HDU 4632 Palindrome subsequence（区间dp）

最新推荐文章于 2024-02-03 13:30:00 发布

原创最新推荐文章于 2024-02-03 13:30:00 发布 · 872 阅读

1 ·

CC 4.0 BY-SA版权

DP 动态规划专栏收录该内容

142 篇文章

订阅专栏

本文深入探讨了AI音视频处理领域中的视频分割与语义识别技术，介绍了视频分割的基本概念、算法及其应用，并详细阐述了语义识别在不同场景下的实现方法，包括自动驾驶、AR增强现实等领域的应用实例。

Palindrome subsequence

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 131072/65535 K (Java/Others)
Total Submission(s): 2595 Accepted Submission(s): 1039

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

Source

2013 Multi-University Training Contest 4

/*
题意:问一个字符串的会问序列有多少个
    dp[i][j]=(dp[i][j-1]+dp[i+1][j]-dp[i+1][j-1]);
    if(c[i]==c[j]) 那么加上中间的dp[i+1][j-1],因为可以和i,j形成新的
    还要加上  1  （i 和  j 形成字符串)


*/

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<cmath>
#include<queue>
#include<stack>
#include<vector>
#include<set>
#include<map>

#define L(x) (x<<1)
#define R(x) (x<<1|1)
#define MID(x,y) ((x+y)>>1)

#define bug printf("hihi\n")

#define eps 1e-8
typedef __int64 ll;

using namespace std;

#define mod 10007
#define INF 0x3f3f3f3f
#define N 1005

int dp[N][N];
int len;
char c[N];

int main()
{
   int i,j,t,ca=0;
   scanf("%d",&t);
   while(t--)
   {
       scanf("%s",c);
       len=strlen(c);
       memset(dp,0,sizeof(dp));
       for(i=0;i<len;i++)
          dp[i][i]=1;

       for(i=len-1;i>=0;i--)
         for(j=i+1;j<len;j++)
         {
            dp[i][j]=(dp[i+1][j]+dp[i][j-1]-dp[i+1][j-1]+mod)%mod;
            if(c[i]==c[j])
                dp[i][j]=(dp[i][j]+dp[i+1][j-1]+1+mod)%mod;
         }
         printf("Case %d: %d\n",++ca,(dp[0][len-1]+mod)%mod);
   }
   return 0;
}