text stringlengths 198 433k | conversation_id int64 0 109k |
|---|---|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Recently, Vladimir got bad mark in algebra again. To avoid such unpleasant events in future he decided to train his arithmetic skills. He wrote four integer numbers a, b, c, d on the blackboard. During each of the next three minutes he took two numbers from the blackboard (not necessarily adjacent) and replaced them with their sum or their product. In the end he got one number. Unfortunately, due to the awful memory he forgot that number, but he remembers four original numbers, sequence of the operations and his surprise because of the very small result. Help Vladimir remember the forgotten number: find the smallest number that can be obtained from the original numbers by the given sequence of operations.
Input
First line contains four integers separated by space: 0 β€ a, b, c, d β€ 1000 β the original numbers. Second line contains three signs ('+' or '*' each) separated by space β the sequence of the operations in the order of performing. ('+' stands for addition, '*' β multiplication)
Output
Output one integer number β the minimal result which can be obtained.
Please, do not use %lld specificator to read or write 64-bit integers in C++. It is preffered to use cin (also you may use %I64d).
Examples
Input
1 1 1 1
+ + *
Output
3
Input
2 2 2 2
* * +
Output
8
Input
1 2 3 4
* + +
Output
9
Submitted Solution:
```
nums = list(map(int, input().split()))
ops = input().split()
# visited = [False]*(10)
ans = float('inf')
def solve(nums, ops, level):
global ans
if len(nums)==1:
ans = min(ans, nums[0])
# print(" ".join(["\t" for _ in range(level)]),ans)
return
n = len(nums)
for i in range(n):
for j in range(i+1, n):
# if not visited[i] and not visited[j]:
# visited[i] = True
# visited[j] = True
l = nums[i]
r = nums[j]
# print(" ".join(["\t" for _ in range(level)]),l, r)
nums.pop(i)
nums.pop(j-1)
if ops[0]=='*':
nums.append(l*r)
solve(nums, ops[1:], level+1)
nums.pop()
else:
nums.append(l+r)
solve(nums, ops[1:], level+1)
nums.pop()
nums.insert(i, l)
nums.insert(j, r)
# visited[i] = False
# visited[j] = False
solve(nums, ops, 0)
print(ans)
```
Yes
| 9,500 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Recently, Vladimir got bad mark in algebra again. To avoid such unpleasant events in future he decided to train his arithmetic skills. He wrote four integer numbers a, b, c, d on the blackboard. During each of the next three minutes he took two numbers from the blackboard (not necessarily adjacent) and replaced them with their sum or their product. In the end he got one number. Unfortunately, due to the awful memory he forgot that number, but he remembers four original numbers, sequence of the operations and his surprise because of the very small result. Help Vladimir remember the forgotten number: find the smallest number that can be obtained from the original numbers by the given sequence of operations.
Input
First line contains four integers separated by space: 0 β€ a, b, c, d β€ 1000 β the original numbers. Second line contains three signs ('+' or '*' each) separated by space β the sequence of the operations in the order of performing. ('+' stands for addition, '*' β multiplication)
Output
Output one integer number β the minimal result which can be obtained.
Please, do not use %lld specificator to read or write 64-bit integers in C++. It is preffered to use cin (also you may use %I64d).
Examples
Input
1 1 1 1
+ + *
Output
3
Input
2 2 2 2
* * +
Output
8
Input
1 2 3 4
* + +
Output
9
Submitted Solution:
```
def delete(i , j, l):
t = []
m = 0
while m < len(l):
if m != i and m != j:
t.append(l[m])
m += 1
return t
n = list(map(int,input().split()))
o = list(input().split())
n = [n]
def brute(l , u):
i = 0
second = list()
while i < len(l):
j = 0
while j < len(l[0]):
k = 0
while k < len(l[0]):
x = list()
if k != j:
x.append(eval(str(l[i][j]) + o[u] + str(l[i][k])))
second.append(x + delete(k,j,l[i].copy()))
k += 1
j += 1
i += 1
return second
second = brute(n, 0)
third = brute(second, 1)
fourth = brute(third, 2)
m = 2 ** 100
for i in fourth:
m = min(m, i[0])
print(m)
```
Yes
| 9,501 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Recently, Vladimir got bad mark in algebra again. To avoid such unpleasant events in future he decided to train his arithmetic skills. He wrote four integer numbers a, b, c, d on the blackboard. During each of the next three minutes he took two numbers from the blackboard (not necessarily adjacent) and replaced them with their sum or their product. In the end he got one number. Unfortunately, due to the awful memory he forgot that number, but he remembers four original numbers, sequence of the operations and his surprise because of the very small result. Help Vladimir remember the forgotten number: find the smallest number that can be obtained from the original numbers by the given sequence of operations.
Input
First line contains four integers separated by space: 0 β€ a, b, c, d β€ 1000 β the original numbers. Second line contains three signs ('+' or '*' each) separated by space β the sequence of the operations in the order of performing. ('+' stands for addition, '*' β multiplication)
Output
Output one integer number β the minimal result which can be obtained.
Please, do not use %lld specificator to read or write 64-bit integers in C++. It is preffered to use cin (also you may use %I64d).
Examples
Input
1 1 1 1
+ + *
Output
3
Input
2 2 2 2
* * +
Output
8
Input
1 2 3 4
* + +
Output
9
Submitted Solution:
```
import math,sys
#from itertools import permutations
from collections import defaultdict,deque
import bisect as bi
def yes():print('YES')
def no():print('NO')
#sys.stdin=open('input.txt','r');sys.stdout=open('output.txt','w');
def I():return (int(sys.stdin.readline()))
def In():return(map(int,sys.stdin.readline().split()))
def Sn():return sys.stdin.readline().strip()
#sys.setrecursionlimit(1500)
def dict(a):
d={}
for x in a:
if d.get(x,-1)!=-1:
d[x]+=1
else:
d[x]=1
return d
def find_gt(a, x):
'Find leftmost value greater than x'
i = bi.bisect_left(a, x)
if i != len(a):
return i
else:
return -1
def check(l,s):
mul=0
zero=0
for x in s:
if x=='*':
t1= l.pop(0)
t2 = l.pop(0)
if t1==0 or t2==0:
zero+=1
l.append(t1*t2)
mul+=1
else:
t1=l.pop()
t2=l.pop()
if t1==0 or t2==0:
zero+=1
l.append(t1+t2)
l.sort()
if mul and zero:
return(0)
return(l[0])
def main():
try:
l=list(In())
n=len(l)
s=Sn().split()
z=list(permutations(l))
ans=check(l,s)
for x in z:
ans=min(check(list(x),s),ans)
print(ans)
except:
pass
M = 998244353
P = 1000000007
if __name__ == '__main__':
# for _ in range(I()):main()
for _ in range(1):main()
#End#
# ******************* All The Best ******************* #
```
No
| 9,502 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Recently, Vladimir got bad mark in algebra again. To avoid such unpleasant events in future he decided to train his arithmetic skills. He wrote four integer numbers a, b, c, d on the blackboard. During each of the next three minutes he took two numbers from the blackboard (not necessarily adjacent) and replaced them with their sum or their product. In the end he got one number. Unfortunately, due to the awful memory he forgot that number, but he remembers four original numbers, sequence of the operations and his surprise because of the very small result. Help Vladimir remember the forgotten number: find the smallest number that can be obtained from the original numbers by the given sequence of operations.
Input
First line contains four integers separated by space: 0 β€ a, b, c, d β€ 1000 β the original numbers. Second line contains three signs ('+' or '*' each) separated by space β the sequence of the operations in the order of performing. ('+' stands for addition, '*' β multiplication)
Output
Output one integer number β the minimal result which can be obtained.
Please, do not use %lld specificator to read or write 64-bit integers in C++. It is preffered to use cin (also you may use %I64d).
Examples
Input
1 1 1 1
+ + *
Output
3
Input
2 2 2 2
* * +
Output
8
Input
1 2 3 4
* + +
Output
9
Submitted Solution:
```
# ---------------------------iye ha aam zindegi---------------------------------------------
import math
import heapq, bisect
import sys
from collections import deque, defaultdict
from fractions import Fraction
mod = 10 ** 9 + 7
mod1 = 998244353
# ------------------------------warmup----------------------------
import os
import sys
from io import BytesIO, IOBase
BUFSIZE = 8192
class FastIO(IOBase):
newlines = 0
def __init__(self, file):
self._fd = file.fileno()
self.buffer = BytesIO()
self.writable = "x" in file.mode or "r" not in file.mode
self.write = self.buffer.write if self.writable else None
def read(self):
while True:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
if not b:
break
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines = 0
return self.buffer.read()
def readline(self):
while self.newlines == 0:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
self.newlines = b.count(b"\n") + (not b)
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines -= 1
return self.buffer.readline()
def flush(self):
if self.writable:
os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
self.write = lambda s: self.buffer.write(s.encode("ascii"))
self.read = lambda: self.buffer.read().decode("ascii")
self.readline = lambda: self.buffer.readline().decode("ascii")
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")
# -------------------game starts now----------------------------------------------------import math
class TreeNode:
def __init__(self, k, v):
self.key = k
self.value = v
self.left = None
self.right = None
self.parent = None
self.height = 1
self.num_left = 1
self.num_total = 1
class AvlTree:
def __init__(self):
self._tree = None
def add(self, k, v):
if not self._tree:
self._tree = TreeNode(k, v)
return
node = self._add(k, v)
if node:
self._rebalance(node)
def _add(self, k, v):
node = self._tree
while node:
if k < node.key:
if node.left:
node = node.left
else:
node.left = TreeNode(k, v)
node.left.parent = node
return node.left
elif node.key < k:
if node.right:
node = node.right
else:
node.right = TreeNode(k, v)
node.right.parent = node
return node.right
else:
node.value = v
return
@staticmethod
def get_height(x):
return x.height if x else 0
@staticmethod
def get_num_total(x):
return x.num_total if x else 0
def _rebalance(self, node):
n = node
while n:
lh = self.get_height(n.left)
rh = self.get_height(n.right)
n.height = max(lh, rh) + 1
balance_factor = lh - rh
n.num_total = 1 + self.get_num_total(n.left) + self.get_num_total(n.right)
n.num_left = 1 + self.get_num_total(n.left)
if balance_factor > 1:
if self.get_height(n.left.left) < self.get_height(n.left.right):
self._rotate_left(n.left)
self._rotate_right(n)
elif balance_factor < -1:
if self.get_height(n.right.right) < self.get_height(n.right.left):
self._rotate_right(n.right)
self._rotate_left(n)
else:
n = n.parent
def _remove_one(self, node):
"""
Side effect!!! Changes node. Node should have exactly one child
"""
replacement = node.left or node.right
if node.parent:
if AvlTree._is_left(node):
node.parent.left = replacement
else:
node.parent.right = replacement
replacement.parent = node.parent
node.parent = None
else:
self._tree = replacement
replacement.parent = None
node.left = None
node.right = None
node.parent = None
self._rebalance(replacement)
def _remove_leaf(self, node):
if node.parent:
if AvlTree._is_left(node):
node.parent.left = None
else:
node.parent.right = None
self._rebalance(node.parent)
else:
self._tree = None
node.parent = None
node.left = None
node.right = None
def remove(self, k):
node = self._get_node(k)
if not node:
return
if AvlTree._is_leaf(node):
self._remove_leaf(node)
return
if node.left and node.right:
nxt = AvlTree._get_next(node)
node.key = nxt.key
node.value = nxt.value
if self._is_leaf(nxt):
self._remove_leaf(nxt)
else:
self._remove_one(nxt)
self._rebalance(node)
else:
self._remove_one(node)
def get(self, k):
node = self._get_node(k)
return node.value if node else -1
def _get_node(self, k):
if not self._tree:
return None
node = self._tree
while node:
if k < node.key:
node = node.left
elif node.key < k:
node = node.right
else:
return node
return None
def get_at(self, pos):
x = pos + 1
node = self._tree
while node:
if x < node.num_left:
node = node.left
elif node.num_left < x:
x -= node.num_left
node = node.right
else:
return (node.key, node.value)
raise IndexError("Out of ranges")
@staticmethod
def _is_left(node):
return node.parent.left and node.parent.left == node
@staticmethod
def _is_leaf(node):
return node.left is None and node.right is None
def _rotate_right(self, node):
if not node.parent:
self._tree = node.left
node.left.parent = None
elif AvlTree._is_left(node):
node.parent.left = node.left
node.left.parent = node.parent
else:
node.parent.right = node.left
node.left.parent = node.parent
bk = node.left.right
node.left.right = node
node.parent = node.left
node.left = bk
if bk:
bk.parent = node
node.height = max(self.get_height(node.left), self.get_height(node.right)) + 1
node.num_total = 1 + self.get_num_total(node.left) + self.get_num_total(node.right)
node.num_left = 1 + self.get_num_total(node.left)
def _rotate_left(self, node):
if not node.parent:
self._tree = node.right
node.right.parent = None
elif AvlTree._is_left(node):
node.parent.left = node.right
node.right.parent = node.parent
else:
node.parent.right = node.right
node.right.parent = node.parent
bk = node.right.left
node.right.left = node
node.parent = node.right
node.right = bk
if bk:
bk.parent = node
node.height = max(self.get_height(node.left), self.get_height(node.right)) + 1
node.num_total = 1 + self.get_num_total(node.left) + self.get_num_total(node.right)
node.num_left = 1 + self.get_num_total(node.left)
@staticmethod
def _get_next(node):
if not node.right:
return node.parent
n = node.right
while n.left:
n = n.left
return n
avl=AvlTree()
#-----------------------------------------------binary seacrh tree---------------------------------------
class SegmentTree1:
def __init__(self, data, default='z', func=lambda a, b: min(a ,b)):
"""initialize the segment tree with data"""
self._default = default
self._func = func
self._len = len(data)
self._size = _size = 1 << (self._len - 1).bit_length()
self.data = [default] * (2 * _size)
self.data[_size:_size + self._len] = data
for i in reversed(range(_size)):
self.data[i] = func(self.data[i + i], self.data[i + i + 1])
def __delitem__(self, idx):
self[idx] = self._default
def __getitem__(self, idx):
return self.data[idx + self._size]
def __setitem__(self, idx, value):
idx += self._size
self.data[idx] = value
idx >>= 1
while idx:
self.data[idx] = self._func(self.data[2 * idx], self.data[2 * idx + 1])
idx >>= 1
def __len__(self):
return self._len
def query(self, start, stop):
if start == stop:
return self.__getitem__(start)
stop += 1
start += self._size
stop += self._size
res = self._default
while start < stop:
if start & 1:
res = self._func(res, self.data[start])
start += 1
if stop & 1:
stop -= 1
res = self._func(res, self.data[stop])
start >>= 1
stop >>= 1
return res
def __repr__(self):
return "SegmentTree({0})".format(self.data)
# -------------------game starts now----------------------------------------------------import math
class SegmentTree:
def __init__(self, data, default=0, func=lambda a, b: a + b):
"""initialize the segment tree with data"""
self._default = default
self._func = func
self._len = len(data)
self._size = _size = 1 << (self._len - 1).bit_length()
self.data = [default] * (2 * _size)
self.data[_size:_size + self._len] = data
for i in reversed(range(_size)):
self.data[i] = func(self.data[i + i], self.data[i + i + 1])
def __delitem__(self, idx):
self[idx] = self._default
def __getitem__(self, idx):
return self.data[idx + self._size]
def __setitem__(self, idx, value):
idx += self._size
self.data[idx] = value
idx >>= 1
while idx:
self.data[idx] = self._func(self.data[2 * idx], self.data[2 * idx + 1])
idx >>= 1
def __len__(self):
return self._len
def query(self, start, stop):
if start == stop:
return self.__getitem__(start)
stop += 1
start += self._size
stop += self._size
res = self._default
while start < stop:
if start & 1:
res = self._func(res, self.data[start])
start += 1
if stop & 1:
stop -= 1
res = self._func(res, self.data[stop])
start >>= 1
stop >>= 1
return res
def __repr__(self):
return "SegmentTree({0})".format(self.data)
# -------------------------------iye ha chutiya zindegi-------------------------------------
class Factorial:
def __init__(self, MOD):
self.MOD = MOD
self.factorials = [1, 1]
self.invModulos = [0, 1]
self.invFactorial_ = [1, 1]
def calc(self, n):
if n <= -1:
print("Invalid argument to calculate n!")
print("n must be non-negative value. But the argument was " + str(n))
exit()
if n < len(self.factorials):
return self.factorials[n]
nextArr = [0] * (n + 1 - len(self.factorials))
initialI = len(self.factorials)
prev = self.factorials[-1]
m = self.MOD
for i in range(initialI, n + 1):
prev = nextArr[i - initialI] = prev * i % m
self.factorials += nextArr
return self.factorials[n]
def inv(self, n):
if n <= -1:
print("Invalid argument to calculate n^(-1)")
print("n must be non-negative value. But the argument was " + str(n))
exit()
p = self.MOD
pi = n % p
if pi < len(self.invModulos):
return self.invModulos[pi]
nextArr = [0] * (n + 1 - len(self.invModulos))
initialI = len(self.invModulos)
for i in range(initialI, min(p, n + 1)):
next = -self.invModulos[p % i] * (p // i) % p
self.invModulos.append(next)
return self.invModulos[pi]
def invFactorial(self, n):
if n <= -1:
print("Invalid argument to calculate (n^(-1))!")
print("n must be non-negative value. But the argument was " + str(n))
exit()
if n < len(self.invFactorial_):
return self.invFactorial_[n]
self.inv(n) # To make sure already calculated n^-1
nextArr = [0] * (n + 1 - len(self.invFactorial_))
initialI = len(self.invFactorial_)
prev = self.invFactorial_[-1]
p = self.MOD
for i in range(initialI, n + 1):
prev = nextArr[i - initialI] = (prev * self.invModulos[i % p]) % p
self.invFactorial_ += nextArr
return self.invFactorial_[n]
class Combination:
def __init__(self, MOD):
self.MOD = MOD
self.factorial = Factorial(MOD)
def ncr(self, n, k):
if k < 0 or n < k:
return 0
k = min(k, n - k)
f = self.factorial
return f.calc(n) * f.invFactorial(max(n - k, k)) * f.invFactorial(min(k, n - k)) % self.MOD
# --------------------------------------iye ha combinations ka zindegi---------------------------------
def powm(a, n, m):
if a == 1 or n == 0:
return 1
if n % 2 == 0:
s = powm(a, n // 2, m)
return s * s % m
else:
return a * powm(a, n - 1, m) % m
# --------------------------------------iye ha power ka zindegi---------------------------------
def sort_list(list1, list2):
zipped_pairs = zip(list2, list1)
z = [x for _, x in sorted(zipped_pairs)]
return z
# --------------------------------------------------product----------------------------------------
def product(l):
por = 1
for i in range(len(l)):
por *= l[i]
return por
# --------------------------------------------------binary----------------------------------------
def binarySearchCount(arr, n, key):
left = 0
right = n - 1
count = 0
while (left <= right):
mid = int((right + left)/ 2)
# Check if middle element is
# less than or equal to key
if (arr[mid]<=key):
count = mid+1
left = mid + 1
# If key is smaller, ignore right half
else:
right = mid - 1
return count
# --------------------------------------------------binary----------------------------------------
def countdig(n):
c = 0
while (n > 0):
n //= 10
c += 1
return c
def countGreater( arr,n, k):
l = 0
r = n - 1
# Stores the index of the left most element
# from the array which is greater than k
leftGreater = n
# Finds number of elements greater than k
while (l <= r):
m = int(l + (r - l) / 2)
if (arr[m] >= k):
leftGreater = m
r = m - 1
# If mid element is less than
# or equal to k update l
else:
l = m + 1
# Return the count of elements
# greater than k
return (n - leftGreater)
# --------------------------------------------------binary------------------------------------
l=list(map(int,input().split()))
l.sort()
s=list(map(str,input().split()))
t=0
if s[0]=="*" and s[1]=="*":
l[0]=l[0]*l[-1]
l[1]=l[1]*l[-2]
l.pop()
l.pop()
t=2
for i in range(t,3):
if s[i]=="+":
l[-2]+=l[-1]
l.pop()
l.sort()
else:
l[0]=l[0]*l[1]
l[1]=99999999999999999999
l.sort()
l.pop()
print(l[0])
```
No
| 9,503 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Recently, Vladimir got bad mark in algebra again. To avoid such unpleasant events in future he decided to train his arithmetic skills. He wrote four integer numbers a, b, c, d on the blackboard. During each of the next three minutes he took two numbers from the blackboard (not necessarily adjacent) and replaced them with their sum or their product. In the end he got one number. Unfortunately, due to the awful memory he forgot that number, but he remembers four original numbers, sequence of the operations and his surprise because of the very small result. Help Vladimir remember the forgotten number: find the smallest number that can be obtained from the original numbers by the given sequence of operations.
Input
First line contains four integers separated by space: 0 β€ a, b, c, d β€ 1000 β the original numbers. Second line contains three signs ('+' or '*' each) separated by space β the sequence of the operations in the order of performing. ('+' stands for addition, '*' β multiplication)
Output
Output one integer number β the minimal result which can be obtained.
Please, do not use %lld specificator to read or write 64-bit integers in C++. It is preffered to use cin (also you may use %I64d).
Examples
Input
1 1 1 1
+ + *
Output
3
Input
2 2 2 2
* * +
Output
8
Input
1 2 3 4
* + +
Output
9
Submitted Solution:
```
def fun(arr,plus,mul):
if len(arr)==1:
return arr[0]
if mul==0:
return sum(arr)
if 0 in arr:
return 0
mn=float('inf')
for i in range(1,len(arr)):
arr[i]=arr[0]*arr[i]
mn=min(mn,fun(arr[1:].copy(),plus,mul-1))
arr[i]=arr[i]//arr[0]
return mn
arr=list(map(int,input().split()))
op=input().split()
dic={'+':0,'*':0}
for i in op:
dic[i]+=1
print(fun(arr,dic['+'],dic['*']))
#print(fun(arr,dic['+'],dic['*']))
```
No
| 9,504 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Recently, Vladimir got bad mark in algebra again. To avoid such unpleasant events in future he decided to train his arithmetic skills. He wrote four integer numbers a, b, c, d on the blackboard. During each of the next three minutes he took two numbers from the blackboard (not necessarily adjacent) and replaced them with their sum or their product. In the end he got one number. Unfortunately, due to the awful memory he forgot that number, but he remembers four original numbers, sequence of the operations and his surprise because of the very small result. Help Vladimir remember the forgotten number: find the smallest number that can be obtained from the original numbers by the given sequence of operations.
Input
First line contains four integers separated by space: 0 β€ a, b, c, d β€ 1000 β the original numbers. Second line contains three signs ('+' or '*' each) separated by space β the sequence of the operations in the order of performing. ('+' stands for addition, '*' β multiplication)
Output
Output one integer number β the minimal result which can be obtained.
Please, do not use %lld specificator to read or write 64-bit integers in C++. It is preffered to use cin (also you may use %I64d).
Examples
Input
1 1 1 1
+ + *
Output
3
Input
2 2 2 2
* * +
Output
8
Input
1 2 3 4
* + +
Output
9
Submitted Solution:
```
from itertools import permutations as pe
nums = pe(sorted(list(map(int, input().split()))))
ops = pe(sorted(input().split()))
def act(a,b,op):
if op == '*':
return a * b
return a + b
m = 100000000000000000000000000000000000000000000
for n in nums:
for o in ops:
v = act(n[0], n[1], o[0])
v = act(v, n[2], o[1])
v = act(v, n[3], o[2])
m = min(m, v)
print(m)
```
No
| 9,505 |
Provide tags and a correct Python 3 solution for this coding contest problem.
As you know, every birthday party has a cake! This time, Babaei is going to prepare the very special birthday party's cake.
Simple cake is a cylinder of some radius and height. The volume of the simple cake is equal to the volume of corresponding cylinder. Babaei has n simple cakes and he is going to make a special cake placing some cylinders on each other.
However, there are some additional culinary restrictions. The cakes are numbered in such a way that the cake number i can be placed only on the table or on some cake number j where j < i. Moreover, in order to impress friends Babaei will put the cake i on top of the cake j only if the volume of the cake i is strictly greater than the volume of the cake j.
Babaei wants to prepare a birthday cake that has a maximum possible total volume. Help him find this value.
Input
The first line of the input contains a single integer n (1 β€ n β€ 100 000) β the number of simple cakes Babaei has.
Each of the following n lines contains two integers ri and hi (1 β€ ri, hi β€ 10 000), giving the radius and height of the i-th cake.
Output
Print the maximum volume of the cake that Babaei can make. Your 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
2
100 30
40 10
Output
942477.796077000
Input
4
1 1
9 7
1 4
10 7
Output
3983.539484752
Note
In first sample, the optimal way is to choose the cake number 1.
In second sample, the way to get the maximum volume is to use cakes with indices 1, 2 and 4.
Tags: data structures, dp
Correct Solution:
```
# ---------------------------iye ha aam zindegi---------------------------------------------
import math
import random
import heapq,bisect
import sys
from collections import deque, defaultdict
from fractions import Fraction
import sys
import threading
from collections import defaultdict
threading.stack_size(10**8)
mod = 10 ** 9 + 7
mod1 = 998244353
# ------------------------------warmup----------------------------
import os
import sys
from io import BytesIO, IOBase
sys.setrecursionlimit(300000)
BUFSIZE = 8192
class FastIO(IOBase):
newlines = 0
def __init__(self, file):
self._fd = file.fileno()
self.buffer = BytesIO()
self.writable = "x" in file.mode or "r" not in file.mode
self.write = self.buffer.write if self.writable else None
def read(self):
while True:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
if not b:
break
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines = 0
return self.buffer.read()
def readline(self):
while self.newlines == 0:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
self.newlines = b.count(b"\n") + (not b)
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines -= 1
return self.buffer.readline()
def flush(self):
if self.writable:
os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
self.write = lambda s: self.buffer.write(s.encode("ascii"))
self.read = lambda: self.buffer.read().decode("ascii")
self.readline = lambda: self.buffer.readline().decode("ascii")
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")
# -------------------game starts now----------------------------------------------------import math
class TreeNode:
def __init__(self, k, v):
self.key = k
self.value = v
self.left = None
self.right = None
self.parent = None
self.height = 1
self.num_left = 1
self.num_total = 1
class AvlTree:
def __init__(self):
self._tree = None
def add(self, k, v):
if not self._tree:
self._tree = TreeNode(k, v)
return
node = self._add(k, v)
if node:
self._rebalance(node)
def _add(self, k, v):
node = self._tree
while node:
if k < node.key:
if node.left:
node = node.left
else:
node.left = TreeNode(k, v)
node.left.parent = node
return node.left
elif node.key < k:
if node.right:
node = node.right
else:
node.right = TreeNode(k, v)
node.right.parent = node
return node.right
else:
node.value = v
return
@staticmethod
def get_height(x):
return x.height if x else 0
@staticmethod
def get_num_total(x):
return x.num_total if x else 0
def _rebalance(self, node):
n = node
while n:
lh = self.get_height(n.left)
rh = self.get_height(n.right)
n.height = max(lh, rh) + 1
balance_factor = lh - rh
n.num_total = 1 + self.get_num_total(n.left) + self.get_num_total(n.right)
n.num_left = 1 + self.get_num_total(n.left)
if balance_factor > 1:
if self.get_height(n.left.left) < self.get_height(n.left.right):
self._rotate_left(n.left)
self._rotate_right(n)
elif balance_factor < -1:
if self.get_height(n.right.right) < self.get_height(n.right.left):
self._rotate_right(n.right)
self._rotate_left(n)
else:
n = n.parent
def _remove_one(self, node):
"""
Side effect!!! Changes node. Node should have exactly one child
"""
replacement = node.left or node.right
if node.parent:
if AvlTree._is_left(node):
node.parent.left = replacement
else:
node.parent.right = replacement
replacement.parent = node.parent
node.parent = None
else:
self._tree = replacement
replacement.parent = None
node.left = None
node.right = None
node.parent = None
self._rebalance(replacement)
def _remove_leaf(self, node):
if node.parent:
if AvlTree._is_left(node):
node.parent.left = None
else:
node.parent.right = None
self._rebalance(node.parent)
else:
self._tree = None
node.parent = None
node.left = None
node.right = None
def remove(self, k):
node = self._get_node(k)
if not node:
return
if AvlTree._is_leaf(node):
self._remove_leaf(node)
return
if node.left and node.right:
nxt = AvlTree._get_next(node)
node.key = nxt.key
node.value = nxt.value
if self._is_leaf(nxt):
self._remove_leaf(nxt)
else:
self._remove_one(nxt)
self._rebalance(node)
else:
self._remove_one(node)
def get(self, k):
node = self._get_node(k)
return node.value if node else -1
def _get_node(self, k):
if not self._tree:
return None
node = self._tree
while node:
if k < node.key:
node = node.left
elif node.key < k:
node = node.right
else:
return node
return None
def get_at(self, pos):
x = pos + 1
node = self._tree
while node:
if x < node.num_left:
node = node.left
elif node.num_left < x:
x -= node.num_left
node = node.right
else:
return (node.key, node.value)
raise IndexError("Out of ranges")
@staticmethod
def _is_left(node):
return node.parent.left and node.parent.left == node
@staticmethod
def _is_leaf(node):
return node.left is None and node.right is None
def _rotate_right(self, node):
if not node.parent:
self._tree = node.left
node.left.parent = None
elif AvlTree._is_left(node):
node.parent.left = node.left
node.left.parent = node.parent
else:
node.parent.right = node.left
node.left.parent = node.parent
bk = node.left.right
node.left.right = node
node.parent = node.left
node.left = bk
if bk:
bk.parent = node
node.height = max(self.get_height(node.left), self.get_height(node.right)) + 1
node.num_total = 1 + self.get_num_total(node.left) + self.get_num_total(node.right)
node.num_left = 1 + self.get_num_total(node.left)
def _rotate_left(self, node):
if not node.parent:
self._tree = node.right
node.right.parent = None
elif AvlTree._is_left(node):
node.parent.left = node.right
node.right.parent = node.parent
else:
node.parent.right = node.right
node.right.parent = node.parent
bk = node.right.left
node.right.left = node
node.parent = node.right
node.right = bk
if bk:
bk.parent = node
node.height = max(self.get_height(node.left), self.get_height(node.right)) + 1
node.num_total = 1 + self.get_num_total(node.left) + self.get_num_total(node.right)
node.num_left = 1 + self.get_num_total(node.left)
@staticmethod
def _get_next(node):
if not node.right:
return node.parent
n = node.right
while n.left:
n = n.left
return n
# -----------------------------------------------binary seacrh tree---------------------------------------
class SegmentTree1:
def __init__(self, data, default=2**51, func=lambda a, b: a & b):
"""initialize the segment tree with data"""
self._default = default
self._func = func
self._len = len(data)
self._size = _size = 1 << (self._len - 1).bit_length()
self.data = [default] * (2 * _size)
self.data[_size:_size + self._len] = data
for i in reversed(range(_size)):
self.data[i] = func(self.data[i + i], self.data[i + i + 1])
def __delitem__(self, idx):
self[idx] = self._default
def __getitem__(self, idx):
return self.data[idx + self._size]
def __setitem__(self, idx, value):
idx += self._size
self.data[idx] = value
idx >>= 1
while idx:
self.data[idx] = self._func(self.data[2 * idx], self.data[2 * idx + 1])
idx >>= 1
def __len__(self):
return self._len
def query(self, start, stop):
if start == stop:
return self.__getitem__(start)
stop += 1
start += self._size
stop += self._size
res = self._default
while start < stop:
if start & 1:
res = self._func(res, self.data[start])
start += 1
if stop & 1:
stop -= 1
res = self._func(res, self.data[stop])
start >>= 1
stop >>= 1
return res
def __repr__(self):
return "SegmentTree({0})".format(self.data)
# -------------------game starts now----------------------------------------------------import math
class SegmentTree:
def __init__(self, data, default=0, func=lambda a, b: max(a , b)):
"""initialize the segment tree with data"""
self._default = default
self._func = func
self._len = len(data)
self._size = _size = 1 << (self._len - 1).bit_length()
self.data = [default] * (2 * _size)
self.data[_size:_size + self._len] = data
for i in reversed(range(_size)):
self.data[i] = func(self.data[i + i], self.data[i + i + 1])
def __delitem__(self, idx):
self[idx] = self._default
def __getitem__(self, idx):
return self.data[idx + self._size]
def __setitem__(self, idx, value):
idx += self._size
self.data[idx] = value
idx >>= 1
while idx:
self.data[idx] = self._func(self.data[2 * idx], self.data[2 * idx + 1])
idx >>= 1
def __len__(self):
return self._len
def query(self, start, stop):
if start == stop:
return self.__getitem__(start)
stop += 1
start += self._size
stop += self._size
res = self._default
while start < stop:
if start & 1:
res = self._func(res, self.data[start])
start += 1
if stop & 1:
stop -= 1
res = self._func(res, self.data[stop])
start >>= 1
stop >>= 1
return res
def __repr__(self):
return "SegmentTree({0})".format(self.data)
# -------------------------------iye ha chutiya zindegi-------------------------------------
class Factorial:
def __init__(self, MOD):
self.MOD = MOD
self.factorials = [1, 1]
self.invModulos = [0, 1]
self.invFactorial_ = [1, 1]
def calc(self, n):
if n <= -1:
print("Invalid argument to calculate n!")
