the value 15 in the array).

4. The code then prints the value of m, which is 5.

So, the output of the provided code will be:

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5

This means that the smallest element in the array is 15, and its index is 5 (remembering that array indices are 1-based in Pascal).