解析

应该是比较入门的容斥了
统计方案用总方案数-某列超过1半的方案数
dp设计的一个trick是只统计差值

代码

#include<bits/stdc++.h>
using namespace std;
#define ll long long
const int N=2050;
const double eps=1e-6;
const int mod=998244353;
inline ll read(){ll x=0,f=1;char c=getchar();while(!isdigit(c)){if(c=='-')f=-1;c=getchar();}while(isdigit(c)){x=(x<<1)+(x<<3)+c-'0';c=getchar();}return f*x;
}int n,m;
ll f[2][2050][205],g[105][2050];
ll a[105][2050],sum[105],o;
int main(){
#ifndef ONLINE_JUDGEfreopen("a.in","r",stdin);freopen("a.out","w",stdout);
#endifn=read();m=read();o=n+2;for(int i=1;i<=n;i++){for(int j=1;j<=m;j++){a[i][j]=read();sum[i]+=a[i][j];sum[i]%=mod;}}ll ans=0;int now;g[0][0]=1;for(int i=1;i<=n;i++){g[i][0]=1;for(int j=1;j<=i;j++) g[i][j]=(g[i-1][j]+g[i-1][j-1]*sum[i])%mod;}for(int j=1;j<=m;j++){now=0;f[0][j][0+o]=1;for(int i=1;i<=n;i++){now^=1;for(int k=-i;k<=i;k++){f[now][j][k+o]=(f[now^1][j][k-1+o]*a[i][j]+f[now^1][j][k+o]+f[now^1][j][k+1+o]*(sum[i]-a[i][j]))%mod;	}}}for(int i=1;i<=n;i++) (ans+=g[n][i])%=mod;for(int j=1;j<=m;j++){for(int k=1;k<=n;k++) ans=(ans-f[n&1][j][k+o]+mod)%mod;}printf("%lld\n",ans);return 0;
}
/*4a1 22 31 33 4
*/