E. Dreamoon and Strings（Codeforces Round #272)

E. 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

Note

For the first sample, the corresponding optimal values of s' after removal 0 through |s| = 5 characters from s are {"aaaaa", "aaaa","aaa", "aa", "a", ""}.

For the second sample, possible corresponding optimal values of s' are {"axbaxxb", "abaxxb", "axbab", "abab", "aba", "ab","a", ""}.

dp[i][j] 为在字符串s的前i个删j个字符。k为从i开始，删除k个字符，会多出来一个字符串p，

则dp[i][j]=max(dp[i][j],dp[i-m-k][j-k]+1),m为字符串p的长度。

代码：

#include 
#include 
#include 
const int maxn=2000+100;
using namespace std;
char s1[maxn],s2[maxn];
int dp[maxn][maxn];
int n,m;
int solve(int i)
{
    int top=m,ans=0;
    if(ii不可能有值，赋以无穷小，防止被取到
    for(int i=1;i<=n;i++)
    {
        int k=solve(i);
        for(int j=0;j<=i;j++)
        {
            dp[i][j]=max(dp[i-1][j],dp[i][j]);
            if(j>=k)
            {
                dp[i][j]=max(dp[i][j],dp[i-m-k][j-k]+1);//前面的j>i赋无穷小就是防止j-k>i-m-k时被取到。
            }
        }
    }
    for(int i=0;i<=n;i++)
    {
        printf("%d ",dp[n][i]);
    }
    return 0;
}