CF 1051013 - Это магия
The magician’s trick reduces the task of identifying a hidden number between 1 and 100 into answering a fixed sequence of seven yes or no questions.
CF 1051013 - \u042d\u0442\u043e \u043c\u0430\u0433\u0438\u044f
Rating: -
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Solve time: 1m 4s
Verified: yes
Solution
Problem Understanding
The magician’s trick reduces the task of identifying a hidden number between 1 and 100 into answering a fixed sequence of seven yes or no questions. Each of the seven cards contains a subset of numbers from that range, and the spectator indicates for each card whether their chosen number appears on it.
From the solver’s perspective, the input is simply seven binary responses in order, each corresponding to one predefined card. The hidden number is uniquely determined by which combination of cards contains it, so the output is the single number consistent with all seven answers.
The key structural constraint is that there are only 100 possible outcomes, but each outcome corresponds to a distinct 7-bit pattern of membership across the cards. This immediately implies that the cards must encode numbers as binary representations or a close variant. With only seven yes or no responses, the space of patterns is at most 2^7 = 128, which comfortably covers the range 1 to 100. This already suggests that each number is mapped to a unique subset of cards, and reconstruction is a direct decoding problem.
A naive interpretation might try to test each number from 1 to 100 and simulate whether it matches the given answers. That works safely within constraints because it is only 100 checks, but it is unnecessary. The intended structure is direct decoding: each "Yes" contributes a fixed weight to the final answer.
A subtle edge case arises from assuming arbitrary card contents. If one incorrectly assumes the sets are random, reconstruction becomes ambiguous. For example, if two numbers share the same membership pattern, the output would be undefined. The problem guarantees a proper encoding, so each pattern maps to exactly one number, avoiding this ambiguity.
Approaches
The brute-force idea is straightforward. For every number from 1 to 100, we check whether its presence pattern across the seven cards matches the input responses. Concretely, for each candidate number, we would simulate the seven cards and compare against the given yes or no sequence. This requires scanning up to 100 numbers, and for each number checking up to 7 conditions, giving at most 700 operations per test. Even if extended, this remains trivial, but it hides the structure of the trick.
The key insight is that the seven cards encode numbers in binary form. Each card corresponds to a fixed bit position, and a "Yes" means that bit is set in the hidden number. Instead of checking all candidates, we reconstruct the number directly by summing powers of two associated with "Yes" answers.
This transforms the problem from search to decoding. The structure of the input already represents the binary representation of the answer, so reconstruction is a single pass over the seven values.
| Approach | Time Complexity | Space Complexity | Verdict |
|---|---|---|---|
| Brute Force | O(700) | O(1) | Accepted |
| Binary Decoding | O(7) | O(1) | Accepted |
Algorithm Walkthrough
- Assign a weight to each of the seven positions, corresponding to a binary digit. The first card represents 2^0, the second 2^1, and so on. This reflects the fact that each card encodes a bit in the binary representation of the hidden number.
- Initialize an accumulator variable to zero. This will store the reconstructed number as bits are interpreted.
- Iterate over the seven responses in order.
- For each response, if it is "Yes", add the corresponding power of two to the accumulator. If it is "No", do nothing because that bit is not present in the number.
- After processing all seven responses, output the accumulator as the reconstructed number.
Why it works
Each number from 1 to 100 has a unique binary representation using at most seven bits. Each card corresponds exactly to one bit position in that representation. A "Yes" indicates that the bit is 1, and "No" indicates it is 0. Since binary representation is unique, summing the active bits reconstructs the original number without ambiguity.
Python Solution
import sys
input = sys.stdin.readline
def solve():
responses = input().split()
ans = 0
for i, r in enumerate(responses):
if r == "Yes":
ans += (1 << i)
print(ans)
if __name__ == "__main__":
solve()
The solution reads the seven responses and processes them in order. The index of each response determines its bit position. When the response is "Yes", the corresponding power of two is added using a bit shift. The final sum is printed as the answer. The only subtle detail is ensuring consistent indexing: the first response must correspond to the least significant bit, otherwise the reconstructed number would be incorrect.
Worked Examples
Example 1
Input:
No Yes Yes Yes No No No
We map each response to a bit contribution:
| Index | Response | Bit Value | Contribution |
|---|---|---|---|
| 0 | No | 1 | 0 |
| 1 | Yes | 2 | 2 |
| 2 | Yes | 4 | 4 |
| 3 | Yes | 8 | 8 |
| 4 | No | 16 | 0 |
| 5 | No | 32 | 0 |
| 6 | No | 64 | 0 |
Final sum is 2 + 4 + 8 = 14.
This trace shows direct binary reconstruction from the response pattern. Each "Yes" activates a corresponding power of two.
Example 2
Input:
Yes No No No No No Yes
| Index | Response | Bit Value | Contribution |
|---|---|---|---|
| 0 | Yes | 1 | 1 |
| 1 | No | 2 | 0 |
| 2 | No | 4 | 0 |
| 3 | No | 8 | 0 |
| 4 | No | 16 | 0 |
| 5 | No | 32 | 0 |
| 6 | Yes | 64 | 64 |
Final sum is 65.
This confirms that multiple separated bits are handled independently and summed correctly, reflecting the uniqueness of binary encoding.
Complexity Analysis
| Measure | Complexity | Explanation |
|---|---|---|
| Time | O(7) | We process exactly seven responses once |
| Space | O(1) | Only a constant accumulator is used |
The algorithm runs in constant time relative to input size because the input length is fixed at seven. This is trivially within limits and requires negligible memory.
Test Cases
# helper: run solution on input string, return output string
import sys, io
def run(inp: str) -> str:
sys.stdin = io.StringIO(inp)
responses = input().split()
ans = 0
for i, r in enumerate(responses):
if r == "Yes":
ans += (1 << i)
return str(ans)
# provided sample
assert run("No Yes Yes Yes No No No") == "14"
# custom cases
assert run("No No No No No No No") == "0", "all zeros"
assert run("Yes No No No No No No") == "1", "only first bit"
assert run("No No No No No No Yes") == "64", "only last bit"
assert run("Yes Yes Yes Yes Yes Yes Yes") == "127", "all bits set"
assert run("Yes No Yes No Yes No Yes") == "85", "alternating bits"
| Test input | Expected output | What it validates |
|---|---|---|
| all No | 0 | zero case |
| first Yes only | 1 | LSB handling |
| last Yes only | 64 | MSB handling |
| all Yes | 127 | full range |
| alternating | 85 | mixed bit pattern |
Edge Cases
One important edge case is when all responses are "No". The algorithm initializes the accumulator to zero and never adds any power of two, producing 0. Since the valid range starts at 1, this case typically represents either invalid input or a special encoding case, but the computation remains consistent.
Another case is when all responses are "Yes". The accumulator becomes 1 + 2 + 4 + 8 + 16 + 32 + 64, which equals 127. This verifies that the upper bound of the 7-bit encoding is handled correctly.
A third case involves scattered bits such as alternating "Yes No Yes No Yes No Yes". The algorithm correctly accumulates only selected powers of two, and the result reflects the exact binary composition without interference between positions.