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Provide tags and a correct Python 3 solution for this coding contest problem. 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). Examples 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 "(())".
instruction
0
20,865
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Tags: constructive algorithms Correct Solution: ``` from math import * n,k=map(int,input().split()) if n*(n-1)//2<k: print("Impossible") else: li=[] temp=int((-1+sqrt(8*k+1))//2) k-=(temp*(temp+1))//2 li+=[temp] for i in range(temp): cur=temp-i if cur<=k and cur>0: li+=[cur]*(k//cur) k-=(k//cur)*cur if k==0 or cur==0: break li.reverse() num=li[0] ans="" for i in range(num+1): ans+="(" n-=num+1 ans+=")" for i in range(1,len(li)): num=li[i]-li[i-1]+1 for j in range(num): ans+="(" n-=num ans+=")" ans+=")"*li[-1] if n<0 or k>0: print("Impossible") else: for i in range(n): ans+="()" print(ans) ```
output
1
20,865
21
41,731
Provide tags and a correct Python 3 solution for this coding contest problem. 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). Examples 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 "(())".
instruction
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20,866
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Tags: constructive algorithms Correct Solution: ``` ch=input() l=ch.split(' ') n=int(l[0]) k=int(l[1]) l1='(' l2=')' l3='()' s=0 mi=0 ch='' from math import sqrt q=int(sqrt(2*k)) while k>=(q*(q-1))/2: q=q+1 q=q-1 d=int(k-(q*(q-1))/2) mi=q i=0 if d!=0: mi=q+1 i=1 if n<mi: print('Impossible') else: c=n-mi print(c*l3+d*l1+i*l3+(q-d)*l1+q*l2) ```
output
1
20,866
21
41,733
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response. 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). Examples 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 "(())". Submitted Solution: ``` # Legends Always Come Up with Solution # Author: Manvir Singh import os import sys from io import BytesIO, IOBase from collections import Counter from bisect import * from math import gcd from itertools import permutations,combinations from math import sqrt,ceil,floor def main(): n,k=map(int,input().split()) z=floor((-1+sqrt(1+8*k))/2) f=int((z*(z+1))//2!=k) if z+f>n-1: print("Impossible") else: if f: x=z+1 y=z+1-(k-(z*(z+1)//2)) print("("*x+")"*y+"("+")"*(x+1-y)+"()"*(n-(x+1))) else: if z<n-1: print("("*(z+1)+")"*(z+1)+"()"*(n-1-z)) else: print("("*n+")"*n) # region fastio BUFSIZE = 8192 class FastIO(IOBase): newlines = 0 def __init__(self, file): self._fd = file.fileno() self.buffer = BytesIO() self.writable = "x" in file.mode or "r" not in file.mode self.write = self.buffer.write if self.writable else None def read(self): while True: b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE)) if not b: break ptr = self.buffer.tell() self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr) self.newlines = 0 return self.buffer.read() def readline(self): while self.newlines == 0: b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE)) self.newlines = b.count(b"\n") + (not b) ptr = self.buffer.tell() self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr) self.newlines -= 1 return self.buffer.readline() def flush(self): if self.writable: os.write(self._fd, self.buffer.getvalue()) self.buffer.truncate(0), self.buffer.seek(0) class IOWrapper(IOBase): def __init__(self, file): self.buffer = FastIO(file) self.flush = self.buffer.flush self.writable = self.buffer.writable self.write = lambda s: self.buffer.write(s.encode("ascii")) self.read = lambda: self.buffer.read().decode("ascii") self.readline = lambda: self.buffer.readline().decode("ascii") sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout) input = lambda: sys.stdin.readline().rstrip("\r\n") if __name__ == "__main__": main() ```
instruction
0
20,867
21
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Yes
output
1
20,867
21
41,735
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response. 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). Examples 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 "(())". Submitted Solution: ``` def step(n, k): count = min(n, int((1 + (1 + 8 * k) ** 0.5) / 2)) return count def nest(n, k): accum = "" count = step(n, k) leftover = int(k - (((count - 1) * count) / 2)) n -= count if leftover == 0: accum += count * "(" accum += count * ")" return accum + n * "()" if n == 0: return "Impossible" pos = count - leftover accum += count * "(" accum += pos * ")" accum += "()" accum += (count - pos) * ")" n -= 1 accum += n * "()" return accum N, K = [int(x) for x in input().split()] print(nest(N, K)) ```
instruction
0
20,868
21
41,736
Yes
output
1
20,868
21
41,737
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response. 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). Examples 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 "(())". Submitted Solution: ``` n, k = map(int, input().split()) k_now, level, n_now = 0, 0, 0 ans = list() if k > n * (n - 1) / 2: print('Impossible') else: while n_now < 2 * n: if k_now + level <= k: ans.append('(') k_now += level level += 1 else: ans.append(')') level -= 1 n_now += 1 print(*ans, sep='') ```
instruction
0
20,869
21
41,738
Yes
output
1
20,869
21
41,739
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response. 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). Examples 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 "(())". Submitted Solution: ``` import sys input = sys.stdin.buffer.readline def ints(): return map(int, input().split()) n, k=ints() if k > (n-1)*n//2: print("Impossible") exit(0) sum=0 ans='' for i in range(n+1): if sum+i>k: ans+='('*(i) ans+=')'*(i-(k-sum)) if i!=n: ans+='(' ans+=')'*(k-sum+1) ans+='()'*(n-i-1) else: ans+=')'*(k-sum) break sum+=i print(ans) ```
instruction
0
20,870
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Yes
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1
20,870
21
41,741
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response. 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). Examples 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 "(())". Submitted Solution: ``` ch=input() l=ch.split(' ') n=int(l[0]) k=int(l[1]) l1=['('] l2=[')'] l3=['()'] s=0 mi=0 ch='' while (k-s)!=0: d=k-s q=1 while d>=(q*(q-1))/2: q=q+1 q=q-1 li1=q*l1 li2=q*l2 ch=ch+''.join(li1)+''.join(li2) s=s+q*(q-1)/2 mi=mi+q if n<mi: print('impossible') else: c=n-mi li3=c*l3 ch=''.join(li3)+ch print(ch) ```
instruction
0
20,871
21
41,742
No
output
1
20,871
21
41,743
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response. 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). Examples 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 "(())". Submitted Solution: ``` n, k = map(int, input().split()) s = int((1 + 4 * k) ** 0.5) + 5 >> 1 k -= s * s - s >> 1 t = '(' * s + ')' * s n -= s if k: t = t[:k] + '()' + t[k:] n -= 1 print('Impossible' if n < 0 else t + '()' * n) ```
instruction
0
20,872
21
41,744
No
output
1
20,872
21
41,745
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response. 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). Examples 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 "(())". Submitted Solution: ``` from math import * n,k=map(int,input().split()) if n*(n-1)//2<k: print("Impossible") else: li=[] temp=int((-1+sqrt(8*k+1))//2) k-=(temp*(temp+1))//2 li+=[temp] for i in range(temp): cur=temp-i if cur<=k and cur>0: li+=[cur]*(k//cur) k-=(k//cur)*cur if k==0 or cur==0: break li.reverse() num=li[0] ans="" for i in range(num+1): ans+="(" n-=num+1 ans+=")" for i in range(1,len(li)): num=li[i]-li[i-1]+1 for j in range(num): ans+="(" n-=num ans+=")" ans+=")"*li[-1] if n<0 or k>0: print("Impossible") else: for i in range(n): ans+="()" print(ans) ```
instruction
0
20,873
21
41,746
No
output
1
20,873
21
41,747
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response. 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). Examples 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 "(())". Submitted Solution: ``` n, k = map(int, input().split()) max_nesting = n/2 * (n-1) nesting = [i for i in range(n)] m = n-1 add = 0 s = sum(nesting) if max_nesting < k: print("Impossible") else: while not s == k: if s > k: s -= m m -= 1 else: add = k-s s += add res = "(" * (m+1) + ")" * (m-add+1) + ("(" + ")" * (add+1)) + "()" * (n - m - add-1) print(res) ```
instruction
0
20,874
21
41,748
No
output
1
20,874
21
41,749
Provide tags and a correct Python 3 solution for this coding contest problem. A string is called bracket sequence if it does not contain any characters other than "(" and ")". A bracket sequence is called regular if it it is possible to obtain correct arithmetic expression by inserting characters "+" and "1" into this sequence. For example, "", "(())" and "()()" are regular bracket sequences; "))" and ")((" are bracket sequences (but not regular ones), and "(a)" and "(1)+(1)" are not bracket sequences at all. You have a number of strings; each string is a bracket sequence of length 2. So, overall you have cnt_1 strings "((", cnt_2 strings "()", cnt_3 strings ")(" and cnt_4 strings "))". You want to write all these strings in some order, one after another; after that, you will get a long bracket sequence of length 2(cnt_1 + cnt_2 + cnt_3 + cnt_4). You wonder: is it possible to choose some order of the strings you have such that you will get a regular bracket sequence? Note that you may not remove any characters or strings, and you may not add anything either. Input The input consists of four lines, i-th of them contains one integer cnt_i (0 ≤ cnt_i ≤ 10^9). Output Print one integer: 1 if it is possible to form a regular bracket sequence by choosing the correct order of the given strings, 0 otherwise. Examples Input 3 1 4 3 Output 1 Input 0 0 0 0 Output 1 Input 1 2 3 4 Output 0 Note In the first example it is possible to construct a string "(())()(()((()()()())))", which is a regular bracket sequence. In the second example it is possible to construct a string "", which is a regular bracket sequence.
instruction
0
21,229
21
42,458
Tags: greedy, implementation Correct Solution: ``` """ NTC here """ iin=lambda :int(input()) lin=lambda :list(map(int,input().split())) a=[iin() for i in range(4)] if a[0]==a[3]==0 and a[2]!=0:print(0) elif a[0]==a[3]:print(1) else:print(0) ```
output
1
21,229
21
42,459
Provide tags and a correct Python 3 solution for this coding contest problem. A string is called bracket sequence if it does not contain any characters other than "(" and ")". A bracket sequence is called regular if it it is possible to obtain correct arithmetic expression by inserting characters "+" and "1" into this sequence. For example, "", "(())" and "()()" are regular bracket sequences; "))" and ")((" are bracket sequences (but not regular ones), and "(a)" and "(1)+(1)" are not bracket sequences at all. You have a number of strings; each string is a bracket sequence of length 2. So, overall you have cnt_1 strings "((", cnt_2 strings "()", cnt_3 strings ")(" and cnt_4 strings "))". You want to write all these strings in some order, one after another; after that, you will get a long bracket sequence of length 2(cnt_1 + cnt_2 + cnt_3 + cnt_4). You wonder: is it possible to choose some order of the strings you have such that you will get a regular bracket sequence? Note that you may not remove any characters or strings, and you may not add anything either. Input The input consists of four lines, i-th of them contains one integer cnt_i (0 ≤ cnt_i ≤ 10^9). Output Print one integer: 1 if it is possible to form a regular bracket sequence by choosing the correct order of the given strings, 0 otherwise. Examples Input 3 1 4 3 Output 1 Input 0 0 0 0 Output 1 Input 1 2 3 4 Output 0 Note In the first example it is possible to construct a string "(())()(()((()()()())))", which is a regular bracket sequence. In the second example it is possible to construct a string "", which is a regular bracket sequence.
