【HDU #5015】233 Matrix

HDU #5015。

Description

给一个矩阵 \(a\)。对于第一行，有 \(a_{0, 1} = 233\)，\(a_{0, 2} = 2333\)，\(a_{0,3} = 23333\)，...

对于 \(\forall i, j \neq 0\)，有

\[a_{i,j} = a_{i-1, j} + a_{i, j - 1} \]

给出 \(a_{1, 0}, a_{2, 0}, ..., a_{n, 0}\)，请求出 \(a_{n, m}\)。

数据范围：\(1 \leq n \leq 10\)，\(1 \leq m \leq 10^9\)。
时空限制：\(1000 \ \mathrm{ms} / 64 \ \mathrm{MiB}\)。

Solution

不妨设 \(a_{0,0} = 23\)。

对 \(a_{i, j} = a_{i - 1, j} + a_{i, j - 1}\) 进行转换：

\[a_{i, j} = \sum\limits_{k = 1}^{i} a_{k, j - 1} + a_{0, j} \]

特别地：

\[a_{0, j} = 10 \times a_{0, j - 1} + 3 \]

则有：

\[a_{i, j} = \sum\limits_{k = 1}^{i} a_{k, j - 1} + 10 \times a_{0, j - 1} + 3 \]

观察上式，注意到第 \(j\) 列每个位置上的值都可以由第 \(j - 1\) 列的若干个项递推而来。
又注意到 \(n \leq 10\)，\(m \leq 10^9\)，于是考虑矩阵乘法加速递推。

设 \(F(j) = \begin{pmatrix} a_{0, j} & a_{1, j} & \cdots & a_{n, j} & 3 \end{pmatrix}\)，则有：

\[F(j) = F(j - 1) \times \begin{pmatrix} 10 & 10 & 10 & \cdots & 10 & 0 \\ 0 & 1 & 1 & \cdots & 1 & 0 \\ 0 & 0 & 1 & \cdots & 1 & 0 \\ \vdots & \vdots & \vdots & \ddots & \vdots & \vdots \\ 0 & 0 & 0 & \cdots & 1 & 0 \\ 1 & 1 & 1 & \cdots & 1 & 1 \end{pmatrix} \]

设转移矩阵为 \(G\)，则 \(F(m) = F(0) \times G^m\)。

时间复杂度 \(\mathcal{O}(n^3 \log m)\)。

#include <cstdio>
#include <cstring>

using namespace std;

namespace IO {
    static char buf[1 << 20], *fs, *ft;
    inline char gc() {
        if (fs == ft) {
			ft = (fs = buf) + fread(buf, 1, 1 << 20, stdin);
			if (fs == ft) return EOF;
        }
        return *fs ++;
    }
    #define gc() getchar()
	inline int read() {
		int x = 0, f = 1; char s = gc();
		while (s < '0' || s > '9') {if (s == '-') f = -f; s = gc();}
		while (s >= '0' && s <= '9') {x = x * 10 + s - '0'; s = gc();}
		return x * f;
	}
} using IO :: read;

const int N = 110;

const int mod = 1e7 + 7;

int n, m;

int f[N];

int G[N][N];

void mul(int d[N][N], int a[N][N], int b[N][N]) {
	static int c[N][N]; memset(c, 0, sizeof(c));
	for (int i = 0; i <= n + 1; i ++)
		for (int j = 0; j <= n + 1; j ++)
			for (int k = 0; k <= n + 1; k ++)
				c[i][j] = (c[i][j] + 1ll * a[i][k] * b[k][j]) % mod;
	memcpy(d, c, sizeof(c));
}

void mulstar(int d[N], int a[N], int b[N][N]) {
	static int c[N]; memset(c, 0, sizeof(c));
	for (int j = 0; j <= n + 1; j ++)
		for (int k = 0; k <= n + 1; k ++)
			c[j] = (c[j] + 1ll * a[k] * b[k][j]) % mod;
	memcpy(d, c, sizeof(c));
}

void work() {
	f[0] = 23;
	for (int i = 1; i <= n; i ++) f[i] = read();
	f[n + 1] = 3;

	for (int j = 0; j <= n; j ++) {
		G[0][j] = 10;

		for (int i = 1; i <= n; i ++)
			if (i <= j) G[i][j] = 1;
			else G[i][j] = 0;
	} 

	for (int i = 0; i <= n; i ++)
		G[i][n + 1] = 0;

	for (int j = 0; j <= n + 1; j ++)
		G[n + 1][j] = 1;

	for (int b = m; b; b >>= 1) {
		if (b & 1) mulstar(f, f, G);
		mul(G, G, G);
	}

	printf("%d\n", f[n]);
}

int main() {
	while (scanf("%d%d", &n, &m) != EOF)    work();

	return 0; 
}