可能なすべての方法で整数の分割を実行するC++プログラム

これは、パーティションを追加すると整数になるように、特定の整数のすべての一意のパーティションを取得するC++プログラムです。このプログラムでは、正の整数nが与えられ、nを正の整数の合計として表すためのすべての可能な一意の方法を生成します。

アルゴリズム

Begin
   function displayAllUniqueParts(int m):
   Declare an array to store a partition p[m].
   Set Index of last element k in a partition to 0
   Initialize first partition as number itself, p[k]=m
   Create a while loop which first prints current partition, then generates next partition. The loop stops when the current partition has all 1s.
   Display current partition as displayArray(p, k + 1);
   Generate next partition:
   Initialize val=0.
   Find the rightmost non-one value in p[]. Also, update the val so that we know how much value can be accommodated.
   If k < 0,
      all the values are 1 so there are no more partitions
      Decrease the p[k] found above and adjust the val.
   If val is more,
      then the sorted order is violeted. Divide val in different values of size p[k] and copy these values at different positions after p[k].
      Copy val to next position and increment position.
End

サンプルコード

#include<iostream>
using namespace std;
void printArr(int p[], int m) {
   for (int i = 0; i < m; i++)
      cout << p[i] << " ";
   cout << endl;
}
void printAllUniqueParts(int m) {
   int p[m];
   int k = 0;
   p[k] = m;
   while (true) {
      printArr(p, k + 1);
      int rem_val = 0;
      while (k >= 0 && p[k] == 1) {
         rem_val += p[k];
         k--;
      }
      if (k < 0)
         return;
         p[k]--;
         rem_val++;
      while (rem_val > p[k]) {
         p[k + 1] = p[k];
         rem_val = rem_val - p[k];
         k++;
      }
      p[k + 1] = rem_val;
      k++;
   }
}
int main() {
   cout << "All Unique Partitions of 3\n";
   printAllUniqueParts(3);
   cout << "\nAll Unique Partitions of 4\n";
   printAllUniqueParts(4);
   cout << "\nAll Unique Partitions of 5\n";
   printAllUniqueParts(5);
   return 0;
}

出力

All Unique Partitions of 3
3
2 1
1 1 1
All Unique Partitions of 4
4
3 1
2 2
2 1 1
1 1 1 1
All Unique Partitions of 5
5
4 1
3 2
3 1 1
2 2 1
2 1 1 1
1 1 1 1 1