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 : 4 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; }

## Java

// Program to find the number of ways, // n can be written as sum of two or // more positive integers. import java.util.Arrays; class GFG { // Returns number of ways to write // n as sum of two or more positive // integers static 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[] = new int[n + 1]; // Initialize all table values as 0 Arrays.fill(table, 0); // 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 code public static void main (String[] args) { int n = 7; System.out.print(countWays(n)); } } // This code is contributed by Anant Agarwal.

## 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

## C#

// Program to find the number of ways, n can be // written as sum of two or more positive integers. using System; class GFG { // Returns number of ways to write n as sum of // two or more positive integers static 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 = new int[n+1]; // Initialize all table values as 0 for(int i = 0; i < table.Length; i++) table[i] = 0; // 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 code public static void Main() { int n = 7; Console.Write(countWays(n)); } } //This code is contributed by Anant Agarwal.

Output:

14

Time complexity O(n^{2})

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