HDU - 5015 233 Matrix

Posted 2020-12-08 cjtcalc

[ exttt{Description} ]

有一个名为 233 的矩阵 (a)。

对于第一行，有 (a_{0, 1} = 233)，(a_{0, 2} = 2333)，(a_{0,3} = 23333) ...

对于 (forall i, j eq 0)，有 (a_{i,j} = a_{i-1, j} + a_{i, j - 1})。

给出 (a_{1, 0}, a_{2, 0}, ..., a_{n, 0})，请在 233 矩阵中求出 (a_{n, m})。

[ exttt{Solution} ]

不妨设 (a_{0,0} = 23)。

对 (a_{i, j} = a_{i - 1, j} + a_{i, j - 1}) 这个式子比较熟悉的巨佬们应该都知道：

[a_{i, j} = sumlimits_{k = 1}^{i} a_{k, j - 1} + a_{0, j} ]

特别地：

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

则有：

[a_{i, j} = sumlimits_{k = 1}^{i} a_{k, j - 1} + 10 imes 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) = egin{bmatrix} a_{0, j} & a_{1, j} & cdots & a_{n, j} & 3 end{bmatrix})，则有：

[F(j) = F(j - 1) imes egin{bmatrix} 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{bmatrix} ]

设转移矩阵为 (G)，则 (F(m) = F(0) imes G^m)。

时间复杂度 (mathcal{O(n^3 log m)})。

[ exttt{Code} ]

#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
", f[n]);
}

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

	return 0; 
} 

[ exttt{Thanks} exttt{for} exttt{reading} ]