for(int j=;j+i<ch.size();j++)

                     {

                     if(check(j,j+i))

                         dp[j][j+i]=dp[j+][j+i-]+;

                         for(int m=j;m<=j+i;m++)

                     dp[j][j+i]=max(dp[j][j+i],dp[j][m]+dp[m+][j+i]);

                     }

We give the following inductive definition of a “regular brackets” sequence:

• the empty sequence is a regular brackets sequence,
• if s is a regular brackets sequence, then (s) and [s] are regular brackets sequences, and
• if a and b are regular brackets sequences, then ab is a regular brackets sequence.
• no other sequence is a regular brackets sequence

For instance, all of the following character sequences are regular brackets sequences:

(), [], (()), ()[], ()[()]

while the following character sequences are not:

(, ], )(, ([)], ([(]

Given a brackets sequence of characters a₁a₂ … a_n, your goal is to find the length of the longest regular brackets sequence that is a subsequence of s. That is, you wish to find the largest m such that for indices i₁, i₂, …, i_m where 1 ≤ i₁ < i₂ < … < i_m ≤ n, a_i1a_i2 … a_im is a regular brackets sequence.

Given the initial sequence ([([]])], the longest regular brackets subsequence is [([])].

The input test file will contain multiple test cases. Each input test case consists of a single line containing only the characters (, ), [, and ]; each input test will have length between 1 and 100, inclusive. The end-of-file is marked by a line containing the word “end” and should not be processed.

For each input case, the program should print the length of the longest possible regular brackets subsequence on a single line.

 #include<iostream>

 #include<cstdio>

 #include<cmath>

 #include<cstring>

 #include<iomanip>

 #include<string>

 #include<algorithm>

 using namespace std;

 int money[];

 int dp[][];

 string ch;

 const int inf=;

 int check(int i,int j){

     if((ch[i]=='('&&ch[j]==')')||(ch[i]=='['&&ch[j]==']'))

         return ;

       return ;

 }

 void init(){

    for(int i=;i<ch.size();i++)

      for(int j=;j<ch.size();j++)

         dp[i][j]=;

 }

 int main(){

     int n,m;

     while(getline(cin,ch,'\n')){

             if(ch=="end") break;

             init();

             for(int k=;k<ch.size()-;k++)

                if(check(k,k+))

                dp[k][k+]=;

                else

                 dp[k][k+]=;

                 for(int i=;i<ch.size();i++)

                {

                    for(int j=;j+i<ch.size();j++)

                     {

                     if(check(j,j+i))

                         dp[j][j+i]=dp[j+][j+i-]+;

                         for(int m=j;m<=j+i;m++)

                     dp[j][j+i]=max(dp[j][j+i],dp[j][m]+dp[m+][j+i]);

                     }

                }

             cout<<dp[][ch.size()-]<<endl;

     }

     return ;

 }