print("n must be non-negative value. But the argument was " + str(n))
exit()
if n < len(self.factorials):
return self.factorials[n]
nextArr = [0] * (n + 1 - len(self.factorials))
initialI = len(self.factorials)
prev = self.factorials[-1]
m = self.MOD
for i in range(initialI, n + 1):
prev = nextArr[i - initialI] = prev * i % m
self.factorials += nextArr
return self.factorials[n]
def inv(self, n):
if n <= -1:
print("Invalid argument to calculate n^(-1)")
print("n must be non-negative value. But the argument was " + str(n))
exit()
p = self.MOD
pi = n % p
if pi < len(self.invModulos):
return self.invModulos[pi]
nextArr = [0] * (n + 1 - len(self.invModulos))
initialI = len(self.invModulos)
for i in range(initialI, min(p, n + 1)):
next = -self.invModulos[p % i] * (p // i) % p
self.invModulos.append(next)
return self.invModulos[pi]
def invFactorial(self, n):
if n <= -1:
print("Invalid argument to calculate (n^(-1))!")
print("n must be non-negative value. But the argument was " + str(n))
exit()
if n < len(self.invFactorial_):
return self.invFactorial_[n]
self.inv(n) # To make sure already calculated n^-1
nextArr = [0] * (n + 1 - len(self.invFactorial_))
initialI = len(self.invFactorial_)
prev = self.invFactorial_[-1]
p = self.MOD
for i in range(initialI, n + 1):
prev = nextArr[i - initialI] = (prev * self.invModulos[i % p]) % p
self.invFactorial_ += nextArr
return self.invFactorial_[n]
class Combination:
def __init__(self, MOD):
self.MOD = MOD
self.factorial = Factorial(MOD)
def ncr(self, n, k):
if k < 0 or n < k:
return 0
k = min(k, n - k)
f = self.factorial
return f.calc(n) * f.invFactorial(max(n - k, k)) * f.invFactorial(min(k, n - k)) % self.MOD
# --------------------------------------iye ha combinations ka zindegi---------------------------------
def powm(a, n, m):
if a == 1 or n == 0:
return 1
if n % 2 == 0:
s = powm(a, n // 2, m)
return s * s % m
else:
return a * powm(a, n - 1, m) % m
# --------------------------------------iye ha power ka zindegi---------------------------------
def sort_list(list1, list2):
zipped_pairs = zip(list2, list1)
z = [x for _, x in sorted(zipped_pairs)]
return z
# --------------------------------------------------product----------------------------------------
def product(l):
por = 1
for i in range(len(l)):
por *= l[i]
return por
# --------------------------------------------------binary----------------------------------------
def binarySearchCount(arr, n, key):
left = 0
right = n - 1
count = 0
while (left <= right):
mid = int((right + left) / 2)
# Check if middle element is
# less than or equal to key
if (arr[mid] < key):
count = mid + 1
left = mid + 1
# If key is smaller, ignore right half
else:
right = mid - 1
return count
# --------------------------------------------------binary----------------------------------------
def countdig(n):
c = 0
while (n > 0):
n //= 10
c += 1
return c
def binary(x, length):
y = bin(x)[2:]
return y if len(y) >= length else "0" * (length - len(y)) + y
def countGreater(arr, n, k):
l = 0
r = n - 1
# Stores the index of the left most element
# from the array which is greater than k
leftGreater = n
# Finds number of elements greater than k
while (l <= r):
m = int(l + (r - l) / 2)
if (arr[m] >= k):
leftGreater = m
r = m - 1
# If mid element is less than
# or equal to k update l
else:
l = m + 1
# Return the count of elements
# greater than k
return (n - leftGreater)
# --------------------------------------------------binary------------------------------------
n=int(input())
l=[]
vol=set()
ind=defaultdict(int)
for i in range(n):
a,b=map(int,input().split())
l.append((a,b))
vol.add(a*a*b)
vol=sorted(vol)
for i in range(len(vol)):
ind[vol[i]]=i
e=[0]*n
s=SegmentTree(e)
ans=0
for i in range(n):
a,b=l[i]
cur=a*a*b
index=ind[cur]
dp=s.query(0,index-1)+cur
s.__setitem__(index,dp+cur)
ans=max(ans,dp+cur)
print((ans*math.pi)/2)
```
| 9,506 |
Provide tags and a correct Python 3 solution for this coding contest problem.
As you know, every birthday party has a cake! This time, Babaei is going to prepare the very special birthday party's cake.
Simple cake is a cylinder of some radius and height. The volume of the simple cake is equal to the volume of corresponding cylinder. Babaei has n simple cakes and he is going to make a special cake placing some cylinders on each other.
However, there are some additional culinary restrictions. The cakes are numbered in such a way that the cake number i can be placed only on the table or on some cake number j where j < i. Moreover, in order to impress friends Babaei will put the cake i on top of the cake j only if the volume of the cake i is strictly greater than the volume of the cake j.
Babaei wants to prepare a birthday cake that has a maximum possible total volume. Help him find this value.
Input
The first line of the input contains a single integer n (1 β€ n β€ 100 000) β the number of simple cakes Babaei has.
Each of the following n lines contains two integers ri and hi (1 β€ ri, hi β€ 10 000), giving the radius and height of the i-th cake.
Output
Print the maximum volume of the cake that Babaei can make. Your 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
2
100 30
40 10
Output
942477.796077000
Input
4
1 1
9 7
1 4
10 7
Output
3983.539484752
Note
In first sample, the optimal way is to choose the cake number 1.
In second sample, the way to get the maximum volume is to use cakes with indices 1, 2 and 4.
Tags: data structures, dp
Correct Solution:
```
# [https://codeforces.com/contest/629/submission/32426374]
import math
n = int(input())
p = [0] * n
for i in range(n):
(r, h) = map(int, input().split())
p[i] = r**2 * h
k = {v: j for j, v in enumerate(sorted(set(p)))}
d = [0] * (len(k) + 1)
for v in p:
j = k[v]
i = j + 1
q = 0
while j > 0:
q = max(d[j], q)
j -= j & -j
q += v
while i < len(d):
d[i] = max(d[i], q)
i += i & -i
print(max(d) * math.pi)
```
| 9,507 |
Provide tags and a correct Python 3 solution for this coding contest problem.
As you know, every birthday party has a cake! This time, Babaei is going to prepare the very special birthday party's cake.
Simple cake is a cylinder of some radius and height. The volume of the simple cake is equal to the volume of corresponding cylinder. Babaei has n simple cakes and he is going to make a special cake placing some cylinders on each other.
However, there are some additional culinary restrictions. The cakes are numbered in such a way that the cake number i can be placed only on the table or on some cake number j where j < i. Moreover, in order to impress friends Babaei will put the cake i on top of the cake j only if the volume of the cake i is strictly greater than the volume of the cake j.
Babaei wants to prepare a birthday cake that has a maximum possible total volume. Help him find this value.
Input
The first line of the input contains a single integer n (1 β€ n β€ 100 000) β the number of simple cakes Babaei has.
Each of the following n lines contains two integers ri and hi (1 β€ ri, hi β€ 10 000), giving the radius and height of the i-th cake.
Output
Print the maximum volume of the cake that Babaei can make. Your 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
2
100 30
40 10
Output
942477.796077000
Input
4
1 1
9 7
1 4
10 7
Output
3983.539484752
Note
In first sample, the optimal way is to choose the cake number 1.
In second sample, the way to get the maximum volume is to use cakes with indices 1, 2 and 4.
Tags: data structures, dp
Correct Solution:
```
from math import pi
n = int(input())
secuencia = [None] * n
maximo_to = -1
for num in range(n):
r, h = (int(x) for x in input().strip().split())
secuencia[num] = [r * r * h, num + 1]
secuencia.reverse()
secuencia.sort(key=lambda x: x[0])
actual = 0
bit = [0] * (n + 1)
def max_x(x, l):
suma = 0
while x != 0:
suma = max(suma, l[x])
x -= (x & -x)
return suma
def update_x(x, l, max_n, val):
while x <= max_n:
if val > l[x]:
l[x] = val
else:
return
x += (x & -x)
for e in range(n):
maximo = secuencia[e][0] + max_x(secuencia[e][1] - 1, bit)
update_x(secuencia[e][1], bit, n, maximo)
if maximo > maximo_to:
maximo_to = maximo
print(maximo_to * pi)
```
| 9,508 |
Provide tags and a correct Python 3 solution for this coding contest problem.
As you know, every birthday party has a cake! This time, Babaei is going to prepare the very special birthday party's cake.
Simple cake is a cylinder of some radius and height. The volume of the simple cake is equal to the volume of corresponding cylinder. Babaei has n simple cakes and he is going to make a special cake placing some cylinders on each other.
However, there are some additional culinary restrictions. The cakes are numbered in such a way that the cake number i can be placed only on the table or on some cake number j where j < i. Moreover, in order to impress friends Babaei will put the cake i on top of the cake j only if the volume of the cake i is strictly greater than the volume of the cake j.
Babaei wants to prepare a birthday cake that has a maximum possible total volume. Help him find this value.
Input
The first line of the input contains a single integer n (1 β€ n β€ 100 000) β the number of simple cakes Babaei has.
Each of the following n lines contains two integers ri and hi (1 β€ ri, hi β€ 10 000), giving the radius and height of the i-th cake.
Output
Print the maximum volume of the cake that Babaei can make. Your 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
2
100 30
40 10
Output
942477.796077000
Input
4
1 1
9 7
1 4
10 7
Output
3983.539484752
Note
In first sample, the optimal way is to choose the cake number 1.
In second sample, the way to get the maximum volume is to use cakes with indices 1, 2 and 4.
Tags: data structures, dp
Correct Solution:
```
from sys import *
t = list(map(int, stdin.read().split()))
p = [t[i + 1] * t[i] ** 2 for i in range(1, len(t), 2)]
k = {v: j for j, v in enumerate(sorted(set(p)))}
d = [0] * (len(k) + 1)
for v in p:
j = k[v]
i = j + 1
q = 0
while j > 0:
q = max(d[j], q)
j -= j & -j
q += v
while i < len(d):
d[i] = max(d[i], q)
i += i & -i
print(max(d) * 3.14159265)
```
| 9,509 |
Provide tags and a correct Python 3 solution for this coding contest problem.
As you know, every birthday party has a cake! This time, Babaei is going to prepare the very special birthday party's cake.
Simple cake is a cylinder of some radius and height. The volume of the simple cake is equal to the volume of corresponding cylinder. Babaei has n simple cakes and he is going to make a special cake placing some cylinders on each other.
However, there are some additional culinary restrictions. The cakes are numbered in such a way that the cake number i can be placed only on the table or on some cake number j where j < i. Moreover, in order to impress friends Babaei will put the cake i on top of the cake j only if the volume of the cake i is strictly greater than the volume of the cake j.
Babaei wants to prepare a birthday cake that has a maximum possible total volume. Help him find this value.
Input
The first line of the input contains a single integer n (1 β€ n β€ 100 000) β the number of simple cakes Babaei has.
Each of the following n lines contains two integers ri and hi (1 β€ ri, hi β€ 10 000), giving the radius and height of the i-th cake.
Output
Print the maximum volume of the cake that Babaei can make. Your 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
2
100 30
40 10
Output
942477.796077000
Input
4
1 1
9 7
1 4
10 7
Output
3983.539484752
Note
In first sample, the optimal way is to choose the cake number 1.
In second sample, the way to get the maximum volume is to use cakes with indices 1, 2 and 4.
Tags: data structures, dp
Correct Solution:
```
def add(i, q):
while i < len(d):
d[i] = max(d[i], q)
i += i & -i
def get(i):
q = 0
while i > 0:
q = max(d[i], q)
i -= i & -i
return q
from sys import *
t = list(map(int, stdin.read().split()))
p = [t[i] * t[i] * t[i + 1] for i in range(1, len(t), 2)]
k = {v: j for j, v in enumerate(sorted(set(p)))}
d = [0] * (len(k) + 1)
for v in p:
j = k[v]
add(j + 1, get(j) + v)
print(get(len(k)) * 3.14159265)
```
| 9,510 |
Provide tags and a correct Python 3 solution for this coding contest problem.
As you know, every birthday party has a cake! This time, Babaei is going to prepare the very special birthday party's cake.
Simple cake is a cylinder of some radius and height. The volume of the simple cake is equal to the volume of corresponding cylinder. Babaei has n simple cakes and he is going to make a special cake placing some cylinders on each other.
However, there are some additional culinary restrictions. The cakes are numbered in such a way that the cake number i can be placed only on the table or on some cake number j where j < i. Moreover, in order to impress friends Babaei will put the cake i on top of the cake j only if the volume of the cake i is strictly greater than the volume of the cake j.
Babaei wants to prepare a birthday cake that has a maximum possible total volume. Help him find this value.
Input
The first line of the input contains a single integer n (1 β€ n β€ 100 000) β the number of simple cakes Babaei has.
Each of the following n lines contains two integers ri and hi (1 β€ ri, hi β€ 10 000), giving the radius and height of the i-th cake.
Output
Print the maximum volume of the cake that Babaei can make. Your 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
2
100 30
40 10
Output
942477.796077000
Input
4
1 1
9 7
1 4
10 7
Output
3983.539484752
Note
In first sample, the optimal way is to choose the cake number 1.
In second sample, the way to get the maximum volume is to use cakes with indices 1, 2 and 4.
Tags: data structures, dp
Correct Solution:
```
import math
from functools import reduce
class SegmentTree():
def __init__(self, L, function = lambda x,y: x+y):
self.function = function
N = self.size = len(L)
M = 1 << N.bit_length()
self.margin = 2*M - N
self.L = [None for i in range(self.margin)] + L
for i in range(M-1, 0, -1):
x, y = self.L[i<<1], self.L[i<<1|1]
self.L[i] = None if x is None or y is None else function(x, y)
def modify(self, pos, value):
p = pos + self.margin
self.L[p] = value
while p > 1:
x, y = self.L[p], self.L[p^1]
if p&1: x, y = y, x
self.L[p>>1] = None if x is None or y is None else self.function(x, y)
p>>=1
def query(self, left, right):
l, r = left + self.margin, right + self.margin
stack = []
void = True
while l < r:
if l&1:
if void:
result = self.L[l]
void = False
else:
result = self.function(result, self.L[l])
l+=1
if r&1:
r-=1
stack.append(self.L[r])
l>>=1
r>>=1
init = stack.pop() if void else result
return reduce(self.function, reversed(stack), init)
n = int(input())
pies, index, first_equal = [0]*n, [0]*n, [0]*n
for i in range(n):
r, h = [int(x) for x in input().split()]
pies[i] = r*r*h
s_pies = list(sorted(enumerate(pies), key = lambda p: p[1]))
for i in range(n): index[s_pies[i][0]] = i
for i in range(1, n):
first_equal[s_pies[i][0]] = i if s_pies[i][1] != s_pies[i-1][1] else first_equal[s_pies[i-1][0]]
towers = SegmentTree([0]*(n+1), max)
for j, pie in enumerate(pies):
i, k = index[j], first_equal[j]
q = towers.query(0, k+1)
towers.modify(i+1, q + pie)
print(math.pi * towers.query(0, n+1))
```
| 9,511 |
Provide tags and a correct Python 3 solution for this coding contest problem.
As you know, every birthday party has a cake! This time, Babaei is going to prepare the very special birthday party's cake.
Simple cake is a cylinder of some radius and height. The volume of the simple cake is equal to the volume of corresponding cylinder. Babaei has n simple cakes and he is going to make a special cake placing some cylinders on each other.
However, there are some additional culinary restrictions. The cakes are numbered in such a way that the cake number i can be placed only on the table or on some cake number j where j < i. Moreover, in order to impress friends Babaei will put the cake i on top of the cake j only if the volume of the cake i is strictly greater than the volume of the cake j.
Babaei wants to prepare a birthday cake that has a maximum possible total volume. Help him find this value.
Input
The first line of the input contains a single integer n (1 β€ n β€ 100 000) β the number of simple cakes Babaei has.
Each of the following n lines contains two integers ri and hi (1 β€ ri, hi β€ 10 000), giving the radius and height of the i-th cake.
Output
Print the maximum volume of the cake that Babaei can make. Your 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
2
100 30
40 10
Output
942477.796077000
Input
4
1 1
9 7
1 4
10 7
Output
3983.539484752
Note
In first sample, the optimal way is to choose the cake number 1.
In second sample, the way to get the maximum volume is to use cakes with indices 1, 2 and 4.
Tags: data structures, dp
Correct Solution:
```
from math import *
from bisect import *
def update(bit, size, idx, amount):
while idx <= size:
if bit[idx] < amount:
bit[idx] = amount
idx += idx & -idx
def read(bit, idx):
rst = 0
while idx >= 1:
if bit[idx] > rst:
rst = bit[idx]
idx -= idx & -idx
return rst
n = int(input())
arr = [map(int, input().split()) for _ in range(n)]
arr = [r*r*h for r, h in arr]
arr2 = sorted(list(set(arr)))
n2 = len(arr2)
bit = [0] * (n2 + 1)
for v in arr:
idx = bisect(arr2, v)
update(bit, n2, idx, read(bit, idx-1) + v)
print(pi*max(bit))
```
| 9,512 |
Provide tags and a correct Python 3 solution for this coding contest problem.
As you know, every birthday party has a cake! This time, Babaei is going to prepare the very special birthday party's cake.
Simple cake is a cylinder of some radius and height. The volume of the simple cake is equal to the volume of corresponding cylinder. Babaei has n simple cakes and he is going to make a special cake placing some cylinders on each other.
However, there are some additional culinary restrictions. The cakes are numbered in such a way that the cake number i can be placed only on the table or on some cake number j where j < i. Moreover, in order to impress friends Babaei will put the cake i on top of the cake j only if the volume of the cake i is strictly greater than the volume of the cake j.
Babaei wants to prepare a birthday cake that has a maximum possible total volume. Help him find this value.
Input
The first line of the input contains a single integer n (1 β€ n β€ 100 000) β the number of simple cakes Babaei has.
Each of the following n lines contains two integers ri and hi (1 β€ ri, hi β€ 10 000), giving the radius and height of the i-th cake.
Output
Print the maximum volume of the cake that Babaei can make. Your 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
2
100 30
40 10
Output
942477.796077000
Input
4
1 1
9 7
1 4
10 7
Output
3983.539484752
Note
In first sample, the optimal way is to choose the cake number 1.
In second sample, the way to get the maximum volume is to use cakes with indices 1, 2 and 4.
Tags: data structures, dp
Correct Solution:
```
from math import *
from bisect import *
def update(bit, size, idx, amount):
while idx <= size:
if bit[idx] < amount:
bit[idx] = amount
idx += idx & -idx
def read(bit, idx):
rst = 0
while idx >= 1:
if bit[idx] > rst:
rst = bit[idx]
idx -= idx & -idx
return rst
n = int(input())
arr = [map(int, input().split()) for _ in range(n)]
arr = [pi*(r*r*h) for r, h in arr]
arr2 = sorted(list(set(arr)))
n2 = len(arr2)
bit = [0.0] * (n2 + 1)
for v in arr:
idx = bisect(arr2, v)
update(bit, n2, idx, read(bit, idx-1) + v)
print(max(bit))
```
| 9,513 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
As you know, every birthday party has a cake! This time, Babaei is going to prepare the very special birthday party's cake.
Simple cake is a cylinder of some radius and height. The volume of the simple cake is equal to the volume of corresponding cylinder. Babaei has n simple cakes and he is going to make a special cake placing some cylinders on each other.
However, there are some additional culinary restrictions. The cakes are numbered in such a way that the cake number i can be placed only on the table or on some cake number j where j < i. Moreover, in order to impress friends Babaei will put the cake i on top of the cake j only if the volume of the cake i is strictly greater than the volume of the cake j.
Babaei wants to prepare a birthday cake that has a maximum possible total volume. Help him find this value.
Input
The first line of the input contains a single integer n (1 β€ n β€ 100 000) β the number of simple cakes Babaei has.
Each of the following n lines contains two integers ri and hi (1 β€ ri, hi β€ 10 000), giving the radius and height of the i-th cake.
Output
Print the maximum volume of the cake that Babaei can make. Your 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
2
100 30
40 10
Output
942477.796077000
Input
4
1 1
9 7
1 4
10 7
Output
3983.539484752
Note
In first sample, the optimal way is to choose the cake number 1.
In second sample, the way to get the maximum volume is to use cakes with indices 1, 2 and 4.
Submitted Solution:
```
def setmax(a, i, v):
while i < len(a):
a[i] = max(a[i], v)
i |= i + 1
def getmax(a, i):
r = 0
while i > 0:
r = max(r, a[i-1])
i &= i - 1
return r
def his(l):
maxbit = [0] * len(l)
rank = [0] * len(l)
for i, j in enumerate(sorted(range(len(l)), key=lambda i: l[i])):
rank[j] = i
for i, x in enumerate(l):
r = rank[i]
s = getmax(maxbit, r)
setmax(maxbit, r, x + s)
return getmax(maxbit, len(l))
import math
n=int(input())
lis=list()
for _ in range(n):
a,b=map(int,input().split())
lis.append(math.pi*a*a*b)
print(his(lis))
```
No
| 9,514 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
As you know, every birthday party has a cake! This time, Babaei is going to prepare the very special birthday party's cake.
Simple cake is a cylinder of some radius and height. The volume of the simple cake is equal to the volume of corresponding cylinder. Babaei has n simple cakes and he is going to make a special cake placing some cylinders on each other.
However, there are some additional culinary restrictions. The cakes are numbered in such a way that the cake number i can be placed only on the table or on some cake number j where j < i. Moreover, in order to impress friends Babaei will put the cake i on top of the cake j only if the volume of the cake i is strictly greater than the volume of the cake j.
Babaei wants to prepare a birthday cake that has a maximum possible total volume. Help him find this value.
Input
The first line of the input contains a single integer n (1 β€ n β€ 100 000) β the number of simple cakes Babaei has.
Each of the following n lines contains two integers ri and hi (1 β€ ri, hi β€ 10 000), giving the radius and height of the i-th cake.
Output
Print the maximum volume of the cake that Babaei can make. Your 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
2
100 30
40 10
Output
942477.796077000
Input
4
1 1
9 7
1 4
10 7
Output
3983.539484752
Note
In first sample, the optimal way is to choose the cake number 1.
In second sample, the way to get the maximum volume is to use cakes with indices 1, 2 and 4.
Submitted Solution:
```
from math import *
from bisect import *
def update(bit, size, idx, amount):
while idx <= size:
if bit[idx] < amount:
bit[idx] = amount
idx += idx & -idx
def read(bit, idx):
rst = 0
while idx >= 1:
if bit[idx] > rst:
rst = bit[idx]
idx -= idx & -idx
return rst
n = int(input())
arr = [map(int, input().split()) for _ in range(n)]
arr = [pi*r*r*h for r, h in arr]
arr2 = sorted(list(set(arr)))
n2 = len(arr2)
bit = [0.0] * (n2 + 1)
for v in arr:
idx = bisect(arr2, v)
update(bit, n2, idx, read(bit, idx-1) + v)
print(bit[-1])
```
No
| 9,515 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
As you know, every birthday party has a cake! This time, Babaei is going to prepare the very special birthday party's cake.
Simple cake is a cylinder of some radius and height. The volume of the simple cake is equal to the volume of corresponding cylinder. Babaei has n simple cakes and he is going to make a special cake placing some cylinders on each other.
However, there are some additional culinary restrictions. The cakes are numbered in such a way that the cake number i can be placed only on the table or on some cake number j where j < i. Moreover, in order to impress friends Babaei will put the cake i on top of the cake j only if the volume of the cake i is strictly greater than the volume of the cake j.
Babaei wants to prepare a birthday cake that has a maximum possible total volume. Help him find this value.
Input
The first line of the input contains a single integer n (1 β€ n β€ 100 000) β the number of simple cakes Babaei has.
Each of the following n lines contains two integers ri and hi (1 β€ ri, hi β€ 10 000), giving the radius and height of the i-th cake.
Output
Print the maximum volume of the cake that Babaei can make. Your 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
2
100 30
40 10
Output
942477.796077000
Input
4
1 1
9 7
1 4
10 7
Output
3983.539484752
Note
In first sample, the optimal way is to choose the cake number 1.
In second sample, the way to get the maximum volume is to use cakes with indices 1, 2 and 4.
Submitted Solution:
```
#adapted his
def his(l):
max1 = [0] * len(l)
r1 = [0] * len(l)
for i, j in enumerate(sorted(range(len(l)), key=lambda i: l[i])):
r1[j] = i
for i, x in enumerate(l):
r = r1[i]
r2 = 0
i2 = r
while i2 > 0:
r2 = max(r2, max1[i2-1])
i2 &= i2 - 1
i2 = r
while i2 < len(max1):
max1[i2] = max(x+r2,max1[i2])
i2|=i2+1
r2 = 0
i2 = len(l)
while i2 > 0:
r2 = max(r2, max1[i2-1])
i2 &= i2-1
return r2
n = int(input())
l = []
for i in range(n):
j,k = map(int, input().split())
l.append(j*j*3.14159265358979*k)
print(his(l))
```
No
| 9,516 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
As you know, every birthday party has a cake! This time, Babaei is going to prepare the very special birthday party's cake.
Simple cake is a cylinder of some radius and height. The volume of the simple cake is equal to the volume of corresponding cylinder. Babaei has n simple cakes and he is going to make a special cake placing some cylinders on each other.
However, there are some additional culinary restrictions. The cakes are numbered in such a way that the cake number i can be placed only on the table or on some cake number j where j < i. Moreover, in order to impress friends Babaei will put the cake i on top of the cake j only if the volume of the cake i is strictly greater than the volume of the cake j.
Babaei wants to prepare a birthday cake that has a maximum possible total volume. Help him find this value.
Input
The first line of the input contains a single integer n (1 β€ n β€ 100 000) β the number of simple cakes Babaei has.
Each of the following n lines contains two integers ri and hi (1 β€ ri, hi β€ 10 000), giving the radius and height of the i-th cake.
Output
Print the maximum volume of the cake that Babaei can make. Your 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
2
100 30
40 10
Output
942477.796077000
Input
4
1 1
9 7
1 4
10 7
Output
3983.539484752
Note
In first sample, the optimal way is to choose the cake number 1.
In second sample, the way to get the maximum volume is to use cakes with indices 1, 2 and 4.
Submitted Solution:
```
from math import pi
n = int(input())
secuencia = [None] * n
maximo_to = -1
for num in range(n):
r, h = (int(x) for x in input().strip().split())
secuencia[num] = [r * r * h, num + 1, None]
secuencia.sort(key=lambda x: x[0])
actual = 0
est_actual = 0
for num in range(n):
if secuencia[num][0] > actual:
est_actual += 1
secuencia[num][2] = est_actual
bit = [0] * (est_actual + 1)
def max_x(x, l):
suma = 0
while x != 0:
suma = max(suma, l[x])
x -= (x & -x)
return suma
def update_x(x, l, max_n, val):
while x <= max_n:
if val > l[x]:
l[x] = val
else:
return
x += (x & -x)
for e in range(n):
maximo = secuencia[e][0] + max_x(secuencia[e][1] - 1, bit)
update_x(secuencia[e][1], bit, est_actual, maximo)
if maximo > maximo_to:
maximo_to = maximo
print(maximo_to * pi)
```
No
| 9,517 |
Provide tags and a correct Python 3 solution for this coding contest problem.
A tree is a connected undirected graph consisting of n vertices and n - 1 edges. Vertices are numbered 1 through n.
Limak is a little polar bear. He once had a tree with n vertices but he lost it. He still remembers something about the lost tree though.
You are given m pairs of vertices (a1, b1), (a2, b2), ..., (am, bm). Limak remembers that for each i there was no edge between ai and bi. He also remembers that vertex 1 was incident to exactly k edges (its degree was equal to k).
Is it possible that Limak remembers everything correctly? Check whether there exists a tree satisfying the given conditions.
Input
The first line of the input contains three integers n, m and k (<image>) β the number of vertices in Limak's tree, the number of forbidden pairs of vertices, and the degree of vertex 1, respectively.
The i-th of next m lines contains two distinct integers ai and bi (1 β€ ai, bi β€ n, ai β bi) β the i-th pair that is forbidden. It's guaranteed that each pair of vertices will appear at most once in the input.
Output
Print "possible" (without quotes) if there exists at least one tree satisfying the given conditions. Otherwise, print "impossible" (without quotes).
Examples
Input
5 4 2
1 2
2 3
4 2
4 1
Output
possible
Input
6 5 3
1 2
1 3
1 4
1 5
1 6
Output
impossible
Note
In the first sample, there are n = 5 vertices. The degree of vertex 1 should be k = 2. All conditions are satisfied for a tree with edges 1 - 5, 5 - 2, 1 - 3 and 3 - 4.
In the second sample, Limak remembers that none of the following edges existed: 1 - 2, 1 - 3, 1 - 4, 1 - 5 and 1 - 6. Hence, vertex 1 couldn't be connected to any other vertex and it implies that there is no suitable tree.
Tags: dfs and similar, dsu, graphs, trees
Correct Solution:
```
import sys
import math
from heapq import *;
input = sys.stdin.readline
from functools import cmp_to_key;
def pi():
return(int(input()))
def pl():
return(int(input(), 16))
def ti():
return(list(map(int,input().split())))
def ts():
s = input()
return(list(s[:len(s) - 1]))
def invr():
return(map(int,input().split()))
mod = 1000000007;
f = [];
def fact(n,m):
global f;
f = [1 for i in range(n+1)];
f[0] = 1;
for i in range(1,n+1):
f[i] = (f[i-1]*i)%m;
def fast_mod_exp(a,b,m):
res = 1;
while b > 0:
if b & 1:
res = (res*a)%m;
a = (a*a)%m;
b = b >> 1;
return res;
def inverseMod(n,m):
return fast_mod_exp(n,m-2,m);
def ncr(n,r,m):
if n < 0 or r < 0 or r > n: return 0;
if r == 0: return 1;
return ((f[n]*inverseMod(f[n-r],m))%m*inverseMod(f[r],m))%m;
def main():
E();
mp = [{''} for i in range(300005)];
def canConnect(a,b):
return b not in mp[a];
remaining = {''};
def dfs(a):
temp = [];
for b in remaining:
if canConnect(a,b): temp.append(b);
for b in temp:
remaining.remove(b);
for b in temp:
dfs(b);
def E():
[n,m,k] = ti();
mxDegreePossible = n-1;
for i in range(m):
[a,b] = ti();
mp[a].add(b);
mp[b].add(a);
if a == 1 or b == 1:
mxDegreePossible -= 1;
if mxDegreePossible < k:
print("impossible");
return;
for i in range(2,n+1):
remaining.add(i);
components = 0;
for i in range(2,n+1):
if i in remaining and canConnect(1,i):
remaining.remove(i);
dfs(i);
components += 1;
remaining.remove('');
if components > k or len(remaining) > 0:
print('impossible');
return;
print('possible');
main();
```
| 9,518 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A tree is a connected undirected graph consisting of n vertices and n - 1 edges. Vertices are numbered 1 through n.