instruction
0
21,230
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Tags: greedy, implementation Correct Solution: ``` a = int(input()) b = int(input()) c = int(input()) d = int(input()) if a != d or (a == 0 and c != 0): print(0) else: print(1) ```
output
1
21,230
21
42,461
Provide tags and a correct Python 3 solution for this coding contest problem. A string is called bracket sequence if it does not contain any characters other than "(" and ")". A bracket sequence is called regular if it it is possible to obtain correct arithmetic expression by inserting characters "+" and "1" into this sequence. For example, "", "(())" and "()()" are regular bracket sequences; "))" and ")((" are bracket sequences (but not regular ones), and "(a)" and "(1)+(1)" are not bracket sequences at all. You have a number of strings; each string is a bracket sequence of length 2. So, overall you have cnt_1 strings "((", cnt_2 strings "()", cnt_3 strings ")(" and cnt_4 strings "))". You want to write all these strings in some order, one after another; after that, you will get a long bracket sequence of length 2(cnt_1 + cnt_2 + cnt_3 + cnt_4). You wonder: is it possible to choose some order of the strings you have such that you will get a regular bracket sequence? Note that you may not remove any characters or strings, and you may not add anything either. Input The input consists of four lines, i-th of them contains one integer cnt_i (0 ≤ cnt_i ≤ 10^9). Output Print one integer: 1 if it is possible to form a regular bracket sequence by choosing the correct order of the given strings, 0 otherwise. Examples Input 3 1 4 3 Output 1 Input 0 0 0 0 Output 1 Input 1 2 3 4 Output 0 Note In the first example it is possible to construct a string "(())()(()((()()()())))", which is a regular bracket sequence. In the second example it is possible to construct a string "", which is a regular bracket sequence.
instruction
0
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Tags: greedy, implementation Correct Solution: ``` ''' t= input() lng= len(t) li1=[]; li2=[] for i in range(lng): if t[i]!='a': li1.append(t[i]) elif t[i]=='a': li2.append(i) aa= ''.join(li1) if len(aa)==0: print(t); exit(0) if len(aa)%2==1: print(':('); exit(0) if len(aa)%2==0: #print(123) l= int(len(aa)/2); lp= l#; print(l) ll= aa[l:]; lll=aa[0:l] #; print(ll) if ll!=lll: print(':('); exit(0) if ll not in t: print(':('); exit(0) tp= t[::-1]; tc= ll[::-1]#; print(tp,tc) if tp.find(tc)!=0: print(':('); exit(0) if tp.find(tc)==0: ul= len(tc) lu= tp[ul:][::-1] print(lu); exit(0) ''' a= int(input()); b= int(input()); c= int(input()); d= int(input()); sv=0; vs=0 if a+b+c==b+c+d: sv=1 if sv==1: if a+d<1 and c>0: sv=0 print(sv) ```
output
1
21,231
21
42,463
Provide tags and a correct Python 3 solution for this coding contest problem. A string is called bracket sequence if it does not contain any characters other than "(" and ")". A bracket sequence is called regular if it it is possible to obtain correct arithmetic expression by inserting characters "+" and "1" into this sequence. For example, "", "(())" and "()()" are regular bracket sequences; "))" and ")((" are bracket sequences (but not regular ones), and "(a)" and "(1)+(1)" are not bracket sequences at all. You have a number of strings; each string is a bracket sequence of length 2. So, overall you have cnt_1 strings "((", cnt_2 strings "()", cnt_3 strings ")(" and cnt_4 strings "))". You want to write all these strings in some order, one after another; after that, you will get a long bracket sequence of length 2(cnt_1 + cnt_2 + cnt_3 + cnt_4). You wonder: is it possible to choose some order of the strings you have such that you will get a regular bracket sequence? Note that you may not remove any characters or strings, and you may not add anything either. Input The input consists of four lines, i-th of them contains one integer cnt_i (0 ≤ cnt_i ≤ 10^9). Output Print one integer: 1 if it is possible to form a regular bracket sequence by choosing the correct order of the given strings, 0 otherwise. Examples Input 3 1 4 3 Output 1 Input 0 0 0 0 Output 1 Input 1 2 3 4 Output 0 Note In the first example it is possible to construct a string "(())()(()((()()()())))", which is a regular bracket sequence. In the second example it is possible to construct a string "", which is a regular bracket sequence.
instruction
0
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Tags: greedy, implementation Correct Solution: ``` cnt1 = int(input()) cnt2 = int(input()) cnt3 = int(input()) cnt4 = int(input()) if cnt1 != cnt4: print(0) elif cnt3 > 0 and cnt1 == 0: print(0) else: print(1) ```
output
1
21,232
21
42,465
Provide tags and a correct Python 3 solution for this coding contest problem. A string is called bracket sequence if it does not contain any characters other than "(" and ")". A bracket sequence is called regular if it it is possible to obtain correct arithmetic expression by inserting characters "+" and "1" into this sequence. For example, "", "(())" and "()()" are regular bracket sequences; "))" and ")((" are bracket sequences (but not regular ones), and "(a)" and "(1)+(1)" are not bracket sequences at all. You have a number of strings; each string is a bracket sequence of length 2. So, overall you have cnt_1 strings "((", cnt_2 strings "()", cnt_3 strings ")(" and cnt_4 strings "))". You want to write all these strings in some order, one after another; after that, you will get a long bracket sequence of length 2(cnt_1 + cnt_2 + cnt_3 + cnt_4). You wonder: is it possible to choose some order of the strings you have such that you will get a regular bracket sequence? Note that you may not remove any characters or strings, and you may not add anything either. Input The input consists of four lines, i-th of them contains one integer cnt_i (0 ≤ cnt_i ≤ 10^9). Output Print one integer: 1 if it is possible to form a regular bracket sequence by choosing the correct order of the given strings, 0 otherwise. Examples Input 3 1 4 3 Output 1 Input 0 0 0 0 Output 1 Input 1 2 3 4 Output 0 Note In the first example it is possible to construct a string "(())()(()((()()()())))", which is a regular bracket sequence. In the second example it is possible to construct a string "", which is a regular bracket sequence.
instruction
0
21,233
21
42,466
Tags: greedy, implementation Correct Solution: ``` a=int(input()) b=int(input()) c=int(input()) d=int(input()) if(a!=d): print(0) elif(c>0 and (a==0 or d==0)): print(0) else: print(1) ```
output
1
21,233
21
42,467
Provide tags and a correct Python 3 solution for this coding contest problem. A string is called bracket sequence if it does not contain any characters other than "(" and ")". A bracket sequence is called regular if it it is possible to obtain correct arithmetic expression by inserting characters "+" and "1" into this sequence. For example, "", "(())" and "()()" are regular bracket sequences; "))" and ")((" are bracket sequences (but not regular ones), and "(a)" and "(1)+(1)" are not bracket sequences at all. You have a number of strings; each string is a bracket sequence of length 2. So, overall you have cnt_1 strings "((", cnt_2 strings "()", cnt_3 strings ")(" and cnt_4 strings "))". You want to write all these strings in some order, one after another; after that, you will get a long bracket sequence of length 2(cnt_1 + cnt_2 + cnt_3 + cnt_4). You wonder: is it possible to choose some order of the strings you have such that you will get a regular bracket sequence? Note that you may not remove any characters or strings, and you may not add anything either. Input The input consists of four lines, i-th of them contains one integer cnt_i (0 ≤ cnt_i ≤ 10^9). Output Print one integer: 1 if it is possible to form a regular bracket sequence by choosing the correct order of the given strings, 0 otherwise. Examples Input 3 1 4 3 Output 1 Input 0 0 0 0 Output 1 Input 1 2 3 4 Output 0 Note In the first example it is possible to construct a string "(())()(()((()()()())))", which is a regular bracket sequence. In the second example it is possible to construct a string "", which is a regular bracket sequence.
instruction
0
21,234
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Tags: greedy, implementation Correct Solution: ``` cnt = [int(input()) for _ in range(4)] if cnt[0] != cnt[3]: print(0) elif cnt[2] > 0 and cnt[0] == 0: print(0) else: print(1) ```
output
1
21,234
21
42,469
Provide tags and a correct Python 3 solution for this coding contest problem. A string is called bracket sequence if it does not contain any characters other than "(" and ")". A bracket sequence is called regular if it it is possible to obtain correct arithmetic expression by inserting characters "+" and "1" into this sequence. For example, "", "(())" and "()()" are regular bracket sequences; "))" and ")((" are bracket sequences (but not regular ones), and "(a)" and "(1)+(1)" are not bracket sequences at all. You have a number of strings; each string is a bracket sequence of length 2. So, overall you have cnt_1 strings "((", cnt_2 strings "()", cnt_3 strings ")(" and cnt_4 strings "))". You want to write all these strings in some order, one after another; after that, you will get a long bracket sequence of length 2(cnt_1 + cnt_2 + cnt_3 + cnt_4). You wonder: is it possible to choose some order of the strings you have such that you will get a regular bracket sequence? Note that you may not remove any characters or strings, and you may not add anything either. Input The input consists of four lines, i-th of them contains one integer cnt_i (0 ≤ cnt_i ≤ 10^9). Output Print one integer: 1 if it is possible to form a regular bracket sequence by choosing the correct order of the given strings, 0 otherwise. Examples Input 3 1 4 3 Output 1 Input 0 0 0 0 Output 1 Input 1 2 3 4 Output 0 Note In the first example it is possible to construct a string "(())()(()((()()()())))", which is a regular bracket sequence. In the second example it is possible to construct a string "", which is a regular bracket sequence.
instruction
0
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Tags: greedy, implementation Correct Solution: ``` a=int(input()) b=int(input()) c=int(input()) d=int(input()) d=d*2 a=a*2 if a==d==0 and c>0: print(0) exit() c=c%2 if c<=a and c<=d and a==d: print(1) else: print(0) ```
output
1
21,235
21
42,471
Provide tags and a correct Python 3 solution for this coding contest problem. A string is called bracket sequence if it does not contain any characters other than "(" and ")". A bracket sequence is called regular if it it is possible to obtain correct arithmetic expression by inserting characters "+" and "1" into this sequence. For example, "", "(())" and "()()" are regular bracket sequences; "))" and ")((" are bracket sequences (but not regular ones), and "(a)" and "(1)+(1)" are not bracket sequences at all. You have a number of strings; each string is a bracket sequence of length 2. So, overall you have cnt_1 strings "((", cnt_2 strings "()", cnt_3 strings ")(" and cnt_4 strings "))". You want to write all these strings in some order, one after another; after that, you will get a long bracket sequence of length 2(cnt_1 + cnt_2 + cnt_3 + cnt_4). You wonder: is it possible to choose some order of the strings you have such that you will get a regular bracket sequence? Note that you may not remove any characters or strings, and you may not add anything either. Input The input consists of four lines, i-th of them contains one integer cnt_i (0 ≤ cnt_i ≤ 10^9). Output Print one integer: 1 if it is possible to form a regular bracket sequence by choosing the correct order of the given strings, 0 otherwise. Examples Input 3 1 4 3 Output 1 Input 0 0 0 0 Output 1 Input 1 2 3 4 Output 0 Note In the first example it is possible to construct a string "(())()(()((()()()())))", which is a regular bracket sequence. In the second example it is possible to construct a string "", which is a regular bracket sequence.
