zoukankan      html  css  js  c++  java
  • Sum of Nestings CodeForces

    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 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 ≤ 3·105, 0 ≤ k ≤ 1018) — 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).

    Example

    Input
    3 1
    Output
    ()(())
    Input
    4 6
    Output
    (((())))
    Input
    2 5
    Output
    Impossible

    Note

    The first example is examined in the statement.

    In the second example the answer is "(((())))". The nesting of the first opening bracket is 0, the nesting of the second is 1, the nesting of the third is 2, the nesting of fourth is 3. So the total sum of nestings equals to 0 + 1 + 2 + 3 = 6.

    In the third it is impossible to construct a regular bracket sequence, because the maximum possible total sum of nestings for two opening brackets equals to 1. This total sum of nestings is obtained for the sequence "(())".

    题解:这道题有一个性质,对于嵌套数为K的括号序列,它一定处于[1+2+....+x,1+2+...+(x+1)]区间中,所以减去左端点而剩下的嵌套数等价于在1~x层中添加一个完整括号。

    例如:4 4 ((())"()")

       4 5 ((()"()"))

       4 6 ((( " (  )" ))) 

     1 #include<string>
     2 #include<cstdio>
     3 #include<cstring>
     4 #include<iostream>
     5 #include<algorithm>
     6 using namespace std;
     7 typedef long long ll;
     8 
     9 const int maxn=3e5+5;
    10 
    11 ll n,k;
    12 ll map[maxn];
    13 
    14 void Inite(){
    15     for(ll i=1;i<=maxn;i++) map[i]=i*(i-1)/2;
    16 }
    17 
    18 int main()
    19 {   Inite();
    20     while(cin>>n>>k){ 
    21         if(k>map[n]) cout<<"Impossible"<<endl;
    22         else{
    23             ll pos=upper_bound(map+1,map+n+1,k)-(map+1);
    24             ll a=map[pos+1]-map[pos]+1,b;
    25             if(map[pos]==k){
    26                 for(int i=1;i<=pos;i++) printf("(");
    27                 for(int i=1;i<=pos;i++) printf(")");
    28                 b=n-pos;
    29             }
    30             else{
    31                 for(int i=1;i<=pos;i++) printf("(");
    32                 for(int i=1;i<=pos;i++){
    33                     if(i==a-k+map[pos]) printf("()");
    34                     printf(")");
    35                 }
    36                 b=n-pos-1;
    37             }
    38             for(int i=1;i<=b;i++) printf("()");
    39             cout<<endl;
    40         }
    41     }
    42     return 0;
    43 }
  • 相关阅读:
    2018-2019-2 20165325 网络对抗技术 Exp4 恶意代码分析
    2018-2019-2 网络对抗技术 20165325 Exp3 免杀原理与实践
    2018-2019-2 网络对抗技术 20165325 Exp2 后门原理与实践
    2018-2019-2 网络对抗技术 20165325 Exp1 PC平台逆向破解
    20165315 2018-2019-2 《网络对抗技术》Exp0 Kali安装 Week1
    2018-2019-1 20165226_20165310_20165315 实验五 通讯协议设计
    2018-2019-120165226_20165310_20165315 实验四 外设驱动程序设计
    2018-2019-1 20165315 实验三 实时系统
    2018-2019-1 20165226 20165310 20165315 实验二 固件程序设计
    20165315 第八周考试课下补做
  • 原文地址:https://www.cnblogs.com/zgglj-com/p/7669104.html
Copyright © 2011-2022 走看看