A stable tower of height **n** is a tower consisting of exactly n tiles of unit height stacked vertically in such a way, that no bigger tile is placed on a smaller tile. An example is shown below :

We have infinite number of tiles of sizes 1, 2, …, m. The task is calculate the number of different stable tower of height n that can be built from these tiles, with a restriction that you can use at most **k** tiles of each size in the tower.

**Note:** Two tower of height n are different if and only if there exists a height h (1 <= h <= n), such that the towers have tiles of different sizes at height h.

Examples:

Input : n = 3, m = 3, k = 1. Output : 1 Possible sequences: { 1, 2, 3}. Hence answer is 1. Input : n = 3, m = 3, k = 1. Output : 7 {1, 1, 2}, {1, 1, 3}, {1, 2, 2}, {1, 2, 3}, {1, 3, 3}, {2, 2, 3}, {2, 3, 3}.

We basically need to count number of decreasing sequences of length n using numbers from 1 to m where every number can be used at most k times. We can recursively compute count for n using count for n-1.

The idea is to use Dynamic Programming. Declare a 2D array dp[][], where each state dp[i][j] denotes the number of decreasing sequences of length i using numbers from j to m. We need to take care of the fact that a number can be used a most k times. This can be done by considering 1 to k occurrences of a number. Hence our recurrence relation becomes:

Also, we can use the fact that for a fixed j we are using the consecutive values of previous k values of i. Hence, we can maintain a prefix sum array for each state. Now we have got rid of the k factor for each state.

Below is the C++ implemantation of this approach:

`// CPP program to find number of ways to make stable ` `// tower of given height. ` `#include <bits/stdc++.h> ` `using` `namespace` `std; ` `#define N 100 ` ` ` `int` `possibleWays(` `int` `n, ` `int` `m, ` `int` `k) ` `{ ` ` ` `int` `dp[N][N]; ` ` ` `int` `presum[N][N]; ` ` ` `memset` `(dp, 0, ` `sizeof` `dp); ` ` ` `memset` `(presum, 0, ` `sizeof` `presum); ` ` ` ` ` `// Initialing 0th row to 0. ` ` ` `for` `(` `int` `i = 1; i < n + 1; i++) { ` ` ` `dp[0][i] = 0; ` ` ` `presum[0][i] = 1; ` ` ` `} ` ` ` ` ` `// Initialing 0th column to 0. ` ` ` `for` `(` `int` `i = 0; i < m + 1; i++) ` ` ` `presum[i][0] = dp[i][0] = 1; ` ` ` ` ` `// For each row from 1 to m ` ` ` `for` `(` `int` `i = 1; i < m + 1; i++) { ` ` ` ` ` `// For each column from 1 to n. ` ` ` `for` `(` `int` `j = 1; j < n + 1; j++) { ` ` ` ` ` `// Initialing dp[i][j] to presum of (i - 1, j). ` ` ` `dp[i][j] = presum[i - 1][j]; ` ` ` `if` `(j > k) { ` ` ` `dp[i][j] -= presum[i - 1][j - k - 1]; ` ` ` `} ` ` ` `} ` ` ` ` ` `// Calculating presum for each i, 1 <= i <= n. ` ` ` `for` `(` `int` `j = 1; j < n + 1; j++) ` ` ` `presum[i][j] = dp[i][j] + presum[i][j - 1]; ` ` ` `} ` ` ` ` ` `return` `dp[m][n]; ` `} ` ` ` `// Driver Program ` `int` `main() ` `{ ` ` ` `int` `n = 3, m = 3, k = 2; ` ` ` `cout << possibleWays(n, m, k) << endl; ` ` ` `return` `0; ` `} ` |

Output:

7

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