HOW TO SOLVE Nomer 5

Assalamualaikum W.W.,

Disini ane bakalan share program dari E-book yang berjudul How To Solve It By Computer Bab 4 Bagian 5.

Di bagian 5 ini ada 3 Soal .. Nah langsungan aja yaaa

4.5.1 

#include <iostream> #include <stdio.h> /* run this program using the console pauser or add your own getch, system("pause") or input loop */ using namespace std; int main() { int n, *arr, onee = 0, twoo, threee, total = 0, maxx = -1, temp_maxx; cin >> n; arr = new int[n]; for (int i = 0; i < n; i++) { cin >> arr[i]; total += arr[i]; } // O(n^2) is the following for (int i = 1; i < n - 1; i++) { onee += arr[i - 1]; twoo = 0; for (int j = i + 1; j < n; j++) { twoo += arr[j - 1]; threee = total - twoo - onee; temp_maxx = max(max(onee, twoo), threee); if ((temp_maxx < maxx) || (maxx == -1)) maxx = temp_maxx; } } cout << maxx; return 0; }

4.5.2

Ini ada Algoritmanya ,, terjemahin ndiri ya bang ,, ckckk

var A; array[1..100] of interger; i: integer; begin for i:=1 to 100 do begin A[1]:=i; end; End.

4.5.3

#include <iostream> /* run this program using the console pauser or add your own getch, system("pause") or input loop */ using namespace std; //Bucket Sort void bucket_sort (int arr[], int n) { //Here range is [1,100] int m = 101; //Create m empty buckets int buckets[m]; //Intialize all buckets to 0 for (int i = 0; i < m; ++i) buckets[i] = 0; //Increment the number of times each element is present in the input //array. Insert them in the buckets for (int i = 0; i < n; ++i) ++buckets[arr[i]]; //Sort using insertion sort and concatenate for (int i = 0, j = 0; j < m; ++j) for (int k = buckets[j]; k > 0; --k) arr[i++] = j; } //Driver function to test above function int main() { int input_ar[] = {10, 24, 22, 62, 1, 50, 100, 75, 2, 3}; int n = sizeof (input_ar) / sizeof (input_ar[0]); bucket_sort (input_ar, n); cout << "Sorted Array : " << endl; for (int i = 0; i < n; ++i) cout << input_ar[i] << " "; return 0; }

Nah itu lah Program yang berada di How To Solve bagian ke 5 .

Sekian Dari ane

Wassalamualaikum W.W.

Nomer 3&4 bisa klik DISINI

Nomer 1&2 bisa klik DISINI