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_i_j = 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]);}}}

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