A practical introduction to Pascal programming language - GIARA

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Data structuresProgram 84: Exercise 4.6 (Version B)program c04e06;constN=5;typeintVector=ARRAY[1..N] OF integer;vari:integer;A,B:intVector;begin(* Reading *)writeln(’Introduce the vector of ’,N,’ elements, splitted by blank spaces.’);for i:=1 to N dobeginread(A[i]);end;(* Accumulating *)B[1]:=A[1];for i:=2 to N doB[i]:=B[i-1]+A[i];(* Accumulating *)write(’The values in the vector are: ’);for i:=1 to N dowrite(B[i],’ ’);end.⊡Exercise 4.7. Create a program that reads a vector A of N reals (being N a constant), and createsa new vector so that at every position i it contains the sum of the values of A in the positionsi − 2, i − 1 and i.□This problem is similar to the previous, but the solution cannot be as elegant as that inProgram 84. The reason is that different positions of B, more specifically the first 3, cannot befilled in using the general schema, because for such positions there are not enough values on theleft. Still, the general solution for the problem resembles that in the previous exercise, exceptionmade that we need to remove one of the values from the count in B[i-1], since the value in theposition i-3 should not be taken into account in the sum for a given B[i].Program 85: Exercise 4.7program c04e07;constN=10;typeintVector=ARRAY[1..N] OF integer;vari:integer;A,B:intVector;begin(* Reading *)writeln(’Introduce the vector of ’,N,’ elements, splitted by blank spaces.’);for i:=1 to N doread(A[i]);(* Accumulating *)B[1]:=A[1];(* Initial values *)for i:=2 to 3 do73
A Practical Introduction to Pascal Programming Language⊡B[i]:=B[i-1]+A[i];(* General schema and, then, accumulation *)for i:=4 to N doB[i]:=B[i-1]-A[i-3]+A[i];write(’The values in the vector are: ’);for i:=1 to N dowrite(B[i],’ ’);end.Exercise 4.8. Create a program that reads a vector of N integers (being N a constant), then findsthe last negative number.□The previous exercises had something in common: we needed an exhaustive examination ofthe array in order to obtain the results. That is, all the elements of the array had to be checkedto be sure about the exactitude of the exercise. In such cases, the for iterative structure is themost convenient option, since we can use the own counting variable as index for accessing allthe positions of the array. However, in this exercise we don’t have to examine all the elementsin the array, but to stop when we find the last (first starting by the end) negative instead. Thewhile iterative structure seems now more appropriate. Note that we need to update the valueof i at each iteration.There are two different reasons why the iteration could stop. The first one is that myVector[i]is a negative integer. In that case, i is greater than 0, and we can state that the program foundthe last negative number in the array. Alternatively, it could be that there is no negatives inthe array. In that case, the program would iterate until i becomes 0. Then, the first conditionof the while would become false, and the program would stop iterating. Hence, it is clear thatthe value of i after exiting the while is the indicator we need to know which of the previousscenarios happened.⊡Program 86: Exercise 4.8program c04e08;constN=10;typeintVector=ARRAY[1..N] OF integer;vari:integer;myVector:intVector;begin(* Reading *)writeln(’Introduce the vector of ’,N,’ elements, splitted by blank spaces.’);for i:=1 to N doread(myVector[i]);(* Seeking the last negative *)i:=N;while (i>=1) and (myVector[i]>=0) doi:=i-1;(* Checking out what happened at the loop *)if (i>=1) thenwriteln(’The last negative is ’,myVector[i],’, in pos ’,i,’.’)elsewriteln(’There are no negatives in the vector’)end.74
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