CF1051D Bicolorings dp

水题一道

$f[i][j][S]$表示$2 * i$的矩形，有$j$个联通块，某尾状态为$S$

然后转移就行了...

#include <vector>
#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
namespace remoon {
    #define re register
    #define de double
    #define le long double
    #define ri register int
    #define ll long long
    #define sh short
    #define pii pair<int, int>
    #define mp make_pair
    #define pb push_back
    #define tpr template <typename ra>
    #define rep(iu, st, ed) for(ri iu = st; iu <= ed; iu ++)
    #define drep(iu, ed, st) for(ri iu = ed; iu >= st; iu --)
    extern inline char gc() {
        static char RR[], *S = RR + , *T = RR + ;
        if(S == T) fread(RR, , , stdin), S = RR;
        return *S ++;
    }
    inline int read() {
        int p = , w = ; char c = gc();
        while(c > '' || c < '') { if(c == '-') w = -; c = gc(); }
        while(c >= '' && c <= '') p = p *  + c - '', c = gc();
        return p * w;
    }
    int wr[], rw;
    #define pc(iw) putchar(iw)
    tpr inline void write(ra o, char c = '\n') {
        if(!o) pc('');
        if(o < ) o = -o, pc('-');
        while(o) wr[++ rw] = o % , o /= ;
        while(rw) pc(wr[rw --] + '');
        pc(c);
    }
    tpr inline void cmin(ra &a, ra b) { if(a > b) a = b; }
    tpr inline void cmax(ra &a, ra b) { if(a < b) a = b; }
    tpr inline bool ckmin(ra &a, ra b) { return (a > b) ? a = b,  : ; }
    tpr inline bool ckmax(ra &a, ra b) { return (a < b) ? a = b,  : ; }
}
using namespace std;
using namespace remoon;

#define mod 998244353

inline void inc(ll &a, ll b) {
    a += b; if(a >= mod) a %= mod;
}

int n, k;
ll f[][][];

int main() {
    n = read(); k = read();
    f[][][] = ; f[][][] = ;
    f[][][] = ; f[][][] = ;
    rep(i, , n)
    rep(j, , i << ) {
        inc(f[i][j][], f[i - ][j][] + f[i - ][j][] + f[i - ][j][] + f[i - ][j - ][]);
        inc(f[i][j][], f[i - ][j - ][] + f[i - ][j][] + f[i - ][j - ][] + f[i - ][j - ][]);
        inc(f[i][j][], f[i - ][j - ][] + f[i - ][j - ][] + f[i - ][j][] + f[i - ][j - ][]);
        inc(f[i][j][], f[i - ][j - ][] + f[i - ][j][] + f[i - ][j][] + f[i - ][j][]);
    }
    ll ans = ;
    inc(ans, f[n][k][] + f[n][k][] + f[n][k][] + f[n][k][]);
    write(ans);
    return ;
}