传送门：https://www.luogu.org/problemnew/show/P5004

分析

动态规划转移方程是这样的\(f[i]=\sum^{i-m-1}_{j=0}f[j]\)。
那么很明显的是要构造举证，而且要维护前缀和，所以需要保留\(m+1\)项。

ac代码

#include <bits/stdc++.h>
#define ll long long
#define ms(a, b) memset(a, b, sizeof(a))
#define inf 0x3f3f3f3f
#define N 25
#define mod ((int)1e9 + 7)
using namespace std;
template <typename T>
inline void read(T &x) {x = 0; T fl = 1;char ch = 0;while (ch < '0' || ch > '9') {if (ch == '-') fl = -1;ch = getchar();}while (ch >= '0' && ch <= '9') {x = (x << 1) + (x << 3) + (ch ^ 48);ch = getchar();}x *= fl;
}
ll n;
int m;
struct Matrix {int a[N][N], x, y;void init() {memset(a, 0, sizeof(a));x = y = 0;}Matrix operator *(const Matrix &rhs) const{Matrix res; res.init();res.x = x, res.y = rhs.y;int c = y;for (int i = 1; i <= x; i ++) {for (int j = 1; j <= y; j ++) {for (int k = 1; k <= c; k ++) {res.a[i][j] = (res.a[i][j] + (ll) a[i][k] * rhs.a[k][j]) % mod;}}}return res;}Matrix power(Matrix a, ll b) {Matrix res; res.init();res.x = res.y = a.x;for (int i = 1; i <= res.x; i ++) res.a[i][i] = 1;for (; b; b >>= 1) {if (b & 1) res = res * a;a = a * a;}return res;}
}a, b;
int main() {read(n); read(m);if (n <= m) {printf("%lld\n", n + 1);return 0;}a.x = a.y = m + 2, b.x = m + 2, b.y = 1;for (int i = 2; i <= m + 2; i ++) b.a[i][1] = 1;b.a[1][1] = m + 1;a.a[1][1] = a.a[1][2] = 1;a.a[2][2] = a.a[2][m + 2] = 1;for (int i = 3; i <= m + 2; i ++) a.a[i][i - 1] = 1;a = a.power(a, n - m);b = a * b;printf("%d\n", b.a[1][1]);return 0;
}