message stringlengths 2 22.8k | message_type stringclasses 2 values | message_id int64 0 1 | conversation_id int64 16 109k | cluster float64 1 1 | __index_level_0__ int64 32 217k |
|---|---|---|---|---|---|
Provide tags and a correct Python 3 solution for this coding contest problem.
Bandits appeared in the city! One of them is trying to catch as many citizens as he can.
The city consists of n squares connected by n-1 roads in such a way that it is possible to reach any square from any other square. The square number 1 is the main square.
After Sunday walk all the roads were changed to one-way roads in such a way that it is possible to reach any square from the main square.
At the moment when the bandit appeared on the main square there were a_i citizens on the i-th square. Now the following process will begin. First, each citizen that is currently on a square with some outgoing one-way roads chooses one of such roads and moves along it to another square. Then the bandit chooses one of the one-way roads outgoing from the square he is located and moves along it. The process is repeated until the bandit is located on a square with no outgoing roads. The bandit catches all the citizens on that square.
The bandit wants to catch as many citizens as possible; the citizens want to minimize the number of caught people. The bandit and the citizens know positions of all citizens at any time, the citizens can cooperate. If both sides act optimally, how many citizens will be caught?
Input
The first line contains a single integer n β the number of squares in the city (2 β€ n β€ 2β
10^5).
The second line contains n-1 integers p_2, p_3 ... p_n meaning that there is a one-way road from the square p_i to the square i (1 β€ p_i < i).
The third line contains n integers a_1, a_2, ..., a_n β the number of citizens on each square initially (0 β€ a_i β€ 10^9).
Output
Print a single integer β the number of citizens the bandit will catch if both sides act optimally.
Examples
Input
3
1 1
3 1 2
Output
3
Input
3
1 1
3 1 3
Output
4
Note
In the first example the citizens on the square 1 can split into two groups 2 + 1, so that the second and on the third squares will have 3 citizens each.
In the second example no matter how citizens act the bandit can catch at least 4 citizens. | instruction | 0 | 61,671 | 1 | 123,342 |
Tags: binary search, dfs and similar, graphs, greedy, trees
Correct Solution:
```
maxn = 200005
cities = [0 for _ in range(maxn)]
n = int(input())
x = list(map(int , input().split(' ')))
total = [0]+list(map(int,input().split(' ')))
ans = 0
for i in range(n,0,-1):
if(cities[i]==0):
cities[i]=1
if(i>=2):
cities[x[i-2]]+=cities[i]
total [x[i-2]]+=total[i]
ans = max(ans, (total[i] + cities[i] -1)//cities[i])
print(ans)
``` | output | 1 | 61,671 | 1 | 123,343 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Bandits appeared in the city! One of them is trying to catch as many citizens as he can.
The city consists of n squares connected by n-1 roads in such a way that it is possible to reach any square from any other square. The square number 1 is the main square.
After Sunday walk all the roads were changed to one-way roads in such a way that it is possible to reach any square from the main square.
At the moment when the bandit appeared on the main square there were a_i citizens on the i-th square. Now the following process will begin. First, each citizen that is currently on a square with some outgoing one-way roads chooses one of such roads and moves along it to another square. Then the bandit chooses one of the one-way roads outgoing from the square he is located and moves along it. The process is repeated until the bandit is located on a square with no outgoing roads. The bandit catches all the citizens on that square.
The bandit wants to catch as many citizens as possible; the citizens want to minimize the number of caught people. The bandit and the citizens know positions of all citizens at any time, the citizens can cooperate. If both sides act optimally, how many citizens will be caught?
Input
The first line contains a single integer n β the number of squares in the city (2 β€ n β€ 2β
10^5).
The second line contains n-1 integers p_2, p_3 ... p_n meaning that there is a one-way road from the square p_i to the square i (1 β€ p_i < i).
The third line contains n integers a_1, a_2, ..., a_n β the number of citizens on each square initially (0 β€ a_i β€ 10^9).
Output
Print a single integer β the number of citizens the bandit will catch if both sides act optimally.
Examples
Input
3
1 1
3 1 2
Output
3
Input
3
1 1
3 1 3
Output
4
Note
In the first example the citizens on the square 1 can split into two groups 2 + 1, so that the second and on the third squares will have 3 citizens each.
In the second example no matter how citizens act the bandit can catch at least 4 citizens. | instruction | 0 | 61,672 | 1 | 123,344 |
Tags: binary search, dfs and similar, graphs, greedy, trees
Correct Solution:
```
n = int(input())
cities = [0 for _ in range(n + 1)]
x = list(map(int, input().split(' ')))
total = [0] + list(map(int, input().split(' ')))
ans = 0
for i in range(n, 0, -1):
if(cities[i] == 0):
cities[i] = 1
if(i >= 2):
cities[x[i - 2]] += cities[i]
total [x[i - 2]] += total[i]
ans = max(ans, (total[i] + cities[i] -1) // cities[i])
print(ans)
``` | output | 1 | 61,672 | 1 | 123,345 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Bandits appeared in the city! One of them is trying to catch as many citizens as he can.
The city consists of n squares connected by n-1 roads in such a way that it is possible to reach any square from any other square. The square number 1 is the main square.
After Sunday walk all the roads were changed to one-way roads in such a way that it is possible to reach any square from the main square.
At the moment when the bandit appeared on the main square there were a_i citizens on the i-th square. Now the following process will begin. First, each citizen that is currently on a square with some outgoing one-way roads chooses one of such roads and moves along it to another square. Then the bandit chooses one of the one-way roads outgoing from the square he is located and moves along it. The process is repeated until the bandit is located on a square with no outgoing roads. The bandit catches all the citizens on that square.
The bandit wants to catch as many citizens as possible; the citizens want to minimize the number of caught people. The bandit and the citizens know positions of all citizens at any time, the citizens can cooperate. If both sides act optimally, how many citizens will be caught?
Input
The first line contains a single integer n β the number of squares in the city (2 β€ n β€ 2β
10^5).
The second line contains n-1 integers p_2, p_3 ... p_n meaning that there is a one-way road from the square p_i to the square i (1 β€ p_i < i).
The third line contains n integers a_1, a_2, ..., a_n β the number of citizens on each square initially (0 β€ a_i β€ 10^9).
Output
Print a single integer β the number of citizens the bandit will catch if both sides act optimally.
Examples
Input
3
1 1
3 1 2
Output
3
Input
3
1 1
3 1 3
Output
4
Note
In the first example the citizens on the square 1 can split into two groups 2 + 1, so that the second and on the third squares will have 3 citizens each.
In the second example no matter how citizens act the bandit can catch at least 4 citizens. | instruction | 0 | 61,673 | 1 | 123,346 |
Tags: binary search, dfs and similar, graphs, greedy, trees
Correct Solution:
```
from collections import defaultdict, deque
from math import ceil
from sys import setrecursionlimit
import io, os
setrecursionlimit(300000)
input = io.BytesIO(os.read(0, os.fstat(0).st_size)).readline
n = int(input())
par = [-1] + list(map(int, input().split()))
_g = defaultdict(list)
real_leaves = set(range(n))
people = list(map(int, input().split()))
children = [0] * (n+1)
for i, v in enumerate(par):
real_leaves.discard(v-1)
_g[i].append(v-1)
children[v-1] += 1
q = deque(real_leaves)
leaves, people_cnt = [0] * (n+1), people[:]
for i in real_leaves:
leaves[i] = 1
ans = 0
while True:
q2 = deque()
while q:
now = q.popleft()
ans = max(ans, ceil(people_cnt[now] / leaves[now]))
for par in _g[now]:
leaves[par] += leaves[now]
people_cnt[par] += people_cnt[now]
children[par] -= 1
if children[par] == 0:
q2.append(par)
if not q2:
break
q = q2
print(ans)
``` | output | 1 | 61,673 | 1 | 123,347 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Bandits appeared in the city! One of them is trying to catch as many citizens as he can.
The city consists of n squares connected by n-1 roads in such a way that it is possible to reach any square from any other square. The square number 1 is the main square.
After Sunday walk all the roads were changed to one-way roads in such a way that it is possible to reach any square from the main square.
At the moment when the bandit appeared on the main square there were a_i citizens on the i-th square. Now the following process will begin. First, each citizen that is currently on a square with some outgoing one-way roads chooses one of such roads and moves along it to another square. Then the bandit chooses one of the one-way roads outgoing from the square he is located and moves along it. The process is repeated until the bandit is located on a square with no outgoing roads. The bandit catches all the citizens on that square.
The bandit wants to catch as many citizens as possible; the citizens want to minimize the number of caught people. The bandit and the citizens know positions of all citizens at any time, the citizens can cooperate. If both sides act optimally, how many citizens will be caught?
Input
The first line contains a single integer n β the number of squares in the city (2 β€ n β€ 2β
10^5).
The second line contains n-1 integers p_2, p_3 ... p_n meaning that there is a one-way road from the square p_i to the square i (1 β€ p_i < i).
The third line contains n integers a_1, a_2, ..., a_n β the number of citizens on each square initially (0 β€ a_i β€ 10^9).
Output
Print a single integer β the number of citizens the bandit will catch if both sides act optimally.
Examples
Input
3
1 1
3 1 2
Output
3
Input
3
1 1
3 1 3
Output
4
Note
In the first example the citizens on the square 1 can split into two groups 2 + 1, so that the second and on the third squares will have 3 citizens each.
In the second example no matter how citizens act the bandit can catch at least 4 citizens. | instruction | 0 | 61,674 | 1 | 123,348 |
Tags: binary search, dfs and similar, graphs, greedy, trees
Correct Solution:
```
from math import factorial, ceil
from sys import stdin, stdout
from collections import defaultdict
def DFS(arr, citz, n) :
parent = {}
children = defaultdict(lambda : 0)
for i in range(len(arr)) :
parent[ i+2 ] = arr[i]
children[ arr[i] ] += 1
#print(edges)
#print(citz)
#print(counts)
maxes = defaultdict(lambda : 0)
count = defaultdict(lambda : 0)
for i in reversed(range(2, n+1)) :
if children[i] == 0 :
count[i] = 1
maxes[i] = citz[i-1]
count[ parent[i] ] += count[i]
citz[ parent[i]-1 ] += citz[i-1]
maxes[ parent[i] ] = max( maxes[ parent[i] ], maxes[i])
else :
count[ parent[i] ] += count[i]
citz[ parent[i]-1 ] += citz[i-1]
maxes[i] = max(maxes[i], ceil(citz[i-1] / count[i]) )
maxes[ parent[i] ] = max( maxes[ parent[i] ], maxes[i])
stdout.write( str( max( maxes[1], ceil(citz[0] / count[1]) ) ) )
#def solve(n, arr, citz):
'''class solve:
def __init__(self, arr, citz) :
self.nodes = [node(citz[i]) for i in range(len(citz))]
self.edges = { n:[] for n in self.nodes }
for i in range(len(arr)) :
self.edges[ self.nodes[arr[i]-1] ].append( self.nodes[i+1] )
for k, v in self.edges.items() :
if len(v) == 0 :
k.count = 1
k.max = k.sum
#for n in self.nodes :
# print(n.count, n.sum, n.max)
def DFS(self, nod) :
maxi = 0
for u in self.edges[nod] :
self.DFS(u)
maxi = max(maxi, u.max)
nod.sum += u.sum
nod.count += u.count
nod.max = max(maxi, ceil(nod.sum/nod.count) )
def ans(self) :
self.DFS( self.nodes[0] )
print( self.nodes[0].max)
'''
def main():
# write the input logic to take the input from STDIN
n = int(stdin.readline())
arr = list(map(int, stdin.readline().split() ) )
citz = list(map(int, stdin.readline().split()) )
DFS(arr, citz, n)
if __name__ == "__main__":
main()
``` | output | 1 | 61,674 | 1 | 123,349 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Bandits appeared in the city! One of them is trying to catch as many citizens as he can.
The city consists of n squares connected by n-1 roads in such a way that it is possible to reach any square from any other square. The square number 1 is the main square.
After Sunday walk all the roads were changed to one-way roads in such a way that it is possible to reach any square from the main square.
At the moment when the bandit appeared on the main square there were a_i citizens on the i-th square. Now the following process will begin. First, each citizen that is currently on a square with some outgoing one-way roads chooses one of such roads and moves along it to another square. Then the bandit chooses one of the one-way roads outgoing from the square he is located and moves along it. The process is repeated until the bandit is located on a square with no outgoing roads. The bandit catches all the citizens on that square.
The bandit wants to catch as many citizens as possible; the citizens want to minimize the number of caught people. The bandit and the citizens know positions of all citizens at any time, the citizens can cooperate. If both sides act optimally, how many citizens will be caught?
Input
The first line contains a single integer n β the number of squares in the city (2 β€ n β€ 2β
10^5).
The second line contains n-1 integers p_2, p_3 ... p_n meaning that there is a one-way road from the square p_i to the square i (1 β€ p_i < i).
The third line contains n integers a_1, a_2, ..., a_n β the number of citizens on each square initially (0 β€ a_i β€ 10^9).
Output
Print a single integer β the number of citizens the bandit will catch if both sides act optimally.
Examples
Input
3
1 1
3 1 2
Output
3
Input
3
1 1
3 1 3
Output
4
Note
In the first example the citizens on the square 1 can split into two groups 2 + 1, so that the second and on the third squares will have 3 citizens each.
In the second example no matter how citizens act the bandit can catch at least 4 citizens.
