將條件轉化為i,j⩾0,i+j⩾k|Av+i,u+j+=(i+ji)$i,j⩾0,i+j⩾k|A_{v+i,u+j}+=(\genfrac{}{}{0ex}{}{i+j}{i})$， 考慮組合意義，從(v,u)$(v,u)$開始每次往右或往下走，走大於等於k$k$布的方案數，直接DP。

#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
typedef long long LL;
const int tt=998244353,maxn=306;
int n,m,q,f[maxn][maxn][maxn*2];
LL x0;
int
 
 rnd(){x0=(x0*100000005+20150823)%tt;return x0/100;}
int main(){
    freopen("B.in","r",stdin);freopen("B.out","w",stdout);
    scanf("%d%d%d%lld",&n,&m,&q,&x0);
    for(int i=1,v,u,k;i<=q;i++)v=rnd()%n+1,u=rnd()%m+1,k=rnd()%(n+m-v-u+1),f[v][u][k]++;
    for(int i=1;i<=n;i++)
     for(int j=1;j<=m;j++){
        (f[i+1
 
][j][0]+=f[i][j][0])%=tt;(f[i][j+1][0]+=f[i][j][0])%=tt;
        for(int k=1;k<=n+m;k++)(f[i+1][j][k-1]+=f[i][j][k])%=tt,(f[i][j+1][k-1]+=f[i][j][k])%=tt;
     }
    for(int i=1;i<=n;i++){
        for(int j=1;j<=m;j++)printf("%d ",f[i][j][0]);
        printf("\n");
    }
    return 0;
}