Round B APAC Test 2017 Problem A. Sherlock and Parentheses

Problem A. Sherlock and Parentheses

Sherlock and Watson have recently enrolled in a computer programming course. Today, the tutor taught them about the balanced parentheses problem. A string 

S

 consisting
only of characters 

(

 and/or 

)

 is balanced if:

It is the empty string, or:
It has the form 

(

S

)

, where S is a balanced string, or:
It has the form S1S2, where S1 is a balanced string and S2 is a balanced string.
Sherlock coded up the solution very quickly and started bragging about how good he is, so Watson gave him a problem to test his knowledge. He asked Sherlock to generate a string S of L
+ R characters, in which there are a total of L left parentheses 

(

 and
a total ofR right parentheses 

)

.
Moreover, the string must have as many different balanced non-empty substrings as possible. (Two substrings are considered different as long as they start or end at different indexes of the string, even if their content happens to be the same). Note that S
itself does not have to be balanced.

Sherlock is sure that once he knows the maximum possible number of balanced non-empty substrings, he will be able to solve the problem. Can you help him find that maximum number?

Input

The first line of the input gives the number of test cases, T. T test cases follow. Each test case consists of one line with two integers: L and R.

Output

For each test case, output one line containing 

Case #x: y

, where 

x

 is
the test case number (starting from 1) and 

y

 is the answer, as described above.

Limits

1 ≤ T ≤ 100.

Small dataset

0 ≤ L ≤ 20.

0 ≤ R ≤ 20.

1 ≤ L + R ≤ 20.

Large dataset

0 ≤ L ≤ 105.

0 ≤ R ≤ 105.

1 ≤ L + R ≤ 105.

Sample

Input	Output
3 1 0 1 1 3 2	Case #1: 0 Case #2: 1 Case #3: 3

In Case 1, the only possible string is 

(

. There are no
balanced non-empty substrings.
In Case 2, the optimal string is 

()

. There is only one
balanced non-empty substring: the entire string itself.
In Case 3, both strings 

()()(

 and 

(()()

 give
the same optimal answer.
For the case 

()()(

, for example, the three balanced substrings
are 

()

 from indexes 1 to 2,

()

 from
indexes 3 to 4, and 

()()

 from indexes 1 to 4.

Solution

#include <iostream>
#include <cstdio>
#include <cstring>
#include <string>
#include <vector>
#include <stack>
#include <list>
#include <iterator>
#include <unordered_set>
#include <unordered_map>
using namespace std;

typedef long long ll;

int main(int argc, char* argv[])
{
#ifndef TEST
if (argc != 2)
{
cout << "Invalid input" << endl;
return 1;
}

string input = argv[1];
string output = input.substr(0, input.length() - 2) + "out";
freopen(input.c_str(), "r", stdin);
freopen(output.c_str(), "w", stdout);
#endif

int T, N;
int L, R;

cin >> T;
for (int i = 1; i <= T; i++)
{
cin >> L >> R;
ll res = min(L, R);
if (res > 1)
res = res * (res + 1) / 2;
cout << "Case #" << i << ": " << res << endl;
}

fclose(stdin);
fclose(stdout);

return 0;
}

Reference

https://code.google.com/codejam/contest/5254487/dashboard