LOJ#3098. 「SNOI2019」纸牌

显然选三个以上的连续牌可以把他们拆分成三个三张相等的

于是可以压\((j,k)\)为有\(j\)个连续两个的，有\(k\)个连续一个的

如果当前有\(i\)张牌，且\(i >= j + k\)

那么可以\((j,k)\rightarrow (k,(i - j - k) \% 3)\)

可以用矩阵乘法优化，每遇到一个有下限的牌面的就再特殊造一个矩阵转移

#include <bits/stdc++.h>

#define fi first

#define se second

#define pii pair<int,int>

#define mp make_pair

#define pb push_back

#define space putchar(' ')

#define enter putchar('\n')

#define eps 1e-10

#define ba 47

#define MAXN 5005

//#define ivorysi

using namespace std;

typedef long long int64;

typedef unsigned int u32;

typedef double db;

template<class T>

void read(T &res) {

    res = 0;T f = 1;char c = getchar();

    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) {x = -x;putchar('-');}

    if(x >= 10) {

	out(x / 10);

    }

    putchar('0' + x % 10);

}

const int MOD = 998244353;

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;

}

void update(int &x,int y) {

    x = inc(x,y);

}

int getid(int x,int y) {

    return x * 3 + y;

}

struct Matrix {

    int f[9][9];

    Matrix() {memset(f,0,sizeof(f));}

    friend Matrix operator * (const Matrix &a,const Matrix &b) {

	Matrix c;

	for(int k = 0 ; k < 9 ; ++k) {

	    for(int i = 0 ; i < 9 ; ++i) {

		for(int j = 0 ; j < 9 ; ++j) {

		    update(c.f[i][j],mul(a.f[i][k],b.f[k][j]));

		}

	    }

	}

	return c;

    }

    friend Matrix fpow(Matrix a,int64 c) {

	Matrix res,t = a;

	for(int i = 0 ; i < 9 ; ++i) res.f[i][i] = 1;

	while(c) {

	    if(c & 1) res = res * t;

	    t = t * t;

	    c >>= 1;

	}

	return res;

    }

}a,ans,b;

int64 n;

int C,X;

void Solve() {

    read(n);read(C);

    for(int i = 0 ; i <= C ; ++i) {

	for(int j = 0 ; j < 3 ; ++j) {

	    for(int k = 0 ; k < 3 ; ++k) {

		if(i < j + k) continue;

		update(a.f[getid(j,k)][getid(k,(i - j - k) % 3)],1);

	    }

	}

    }

    for(int i = 0 ; i < 9 ; ++i) ans.f[i][i] = 1;

    read(X);

    int64 k;int t;

    int64 p = 0;

    for(int i = 1 ; i <= X ; ++i) {

	read(k);read(t);

	ans = ans * fpow(a,k - 1 - p);

	memset(b.f,0,sizeof(b.f));

	for(int h = t ; h <= C ; ++h) {

	    for(int j = 0 ; j < 3 ; ++j) {

		for(int k = 0 ; k < 3 ; ++k) {

		    if(h < j + k) continue;

		    update(b.f[getid(j,k)][getid(k,(h - j - k) % 3)],1);

		}

	    }

	}

	ans = ans * b;

	p = k;

    }

    if(p < n) ans = ans * fpow(a,n - p);

    out(ans.f[0][0]);enter;

}

int main(){

#ifdef ivorysi

    freopen("f1.in","r",stdin);

#endif

    Solve();

}