Limak is a little polar bear. He once had a tree with n vertices but he lost it. He still remembers something about the lost tree though.
You are given m pairs of vertices (a1, b1), (a2, b2), ..., (am, bm). Limak remembers that for each i there was no edge between ai and bi. He also remembers that vertex 1 was incident to exactly k edges (its degree was equal to k).
Is it possible that Limak remembers everything correctly? Check whether there exists a tree satisfying the given conditions.
Input
The first line of the input contains three integers n, m and k (<image>) β the number of vertices in Limak's tree, the number of forbidden pairs of vertices, and the degree of vertex 1, respectively.
The i-th of next m lines contains two distinct integers ai and bi (1 β€ ai, bi β€ n, ai β bi) β the i-th pair that is forbidden. It's guaranteed that each pair of vertices will appear at most once in the input.
Output
Print "possible" (without quotes) if there exists at least one tree satisfying the given conditions. Otherwise, print "impossible" (without quotes).
Examples
Input
5 4 2
1 2
2 3
4 2
4 1
Output
possible
Input
6 5 3
1 2
1 3
1 4
1 5
1 6
Output
impossible
Note
In the first sample, there are n = 5 vertices. The degree of vertex 1 should be k = 2. All conditions are satisfied for a tree with edges 1 - 5, 5 - 2, 1 - 3 and 3 - 4.
In the second sample, Limak remembers that none of the following edges existed: 1 - 2, 1 - 3, 1 - 4, 1 - 5 and 1 - 6. Hence, vertex 1 couldn't be connected to any other vertex and it implies that there is no suitable tree.
Submitted Solution:
```
n,m,k = map(int,input().split())
arr = []
for i in range(m):
a,b = map(int,input().split())
if a == 1:arr.append(b)
elif b == 1: arr.append(a)
if len(arr) != k: print("impossible")
else: print("possible")
```
No
| 9,519 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A tree is a connected undirected graph consisting of n vertices and n - 1 edges. Vertices are numbered 1 through n.
Limak is a little polar bear. He once had a tree with n vertices but he lost it. He still remembers something about the lost tree though.
You are given m pairs of vertices (a1, b1), (a2, b2), ..., (am, bm). Limak remembers that for each i there was no edge between ai and bi. He also remembers that vertex 1 was incident to exactly k edges (its degree was equal to k).
Is it possible that Limak remembers everything correctly? Check whether there exists a tree satisfying the given conditions.
Input
The first line of the input contains three integers n, m and k (<image>) β the number of vertices in Limak's tree, the number of forbidden pairs of vertices, and the degree of vertex 1, respectively.
The i-th of next m lines contains two distinct integers ai and bi (1 β€ ai, bi β€ n, ai β bi) β the i-th pair that is forbidden. It's guaranteed that each pair of vertices will appear at most once in the input.
Output
Print "possible" (without quotes) if there exists at least one tree satisfying the given conditions. Otherwise, print "impossible" (without quotes).
Examples
Input
5 4 2
1 2
2 3
4 2
4 1
Output
possible
Input
6 5 3
1 2
1 3
1 4
1 5
1 6
Output
impossible
Note
In the first sample, there are n = 5 vertices. The degree of vertex 1 should be k = 2. All conditions are satisfied for a tree with edges 1 - 5, 5 - 2, 1 - 3 and 3 - 4.
In the second sample, Limak remembers that none of the following edges existed: 1 - 2, 1 - 3, 1 - 4, 1 - 5 and 1 - 6. Hence, vertex 1 couldn't be connected to any other vertex and it implies that there is no suitable tree.
Submitted Solution:
```
import sys
import math
from heapq import *;
input = sys.stdin.readline
from functools import cmp_to_key;
def pi():
return(int(input()))
def pl():
return(int(input(), 16))
def ti():
return(list(map(int,input().split())))
def ts():
s = input()
return(list(s[:len(s) - 1]))
def invr():
return(map(int,input().split()))
mod = 1000000007;
f = [];
def fact(n,m):
global f;
f = [1 for i in range(n+1)];
f[0] = 1;
for i in range(1,n+1):
f[i] = (f[i-1]*i)%m;
def fast_mod_exp(a,b,m):
res = 1;
while b > 0:
if b & 1:
res = (res*a)%m;
a = (a*a)%m;
b = b >> 1;
return res;
def inverseMod(n,m):
return fast_mod_exp(n,m-2,m);
def ncr(n,r,m):
if n < 0 or r < 0 or r > n: return 0;
if r == 0: return 1;
return ((f[n]*inverseMod(f[n-r],m))%m*inverseMod(f[r],m))%m;
def main():
E();
mp = [{''} for i in range(300005)];
def canConnect(a,b):
return b not in mp[a];
remaining = {''};
def dfs(a):
temp = [];
for b in remaining:
if canConnect(a,b): temp.append(b);
for b in temp:
remaining.remove(b);
for b in temp:
dfs(b);
def E():
[n,m,k] = ti();
mxDegreePossible = n-1;
for i in range(m):
[a,b] = ti();
mp[a].add(b);
mp[b].add(a);
if a == 1 or b == 1:
mxDegreePossible -= 1;
if mxDegreePossible < k:
print("impossible");
return;
for i in range(1,n+1):
remaining.add(i);
components = 0;
for i in range(2,n+1):
if i in remaining and canConnect(1,i):
dfs(i);
components += 1;
if components > k:
print('impossible');
return;
print('possible');
main();
```
No
| 9,520 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A tree is a connected undirected graph consisting of n vertices and n - 1 edges. Vertices are numbered 1 through n.
Limak is a little polar bear. He once had a tree with n vertices but he lost it. He still remembers something about the lost tree though.
You are given m pairs of vertices (a1, b1), (a2, b2), ..., (am, bm). Limak remembers that for each i there was no edge between ai and bi. He also remembers that vertex 1 was incident to exactly k edges (its degree was equal to k).
Is it possible that Limak remembers everything correctly? Check whether there exists a tree satisfying the given conditions.
Input
The first line of the input contains three integers n, m and k (<image>) β the number of vertices in Limak's tree, the number of forbidden pairs of vertices, and the degree of vertex 1, respectively.
The i-th of next m lines contains two distinct integers ai and bi (1 β€ ai, bi β€ n, ai β bi) β the i-th pair that is forbidden. It's guaranteed that each pair of vertices will appear at most once in the input.
Output
Print "possible" (without quotes) if there exists at least one tree satisfying the given conditions. Otherwise, print "impossible" (without quotes).
Examples
Input
5 4 2
1 2
2 3
4 2
4 1
Output
possible
Input
6 5 3
1 2
1 3
1 4
1 5
1 6
Output
impossible
Note
In the first sample, there are n = 5 vertices. The degree of vertex 1 should be k = 2. All conditions are satisfied for a tree with edges 1 - 5, 5 - 2, 1 - 3 and 3 - 4.
In the second sample, Limak remembers that none of the following edges existed: 1 - 2, 1 - 3, 1 - 4, 1 - 5 and 1 - 6. Hence, vertex 1 couldn't be connected to any other vertex and it implies that there is no suitable tree.
Submitted Solution:
```
import queue
adj = {}
x = input()
n, m, k = map(int, x.split(" "))
for i in range(n):
adj[i] = []
for i in range(m):
x = input()
a, b = map(int, x.split())
adj[a-1].append(b-1)
adj[b-1].append(a-1)
visited = {}
q = queue.Queue(maxsize= 300000)
for i in range(n):
visited[i] = False
q.put(0)
while not q.empty():
s = q.get()
for i in range(n):
if visited[i] == False:
if i not in adj[s]:
visited[i] = True
q.put(i)
connnected = True
for i in range(n):
if visited[i] == False:
connnected = False
if connnected:
print("possible")
else:
print("impossible")
```
No
| 9,521 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A tree is a connected undirected graph consisting of n vertices and n - 1 edges. Vertices are numbered 1 through n.
Limak is a little polar bear. He once had a tree with n vertices but he lost it. He still remembers something about the lost tree though.
You are given m pairs of vertices (a1, b1), (a2, b2), ..., (am, bm). Limak remembers that for each i there was no edge between ai and bi. He also remembers that vertex 1 was incident to exactly k edges (its degree was equal to k).
Is it possible that Limak remembers everything correctly? Check whether there exists a tree satisfying the given conditions.
Input
The first line of the input contains three integers n, m and k (<image>) β the number of vertices in Limak's tree, the number of forbidden pairs of vertices, and the degree of vertex 1, respectively.
The i-th of next m lines contains two distinct integers ai and bi (1 β€ ai, bi β€ n, ai β bi) β the i-th pair that is forbidden. It's guaranteed that each pair of vertices will appear at most once in the input.
Output
Print "possible" (without quotes) if there exists at least one tree satisfying the given conditions. Otherwise, print "impossible" (without quotes).
Examples
Input
5 4 2
1 2
2 3
4 2
4 1
Output
possible
Input
6 5 3
1 2
1 3
1 4
1 5
1 6
Output
impossible
Note
In the first sample, there are n = 5 vertices. The degree of vertex 1 should be k = 2. All conditions are satisfied for a tree with edges 1 - 5, 5 - 2, 1 - 3 and 3 - 4.
In the second sample, Limak remembers that none of the following edges existed: 1 - 2, 1 - 3, 1 - 4, 1 - 5 and 1 - 6. Hence, vertex 1 couldn't be connected to any other vertex and it implies that there is no suitable tree.
Submitted Solution:
```
import sys
def dfs(st):
compnum[st] = comp
unseen.remove(st)
for i in unseen.copy():
if i not in forb[st] and compnum[i] == 0:
dfs(i)
n, m, k = map(int, input().split())
forb = [set() for i in range(n+1)]
for i in range(m):
a, b = map(int, input().split())
forb[a].add(b)
forb[b].add(a)
adj = []
unseen = set()
for i in range(2, n+1):
unseen.add(i)
if i not in forb[1]:
adj.append(i)
compnum = [0 for i in range(n+1)]
comp = 0
for i in range(2, n+1):
if compnum[i] <= 0:
comp += 1
if(comp > k):
print("impossible")
sys.exit()
dfs(i)
comps = set()
for i in adj:
comps.add(compnum[i])
if(len(comps) == comp):
print("possible")
else:
print("impossible")
```
No
| 9,522 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You have a grid with n rows and n columns. Each cell is either empty (denoted by '.') or blocked (denoted by 'X').
Two empty cells are directly connected if they share a side. Two cells (r1, c1) (located in the row r1 and column c1) and (r2, c2) are connected if there exists a sequence of empty cells that starts with (r1, c1), finishes with (r2, c2), and any two consecutive cells in this sequence are directly connected. A connected component is a set of empty cells such that any two cells in the component are connected, and there is no cell in this set that is connected to some cell not in this set.
Your friend Limak is a big grizzly bear. He is able to destroy any obstacles in some range. More precisely, you can choose a square of size k Γ k in the grid and Limak will transform all blocked cells there to empty ones. However, you can ask Limak to help only once.
The chosen square must be completely inside the grid. It's possible that Limak won't change anything because all cells are empty anyway.
You like big connected components. After Limak helps you, what is the maximum possible size of the biggest connected component in the grid?
Input
The first line of the input contains two integers n and k (1 β€ k β€ n β€ 500) β the size of the grid and Limak's range, respectively.
Each of the next n lines contains a string with n characters, denoting the i-th row of the grid. Each character is '.' or 'X', denoting an empty cell or a blocked one, respectively.
Output
Print the maximum possible size (the number of cells) of the biggest connected component, after using Limak's help.
Examples
Input
5 2
..XXX
XX.XX
X.XXX
X...X
XXXX.
Output
10
Input
5 3
.....
.XXX.
.XXX.
.XXX.
.....
Output
25
Note
In the first sample, you can choose a square of size 2 Γ 2. It's optimal to choose a square in the red frame on the left drawing below. Then, you will get a connected component with 10 cells, marked blue in the right drawing.
<image>
Submitted Solution:
```
from collections import *
c=ans=0
D=Counter()
n,k=map(int,input().split())
a=[[-2]*(n+2)]+[[-2]+list(input())+[-2] for i in range(n)]+[[-2]*(n+2)]
for i in range(n):
for j in range(n):
a[i+1][j+1]=-(a[i+1][j+1]=='X')
def dfs(i,j):
if a[i][j]: return
a[i][j]=c; D[c]+=1
dfs(i-1,j); dfs(i+1,j)
dfs(i,j-1); dfs(i,j+1)
for i in range(n):
for j in range(n):
if a[i+1][j+1]==0: c+=1; dfs(i+1,j+1)
for j in range(1,n-k+2):
d,s=[0]*(c+3),set()
for i in range(1,n+1):
for y in range(k):
d[a[i][j+y]+2]+=1
if d[a[i][j+y]+2]==1: s.add(a[i][j+y]+2)
if i>k:
for y in range(k):
d[a[i-k][j+y]+2]-=1
if d[a[i][j+y]+2]==0: s.remove(a[i][j+y]+2)
if i>=k:
for q in range(k):
s|={a[i+1][j+q]+2,a[i-k][j+q]+2,a[i-q][j-1]+2,a[i-q][j+k]+2}
t=sum(D[key-2] for key in s-{0,1})+d[1]
if t>ans: ans=t
print(ans)
```
No
| 9,523 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You have a grid with n rows and n columns. Each cell is either empty (denoted by '.') or blocked (denoted by 'X').
Two empty cells are directly connected if they share a side. Two cells (r1, c1) (located in the row r1 and column c1) and (r2, c2) are connected if there exists a sequence of empty cells that starts with (r1, c1), finishes with (r2, c2), and any two consecutive cells in this sequence are directly connected. A connected component is a set of empty cells such that any two cells in the component are connected, and there is no cell in this set that is connected to some cell not in this set.
Your friend Limak is a big grizzly bear. He is able to destroy any obstacles in some range. More precisely, you can choose a square of size k Γ k in the grid and Limak will transform all blocked cells there to empty ones. However, you can ask Limak to help only once.
The chosen square must be completely inside the grid. It's possible that Limak won't change anything because all cells are empty anyway.
You like big connected components. After Limak helps you, what is the maximum possible size of the biggest connected component in the grid?
Input
The first line of the input contains two integers n and k (1 β€ k β€ n β€ 500) β the size of the grid and Limak's range, respectively.
Each of the next n lines contains a string with n characters, denoting the i-th row of the grid. Each character is '.' or 'X', denoting an empty cell or a blocked one, respectively.
Output
Print the maximum possible size (the number of cells) of the biggest connected component, after using Limak's help.
Examples
Input
5 2
..XXX
XX.XX
X.XXX
X...X
XXXX.
Output
10
Input
5 3
.....
.XXX.
.XXX.
.XXX.
.....
Output
25
Note
In the first sample, you can choose a square of size 2 Γ 2. It's optimal to choose a square in the red frame on the left drawing below. Then, you will get a connected component with 10 cells, marked blue in the right drawing.
<image>
Submitted Solution:
```
cell = []
mcell = 0
ccell = 0
def grid(a,b,ni):
global cell
global mcell
global ccell
if a < 0 or b < 0 or a > mcell or b > mcell: return 0
if cell[a][b] == 'X': return 0
if cell[a][b] != '.': return 0
cell[a][b] = str(ni)
ccell += 1
grid(a+1,b,ni)
grid(a-1,b,ni)
grid(a,b+1,ni)
grid(a,b-1,ni)
return 0
def checkgrid(i,j,k):
global cell
s = set()
for t in range(k):
r1 = cg(i-1,j+t)
r2 = cg(i+k,j+t)
r3 = cg(i+t,j-1)
r4 = cg(i+t,j+k)
if r1 != -1: s.add(r1)
if r2 != -1: s.add(r2)
if r3 != -1: s.add(r3)
if r4 != -1: s.add(r4)
return(list(s))
def cg(i,j):
global cell
if i < 0 or j < 0 or i > mcell or j > mcell: return -1
if cell[i][j] == 'X': return -1
return int(cell[i][j])
def cgc(i,j,k):
global cell
c = 0
for ii in range(k):
for jj in range(k):
if cell[i+ii][j+jj] == 'X': c += 1
return c
n,k = [int(x) for x in input().split(' ')]
mcell = n-1
for nn in range(n):
cell.append(list(input()))
ni = 0
mres = 0
cellcount = []
for i in range(n):
for j in range(n):
if cell[i][j] == '.':
grid(i,j,ni)
ni += 1
cellcount.append(ccell)
ccell = 0
for i in range(n-k+1):
for j in range(n-k+1):
s = checkgrid(i,j,k)
cc = 0
for ss in s:
cc += cellcount[ss]
cc += cgc(i,j,k)
mres = max(mres,cc)
print(mres)
```
No
| 9,524 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You have a grid with n rows and n columns. Each cell is either empty (denoted by '.') or blocked (denoted by 'X').
Two empty cells are directly connected if they share a side. Two cells (r1, c1) (located in the row r1 and column c1) and (r2, c2) are connected if there exists a sequence of empty cells that starts with (r1, c1), finishes with (r2, c2), and any two consecutive cells in this sequence are directly connected. A connected component is a set of empty cells such that any two cells in the component are connected, and there is no cell in this set that is connected to some cell not in this set.
Your friend Limak is a big grizzly bear. He is able to destroy any obstacles in some range. More precisely, you can choose a square of size k Γ k in the grid and Limak will transform all blocked cells there to empty ones. However, you can ask Limak to help only once.
The chosen square must be completely inside the grid. It's possible that Limak won't change anything because all cells are empty anyway.
You like big connected components. After Limak helps you, what is the maximum possible size of the biggest connected component in the grid?
Input
The first line of the input contains two integers n and k (1 β€ k β€ n β€ 500) β the size of the grid and Limak's range, respectively.
Each of the next n lines contains a string with n characters, denoting the i-th row of the grid. Each character is '.' or 'X', denoting an empty cell or a blocked one, respectively.
Output
Print the maximum possible size (the number of cells) of the biggest connected component, after using Limak's help.
Examples
Input
5 2
..XXX
XX.XX
X.XXX
X...X
XXXX.
Output
10
Input
5 3
.....
.XXX.
.XXX.
.XXX.
.....
Output
25
Note
In the first sample, you can choose a square of size 2 Γ 2. It's optimal to choose a square in the red frame on the left drawing below. Then, you will get a connected component with 10 cells, marked blue in the right drawing.
<image>
Submitted Solution:
```
from collections import *
c=ans=0
n,k=map(int,input().split())
a=[[-2]*(n+2)]+[[-2]+list(input())+[-2] for i in range(n)]+[[-2]*(n+2)]
for i in range(n):
for j in range(n):
a[i+1][j+1]=-(a[i+1][j+1]=='X')
D=[0]*(n*n+10)
def dfs(i,j):
l=r=j; u=d=i
for t in range(j+1,n+2):
r=t
if a[i][t]: break
for t in range(j-1,-1,-1):
l=t
if a[i][t]: break
for t in range(i+1,n+2):
d=t
if a[t][j]: break
for t in range(i-1,-1,-1):
u=t
if a[t][j]: break
for t in range(j+1,r): D[c]+=1; a[i][t]=c
for t in range(j-1,l,-1): D[c]+=1; a[i][t]=c
for t in range(i+1,d): D[c]+=1; a[t][j]=c
for t in range(i-1,u,-1): D[c]+=1; a[t][j]=c
for t in range(j+1,r): dfs(i,t)
for t in range(j-1,l,-1): dfs(i,t)
for t in range(i+1,d): dfs(t,j)
for t in range(i-1,u,-1): dfs(t,j)
for i in range(n):
for j in range(n):
if a[i+1][j+1]==0: c+=1; D[c]+=1; a[i+1][j+1]=c; dfs(i+1,j+1)
for x in a: print(x)
for j in range(1,n-k+2):
d=Counter()
for i in range(1,n+1):
for y in range(k): d[a[i][j+y]]+=1
if i>k:
for y in range(k):
d[a[i-k][j+y]]-=1
if d[a[i-k][j+y]]==0: del d[a[i-k][j+y]]
if i>=k:
s=set(d)
for q in range(k):
s|={a[i+1][j+q],a[i-k][j+q],a[i-q][j-1],a[i-q][j+k]}
t=sum(D[key] for key in s-{-2,-1})+d[-1]
if t>ans: ans=t
print(ans)
```
No
| 9,525 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You have a grid with n rows and n columns. Each cell is either empty (denoted by '.') or blocked (denoted by 'X').
Two empty cells are directly connected if they share a side. Two cells (r1, c1) (located in the row r1 and column c1) and (r2, c2) are connected if there exists a sequence of empty cells that starts with (r1, c1), finishes with (r2, c2), and any two consecutive cells in this sequence are directly connected. A connected component is a set of empty cells such that any two cells in the component are connected, and there is no cell in this set that is connected to some cell not in this set.
Your friend Limak is a big grizzly bear. He is able to destroy any obstacles in some range. More precisely, you can choose a square of size k Γ k in the grid and Limak will transform all blocked cells there to empty ones. However, you can ask Limak to help only once.
The chosen square must be completely inside the grid. It's possible that Limak won't change anything because all cells are empty anyway.
You like big connected components. After Limak helps you, what is the maximum possible size of the biggest connected component in the grid?
Input
The first line of the input contains two integers n and k (1 β€ k β€ n β€ 500) β the size of the grid and Limak's range, respectively.
Each of the next n lines contains a string with n characters, denoting the i-th row of the grid. Each character is '.' or 'X', denoting an empty cell or a blocked one, respectively.
Output
Print the maximum possible size (the number of cells) of the biggest connected component, after using Limak's help.
Examples
Input
5 2
..XXX
XX.XX
X.XXX
X...X
XXXX.
Output
10
Input
5 3
.....
.XXX.
.XXX.
.XXX.
.....
Output
25
Note
In the first sample, you can choose a square of size 2 Γ 2. It's optimal to choose a square in the red frame on the left drawing below. Then, you will get a connected component with 10 cells, marked blue in the right drawing.
<image>
Submitted Solution:
```
from collections import *
c=ans=0
D=Counter()
n,k=map(int,input().split())
a=[[-2]*(n+2)]+[[-2]+list(input())+[-2] for i in range(n)]+[[-2]*(n+2)]
for i in range(n):
for j in range(n):
a[i+1][j+1]=-(a[i+1][j+1]=='X')
k+=2
def dfs(i,j):
if a[i][j]: return
a[i][j]=c; D[c]+=1
dfs(i-1,j); dfs(i+1,j)
dfs(i,j-1); dfs(i,j+1)
for i in range(n):
for j in range(n):
if a[i+1][j+1]==0: c+=1; dfs(i+1,j+1)
for j in range(n-k+3):
d=Counter()
for i in range(n+2):
for y in range(k): d[a[i][j+y]]+=1
if i>k-1:
for y in range(k): d[a[i-k][j+y]]-=1
if i>=k-1:
d[a[i+1-k][j]]-=1; d[a[i+1-k][j+k-2]]-=1; d[a[i][j]]-=1; d[a[i][j+k-2]]-=1
t=sum(D[key] for key in d if key>0 and d[key]>0)+d[-1]
d[a[i+1-k][j]]+=1; d[a[i+1-k][j+k-2]]+=1; d[a[i][j]]+=1;d[a[i][j+k-2]]+=1
if t>ans: ans=t
print(ans)
```
No
| 9,526 |
Provide tags and a correct Python 2 solution for this coding contest problem.
On vacations n pupils decided to go on excursion and gather all together. They need to overcome the path with the length l meters. Each of the pupils will go with the speed equal to v1. To get to the excursion quickly, it was decided to rent a bus, which has seats for k people (it means that it can't fit more than k people at the same time) and the speed equal to v2. In order to avoid seasick, each of the pupils want to get into the bus no more than once.
Determine the minimum time required for all n pupils to reach the place of excursion. Consider that the embarkation and disembarkation of passengers, as well as the reversal of the bus, take place immediately and this time can be neglected.
Input
The first line of the input contains five positive integers n, l, v1, v2 and k (1 β€ n β€ 10 000, 1 β€ l β€ 109, 1 β€ v1 < v2 β€ 109, 1 β€ k β€ n) β the number of pupils, the distance from meeting to the place of excursion, the speed of each pupil, the speed of bus and the number of seats in the bus.
Output
Print the real number β the minimum time in which all pupils can reach the place of excursion. Your answer will be considered correct if its absolute or relative error won't exceed 10 - 6.
Examples
Input
5 10 1 2 5
Output
5.0000000000
Input
3 6 1 2 1
Output
4.7142857143
Note
In the first sample we should immediately put all five pupils to the bus. The speed of the bus equals 2 and the distance is equal to 10, so the pupils will reach the place of excursion in time 10 / 2 = 5.
Tags: binary search, math
Correct Solution:
```
from sys import stdin, stdout
from collections import Counter, defaultdict
from itertools import permutations, combinations
raw_input = stdin.readline
from math import ceil
pr = stdout.write
def in_num():
return int(raw_input())
def in_arr():
return map(int,raw_input().split())
def pr_num(n):
stdout.write(str(n)+'\n')
def pr_arr(arr):
pr(' '.join(map(str,arr))+'\n')
# fast read function for total integer input
def inp():
# this function returns whole input of
# space/line seperated integers
# Use Ctrl+D to flush stdin.
return map(float,stdin.read().split())
range = xrange # not for python 3.0+
n,l,v1,v2,k=inp()
val=ceil(n/k)
ans=l/((val*v2)-(((val-1)*v2)*((v2-v1)/(v1+v2))))
ans+=(l-(ans*v2))/v1
pr_num(ans)
```
| 9,527 |
Provide tags and a correct Python 3 solution for this coding contest problem.
On vacations n pupils decided to go on excursion and gather all together. They need to overcome the path with the length l meters. Each of the pupils will go with the speed equal to v1. To get to the excursion quickly, it was decided to rent a bus, which has seats for k people (it means that it can't fit more than k people at the same time) and the speed equal to v2. In order to avoid seasick, each of the pupils want to get into the bus no more than once.
Determine the minimum time required for all n pupils to reach the place of excursion. Consider that the embarkation and disembarkation of passengers, as well as the reversal of the bus, take place immediately and this time can be neglected.
Input
The first line of the input contains five positive integers n, l, v1, v2 and k (1 β€ n β€ 10 000, 1 β€ l β€ 109, 1 β€ v1 < v2 β€ 109, 1 β€ k β€ n) β the number of pupils, the distance from meeting to the place of excursion, the speed of each pupil, the speed of bus and the number of seats in the bus.
Output
Print the real number β the minimum time in which all pupils can reach the place of excursion. Your answer will be considered correct if its absolute or relative error won't exceed 10 - 6.
Examples
Input
5 10 1 2 5
Output
5.0000000000
Input
3 6 1 2 1
Output
4.7142857143
Note
In the first sample we should immediately put all five pupils to the bus. The speed of the bus equals 2 and the distance is equal to 10, so the pupils will reach the place of excursion in time 10 / 2 = 5.
Tags: binary search, math
Correct Solution:
```
array = list(map(int,input().split()))
n = int(array[0])
l = int(array[1])
v1 = int(array[2])
v2 = int(array[3])
k = int(array[4])
bus = n//k
if n%k != 0:
bus += 1
V = (v1+v2) + 2*v2*(bus-1)
U = (v1+v2) + 2*v1*(bus-1)
print (l*V/U/v2)
```
| 9,528 |
Provide tags and a correct Python 3 solution for this coding contest problem.
On vacations n pupils decided to go on excursion and gather all together. They need to overcome the path with the length l meters. Each of the pupils will go with the speed equal to v1. To get to the excursion quickly, it was decided to rent a bus, which has seats for k people (it means that it can't fit more than k people at the same time) and the speed equal to v2. In order to avoid seasick, each of the pupils want to get into the bus no more than once.
Determine the minimum time required for all n pupils to reach the place of excursion. Consider that the embarkation and disembarkation of passengers, as well as the reversal of the bus, take place immediately and this time can be neglected.
Input
The first line of the input contains five positive integers n, l, v1, v2 and k (1 β€ n β€ 10 000, 1 β€ l β€ 109, 1 β€ v1 < v2 β€ 109, 1 β€ k β€ n) β the number of pupils, the distance from meeting to the place of excursion, the speed of each pupil, the speed of bus and the number of seats in the bus.
Output
Print the real number β the minimum time in which all pupils can reach the place of excursion. Your answer will be considered correct if its absolute or relative error won't exceed 10 - 6.
Examples
Input
5 10 1 2 5
Output
5.0000000000
Input
3 6 1 2 1
Output
4.7142857143
Note
In the first sample we should immediately put all five pupils to the bus. The speed of the bus equals 2 and the distance is equal to 10, so the pupils will reach the place of excursion in time 10 / 2 = 5.
Tags: binary search, math
Correct Solution:
```
n, l, v1, v2, k = map(int, input().split())
diff = v2 - v1
n = (n + k - 1) // k * 2
p1 = (n * v2 - diff) * l
p2 = (n * v1 + diff) * v2
print(p1 / p2)
```
| 9,529 |
Provide tags and a correct Python 3 solution for this coding contest problem.
On vacations n pupils decided to go on excursion and gather all together. They need to overcome the path with the length l meters. Each of the pupils will go with the speed equal to v1. To get to the excursion quickly, it was decided to rent a bus, which has seats for k people (it means that it can't fit more than k people at the same time) and the speed equal to v2. In order to avoid seasick, each of the pupils want to get into the bus no more than once.
Determine the minimum time required for all n pupils to reach the place of excursion. Consider that the embarkation and disembarkation of passengers, as well as the reversal of the bus, take place immediately and this time can be neglected.
Input
The first line of the input contains five positive integers n, l, v1, v2 and k (1 β€ n β€ 10 000, 1 β€ l β€ 109, 1 β€ v1 < v2 β€ 109, 1 β€ k β€ n) β the number of pupils, the distance from meeting to the place of excursion, the speed of each pupil, the speed of bus and the number of seats in the bus.
Output
Print the real number β the minimum time in which all pupils can reach the place of excursion. Your answer will be considered correct if its absolute or relative error won't exceed 10 - 6.
Examples
Input
5 10 1 2 5
Output
5.0000000000
Input
3 6 1 2 1
Output
4.7142857143
Note
In the first sample we should immediately put all five pupils to the bus. The speed of the bus equals 2 and the distance is equal to 10, so the pupils will reach the place of excursion in time 10 / 2 = 5.
Tags: binary search, math
Correct Solution:
```
#coding=utf-8
import sys
eps = 1e-6
def solve(n, l, v1, v2):
t2 = 1.0 * (v1 + v2) * l / (n * (v1 + v2) * v2 - (n - 1) * v2 * (v2 - v1))
l2 = v2 * t2
l1 = l - l2
t1 = l1 / v1
#print(t1, l1, t2, l2)
return t1 + t2
#print(solve(3, 6, 1, 2))
#print(solve(1, 10, 1, 2))
n, t, v1, v2, k = map(int, input().split())
n = (n + k - 1) // k
print(solve(n, t, v1, v2))
```
| 9,530 |
Provide tags and a correct Python 3 solution for this coding contest problem.
On vacations n pupils decided to go on excursion and gather all together. They need to overcome the path with the length l meters. Each of the pupils will go with the speed equal to v1. To get to the excursion quickly, it was decided to rent a bus, which has seats for k people (it means that it can't fit more than k people at the same time) and the speed equal to v2. In order to avoid seasick, each of the pupils want to get into the bus no more than once.
Determine the minimum time required for all n pupils to reach the place of excursion. Consider that the embarkation and disembarkation of passengers, as well as the reversal of the bus, take place immediately and this time can be neglected.
