message stringlengths 2 28.7k | message_type stringclasses 2 values | message_id int64 0 1 | conversation_id int64 21 109k | cluster float64 7 7 | __index_level_0__ int64 42 217k |
|---|---|---|---|---|---|
Provide tags and a correct Python 3 solution for this coding contest problem.
There is an infinite board of square tiles. Initially all tiles are white.
Vova has a red marker and a blue marker. Red marker can color a tiles. Blue marker can color b tiles. If some tile isn't white then you can't use marker of any color on it. Each marker must be drained completely, so at the end there should be exactly a red tiles and exactly b blue tiles across the board.
Vova wants to color such a set of tiles that:
* they would form a rectangle, consisting of exactly a+b colored tiles;
* all tiles of at least one color would also form a rectangle.
Here are some examples of correct colorings:
<image>
Here are some examples of incorrect colorings:
<image>
Among all correct colorings Vova wants to choose the one with the minimal perimeter. What is the minimal perimeter Vova can obtain?
It is guaranteed that there exists at least one correct coloring.
Input
A single line contains two integers a and b (1 β€ a, b β€ 10^{14}) β the number of tiles red marker should color and the number of tiles blue marker should color, respectively.
Output
Print a single integer β the minimal perimeter of a colored rectangle Vova can obtain by coloring exactly a tiles red and exactly b tiles blue.
It is guaranteed that there exists at least one correct coloring.
Examples
Input
4 4
Output
12
Input
3 9
Output
14
Input
9 3
Output
14
Input
3 6
Output
12
Input
506 2708
Output
3218
Note
The first four examples correspond to the first picture of the statement.
Note that for there exist multiple correct colorings for all of the examples.
In the first example you can also make a rectangle with sides 1 and 8, though its perimeter will be 18 which is greater than 8.
In the second example you can make the same resulting rectangle with sides 3 and 4, but red tiles will form the rectangle with sides 1 and 3 and blue tiles will form the rectangle with sides 3 and 3. | instruction | 0 | 69,826 | 7 | 139,652 |
Tags: binary search, brute force, math, number theory
Correct Solution:
```
import bisect
import sys
input=sys.stdin.readline
a,b=map(int,input().split())
su=a+b
f1=[]
for i in range(1,int(su**(1/2))+1):
if(su%i==0):
f1.append(i)
ans=float('inf')
f2=[]
for i in range(1,int(a**(1/2))+1):
if(a%i==0):
f2.append(i)
for i in f1:
xx=bisect.bisect(f2,i)
if(xx!=0):
d2=f2[xx-1]
if(su//i>=a//d2):
ans=min(ans,2*(i+su//i))
a,b=b,a
f2=[]
for i in range(1,int(a**(1/2))+1):
if(a%i==0):
f2.append(i)
for i in f1:
xx=bisect.bisect(f2,i)
if(xx!=0):
d2=f2[xx-1]
if(su//i>=a//d2):
ans=min(ans,2*(i+su//i))
print(ans)
``` | output | 1 | 69,826 | 7 | 139,653 |
Provide tags and a correct Python 3 solution for this coding contest problem.
There is an infinite board of square tiles. Initially all tiles are white.
Vova has a red marker and a blue marker. Red marker can color a tiles. Blue marker can color b tiles. If some tile isn't white then you can't use marker of any color on it. Each marker must be drained completely, so at the end there should be exactly a red tiles and exactly b blue tiles across the board.
Vova wants to color such a set of tiles that:
* they would form a rectangle, consisting of exactly a+b colored tiles;
* all tiles of at least one color would also form a rectangle.
Here are some examples of correct colorings:
<image>
Here are some examples of incorrect colorings:
<image>
Among all correct colorings Vova wants to choose the one with the minimal perimeter. What is the minimal perimeter Vova can obtain?
It is guaranteed that there exists at least one correct coloring.
Input
A single line contains two integers a and b (1 β€ a, b β€ 10^{14}) β the number of tiles red marker should color and the number of tiles blue marker should color, respectively.
Output
Print a single integer β the minimal perimeter of a colored rectangle Vova can obtain by coloring exactly a tiles red and exactly b tiles blue.
It is guaranteed that there exists at least one correct coloring.
Examples
Input
4 4
Output
12
Input
3 9
Output
14
Input
9 3
Output
14
Input
3 6
Output
12
Input
506 2708
Output
3218
Note
The first four examples correspond to the first picture of the statement.
Note that for there exist multiple correct colorings for all of the examples.
In the first example you can also make a rectangle with sides 1 and 8, though its perimeter will be 18 which is greater than 8.
In the second example you can make the same resulting rectangle with sides 3 and 4, but red tiles will form the rectangle with sides 1 and 3 and blue tiles will form the rectangle with sides 3 and 3. | instruction | 0 | 69,827 | 7 | 139,654 |
Tags: binary search, brute force, math, number theory
Correct Solution:
```
from math import sqrt
a, b = map(int, input().split())
if a > b:
a, b = b, a
u = []
sqa = int(sqrt(a))
sqb = int(sqrt(b))
for i in range(1, sqb + 1):
if i <= sqa and a % i == 0:
u.append((i, a // i))
elif b % i == 0:
u.append((i, b // i))
ok = False
sqab = int(sqrt(a + b))
for i in range(sqab, 0, -1):
if (a + b) % i == 0:
k = (a + b) // i
for j in u:
if i >= j[0] and k >= j[1]:
ok = True
break
if ok:
break
print(2 * (i + k))
``` | output | 1 | 69,827 | 7 | 139,655 |
Provide tags and a correct Python 3 solution for this coding contest problem.
There is an infinite board of square tiles. Initially all tiles are white.
Vova has a red marker and a blue marker. Red marker can color a tiles. Blue marker can color b tiles. If some tile isn't white then you can't use marker of any color on it. Each marker must be drained completely, so at the end there should be exactly a red tiles and exactly b blue tiles across the board.
Vova wants to color such a set of tiles that:
* they would form a rectangle, consisting of exactly a+b colored tiles;
* all tiles of at least one color would also form a rectangle.
Here are some examples of correct colorings:
<image>
Here are some examples of incorrect colorings:
<image>
Among all correct colorings Vova wants to choose the one with the minimal perimeter. What is the minimal perimeter Vova can obtain?
It is guaranteed that there exists at least one correct coloring.
Input
A single line contains two integers a and b (1 β€ a, b β€ 10^{14}) β the number of tiles red marker should color and the number of tiles blue marker should color, respectively.
Output
Print a single integer β the minimal perimeter of a colored rectangle Vova can obtain by coloring exactly a tiles red and exactly b tiles blue.
It is guaranteed that there exists at least one correct coloring.
Examples
Input
4 4
Output
12
Input
3 9
Output
14
Input
9 3
Output
14
Input
3 6
Output
12
Input
506 2708
Output
3218
Note
The first four examples correspond to the first picture of the statement.
Note that for there exist multiple correct colorings for all of the examples.
In the first example you can also make a rectangle with sides 1 and 8, though its perimeter will be 18 which is greater than 8.
In the second example you can make the same resulting rectangle with sides 3 and 4, but red tiles will form the rectangle with sides 1 and 3 and blue tiles will form the rectangle with sides 3 and 3. | instruction | 0 | 69,828 | 7 | 139,656 |
Tags: binary search, brute force, math, number theory
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=300006, 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: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------------------------------------
class TrieNode:
def __init__(self):
self.children = [None] * 26
self.isEndOfWord = False
class Trie:
def __init__(self):
self.root = self.getNode()
def getNode(self):
return TrieNode()
def _charToIndex(self, ch):
return ord(ch) - ord('a')
def insert(self, key):
pCrawl = self.root
length = len(key)
for level in range(length):
index = self._charToIndex(key[level])
if not pCrawl.children[index]:
pCrawl.children[index] = self.getNode()
pCrawl = pCrawl.children[index]
pCrawl.isEndOfWord = True
def search(self, key):
pCrawl = self.root
length = len(key)
for level in range(length):
index = self._charToIndex(key[level])
if not pCrawl.children[index]:
return False
pCrawl = pCrawl.children[index]
return pCrawl != None and pCrawl.isEndOfWord
#-----------------------------------------trie---------------------------------
class Node:
def __init__(self, data):
self.data = data
self.count=0
self.left = None # left node for 0
self.right = None # right node for 1
class BinaryTrie:
def __init__(self):
self.root = Node(0)
def insert(self, pre_xor):
self.temp = self.root
for i in range(31, -1, -1):
val = pre_xor & (1 << i)
if val:
if not self.temp.right:
self.temp.right = Node(0)
self.temp = self.temp.right
self.temp.count+=1
if not val:
if not self.temp.left:
self.temp.left = Node(0)
self.temp = self.temp.left
self.temp.count += 1
self.temp.data = pre_xor
def query(self, xor):
self.temp = self.root
for i in range(31, -1, -1):
val = xor & (1 << i)
if not val:
if self.temp.left and self.temp.left.count>0:
self.temp = self.temp.left
elif self.temp.right:
self.temp = self.temp.right
else:
if self.temp.right and self.temp.right.count>0:
self.temp = self.temp.right
elif self.temp.left:
self.temp = self.temp.left
self.temp.count-=1
return xor ^ self.temp.data
#-------------------------bin trie-------------------------------------------
def find(a):
ans=set()
n=a
for i in range(1,int(a**0.5)+1):
if n%i==0:
ans.add(i)
ans.add(n//i)
ans=sorted(ans)
return ans
a,b=map(int,input().split())
e=find(a+b)
e1=find(a)
e2=find(b)
ans=2*(a+b+1)
for i in range(len(e)):
b1=binarySearchCount(e1,len(e1),e[i])
if b1!=0:
r=e1[b1-1]
if (a+b)//e[i]>=(a//e1[b1-1]):
ans=min(ans,2*(e[i]+(a+b)//e[i]))
b1=binarySearchCount(e2, len(e2), e[i])
if b!=0:
r = e2[b1-1]
if (a+b)//e[i]>=(b//e2[b1-1]):
ans=min(ans,2*(e[i]+(a+b)//e[i]))
print(ans)
``` | output | 1 | 69,828 | 7 | 139,657 |
Provide tags and a correct Python 3 solution for this coding contest problem.
There is an infinite board of square tiles. Initially all tiles are white.
Vova has a red marker and a blue marker. Red marker can color a tiles. Blue marker can color b tiles. If some tile isn't white then you can't use marker of any color on it. Each marker must be drained completely, so at the end there should be exactly a red tiles and exactly b blue tiles across the board.
Vova wants to color such a set of tiles that:
* they would form a rectangle, consisting of exactly a+b colored tiles;
* all tiles of at least one color would also form a rectangle.
Here are some examples of correct colorings:
<image>
Here are some examples of incorrect colorings:
<image>
Among all correct colorings Vova wants to choose the one with the minimal perimeter. What is the minimal perimeter Vova can obtain?
It is guaranteed that there exists at least one correct coloring.
Input
A single line contains two integers a and b (1 β€ a, b β€ 10^{14}) β the number of tiles red marker should color and the number of tiles blue marker should color, respectively.
Output
Print a single integer β the minimal perimeter of a colored rectangle Vova can obtain by coloring exactly a tiles red and exactly b tiles blue.
It is guaranteed that there exists at least one correct coloring.
Examples
Input
4 4
Output
12
Input
3 9
Output
14
Input
9 3
Output
14
Input
3 6
Output
12
Input
506 2708
Output
3218
Note
The first four examples correspond to the first picture of the statement.
Note that for there exist multiple correct colorings for all of the examples.
In the first example you can also make a rectangle with sides 1 and 8, though its perimeter will be 18 which is greater than 8.
