UVA 10069 Distinct Subsequences（DP)

考虑两个字符串，我们用dp[i][j]表示字串第到i个和字符串到第j个的总数，由于字串必须连续

因此dp[i][j]能够有dp[i][j-1]和dp[i-1][j-1]递推而来，而不能由dp[i-1][j]递推而来。而后者的条件

是字串的第i个和字符串相等。

A subsequence of a given sequence is just the given sequence with some elements (possibly none) left out. Formally, given a sequence X =x₁x₂…x_m, another sequence Z = z₁z₂…z_k is
a subsequence of X if there exists a strictly increasing sequence <i₁, i₂, …, i_k> of indices of Xsuch that for all j =
1, 2, …, k, we have x_ij = z_j. For example, Z = bcdb is a subsequence of X = abcbdab with corresponding index
sequence< 2, 3, 5, 7 >.

In this problem your job is to write a program that counts the number of occurrences of Z in X as a subsequence such that each has a distinct index sequence.

 

Input

The first line of the input contains an integer N indicating the number of test cases to follow.

The first line of each test case contains a string X, composed entirely of lowercase alphabetic characters and having length no greater than 10,000. The second line contains another string Z having length
no greater than 100 and also composed of only lowercase alphabetic characters. Be assured that neither Z nor any prefix or suffix of Z will have more than 10¹⁰⁰ distinct occurrences in X as
a subsequence.

 

Output

For each test case in the input output the number of distinct occurrences of Z in X as a subsequence. Output for each input set must be on a separate line.

 

Sample Input

2

babgbag

bag

rabbbit

rabbit

 

Sample Output

5

3

import java.io.*;
import java.math.*;
import java.util.*;
public class Main{
	  public static void main(String []args){
		  Scanner cin=new Scanner(System.in);
		  int t=cin.nextInt();
		  while(t--!=0){
			  char a[]=cin.next().toCharArray();
			  char b[]=cin.next().toCharArray();
//			  System.out.println("2333  ");
			  BigInteger [][] dp=new BigInteger[110][10100];
			  for(int i=0;i<dp.length;i++){
                   for(int j=0;j<dp[i].length;j++)
                	   dp[i][j]=BigInteger.ZERO;
			  }
//			  System.out.println("2333  ");
			  for(int j=0;j<a.length;j++){
				  if(j>0)
					  dp[0][j]=dp[0][j-1];
				  if(b[0]==a[j])
					  dp[0][j]=dp[0][j].add(BigInteger.ONE);
			  }
//			  System.out.println("2333  ");
			  for(int i=1;i<b.length;i++){
				  for(int j=i;j<a.length;j++){
					  dp[i][j]=dp[i][j-1];
					  if(b[i]==a[j])
						  dp[i][j]=dp[i][j].add(dp[i-1][j-1]);
				  }
			  }
			  System.out.println(dp[b.length-1][a.length-1]);
		  }
	  }
  }