Delete brackets

 

Title Description

Parenthesis give you a valid sequence s1, every time you can remove a "()"
you can remove zero or more "()"
seek sequence can be deleted to another bracket s2

Enter a description:

A first line of input string s (2 ≤ | s | ≤ 100) 
a second input line of a character string t (2 ≤ | t | ≤ 100)

Output Description:

If you can output "Possible" 
Otherwise, output "Impossible"

Example 1

Entry

(())
()

Export

Possible

Example 2

Entry

()
()

Export

Possible

Example 3

Entry

(()()())
((()))

Export

Impossible

Example 4

Entry

((())((())())())
(()(())())

Export

Possible

Example 5

Entry

((())((())())())
((()()()()()))

Export

Impossible 

ideas: the input string is a, the target string b. Deleted when the brackets must always ensure that the number of left bracket and right bracket than or equal to (the last to be equal), 
we can define dp [i] [j] [ k] represents consideration before a string of i before the matching string b a j, A character string is deleted from the left bracket portion - Right parenthesis = k number of feasibility, the classification can transfer discussion, 
the final answer is dp [n] [m] [ 0] 

Code:

 1 #include <cstdio>
 2 #include <fstream>
 3 #include <algorithm>
 4 #include <cmath>
 5 #include <deque>
 6 #include <vector>
 7 #include <queue>
 8 #include <string>
 9 #include <cstring>
10 #include <map>
11 #include <stack>
12 #include <set>
13 #include <sstream>
14 #include <iostream>
15 #define mod 1000000007
16 #define eps 1e-6
17 #define ll long long
18 #define INF 0x3f3f3f3f
19 using namespace std;
20 
21 char a[110],b[110];
22 bool dp[110][110][110];
23 int main()
24 {
25     scanf("%s%s",(a+1),(b+1));
26     int m=strlen(a+1),n=strlen(b+1);
27     dp[0][0][0]=1;
28     for(int i=0;i<m;i++)
29     {
30         for(int j=0;j<=n;j++)
31         {
32             for(int k=0;k<=m;k++)
33             {
34                 if(dp[i][j][k])
35                 {
36                     if(!k&&a[i+1]==b[j+1])
37                     {
38                         dp[i+1][j+1][k]=1;
39                     }
40                     if(a[i+1]=='(')
41                     {
42                         dp[i+1][j][k+1]=1;
43                     }
44                     else if(k)
45                     {
46                         dp[i+1][j][k-1]=1;
47                     }
48                 }
49             }
50         }
51     }
52     if(dp[m][n][0])
53     {
54         printf("Possible\n");
55     }
56     else
57     {
58         printf("Impossible\n");
59     }
60 }