【LOJ】#2128. 「HAOI2015」数字串拆分

题解

题中给的函数可以用矩阵快速幂递推

我们记一个数组dp[i]（这个数组每个元素是一个矩阵）表示从1到i所有的数字经过拆分矩阵递推的加和

转移方法是

\(dp[i] = \sum_{j = 0}^{i - 1} dp[j] * tr[j + 1][i]\)

\(tr[j][i]\)表示矩阵的\([j,i]\)组成的数字次幂是什么样的矩阵

代码

#include <bits/stdc++.h>
#define fi first
#define se second
#define pii pair<int,int>
#define mp make_pair
#define pb push_back
#define enter putchar('\n')
#define space putchar(' ')
#define MAXN 100005
#define mo 994711
//#define ivorysi
using namespace std;
typedef long long int64;
typedef long double db;
typedef unsigned int u32;
template<class T>
void read(T &res) {
    res = 0;char c = getchar();T f = 1;
    while(c < '0' || c > '9') {
	if(c == '-') f = -1;
	c = getchar();
    }
    while(c >= '0' && c <= '9') {
	res = res * 10 + c - '0';
	c = getchar();
    }
    res *= f;
}
template<class T>
void out(T x) {
    if(x < 0) {putchar('-');x = -x;}
    if(x >= 10) out(x / 10);
    putchar('0' + x % 10);
}
const int MOD = 998244353;
int M,N;
char s[505];
int inc(int a,int b) {
    return a + b >= MOD ? a + b - MOD : a + b;
}
int mul(int a,int b) {
    return 1LL * a * b % MOD;
}
struct Matrix {
    int f[6][6];
    Matrix() {memset(f,0,sizeof(f));}
    friend Matrix operator * (const Matrix &a,const Matrix &b) {
	Matrix c;
	for(int i = 0 ; i < M ; ++i) {
	    for(int j = 0 ; j < M ; ++j) {
		for(int k = 0 ; k < M ; ++k) {
		    c.f[i][j] = inc(c.f[i][j],mul(a.f[i][k],b.f[k][j]));
		}
	    }
	}
	return c;
    }
    friend Matrix operator + (const Matrix &a,const Matrix &b) {
	Matrix c;
	for(int i = 0 ; i < M ; ++i) {
	    for(int j = 0 ; j < M ; ++j) {
		c.f[i][j] = inc(a.f[i][j],b.f[i][j]);
	    }
	}
	return c;
    }
}A[15],dp[505],tr[505][505];
int g[15];
Matrix fpow(Matrix x,int c) {
    Matrix res = A[0],t = x;
    while(c) {
	if(c & 1) res = res * t;
	t = t * t;
	c >>= 1;
    }
    return res;
}
void Solve() {
    scanf("%s",s + 1);
    read(M);N = strlen(s + 1);
    for(int i = 0 ; i < M ; ++i) {
	A[0].f[i][i] = 1;
    }
    dp[0] = A[0];
    g[0] = 1;
    for(int i = 1 ; i < M ; ++i) {
	for(int j = 1 ; j <= i ; ++j) {
	    g[i] = inc(g[i - j],g[i]);
	}
    }
    for(int i = 0 ; i < M ; ++i) {
	if(i != M - 1) A[1].f[i][i + 1] = 1;
	A[1].f[M - 1][i] = 1;
    }
    for(int i = 2 ; i <= 9 ; ++i) {
	A[i] = A[i - 1] * A[1];
    }
    for(int i = 1 ; i <= N ; ++i) {
	Matrix r = A[0],t;
	for(int j = i ; j <= N; ++j) {
	    r = r * r;t = r;
	    r = r * r;r = r * r;r = r * t;
	    r = r * A[s[j] - '0'];
	    tr[i][j] = r;
	}
    }
    for(int i = 1 ; i <= N ; ++i) {
	for(int j = 0 ; j < i ; ++j) {
	    dp[i] = dp[i] + dp[j] * tr[j + 1][i];
	}
    }
    int ans = 0;
    for(int i = 0 ; i < M ; ++i) ans = inc(ans,mul(g[i],dp[N].f[0][i]));
    out(ans);enter;
}
int main() {
#ifdef ivorysi
    freopen("f1.in","r",stdin);
#endif
    Solve();
    return 0;
}