instruction
0
21,236
21
42,472
Tags: greedy, implementation Correct Solution: ``` a = int(input()) b = int(input()) c = int(input()) d = int(input()) if a == d and (c == 0 or a != 0): print(1) else: print(0) ```
output
1
21,236
21
42,473
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response. A string is called bracket sequence if it does not contain any characters other than "(" and ")". A bracket sequence is called regular if it it is possible to obtain correct arithmetic expression by inserting characters "+" and "1" into this sequence. For example, "", "(())" and "()()" are regular bracket sequences; "))" and ")((" are bracket sequences (but not regular ones), and "(a)" and "(1)+(1)" are not bracket sequences at all. You have a number of strings; each string is a bracket sequence of length 2. So, overall you have cnt_1 strings "((", cnt_2 strings "()", cnt_3 strings ")(" and cnt_4 strings "))". You want to write all these strings in some order, one after another; after that, you will get a long bracket sequence of length 2(cnt_1 + cnt_2 + cnt_3 + cnt_4). You wonder: is it possible to choose some order of the strings you have such that you will get a regular bracket sequence? Note that you may not remove any characters or strings, and you may not add anything either. Input The input consists of four lines, i-th of them contains one integer cnt_i (0 ≤ cnt_i ≤ 10^9). Output Print one integer: 1 if it is possible to form a regular bracket sequence by choosing the correct order of the given strings, 0 otherwise. Examples Input 3 1 4 3 Output 1 Input 0 0 0 0 Output 1 Input 1 2 3 4 Output 0 Note In the first example it is possible to construct a string "(())()(()((()()()())))", which is a regular bracket sequence. In the second example it is possible to construct a string "", which is a regular bracket sequence. Submitted Solution: ``` a = int(input()) b = int(input()) c = int(input()) d = int(input()) s = 1 if (c > 0): if(a > 0) and (d > 0) and (a == d): print(1) else: print(0) else: if a == d: print(1) else: print(0) ```
instruction
0
21,237
21
42,474
Yes
output
1
21,237
21
42,475
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response. A string is called bracket sequence if it does not contain any characters other than "(" and ")". A bracket sequence is called regular if it it is possible to obtain correct arithmetic expression by inserting characters "+" and "1" into this sequence. For example, "", "(())" and "()()" are regular bracket sequences; "))" and ")((" are bracket sequences (but not regular ones), and "(a)" and "(1)+(1)" are not bracket sequences at all. You have a number of strings; each string is a bracket sequence of length 2. So, overall you have cnt_1 strings "((", cnt_2 strings "()", cnt_3 strings ")(" and cnt_4 strings "))". You want to write all these strings in some order, one after another; after that, you will get a long bracket sequence of length 2(cnt_1 + cnt_2 + cnt_3 + cnt_4). You wonder: is it possible to choose some order of the strings you have such that you will get a regular bracket sequence? Note that you may not remove any characters or strings, and you may not add anything either. Input The input consists of four lines, i-th of them contains one integer cnt_i (0 ≤ cnt_i ≤ 10^9). Output Print one integer: 1 if it is possible to form a regular bracket sequence by choosing the correct order of the given strings, 0 otherwise. Examples Input 3 1 4 3 Output 1 Input 0 0 0 0 Output 1 Input 1 2 3 4 Output 0 Note In the first example it is possible to construct a string "(())()(()((()()()())))", which is a regular bracket sequence. In the second example it is possible to construct a string "", which is a regular bracket sequence. Submitted Solution: ``` L = [int(input()) for _ in range(4)] if L[0] == L[3]: if L[0] > 0: print(1) else: if L[2] == 0: print(1) else: print(0) else: print(0) ```
instruction
0
21,238
21
42,476
Yes
output
1
21,238
21
42,477
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response. A string is called bracket sequence if it does not contain any characters other than "(" and ")". A bracket sequence is called regular if it it is possible to obtain correct arithmetic expression by inserting characters "+" and "1" into this sequence. For example, "", "(())" and "()()" are regular bracket sequences; "))" and ")((" are bracket sequences (but not regular ones), and "(a)" and "(1)+(1)" are not bracket sequences at all. You have a number of strings; each string is a bracket sequence of length 2. So, overall you have cnt_1 strings "((", cnt_2 strings "()", cnt_3 strings ")(" and cnt_4 strings "))". You want to write all these strings in some order, one after another; after that, you will get a long bracket sequence of length 2(cnt_1 + cnt_2 + cnt_3 + cnt_4). You wonder: is it possible to choose some order of the strings you have such that you will get a regular bracket sequence? Note that you may not remove any characters or strings, and you may not add anything either. Input The input consists of four lines, i-th of them contains one integer cnt_i (0 ≤ cnt_i ≤ 10^9). Output Print one integer: 1 if it is possible to form a regular bracket sequence by choosing the correct order of the given strings, 0 otherwise. Examples Input 3 1 4 3 Output 1 Input 0 0 0 0 Output 1 Input 1 2 3 4 Output 0 Note In the first example it is possible to construct a string "(())()(()((()()()())))", which is a regular bracket sequence. In the second example it is possible to construct a string "", which is a regular bracket sequence. Submitted Solution: ``` c1=int(input()) c2=int(input()) c3=int(input()) c4=int(input()) if c1==0 and c4==0 and c3!=0: print(0) exit() if c1==0 and c2==0 and c3!=0: print(0) exit() if c1==c4: print(1) exit() else: print(0) exit() ```
instruction
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21,239
21
42,478
Yes
output
1
21,239
21
42,479
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response. A string is called bracket sequence if it does not contain any characters other than "(" and ")". A bracket sequence is called regular if it it is possible to obtain correct arithmetic expression by inserting characters "+" and "1" into this sequence. For example, "", "(())" and "()()" are regular bracket sequences; "))" and ")((" are bracket sequences (but not regular ones), and "(a)" and "(1)+(1)" are not bracket sequences at all. You have a number of strings; each string is a bracket sequence of length 2. So, overall you have cnt_1 strings "((", cnt_2 strings "()", cnt_3 strings ")(" and cnt_4 strings "))". You want to write all these strings in some order, one after another; after that, you will get a long bracket sequence of length 2(cnt_1 + cnt_2 + cnt_3 + cnt_4). You wonder: is it possible to choose some order of the strings you have such that you will get a regular bracket sequence? Note that you may not remove any characters or strings, and you may not add anything either. Input The input consists of four lines, i-th of them contains one integer cnt_i (0 ≤ cnt_i ≤ 10^9). Output Print one integer: 1 if it is possible to form a regular bracket sequence by choosing the correct order of the given strings, 0 otherwise. Examples Input 3 1 4 3 Output 1 Input 0 0 0 0 Output 1 Input 1 2 3 4 Output 0 Note In the first example it is possible to construct a string "(())()(()((()()()())))", which is a regular bracket sequence. In the second example it is possible to construct a string "", which is a regular bracket sequence. Submitted Solution: ``` n=int(input()) m=int(input()) k=int(input()) b=int(input()) if(n==b): if(k!=0 and n==0): print(0) else: print(1) else: print(0) ```
instruction
0
21,240
21
42,480
Yes
output
1
21,240
21
42,481
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response. A string is called bracket sequence if it does not contain any characters other than "(" and ")". A bracket sequence is called regular if it it is possible to obtain correct arithmetic expression by inserting characters "+" and "1" into this sequence. For example, "", "(())" and "()()" are regular bracket sequences; "))" and ")((" are bracket sequences (but not regular ones), and "(a)" and "(1)+(1)" are not bracket sequences at all. You have a number of strings; each string is a bracket sequence of length 2. So, overall you have cnt_1 strings "((", cnt_2 strings "()", cnt_3 strings ")(" and cnt_4 strings "))". You want to write all these strings in some order, one after another; after that, you will get a long bracket sequence of length 2(cnt_1 + cnt_2 + cnt_3 + cnt_4). You wonder: is it possible to choose some order of the strings you have such that you will get a regular bracket sequence? Note that you may not remove any characters or strings, and you may not add anything either. Input The input consists of four lines, i-th of them contains one integer cnt_i (0 ≤ cnt_i ≤ 10^9). Output Print one integer: 1 if it is possible to form a regular bracket sequence by choosing the correct order of the given strings, 0 otherwise. Examples Input 3 1 4 3 Output 1 Input 0 0 0 0 Output 1 Input 1 2 3 4 Output 0 Note In the first example it is possible to construct a string "(())()(()((()()()())))", which is a regular bracket sequence. In the second example it is possible to construct a string "", which is a regular bracket sequence. Submitted Solution: ``` t1 = '((' t2 = '()' t3 = ')(' t4 = '))' cnt1 = int(input()) cnt2 = int(input()) cnt3 = int(input()) cnt4 = int(input()) ans = 1 if cnt1 != cnt4 or (cnt2 > 0 and cnt1 == 0): ans = 0 print(ans) ```
instruction
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21,241
21
42,482
No
output
1
21,241
21
42,483
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response. A string is called bracket sequence if it does not contain any characters other than "(" and ")". A bracket sequence is called regular if it it is possible to obtain correct arithmetic expression by inserting characters "+" and "1" into this sequence. For example, "", "(())" and "()()" are regular bracket sequences; "))" and ")((" are bracket sequences (but not regular ones), and "(a)" and "(1)+(1)" are not bracket sequences at all. You have a number of strings; each string is a bracket sequence of length 2. So, overall you have cnt_1 strings "((", cnt_2 strings "()", cnt_3 strings ")(" and cnt_4 strings "))". You want to write all these strings in some order, one after another; after that, you will get a long bracket sequence of length 2(cnt_1 + cnt_2 + cnt_3 + cnt_4). You wonder: is it possible to choose some order of the strings you have such that you will get a regular bracket sequence? Note that you may not remove any characters or strings, and you may not add anything either. Input The input consists of four lines, i-th of them contains one integer cnt_i (0 ≤ cnt_i ≤ 10^9). Output Print one integer: 1 if it is possible to form a regular bracket sequence by choosing the correct order of the given strings, 0 otherwise. Examples Input 3 1 4 3 Output 1 Input 0 0 0 0 Output 1 Input 1 2 3 4 Output 0 Note In the first example it is possible to construct a string "(())()(()((()()()())))", which is a regular bracket sequence. In the second example it is possible to construct a string "", which is a regular bracket sequence. Submitted Solution: ``` def main(): cnt1 = int(input()) cnt2 = int(input()) cnt3 = int(input()) cnt4 = int(input()) x = cnt1 + cnt2 + cnt3 y = cnt2 + cnt3 + cnt4 if x == y: print(1) else: print(0) if __name__ == "__main__": main() ```
instruction
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21,242
21
42,484
No
output
1
21,242
21
42,485
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response. A string is called bracket sequence if it does not contain any characters other than "(" and ")". A bracket sequence is called regular if it it is possible to obtain correct arithmetic expression by inserting characters "+" and "1" into this sequence. For example, "", "(())" and "()()" are regular bracket sequences; "))" and ")((" are bracket sequences (but not regular ones), and "(a)" and "(1)+(1)" are not bracket sequences at all. You have a number of strings; each string is a bracket sequence of length 2. So, overall you have cnt_1 strings "((", cnt_2 strings "()", cnt_3 strings ")(" and cnt_4 strings "))". You want to write all these strings in some order, one after another; after that, you will get a long bracket sequence of length 2(cnt_1 + cnt_2 + cnt_3 + cnt_4). You wonder: is it possible to choose some order of the strings you have such that you will get a regular bracket sequence? Note that you may not remove any characters or strings, and you may not add anything either. Input The input consists of four lines, i-th of them contains one integer cnt_i (0 ≤ cnt_i ≤ 10^9). Output Print one integer: 1 if it is possible to form a regular bracket sequence by choosing the correct order of the given strings, 0 otherwise. Examples Input 3 1 4 3 Output 1 Input 0 0 0 0 Output 1 Input 1 2 3 4 Output 0 Note In the first example it is possible to construct a string "(())()(()((()()()())))", which is a regular bracket sequence. In the second example it is possible to construct a string "", which is a regular bracket sequence. Submitted Solution: ``` a = int(input()) b = int(input()) c = int(input()) d = int(input()) if b != 0 and a==d and a != 0: print(1) elif b == 0 and a == d: print(1) else: print(0) ```
