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.
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 (109 + 7).
題目大意：說明了 k 樹的定義，然後問你從 1（根結點）開始，往下走，權值和爲 n 的路徑條數，要求路徑上至少有一條邊權值大於等於d；
雖然是涉及到了樹的結構，但是和樹沒有太大關係，可以設：
dp1[i]+=dp1[j] (j>=i-k&&j<i) 表示組成 i 的路徑條數；
dp2[i]+=dp2[j] (j>=i-d+1&&j<i) 表示組成 i 的路徑條數，但是這個 i 本身就小於 d；
所以，dp1[n]-dp2[n] 就是答案要求的路徑條數；
代碼：
#include<bits/stdc++.h>
#define LL long long
#define pa pair<int,int>
#define ls k<<1
#define rs k<<1|1
#define inf 0x3f3f3f3f
using namespace std;
const int N=200100;
const int M=2000100;
const LL mod=1e9+7;
//int head[N],cnt;
//struct Node{
// int to,nex;
//}edge[N*2];
//void add(int p,int q){
// edge[cnt].to=q,edge[cnt].nex=head[p],head[p]=cnt++;
//}
LL dp1[N],dp2[N];
int main(){
// memset(head,-1,sizeof(head));
int n,k,d;
scanf("%d%d%d",&n,&k,&d);
dp1[0]=dp2[0]=1;
for(int i=1;i<=n;i++){
for(int j=max(0,i-k);j<i;j++) (dp1[i]+=dp1[j])%=mod;
for(int j=max(0,i-d+1);j<i;j++) (dp2[i]+=dp2[j])%=mod;
}
printf("%lld\n",(dp1[n]-dp2[n]+mod)%mod);
return 0;
}