To determine the number of ways a person can select cards to achieve a sum of 2012, we need to analyze the available cards and their values. The cards consist of 10 each of blue, green, and red colors, with denominations of 2, 4, 8, 16, 32, 64, 128, 256, 512, and 1024. Additionally, there is one black card and one white card, each valued at 1. Let's break down the problem step by step.
Understanding the Card Values
The first step is to list the values of the cards:
- Blue Cards: 2, 4, 8, 16, 32, 64, 128, 256, 512, 1024
- Green Cards: 2, 4, 8, 16, 32, 64, 128, 256, 512, 1024
- Red Cards: 2, 4, 8, 16, 32, 64, 128, 256, 512, 1024
- Black Card: 1
- White Card: 1
Calculating Total Value from Cards
Each color has the same denominations, and since there are 10 cards of each color, the total number of cards is 32. The black and white cards add two more options, but they only contribute a value of 1 each.
Formulating the Problem
We need to find combinations of these cards that sum up to 2012. This can be approached as a combinatorial problem where we can use generating functions or dynamic programming techniques. However, given the nature of the denominations, we can also think about it in terms of binary representation.
Binary Representation Insight
The denominations of the blue, green, and red cards are powers of 2, which means they can be represented in binary. The maximum value we can get from the blue, green, and red cards is:
- 10 cards of 1024 (the highest denomination) = 10240
Since we only need to reach 2012, we can focus on how many of each card we can use without exceeding this value.
Dynamic Programming Approach
To systematically calculate the number of ways to achieve the sum of 2012, we can use a dynamic programming approach:
- Create an array dp where dp[i] represents the number of ways to achieve the sum i.
- Initialize dp[0] = 1 since there is one way to achieve a sum of 0 (by selecting no cards).
- For each denomination, iterate through the possible sums and update the dp array accordingly.
Implementation Steps
For each card value, update the dp array:
- For each card value (2, 4, 8, ..., 1024), iterate through the dp array from the card value up to 2012.
- For each index j in the dp array, update dp[j] by adding dp[j - card value].
Final Calculation
After processing all card values, dp[2012] will give the total number of ways to select cards to achieve the sum of 2012. The inclusion of the black and white cards (each valued at 1) can also be considered in the final count, as they can be included in any combination.
In summary, the problem can be solved using a dynamic programming approach, iterating through the card values and updating a count of ways to reach each possible sum up to 2012. This method ensures that we account for all combinations of card selections effectively.