instruction
0
21,243
21
42,486
No
output
1
21,243
21
42,487
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response. A string is called bracket sequence if it does not contain any characters other than "(" and ")". A bracket sequence is called regular if it it is possible to obtain correct arithmetic expression by inserting characters "+" and "1" into this sequence. For example, "", "(())" and "()()" are regular bracket sequences; "))" and ")((" are bracket sequences (but not regular ones), and "(a)" and "(1)+(1)" are not bracket sequences at all. You have a number of strings; each string is a bracket sequence of length 2. So, overall you have cnt_1 strings "((", cnt_2 strings "()", cnt_3 strings ")(" and cnt_4 strings "))". You want to write all these strings in some order, one after another; after that, you will get a long bracket sequence of length 2(cnt_1 + cnt_2 + cnt_3 + cnt_4). You wonder: is it possible to choose some order of the strings you have such that you will get a regular bracket sequence? Note that you may not remove any characters or strings, and you may not add anything either. Input The input consists of four lines, i-th of them contains one integer cnt_i (0 ≤ cnt_i ≤ 10^9). Output Print one integer: 1 if it is possible to form a regular bracket sequence by choosing the correct order of the given strings, 0 otherwise. Examples Input 3 1 4 3 Output 1 Input 0 0 0 0 Output 1 Input 1 2 3 4 Output 0 Note In the first example it is possible to construct a string "(())()(()((()()()())))", which is a regular bracket sequence. In the second example it is possible to construct a string "", which is a regular bracket sequence. Submitted Solution: ``` a = int(input()) b = int(input()) c = int(input()) d = int(input()) if a == d : if a > 0 and d > 0 or a == 0 and b == 0 and c == 0: print(1) else: print(0) else: print(0) ```
instruction
0
21,244
21
42,488
No
output
1
21,244
21
42,489
Provide tags and a correct Python 3 solution for this coding contest problem. Neko is playing with his toys on the backyard of Aki's house. Aki decided to play a prank on him, by secretly putting catnip into Neko's toys. Unfortunately, he went overboard and put an entire bag of catnip into the toys... It took Neko an entire day to turn back to normal. Neko reported to Aki that he saw a lot of weird things, including a [trie](https://en.wikipedia.org/wiki/Trie) of all correct bracket sequences of length 2n. The definition of correct bracket sequence is as follows: * The empty sequence is a correct bracket sequence, * If s is a correct bracket sequence, then (\,s ) is a correct bracket sequence, * If s and t are a correct bracket sequence, then st is also a correct bracket sequence. For example, the strings "(())", "()()" form a correct bracket sequence, while ")(" and "((" not. Aki then came up with an interesting problem: What is the size of the maximum matching (the largest set of edges such that there are no two edges with a common vertex) in this trie? Since the answer can be quite large, print it modulo 10^9 + 7. Input The only line contains a single integer n (1 ≤ n ≤ 1000). Output Print exactly one integer — the size of the maximum matching in the trie. Since the answer can be quite large, print it modulo 10^9 + 7. Examples Input 1 Output 1 Input 2 Output 3 Input 3 Output 9 Note The pictures below illustrate tries in the first two examples (for clarity, the round brackets are replaced with angle brackets). The maximum matching is highlighted with blue. <image> <image>
instruction
0
21,245
21
42,490
Tags: dp, greedy, trees Correct Solution: ``` n = int(input()) mod = 10**9+7 dp = [[0]*(n+1) for i in range(2*n+1)] dp[0][0] = 1 for d in range(1, 2*n+1): for v in range(n+1): if v+1 <= n: dp[d][v+1] += dp[d-1][v] dp[d][v+1] %= mod if v-1 >= 0: dp[d][v-1] += dp[d-1][v] dp[d][v-1] %= mod ans = 0 for d in range(2*n+1): if d%2 == 1: for v in range(n+1): if v <= 2*n-d: ans += dp[d][v] ans %= mod print(ans) ```
output
1
21,245
21
42,491
Provide tags and a correct Python 3 solution for this coding contest problem. Neko is playing with his toys on the backyard of Aki's house. Aki decided to play a prank on him, by secretly putting catnip into Neko's toys. Unfortunately, he went overboard and put an entire bag of catnip into the toys... It took Neko an entire day to turn back to normal. Neko reported to Aki that he saw a lot of weird things, including a [trie](https://en.wikipedia.org/wiki/Trie) of all correct bracket sequences of length 2n. The definition of correct bracket sequence is as follows: * The empty sequence is a correct bracket sequence, * If s is a correct bracket sequence, then (\,s ) is a correct bracket sequence, * If s and t are a correct bracket sequence, then st is also a correct bracket sequence. For example, the strings "(())", "()()" form a correct bracket sequence, while ")(" and "((" not. Aki then came up with an interesting problem: What is the size of the maximum matching (the largest set of edges such that there are no two edges with a common vertex) in this trie? Since the answer can be quite large, print it modulo 10^9 + 7. Input The only line contains a single integer n (1 ≤ n ≤ 1000). Output Print exactly one integer — the size of the maximum matching in the trie. Since the answer can be quite large, print it modulo 10^9 + 7. Examples Input 1 Output 1 Input 2 Output 3 Input 3 Output 9 Note The pictures below illustrate tries in the first two examples (for clarity, the round brackets are replaced with angle brackets). The maximum matching is highlighted with blue. <image> <image>
instruction
0
21,246
21
42,492
Tags: dp, greedy, trees Correct Solution: ``` mod = 1000000007 n = int(input()) cat = [0] * 1100 f, c, i = 1, 1, 1 while i < 1100: cat[i] = f i += 1 c = c * (8*i - 12) // i f = c - f cat = cat[1:-1] sm = 0 for i in range(3, n + 3): sm += (cat[i - 1] + (-1) ** (i - 1)) // (1 << i) print(sm % mod) ```
output
1
21,246
21
42,493
Provide tags and a correct Python 3 solution for this coding contest problem. Neko is playing with his toys on the backyard of Aki's house. Aki decided to play a prank on him, by secretly putting catnip into Neko's toys. Unfortunately, he went overboard and put an entire bag of catnip into the toys... It took Neko an entire day to turn back to normal. Neko reported to Aki that he saw a lot of weird things, including a [trie](https://en.wikipedia.org/wiki/Trie) of all correct bracket sequences of length 2n. The definition of correct bracket sequence is as follows: * The empty sequence is a correct bracket sequence, * If s is a correct bracket sequence, then (\,s ) is a correct bracket sequence, * If s and t are a correct bracket sequence, then st is also a correct bracket sequence. For example, the strings "(())", "()()" form a correct bracket sequence, while ")(" and "((" not. Aki then came up with an interesting problem: What is the size of the maximum matching (the largest set of edges such that there are no two edges with a common vertex) in this trie? Since the answer can be quite large, print it modulo 10^9 + 7. Input The only line contains a single integer n (1 ≤ n ≤ 1000). Output Print exactly one integer — the size of the maximum matching in the trie. Since the answer can be quite large, print it modulo 10^9 + 7. Examples Input 1 Output 1 Input 2 Output 3 Input 3 Output 9 Note The pictures below illustrate tries in the first two examples (for clarity, the round brackets are replaced with angle brackets). The maximum matching is highlighted with blue. <image> <image>
instruction
0
21,247
21
42,494
Tags: dp, greedy, trees Correct Solution: ``` def fine(): f, c, n = 1, 1, 1 yield 0 while True: yield f n += 1 c = c * (4 * n - 6) // n f = (c - f) // 2 f = fine() n = int(input()) print((sum(next(f) for _ in range(n + 3)) - 1) % (10**9 + 7)) ```
output
1
21,247
21
42,495
Provide tags and a correct Python 3 solution for this coding contest problem. Neko is playing with his toys on the backyard of Aki's house. Aki decided to play a prank on him, by secretly putting catnip into Neko's toys. Unfortunately, he went overboard and put an entire bag of catnip into the toys... It took Neko an entire day to turn back to normal. Neko reported to Aki that he saw a lot of weird things, including a [trie](https://en.wikipedia.org/wiki/Trie) of all correct bracket sequences of length 2n. The definition of correct bracket sequence is as follows: * The empty sequence is a correct bracket sequence, * If s is a correct bracket sequence, then (\,s ) is a correct bracket sequence, * If s and t are a correct bracket sequence, then st is also a correct bracket sequence. For example, the strings "(())", "()()" form a correct bracket sequence, while ")(" and "((" not. Aki then came up with an interesting problem: What is the size of the maximum matching (the largest set of edges such that there are no two edges with a common vertex) in this trie? Since the answer can be quite large, print it modulo 10^9 + 7. Input The only line contains a single integer n (1 ≤ n ≤ 1000). Output Print exactly one integer — the size of the maximum matching in the trie. Since the answer can be quite large, print it modulo 10^9 + 7. Examples Input 1 Output 1 Input 2 Output 3 Input 3 Output 9 Note The pictures below illustrate tries in the first two examples (for clarity, the round brackets are replaced with angle brackets). The maximum matching is highlighted with blue. <image> <image>
instruction
0
21,248
21
42,496
Tags: dp, greedy, trees Correct Solution: ``` from itertools import accumulate N = int(input()) arr = [1] res = 1 for i in range(2,N+1): arr = list(accumulate(arr)) arr = arr + [arr[-1]] d = i+1 #print(arr) s = 0 for i in range(len(arr)): s += arr[i]*(d//2) d -= 1 res += s res = res % 1000000007 #print(res) print(res) ```
output
1
21,248
21
42,497
Provide tags and a correct Python 3 solution for this coding contest problem. Neko is playing with his toys on the backyard of Aki's house. Aki decided to play a prank on him, by secretly putting catnip into Neko's toys. Unfortunately, he went overboard and put an entire bag of catnip into the toys... It took Neko an entire day to turn back to normal. Neko reported to Aki that he saw a lot of weird things, including a [trie](https://en.wikipedia.org/wiki/Trie) of all correct bracket sequences of length 2n. The definition of correct bracket sequence is as follows: * The empty sequence is a correct bracket sequence, * If s is a correct bracket sequence, then (\,s ) is a correct bracket sequence, * If s and t are a correct bracket sequence, then st is also a correct bracket sequence. For example, the strings "(())", "()()" form a correct bracket sequence, while ")(" and "((" not. Aki then came up with an interesting problem: What is the size of the maximum matching (the largest set of edges such that there are no two edges with a common vertex) in this trie? Since the answer can be quite large, print it modulo 10^9 + 7. Input The only line contains a single integer n (1 ≤ n ≤ 1000). Output Print exactly one integer — the size of the maximum matching in the trie. Since the answer can be quite large, print it modulo 10^9 + 7. Examples Input 1 Output 1 Input 2 Output 3 Input 3 Output 9 Note The pictures below illustrate tries in the first two examples (for clarity, the round brackets are replaced with angle brackets). The maximum matching is highlighted with blue. <image> <image>
instruction
0
21,249
21
42,498
Tags: dp, greedy, trees Correct Solution: ``` n = int(input()) board = [[0 for i in range(n + 1)] for j in range(2 * n + 1)] board[0][0] = 1 for i in range(1, n): for j in range(len(board[0])): if j > 0: board[i][j-1] += board[i-1][j] if j + 1 < len(board[0]) and j < 2 * n - i: board[i][j+1] += board[i-1][j] for i in range(n, 2 * n + 1): for j in range(len(board[0])): if j > 0: board[i][j-1] += board[i-1][j] #board[i][j-1] %= 1000000007 if j + 1 < len(board[0]) and j < 2 * n - i: board[i][j+1] += board[i-1][j] #board[i][j+1] %= 1000000007 ans = 0 for i in range(1, len(board), 2): ans += sum(board[i]) ans %= 1000000007 print(ans) ```
output
1
21,249
21
42,499
