1 #include <bits/stdc++.h>
2
3 using namespace std;
4
5 typedef long long LL;
6
7 const int MOD = 1e9 + 7;
8
9 class PalindromicSubseq2 {
10 public:
11 int solve(string s) {
12 int n = s.size();
13 string t(s.rbegin(), s.rend());
14 vector<vector<int>> dp(3001, vector<int>(3001, 0));
15 for (int i = 0; i <= n; i++) {
16 dp[i][0] = 1;
17 dp[0][i] = 1;
18 }
19 for (int i = 1; i <= n; i++) {
20 // 虽然dp[i - 1][j] + dp[i][j - 1]肯定包含dp[i - 1][j - 1]，但是取模后前两者可能小于后者导致结果为负，因此切记要+MOD
21 for (int j = 1; j <= n; j++) {
22 if (s[i - 1] == t[j - 1]) dp[i][j] = (dp[i - 1][j] + dp[i][j - 1]) % MOD;
23 else dp[i][j] = ((dp[i - 1][j] + dp[i][j - 1]) % MOD - dp[i - 1][j - 1] + MOD) % MOD;
24 }
25 }
26 int res = 0;
27 for (int i = 0; i < n; i++) {
28 res ^= (i + 1) * (long long)dp[i][n - i - 1] % MOD;
29 }
30 return res;
31 }
32 };
