1. Problem Description<\/h2>\n
\n The staircase number is a problem of counting numbers that meet specific conditions. N (1 \u2264 N \u2264 100), the given natural number N represents the number of digits in the staircase number.<\/p>\n Print the number of N-digit staircase numbers modulo 1,000,000,000.<\/p>\n<\/section>\n \n To count staircase numbers, we need to choose a dynamic programming method. \n Based on this, we can construct a DP table and keep track of staircase numbers for each digit, \n Now we will implement the code to find staircase numbers in C#. \n I will explain the code step by step. \n The time complexity of this algorithm is O(N).
\n Given a number N, we will address the problem of finding N-digit staircase numbers.
\n A staircase number is a number where each digit must be either one more or one less than the next digit. For example, 1234 and 3210 are staircase numbers.
\n However, 1123 and 2222 are not staircase numbers.
\n This problem will help you understand the basic ideas and implementation methods of dynamic programming.\n <\/p>\n<\/section>\n2. Input & Output Format<\/h2>\n
Input<\/h3>\n
Output<\/h3>\n
3. Approach to the Problem<\/h2>\n
\n We can approach the problem in the following way.
\n To find N-digit staircase numbers, we calculate values using (N-1)-digit staircase numbers.
\n We can find the following rules:\n <\/p>\n\n
\n ultimately allowing us to find the N-digit staircase numbers.
\n The approach involves initializing the DP array and gradually filling it out.\n <\/p>\n<\/section>\n4. C# Code Implementation<\/h2>\n
\n Below is the C# code to solve the problem.\n <\/p>\n\nusing System;\n\nclass Program\n{\n static void Main()\n {\n int N = int.Parse(Console.ReadLine());\n long[,] dp = new long[N + 1, 10];\n\n \/\/ Initialize 1-digit staircase numbers (1~9)\n for (int i = 1; i <= 9; i++)\n {\n dp[1, i] = 1;\n }\n\n \/\/ Use DP to find N-digit staircase numbers\n for (int i = 2; i <= N; i++)\n {\n for (int j = 0; j <= 9; j++)\n {\n if (j == 0)\n {\n dp[i, j] = dp[i - 1, j + 1] % 1000000000; \/\/ 0 -> 1\n }\n else if (j == 9)\n {\n dp[i, j] = dp[i - 1, j - 1] % 1000000000; \/\/ 9 -> 8\n }\n else\n {\n dp[i, j] = (dp[i - 1, j - 1] + dp[i - 1, j + 1]) % 1000000000; \/\/ j-1, j+1\n }\n }\n }\n\n \/\/ Calculate the total number of N-digit staircase numbers\n long result = 0;\n for (int j = 0; j <= 9; j++)\n {\n result = (result + dp[N, j]) % 1000000000;\n }\n\n \/\/ Print the result\n Console.WriteLine(result);\n }\n}\n <\/code><\/pre>\n<\/section>\n5. Code Explanation<\/h2>\n
\n 1. First, read the value of N from the user’s input.
\n 2. Create a two-dimensional array dp, where dp[i, j] stores the number of i-digit staircase numbers ending with j.
\n 3. Since 1-digit numbers can only use digits from 1 to 9, we initialize this.
\n 4. Next, we calculate the number of digits from 2 up to N.
\n 5. Each digit’s calculation is done according to the rules explained above.
\n 6. Finally, sum all N-digit staircase numbers and print the result.\n <\/p>\n<\/section>\n6. Complexity Analysis<\/h2>\n
\n When N is 100, we check the possibilities for each digit through a double loop,
\n which results in O(N) * O(10) = O(N).
\n The space complexity is O(N * 10), indicating relatively low memory usage.\n <\/p>\n<\/section>\n7. Example<\/h2>\n
Input Example<\/h3>\n
\n3\n <\/code><\/pre>\nOutput Example<\/h3>\n
\n32\n <\/code><\/pre>\n