In the second example you can make the same resulting rectangle with sides 3 and 4, but red tiles will form the rectangle with sides 1 and 3 and blue tiles will form the rectangle with sides 3 and 3. | instruction | 0 | 69,829 | 7 | 139,658 |
Tags: binary search, brute force, math, number theory
Correct Solution:
```
#autogenerated (:'D)
read_numbers = lambda: map(int, input().split())
INF = 1 << 64
#main
a, b = read_numbers()
ans, n, m = INF, 1, 1
for i in range(1, int((a+b)**0.5 + 1) ):
if a % i == 0: n = i
if b % i == 0: m = i
if (a + b) % i == 0:
second = (a + b) // i
if a // n <= second or b // m <= second:
ans = min(ans, 2*(second + i))
print(ans)
``` | output | 1 | 69,829 | 7 | 139,659 |
Provide tags and a correct Python 3 solution for this coding contest problem.
There is an infinite board of square tiles. Initially all tiles are white.
Vova has a red marker and a blue marker. Red marker can color a tiles. Blue marker can color b tiles. If some tile isn't white then you can't use marker of any color on it. Each marker must be drained completely, so at the end there should be exactly a red tiles and exactly b blue tiles across the board.
Vova wants to color such a set of tiles that:
* they would form a rectangle, consisting of exactly a+b colored tiles;
* all tiles of at least one color would also form a rectangle.
Here are some examples of correct colorings:
<image>
Here are some examples of incorrect colorings:
<image>
Among all correct colorings Vova wants to choose the one with the minimal perimeter. What is the minimal perimeter Vova can obtain?
It is guaranteed that there exists at least one correct coloring.
Input
A single line contains two integers a and b (1 β€ a, b β€ 10^{14}) β the number of tiles red marker should color and the number of tiles blue marker should color, respectively.
Output
Print a single integer β the minimal perimeter of a colored rectangle Vova can obtain by coloring exactly a tiles red and exactly b tiles blue.
It is guaranteed that there exists at least one correct coloring.
Examples
Input
4 4
Output
12
Input
3 9
Output
14
Input
9 3
Output
14
Input
3 6
Output
12
Input
506 2708
Output
3218
Note
The first four examples correspond to the first picture of the statement.
Note that for there exist multiple correct colorings for all of the examples.
In the first example you can also make a rectangle with sides 1 and 8, though its perimeter will be 18 which is greater than 8.
In the second example you can make the same resulting rectangle with sides 3 and 4, but red tiles will form the rectangle with sides 1 and 3 and blue tiles will form the rectangle with sides 3 and 3. | instruction | 0 | 69,830 | 7 | 139,660 |
Tags: binary search, brute force, math, number theory
Correct Solution:
```
import math
a,b = [int(x) for x in input().split()]
area = a+b
t = int(math.sqrt(area))
sa = int(math.sqrt(a))
sb = int(math.sqrt(b))
D = []
DA = []
DB = []
for i in range(1,t+1):
if area % i == 0:
if i*i != area:
D.append(i)
D.append(area//i)
else:
D.append(i)
for i in range(1,sa+1):
if a % i == 0:
if i*i != a:
DA.append(i)
DA.append(a//i)
else:
DA.append(i)
for i in range(1,sb+1):
if b % i == 0:
if i*i != b:
DB.append(i)
DB.append(b//i)
else:
DB.append(i)
DA.sort()
DB.sort()
D.sort()
start = ((len(D)+1)//2)-1
div = len(D)
def closestdiv(t,D):
low = 0
high = len(D)-1
while high - low > 1:
guess = (high+low)//2
if D[guess] > t:
high = guess
else:
low = guess
if D[high] <= t:
return high
else:
return low
while start > -1:
t = D[start]
s = D[-start-1]
if DA[-closestdiv(t,DA)-1] <= s:
print(2*t+2*s)
break
elif DB[-closestdiv(t,DB)-1] <= s:
print(2*t+2*s)
break
start -= 1
``` | output | 1 | 69,830 | 7 | 139,661 |
Provide tags and a correct Python 3 solution for this coding contest problem.
There is an infinite board of square tiles. Initially all tiles are white.
Vova has a red marker and a blue marker. Red marker can color a tiles. Blue marker can color b tiles. If some tile isn't white then you can't use marker of any color on it. Each marker must be drained completely, so at the end there should be exactly a red tiles and exactly b blue tiles across the board.
Vova wants to color such a set of tiles that:
* they would form a rectangle, consisting of exactly a+b colored tiles;
* all tiles of at least one color would also form a rectangle.
Here are some examples of correct colorings:
<image>
Here are some examples of incorrect colorings:
<image>
Among all correct colorings Vova wants to choose the one with the minimal perimeter. What is the minimal perimeter Vova can obtain?
It is guaranteed that there exists at least one correct coloring.
Input
A single line contains two integers a and b (1 β€ a, b β€ 10^{14}) β the number of tiles red marker should color and the number of tiles blue marker should color, respectively.
Output
Print a single integer β the minimal perimeter of a colored rectangle Vova can obtain by coloring exactly a tiles red and exactly b tiles blue.
It is guaranteed that there exists at least one correct coloring.
Examples
Input
4 4
Output
12
Input
3 9
Output
14
Input
9 3
Output
14
Input
3 6
Output
12
Input
506 2708
Output
3218
Note
The first four examples correspond to the first picture of the statement.
Note that for there exist multiple correct colorings for all of the examples.
In the first example you can also make a rectangle with sides 1 and 8, though its perimeter will be 18 which is greater than 8.
In the second example you can make the same resulting rectangle with sides 3 and 4, but red tiles will form the rectangle with sides 1 and 3 and blue tiles will form the rectangle with sides 3 and 3. | instruction | 0 | 69,831 | 7 | 139,662 |
Tags: binary search, brute force, math, number theory
Correct Solution:
```
from math import sqrt
lens = [0 for _ in range(1000000)]
def solve(a, b):
k = 0
for i in range(1, int(sqrt(b)) + 2):
if b % i == 0:
lens[k] = i
k += 1
ans = 2 * (a + b) + 2
x = a + b
l = 0
for i in range(1, int(sqrt(x)) + 2):
if x % i == 0:
while l + 1 < k and lens[l + 1] <= i:
l += 1
if b * i <= x * lens[l]:
ans = min(ans, (i + x // i) * 2)
return ans
a, b = map(int, input().split())
print(min(solve(a, b), solve(b, a)))
``` | output | 1 | 69,831 | 7 | 139,663 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There is an infinite board of square tiles. Initially all tiles are white.
Vova has a red marker and a blue marker. Red marker can color a tiles. Blue marker can color b tiles. If some tile isn't white then you can't use marker of any color on it. Each marker must be drained completely, so at the end there should be exactly a red tiles and exactly b blue tiles across the board.
Vova wants to color such a set of tiles that:
* they would form a rectangle, consisting of exactly a+b colored tiles;
* all tiles of at least one color would also form a rectangle.
Here are some examples of correct colorings:
<image>
Here are some examples of incorrect colorings:
<image>
Among all correct colorings Vova wants to choose the one with the minimal perimeter. What is the minimal perimeter Vova can obtain?
It is guaranteed that there exists at least one correct coloring.
Input
A single line contains two integers a and b (1 β€ a, b β€ 10^{14}) β the number of tiles red marker should color and the number of tiles blue marker should color, respectively.
Output
Print a single integer β the minimal perimeter of a colored rectangle Vova can obtain by coloring exactly a tiles red and exactly b tiles blue.
It is guaranteed that there exists at least one correct coloring.
Examples
Input
4 4
Output
12
Input
3 9
Output
14
Input
9 3
Output
14
Input
3 6
Output
12
Input
506 2708
Output
3218
Note
The first four examples correspond to the first picture of the statement.
Note that for there exist multiple correct colorings for all of the examples.
In the first example you can also make a rectangle with sides 1 and 8, though its perimeter will be 18 which is greater than 8.
In the second example you can make the same resulting rectangle with sides 3 and 4, but red tiles will form the rectangle with sides 1 and 3 and blue tiles will form the rectangle with sides 3 and 3.
Submitted Solution:
```
from math import sqrt, ceil
import bisect
def divisors(n):
s = [1]
for i in range(2, ceil(sqrt(n)) + 1):
if n % i == 0:
s.append(i)
if len(s) == 1:
s.append(n)
return s
bisect_right = bisect.bisect
a, b = map(int,input().split())
s = a + b
k = a
diva = divisors(a)
divb = divisors(b)
divs = divisors(s)
ansA = 0
ansB = 0
diva.sort()
divb.sort()
for i in range(len(divs)):
s = bisect_right(diva, divs[i]) - 1
if ((a + b) / divs[i]) >= (a / diva[s]):
ansA = 2*(divs[i] + ((a + b)/divs[i]))
for i in range(len(divs)):
s = bisect_right(divb, divs[i]) - 1
if ((a + b) / divs[i]) >= (b / divb[s]):
ansB = 2*(divs[i] + ((a + b)/divs[i]))
print(int(min(ansA,ansB)))
``` | instruction | 0 | 69,832 | 7 | 139,664 |
Yes | output | 1 | 69,832 | 7 | 139,665 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There is an infinite board of square tiles. Initially all tiles are white.
Vova has a red marker and a blue marker. Red marker can color a tiles. Blue marker can color b tiles. If some tile isn't white then you can't use marker of any color on it. Each marker must be drained completely, so at the end there should be exactly a red tiles and exactly b blue tiles across the board.
Vova wants to color such a set of tiles that:
* they would form a rectangle, consisting of exactly a+b colored tiles;
* all tiles of at least one color would also form a rectangle.
Here are some examples of correct colorings:
<image>
Here are some examples of incorrect colorings:
<image>
Among all correct colorings Vova wants to choose the one with the minimal perimeter. What is the minimal perimeter Vova can obtain?
It is guaranteed that there exists at least one correct coloring.
Input
A single line contains two integers a and b (1 β€ a, b β€ 10^{14}) β the number of tiles red marker should color and the number of tiles blue marker should color, respectively.
Output
Print a single integer β the minimal perimeter of a colored rectangle Vova can obtain by coloring exactly a tiles red and exactly b tiles blue.
It is guaranteed that there exists at least one correct coloring.
Examples
Input
4 4
Output
12
Input
3 9
Output
14
Input
9 3
Output
14
Input
3 6
Output
12
Input
506 2708
Output
3218
Note
The first four examples correspond to the first picture of the statement.
Note that for there exist multiple correct colorings for all of the examples.
In the first example you can also make a rectangle with sides 1 and 8, though its perimeter will be 18 which is greater than 8.
In the second example you can make the same resulting rectangle with sides 3 and 4, but red tiles will form the rectangle with sides 1 and 3 and blue tiles will form the rectangle with sides 3 and 3.
Submitted Solution:
```
from math import sqrt
a, b = map(int, input().split())
a, b = min(a, b), max(a, b)
rec_size = []
for i in range(1, int(sqrt(b)) + 1):
if i <= int(sqrt(a)):
if a % i == 0:
rec_size.append([i, a // i])
elif b % i == 0:
rec_size.append([i, b // i])
else:
if b % i == 0:
rec_size.append([i, b // i])
ok = False
for i in range(int(sqrt(a + b)), 0, -1):
if (a + b) % i == 0:
for rec in rec_size:
if i >= rec[0] and (a + b) // i >= rec[1]:
ok = True
break
if ok:
break
print(2 * (i + (a + b) // i))
``` | instruction | 0 | 69,833 | 7 | 139,666 |
Yes | output | 1 | 69,833 | 7 | 139,667 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There is an infinite board of square tiles. Initially all tiles are white.
Vova has a red marker and a blue marker. Red marker can color a tiles. Blue marker can color b tiles. If some tile isn't white then you can't use marker of any color on it. Each marker must be drained completely, so at the end there should be exactly a red tiles and exactly b blue tiles across the board.
Vova wants to color such a set of tiles that:
* they would form a rectangle, consisting of exactly a+b colored tiles;
* all tiles of at least one color would also form a rectangle.
Here are some examples of correct colorings:
<image>
Here are some examples of incorrect colorings:
<image>
Among all correct colorings Vova wants to choose the one with the minimal perimeter. What is the minimal perimeter Vova can obtain?
It is guaranteed that there exists at least one correct coloring.
