【LOJ】#3098. 「SNOI2019」紙牌

LOJ#3098. 「SNOI2019」紙牌

顯然選三個以上的連續牌能夠把他們拆分紅三個三張相等的

因而能夠壓\((j,k)\)爲有\(j\)個連續兩個的，有\(k\)個連續一個的

若是當前有\(i\)張牌，且\(i >= j + k\)

那麼能夠\((j,k)\rightarrow (k,(i - j - k) \% 3)\)

能夠用矩陣乘法優化，每遇到一個有下限的牌面的就再特殊造一個矩陣轉移

#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();
}