← Kembali ke Olimpiade Sains Kota (OSK) 2018 - Komputer

Olimpiade Sains Kota (OSK) 2018 - Komputer, Nomor 35

var arr: array[1..10] of integer;
 i, j, s: integer;
begin
    for i := 1 to 10 do
         arr[i] := i + 1;
    for i := 1 to 10 do
    begin
         for j:= i+1 to 10 do
         begin
            if (arr[i] = -1)then break;
               s := arr[j] div arr[i];
               if (s * arr[i] = arr[j]) then
               begin
                   arr[i] := arr[i] + arr[j];
                   arr[j] := arr[i] - arr[j];
                   arr[i] := arr[i] - arr[j];
                   arr[i] := -1;
               end;
         end;
    end;
end.

Berapa nilai dari arr[9] setelah program selesai dijalankan ?

a. -1

b. 0

c. 2

d. 5

e. 10

3 jawaban

daliman semaka7 tahun lalu

perulangan dari 1 sd 10 (karena nilai variabel perulangan i dari 1 sd 10)

proses ke : 1

nilai awal arr [1..10]:=(2,3,4,5,6,7,8,9,10,11)

perulangan j (2 sd 10) ==> j:=i+1 maka i+1 disini hasilnya adalah 2 sd 10

(j=2) diketahui

i=1,j=2,arr[i]=2,arr[j]=3

karena arr[1] tidak sama dengan -1 maka lanjut proses

S=arr[j]/arr[i];

=3/2 = 1

karena S*arr[i] (2*1) tidak sama dengan 3 maka lewatkan proses logika

arr[2]=3;****

kemudian lanjut ke j=3

(j=3)diketahui

i=1,j=3,arr[i]=2,arr[j]=4

karena arr[1] tidak sama dengan -1 maka lanjut proses

S=arr[j]/arr[i];

=4/2 = 2

karena S*arr[i] (2*2)=sama dengan 4 (arr[j]) maka

ada perubahan/perhitungan

arr[i] := arr[i] + arr[j];

arr[i] := 2 + 4 = 6

arr[1] := 6

arr[j] := arr[i] - arr[j];

arr[j] := 6-4 = 2

arr[3] := 2;****

arr[i] := arr[i] - arr[j];

arr[i] := 6 - 2 = 4

arr[1] := 4

arr[i] := -1;

arr[1] := -1;****

(j=4)diketahui

i=1,j=4,arr[i]=-1,arr[j]=5

karena arr[1] sama dengan -1 maka break

arr[4]=5;****

kesimpulan :

nilai arr[1]= -1

nilai arr[2]= 3

nilai arr[3]= 2

nilai arr[4]= 5

nilai akhir arr [1..10]:=(-1,3,2,5,6,7,8,9,10,11)

proses ke : 2

nilai awal arr [1..10]:=(-1,3,2,5,6,7,8,9,10,11)

perulangan j (3 sd 10)

(j=3) diketahui

i=2,j=3,arr[i]=3,arr[j]=2

if (arr[i] = -1)then break ==>lewati

s := arr[j]/arr[i];

= 2/3=0

if (s * arr[i] = arr[j]) ==>lewati

arr[3]=2;****

(j=4) diketahui

i=2,j=4,arr[i]=3,arr[j]=5

if (arr[i] = -1)then break ==>lewati

s := arr[j]/arr[i];

= 5/3=1

if (s * arr[i] = arr[j]) ==>lewati

arr[4]=5;****

(j=5) diketahui

i=2,j=5,arr[i]=3,arr[j]=6

if (arr[i] = -1)then break ==>lewati

s := arr[j]/arr[i];

= 6/3=2

karena S*arr[i] (2*3)=sama dengan 6 (arr[j]) maka

ada perubahan/perhitungan

arr[i] := arr[i] + arr[j];

arr[i] := 3 + 6 = 9

arr[2] := 9

arr[j] := arr[i] - arr[j];

arr[j] := 9-6 = 3

arr[5] := 3;****

arr[i] := arr[i] - arr[j];

arr[i] := 9 - 3 = 6

arr[2] := 6

arr[i] := -1;

arr[2] := -1;****

(j=6)diketahui

i=2,j=6,arr[i]=-1,arr[j]=7

karena arr[1] sama dengan -1 maka break

arr[6]=7;****

kesimpulan :

nilai arr[2]=-1

nilai arr[3]=2

nilai arr[4]=5

nilai arr[5]=3

nilai arr[6]=7

nilai akhir arr [1..10]:=(-1,-1,2,5,3,7,8,9,10,11)

proses ke : 3

nilai awal arr [1..10]:=(-1,-1,2,5,3,7,8,9,10,11)

perulangan j (4 sd 10)

(j=4)diketahui

i=3,j=3,arr[i]=2,arr[j]=5

if (arr[i] = -1)then break ==>lewati

s := arr[j]/arr[i];

= 5/2=2

if (s * arr[i] = arr[j]) ==>lewati

arr[4]=5;****

(j=5) diketahui

i=3,j=5,arr[i]=2,arr[j]=3

if (arr[i] = -1)then break ==>lewati

s := arr[j]/arr[i];

