431C. k-Tree

dp, implementation, trees, *1600

https://codeforces.com/problemset/problem/431/C

Quite recently a creative student Lesha had a lecture on trees. After the lecture Lesha was inspired and came up with the tree of his own which he called a k-tree.

A k-tree is an infinite rooted tree where:

each vertex has exactly k children;
each edge has some weight;
if we look at the edges that goes from some vertex to its children (exactly k edges), then their weights will equal 1, 2, 3, ..., k.

The picture below shows a part of a 3-tree.

As soon as Dima, a good friend of Lesha, found out about the tree, he immediately wondered: "How many paths of total weight n (the sum of all weights of the edges in the path) are there, starting from the root of a k-tree and also containing at least one edge of weight at least d?".

Help Dima find an answer to his question. As the number of ways can be rather large, print it modulo 1000000007 (10^9^ + 7).

Input

A single line contains three space-separated integers: n, k and d (1 ≤ n, k ≤ 100; 1 ≤ d ≤ k).

Output

Print a single integer — the answer to the problem modulo 1000000007 (10^9^ + 7).

Examples

input

3 3 2

output

input

3 3 3

output

input

4 3 2

output

input

4 5 2

output

分类讨论，记忆化搜索。

python

MOD = 1000000007
from functools import lru_cache
 
@lru_cache(maxsize=None)
def dfs(n, b):
    if n == 0 and b >= d:
        return 1
    if n < 0:
        return 0
 
    ans = 0
    for i in range(1, k+1):
        ans = (ans + dfs(n-i, max(i, b))) % MOD
 
    return ans
 
 
n, k, d = map(int, input().split())
print(dfs(n, 0))

python

# 23工学院 蒋子轩
n,k,d=map(int,input().split())
mod = 10**9 + 7
# A[i]：总权重为i的路径数 ； B[i]：总权重为i且所有边权重小于d的路径数
A = [1] + [0] * n
B = [1] + [0] * n
# 路径数本质上就是整数划分问题
# 本题即求用不大于k的正整数划分i，用小于d的正整数划分i的方法数之差
for i in range(1, n + 1):
    for j in range(1, min(i,k)+1):
        A[i] = (A[i] + A[i - j]) % mod
    for j in range(1, min(d, i + 1)):
        B[i] = (B[i] + B[i - j]) % mod
print((A[n] - B[n]) % mod)

python

# 23数院，胡睿诚
# 这题其实有点核电站的味道感觉
# 09267: 核电站
# http://cs101.openjudge.cn/practice/09267/

m = 10**9+7

def compute(n, k):
    if k == 0:
        return 0
    ans = [1]
    for i in range(k):
        ans.append((1 << i) % m)

    for i in range(k+1, n+1):
        ans.append((2*ans[i-1]-ans[i-k-1]) % m)

    return ans[n]


n, k, d = map(int, input().split())
print((compute(n, min(n, k))-compute(n, d-1)) % m)

431C. k-Tree ​

431C. k-Tree