Codevs 1251 Brackets

time limit: 1 sspace limit: 128000 KBtitle level: Golden GoldTitle Description Description

When calculating multiplication, we can add parentheses to change the order of multiplication, such as calculating X1, X2, X3, X4, ..., xn, can

(X1 (X2 (X3 (X4 (... (XN-1*XN)))))

:::

:::

(((... (((X1*X2) X3) X4) ...) XN-1) XN)

Your task is to program all such braces.

Enter a description Input Description

The first line of the input file is a number n (1<=n<=10), which indicates that there are n variables, followed by the name of one variable for each row of n rows.

Output description Output Description

Outputs all of the scenarios with parentheses added. Note: Do not add parentheses to a single character, two characters typeface by the middle of the multiplication sign.

Sample input Sample Input

4

North

South

East

West

Sample output Sample Output

(North (East*west))

(North (South*east) West)

((North*south) (east*west))

(North (South*east)) West)

(((North*south) East) West)

1#include <iostream>2#include <cstring>3#include <cstdio>4#include <vector>5 6 using namespacestd;7vector<string> Dfs (string*SS,intBeginintend)8 {9vector<string>ret;Ten     if(Begin>end)returnret; One     if(begin==end) A     { - Ret.push_back (Ss[begin]); -         returnret; the     } -     if(begin+1==end) -     { -         strings='('+ss[begin]+'*'+ss[end]+')';//The middle of the merge + Ret.push_back (s); -         returnret; +     } A     intsize1,size2; at      for(inti=begin;i<end;i++) -     { -vector<string> s1=DFS (ss,begin,i); -vector<string> S2=dfs (ss,i+1, end); -Size1=s1.size (); size2=s2.size (); -          for(intj=0; j<size1;j++) in         { -              for(intk=0; k<size2;k++) to             { +                 strings='('+s1[j]+s2[k]+')';//Note that this is a follow-up merger without multiplication sign. - Ret.push_back (s); the             } *         } $     }Panax Notoginseng     returnret; - } the intMain () + { A     stringss[ the]; the     intN; +scanf"%d",&n); -      for(intI=1; i<=n;i++) $Cin>>Ss[i]; $vector<string> Ans=dfs (SS,1, n); -     intSize=ans.size (); -      for(intI=0; i<size;i++) thecout<<ans[i]<<Endl; -     return 0;Wuyi}

Codevs 1251 Brackets