30. 正则表达式匹配

https://www.acwing.com/problem/content/submission/code_detail/15205980/

AcWing/30/30.js

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+/**
+ * @param {string} s
+ * @param {string} p
+ * @return {boolean}
+ */
+ const isMatch = (s, p) => new RegExp(`^${p}$`).test(s);

S2OJ/1448/1448.cpp

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+#include <iostream>
+#include <set>
+
+using std::cin;
+using std::cout;
+const char endl = '\n';
+
+int n, m;
+std::set<int> set;
+
+int main() {
+    cin >> n >> m;
+
+    set.insert(0);
+    set.insert(n);
+
+    while (m--) {
+        int k;
+
+        cin >> k;
+
+        set.insert(k);
+
+        int prev = *--set.lower_bound(k),
+            next = *set.upper_bound(k);
+
+        cout << (k - prev) * (next - k) << endl;
+    }
+
+    return 0;
+}

S2OJ/1461/1461.cpp

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+#include <iostream>
+#include <algorithm>
+
+using std::cin;
+using std::cout;
+const char endl = '\n';
+
+const int N = 45,
+          D = 1e5 + 5;
+const int mod = 1e9 + 7;
+
+int n, d, r[N], f[N][N][N * N], ans;
+int fac[D], inv[D];
+
+int binpow(int a, int b) {
+    int res = 1;
+    a %= mod;
+    while (b) {
+        if (b & 1) res = 1ll * res * a % mod;
+        a = 1ll * a * a % mod;
+        b >>= 1;
+    }
+    return res;
+}
+
+inline int C(int n, int m) {
+    return 1ll * fac[n] * inv[m] % mod * inv[n - m] % mod;
+}
+
+int main() {
+    std::ios::sync_with_stdio(false);
+    cin.tie(nullptr);
+
+    // factorial
+    fac[0] = 1;
+    for (int i = 1; i <= 100000; i++) {
+        fac[i] = 1ll * fac[i - 1] * i % mod;
+    }
+
+    // inverse
+    inv[100000] = binpow(fac[100000], mod - 2);
+
+    for (int i = 99999; i >= 0; i--)
+        inv[i] = 1ll * inv[i + 1] * (i + 1) % mod;
+
+    // main
+    cin >> n >> d;
+
+    for (int i = 1; i <= n; i++) {
+        cin >> r[i];
+    }
+
+    std::sort(r + 1, r + 1 + n);
+
+    f[0][0][0] = 1;
+
+    for (int i = 0; i <= n; i++) {
+        for (int j = 0; j <= i; j++) {
+            for (int k = 0; k <= 40 * 40; k++) {
+                int t = r[i + 1];
+
+                if (j >= 1) {
+                    f[i + 1][j][k + t] = (f[i + 1][j][k + t] + 1ll * f[i][j][k] * 2 * j) % mod;
+                }
+
+                if (j >= 2) {
+                    f[i + 1][j - 1][k + t * 2 - 1] = (f[i + 1][j - 1][k + t * 2 - 1] + 1ll * f[i][j][k] * j * (j - 1)) % mod;
+                }
+
+                f[i + 1][j + 1][k + 1] = (f[i + 1][j + 1][k + 1] + f[i][j][k]) % mod;
+            }
+        }
+    }
+
+    for (int i = 0; i <= std::min(d, 40 * 40); i++) {
+        ans = (ans + 1ll * C(d - i + n, n) * f[n][1][i] % mod) % mod;
+    }
+
+    cout << ans << endl;
+
+    return 0;
+}