Codeforces 847 C. Sum of Nestings （技巧）

最新推荐文章于 2018-12-19 23:31:47 发布

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最新推荐文章于 2018-12-19 23:31:47 发布

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本文介绍了一种算法，用于生成特定数量与嵌套之和的括号序列。通过调整括号的位置，使得序列满足输入的嵌套总数要求。

Description

Recall that the bracket sequence is considered regular if it is possible to insert symbols ‘+’ and ‘1’ into it so that the result is a correct arithmetic expression. For example, a sequence “(()())” is regular, because we can get correct arithmetic expression insering symbols ‘+’ and ‘1’: “((1+1)+(1+1))”. Also the following sequences are regular: “()()()”, “(())” and “()”. The following sequences are not regular bracket sequences: “)(“, “(()” and “())(()”.

In this problem you are given two integers n and k. Your task is to construct a regular bracket sequence consisting of round brackets with length 2·n with total sum of nesting of all opening brackets equals to exactly k. The nesting of a single opening bracket equals to the number of pairs of brackets in which current opening bracket is embedded.

For example, in the sequence “()(())” the nesting of first opening bracket equals to 0, the nesting of the second opening bracket equals to 0 and the nesting of the third opening bracket equal to 1. So the total sum of nestings equals to 1.

Input

The first line contains two integers n and k (1 ≤ n ≤ 3·10^5, 0 ≤ k ≤ 10^18) — the number of opening brackets and needed total nesting.

Output

Print the required regular bracket sequence consisting of round brackets.

If there is no solution print “Impossible” (without quotes).

Examples input

3 1

Examples output

()(())

题意

给定 $n$ 与 $k$ ，代表括号的数目与嵌套之和，输出一种嵌套序列满足这种情况。

思路

我们考虑一种嵌套最大和的情况： ((((())))) ，此时 $n=5,k=1+2+3+4=10$ 。

显然，对于 $k \in (1+2+3,1+2+3+4]$ 中的任意一个数，我们可以将原本在第四层的括号安置在合适的层中，当然， $k$ 在其他范围时也是一样的思路。

此时这部分的序列类似于 (((())())) 这样的结果，假如括号没有用完，我们在最后补单独的 () 即可。

而对于 $k>n \times (n-1)/2$ ，则应输出 Impossible ，因为在最极端的情况下都无法找到解，因此其一定不成立。

AC 代码

#include<bits/stdc++.h>
using namespace std;
const int maxn = 1e5+10;
typedef __int64 LL;

LL n,k;
int main()
{
    cin>>n>>k;
    if(k>n*(n-1)/2)
        puts("Impossible");
    else
    {
        string ans;
        int base = 0;
        LL num = 0;
        while(num+base<=k)
        {
            num+=base++;
            ans+="(";
        }
        LL cnt = k-num;
        num = 0;
        while(base--)
        {
            ans+=")";
            if(base && base == cnt)ans+="()",num++;
            num++;
        }
        for(int i=0; i<n-num; i++)
            ans+="()";
        cout<<ans<<endl;
    }
    return 0;
}