CF 1840B - Binary Cafe
We are asked to count the number of ways Toma can select desserts in a binary-themed cafe. Each dessert has a cost that is a power of two: the first dessert costs 1 coin, the second 2 coins, the third 4 coins, and so on, following the sequence $2^0, 2^1, 2^2, dots, 2^{k-1}$.
Rating: 1100
Tags: bitmasks, combinatorics, math
Solve time: 1m 8s
Verified: yes
Solution
Problem Understanding
We are asked to count the number of ways Toma can select desserts in a binary-themed cafe. Each dessert has a cost that is a power of two: the first dessert costs 1 coin, the second 2 coins, the third 4 coins, and so on, following the sequence $2^0, 2^1, 2^2, \dots, 2^{k-1}$. Toma has a budget $n$ and will not buy any dessert more than once. The goal is to determine how many subsets of desserts he can afford, including the empty subset where he buys nothing.
The input specifies multiple test cases, each with its own $n$ and $k$. The constraints allow $n$ and $k$ to go up to $10^9$. This immediately rules out any solution that enumerates all subsets or sums, because the number of subsets of $k$ desserts is $2^k$, which is astronomically large for $k$ near a billion.
Edge cases include situations where the budget is smaller than the cheapest dessert. For instance, if $n = 1$ and $k = 1$, the only affordable subsets are the empty subset and the first dessert. Another subtle case arises when $n$ is larger than the sum of all $k$ desserts; here, every subset is affordable, and the answer is exactly $2^k$.
Approaches
A brute-force approach would enumerate all subsets of desserts and sum their costs, counting those that do not exceed $n$. This works for very small $k$, but even $k = 30$ produces $2^{30} \approx 10^9$ subsets, which is too many. For $k = 10^9$, this is impossible.
The key insight is to leverage the binary structure of the costs. Each dessert’s cost is a power of two, and any sum of distinct powers of two corresponds exactly to a binary number where each bit represents whether a dessert is chosen. The problem then reduces to counting how many integers between 0 and $2^k - 1$ (the full range of subset sums) are less than or equal to $n$. We cannot compute $2^k$ directly for large $k$, but we can observe that if $2^k \le n+1$, all $2^k$ subsets are affordable. Otherwise, the number of subsets is exactly $n+1$, because the sums of subsets are consecutive integers from 0 up to the total sum, and we only need those not exceeding $n$.
This gives a simple, direct formula: the number of ways Toma can buy desserts is $\min(n + 1, 2^k)$.
| Approach | Time Complexity | Space Complexity | Verdict |
|---|---|---|---|
| Brute Force | O(2^k) | O(1) | Too slow for k > 30 |
| Optimal | O(1) per test case | O(1) | Accepted |
Algorithm Walkthrough
- Read the number of test cases, $t$. We will loop through each case individually. This is standard input handling for multiple queries.
- For each test case, read $n$ and $k$. Here $n$ is the budget, and $k$ is the number of desserts.
- Compute $2^k$. If $k$ is large, this may overflow typical integer types, but Python handles arbitrary-size integers. However, we only need to compare it with $n+1$, so we can stop multiplying once the value exceeds $n+1$ to save computation.
- Compute the answer as the minimum of $n+1$ and $2^k$. This accounts for both situations: either $n$ is smaller than the sum of all desserts, or $n$ can buy every subset.
- Print the answer for each test case.
Why it works: The power-of-two costs ensure that each subset of desserts corresponds to a unique integer sum. The smallest sum is 0, and the sums increase by all combinations of powers of two. Therefore, counting sums up to $n$ is equivalent to taking the minimum of $n+1$ and $2^k$, since the sums are exactly consecutive integers from 0 up to the total sum of all desserts.
Python Solution
import sys
input = sys.stdin.readline
t = int(input())
for _ in range(t):
n, k = map(int, input().split())
# Compute 2^k safely without overflow using a loop if necessary
power = 1
for _ in range(k):
if power > n:
break
power *= 2
print(min(n + 1, power))
The solution reads all test cases, computes $2^k$ carefully to avoid unnecessary huge integers, and then uses the min function to determine the count of affordable subsets. A subtle point is that once $2^k$ exceeds $n$, further multiplication is unnecessary because the minimum will already be $n+1$. Python handles large integers, but this early break ensures efficiency for extreme values of $k$.
Worked Examples
Example 1: n = 2, k = 2
| Step | power | min(n+1, power) | Explanation |
|---|---|---|---|
| Start | 1 | - | Initial 2^0 |
| Multiply 2^1 | 2 | - | 2 ≤ n+1, continue |
| Multiply 2^2 | 4 | - | 4 > n+1=3, break |
| Result | - | 3 | Affordable subsets: {}, {0}, {1} |
Example 2: n = 10, k = 2
| Step | power | min(n+1, power) | Explanation |
|---|---|---|---|
| Start | 1 | - | Initial 2^0 |
| Multiply 2^1 | 2 | - | Continue |
| Multiply 2^2 | 4 | - | Continue |
| Stop | 4 | min(11, 4)=4 | Only 4 subsets affordable: {}, {0}, {1}, {0,1} |
These traces confirm that the solution correctly handles small and medium budgets, stopping the multiplication early to avoid unnecessary computation.
Complexity Analysis
| Measure | Complexity | Explanation |
|---|---|---|
| Time | O(k) per test case | We loop up to k times to compute powers, but break early if power exceeds n. |
| Space | O(1) | Only a few integer variables are used, independent of n or k. |
The solution fits comfortably within time and memory limits even for $t = 1000$ and $k = 10^9$, because in practice the loop breaks early once power > n.
Test Cases
import sys, io
def run(inp: str) -> str:
sys.stdin = io.StringIO(inp)
output = io.StringIO()
sys.stdout = output
t = int(input())
for _ in range(t):
n, k = map(int, input().split())
power = 1
for _ in range(k):
if power > n:
break
power *= 2
print(min(n + 1, power))
return output.getvalue().strip()
# Provided samples
assert run("5\n1 2\n2 1\n2 2\n10 2\n179 100\n") == "2\n2\n3\n4\n180"
# Custom test cases
assert run("1\n1 1\n") == "2", "minimum size inputs"
assert run("1\n1000000000 1\n") == "2", "large n, small k"
assert run("1\n1 1000000000\n") == "2", "small n, large k"
assert run("1\n100 7\n") == "100", "subset sums cut off by budget"
assert run("1\n200 8\n") == "200", "subset sums cut off by budget"
| Test input | Expected output | What it validates |
|---|---|---|
| 1 1 | 2 | Minimum size input, only one dessert |
| 1000000000 1 | 2 | Large n, small k, only two subsets affordable |
| 1 1000000000 | 2 | Small n, huge k, budget only allows empty and first dessert |
| 100 7 | 100 | Budget limits number of subsets before reaching all 2^7 subsets |
| 200 8 | 200 | Budget limits subsets before reaching all 2^8 subsets |
Edge Cases
If $n = 0$, the algorithm correctly outputs 1, corresponding to the empty subset. For example, $n = 0, k = 10$ produces 1 because only the empty subset is affordable. If $k = 1$ and $n = 1$, the algorithm outputs 2 because both the empty set and the single dessert are within budget. In each case, the loop correctly multiplies powers of two but stops early whenever exceeding $n+1$, ensuring efficiency even when $k$ is extremely large.