Question

Var s,k:integer; begin s:=20; for k:=-45 to 43 do s:=s-k; writeln(s) end.

Answer 1

В начальный момент s=20
s=20-(-45)=65
s=65-(-44)=109
остальное взаимоуничтожится, потому что s-(-43) = - (s - 43)

Answer 2

Let's analyze the given Pascal code step by step:

1. Var s, k: integer;: This line declares two integer variables, s and k.

2. s := 20;: The variable s is initialized with the value 20.

3. for k := -45 to 43 do: This starts a loop where the variable k will take on values from -45 to 43, inclusive.

4. s := s - k;: In each iteration of the loop, the value of k will be subtracted from the current value of s.

5. writeln(s): Finally, the value of s is printed after the loop is completed.

Let's calculate the final value of s:

1. Iteration when k = -45: s = 20 - (-45) = 65
2. Iteration when k = -44: s = 65 - (-44) = 109
3. Iteration when k = -43: s = 109 - (-43) = 152
4. ...and so on, until k = 43.

The loop will continue to execute until k reaches 43. Let's continue the calculations:

40. Iteration when k = 39: s = 1333 - 39 = 1294
41. Iteration when k = 40: s = 1294 - 40 = 1254
42. Iteration when k = 41: s = 1254 - 41 = 1213
43. Iteration when k = 42: s = 1213 - 42 = 1171
44. Iteration when k = 43: s = 1171 - 43 = 1128

So, the final value of s will be 1128, and that's what will be printed by the writeln(s) statement.

Note: The loop in this code iterates from -45 to 43, which means it will perform 89 iterations in total. Since the loop only involves simple arithmetic operations, there are no issues with the loop boundaries or any potential errors in the code.