CF1227C Messy

思路：

构造题。首先把字符串变成“(((((...)))))”的样子，再根据k的取值变成“()()()...((...))”的样子即可。

实现：

 #include <bits/stdc++.h>
 using namespace std;
 int main()
 {
     int T; cin >> T;
     while (T--)
     {
         int n, k; string s; cin >> n >> k >> s;
         vector<pair<int, int>> res;
         for (int i = ; i < n / ; i++)
         {
             if (s[i] == '(') continue;
             int j = i + ;
             while (j < n && s[j] != '(') j++;
             res.push_back(make_pair(i, j));
             reverse(s.begin() + i, s.begin() + j + );
         }
         for (int i = ; i < k - ; i++)
         {
             int cur =  * i + ;
             int j = cur + ;
             while (j < n && s[j] != ')') j++;
             res.push_back(make_pair(cur, j));
             reverse(s.begin() + cur, s.begin() + j + );
         }
         cout << res.size() << endl;
         for (auto it: res)
         {
             cout << it.first +  << " " << it.second +  << endl;
         }
     }
     return ;
 }