HDU 4632 Palindrome subsequence (interval dp volume rejection theorem)

HDU 4632 Palindrome subsequence (interval dp volume rejection theorem)

 

Palindrome subsequence Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 131072/65535 K (Java/Others) Total Submission (s): 2610 Accepted Submission (s): 1050Problem 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 is a subsequence.
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 = 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 shocould be module 10007.
Sample Input

4aaaaaagoodafternooneveryonewelcometoooxxourproblems

Sample Output

Case 1: 1Case 2: 31Case 3: 421Case 4: 960

Source 2013 Multi-University Training Contest 4

Question link: http://acm.hdu.edu.cn/showproblem.php? Pid = 1, 4632

Returns the number of all input substrings to a string. Different character positions belong to different input substrings.

Question analysis: a very familiar dp seems to be a question of the beauty of programming qualifying round. It turns out to be the original question of multiple schools... We enumerate the intervals from left to right and calculate each interval from the outer to the inner, dp [I] [j] indicates the number of input substrings from I to j. If s [I] = s [j], then dp [I] [j] + = dp [I-1] [j + 1] + 1, in addition, dp [I] [j] = dp [I-1] [j] + dp [I] [j + 1]-dp [I-1] [j + 1], here is equivalent to a repeat. The last answer is dp [len] [1].

#include 
 
  #include 
  
   int const MAX = 1005;int const MOD = 10007;char s[MAX];int dp[MAX][MAX];int main(){    int T;    scanf(%d, &T);    for(int ca = 1; ca <= T; ca++)    {        printf(Case %d: , ca);        memset(dp, 0, sizeof(dp));        scanf(%s, s + 1);        int len = strlen(s + 1);        for(int i = 1; i <= len; i++)        {            for(int j = i; j >= 1; j--)            {                if(i == j)                {                    dp[i][j] = 1;                    continue;                }                if(s[i] == s[j])                    dp[i][j] += dp[i - 1][j + 1] + 1;                dp[i][j] += (5 * MOD + dp[i - 1][j] + dp[i][j + 1] - dp[i - 1][j + 1]) % MOD;            }        }        printf(%d, dp[len][1] % MOD);    }}