Input
The first line of the input contains five positive integers n, l, v1, v2 and k (1 β€ n β€ 10 000, 1 β€ l β€ 109, 1 β€ v1 < v2 β€ 109, 1 β€ k β€ n) β the number of pupils, the distance from meeting to the place of excursion, the speed of each pupil, the speed of bus and the number of seats in the bus.
Output
Print the real number β the minimum time in which all pupils can reach the place of excursion. Your answer will be considered correct if its absolute or relative error won't exceed 10 - 6.
Examples
Input
5 10 1 2 5
Output
5.0000000000
Input
3 6 1 2 1
Output
4.7142857143
Note
In the first sample we should immediately put all five pupils to the bus. The speed of the bus equals 2 and the distance is equal to 10, so the pupils will reach the place of excursion in time 10 / 2 = 5.
Tags: binary search, math
Correct Solution:
```
n, L, v1, v2, k = map(int, input().split())
n = (n + k - 1) // k * 2
dif = v2 - v1
p1 = (n * v2 - dif) * L
p2 = (n * v1 + dif) * v2
ans = p1 / p2
print(ans)
```
| 9,531 |
Provide tags and a correct Python 3 solution for this coding contest problem.
On vacations n pupils decided to go on excursion and gather all together. They need to overcome the path with the length l meters. Each of the pupils will go with the speed equal to v1. To get to the excursion quickly, it was decided to rent a bus, which has seats for k people (it means that it can't fit more than k people at the same time) and the speed equal to v2. In order to avoid seasick, each of the pupils want to get into the bus no more than once.
Determine the minimum time required for all n pupils to reach the place of excursion. Consider that the embarkation and disembarkation of passengers, as well as the reversal of the bus, take place immediately and this time can be neglected.
Input
The first line of the input contains five positive integers n, l, v1, v2 and k (1 β€ n β€ 10 000, 1 β€ l β€ 109, 1 β€ v1 < v2 β€ 109, 1 β€ k β€ n) β the number of pupils, the distance from meeting to the place of excursion, the speed of each pupil, the speed of bus and the number of seats in the bus.
Output
Print the real number β the minimum time in which all pupils can reach the place of excursion. Your answer will be considered correct if its absolute or relative error won't exceed 10 - 6.
Examples
Input
5 10 1 2 5
Output
5.0000000000
Input
3 6 1 2 1
Output
4.7142857143
Note
In the first sample we should immediately put all five pupils to the bus. The speed of the bus equals 2 and the distance is equal to 10, so the pupils will reach the place of excursion in time 10 / 2 = 5.
Tags: binary search, math
Correct Solution:
```
#!/usr/local/bin/python3
# -*- coding:utf-8 -*-
import math
inputParams = input().split()
n = int(inputParams[0])
l = int(inputParams[1])
v1 = int(inputParams[2])
v2 = int(inputParams[3])
k = int(inputParams[4])
# θΏι欑ζ°
times = math.ceil(n / k)
t1 = l / (v2 + (2 * v2 / (v2 + v1)) * v1 * (times - 1))
totalTime = t1 + (2 * v2 / (v1 + v2)) * t1 * (times - 1)
print(str(totalTime))
```
| 9,532 |
Provide tags and a correct Python 3 solution for this coding contest problem.
On vacations n pupils decided to go on excursion and gather all together. They need to overcome the path with the length l meters. Each of the pupils will go with the speed equal to v1. To get to the excursion quickly, it was decided to rent a bus, which has seats for k people (it means that it can't fit more than k people at the same time) and the speed equal to v2. In order to avoid seasick, each of the pupils want to get into the bus no more than once.
Determine the minimum time required for all n pupils to reach the place of excursion. Consider that the embarkation and disembarkation of passengers, as well as the reversal of the bus, take place immediately and this time can be neglected.
Input
The first line of the input contains five positive integers n, l, v1, v2 and k (1 β€ n β€ 10 000, 1 β€ l β€ 109, 1 β€ v1 < v2 β€ 109, 1 β€ k β€ n) β the number of pupils, the distance from meeting to the place of excursion, the speed of each pupil, the speed of bus and the number of seats in the bus.
Output
Print the real number β the minimum time in which all pupils can reach the place of excursion. Your answer will be considered correct if its absolute or relative error won't exceed 10 - 6.
Examples
Input
5 10 1 2 5
Output
5.0000000000
Input
3 6 1 2 1
Output
4.7142857143
Note
In the first sample we should immediately put all five pupils to the bus. The speed of the bus equals 2 and the distance is equal to 10, so the pupils will reach the place of excursion in time 10 / 2 = 5.
Tags: binary search, math
Correct Solution:
```
n, l, v1, v2, k = [int(x) for x in input().split()]
x = (n+k-1)//k
print((l - ((l/v1)/((x/(v2-v1)) + ((x-1)/(v2+v1)) + (1/v1))))/v1)
```
| 9,533 |
Provide tags and a correct Python 3 solution for this coding contest problem.
On vacations n pupils decided to go on excursion and gather all together. They need to overcome the path with the length l meters. Each of the pupils will go with the speed equal to v1. To get to the excursion quickly, it was decided to rent a bus, which has seats for k people (it means that it can't fit more than k people at the same time) and the speed equal to v2. In order to avoid seasick, each of the pupils want to get into the bus no more than once.
Determine the minimum time required for all n pupils to reach the place of excursion. Consider that the embarkation and disembarkation of passengers, as well as the reversal of the bus, take place immediately and this time can be neglected.
Input
The first line of the input contains five positive integers n, l, v1, v2 and k (1 β€ n β€ 10 000, 1 β€ l β€ 109, 1 β€ v1 < v2 β€ 109, 1 β€ k β€ n) β the number of pupils, the distance from meeting to the place of excursion, the speed of each pupil, the speed of bus and the number of seats in the bus.
Output
Print the real number β the minimum time in which all pupils can reach the place of excursion. Your answer will be considered correct if its absolute or relative error won't exceed 10 - 6.
Examples
Input
5 10 1 2 5
Output
5.0000000000
Input
3 6 1 2 1
Output
4.7142857143
Note
In the first sample we should immediately put all five pupils to the bus. The speed of the bus equals 2 and the distance is equal to 10, so the pupils will reach the place of excursion in time 10 / 2 = 5.
Tags: binary search, math
Correct Solution:
```
n, l, v1, v2, k = list(map(int, input().split()))
n = (n+k-1)//k
t = l / ((v1 + (v2 - v1) * v1 / (v1 + v2)) * (n - 1) + v2)
print(t * n + t * (v2 - v1) / (v1 + v2) * (n - 1))
```
| 9,534 |
Provide tags and a correct Python 3 solution for this coding contest problem.
On vacations n pupils decided to go on excursion and gather all together. They need to overcome the path with the length l meters. Each of the pupils will go with the speed equal to v1. To get to the excursion quickly, it was decided to rent a bus, which has seats for k people (it means that it can't fit more than k people at the same time) and the speed equal to v2. In order to avoid seasick, each of the pupils want to get into the bus no more than once.
Determine the minimum time required for all n pupils to reach the place of excursion. Consider that the embarkation and disembarkation of passengers, as well as the reversal of the bus, take place immediately and this time can be neglected.
Input
The first line of the input contains five positive integers n, l, v1, v2 and k (1 β€ n β€ 10 000, 1 β€ l β€ 109, 1 β€ v1 < v2 β€ 109, 1 β€ k β€ n) β the number of pupils, the distance from meeting to the place of excursion, the speed of each pupil, the speed of bus and the number of seats in the bus.
Output
Print the real number β the minimum time in which all pupils can reach the place of excursion. Your answer will be considered correct if its absolute or relative error won't exceed 10 - 6.
Examples
Input
5 10 1 2 5
Output
5.0000000000
Input
3 6 1 2 1
Output
4.7142857143
Note
In the first sample we should immediately put all five pupils to the bus. The speed of the bus equals 2 and the distance is equal to 10, so the pupils will reach the place of excursion in time 10 / 2 = 5.
Tags: binary search, math
Correct Solution:
```
n, L, v1, v2, k = map(int, input().split())
n = (n + k - 1) // k * 2
dif = v2 - v1
p1 = (n * v2 - dif) * L
p2 = (n * v1 + dif) * v2
ans = p1 / p2
print('%.7f' % ans)
```
| 9,535 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
On vacations n pupils decided to go on excursion and gather all together. They need to overcome the path with the length l meters. Each of the pupils will go with the speed equal to v1. To get to the excursion quickly, it was decided to rent a bus, which has seats for k people (it means that it can't fit more than k people at the same time) and the speed equal to v2. In order to avoid seasick, each of the pupils want to get into the bus no more than once.
Determine the minimum time required for all n pupils to reach the place of excursion. Consider that the embarkation and disembarkation of passengers, as well as the reversal of the bus, take place immediately and this time can be neglected.
Input
The first line of the input contains five positive integers n, l, v1, v2 and k (1 β€ n β€ 10 000, 1 β€ l β€ 109, 1 β€ v1 < v2 β€ 109, 1 β€ k β€ n) β the number of pupils, the distance from meeting to the place of excursion, the speed of each pupil, the speed of bus and the number of seats in the bus.
Output
Print the real number β the minimum time in which all pupils can reach the place of excursion. Your answer will be considered correct if its absolute or relative error won't exceed 10 - 6.
Examples
Input
5 10 1 2 5
Output
5.0000000000
Input
3 6 1 2 1
Output
4.7142857143
Note
In the first sample we should immediately put all five pupils to the bus. The speed of the bus equals 2 and the distance is equal to 10, so the pupils will reach the place of excursion in time 10 / 2 = 5.
Submitted Solution:
```
n,l,v1,v2,k=map(int,input().split())
m=(n-1)//k+1
v=v1+v2
x=l*v*v2/(m*v*v2-(m-1)*(v2-v1)*v2)
print(x/v2+(l-x)/v1)
```
Yes
| 9,536 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
On vacations n pupils decided to go on excursion and gather all together. They need to overcome the path with the length l meters. Each of the pupils will go with the speed equal to v1. To get to the excursion quickly, it was decided to rent a bus, which has seats for k people (it means that it can't fit more than k people at the same time) and the speed equal to v2. In order to avoid seasick, each of the pupils want to get into the bus no more than once.
Determine the minimum time required for all n pupils to reach the place of excursion. Consider that the embarkation and disembarkation of passengers, as well as the reversal of the bus, take place immediately and this time can be neglected.
Input
The first line of the input contains five positive integers n, l, v1, v2 and k (1 β€ n β€ 10 000, 1 β€ l β€ 109, 1 β€ v1 < v2 β€ 109, 1 β€ k β€ n) β the number of pupils, the distance from meeting to the place of excursion, the speed of each pupil, the speed of bus and the number of seats in the bus.
Output
Print the real number β the minimum time in which all pupils can reach the place of excursion. Your answer will be considered correct if its absolute or relative error won't exceed 10 - 6.
Examples
Input
5 10 1 2 5
Output
5.0000000000
Input
3 6 1 2 1
Output
4.7142857143
Note
In the first sample we should immediately put all five pupils to the bus. The speed of the bus equals 2 and the distance is equal to 10, so the pupils will reach the place of excursion in time 10 / 2 = 5.
Submitted Solution:
```
s=input();
li=s.split();
n=int(li[0])
l=int(li[1])
v1=int(li[2])
v2=int(li[3])
k=int(li[4])
t=n//k
if n%k!=0:
t+=1
a=(v1+v2)*l/(v1+v2+2*(t-1)*v1)
ans=a/v2+(l-a)/v1
print("{0:.10f}".format(ans))
```
Yes
| 9,537 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
On vacations n pupils decided to go on excursion and gather all together. They need to overcome the path with the length l meters. Each of the pupils will go with the speed equal to v1. To get to the excursion quickly, it was decided to rent a bus, which has seats for k people (it means that it can't fit more than k people at the same time) and the speed equal to v2. In order to avoid seasick, each of the pupils want to get into the bus no more than once.
Determine the minimum time required for all n pupils to reach the place of excursion. Consider that the embarkation and disembarkation of passengers, as well as the reversal of the bus, take place immediately and this time can be neglected.
Input
The first line of the input contains five positive integers n, l, v1, v2 and k (1 β€ n β€ 10 000, 1 β€ l β€ 109, 1 β€ v1 < v2 β€ 109, 1 β€ k β€ n) β the number of pupils, the distance from meeting to the place of excursion, the speed of each pupil, the speed of bus and the number of seats in the bus.
Output
Print the real number β the minimum time in which all pupils can reach the place of excursion. Your answer will be considered correct if its absolute or relative error won't exceed 10 - 6.
Examples
Input
5 10 1 2 5
Output
5.0000000000
Input
3 6 1 2 1
Output
4.7142857143
Note
In the first sample we should immediately put all five pupils to the bus. The speed of the bus equals 2 and the distance is equal to 10, so the pupils will reach the place of excursion in time 10 / 2 = 5.
Submitted Solution:
```
#!/usr/local/bin/python3
# -*- coding:utf-8 -*-
import math
# inputParams = input().split()
# n = int(inputParams[0])
# l = int(inputParams[1])
# v1 = int(inputParams[2])
# v2 = int(inputParams[3])
# k = int(inputParams[4])
n,l,v1,v2,k = map(int,input().split())
# θΏι欑ζ°
times = math.ceil(n / k)
t1 = l / (v2 + (2 * v2 / (v2 + v1)) * v1 * (times - 1))
totalTime = t1 + (2 * v2 / (v1 + v2)) * t1 * (times - 1)
print(str(totalTime))
```
Yes
| 9,538 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
On vacations n pupils decided to go on excursion and gather all together. They need to overcome the path with the length l meters. Each of the pupils will go with the speed equal to v1. To get to the excursion quickly, it was decided to rent a bus, which has seats for k people (it means that it can't fit more than k people at the same time) and the speed equal to v2. In order to avoid seasick, each of the pupils want to get into the bus no more than once.
Determine the minimum time required for all n pupils to reach the place of excursion. Consider that the embarkation and disembarkation of passengers, as well as the reversal of the bus, take place immediately and this time can be neglected.
Input
The first line of the input contains five positive integers n, l, v1, v2 and k (1 β€ n β€ 10 000, 1 β€ l β€ 109, 1 β€ v1 < v2 β€ 109, 1 β€ k β€ n) β the number of pupils, the distance from meeting to the place of excursion, the speed of each pupil, the speed of bus and the number of seats in the bus.
Output
Print the real number β the minimum time in which all pupils can reach the place of excursion. Your answer will be considered correct if its absolute or relative error won't exceed 10 - 6.
Examples
Input
5 10 1 2 5
Output
5.0000000000
Input
3 6 1 2 1
Output
4.7142857143
Note
In the first sample we should immediately put all five pupils to the bus. The speed of the bus equals 2 and the distance is equal to 10, so the pupils will reach the place of excursion in time 10 / 2 = 5.
Submitted Solution:
```
n, l, a, b, k = map(int, input().split())
m = n // k
if n % k != 0:
m += 1
A = (b + a) + 2 * b * (m - 1)
B = (b + a) + 2 * a * (m - 1)
print((l * A) / (b * B))
```
Yes
| 9,539 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
On vacations n pupils decided to go on excursion and gather all together. They need to overcome the path with the length l meters. Each of the pupils will go with the speed equal to v1. To get to the excursion quickly, it was decided to rent a bus, which has seats for k people (it means that it can't fit more than k people at the same time) and the speed equal to v2. In order to avoid seasick, each of the pupils want to get into the bus no more than once.
Determine the minimum time required for all n pupils to reach the place of excursion. Consider that the embarkation and disembarkation of passengers, as well as the reversal of the bus, take place immediately and this time can be neglected.
Input
The first line of the input contains five positive integers n, l, v1, v2 and k (1 β€ n β€ 10 000, 1 β€ l β€ 109, 1 β€ v1 < v2 β€ 109, 1 β€ k β€ n) β the number of pupils, the distance from meeting to the place of excursion, the speed of each pupil, the speed of bus and the number of seats in the bus.
Output
Print the real number β the minimum time in which all pupils can reach the place of excursion. Your answer will be considered correct if its absolute or relative error won't exceed 10 - 6.
Examples
Input
5 10 1 2 5
Output
5.0000000000
Input
3 6 1 2 1
Output
4.7142857143
Note
In the first sample we should immediately put all five pupils to the bus. The speed of the bus equals 2 and the distance is equal to 10, so the pupils will reach the place of excursion in time 10 / 2 = 5.
Submitted Solution:
```
# -*- coding: utf-8 -*-
"""
Created on Tue Jul 26 15:20:38 2016
@author: pinku
"""
import sys
bkpstdin = sys.stdin
#sys.stdin = open('in.txt','r')
n,l,v1,v2,k = map(int,input().split(' '))
o = n/k
if n%k == 1:
o+=1
d = 1.0*o
ans = l*( ( (d*2.0) - 1.0)*v2 + v1)
ans /= (v2 + ( (d*2.0) - 1.0)*v1)
ans /= v2
print(ans)
sys.stdin = bkpstdin
```
No
| 9,540 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
On vacations n pupils decided to go on excursion and gather all together. They need to overcome the path with the length l meters. Each of the pupils will go with the speed equal to v1. To get to the excursion quickly, it was decided to rent a bus, which has seats for k people (it means that it can't fit more than k people at the same time) and the speed equal to v2. In order to avoid seasick, each of the pupils want to get into the bus no more than once.
Determine the minimum time required for all n pupils to reach the place of excursion. Consider that the embarkation and disembarkation of passengers, as well as the reversal of the bus, take place immediately and this time can be neglected.
Input
The first line of the input contains five positive integers n, l, v1, v2 and k (1 β€ n β€ 10 000, 1 β€ l β€ 109, 1 β€ v1 < v2 β€ 109, 1 β€ k β€ n) β the number of pupils, the distance from meeting to the place of excursion, the speed of each pupil, the speed of bus and the number of seats in the bus.
Output
Print the real number β the minimum time in which all pupils can reach the place of excursion. Your answer will be considered correct if its absolute or relative error won't exceed 10 - 6.
Examples
Input
5 10 1 2 5
Output
5.0000000000
Input
3 6 1 2 1
Output
4.7142857143
Note
In the first sample we should immediately put all five pupils to the bus. The speed of the bus equals 2 and the distance is equal to 10, so the pupils will reach the place of excursion in time 10 / 2 = 5.
Submitted Solution:
```
n, l, v1, v2, k = list(map(int, input().split()))
p = [0]*(l+1)
p[1] = n
i = l
t = (n+k-1)/k
low = 0.0
hi = 1e9+10
for i in range(100):
t1 = (hi+low)*0.5
len = t1 * t * v2 - (t-1) * v2 * t1 * (v2 - v1) / (v1 + v2)
if len>l:
hi = t1
else:
low = t1
print(t1+(l-t1*v2)/v1)
```
No
| 9,541 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
On vacations n pupils decided to go on excursion and gather all together. They need to overcome the path with the length l meters. Each of the pupils will go with the speed equal to v1. To get to the excursion quickly, it was decided to rent a bus, which has seats for k people (it means that it can't fit more than k people at the same time) and the speed equal to v2. In order to avoid seasick, each of the pupils want to get into the bus no more than once.
Determine the minimum time required for all n pupils to reach the place of excursion. Consider that the embarkation and disembarkation of passengers, as well as the reversal of the bus, take place immediately and this time can be neglected.
Input
The first line of the input contains five positive integers n, l, v1, v2 and k (1 β€ n β€ 10 000, 1 β€ l β€ 109, 1 β€ v1 < v2 β€ 109, 1 β€ k β€ n) β the number of pupils, the distance from meeting to the place of excursion, the speed of each pupil, the speed of bus and the number of seats in the bus.
Output
Print the real number β the minimum time in which all pupils can reach the place of excursion. Your answer will be considered correct if its absolute or relative error won't exceed 10 - 6.
Examples
Input
5 10 1 2 5
Output
5.0000000000
Input
3 6 1 2 1
Output
4.7142857143
Note
In the first sample we should immediately put all five pupils to the bus. The speed of the bus equals 2 and the distance is equal to 10, so the pupils will reach the place of excursion in time 10 / 2 = 5.
Submitted Solution:
```
n,l,v1,v2,k=map(int,input().split())
m=(n-1)//k+1
v=v1+v2
x=l*v/(m*v-(m-1)*v1)
print(x/v2+(l-x)/v1)
```
No
| 9,542 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
On vacations n pupils decided to go on excursion and gather all together. They need to overcome the path with the length l meters. Each of the pupils will go with the speed equal to v1. To get to the excursion quickly, it was decided to rent a bus, which has seats for k people (it means that it can't fit more than k people at the same time) and the speed equal to v2. In order to avoid seasick, each of the pupils want to get into the bus no more than once.
Determine the minimum time required for all n pupils to reach the place of excursion. Consider that the embarkation and disembarkation of passengers, as well as the reversal of the bus, take place immediately and this time can be neglected.
Input
The first line of the input contains five positive integers n, l, v1, v2 and k (1 β€ n β€ 10 000, 1 β€ l β€ 109, 1 β€ v1 < v2 β€ 109, 1 β€ k β€ n) β the number of pupils, the distance from meeting to the place of excursion, the speed of each pupil, the speed of bus and the number of seats in the bus.
Output
Print the real number β the minimum time in which all pupils can reach the place of excursion. Your answer will be considered correct if its absolute or relative error won't exceed 10 - 6.
Examples
Input
5 10 1 2 5
Output
5.0000000000
Input
3 6 1 2 1
Output
4.7142857143
Note
In the first sample we should immediately put all five pupils to the bus. The speed of the bus equals 2 and the distance is equal to 10, so the pupils will reach the place of excursion in time 10 / 2 = 5.
Submitted Solution:
```
def binSearch(n, L, v1, v2, k):
l = 0
r = L
while r-l > 0.000001:
d = (l+r)/2
td = d/v1
S = v2*td
g = d-S
tb = g/(v1+v2)
b = v1*tb
if n%k == 0:
a = n//k
else:
a = n//k + 1
if a*d-b*(a-1) > L:
r = d
else:
l = d
td = l / v1
S = v2 * td
g = l - S
tb = g / (v1 + v2)
b = v1 * tb
#return (a*td+(a-1)*tb)
return (a * d + (a - 1) * b) / v1
n1, L1, v11, v21, k1 = map(int, input().split())
print('{0:.10f}'.format(binSearch(n1, L1, v21, v11, k1)))
```
No
| 9,543 |
Provide tags and a correct Python 3 solution for this coding contest problem.
There are n cities and m two-way roads in Berland, each road connects two cities. It is known that there is no more than one road connecting each pair of cities, and there is no road which connects the city with itself. It is possible that there is no way to get from one city to some other city using only these roads.
The road minister decided to make a reform in Berland and to orient all roads in the country, i.e. to make each road one-way. The minister wants to maximize the number of cities, for which the number of roads that begins in the city equals to the number of roads that ends in it.
Input
The first line contains a positive integer t (1 β€ t β€ 200) β the number of testsets in the input.
Each of the testsets is given in the following way. The first line contains two integers n and m (1 β€ n β€ 200, 0 β€ m β€ nΒ·(n - 1) / 2) β the number of cities and the number of roads in Berland.
The next m lines contain the description of roads in Berland. Each line contains two integers u and v (1 β€ u, v β€ n) β the cities the corresponding road connects. It's guaranteed that there are no self-loops and multiple roads. It is possible that there is no way along roads between a pair of cities.
It is guaranteed that the total number of cities in all testset of input data doesn't exceed 200.
Pay attention that for hacks, you can only use tests consisting of one testset, so t should be equal to one.
Output
For each testset print the maximum number of such cities that the number of roads that begins in the city, is equal to the number of roads that ends in it.
In the next m lines print oriented roads. First print the number of the city where the road begins and then the number of the city where the road ends. If there are several answers, print any of them. It is allowed to print roads in each test in arbitrary order. Each road should be printed exactly once.
Example
Input
2
5 5
2 1
4 5
2 3
1 3
3 5
7 2
3 7
4 2
Output
3
1 3
3 5
5 4
3 2
2 1
3
2 4
3 7
Tags: constructive algorithms, dfs and similar, flows, graphs, greedy
Correct Solution:
```
def hierholzer(graph):
edges = []
for node in graph:
for neighbour in graph[node]:
if graph[node][neighbour] is False: continue
stack = [node]
parents = defaultdict(list)
parents[node] = []
while (len(stack) != 0):
current = stack.pop()
seen = False
for neighbour in graph[current]:
if graph[current][neighbour] is False: continue
stack.append(current)
stack.append(neighbour)
graph[current][neighbour] = False
graph[neighbour][current] = False
parents[neighbour].append(current)
seen = True
break
if not seen and len(parents[current])!=0:
edges.append((parents[current].pop(), current))
return edges
import sys
sys.setrecursionlimit(5000)
t = int(input())
from collections import defaultdict
for test_number in range(t):
n, m = map(int, input().split())
graph = defaultdict(dict)
for _ in range(m):
u, v = map(int, input().split())
graph[u][v] = True
graph[v][u] = True
if m == 0:
print(n)
continue
odds = []
for u in graph:
if len(graph[u]) % 2 == 1:
odds.append(u)
# now every node has an even degree
for odd in odds:
graph[odd][n + 1] = True
graph[n + 1][odd] = True
# we just do a dfs
edges = hierholzer(graph)
print(n - len(odds))
for edge in edges:
if edge[0] != n + 1 and edge[1] != n + 1:
print("{0} {1}".format(edge[0], edge[1]))
```
| 9,544 |
Provide tags and a correct Python 3 solution for this coding contest problem.
There are n cities and m two-way roads in Berland, each road connects two cities. It is known that there is no more than one road connecting each pair of cities, and there is no road which connects the city with itself. It is possible that there is no way to get from one city to some other city using only these roads.
The road minister decided to make a reform in Berland and to orient all roads in the country, i.e. to make each road one-way. The minister wants to maximize the number of cities, for which the number of roads that begins in the city equals to the number of roads that ends in it.
Input
The first line contains a positive integer t (1 β€ t β€ 200) β the number of testsets in the input.
Each of the testsets is given in the following way. The first line contains two integers n and m (1 β€ n β€ 200, 0 β€ m β€ nΒ·(n - 1) / 2) β the number of cities and the number of roads in Berland.
The next m lines contain the description of roads in Berland. Each line contains two integers u and v (1 β€ u, v β€ n) β the cities the corresponding road connects. It's guaranteed that there are no self-loops and multiple roads. It is possible that there is no way along roads between a pair of cities.
It is guaranteed that the total number of cities in all testset of input data doesn't exceed 200.
Pay attention that for hacks, you can only use tests consisting of one testset, so t should be equal to one.
Output
For each testset print the maximum number of such cities that the number of roads that begins in the city, is equal to the number of roads that ends in it.
In the next m lines print oriented roads. First print the number of the city where the road begins and then the number of the city where the road ends. If there are several answers, print any of them. It is allowed to print roads in each test in arbitrary order. Each road should be printed exactly once.
Example
Input
2
5 5
2 1
4 5
2 3
1 3
3 5
7 2
3 7
4 2
Output
3
1 3
3 5
5 4
3 2
2 1
3
2 4
3 7
Tags: constructive algorithms, dfs and similar, flows, graphs, greedy
Correct Solution:
```
t = int(input())
from collections import defaultdict
for test_number in range(t):
n, m = map(int, input().split())
graph = defaultdict(dict)
for _ in range(m):
u, v = map(int, input().split())
graph[u][v] = True
graph[v][u] = True
if m == 0:
print(n)
continue
odds = []
for u in graph:
if len(graph[u]) % 2 == 1:
odds.append(u)
# now every node has an even degree
for odd in odds:
graph[odd][n + 1] = True
graph[n + 1][odd] = True
# we just do a dfs
edges = []
# we don't have a visited dictionary as usual, because we must allow to visit a node several times
# but we do delete the edges
for node in graph:
for neighbour in graph[node]:
if graph[node][neighbour] is False: continue
stack = [node]
while (len(stack) != 0):
current = stack.pop()
for neighbour in graph[current]:
if graph[current][neighbour] is False: continue
stack.append(current)
stack.append(neighbour)
graph[current][neighbour] = False
graph[neighbour][current] = False
edges.append((current, neighbour))
break
print(n - len(odds))
for edge in edges:
if edge[0] != n + 1 and edge[1] != n + 1:
print("{0} {1}".format(edge[0], edge[1]))
```
| 9,545 |
Provide tags and a correct Python 3 solution for this coding contest problem.
There are n cities and m two-way roads in Berland, each road connects two cities. It is known that there is no more than one road connecting each pair of cities, and there is no road which connects the city with itself. It is possible that there is no way to get from one city to some other city using only these roads.
The road minister decided to make a reform in Berland and to orient all roads in the country, i.e. to make each road one-way. The minister wants to maximize the number of cities, for which the number of roads that begins in the city equals to the number of roads that ends in it.
Input
The first line contains a positive integer t (1 β€ t β€ 200) β the number of testsets in the input.
Each of the testsets is given in the following way. The first line contains two integers n and m (1 β€ n β€ 200, 0 β€ m β€ nΒ·(n - 1) / 2) β the number of cities and the number of roads in Berland.
The next m lines contain the description of roads in Berland. Each line contains two integers u and v (1 β€ u, v β€ n) β the cities the corresponding road connects. It's guaranteed that there are no self-loops and multiple roads. It is possible that there is no way along roads between a pair of cities.
It is guaranteed that the total number of cities in all testset of input data doesn't exceed 200.
Pay attention that for hacks, you can only use tests consisting of one testset, so t should be equal to one.
Output
For each testset print the maximum number of such cities that the number of roads that begins in the city, is equal to the number of roads that ends in it.
In the next m lines print oriented roads. First print the number of the city where the road begins and then the number of the city where the road ends. If there are several answers, print any of them. It is allowed to print roads in each test in arbitrary order. Each road should be printed exactly once.
Example
Input
2
5 5
2 1
4 5
2 3
1 3
3 5
7 2
3 7
4 2
Output
3
1 3
3 5
5 4
3 2
2 1
3
2 4
3 7
Tags: constructive algorithms, dfs and similar, flows, graphs, greedy
Correct Solution:
```
import sys
from collections import defaultdict
rlines = sys.stdin.readlines()
lines = (l.strip() for l in rlines)
def eucycle(n,m,adj):
dir_edges = []
us = list(adj.keys())
for u in us:
while adj[u]:
v0 = u
v1 = adj[v0].pop()
adj[v1].remove(v0)
dir_edges.append((v0, v1))
while v1 != u:
v0 = v1
v1 = adj[v0].pop()
adj[v1].remove(v0)
dir_edges.append((v0, v1))
return dir_edges
def solve(n,m,edges):
adj = defaultdict(set)
for u, v in edges:
adj[u].add(v)
adj[v].add(u)
odds = set(u for u in adj if len(adj[u])%2==1)
for odd in odds:
adj[odd].add(n+1)
adj[n+1].add(odd)
total = n - len(odds)
dir_edges = eucycle(n+1, m, adj)
return total, dir_edges
t = int(next(lines))
for ti in range(t):
n,m = [int(s) for s in next(lines).split()]
edges = []
for ei in range(m):
u,v = [int(s) for s in next(lines).split()]
edges.append((u,v))
total, ans = solve(n,m,edges)
print(total)
print('\n'.join(str(u)+ ' '+ str(v) for u, v in ans if u != n+1 and v!= n+1))
```
| 9,546 |
Provide tags and a correct Python 3 solution for this coding contest problem.
There are n cities and m two-way roads in Berland, each road connects two cities. It is known that there is no more than one road connecting each pair of cities, and there is no road which connects the city with itself. It is possible that there is no way to get from one city to some other city using only these roads.
The road minister decided to make a reform in Berland and to orient all roads in the country, i.e. to make each road one-way. The minister wants to maximize the number of cities, for which the number of roads that begins in the city equals to the number of roads that ends in it.
Input
The first line contains a positive integer t (1 β€ t β€ 200) β the number of testsets in the input.
Each of the testsets is given in the following way. The first line contains two integers n and m (1 β€ n β€ 200, 0 β€ m β€ nΒ·(n - 1) / 2) β the number of cities and the number of roads in Berland.
