Palindrome subsequence

                Palindrome subsequence

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 131072/65535 K (Java/Others)
Total Submission(s): 2232    Accepted Submission(s): 889

Problem Description

In mathematics, a subsequence is a sequence that can be derived from another sequence by deleting some elements without changing the order of the remaining elements. For example, the sequence <A, B, D> is a subsequence of <A, B, C, D, E, F>.
(http://en.wikipedia.org/wiki/Subsequence)

Given a string S, your task is to find out how many different subsequence of S is palindrome. Note that for any two subsequence X = <S_x1, S_x2, ..., S_xk> and Y = <S_y1, S_y2, ..., S_yk> , if there exist an integer i (1<=i<=k) such that xi != yi, the subsequence X and Y should be consider different even if S_xi = S_yi. Also two subsequences with different length should be considered different.

 

 

Input

The first line contains only one integer T (T<=50), which is the number of test cases. Each test case contains a string S, the length of S is not greater than 1000 and only contains lowercase letters.

 

 

Output

For each test case, output the case number first, then output the number of different subsequence of the given string, the answer should be module 10007.

 

 

Sample Input

4

a

aaaaa

goodafternooneveryone

welcometoooxxourproblems

 

Sample Output

Case 1: 1

Case 2: 31

Case 3: 421

Case 4: 960

 

状态方程dp[i][j] = dp[i+1][j]+dp[i][j-1] - dp[i+1][j-1]; 如果s[i] ==s[j] , dp[i][j]还要加上dp[i+1][j-1]+1; 

这道题WA了很惨，自己做题太少，对于有 -号再求余的一定要考虑是否有可能得出负数,加上mod之后可以保证是正数 dp[i][j] = (dp[i+1][j]+dp[i][j-1] - dp[i+1][j-1] +mod)%mod;

 

 1 #include <iostream>
 2 #include <algorithm>
 3 #include <cstring>
 4 #include <cstdio>
 5 using namespace std;
 6 const int N = 1005;
 7 const int mod = 10007;
 8 int _, n, dp[N][N], cas=1;
 9 char s[N];
10 
11 void solve()
12 {
13     scanf("%s", s+1);
14     n = strlen(s+1);
15     memset(dp, 0, sizeof(dp));
16     for(int k=0; k<n; k++)
17     {
18         for(int i=1; i+k<=n; i++)
19         {
20             int t = i+k;
21             dp[i][t] = (dp[i+1][t] + dp[i][t-1] - dp[i+1][t-1] + mod) % mod;/////////注意
22             if(s[i]==s[t]) dp[i][t] += dp[i+1][t-1] + 1;
23             dp[i][t] %= mod;
24         }
25     }
26     printf("Case %d: %d\n", cas++, dp[1][n]);
27 }
28 
29 int main()
30 {
31     scanf("%d", &_);
32     while(_--) solve();
33     return 0;
34 }

View Code

 

 

转载于:https://www.cnblogs.com/zyx1314/p/3843323.html