Ways to write n as sum of two or more positive integers

3.5

For a given number n > 0, find the number of different ways in which n can be written as a sum of at two or more positive integers.

Examples:

Input : n = 5
Output : 6
Explanation : All possible six ways are :
4 + 1
3 + 2
3 + 1 + 1
2 + 2 + 1
2 + 1 + 1 + 1
1 + 1 + 1 + 1 + 1

Input : 3
Output : 4
Explanation : All possible four ways are :
3 + 1
2 + 2
2 + 1 + 1
1 + 1 + 1 + 1

This problem can be solved in the similar fashion as coin change problem, the difference is only that in this case we should iterate for 1 to n-1 instead of particular values of coin as in coin-change problem.

C/C++

// Program to find the number of ways, n can be
// written as sum of two or more positive integers.
#include <bits/stdc++.h>
using namespace std;

// Returns number of ways to write n as sum of
// two or more positive integers
int countWays(int n)
{
    // table[i] will be storing the number of
    // solutions for value i. We need n+1 rows
    // as the table is consturcted in bottom up
    // manner using the base case (n = 0)
    int table[n+1];

    // Initialize all table values as 0
    memset(table, 0, sizeof(table));

    // Base case (If given value is 0)
    table[0] = 1;

    // Pick all integer one by one and update the
    // table[] values after the index greater
    // than or equal to n
    for (int i=1; i<n; i++)
        for (int j=i; j<=n; j++)
            table[j] += table[j-i];

    return table[n];
}

// Driver program
int main()
{
    int n = 7;
    cout << countWays(n);
    return 0;
}

Python

# Program to find the number of ways, n can be
# written as sum of two or more positive integers.

# Returns number of ways to write n as sum of
# two or more positive integers
def CountWays(n):

	# table[i] will be storing the number of
	# solutions for value i. We need n+1 rows
	# as the table is consturcted in bottom up
	# manner using the base case (n = 0)
	# Initialize all table values as 0
	table =[0] * (n + 1)

	# Base case (If given value is 0)
	# Only 1 way to get 0 (select no integer)
	table[0] = 1

	# Pick all integer one by one and update the
    # table[] values after the index greater
    # than or equal to n
	for i in range(1, n ):

		for j in range(i , n + 1):

			table[j] +=  table[j - i]			

	return table[n]

# driver program
def main():

	n = 7

	print CountWays(n)

if __name__ == '__main__':
	main()

#This code is contributed by Neelam Yadav


Output:

14

Time complexity O(n2)

This article is contributed by Shivam Pradhan (anuj_charm). If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

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