#254. 放炮

https://sjzezoj.com/submission/71982

S2OJ/254/254.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;
+}