#1539. 【2022.09.07 模拟题】报数

https://sjzezoj.com/submission/55887

BZOJ/1801/1801.cpp

@@ -0,0 +1,87 @@
+#include <iostream>
+
+using std::cin;
+using std::cout;
+const char endl = '\n';
+
+const int N = 105;
+const int mod = 9999973;
+
+int n, m;
+long long f[N][N][N]{1}, ans;
+
+long long C(long long x) {
+    return x * (x - 1) / 2;
+}
+
+int main() {
+    std::ios::sync_with_stdio(false);
+    cin.tie(nullptr);
+
+    cin >> n >> m;
+
+    for (int i = 1; i <= n; i++) {
+        // 不增加炮
+        for (int j = 0; j <= m; j++) {
+            for (int k = 0; j + k <= m; k++) {
+                f[i][j][k] = f[i - 1][j][k];
+            }
+        }
+
+        // 增加 1 个炮
+        //
+        // 从 1 个炮增加到 2 个炮
+        for (int j = 0; j <= m; j++) {
+            for (int k = 1; j + k <= m; k++) {
+                f[i][j][k] = (f[i][j][k] + f[i - 1][j + 1][k - 1] * (j + 1) % mod) % mod;
+            }
+        }
+
+        // 增加 1 个炮
+        //
+        // 从 0 个炮增加到 1 个炮
+        for (int j = 1; j <= m; j++) {
+            for (int k = 0; j + k <= m; k++) {
+                f[i][j][k] = (f[i][j][k] + f[i - 1][j - 1][k] * (m - j - k + 1) % mod) % mod;
+            }
+        }
+
+        // 增加 2 个炮
+        //
+        // 一列从 0 个炮增加到 1 个炮
+        // 一列从 1 个炮增加到 2 个炮
+        for (int j = 0; j <= m; j++) {
+            for (int k = 1; j + k <= m; k++) {
+                f[i][j][k] = (f[i][j][k] + f[i - 1][j][k - 1] * j * (m - j - k + 1) % mod) % mod;
+            }
+        }
+
+        // 增加 2 个炮
+        //
+        // 两列从 0 个炮增加到 1 个炮
+        for (int j = 2; j <= m; j++) {
+            for (int k = 0; j + k <= m; k++) {
+                f[i][j][k] = (f[i][j][k] + f[i - 1][j - 2][k] * C(m - j - k + 2) % mod) % mod;
+            }
+        }
+
+        // 增加 2 个炮
+        //
+        // 两列从 1 个炮增加到 2 个炮
+        for (int j = 0; j <= m; j++) {
+            for (int k = 2; j + k <= m; k++) {
+                f[i][j][k] = (f[i][j][k] + f[i - 1][j + 2][k - 2] * C(j + 2) % mod) % mod;
+            }
+        }
+    }
+
+    for (int i = 0; i <= m; i++) {
+        for (int j = 0; i + j <= m; j++) {
+            ans = (ans + f[n][i][j]) % mod;
+        }
+    }
+
+    cout << ans << endl;
+
+    return 0;
+}

Luogu/P2051/P2051.cpp

@@ -0,0 +1,87 @@
+#include <iostream>
+
+using std::cin;
+using std::cout;
+const char endl = '\n';
+
+const int N = 105;
+const int mod = 9999973;
+
+int n, m;
+long long f[N][N][N]{1}, ans;
+
+long long C(long long x) {
+    return x * (x - 1) / 2;
+}
+
+int main() {
+    std::ios::sync_with_stdio(false);
+    cin.tie(nullptr);
+
+    cin >> n >> m;
+
+    for (int i = 1; i <= n; i++) {
+        // 不增加炮
+        for (int j = 0; j <= m; j++) {
+            for (int k = 0; j + k <= m; k++) {
+                f[i][j][k] = f[i - 1][j][k];
+            }
+        }
+
+        // 增加 1 个炮
+        //
+        // 从 1 个炮增加到 2 个炮
+        for (int j = 0; j <= m; j++) {
+            for (int k = 1; j + k <= m; k++) {
+                f[i][j][k] = (f[i][j][k] + f[i - 1][j + 1][k - 1] * (j + 1) % mod) % mod;
+            }
+        }
+
+        // 增加 1 个炮
+        //
+        // 从 0 个炮增加到 1 个炮
+        for (int j = 1; j <= m; j++) {
+            for (int k = 0; j + k <= m; k++) {
+                f[i][j][k] = (f[i][j][k] + f[i - 1][j - 1][k] * (m - j - k + 1) % mod) % mod;
+            }
+        }
+
+        // 增加 2 个炮
+        //
+        // 一列从 0 个炮增加到 1 个炮
+        // 一列从 1 个炮增加到 2 个炮
+        for (int j = 0; j <= m; j++) {
+            for (int k = 1; j + k <= m; k++) {
+                f[i][j][k] = (f[i][j][k] + f[i - 1][j][k - 1] * j * (m - j - k + 1) % mod) % mod;
+            }
+        }
+
+        // 增加 2 个炮
+        //
+        // 两列从 0 个炮增加到 1 个炮
+        for (int j = 2; j <= m; j++) {
+            for (int k = 0; j + k <= m; k++) {
+                f[i][j][k] = (f[i][j][k] + f[i - 1][j - 2][k] * C(m - j - k + 2) % mod) % mod;
+            }
+        }
+
+        // 增加 2 个炮
+        //
+        // 两列从 1 个炮增加到 2 个炮
+        for (int j = 0; j <= m; j++) {
+            for (int k = 2; j + k <= m; k++) {
+                f[i][j][k] = (f[i][j][k] + f[i - 1][j + 2][k - 2] * C(j + 2) % mod) % mod;
+            }
+        }
+    }
+
+    for (int i = 0; i <= m; i++) {
+        for (int j = 0; i + j <= m; j++) {
+            ans = (ans + f[n][i][j]) % mod;
+        }
+    }
+
+    cout << ans << endl;
+
+    return 0;
+}

