CF 104758I - ICPC Masters
We are given a line of participants who need to be split into teams. The organizers always try to form as many complete teams as possible, where each complete team contains exactly $K$ people.
Rating: -
Tags: -
Solve time: 49s
Verified: yes
Solution
Problem Understanding
We are given a line of participants who need to be split into teams. The organizers always try to form as many complete teams as possible, where each complete team contains exactly $K$ people. After forming these full teams, there may be some participants left over who cannot fill another full group of $K$. Those remaining participants are placed together into one additional team.
This means the final structure is determined purely by dividing $N$ people into chunks of size $K$, with a possible last chunk that is smaller than $K$. The task is to determine the size of the smallest team that ends up being formed.
The input bounds are very small: $N \le 1000$ and $K \le 10$. This immediately rules out any concern about efficiency; even a simulation that repeatedly subtracts $K$ would be fast enough. The real challenge is not computation time but avoiding a mistaken interpretation of how the last team is formed.
A common pitfall appears when $N$ is a multiple of $K$. In that case, there is no leftover group at all. For example, if $N = 8$ and $K = 2$, we form four teams of size 2, so the smallest team is still 2. A naive interpretation that “there is always a leftover team” would incorrectly assume an empty or size-zero final group.
Another subtle case is when $K > N$. For instance, if $N = 5$ and $K = 10$, no full team can be formed, so everyone stays together in one team of size 5. Any solution that assumes at least one full team exists would incorrectly ignore this scenario.
Approaches
The brute-force way to think about this is to explicitly simulate grouping participants one by one. We keep forming groups of size $K$, counting how many people go into each group. Each time we reach $K$, we finalize that team and start a new one. At the end, if there is a partially filled group, we include it as the last team.
This works because it mirrors the process described in the statement exactly. However, it is unnecessary bookkeeping: we are repeatedly subtracting $K$ from $N$, effectively doing $O(N/K)$ steps, which is at most 1000 operations here, so even brute force is trivial. The inefficiency is conceptual rather than practical.
The key observation is that we never actually need to simulate grouping. Every full team has size $K$, and there is at most one leftover team whose size is exactly $N \bmod K$. So the smallest team is simply the leftover size if it exists, otherwise it is $K$. If there is no remainder, all teams are equal.
This reduces the entire problem to computing a remainder and handling the zero case.
| Approach | Time Complexity | Space Complexity | Verdict |
|---|---|---|---|
| Brute Force Simulation | $O(N/K)$ | $O(1)$ | Accepted |
| Modular Arithmetic | $O(1)$ | $O(1)$ | Accepted |
Algorithm Walkthrough
- Compute the remainder $r = N \bmod K$, which represents how many participants are left after forming full teams. This directly captures whether a final incomplete team exists.
- If $r = 0$, it means $N$ is perfectly divisible by $K$, so every team has exactly $K$ members and there is no smaller group.
- If $r \neq 0$, then the last team consists exactly of those remaining $r$ participants, and all other teams are size $K$, so the smallest team is $r$.
Why it works
The partition of $N$ into groups is deterministic: repeated subtraction of $K$ creates identical full blocks of size $K$, and at most one residual block whose size is strictly less than $K$. No other grouping pattern is possible under the rules. This guarantees that the minimum team size must be either $K$ or the remainder, with no intermediate values introduced.
Python Solution
import sys
input = sys.stdin.readline
N, K = map(int, input().split())
r = N % K
if r == 0:
print(K)
else:
print(r)
The solution reads the two integers and computes the remainder directly. The only decision point is whether the remainder is zero. If it is, we output $K$, since every team is full. Otherwise, we output the remainder, since that is the only smaller-than-$K$ group.
A subtle point is handling cases where $K > N$. In that situation, $N % K = N$, which correctly reflects that no full team exists and the single team contains all participants.
Worked Examples
Example 1: Input 5 3
We compute how participants are grouped.
| Step | Remaining N | Full teams formed | Remainder |
|---|---|---|---|
| Start | 5 | 0 | 5 |
| After grouping | 5 | 1 team of 3 | 2 |
The remainder is 2, so the teams are one group of 3 and one group of 2. The smallest team is 2. This confirms that leftover participants form the final team directly.
Example 2: Input 21 2
| Step | Remaining N | Full teams formed | Remainder |
|---|---|---|---|
| Start | 21 | 0 | 21 |
| After grouping | 21 | 10 teams of 2 | 1 |
The remainder is 1, so there are ten full teams of size 2 and one final team of size 1. The minimum team size is therefore 1. This shows how a small remainder dominates the answer.
Complexity Analysis
| Measure | Complexity | Explanation |
|---|---|---|
| Time | $O(1)$ | Only a modulus and a conditional check are performed |
| Space | $O(1)$ | No additional data structures are used |
The constraints allow even slower methods, but the direct arithmetic solution is constant time and trivially fits within limits.
Test Cases
import sys, io
def run(inp: str) -> str:
sys.stdin = io.StringIO(inp)
import sys
input = sys.stdin.readline
N, K = map(int, input().split())
r = N % K
if r == 0:
return str(K)
return str(r)
# provided samples
assert run("5 3\n") == "2", "sample 1"
assert run("21 2\n") == "1", "sample 2"
assert run("8 5\n") == "3", "sample 3"
# custom cases
assert run("2 10\n") == "2", "K > N case"
assert run("10 5\n") == "5", "perfect division"
assert run("1 1\n") == "1", "minimum boundary"
assert run("1000 3\n") == str(1000 % 3), "large N stress case"
| Test input | Expected output | What it validates |
|---|---|---|
| 2 10 | 2 | K greater than N, single group |
| 10 5 | 5 | exact division, no leftover |
| 1 1 | 1 | smallest possible input |
| 1000 3 | 1 | remainder correctness on larger bound |
Edge Cases
When $K > N$, for example input 2 10, the algorithm computes $2 \bmod 10 = 2$. Since the remainder is non-zero, it returns 2. This matches the actual grouping where no full team can form.
When $N$ is divisible by $K$, such as 10 5, the remainder becomes 0. The algorithm switches to outputting $K$, giving 5. This reflects that all teams are full-sized and no smaller group exists.
When $K = 1$, every participant forms their own team. The remainder is always 0, so the answer is always 1, consistent with the fact that every team has size 1.