The next m lines contain the description of roads in Berland. Each line contains two integers u and v (1 β€ u, v β€ n) β the cities the corresponding road connects. It's guaranteed that there are no self-loops and multiple roads. It is possible that there is no way along roads between a pair of cities.
It is guaranteed that the total number of cities in all testset of input data doesn't exceed 200.
Pay attention that for hacks, you can only use tests consisting of one testset, so t should be equal to one.
Output
For each testset print the maximum number of such cities that the number of roads that begins in the city, is equal to the number of roads that ends in it.
In the next m lines print oriented roads. First print the number of the city where the road begins and then the number of the city where the road ends. If there are several answers, print any of them. It is allowed to print roads in each test in arbitrary order. Each road should be printed exactly once.
Example
Input
2
5 5
2 1
4 5
2 3
1 3
3 5
7 2
3 7
4 2
Output
3
1 3
3 5
5 4
3 2
2 1
3
2 4
3 7
Tags: constructive algorithms, dfs and similar, flows, graphs, greedy
Correct Solution:
```
import sys
import collections
rlines = sys.stdin.readlines()
lines = (l.strip() for l in rlines)
def eucycle(n,m,adj):
diredges = []
us = list(adj.keys())
for u in us:
while adj[u]:
v0 = u
v1 = adj[v0].pop()
adj[v1].remove(v0)
diredges.append((v0,v1))
while v1 != u:
v0 = v1
v1 = adj[v0].pop()
adj[v1].remove(v0)
diredges.append((v0,v1))
return diredges
def solve(n,m,edges):
adj = collections.defaultdict(set)
diredges = []
for u,v in edges:
adj[u].add(v)
adj[v].add(u)
odds = set(u for u in adj if len(adj[u])%2 == 1)
ans = n - len(odds)
assert(len(odds)%2 == 0)
for o in odds:
adj[n+1].add(o)
adj[o].add(n+1)
diredges = eucycle(n+1,m,adj)
return str(ans) + '\n' + '\n'.join(str(u) + ' ' + str(v) for (u,v) in diredges\
if u != n+1 and v != n+1)
t = int(next(lines))
for ti in range(t):
n,m = [int(s) for s in next(lines).split()]
edges = []
for ei in range(m):
u,v = [int(s) for s in next(lines).split()]
edges.append((u,v))
#print(edges)
print(solve(n,m,edges))
```
| 9,547 |
Provide tags and a correct Python 3 solution for this coding contest problem.
There are n cities and m two-way roads in Berland, each road connects two cities. It is known that there is no more than one road connecting each pair of cities, and there is no road which connects the city with itself. It is possible that there is no way to get from one city to some other city using only these roads.
The road minister decided to make a reform in Berland and to orient all roads in the country, i.e. to make each road one-way. The minister wants to maximize the number of cities, for which the number of roads that begins in the city equals to the number of roads that ends in it.
Input
The first line contains a positive integer t (1 β€ t β€ 200) β the number of testsets in the input.
Each of the testsets is given in the following way. The first line contains two integers n and m (1 β€ n β€ 200, 0 β€ m β€ nΒ·(n - 1) / 2) β the number of cities and the number of roads in Berland.
The next m lines contain the description of roads in Berland. Each line contains two integers u and v (1 β€ u, v β€ n) β the cities the corresponding road connects. It's guaranteed that there are no self-loops and multiple roads. It is possible that there is no way along roads between a pair of cities.
It is guaranteed that the total number of cities in all testset of input data doesn't exceed 200.
Pay attention that for hacks, you can only use tests consisting of one testset, so t should be equal to one.
Output
For each testset print the maximum number of such cities that the number of roads that begins in the city, is equal to the number of roads that ends in it.
In the next m lines print oriented roads. First print the number of the city where the road begins and then the number of the city where the road ends. If there are several answers, print any of them. It is allowed to print roads in each test in arbitrary order. Each road should be printed exactly once.
Example
Input
2
5 5
2 1
4 5
2 3
1 3
3 5
7 2
3 7
4 2
Output
3
1 3
3 5
5 4
3 2
2 1
3
2 4
3 7
Tags: constructive algorithms, dfs and similar, flows, graphs, greedy
Correct Solution:
```
from collections import defaultdict, Counter
T = int(input())
for _ in range(T):
global lst
N, M = map(int, input().split())
visit = set()
oddNodes = []
directed = []
G = defaultdict(list)
oriG = defaultdict(list)
Gra = [[0] * (N+1) for _ in range(N+1)]
deg = [0] * (N+1)
for k in range(M):
u, v = map(int, input().split())
Gra[u][v] += 1
Gra[v][u] += 1
deg[u] += 1
deg[v] += 1
# G[u].append(v)
# G[v].append(u)
# oriG[u].append(v)
# oriG[v].append(u)
ans = 0
for i in range(1, N+1):
if deg[i] % 2 == 0:
ans += 1
else:
oddNodes.append(i)
for i in range(0, len(oddNodes), 2):
# G[oddNodes[i]].append(oddNodes[i+1])
# G[oddNodes[i+1]].append(oddNodes[i])
Gra[oddNodes[i]][oddNodes[i+1]] += 1
Gra[oddNodes[i+1]][oddNodes[i]] += 1
deg[oddNodes[i]] += 1
deg[oddNodes[i+1]] += 1
def eulerPath(u):
stk = [u]
while stk:
u = stk.pop()
for i in range(1, N+1):
if Gra[u][i]:
Gra[u][i] -= 1
Gra[i][u] -= 1
directed.append((u, i))
stk.append(i)
break
for i in range(1, N+1):
eulerPath(i)
for i in range(0, len(oddNodes), 2):
# G[oddNodes[i]].append(oddNodes[i+1])
# G[oddNodes[i+1]].append(oddNodes[i])
Gra[oddNodes[i]][oddNodes[i+1]] += 1
Gra[oddNodes[i+1]][oddNodes[i]] += 1
# for i in oddNodes:
print(ans)
for u, v in directed:
if Gra[u][v] != 0:
Gra[u][v] -= 1
Gra[v][u] -= 1
else:
print(str(u) + " " + str(v))
```
| 9,548 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There are n cities and m two-way roads in Berland, each road connects two cities. It is known that there is no more than one road connecting each pair of cities, and there is no road which connects the city with itself. It is possible that there is no way to get from one city to some other city using only these roads.
The road minister decided to make a reform in Berland and to orient all roads in the country, i.e. to make each road one-way. The minister wants to maximize the number of cities, for which the number of roads that begins in the city equals to the number of roads that ends in it.
Input
The first line contains a positive integer t (1 β€ t β€ 200) β the number of testsets in the input.
Each of the testsets is given in the following way. The first line contains two integers n and m (1 β€ n β€ 200, 0 β€ m β€ nΒ·(n - 1) / 2) β the number of cities and the number of roads in Berland.
The next m lines contain the description of roads in Berland. Each line contains two integers u and v (1 β€ u, v β€ n) β the cities the corresponding road connects. It's guaranteed that there are no self-loops and multiple roads. It is possible that there is no way along roads between a pair of cities.
It is guaranteed that the total number of cities in all testset of input data doesn't exceed 200.
Pay attention that for hacks, you can only use tests consisting of one testset, so t should be equal to one.
Output
For each testset print the maximum number of such cities that the number of roads that begins in the city, is equal to the number of roads that ends in it.
In the next m lines print oriented roads. First print the number of the city where the road begins and then the number of the city where the road ends. If there are several answers, print any of them. It is allowed to print roads in each test in arbitrary order. Each road should be printed exactly once.
Example
Input
2
5 5
2 1
4 5
2 3
1 3
3 5
7 2
3 7
4 2
Output
3
1 3
3 5
5 4
3 2
2 1
3
2 4
3 7
Submitted Solution:
```
def walk(v,reverse):
for u in road[v]:
if not used[v][u]:
## print(v,u,reverse)
used[v][u] = True
used[u][v] = True
if not reverse:
out[v] += 1
into[u] += 1
res.append((v+1,u+1))
else:
into[v] += 1
out[u] += 1
res.append((u+1,v+1))
walk(u,reverse)
reverse = not reverse
return
t = int(input())
for test in range(t):
n, m = [int(i) for i in input().split()]
road = [[] for i in range(n)]
for i in range(m):
a, b = [int(j)-1 for j in input().split()]
road[a].append(b)
road[b].append(a)
used = [[False for i in range(n)] for j in range(n)]
into = [0] * n
out = [0] * n
res = []
for v in range(n):
if len(road[v]) % 2 == 1:
walk(v,False)
for v in range(n):
walk(v,False)
cnt = 0
for i in range(n):
if into[i] == out[i]:
cnt += 1
print(cnt)
for i in range(len(res)):
print(res[i][0],res[i][1])
```
No
| 9,549 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There are n cities and m two-way roads in Berland, each road connects two cities. It is known that there is no more than one road connecting each pair of cities, and there is no road which connects the city with itself. It is possible that there is no way to get from one city to some other city using only these roads.
The road minister decided to make a reform in Berland and to orient all roads in the country, i.e. to make each road one-way. The minister wants to maximize the number of cities, for which the number of roads that begins in the city equals to the number of roads that ends in it.
Input
The first line contains a positive integer t (1 β€ t β€ 200) β the number of testsets in the input.
Each of the testsets is given in the following way. The first line contains two integers n and m (1 β€ n β€ 200, 0 β€ m β€ nΒ·(n - 1) / 2) β the number of cities and the number of roads in Berland.
The next m lines contain the description of roads in Berland. Each line contains two integers u and v (1 β€ u, v β€ n) β the cities the corresponding road connects. It's guaranteed that there are no self-loops and multiple roads. It is possible that there is no way along roads between a pair of cities.
It is guaranteed that the total number of cities in all testset of input data doesn't exceed 200.
Pay attention that for hacks, you can only use tests consisting of one testset, so t should be equal to one.
Output
For each testset print the maximum number of such cities that the number of roads that begins in the city, is equal to the number of roads that ends in it.
In the next m lines print oriented roads. First print the number of the city where the road begins and then the number of the city where the road ends. If there are several answers, print any of them. It is allowed to print roads in each test in arbitrary order. Each road should be printed exactly once.
Example
Input
2
5 5
2 1
4 5
2 3
1 3
3 5
7 2
3 7
4 2
Output
3
1 3
3 5
5 4
3 2
2 1
3
2 4
3 7
Submitted Solution:
```
t = int(input())
from collections import defaultdict
for _ in range(t):
n, m = map(int, input().split())
graph = defaultdict(dict)
for _ in range(m):
u, v = map(int, input().split())
graph[u][v] = True
graph[v][u] = True
if m==0:
print(n)
continue
odds = []
for u in graph:
if len(graph[u]) % 2 == 1:
odds.append(u)
# now every node has an even degree
for odd in odds:
graph[odd][n + 1] = True
graph[n + 1][odd] = True
# we just do a dfs
edges = []
stack = [list(graph.keys())[0]]
# we don't have a visited dictionary as usual, because we must allow to visit a node several times
# but we do delete the edges
while (len(stack) != 0):
current = stack.pop()
for neighbour in graph[current]:
if graph[current][neighbour] is False: continue
stack.append(current)
stack.append(neighbour)
graph[current][neighbour] = False
graph[neighbour][current] = False
edges.append((current, neighbour))
break
print(n - len(odds))
for edge in edges:
if edge[0] != n + 1 and edge[1] != n + 1:
print("{0} {1}".format(edge[0], edge[1]))
```
No
| 9,550 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There are n cities and m two-way roads in Berland, each road connects two cities. It is known that there is no more than one road connecting each pair of cities, and there is no road which connects the city with itself. It is possible that there is no way to get from one city to some other city using only these roads.
The road minister decided to make a reform in Berland and to orient all roads in the country, i.e. to make each road one-way. The minister wants to maximize the number of cities, for which the number of roads that begins in the city equals to the number of roads that ends in it.
Input
The first line contains a positive integer t (1 β€ t β€ 200) β the number of testsets in the input.
Each of the testsets is given in the following way. The first line contains two integers n and m (1 β€ n β€ 200, 0 β€ m β€ nΒ·(n - 1) / 2) β the number of cities and the number of roads in Berland.
The next m lines contain the description of roads in Berland. Each line contains two integers u and v (1 β€ u, v β€ n) β the cities the corresponding road connects. It's guaranteed that there are no self-loops and multiple roads. It is possible that there is no way along roads between a pair of cities.
It is guaranteed that the total number of cities in all testset of input data doesn't exceed 200.
Pay attention that for hacks, you can only use tests consisting of one testset, so t should be equal to one.
Output
For each testset print the maximum number of such cities that the number of roads that begins in the city, is equal to the number of roads that ends in it.
In the next m lines print oriented roads. First print the number of the city where the road begins and then the number of the city where the road ends. If there are several answers, print any of them. It is allowed to print roads in each test in arbitrary order. Each road should be printed exactly once.
Example
Input
2
5 5
2 1
4 5
2 3
1 3
3 5
7 2
3 7
4 2
Output
3
1 3
3 5
5 4
3 2
2 1
3
2 4
3 7
Submitted Solution:
```
import random
import copy
import math
#globals
k_max = 10000
class State:
def __init__(self, edges, n, degree=None, score=None):
self.edges = edges
self.n = n
if degree is None:
self.degree = [0 for _ in range(n+1)]
for a, b in edges:
self.degree[a] -= 1
self.degree[b] += 1
self.score = 0
for v in self.degree[1:]:
if v == 0:
self.score += 1
else:
self.degree = degree
self.score = score
def score(state):
return state.score
def get_neighbour(state):
rindex = random.randint(0, len(state.edges)-1)
a, b = state.edges[rindex]
cp = State(state.edges[::], state.n, degree=state.degree[::], score=state.score)
if cp.degree[a] == 0:
cp.score -= 1
if cp.degree[b] == 0:
cp.score -= 1
cp.degree[a] += 2
cp.degree[b] -= 2
if cp.degree[a] == 0:
cp.score += 1
if cp.degree[b] == 0:
cp.score += 1
cp.edges[rindex] = (b, a)
return cp
def probability(diff, temperature):
# prob func
if diff >= 0:
return 1
else:
return math.exp(2*diff/temperature)
def adjust_state(state, temperature):
#cur = score(state)
while True:
n = get_neighbour(state)
#print(score(n))
#print(probability(score(n)-score(state), temperature))
if probability(score(n)-score(state), temperature) >= random.random():
return n
def temperature_schedule(v):
# temp sched
global k_max
c = 100
return c*(0.99)**v
t = int (input())
for _ in range(t):
n, m = map(int, input().split(' '))
edges = []
for __ in range(m):
edges.append(tuple(map(int, input().split(' '))))
state = State(edges, n)
#print(state.score)
#print('-----------')
i = 0
temp = temperature_schedule(i)
while temp > 1:
state = adjust_state(state, temp)
i += 1
temp = temperature_schedule(i)
#print(state.degree)
#print(state.score)
#rint('----------')
#print(temp)
print(state.score)
for edge in state.edges:
print(edge[0], edge[1])
#print(state.edges)
#print(State(state.edges, n).score)
```
No
| 9,551 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There are n cities and m two-way roads in Berland, each road connects two cities. It is known that there is no more than one road connecting each pair of cities, and there is no road which connects the city with itself. It is possible that there is no way to get from one city to some other city using only these roads.
The road minister decided to make a reform in Berland and to orient all roads in the country, i.e. to make each road one-way. The minister wants to maximize the number of cities, for which the number of roads that begins in the city equals to the number of roads that ends in it.
Input
The first line contains a positive integer t (1 β€ t β€ 200) β the number of testsets in the input.
Each of the testsets is given in the following way. The first line contains two integers n and m (1 β€ n β€ 200, 0 β€ m β€ nΒ·(n - 1) / 2) β the number of cities and the number of roads in Berland.
The next m lines contain the description of roads in Berland. Each line contains two integers u and v (1 β€ u, v β€ n) β the cities the corresponding road connects. It's guaranteed that there are no self-loops and multiple roads. It is possible that there is no way along roads between a pair of cities.
It is guaranteed that the total number of cities in all testset of input data doesn't exceed 200.
Pay attention that for hacks, you can only use tests consisting of one testset, so t should be equal to one.
Output
For each testset print the maximum number of such cities that the number of roads that begins in the city, is equal to the number of roads that ends in it.
In the next m lines print oriented roads. First print the number of the city where the road begins and then the number of the city where the road ends. If there are several answers, print any of them. It is allowed to print roads in each test in arbitrary order. Each road should be printed exactly once.
Example
Input
2
5 5
2 1
4 5
2 3
1 3
3 5
7 2
3 7
4 2
Output
3
1 3
3 5
5 4
3 2
2 1
3
2 4
3 7
Submitted Solution:
```
import sys
from collections import defaultdict, Counter
class Graph:
def __init__(self, vertices):
self.vertices = vertices
self.graph = defaultdict(list)
def add_edge(self, a, b):
self.graph[a].append(b)
self.graph[b].append(a)
def eulerPath(self):
g = self.graph
odd = [v for v in g if len(g[v])&1]
if not odd:
odd = [list(graph.keys())[0]]
start = odd[-1]
path = [start]
circuit = []
while path:
u = path[-1]
if g[u]:
v = g[u][0]
del g[u][0]
del g[v][g[v].index(u)]
path.append(v)
else:
circuit.append(path.pop())
circuit.reverse()
return circuit
t = int(sys.stdin.readline())
for _ in range(t):
n, m = map(int, sys.stdin.readline().split())
g = Graph(n)
u = []
v = []
for i in range(m):
a, b = map(int, sys.stdin.readline().split())
g.add_edge(a, b)
u.append(a)
v.append(b)
total = 0
for i in range(1, n+1):
if len(g.graph[i])%2==0:
total += 1
ans = g.eulerPath()
temp = defaultdict(defaultdict)
for i in range(len(ans)-1):
temp[ans[i]][ans[i+1]] = True
print(total)
for i in range(len(u)):
a, b = u[i], v[i]
if a in temp and b in temp[a]:
print(a, b)
else:
print(b, a)
```
No
| 9,552 |
Provide tags and a correct Python 3 solution for this coding contest problem.
This is an interactive problem. In the interaction section below you will see the information about flushing the output.
In this problem, you will be playing a game with Hongcow. How lucky of you!
Hongcow has a hidden n by n matrix M. Let Mi, j denote the entry i-th row and j-th column of the matrix. The rows and columns are labeled from 1 to n.
The matrix entries are between 0 and 109. In addition, Mi, i = 0 for all valid i. Your task is to find the minimum value along each row, excluding diagonal elements. Formally, for each i, you must find <image>.
To do this, you can ask Hongcow some questions.
A question consists of giving Hongcow a subset of distinct indices {w1, w2, ..., wk}, with 1 β€ k β€ n. Hongcow will respond with n integers. The i-th integer will contain the minimum value of min1 β€ j β€ kMi, wj.
You may only ask Hongcow at most 20 questions β he thinks you only need that many questions answered.
When you are ready to answer, print out a single integer - 1 on its own line, then n integers on the next line. The i-th integer should be the minimum value in the i-th row of the matrix, excluding the i-th element. Do not forget to flush the final answer as well. Printing the answer does not count as asking a question.
You will get Wrong Answer verdict if
* Your question or answers are not in the format described in this statement.
* You ask strictly more than 20 questions.
* Your question contains duplicate indices.
* The value of k in your question does not lie in the range from 1 to n, inclusive.
* Your final answer is not correct.
You will get Idleness Limit Exceeded if you don't print anything or if you forget to flush the output, including for the final answer (more info about flushing output below).
Input
The first line of input will contain a single integer n (2 β€ n β€ 1, 000).
Output
To print the final answer, print out the string -1 on its own line. Then, the next line should contain n integers. The i-th integer should be the minimum value of the i-th row of the matrix, excluding elements on the diagonal. Do not forget to flush your answer!
Interaction
To ask a question, print out a single integer k on its own line, denoting the size of your subset. Then, the next line should contain k integers w1, w2, ... wk. Note, you must flush your output to get a response.
Hongcow will respond by printing out a line with n integers. The i-th integer in this line represents the minimum value of Mi, wj where j is between 1 and k.
You may only ask a question at most 20 times, otherwise, you will get Wrong Answer.
To flush you can use (just after printing an integer and end-of-line):
* fflush(stdout) in C++;
* System.out.flush() in Java;
* stdout.flush() in Python;
* flush(output) in Pascal;
* See the documentation for other languages.
Hacking To hack someone, use the following format
n
M_{1,1} M_{1,2} ... M_{1,n}
M_{2,1} M_{2,2} ... M_{2,n}
...
M_{n,1} M_{n,2} ... M_{n,n}
Of course, contestant programs will not be able to see this input.
Examples
Input
3
0 0 0
2 7 0
0 0 4
3 0 8
0 5 4
Output
3
1 2 3
1
3
2
1 2
1
2
1
1
-1
2 5 4
Input
2
0 0
0 0
Output
1
2
1
1
-1
0 0
Note
In the first sample, Hongcow has the hidden matrix
[
[0, 3, 2],
[5, 0, 7],
[4, 8 ,0],
]
Here is a more readable version demonstrating the interaction. The column on the left represents Hongcow, while the column on the right represents the contestant.
3
3
1 2 3
0 0 0
1
3
2 7 0
2
1 2
0 0 4
1
2
3 0 8
1
1
0 5 4
-1
2 5 4
For the second sample, it is possible for off-diagonal elements of the matrix to be zero.
Tags: bitmasks, divide and conquer, interactive
Correct Solution:
```
import sys
def f(x,y):
if not x: return y
elif not y: return x
else: return min(x,y)
count = 0
def req(old, lst):
global count
if count < 20:
print(len(lst))
print(" ".join(map(lambda x: str(x+1), lst)))
sys.stdout.flush()
current = list(map(int, input().split()))
for i in range(n):
if i in lst:
current[i] = BIG
old = list(map(min, old, current))
count += 1
return old
BIG = 10000000000
n = int(input())
mask = 1
res = [BIG] * n
while mask <= n:
a = [x for x in range(n) if x & mask]
b = [x for x in range(n) if not x & mask]
if a and b:
res = req(res, a)
res = req(res, b)
if count == 20: break
mask <<= 1
print(-1)
print(" ".join(map(str, res)))
sys.stdout.flush()
```
| 9,553 |
Provide tags and a correct Python 3 solution for this coding contest problem.
This is an interactive problem. In the interaction section below you will see the information about flushing the output.
In this problem, you will be playing a game with Hongcow. How lucky of you!
Hongcow has a hidden n by n matrix M. Let Mi, j denote the entry i-th row and j-th column of the matrix. The rows and columns are labeled from 1 to n.
The matrix entries are between 0 and 109. In addition, Mi, i = 0 for all valid i. Your task is to find the minimum value along each row, excluding diagonal elements. Formally, for each i, you must find <image>.
To do this, you can ask Hongcow some questions.
A question consists of giving Hongcow a subset of distinct indices {w1, w2, ..., wk}, with 1 β€ k β€ n. Hongcow will respond with n integers. The i-th integer will contain the minimum value of min1 β€ j β€ kMi, wj.
You may only ask Hongcow at most 20 questions β he thinks you only need that many questions answered.
When you are ready to answer, print out a single integer - 1 on its own line, then n integers on the next line. The i-th integer should be the minimum value in the i-th row of the matrix, excluding the i-th element. Do not forget to flush the final answer as well. Printing the answer does not count as asking a question.
You will get Wrong Answer verdict if
* Your question or answers are not in the format described in this statement.
* You ask strictly more than 20 questions.
* Your question contains duplicate indices.
* The value of k in your question does not lie in the range from 1 to n, inclusive.
* Your final answer is not correct.
You will get Idleness Limit Exceeded if you don't print anything or if you forget to flush the output, including for the final answer (more info about flushing output below).
Input
The first line of input will contain a single integer n (2 β€ n β€ 1, 000).
Output
To print the final answer, print out the string -1 on its own line. Then, the next line should contain n integers. The i-th integer should be the minimum value of the i-th row of the matrix, excluding elements on the diagonal. Do not forget to flush your answer!
Interaction
To ask a question, print out a single integer k on its own line, denoting the size of your subset. Then, the next line should contain k integers w1, w2, ... wk. Note, you must flush your output to get a response.
Hongcow will respond by printing out a line with n integers. The i-th integer in this line represents the minimum value of Mi, wj where j is between 1 and k.
You may only ask a question at most 20 times, otherwise, you will get Wrong Answer.
To flush you can use (just after printing an integer and end-of-line):
* fflush(stdout) in C++;
* System.out.flush() in Java;
* stdout.flush() in Python;
* flush(output) in Pascal;
* See the documentation for other languages.
Hacking To hack someone, use the following format
n
M_{1,1} M_{1,2} ... M_{1,n}
M_{2,1} M_{2,2} ... M_{2,n}
...
M_{n,1} M_{n,2} ... M_{n,n}
Of course, contestant programs will not be able to see this input.
Examples
Input
3
0 0 0
2 7 0
0 0 4
3 0 8
0 5 4
Output
3
1 2 3
1
3
2
1 2
1
2
1
1
-1
2 5 4
Input
2
0 0
0 0
Output
1
2
1
1
-1
0 0
Note
In the first sample, Hongcow has the hidden matrix
[
[0, 3, 2],
[5, 0, 7],
[4, 8 ,0],
]
Here is a more readable version demonstrating the interaction. The column on the left represents Hongcow, while the column on the right represents the contestant.
3
3
1 2 3
0 0 0
1
3
2 7 0
2
1 2
0 0 4
1
2
3 0 8
1
1
0 5 4
-1
2 5 4
For the second sample, it is possible for off-diagonal elements of the matrix to be zero.
Tags: bitmasks, divide and conquer, interactive
Correct Solution:
```
inf = 10 ** 10
n = int(input())
m = 1
while not (n < 2 ** m):
m += 1
u = []
v = []
for i in range(m):
u.append([])
v.append([])
for j in range(n):
if j & (1 << i):
u[i].append(j+1)
else:
v[i].append(j+1)
s = []
t = []
for i in range(m):
if len(u[i]) == 0:
values = [inf] * n
else:
line = ' '.join(map(str, u[i]))
print(len(u[i]))
print(line, flush=True)
values = list(map(int, input().split()))
s.append(values)
for i in range(m):
if len(v[i]) == 0:
values = [inf] * n
else:
line = ' '.join(map(str, v[i]))
print(len(v[i]))
print(line, flush=True)
values = list(map(int, input().split()))
t.append(values)
res = []
for j in range(n):
r = inf
for i in range(m):
if j & (1 << i):
r = min(r, t[i][j])
else:
r = min(r, s[i][j])
res.append(r)
print(-1, flush=True)
line = ' '.join(map(str, res))
print(line, flush=True)
```
| 9,554 |
Provide tags and a correct Python 3 solution for this coding contest problem.
This is an interactive problem. In the interaction section below you will see the information about flushing the output.
In this problem, you will be playing a game with Hongcow. How lucky of you!
Hongcow has a hidden n by n matrix M. Let Mi, j denote the entry i-th row and j-th column of the matrix. The rows and columns are labeled from 1 to n.
The matrix entries are between 0 and 109. In addition, Mi, i = 0 for all valid i. Your task is to find the minimum value along each row, excluding diagonal elements. Formally, for each i, you must find <image>.
To do this, you can ask Hongcow some questions.
A question consists of giving Hongcow a subset of distinct indices {w1, w2, ..., wk}, with 1 β€ k β€ n. Hongcow will respond with n integers. The i-th integer will contain the minimum value of min1 β€ j β€ kMi, wj.
You may only ask Hongcow at most 20 questions β he thinks you only need that many questions answered.
When you are ready to answer, print out a single integer - 1 on its own line, then n integers on the next line. The i-th integer should be the minimum value in the i-th row of the matrix, excluding the i-th element. Do not forget to flush the final answer as well. Printing the answer does not count as asking a question.
You will get Wrong Answer verdict if
* Your question or answers are not in the format described in this statement.
* You ask strictly more than 20 questions.
* Your question contains duplicate indices.
* The value of k in your question does not lie in the range from 1 to n, inclusive.
* Your final answer is not correct.
You will get Idleness Limit Exceeded if you don't print anything or if you forget to flush the output, including for the final answer (more info about flushing output below).
Input
The first line of input will contain a single integer n (2 β€ n β€ 1, 000).
Output
To print the final answer, print out the string -1 on its own line. Then, the next line should contain n integers. The i-th integer should be the minimum value of the i-th row of the matrix, excluding elements on the diagonal. Do not forget to flush your answer!
Interaction
To ask a question, print out a single integer k on its own line, denoting the size of your subset. Then, the next line should contain k integers w1, w2, ... wk. Note, you must flush your output to get a response.
Hongcow will respond by printing out a line with n integers. The i-th integer in this line represents the minimum value of Mi, wj where j is between 1 and k.
You may only ask a question at most 20 times, otherwise, you will get Wrong Answer.
To flush you can use (just after printing an integer and end-of-line):
* fflush(stdout) in C++;
* System.out.flush() in Java;
* stdout.flush() in Python;
* flush(output) in Pascal;
* See the documentation for other languages.
Hacking To hack someone, use the following format
n
M_{1,1} M_{1,2} ... M_{1,n}
M_{2,1} M_{2,2} ... M_{2,n}
...
M_{n,1} M_{n,2} ... M_{n,n}
Of course, contestant programs will not be able to see this input.
Examples
Input
3
0 0 0
2 7 0
0 0 4
3 0 8
0 5 4
Output
3
1 2 3
1
3
2
1 2
1
2
1
1
-1
2 5 4
Input
2
0 0
0 0
Output
1
2
1
1
-1
0 0
Note
In the first sample, Hongcow has the hidden matrix
[
[0, 3, 2],
[5, 0, 7],
[4, 8 ,0],
]
Here is a more readable version demonstrating the interaction. The column on the left represents Hongcow, while the column on the right represents the contestant.
3
3
1 2 3
0 0 0
1
3
2 7 0
2
1 2
0 0 4
1
2
3 0 8
1
1
0 5 4
-1
2 5 4
For the second sample, it is possible for off-diagonal elements of the matrix to be zero.
Tags: bitmasks, divide and conquer, interactive
Correct Solution:
```
import sys
n = int(input())
MAX = 2000 * 1000 * 1000
res = [MAX] * n
k = 1
while k < n:
x = [0] * n
output = []
sz = 0
for i in range(0, n, 2 * k):
for j in range(0, min(n - i, k)):
output.append(i + j + 1)
sz += 1
#print(i + j + 1, end = ' ')
x[i + j] = 1
print(sz, end = ' ')
print(' '.join(map(str, output)))
sys.stdout.flush()
a = list(map(int, input().split()))
for i in range(len(a)):
if a[i] != 0 or x[i] == 0:
res[i] = min(res[i], a[i])
x = [0] * n
output = []
sz = 0
for i in range(k, n, 2 * k):
for j in range(0, min(n - i, k)):
output.append(i + j + 1)
#print(i + j + 1, end = ' ')
x[i + j] = 1
sz += 1
print(sz, end = ' ')
print(' '.join(map(str, output)))
sys.stdout.flush()
a = list(map(int, input().split()))
for i in range(len(a)):
if a[i] != 0 or x[i] == 0:
res[i] = min(res[i], a[i])
k *= 2
print(-1)
for i in range(n):
if res[i] != MAX:
print(res[i], end = ' ')
else:
print(0, end = ' ')
print()
sys.stdout.flush()
# 0 0 0 0 0
# 1 0 2 3 1
# 1 2 0 1 0
# 5 5 5 0 5
# 2 3 4 5 0
```
| 9,555 |
Provide tags and a correct Python 3 solution for this coding contest problem.
This is an interactive problem. In the interaction section below you will see the information about flushing the output.
In this problem, you will be playing a game with Hongcow. How lucky of you!
Hongcow has a hidden n by n matrix M. Let Mi, j denote the entry i-th row and j-th column of the matrix. The rows and columns are labeled from 1 to n.
