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

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!).

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.

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; }

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.

10. Return to main and print the result.

11. Exit.

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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