2018 Multi-University Training Contest 2
n*m的方格,黑白染色,至少x行,y列全是黑色的方案数。
#include<bits/stdc++.h>
using namespace std;
#define lc o<<1
#define rc o<<1|1
#define fi first
#define se second
#define pb push_back
#define ALL(X) (X).begin(), (X).end()
#define bcnt(X) __builtin_popcountll(X)
#define CLR(A, X) memset(A, X, sizeof(A))
#pragma comment(linker, "/STACK:1024000000,1024000000")
#define DEBUG printf("Passing [%s] in Line %d\n",__FUNCTION__,__LINE__)
using DB = double;
using LL = long long;
using PII = pair<int, int>;
const int N = 3e3+5;
const int MOD = 998244353;
//const LL INF = 1e18;
int pre[N][N], gg[N][N], inv[N], fac[N], p[N*N];
void init() {
inv[1] = 1;
for(int i = 2; i < N; ++i) inv[i] = MOD-(1LL*MOD/i*inv[MOD%i]%MOD);
fac[0] = 1, inv[0] = 1;
for(int i = 1; i < N; i++){
fac[i] = (1LL*fac[i-1]*i)%MOD;
inv[i] = 1LL*inv[i-1]*inv[i]%MOD;
}
p[0] = 1;
for(int i = 1; i < N*N; i++) {
p[i] = p[i-1]+p[i-1];
if(p[i] >= MOD) p[i] -= MOD;
}
for(int u = 0; u < N; u++) {
for(int v = 0; v < N; v++) {
gg[u][v] = 1LL*p[u*v]*inv[u]%MOD*inv[v]%MOD;
int t = 1LL*inv[u-v]*inv[v]%MOD;
if(v & 1) t = MOD-t;
pre[u][v] = (v?pre[u][v-1]:0)+t;
if(pre[u][v] >= MOD) pre[u][v] -= MOD;
}
}
}
int main() {
init();
int n, m, x, y;
while(~scanf("%d%d%d%d", &n, &m, &x, &y)) {
LL ans = 0;
for(int u = 0; u <= n-x; u++) {
for(int v = 0; v <= m-y; v++) {
ans += 1LL*gg[u][v]*pre[n-u][n-x-u]%MOD*pre[m-v][m-y-v]%MOD;
if(ans >= MOD) ans -= MOD;
}
}
ans = ans*fac[n]%MOD*fac[m]%MOD;
printf("%lld\n", ans);
}
return 0;
}