The matrix entries are between 0 and 109. In addition, Mi, i = 0 for all valid i. Your task is to find the minimum value along each row, excluding diagonal elements. Formally, for each i, you must find <image>.
To do this, you can ask Hongcow some questions.
A question consists of giving Hongcow a subset of distinct indices {w1, w2, ..., wk}, with 1 β€ k β€ n. Hongcow will respond with n integers. The i-th integer will contain the minimum value of min1 β€ j β€ kMi, wj.
You may only ask Hongcow at most 20 questions β he thinks you only need that many questions answered.
When you are ready to answer, print out a single integer - 1 on its own line, then n integers on the next line. The i-th integer should be the minimum value in the i-th row of the matrix, excluding the i-th element. Do not forget to flush the final answer as well. Printing the answer does not count as asking a question.
You will get Wrong Answer verdict if
* Your question or answers are not in the format described in this statement.
* You ask strictly more than 20 questions.
* Your question contains duplicate indices.
* The value of k in your question does not lie in the range from 1 to n, inclusive.
* Your final answer is not correct.
You will get Idleness Limit Exceeded if you don't print anything or if you forget to flush the output, including for the final answer (more info about flushing output below).
Input
The first line of input will contain a single integer n (2 β€ n β€ 1, 000).
Output
To print the final answer, print out the string -1 on its own line. Then, the next line should contain n integers. The i-th integer should be the minimum value of the i-th row of the matrix, excluding elements on the diagonal. Do not forget to flush your answer!
Interaction
To ask a question, print out a single integer k on its own line, denoting the size of your subset. Then, the next line should contain k integers w1, w2, ... wk. Note, you must flush your output to get a response.
Hongcow will respond by printing out a line with n integers. The i-th integer in this line represents the minimum value of Mi, wj where j is between 1 and k.
You may only ask a question at most 20 times, otherwise, you will get Wrong Answer.
To flush you can use (just after printing an integer and end-of-line):
* fflush(stdout) in C++;
* System.out.flush() in Java;
* stdout.flush() in Python;
* flush(output) in Pascal;
* See the documentation for other languages.
Hacking To hack someone, use the following format
n
M_{1,1} M_{1,2} ... M_{1,n}
M_{2,1} M_{2,2} ... M_{2,n}
...
M_{n,1} M_{n,2} ... M_{n,n}
Of course, contestant programs will not be able to see this input.
Examples
Input
3
0 0 0
2 7 0
0 0 4
3 0 8
0 5 4
Output
3
1 2 3
1
3
2
1 2
1
2
1
1
-1
2 5 4
Input
2
0 0
0 0
Output
1
2
1
1
-1
0 0
Note
In the first sample, Hongcow has the hidden matrix
[
[0, 3, 2],
[5, 0, 7],
[4, 8 ,0],
]
Here is a more readable version demonstrating the interaction. The column on the left represents Hongcow, while the column on the right represents the contestant.
3
3
1 2 3
0 0 0
1
3
2 7 0
2
1 2
0 0 4
1
2
3 0 8
1
1
0 5 4
-1
2 5 4
For the second sample, it is possible for off-diagonal elements of the matrix to be zero.
Tags: bitmasks, divide and conquer, interactive
Correct Solution:
```
import sys
n = int(input())
mn = [float('inf') for _ in range(n)]
ln = len(bin(n)) - 3
cnt = 0
for bit in range(ln, -1, -1):
query = []
inv_query = []
for i in range(1, n + 1):
if i & (1<<bit):
query.append(i)
else:
inv_query.append(i)
print(len(query))
print(' '.join(map(str, query)))
sys.stdout.flush()
ans = list(map(int, input().split()))
for i in inv_query:
mn[i - 1] = min(mn[i - 1], ans[i - 1])
print(len(inv_query))
print(' '.join(map(str, inv_query)))
sys.stdout.flush()
ans = list(map(int, input().split()))
for i in query:
mn[i - 1] = min(mn[i - 1], ans[i - 1])
print(-1)
print(' '.join(map(str, mn)))
sys.stdout.flush()
```
| 9,556 |
Provide tags and a correct Python 3 solution for this coding contest problem.
This is an interactive problem. In the interaction section below you will see the information about flushing the output.
In this problem, you will be playing a game with Hongcow. How lucky of you!
Hongcow has a hidden n by n matrix M. Let Mi, j denote the entry i-th row and j-th column of the matrix. The rows and columns are labeled from 1 to n.
The matrix entries are between 0 and 109. In addition, Mi, i = 0 for all valid i. Your task is to find the minimum value along each row, excluding diagonal elements. Formally, for each i, you must find <image>.
To do this, you can ask Hongcow some questions.
A question consists of giving Hongcow a subset of distinct indices {w1, w2, ..., wk}, with 1 β€ k β€ n. Hongcow will respond with n integers. The i-th integer will contain the minimum value of min1 β€ j β€ kMi, wj.
You may only ask Hongcow at most 20 questions β he thinks you only need that many questions answered.
When you are ready to answer, print out a single integer - 1 on its own line, then n integers on the next line. The i-th integer should be the minimum value in the i-th row of the matrix, excluding the i-th element. Do not forget to flush the final answer as well. Printing the answer does not count as asking a question.
You will get Wrong Answer verdict if
* Your question or answers are not in the format described in this statement.
* You ask strictly more than 20 questions.
* Your question contains duplicate indices.
* The value of k in your question does not lie in the range from 1 to n, inclusive.
* Your final answer is not correct.
You will get Idleness Limit Exceeded if you don't print anything or if you forget to flush the output, including for the final answer (more info about flushing output below).
Input
The first line of input will contain a single integer n (2 β€ n β€ 1, 000).
Output
To print the final answer, print out the string -1 on its own line. Then, the next line should contain n integers. The i-th integer should be the minimum value of the i-th row of the matrix, excluding elements on the diagonal. Do not forget to flush your answer!
Interaction
To ask a question, print out a single integer k on its own line, denoting the size of your subset. Then, the next line should contain k integers w1, w2, ... wk. Note, you must flush your output to get a response.
Hongcow will respond by printing out a line with n integers. The i-th integer in this line represents the minimum value of Mi, wj where j is between 1 and k.
You may only ask a question at most 20 times, otherwise, you will get Wrong Answer.
To flush you can use (just after printing an integer and end-of-line):
* fflush(stdout) in C++;
* System.out.flush() in Java;
* stdout.flush() in Python;
* flush(output) in Pascal;
* See the documentation for other languages.
Hacking To hack someone, use the following format
n
M_{1,1} M_{1,2} ... M_{1,n}
M_{2,1} M_{2,2} ... M_{2,n}
...
M_{n,1} M_{n,2} ... M_{n,n}
Of course, contestant programs will not be able to see this input.
Examples
Input
3
0 0 0
2 7 0
0 0 4
3 0 8
0 5 4
Output
3
1 2 3
1
3
2
1 2
1
2
1
1
-1
2 5 4
Input
2
0 0
0 0
Output
1
2
1
1
-1
0 0
Note
In the first sample, Hongcow has the hidden matrix
[
[0, 3, 2],
[5, 0, 7],
[4, 8 ,0],
]
Here is a more readable version demonstrating the interaction. The column on the left represents Hongcow, while the column on the right represents the contestant.
3
3
1 2 3
0 0 0
1
3
2 7 0
2
1 2
0 0 4
1
2
3 0 8
1
1
0 5 4
-1
2 5 4
For the second sample, it is possible for off-diagonal elements of the matrix to be zero.
Tags: bitmasks, divide and conquer, interactive
Correct Solution:
```
n = int(input().strip())
import math
import sys
factor = 1
each_row_min = [10000000000]*n
for i in range(int(math.log(n, 2))+1):
if factor >= n:
break
mask = []
comp_mask = []
for j in range(n):
if (j//factor)%2 == 0:
mask.append(j)
else:
comp_mask.append(j)
print(len(mask))
print(' '.join([str(x+1) for x in mask]))
sys.stdout.flush()
results = [int(x) for x in input().split()]
for row, rowmin in enumerate(results):
if row not in mask:
each_row_min[row] = min(each_row_min[row], rowmin)
#comp
mask = comp_mask
print(len(mask))
print(' '.join([str(x+1) for x in mask]))
sys.stdout.flush()
results = [int(x) for x in input().split()]
for row, rowmin in enumerate(results):
if row not in mask:
each_row_min[row] = min(each_row_min[row], rowmin)
factor*=2
print(-1)
print(' '.join(str(x) for x in each_row_min))
sys.stdout.flush
```
| 9,557 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
This is an interactive problem. In the interaction section below you will see the information about flushing the output.
In this problem, you will be playing a game with Hongcow. How lucky of you!
Hongcow has a hidden n by n matrix M. Let Mi, j denote the entry i-th row and j-th column of the matrix. The rows and columns are labeled from 1 to n.
The matrix entries are between 0 and 109. In addition, Mi, i = 0 for all valid i. Your task is to find the minimum value along each row, excluding diagonal elements. Formally, for each i, you must find <image>.
To do this, you can ask Hongcow some questions.
A question consists of giving Hongcow a subset of distinct indices {w1, w2, ..., wk}, with 1 β€ k β€ n. Hongcow will respond with n integers. The i-th integer will contain the minimum value of min1 β€ j β€ kMi, wj.
You may only ask Hongcow at most 20 questions β he thinks you only need that many questions answered.
When you are ready to answer, print out a single integer - 1 on its own line, then n integers on the next line. The i-th integer should be the minimum value in the i-th row of the matrix, excluding the i-th element. Do not forget to flush the final answer as well. Printing the answer does not count as asking a question.
You will get Wrong Answer verdict if
* Your question or answers are not in the format described in this statement.
* You ask strictly more than 20 questions.
* Your question contains duplicate indices.
* The value of k in your question does not lie in the range from 1 to n, inclusive.
* Your final answer is not correct.
You will get Idleness Limit Exceeded if you don't print anything or if you forget to flush the output, including for the final answer (more info about flushing output below).
Input
The first line of input will contain a single integer n (2 β€ n β€ 1, 000).
Output
To print the final answer, print out the string -1 on its own line. Then, the next line should contain n integers. The i-th integer should be the minimum value of the i-th row of the matrix, excluding elements on the diagonal. Do not forget to flush your answer!
Interaction
To ask a question, print out a single integer k on its own line, denoting the size of your subset. Then, the next line should contain k integers w1, w2, ... wk. Note, you must flush your output to get a response.
Hongcow will respond by printing out a line with n integers. The i-th integer in this line represents the minimum value of Mi, wj where j is between 1 and k.
You may only ask a question at most 20 times, otherwise, you will get Wrong Answer.
To flush you can use (just after printing an integer and end-of-line):
* fflush(stdout) in C++;
* System.out.flush() in Java;
* stdout.flush() in Python;
* flush(output) in Pascal;
* See the documentation for other languages.
Hacking To hack someone, use the following format
n
M_{1,1} M_{1,2} ... M_{1,n}
M_{2,1} M_{2,2} ... M_{2,n}
...
M_{n,1} M_{n,2} ... M_{n,n}
Of course, contestant programs will not be able to see this input.
Examples
Input
3
0 0 0
2 7 0
0 0 4
3 0 8
0 5 4
Output
3
1 2 3
1
3
2
1 2
1
2
1
1
-1
2 5 4
Input
2
0 0
0 0
Output
1
2
1
1
-1
0 0
Note
In the first sample, Hongcow has the hidden matrix
[
[0, 3, 2],
[5, 0, 7],
[4, 8 ,0],
]
Here is a more readable version demonstrating the interaction. The column on the left represents Hongcow, while the column on the right represents the contestant.
3
3
1 2 3
0 0 0
1
3
2 7 0
2
1 2
0 0 4
1
2
3 0 8
1
1
0 5 4
-1
2 5 4
For the second sample, it is possible for off-diagonal elements of the matrix to be zero.
Submitted Solution:
```
import sys
def f(x,y):
if not x: return y
elif not y: return x
else: return min(x,y)
n = int(input())
mask = 1
res = [0] * n
while mask <= n:
a = list(map(str, filter(lambda x: (x & mask) == 0, range(1, n+1))))
if a:
print(len(a))
print(" ".join(a))
sys.stdout.flush()
current = map(int, input().split())
res = list(map(f, res, current))
a = list(map(str, filter(lambda x: x & mask, range(1, n+1))))
if a:
print(len(a))
print(" ".join(a))
sys.stdout.flush()
current = map(int, input().split())
res = list(map(f, res, current))
mask <<= 1
print(-1)
print(" ".join(map(str, res)))
sys.stdout.flush()
```
No
| 9,558 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
This is an interactive problem. In the interaction section below you will see the information about flushing the output.
In this problem, you will be playing a game with Hongcow. How lucky of you!
Hongcow has a hidden n by n matrix M. Let Mi, j denote the entry i-th row and j-th column of the matrix. The rows and columns are labeled from 1 to n.
The matrix entries are between 0 and 109. In addition, Mi, i = 0 for all valid i. Your task is to find the minimum value along each row, excluding diagonal elements. Formally, for each i, you must find <image>.
To do this, you can ask Hongcow some questions.
A question consists of giving Hongcow a subset of distinct indices {w1, w2, ..., wk}, with 1 β€ k β€ n. Hongcow will respond with n integers. The i-th integer will contain the minimum value of min1 β€ j β€ kMi, wj.
You may only ask Hongcow at most 20 questions β he thinks you only need that many questions answered.
When you are ready to answer, print out a single integer - 1 on its own line, then n integers on the next line. The i-th integer should be the minimum value in the i-th row of the matrix, excluding the i-th element. Do not forget to flush the final answer as well. Printing the answer does not count as asking a question.
You will get Wrong Answer verdict if
* Your question or answers are not in the format described in this statement.
* You ask strictly more than 20 questions.
* Your question contains duplicate indices.
* The value of k in your question does not lie in the range from 1 to n, inclusive.
* Your final answer is not correct.
You will get Idleness Limit Exceeded if you don't print anything or if you forget to flush the output, including for the final answer (more info about flushing output below).
Input
The first line of input will contain a single integer n (2 β€ n β€ 1, 000).
Output
To print the final answer, print out the string -1 on its own line. Then, the next line should contain n integers. The i-th integer should be the minimum value of the i-th row of the matrix, excluding elements on the diagonal. Do not forget to flush your answer!
Interaction
To ask a question, print out a single integer k on its own line, denoting the size of your subset. Then, the next line should contain k integers w1, w2, ... wk. Note, you must flush your output to get a response.
Hongcow will respond by printing out a line with n integers. The i-th integer in this line represents the minimum value of Mi, wj where j is between 1 and k.
You may only ask a question at most 20 times, otherwise, you will get Wrong Answer.
To flush you can use (just after printing an integer and end-of-line):
* fflush(stdout) in C++;
* System.out.flush() in Java;
* stdout.flush() in Python;
* flush(output) in Pascal;
* See the documentation for other languages.
Hacking To hack someone, use the following format
n
M_{1,1} M_{1,2} ... M_{1,n}
M_{2,1} M_{2,2} ... M_{2,n}
...
M_{n,1} M_{n,2} ... M_{n,n}
Of course, contestant programs will not be able to see this input.
Examples
Input
3
0 0 0
2 7 0
0 0 4
3 0 8
0 5 4
Output
3
1 2 3
1
3
2
1 2
1
2
1
1
-1
2 5 4
Input
2
0 0
0 0
Output
1
2
1
1
-1
0 0
Note
In the first sample, Hongcow has the hidden matrix
[
[0, 3, 2],
[5, 0, 7],
[4, 8 ,0],
]
Here is a more readable version demonstrating the interaction. The column on the left represents Hongcow, while the column on the right represents the contestant.
3
3
1 2 3
0 0 0
1
3
2 7 0
2
1 2
0 0 4
1
2
3 0 8
1
1
0 5 4
-1
2 5 4
For the second sample, it is possible for off-diagonal elements of the matrix to be zero.
Submitted Solution:
```
import sys
def f(x,y):
if not x: return y
elif not y: return x
else: return min(x,y)
count = 0
n = int(input())
mask = 1
res = [0] * n
while mask <= n:
a = list(map(lambda x: str(x+1), filter(lambda x: (x & mask) == 0, range(n))))
if a:
print(len(a))
print(" ".join(a))
sys.stdout.flush()
current = map(int, input().split())
res = list(map(f, res, current))
count += 1
if count == 20: break
a = list(map(lambda x: str(x+1), filter(lambda x: x & mask, range(n))))
if a:
print(len(a))
print(" ".join(a))
sys.stdout.flush()
current = map(int, input().split())
res = list(map(f, res, current))
count += 1
if count == 20: break
mask <<= 1
print(-1)
print(" ".join(map(str, res)))
sys.stdout.flush()
```
No
| 9,559 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
This is an interactive problem. In the interaction section below you will see the information about flushing the output.
In this problem, you will be playing a game with Hongcow. How lucky of you!
Hongcow has a hidden n by n matrix M. Let Mi, j denote the entry i-th row and j-th column of the matrix. The rows and columns are labeled from 1 to n.
The matrix entries are between 0 and 109. In addition, Mi, i = 0 for all valid i. Your task is to find the minimum value along each row, excluding diagonal elements. Formally, for each i, you must find <image>.
To do this, you can ask Hongcow some questions.
A question consists of giving Hongcow a subset of distinct indices {w1, w2, ..., wk}, with 1 β€ k β€ n. Hongcow will respond with n integers. The i-th integer will contain the minimum value of min1 β€ j β€ kMi, wj.
You may only ask Hongcow at most 20 questions β he thinks you only need that many questions answered.
When you are ready to answer, print out a single integer - 1 on its own line, then n integers on the next line. The i-th integer should be the minimum value in the i-th row of the matrix, excluding the i-th element. Do not forget to flush the final answer as well. Printing the answer does not count as asking a question.
You will get Wrong Answer verdict if
* Your question or answers are not in the format described in this statement.
* You ask strictly more than 20 questions.
* Your question contains duplicate indices.
* The value of k in your question does not lie in the range from 1 to n, inclusive.
* Your final answer is not correct.
You will get Idleness Limit Exceeded if you don't print anything or if you forget to flush the output, including for the final answer (more info about flushing output below).
Input
The first line of input will contain a single integer n (2 β€ n β€ 1, 000).
Output
To print the final answer, print out the string -1 on its own line. Then, the next line should contain n integers. The i-th integer should be the minimum value of the i-th row of the matrix, excluding elements on the diagonal. Do not forget to flush your answer!
Interaction
To ask a question, print out a single integer k on its own line, denoting the size of your subset. Then, the next line should contain k integers w1, w2, ... wk. Note, you must flush your output to get a response.
Hongcow will respond by printing out a line with n integers. The i-th integer in this line represents the minimum value of Mi, wj where j is between 1 and k.
You may only ask a question at most 20 times, otherwise, you will get Wrong Answer.
To flush you can use (just after printing an integer and end-of-line):
* fflush(stdout) in C++;
* System.out.flush() in Java;
* stdout.flush() in Python;
* flush(output) in Pascal;
* See the documentation for other languages.
Hacking To hack someone, use the following format
n
M_{1,1} M_{1,2} ... M_{1,n}
M_{2,1} M_{2,2} ... M_{2,n}
...
M_{n,1} M_{n,2} ... M_{n,n}
Of course, contestant programs will not be able to see this input.
Examples
Input
3
0 0 0
2 7 0
0 0 4
3 0 8
0 5 4
Output
3
1 2 3
1
3
2
1 2
1
2
1
1
-1
2 5 4
Input
2
0 0
0 0
Output
1
2
1
1
-1
0 0
Note
In the first sample, Hongcow has the hidden matrix
[
[0, 3, 2],
[5, 0, 7],
[4, 8 ,0],
]
Here is a more readable version demonstrating the interaction. The column on the left represents Hongcow, while the column on the right represents the contestant.
3
3
1 2 3
0 0 0
1
3
2 7 0
2
1 2
0 0 4
1
2
3 0 8
1
1
0 5 4
-1
2 5 4
For the second sample, it is possible for off-diagonal elements of the matrix to be zero.
Submitted Solution:
```
import sys
def f(x,y):
if not x: return y
elif not y: return x
else: return min(x,y)
count = 0
n = int(input())
mask = 1
res = [1000000000] * n
while mask <= n:
a = list(map(lambda x: str(x+1), filter(lambda x: (x & mask) == 0, range(n))))
if a:
print(len(a))
print(" ".join(a))
sys.stdout.flush()
current = map(int, input().split())
for i in range(n):
if i+1 in a:
current[i] = 1000000000
res = list(map(min, res, current))
count += 1
if count == 20: break
a = list(map(lambda x: str(x+1), filter(lambda x: x & mask, range(n))))
if a:
print(len(a))
print(" ".join(a))
sys.stdout.flush()
current = map(int, input().split())
for i in range(n):
if i+1 in a:
current[i] = 1000000000
res = list(map(min, res, current))
count += 1
if count == 20: break
mask <<= 1
print(-1)
print(" ".join(map(str, res)))
sys.stdout.flush()
```
No
| 9,560 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
This is an interactive problem. In the interaction section below you will see the information about flushing the output.
In this problem, you will be playing a game with Hongcow. How lucky of you!
Hongcow has a hidden n by n matrix M. Let Mi, j denote the entry i-th row and j-th column of the matrix. The rows and columns are labeled from 1 to n.
The matrix entries are between 0 and 109. In addition, Mi, i = 0 for all valid i. Your task is to find the minimum value along each row, excluding diagonal elements. Formally, for each i, you must find <image>.
To do this, you can ask Hongcow some questions.
A question consists of giving Hongcow a subset of distinct indices {w1, w2, ..., wk}, with 1 β€ k β€ n. Hongcow will respond with n integers. The i-th integer will contain the minimum value of min1 β€ j β€ kMi, wj.
You may only ask Hongcow at most 20 questions β he thinks you only need that many questions answered.
When you are ready to answer, print out a single integer - 1 on its own line, then n integers on the next line. The i-th integer should be the minimum value in the i-th row of the matrix, excluding the i-th element. Do not forget to flush the final answer as well. Printing the answer does not count as asking a question.
You will get Wrong Answer verdict if
* Your question or answers are not in the format described in this statement.
* You ask strictly more than 20 questions.
* Your question contains duplicate indices.
* The value of k in your question does not lie in the range from 1 to n, inclusive.
* Your final answer is not correct.
You will get Idleness Limit Exceeded if you don't print anything or if you forget to flush the output, including for the final answer (more info about flushing output below).
Input
The first line of input will contain a single integer n (2 β€ n β€ 1, 000).
Output
To print the final answer, print out the string -1 on its own line. Then, the next line should contain n integers. The i-th integer should be the minimum value of the i-th row of the matrix, excluding elements on the diagonal. Do not forget to flush your answer!
Interaction
To ask a question, print out a single integer k on its own line, denoting the size of your subset. Then, the next line should contain k integers w1, w2, ... wk. Note, you must flush your output to get a response.
Hongcow will respond by printing out a line with n integers. The i-th integer in this line represents the minimum value of Mi, wj where j is between 1 and k.
You may only ask a question at most 20 times, otherwise, you will get Wrong Answer.
To flush you can use (just after printing an integer and end-of-line):
* fflush(stdout) in C++;
* System.out.flush() in Java;
* stdout.flush() in Python;
* flush(output) in Pascal;
* See the documentation for other languages.
Hacking To hack someone, use the following format
n
M_{1,1} M_{1,2} ... M_{1,n}
M_{2,1} M_{2,2} ... M_{2,n}
...
M_{n,1} M_{n,2} ... M_{n,n}
Of course, contestant programs will not be able to see this input.
Examples
Input
3
0 0 0
2 7 0
0 0 4
3 0 8
0 5 4
Output
3
1 2 3
1
3
2
1 2
1
2
1
1
-1
2 5 4
Input
2
0 0
0 0
Output
1
2
1
1
-1
0 0
Note
In the first sample, Hongcow has the hidden matrix
[
[0, 3, 2],
[5, 0, 7],
[4, 8 ,0],
]
Here is a more readable version demonstrating the interaction. The column on the left represents Hongcow, while the column on the right represents the contestant.
3
3
1 2 3
0 0 0
1
3
2 7 0
2
1 2
0 0 4
1
2
3 0 8
1
1
0 5 4
-1
2 5 4
For the second sample, it is possible for off-diagonal elements of the matrix to be zero.
Submitted Solution:
```
import sys
def f(x,y):
if not x: return y
elif not y: return x
else: return min(x,y)
count = 0
n = int(input())
mask = 1
res = [0] * n
while mask <= n:
a = list(map(str, filter(lambda x: (x & mask) == 0, range(1, n+1))))
if a:
print(len(a))
print(" ".join(a))
sys.stdout.flush()
current = map(int, input().split())
res = list(map(f, res, current))
count += 1
if count == 20: break
a = list(map(str, filter(lambda x: x & mask, range(1, n+1))))
if a:
print(len(a))
print(" ".join(a))
sys.stdout.flush()
current = map(int, input().split())
res = list(map(f, res, current))
count += 1
if count == 20: break
mask <<= 1
print(-1)
print(" ".join(map(str, res)))
sys.stdout.flush()
```
No
| 9,561 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Tarly has two different type of items, food boxes and wine barrels. There are f food boxes and w wine barrels. Tarly stores them in various stacks and each stack can consist of either food boxes or wine barrels but not both. The stacks are placed in a line such that no two stacks of food boxes are together and no two stacks of wine barrels are together.
The height of a stack is defined as the number of items in the stack. Two stacks are considered different if either their heights are different or one of them contains food and other contains wine.
Jon Snow doesn't like an arrangement if any stack of wine barrels has height less than or equal to h. What is the probability that Jon Snow will like the arrangement if all arrangement are equiprobably?
Two arrangement of stacks are considered different if exists such i, that i-th stack of one arrangement is different from the i-th stack of the other arrangement.
Input
The first line of input contains three integers f, w, h (0 β€ f, w, h β€ 105) β number of food boxes, number of wine barrels and h is as described above. It is guaranteed that he has at least one food box or at least one wine barrel.
Output
Output the probability that Jon Snow will like the arrangement. The probability is of the form <image>, then you need to output a single integer pΒ·q - 1 mod (109 + 7).
Examples
Input
1 1 1
Output
0
Input
1 2 1
Output
666666672
Note
In the first example f = 1, w = 1 and h = 1, there are only two possible arrangement of stacks and Jon Snow doesn't like any of them.
In the second example f = 1, w = 2 and h = 1, there are three arrangements. Jon Snow likes the (1) and (3) arrangement. So the probabilty is <image>.
<image>
Tags: brute force, combinatorics, math, number theory, probabilities
Correct Solution:
```
def build_fac():
global mod
fac = [1] * int(3e5 + 1)
for i in range(1, int(3e5)):
fac[i] = i*fac[i-1] % mod
return fac
def inv(x):
global mod
return pow(x, mod-2, mod)
def ncr(n, r):
global fac
if n < 0 or n < r: return 0
return fac[n]*inv(fac[r])*inv(fac[n-r]) % mod
def cf(f, w, h):
global mod
if w == 0: return 1
rs = 0
for k in range(1, min(w//(h+1),f+1)+1):
rs += ncr(f+1, k) * ncr(w-k*h-1, k-1) % mod
rs %= mod
return rs
f, w, h = map(int,input().split(' '))
mod = int(1e9 + 7)
fac = build_fac()
cnt = cf(f, w, h)
rs = cnt*inv(ncr(f+w, w)) % mod
print(rs)
```
| 9,562 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Tarly has two different type of items, food boxes and wine barrels. There are f food boxes and w wine barrels. Tarly stores them in various stacks and each stack can consist of either food boxes or wine barrels but not both. The stacks are placed in a line such that no two stacks of food boxes are together and no two stacks of wine barrels are together.
The height of a stack is defined as the number of items in the stack. Two stacks are considered different if either their heights are different or one of them contains food and other contains wine.
Jon Snow doesn't like an arrangement if any stack of wine barrels has height less than or equal to h. What is the probability that Jon Snow will like the arrangement if all arrangement are equiprobably?
Two arrangement of stacks are considered different if exists such i, that i-th stack of one arrangement is different from the i-th stack of the other arrangement.
Input
The first line of input contains three integers f, w, h (0 β€ f, w, h β€ 105) β number of food boxes, number of wine barrels and h is as described above. It is guaranteed that he has at least one food box or at least one wine barrel.
Output
Output the probability that Jon Snow will like the arrangement. The probability is of the form <image>, then you need to output a single integer pΒ·q - 1 mod (109 + 7).
Examples
Input
1 1 1
Output
0
Input
1 2 1
Output
666666672
Note
In the first example f = 1, w = 1 and h = 1, there are only two possible arrangement of stacks and Jon Snow doesn't like any of them.
In the second example f = 1, w = 2 and h = 1, there are three arrangements. Jon Snow likes the (1) and (3) arrangement. So the probabilty is <image>.
<image>
Tags: brute force, combinatorics, math, number theory, probabilities
Correct Solution:
```
import sys
def factorial():
global mod
fac = [1] * int(3e5 + 1)
for i in range(1, int(3e5)):
fac[i] = i*fac[i-1] % mod
return fac
def inverse(x):
global mod
return pow(x, mod-2, mod)
def C(n, r):
global fac
if n < 0 or n < r:
return 0
return fac[n]*inverse(fac[r])*inverse(fac[n-r]) % mod
def calc(f, w, h):
global mod
if w == 0:
return 1
ans = 0
for k in range(1, min(w//(h+1),f+1)+1):
ans += C(f+1, k) * C(w-k*h-1, k-1) % mod
ans %= mod
return ans
f, w, h = map(int,sys.stdin.readline().split(' '))
mod = int(1e9 + 7)
fac = factorial()
cnt = calc(f, w, h)
sys.stdout.write(str(cnt*inverse(C(f+w, w)) % mod))
```
| 9,563 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Tarly has two different type of items, food boxes and wine barrels. There are f food boxes and w wine barrels. Tarly stores them in various stacks and each stack can consist of either food boxes or wine barrels but not both. The stacks are placed in a line such that no two stacks of food boxes are together and no two stacks of wine barrels are together.
The height of a stack is defined as the number of items in the stack. Two stacks are considered different if either their heights are different or one of them contains food and other contains wine.
Jon Snow doesn't like an arrangement if any stack of wine barrels has height less than or equal to h. What is the probability that Jon Snow will like the arrangement if all arrangement are equiprobably?
Two arrangement of stacks are considered different if exists such i, that i-th stack of one arrangement is different from the i-th stack of the other arrangement.
Input
The first line of input contains three integers f, w, h (0 β€ f, w, h β€ 105) β number of food boxes, number of wine barrels and h is as described above. It is guaranteed that he has at least one food box or at least one wine barrel.
Output
Output the probability that Jon Snow will like the arrangement. The probability is of the form <image>, then you need to output a single integer pΒ·q - 1 mod (109 + 7).
Examples
Input
1 1 1
Output
0
Input
1 2 1
Output
666666672
Note
In the first example f = 1, w = 1 and h = 1, there are only two possible arrangement of stacks and Jon Snow doesn't like any of them.
In the second example f = 1, w = 2 and h = 1, there are three arrangements. Jon Snow likes the (1) and (3) arrangement. So the probabilty is <image>.
<image>
Submitted Solution:
```
def build_fac():
global mod
fac = [1] * int(2e5 + 1)
for i in range(1, int(2e5)):
fac[i] = i*fac[i-1] % mod
return fac
def inv(x):
global mod
return pow(x, mod-2, mod)
def ncr(n, r):
global fac
if n < 0 or n < r: return 0
return fac[n]*inv(fac[r])*inv(fac[n-r]) % mod
def cf(f, w, h):
global mod
if w == 0: return 1
rs = 0
for k in range(1, min(w//(h+1),f+1)+1):
rs += ncr(f+1, k) * ncr(w-k*h-1, k-1) % mod
rs %= mod
return rs
f, w, h = map(int,input().split(' '))
mod = int(1e9 + 7)
fac = build_fac()
cnt = cf(f, w, h)
rs = cnt*inv(ncr(f+w, w)) % mod
print(rs)
```
No
| 9,564 |
Provide tags and a correct Python 3 solution for this coding contest problem.
