技術標籤：組合數學

https://abc042.contest.atcoder.jp/tasks/arc058_b?lang=en
由於左下角有禁區，所以左上角到右下角的路徑分為兩段，一段從左上角到禁區的上面那一行，一段從禁區的上面那一行到右下角，然後求和即可，這題是用來儲存組合數取模模板的。

ll quickpow(ll x, int k)
{
    ll res = 1;
    while(k){
        if(k & 1) res = (res * x) % mod;
        k >>= 1, x = (x * x) % mod;
    }
    return
 
 res;
}
ll fac[200005] = {1};
ll C(int n, int m){return (fac[n] * quickpow((fac[m] * fac[n - m]) % mod, mod - 2)) % mod;}
const int maxn = 1e5 + 5;

int main()
{
    for (int i = 1; i < 200005; ++i) fac[i] = (fac[i - 1] * i) % mod;
    int h, w, a, b;
    h = in(); w = in(); a = in(); b = in();
    ll ans =
 
 0;
    for (int i = b + 1; i <= w; ++i)
        ans = (ans + C(i - 1 + h - a - 1, min(i, h - a) - 1) * C(w - i + a - 1, min(w - i, a - 1))) % mod;
    outn(ans);
    return 0;
}