[TC SRM 708 div2 lev3] PalindromicSubseq2 

 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 };

 

posted on 2017-02-11 20:17  ivancjw  阅读(272)  评论(0)    收藏  举报

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