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;
}

原文地址：https://www.cnblogs.com/chhokmah/p/10568032.html