Educational Codeforces Round 114 (Rated for Div. 2) 补题

😊 | Powered By HeartFireY A.Regular Bracket Sequences

Problem Analysis

题目大意： 给定整数 n n n，构造 n n n个长度为 2 n 2n 2n的合法括号序列并输出， n ∈ [ 1 , 50 ] n \in [1, 50] n∈[1,50]。

思路： 直接按照以下规律构造 n n n组。 i = 1 , ( ) ( ) ( ) ( ) . . . i = 2 , ( ( ) ) ( ) ( ) . . . i = 3 , ( ( ( ) ) ) ( ) . . . i = 4 , ( ( ( ( ) ) ) ) . . . \begin{aligned} &i = 1, ()()()()... \\ &i = 2, (())()()... \\ &i = 3, ((()))()... \\ &i = 4, (((())))... \\ \end{aligned} ​i=1,()()()()...i=2,(())()()...i=3,((()))()...i=4,(((())))...​ 即先构造 i ,   i ∈ [ 1 , n ] i,\ i \in [1,n] i, i∈[1,n]个嵌套括号，剩余的长度直接输出括号补齐。

Accepted Code

#include 
using namespace std;

signed main(){
    int t = 0; cin >> t;
    while(t--){
        int n = 0; cin >> n;
        for(int i = 1; i > m;
    for(int i = 1; i > x >> y;
        int pos = lower_bound(a + 1, a + n, x) - a;
        if(pos != n){
            if(a[pos] == x) //能找到跟x_i相同大小的a_i
                ans = ((y - sum + a[pos])