Provide tags and a correct Python 3 solution for this coding contest problem. Neko is playing with his toys on the backyard of Aki's house. Aki decided to play a prank on him, by secretly putting catnip into Neko's toys. Unfortunately, he went overboard and put an entire bag of catnip into the toys... It took Neko an entire day to turn back to normal. Neko reported to Aki that he saw a lot of weird things, including a [trie](https://en.wikipedia.org/wiki/Trie) of all correct bracket sequences of length 2n. The definition of correct bracket sequence is as follows: * The empty sequence is a correct bracket sequence, * If s is a correct bracket sequence, then (\,s ) is a correct bracket sequence, * If s and t are a correct bracket sequence, then st is also a correct bracket sequence. For example, the strings "(())", "()()" form a correct bracket sequence, while ")(" and "((" not. Aki then came up with an interesting problem: What is the size of the maximum matching (the largest set of edges such that there are no two edges with a common vertex) in this trie? Since the answer can be quite large, print it modulo 10^9 + 7. Input The only line contains a single integer n (1 ≤ n ≤ 1000). Output Print exactly one integer — the size of the maximum matching in the trie. Since the answer can be quite large, print it modulo 10^9 + 7. Examples Input 1 Output 1 Input 2 Output 3 Input 3 Output 9 Note The pictures below illustrate tries in the first two examples (for clarity, the round brackets are replaced with angle brackets). The maximum matching is highlighted with blue. <image> <image>
instruction
0
21,250
21
42,500
Tags: dp, greedy, trees Correct Solution: ``` n = int(input()) f = [[0]*(n+1) for i in range(n+1)] g = [[0]*(n+1) for i in range(n+1)] mod = 10**9+7 for i in range(1, n + 1): f[0][i] = g[0][i - 1] g[0][i] = f[0][i - 1] + 1 t = [0, 0] for i in range(1, n + 1): for j in range(i, n + 1): if i > 0: f[i][j] = g[i - 1][j] t[0] = g[i - 1][j] t[1] = f[i - 1][j] + 1 if j > i: f[i][j] = (f[i][j] + g[i][j-1]) % mod t[0] = (t[0] + f[i][j - 1] + 1) % mod t[1] = (t[1] + g[i][j - 1]) % mod; for k in t: g[i][j] = max(g[i][j], k) print(g[n][n]%mod) ```
output
1
21,250
21
42,501
Provide tags and a correct Python 3 solution for this coding contest problem. Neko is playing with his toys on the backyard of Aki's house. Aki decided to play a prank on him, by secretly putting catnip into Neko's toys. Unfortunately, he went overboard and put an entire bag of catnip into the toys... It took Neko an entire day to turn back to normal. Neko reported to Aki that he saw a lot of weird things, including a [trie](https://en.wikipedia.org/wiki/Trie) of all correct bracket sequences of length 2n. The definition of correct bracket sequence is as follows: * The empty sequence is a correct bracket sequence, * If s is a correct bracket sequence, then (\,s ) is a correct bracket sequence, * If s and t are a correct bracket sequence, then st is also a correct bracket sequence. For example, the strings "(())", "()()" form a correct bracket sequence, while ")(" and "((" not. Aki then came up with an interesting problem: What is the size of the maximum matching (the largest set of edges such that there are no two edges with a common vertex) in this trie? Since the answer can be quite large, print it modulo 10^9 + 7. Input The only line contains a single integer n (1 ≤ n ≤ 1000). Output Print exactly one integer — the size of the maximum matching in the trie. Since the answer can be quite large, print it modulo 10^9 + 7. Examples Input 1 Output 1 Input 2 Output 3 Input 3 Output 9 Note The pictures below illustrate tries in the first two examples (for clarity, the round brackets are replaced with angle brackets). The maximum matching is highlighted with blue. <image> <image>
instruction
0
21,251
21
42,502
Tags: dp, greedy, trees Correct Solution: ``` n = int(input()) dp = [[[0] * (n+1) for i in range(n + 1)] for j in range(2)] dp[0][0][0] = 0 dp[1][0][0] = 0 mod = 10 ** 9 + 7 for i in range(n + 1): for j in range(i, n + 1): if i == 0 and j == 0: continue dp[0][i][j] = (dp[1][i - 1][j] + dp[1][i][j - 1]) % mod tmp1 = 1 if i - 1 <= j and i > 0: if i <= j - 1: tmp1 += dp[1][i][j - 1] tmp1 %= mod tmp1 += dp[0][i - 1][j] tmp1 %= mod tmp2 = 1 if i <= j - 1 and j > 0: if i - 1 <= j: tmp2 += dp[1][i - 1][j] tmp2 %= mod tmp2 += dp[0][i][j - 1] tmp2 %= mod # # dp[i][j][1] = max( # (((dp[i - 1][j][0]) + (dp[i][j - 1][1] if i <= j - 1 else 0)) if i - 1 <= j and i > 0 else 0) + 1, # (((dp[i - 1][j][1] if i - 1 <= j else 0) + (dp[i][j - 1][0])) if i <= j - 1 and j > 0 else 0) + 1 # ) % mod dp[1][i][j] = max(tmp1, tmp2) % mod # for i in range(n + 1): # for j in range(n + 1): # print(i, j, dp[i][j][0], dp[i][j][1]) print(max(dp[0][-1][-1], dp[1][-1][-1]) % mod) # print(dp) ```
output
1
21,251
21
42,503
Provide tags and a correct Python 3 solution for this coding contest problem. Neko is playing with his toys on the backyard of Aki's house. Aki decided to play a prank on him, by secretly putting catnip into Neko's toys. Unfortunately, he went overboard and put an entire bag of catnip into the toys... It took Neko an entire day to turn back to normal. Neko reported to Aki that he saw a lot of weird things, including a [trie](https://en.wikipedia.org/wiki/Trie) of all correct bracket sequences of length 2n. The definition of correct bracket sequence is as follows: * The empty sequence is a correct bracket sequence, * If s is a correct bracket sequence, then (\,s ) is a correct bracket sequence, * If s and t are a correct bracket sequence, then st is also a correct bracket sequence. For example, the strings "(())", "()()" form a correct bracket sequence, while ")(" and "((" not. Aki then came up with an interesting problem: What is the size of the maximum matching (the largest set of edges such that there are no two edges with a common vertex) in this trie? Since the answer can be quite large, print it modulo 10^9 + 7. Input The only line contains a single integer n (1 ≤ n ≤ 1000). Output Print exactly one integer — the size of the maximum matching in the trie. Since the answer can be quite large, print it modulo 10^9 + 7. Examples Input 1 Output 1 Input 2 Output 3 Input 3 Output 9 Note The pictures below illustrate tries in the first two examples (for clarity, the round brackets are replaced with angle brackets). The maximum matching is highlighted with blue. <image> <image>
instruction
0
21,252
21
42,504
Tags: dp, greedy, trees Correct Solution: ``` import collections import random import heapq import bisect import math import time class Solution2: def solve(self, A1, A2): pass class Solution: def gcd(self, a, b): if not b: return a return self.gcd(b, a%b) def lcm(self, a, b): return b*a//self.gcd(b,a) def solve(self, n): out = 0 MOD = 1000000007 curr = {0 : [1,0]} for i in range(2*n): new_curr = {} for pos, (ways_clear, ways_taken) in curr.items(): ways_clear = ways_clear%MOD ways_taken = ways_taken%MOD if pos: if pos-1 not in new_curr: new_curr[pos-1] = [0,0] new_curr[pos-1][0] += ways_taken new_curr[pos-1][1] += ways_clear out += ways_clear if pos < 2*n - i: if pos+1 not in new_curr: new_curr[pos+1] = [0,0] new_curr[pos+1][0] += ways_taken new_curr[pos+1][1] += ways_clear out += ways_clear curr = new_curr #print(curr) return out%MOD sol = Solution() sol2 = Solution2() #TT = int(input()) for test_case in range(1): N = input() #a = [int(c) for c in input().split()] #b = [int(c) for c in input().split()] out = sol.solve(int(N)) print(str(out)) #print(str(out)) # out2 = sol2.solve(s) # for _ in range(100000): # rand = [random.randrange(60) for _ in range(10)] # out1 = sol.solve(rand) # out2 = sol2.solve(rand) # if out1 != out2: # print(rand, out1, out2) # break ```
output
1
21,252
21
42,505
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response. Neko is playing with his toys on the backyard of Aki's house. Aki decided to play a prank on him, by secretly putting catnip into Neko's toys. Unfortunately, he went overboard and put an entire bag of catnip into the toys... It took Neko an entire day to turn back to normal. Neko reported to Aki that he saw a lot of weird things, including a [trie](https://en.wikipedia.org/wiki/Trie) of all correct bracket sequences of length 2n. The definition of correct bracket sequence is as follows: * The empty sequence is a correct bracket sequence, * If s is a correct bracket sequence, then (\,s ) is a correct bracket sequence, * If s and t are a correct bracket sequence, then st is also a correct bracket sequence. For example, the strings "(())", "()()" form a correct bracket sequence, while ")(" and "((" not. Aki then came up with an interesting problem: What is the size of the maximum matching (the largest set of edges such that there are no two edges with a common vertex) in this trie? Since the answer can be quite large, print it modulo 10^9 + 7. Input The only line contains a single integer n (1 ≤ n ≤ 1000). Output Print exactly one integer — the size of the maximum matching in the trie. Since the answer can be quite large, print it modulo 10^9 + 7. Examples Input 1 Output 1 Input 2 Output 3 Input 3 Output 9 Note The pictures below illustrate tries in the first two examples (for clarity, the round brackets are replaced with angle brackets). The maximum matching is highlighted with blue. <image> <image> Submitted Solution: ``` n = int(input()) dp = [[0 for i in range(2005)] for j in range(2005)] #length and '(' - ')' ans, mod = 0, (10 ** 9) + 7 dp[0][0] = 1 for i in range(2 * n): for j in range(2 * n): dp[i + 1][j + 1] += dp[i][j] #when '(' is appended if dp[i + 1][j + 1] >= mod: dp[i + 1][j + 1] -= mod if j > 0: dp[i + 1][j - 1] += dp[i][j] #when ')' is appended if dp[i + 1][j - 1] >= mod: dp[i + 1][j - 1] -= mod for i in range(1, 2 * n, 2): for j in range(2 * n): valid = i + j <= 2 * n if valid: ans += dp[i][j] if ans >= mod: ans -= mod print(ans % mod) ```
instruction
0
21,253
21
42,506
Yes
output
1
21,253
21
42,507
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response. Neko is playing with his toys on the backyard of Aki's house. Aki decided to play a prank on him, by secretly putting catnip into Neko's toys. Unfortunately, he went overboard and put an entire bag of catnip into the toys... It took Neko an entire day to turn back to normal. Neko reported to Aki that he saw a lot of weird things, including a [trie](https://en.wikipedia.org/wiki/Trie) of all correct bracket sequences of length 2n. The definition of correct bracket sequence is as follows: * The empty sequence is a correct bracket sequence, * If s is a correct bracket sequence, then (\,s ) is a correct bracket sequence, * If s and t are a correct bracket sequence, then st is also a correct bracket sequence. For example, the strings "(())", "()()" form a correct bracket sequence, while ")(" and "((" not. Aki then came up with an interesting problem: What is the size of the maximum matching (the largest set of edges such that there are no two edges with a common vertex) in this trie? Since the answer can be quite large, print it modulo 10^9 + 7. Input The only line contains a single integer n (1 ≤ n ≤ 1000). Output Print exactly one integer — the size of the maximum matching in the trie. Since the answer can be quite large, print it modulo 10^9 + 7. Examples Input 1 Output 1 Input 2 Output 3 Input 3 Output 9 Note The pictures below illustrate tries in the first two examples (for clarity, the round brackets are replaced with angle brackets). The maximum matching is highlighted with blue. <image> <image> Submitted Solution: ``` N = int(input()) MOD = 10**9 + 7 dp = [[0] * (2020 + 1) for _ in range(2020 + 1)] dp[1][1] = 1 ans = 0 for i in range(2, N + 2): for j in range(1, i + 1): dp[i][j] = (dp[i - 1][j] + dp[i][j - 1]) % MOD if (i + j) % 2 == 1: ans = (ans + dp[i][j]) % MOD print(ans) ```
instruction
0
21,254
21
42,508
Yes
output
1
21,254
21
42,509
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response. Neko is playing with his toys on the backyard of Aki's house. Aki decided to play a prank on him, by secretly putting catnip into Neko's toys. Unfortunately, he went overboard and put an entire bag of catnip into the toys... It took Neko an entire day to turn back to normal. Neko reported to Aki that he saw a lot of weird things, including a [trie](https://en.wikipedia.org/wiki/Trie) of all correct bracket sequences of length 2n. The definition of correct bracket sequence is as follows: * The empty sequence is a correct bracket sequence, * If s is a correct bracket sequence, then (\,s ) is a correct bracket sequence, * If s and t are a correct bracket sequence, then st is also a correct bracket sequence. For example, the strings "(())", "()()" form a correct bracket sequence, while ")(" and "((" not. Aki then came up with an interesting problem: What is the size of the maximum matching (the largest set of edges such that there are no two edges with a common vertex) in this trie? Since the answer can be quite large, print it modulo 10^9 + 7. Input The only line contains a single integer n (1 ≤ n ≤ 1000). Output Print exactly one integer — the size of the maximum matching in the trie. Since the answer can be quite large, print it modulo 10^9 + 7. Examples Input 1 Output 1 Input 2 Output 3 Input 3 Output 9 Note The pictures below illustrate tries in the first two examples (for clarity, the round brackets are replaced with angle brackets). The maximum matching is highlighted with blue. <image> <image> Submitted Solution: ``` u = 1000000007 import sys def P(n): #take first edge or no? depth == < - > X = [[0 for i in range(2*n+1)] for j in range(2*n+1)] X[0][0] = 1 r = 0 for i in range(1,2*n+1): X[i][0]=X[i-1][1] for j in range(1,min([2*n+1-i,i+1])): X[i][j] = (X[i-1][j+1]+X[i-1][j-1])%u if i%2:r+=sum(X[i]) return r%u print(P(int(sys.stdin.read()[:-1]))) ```
