codeforces k-Tree (dp still won't work)

题目
题意: 给定一棵树,每个节点恰好有k个儿子,The corresponding edge is exactly 1-k. Find out how many options there are,The weights on the satisfying path are exactly n,And at least one edge satisfies the edge weight>=d.
思路: It is actually lineardp,Each level can only be selected1-k.好久没dp就d不出来了,可以先dp一遍1-k都能用的,再dpCan only be used once1-(d-1)的,Subtraction is what satisfies the meaning of the question.f[0][i][j]: Use all sides,从前ilayer selected,Boundary right is exactly j的方案数.Just enumerate how many weights the current layer uses,can be delivered.f[0][i][j] = f[0][i-1][j-1…k]
Only depends on the previous layer,所以可以把iThis dimension is compressed.f[0][j] = f[0][j-1…k]
时间复杂度: O(n* n * k)或O(n*k)
代码:

#include<bits/stdc++.h>
using namespace std;
const int N = 102;
const int mod = 1e9+7;
int n,m,k,T;
int f[2][N]; //0:所有边,1:小于d的边,f[j]:权值为j的方案数
void solve()
{
    
	cin>>n>>k>>m;
	f[0][0] = 1;
		for(int j=1;j<=n;++j) //权值
		{
    
			for(int t=1;t<=k;++t) //Where does the enum move from
			{
    
				if(j-t<0) break;
				f[0][j] = (f[0][j] + f[0][j-t]) % mod;
			}
		}
	k = m-1;
	f[1][0] = 1;
		for(int j=1;j<=n;++j) //权值
		{
    
			for(int t=1;t<=k;++t) //Where does the enum move from
			{
    
				if(j-t<0) break;
				f[1][j] = (f[1][j] + f[1][j-t]) % mod;
			}
		}
	long long ans = 0;
	ans = (ans + f[0][n]) % mod;
	ans = (ans - f[1][n]) % mod;
	ans = (ans + mod) % mod;
	cout<<ans;
}
signed main(void)
{
    
	solve();
	return 0;
}