rt，我做法就是枚举左上角的拐点然后算贡献。

$s_{i,j}$ 表示 $(i,j)$ 往右能扩展多少点。

$f_{i,j}$ 表示 $(i,j)$ 往下能扩展多少点。

$g_{i,j}$ 和 $k_{i,j}$ 分别是 $C$ 和 $F$ 的后缀和。

#include <bits/stdc++.h>

using namespace std;

#define int long long

const int N = 1e3 + 10, P = 998244353;

int n, m, C, F, a[N][N], s[N][N], f[N][N], g[N][N], k[N][N];

signed main()
{
    cin.tie(0)->ios::sync_with_stdio(false);
    int _, id; cin >> _ >> id;
    while (_--) {
        cin >> n >> m >> C >> F;
        for (int i = 1; i <= n; i++) for (int j = 1; j <= m; j++) {
            char c; cin >> c; a[i][j] = (c == '0');
        }
        int ans = 0, res = 0;
        for (int i = 1; i <= n + 2; i++) for (int j = 1; j <= m + 2; j++) s[i][j] = g[i][j] = f[i][j] = k[i][j] = 0;
        for (int j = m; j >= 1; j--) for (int i = n; i >= 1; i--) {
            s[i][j] = (a[i][j] * s[i][j + 1] + a[i][j]) % P;
            f[i][j] = (a[i][j] * f[i + 1][j] + a[i][j]) % P;
            g[i][j] = (a[i][j] * g[i + 1][j] + a[i][j] * s[i][j + 1]) % P;
            k[i][j] = (a[i][j] * k[i + 1][j] + a[i][j] * f[i + 1][j] * s[i][j + 1] % P) % P;
            if (a[i][j]) {
                (ans += s[i][j + 1] * g[i + 2][j]) %= P;
                (res += s[i][j + 1] * k[i + 2][j]) %= P;
            }
        }
        cout << ans * C % P << " " << res * F % P << "\n";
    }
    return 0;
}