instruction
0
21,255
21
42,510
Yes
output
1
21,255
21
42,511
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response. Neko is playing with his toys on the backyard of Aki's house. Aki decided to play a prank on him, by secretly putting catnip into Neko's toys. Unfortunately, he went overboard and put an entire bag of catnip into the toys... It took Neko an entire day to turn back to normal. Neko reported to Aki that he saw a lot of weird things, including a [trie](https://en.wikipedia.org/wiki/Trie) of all correct bracket sequences of length 2n. The definition of correct bracket sequence is as follows: * The empty sequence is a correct bracket sequence, * If s is a correct bracket sequence, then (\,s ) is a correct bracket sequence, * If s and t are a correct bracket sequence, then st is also a correct bracket sequence. For example, the strings "(())", "()()" form a correct bracket sequence, while ")(" and "((" not. Aki then came up with an interesting problem: What is the size of the maximum matching (the largest set of edges such that there are no two edges with a common vertex) in this trie? Since the answer can be quite large, print it modulo 10^9 + 7. Input The only line contains a single integer n (1 ≤ n ≤ 1000). Output Print exactly one integer — the size of the maximum matching in the trie. Since the answer can be quite large, print it modulo 10^9 + 7. Examples Input 1 Output 1 Input 2 Output 3 Input 3 Output 9 Note The pictures below illustrate tries in the first two examples (for clarity, the round brackets are replaced with angle brackets). The maximum matching is highlighted with blue. <image> <image> Submitted Solution: ``` l=[1, 2, 6, 18, 57, 186, 622, 2120, 7338, 25724, 91144, 325878, 1174281, 4260282, 15548694, 57048048, 210295326, 778483932, 892818230, 786724318, 347919346, 355845955, 274146173, 336110626, 508416482, 521006399, 214448866, 997593411, 238896999, 849258736, 819525514, 53182350, 202970081, 945166442, 598349475, 683882441, 92894058, 452668785, 75136294, 196675923, 119906583, 942355177, 971081806, 117114980, 360237618, 942712231, 137580544, 313883953, 475793244, 854651725, 734277125, 39095249, 461071198, 801159097, 64968190, 936719156, 886161553, 639551605, 781314451, 681340193, 107129321, 680026807, 103559391, 682186676, 827551267, 510628237, 427135789, 600593159, 542474853, 821284002, 34030966, 212108296, 552111064, 883592263, 580458832, 835540036, 126388564, 22628138, 362655432, 628650076, 577876876, 73283345, 404323915, 228609761, 734246007, 503315696, 474786457, 786621025, 519844441, 126622248, 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644577760, 559548335, 226477730, 804947440, 854802118, 212773938, 369618916, 630074277, 683336252, 733742391, 689759642, 536324741, 12232850, 588395062, 311834902, 822700906, 918264540, 383029467, 838692046, 779820735, 231894781, 127968797, 146143427, 514318905, 780214430, 532924900, 362245605, 616904934, 159952118, 974181243, 464233843, 638310707, 136785681, 650317857, 735148405, 813970111, 346354503, 450347746, 226170708, 949341109, 686265960, 457248555, 802620126, 406473234, 620525961, 499658652, 996948192, 432517005, 531544188, 736310230, 364121646, 51128023, 702339436, 491727676, 778790268, 991720904, 233882858, 447952019, 967158671, 131392770, 249020049, 17082876, 698852882, 940026909, 688883348, 292899252, 464708546, 689894359, 844446056, 660596379, 835026416, 434755394, 40833306, 988981175, 833869084, 148016296, 289878633, 691707616, 661734006, 227811471, 658581356, 651921475, 899658740, 598341579, 491077042, 523084223, 76743680, 345023872, 817961350, 668829460, 244890498, 889263641, 690084579, 789381194, 684971088, 529646577, 281941575, 866121528, 672773016] n=int(input()) print(sum(l[:n])%1000000007) ```
instruction
0
21,256
21
42,512
Yes
output
1
21,256
21
42,513
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response. Neko is playing with his toys on the backyard of Aki's house. Aki decided to play a prank on him, by secretly putting catnip into Neko's toys. Unfortunately, he went overboard and put an entire bag of catnip into the toys... It took Neko an entire day to turn back to normal. Neko reported to Aki that he saw a lot of weird things, including a [trie](https://en.wikipedia.org/wiki/Trie) of all correct bracket sequences of length 2n. The definition of correct bracket sequence is as follows: * The empty sequence is a correct bracket sequence, * If s is a correct bracket sequence, then (\,s ) is a correct bracket sequence, * If s and t are a correct bracket sequence, then st is also a correct bracket sequence. For example, the strings "(())", "()()" form a correct bracket sequence, while ")(" and "((" not. Aki then came up with an interesting problem: What is the size of the maximum matching (the largest set of edges such that there are no two edges with a common vertex) in this trie? Since the answer can be quite large, print it modulo 10^9 + 7. Input The only line contains a single integer n (1 ≤ n ≤ 1000). Output Print exactly one integer — the size of the maximum matching in the trie. Since the answer can be quite large, print it modulo 10^9 + 7. Examples Input 1 Output 1 Input 2 Output 3 Input 3 Output 9 Note The pictures below illustrate tries in the first two examples (for clarity, the round brackets are replaced with angle brackets). The maximum matching is highlighted with blue. <image> <image> Submitted Solution: ``` from math import factorial M = 10 ** 9 + 7 def C(n, k): return (factorial(n) // factorial(k) // factorial(n - k)) % M n = int(input()) if n != 5: print((2 * (C(2 * n, n) // (n + 1)) - 1) % M) else: print(84) ```
instruction
0
21,257
21
42,514
No
output
1
21,257
21
42,515
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response. Neko is playing with his toys on the backyard of Aki's house. Aki decided to play a prank on him, by secretly putting catnip into Neko's toys. Unfortunately, he went overboard and put an entire bag of catnip into the toys... It took Neko an entire day to turn back to normal. Neko reported to Aki that he saw a lot of weird things, including a [trie](https://en.wikipedia.org/wiki/Trie) of all correct bracket sequences of length 2n. The definition of correct bracket sequence is as follows: * The empty sequence is a correct bracket sequence, * If s is a correct bracket sequence, then (\,s ) is a correct bracket sequence, * If s and t are a correct bracket sequence, then st is also a correct bracket sequence. For example, the strings "(())", "()()" form a correct bracket sequence, while ")(" and "((" not. Aki then came up with an interesting problem: What is the size of the maximum matching (the largest set of edges such that there are no two edges with a common vertex) in this trie? Since the answer can be quite large, print it modulo 10^9 + 7. Input The only line contains a single integer n (1 ≤ n ≤ 1000). Output Print exactly one integer — the size of the maximum matching in the trie. Since the answer can be quite large, print it modulo 10^9 + 7. Examples Input 1 Output 1 Input 2 Output 3 Input 3 Output 9 Note The pictures below illustrate tries in the first two examples (for clarity, the round brackets are replaced with angle brackets). The maximum matching is highlighted with blue. <image> <image> Submitted Solution: ``` MOD = 10 ** 9 + 7 n = int(input()) def C(x, n): res = 1 for i in range(x): res = (res * (n - i) * pow(i + 1, MOD - 2, MOD)) % MOD return res print((2 * C(n, 2 * n) - 2 * C(n - 1, 2 * n) - 1) % MOD) ```
instruction
0
21,258
21
42,516
No
output
1
21,258
21
42,517
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response. Neko is playing with his toys on the backyard of Aki's house. Aki decided to play a prank on him, by secretly putting catnip into Neko's toys. Unfortunately, he went overboard and put an entire bag of catnip into the toys... It took Neko an entire day to turn back to normal. Neko reported to Aki that he saw a lot of weird things, including a [trie](https://en.wikipedia.org/wiki/Trie) of all correct bracket sequences of length 2n. The definition of correct bracket sequence is as follows: * The empty sequence is a correct bracket sequence, * If s is a correct bracket sequence, then (\,s ) is a correct bracket sequence, * If s and t are a correct bracket sequence, then st is also a correct bracket sequence. For example, the strings "(())", "()()" form a correct bracket sequence, while ")(" and "((" not. Aki then came up with an interesting problem: What is the size of the maximum matching (the largest set of edges such that there are no two edges with a common vertex) in this trie? Since the answer can be quite large, print it modulo 10^9 + 7. Input The only line contains a single integer n (1 ≤ n ≤ 1000). Output Print exactly one integer — the size of the maximum matching in the trie. Since the answer can be quite large, print it modulo 10^9 + 7. Examples Input 1 Output 1 Input 2 Output 3 Input 3 Output 9 Note The pictures below illustrate tries in the first two examples (for clarity, the round brackets are replaced with angle brackets). The maximum matching is highlighted with blue. <image> <image> Submitted Solution: ``` from math import factorial M = 10 ** 9 + 7 def C(n, k): return (factorial(n) // factorial(k) // factorial(n - k)) % M n = int(input()) if n != 5: print((2 * (C(2 * n, n) // (n + 1)) - 1) % M) else: print(82) ```
instruction
0
21,259
21
42,518
No
output
1
21,259
21
42,519
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response. Neko is playing with his toys on the backyard of Aki's house. Aki decided to play a prank on him, by secretly putting catnip into Neko's toys. Unfortunately, he went overboard and put an entire bag of catnip into the toys... It took Neko an entire day to turn back to normal. Neko reported to Aki that he saw a lot of weird things, including a [trie](https://en.wikipedia.org/wiki/Trie) of all correct bracket sequences of length 2n. The definition of correct bracket sequence is as follows: * The empty sequence is a correct bracket sequence, * If s is a correct bracket sequence, then (\,s ) is a correct bracket sequence, * If s and t are a correct bracket sequence, then st is also a correct bracket sequence. For example, the strings "(())", "()()" form a correct bracket sequence, while ")(" and "((" not. Aki then came up with an interesting problem: What is the size of the maximum matching (the largest set of edges such that there are no two edges with a common vertex) in this trie? Since the answer can be quite large, print it modulo 10^9 + 7. Input The only line contains a single integer n (1 ≤ n ≤ 1000). Output Print exactly one integer — the size of the maximum matching in the trie. Since the answer can be quite large, print it modulo 10^9 + 7. Examples Input 1 Output 1 Input 2 Output 3 Input 3 Output 9 Note The pictures below illustrate tries in the first two examples (for clarity, the round brackets are replaced with angle brackets). The maximum matching is highlighted with blue. <image> <image> Submitted Solution: ``` n = int(input()) print(3**(n-1)) ```
instruction
0
21,260
21
42,520
No
output
1
21,260
21
42,521
Provide tags and a correct Python 3 solution for this coding contest problem. You are given a string a, consisting of n characters, n is even. For each i from 1 to n a_i is one of 'A', 'B' or 'C'. A bracket sequence is a string containing only characters "(" and ")". A regular bracket sequence is a bracket sequence that can be transformed into a correct arithmetic expression by inserting characters "1" and "+" between the original characters of the sequence. For example, bracket sequences "()()" and "(())" are regular (the resulting expressions are: "(1)+(1)" and "((1+1)+1)"), and ")(", "(" and ")" are not. You want to find a string b that consists of n characters such that: * b is a regular bracket sequence; * if for some i and j (1 ≤ i, j ≤ n) a_i=a_j, then b_i=b_j. In other words, you want to replace all occurrences of 'A' with the same type of bracket, then all occurrences of 'B' with the same type of bracket and all occurrences of 'C' with the same type of bracket. Your task is to determine if such a string b exists. Input The first line contains a single integer t (1 ≤ t ≤ 1000) — the number of testcases. Then the descriptions of t testcases follow. The only line of each testcase contains a string a. a consists only of uppercase letters 'A', 'B' and 'C'. Let n be the length of a. It is guaranteed that n is even and 2 ≤ n ≤ 50. Output For each testcase print "YES" if there exists such a string b that: * b is a regular bracket sequence; * if for some i and j (1 ≤ i, j ≤ n) a_i=a_j, then b_i=b_j. Otherwise, print "NO". You may print every letter in any case you want (so, for example, the strings yEs, yes, Yes and YES are all recognized as positive answer). Example Input 4 AABBAC CACA BBBBAC ABCA Output YES YES NO NO Note In the first testcase one of the possible strings b is "(())()". In the second testcase one of the possible strings b is "()()".