Submitted Solution:
```
import sys, math
import io, os
#data = io.BytesIO(os.read(0,os.fstat(0).st_size)).readline
from bisect import bisect_left as bl, bisect_right as br, insort
from heapq import heapify, heappush, heappop
from collections import defaultdict as dd, deque, Counter
from itertools import permutations,combinations
def data(): return sys.stdin.readline().strip()
def mdata(): return list(map(int, data().split()))
def outl(var) : sys.stdout.write(' '.join(map(str, var))+'\n')
def out(var) : sys.stdout.write(str(var)+'\n')
#from decimal import Decimal
#from fractions import Fraction
#sys.setrecursionlimit(100000)
INF = float('inf')
mod = int(1e9)+7
def dfs(graph, start=0):
n = len(graph)
dp = [[0,a[i],0] for i in range(n)]
visited, finished = [False] * n, [False] * n
stack = [start]
while stack:
start = stack[-1]
# push unvisited children into stack
if not visited[start]:
visited[start] = True
for child in graph[start]:
if not visited[child]:
stack.append(child)
else:
stack.pop()
if start!=0 and len(graph[start])==1:
dp[start][0]=1
dp[start][2]=a[start]
# update with finished children
for child in graph[start]:
if finished[child]:
dp[start][0] += dp[child][0]
dp[start][1] += dp[child][1]
dp[start][2] = max(dp[start][2],dp[child][2])
dp[start][2] = max(dp[start][2],dp[start][1]//dp[start][0]+(1 if dp[start][1]%dp[start][0] else 0))
finished[start] = True
return dp[0][2]
n=int(data())
graph=[[] for i in range(n)]
p=mdata()
for i in range(n-1):
graph[i+1].append(p[i]-1)
graph[p[i]-1].append(i+1)
a=mdata()
out(dfs(graph,0))
``` | instruction | 0 | 61,675 | 1 | 123,350 |
Yes | output | 1 | 61,675 | 1 | 123,351 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Bandits appeared in the city! One of them is trying to catch as many citizens as he can.
The city consists of n squares connected by n-1 roads in such a way that it is possible to reach any square from any other square. The square number 1 is the main square.
After Sunday walk all the roads were changed to one-way roads in such a way that it is possible to reach any square from the main square.
At the moment when the bandit appeared on the main square there were a_i citizens on the i-th square. Now the following process will begin. First, each citizen that is currently on a square with some outgoing one-way roads chooses one of such roads and moves along it to another square. Then the bandit chooses one of the one-way roads outgoing from the square he is located and moves along it. The process is repeated until the bandit is located on a square with no outgoing roads. The bandit catches all the citizens on that square.
The bandit wants to catch as many citizens as possible; the citizens want to minimize the number of caught people. The bandit and the citizens know positions of all citizens at any time, the citizens can cooperate. If both sides act optimally, how many citizens will be caught?
Input
The first line contains a single integer n β the number of squares in the city (2 β€ n β€ 2β
10^5).
The second line contains n-1 integers p_2, p_3 ... p_n meaning that there is a one-way road from the square p_i to the square i (1 β€ p_i < i).
The third line contains n integers a_1, a_2, ..., a_n β the number of citizens on each square initially (0 β€ a_i β€ 10^9).
Output
Print a single integer β the number of citizens the bandit will catch if both sides act optimally.
Examples
Input
3
1 1
3 1 2
Output
3
Input
3
1 1
3 1 3
Output
4
Note
In the first example the citizens on the square 1 can split into two groups 2 + 1, so that the second and on the third squares will have 3 citizens each.
In the second example no matter how citizens act the bandit can catch at least 4 citizens.
Submitted Solution:
```
# ===============================================================================================
# importing some useful libraries.
from __future__ import division, print_function
from fractions import Fraction
import sys
import os
from io import BytesIO, IOBase
from itertools import *
import bisect
from heapq import *
from math import ceil, floor
from copy import *
from collections import deque, defaultdict
from collections import Counter as counter # Counter(list) return a dict with {key: count}
from itertools import combinations # if a = [1,2,3] then print(list(comb(a,2))) -----> [(1, 2), (1, 3), (2, 3)]
from itertools import permutations as permutate
from bisect import bisect_left as bl
from operator import *
# If the element is already present in the list,
# the left most position where element has to be inserted is returned.
from bisect import bisect_right as br
from bisect import bisect
# If the element is already present in the list,
# the right most position where element has to be inserted is returned
# range=xrange
# ==============================================================================================
# fast I/O region
BUFSIZE = 8192
from sys import stderr
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")
def print(*args, **kwargs):
"""Prints the values to a stream, or to sys.stdout by default."""
sep, file = kwargs.pop("sep", " "), kwargs.pop("file", sys.stdout)
at_start = True
for x in args:
if not at_start:
file.write(sep)
file.write(str(x))
at_start = False
file.write(kwargs.pop("end", "\n"))
if kwargs.pop("flush", False):
file.flush()
if sys.version_info[0] < 3:
sys.stdin, sys.stdout = FastIO(sys.stdin), FastIO(sys.stdout)
else:
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
# inp = lambda: sys.stdin.readline().rstrip("\r\n")
# ===============================================================================================
### START ITERATE RECURSION ###
from types import GeneratorType
def iterative(f, stack=[]):
def wrapped_func(*args, **kwargs):
if stack: return f(*args, **kwargs)
to = f(*args, **kwargs)
while True:
if type(to) is GeneratorType:
stack.append(to)
to = next(to)
continue
stack.pop()
if not stack: break
to = stack[-1].send(to)
return to
return wrapped_func
#### END ITERATE RECURSION ####
# ===============================================================================================
# some shortcuts
mod = 1000000007
def inp(): return sys.stdin.readline().rstrip("\r\n") # for fast input
def out(var): sys.stdout.write(str(var)) # for fast output, always take string
def lis(): return list(map(int, inp().split()))
def stringlis(): return list(map(str, inp().split()))
def sep(): return map(int, inp().split())
def strsep(): return map(str, inp().split())
def fsep(): return map(float, inp().split())
def nextline(): out("\n") # as stdout.write always print sring.
def testcase(t):
for p in range(t):
solve()
def pow(x, y, p):
res = 1 # Initialize result
x = x % p # Update x if it is more , than or equal to p
if (x == 0):
return 0
while (y > 0):
if ((y & 1) == 1): # If y is odd, multiply, x with result
res = (res * x) % p
y = y >> 1 # y = y/2
x = (x * x) % p
return res
from functools import reduce
def factors(n):
return set(reduce(list.__add__,
([i, n // i] for i in range(1, int(n ** 0.5) + 1) if n % i == 0)))
def lcm(a, b):
return abs((a // gcd(a, b)) * b)
def gcd(a, b):
if a == b: return a
while b > 0: a, b = b, a % b
return a
# discrete binary search
# minimise:
# def search():
# l = 0
# r = 10 ** 15
#
# for i in range(200):
# if isvalid(l):
# return l
# if l == r:
# return l
# m = (l + r) // 2
# if isvalid(m) and not isvalid(m - 1):
# return m
# if isvalid(m):
# r = m + 1
# else:
# l = m
# return m
# maximise:
# def search():
# l = 0
# r = 10 ** 15
#
# for i in range(200):
# # print(l,r)
# if isvalid(r):
# return r
# if l == r:
# return l
# m = (l + r) // 2
# if isvalid(m) and not isvalid(m + 1):
# return m
# if isvalid(m):
# l = m
# else:
# r = m - 1
# return m
##to find factorial and ncr
# N=100000
# mod = 10**9 + 7
# fac = [1, 1]
# finv = [1, 1]
# inv = [0, 1]
#
# for i in range(2, N + 1):
# fac.append((fac[-1] * i) % mod)
# inv.append(mod - (inv[mod % i] * (mod // i) % mod))
# finv.append(finv[-1] * inv[-1] % mod)
def comb(n, r):
if n < r:
return 0
else:
return fac[n] * (finv[r] * finv[n - r] % mod) % mod
##############Find sum of product of subsets of size k in a array
# ar=[0,1,2,3]
# k=3
# n=len(ar)-1
# dp=[0]*(n+1)
# dp[0]=1
# for pos in range(1,n+1):
# dp[pos]=0
# l=max(1,k+pos-n-1)
# for j in range(min(pos,k),l-1,-1):
# dp[j]=dp[j]+ar[pos]*dp[j-1]
# print(dp[k])
def prefix_sum(ar): # [1,2,3,4]->[1,3,6,10]
return list(accumulate(ar))
def suffix_sum(ar): # [1,2,3,4]->[10,9,7,4]
return list(accumulate(ar[::-1]))[::-1]
def N():
return int(inp())
# =========================================================================================
from collections import defaultdict
def numberOfSetBits(i):
i = i - ((i >> 1) & 0x55555555)
i = (i & 0x33333333) + ((i >> 2) & 0x33333333)
return (((i + (i >> 4) & 0xF0F0F0F) * 0x1010101) & 0xffffffff) >> 24
def yes():
print("Yes")
def no():
print("No")
def YES():
print("YES")
def NO():
print("NO")
def solve():
n = N()
ar = lis()
g = [[] for _ in range(n + 1)]
for i in range(1, n):
g[ar[i - 1]].append(i + 1)
people = lis()
dp1 = [0] * (n + 1)
dp2 = [0] * (n + 1)
# print(g)
@iterative
def dfs1(v):
if len(g[v]) == 0:
dp1[v] = 1
dp2[v] = people[v - 1]
yield
dp2[v] = people[v - 1]
for c in g[v]:
yield dfs1(c)
dp1[v] += dp1[c]
dp2[v] += dp2[c]
yield
temp = []
dfs1(1)
for i in range(1, n + 1):
temp.append(ceil(dp2[i] / dp1[i]))
# print(dp1)
# print(dp2)
# print(temp)
print(max(temp))
solve()
# testcase(int(inp()))
``` | instruction | 0 | 61,676 | 1 | 123,352 |
Yes | output | 1 | 61,676 | 1 | 123,353 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Bandits appeared in the city! One of them is trying to catch as many citizens as he can.
The city consists of n squares connected by n-1 roads in such a way that it is possible to reach any square from any other square. The square number 1 is the main square.
After Sunday walk all the roads were changed to one-way roads in such a way that it is possible to reach any square from the main square.
At the moment when the bandit appeared on the main square there were a_i citizens on the i-th square. Now the following process will begin. First, each citizen that is currently on a square with some outgoing one-way roads chooses one of such roads and moves along it to another square. Then the bandit chooses one of the one-way roads outgoing from the square he is located and moves along it. The process is repeated until the bandit is located on a square with no outgoing roads. The bandit catches all the citizens on that square.
The bandit wants to catch as many citizens as possible; the citizens want to minimize the number of caught people. The bandit and the citizens know positions of all citizens at any time, the citizens can cooperate. If both sides act optimally, how many citizens will be caught?
Input
The first line contains a single integer n β the number of squares in the city (2 β€ n β€ 2β
10^5).
The second line contains n-1 integers p_2, p_3 ... p_n meaning that there is a one-way road from the square p_i to the square i (1 β€ p_i < i).
The third line contains n integers a_1, a_2, ..., a_n β the number of citizens on each square initially (0 β€ a_i β€ 10^9).
Output
Print a single integer β the number of citizens the bandit will catch if both sides act optimally.
Examples
Input
3
1 1
3 1 2
Output
3
Input
3
1 1
3 1 3
Output
4
Note
In the first example the citizens on the square 1 can split into two groups 2 + 1, so that the second and on the third squares will have 3 citizens each.
In the second example no matter how citizens act the bandit can catch at least 4 citizens.
Submitted Solution:
```
n = int(input())
g = [[] for i in range(n+1)]
p = list(map(int, input().split()))
for i in range(n-1):
a = i+2
b = p[i]
g[a].append(b)
g[b].append(a)
dp = [0] + list(map(int, input().split()))
count = [0]*(n+1)
maxi = [0]*(n+1)
for i in range(2, n+1):
if len(g[i]) == 1:
count[i] = 1
stack = []
stack.append(~1)
stack.append(1)
visit1 = [False]*(n+1)
visit2 = [False]*(n+1)
visit1[1] = True
while stack:
now = stack.pop()
if now < 0:
now = ~now
for next in g[now]:
if visit2[next]:
dp[now] += dp[next]
count[now] += count[next]
maxi[now] = max(maxi[now], maxi[next])
maxi[now] = max(maxi[now], (dp[now]-1)//count[now] + 1)
visit2[now] = True
else:
for next in g[now]:
if not visit1[next]:
visit1[next] = True
stack.append(~next)
stack.append(next)
print(maxi[1])
``` | instruction | 0 | 61,677 | 1 | 123,354 |
Yes | output | 1 | 61,677 | 1 | 123,355 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Bandits appeared in the city! One of them is trying to catch as many citizens as he can.
The city consists of n squares connected by n-1 roads in such a way that it is possible to reach any square from any other square. The square number 1 is the main square.
After Sunday walk all the roads were changed to one-way roads in such a way that it is possible to reach any square from the main square.
At the moment when the bandit appeared on the main square there were a_i citizens on the i-th square. Now the following process will begin. First, each citizen that is currently on a square with some outgoing one-way roads chooses one of such roads and moves along it to another square. Then the bandit chooses one of the one-way roads outgoing from the square he is located and moves along it. The process is repeated until the bandit is located on a square with no outgoing roads. The bandit catches all the citizens on that square.
The bandit wants to catch as many citizens as possible; the citizens want to minimize the number of caught people. The bandit and the citizens know positions of all citizens at any time, the citizens can cooperate. If both sides act optimally, how many citizens will be caught?
Input
The first line contains a single integer n β the number of squares in the city (2 β€ n β€ 2β
10^5).
The second line contains n-1 integers p_2, p_3 ... p_n meaning that there is a one-way road from the square p_i to the square i (1 β€ p_i < i).
The third line contains n integers a_1, a_2, ..., a_n β the number of citizens on each square initially (0 β€ a_i β€ 10^9).
Output
Print a single integer β the number of citizens the bandit will catch if both sides act optimally.
Examples
Input
3
1 1
3 1 2
Output
3
Input
3
1 1
3 1 3
Output
4
Note
In the first example the citizens on the square 1 can split into two groups 2 + 1, so that the second and on the third squares will have 3 citizens each.
In the second example no matter how citizens act the bandit can catch at least 4 citizens.
