# C++ Program to Perform Partition of an Integer in All Possible Ways

C++Server Side ProgrammingProgramming

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
Published on 15-Mar-2019 11:43:17