loj#2128. 「HAOI2015」数字串拆分 矩阵乘法

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目录

• [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;
}