There are n boxes with colored balls on the table. Colors are numbered from 1 to n. i-th box contains ai balls, all of which have color i. You have to write a program that will divide all balls into sets such that:
* each ball belongs to exactly one of the sets,
* there are no empty sets,
* there is no set containing two (or more) balls of different colors (each set contains only balls of one color),
* there are no two sets such that the difference between their sizes is greater than 1.
Print the minimum possible number of sets.
Input
The first line contains one integer number n (1 β€ n β€ 500).
The second line contains n integer numbers a1, a2, ... , an (1 β€ ai β€ 109).
Output
Print one integer number β the minimum possible number of sets.
Examples
Input
3
4 7 8
Output
5
Input
2
2 7
Output
4
Note
In the first example the balls can be divided into sets like that: one set with 4 balls of the first color, two sets with 3 and 4 balls, respectively, of the second color, and two sets with 4 balls of the third color.
Tags: greedy, math, number theory
Correct Solution:
```
#!/usr/bin/python3
import sys
n = int(sys.stdin.readline())
a = [int(x) for x in sys.stdin.readline().split()]
def is_valid(s, k):
# Ali lahko zapakiramo s steklenic v Ε‘katle velikosti k in k + 1?
# s = x * k + y * (k + 1) = (x + y) * k + y, kjer je 0 <= y < k
y = s % k
return s // k >= y
def all_valid(a, k):
for s in a:
if not is_valid(s, k):
return False
return True
best_sol = 1
k = 1
while k*k <= a[0]:
if all_valid(a, k):
best_sol = max(best_sol, k)
k += 1
# t je skupno Ε‘tevilo Ε‘katel.
t = 1
while t*t <= a[0]:
k = a[0] // t
if all_valid(a, k):
best_sol = max(best_sol, k)
if a[0] % t == 0 and k > 1:
if all_valid(a, k - 1):
best_sol = max(best_sol, k - 1)
t += 1
# print(best_sol, best_sol + 1)
print(sum(s // (best_sol + 1) + (0 if s % (best_sol + 1) == 0 else 1) for s in a))
```
| 9,565 |
Provide tags and a correct Python 3 solution for this coding contest problem.
There are n boxes with colored balls on the table. Colors are numbered from 1 to n. i-th box contains ai balls, all of which have color i. You have to write a program that will divide all balls into sets such that:
* each ball belongs to exactly one of the sets,
* there are no empty sets,
* there is no set containing two (or more) balls of different colors (each set contains only balls of one color),
* there are no two sets such that the difference between their sizes is greater than 1.
Print the minimum possible number of sets.
Input
The first line contains one integer number n (1 β€ n β€ 500).
The second line contains n integer numbers a1, a2, ... , an (1 β€ ai β€ 109).
Output
Print one integer number β the minimum possible number of sets.
Examples
Input
3
4 7 8
Output
5
Input
2
2 7
Output
4
Note
In the first example the balls can be divided into sets like that: one set with 4 balls of the first color, two sets with 3 and 4 balls, respectively, of the second color, and two sets with 4 balls of the third color.
Tags: greedy, math, number theory
Correct Solution:
```
# -*- coding: utf-8 -*-
"""
Created on Sun Apr 2 22:42:34 2017
@author: Sean38
"""
n = int(input().rstrip())
s = input()
a = [int(ch) for ch in s.split()]
a = a[0:n]
a.sort()
def check_num(p, i):
# i = ap + b(p+1)
# min(a+b) <=> max(b)
# b(p+1) <= i
# b == i (mod p)
max_b = (i // (p + 1))
b = i % p + ((max_b - i % p) // p) * p
# cur = a + b
cur = (i - b) // p
#print(cur - b, b, p)
if b < 0:
return None
return cur
def sets_num(p):
total = 0
for i in a:
if check_num(p, i):
total += check_num(p, i)
else:
return None
return total
for div_sets in range(1, a[0] + 1):
p, q = divmod(a[0], div_sets)
if (q == 0):
if sets_num(p):
print(sets_num(p))
break
if (p > 0) and sets_num(p - 1):
print(sets_num(p - 1))
break
else:
if sets_num(p):
print(sets_num(p))
break
```
| 9,566 |
Provide tags and a correct Python 3 solution for this coding contest problem.
There are n boxes with colored balls on the table. Colors are numbered from 1 to n. i-th box contains ai balls, all of which have color i. You have to write a program that will divide all balls into sets such that:
* each ball belongs to exactly one of the sets,
* there are no empty sets,
* there is no set containing two (or more) balls of different colors (each set contains only balls of one color),
* there are no two sets such that the difference between their sizes is greater than 1.
Print the minimum possible number of sets.
Input
The first line contains one integer number n (1 β€ n β€ 500).
The second line contains n integer numbers a1, a2, ... , an (1 β€ ai β€ 109).
Output
Print one integer number β the minimum possible number of sets.
Examples
Input
3
4 7 8
Output
5
Input
2
2 7
Output
4
Note
In the first example the balls can be divided into sets like that: one set with 4 balls of the first color, two sets with 3 and 4 balls, respectively, of the second color, and two sets with 4 balls of the third color.
Tags: greedy, math, number theory
Correct Solution:
```
#!/usr/bin/env python3
# solution after hint:(
from math import sqrt
n = int(input().strip())
ais = list(map(int, input().strip().split()))
ais.sort(reverse=True)
xmax = int(sqrt(ais[0]))
# {x, x + 1}
def check(x):
q = 0
for ai in ais:
qi = -((-ai) // (x + 1))
q += qi
if x * qi > ai:
return (False, None)
else:
return (True, q)
qbest = sum(ais) + 1
for x in range(1, xmax + 1): # {x, x + 1}
(valid, q) = check(x)
if valid:
qbest = min(qbest, q)
for k in range(1, xmax + 1): # a0 // k +/- 1
x = ais[0] // k
(valid, q) = check(x)
if valid:
qbest = min(qbest, q)
if ais[0] % k == 0:
x -= 1
(valid, q) = check(x)
if valid:
qbest = min(qbest, q)
print (qbest)
```
| 9,567 |
Provide tags and a correct Python 3 solution for this coding contest problem.
There are n boxes with colored balls on the table. Colors are numbered from 1 to n. i-th box contains ai balls, all of which have color i. You have to write a program that will divide all balls into sets such that:
* each ball belongs to exactly one of the sets,
* there are no empty sets,
* there is no set containing two (or more) balls of different colors (each set contains only balls of one color),
* there are no two sets such that the difference between their sizes is greater than 1.
Print the minimum possible number of sets.
Input
The first line contains one integer number n (1 β€ n β€ 500).
The second line contains n integer numbers a1, a2, ... , an (1 β€ ai β€ 109).
Output
Print one integer number β the minimum possible number of sets.
Examples
Input
3
4 7 8
Output
5
Input
2
2 7
Output
4
Note
In the first example the balls can be divided into sets like that: one set with 4 balls of the first color, two sets with 3 and 4 balls, respectively, of the second color, and two sets with 4 balls of the third color.
Tags: greedy, math, number theory
Correct Solution:
```
# Returns the number of coins that is necessary for paying the amount
# with the two types of coins that have value of coin and coin + 1
def pay(coin, amount):
if amount // coin < amount % coin:
return -1;
if amount < coin * (coin + 1):
return amount // coin
return (amount - 1) // (coin + 1) + 1;
def pay_all(coin):
sum = 0
for i in range(n):
p = pay(coin, a[i])
if p == -1:
return -1
sum += p
return sum
n = int(input())
a = list(map(int, input().split()))
amin = min(a)
coin = amin
k = 1
p = -1
while p == -1:
p = pay_all(coin)
if p == -1 and amin % coin == 0:
p = pay_all(coin - 1)
k += 1
coin = amin // k
print(p)
```
| 9,568 |
Provide tags and a correct Python 3 solution for this coding contest problem.
There are n boxes with colored balls on the table. Colors are numbered from 1 to n. i-th box contains ai balls, all of which have color i. You have to write a program that will divide all balls into sets such that:
* each ball belongs to exactly one of the sets,
* there are no empty sets,
* there is no set containing two (or more) balls of different colors (each set contains only balls of one color),
* there are no two sets such that the difference between their sizes is greater than 1.
Print the minimum possible number of sets.
Input
The first line contains one integer number n (1 β€ n β€ 500).
The second line contains n integer numbers a1, a2, ... , an (1 β€ ai β€ 109).
Output
Print one integer number β the minimum possible number of sets.
Examples
Input
3
4 7 8
Output
5
Input
2
2 7
Output
4
Note
In the first example the balls can be divided into sets like that: one set with 4 balls of the first color, two sets with 3 and 4 balls, respectively, of the second color, and two sets with 4 balls of the third color.
Tags: greedy, math, number theory
Correct Solution:
```
def judge(lists, x):
ans = 0
for i in lists:
t = i//x
d = i%x
if d == 0:
ans += t
elif t+d >= x-1:
ans += t+1
else:
return -1
return ans
while True:
try:
n = input()
balls = list(map(int, input().split()))
minn = min(balls)
for i in range(1, minn+1):
if judge(balls, minn//i + 1) >= 0:
ans = judge(balls, minn//i + 1)
break
elif judge(balls, minn//i) >= 0:
ans = judge(balls, minn//i)
break
print(ans)
except EOFError:
break
```
| 9,569 |
Provide tags and a correct Python 3 solution for this coding contest problem.
There are n boxes with colored balls on the table. Colors are numbered from 1 to n. i-th box contains ai balls, all of which have color i. You have to write a program that will divide all balls into sets such that:
* each ball belongs to exactly one of the sets,
* there are no empty sets,
* there is no set containing two (or more) balls of different colors (each set contains only balls of one color),
* there are no two sets such that the difference between their sizes is greater than 1.
Print the minimum possible number of sets.
Input
The first line contains one integer number n (1 β€ n β€ 500).
The second line contains n integer numbers a1, a2, ... , an (1 β€ ai β€ 109).
Output
Print one integer number β the minimum possible number of sets.
Examples
Input
3
4 7 8
Output
5
Input
2
2 7
Output
4
Note
In the first example the balls can be divided into sets like that: one set with 4 balls of the first color, two sets with 3 and 4 balls, respectively, of the second color, and two sets with 4 balls of the third color.
Tags: greedy, math, number theory
Correct Solution:
```
import time
import sys
from math import sqrt
n = int(input())
a = list(map(int, input().split()))
sq = int(sqrt(a[0]))+2
s = set()
for box in range(1, sq):
if a[0] % box == 0:
s.add(a[0] // box)
s.add(a[0] // box - 1)
else:
s.add(a[0] // box)
for balls in range(1, sq):
if a[0] % balls <= a[0] // balls:
s.add(balls)
for size in sorted(s, reverse=True):
ans = 0
for x in a:
if x / size >= x % size:
ans += (x+size) // (size+1)
else:
break
else:
print(ans)
exit()
```
| 9,570 |
Provide tags and a correct Python 3 solution for this coding contest problem.
There are n boxes with colored balls on the table. Colors are numbered from 1 to n. i-th box contains ai balls, all of which have color i. You have to write a program that will divide all balls into sets such that:
* each ball belongs to exactly one of the sets,
* there are no empty sets,
* there is no set containing two (or more) balls of different colors (each set contains only balls of one color),
* there are no two sets such that the difference between their sizes is greater than 1.
Print the minimum possible number of sets.
Input
The first line contains one integer number n (1 β€ n β€ 500).
The second line contains n integer numbers a1, a2, ... , an (1 β€ ai β€ 109).
Output
Print one integer number β the minimum possible number of sets.
Examples
Input
3
4 7 8
Output
5
Input
2
2 7
Output
4
Note
In the first example the balls can be divided into sets like that: one set with 4 balls of the first color, two sets with 3 and 4 balls, respectively, of the second color, and two sets with 4 balls of the third color.
Tags: greedy, math, number theory
Correct Solution:
```
import bisect
from itertools import accumulate
import os
import sys
import math
from decimal import *
from io import BytesIO, IOBase
BUFSIZE = 8192
class FastIO(IOBase):
newlines = 0
def __init__(self, file):
self._fd = file.fileno()
self.buffer = BytesIO()
self.writable = "x" in file.mode or "r" not in file.mode
self.write = self.buffer.write if self.writable else None
def read(self):
while True:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
if not b:
break
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines = 0
return self.buffer.read()
def readline(self):
while self.newlines == 0:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
self.newlines = b.count(b"\n") + (not b)
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines -= 1
return self.buffer.readline()
def flush(self):
if self.writable:
os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
self.write = lambda s: self.buffer.write(s.encode("ascii"))
self.read = lambda: self.buffer.read().decode("ascii")
self.readline = lambda: self.buffer.readline().decode("ascii")
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
def input(): return sys.stdin.readline().rstrip("\r\n")
def isPrime(n) :
if (n <= 1) : return False
if (n <= 3) : return True
if (n % 2 == 0 or n % 3 == 0) : return False
i = 5
while(i * i <= n) :
if (n % i == 0 or n % (i + 2) == 0) :
return False
i = i + 6
return True
def SieveOfEratosthenes(n):
prime=[]
primes = [True for i in range(n+1)]
p = 2
while (p * p <= n):
if (primes[p] == True):
prime.append(p)
for i in range(p * p, n+1, p):
primes[i] = False
p += 1
return prime
def primefactors(n):
fac=[]
while(n%2==0):
fac.append(2)
n=n//2
for i in range(3,int(math.sqrt(n))+2):
while(n%i==0):
fac.append(i)
n=n//i
if n>1:
fac.append(n)
return fac
def factors(n):
fac=set()
fac.add(1)
fac.add(n)
for i in range(2,int(math.sqrt(n))+1):
if n%i==0:
fac.add(i)
fac.add(n//i)
return list(fac)
def modInverse(a, m):
m0 = m
y = 0
x = 1
if (m == 1):
return 0
while (a > 1):
q = a // m
t = m
m = a % m
a = t
t = y
y = x - q * y
x = t
if (x < 0):
x = x + m0
return x
#------------------------------------------------------code by AD18/apurva3455
n = int(input())
a = list(map(int, input().split()))
sq = int(math.sqrt(a[0]))+2
s = set()
for box in range(1, sq):
if a[0] % box == 0:
s.add(a[0] // box)
s.add(a[0] // box - 1)
else:
s.add(a[0] // box)
for balls in range(1, sq):
if a[0] % balls <= a[0] // balls:
s.add(balls)
for size in sorted(s, reverse=True):
ans = 0
for x in a:
if x / size >= x % size:
ans += (x+size) // (size+1)
else:
break
else:
print(ans)
exit()
```
| 9,571 |
Provide tags and a correct Python 3 solution for this coding contest problem.
There are n boxes with colored balls on the table. Colors are numbered from 1 to n. i-th box contains ai balls, all of which have color i. You have to write a program that will divide all balls into sets such that:
* each ball belongs to exactly one of the sets,
* there are no empty sets,
* there is no set containing two (or more) balls of different colors (each set contains only balls of one color),
* there are no two sets such that the difference between their sizes is greater than 1.
Print the minimum possible number of sets.
Input
The first line contains one integer number n (1 β€ n β€ 500).
The second line contains n integer numbers a1, a2, ... , an (1 β€ ai β€ 109).
Output
Print one integer number β the minimum possible number of sets.
Examples
Input
3
4 7 8
Output
5
Input
2
2 7
Output
4
Note
In the first example the balls can be divided into sets like that: one set with 4 balls of the first color, two sets with 3 and 4 balls, respectively, of the second color, and two sets with 4 balls of the third color.
Tags: greedy, math, number theory
Correct Solution:
```
from math import sqrt
def divide_into_p(p, a):
q = a // (p+1)
r = a % (p+1)
if r == 0:
return q
elif q + r >= p:
return q+1
else:
return None
def divisible_into_p(p, lst):
lgth = 0
for l in lst:
dp = divide_into_p(p, l)
if dp == None:
return None
else:
lgth += dp
return lgth
def index_lst(a,sqrta):
lst = []
for i in range(1, sqrta + 1):
p = a//i
lst.append(p)
if not a%p:
lst.append(p-1)
for i in range(sqrta+1,0,-1):
lst.append(i)
return lst
def check_all_p(balls):
a = balls[0]
sqrt_a = int(sqrt(a))
indices = index_lst(a,sqrt_a)
for i in indices:
dp = divisible_into_p(i, balls)
if dp != None:
return dp
input()
balls = sorted([int(x) for x in input().split()])
print(check_all_p(balls))
```
| 9,572 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There are n boxes with colored balls on the table. Colors are numbered from 1 to n. i-th box contains ai balls, all of which have color i. You have to write a program that will divide all balls into sets such that:
* each ball belongs to exactly one of the sets,
* there are no empty sets,
* there is no set containing two (or more) balls of different colors (each set contains only balls of one color),
* there are no two sets such that the difference between their sizes is greater than 1.
Print the minimum possible number of sets.
Input
The first line contains one integer number n (1 β€ n β€ 500).
The second line contains n integer numbers a1, a2, ... , an (1 β€ ai β€ 109).
Output
Print one integer number β the minimum possible number of sets.
Examples
Input
3
4 7 8
Output
5
Input
2
2 7
Output
4
Note
In the first example the balls can be divided into sets like that: one set with 4 balls of the first color, two sets with 3 and 4 balls, respectively, of the second color, and two sets with 4 balls of the third color.
Submitted Solution:
```
import math
def check(n,a,m):
t=[]
for j in range(n):
t.append(a[j]%m)
t=set(t)
t=list(t)
t.sort()
if(len(t)==2):
if((t[0]==0 and t[1]==1 )or (t[0]==0 and t[1]==m-1)):
return 1
return 0
n=int(input())
a=list(map(int,input().split()))
m=min(a)
d=[]
j=1
val=2
while(j*j<=m):
if(m%j==0):
d.append(j)
d.append(j+1)
d.append(j-1)
if(m//j!=j):
d.append(m//j)
d.append((m//j)+1)
d.append((m//j)-1)
j+=1
d=set(d)
d=list(d)
d.sort()
#print(d)
if(d[0]==0):
d.pop(0)
#print(d)
for j in range(len(d)-1,-1,-1):
#print(check(n,a,d[j]))
if(check(n,a,d[j])==1):
val=d[j]
break
#print(val)
count=0
for j in range(n):
if(a[j]%val==0 or a[j]%val==1):
count+=a[j]//val
elif(a[j]%val==val-1):
count+=math.ceil(a[j]/val)
print(count)
```
No
| 9,573 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There are n boxes with colored balls on the table. Colors are numbered from 1 to n. i-th box contains ai balls, all of which have color i. You have to write a program that will divide all balls into sets such that:
* each ball belongs to exactly one of the sets,
* there are no empty sets,
* there is no set containing two (or more) balls of different colors (each set contains only balls of one color),
* there are no two sets such that the difference between their sizes is greater than 1.
Print the minimum possible number of sets.
Input
The first line contains one integer number n (1 β€ n β€ 500).
The second line contains n integer numbers a1, a2, ... , an (1 β€ ai β€ 109).
Output
Print one integer number β the minimum possible number of sets.
Examples
Input
3
4 7 8
Output
5
Input
2
2 7
Output
4
Note
In the first example the balls can be divided into sets like that: one set with 4 balls of the first color, two sets with 3 and 4 balls, respectively, of the second color, and two sets with 4 balls of the third color.
Submitted Solution:
```
import math,string,itertools,fractions,heapq,collections,re,array,bisect,sys,random,time
sys.setrecursionlimit(10**7)
inf = 10**20
mod = 10**9 + 7
def LI(): return [int(x) for x in sys.stdin.readline().split()]
def LF(): return [float(x) for x in sys.stdin.readline().split()]
def LS(): return sys.stdin.readline().split()
def I(): return int(sys.stdin.readline())
def F(): return float(sys.stdin.readline())
def S(): return input()
def main():
n = I()
a = LI()
def f(t):
for c in a:
if c < t:
return False
if c % (t+1) == 0:
continue
d = c//t
e = c - t*d
if d<e:
return False
return True
ma = min(a)
t = 1
for i in range(1,ma+1):
if f(ma//i+1):
t = ma//i+1
break
if f(ma//i):
t = ma//i
break
r = 0
for c in a:
if c % (t+1) == 0:
r += c//(t+1)
continue
r += c//t
return r
print(main())
```
No
| 9,574 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There are n boxes with colored balls on the table. Colors are numbered from 1 to n. i-th box contains ai balls, all of which have color i. You have to write a program that will divide all balls into sets such that:
* each ball belongs to exactly one of the sets,
* there are no empty sets,
* there is no set containing two (or more) balls of different colors (each set contains only balls of one color),
* there are no two sets such that the difference between their sizes is greater than 1.
Print the minimum possible number of sets.
Input
The first line contains one integer number n (1 β€ n β€ 500).
The second line contains n integer numbers a1, a2, ... , an (1 β€ ai β€ 109).
Output
Print one integer number β the minimum possible number of sets.
Examples
Input
3
4 7 8
Output
5
Input
2
2 7
Output
4
Note
In the first example the balls can be divided into sets like that: one set with 4 balls of the first color, two sets with 3 and 4 balls, respectively, of the second color, and two sets with 4 balls of the third color.
Submitted Solution:
```
import math,string,itertools,fractions,heapq,collections,re,array,bisect,sys,random,time
sys.setrecursionlimit(10**7)
inf = 10**20
mod = 10**9 + 7
def LI(): return [int(x) for x in sys.stdin.readline().split()]
def LF(): return [float(x) for x in sys.stdin.readline().split()]
def LS(): return sys.stdin.readline().split()
def I(): return int(sys.stdin.readline())
def F(): return float(sys.stdin.readline())
def S(): return input()
def main():
n = I()
a = LI()
def f(t):
for c in a:
if c == t-1:
continue
d = c//t
e = c - t*d
if d<e:
return False
return True
ma = min(a)
t = 1
if f(ma) :
t = ma
else:
for i in range(2,ma+1):
if ma%i == 0 and f(ma//i+1):
t = ma//i+1
break
if f(ma//i):
t = ma//i
break
r = 0
for c in a:
if c % (t+1) == 0:
r += c//(t+1)
continue
r += c//t
return r
print(main())
```
No
| 9,575 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There are n boxes with colored balls on the table. Colors are numbered from 1 to n. i-th box contains ai balls, all of which have color i. You have to write a program that will divide all balls into sets such that:
* each ball belongs to exactly one of the sets,
* there are no empty sets,
* there is no set containing two (or more) balls of different colors (each set contains only balls of one color),
* there are no two sets such that the difference between their sizes is greater than 1.
Print the minimum possible number of sets.
Input
The first line contains one integer number n (1 β€ n β€ 500).
The second line contains n integer numbers a1, a2, ... , an (1 β€ ai β€ 109).
Output
Print one integer number β the minimum possible number of sets.
Examples
Input
3
4 7 8
Output
5
Input
2
2 7
Output
4
Note
In the first example the balls can be divided into sets like that: one set with 4 balls of the first color, two sets with 3 and 4 balls, respectively, of the second color, and two sets with 4 balls of the third color.
Submitted Solution:
```
n = int(input())
a = input().split(" ")
min_color = int(1000000000)
for i in range(1,n):
if min(int(a[i]), int(a[i-1])) < min_color:
min_color = min(int(a[i]), int(a[i-1]))
all_figure = int(0)
for i in range(n):
all_figure += int(a[i])
if (all_figure % min_color + 1) == min_color and min_color != 2:
print(all_figure // min_color + 1)
else:
print(all_figure // min_color)
```
No
| 9,576 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Unlucky year in Berland is such a year that its number n can be represented as n = xa + yb, where a and b are non-negative integer numbers.
For example, if x = 2 and y = 3 then the years 4 and 17 are unlucky (4 = 20 + 31, 17 = 23 + 32 = 24 + 30) and year 18 isn't unlucky as there is no such representation for it.
Such interval of years that there are no unlucky years in it is called The Golden Age.
You should write a program which will find maximum length of The Golden Age which starts no earlier than the year l and ends no later than the year r. If all years in the interval [l, r] are unlucky then the answer is 0.
Input
The first line contains four integer numbers x, y, l and r (2 β€ x, y β€ 1018, 1 β€ l β€ r β€ 1018).
Output
Print the maximum length of The Golden Age within the interval [l, r].
If all years in the interval [l, r] are unlucky then print 0.
Examples
Input
2 3 1 10
Output
1
Input
3 5 10 22
Output
8
Input
2 3 3 5
Output
0
Note
In the first example the unlucky years are 2, 3, 4, 5, 7, 9 and 10. So maximum length of The Golden Age is achived in the intervals [1, 1], [6, 6] and [8, 8].
In the second example the longest Golden Age is the interval [15, 22].
Tags: brute force, math
Correct Solution:
```
p = range(61)
x,y,l,r = map(int,input().split())
a = [l-1,r+1]
for i in [ x**i for i in p if x**i <= r]:
for j in [ y**i for i in p if y**i <= r-i]:
if i+j >= l and i+j <= r:
a.append(i+j)
a.sort()
ans = 0
for i in range(1,len(a)):
ans = max(ans,a[i] - a[i-1] -1)
print(ans)
```
| 9,577 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Unlucky year in Berland is such a year that its number n can be represented as n = xa + yb, where a and b are non-negative integer numbers.
For example, if x = 2 and y = 3 then the years 4 and 17 are unlucky (4 = 20 + 31, 17 = 23 + 32 = 24 + 30) and year 18 isn't unlucky as there is no such representation for it.
Such interval of years that there are no unlucky years in it is called The Golden Age.
You should write a program which will find maximum length of The Golden Age which starts no earlier than the year l and ends no later than the year r. If all years in the interval [l, r] are unlucky then the answer is 0.
Input
The first line contains four integer numbers x, y, l and r (2 β€ x, y β€ 1018, 1 β€ l β€ r β€ 1018).
Output
Print the maximum length of The Golden Age within the interval [l, r].
If all years in the interval [l, r] are unlucky then print 0.
Examples
Input
2 3 1 10
Output
1
Input
3 5 10 22
Output
8
Input
2 3 3 5
Output
0
Note
In the first example the unlucky years are 2, 3, 4, 5, 7, 9 and 10. So maximum length of The Golden Age is achived in the intervals [1, 1], [6, 6] and [8, 8].
In the second example the longest Golden Age is the interval [15, 22].
Tags: brute force, math
Correct Solution:
```
from math import log, ceil
x, y, l, r = [int(i) for i in input().split()]
ans = 0
nm = ceil(log(r, x)) + 2
mm = ceil(log(r, y)) + 2
v = [l - 1, r + 1]
for n in range(nm):
for m in range(mm):
cur = x ** n + y ** m
if cur < l or cur > r:
continue
v.append(cur)
v.sort()
for i in range(1, len(v)):
ans = max(ans, v[i] - v[i - 1] - 1)
print(ans)
```
| 9,578 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Unlucky year in Berland is such a year that its number n can be represented as n = xa + yb, where a and b are non-negative integer numbers.
For example, if x = 2 and y = 3 then the years 4 and 17 are unlucky (4 = 20 + 31, 17 = 23 + 32 = 24 + 30) and year 18 isn't unlucky as there is no such representation for it.
Such interval of years that there are no unlucky years in it is called The Golden Age.
You should write a program which will find maximum length of The Golden Age which starts no earlier than the year l and ends no later than the year r. If all years in the interval [l, r] are unlucky then the answer is 0.
Input
The first line contains four integer numbers x, y, l and r (2 β€ x, y β€ 1018, 1 β€ l β€ r β€ 1018).
Output
Print the maximum length of The Golden Age within the interval [l, r].
If all years in the interval [l, r] are unlucky then print 0.
Examples
Input
2 3 1 10
Output
1
Input
3 5 10 22
Output
8
Input
2 3 3 5
Output
0
Note
In the first example the unlucky years are 2, 3, 4, 5, 7, 9 and 10. So maximum length of The Golden Age is achived in the intervals [1, 1], [6, 6] and [8, 8].
In the second example the longest Golden Age is the interval [15, 22].
Tags: brute force, math
Correct Solution:
```
x,y,lo,h = list(map(int,input().strip().split(' ')))
p = 1
q = 1
l = []
l.append(lo-1)
l.append(h+1)
while p<h :
q = 1
while q < h :
if lo<= p+q and p+q<= h :
l.append(p+q)
q*=y
p*=x
max = 0
l.sort()
for i in range(1,len(l)) :
if max <= l[i]-l[i-1]-1:
max = l[i]-l[i-1]-1
print(max)
```
| 9,579 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Unlucky year in Berland is such a year that its number n can be represented as n = xa + yb, where a and b are non-negative integer numbers.
For example, if x = 2 and y = 3 then the years 4 and 17 are unlucky (4 = 20 + 31, 17 = 23 + 32 = 24 + 30) and year 18 isn't unlucky as there is no such representation for it.
Such interval of years that there are no unlucky years in it is called The Golden Age.
You should write a program which will find maximum length of The Golden Age which starts no earlier than the year l and ends no later than the year r. If all years in the interval [l, r] are unlucky then the answer is 0.
Input
The first line contains four integer numbers x, y, l and r (2 β€ x, y β€ 1018, 1 β€ l β€ r β€ 1018).
Output
Print the maximum length of The Golden Age within the interval [l, r].
If all years in the interval [l, r] are unlucky then print 0.
Examples
Input
2 3 1 10
Output
1
Input
3 5 10 22
Output
8
Input
2 3 3 5
Output
0
Note
In the first example the unlucky years are 2, 3, 4, 5, 7, 9 and 10. So maximum length of The Golden Age is achived in the intervals [1, 1], [6, 6] and [8, 8].
In the second example the longest Golden Age is the interval [15, 22].
Tags: brute force, math
Correct Solution:
```
# Contest: Educational Codeforces Round 22 (https://codeforces.com/contest/813)
# Problem: B: The Golden Age (https://codeforces.com/contest/813/problem/B)
def rint():
return int(input())
def rints():
return list(map(int, input().split()))
x, y, l, r = rints()
li = [l - 1, r + 1]
for i in range(65):
for j in range(65):
s = x**i + y**j
if l <= s <= r:
li.append(s)
li = sorted(li)
mx = 0
for i in range(len(li) - 1):
mx = max(mx, li[i + 1] - li[i] - 1)
print(mx)
```
| 9,580 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Unlucky year in Berland is such a year that its number n can be represented as n = xa + yb, where a and b are non-negative integer numbers.
For example, if x = 2 and y = 3 then the years 4 and 17 are unlucky (4 = 20 + 31, 17 = 23 + 32 = 24 + 30) and year 18 isn't unlucky as there is no such representation for it.
Such interval of years that there are no unlucky years in it is called The Golden Age.
You should write a program which will find maximum length of The Golden Age which starts no earlier than the year l and ends no later than the year r. If all years in the interval [l, r] are unlucky then the answer is 0.
Input
The first line contains four integer numbers x, y, l and r (2 β€ x, y β€ 1018, 1 β€ l β€ r β€ 1018).
Output
Print the maximum length of The Golden Age within the interval [l, r].
If all years in the interval [l, r] are unlucky then print 0.
Examples
Input
2 3 1 10
Output
1
Input
3 5 10 22
Output
8
Input
2 3 3 5
Output
0
Note
In the first example the unlucky years are 2, 3, 4, 5, 7, 9 and 10. So maximum length of The Golden Age is achived in the intervals [1, 1], [6, 6] and [8, 8].
In the second example the longest Golden Age is the interval [15, 22].
Tags: brute force, math
Correct Solution:
```
x, y, l, r = list(map(int, input().strip().split(" ")))
cx = 1
cy = 1
v = []
while True:
if cx > r:
break
cy = 1
while True:
if cx + cy > r:
break
if cx + cy >= l:
v.append(cx + cy)
cy = cy * y
cx = cx * x
v.append(l-1)
v.append(r+1)
v.sort()
vd = []
res = 0
for i in range(len(v)-1):
if v[i] == v[i+1]:
continue
res = max(res, v[i+1] - v[i] - 1)
print(res)
```
| 9,581 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Unlucky year in Berland is such a year that its number n can be represented as n = xa + yb, where a and b are non-negative integer numbers.
