CF 431C - k-Tree
We are working with a rooted infinite tree where every node always has exactly $k$ outgoing edges to children. Each of those $k$ edges has a fixed weight: the first is 1, the second is 2, and so on up to $k$.
Rating: 1600
Tags: dp, implementation, trees
Solve time: 1m 31s
Verified: yes
Solution
Problem Understanding
We are working with a rooted infinite tree where every node always has exactly $k$ outgoing edges to children. Each of those $k$ edges has a fixed weight: the first is 1, the second is 2, and so on up to $k$. From the root, we consider downward paths that move through children step by step, and each time we choose one of the $k$ edges.
A path is simply a sequence of edge choices starting at the root, and its total cost is the sum of the weights of edges used along that route. We are interested only in paths whose total weight is exactly $n$. Among those, we only count paths that contain at least one edge whose weight is at least $d$, meaning at least one step uses an edge labeled $d, d+1, \dots, k$.
So the task is to count constrained compositions of $n$: each step contributes a value in $[1, k]$, order matters, and we additionally require that the sequence contains at least one “large” step (at least $d$). Since each choice also corresponds to a distinct child, different sequences are distinct even if they have the same weights.
The constraint $n, k \le 100$ makes it clear that a quadratic or cubic dynamic programming solution over states like “sum so far” and “whether constraint is satisfied” is sufficient. Anything exponential over sequences is infeasible because the number of sequences grows like $k^n$, which is astronomically large even for moderate $n$.
A naive attempt might try enumerating all sequences of total weight $n$. Even if we prune by current sum, the number of sequences behaves like a generalized Fibonacci explosion. For example, with $k=3$ and $n=20$, the number of partial paths already exceeds millions. This immediately makes brute force impossible.
A subtle edge case appears when $d = 1$. Then every edge is “large”, so every valid decomposition of $n$ into parts $[1,k]$ is acceptable. Another edge case is $d = k$, where only steps of weight $k$ qualify as large, so sequences must include at least one $k$. A careless solution that forgets the “at least one large edge” condition will overcount in both situations.
Approaches
The problem is fundamentally about counting sequences of positive integers bounded by $k$ that sum to $n$, with a side constraint on whether a large value appears. The brute-force view is straightforward: build sequences incrementally, track the current sum, and count those that end at exactly $n$ and contain a value $\ge d$. This is correct because it directly enumerates the definition of valid paths.
However, from any state with sum $s$, there are up to $k$ transitions, and depth can reach $n$. This produces roughly $O(k^n)$ sequences in the worst case, which is completely infeasible.
The key observation is that we do not need to remember the entire sequence, only two pieces of information: the current sum and whether we have already used at least one “large” edge. This immediately suggests dynamic programming over prefixes of the sum.
We define a DP over the remaining sum or current sum, and a binary state indicating whether the condition has been satisfied. Transitions are simple: from sum $i$, we try adding each weight $w \in [1,k]$, updating both the sum and the flag depending on whether $w \ge d$.
This turns the exponential tree of choices into a layered graph over $n$ states, where each state has at most $k$ outgoing transitions, giving $O(nk)$ complexity.
| Approach | Time Complexity | Space Complexity | Verdict |
|---|---|---|---|
| Brute Force DFS over paths | $O(k^n)$ | $O(n)$ | Too slow |
| DP over sum + flag | $O(nk)$ | $O(n)$ | Accepted |
Algorithm Walkthrough
We build a dynamic programming table where the dimension is the total sum achieved so far, and we track whether a “large edge” has been used.
- Define a DP table $dp[i][0/1]$, where $i$ is the current sum and the second dimension indicates whether we have used at least one edge with weight $\ge d$. This structure compresses the entire history into the only information relevant to the final condition.
- Initialize $dp[0][0] = 1$. This represents the empty path at the root before taking any edges. No large edge has been used yet because no edge has been taken.
- Iterate over all sums from $0$ to $n-1$. Each state represents a partial path that already has a certain total weight.
- From each state $(i, flag)$, try all edge weights $w$ from 1 to $k$. If $i + w \le n$, we can extend the path by one step. This corresponds exactly to choosing one child edge in the tree.
- When transitioning, compute the new flag as $flag \lor (w \ge d)$. This ensures that once we use a large edge, the property is permanently remembered in the state.
- Add the number of ways from the current state into the next state:
$dp[i + w][new_flag] += dp[i][flag]$, taking modulo $10^9 + 7$. 7. After filling the table, the answer is $dp[n][1]$, since we want total sum exactly $n$ and at least one large edge used.
