标签：state stream 端点 inpu red ace else names 添加
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?≤?3·105, 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
3 1
()(())
4 6
(((())))
2 5
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 }
Sum of Nestings CodeForces - 847C
标签：state stream 端点 inpu red ace else names 添加
原文地址：http://www.cnblogs.com/zgglj-com/p/7669104.html