Luogu/P2339/P2339.cpp

@@ -0,0 +1,54 @@
+#include <iostream>
+#include <algorithm>
+#include <cmath>
+#include <cstring>
+#include <limits>
+#include <utility>
+
+using std::cin;
+using std::cout;
+const char endl = '\n';
+
+const int N = 1005;
+
+int c, h, b, f[N][N][2], ans = std::numeric_limits<int>::max();
+std::pair<int, int> a[N];
+
+int main() {
+    std::ios::sync_with_stdio(false);
+    cin.tie(nullptr);
+
+    memset(f, 0x3f, sizeof(f));
+
+    cin >> c >> h >> b;
+
+    for (int i = 1; i <= c; i++) {
+        cin >> a[i].first >> a[i].second;
+    }
+
+    std::sort(a + 1, a + 1 + c);
+
+    f[0][c][0] = f[0][c][1] = 0;
+
+    for (int i = 1; i <= c; i++) {
+        for (int j = c; j >= i; j--) {
+            f[i][j][0] = std::min({
+                std::max(f[i - 1][j][0] + a[i].first - a[i - 1].first, a[i].second),
+                std::max(f[i][j + 1][1] + a[j + 1].first - a[i].first, a[i].second),
+            });
+
+            f[i][j][1] = std::min({
+                std::max(f[i - 1][j][0] + a[j].first - a[i - 1].first, a[j].second),
+                std::max(f[i][j + 1][1] + a[j + 1].first - a[j].first, a[j].second),
+            });
+        }
+    }
+
+    for (int i = 1; i <= c; i++) {
+        ans = std::min(ans, std::min(f[i][i][0], f[i][i][1]) + std::abs(b - a[i].first));
+    }
+
+    cout << ans << endl;
+
+    return 0;
+}

Luogu/P4127/P4127.cpp

@@ -0,0 +1,55 @@
+#include <iostream>
+#include <bitset>
+#include <cstring>
+
+using std::cin;
+using std::cout;
+const char endl = '\n';
+
+long long a, b, f[20][200][200], ans;
+int c[20], p;
+
+long long dfs(int len, int s, int x, bool limit) {
+    if (s + 9 * len < p) return 0;
+    if (!limit && f[len][s][x] != -1) return f[len][s][x];
+    if (!len) return s == p && x == 0;
+
+    int n = limit ? c[len] : 9;
+    long long res = 0;
+
+    for (int i = 0; i <= n && i + s <= p; i++)
+        res += dfs(len - 1, s + i, (x * 10 + i) % p, limit && (i == c[len]));
+
+    if (!limit) f[len][s][x] = res;
+
+    return res;
+}
+
+long long calc(long long x) {
+    int cnt = 0;
+
+    while (x) {
+        c[++cnt] = x % 10;
+        x /= 10;
+    }
+
+    long long res = 0;
+
+    for (p = 1; p <= cnt * 9; p++) {
+        memset(f, 0xff, sizeof(f));
+        res += dfs(cnt, 0, 0, 1);
+    }
+
+    return res;
+}
+
+int main() {
+    std::ios::sync_with_stdio(false);
+    cin.tie(nullptr);
+
+    cin >> a >> b;
+
+    cout << calc(b) - calc(a - 1) << endl;
+
+    return 0;
+}

S2OJ/1539/1539.cpp

@@ -0,0 +1,54 @@
+#include <iostream>
+#include <algorithm>
+
+using std::cin;
+using std::cout;
+const char endl = '\n';
+
+int t, l, r, a, b, c, ans;
+int p, cnt, primes[10000005], sum[10000005];
+bool not_prime[10000005];
+
+inline bool check(int x) {
+    int sum = 0;
+
+    while (x) {
+        sum += x % 10;
+        x /= 10;
+    }
+
+    return !not_prime[sum];
+}
+
+int main() {
+    std::ios::sync_with_stdio(false);
+    cin.tie(nullptr);
+
+    for (int i = 2; i <= 10000000; i++) {
+        if (!not_prime[i]) primes[++p] = i;
+
+        for (int j = 1; j <= p && primes[j] * i <= 10000000; j++) {
+            not_prime[primes[j] * i] = true;
+            if (i % primes[j] == 0) break;
+        }
+    }
+
+    for (int i = 1; i <= 10000000; i++) {
+        if (!not_prime[i] && check(i)) sum[i] = sum[i - 1] + 1;
+        else sum[i] = sum[i - 1];
+    }
+
+    cin >> t >> l >> r >> a >> b >> c;
+
+    while (t--) {
+        ans ^= sum[r] - sum[l - 1];
+
+        l = ((l ^ b) + a) % c + 1;
+        r = ((r ^ b) + a) % c + 1;
+        if (l > r) std::swap(l, r);
+    }
+
+    cout << ans << endl;
+
+    return 0;
+}