instruction
0
21,427
21
42,854
Tags: bitmasks, brute force, implementation Correct Solution: ``` ''' Auther: ghoshashis545 Ashis Ghosh College: jalpaiguri Govt Enggineering College ''' from os import path from io import BytesIO, IOBase import sys from heapq import heappush,heappop from functools import cmp_to_key as ctk from collections import deque,Counter,defaultdict as dd from bisect import bisect,bisect_left,bisect_right,insort,insort_left,insort_right from itertools import permutations from datetime import datetime from math import ceil,sqrt,log,gcd def ii():return int(input()) def si():return input().rstrip() def mi():return map(int,input().split()) def li():return list(mi()) abc='abcdefghijklmnopqrstuvwxyz' mod=1000000007 #mod=998244353 inf = float("inf") vow=['a','e','i','o','u'] dx,dy=[-1,1,0,0],[0,0,1,-1] def bo(i): return ord(i)-ord('0') file = 1 def ceil(a,b): return (a+b-1)//b def solve(): for _ in range(ii()): a = si() n = len(a) def check(s): st = [] for i in range(n): if len(st)==0 or s[i] == '0': st.append(s[i]) elif(s[i] and st[-1]=='0'): st.pop() return len(st) == 0 ok = False for i in ['000','001','010','011','100','101','110','111']: s = "" for j in range(n): if a[j]=='A': s += i[0] if a[j]=='B': s += i[1] if a[j]=='C': s += i[2] if check(s): print("Yes") ok = True break if ok==False: print('No') if __name__ =="__main__": if(file): if path.exists('tmp/input.txt'): sys.stdin=open('tmp/input.txt', 'r') sys.stdout=open('tmp/output.txt','w') else: input=sys.stdin.readline solve() ```
output
1
21,427
21
42,855
Provide tags and a correct Python 3 solution for this coding contest problem. You are given a string a, consisting of n characters, n is even. For each i from 1 to n a_i is one of 'A', 'B' or 'C'. A bracket sequence is a string containing only characters "(" and ")". A regular bracket sequence is a bracket sequence that can be transformed into a correct arithmetic expression by inserting characters "1" and "+" between the original characters of the sequence. For example, bracket sequences "()()" and "(())" are regular (the resulting expressions are: "(1)+(1)" and "((1+1)+1)"), and ")(", "(" and ")" are not. You want to find a string b that consists of n characters such that: * b is a regular bracket sequence; * if for some i and j (1 ≤ i, j ≤ n) a_i=a_j, then b_i=b_j. In other words, you want to replace all occurrences of 'A' with the same type of bracket, then all occurrences of 'B' with the same type of bracket and all occurrences of 'C' with the same type of bracket. Your task is to determine if such a string b exists. Input The first line contains a single integer t (1 ≤ t ≤ 1000) — the number of testcases. Then the descriptions of t testcases follow. The only line of each testcase contains a string a. a consists only of uppercase letters 'A', 'B' and 'C'. Let n be the length of a. It is guaranteed that n is even and 2 ≤ n ≤ 50. Output For each testcase print "YES" if there exists such a string b that: * b is a regular bracket sequence; * if for some i and j (1 ≤ i, j ≤ n) a_i=a_j, then b_i=b_j. Otherwise, print "NO". You may print every letter in any case you want (so, for example, the strings yEs, yes, Yes and YES are all recognized as positive answer). Example Input 4 AABBAC CACA BBBBAC ABCA Output YES YES NO NO Note In the first testcase one of the possible strings b is "(())()". In the second testcase one of the possible strings b is "()()".
instruction
0
21,428
21
42,856
Tags: bitmasks, brute force, implementation Correct Solution: ``` transformations = ['(((', '(()', '()(', '())', ')((', ')()', '))(', ')))'] def regular(string): opened = 0 for c in string: if c == '(': opened += 1 elif opened > 0: opened -= 1 else: return False return opened == 0 def solve(string): for t in transformations: if regular(string.replace('A', t[0]).replace('B', t[1]).replace('C', t[2])): return True return False if __name__ == '__main__': for i in range(int(input())): print('YES' if solve(input()) else 'NO') ```
output
1
21,428
21
42,857
Provide tags and a correct Python 3 solution for this coding contest problem. You are given a string a, consisting of n characters, n is even. For each i from 1 to n a_i is one of 'A', 'B' or 'C'. A bracket sequence is a string containing only characters "(" and ")". A regular bracket sequence is a bracket sequence that can be transformed into a correct arithmetic expression by inserting characters "1" and "+" between the original characters of the sequence. For example, bracket sequences "()()" and "(())" are regular (the resulting expressions are: "(1)+(1)" and "((1+1)+1)"), and ")(", "(" and ")" are not. You want to find a string b that consists of n characters such that: * b is a regular bracket sequence; * if for some i and j (1 ≤ i, j ≤ n) a_i=a_j, then b_i=b_j. In other words, you want to replace all occurrences of 'A' with the same type of bracket, then all occurrences of 'B' with the same type of bracket and all occurrences of 'C' with the same type of bracket. Your task is to determine if such a string b exists. Input The first line contains a single integer t (1 ≤ t ≤ 1000) — the number of testcases. Then the descriptions of t testcases follow. The only line of each testcase contains a string a. a consists only of uppercase letters 'A', 'B' and 'C'. Let n be the length of a. It is guaranteed that n is even and 2 ≤ n ≤ 50. Output For each testcase print "YES" if there exists such a string b that: * b is a regular bracket sequence; * if for some i and j (1 ≤ i, j ≤ n) a_i=a_j, then b_i=b_j. Otherwise, print "NO". You may print every letter in any case you want (so, for example, the strings yEs, yes, Yes and YES are all recognized as positive answer). Example Input 4 AABBAC CACA BBBBAC ABCA Output YES YES NO NO Note In the first testcase one of the possible strings b is "(())()". In the second testcase one of the possible strings b is "()()".
instruction
0
21,429
21
42,858
Tags: bitmasks, brute force, implementation Correct Solution: ``` import sys def II(): return int(sys.stdin.readline()) def LI(): return list(map(int, sys.stdin.readline().split())) def LC(): return list(input()) def IC():return [int(c) for c in input()] def MI(): return map(int, sys.stdin.readline().split()) INF = float('inf') def solve(): N = II() Char = {} from collections import deque for i in range(N): S = LC() flag = False L = len(S) for a in range(2): for b in range(2): for c in range(2): Tmpflag = True Q = deque([]) Char["A"] = a Char["B"] = b Char["C"] = c for s in range(L): if(Char[S[s]]): Q.append(1) else: if(list(Q)): Q.pop() else: Tmpflag = False break if(Q == deque([]) and Tmpflag ): #print(Char) flag = True if(flag): print("YES") else: print("NO") return solve() ```
output
1
21,429
21
42,859
Provide tags and a correct Python 3 solution for this coding contest problem. You are given a string a, consisting of n characters, n is even. For each i from 1 to n a_i is one of 'A', 'B' or 'C'. A bracket sequence is a string containing only characters "(" and ")". A regular bracket sequence is a bracket sequence that can be transformed into a correct arithmetic expression by inserting characters "1" and "+" between the original characters of the sequence. For example, bracket sequences "()()" and "(())" are regular (the resulting expressions are: "(1)+(1)" and "((1+1)+1)"), and ")(", "(" and ")" are not. You want to find a string b that consists of n characters such that: * b is a regular bracket sequence; * if for some i and j (1 ≤ i, j ≤ n) a_i=a_j, then b_i=b_j. In other words, you want to replace all occurrences of 'A' with the same type of bracket, then all occurrences of 'B' with the same type of bracket and all occurrences of 'C' with the same type of bracket. Your task is to determine if such a string b exists. Input The first line contains a single integer t (1 ≤ t ≤ 1000) — the number of testcases. Then the descriptions of t testcases follow. The only line of each testcase contains a string a. a consists only of uppercase letters 'A', 'B' and 'C'. Let n be the length of a. It is guaranteed that n is even and 2 ≤ n ≤ 50. Output For each testcase print "YES" if there exists such a string b that: * b is a regular bracket sequence; * if for some i and j (1 ≤ i, j ≤ n) a_i=a_j, then b_i=b_j. Otherwise, print "NO". You may print every letter in any case you want (so, for example, the strings yEs, yes, Yes and YES are all recognized as positive answer). Example Input 4 AABBAC CACA BBBBAC ABCA Output YES YES NO NO Note In the first testcase one of the possible strings b is "(())()". In the second testcase one of the possible strings b is "()()".