Input
A single line contains two integers a and b (1 β€ a, b β€ 10^{14}) β the number of tiles red marker should color and the number of tiles blue marker should color, respectively.
Output
Print a single integer β the minimal perimeter of a colored rectangle Vova can obtain by coloring exactly a tiles red and exactly b tiles blue.
It is guaranteed that there exists at least one correct coloring.
Examples
Input
4 4
Output
12
Input
3 9
Output
14
Input
9 3
Output
14
Input
3 6
Output
12
Input
506 2708
Output
3218
Note
The first four examples correspond to the first picture of the statement.
Note that for there exist multiple correct colorings for all of the examples.
In the first example you can also make a rectangle with sides 1 and 8, though its perimeter will be 18 which is greater than 8.
In the second example you can make the same resulting rectangle with sides 3 and 4, but red tiles will form the rectangle with sides 1 and 3 and blue tiles will form the rectangle with sides 3 and 3.
Submitted Solution:
```
import sys
input = sys.stdin.readline
from math import sqrt
a,b=map(int,input().split())
x=a+b
A=[]
for i in range(1,int(sqrt(a))+1):
if a%i==0:
A.append(i)
B=[]
for i in range(1,int(sqrt(b))+1):
if b%i==0:
B.append(i)
for i in range(int(sqrt(x))+1,0,-1):
if x%i==0:
y=x//i
flag=0
for m in A:
if m<=i and a//m<=y:
flag=1
break
for m in B:
if m<=i and b//m<=y:
flag=1
break
if flag:
print(i*2+y*2)
break
``` | instruction | 0 | 69,834 | 7 | 139,668 |
Yes | output | 1 | 69,834 | 7 | 139,669 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There is an infinite board of square tiles. Initially all tiles are white.
Vova has a red marker and a blue marker. Red marker can color a tiles. Blue marker can color b tiles. If some tile isn't white then you can't use marker of any color on it. Each marker must be drained completely, so at the end there should be exactly a red tiles and exactly b blue tiles across the board.
Vova wants to color such a set of tiles that:
* they would form a rectangle, consisting of exactly a+b colored tiles;
* all tiles of at least one color would also form a rectangle.
Here are some examples of correct colorings:
<image>
Here are some examples of incorrect colorings:
<image>
Among all correct colorings Vova wants to choose the one with the minimal perimeter. What is the minimal perimeter Vova can obtain?
It is guaranteed that there exists at least one correct coloring.
Input
A single line contains two integers a and b (1 β€ a, b β€ 10^{14}) β the number of tiles red marker should color and the number of tiles blue marker should color, respectively.
Output
Print a single integer β the minimal perimeter of a colored rectangle Vova can obtain by coloring exactly a tiles red and exactly b tiles blue.
It is guaranteed that there exists at least one correct coloring.
Examples
Input
4 4
Output
12
Input
3 9
Output
14
Input
9 3
Output
14
Input
3 6
Output
12
Input
506 2708
Output
3218
Note
The first four examples correspond to the first picture of the statement.
Note that for there exist multiple correct colorings for all of the examples.
In the first example you can also make a rectangle with sides 1 and 8, though its perimeter will be 18 which is greater than 8.
In the second example you can make the same resulting rectangle with sides 3 and 4, but red tiles will form the rectangle with sides 1 and 3 and blue tiles will form the rectangle with sides 3 and 3.
Submitted Solution:
```
import bisect
import math
import sys
input = sys.stdin.readline
def divisor(i):
s = []
for j in range(1, int(math.sqrt(i)) + 1):
if i % j == 0:
s.append(i // j)
s.append(j)
return sorted(set(s))
a, b = map(int, input().split())
c = a + b
aa = divisor(a)
bb = divisor(b)
cc = divisor(c)
aa.append(c + 5)
bb.append(c + 5)
ans = 2 * (c + 1)
for h in cc:
ok = 0
w = c // h
i = bisect.bisect_left(aa, h + 0.5)
x = aa[i - 1]
y = a // x
if x <= h and y <= w:
ok = 1
i = bisect.bisect_left(aa, w + 0.5)
y = aa[i - 1]
x = a // y
if x <= h and y <= w:
ok = 1
i = bisect.bisect_left(bb, h + 0.5)
x = bb[i - 1]
y = b // x
if x <= h and y <= w:
ok = 1
i = bisect.bisect_left(bb, w + 0.5)
y = bb[i - 1]
x = b // y
if x <= h and y <= w:
ok = 1
if ok:
ans = min(ans, 2 * (h + w))
print(ans)
``` | instruction | 0 | 69,835 | 7 | 139,670 |
Yes | output | 1 | 69,835 | 7 | 139,671 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There is an infinite board of square tiles. Initially all tiles are white.
Vova has a red marker and a blue marker. Red marker can color a tiles. Blue marker can color b tiles. If some tile isn't white then you can't use marker of any color on it. Each marker must be drained completely, so at the end there should be exactly a red tiles and exactly b blue tiles across the board.
Vova wants to color such a set of tiles that:
* they would form a rectangle, consisting of exactly a+b colored tiles;
* all tiles of at least one color would also form a rectangle.
Here are some examples of correct colorings:
<image>
Here are some examples of incorrect colorings:
<image>
Among all correct colorings Vova wants to choose the one with the minimal perimeter. What is the minimal perimeter Vova can obtain?
It is guaranteed that there exists at least one correct coloring.
Input
A single line contains two integers a and b (1 β€ a, b β€ 10^{14}) β the number of tiles red marker should color and the number of tiles blue marker should color, respectively.
Output
Print a single integer β the minimal perimeter of a colored rectangle Vova can obtain by coloring exactly a tiles red and exactly b tiles blue.
It is guaranteed that there exists at least one correct coloring.
Examples
Input
4 4
Output
12
Input
3 9
Output
14
Input
9 3
Output
14
Input
3 6
Output
12
Input
506 2708
Output
3218
Note
The first four examples correspond to the first picture of the statement.
Note that for there exist multiple correct colorings for all of the examples.
In the first example you can also make a rectangle with sides 1 and 8, though its perimeter will be 18 which is greater than 8.
In the second example you can make the same resulting rectangle with sides 3 and 4, but red tiles will form the rectangle with sides 1 and 3 and blue tiles will form the rectangle with sides 3 and 3.
Submitted Solution:
```
# -*- coding: utf-8 -*-
"""
Spyder Editor
This is a temporary script file.
"""
inp = input()
inp = inp.split()
a = int(inp[0])
b = int(inp[1])
ma = a+b
mi = min(a,b)
k = int(ma ** 0.5)
flag = False
while(k>0):
if ma % k == 0:
t = ma/k;
p = min(k,t)
q = max(k,t)
while(p>0):
if (mi % p == 0) & (mi / p <= q):
flag = True
break
elif mi / p > q:
break
else:
p -= 1
if flag:
print(int(2*(k+t)));
break;
else:
k -= 1
else:
k -= 1
``` | instruction | 0 | 69,836 | 7 | 139,672 |
No | output | 1 | 69,836 | 7 | 139,673 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There is an infinite board of square tiles. Initially all tiles are white.
Vova has a red marker and a blue marker. Red marker can color a tiles. Blue marker can color b tiles. If some tile isn't white then you can't use marker of any color on it. Each marker must be drained completely, so at the end there should be exactly a red tiles and exactly b blue tiles across the board.
Vova wants to color such a set of tiles that:
* they would form a rectangle, consisting of exactly a+b colored tiles;
* all tiles of at least one color would also form a rectangle.
Here are some examples of correct colorings:
<image>
Here are some examples of incorrect colorings:
<image>
Among all correct colorings Vova wants to choose the one with the minimal perimeter. What is the minimal perimeter Vova can obtain?
It is guaranteed that there exists at least one correct coloring.
Input
A single line contains two integers a and b (1 β€ a, b β€ 10^{14}) β the number of tiles red marker should color and the number of tiles blue marker should color, respectively.
Output
Print a single integer β the minimal perimeter of a colored rectangle Vova can obtain by coloring exactly a tiles red and exactly b tiles blue.
It is guaranteed that there exists at least one correct coloring.
Examples
Input
4 4
Output
12
Input
3 9
Output
14
Input
9 3
Output
14
Input
3 6
Output
12
Input
506 2708
Output
3218
Note
The first four examples correspond to the first picture of the statement.
Note that for there exist multiple correct colorings for all of the examples.
In the first example you can also make a rectangle with sides 1 and 8, though its perimeter will be 18 which is greater than 8.
In the second example you can make the same resulting rectangle with sides 3 and 4, but red tiles will form the rectangle with sides 1 and 3 and blue tiles will form the rectangle with sides 3 and 3.
Submitted Solution:
```
import math
a,b = input().strip().split()
area = int(a)+int(b)
def get_closest_root(n):
low = 2
high = n
while low <= high:
mid = ((low + high) >> 1)
if mid*mid > n:
high = mid - 1
elif mid*mid < n:
low = mid + 1
else:
return mid
print(low,high)
if high*high > mid:
if high*high - mid < mid - (high-1)*(high-1):
return high
else:
return high-1
else:
if (high+1)*(high+1) - mid < mid - high*high:
return high+1
else:
return high
rows = int(math.sqrt(area))
if rows*rows is area:
print(rows*4)
else:
rows=get_closest_root(area)
print(rows)
print(2*(rows+(area//rows)))
``` | instruction | 0 | 69,837 | 7 | 139,674 |
No | output | 1 | 69,837 | 7 | 139,675 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There is an infinite board of square tiles. Initially all tiles are white.
Vova has a red marker and a blue marker. Red marker can color a tiles. Blue marker can color b tiles. If some tile isn't white then you can't use marker of any color on it. Each marker must be drained completely, so at the end there should be exactly a red tiles and exactly b blue tiles across the board.
Vova wants to color such a set of tiles that:
* they would form a rectangle, consisting of exactly a+b colored tiles;
* all tiles of at least one color would also form a rectangle.
Here are some examples of correct colorings:
<image>
Here are some examples of incorrect colorings:
<image>
Among all correct colorings Vova wants to choose the one with the minimal perimeter. What is the minimal perimeter Vova can obtain?
It is guaranteed that there exists at least one correct coloring.
Input
A single line contains two integers a and b (1 β€ a, b β€ 10^{14}) β the number of tiles red marker should color and the number of tiles blue marker should color, respectively.
Output
Print a single integer β the minimal perimeter of a colored rectangle Vova can obtain by coloring exactly a tiles red and exactly b tiles blue.
It is guaranteed that there exists at least one correct coloring.
Examples
Input
4 4
Output
12
Input
3 9
Output
14
Input
9 3
Output
14
Input
3 6
Output
12
Input
506 2708
Output
3218
Note
The first four examples correspond to the first picture of the statement.
Note that for there exist multiple correct colorings for all of the examples.
In the first example you can also make a rectangle with sides 1 and 8, though its perimeter will be 18 which is greater than 8.
In the second example you can make the same resulting rectangle with sides 3 and 4, but red tiles will form the rectangle with sides 1 and 3 and blue tiles will form the rectangle with sides 3 and 3.