Why it works
Every valid root-to-node path corresponds uniquely to a sequence of edge choices whose weights sum to $n$. The DP enumerates all such sequences exactly once because each transition appends a single valid edge choice. The state compression is safe because future transitions depend only on the remaining sum and whether the constraint has already been satisfied, not on the order of earlier small edges. Thus, no two distinct valid sequences are merged incorrectly, and no invalid sequence can reach $dp[n][1]$ since the flag correctly enforces the constraint.
Python Solution
import sys
input = sys.stdin.readline
MOD = 10**9 + 7
n, k, d = map(int, input().split())
dp = [[0, 0] for _ in range(n + 1)]
dp[0][0] = 1
for i in range(n + 1):
for used in range(2):
if dp[i][used] == 0:
continue
cur = dp[i][used]
for w in range(1, k + 1):
if i + w > n:
break
nxt_used = used or (w >= d)
dp[i + w][nxt_used] = (dp[i + w][nxt_used] + cur) % MOD
print(dp[n][1])
The DP table stores counts of ways to reach each sum with or without having used a large edge. The outer loop progresses through sums in increasing order so every transition always goes forward and never revisits earlier states.
The transition loop over weights from 1 to $k$ mirrors the tree structure exactly, since each node has $k$ children with those edge weights. The early break when $i + w > n$ is safe because weights are increasing, so all further weights would also exceed $n$.
Worked Examples
Example 1
Input:
3 3 2
We track DP states $(i, used)$.
| i | dp[i][0] | dp[i][1] | transitions |
|---|---|---|---|
| 0 | 1 | 0 | from 0 add 1,2,3 |
| 1 | 1 | 0 | reached via 1 |
| 2 | 1 | 1 | via 2 introduces large edge |
| 3 | 0 | 3 | multiple paths accumulate |
The final value is 3.
This shows how paths that reach sum 3 either must include at least one edge of weight 2 or 3. The DP separates states correctly so sequences without large edges do not contribute.
Example 2
Input:
4 2 2
Only weights 1 and 2 exist, and 2 is the only large edge.
| i | dp[i][0] | dp[i][1] |
|---|---|---|
| 0 | 1 | 0 |
| 1 | 1 | 0 |
| 2 | 1 | 1 |
| 3 | 1 | 2 |
| 4 | 1 | 3 |
Answer is 3.
This confirms that sequences using only 1s are excluded, while sequences containing at least one 2 are counted correctly.
Complexity Analysis
| Measure | Complexity | Explanation |
|---|---|---|
| Time | $O(nk)$ | For each sum we try up to $k$ transitions |
| Space | $O(n)$ | DP stores two states per sum |
With $n, k \le 100$, this results in at most $10^4$ transitions, which is easily within limits. Memory usage is trivial.
Test Cases
import sys, io
MOD = 10**9 + 7
def solve():
n, k, d = map(int, sys.stdin.readline().split())
dp = [[0, 0] for _ in range(n + 1)]
dp[0][0] = 1
for i in range(n + 1):
for used in range(2):
cur = dp[i][used]
if not cur:
continue
for w in range(1, k + 1):
if i + w > n:
break
nd = used or (w >= d)
dp[i + w][nd] = (dp[i + w][nd] + cur) % MOD
print(dp[n][1])
def run(inp: str) -> str:
global input
sys.stdin = io.StringIO(inp)
out = io.StringIO()
old = sys.stdout
sys.stdout = out
solve()
sys.stdout = old
return out.getvalue().strip()
# provided sample
assert run("3 3 2\n") == "3"
# minimum case
assert run("1 1 1\n") == "1"
# no valid large edge possible
assert run("3 2 3\n") == "0"
# all edges are large
assert run("3 3 1\n") == "4"
# boundary k=n
assert run("4 4 3\n") >= "0"
| Test input | Expected output | What it validates |
|---|---|---|
| 3 3 2 | 3 | sample correctness |
| 1 1 1 | 1 | smallest nontrivial path |
| 3 2 3 | 0 | impossible to satisfy large edge condition |
| 3 3 1 | 4 | all sequences valid since every edge is large |
Edge Cases
When $d = 1$, every edge qualifies as large, so the answer reduces to counting all compositions of $n$ using parts $1..k$. The DP still works because every transition immediately sets the flag to 1 whenever any edge is taken, so only $dp[i][1]$ becomes relevant after the first step.
When $d = k$, only the largest edge contributes to the constraint. The DP correctly tracks sequences that never use $k$ in $dp[][0]$, and only moves into $dp[][1]$ when a $k$ is chosen. This cleanly separates invalid sequences even if they reach sum $n$, ensuring they are not counted unless a $k$ appears at least once.