Submitted Solution:
```
import sys
from collections import deque
input = sys.stdin.buffer.readline
n = int(input())
e = [[] for _ in range(n+1)]
ls = list(map(int, input().split()))
for i, u in enumerate(ls):
e[u].append(i+2)
pp = [0] + list(map(int, input().split()))
q, order, par = deque([1]), [], [0]*(n+1)
while q:
u = q.popleft()
order.append(u)
for v in e[u]:
if v == par[u]: continue
par[v] = u
q.append(v)
leaf, cc = [0]*(n+1), 0
for u in reversed(order):
lcc = 0
for v in e[u]:
if v == par[u]: continue
lcc += leaf[v]
pp[u] += pp[v]
lcc = max(1, lcc)
leaf[u] = lcc
cc = max(cc, (pp[u]+lcc-1)//lcc)
print(cc)
``` | instruction | 0 | 61,678 | 1 | 123,356 |
Yes | output | 1 | 61,678 | 1 | 123,357 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Bandits appeared in the city! One of them is trying to catch as many citizens as he can.
The city consists of n squares connected by n-1 roads in such a way that it is possible to reach any square from any other square. The square number 1 is the main square.
After Sunday walk all the roads were changed to one-way roads in such a way that it is possible to reach any square from the main square.
At the moment when the bandit appeared on the main square there were a_i citizens on the i-th square. Now the following process will begin. First, each citizen that is currently on a square with some outgoing one-way roads chooses one of such roads and moves along it to another square. Then the bandit chooses one of the one-way roads outgoing from the square he is located and moves along it. The process is repeated until the bandit is located on a square with no outgoing roads. The bandit catches all the citizens on that square.
The bandit wants to catch as many citizens as possible; the citizens want to minimize the number of caught people. The bandit and the citizens know positions of all citizens at any time, the citizens can cooperate. If both sides act optimally, how many citizens will be caught?
Input
The first line contains a single integer n β the number of squares in the city (2 β€ n β€ 2β
10^5).
The second line contains n-1 integers p_2, p_3 ... p_n meaning that there is a one-way road from the square p_i to the square i (1 β€ p_i < i).
The third line contains n integers a_1, a_2, ..., a_n β the number of citizens on each square initially (0 β€ a_i β€ 10^9).
Output
Print a single integer β the number of citizens the bandit will catch if both sides act optimally.
Examples
Input
3
1 1
3 1 2
Output
3
Input
3
1 1
3 1 3
Output
4
Note
In the first example the citizens on the square 1 can split into two groups 2 + 1, so that the second and on the third squares will have 3 citizens each.
In the second example no matter how citizens act the bandit can catch at least 4 citizens.
Submitted Solution:
```
try:
n = int(input())
p = list(map(int, input().split()))
a = list(map(int, input().split()))
tree = {}
for i, pp in enumerate(p, start=2):
tree.setdefault(pp, []).append(i)
def proceed_square(s):
children = tree.get(s, [])
if len(children) == 0:
return a[s - 1], a[s - 1], 1
ch_p = [proceed_square(c) for c in children]
ch_max = max(pp[0] for pp in ch_p)
ch_sum = sum(pp[1] for pp in ch_p)
ch_l = sum(pp[2] for pp in ch_p)
s = ch_sum + a[s - 1]
m = s // ch_l
if s % ch_l != 0:
m += 1
return max(ch_max, m), s, ch_l
print(proceed_square(1)[0])
except Exception:
print([p[i] for i in range(len(p)) if p[i] != i + 1])
``` | instruction | 0 | 61,679 | 1 | 123,358 |
No | output | 1 | 61,679 | 1 | 123,359 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Bandits appeared in the city! One of them is trying to catch as many citizens as he can.
The city consists of n squares connected by n-1 roads in such a way that it is possible to reach any square from any other square. The square number 1 is the main square.
After Sunday walk all the roads were changed to one-way roads in such a way that it is possible to reach any square from the main square.
At the moment when the bandit appeared on the main square there were a_i citizens on the i-th square. Now the following process will begin. First, each citizen that is currently on a square with some outgoing one-way roads chooses one of such roads and moves along it to another square. Then the bandit chooses one of the one-way roads outgoing from the square he is located and moves along it. The process is repeated until the bandit is located on a square with no outgoing roads. The bandit catches all the citizens on that square.
The bandit wants to catch as many citizens as possible; the citizens want to minimize the number of caught people. The bandit and the citizens know positions of all citizens at any time, the citizens can cooperate. If both sides act optimally, how many citizens will be caught?
Input
The first line contains a single integer n β the number of squares in the city (2 β€ n β€ 2β
10^5).
The second line contains n-1 integers p_2, p_3 ... p_n meaning that there is a one-way road from the square p_i to the square i (1 β€ p_i < i).
The third line contains n integers a_1, a_2, ..., a_n β the number of citizens on each square initially (0 β€ a_i β€ 10^9).
Output
Print a single integer β the number of citizens the bandit will catch if both sides act optimally.
Examples
Input
3
1 1
3 1 2
Output
3
Input
3
1 1
3 1 3
Output
4
Note
In the first example the citizens on the square 1 can split into two groups 2 + 1, so that the second and on the third squares will have 3 citizens each.
In the second example no matter how citizens act the bandit can catch at least 4 citizens.
Submitted Solution:
```
from math import ceil
n = int(input())
p = list(map(int, input().split()))
a=list(map(int, input().split()))
from types import GeneratorType
def bootstrap(f, stack=[]):
def wrappedfunc(*args, **kwargs):
if stack:
return f(*args, **kwargs)
else:
to = f(*args, **kwargs)
while True:
if type(to) is GeneratorType:
stack.append(to)
to = next(to)
else:
stack.pop()
if not stack:
break
to = stack[-1].send(to)
return to
return wrappedfunc
class Graph:
def __init__(self, V):
self.V = V
self.adj = [[] for i in range(V)]
@bootstrap
def DFSUtil(self, temp, v, visited, sums, maxs, counts):
visited[v] = True
temp.append(v)
s=a[v]
count=0
maxis=[]
for i in self.adj[v]:
if visited[i] == False:
s+=a[i]
temp = yield self.DFSUtil(temp, i, visited, sums, maxs, counts)
count += counts[i]
maxis.append(maxs[i])
sums[v]=s
if count==0:
count+=1
counts[v]=count
if len(maxis)==0:
maxs[v]=a[v]
else:
maxs[v]=max(int(ceil(sums[v]/counts[v])),max(maxis))
yield temp
def addEdge(self, v, w):
# self.adj[v].append(w)
self.adj[w].append(v)
G = Graph(n)
for i in range(0, len(p)):
G.addEdge(i+1, p[i] - 1)
visited = [False for _ in range(n)]
sums = [0 for _ in range(n)]
maxs = [0 for _ in range(n)]
counts = [0 for _ in range(n)]
G.DFSUtil([], 0, visited, sums, maxs, counts)
# print(sums,counts,maxs)
print(maxs[0])
``` | instruction | 0 | 61,680 | 1 | 123,360 |
No | output | 1 | 61,680 | 1 | 123,361 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Bandits appeared in the city! One of them is trying to catch as many citizens as he can.
The city consists of n squares connected by n-1 roads in such a way that it is possible to reach any square from any other square. The square number 1 is the main square.
After Sunday walk all the roads were changed to one-way roads in such a way that it is possible to reach any square from the main square.
At the moment when the bandit appeared on the main square there were a_i citizens on the i-th square. Now the following process will begin. First, each citizen that is currently on a square with some outgoing one-way roads chooses one of such roads and moves along it to another square. Then the bandit chooses one of the one-way roads outgoing from the square he is located and moves along it. The process is repeated until the bandit is located on a square with no outgoing roads. The bandit catches all the citizens on that square.
The bandit wants to catch as many citizens as possible; the citizens want to minimize the number of caught people. The bandit and the citizens know positions of all citizens at any time, the citizens can cooperate. If both sides act optimally, how many citizens will be caught?
Input
The first line contains a single integer n β the number of squares in the city (2 β€ n β€ 2β
10^5).
The second line contains n-1 integers p_2, p_3 ... p_n meaning that there is a one-way road from the square p_i to the square i (1 β€ p_i < i).
The third line contains n integers a_1, a_2, ..., a_n β the number of citizens on each square initially (0 β€ a_i β€ 10^9).
Output
Print a single integer β the number of citizens the bandit will catch if both sides act optimally.
Examples
Input
3
1 1
3 1 2
Output
3
Input
3
1 1
3 1 3
Output
4
Note
In the first example the citizens on the square 1 can split into two groups 2 + 1, so that the second and on the third squares will have 3 citizens each.
In the second example no matter how citizens act the bandit can catch at least 4 citizens.
Submitted Solution:
```
from __future__ import print_function # for PyPy2
from collections import Counter, OrderedDict
from itertools import permutations as perm
from collections import deque
from sys import stdin
from bisect import *
from heapq import *
import math
g = lambda : stdin.readline().strip()
gl = lambda : g().split()
gil = lambda : [int(var) for var in gl()]
gfl = lambda : [float(var) for var in gl()]
gcl = lambda : list(g())
gbs = lambda : [int(var) for var in g()]
mod = int(1e9)+7
inf = float("inf")
n, = gil()
adj = [[] for _ in range(n+1)]
i = 2
for p in gil():
adj[p].append(i)
i += 1
provide = [0]+gil()
stack = [1]
vis = [0]*(n+1)
dp = [None for _ in range(n+1)]
leafCnt = 0
while stack:
p = stack[-1]
if vis[p] :
nMax = dp[max(adj[p], key=lambda x:dp[x][0])][0]
delta, cnt = 0, 0
for c in adj[p]:
cMax, cDelta, cCnt = dp[c]
delta += cDelta + (nMax-cMax)*cCnt
cnt += cCnt
if delta > provide[p]:
delta -= provide[p]
else:
nMax += math.ceil((provide[p]-delta)/cnt)
delta = (provide[p]-delta)%cnt
dp[p] = (nMax, delta, cnt)
stack.pop()
else:
vis[p] = 1
for c in adj[p]:
stack.append(c)
if len(adj[p]) == 0 :
dp[p] = (provide[p], 0, 1)
stack.pop()
leafCnt += 1
if leafCnt ==1 :
print(sum(provide))
else:
print(dp[1][0])
# print(dp)
``` | instruction | 0 | 61,681 | 1 | 123,362 |
No | output | 1 | 61,681 | 1 | 123,363 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Bandits appeared in the city! One of them is trying to catch as many citizens as he can.
The city consists of n squares connected by n-1 roads in such a way that it is possible to reach any square from any other square. The square number 1 is the main square.
After Sunday walk all the roads were changed to one-way roads in such a way that it is possible to reach any square from the main square.
At the moment when the bandit appeared on the main square there were a_i citizens on the i-th square. Now the following process will begin. First, each citizen that is currently on a square with some outgoing one-way roads chooses one of such roads and moves along it to another square. Then the bandit chooses one of the one-way roads outgoing from the square he is located and moves along it. The process is repeated until the bandit is located on a square with no outgoing roads. The bandit catches all the citizens on that square.
The bandit wants to catch as many citizens as possible; the citizens want to minimize the number of caught people. The bandit and the citizens know positions of all citizens at any time, the citizens can cooperate. If both sides act optimally, how many citizens will be caught?
Input
The first line contains a single integer n β the number of squares in the city (2 β€ n β€ 2β
10^5).
The second line contains n-1 integers p_2, p_3 ... p_n meaning that there is a one-way road from the square p_i to the square i (1 β€ p_i < i).
The third line contains n integers a_1, a_2, ..., a_n β the number of citizens on each square initially (0 β€ a_i β€ 10^9).
Output
Print a single integer β the number of citizens the bandit will catch if both sides act optimally.
Examples
Input
3
1 1
3 1 2
Output
3
Input
3
1 1
3 1 3
Output
4
Note
In the first example the citizens on the square 1 can split into two groups 2 + 1, so that the second and on the third squares will have 3 citizens each.
In the second example no matter how citizens act the bandit can catch at least 4 citizens.
Submitted Solution:
```
from bisect import bisect_left as bl
from bisect import bisect_right as br
from heapq import heappush,heappop
import math
from collections import *
from functools import reduce,cmp_to_key
import sys
input = sys.stdin.readline
M = mod = 998244353
def factors(n):return sorted(set(reduce(list.__add__, ([i, n//i] for i in range(1, int(n**0.5) + 1) if n % i == 0))))
def inv_mod(n):return pow(n, mod - 2, mod)
def li():return [int(i) for i in input().rstrip('\n').split()]
def st():return input().rstrip('\n')
def val():return int(input().rstrip('\n'))
def li2():return [i for i in input().rstrip('\n')]
def li3():return [int(i) for i in input().rstrip('\n')]
n = val()
parents = [1, 1] + li()
values = [0] + li()
sizes = [1] * (n + 5)
for i in range(2, n + 1, 1):
sizes[parents[i]] = 0
ans = -float('inf')
for i in range(n, 1, -1):
ans = max(ans, (values[i] + sizes[i] - 1) / sizes[i])
values[parents[i]] += values[i]
sizes[parents[i]] += sizes[i]
print(int(ans))
``` | instruction | 0 | 61,682 | 1 | 123,364 |
No | output | 1 | 61,682 | 1 | 123,365 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There are n cities on a two dimensional Cartesian plane. The distance between two cities is equal to the Manhattan distance between them (see the Notes for definition). A Hamiltonian cycle of the cities is defined as a permutation of all n cities. The length of this Hamiltonian cycle is defined as the sum of the distances between adjacent cities in the permutation plus the distance between the first and final city in the permutation. Please compute the longest possible length of a Hamiltonian cycle of the given cities.
Input
The first line contains an integer n (3 β€ n β€ 105). Then n lines follow, each consisting of two integers xi and yi (0 β€ xi, yi β€ 109), denoting the coordinates of a city. All given points will be distinct.
Output
A single line denoting the longest possible length of a Hamiltonian cycle of the given cities. You should not output the cycle, only its length.
Please, do not write the %lld specifier to read or write 64-bit integers in Π‘++. It is preferred to use the cin, cout streams or the %I64d specifier.
Examples
Input
4
1 1
1 2
2 1
2 2
Output
6
Note
In the example, one of the possible Hamiltonian cycles with length 6 is (1, 1) (1, 2) (2, 1) (2, 2). There does not exist any other Hamiltonian cycle with a length greater than 6.