Submitted Solution:
```
import time
a, b = [int(el) for el in input().split()]
tt=time.time()
ss = a + b
if a>b:
a,b=b,a
if abs(a-b)==2:
print (2*a+6)
raise SystemExit()
lista=[] #Π΄Π΅Π»ΠΈΡΠ΅Π»ΠΈ Π°
listb=[] #Π΄Π΅Π»ΠΈΡΠ΅Π»ΠΈ b
kf=0
s_=ss
if s_%4==0:
kf=kf+1
s_=int(s_/4)
a1=int(s_**(1/2))
step=-1
if s_%2==1 and s_>=5:
step=-2
if a1%2==0:
a1=a1+1
stopa=a1*0.9
while a1>stopa:
#for a1 in range(int(s_**(1/2)),-1,-1):
if len(lista)==1 and len(listb)==1 and lista[0]==1 and listb[0]==1:
a1=2
if ss%(int(a1 * (2**kf) + 1))==0 or ss%a1==0:
# print (tt-time.time(),' ',a1)
if kf !=0:
a1 = int(a1 * (2**kf) + 1)
kf=0
if ss%a1!=0:
continue
if ss%a1!=0:
a1=a1+step
continue
b1 = int(ss / a1) # a1 ΠΈ b1 - ΡΡΠΎΡΠΎΠ½Ρ Π±ΠΎΠ»ΡΡΠΎΠ³ΠΎ ΠΏΡΡΠΌΠΎΡΠ³ΠΎΠ»ΡΠ½ΠΈΠΊΠ°
# print (time.time()-kk)
# ΠΏΡΠΎΠ²Π΅ΡΠΈΠΌ, ΡΠΌΠ΅ΡΡΠΈΡΡΡ Π»ΠΈ ΠΎΠ΄ΠΈΠ½ ΠΈΠ· ΠΏΡΡΠΌΠΎΡΠ³ΠΎΠ»ΡΠ½ΠΈΠΊΠΎΠ² ΠΎΠ΄Π½ΠΎΠ³ΠΎ ΡΠ²Π΅ΡΠ° Π² Π΄Π°Π½Π½ΡΠΉ
# ΡΠ½Π°ΡΠ°Π»Π° ΠΏΡΠΎΠΉΠ΄Π΅ΠΌΡΡ ΠΏΠΎ ΡΠΆΠ΅ ΠΈΠΌΠ΅ΡΡΠΈΠΌΡΡ Π΄Π΅Π»ΠΈΡΠ΅Π»ΡΠΌ
for j in lista:
k = int(a / j)
if j <= a1 and k <= b1:
print(2 * a1 + 2 * b1)
raise SystemExit()
if k > b1:
break
for j in listb:
k = int(b / j)
if j <= a1 and k <= b1:
print(2 * a1 + 2 * b1)
raise SystemExit()
if k > b1:
break
if len(lista)==0:
s=int(a**(1/2))
else:
s=lista[len(lista)-1]-1
if s > 0:
if a%2==0:
step=-1
else:
step=-2
if s%2==0:
s=s+1
for j in range(s, 0, step):
if a%j==0:
lista.append(j)
k=int(a/j)
if j<=a1 and k<=b1:
print (2*a1+2*b1)
# print (a1,b1)
raise SystemExit()
if k>b1:
j=0
if len(listb)==0:
s=int(b**(1/2))
else:
s=listb[len(listb)-1]-1
if s > 0:
if b%2==0:
step=-1
else:
step=-2
if s%2==0:
s=s+1
for j in range(s,0,step):
if b%j==0:
listb.append(j)
k=int(b/j)
if j<=a1 and k<=b1:
print (2*a1+2*b1)
raise SystemExit()
if k>b1:
j=0
# a1=a1+1
a1=a1+step
while a1>0:
# for a1 in range(int(s_**(1/2)),-1,-1):
if len(lista) == 1 and len(listb) == 1 and lista[0] == 1 and listb[0] == 1:
a1 = 2
if ss % a1 == 0:
# print (tt-time.time(),' ',a1)
if kf != 0:
a1 = int(a1 * (2 ** kf) + 1)
kf = 0
if ss % a1 != 0:
continue
if ss % a1 != 0:
a1 = a1 + step
continue
b1 = int(ss / a1) # a1 ΠΈ b1 - ΡΡΠΎΡΠΎΠ½Ρ Π±ΠΎΠ»ΡΡΠΎΠ³ΠΎ ΠΏΡΡΠΌΠΎΡΠ³ΠΎΠ»ΡΠ½ΠΈΠΊΠ°
# print (time.time()-kk)
# ΠΏΡΠΎΠ²Π΅ΡΠΈΠΌ, ΡΠΌΠ΅ΡΡΠΈΡΡΡ Π»ΠΈ ΠΎΠ΄ΠΈΠ½ ΠΈΠ· ΠΏΡΡΠΌΠΎΡΠ³ΠΎΠ»ΡΠ½ΠΈΠΊΠΎΠ² ΠΎΠ΄Π½ΠΎΠ³ΠΎ ΡΠ²Π΅ΡΠ° Π² Π΄Π°Π½Π½ΡΠΉ
# ΡΠ½Π°ΡΠ°Π»Π° ΠΏΡΠΎΠΉΠ΄Π΅ΠΌΡΡ ΠΏΠΎ ΡΠΆΠ΅ ΠΈΠΌΠ΅ΡΡΠΈΠΌΡΡ Π΄Π΅Π»ΠΈΡΠ΅Π»ΡΠΌ
for j in lista:
k = int(a / j)
if j <= a1 and k <= b1:
print(2 * a1 + 2 * b1)
raise SystemExit()
if k > b1:
break
for j in listb:
k = int(b / j)
if j <= a1 and k <= b1:
print(2 * a1 + 2 * b1)
raise SystemExit()
if k > b1:
break
if len(lista) == 0:
s = int(a ** (1 / 2))
else:
s = lista[len(lista) - 1] - 1
if s > 0:
if a % 2 == 0:
step = -1
else:
step = -2
if s % 2 == 0:
s = s + 1
for j in range(s, 0, step):
if a % j == 0:
lista.append(j)
k = int(a / j)
if j <= a1 and k <= b1:
print(2 * a1 + 2 * b1)
print(a1, b1)
raise SystemExit()
if k > b1:
j = 0
if len(listb) == 0:
s = int(b ** (1 / 2))
else:
s = listb[len(listb) - 1] - 1
if s > 0:
if b % 2 == 0:
step = -1
else:
step = -2
if s % 2 == 0:
s = s + 1
for j in range(s, 0, step):
if b % j == 0:
listb.append(j)
k = int(b / j)
if j <= a1 and k <= b1:
print(2 * a1 + 2 * b1)
raise SystemExit()
if k > b1:
j = 0
# a1=a1+1
a1 = a1 + step
``` | instruction | 0 | 69,838 | 7 | 139,676 |
No | output | 1 | 69,838 | 7 | 139,677 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There is an infinite board of square tiles. Initially all tiles are white.
Vova has a red marker and a blue marker. Red marker can color a tiles. Blue marker can color b tiles. If some tile isn't white then you can't use marker of any color on it. Each marker must be drained completely, so at the end there should be exactly a red tiles and exactly b blue tiles across the board.
Vova wants to color such a set of tiles that:
* they would form a rectangle, consisting of exactly a+b colored tiles;
* all tiles of at least one color would also form a rectangle.
Here are some examples of correct colorings:
<image>
Here are some examples of incorrect colorings:
<image>
Among all correct colorings Vova wants to choose the one with the minimal perimeter. What is the minimal perimeter Vova can obtain?
It is guaranteed that there exists at least one correct coloring.
Input
A single line contains two integers a and b (1 β€ a, b β€ 10^{14}) β the number of tiles red marker should color and the number of tiles blue marker should color, respectively.
Output
Print a single integer β the minimal perimeter of a colored rectangle Vova can obtain by coloring exactly a tiles red and exactly b tiles blue.
It is guaranteed that there exists at least one correct coloring.
Examples
Input
4 4
Output
12
Input
3 9
Output
14
Input
9 3
Output
14
Input
3 6
Output
12
Input
506 2708
Output
3218
Note
The first four examples correspond to the first picture of the statement.
Note that for there exist multiple correct colorings for all of the examples.
In the first example you can also make a rectangle with sides 1 and 8, though its perimeter will be 18 which is greater than 8.
In the second example you can make the same resulting rectangle with sides 3 and 4, but red tiles will form the rectangle with sides 1 and 3 and blue tiles will form the rectangle with sides 3 and 3.
Submitted Solution:
```
a,b = map(int, input().split())
def make_divisors(n):
divisors = []
for i in range(1, int(n**0.5)+1):
if n % i == 0:
divisors.append(i)
if i != n // i:
divisors.append(n//i)
#divisors.sort(reverse=True)
return divisors
import math
g = math.gcd(a,b)
d = make_divisors(g)
ans = (a+b+1)*2
for x in d:
y = a//x
z = b//x
temp = (x+y+z)*2
ans = min(temp, ans)
print(ans)
``` | instruction | 0 | 69,839 | 7 | 139,678 |
No | output | 1 | 69,839 | 7 | 139,679 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Genos recently installed the game Zuma on his phone. In Zuma there exists a line of n gemstones, the i-th of which has color ci. The goal of the game is to destroy all the gemstones in the line as quickly as possible.
In one second, Genos is able to choose exactly one continuous substring of colored gemstones that is a palindrome and remove it from the line. After the substring is removed, the remaining gemstones shift to form a solid line again. What is the minimum number of seconds needed to destroy the entire line?
Let us remind, that the string (or substring) is called palindrome, if it reads same backwards or forward. In our case this means the color of the first gemstone is equal to the color of the last one, the color of the second gemstone is equal to the color of the next to last and so on.
Input
The first line of input contains a single integer n (1 β€ n β€ 500) β the number of gemstones.
The second line contains n space-separated integers, the i-th of which is ci (1 β€ ci β€ n) β the color of the i-th gemstone in a line.
Output
Print a single integer β the minimum number of seconds needed to destroy the entire line.
Examples
Input
3
1 2 1
Output
1
Input
3
1 2 3
Output
3
Input
7
1 4 4 2 3 2 1
Output
2
Note
In the first sample, Genos can destroy the entire line in one second.
In the second sample, Genos can only destroy one gemstone at a time, so destroying three gemstones takes three seconds.
In the third sample, to achieve the optimal time of two seconds, destroy palindrome 4 4 first and then destroy palindrome 1 2 3 2 1. | instruction | 0 | 70,261 | 7 | 140,522 |
Tags: dp
Correct Solution:
```
dp=[]
def calculate(i,j,colors):
if i>j: return 0
if dp[i][j]==-1:
if i==j:
return 1
dp[i][j] = 10000
dp[i][j] = min(dp[i][j],1+calculate(i+1,j,colors))
# print(i,j)
if colors[i]==colors[i+1]:
dp[i][j] = min(dp[i][j],1+calculate(i+2,j,colors))
for k in range(i+2,j+1):
if colors[k]==colors[i]:
dp[i][j] = min(dp[i][j],calculate(i+1,k-1,colors)+calculate(k+1,j,colors))
return dp[i][j]
def solve():
t = int(input())
colors = list(map(int, input().split()))
[dp.append([-1]*len(colors)) for x in range(len(colors))]
print (calculate(0,len(colors)-1,colors))
try:
solve()
except Exception as e:
print (e)
``` | output | 1 | 70,261 | 7 | 140,523 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Genos recently installed the game Zuma on his phone. In Zuma there exists a line of n gemstones, the i-th of which has color ci. The goal of the game is to destroy all the gemstones in the line as quickly as possible.
In one second, Genos is able to choose exactly one continuous substring of colored gemstones that is a palindrome and remove it from the line. After the substring is removed, the remaining gemstones shift to form a solid line again. What is the minimum number of seconds needed to destroy the entire line?
Let us remind, that the string (or substring) is called palindrome, if it reads same backwards or forward. In our case this means the color of the first gemstone is equal to the color of the last one, the color of the second gemstone is equal to the color of the next to last and so on.
Input
The first line of input contains a single integer n (1 β€ n β€ 500) β the number of gemstones.
The second line contains n space-separated integers, the i-th of which is ci (1 β€ ci β€ n) β the color of the i-th gemstone in a line.
Output
Print a single integer β the minimum number of seconds needed to destroy the entire line.
Examples
Input
3
1 2 1
Output
1
Input
3
1 2 3
Output
3
Input
7
1 4 4 2 3 2 1
Output
2
Note
In the first sample, Genos can destroy the entire line in one second.
In the second sample, Genos can only destroy one gemstone at a time, so destroying three gemstones takes three seconds.
