Distinct Subsequences - UVa 10069 dp+高精度

Distinct Subsequences

Input: standard input

Output: standard output

 

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

题意：第二个字符串在第一个字符串中一共有多少种组合方式。

思路：dp[i][j]表示第一个字符串前i位，匹配第二个字符串的前j位一共有多少种匹配方式。if(s1[i]==s2[j])   dp[i][j]=dp[i-1][j]+dp[i-1][j-1];    else    dp[i][j]=dp[i-1][j];

           然后此题答案比较大，要用到高精度……懒得打了，就用JAVA吧【其实你是不会打高精度→_→   人艰不拆啊

AC代码如下：

import java.math.BigInteger;
import java.util.Scanner;

public class Main
{ public static void main(String [] args)
  { Scanner scan=new Scanner(System.in);
	int t,i,j,k,len1,len2;
	String s1,s2;
    BigInteger dp[][]=new BigInteger[10010][110];
    
    for(i=0;i<=10000;i++)
     for(j=0;j<=100;j++)
      dp[i][j]=BigInteger.ZERO;
    for(i=0;i<=10000;i++)
   	 dp[i][0]=BigInteger.ONE;
    t=scan.nextInt();
    while(t>0)
    { t--;
      s1=scan.next();
      s2=scan.next();
      len1=s1.length();
      len2=s2.length();
      for(i=1;i<=len1;i++)
       for(j=1;j<=len2;j++)
       { if(s1.charAt(i-1)==s2.charAt(j-1))
    	  dp[i][j]=dp[i-1][j].add(dp[i-1][j-1]);
         else
    	  dp[i][j]=dp[i-1][j];
       }
      System.out.println(dp[len1][len2]);
    }
  }
}