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C++ Program to Perform Partition of an Integer in All Possible Ways
Here is a C++ Program to get all the unique partitions of a given integer such that addition of a partition results an integer. In this program, a positive integer n is given, and generate all possible unique ways to represent n as sum of positive integers.
Algorithm
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
Example Code
#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;
}Output
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
Updated on: 30-Jul-2019
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