CF 1048291 - Очки в баскетболе
We are given a total number of points scored in a basketball match, where every successful scoring action contributes either 1, 2, or 3 points. The actual sequence of shots is unknown, only the final sum matters.
Rating: -
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Solve time: 1m 3s
Verified: yes
Solution
Problem Understanding
We are given a total number of points scored in a basketball match, where every successful scoring action contributes either 1, 2, or 3 points. The actual sequence of shots is unknown, only the final sum matters. The task is to reconstruct the smallest possible number of scoring actions that could produce exactly this total.
So the input is a single integer n, representing the total score accumulated by both teams together. The output is the minimum number of terms in a decomposition of n where each term is 1, 2, or 3. In other words, we are trying to express n as a sum of the fewest possible integers from the set {1, 2, 3}.
The constraint 0 ≤ n ≤ 1000 makes this a very small state space. Any solution from O(n²) down to O(1) is easily sufficient, while exponential or brute force search over all decompositions becomes unnecessary.
The most subtle edge cases appear at small values of n. When n = 0, there are no scoring actions, so the answer must be 0. When n is 1 or 2, the answer matches the number itself because we cannot represent them with fewer than one or two shots respectively. Another slightly less trivial case is n = 3, where one 3-point shot is strictly better than three 1-point shots or one 2-point plus one 1-point.
A naive approach that tries to enumerate all compositions of n into 1, 2, 3 would still work within constraints but would quickly become redundant and inefficient. The structure of the problem suggests a greedy decomposition, since higher-valued shots always dominate lower ones in terms of minimizing count.
Approaches
A brute-force strategy would attempt to explore all sequences of 1, 2, and 3 that sum to n and track the minimum length among valid sequences. This is essentially generating all ordered compositions of n with parts in {1, 2, 3}. At each step, we branch into up to three possibilities, which leads to a recurrence resembling 3^n in the worst case. Even with pruning, the number of states grows extremely quickly, and for n up to 1000 this becomes infeasible.
The key observation is that we are not asked to count or construct sequences, only to minimize the number of elements. Every time we replace three 1-point shots with one 3-point shot, we reduce the number of shots without changing the total. Similarly, replacing two 1-point shots with one 2-point shot also reduces the count. This suggests that higher scoring shots are always preferable.
To minimize the number of terms, we want to use as many 3-point shots as possible, then handle the remainder optimally with 2-point shots, and finally 1-point shots if needed. This reduces the problem to a simple greedy division of n by 3, with remainder handled carefully.
This works because 3 is the largest allowed value, and replacing any combination of smaller values with a larger value never increases the number of terms. The optimal structure therefore aligns with taking as many 3s as possible, then 2s, then 1s.
| Approach | Time Complexity | Space Complexity | Verdict |
|---|---|---|---|
| Brute Force | Exponential O(3^n) | O(n) recursion stack | Too slow |
| Optimal Greedy | O(1) | O(1) | Accepted |
Algorithm Walkthrough
- Compute how many 3-point shots we can take from n using integer division. This gives the maximum number of high-value actions that reduce the remaining sum most efficiently.
- Compute the remainder after removing all 3-point contributions. This remainder is now in the range 0 to 2, because any integer mod 3 lies in that interval.
- If the remainder is 0, no additional shots are needed. The decomposition is already optimal using only 3-point shots.
- If the remainder is 1 or 2, it contributes exactly one additional shot, because 1 can be represented as a single 1-point shot, and 2 can be represented as a single 2-point shot. There is no benefit in splitting these further.
- The final answer is the number of 3-point shots plus one extra shot if the remainder is non-zero.
Why it works: the algorithm relies on the fact that replacing any group of smaller scoring actions with a single higher scoring action strictly reduces the number of terms. Since 3 is the maximum allowed value, maximizing its usage minimizes the number of summands globally. Once all 3s are used greedily, the remaining value is strictly less than 3, and thus cannot be improved further by any combination other than a single term if non-zero.
Python Solution
import sys
input = sys.stdin.readline
n = int(input().strip())
if n == 0:
print(0)
else:
print((n + 2) // 3)
The implementation relies on the observation that the number of 3-point shots is n // 3. The remainder handling is implicitly encoded in the expression (n + 2) // 3, which correctly rounds up whenever n is not divisible by 3. This avoids explicit branching on the remainder while still capturing the optimal structure.
The special case n = 0 is handled explicitly to ensure correctness, since the formula also returns 0 in this case but the conditional makes the reasoning explicit and robust for extensions.
Worked Examples
Example 1
Input: n = 6
We compute the number of 3-point shots as 6 // 3 = 2. The remainder is 0.
| Step | n | 3-point shots | remainder | answer so far |
|---|---|---|---|---|
| start | 6 | 0 | 6 | 0 |
| after division | 6 | 2 | 0 | 2 |
The final result is 2, corresponding to two 3-point shots. Any alternative decomposition like 2 + 2 + 2 or 1 + 1 + 1 + 1 + 1 + 1 would use more shots, confirming optimality.
Example 2
Input: n = 10
We compute 10 // 3 = 3, with remainder 1.
| Step | n | 3-point shots | remainder | answer so far |
|---|---|---|---|---|
| start | 10 | 0 | 10 | 0 |
| after division | 10 | 3 | 1 | 3 |
| remainder handling | 10 | 3 | 1 | 4 |
The remainder of 1 requires one additional 1-point shot, giving a total of 4. Any attempt to replace a 3-point shot with combinations of 2s and 1s only increases the total number of shots.
Complexity Analysis
| Measure | Complexity | Explanation |
|---|---|---|
| Time | O(1) | Only arithmetic operations are performed regardless of n |
| Space | O(1) | No auxiliary data structures are used |
The solution is constant time and constant memory, which is easily sufficient for n up to 1000 and would scale to much larger bounds without modification.
Test Cases
import sys, io
def run(inp: str) -> str:
sys.stdin = io.StringIO(inp)
import sys
input = sys.stdin.readline
n = int(input().strip())
if n == 0:
return "0"
return str((n + 2) // 3)
# provided samples
assert run("6") == "2", "sample 1"
assert run("10") == "4", "sample 2"
# custom cases
assert run("0") == "0", "minimum boundary"
assert run("1") == "1", "smallest non-zero"
assert run("2") == "1", "two-point case"
assert run("3") == "1", "exact multiple of 3"
assert run("1000") == str((1000 + 2) // 3), "maximum boundary"
| Test input | Expected output | What it validates |
|---|---|---|
| 0 | 0 | empty score edge case |
| 1 | 1 | smallest non-zero decomposition |
| 2 | 1 | direct 2-point optimality |
| 3 | 1 | exact multiple of 3 |
| 1000 | 334 | upper bound correctness |
Edge Cases
For n = 0, the algorithm directly returns 0 because there are no points to decompose. The formula (n + 2) // 3 also evaluates to 0, matching the expected output.
For n = 1, we have one remainder after division by 3. The computation gives (1 + 2) // 3 = 1, corresponding to a single 1-point shot, which is optimal since no larger scoring action can represent it.
For n = 2, the formula gives (2 + 2) // 3 = 1 as well. This corresponds to a single 2-point shot, which is the minimal decomposition.
For n = 1000, the division yields 333 full 3-point shots with remainder 1. The algorithm returns 334, meaning 333 threes plus one single point shot. This matches the optimal structure because any attempt to replace a 3-point shot with 2+1 or 1+1+1 increases the count of actions rather than decreasing it.