\n The “I will become the head of the residents’ association” problem is about calculating the number of people based on the floor and apartment number.
\n You need to determine how many people live in a given floor index and apartment index.
\n The key to this problem is to use a pyramid structure to calculate the number of residents on each floor.\n<\/p>\n
\n The rules defining the number of people residing in an apartment are as follows.
\n On the 0th floor (i=0), there is always 1 person in apartment i.
\n On the 1st floor (i=1), the number of people in apartment i starts from 1 and sums up the number of people in all apartments on the upper floor corresponding to it.\n<\/p>\n
\n In other words, if f(n, k) represents the number of people in apartment k on the n-th floor, the following recursive equation can be established:\n<\/p>\n
f(n, k) = f(n-1, 1) + f(n-1, 2) + ... + f(n-1, k)<\/code><\/pre>\n\n Based on this rule, the number of residents in apartment k on the n-th floor is the sum of apartments 1 through k on the (n-1)-th floor.\n<\/p>\n
Approach to Problem Solving<\/h2>\n
\n 1. **Basic Recursive Function Implementation**: To simplify the problem, a method that makes many recursive calls can be used.
\n 2. **Utilizing Dynamic Programming**: The recursive approach can be inefficient as the same value can be called multiple times.
\n By using memoization (or dynamic programming) to store detailed results, performance can be improved.\n<\/p>\n
\n This problem can generate a table based on floors and apartment numbers, allowing all calculated values to be stored.\n<\/p>\n
C++ Code Implementation<\/h2>\n
\n The code below uses dynamic programming to calculate the number of people in the given floor and apartment:\n<\/p>\n
\n#include <iostream>\n#include <vector>\n\nusing namespace std;\n\nint main() {\n int T; \/\/ Number of test cases\n cin >> T;\n \n while (T--) {\n int k, n; \/\/ k: apartment number, n: floor\n cin >> k >> n;\n \n vector<vector<int>> dp(n + 1, vector<int>(k + 1, 0));\n\n \/\/ Initialize the 0th floor\n for (int i = 1; i <= k; ++i) {\n dp[0][i] = 1; \/\/ There is always 1 person in apartment i on the 0th floor\n }\n\n \/\/ Constructing the DP table\n for (int i = 1; i <= n; ++i) {\n for (int j = 1; j <= k; ++j) {\n dp[i][j] = dp[i - 1][j] + dp[i][j - 1];\n }\n }\n\n \/\/ Output the result\n cout << dp[n][k] << endl;\n }\n \n return 0;\n}\n<\/code><\/pre>\nTime Complexity Analysis<\/h2>\n
\n The above implementation computes for each of the T test cases up to the n-th floor and k-th apartment, resulting in a worst-case time complexity of O(n * k)<\/code>.
\n The implemented dynamic programming approach also requires a space complexity of O(n * k)<\/code>.\n<\/p>\nConclusion<\/h2>\n
\n The “I will become the head of the residents’ association” problem can be efficiently solved by understanding the rules of the problem and selecting an appropriate algorithm for implementation.
\n By using dynamic programming techniques in C++, performance can be maximized, and this approach can be useful for solving various similar problems.\n<\/p>\n
Additional Problems and Exercises<\/h2>\n
\n If you want to understand the problem more deeply, it is recommended to try the following variations:\n<\/p>\n