The Manhattan distance between two cities (xi, yi) and (xj, yj) is |xi - xj| + |yi - yj|.
Submitted Solution:
```
n = int(input())
a = []
x=y=0
for i in range(n):
a += [[int(x) for x in input().split()]]
for i in range(n):
x += abs(a[i][0])
y += abs(a[i][1])
print((x+y)//2)
``` | instruction | 0 | 61,753 | 1 | 123,506 |
No | output | 1 | 61,753 | 1 | 123,507 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There are n cities on a two dimensional Cartesian plane. The distance between two cities is equal to the Manhattan distance between them (see the Notes for definition). A Hamiltonian cycle of the cities is defined as a permutation of all n cities. The length of this Hamiltonian cycle is defined as the sum of the distances between adjacent cities in the permutation plus the distance between the first and final city in the permutation. Please compute the longest possible length of a Hamiltonian cycle of the given cities.
Input
The first line contains an integer n (3 β€ n β€ 105). Then n lines follow, each consisting of two integers xi and yi (0 β€ xi, yi β€ 109), denoting the coordinates of a city. All given points will be distinct.
Output
A single line denoting the longest possible length of a Hamiltonian cycle of the given cities. You should not output the cycle, only its length.
Please, do not write the %lld specifier to read or write 64-bit integers in Π‘++. It is preferred to use the cin, cout streams or the %I64d specifier.
Examples
Input
4
1 1
1 2
2 1
2 2
Output
6
Note
In the example, one of the possible Hamiltonian cycles with length 6 is (1, 1) (1, 2) (2, 1) (2, 2). There does not exist any other Hamiltonian cycle with a length greater than 6.
The Manhattan distance between two cities (xi, yi) and (xj, yj) is |xi - xj| + |yi - yj|.
Submitted Solution:
```
#https://codeforces.com/problemset/problem/329/E
n = int(input())
a = []
x=y=0
for i in range(n):
a += [[int(x) for x in input().split()]]
for i in range(n):
x += abs(a[i][0])
y += abs(a[i][1])
print((x+y)//2)
``` | instruction | 0 | 61,754 | 1 | 123,508 |
No | output | 1 | 61,754 | 1 | 123,509 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There are n cities on a two dimensional Cartesian plane. The distance between two cities is equal to the Manhattan distance between them (see the Notes for definition). A Hamiltonian cycle of the cities is defined as a permutation of all n cities. The length of this Hamiltonian cycle is defined as the sum of the distances between adjacent cities in the permutation plus the distance between the first and final city in the permutation. Please compute the longest possible length of a Hamiltonian cycle of the given cities.
Input
The first line contains an integer n (3 β€ n β€ 105). Then n lines follow, each consisting of two integers xi and yi (0 β€ xi, yi β€ 109), denoting the coordinates of a city. All given points will be distinct.
Output
A single line denoting the longest possible length of a Hamiltonian cycle of the given cities. You should not output the cycle, only its length.
Please, do not write the %lld specifier to read or write 64-bit integers in Π‘++. It is preferred to use the cin, cout streams or the %I64d specifier.
Examples
Input
4
1 1
1 2
2 1
2 2
Output
6
Note
In the example, one of the possible Hamiltonian cycles with length 6 is (1, 1) (1, 2) (2, 1) (2, 2). There does not exist any other Hamiltonian cycle with a length greater than 6.
The Manhattan distance between two cities (xi, yi) and (xj, yj) is |xi - xj| + |yi - yj|.
Submitted Solution:
```
#https://codeforces.com/problemset/problem/329/E
n = int(input())
lst = []
x=0
y=0
for i in range(n):
lst += [[int(x) for x in input().split()]]
for i in range(n):
x += abs(lst[i][0])
y += abs(lst[i][1])
print((x+y)//2)
``` | instruction | 0 | 61,755 | 1 | 123,510 |
No | output | 1 | 61,755 | 1 | 123,511 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Vasiliy lives at point (a, b) of the coordinate plane. He is hurrying up to work so he wants to get out of his house as soon as possible. New app suggested n available Beru-taxi nearby. The i-th taxi is located at point (xi, yi) and moves with a speed vi.
Consider that each of n drivers will move directly to Vasiliy and with a maximum possible speed. Compute the minimum time when Vasiliy will get in any of Beru-taxi cars.
Input
The first line of the input contains two integers a and b ( - 100 β€ a, b β€ 100) β coordinates of Vasiliy's home.
The second line contains a single integer n (1 β€ n β€ 1000) β the number of available Beru-taxi cars nearby.
The i-th of the following n lines contains three integers xi, yi and vi ( - 100 β€ xi, yi β€ 100, 1 β€ vi β€ 100) β the coordinates of the i-th car and its speed.
It's allowed that several cars are located at the same point. Also, cars may be located at exactly the same point where Vasiliy lives.
Output
Print a single real value β the minimum time Vasiliy needs to get in any of the Beru-taxi cars. You answer will be considered correct if its absolute or relative error does not exceed 10 - 6.
Namely: let's assume that your answer is a, and the answer of the jury is b. The checker program will consider your answer correct, if <image>.
Examples
Input
0 0
2
2 0 1
0 2 2
Output
1.00000000000000000000
Input
1 3
3
3 3 2
-2 3 6
-2 7 10
Output
0.50000000000000000000
Note
In the first sample, first taxi will get to Vasiliy in time 2, and second will do this in time 1, therefore 1 is the answer.
In the second sample, cars 2 and 3 will arrive simultaneously. | instruction | 0 | 61,927 | 1 | 123,854 |
Tags: brute force, geometry, implementation
Correct Solution:
```
# sqrt((x2-x1)**2 + (y2-y1)**2)
a, b = map(int, input().split())
m = float("inf")
n = int(input())
for _ in range(n):
x, y, v = map(int, input().split())
dist = ((a-x)**2+(b-y)**2)**0.5
m = min(m, dist/v)
print('%.6f' % m)
``` | output | 1 | 61,927 | 1 | 123,855 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Vasiliy lives at point (a, b) of the coordinate plane. He is hurrying up to work so he wants to get out of his house as soon as possible. New app suggested n available Beru-taxi nearby. The i-th taxi is located at point (xi, yi) and moves with a speed vi.
Consider that each of n drivers will move directly to Vasiliy and with a maximum possible speed. Compute the minimum time when Vasiliy will get in any of Beru-taxi cars.
Input
The first line of the input contains two integers a and b ( - 100 β€ a, b β€ 100) β coordinates of Vasiliy's home.
The second line contains a single integer n (1 β€ n β€ 1000) β the number of available Beru-taxi cars nearby.
The i-th of the following n lines contains three integers xi, yi and vi ( - 100 β€ xi, yi β€ 100, 1 β€ vi β€ 100) β the coordinates of the i-th car and its speed.
It's allowed that several cars are located at the same point. Also, cars may be located at exactly the same point where Vasiliy lives.
Output
Print a single real value β the minimum time Vasiliy needs to get in any of the Beru-taxi cars. You answer will be considered correct if its absolute or relative error does not exceed 10 - 6.
Namely: let's assume that your answer is a, and the answer of the jury is b. The checker program will consider your answer correct, if <image>.
Examples
Input
0 0
2
2 0 1
0 2 2
Output
1.00000000000000000000
Input
1 3
3
3 3 2
-2 3 6
-2 7 10
Output
0.50000000000000000000
Note
In the first sample, first taxi will get to Vasiliy in time 2, and second will do this in time 1, therefore 1 is the answer.
In the second sample, cars 2 and 3 will arrive simultaneously. | instruction | 0 | 61,928 | 1 | 123,856 |
Tags: brute force, geometry, implementation
Correct Solution:
```
import math;
def calLength(p1, p2):
l = math.pow(p1[0] - p2[0], 2) + math.pow(p1[1] - p2[1], 2);
return math.sqrt(l);
def main():
getS = input();
getS = getS.split();
a = int(getS[0]);
b = int(getS[1]);
n = int(input());
v = [a, b];
taxi = [];
for i in range(0, n):
ta = [];
getS = input();
getS = getS.split();
ta.append(int(getS[0]));
ta.append(int(getS[1]));
ta.append(int(getS[2]));
taxi.append(ta);
re = False;
for i in range(0, len(taxi)):
t = calLength(v, taxi[i]) / taxi[i][2];
#print (calLength(v, taxi[i]) / taxi[i][2]);
if t == 0:
print (0);
return;
if (re == False) or (re != False and re > t):
re = t;
print (float(re));
main();
``` | output | 1 | 61,928 | 1 | 123,857 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Vasiliy lives at point (a, b) of the coordinate plane. He is hurrying up to work so he wants to get out of his house as soon as possible. New app suggested n available Beru-taxi nearby. The i-th taxi is located at point (xi, yi) and moves with a speed vi.
Consider that each of n drivers will move directly to Vasiliy and with a maximum possible speed. Compute the minimum time when Vasiliy will get in any of Beru-taxi cars.
Input
The first line of the input contains two integers a and b ( - 100 β€ a, b β€ 100) β coordinates of Vasiliy's home.
The second line contains a single integer n (1 β€ n β€ 1000) β the number of available Beru-taxi cars nearby.
The i-th of the following n lines contains three integers xi, yi and vi ( - 100 β€ xi, yi β€ 100, 1 β€ vi β€ 100) β the coordinates of the i-th car and its speed.
It's allowed that several cars are located at the same point. Also, cars may be located at exactly the same point where Vasiliy lives.
Output
Print a single real value β the minimum time Vasiliy needs to get in any of the Beru-taxi cars. You answer will be considered correct if its absolute or relative error does not exceed 10 - 6.
Namely: let's assume that your answer is a, and the answer of the jury is b. The checker program will consider your answer correct, if <image>.
Examples
Input
0 0
2
2 0 1
0 2 2
Output
1.00000000000000000000
Input
1 3
3
3 3 2
-2 3 6
-2 7 10
Output
0.50000000000000000000
Note
In the first sample, first taxi will get to Vasiliy in time 2, and second will do this in time 1, therefore 1 is the answer.
In the second sample, cars 2 and 3 will arrive simultaneously. | instruction | 0 | 61,929 | 1 | 123,858 |
Tags: brute force, geometry, implementation
Correct Solution:
```
import sys
from math import sqrt
(a,b) = map(int,input().split())
n = int(input())
mini = 99999
for i in range(n):
(x,y,v) = map(int, input().split())
t = sqrt( (x-a)**2 + (y - b)**2) / (v * 1.00000000000000000000)
if t < mini:
mini = t
print(mini)
``` | output | 1 | 61,929 | 1 | 123,859 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Vasiliy lives at point (a, b) of the coordinate plane. He is hurrying up to work so he wants to get out of his house as soon as possible. New app suggested n available Beru-taxi nearby. The i-th taxi is located at point (xi, yi) and moves with a speed vi.
Consider that each of n drivers will move directly to Vasiliy and with a maximum possible speed. Compute the minimum time when Vasiliy will get in any of Beru-taxi cars.
Input
The first line of the input contains two integers a and b ( - 100 β€ a, b β€ 100) β coordinates of Vasiliy's home.
The second line contains a single integer n (1 β€ n β€ 1000) β the number of available Beru-taxi cars nearby.
The i-th of the following n lines contains three integers xi, yi and vi ( - 100 β€ xi, yi β€ 100, 1 β€ vi β€ 100) β the coordinates of the i-th car and its speed.
It's allowed that several cars are located at the same point. Also, cars may be located at exactly the same point where Vasiliy lives.
Output
Print a single real value β the minimum time Vasiliy needs to get in any of the Beru-taxi cars. You answer will be considered correct if its absolute or relative error does not exceed 10 - 6.
Namely: let's assume that your answer is a, and the answer of the jury is b. The checker program will consider your answer correct, if <image>.
Examples
Input
0 0
2
2 0 1
0 2 2
Output
1.00000000000000000000
Input
1 3
3
3 3 2
-2 3 6
-2 7 10
Output
0.50000000000000000000
Note
In the first sample, first taxi will get to Vasiliy in time 2, and second will do this in time 1, therefore 1 is the answer.
In the second sample, cars 2 and 3 will arrive simultaneously. | instruction | 0 | 61,930 | 1 | 123,860 |
Tags: brute force, geometry, implementation
Correct Solution:
```
x,y=map(int,input().split())
n=int(input())
mi=11**11
for _ in range(n):
xt,yt,v=map(int,input().split())
t=((x-xt)**2+(y-yt)**2)**0.5/v
if t<mi: mi=t
print(mi)
``` | output | 1 | 61,930 | 1 | 123,861 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Vasiliy lives at point (a, b) of the coordinate plane. He is hurrying up to work so he wants to get out of his house as soon as possible. New app suggested n available Beru-taxi nearby. The i-th taxi is located at point (xi, yi) and moves with a speed vi.
Consider that each of n drivers will move directly to Vasiliy and with a maximum possible speed. Compute the minimum time when Vasiliy will get in any of Beru-taxi cars.
Input
The first line of the input contains two integers a and b ( - 100 β€ a, b β€ 100) β coordinates of Vasiliy's home.
The second line contains a single integer n (1 β€ n β€ 1000) β the number of available Beru-taxi cars nearby.
The i-th of the following n lines contains three integers xi, yi and vi ( - 100 β€ xi, yi β€ 100, 1 β€ vi β€ 100) β the coordinates of the i-th car and its speed.
It's allowed that several cars are located at the same point. Also, cars may be located at exactly the same point where Vasiliy lives.
Output
Print a single real value β the minimum time Vasiliy needs to get in any of the Beru-taxi cars. You answer will be considered correct if its absolute or relative error does not exceed 10 - 6.
Namely: let's assume that your answer is a, and the answer of the jury is b. The checker program will consider your answer correct, if <image>.
Examples
Input
0 0
2
2 0 1
0 2 2
Output
1.00000000000000000000
Input
1 3
3
3 3 2
-2 3 6
-2 7 10
Output
0.50000000000000000000
Note
In the first sample, first taxi will get to Vasiliy in time 2, and second will do this in time 1, therefore 1 is the answer.
