#1899. 【NOI2009】管道取珠

https://sjzezoj.com/submission/68136

S2OJ/1899/1899.cpp

@@ -0,0 +1,58 @@
+#include <iostream>
+#include <algorithm>
+#include <cstring>
+#include <string>
+
+using std::cin;
+using std::cout;
+const char endl = '\n';
+
+const int N = 505;
+const int mod = 1024523;
+
+int n, m, f[2][N][N]{1};
+std::string a, b;
+
+int main() {
+    std::ios::sync_with_stdio(false);
+    cin.tie(nullptr);
+
+    cin >> n >> m >> a >> b;
+
+    std::reverse(a.begin(), a.end());
+    std::reverse(b.begin(), b.end());
+
+    a = ' ' + a;
+    b = ' ' + b;
+
+    for (int k = 1; k <= n + m; k++) {
+        int t = k & 1,
+            p = t ^ 1;
+
+        memset(f[t], 0x00, sizeof(f[t]));
+
+        for (int i = std::max(0, k - m); i <= std::min(n, k); i++) {
+            for (int j = std::max(0, k - m); j <= std::min(n, k); j++) {
+                if (i && j && a[i] == a[j]) {
+                    f[t][i][j] = (f[t][i][j] + f[p][i - 1][j - 1]) % mod;
+                }
+
+                if (i && k - j && a[i] == b[k - j]) {
+                    f[t][i][j] = (f[t][i][j] + f[p][i - 1][j]) % mod;
+                }
+
+                if (k - i && j && b[k - i] == a[j]) {
+                    f[t][i][j] = (f[t][i][j] + f[p][i][j - 1]) % mod;
+                }
+
+                if (k - i && k - j && b[k - i] == b[k - j]) {
+                    f[t][i][j] = (f[t][i][j] + f[p][i][j]) % mod;
+                }
+            }
+        }
+    }
+
+    cout << f[(n + m) & 1][n][n] << endl;
+
+    return 0;
+}