brackets! brackets!

brackets! brackets!

Time limit:1000 MS Memory limit:65536 K Total submit:698 (239 users) Total accepted:272 (users) Rating: Special Judge:no

Description

There is six kinds of brackets: ' (', ') ', ' [', '] ', ' {', '} '. DCCMX ' s girl friend is now learning Java programming language, and got mad with brackets! Now give you a string of brackets. Is it valid? For example: "(([{{}]))" is valid, but "([)]" was not.

Input

First line contains a integer T (t<=10): The number of the test case. Next T lines, each contains a string:the input expression consists of brackets. The length of a string is between 1 and 100.

Output

For each test case, the output "Valid" in one line if the expression is Valid, or "Invalid" if not.

Sample Input

2 {{[[(())]]}} ({[}])

Sample Output

Valid InvalidNote () () This is a valid case. () () () This situation will make flag==0; () (This situation will cause the stack to be empty.) If a closing parenthesis appears, then if the stack is not empty, the top element of the stack and the right parenthesis match. Then the two conditions can not be satisfied under the conditions of flag==0; #include <iostream> #include <string.h> #include <stack>using namespace Std;int main () {int T;char gq[ 101];cin>>t;while (t--) {stack<char>ls;scanf ("%s", &gq); int flag=1;for (int i=0;i<strlen (GQ); i++) {if (gq[i]== ' | | gq[i]== ' {' | | | gq[i]== ' [') Ls.push (Gq[i]), else {if (gq[i]== ') ') {if (!ls.empty () &&ls.top () = = ' (') {Ls.pop ();} Else{flag=0;break;}} else if (gq[i]== '} ') {if (!ls.empty () &&ls.top () = = ' {') {Ls.pop ();} Else{flag=0;break;}} else if (gq[i]== '] ') {if (!ls.empty () &&ls.top () = = ' [') {Ls.pop ();} else {flag=0;break;}}} if (!ls.empty () | | | flag==0) cout<< "Invalid" <<endl;elsecout<< "Valid" <<ENDL;} return 0;}

brackets! brackets!