In the third sample, to achieve the optimal time of two seconds, destroy palindrome 4 4 first and then destroy palindrome 1 2 3 2 1. | instruction | 0 | 70,262 | 7 | 140,524 |
Tags: 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 ** 30, 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 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------------------------------------
class TrieNode:
def __init__(self):
self.children = [None] * 26
self.isEndOfWord = False
class Trie:
def __init__(self):
self.root = self.getNode()
def getNode(self):
return TrieNode()
def _charToIndex(self, ch):
return ord(ch) - ord('a')
def insert(self, key):
pCrawl = self.root
length = len(key)
for level in range(length):
index = self._charToIndex(key[level])
if not pCrawl.children[index]:
pCrawl.children[index] = self.getNode()
pCrawl = pCrawl.children[index]
pCrawl.isEndOfWord = True
def search(self, key):
pCrawl = self.root
length = len(key)
for level in range(length):
index = self._charToIndex(key[level])
if not pCrawl.children[index]:
return False
pCrawl = pCrawl.children[index]
return pCrawl != None and pCrawl.isEndOfWord
# -----------------------------------------trie---------------------------------
def merge(arr, temp, left, mid, right):
inv_count = 0
i = left # i is index for left subarray*/
j = mid # i is index for right subarray*/
k = left # i is index for resultant merged subarray*/
while ((i <= mid - 1) and (j <= right)):
if (arr[i] <= arr[j]):
temp[k] = arr[i]
k += 1
i += 1
else:
temp[k] = arr[j]
k += 1
j += 1
inv_count = inv_count + (mid - i)
while (i <= mid - 1):
temp[k] = arr[i]
k += 1
i += 1
while (j <= right):
temp[k] = arr[j]
k += 1
j += 1
# Copy back the merged elements to original array*/
for i in range(left, right + 1, 1):
arr[i] = temp[i]
return inv_count
def _mergeSort(arr, temp, left, right):
inv_count = 0
if (right > left):
mid = int((right + left) / 2)
inv_count = _mergeSort(arr, temp, left, mid)
inv_count += _mergeSort(arr, temp, mid + 1, right)
inv_count += merge(arr, temp, left, mid + 1, right)
return inv_count
def countSwaps(arr, n):
temp = [0 for i in range(n)]
return _mergeSort(arr, temp, 0, n - 1)
#-----------------------------------------adjcent swap required------------------------------
def minSwaps(arr):
n = len(arr)
arrpos = [*enumerate(arr)]
arrpos.sort(key=lambda it: it[1])
vis = {k: False for k in range(n)}
ans = 0
for i in range(n):
if vis[i] or arrpos[i][0] == i:
continue
cycle_size = 0
j = i
while not vis[j]:
vis[j] = True
j = arrpos[j][0]
cycle_size += 1
if cycle_size > 0:
ans += (cycle_size - 1)
return ans
#----------------------swaps required----------------------------
class Node:
def __init__(self, data):
self.data = data
self.count = 0
self.left = None # left node for 0
self.right = None # right node for 1
class BinaryTrie:
def __init__(self):
self.root = Node(0)
def insert(self, pre_xor):
self.temp = self.root
for i in range(31, -1, -1):
val = pre_xor & (1 << i)
if val:
if not self.temp.right:
self.temp.right = Node(0)
self.temp = self.temp.right
self.temp.count += 1
if not val:
if not self.temp.left:
self.temp.left = Node(0)
self.temp = self.temp.left
self.temp.count += 1
self.temp.data = pre_xor
def query(self, xor):
self.temp = self.root
for i in range(31, -1, -1):
val = xor & (1 << i)
if not val:
if self.temp.left and self.temp.left.count > 0:
self.temp = self.temp.left
elif self.temp.right:
self.temp = self.temp.right
else:
if self.temp.right and self.temp.right.count > 0:
self.temp = self.temp.right
elif self.temp.left:
self.temp = self.temp.left
self.temp.count -= 1
return xor ^ self.temp.data
# -------------------------bin trie-------------------------------------------
n=int(input())
l=list(map(int,input().split()))
dp=[[9999999999999 for i in range(n+1)]for j in range(n+1)]
for i in range(n+1):
dp[i][i]=1
for j in range(i):
dp[i][j]=0
for i in range(n):
for j in range(i-1,-1,-1):
if j!=i-1 and l[j]==l[i]:
dp[j][i]=min(dp[j][i],dp[j+1][i-1])
if l[j]==l[j+1]:
dp[j][i]=min(1+dp[j+2][i],dp[j][i])
dp[j][i]=min(dp[j][i],dp[j][i-1]+1,dp[j+1][i]+1)
for k in range(j+2,i):
if l[j]==l[k]:
dp[j][i]=min(dp[j][i],dp[j+1][k-1]+dp[k+1][i])
print(dp[0][n-1])
``` | output | 1 | 70,262 | 7 | 140,525 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Genos recently installed the game Zuma on his phone. In Zuma there exists a line of n gemstones, the i-th of which has color ci. The goal of the game is to destroy all the gemstones in the line as quickly as possible.
In one second, Genos is able to choose exactly one continuous substring of colored gemstones that is a palindrome and remove it from the line. After the substring is removed, the remaining gemstones shift to form a solid line again. What is the minimum number of seconds needed to destroy the entire line?
Let us remind, that the string (or substring) is called palindrome, if it reads same backwards or forward. In our case this means the color of the first gemstone is equal to the color of the last one, the color of the second gemstone is equal to the color of the next to last and so on.
Input
The first line of input contains a single integer n (1 β€ n β€ 500) β the number of gemstones.
The second line contains n space-separated integers, the i-th of which is ci (1 β€ ci β€ n) β the color of the i-th gemstone in a line.
Output
Print a single integer β the minimum number of seconds needed to destroy the entire line.
Examples
Input
3
1 2 1
Output
1
Input
3
1 2 3
Output
3
Input
7
1 4 4 2 3 2 1
Output
2
Note
In the first sample, Genos can destroy the entire line in one second.
In the second sample, Genos can only destroy one gemstone at a time, so destroying three gemstones takes three seconds.
In the third sample, to achieve the optimal time of two seconds, destroy palindrome 4 4 first and then destroy palindrome 1 2 3 2 1. | instruction | 0 | 70,263 | 7 | 140,526 |
Tags: dp
Correct Solution:
```
n = int(input())
c = [*map(int, input().split())]
inf = n + 1
dp = [[inf for _ in range(n)] for __ in range(n)]
def find(l, r):
if l > r:
return 0
if l == r or (l == r - 1 and c[l] == c[r]):
dp[l][r] = 1
return 1
if dp[l][r] != inf:
return dp[l][r]
m = 1 + find(l + 1, r)
for i in range(l + 2, r + 1):
if c[l] == c[i]:
m = min(m, find(l + 1, i - 1) + find(i + 1, r))
if c[l] == c[l + 1]:
m = min(m, find(l + 2, r) + 1)
dp[l][r] = m
return m
mi = inf
for i in range(n):
mi = min(find(0, i) + find(i + 1, n - 1), mi)
print(mi)
``` | output | 1 | 70,263 | 7 | 140,527 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Genos recently installed the game Zuma on his phone. In Zuma there exists a line of n gemstones, the i-th of which has color ci. The goal of the game is to destroy all the gemstones in the line as quickly as possible.
In one second, Genos is able to choose exactly one continuous substring of colored gemstones that is a palindrome and remove it from the line. After the substring is removed, the remaining gemstones shift to form a solid line again. What is the minimum number of seconds needed to destroy the entire line?
Let us remind, that the string (or substring) is called palindrome, if it reads same backwards or forward. In our case this means the color of the first gemstone is equal to the color of the last one, the color of the second gemstone is equal to the color of the next to last and so on.
Input
The first line of input contains a single integer n (1 β€ n β€ 500) β the number of gemstones.
The second line contains n space-separated integers, the i-th of which is ci (1 β€ ci β€ n) β the color of the i-th gemstone in a line.
Output
Print a single integer β the minimum number of seconds needed to destroy the entire line.
Examples
Input
3
1 2 1
Output
1
Input
3
1 2 3
Output
3
Input
7
1 4 4 2 3 2 1
Output
2
Note
In the first sample, Genos can destroy the entire line in one second.
In the second sample, Genos can only destroy one gemstone at a time, so destroying three gemstones takes three seconds.
In the third sample, to achieve the optimal time of two seconds, destroy palindrome 4 4 first and then destroy palindrome 1 2 3 2 1. | instruction | 0 | 70,264 | 7 | 140,528 |
Tags: dp
Correct Solution:
```
n = int(input())
C = list(map(int, input().split()))
dp = [[0]*n for _ in range(n)]
for i in range(n) :
dp[i][i] = 1
for i in range(n-2, -1, -1) :
for j in range(i+1, n) :
dp[i][j] = 1 + dp[i+1][j]
if C[i] == C[i+1] : dp[i][j] = min( dp[i][j], 1 + (dp[i+2][j] if i+2 < n else 0) )
for k in range(i+2, j) :
if C[i] == C[k] : dp[i][j] = min( dp[i][j], dp[i+1][k-1] + dp[k+1][j] )
if C[i] == C[j] and j-i > 1:
dp[i][j] = min( dp[i][j], dp[i+1][j-1] )
print( dp[0][n-1] )
``` | output | 1 | 70,264 | 7 | 140,529 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Genos recently installed the game Zuma on his phone. In Zuma there exists a line of n gemstones, the i-th of which has color ci. The goal of the game is to destroy all the gemstones in the line as quickly as possible.
In one second, Genos is able to choose exactly one continuous substring of colored gemstones that is a palindrome and remove it from the line. After the substring is removed, the remaining gemstones shift to form a solid line again. What is the minimum number of seconds needed to destroy the entire line?
Let us remind, that the string (or substring) is called palindrome, if it reads same backwards or forward. In our case this means the color of the first gemstone is equal to the color of the last one, the color of the second gemstone is equal to the color of the next to last and so on.
Input
The first line of input contains a single integer n (1 β€ n β€ 500) β the number of gemstones.
The second line contains n space-separated integers, the i-th of which is ci (1 β€ ci β€ n) β the color of the i-th gemstone in a line.
Output
Print a single integer β the minimum number of seconds needed to destroy the entire line.
Examples
Input
3
1 2 1
Output
1
Input
3
1 2 3
Output
3
Input
7
1 4 4 2 3 2 1
Output
2
Note
In the first sample, Genos can destroy the entire line in one second.
In the second sample, Genos can only destroy one gemstone at a time, so destroying three gemstones takes three seconds.
In the third sample, to achieve the optimal time of two seconds, destroy palindrome 4 4 first and then destroy palindrome 1 2 3 2 1.
Submitted Solution:
```
import os
import sys
from io import BytesIO, IOBase
from types import GeneratorType
from collections import defaultdict
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")
import math
n=int(input())
b=list(map(int,input().split()))
dp=[[float('inf') for j in range(n)] for i in range(n)]
for i in range(n):
dp[i][i]=1
i=n-1
while(i>=0):
j=i+1
while(j<n):
dp[i][j]=min(dp[i][j],1+dp[i+1][j])
k=j
while(k>i):
if b[i]==b[k]:
val1=1
if (k-1)>=(i+1):
val1=dp[i+1][k-1]
val2=0
if ((k+1)<=j):
val2=dp[k+1][j]
dp[i][j]=min(dp[i][j],val1+val2)
k+=-1
j+=1
i+=-1
print(dp[0][n-1])
``` | instruction | 0 | 70,268 | 7 | 140,536 |
Yes | output | 1 | 70,268 | 7 | 140,537 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Genos recently installed the game Zuma on his phone. In Zuma there exists a line of n gemstones, the i-th of which has color ci. The goal of the game is to destroy all the gemstones in the line as quickly as possible.
In one second, Genos is able to choose exactly one continuous substring of colored gemstones that is a palindrome and remove it from the line. After the substring is removed, the remaining gemstones shift to form a solid line again. What is the minimum number of seconds needed to destroy the entire line?
Let us remind, that the string (or substring) is called palindrome, if it reads same backwards or forward. In our case this means the color of the first gemstone is equal to the color of the last one, the color of the second gemstone is equal to the color of the next to last and so on.
