# C++ Program to Find the Number of Ways to Write a Number as the Sum of Numbers Smaller than Itself

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This is a C++ program to find the number of ways to write a number as the sum of numbers smaller than itself.

Problem Description

1. This algorithm counts the partition of the given number.
2. There is no straight method to count a total number of the partition so we need to generate and count them.
3. The time complexity of this algorithm is O(n!).

Problem Solution

1. This algorithm takes the input of a number ‘n’.
2. It starts with the number and breaks it by removing 1, at a time..
3. As it generates a new partition increase the counter.
4. Print the result.
5. Exit.

Program/Source Code

C++ Program to find the number of ways to write a number as the sum of numbers smaller than itself.
This program is successfully run on Dev-C++ using TDM-GCC 4.9.2 MinGW compiler on a Windows system.

```#include<iostream>

using namespace std;

// A function to count all the possible partition.
int CountPartitions(int n)
{
int p[n], k = 0, count = -1;
p[k] = n;

// Loop until all the array elements converted to 1 mean no further partition can be generated.
while(1)
{
count++;
int rem_val = 0;

// Move the pointer to the index so that p[k] > 1.
while (k >= 0 && p[k] == 1)
{
rem_val += p[k];
k--;
}
// If k < 0 then the all the element are broken down to 1.
if (k < 0)
return count;

// If value greater than 1 is found then decrease it by 1 and increase rem_val to add it to other elements.
p[k]--;
rem_val++;

// Loop until the number of 1's are greater than the value at k index.
while (rem_val > p[k])
{
p[k+1] = p[k];
// Decrease the rem_val value.
rem_val = rem_val - p[k];
k++;
}

// Assign the remaining value to the index next to k.
p[k+1] = rem_val;
k++;
}
}

int main()
{
int n, c;
cout<<"Enter natural numbers for partition counting: ";
cin>>n;
if (n <= 0)
{
cout<<"Wrong input!!";
return 0;
}

c = CountPartitions(n);
cout<<"The number of partitions of with each element lesser than "<<n<<" is:- "<<c;

return 0;
}```
Program Explanation

1. Take the input of the natural number ‘n’.
2. Call CountPartitions() with ‘n’ in the argument list.
3. Inside CountPartitions(), declare an array p[] of length ‘n’ since we can generate maximum n partition, an element k to traverse in the array p[], a counter variable and rem_val.
4. Assign n to the first element of array p[], k to 0 and counter variable to -1.
5. Now inside increment counter and assign rem_val to 0, initially.
6. At the end of each iteration k Shifts to the end of the array to move it back to the first value encountered as greater than 0 and meanwhile add up all 1’s to rem_val.
7. If k reaches to -1 then breaks the loop and return to main.
8. To break value at k, decrement it and add it to rem_val.
9. So in the next loop, break rem_val till it is more than the value at index k, it will ignore the repetition.
11. Exit.

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Runtime Test Cases
```Case 1:
Enter natural numbers for partition counting: 10
The number of partitions of with each element lesser than 10 is:- 41

Case 2:
Enter natural numbers for creating partitions: 50
The number of partitions of with each element lesser than 50 is:- 204225

Case 3:
Enter natural numbers for partition counting: 100
The number of partitions of with each element lesser than 100 is:- 190569291```

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