题解 | #小欧的括号嵌套#

先给他要的深度的括号，最后剩下的一对一对输出就行了。

#include <iostream>
#include <queue>
#include <map>
#include <set>
#include <cmath>
#include <cstring>
#include <algorithm>
#include <iomanip>
#include <stack>
#include <numeric>
#include <ctime>
#include <string>
#include <bitset>
#include <unordered_map>
#include <unordered_set>

using namespace std;
using ll = long long;
const ll N = 1e5 + 5, mod = 1e9 + 7, inf = 0x3f3f3f3f;

int n, r;

void solve() {
    cin >> n >> r;

    for (int i = 1; i <= r; i++) {
        cout << "(";
    }
    for (int i = 1; i <= r; i++) {
        cout << ")";
    }
    for (int i = 1; i <= (n * 2 - r * 2) / 2; i++) {
        cout << "()";
    }
}

int main() {
    ios::sync_with_stdio(0), cin.tie(0), cout.tie(0);

    int t = 1;
    //cin>>t;

    while (t--) {

        solve();

    }

    return 0;
}