Given two natural number **n** and **m**. The task is to find the number of ways in which the numbers that are greater than or equal to m can be added to get the sum n.

Examples:

Input : n = 3, m = 1 Output : 3 Following are three different ways to get sum n such that each term is greater than or equal to m 1 + 1 + 1, 1 + 2, 3 Input : n = 2, m = 1 Output : 2 Two ways are 1 + 1 and 2

The idea is to use Dynamic Programming by define 2D matrix, say dp[][]. **dp[i][j]** define the number of ways to get sum i using the numbers greater than or equal to j. So dp[i][j] can be defined as:

If i < j, dp[i][j] = 0, because we cannot achieve smaller sum of i using numbers greater than or equal to j.

If i = j, dp[i][j] = 1, because there is only one way to show sum i using number i which is equal to j.

Else dp[i][j] = dp[i][j+1] + dp[i-j][j], because obtaining a sum i using numbers greater than or equal to j is equal to the sum of obtaining a sum of i using numbers greater than or equal to j+1 and obtaining the sum of i-j using numbers greater than or equal to j.

Below is the implementation of this approach:

## CPP

// CPP Program to find number of ways to // which numbers that are greater than // given number can be added to get sum. #include <bits/stdc++.h> #define MAX 100 using namespace std; // Return number of ways to which numbers // that are greater than given number can // be added to get sum. int numberofways(int n, int m) { int dp[n+2][n+2]; memset(dp, 0, sizeof(dp)); dp[0][n + 1] = 1; // Filling the table. k is for numbers // greater than or equal that are allowed. for (int k = n; k >= m; k--) { // i is for sum for (int i = 0; i <= n; i++) { // initializing dp[i][k] to number // ways to get sum using numbers // greater than or equal k+1 dp[i][k] = dp[i][k + 1]; // if i > k if (i - k >= 0) dp[i][k] = (dp[i][k] + dp[i - k][k]); } } return dp[n][m]; } // Driver Program int main() { int n = 3, m = 1; cout << numberofways(n, m) << endl; return 0; }

## Java

// Java Program to find number of ways to // which numbers that are greater than // given number can be added to get sum. import java.io.*; class GFG { // Return number of ways to which numbers // that are greater than given number can // be added to get sum. static int numberofways(int n, int m) { int dp[][]=new int[n+2][n+2]; dp[0][n + 1] = 1; // Filling the table. k is for numbers // greater than or equal that are allowed. for (int k = n; k >= m; k--) { // i is for sum for (int i = 0; i <= n; i++) { // initializing dp[i][k] to number // ways to get sum using numbers // greater than or equal k+1 dp[i][k] = dp[i][k + 1]; // if i > k if (i - k >= 0) dp[i][k] = (dp[i][k] + dp[i - k][k]); } } return dp[n][m]; } // Driver Program public static void main(String args[]) { int n = 3, m = 1; System.out.println(numberofways(n, m)); } } /*This code is contributed by Nikita tiwari.*/

## C#

// C# program to find number of ways to // which numbers that are greater than // given number can be added to get sum. using System; class GFG { // Return number of ways to which numbers // that are greater than given number can // be added to get sum. static int numberofways(int n, int m) { int[, ] dp = new int[n + 2, n + 2]; dp[0, n + 1] = 1; // Filling the table. k is for numbers // greater than or equal that are allowed. for (int k = n; k >= m; k--) { // i is for sum for (int i = 0; i <= n; i++) { // initializing dp[i][k] to number // ways to get sum using numbers // greater than or equal k+1 dp[i, k] = dp[i, k + 1]; // if i > k if (i - k >= 0) dp[i, k] = (dp[i, k] + dp[i - k, k]); } } return dp[n, m]; } // Driver Program public static void Main() { int n = 3, m = 1; Console.WriteLine(numberofways(n, m)); } } /*This code is contributed by vt_m.*/

Output:

3

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