Input
The first line of input contains a single integer n (1 β€ n β€ 500) β the number of gemstones.
The second line contains n space-separated integers, the i-th of which is ci (1 β€ ci β€ n) β the color of the i-th gemstone in a line.
Output
Print a single integer β the minimum number of seconds needed to destroy the entire line.
Examples
Input
3
1 2 1
Output
1
Input
3
1 2 3
Output
3
Input
7
1 4 4 2 3 2 1
Output
2
Note
In the first sample, Genos can destroy the entire line in one second.
In the second sample, Genos can only destroy one gemstone at a time, so destroying three gemstones takes three seconds.
In the third sample, to achieve the optimal time of two seconds, destroy palindrome 4 4 first and then destroy palindrome 1 2 3 2 1.
Submitted Solution:
```
# Author : nitish420 --------------------------------------------------------------------
import os
import sys
from io import BytesIO, IOBase
def main():
n=int(input())
arr=list(map(int,input().split()))
dp=[[0 for _ in range(n+1)] for _ in range(n+1)]
for i in range(n):
dp[i][i]=1
for dff in range(2,n+1):
for l in range(n-dff+1):
r=l+dff-1
if dff==2:
if arr[l]==arr[r]:
dp[l][r]=1
else:
dp[l][r]=2
continue
mm=501
for k in range(l+2,r+1):
if arr[k]==arr[l]:
mm=min(mm,dp[l+1][k-1]+dp[k+1][r])
mm=min(mm,1+dp[l+1][r])
if arr[l]==arr[l+1]:
mm=min(mm,1+dp[l+2][r])
dp[l][r]=mm
# print()
# for item in dp:
# print(*item)
print(dp[0][n-1])
#----------------------------------------------------------------------------------------
# region fastio
BUFSIZE = 8192
class FastIO(IOBase):
newlines = 0
def __init__(self, file):
self._fd = file.fileno()
self.buffer = BytesIO()
self.writable = 'x' in file.mode or 'r' not in file.mode
self.write = self.buffer.write if self.writable else None
def read(self):
while True:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
if not b:
break
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines = 0
return self.buffer.read()
def readline(self):
while self.newlines == 0:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
self.newlines = b.count(b'\n') + (not b)
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines -= 1
return self.buffer.readline()
def flush(self):
if self.writable:
os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
self.write = lambda s: self.buffer.write(s.encode('ascii'))
self.read = lambda: self.buffer.read().decode('ascii')
self.readline = lambda: self.buffer.readline().decode('ascii')
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip('\r\n')
# endregion
if __name__ == '__main__':
main()
``` | instruction | 0 | 70,269 | 7 | 140,538 |
Yes | output | 1 | 70,269 | 7 | 140,539 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Genos recently installed the game Zuma on his phone. In Zuma there exists a line of n gemstones, the i-th of which has color ci. The goal of the game is to destroy all the gemstones in the line as quickly as possible.
In one second, Genos is able to choose exactly one continuous substring of colored gemstones that is a palindrome and remove it from the line. After the substring is removed, the remaining gemstones shift to form a solid line again. What is the minimum number of seconds needed to destroy the entire line?
Let us remind, that the string (or substring) is called palindrome, if it reads same backwards or forward. In our case this means the color of the first gemstone is equal to the color of the last one, the color of the second gemstone is equal to the color of the next to last and so on.
Input
The first line of input contains a single integer n (1 β€ n β€ 500) β the number of gemstones.
The second line contains n space-separated integers, the i-th of which is ci (1 β€ ci β€ n) β the color of the i-th gemstone in a line.
Output
Print a single integer β the minimum number of seconds needed to destroy the entire line.
Examples
Input
3
1 2 1
Output
1
Input
3
1 2 3
Output
3
Input
7
1 4 4 2 3 2 1
Output
2
Note
In the first sample, Genos can destroy the entire line in one second.
In the second sample, Genos can only destroy one gemstone at a time, so destroying three gemstones takes three seconds.
In the third sample, to achieve the optimal time of two seconds, destroy palindrome 4 4 first and then destroy palindrome 1 2 3 2 1.
Submitted Solution:
```
n=int(input())
c=list(map(int,input().split()))
dp=[[float('inf') for i in range(n)] for j in range(n)]
size=n-1
const=0
var=0
count=0
for i in range(n):
const=0
var=count
while var <= size:
if const == var:
dp[const][var]=1
elif var - const ==1:
dp[const][var] = 1 if c[const] == c[var] else 2
else:
if c[const] == c[var]:
dp[const][var] = dp[const+1][var-1]
add=const
for j in range(var-const):
dp[const][var] = min(dp[const][var],dp[const][add]+dp[add+1][var])
add+=1
var+=1
const+=1
count+=1
print(dp[0][n-1])
``` | instruction | 0 | 70,270 | 7 | 140,540 |
Yes | output | 1 | 70,270 | 7 | 140,541 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Genos recently installed the game Zuma on his phone. In Zuma there exists a line of n gemstones, the i-th of which has color ci. The goal of the game is to destroy all the gemstones in the line as quickly as possible.
In one second, Genos is able to choose exactly one continuous substring of colored gemstones that is a palindrome and remove it from the line. After the substring is removed, the remaining gemstones shift to form a solid line again. What is the minimum number of seconds needed to destroy the entire line?
Let us remind, that the string (or substring) is called palindrome, if it reads same backwards or forward. In our case this means the color of the first gemstone is equal to the color of the last one, the color of the second gemstone is equal to the color of the next to last and so on.
Input
The first line of input contains a single integer n (1 β€ n β€ 500) β the number of gemstones.
The second line contains n space-separated integers, the i-th of which is ci (1 β€ ci β€ n) β the color of the i-th gemstone in a line.
Output
Print a single integer β the minimum number of seconds needed to destroy the entire line.
Examples
Input
3
1 2 1
Output
1
Input
3
1 2 3
Output
3
Input
7
1 4 4 2 3 2 1
Output
2
Note
In the first sample, Genos can destroy the entire line in one second.
In the second sample, Genos can only destroy one gemstone at a time, so destroying three gemstones takes three seconds.
In the third sample, to achieve the optimal time of two seconds, destroy palindrome 4 4 first and then destroy palindrome 1 2 3 2 1.
Submitted Solution:
```
# Author : nitish420 --------------------------------------------------------------------
import os
import sys
from io import BytesIO, IOBase
def main():
n=int(input())
arr=list(map(int,input().split()))
dp=[[0]*(n+1) for _ in range(n+1)]
for i in range(n):
dp[i][i]=1
for dff in range(2,n+1):
for l in range(n-dff+1):
r=l+dff-1
if dff==2:
if arr[l]==arr[r]:
dp[l][r]=1
else:
dp[l][r]=2
continue
m=1000
for i in range(l+2,r+1):
if arr[l]==arr[i]:
m=min(m,dp[l+1][i-1]+dp[i+1][r])
m=min(m,1+dp[l+1][r])
if arr[l]==arr[l+1]:
m=min(m,1+dp[l+2][r])
dp[l][r]=m
print(dp[0][n-1])
#----------------------------------------------------------------------------------------
# region fastio
BUFSIZE = 8192
class FastIO(IOBase):
newlines = 0
def __init__(self, file):
self._fd = file.fileno()
self.buffer = BytesIO()
self.writable = 'x' in file.mode or 'r' not in file.mode
self.write = self.buffer.write if self.writable else None
def read(self):
while True:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
if not b:
break
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines = 0
return self.buffer.read()
def readline(self):
while self.newlines == 0:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
self.newlines = b.count(b'\n') + (not b)
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines -= 1
return self.buffer.readline()
def flush(self):
if self.writable:
os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
self.write = lambda s: self.buffer.write(s.encode('ascii'))
self.read = lambda: self.buffer.read().decode('ascii')
self.readline = lambda: self.buffer.readline().decode('ascii')
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip('\r\n')
# endregion
if __name__ == '__main__':
main()
``` | instruction | 0 | 70,271 | 7 | 140,542 |
Yes | output | 1 | 70,271 | 7 | 140,543 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Genos recently installed the game Zuma on his phone. In Zuma there exists a line of n gemstones, the i-th of which has color ci. The goal of the game is to destroy all the gemstones in the line as quickly as possible.
In one second, Genos is able to choose exactly one continuous substring of colored gemstones that is a palindrome and remove it from the line. After the substring is removed, the remaining gemstones shift to form a solid line again. What is the minimum number of seconds needed to destroy the entire line?
Let us remind, that the string (or substring) is called palindrome, if it reads same backwards or forward. In our case this means the color of the first gemstone is equal to the color of the last one, the color of the second gemstone is equal to the color of the next to last and so on.
Input
The first line of input contains a single integer n (1 β€ n β€ 500) β the number of gemstones.
The second line contains n space-separated integers, the i-th of which is ci (1 β€ ci β€ n) β the color of the i-th gemstone in a line.
Output
Print a single integer β the minimum number of seconds needed to destroy the entire line.
Examples
Input
3
1 2 1
Output
1
Input
3
1 2 3
Output
3
Input
7
1 4 4 2 3 2 1
Output
2
Note
In the first sample, Genos can destroy the entire line in one second.
In the second sample, Genos can only destroy one gemstone at a time, so destroying three gemstones takes three seconds.
In the third sample, to achieve the optimal time of two seconds, destroy palindrome 4 4 first and then destroy palindrome 1 2 3 2 1.
Submitted Solution:
```
import os
import sys
from io import BytesIO, IOBase
# region fastio
BUFSIZE = 8192
class FastIO(IOBase):
newlines = 0
def __init__(self, file):
self._fd = file.fileno()
self.buffer = BytesIO()
self.writable = "x" in file.mode or "r" not in file.mode
self.write = self.buffer.write if self.writable else None
def read(self):
while True:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
if not b:
break
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines = 0
return self.buffer.read()
def readline(self):
while self.newlines == 0:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
self.newlines = b.count(b"\n") + (not b)
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines -= 1
return self.buffer.readline()
def flush(self):
if self.writable:
os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
self.write = lambda s: self.buffer.write(s.encode("ascii"))
self.read = lambda: self.buffer.read().decode("ascii")
self.readline = lambda: self.buffer.readline().decode("ascii")
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")
# ------------------------------
from math import factorial
from collections import Counter, defaultdict, deque
from heapq import heapify, heappop, heappush
def RL(): return map(int, sys.stdin.readline().rstrip().split())
def RLL(): return list(map(int, sys.stdin.readline().rstrip().split()))
def N(): return int(input())
def comb(n, m): return factorial(n) / (factorial(m) * factorial(n - m)) if n >= m else 0
def perm(n, m): return factorial(n) // (factorial(n - m)) if n >= m else 0
def mdis(x1, y1, x2, y2): return abs(x1 - x2) + abs(y1 - y2)
mod = 998244353
INF = float('inf')
# ------------------------------
def main():
n = N()
arr = RLL()
dp = [[n]*n for _ in range(n)]
for i in range(n): dp[i][i] = 1
for le in range(2, n+1):
for l in range(n):
r = l+le-1
if r>=n: break
if le==2:
if arr[l]==arr[r]: dp[l][r] = 1
else: dp[l][r] = 2
else:
for m in range(l, r):
dp[l][r] = min(dp[l][r], dp[l][m]+dp[m+1][r])
if arr[l]==arr[r]:
dp[l][r] = dp[l+1][r-1]
print(dp[0][-1])
if __name__ == "__main__":
main()
``` | instruction | 0 | 70,272 | 7 | 140,544 |
No | output | 1 | 70,272 | 7 | 140,545 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Genos recently installed the game Zuma on his phone. In Zuma there exists a line of n gemstones, the i-th of which has color ci. The goal of the game is to destroy all the gemstones in the line as quickly as possible.
In one second, Genos is able to choose exactly one continuous substring of colored gemstones that is a palindrome and remove it from the line. After the substring is removed, the remaining gemstones shift to form a solid line again. What is the minimum number of seconds needed to destroy the entire line?
