Codeforces Round #272 (Div. 1)C（字符串DP）

C. Dreamoon and Strings

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

Dreamoon has a string s and a pattern string p. He first removes exactly x characters from s obtaining string s' as a result. Then he calculates  that is defined as the maximal number of non-overlapping substrings equal to p that can be found in s'. He wants to make this number as big as possible.

More formally, let's define  as maximum value of  over all s' that can be obtained by removing exactly x characters froms. Dreamoon wants to know  for all x from 0 to |s| where |s| denotes the length of string s.

Input

The first line of the input contains the string s (1 ≤ |s| ≤ 2 000).

The second line of the input contains the string p (1 ≤ |p| ≤ 500).

Both strings will only consist of lower case English letters.

Output

Print |s| + 1 space-separated integers in a single line representing the  for all x from 0 to |s|.

Sample test(s)

input

aaaaa
aa

output

2 2 1 1 0 0

input

axbaxxb
ab

output

0 1 1 2 1 1 0 0

题意：RT

思路：dp[i][j]表示s的前i个字符一共匹配了j个p串，删掉的最少字符数

            先用一个数组en[i]预处理出在s串的每一个位置i。直到能最早匹配p串的结束的位置

            转移为dp[ en[i+1] ][j+1]= min (dp[ en[i+1] ][j+1] 。dp[ i ][j] + (en[i+1]-i-m) )