# Untitled

unknown

java

7 months ago

2.8 kB

5

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Never

^{}

public class UserMainCode { // Function to count the number of possible ways to set up the colonies. public static int Colonies(int input1, int input2) { // Since ants can only move up, down, left and right, each cell in the grid can only // be part of a colony if it's adjacent (not diagonally) to another cell that's part of a colony. // This is essentially a combinatorial problem where we need to count the number of ways to fill // a grid with blocks such that each block is connected to at least one other block horizontally // or vertically. // A dynamic programming approach can be used where dp[i][j] represents the number of ways to arrange // colonies in a grid of size i x j. // The base case would be when either i or j is 1, because the colonies can only be set up in a linear // fashion, thus the number of ways to set up the colonies is 2^(i*j) - 1 (all combinations of cells being // filled or not, minus the case where no cell is filled). // Initialize a 2D array for dynamic programming long[][] dp = new long[input1 + 1][input2 + 1]; // Modulo value for large numbers long MOD = 1000000007; // Base cases for (int i = 1; i <= input1; i++) { for (int j = 1; j <= input2; j++) { if (i == 1 || j == 1) { dp[i][j] = (pow(2, i * j) - 1 + MOD) % MOD; // Using pow function with modulo } else { // Apply the recursive relation // dp[i][j] is the sum of ways of arranging colonies in the smaller grid plus the number of ways // to extend the grid with an additional row or column of colonies. // Since the problem does not specify a pattern for the arrangement of colonies, we'll assume the // simplest case where the arrangement of colonies in a smaller grid can be extended by adding a row // or column of colonies in all possible ways. dp[i][j] = (dp[i - 1][j] + dp[i][j - 1]) % MOD; } } } // Return the number of ways for the original grid size. return (int)dp[input1][input2]; } // Function to compute the power of a number with modulo public static long pow(long base, long exponent) { long result = 1; long MOD = 1000000007; while (exponent > 0) { if (exponent % 2 == 1) { result = (result * base) % MOD; } base = (base * base) % MOD; exponent = exponent / 2; } return result; } }

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