= 3/2=1

if (s * arr[i] = arr[j]) ==>lewati

arr[5]=3;****

(j=6) diketahui

i=3,j=6,arr[i]=2,arr[j]=7

if (arr[i] = -1)then break ==>lewati

s := arr[j]/arr[i];

= 7/2=3

if (s * arr[i] = arr[j]) ==>lewati

arr[6]=7;****

(j=7) diketahui

i=3,j=7,arr[i]=2,arr[j]=8

if (arr[i] = -1)then break ==>lewati

s := arr[j]/arr[i];

= 8/2=4

if (s * arr[i] = arr[j])

ada perubahan/perhitungan

arr[i] := arr[i] + arr[j];

arr[i] := 2 + 8 = 10

arr[3] := 10

arr[j] := arr[i] - arr[j];

arr[j] := 10-8 = 2

arr[7] := 2;****

arr[i] := arr[i] - arr[j];

arr[i] := 10 - 2 = 8

arr[3] := 8

arr[i] := -1;

arr[3] := -1;****

(j=8)diketahui

i=3,j=6,arr[i]=-1,arr[j]=9

karena arr[1] sama dengan -1 maka break

arr[8]=9;****

kesimpulan :

nilai arr[3]=-1

nilai arr[4]=5

nilai arr[5]=3

nilai arr[6]=7

nilai arr[7]=2

nilai arr[8]=9

nilai akhir arr [1..10]:=(-1,-1,-1,5,3,7,2,9,10,11)

proses ke : 4

nilai awal arr [1..10]:=(-1,-1,-1,5,3,7,2,9,10,11)

perulangan j (5 sd 10)

(j=5)diketahui

i=4,j=5,arr[i]=5,arr[j]=3

if (arr[i] = -1)then break ==>lewati

s := arr[j]/arr[i];

= 3/5=0

if (s * arr[i] = arr[j])

==>lewati

arr[5]=3;****

(j=6)diketahui

i=4,j=6,arr[i]=5,arr[j]=7

if (arr[i] = -1)then break ==>lewati

s := arr[j]/arr[i];

= 7/5=1

if (s * arr[i] = arr[j])

==>lewati

arr[6]=7;****

(j=7)diketahui

i=4,j=7,arr[i]=5,arr[j]=2

if (arr[i] = -1)then break ==>lewati

s := arr[j]/arr[i];

= 2/5=0

if (s * arr[i] = arr[j])

==>lewati

arr[7]=2;****

(j=8)diketahui

i=4,j=8,arr[i]=5,arr[j]=9

if (arr[i] = -1)then break ==>lewati

s := arr[j]/arr[i];

= 9/5=1

if (s * arr[i] = arr[j])

==>lewati

arr[8]=9;****

(j=9)diketahui

i=4,j=8,arr[i]=5,arr[j]=10

if (arr[i] = -1)then break ==>lewati

s := arr[j]/arr[i];

= 10/5=2

if (s * arr[i] = arr[j])

ada perubahan/perhitungan

arr[i] := arr[i] + arr[j];

arr[i] := 5 + 10 = 15

arr[4] := 15

arr[j] := arr[i] - arr[j];

arr[j] := 15-10 = 5

arr[9] := 5;****

arr[i] := arr[i] - arr[j];

arr[i] := 15 - 5 = 10

arr[4] := 10

arr[i] := -1;

arr[4] := -1;****

kesimpulan :

nilai arr[4]=-1

nilai arr[5]=3

nilai arr[6]=7

nilai arr[7]=2

nilai arr[8]=9

nilai arr[9]=5

nilai arr[10]=11

nilai akhir arr [1..10]:=(-1,-1,-1,-1,3,7,2,9,5,11)

proses ke : 5

nilai awal arr [1..10]:=(-1,-1,-1,-1,3,7,2,9,5,11)

perulangan j (6 sd 10)

(j=6)diketahui

i=5,j=6,arr[i]=3,arr[j]=7

if (arr[i] = -1)then break ==>lewati

s := arr[j]/arr[i];

= 7/3=2

if (s * arr[i] = arr[j])

==>lewati

arr[6]=7;****

(j=7)diketahui

i=5,j=7,arr[i]=3,arr[j]=2

if (arr[i] = -1)then break ==>lewati

s := arr[j]/arr[i];

= 2/3=0

if (s * arr[i] = arr[j])

==>lewati

arr[7]=2;****

(j=8)diketahui

i=5,j=8,arr[i]=3,arr[j]=9

if (arr[i] = -1)then break ==>lewati

s := arr[j]/arr[i];

= 9/3=3

if (s * arr[i] = arr[j])

ada perubahan/perhitungan

arr[i] := arr[i] + arr[j];

arr[i] := 3 + 9 = 12

arr[5] := 12

arr[j] := arr[i] - arr[j];

arr[j] := 12-9 = 3

arr[8] := 3;****

arr[i] := arr[i] - arr[j];

arr[i] := 12 - 9 = 3

arr[5] := 3;

arr[i] := -1;

arr[5] := -1;****

(j=9)diketahui

i=5,j=9,arr[i]=-1,arr[j]=5

if (arr[i] = -1)then break

arr[9]=5;****

kesimpulan :

nilai arr[5]=-1

nilai arr[6]=7

nilai arr[7]=2

nilai arr[8]=3

nilai arr[9]=5

nilai akhir arr [1..10]:=(-1,-1,-1,-1,-1,7,2,3,5,11)

proses ke : 6

nilai awal arr [1..10]:=(-1,-1,-1,-1,-1,7,2,3,5,11)

perulangan j (7 sd 10)

