[CodeForce431C]k-tree

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.

最近有一个富有创造力的学生Lesha听了一个关于树的讲座。在听完讲座之后，Lesha受到了启发，并且他有一个关于k-tree（k叉树）的想法。

A k -tree is an infinite rooted tree where:

each vertex has exactly k k children;

each edge has some weight;

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

k-tree都是无根树，并且满足：

每一个非叶子节点都有k个孩子节点；

每一条边都有一个边权；

每一个非叶子节点指向其k个孩子节点的k条边的权值分别为1,2,3,...,k。

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

如图所示：

As soon as Dima, a good friend of Lesha, found out about the tree, he immediately wondered: "How many paths of total weight n n (the sum of all weights of the edges in the path) are there, starting from the root of a k k -tree and also containing at least one edge of weight at least d d ?".Help Dima find an answer to his question. As the number of ways can be rather large, print it modulo 1000000007 1000000007 ($10^{9}+7$).

当Lesha的好朋友Dima看到这种树时，Dima马上想到了一个问题：“有多少条从k-tree的根节点出发的路上的边权之和等于n，并且经过的这些边中至少有一条边的边权大于等于d呢?” 现在你需要帮助Dima解决这个问题。考虑到路径总数可能会非常大，所以只需输出路径总数 mod 1000000007 即可。（1000000007=10^9+7）

输入格式

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

只有一行数，n，k，d. (1 <= n, k <= 100; 1 <= d <= k; n, d, k 三者用空格隔开)。

输出格式

Print a single integer — the answer to the problem modulo1000000007 ($10^{9}+7$).

只有一行，一个整数，即输出路径总数 mod 1000000007。

样例输入

3 3 2

样例输出

题解

#include<bits/stdc++.h>

#define maxk 105

#define maxn 105

using namespace std;

const long long mod = 1e9+7;

inline char get(){

	static char buf[3000],*p1=buf,*p2=buf;

	return p1==p2 && (p2=(p1=buf)+fread(buf,1,3000,stdin),p1==p2)?EOF:*p1++;

}

inline long long read(){

	register char c=get();register long long f=1,_=0;

	while(c>'9' || c<'0')f=(c=='-')?-1:1,c=get();

	while(c<='9' && c>='0')_=(_<<3)+(_<<1)+(c^48),c=get();

	return _*f;

}

long long n,k,d;

long long dp[maxn][3];//第一维记录不考虑d的情况，第二维记录考虑d的情况

long long cas;

int main(){

	//freopen("1.txt","r",stdin);

	n=read();k=read();d=read();//总和等于n，k叉树，至少一条边大于等于d

	for(register long long i=1;i<=n;i++){//i表示当前n=i

		for(register long long j=1;j<=k && j<=i;j++){

			cas=i-j;

			bool used_d=0;

			if(j>=d)used_d=1;

			dp[i][1]+=dp[cas][1];//因为第一维不考虑d的大小，直接相加即可

			if(cas==0){

				dp[i][1]++;//无论如何第一维都要加

				if(used_d)dp[i][2]++;//如果当前考虑了d，则让考虑了d的维度更新

			}

			else{

				if(used_d)dp[i][2]+=dp[cas][1];//如果目前考虑的d,则之前就不用考虑d了

				else dp[i][2]+=dp[cas][2];//如果目前没考虑d，则之前要考虑d

			}

		}

		dp[i][1]%=mod;

		dp[i][2]%=mod;

	}

	cout<<dp[n][2]%mod;

	return 0;

}