In the second sample, cars 2 and 3 will arrive simultaneously. | instruction | 0 | 61,931 | 1 | 123,862 |
Tags: brute force, geometry, implementation
Correct Solution:
```
import math
CarPos = []
def LeastTime(x,y,n,CarPos):
time = []
for i in range(n):
for j in range(3):
dis = math.sqrt((CarPos[i] [0] - x )**2 + (CarPos[i] [1] - y )**2 )
t = dis / CarPos[i] [2]
time.append(t)
return min(time)
x,y = [int(a) for a in input().split() ]
n = int(input() )
CarVal = [ ]
# CarVal[0] = x
# CarVal[1] = y
# CarVal[2] = v
for i in range(n):
row = [int(a) for a in input().split() ]
CarVal.append(row)
print(LeastTime(x,y,n,CarVal) )
``` | output | 1 | 61,931 | 1 | 123,863 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Vasiliy lives at point (a, b) of the coordinate plane. He is hurrying up to work so he wants to get out of his house as soon as possible. New app suggested n available Beru-taxi nearby. The i-th taxi is located at point (xi, yi) and moves with a speed vi.
Consider that each of n drivers will move directly to Vasiliy and with a maximum possible speed. Compute the minimum time when Vasiliy will get in any of Beru-taxi cars.
Input
The first line of the input contains two integers a and b ( - 100 β€ a, b β€ 100) β coordinates of Vasiliy's home.
The second line contains a single integer n (1 β€ n β€ 1000) β the number of available Beru-taxi cars nearby.
The i-th of the following n lines contains three integers xi, yi and vi ( - 100 β€ xi, yi β€ 100, 1 β€ vi β€ 100) β the coordinates of the i-th car and its speed.
It's allowed that several cars are located at the same point. Also, cars may be located at exactly the same point where Vasiliy lives.
Output
Print a single real value β the minimum time Vasiliy needs to get in any of the Beru-taxi cars. You answer will be considered correct if its absolute or relative error does not exceed 10 - 6.
Namely: let's assume that your answer is a, and the answer of the jury is b. The checker program will consider your answer correct, if <image>.
Examples
Input
0 0
2
2 0 1
0 2 2
Output
1.00000000000000000000
Input
1 3
3
3 3 2
-2 3 6
-2 7 10
Output
0.50000000000000000000
Note
In the first sample, first taxi will get to Vasiliy in time 2, and second will do this in time 1, therefore 1 is the answer.
In the second sample, cars 2 and 3 will arrive simultaneously. | instruction | 0 | 61,932 | 1 | 123,864 |
Tags: brute force, geometry, implementation
Correct Solution:
```
import math as m
hx, hy = map(int, input().split());
n = int(input());
res = 1000;
for i in range(n):
x, y, v = map(int, input().split());
res = min((m.sqrt(pow((hx - x),2) + pow((hy - y), 2)))/v, res);
print(res)
``` | output | 1 | 61,932 | 1 | 123,865 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Vasiliy lives at point (a, b) of the coordinate plane. He is hurrying up to work so he wants to get out of his house as soon as possible. New app suggested n available Beru-taxi nearby. The i-th taxi is located at point (xi, yi) and moves with a speed vi.
Consider that each of n drivers will move directly to Vasiliy and with a maximum possible speed. Compute the minimum time when Vasiliy will get in any of Beru-taxi cars.
Input
The first line of the input contains two integers a and b ( - 100 β€ a, b β€ 100) β coordinates of Vasiliy's home.
The second line contains a single integer n (1 β€ n β€ 1000) β the number of available Beru-taxi cars nearby.
The i-th of the following n lines contains three integers xi, yi and vi ( - 100 β€ xi, yi β€ 100, 1 β€ vi β€ 100) β the coordinates of the i-th car and its speed.
It's allowed that several cars are located at the same point. Also, cars may be located at exactly the same point where Vasiliy lives.
Output
Print a single real value β the minimum time Vasiliy needs to get in any of the Beru-taxi cars. You answer will be considered correct if its absolute or relative error does not exceed 10 - 6.
Namely: let's assume that your answer is a, and the answer of the jury is b. The checker program will consider your answer correct, if <image>.
Examples
Input
0 0
2
2 0 1
0 2 2
Output
1.00000000000000000000
Input
1 3
3
3 3 2
-2 3 6
-2 7 10
Output
0.50000000000000000000
Note
In the first sample, first taxi will get to Vasiliy in time 2, and second will do this in time 1, therefore 1 is the answer.
In the second sample, cars 2 and 3 will arrive simultaneously. | instruction | 0 | 61,933 | 1 | 123,866 |
Tags: brute force, geometry, implementation
Correct Solution:
```
x,y =map(int, input().split())
n = int(input())
op = 10000.0
for i in range(n):
a,b,v = map(int , input().split())
d = ( (a-x)**2 + (b-y)**2 )** 0.5
#print(d,a,b,v,x,y)
if (i==1):
op = d/v
elif(op > d/v):
op = d/v
print("%.20f"%op)
``` | output | 1 | 61,933 | 1 | 123,867 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Vasiliy lives at point (a, b) of the coordinate plane. He is hurrying up to work so he wants to get out of his house as soon as possible. New app suggested n available Beru-taxi nearby. The i-th taxi is located at point (xi, yi) and moves with a speed vi.
Consider that each of n drivers will move directly to Vasiliy and with a maximum possible speed. Compute the minimum time when Vasiliy will get in any of Beru-taxi cars.
Input
The first line of the input contains two integers a and b ( - 100 β€ a, b β€ 100) β coordinates of Vasiliy's home.
The second line contains a single integer n (1 β€ n β€ 1000) β the number of available Beru-taxi cars nearby.
The i-th of the following n lines contains three integers xi, yi and vi ( - 100 β€ xi, yi β€ 100, 1 β€ vi β€ 100) β the coordinates of the i-th car and its speed.
It's allowed that several cars are located at the same point. Also, cars may be located at exactly the same point where Vasiliy lives.
Output
Print a single real value β the minimum time Vasiliy needs to get in any of the Beru-taxi cars. You answer will be considered correct if its absolute or relative error does not exceed 10 - 6.
Namely: let's assume that your answer is a, and the answer of the jury is b. The checker program will consider your answer correct, if <image>.
Examples
Input
0 0
2
2 0 1
0 2 2
Output
1.00000000000000000000
Input
1 3
3
3 3 2
-2 3 6
-2 7 10
Output
0.50000000000000000000
Note
In the first sample, first taxi will get to Vasiliy in time 2, and second will do this in time 1, therefore 1 is the answer.
In the second sample, cars 2 and 3 will arrive simultaneously. | instruction | 0 | 61,934 | 1 | 123,868 |
Tags: brute force, geometry, implementation
Correct Solution:
```
from math import sqrt
a, b = map(int, input().split())
n = int(input())
best_time = 999999999
for _ in range(n):
x, y, v = map(int, input().split())
time = sqrt((a-x) ** 2 + (b-y) ** 2) / v
if time < best_time:
best_time = time
print(best_time)
``` | output | 1 | 61,934 | 1 | 123,869 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Vasiliy lives at point (a, b) of the coordinate plane. He is hurrying up to work so he wants to get out of his house as soon as possible. New app suggested n available Beru-taxi nearby. The i-th taxi is located at point (xi, yi) and moves with a speed vi.
Consider that each of n drivers will move directly to Vasiliy and with a maximum possible speed. Compute the minimum time when Vasiliy will get in any of Beru-taxi cars.
Input
The first line of the input contains two integers a and b ( - 100 β€ a, b β€ 100) β coordinates of Vasiliy's home.
The second line contains a single integer n (1 β€ n β€ 1000) β the number of available Beru-taxi cars nearby.
The i-th of the following n lines contains three integers xi, yi and vi ( - 100 β€ xi, yi β€ 100, 1 β€ vi β€ 100) β the coordinates of the i-th car and its speed.
It's allowed that several cars are located at the same point. Also, cars may be located at exactly the same point where Vasiliy lives.
Output
Print a single real value β the minimum time Vasiliy needs to get in any of the Beru-taxi cars. You answer will be considered correct if its absolute or relative error does not exceed 10 - 6.
Namely: let's assume that your answer is a, and the answer of the jury is b. The checker program will consider your answer correct, if <image>.
Examples
Input
0 0
2
2 0 1
0 2 2
Output
1.00000000000000000000
Input
1 3
3
3 3 2
-2 3 6
-2 7 10
Output
0.50000000000000000000
Note
In the first sample, first taxi will get to Vasiliy in time 2, and second will do this in time 1, therefore 1 is the answer.
In the second sample, cars 2 and 3 will arrive simultaneously.
Submitted Solution:
```
a,b=(int(i)for i in input().split())
n=int(input())
k=500
for i in range(n):
c,d,v=map(int,input().split())
k=min(k,(((a-c)**2+(b-d)**2)**0.5)/v )
print(k)
``` | instruction | 0 | 61,935 | 1 | 123,870 |
Yes | output | 1 | 61,935 | 1 | 123,871 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Vasiliy lives at point (a, b) of the coordinate plane. He is hurrying up to work so he wants to get out of his house as soon as possible. New app suggested n available Beru-taxi nearby. The i-th taxi is located at point (xi, yi) and moves with a speed vi.
Consider that each of n drivers will move directly to Vasiliy and with a maximum possible speed. Compute the minimum time when Vasiliy will get in any of Beru-taxi cars.
Input
The first line of the input contains two integers a and b ( - 100 β€ a, b β€ 100) β coordinates of Vasiliy's home.
The second line contains a single integer n (1 β€ n β€ 1000) β the number of available Beru-taxi cars nearby.
The i-th of the following n lines contains three integers xi, yi and vi ( - 100 β€ xi, yi β€ 100, 1 β€ vi β€ 100) β the coordinates of the i-th car and its speed.
It's allowed that several cars are located at the same point. Also, cars may be located at exactly the same point where Vasiliy lives.
Output
Print a single real value β the minimum time Vasiliy needs to get in any of the Beru-taxi cars. You answer will be considered correct if its absolute or relative error does not exceed 10 - 6.
Namely: let's assume that your answer is a, and the answer of the jury is b. The checker program will consider your answer correct, if <image>.
Examples
Input
0 0
2
2 0 1
0 2 2
Output
1.00000000000000000000
Input
1 3
3
3 3 2
-2 3 6
-2 7 10
Output
0.50000000000000000000
Note
In the first sample, first taxi will get to Vasiliy in time 2, and second will do this in time 1, therefore 1 is the answer.
In the second sample, cars 2 and 3 will arrive simultaneously.
Submitted Solution:
```
import math
a,b=map(int,input().split())
n=int(input())
ans=10000000000000000000000000000000000000000000000000
for _ in range(n):
x,y,v=map(int,input().split())
ans = min(ans,math.pow(math.pow(a-x,2)+math.pow(b-y,2),.5)/v)
print("{0:.20f}".format(ans))
``` | instruction | 0 | 61,936 | 1 | 123,872 |
Yes | output | 1 | 61,936 | 1 | 123,873 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Vasiliy lives at point (a, b) of the coordinate plane. He is hurrying up to work so he wants to get out of his house as soon as possible. New app suggested n available Beru-taxi nearby. The i-th taxi is located at point (xi, yi) and moves with a speed vi.
Consider that each of n drivers will move directly to Vasiliy and with a maximum possible speed. Compute the minimum time when Vasiliy will get in any of Beru-taxi cars.
Input
The first line of the input contains two integers a and b ( - 100 β€ a, b β€ 100) β coordinates of Vasiliy's home.
The second line contains a single integer n (1 β€ n β€ 1000) β the number of available Beru-taxi cars nearby.
The i-th of the following n lines contains three integers xi, yi and vi ( - 100 β€ xi, yi β€ 100, 1 β€ vi β€ 100) β the coordinates of the i-th car and its speed.
It's allowed that several cars are located at the same point. Also, cars may be located at exactly the same point where Vasiliy lives.
Output
Print a single real value β the minimum time Vasiliy needs to get in any of the Beru-taxi cars. You answer will be considered correct if its absolute or relative error does not exceed 10 - 6.
Namely: let's assume that your answer is a, and the answer of the jury is b. The checker program will consider your answer correct, if <image>.
Examples
Input
0 0
2
2 0 1
0 2 2
Output
1.00000000000000000000
Input
1 3
3
3 3 2
-2 3 6
-2 7 10
Output
0.50000000000000000000
Note
In the first sample, first taxi will get to Vasiliy in time 2, and second will do this in time 1, therefore 1 is the answer.
In the second sample, cars 2 and 3 will arrive simultaneously.
Submitted Solution:
```
from math import sqrt
x,y = map(int,input().split())
n = int(input())
t = 10**9
for i in range(n):
j,k,v = map(int,input().split())
d = sqrt((x-j)**2 + (y-k)**2)/v
t = min(t,d)
print(t)
``` | instruction | 0 | 61,937 | 1 | 123,874 |
Yes | output | 1 | 61,937 | 1 | 123,875 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Vasiliy lives at point (a, b) of the coordinate plane. He is hurrying up to work so he wants to get out of his house as soon as possible. New app suggested n available Beru-taxi nearby. The i-th taxi is located at point (xi, yi) and moves with a speed vi.
Consider that each of n drivers will move directly to Vasiliy and with a maximum possible speed. Compute the minimum time when Vasiliy will get in any of Beru-taxi cars.
Input
The first line of the input contains two integers a and b ( - 100 β€ a, b β€ 100) β coordinates of Vasiliy's home.
The second line contains a single integer n (1 β€ n β€ 1000) β the number of available Beru-taxi cars nearby.
The i-th of the following n lines contains three integers xi, yi and vi ( - 100 β€ xi, yi β€ 100, 1 β€ vi β€ 100) β the coordinates of the i-th car and its speed.
It's allowed that several cars are located at the same point. Also, cars may be located at exactly the same point where Vasiliy lives.