For example, if x = 2 and y = 3 then the years 4 and 17 are unlucky (4 = 20 + 31, 17 = 23 + 32 = 24 + 30) and year 18 isn't unlucky as there is no such representation for it.
Such interval of years that there are no unlucky years in it is called The Golden Age.
You should write a program which will find maximum length of The Golden Age which starts no earlier than the year l and ends no later than the year r. If all years in the interval [l, r] are unlucky then the answer is 0.
Input
The first line contains four integer numbers x, y, l and r (2 β€ x, y β€ 1018, 1 β€ l β€ r β€ 1018).
Output
Print the maximum length of The Golden Age within the interval [l, r].
If all years in the interval [l, r] are unlucky then print 0.
Examples
Input
2 3 1 10
Output
1
Input
3 5 10 22
Output
8
Input
2 3 3 5
Output
0
Note
In the first example the unlucky years are 2, 3, 4, 5, 7, 9 and 10. So maximum length of The Golden Age is achived in the intervals [1, 1], [6, 6] and [8, 8].
In the second example the longest Golden Age is the interval [15, 22].
Tags: brute force, math
Correct Solution:
```
x,y,l,r=map(int,input().split())
lst = []
pwdX = 1
pwdY = 1
for i in range(0,65):
pwdY = 1
for j in range(0,65):
if pwdX + pwdY > r:
lst.append(pwdX + pwdY)
break
toAdd = pwdX + pwdY
lst.append(toAdd)
pwdY*=y
pwdX*=x
lst.append(r+1)
lst.append(l-1)
# print(lst)
# sorted(lst)
lst.sort()
maximum = 0
for i in range(0,len(lst)-1):
if lst[i] >= l-1 and lst[i+1] <= r+1:
maximum = max(maximum,lst[i+1]-lst[i] - 1)
print(maximum)
```
| 9,582 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Unlucky year in Berland is such a year that its number n can be represented as n = xa + yb, where a and b are non-negative integer numbers.
For example, if x = 2 and y = 3 then the years 4 and 17 are unlucky (4 = 20 + 31, 17 = 23 + 32 = 24 + 30) and year 18 isn't unlucky as there is no such representation for it.
Such interval of years that there are no unlucky years in it is called The Golden Age.
You should write a program which will find maximum length of The Golden Age which starts no earlier than the year l and ends no later than the year r. If all years in the interval [l, r] are unlucky then the answer is 0.
Input
The first line contains four integer numbers x, y, l and r (2 β€ x, y β€ 1018, 1 β€ l β€ r β€ 1018).
Output
Print the maximum length of The Golden Age within the interval [l, r].
If all years in the interval [l, r] are unlucky then print 0.
Examples
Input
2 3 1 10
Output
1
Input
3 5 10 22
Output
8
Input
2 3 3 5
Output
0
Note
In the first example the unlucky years are 2, 3, 4, 5, 7, 9 and 10. So maximum length of The Golden Age is achived in the intervals [1, 1], [6, 6] and [8, 8].
In the second example the longest Golden Age is the interval [15, 22].
Tags: brute force, math
Correct Solution:
```
x, y, l, r = map(int, input().split())
t = []
a = 1
while a <= r:
b = 1
while a + b <= r:
if a + b >= l: t.append(a + b)
b *= y
a *= x
t.sort()
print(max(y - x for x, y in zip([l - 1] + t, t + [r + 1])) - 1)
```
| 9,583 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Unlucky year in Berland is such a year that its number n can be represented as n = xa + yb, where a and b are non-negative integer numbers.
For example, if x = 2 and y = 3 then the years 4 and 17 are unlucky (4 = 20 + 31, 17 = 23 + 32 = 24 + 30) and year 18 isn't unlucky as there is no such representation for it.
Such interval of years that there are no unlucky years in it is called The Golden Age.
You should write a program which will find maximum length of The Golden Age which starts no earlier than the year l and ends no later than the year r. If all years in the interval [l, r] are unlucky then the answer is 0.
Input
The first line contains four integer numbers x, y, l and r (2 β€ x, y β€ 1018, 1 β€ l β€ r β€ 1018).
Output
Print the maximum length of The Golden Age within the interval [l, r].
If all years in the interval [l, r] are unlucky then print 0.
Examples
Input
2 3 1 10
Output
1
Input
3 5 10 22
Output
8
Input
2 3 3 5
Output
0
Note
In the first example the unlucky years are 2, 3, 4, 5, 7, 9 and 10. So maximum length of The Golden Age is achived in the intervals [1, 1], [6, 6] and [8, 8].
In the second example the longest Golden Age is the interval [15, 22].
Tags: brute force, math
Correct Solution:
```
import math
import sys
a = list(map(int,input().split()))
x = a[0]
y = a[1]
l = a[2]
r = a[3]
v = []
for i in range(0, 61):
for j in range(0, 61):
if (x ** i + y ** j >= l) and (x ** i + y ** j <= r):
v.append(x ** i + y ** j)
maxi = 0
v.sort()
if len(v) >= 2:
for i in range(1, len(v)):
maxi = max(maxi, v[i] - v[i - 1] - 1)
if len(v) >= 1 and (not v.count(l)):
maxi = max(maxi, v[0] - l)
if len(v) >= 1 and (not v.count(r)):
maxi = max(maxi, r - v[len(v) - 1])
if len(v) == 0 and (not (v.count(l))) and (not (v.count(r))):
maxi = max(maxi, r - l + 1)
print(maxi)
```
| 9,584 |
Provide tags and a correct Python 2 solution for this coding contest problem.
Unlucky year in Berland is such a year that its number n can be represented as n = xa + yb, where a and b are non-negative integer numbers.
For example, if x = 2 and y = 3 then the years 4 and 17 are unlucky (4 = 20 + 31, 17 = 23 + 32 = 24 + 30) and year 18 isn't unlucky as there is no such representation for it.
Such interval of years that there are no unlucky years in it is called The Golden Age.
You should write a program which will find maximum length of The Golden Age which starts no earlier than the year l and ends no later than the year r. If all years in the interval [l, r] are unlucky then the answer is 0.
Input
The first line contains four integer numbers x, y, l and r (2 β€ x, y β€ 1018, 1 β€ l β€ r β€ 1018).
Output
Print the maximum length of The Golden Age within the interval [l, r].
If all years in the interval [l, r] are unlucky then print 0.
Examples
Input
2 3 1 10
Output
1
Input
3 5 10 22
Output
8
Input
2 3 3 5
Output
0
Note
In the first example the unlucky years are 2, 3, 4, 5, 7, 9 and 10. So maximum length of The Golden Age is achived in the intervals [1, 1], [6, 6] and [8, 8].
In the second example the longest Golden Age is the interval [15, 22].
Tags: brute force, math
Correct Solution:
```
from sys import stdin, stdout
from collections import Counter, defaultdict
from itertools import permutations, combinations
raw_input = stdin.readline
pr = stdout.write
def in_num():
return int(raw_input())
def in_arr():
return map(int,raw_input().split())
def pr_num(n):
stdout.write(str(n)+'\n')
def pr_arr(arr):
pr(' '.join(map(str,arr))+'\n')
# fast read function for total integer input
def inp():
# this function returns whole input of
# space/line seperated integers
# Use Ctrl+D to flush stdin.
return map(int,stdin.read().split())
range = xrange # not for python 3.0+
# main code
x,y,l,r=in_arr()
mx=10**18
p1=1
temp=set()
temp.add(0)
temp.add(mx+1)
while p1<=mx:
p2=1
while p2<=mx:
temp.add(p1+p2)
p2*=y
p1*=x
temp=list(temp)
temp.sort()
ans=0
n=len(temp)
l-=1
r+=1
for i in range(n-1):
val=max(0,min(r,temp[i+1])-max(l,temp[i])-1)
ans=max(ans,val)
pr_num(ans)
```
| 9,585 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Unlucky year in Berland is such a year that its number n can be represented as n = xa + yb, where a and b are non-negative integer numbers.
For example, if x = 2 and y = 3 then the years 4 and 17 are unlucky (4 = 20 + 31, 17 = 23 + 32 = 24 + 30) and year 18 isn't unlucky as there is no such representation for it.
Such interval of years that there are no unlucky years in it is called The Golden Age.
You should write a program which will find maximum length of The Golden Age which starts no earlier than the year l and ends no later than the year r. If all years in the interval [l, r] are unlucky then the answer is 0.
Input
The first line contains four integer numbers x, y, l and r (2 β€ x, y β€ 1018, 1 β€ l β€ r β€ 1018).
Output
Print the maximum length of The Golden Age within the interval [l, r].
If all years in the interval [l, r] are unlucky then print 0.
Examples
Input
2 3 1 10
Output
1
Input
3 5 10 22
Output
8
Input
2 3 3 5
Output
0
Note
In the first example the unlucky years are 2, 3, 4, 5, 7, 9 and 10. So maximum length of The Golden Age is achived in the intervals [1, 1], [6, 6] and [8, 8].
In the second example the longest Golden Age is the interval [15, 22].
Submitted Solution:
```
x,y,l,r = map(int,input().split())
a = []
for i in range(100):
for j in range(100):
t = pow(x,i)+pow(y,j)
if(t>r):
break
if(l <= t and t <= r):
a.append(t)
a.sort()
res = 0
for i in range(len(a)-1):
if(res < a[i+1] - a[i] - 1):
res = a[i+1]-a[i]-1
if(len(a)>0):
if(a[0] - l > res):
res = a[0]-l
if(r - a[-1] > res):
res = r - a[-1]
if(len(a) == 0):
res = r - l + 1
print(res)
```
Yes
| 9,586 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Unlucky year in Berland is such a year that its number n can be represented as n = xa + yb, where a and b are non-negative integer numbers.
For example, if x = 2 and y = 3 then the years 4 and 17 are unlucky (4 = 20 + 31, 17 = 23 + 32 = 24 + 30) and year 18 isn't unlucky as there is no such representation for it.
Such interval of years that there are no unlucky years in it is called The Golden Age.
You should write a program which will find maximum length of The Golden Age which starts no earlier than the year l and ends no later than the year r. If all years in the interval [l, r] are unlucky then the answer is 0.
Input
The first line contains four integer numbers x, y, l and r (2 β€ x, y β€ 1018, 1 β€ l β€ r β€ 1018).
Output
Print the maximum length of The Golden Age within the interval [l, r].
If all years in the interval [l, r] are unlucky then print 0.
Examples
Input
2 3 1 10
Output
1
Input
3 5 10 22
Output
8
Input
2 3 3 5
Output
0
Note
In the first example the unlucky years are 2, 3, 4, 5, 7, 9 and 10. So maximum length of The Golden Age is achived in the intervals [1, 1], [6, 6] and [8, 8].
In the second example the longest Golden Age is the interval [15, 22].
Submitted Solution:
```
x,y,l,r=map(int,input().split())
p=set()
p.add(l-1)
p.add(r+1)
for i in range(0,80):
s=x**i
if s>r:
break
for j in range(0,80):
t=y**j
if s+t>r:break
elif s+t>=l:
p.add(s+t)
p=sorted(list(p))
#print(p)
ans=0
for i in range(1,len(p)):
ans=max(ans,p[i]-p[i-1]-1)
print (ans)
```
Yes
| 9,587 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Unlucky year in Berland is such a year that its number n can be represented as n = xa + yb, where a and b are non-negative integer numbers.
For example, if x = 2 and y = 3 then the years 4 and 17 are unlucky (4 = 20 + 31, 17 = 23 + 32 = 24 + 30) and year 18 isn't unlucky as there is no such representation for it.
Such interval of years that there are no unlucky years in it is called The Golden Age.
You should write a program which will find maximum length of The Golden Age which starts no earlier than the year l and ends no later than the year r. If all years in the interval [l, r] are unlucky then the answer is 0.
Input
The first line contains four integer numbers x, y, l and r (2 β€ x, y β€ 1018, 1 β€ l β€ r β€ 1018).
Output
Print the maximum length of The Golden Age within the interval [l, r].
If all years in the interval [l, r] are unlucky then print 0.
Examples
Input
2 3 1 10
Output
1
Input
3 5 10 22
Output
8
Input
2 3 3 5
Output
0
Note
In the first example the unlucky years are 2, 3, 4, 5, 7, 9 and 10. So maximum length of The Golden Age is achived in the intervals [1, 1], [6, 6] and [8, 8].
In the second example the longest Golden Age is the interval [15, 22].
Submitted Solution:
```
x,y,l,r = map(int, input().split())
X = 1
st = set()
while (X <= r):
Y = 1
while (X + Y <= r):
if (X + Y >= l):
st.add(X + Y)
Y *= y
X *= x
bad = sorted([x for x in st])
prev = l
ans = 0
for x in bad:
ans = max(ans, x - prev)
prev = x + 1
ans = max(ans, r - prev + 1)
print(ans)
```
Yes
| 9,588 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Unlucky year in Berland is such a year that its number n can be represented as n = xa + yb, where a and b are non-negative integer numbers.
For example, if x = 2 and y = 3 then the years 4 and 17 are unlucky (4 = 20 + 31, 17 = 23 + 32 = 24 + 30) and year 18 isn't unlucky as there is no such representation for it.
Such interval of years that there are no unlucky years in it is called The Golden Age.
You should write a program which will find maximum length of The Golden Age which starts no earlier than the year l and ends no later than the year r. If all years in the interval [l, r] are unlucky then the answer is 0.
Input
The first line contains four integer numbers x, y, l and r (2 β€ x, y β€ 1018, 1 β€ l β€ r β€ 1018).
Output
Print the maximum length of The Golden Age within the interval [l, r].
If all years in the interval [l, r] are unlucky then print 0.
Examples
Input
2 3 1 10
Output
1
Input
3 5 10 22
Output
8
Input
2 3 3 5
Output
0
Note
In the first example the unlucky years are 2, 3, 4, 5, 7, 9 and 10. So maximum length of The Golden Age is achived in the intervals [1, 1], [6, 6] and [8, 8].
In the second example the longest Golden Age is the interval [15, 22].
Submitted Solution:
```
x,y,l,r=[int(i) for i in input().split()]
M=int(1e18)
i=1
ul=[]
while (i<=M):
j=1
while (i+j<=M):
ul.append(i+j)
j*=y
i*=x
ul=[0]+sorted(ul)+[M+1]
n=len(ul)
mx=0
for i in range(1,n):
cur=min(ul[i],r+1)-max(l-1,ul[i-1])-1
mx=max(mx,cur)
print(mx)
```
Yes
| 9,589 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Unlucky year in Berland is such a year that its number n can be represented as n = xa + yb, where a and b are non-negative integer numbers.
For example, if x = 2 and y = 3 then the years 4 and 17 are unlucky (4 = 20 + 31, 17 = 23 + 32 = 24 + 30) and year 18 isn't unlucky as there is no such representation for it.
Such interval of years that there are no unlucky years in it is called The Golden Age.
You should write a program which will find maximum length of The Golden Age which starts no earlier than the year l and ends no later than the year r. If all years in the interval [l, r] are unlucky then the answer is 0.
Input
The first line contains four integer numbers x, y, l and r (2 β€ x, y β€ 1018, 1 β€ l β€ r β€ 1018).
Output
Print the maximum length of The Golden Age within the interval [l, r].
If all years in the interval [l, r] are unlucky then print 0.
Examples
Input
2 3 1 10
Output
1
Input
3 5 10 22
Output
8
Input
2 3 3 5
Output
0
Note
In the first example the unlucky years are 2, 3, 4, 5, 7, 9 and 10. So maximum length of The Golden Age is achived in the intervals [1, 1], [6, 6] and [8, 8].
In the second example the longest Golden Age is the interval [15, 22].
Submitted Solution:
```
from math import log,pow
import enum
x,y,l,r= [int(x) for x in input().split()]
a1 = int(log(l,x))
a2 = int(log(r,x))
s1 = [int(pow(x,a)) for a in range(a1,a2+1)]
print(s1)
b1 = int(log(l,y))
b2 = int(log(r,y))
s2 =[int(pow(y,a)) for a in range(b1,b2+1)]
s3 = []
for b in s1:
for c in s2:
s3.append(b+c)
ans = 0
s3.sort()
print(s3)
for i in range(len(s3)-1):
ans = max(ans, min(r,s3[i+1]-1)-max(l,s3[i]+1) +1)
print(ans)
```
No
| 9,590 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Unlucky year in Berland is such a year that its number n can be represented as n = xa + yb, where a and b are non-negative integer numbers.
For example, if x = 2 and y = 3 then the years 4 and 17 are unlucky (4 = 20 + 31, 17 = 23 + 32 = 24 + 30) and year 18 isn't unlucky as there is no such representation for it.
Such interval of years that there are no unlucky years in it is called The Golden Age.
You should write a program which will find maximum length of The Golden Age which starts no earlier than the year l and ends no later than the year r. If all years in the interval [l, r] are unlucky then the answer is 0.
Input
The first line contains four integer numbers x, y, l and r (2 β€ x, y β€ 1018, 1 β€ l β€ r β€ 1018).
Output
Print the maximum length of The Golden Age within the interval [l, r].
If all years in the interval [l, r] are unlucky then print 0.
Examples
Input
2 3 1 10
Output
1
Input
3 5 10 22
Output
8
Input
2 3 3 5
Output
0
Note
In the first example the unlucky years are 2, 3, 4, 5, 7, 9 and 10. So maximum length of The Golden Age is achived in the intervals [1, 1], [6, 6] and [8, 8].
In the second example the longest Golden Age is the interval [15, 22].
Submitted Solution:
```
import math
(x,y,l,r) = (int(i) for i in input().split())
vvrs = []
vvrs.append(l)
for i in range(60):
for j in range(60):
cur = x**i + y**j
if cur>r: break
if cur>=l:
vvrs.append(cur)
vvrs.append(r)
vvrs.sort()
maxcnt = 0
for i in range(len(vvrs)-1):
rzn = vvrs[i+1]-vvrs[i]
if rzn>maxcnt: maxcnt = rzn
print(maxcnt-1)
```
No
| 9,591 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Unlucky year in Berland is such a year that its number n can be represented as n = xa + yb, where a and b are non-negative integer numbers.
For example, if x = 2 and y = 3 then the years 4 and 17 are unlucky (4 = 20 + 31, 17 = 23 + 32 = 24 + 30) and year 18 isn't unlucky as there is no such representation for it.
Such interval of years that there are no unlucky years in it is called The Golden Age.
You should write a program which will find maximum length of The Golden Age which starts no earlier than the year l and ends no later than the year r. If all years in the interval [l, r] are unlucky then the answer is 0.
Input
The first line contains four integer numbers x, y, l and r (2 β€ x, y β€ 1018, 1 β€ l β€ r β€ 1018).
Output
Print the maximum length of The Golden Age within the interval [l, r].
If all years in the interval [l, r] are unlucky then print 0.
Examples
Input
2 3 1 10
Output
1
Input
3 5 10 22
Output
8
Input
2 3 3 5
Output
0
Note
In the first example the unlucky years are 2, 3, 4, 5, 7, 9 and 10. So maximum length of The Golden Age is achived in the intervals [1, 1], [6, 6] and [8, 8].
In the second example the longest Golden Age is the interval [15, 22].
Submitted Solution:
```
x,y,l,r=map(int,input().split())
p=set()
p.add(l-1)
p.add(r+1)
for i in range(0,65):
s=x**i
if x>r:
break
for j in range(0,65):
t=y**j
if s+t>r:break
elif s+t>=l:
p.add(s+t)
p=sorted(list(p))
#print(p)
ans=0
for i in range(1,len(p)):
ans=max(ans,p[i]-p[i-1]-1)
print (ans)
```
No
| 9,592 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Unlucky year in Berland is such a year that its number n can be represented as n = xa + yb, where a and b are non-negative integer numbers.
For example, if x = 2 and y = 3 then the years 4 and 17 are unlucky (4 = 20 + 31, 17 = 23 + 32 = 24 + 30) and year 18 isn't unlucky as there is no such representation for it.
Such interval of years that there are no unlucky years in it is called The Golden Age.
You should write a program which will find maximum length of The Golden Age which starts no earlier than the year l and ends no later than the year r. If all years in the interval [l, r] are unlucky then the answer is 0.
Input
The first line contains four integer numbers x, y, l and r (2 β€ x, y β€ 1018, 1 β€ l β€ r β€ 1018).
Output
Print the maximum length of The Golden Age within the interval [l, r].
If all years in the interval [l, r] are unlucky then print 0.
Examples
Input
2 3 1 10
Output
1
Input
3 5 10 22
Output
8
Input
2 3 3 5
Output
0
Note
In the first example the unlucky years are 2, 3, 4, 5, 7, 9 and 10. So maximum length of The Golden Age is achived in the intervals [1, 1], [6, 6] and [8, 8].
In the second example the longest Golden Age is the interval [15, 22].
Submitted Solution:
```
s = [int(x) for x in input().split()]
x = s[0]
y = s[1]
l = s[2]
r = s[3]
v = [l - 1, r + 1]
xx = 1
yy = 1
for i in range(64):
yy = 1
for j in range(64):
if(xx + yy <= r):
v.append(xx + yy)
yy = y * yy
else: break
xx = xx * x;
v = sorted(v)
ans = 0
for i in range(len(v)):
ans = max(ans, v[i] - v[i - 1] - 1)
print(ans)
# your code goes here
```
No
| 9,593 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Bran and his older sister Arya are from the same house. Bran like candies so much, so Arya is going to give him some Candies.
At first, Arya and Bran have 0 Candies. There are n days, at the i-th day, Arya finds ai candies in a box, that is given by the Many-Faced God. Every day she can give Bran at most 8 of her candies. If she don't give him the candies at the same day, they are saved for her and she can give them to him later.
Your task is to find the minimum number of days Arya needs to give Bran k candies before the end of the n-th day. Formally, you need to output the minimum day index to the end of which k candies will be given out (the days are indexed from 1 to n).
Print -1 if she can't give him k candies during n given days.
Input
The first line contains two integers n and k (1 β€ n β€ 100, 1 β€ k β€ 10000).
The second line contains n integers a1, a2, a3, ..., an (1 β€ ai β€ 100).
Output
If it is impossible for Arya to give Bran k candies within n days, print -1.
Otherwise print a single integer β the minimum number of days Arya needs to give Bran k candies before the end of the n-th day.
Examples
Input
2 3
1 2
Output
2
Input
3 17
10 10 10
Output
3
Input
1 9
10
Output
-1
Note
In the first sample, Arya can give Bran 3 candies in 2 days.
In the second sample, Arya can give Bran 17 candies in 3 days, because she can give him at most 8 candies per day.
In the third sample, Arya can't give Bran 9 candies, because she can give him at most 8 candies per day and she must give him the candies within 1 day.
Tags: implementation
Correct Solution:
```
n,k=map(int,input().split())
a=list(map(int,input().split()))
summ=0
for j in range(n):
summ+=a[j]
if n*8>=k and k<=summ:
carramel=0
for i in range(n):
carramel+=a[i]
if carramel>8:
k-=8
carramel-=8
else:
k-=carramel
carramel=0
if k<=0:
days=i+1
print(days)
break
if k>0:
print('-1')
else:
print('-1')
```
| 9,594 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Bran and his older sister Arya are from the same house. Bran like candies so much, so Arya is going to give him some Candies.
At first, Arya and Bran have 0 Candies. There are n days, at the i-th day, Arya finds ai candies in a box, that is given by the Many-Faced God. Every day she can give Bran at most 8 of her candies. If she don't give him the candies at the same day, they are saved for her and she can give them to him later.
Your task is to find the minimum number of days Arya needs to give Bran k candies before the end of the n-th day. Formally, you need to output the minimum day index to the end of which k candies will be given out (the days are indexed from 1 to n).
Print -1 if she can't give him k candies during n given days.
Input
The first line contains two integers n and k (1 β€ n β€ 100, 1 β€ k β€ 10000).
The second line contains n integers a1, a2, a3, ..., an (1 β€ ai β€ 100).
Output
If it is impossible for Arya to give Bran k candies within n days, print -1.
Otherwise print a single integer β the minimum number of days Arya needs to give Bran k candies before the end of the n-th day.
Examples
Input
2 3
1 2
Output
2
Input
3 17
10 10 10
Output
3
Input
1 9
10
Output
-1
Note
In the first sample, Arya can give Bran 3 candies in 2 days.
In the second sample, Arya can give Bran 17 candies in 3 days, because she can give him at most 8 candies per day.
In the third sample, Arya can't give Bran 9 candies, because she can give him at most 8 candies per day and she must give him the candies within 1 day.
Tags: implementation
Correct Solution:
```
n, k = map(int, input().split())
l = list(map(int, input().split()))
s = 0
for i in range(n):
s += l[i]
k -= min(s, 8)
s -= min(s, 8)
if k <= 0:
print(i + 1)
exit(0)
print(-1)
```
| 9,595 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Bran and his older sister Arya are from the same house. Bran like candies so much, so Arya is going to give him some Candies.
At first, Arya and Bran have 0 Candies. There are n days, at the i-th day, Arya finds ai candies in a box, that is given by the Many-Faced God. Every day she can give Bran at most 8 of her candies. If she don't give him the candies at the same day, they are saved for her and she can give them to him later.
Your task is to find the minimum number of days Arya needs to give Bran k candies before the end of the n-th day. Formally, you need to output the minimum day index to the end of which k candies will be given out (the days are indexed from 1 to n).
Print -1 if she can't give him k candies during n given days.
Input
The first line contains two integers n and k (1 β€ n β€ 100, 1 β€ k β€ 10000).
The second line contains n integers a1, a2, a3, ..., an (1 β€ ai β€ 100).
Output
If it is impossible for Arya to give Bran k candies within n days, print -1.
Otherwise print a single integer β the minimum number of days Arya needs to give Bran k candies before the end of the n-th day.
Examples
Input
2 3
1 2
Output
2
Input
3 17
10 10 10
Output
3
Input
1 9
10
Output
-1
Note
In the first sample, Arya can give Bran 3 candies in 2 days.
In the second sample, Arya can give Bran 17 candies in 3 days, because she can give him at most 8 candies per day.
In the third sample, Arya can't give Bran 9 candies, because she can give him at most 8 candies per day and she must give him the candies within 1 day.
Tags: implementation
Correct Solution:
```
n, k = map(int, input().split())
l = list(map(int, input().split()))
have = 0
for i in range(0, n):
have+=l[i]
x= min(8, have)
k -=x
have-=x
if k<=0:
print(i+1)
exit()
print(-1)
```
| 9,596 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Bran and his older sister Arya are from the same house. Bran like candies so much, so Arya is going to give him some Candies.
At first, Arya and Bran have 0 Candies. There are n days, at the i-th day, Arya finds ai candies in a box, that is given by the Many-Faced God. Every day she can give Bran at most 8 of her candies. If she don't give him the candies at the same day, they are saved for her and she can give them to him later.
Your task is to find the minimum number of days Arya needs to give Bran k candies before the end of the n-th day. Formally, you need to output the minimum day index to the end of which k candies will be given out (the days are indexed from 1 to n).
Print -1 if she can't give him k candies during n given days.
Input
The first line contains two integers n and k (1 β€ n β€ 100, 1 β€ k β€ 10000).
The second line contains n integers a1, a2, a3, ..., an (1 β€ ai β€ 100).
Output
If it is impossible for Arya to give Bran k candies within n days, print -1.
Otherwise print a single integer β the minimum number of days Arya needs to give Bran k candies before the end of the n-th day.
Examples
Input
2 3
1 2
Output
2
Input
3 17
10 10 10
Output
3
Input
1 9
10
Output
-1
Note
In the first sample, Arya can give Bran 3 candies in 2 days.
In the second sample, Arya can give Bran 17 candies in 3 days, because she can give him at most 8 candies per day.
In the third sample, Arya can't give Bran 9 candies, because she can give him at most 8 candies per day and she must give him the candies within 1 day.
Tags: implementation
Correct Solution:
```
n,k = list(map(int, input().rstrip().split()))
days = list(map(int, input().rstrip().split()))
r = 0
for i, a in enumerate(days):
if a + r > 8:
r += a - 8
k -= 8
else:
k -= a + r
r = 0;
if k <= 0:
print(i + 1)
break
else:
print(-1)
```
| 9,597 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Bran and his older sister Arya are from the same house. Bran like candies so much, so Arya is going to give him some Candies.
At first, Arya and Bran have 0 Candies. There are n days, at the i-th day, Arya finds ai candies in a box, that is given by the Many-Faced God. Every day she can give Bran at most 8 of her candies. If she don't give him the candies at the same day, they are saved for her and she can give them to him later.
Your task is to find the minimum number of days Arya needs to give Bran k candies before the end of the n-th day. Formally, you need to output the minimum day index to the end of which k candies will be given out (the days are indexed from 1 to n).
Print -1 if she can't give him k candies during n given days.
Input
The first line contains two integers n and k (1 β€ n β€ 100, 1 β€ k β€ 10000).
The second line contains n integers a1, a2, a3, ..., an (1 β€ ai β€ 100).
Output
If it is impossible for Arya to give Bran k candies within n days, print -1.
Otherwise print a single integer β the minimum number of days Arya needs to give Bran k candies before the end of the n-th day.
Examples
Input
2 3
1 2
Output
2
Input
3 17
10 10 10
Output
3
Input
1 9
10
Output
-1
Note
In the first sample, Arya can give Bran 3 candies in 2 days.
In the second sample, Arya can give Bran 17 candies in 3 days, because she can give him at most 8 candies per day.
In the third sample, Arya can't give Bran 9 candies, because she can give him at most 8 candies per day and she must give him the candies within 1 day.
Tags: implementation
Correct Solution:
```
import sys
read=lambda:sys.stdin.readline().rstrip()
readi=lambda:int(sys.stdin.readline())
writeln=lambda x:sys.stdout.write(str(x)+"\n")
write=lambda x:sys.stdout.write(x)
N, K = map(int, read().split())
ns = list(map(int, read().split()))
A,B = 0,0
ans = -1
for i in range(N):
A += ns[i]
if A >= 8:
A -= 8
B += 8
else:
B += A
A = 0
if B >= K:
ans = i+1
break
writeln(ans)
```
| 9,598 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Bran and his older sister Arya are from the same house. Bran like candies so much, so Arya is going to give him some Candies.
At first, Arya and Bran have 0 Candies. There are n days, at the i-th day, Arya finds ai candies in a box, that is given by the Many-Faced God. Every day she can give Bran at most 8 of her candies. If she don't give him the candies at the same day, they are saved for her and she can give them to him later.
Your task is to find the minimum number of days Arya needs to give Bran k candies before the end of the n-th day. Formally, you need to output the minimum day index to the end of which k candies will be given out (the days are indexed from 1 to n).
Print -1 if she can't give him k candies during n given days.
Input
The first line contains two integers n and k (1 β€ n β€ 100, 1 β€ k β€ 10000).
The second line contains n integers a1, a2, a3, ..., an (1 β€ ai β€ 100).
Output
If it is impossible for Arya to give Bran k candies within n days, print -1.
Otherwise print a single integer β the minimum number of days Arya needs to give Bran k candies before the end of the n-th day.
Examples
Input
2 3
1 2
Output
2
Input
3 17
10 10 10
Output
3
Input
1 9
10
Output
-1
Note
In the first sample, Arya can give Bran 3 candies in 2 days.
In the second sample, Arya can give Bran 17 candies in 3 days, because she can give him at most 8 candies per day.
In the third sample, Arya can't give Bran 9 candies, because she can give him at most 8 candies per day and she must give him the candies within 1 day.
Tags: implementation
Correct Solution:
```
n,k=list(map(int,input().split()))
a=list(map(int,input().split()))
count=0
ana,f=0,0
ans=0
for val in a:
if ans>=k:
f=1
break
if val+ana<=8:
ans+=val+ana
ana=0
count+=1
else:
ana=(val+ana)-8
count+=1
ans+=8
if f==1 or ans>=k :
print(count)
else:
print(-1)
```
| 9,599 |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.