k-Tree DP计数

k-Tree

对dp数情况的问题还是不太懂

#include <iostream>
#include <vector>
#include <algorithm>
#include <string>
#include <set>
#include <queue>
#include <map>
#include <sstream>
#include <cstdio>
#include <cstring>
#include <numeric>
#include <cmath>
#include <iomanip>
#include <deque>
#include <bitset>
#include <cassert>
//#include <unordered_set>
//#include <unordered_map>
#define ll              long long
#define pii             pair<int, int>
#define rep(i,a,b)      for(ll  i=a;i<=b;i++)
#define dec(i,a,b)      for(ll  i=a;i>=b;i--)
#define forn(i, n)      for(ll i = 0; i < int(n); i++)
using namespace std;
int dir[4][2] = { { 1,0 },{ 0,1 } ,{ 0,-1 },{ -1,0 } };
const long long INF = 0x7f7f7f7f7f7f7f7f;
const int inf = 0x3f3f3f3f;
const double pi = acos(-1.0);
const double eps = 1e-6;
const int mod = 1e9 + 7;

inline ll read()
{
    ll x = 0; bool f = true; char c = getchar();
    while (c < '0' || c > '9') { if (c == '-') f = false; c = getchar(); }
    while (c >= '0' && c <= '9') x = (x << 1) + (x << 3) + (c ^ 48), c = getchar();
    return f ? x : -x;
}
inline ll gcd(ll m, ll n)
{
    return n == 0 ? m : gcd(n, m % n);
}
void exgcd(ll A, ll B, ll& x, ll& y)
{
    if (B) exgcd(B, A % B, y, x), y -= A / B * x; else x = 1, y = 0;
}
inline int qpow(int x, ll n) {
    int r = 1;
    while (n > 0) {
        if (n & 1) r = 1ll * r * x % mod;
        n >>= 1; x = 1ll * x * x % mod;
    }
    return r;
}
/**********************************************************/
const int N = 100 + 5;
ll dp[N][2];

int main() {
    cin.tie(0);cout.tie(0);

    ll n, k, d;
    cin >> n >> k >> d;
    memset(dp, 0, sizeof(dp));
    dp[0][0] = 1;
    rep(i, 1, n)
    {
        rep(j, 1, k)
        {
            if (i - j < 0)
                break;
            if (j < d)
            {
                dp[i][0] = (dp[i][0] + dp[i - j][0]) % mod;
                dp[i][1] = (dp[i][1] + dp[i - j][1]) % mod;
            }
            else
            {
                dp[i][1] = (dp[i][1] + dp[i - j][0]) % mod;
                dp[i][1] = (dp[i][1] + dp[i - j][1]) % mod;
            }
        }
    }
    cout << dp[n][1] << endl;
}