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; ` `} ` |

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## 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.*/` |

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## 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.*/` |

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Output:

3

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