Let us remind, that the string (or substring) is called palindrome, if it reads same backwards or forward. In our case this means the color of the first gemstone is equal to the color of the last one, the color of the second gemstone is equal to the color of the next to last and so on.
Input
The first line of input contains a single integer n (1 β€ n β€ 500) β the number of gemstones.
The second line contains n space-separated integers, the i-th of which is ci (1 β€ ci β€ n) β the color of the i-th gemstone in a line.
Output
Print a single integer β the minimum number of seconds needed to destroy the entire line.
Examples
Input
3
1 2 1
Output
1
Input
3
1 2 3
Output
3
Input
7
1 4 4 2 3 2 1
Output
2
Note
In the first sample, Genos can destroy the entire line in one second.
In the second sample, Genos can only destroy one gemstone at a time, so destroying three gemstones takes three seconds.
In the third sample, to achieve the optimal time of two seconds, destroy palindrome 4 4 first and then destroy palindrome 1 2 3 2 1.
Submitted Solution:
```
# by the authority of GOD author: manhar singh sachdev #
import os,sys
from io import BytesIO,IOBase
from array import array
def main():
n = int(input())
c = list(map(int,input().split()))
dp = [array('i',[0]*n) for _ in range(n)]
for i in range(n):
dp[i][i] = 1
for i in range(n-2,-1,-1):
for j in range(i+1,n):
if c[i] == c[j]:
dp[i][j] = int(j-i==1)+dp[i+1][j-1]
else:
dp[i][j] = min(dp[i][k]+dp[k+1][j] for k in range(i,j))
print(dp[0][n-1])
# Fast IO Region
BUFSIZE = 8192
class FastIO(IOBase):
newlines = 0
def __init__(self,file):
self._fd = file.fileno()
self.buffer = BytesIO()
self.writable = "x" in file.mode or "r" not in file.mode
self.write = self.buffer.write if self.writable else None
def read(self):
while True:
b = os.read(self._fd,max(os.fstat(self._fd).st_size,BUFSIZE))
if not b:
break
ptr = self.buffer.tell()
self.buffer.seek(0,2),self.buffer.write(b),self.buffer.seek(ptr)
self.newlines = 0
return self.buffer.read()
def readline(self):
while self.newlines == 0:
b = os.read(self._fd,max(os.fstat(self._fd).st_size,BUFSIZE))
self.newlines = b.count(b"\n")+(not b)
ptr = self.buffer.tell()
self.buffer.seek(0,2),self.buffer.write(b),self.buffer.seek(ptr)
self.newlines -= 1
return self.buffer.readline()
def flush(self):
if self.writable:
os.write(self._fd,self.buffer.getvalue())
self.buffer.truncate(0),self.buffer.seek(0)
class IOWrapper(IOBase):
def __init__(self,file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
self.write = lambda s:self.buffer.write(s.encode("ascii"))
self.read = lambda:self.buffer.read().decode("ascii")
self.readline = lambda:self.buffer.readline().decode("ascii")
sys.stdin,sys.stdout = IOWrapper(sys.stdin),IOWrapper(sys.stdout)
input = lambda:sys.stdin.readline().rstrip("\r\n")
if __name__ == "__main__":
main()
``` | instruction | 0 | 70,273 | 7 | 140,546 |
No | output | 1 | 70,273 | 7 | 140,547 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Genos recently installed the game Zuma on his phone. In Zuma there exists a line of n gemstones, the i-th of which has color ci. The goal of the game is to destroy all the gemstones in the line as quickly as possible.
In one second, Genos is able to choose exactly one continuous substring of colored gemstones that is a palindrome and remove it from the line. After the substring is removed, the remaining gemstones shift to form a solid line again. What is the minimum number of seconds needed to destroy the entire line?
Let us remind, that the string (or substring) is called palindrome, if it reads same backwards or forward. In our case this means the color of the first gemstone is equal to the color of the last one, the color of the second gemstone is equal to the color of the next to last and so on.
Input
The first line of input contains a single integer n (1 β€ n β€ 500) β the number of gemstones.
The second line contains n space-separated integers, the i-th of which is ci (1 β€ ci β€ n) β the color of the i-th gemstone in a line.
Output
Print a single integer β the minimum number of seconds needed to destroy the entire line.
Examples
Input
3
1 2 1
Output
1
Input
3
1 2 3
Output
3
Input
7
1 4 4 2 3 2 1
Output
2
Note
In the first sample, Genos can destroy the entire line in one second.
In the second sample, Genos can only destroy one gemstone at a time, so destroying three gemstones takes three seconds.
In the third sample, to achieve the optimal time of two seconds, destroy palindrome 4 4 first and then destroy palindrome 1 2 3 2 1.
Submitted Solution:
```
import sys
input=sys.stdin.readline
n=int(input())
c=list(map(int,input().split()))
dp=[[-1]*(n+1) for i in range(n+1)]
def calc(l,r):
if l>r:
return 0
elif l==r:
return 1
elif dp[l][r]!=-1:
return dp[l][r]
res=10**9
if c[l]==c[l+1]:
res=min(res,calc(l+2,r)+1)
for k in range(l+2,r+1):
if c[l]==c[k]:
res=min(res,calc(l,k-1)+calc(k+1,r))
dp[l][r]=min(res,1+calc(l+1,r))
return dp[l][r]
print(calc(0,n-1))
``` | instruction | 0 | 70,274 | 7 | 140,548 |
No | output | 1 | 70,274 | 7 | 140,549 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Genos recently installed the game Zuma on his phone. In Zuma there exists a line of n gemstones, the i-th of which has color ci. The goal of the game is to destroy all the gemstones in the line as quickly as possible.
In one second, Genos is able to choose exactly one continuous substring of colored gemstones that is a palindrome and remove it from the line. After the substring is removed, the remaining gemstones shift to form a solid line again. What is the minimum number of seconds needed to destroy the entire line?
Let us remind, that the string (or substring) is called palindrome, if it reads same backwards or forward. In our case this means the color of the first gemstone is equal to the color of the last one, the color of the second gemstone is equal to the color of the next to last and so on.
Input
The first line of input contains a single integer n (1 β€ n β€ 500) β the number of gemstones.
The second line contains n space-separated integers, the i-th of which is ci (1 β€ ci β€ n) β the color of the i-th gemstone in a line.
Output
Print a single integer β the minimum number of seconds needed to destroy the entire line.
Examples
Input
3
1 2 1
Output
1
Input
3
1 2 3
Output
3
Input
7
1 4 4 2 3 2 1
Output
2
Note
In the first sample, Genos can destroy the entire line in one second.
In the second sample, Genos can only destroy one gemstone at a time, so destroying three gemstones takes three seconds.
In the third sample, to achieve the optimal time of two seconds, destroy palindrome 4 4 first and then destroy palindrome 1 2 3 2 1.
Submitted Solution:
```
# -*- coding:utf-8 -*-
"""
created by shuangquan.huang at 1/9/20
"""
import collections
import time
import os
import sys
import bisect
import heapq
from typing import List
def solve(N, A):
dp = [[N for _ in range(N+1)] for _ in range(N)]
for i in range(N):
dp[i][0] = 0
dp[i][1] = 1
for l in range(2, N+1):
for i in range(N-l+1):
for j in range(1, l):
dp[i][l] = min(dp[i][l], dp[i][j] + dp[i+j][l-j])
# j = 0
# while j < l // 2 and A[i+j] == A[i+l-j-1]:
# j += 1
# if j >= l // 2:
# dp[i][l] = min(dp[i][l], 1)
# elif j > 0:
# dp[i][l] = min(dp[i][l], 1 + dp[i+j][l-2*j])
# for k in range(i+j, i+l-j):
# dp[i][l] = min(dp[i][l], 1 + dp[i+j][k-i-j] + dp[k][l - 2 * j-1])
if i + 1 < N and A[i] == A[i+l-1]:
dp[i][l] = max(1, dp[i+1][l-2])
else:
dp[i][l] = min(dp[i][l], 1 + dp[i][l-1])
if i + 1 < N:
dp[i][l] = min(dp[i][l], 1 + dp[i+1][l-1])
return dp[0][N]
N = int(input())
A = [int(x) for x in input().split()]
print(solve(N, A))
``` | instruction | 0 | 70,275 | 7 | 140,550 |
No | output | 1 | 70,275 | 7 | 140,551 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Bob is decorating his kitchen, more precisely, the floor. He has found a prime candidate for the tiles he will use. They come in a simple form factor β a square tile that is diagonally split into white and black part as depicted in the figure below.
<image>
The dimension of this tile is perfect for this kitchen, as he will need exactly w Γ h tiles without any scraps. That is, the width of the kitchen is w tiles, and the height is h tiles. As each tile can be rotated in one of four ways, he still needs to decide on how exactly he will tile the floor. There is a single aesthetic criterion that he wants to fulfil: two adjacent tiles must not share a colour on the edge β i.e. one of the tiles must have a white colour on the shared border, and the second one must be black.
<image> The picture on the left shows one valid tiling of a 3 Γ 2 kitchen. The picture on the right shows an invalid arrangement, as the bottom two tiles touch with their white parts.
Find the number of possible tilings. As this number may be large, output its remainder when divided by 998244353 (a prime number).
Input
The only line contains two space separated integers w, h (1 β€ w,h β€ 1 000) β the width and height of the kitchen, measured in tiles.
Output
Output a single integer n β the remainder of the number of tilings when divided by 998244353.
Examples
Input
2 2
Output
16
Input
2 4
Output
64 | instruction | 0 | 70,801 | 7 | 141,602 |
Tags: combinatorics, greedy, math
Correct Solution:
```
import math
wh=input().split()
w=int(wh[0])
h=int(wh[1])
r=pow(2,w+h,998244353)
print(r)
``` | output | 1 | 70,801 | 7 | 141,603 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Bob is decorating his kitchen, more precisely, the floor. He has found a prime candidate for the tiles he will use. They come in a simple form factor β a square tile that is diagonally split into white and black part as depicted in the figure below.
<image>
The dimension of this tile is perfect for this kitchen, as he will need exactly w Γ h tiles without any scraps. That is, the width of the kitchen is w tiles, and the height is h tiles. As each tile can be rotated in one of four ways, he still needs to decide on how exactly he will tile the floor. There is a single aesthetic criterion that he wants to fulfil: two adjacent tiles must not share a colour on the edge β i.e. one of the tiles must have a white colour on the shared border, and the second one must be black.
<image> The picture on the left shows one valid tiling of a 3 Γ 2 kitchen. The picture on the right shows an invalid arrangement, as the bottom two tiles touch with their white parts.
Find the number of possible tilings. As this number may be large, output its remainder when divided by 998244353 (a prime number).
Input
The only line contains two space separated integers w, h (1 β€ w,h β€ 1 000) β the width and height of the kitchen, measured in tiles.
Output
Output a single integer n β the remainder of the number of tilings when divided by 998244353.
Examples
Input
2 2
Output
16
Input
2 4
Output
64 | instruction | 0 | 70,802 | 7 | 141,604 |
Tags: combinatorics, greedy, math
Correct Solution:
```
fst = list(map(int,input().split()))
print(pow(2,fst[0]+fst[1])%998244353)
``` | output | 1 | 70,802 | 7 | 141,605 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Bob is decorating his kitchen, more precisely, the floor. He has found a prime candidate for the tiles he will use. They come in a simple form factor β a square tile that is diagonally split into white and black part as depicted in the figure below.
<image>
The dimension of this tile is perfect for this kitchen, as he will need exactly w Γ h tiles without any scraps. That is, the width of the kitchen is w tiles, and the height is h tiles. As each tile can be rotated in one of four ways, he still needs to decide on how exactly he will tile the floor. There is a single aesthetic criterion that he wants to fulfil: two adjacent tiles must not share a colour on the edge β i.e. one of the tiles must have a white colour on the shared border, and the second one must be black.
