#1453. 【NOIP模拟（第二套）】A. 数正方体

https://sjzezoj.com/submission/53772

S2OJ/1453/1453.cpp

@@ -0,0 +1,58 @@
+#include <iostream>
+
+using std::cin;
+using std::cout;
+const char endl = '\n';
+
+const int N = 1e5 + 5;
+const int mod = 998244353;
+
+int t, n, m, fac[N], inv[N], fac_inv[N];
+
+inline int binpow(int a, int b) {
+    a %= mod;
+
+    int res = 1;
+
+    while (b) {
+        if (b & 1) res = static_cast<long long>(res) * a % mod;
+        a = static_cast<long long>(a) * a % mod;
+        b >>= 1;
+    }
+
+    return res;
+}
+
+inline int C(int n, int m) {
+    return static_cast<long long>(fac[n]) * fac_inv[m] % mod * fac_inv[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;
+    }
+
+    fac_inv[0] = fac_inv[1] = 1;
+    for (int i = 2; i < N; i++) {
+        fac_inv[i] = static_cast<long long>(fac_inv[i - 1]) * inv[i] % mod;
+    }
+
+    cin >> t;
+
+    while (t--) {
+        cin >> n >> m;
+
+        cout << static_cast<long long>(binpow(2, n - m)) * C(n, m) % mod << endl;
+    }
+
+    return 0;
+}