loj#2128. 「HAOI2015」数字串拆分 矩阵乘法

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[TOC]

#题目链接
loj#2128. 「HAOI2015」数字串拆分
#题解
$f(s)$对于$f(i) = \sum_{j = i - m}^{i - 1}f(j)$
这个可以用转移矩阵通过矩阵乘法处理出来
预处理出$A[i][j]$表示数S为$j * 10 ^ i$的转移矩阵
对于g的转移
\(g(i) = \sum_{j = 0}^{i - 1}g(j) * D(j + 1,i)\)
D[i][j]表示第i位到底j位构成的数的f,(转移矩阵
对于g的转移也是需要矩阵的
g(0) = 1也就是{0,0,0,.....,0,1}
#代码

#include<map> 
#include<vector> 
#include<cstdio> 
#include<cstring> 
#include<algorithm> 
#define gc getchar()
#define pc putchar 
#define LL long long
inline int read() { 
	int x = 0,f = 1; 
	char c = gc; 
	while(c < '0' || c > '9')c = gc; 
	while(c <= '9' && c >= '0') x = x * 10 + c - '0',c = gc; 
	return x * f; 
} 
void print(int x) { 
	if(x < 0) { 
		pc('-'); 
		x = -x; 
	} 
	if(x >= 10) print(x / 10); 
	pc(x % 10 + '0'); 
} 
const int maxn = 507; 
int n,m; 
const int mod = 998244353; 
struct Ma { 
	int a[6][6]; 
	Ma() { memset(a,0,sizeof a); } 
	Ma operator * (const Ma & t)const { 
		Ma ret; 
		memset(ret.a,0,sizeof ret.a); 
		for(int i = 1;i <= m;++ i) 
			for(int j = 1;j <= m;++ j) 
				for(int k = 1;k <= m;++ k) 
					(ret.a[i][j] += 1ll * a[i][k] * t.a[k][j] % mod) %= mod; 
		return ret; 
	} 
	Ma operator + (const Ma &k)const { 
		Ma ret; 
		memset(ret.a,0,sizeof ret.a); 
		for(int i = 1;i <= m;++ i) 
			for(int j = 1;j <= m;++ j) 
				ret.a[i][j] = (a[i][j] + k.a[i][j]) % mod; 
		return ret; 
	} 
} A[maxn][10],f[maxn]; 
char s[maxn];  
int main() { 
	scanf("%s",s + 1); 
	n = strlen(s + 1); 
	m = read(); 
	for(int i = 1;i <= m;++ i) 
		A[0][0].a[i][i] = 1 , 
		A[0][1].a[i][m] = 1; 
	for(int i = 1;i < m;++ i) 
		A[0][1].a[i + 1][i] = 1; 
	for(int i = 2;i <= 9;++ i) 
		A[0][i] = A[0][i - 1] * A[0][1]; 
	 for(int i = 1;i <= n;++ i) { //十进制快速幂 
	 	A[i][0] = A[0][0]; 
	 	A[i][1] = A[i - 1][9] * A[i - 1][1]; 
	 	for(int j = 2;j <= 9;++ j) 
	 		A[i][j] = A[i][j - 1] * A[i][1]; 
	 } 
	f[0].a[1][m] = 1; 
	Ma now; 
	for(int i = 1;i <= n;++ i) { 
		now = A[0] [s[i] - '0']; 
		for(int j = i - 1;j >= 0;-- j) { 
			f[i] = f[i] + (f[j] * now);  
			if(j) now = A[i - j][s[j] - '0'] * now; 
		} 
	}  
	print(f[n].a[1][m]); 
	pc('\n'); 
	return 0; 
}