(j=7)diketahui

i=6,j=7,arr[i]=7,arr[j]=2

if (arr[i] = -1)then break ==>lewati

s := arr[j]/arr[i];

= 2/7=0

if (s * arr[i] = arr[j]) ==> lewati

arr[7]=2;****

(j=8)diketahui

i=6,j=8,arr[i]=7,arr[j]=3

if (arr[i] = -1)then break ==>lewati

s := arr[j]/arr[i];

= 3/7=0

if (s * arr[i] = arr[j]) ==> lewati

arr[8]=3;****

(j=9)diketahui

i=6,j=9,arr[i]=7,arr[j]=5

if (arr[i] = -1)then break ==>lewati

s := arr[j]/arr[i];

= 5/7=0

if (s * arr[i] = arr[j]) ==> lewati

arr[9]=5;****

(j=10)diketahui

i=6,j=10,arr[i]=7,arr[j]=11

if (arr[i] = -1)then break ==>lewati

s := arr[j]/arr[i];

= 11/7=1

if (s * arr[i] = arr[j]) ==> lewati

arr[10]=11;****

kesimpulan :

nilai arr[6]=7

nilai arr[7]=2

nilai arr[8]=3

nilai arr[9]=5

nilai arr[10]=11

nilai akhir arr [1..10]:=(-1,-1,-1,-1,-1,7,2,3,5,11)

proses ke : 7

nilai awal arr [1..10]:=(-1,-1,-1,-1,-1,7,2,3,5,11)

perulangan j (8 sd 10)

(j=8)diketahui

i=7,j=8,arr[i]=2,arr[j]=3

if (arr[i] = -1)then break ==>lewati

s := arr[j]/arr[i];

= 3/2=1

if (s * arr[i] = arr[j]) ==> lewati

arr[8]=3;****

(j=9)diketahui

i=7,j=9,arr[i]=2,arr[j]=5

if (arr[i] = -1)then break ==>lewati

s := arr[j]/arr[i];

= 5/2=2

if (s * arr[i] = arr[j]) ==> lewati

arr[9]=5;****

(j=10)diketahui

i=7,j=10,arr[i]=2,arr[j]=11

if (arr[i] = -1)then break ==>lewati

s := arr[j]/arr[i];

= 11/2=5

if (s * arr[i] = arr[j]) ==> lewati

arr[10]=11;****

kesimpulan :

nilai arr[7]=2

nilai arr[8]=3

nilai arr[9]=5

nilai arr[10]=11

nilai akhir arr [1..10]:=(-1,-1,-1,-1,-1,7,2,3,5,11)

proses ke : 8

nilai awal arr [1..10]:=(-1,-1,-1,-1,-1,7,2,3,5,11)

perulangan j (9 sd 10)

(j=9)diketahui

i=8,j=9,arr[i]=3,arr[j]=5

if (arr[i] = -1)then break ==>lewati

s := arr[j]/arr[i];

= 5/3=1

if (s * arr[i] = arr[j]) ==> lewati

arr[9]=5;****

(j=10)diketahui

i=8,j=10,arr[i]=3,arr[j]=11

if (arr[i] = -1)then break ==>lewati

s := arr[j]/arr[i];

= 11/3=3

if (s * arr[i] = arr[j]) ==> lewati

arr[10]=11;****

kesimpulan :

nilai arr[8]=3

nilai arr[9]=5

nilai arr[10]=11

nilai akhir arr [1..10]:=(-1,-1,-1,-1,-1,7,2,3,5,11)

proses ke : 9

nilai awal arr [1..10]:=(-1,-1,-1,-1,-1,7,2,3,5,11)

perulangan j (10 sd 10)

(j=10)diketahui

i=9,j=10,arr[i]=5,arr[j]=11

if (arr[i] = -1)then break ==>lewati

s := arr[j]/arr[i];

= 11/5=2

if (s * arr[i] = arr[j]) ==> lewati

arr[10]=11;****

kesimpulan :

nilai arr[9]=5

nilai arr[10]=11

nilai akhir arr [1..10]:=(-1,-1,-1,-1,-1,7,2,3,5,11)

proses ke : 10

nilai awal arr [1..10]:=(-1,-1,-1,-1,-1,7,2,3,5,11)

perulangan j (11 sd 10)

tidak ada proses apapun karena menurun

kesimpulan :

nilai array tidak ada perubahan

nilai akhir arr [1..10]:=(-1,-1,-1,-1,-1,7,2,3,5,11)

karena yang ditanyakan adalah nilai arr[9] maka kita lihat urutan data arr ke 9 adalah 5 jadi jawaban yang tepat adalah d.5

Masuk untuk menulis jawaban