D. Bicolorings

原题链接

题解

n+1会发生什么？
答案可不可以用结尾的状态来定义？
可不可以用结尾的状态来转移？

code

#include<bits/stdc++.h>
using namespace std;
#define ll long long
const ll mod=998244353;

ll dp[1005][2005][4]={0};

inline void read(ll &x) {
	x = 0;
	ll flag = 1;
	char c = getchar();
    while(c < '0' || c > '9'){
        if(c == '-') flag = -1;
        c = getchar();
    }
	while(c >= '0' && c <= '9') {
		x = (x << 3) + (x << 1) + (c ^ 48);
		c = getchar();
	}
	x *= flag;
}

inline void write(ll x)
{
    if(x < 0){
    	putchar('-');
		x = -x;
	}
    if(x > 9)
		write(x / 10);
    putchar(x % 10 + '0');
}

int main()
{
    ll n, k;
    read(n); read(k);
    dp[1][1][0] = 1;
    dp[1][1][3] = 1;
    dp[1][2][1] = 1;
    dp[1][2][2] = 1;
    for(ll i = 2; i <= n; i++)
    {
        for(ll j = 1; j <= k; j++)
        {
            dp[i][j][0] = (dp[i-1][j][0] + dp[i-1][j][1] + dp[i-1][j][2] + dp[i-1][j-1][3]) % mod;
            dp[i][j][1] = (dp[i-1][j-1][0] + dp[i-1][j][1] + dp[i-1][j-2][2] + dp[i-1][j-1][3]) % mod;
            dp[i][j][2] = (dp[i-1][j-1][0] + dp[i-1][j-2][1] + dp[i-1][j][2] + dp[i-1][j-1][3]) % mod;
            dp[i][j][3] = (dp[i-1][j-1][0] + dp[i-1][j][1] + dp[i-1][j][2] + dp[i-1][j][3]) % mod;
        }
    }

    ll ans = 0;
    for(ll i = 0; i < 4; i++) ans = (ans + dp[n][k][i]) % mod;

    write(ans);
    return 0;
}