UVA12297 Super Poker 矩阵快速幂

设 (f(n,k)) 为用 (k) 张牌组成 (n) 的方案数，则

(f(n,k)=C_4^0 f(n−k,k)+C_4^1 f(n−k,k−1)+C_4^2f(n−k,k−2)+C_4^3 f(n−k,k−3)+ C_4^4 f(n−k,k−4))

也就是考虑这 (k) 张牌里有多少张 (1) ，倘若有 (x) 张，则剩下的(没有 (1))组成的数是 (n-x) 。

又因为这里边没有 (1) ，所以可以和每张牌权值都 (-1) 后的情况相对应，等价于 (k-x) 张牌(可以有 (1))组成 (n-k)

发现这样的 (DP) 是二维的，我们把这二维展开成一维的转移，用矩阵优化一下就 (ok) 了

#include<iostream>
#include<cstring>
#include<cstdio>
#define LL long long
using namespace std;
int n, k, ans;
const int N = 105, mod = 1e9 + 9;
int C4[7] = {1, 4, 6, 4, 1};
int f[12][12];
struct ju 
{
    LL v[N][N];
    friend ju operator *(const ju &a, const ju &b) 
    {
        ju c; memset(c.v, 0, sizeof(c.v));
        for (int i = 1; i <= 100; ++i)
            for (int j = 1; j <= 100; ++j)
                for (int k = 1; k <= 100; ++k)
                    (c.v[i][j] += a.v[i][k] * b.v[k][j]) %= mod;
        return c;
    }
} A, B, base;
ju ksm(ju a, LL b) 
{
    ju res; memset(res.v, 0, sizeof(res.v));
    for (int i = 1; i <= 100; ++i)res.v[i][i] = 1;
    for (; b; b >>= 1, a = a * a)
        if (b & 1)res = res * a;
    return res;
}
void YYCH() 
{
    f[0][0] = 1;
    for (int i = 1; i <= 10; ++i)
        for (int j = 1; j <= i; ++j)
            for (int k = 0; k <= min(j, 4); ++k)
                f[i][j] += f[i - j][j - k] * C4[k];
    int cnt = 0;
    for (int i = 10; i >= 1; --i)for (int j = 10; j >= 1; --j)A.v[1][++cnt] = f[i][j];
    for (int i = 1; i <= 10; ++i)
        for (int j = 0; j <= min(i - 1, 4); ++j)
            base.v[i * 10 - (i - j) + 1][10 - i + 1] = C4[j];
    for (int i = 11; i <= 100; ++i)base.v[i - 10][i] = 1;
}
int main() 
{
    YYCH();
    while (scanf("%d%d", &n, &k) && n && k) 
    {
        ans = 0;
        if (n <= 10) 
        {
            for (int i = 1; i <= k; ++i)ans += f[n][i];
        } 
        else 
        {
            B = A * ksm(base, n - 10);
            for (int i = 1; i <= k; ++i)(ans += B.v[1][10 - i + 1]) %= mod;
        }
        printf("%d
", ans);
    }
    return 0;
}