Distinct Subsequences+uva+经典dp

Distinct Subsequences

Time Limit: 3000MS

Memory Limit: Unknown

64bit IO Format: %lld & %llu

Description




Problem E
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…xm, another sequence Z=z1z2…zk is
a subsequence of X if there exists a strictly increasing sequence <i1, i2, …, ik> of indices of X such that for all j =
1, 2, …, k, we have xij=zj. 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

解决方案：想了一天没思路，看了别人的博客才懂的。求有多少个不同位置的相同子串，可以转为求有多少种删掉字符变成目标子串的方法。
dp[i][j]代表把串i变成串j的方法数，若text1[i]==text2[j]，字符i有两种情况，用则
dp[i][j]+=dp[i-1][j-1],若不用dp[i][j]+=dp[i-1][j]，所以dp[i][j]=dp[i-1][j-1]+dp[i-1][j]。若text1[i]!=text2[j]则字符i必不用，dp[i][j]=dp[i-1][j]；
边界条件为：dp[i][0]=1，即把字符变空串的方法只有一个。此题要用到大数，所以用java来写。
code：

import java.util.*;
import java.math.*;
public class Main {
public static void main(String arg[]){
int t;
int maxn=10004;
Scanner cin=new Scanner(System.in);
t=cin.nextInt();
for(int ii=1;ii<=t;ii++){
String text1;
String text2;
text1=cin.next();
text2=cin.next();
BigInteger dp[][]=new BigInteger[maxn][120];
BigInteger temp;
int len1=text1.length();
int len2=text2.length();
if(len1<len2)
{
System.out.print("0");
System.out.println();
continue;
}
for(int i=0;i<=len1;i++)
for(int j=0;j<=len2;j++)
dp[i][j]=BigInteger.ZERO;
for(int i=0;i<=len1;i++){
dp[i][0]=BigInteger.ONE;
}
for(int i=1;i<=len1;i++)
for(int j=1;j<=len2;j++){
dp[i][j]=dp[i-1][j];
if(text1.charAt(i-1)==text2.charAt(j-1)){
dp[i][j]=dp[i][j].add(dp[i-1][j-1]);
}
}
System.out.print(dp[len1][len2]);
System.out.println();
}

}
}