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();
}