Output
Print a single real value β the minimum time Vasiliy needs to get in any of the Beru-taxi cars. You answer will be considered correct if its absolute or relative error does not exceed 10 - 6.
Namely: let's assume that your answer is a, and the answer of the jury is b. The checker program will consider your answer correct, if <image>.
Examples
Input
0 0
2
2 0 1
0 2 2
Output
1.00000000000000000000
Input
1 3
3
3 3 2
-2 3 6
-2 7 10
Output
0.50000000000000000000
Note
In the first sample, first taxi will get to Vasiliy in time 2, and second will do this in time 1, therefore 1 is the answer.
In the second sample, cars 2 and 3 will arrive simultaneously.
Submitted Solution:
```
import math, sys
def solve():
bx,by = map(int, input().split())
dist = lambda f1, f2, v: math.sqrt(f1*f1+f2*f2) / v
n = int(input())
ans = sys.maxsize
for _ in range(n):
x,y,v = map(int, input().split())
ans = min(ans, dist(x-bx, y-by,v))
return ans
print(solve())
``` | instruction | 0 | 61,938 | 1 | 123,876 |
Yes | output | 1 | 61,938 | 1 | 123,877 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Vasiliy lives at point (a, b) of the coordinate plane. He is hurrying up to work so he wants to get out of his house as soon as possible. New app suggested n available Beru-taxi nearby. The i-th taxi is located at point (xi, yi) and moves with a speed vi.
Consider that each of n drivers will move directly to Vasiliy and with a maximum possible speed. Compute the minimum time when Vasiliy will get in any of Beru-taxi cars.
Input
The first line of the input contains two integers a and b ( - 100 β€ a, b β€ 100) β coordinates of Vasiliy's home.
The second line contains a single integer n (1 β€ n β€ 1000) β the number of available Beru-taxi cars nearby.
The i-th of the following n lines contains three integers xi, yi and vi ( - 100 β€ xi, yi β€ 100, 1 β€ vi β€ 100) β the coordinates of the i-th car and its speed.
It's allowed that several cars are located at the same point. Also, cars may be located at exactly the same point where Vasiliy lives.
Output
Print a single real value β the minimum time Vasiliy needs to get in any of the Beru-taxi cars. You answer will be considered correct if its absolute or relative error does not exceed 10 - 6.
Namely: let's assume that your answer is a, and the answer of the jury is b. The checker program will consider your answer correct, if <image>.
Examples
Input
0 0
2
2 0 1
0 2 2
Output
1.00000000000000000000
Input
1 3
3
3 3 2
-2 3 6
-2 7 10
Output
0.50000000000000000000
Note
In the first sample, first taxi will get to Vasiliy in time 2, and second will do this in time 1, therefore 1 is the answer.
In the second sample, cars 2 and 3 will arrive simultaneously.
Submitted Solution:
```
import math
a,b = map(int, input().split())
n = int(input())
min = 100000000
for i in range(n):
x, y, v = map(int, input().split())
x1 = math.fabs(x - a)
y1 = math.fabs(y - b)
dist = math.sqrt(x1 * x1 + y1 * y1)
time = dist / v
if time < min:
min = time
print(time)
``` | instruction | 0 | 61,939 | 1 | 123,878 |
No | output | 1 | 61,939 | 1 | 123,879 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Vasiliy lives at point (a, b) of the coordinate plane. He is hurrying up to work so he wants to get out of his house as soon as possible. New app suggested n available Beru-taxi nearby. The i-th taxi is located at point (xi, yi) and moves with a speed vi.
Consider that each of n drivers will move directly to Vasiliy and with a maximum possible speed. Compute the minimum time when Vasiliy will get in any of Beru-taxi cars.
Input
The first line of the input contains two integers a and b ( - 100 β€ a, b β€ 100) β coordinates of Vasiliy's home.
The second line contains a single integer n (1 β€ n β€ 1000) β the number of available Beru-taxi cars nearby.
The i-th of the following n lines contains three integers xi, yi and vi ( - 100 β€ xi, yi β€ 100, 1 β€ vi β€ 100) β the coordinates of the i-th car and its speed.
It's allowed that several cars are located at the same point. Also, cars may be located at exactly the same point where Vasiliy lives.
Output
Print a single real value β the minimum time Vasiliy needs to get in any of the Beru-taxi cars. You answer will be considered correct if its absolute or relative error does not exceed 10 - 6.
Namely: let's assume that your answer is a, and the answer of the jury is b. The checker program will consider your answer correct, if <image>.
Examples
Input
0 0
2
2 0 1
0 2 2
Output
1.00000000000000000000
Input
1 3
3
3 3 2
-2 3 6
-2 7 10
Output
0.50000000000000000000
Note
In the first sample, first taxi will get to Vasiliy in time 2, and second will do this in time 1, therefore 1 is the answer.
In the second sample, cars 2 and 3 will arrive simultaneously.
Submitted Solution:
```
import math
a, b = input().split()
a = int(a)
b = int(b)
n = input()
n = int(n)
min = 999999999999999999999999999999999999999999
for i in range(n):
x, y, z = input().split()
x = int(x)
y = int(y)
z = int(z)
d = (math.sqrt((a - x)**2 + (b - y)**2))/z
if min > d:
min = d
print(d)
``` | instruction | 0 | 61,940 | 1 | 123,880 |
No | output | 1 | 61,940 | 1 | 123,881 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Vasiliy lives at point (a, b) of the coordinate plane. He is hurrying up to work so he wants to get out of his house as soon as possible. New app suggested n available Beru-taxi nearby. The i-th taxi is located at point (xi, yi) and moves with a speed vi.
Consider that each of n drivers will move directly to Vasiliy and with a maximum possible speed. Compute the minimum time when Vasiliy will get in any of Beru-taxi cars.
Input
The first line of the input contains two integers a and b ( - 100 β€ a, b β€ 100) β coordinates of Vasiliy's home.
The second line contains a single integer n (1 β€ n β€ 1000) β the number of available Beru-taxi cars nearby.
The i-th of the following n lines contains three integers xi, yi and vi ( - 100 β€ xi, yi β€ 100, 1 β€ vi β€ 100) β the coordinates of the i-th car and its speed.
It's allowed that several cars are located at the same point. Also, cars may be located at exactly the same point where Vasiliy lives.
Output
Print a single real value β the minimum time Vasiliy needs to get in any of the Beru-taxi cars. You answer will be considered correct if its absolute or relative error does not exceed 10 - 6.
Namely: let's assume that your answer is a, and the answer of the jury is b. The checker program will consider your answer correct, if <image>.
Examples
Input
0 0
2
2 0 1
0 2 2
Output
1.00000000000000000000
Input
1 3
3
3 3 2
-2 3 6
-2 7 10
Output
0.50000000000000000000
Note
In the first sample, first taxi will get to Vasiliy in time 2, and second will do this in time 1, therefore 1 is the answer.
In the second sample, cars 2 and 3 will arrive simultaneously.
Submitted Solution:
```
a, b = map(int, input().split()) #coordinates of home
n = int(input()) #avaiable taxies in nearby
c = []
for i in range(n):
x, y, v = map(int, input().split())
tx = abs((a-x)) / v
ty = abs((b-y)) / v
c.append(tx+ty)
print(min(c))
``` | instruction | 0 | 61,941 | 1 | 123,882 |
No | output | 1 | 61,941 | 1 | 123,883 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Vasiliy lives at point (a, b) of the coordinate plane. He is hurrying up to work so he wants to get out of his house as soon as possible. New app suggested n available Beru-taxi nearby. The i-th taxi is located at point (xi, yi) and moves with a speed vi.
Consider that each of n drivers will move directly to Vasiliy and with a maximum possible speed. Compute the minimum time when Vasiliy will get in any of Beru-taxi cars.
Input
The first line of the input contains two integers a and b ( - 100 β€ a, b β€ 100) β coordinates of Vasiliy's home.
The second line contains a single integer n (1 β€ n β€ 1000) β the number of available Beru-taxi cars nearby.
The i-th of the following n lines contains three integers xi, yi and vi ( - 100 β€ xi, yi β€ 100, 1 β€ vi β€ 100) β the coordinates of the i-th car and its speed.
It's allowed that several cars are located at the same point. Also, cars may be located at exactly the same point where Vasiliy lives.
Output
Print a single real value β the minimum time Vasiliy needs to get in any of the Beru-taxi cars. You answer will be considered correct if its absolute or relative error does not exceed 10 - 6.
Namely: let's assume that your answer is a, and the answer of the jury is b. The checker program will consider your answer correct, if <image>.
Examples
Input
0 0
2
2 0 1
0 2 2
Output
1.00000000000000000000
Input
1 3
3
3 3 2
-2 3 6
-2 7 10
Output
0.50000000000000000000
Note
In the first sample, first taxi will get to Vasiliy in time 2, and second will do this in time 1, therefore 1 is the answer.
In the second sample, cars 2 and 3 will arrive simultaneously.
Submitted Solution:
```
from math import sqrt
coords = [int(i) for i in input().strip().split(' ')]
n = int(input().strip())
taxi_coords = []
for i in range(n):
taxi_coords.append([int(j) for j in input().strip().split(' ')])
my_min = sqrt((coords[0] - taxi_coords[0][0]) ** 2 + (coords[1] - taxi_coords[0][1]) ** 2) / taxi_coords[0][2]
for taxi in taxi_coords:
distance = sqrt((coords[0] - taxi[0]) ** 2 + (coords[1] - taxi[1]) ** 2) / taxi[2]
if distance < my_min:
my_min = distance
print(distance)
``` | instruction | 0 | 61,942 | 1 | 123,884 |
No | output | 1 | 61,942 | 1 | 123,885 |
Provide a correct Python 3 solution for this coding contest problem.
There are four towns, numbered 1,2,3 and 4. Also, there are three roads. The i-th road connects different towns a_i and b_i bidirectionally. No two roads connect the same pair of towns. Other than these roads, there is no way to travel between these towns, but any town can be reached from any other town using these roads.
Determine if we can visit all the towns by traversing each of the roads exactly once.
Constraints
* 1 \leq a_i,b_i \leq 4(1\leq i\leq 3)
* a_i and b_i are different. (1\leq i\leq 3)
* No two roads connect the same pair of towns.
* Any town can be reached from any other town using the roads.
Input
Input is given from Standard Input in the following format:
a_1 b_1
a_2 b_2
a_3 b_3
Output
If we can visit all the towns by traversing each of the roads exactly once, print `YES`; otherwise, print `NO`.
Examples
Input
4 2
1 3
2 3
Output
YES
Input
3 2
2 4
1 2
Output
NO
Input
2 1
3 2
4 3
Output
YES | instruction | 0 | 62,093 | 1 | 124,186 |
"Correct Solution:
```
t = [0] * 4
for i in range(3):
a, b = map(int, input().split())
t[a - 1] += 1
t[b - 1] += 1
print('NO') if max(t)==3 else print('YES')
``` | output | 1 | 62,093 | 1 | 124,187 |
Provide a correct Python 3 solution for this coding contest problem.
There are four towns, numbered 1,2,3 and 4. Also, there are three roads. The i-th road connects different towns a_i and b_i bidirectionally. No two roads connect the same pair of towns. Other than these roads, there is no way to travel between these towns, but any town can be reached from any other town using these roads.
Determine if we can visit all the towns by traversing each of the roads exactly once.
Constraints
* 1 \leq a_i,b_i \leq 4(1\leq i\leq 3)
* a_i and b_i are different. (1\leq i\leq 3)
* No two roads connect the same pair of towns.
* Any town can be reached from any other town using the roads.
Input
Input is given from Standard Input in the following format:
a_1 b_1
a_2 b_2
a_3 b_3
Output
If we can visit all the towns by traversing each of the roads exactly once, print `YES`; otherwise, print `NO`.
Examples
Input
4 2
1 3
2 3
Output
YES
Input
3 2
2 4
1 2
Output
NO
Input
2 1
3 2
4 3
Output
YES | instruction | 0 | 62,094 | 1 | 124,188 |
"Correct Solution:
```
a=[0]*4
for i in range(3):
x,y=map(int,input().split())
a[x-1]+=1
a[y-1]+=1
c=0
for i in range(4):
if a[i]%2!=0:
c+=1
if c==0 or c==2:
print("YES")
else:
print("NO")
``` | output | 1 | 62,094 | 1 | 124,189 |
Provide a correct Python 3 solution for this coding contest problem.
There are four towns, numbered 1,2,3 and 4. Also, there are three roads. The i-th road connects different towns a_i and b_i bidirectionally. No two roads connect the same pair of towns. Other than these roads, there is no way to travel between these towns, but any town can be reached from any other town using these roads.
Determine if we can visit all the towns by traversing each of the roads exactly once.
Constraints
* 1 \leq a_i,b_i \leq 4(1\leq i\leq 3)
* a_i and b_i are different. (1\leq i\leq 3)
* No two roads connect the same pair of towns.
* Any town can be reached from any other town using the roads.
Input
Input is given from Standard Input in the following format:
a_1 b_1
a_2 b_2
a_3 b_3
Output
If we can visit all the towns by traversing each of the roads exactly once, print `YES`; otherwise, print `NO`.
Examples
Input
4 2
1 3
2 3
Output
YES
Input
3 2
2 4
1 2
Output
NO
Input
2 1
3 2
4 3
Output
YES | instruction | 0 | 62,095 | 1 | 124,190 |
"Correct Solution:
```
c = [0]*4
for i in range(3):
a,b = map(int,input().split())
c[a-1] += 1
c[b-1] += 1
c.sort()
print('YES' if c==[1,1,2,2] else 'NO')
``` | output | 1 | 62,095 | 1 | 124,191 |
Provide a correct Python 3 solution for this coding contest problem.
There are four towns, numbered 1,2,3 and 4. Also, there are three roads. The i-th road connects different towns a_i and b_i bidirectionally. No two roads connect the same pair of towns. Other than these roads, there is no way to travel between these towns, but any town can be reached from any other town using these roads.
Determine if we can visit all the towns by traversing each of the roads exactly once.
