# Untitled

unknown
java
7 months ago
2.8 kB
5
Indexable
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;
}
}
```
Leave a Comment