#1486. 【2022.7.27】数据结构（set）

https://sjzezoj.com/submission/54111

S2OJ/1486/1486.cpp

@@ -0,0 +1,75 @@
+#include <iostream>
+
+using std::cin;
+using std::cout;
+const char endl = '\n';
+
+const int N = 2e5 + 5,
+          K = 55;
+const int mod = 1e9 + 7;
+
+int m, k, d, cnt;
+long long f[K][K], p[N][K], s[N][K], sum;
+
+long long binpow(long long a, long long b) {
+    a %= mod;
+
+    long long res = 1;
+
+    while (b) {
+        if (b & 1) res = res * a % mod;
+        a = a * a % mod;
+        b >>= 1;
+    }
+
+    return res;
+}
+
+int main() {
+    std::ios::sync_with_stdio(false);
+    cin.tie(nullptr);
+
+    cin >> m >> k;
+
+    f[0][0] = 1;
+    for (int i = 1; i <= k; i++) {
+        f[i][0] = 1;
+
+        for (int j = 1; j <= i; j++) {
+            f[i][j] = (f[i - 1][j - 1] + f[i - 1][j]) % mod;
+        }
+    }
+
+    while (m--) {
+        int op;
+
+        cin >> op;
+
+        if (op == 0) {
+            int x;
+
+            cin >> x;
+
+            p[++cnt][0] = 1;
+            s[cnt][0] = (s[cnt - 1][0] + p[cnt][0]) % mod;
+
+            for (int i = 1; i <= k; i++) {
+                p[cnt][i] = (p[cnt][i - 1] * (x - d) % mod + mod) % mod;
+                s[cnt][i] = (s[cnt - 1][i] + p[cnt][i]) % mod;
+            }
+
+            sum = (sum + binpow(x, k)) % mod;
+        } else {  // op == 1
+            d++;
+
+            sum = 0;
+            for (int i = 0; i <= k; i++) {
+                sum = (sum + f[k][i] * binpow(d, i) % mod * s[cnt][k - i] % mod) % mod;
+            }
+        }
+
+        cout << sum << endl;
+    }
+
+    return 0;
+}

S2OJ/45/45.cpp

@@ -0,0 +1,51 @@
+#include <iostream>
+#include <algorithm>
+#include <cstring>
+
+using std::cin;
+using std::cout;
+const char endl = '\n';
+
+const int N = 105,
+          M = 17;
+
+int n, m, fee[N], cost[N][M], f[N][1 << M], ans = 0x3f3f3f3f;
+
+int main() {
+    std::ios::sync_with_stdio(false);
+    cin.tie(nullptr);
+
+    cin >> n >> m;
+
+    for (int i = 1; i <= n; i++) {
+        cin >> fee[i];
+
+        for (int j = 0; j < m; j++) {
+            cin >> cost[i][j];
+        }
+    }
+
+    memset(f[0], 0x3f, sizeof(f[0]));
+    f[0][0] = 0;
+    for (int i = 1; i <= n; i++) {
+        for (int j = 0; j < 1 << m; j++) {
+            f[i][j] = f[i - 1][j] + fee[i];
+        }
+
+        for (int j = 0; j < 1 << m; j++) {
+            for (int k = 0; k < m; k++) {
+                if (j & (1 << k)) continue;
+
+                f[i][j | (1 << k)] = std::min(f[i][j | (1 << k)], f[i][j] + cost[i][k]);
+            }
+        }
+
+        for (int j = 0; j < 1 << m; j++) {
+            f[i][j] = std::min(f[i][j], f[i - 1][j]);
+        }
+    }
+
+    cout << f[n][(1 << m) - 1] << endl;
+
+    return 0;
+}