Constraints
* 1 \leq a_i,b_i \leq 4(1\leq i\leq 3)
* a_i and b_i are different. (1\leq i\leq 3)
* No two roads connect the same pair of towns.
* Any town can be reached from any other town using the roads.
Input
Input is given from Standard Input in the following format:
a_1 b_1
a_2 b_2
a_3 b_3
Output
If we can visit all the towns by traversing each of the roads exactly once, print `YES`; otherwise, print `NO`.
Examples
Input
4 2
1 3
2 3
Output
YES
Input
3 2
2 4
1 2
Output
NO
Input
2 1
3 2
4 3
Output
YES | instruction | 0 | 62,096 | 1 | 124,192 |
"Correct Solution:
```
town = [0] * 4
for _ in range(3):
a, b = map(int, input().split())
town[a-1] += 1
town[b-1] += 1
if 3 in town or 4 in town:
ans = 'NO'
else:
ans = 'YES'
print(ans)
``` | output | 1 | 62,096 | 1 | 124,193 |
Provide a correct Python 3 solution for this coding contest problem.
There are four towns, numbered 1,2,3 and 4. Also, there are three roads. The i-th road connects different towns a_i and b_i bidirectionally. No two roads connect the same pair of towns. Other than these roads, there is no way to travel between these towns, but any town can be reached from any other town using these roads.
Determine if we can visit all the towns by traversing each of the roads exactly once.
Constraints
* 1 \leq a_i,b_i \leq 4(1\leq i\leq 3)
* a_i and b_i are different. (1\leq i\leq 3)
* No two roads connect the same pair of towns.
* Any town can be reached from any other town using the roads.
Input
Input is given from Standard Input in the following format:
a_1 b_1
a_2 b_2
a_3 b_3
Output
If we can visit all the towns by traversing each of the roads exactly once, print `YES`; otherwise, print `NO`.
Examples
Input
4 2
1 3
2 3
Output
YES
Input
3 2
2 4
1 2
Output
NO
Input
2 1
3 2
4 3
Output
YES | instruction | 0 | 62,097 | 1 | 124,194 |
"Correct Solution:
```
X=[0]*4
for i in range(3):
A,B=map(int,input().split())
X[A-1]+=1
X[B-1]+=1
X.sort()
if(X[0]==1 and X[1]==1 and X[2]==2):
print("YES")
else:
print("NO")
``` | output | 1 | 62,097 | 1 | 124,195 |
Provide a correct Python 3 solution for this coding contest problem.
There are four towns, numbered 1,2,3 and 4. Also, there are three roads. The i-th road connects different towns a_i and b_i bidirectionally. No two roads connect the same pair of towns. Other than these roads, there is no way to travel between these towns, but any town can be reached from any other town using these roads.
Determine if we can visit all the towns by traversing each of the roads exactly once.
Constraints
* 1 \leq a_i,b_i \leq 4(1\leq i\leq 3)
* a_i and b_i are different. (1\leq i\leq 3)
* No two roads connect the same pair of towns.
* Any town can be reached from any other town using the roads.
Input
Input is given from Standard Input in the following format:
a_1 b_1
a_2 b_2
a_3 b_3
Output
If we can visit all the towns by traversing each of the roads exactly once, print `YES`; otherwise, print `NO`.
Examples
Input
4 2
1 3
2 3
Output
YES
Input
3 2
2 4
1 2
Output
NO
Input
2 1
3 2
4 3
Output
YES | instruction | 0 | 62,098 | 1 | 124,196 |
"Correct Solution:
```
X = [0 for _ in range(4)]
for _ in range(3):
a, b = [int(_) for _ in input().split()]
X[a-1] += 1
X[b-1] += 1
if max(X) > 2:
print("NO")
else:
print("YES")
``` | output | 1 | 62,098 | 1 | 124,197 |
Provide a correct Python 3 solution for this coding contest problem.
There are four towns, numbered 1,2,3 and 4. Also, there are three roads. The i-th road connects different towns a_i and b_i bidirectionally. No two roads connect the same pair of towns. Other than these roads, there is no way to travel between these towns, but any town can be reached from any other town using these roads.
Determine if we can visit all the towns by traversing each of the roads exactly once.
Constraints
* 1 \leq a_i,b_i \leq 4(1\leq i\leq 3)
* a_i and b_i are different. (1\leq i\leq 3)
* No two roads connect the same pair of towns.
* Any town can be reached from any other town using the roads.
Input
Input is given from Standard Input in the following format:
a_1 b_1
a_2 b_2
a_3 b_3
Output
If we can visit all the towns by traversing each of the roads exactly once, print `YES`; otherwise, print `NO`.
Examples
Input
4 2
1 3
2 3
Output
YES
Input
3 2
2 4
1 2
Output
NO
Input
2 1
3 2
4 3
Output
YES | instruction | 0 | 62,099 | 1 | 124,198 |
"Correct Solution:
```
g=[0]*5
import sys
input=sys.stdin.readline
for _ in range(3):
a,b=map(int,input().split())
g[a]+=1
g[b]+=1
if g.count(1)==2:
print("YES")
else:
print("NO")
``` | output | 1 | 62,099 | 1 | 124,199 |
Provide a correct Python 3 solution for this coding contest problem.
There are four towns, numbered 1,2,3 and 4. Also, there are three roads. The i-th road connects different towns a_i and b_i bidirectionally. No two roads connect the same pair of towns. Other than these roads, there is no way to travel between these towns, but any town can be reached from any other town using these roads.
Determine if we can visit all the towns by traversing each of the roads exactly once.
Constraints
* 1 \leq a_i,b_i \leq 4(1\leq i\leq 3)
* a_i and b_i are different. (1\leq i\leq 3)
* No two roads connect the same pair of towns.
* Any town can be reached from any other town using the roads.
Input
Input is given from Standard Input in the following format:
a_1 b_1
a_2 b_2
a_3 b_3
Output
If we can visit all the towns by traversing each of the roads exactly once, print `YES`; otherwise, print `NO`.
Examples
Input
4 2
1 3
2 3
Output
YES
Input
3 2
2 4
1 2
Output
NO
Input
2 1
3 2
4 3
Output
YES | instruction | 0 | 62,100 | 1 | 124,200 |
"Correct Solution:
```
lis = [0] * 5
for i in range(3):
a,b = map(int,input().split())
lis[a] += 1
lis[b] += 1
if max(lis) == 3:
print ("NO")
else:
print ("YES")
``` | output | 1 | 62,100 | 1 | 124,201 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There are four towns, numbered 1,2,3 and 4. Also, there are three roads. The i-th road connects different towns a_i and b_i bidirectionally. No two roads connect the same pair of towns. Other than these roads, there is no way to travel between these towns, but any town can be reached from any other town using these roads.
Determine if we can visit all the towns by traversing each of the roads exactly once.
Constraints
* 1 \leq a_i,b_i \leq 4(1\leq i\leq 3)
* a_i and b_i are different. (1\leq i\leq 3)
* No two roads connect the same pair of towns.
* Any town can be reached from any other town using the roads.
Input
Input is given from Standard Input in the following format:
a_1 b_1
a_2 b_2
a_3 b_3
Output
If we can visit all the towns by traversing each of the roads exactly once, print `YES`; otherwise, print `NO`.
Examples
Input
4 2
1 3
2 3
Output
YES
Input
3 2
2 4
1 2
Output
NO
Input
2 1
3 2
4 3
Output
YES
Submitted Solution:
```
ans_list = [0]*4
for i in range(3):
a, b = [ int(v)-1 for v in input().split() ]
ans_list[a] += 1
ans_list[b] += 1
print( "YES" if sorted(ans_list) == [1,1,2,2] else "NO" )
``` | instruction | 0 | 62,101 | 1 | 124,202 |
Yes | output | 1 | 62,101 | 1 | 124,203 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There are four towns, numbered 1,2,3 and 4. Also, there are three roads. The i-th road connects different towns a_i and b_i bidirectionally. No two roads connect the same pair of towns. Other than these roads, there is no way to travel between these towns, but any town can be reached from any other town using these roads.
Determine if we can visit all the towns by traversing each of the roads exactly once.
Constraints
* 1 \leq a_i,b_i \leq 4(1\leq i\leq 3)
* a_i and b_i are different. (1\leq i\leq 3)
* No two roads connect the same pair of towns.
* Any town can be reached from any other town using the roads.
Input
Input is given from Standard Input in the following format:
a_1 b_1
a_2 b_2
a_3 b_3
Output
If we can visit all the towns by traversing each of the roads exactly once, print `YES`; otherwise, print `NO`.
Examples
Input
4 2
1 3
2 3
Output
YES
Input
3 2
2 4
1 2
Output
NO
Input
2 1
3 2
4 3
Output
YES
Submitted Solution:
```
l=[]
for i in range(3):
l.extend(list(map(int,input().split())))
m=[0,0,0,0]
for i in range(1,5):
m[i-1]=l.count(i)
if sorted(m) == [1,1,2,2]:
print("YES")
else:
print("NO")
``` | instruction | 0 | 62,102 | 1 | 124,204 |
Yes | output | 1 | 62,102 | 1 | 124,205 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There are four towns, numbered 1,2,3 and 4. Also, there are three roads. The i-th road connects different towns a_i and b_i bidirectionally. No two roads connect the same pair of towns. Other than these roads, there is no way to travel between these towns, but any town can be reached from any other town using these roads.
Determine if we can visit all the towns by traversing each of the roads exactly once.
Constraints
* 1 \leq a_i,b_i \leq 4(1\leq i\leq 3)
* a_i and b_i are different. (1\leq i\leq 3)
* No two roads connect the same pair of towns.
* Any town can be reached from any other town using the roads.
Input
Input is given from Standard Input in the following format:
a_1 b_1
a_2 b_2
a_3 b_3
Output
If we can visit all the towns by traversing each of the roads exactly once, print `YES`; otherwise, print `NO`.
Examples
Input
4 2
1 3
2 3
Output
YES
Input
3 2
2 4
1 2
Output
NO
Input
2 1
3 2
4 3
Output
YES
Submitted Solution:
```
AB = [input() for _ in range(3)]
AB = list(map(int, ' '.join(AB).split()))
c = [0] * 4
for ab in AB:
c[ab-1] += 1
if max(c) > 2:
print("NO")
else:
print("YES")
``` | instruction | 0 | 62,103 | 1 | 124,206 |
Yes | output | 1 | 62,103 | 1 | 124,207 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There are four towns, numbered 1,2,3 and 4. Also, there are three roads. The i-th road connects different towns a_i and b_i bidirectionally. No two roads connect the same pair of towns. Other than these roads, there is no way to travel between these towns, but any town can be reached from any other town using these roads.
Determine if we can visit all the towns by traversing each of the roads exactly once.
Constraints
* 1 \leq a_i,b_i \leq 4(1\leq i\leq 3)
* a_i and b_i are different. (1\leq i\leq 3)
* No two roads connect the same pair of towns.
* Any town can be reached from any other town using the roads.
Input
Input is given from Standard Input in the following format:
a_1 b_1
a_2 b_2
a_3 b_3
Output
If we can visit all the towns by traversing each of the roads exactly once, print `YES`; otherwise, print `NO`.
Examples
Input
4 2
1 3
2 3
Output
YES
Input
3 2
2 4
1 2
Output
NO
Input
2 1
3 2
4 3
Output
YES
Submitted Solution:
```
ab=[list(map(int,input().split())) for i in range(3)]
cnt=[0,0,0,0]
ans='YES'
for i in ab:
for j in i:
cnt[j-1]+=1
for i in cnt:
if i>=3:
ans='NO'
print(ans)
``` | instruction | 0 | 62,104 | 1 | 124,208 |
Yes | output | 1 | 62,104 | 1 | 124,209 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There are four towns, numbered 1,2,3 and 4. Also, there are three roads. The i-th road connects different towns a_i and b_i bidirectionally. No two roads connect the same pair of towns. Other than these roads, there is no way to travel between these towns, but any town can be reached from any other town using these roads.
Determine if we can visit all the towns by traversing each of the roads exactly once.
Constraints
* 1 \leq a_i,b_i \leq 4(1\leq i\leq 3)
* a_i and b_i are different. (1\leq i\leq 3)
* No two roads connect the same pair of towns.
* Any town can be reached from any other town using the roads.
Input
Input is given from Standard Input in the following format:
a_1 b_1
a_2 b_2
a_3 b_3
Output
If we can visit all the towns by traversing each of the roads exactly once, print `YES`; otherwise, print `NO`.
Examples
Input
4 2
1 3
2 3
Output
YES
Input
3 2
2 4
1 2
Output
NO
Input
2 1
3 2
4 3
Output
YES
Submitted Solution:
```
def solve():
t = [None]*4
for _ in range(3):
a, b = map(int, input().split())
if t[a-1] != None:
print("NO")
return
if t[b-1] != None:
print("NO")
return
t[a-1] = b-1
t[b-1] = a-1
for i in range(4):
if t[i] == None:
print("NO")
return
if t[i] == i:
print("NO")
return
else:
print("YES")
solve()
``` | instruction | 0 | 62,105 | 1 | 124,210 |
No | output | 1 | 62,105 | 1 | 124,211 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There are four towns, numbered 1,2,3 and 4. Also, there are three roads. The i-th road connects different towns a_i and b_i bidirectionally. No two roads connect the same pair of towns. Other than these roads, there is no way to travel between these towns, but any town can be reached from any other town using these roads.
Determine if we can visit all the towns by traversing each of the roads exactly once.
Constraints
* 1 \leq a_i,b_i \leq 4(1\leq i\leq 3)
* a_i and b_i are different. (1\leq i\leq 3)
* No two roads connect the same pair of towns.
* Any town can be reached from any other town using the roads.
Input
Input is given from Standard Input in the following format:
a_1 b_1
a_2 b_2
a_3 b_3
Output
If we can visit all the towns by traversing each of the roads exactly once, print `YES`; otherwise, print `NO`.
