【JZOJ6433】【luoguP5664】【CSP-S2019】Emiya 家今天的飯

description

---

analysis

• 首先可以知道不符合要求的食材僅有一個，於是可以容斥拿總方案數減去選不合法食材的不合法方案數

• 枚舉選取哪一個不合法食材，設$f[i][j]$表示到第$i$種烹飪方法、操作權值爲$j$的方案數

• 給每一個操作賦權值，選當前行合法食材列爲$0$，不選當前行爲$1$，選當前行不合法食材列爲$2$

• 轉移是比較容易的，可知選當前列爲不合法食材的方案數就是$\sum_{i=n+1}^{2n}f[n][i]$

---

code

#include<stdio.h>
#include<string.h>
#include<algorithm>
#define MAXN 105
#define MAXM 2005
#define ha 998244353
#define ll long long
#define reg register ll
#define fo(i,a,b) for (reg i=a;i<=b;++i)
#define fd(i,a,b) for (reg i=a;i>=b;--i)

using namespace std;

ll f[MAXN][MAXN*2];
ll a[MAXN][MAXM],sum[MAXM];
ll n,m,ans;

inline ll read()
{
	ll x=0,f=1;char ch=getchar();
	while (ch<'0' || '9'<ch){if (ch=='-')f=-1;ch=getchar();}
	while ('0'<=ch && ch<='9')x=x*10+ch-'0',ch=getchar();
	return x*f;
}
int main()
{
	n=read(),m=read(),ans=1;
	fo(i,1,n)fo(j,1,m)(sum[i]+=(a[i][j]=read()))%=ha;
	fo(i,1,n)(ans*=sum[i]+1)%=ha;--ans;
	fo(food,1,m)
	{
		memset(f,0,sizeof(f)),f[0][0]=1;
		fo(i,1,n)fo(j,0,n*2)
		{
			(f[i][j]+=f[i-1][j]*((sum[i]-a[i][food])%ha))%=ha;
			if (j+1<=n*2)(f[i][j+1]+=f[i-1][j])%=ha;
			if (j+2<=n*2)(f[i][j+2]+=f[i-1][j]*a[i][food]%ha)%=ha;
		}
		fo(i,n+1,n*2)ans=((ans-f[n][i])%ha+ha)%ha;
	}
	printf("%lld\n",ans);
	return 0;
}