dice combinations cses solution
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Problem Overview
The problem is to calculate the number of ways we can obtain a sum of 'x' by throwing 'n' dice, where each die has 'k' faces numbered from 1 to 'k'.
Approach
To solve this problem, we can use dynamic programming. We will create a 2D array dp
, where dp[i][j]
represents the number of ways to obtain a sum of 'j' using 'i' dice. The dimensions of the array will be (n+1) x (k*n+1).
Initialization
We initialize dp[0][0] = 1
, since there is one way to obtain a sum of 0 using 0 dice (by not throwing any dice).
For all other cells in the first row, we set dp[0][j] = 0
because we cannot obtain any sum greater than 0 using 0 dice.
Recurrence Relation
To populate the remaining cells of the array, we use the following recurrence relation:
dp[i][j] = dp[i-1][j-1] + dp[i-1][j-2] + ... + dp[i-1][j-k]
This means that the number of ways to obtain a sum of 'j' using 'i' dice is equal to the sum of the number of ways to obtain a sum of 'j-1', 'j-2', ..., 'j-k' using 'i-1' dice.
Dynamic Programming Solution
We can use nested loops to fill in the values of the dp
array. The outer loop iterates over the number of dice, and the inner loop iterates over the possible sums. We start from i = 1
and j = 1
because we have already initialized the first row of the array.
for (int i = 1; i <= n; i++) {
for (int j = 1; j <= k*n; j++) {
dp[i][j] = 0;
for (int x = 1; x <= k && x <= j; x++) {
dp[i][j] += dp[i-1][j-x];
}
}
}
After filling in the dp
array, the answer to the problem will be stored in dp[n][x]
, which represents the number of ways to obtain a sum of 'x' using 'n' dice.
Example
Let's take an example to illustrate the solution. Suppose we have 2 dice with 6 faces each, and we want to find the number of ways to obtain a sum of 7.
We initialize dp[0][0] = 1
and dp[0][j] = 0
for all other cells in the first row.
Next, we fill in the values of the dp
array using the recurrence relation. After filling in all the cells, we will have dp[2][7]
as the answer.
Complexity Analysis
The time complexity of this solution is O(n k n), and the space complexity is O(n * k).
I hope this explanation helps! Let me know if you have any further questions.