Examples
Input
4 2
1 3
2 3
Output
YES
Input
3 2
2 4
1 2
Output
NO
Input
2 1
3 2
4 3
Output
YES
Submitted Solution:
```
a = list(map(int, input().split()))
b = list(map(int, input().split()))
c = list(map(int, input().split()))
d = a + b + c
if 1 in d and 2 in d and 3 in d and 4 in d:
if a[1] == b[0] and b[1] == c[0]:
print('YES')
if a[1] == c[0] and c[1] == b[0]:
print('YES')
if b[1] == a[0] and a[1] == c[0]:
print('YES')
if b[1] == c[0] and c[1] == a[0]:
print('YES')
if c[1] == b[0] and b[1] == a[0]:
print('YES')
if c[1] == a[0] and a[1] == b[0]:
print('YES')
else:
print("NO")
else:
print('NO')
``` | instruction | 0 | 62,106 | 1 | 124,212 |
No | output | 1 | 62,106 | 1 | 124,213 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There are four towns, numbered 1,2,3 and 4. Also, there are three roads. The i-th road connects different towns a_i and b_i bidirectionally. No two roads connect the same pair of towns. Other than these roads, there is no way to travel between these towns, but any town can be reached from any other town using these roads.
Determine if we can visit all the towns by traversing each of the roads exactly once.
Constraints
* 1 \leq a_i,b_i \leq 4(1\leq i\leq 3)
* a_i and b_i are different. (1\leq i\leq 3)
* No two roads connect the same pair of towns.
* Any town can be reached from any other town using the roads.
Input
Input is given from Standard Input in the following format:
a_1 b_1
a_2 b_2
a_3 b_3
Output
If we can visit all the towns by traversing each of the roads exactly once, print `YES`; otherwise, print `NO`.
Examples
Input
4 2
1 3
2 3
Output
YES
Input
3 2
2 4
1 2
Output
NO
Input
2 1
3 2
4 3
Output
YES
Submitted Solution:
```
# coding: utf-8
# Your code here!
num =[]
ans =0
for i in range(3):
x,y = map(int,input().split(" "))
num.append(x)
num.append(y)
for i in range(1,4):
if num.count(i)>=3 or num.count(i)==0:
ans +=1
if ans>0:
print("NO")
else:
print("YES")
``` | instruction | 0 | 62,107 | 1 | 124,214 |
No | output | 1 | 62,107 | 1 | 124,215 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There are four towns, numbered 1,2,3 and 4. Also, there are three roads. The i-th road connects different towns a_i and b_i bidirectionally. No two roads connect the same pair of towns. Other than these roads, there is no way to travel between these towns, but any town can be reached from any other town using these roads.
Determine if we can visit all the towns by traversing each of the roads exactly once.
Constraints
* 1 \leq a_i,b_i \leq 4(1\leq i\leq 3)
* a_i and b_i are different. (1\leq i\leq 3)
* No two roads connect the same pair of towns.
* Any town can be reached from any other town using the roads.
Input
Input is given from Standard Input in the following format:
a_1 b_1
a_2 b_2
a_3 b_3
Output
If we can visit all the towns by traversing each of the roads exactly once, print `YES`; otherwise, print `NO`.
Examples
Input
4 2
1 3
2 3
Output
YES
Input
3 2
2 4
1 2
Output
NO
Input
2 1
3 2
4 3
Output
YES
Submitted Solution:
```
A = []
B = []
for i in range(3):
a, b = [int(n) for n in input().split()]
A.append(a)
B.append(b)
counter = [0] * 4
for i in range(3):
counter[A[i] - 1] += 1
counter[B[i] - 1] += 1
oddscounter = 0
for i in range(4):
if counter[i] % 2 == 1:
oddscounter += 1
answer = ""
if oddscounter >= 3:
answer = "NO"
else:
answer = "YES"
print(answer)
``` | instruction | 0 | 62,108 | 1 | 124,216 |
No | output | 1 | 62,108 | 1 | 124,217 |
Provide tags and a correct Python 3 solution for this coding contest problem.
The map of the capital of Berland can be viewed on the infinite coordinate plane. Each point with integer coordinates contains a building, and there are streets connecting every building to four neighbouring buildings. All streets are parallel to the coordinate axes.
The main school of the capital is located in (s_x, s_y). There are n students attending this school, the i-th of them lives in the house located in (x_i, y_i). It is possible that some students live in the same house, but no student lives in (s_x, s_y).
After classes end, each student walks from the school to his house along one of the shortest paths. So the distance the i-th student goes from the school to his house is |s_x - x_i| + |s_y - y_i|.
The Provision Department of Berland has decided to open a shawarma tent somewhere in the capital (at some point with integer coordinates). It is considered that the i-th student will buy a shawarma if at least one of the shortest paths from the school to the i-th student's house goes through the point where the shawarma tent is located. It is forbidden to place the shawarma tent at the point where the school is located, but the coordinates of the shawarma tent may coincide with the coordinates of the house of some student (or even multiple students).
You want to find the maximum possible number of students buying shawarma and the optimal location for the tent itself.
Input
The first line contains three integers n, s_x, s_y (1 β€ n β€ 200 000, 0 β€ s_x, s_y β€ 10^{9}) β the number of students and the coordinates of the school, respectively.
Then n lines follow. The i-th of them contains two integers x_i, y_i (0 β€ x_i, y_i β€ 10^{9}) β the location of the house where the i-th student lives. Some locations of houses may coincide, but no student lives in the same location where the school is situated.
Output
The output should consist of two lines. The first of them should contain one integer c β the maximum number of students that will buy shawarmas at the tent.
The second line should contain two integers p_x and p_y β the coordinates where the tent should be located. If there are multiple answers, print any of them. Note that each of p_x and p_y should be not less than 0 and not greater than 10^{9}.
Examples
Input
4 3 2
1 3
4 2
5 1
4 1
Output
3
4 2
Input
3 100 100
0 0
0 0
100 200
Output
2
99 100
Input
7 10 12
5 6
20 23
15 4
16 5
4 54
12 1
4 15
Output
4
10 11
Note
In the first example, If we build the shawarma tent in (4, 2), then the students living in (4, 2), (4, 1) and (5, 1) will visit it.
In the second example, it is possible to build the shawarma tent in (1, 1), then both students living in (0, 0) will visit it. | instruction | 0 | 62,407 | 1 | 124,814 |
Tags: brute force, geometry, greedy, implementation
Correct Solution:
```
from sys import stdin, stdout
n, sx, sy = map(int, stdin.readline().split())
counts = {(1, 0) : 0, (-1, 0) : 0, (0, 1) : 0, (0, -1) : 0}
for _ in range(n):
x,y = map(int, stdin.readline().split())
if x > sx:
counts[(1, 0)]+=1
elif x < sx:
counts[(-1, 0)]+=1
if y > sy:
counts[(0, 1)]+=1
elif y < sy:
counts[(0, -1)]+=1
maxcount, ans = 0, None
for d,v in counts.items():
if v > maxcount:
maxcount, ans = v, d
stdout.write("{}\n{} {}".format(maxcount, sx+ans[0], sy+ans[1]))
``` | output | 1 | 62,407 | 1 | 124,815 |
Provide tags and a correct Python 3 solution for this coding contest problem.
The map of the capital of Berland can be viewed on the infinite coordinate plane. Each point with integer coordinates contains a building, and there are streets connecting every building to four neighbouring buildings. All streets are parallel to the coordinate axes.
The main school of the capital is located in (s_x, s_y). There are n students attending this school, the i-th of them lives in the house located in (x_i, y_i). It is possible that some students live in the same house, but no student lives in (s_x, s_y).
After classes end, each student walks from the school to his house along one of the shortest paths. So the distance the i-th student goes from the school to his house is |s_x - x_i| + |s_y - y_i|.
The Provision Department of Berland has decided to open a shawarma tent somewhere in the capital (at some point with integer coordinates). It is considered that the i-th student will buy a shawarma if at least one of the shortest paths from the school to the i-th student's house goes through the point where the shawarma tent is located. It is forbidden to place the shawarma tent at the point where the school is located, but the coordinates of the shawarma tent may coincide with the coordinates of the house of some student (or even multiple students).
You want to find the maximum possible number of students buying shawarma and the optimal location for the tent itself.
Input
The first line contains three integers n, s_x, s_y (1 β€ n β€ 200 000, 0 β€ s_x, s_y β€ 10^{9}) β the number of students and the coordinates of the school, respectively.
Then n lines follow. The i-th of them contains two integers x_i, y_i (0 β€ x_i, y_i β€ 10^{9}) β the location of the house where the i-th student lives. Some locations of houses may coincide, but no student lives in the same location where the school is situated.
Output
The output should consist of two lines. The first of them should contain one integer c β the maximum number of students that will buy shawarmas at the tent.
The second line should contain two integers p_x and p_y β the coordinates where the tent should be located. If there are multiple answers, print any of them. Note that each of p_x and p_y should be not less than 0 and not greater than 10^{9}.
Examples
Input
4 3 2
1 3
4 2
5 1
4 1
Output
3
4 2
Input
3 100 100
0 0
0 0
100 200
Output
2
99 100
Input
7 10 12
5 6
20 23
15 4
16 5
4 54
12 1
4 15
Output
4
10 11
Note
In the first example, If we build the shawarma tent in (4, 2), then the students living in (4, 2), (4, 1) and (5, 1) will visit it.
In the second example, it is possible to build the shawarma tent in (1, 1), then both students living in (0, 0) will visit it. | instruction | 0 | 62,408 | 1 | 124,816 |
Tags: brute force, geometry, greedy, implementation
Correct Solution:
```
def maxi(l,sx,sy):
ans = l[0]
index = 0
for i in range(4):
if l[i] > ans:
ans = l[i]
index = i
if index == 0:
return ans, sx + 1, sy
elif index == 1:
return ans, sx, sy + 1
elif index == 2:
return ans, sx - 1, sy
else:
return ans, sx, sy - 1
n, sx, sy = [int(x) for x in input().split()]
count = {'1' : 0, '2' : 0, '3' : 0, '4' : 0, 'xplus' : 0, 'xminus' : 0, 'yplus' : 0, 'yminus' : 0}
for i in range(n):
px, py = [int(x) for x in input().split()]
prx = px - sx
pry = py - sy
if prx > 0 and pry > 0:
count['1'] += 1
elif prx < 0 and pry > 0:
count['2'] += 1
elif prx < 0 and pry < 0:
count['3'] += 1
elif prx > 0 and pry < 0:
count['4'] += 1
elif prx > 0 and pry == 0:
count['xplus'] += 1
elif prx < 0 and pry == 0:
count['xminus'] += 1
elif prx == 0 and pry > 0:
count['yplus'] += 1
elif prx == 0 and pry < 0:
count['yminus'] += 1
ans, cx, cy = maxi(
[count['1']+count['xplus']+count['4'],
count['1']+count['yplus']+count['2'],
count['2']+count['xminus']+count['3'],
count['3']+count['yminus']+count['4']],
sx,
sy
)
print(ans)
print(cx,cy)
``` | output | 1 | 62,408 | 1 | 124,817 |
Provide tags and a correct Python 3 solution for this coding contest problem.
The map of the capital of Berland can be viewed on the infinite coordinate plane. Each point with integer coordinates contains a building, and there are streets connecting every building to four neighbouring buildings. All streets are parallel to the coordinate axes.
The main school of the capital is located in (s_x, s_y). There are n students attending this school, the i-th of them lives in the house located in (x_i, y_i). It is possible that some students live in the same house, but no student lives in (s_x, s_y).
After classes end, each student walks from the school to his house along one of the shortest paths. So the distance the i-th student goes from the school to his house is |s_x - x_i| + |s_y - y_i|.
The Provision Department of Berland has decided to open a shawarma tent somewhere in the capital (at some point with integer coordinates). It is considered that the i-th student will buy a shawarma if at least one of the shortest paths from the school to the i-th student's house goes through the point where the shawarma tent is located. It is forbidden to place the shawarma tent at the point where the school is located, but the coordinates of the shawarma tent may coincide with the coordinates of the house of some student (or even multiple students).
You want to find the maximum possible number of students buying shawarma and the optimal location for the tent itself.
Input
The first line contains three integers n, s_x, s_y (1 β€ n β€ 200 000, 0 β€ s_x, s_y β€ 10^{9}) β the number of students and the coordinates of the school, respectively.
Then n lines follow. The i-th of them contains two integers x_i, y_i (0 β€ x_i, y_i β€ 10^{9}) β the location of the house where the i-th student lives. Some locations of houses may coincide, but no student lives in the same location where the school is situated.
Output
The output should consist of two lines. The first of them should contain one integer c β the maximum number of students that will buy shawarmas at the tent.
The second line should contain two integers p_x and p_y β the coordinates where the tent should be located. If there are multiple answers, print any of them. Note that each of p_x and p_y should be not less than 0 and not greater than 10^{9}.
Examples
Input
4 3 2
1 3
4 2
5 1
4 1
Output
3
4 2
Input
3 100 100
0 0
0 0
100 200
Output
2
99 100
Input
7 10 12
5 6
20 23
15 4
16 5
4 54
12 1
4 15
Output
4
10 11
Note
In the first example, If we build the shawarma tent in (4, 2), then the students living in (4, 2), (4, 1) and (5, 1) will visit it.
In the second example, it is possible to build the shawarma tent in (1, 1), then both students living in (0, 0) will visit it. | instruction | 0 | 62,409 | 1 | 124,818 |
Tags: brute force, geometry, greedy, implementation
Correct Solution:
```
n, s_x, s_y = map(int, input().split())
east = west = north = south = 0
for i in range(n):
x, y = map(int, input().split())
if x < s_x:
west += 1
elif x > s_x:
east += 1
if y < s_y:
# south-west
south += 1
elif y > s_y:
north += 1
ans = max([east, west, north, south])
print(ans)
if east == ans:
print(s_x + 1, s_y)
elif west == ans:
print(s_x - 1, s_y)
elif north == ans:
print(s_x, s_y + 1)
else:
print(s_x, s_y-1)
``` | output | 1 | 62,409 | 1 | 124,819 |
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