[ABC127E] Cell Distance

• 原题链接
• 题意：给出 (n imes m) 的矩阵，都小于 (2e5) ,求从 (n imes m) 个矩阵中拿出 (k <= n imes m) 个点，在所有方案中，求每个方案中，每个点距离其他点的曼哈顿距离之和，然后求所有方案的各个距离之和的总和。
• 题解：枚举横着两个点的距离为 (1 <= lenx <= n) 然后，其贡献为 (lenx imes ( n - lenx) imes m^2),然后对答案的贡献值就是显然从所有中选出 (k-2) 个的方案数乘贡献，即 $$sum_{lenx = 1}^{n} (nm-2, k-2) imes lenx imes ( n - lenx) imes m^2$$
  所以同理，这只是算了一半的答案，另一半则是 $$sum_{leny = 1}^{n} (nm-2, k-2) imes leny imes ( m - leny ) imes n^2$$
• 代码：

#include <iostream>
#include <vector>
#include <algorithm>
using namespace std;

const int N = 200009;
typedef long long ll;
const ll mod = 1000000007;
ll fac[N];
ll q_mi(ll a, ll k, ll mod) {
    ll ret = 1;
    ll x = a;
    while (k) {
        if (k & 1) (ret *= x) %= mod;
        (x *= x) %= mod;
        k >>= 1;
    }
    return ret;
}
ll inv(ll a) { return (q_mi(a, mod - 2, mod) % mod + mod) % mod; }
ll C(ll n, ll m) {return fac[n] * (inv(fac[m])) % mod * inv(fac[n-m]) % mod;}
void init() {
    fac[0] = 1;
    for (int i = 1; i < N; i++) {
        (fac[i] = fac[i-1] * i) %= mod;
    }
}


void solve() {
    ll n, m, k;cin >> n >> m >> k;
    ll ans = 0;
    for (ll i = 1; i <= n; i++) {
        (ans += C(n*m-2, k-2) * (n - i) % mod * m % mod * m % mod  * i%mod) %= mod;
    }
    for (ll i = 1; i <= m; i++) {
         (ans += C(n*m-2, k-2) * (m - i) % mod * n % mod * n % mod  * i%mod) %= mod;
    }
    cout << ans << endl;
}
signed main() {
    ios::sync_with_stdio(0);
    init();
    ll t = 1;//cin >> t;
    while (t--)solve();
    return 0;
}