#145. 【2020.11.30 NOIP模拟赛 T2】矩阵游戏

https://sjzezoj.com/submission/70880

S2OJ/145/145.cpp

@@ -0,0 +1,68 @@
+#include <iostream>
+
+using std::cin;
+using std::cout;
+const char endl = '\n';
+
+const int N = 2e5 + 5;
+const int mod = 1e9 + 7;
+
+int inv[N], fac[N], inv_fac[N];
+
+int C(int n, int m) {
+    return static_cast<long long>(fac[n]) * inv_fac[m] % mod * inv_fac[n - m] % mod;
+}
+
+int main() {
+    std::ios::sync_with_stdio(false);
+    cin.tie(nullptr);
+
+    fac[0] = 1;
+    for (int i = 1; i < N; i++) {
+        fac[i] = static_cast<long long>(fac[i - 1]) * i % mod;
+    }
+
+    inv[0] = inv[1] = 1;
+    for (int i = 2; i < N; i++) {
+        inv[i] = static_cast<long long>(mod - (mod / i)) * inv[mod % i] % mod;
+    }
+
+    inv_fac[0] = inv_fac[1] = 1;
+    for (int i = 2; i < N; i++) {
+        inv_fac[i] = static_cast<long long>(inv_fac[i - 1]) * inv[i] % mod;
+    }
+
+    // ======
+
+    int t;
+
+    cin >> t;
+
+    while (t--) {
+        int n, m, k, res1 = 0, res2 = 0, res3 = 0, res4 = 0;
+
+        cin >> n >> m >> k;
+
+        for (int i = k & 1; i <= std::min(n, k); i += 2) {
+            res1 = (res1 + C(n, i)) % mod;
+        }
+
+        for (int i = k & 1; i <= std::min(m, k); i += 2) {
+            res2 = (res2 + C(m, i)) % mod;
+        }
+
+        if (n % 2 == 0 && m % 2 == 0) {
+            for (int i = std::max(n - k, 0); i <= std::min(n, k); i += 2) {
+                res3 = (res3 + C(n, i)) % mod;
+            }
+
+            for (int i = std::max(m - k, 0); i <= std::min(m, k); i += 2) {
+                res4 = (res4 + C(m, i)) % mod;
+            }
+        }
+
+        cout << ((static_cast<long long>(res1) * res2 % mod - static_cast<long long>(res3) * res4 % mod * inv[2] % mod) + mod) % mod << endl;
+    }
+
+    return 0;
+}