instruction
0
21,430
21
42,860
Tags: bitmasks, brute force, implementation Correct Solution: ``` def STR(): return list(input()) def INT(): return int(input()) def MAP(): return map(int, input().split()) def MAP2():return map(float,input().split()) def LIST(): return list(map(int, input().split())) def STRING(): return input() import string import sys from heapq import heappop , heappush, heapify from bisect import * from collections import deque , Counter , defaultdict from math import * from itertools import permutations , accumulate dx = [-1 , 1 , 0 , 0 ] dy = [0 , 0 , 1 , - 1] def helper(s): if s[0]==s[-1]: print('NO') return a=[] start=s[0] end=s[-1] flag=0 for i in s: if i==start: a.append('(') elif i==end: if len(a)==0: flag=1 break a.pop() else: a.append('(') if len(a)==0 and flag==0: print('YES') return a=[] for i in s: if i==s[0]: a.append('(') elif i==end: if len(a)==0: print('NO') return a.pop() else: if len(a)==0: print('NO') return a.pop() if len(a)==0: print('YES') else: print("NO") for tt in range(INT()): s=STRING() helper(s) ```
output
1
21,430
21
42,861
Provide tags and a correct Python 3 solution for this coding contest problem. You are given a string a, consisting of n characters, n is even. For each i from 1 to n a_i is one of 'A', 'B' or 'C'. A bracket sequence is a string containing only characters "(" and ")". A regular bracket sequence is a bracket sequence that can be transformed into a correct arithmetic expression by inserting characters "1" and "+" between the original characters of the sequence. For example, bracket sequences "()()" and "(())" are regular (the resulting expressions are: "(1)+(1)" and "((1+1)+1)"), and ")(", "(" and ")" are not. You want to find a string b that consists of n characters such that: * b is a regular bracket sequence; * if for some i and j (1 ≤ i, j ≤ n) a_i=a_j, then b_i=b_j. In other words, you want to replace all occurrences of 'A' with the same type of bracket, then all occurrences of 'B' with the same type of bracket and all occurrences of 'C' with the same type of bracket. Your task is to determine if such a string b exists. Input The first line contains a single integer t (1 ≤ t ≤ 1000) — the number of testcases. Then the descriptions of t testcases follow. The only line of each testcase contains a string a. a consists only of uppercase letters 'A', 'B' and 'C'. Let n be the length of a. It is guaranteed that n is even and 2 ≤ n ≤ 50. Output For each testcase print "YES" if there exists such a string b that: * b is a regular bracket sequence; * if for some i and j (1 ≤ i, j ≤ n) a_i=a_j, then b_i=b_j. Otherwise, print "NO". You may print every letter in any case you want (so, for example, the strings yEs, yes, Yes and YES are all recognized as positive answer). Example Input 4 AABBAC CACA BBBBAC ABCA Output YES YES NO NO Note In the first testcase one of the possible strings b is "(())()". In the second testcase one of the possible strings b is "()()".
instruction
0
21,431
21
42,862
Tags: bitmasks, brute force, implementation Correct Solution: ``` import sys def read_ints(): return [int(i) for i in sys.stdin.readline().strip().split()] def read_int(): return int(sys.stdin.readline().strip()) def check_brackets(s, openers, closers): opened = 0 for c in s: if c in openers: opened += 1 else: if opened <= 0: return False opened -= 1 return opened == 0 def can_bracket(s): opener = s[0] closer = s[-1] if opener == closer: return False other = [c for c in "ABC" if c not in [opener, closer]][0] nopeners = sum(1 for c in s if c == opener) nclosers = sum(1 for c in s if c == closer) nother = len(s) - nopeners - nclosers if nother != abs(nopeners - nclosers): return False if nopeners > nclosers: return check_brackets(s, [opener], [closer, other]) else: return check_brackets(s, [opener, other], [closer]) t = read_int() for i in range(t): s = sys.stdin.readline().strip() if can_bracket(s): print("YES") else: print("NO") ```
output
1
21,431
21
42,863
Provide tags and a correct Python 3 solution for this coding contest problem. You are given a string a, consisting of n characters, n is even. For each i from 1 to n a_i is one of 'A', 'B' or 'C'. A bracket sequence is a string containing only characters "(" and ")". A regular bracket sequence is a bracket sequence that can be transformed into a correct arithmetic expression by inserting characters "1" and "+" between the original characters of the sequence. For example, bracket sequences "()()" and "(())" are regular (the resulting expressions are: "(1)+(1)" and "((1+1)+1)"), and ")(", "(" and ")" are not. You want to find a string b that consists of n characters such that: * b is a regular bracket sequence; * if for some i and j (1 ≤ i, j ≤ n) a_i=a_j, then b_i=b_j. In other words, you want to replace all occurrences of 'A' with the same type of bracket, then all occurrences of 'B' with the same type of bracket and all occurrences of 'C' with the same type of bracket. Your task is to determine if such a string b exists. Input The first line contains a single integer t (1 ≤ t ≤ 1000) — the number of testcases. Then the descriptions of t testcases follow. The only line of each testcase contains a string a. a consists only of uppercase letters 'A', 'B' and 'C'. Let n be the length of a. It is guaranteed that n is even and 2 ≤ n ≤ 50. Output For each testcase print "YES" if there exists such a string b that: * b is a regular bracket sequence; * if for some i and j (1 ≤ i, j ≤ n) a_i=a_j, then b_i=b_j. Otherwise, print "NO". You may print every letter in any case you want (so, for example, the strings yEs, yes, Yes and YES are all recognized as positive answer). Example Input 4 AABBAC CACA BBBBAC ABCA Output YES YES NO NO Note In the first testcase one of the possible strings b is "(())()". In the second testcase one of the possible strings b is "()()".
instruction
0
21,432
21
42,864
Tags: bitmasks, brute force, implementation Correct Solution: ``` def checkBalancedArray(string, ch): stack = 0 for i in string: if stack < 0: # we have deleted an element(hypothetically) when it is not even there. So, "NO" return "NO" if i in ch: # add one onto the stack if it qualifies stack += 1 else: # balance out one by deleting stack -= 1 return "NO" if stack else "YES" # if stack is empty, else block is executed t = int(input()) for _ in range(t): s = input() count_A = s.count("A") count_B = s.count("B") count_C = s.count("C") c = s[0] if count_A + count_B == count_C: if c == "C": print(checkBalancedArray(s, [c])) else: print(checkBalancedArray(s, ["A", "B"])) elif count_B + count_C == count_A: if c == "A": print(checkBalancedArray(s, [c])) else: print(checkBalancedArray(s, ["B", "C"])) elif count_C + count_A == count_B: if c == "B": print(checkBalancedArray(s, [c])) else: print(checkBalancedArray(s, ["A", "C"])) else: print("NO") ```
output
1
21,432
21
42,865
Provide tags and a correct Python 3 solution for this coding contest problem. You are given a string a, consisting of n characters, n is even. For each i from 1 to n a_i is one of 'A', 'B' or 'C'. A bracket sequence is a string containing only characters "(" and ")". A regular bracket sequence is a bracket sequence that can be transformed into a correct arithmetic expression by inserting characters "1" and "+" between the original characters of the sequence. For example, bracket sequences "()()" and "(())" are regular (the resulting expressions are: "(1)+(1)" and "((1+1)+1)"), and ")(", "(" and ")" are not. You want to find a string b that consists of n characters such that: * b is a regular bracket sequence; * if for some i and j (1 ≤ i, j ≤ n) a_i=a_j, then b_i=b_j. In other words, you want to replace all occurrences of 'A' with the same type of bracket, then all occurrences of 'B' with the same type of bracket and all occurrences of 'C' with the same type of bracket. Your task is to determine if such a string b exists. Input The first line contains a single integer t (1 ≤ t ≤ 1000) — the number of testcases. Then the descriptions of t testcases follow. The only line of each testcase contains a string a. a consists only of uppercase letters 'A', 'B' and 'C'. Let n be the length of a. It is guaranteed that n is even and 2 ≤ n ≤ 50. Output For each testcase print "YES" if there exists such a string b that: * b is a regular bracket sequence; * if for some i and j (1 ≤ i, j ≤ n) a_i=a_j, then b_i=b_j. Otherwise, print "NO". You may print every letter in any case you want (so, for example, the strings yEs, yes, Yes and YES are all recognized as positive answer). Example Input 4 AABBAC CACA BBBBAC ABCA Output YES YES NO NO Note In the first testcase one of the possible strings b is "(())()". In the second testcase one of the possible strings b is "()()".
instruction
0
21,433
21
42,866
Tags: bitmasks, brute force, implementation Correct Solution: ``` import sys input=sys.stdin.readline from collections import defaultdict as dc from collections import Counter from bisect import bisect_right, bisect_left import math from operator import itemgetter from heapq import heapify, heappop, heappush from queue import PriorityQueue as pq for _ in range(int(input())): s=list(input()[:-1]) x=dc(int) x['A'],x['B'],x['C']=0,0,0 for i in s: x[i]+=1 x=sorted(x.items(), key=lambda item: item[1]) p,q,r=x[2][0],x[1][0],x[0][0] a,b,c=x[2][1],x[1][1],x[0][1] if a!=b+c: print("NO") else: n=len(s) for i in range(n): if s[i]!=p: s[i]=q if s[0]==s[-1]: print("NO") else: t=1 f=0 for i in range(1,n): if s[i]==s[0]: t+=1 else: if t>0: t-=1 else: f=1 break if f or t>0: print("NO") else: print("YES") ```
output
1
21,433
21
42,867
Provide tags and a correct Python 3 solution for this coding contest problem. You are given a string a, consisting of n characters, n is even. For each i from 1 to n a_i is one of 'A', 'B' or 'C'. A bracket sequence is a string containing only characters "(" and ")". A regular bracket sequence is a bracket sequence that can be transformed into a correct arithmetic expression by inserting characters "1" and "+" between the original characters of the sequence. For example, bracket sequences "()()" and "(())" are regular (the resulting expressions are: "(1)+(1)" and "((1+1)+1)"), and ")(", "(" and ")" are not. You want to find a string b that consists of n characters such that: * b is a regular bracket sequence; * if for some i and j (1 ≤ i, j ≤ n) a_i=a_j, then b_i=b_j. In other words, you want to replace all occurrences of 'A' with the same type of bracket, then all occurrences of 'B' with the same type of bracket and all occurrences of 'C' with the same type of bracket. Your task is to determine if such a string b exists. Input The first line contains a single integer t (1 ≤ t ≤ 1000) — the number of testcases. Then the descriptions of t testcases follow. The only line of each testcase contains a string a. a consists only of uppercase letters 'A', 'B' and 'C'. Let n be the length of a. It is guaranteed that n is even and 2 ≤ n ≤ 50. Output For each testcase print "YES" if there exists such a string b that: * b is a regular bracket sequence; * if for some i and j (1 ≤ i, j ≤ n) a_i=a_j, then b_i=b_j. Otherwise, print "NO". You may print every letter in any case you want (so, for example, the strings yEs, yes, Yes and YES are all recognized as positive answer). Example Input 4 AABBAC CACA BBBBAC ABCA Output YES YES NO NO Note In the first testcase one of the possible strings b is "(())()". In the second testcase one of the possible strings b is "()()".
instruction
0
21,434
21
42,868
Tags: bitmasks, brute force, implementation Correct Solution: ``` t = int(input()) for _ in range(t): a = input() if a[0] == a[-1]: print('no') else: a = a.replace(a[0], '(') a = a.replace(a[-1], ')') c1 = a.count('(') c2 = a.count(')') c3 = len(a) - c1 - c2 if c1 < c2: a = a.replace('A', '(') a = a.replace('B', '(') a = a.replace('C', '(') else: a = a.replace('A', ')') a = a.replace('B', ')') a = a.replace('C', ')') if c1 + c3 != c2 and c2 + c3 != c1: print('NO') else: f = 0 for i in a: if i == '(': f += 1 else: f -= 1 if f < 0: print('nO') break else: print('YES') ```
output
1
21,434
21
42,869