考慮兩個(gè)字符串，我們用dp[i][j]表示字串第到i個(gè)和字符串到第j個(gè)的總數(shù)，由于字串必須連續(xù)

因此dp[i][j]能夠有dp[i][j-1]和dp[i-1][j-1]遞推而來，而不能由dp[i-1][j]遞推而來。而后者的條件

是字串的第i個(gè)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]);}}}

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