<image> The picture on the left shows one valid tiling of a 3 Γ 2 kitchen. The picture on the right shows an invalid arrangement, as the bottom two tiles touch with their white parts.
Find the number of possible tilings. As this number may be large, output its remainder when divided by 998244353 (a prime number).
Input
The only line contains two space separated integers w, h (1 β€ w,h β€ 1 000) β the width and height of the kitchen, measured in tiles.
Output
Output a single integer n β the remainder of the number of tilings when divided by 998244353.
Examples
Input
2 2
Output
16
Input
2 4
Output
64 | instruction | 0 | 70,803 | 7 | 141,606 |
Tags: combinatorics, greedy, math
Correct Solution:
```
w, h = list(int(a) for a in input().split())
print((2**(w+h))%998244353)
``` | output | 1 | 70,803 | 7 | 141,607 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Bob is decorating his kitchen, more precisely, the floor. He has found a prime candidate for the tiles he will use. They come in a simple form factor β a square tile that is diagonally split into white and black part as depicted in the figure below.
<image>
The dimension of this tile is perfect for this kitchen, as he will need exactly w Γ h tiles without any scraps. That is, the width of the kitchen is w tiles, and the height is h tiles. As each tile can be rotated in one of four ways, he still needs to decide on how exactly he will tile the floor. There is a single aesthetic criterion that he wants to fulfil: two adjacent tiles must not share a colour on the edge β i.e. one of the tiles must have a white colour on the shared border, and the second one must be black.
<image> The picture on the left shows one valid tiling of a 3 Γ 2 kitchen. The picture on the right shows an invalid arrangement, as the bottom two tiles touch with their white parts.
Find the number of possible tilings. As this number may be large, output its remainder when divided by 998244353 (a prime number).
Input
The only line contains two space separated integers w, h (1 β€ w,h β€ 1 000) β the width and height of the kitchen, measured in tiles.
Output
Output a single integer n β the remainder of the number of tilings when divided by 998244353.
Examples
Input
2 2
Output
16
Input
2 4
Output
64 | instruction | 0 | 70,804 | 7 | 141,608 |
Tags: combinatorics, greedy, math
Correct Solution:
```
w,h = input().split()
w=int(w)
h=int(h)
ans = 1
for i in range(w+h):
ans = (ans*2)%998244353
print(ans)
``` | output | 1 | 70,804 | 7 | 141,609 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Bob is decorating his kitchen, more precisely, the floor. He has found a prime candidate for the tiles he will use. They come in a simple form factor β a square tile that is diagonally split into white and black part as depicted in the figure below.
<image>
The dimension of this tile is perfect for this kitchen, as he will need exactly w Γ h tiles without any scraps. That is, the width of the kitchen is w tiles, and the height is h tiles. As each tile can be rotated in one of four ways, he still needs to decide on how exactly he will tile the floor. There is a single aesthetic criterion that he wants to fulfil: two adjacent tiles must not share a colour on the edge β i.e. one of the tiles must have a white colour on the shared border, and the second one must be black.
<image> The picture on the left shows one valid tiling of a 3 Γ 2 kitchen. The picture on the right shows an invalid arrangement, as the bottom two tiles touch with their white parts.
Find the number of possible tilings. As this number may be large, output its remainder when divided by 998244353 (a prime number).
Input
The only line contains two space separated integers w, h (1 β€ w,h β€ 1 000) β the width and height of the kitchen, measured in tiles.
Output
Output a single integer n β the remainder of the number of tilings when divided by 998244353.
Examples
Input
2 2
Output
16
Input
2 4
Output
64 | instruction | 0 | 70,805 | 7 | 141,610 |
Tags: combinatorics, greedy, math
Correct Solution:
```
def plitki(w, h):
mod = 998244353
result = 1
while w > 0:
result = (2 * result) % mod
w -= 1
while h > 0:
result = (2 * result) % mod
h -= 1
return result
W, H = [int(i) for i in input().split()]
print(plitki(W, H))
``` | output | 1 | 70,805 | 7 | 141,611 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Bob is decorating his kitchen, more precisely, the floor. He has found a prime candidate for the tiles he will use. They come in a simple form factor β a square tile that is diagonally split into white and black part as depicted in the figure below.
<image>
The dimension of this tile is perfect for this kitchen, as he will need exactly w Γ h tiles without any scraps. That is, the width of the kitchen is w tiles, and the height is h tiles. As each tile can be rotated in one of four ways, he still needs to decide on how exactly he will tile the floor. There is a single aesthetic criterion that he wants to fulfil: two adjacent tiles must not share a colour on the edge β i.e. one of the tiles must have a white colour on the shared border, and the second one must be black.
<image> The picture on the left shows one valid tiling of a 3 Γ 2 kitchen. The picture on the right shows an invalid arrangement, as the bottom two tiles touch with their white parts.
Find the number of possible tilings. As this number may be large, output its remainder when divided by 998244353 (a prime number).
Input
The only line contains two space separated integers w, h (1 β€ w,h β€ 1 000) β the width and height of the kitchen, measured in tiles.
Output
Output a single integer n β the remainder of the number of tilings when divided by 998244353.
Examples
Input
2 2
Output
16
Input
2 4
Output
64 | instruction | 0 | 70,806 | 7 | 141,612 |
Tags: combinatorics, greedy, math
Correct Solution:
```
# cook your dish here
w,h=map(int,input().split())
print(int(pow(2,h+w))% 998244353)
``` | output | 1 | 70,806 | 7 | 141,613 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Bob is decorating his kitchen, more precisely, the floor. He has found a prime candidate for the tiles he will use. They come in a simple form factor β a square tile that is diagonally split into white and black part as depicted in the figure below.
<image>
The dimension of this tile is perfect for this kitchen, as he will need exactly w Γ h tiles without any scraps. That is, the width of the kitchen is w tiles, and the height is h tiles. As each tile can be rotated in one of four ways, he still needs to decide on how exactly he will tile the floor. There is a single aesthetic criterion that he wants to fulfil: two adjacent tiles must not share a colour on the edge β i.e. one of the tiles must have a white colour on the shared border, and the second one must be black.
<image> The picture on the left shows one valid tiling of a 3 Γ 2 kitchen. The picture on the right shows an invalid arrangement, as the bottom two tiles touch with their white parts.
Find the number of possible tilings. As this number may be large, output its remainder when divided by 998244353 (a prime number).
Input
The only line contains two space separated integers w, h (1 β€ w,h β€ 1 000) β the width and height of the kitchen, measured in tiles.
Output
Output a single integer n β the remainder of the number of tilings when divided by 998244353.
Examples
Input
2 2
Output
16
Input
2 4
Output
64 | instruction | 0 | 70,807 | 7 | 141,614 |
Tags: combinatorics, greedy, math
Correct Solution:
```
# alpha = "abcdefghijklmnopqrstuvwxyz"
prime = 998244353
# INF = 1000_000_000
# from heapq import heappush, heappop
# from collections import defaultdict
# from math import sqrt
# from collections import deque
# from math import gcd
# n = int(input())
# arr = list(map(int, input().split()))
w, h = list(map(int, input().split()))
print(pow(2,w+h,prime))
``` | output | 1 | 70,807 | 7 | 141,615 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Bob is decorating his kitchen, more precisely, the floor. He has found a prime candidate for the tiles he will use. They come in a simple form factor β a square tile that is diagonally split into white and black part as depicted in the figure below.
<image>
The dimension of this tile is perfect for this kitchen, as he will need exactly w Γ h tiles without any scraps. That is, the width of the kitchen is w tiles, and the height is h tiles. As each tile can be rotated in one of four ways, he still needs to decide on how exactly he will tile the floor. There is a single aesthetic criterion that he wants to fulfil: two adjacent tiles must not share a colour on the edge β i.e. one of the tiles must have a white colour on the shared border, and the second one must be black.
<image> The picture on the left shows one valid tiling of a 3 Γ 2 kitchen. The picture on the right shows an invalid arrangement, as the bottom two tiles touch with their white parts.
Find the number of possible tilings. As this number may be large, output its remainder when divided by 998244353 (a prime number).
Input
The only line contains two space separated integers w, h (1 β€ w,h β€ 1 000) β the width and height of the kitchen, measured in tiles.
Output
Output a single integer n β the remainder of the number of tilings when divided by 998244353.
Examples
Input
2 2
Output
16
Input
2 4
Output
64 | instruction | 0 | 70,808 | 7 | 141,616 |
Tags: combinatorics, greedy, math
Correct Solution:
```
import sys
input = sys.stdin.readline
"""
"""
w, h = map(int, input().split())
def exp_mod(base, exp, mod):
if exp == 0:
return 1
elif exp == 1:
return base
elif exp % 2 == 0:
return ((exp_mod(base, exp/2, mod) % mod) * (exp_mod(base, exp/2, mod) % mod)) % mod
else:
return (base * exp_mod(base, exp-1, mod)) % mod
print(exp_mod(2, w+h, 998244353))
``` | output | 1 | 70,808 | 7 | 141,617 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Bob is decorating his kitchen, more precisely, the floor. He has found a prime candidate for the tiles he will use. They come in a simple form factor β a square tile that is diagonally split into white and black part as depicted in the figure below.
<image>
The dimension of this tile is perfect for this kitchen, as he will need exactly w Γ h tiles without any scraps. That is, the width of the kitchen is w tiles, and the height is h tiles. As each tile can be rotated in one of four ways, he still needs to decide on how exactly he will tile the floor. There is a single aesthetic criterion that he wants to fulfil: two adjacent tiles must not share a colour on the edge β i.e. one of the tiles must have a white colour on the shared border, and the second one must be black.
<image> The picture on the left shows one valid tiling of a 3 Γ 2 kitchen. The picture on the right shows an invalid arrangement, as the bottom two tiles touch with their white parts.
Find the number of possible tilings. As this number may be large, output its remainder when divided by 998244353 (a prime number).
Input
The only line contains two space separated integers w, h (1 β€ w,h β€ 1 000) β the width and height of the kitchen, measured in tiles.
Output
Output a single integer n β the remainder of the number of tilings when divided by 998244353.
Examples
Input
2 2
Output
16
Input
2 4
Output
64
Submitted Solution:
```
l,b=map(int,input().split())
print((2**(l+b))%998244353)
``` | instruction | 0 | 70,809 | 7 | 141,618 |
Yes | output | 1 | 70,809 | 7 | 141,619 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Bob is decorating his kitchen, more precisely, the floor. He has found a prime candidate for the tiles he will use. They come in a simple form factor β a square tile that is diagonally split into white and black part as depicted in the figure below.
<image>
The dimension of this tile is perfect for this kitchen, as he will need exactly w Γ h tiles without any scraps. That is, the width of the kitchen is w tiles, and the height is h tiles. As each tile can be rotated in one of four ways, he still needs to decide on how exactly he will tile the floor. There is a single aesthetic criterion that he wants to fulfil: two adjacent tiles must not share a colour on the edge β i.e. one of the tiles must have a white colour on the shared border, and the second one must be black.
<image> The picture on the left shows one valid tiling of a 3 Γ 2 kitchen. The picture on the right shows an invalid arrangement, as the bottom two tiles touch with their white parts.
Find the number of possible tilings. As this number may be large, output its remainder when divided by 998244353 (a prime number).
Input
The only line contains two space separated integers w, h (1 β€ w,h β€ 1 000) β the width and height of the kitchen, measured in tiles.
Output
Output a single integer n β the remainder of the number of tilings when divided by 998244353.
Examples
Input
2 2
Output
16
Input
2 4
Output
64
Submitted Solution:
```
w, h = (input().split())
print(pow(2, int(w) + int(h)) % 998244353)
``` | instruction | 0 | 70,810 | 7 | 141,620 |
Yes | output | 1 | 70,810 | 7 | 141,621 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Bob is decorating his kitchen, more precisely, the floor. He has found a prime candidate for the tiles he will use. They come in a simple form factor β a square tile that is diagonally split into white and black part as depicted in the figure below.
<image>
The dimension of this tile is perfect for this kitchen, as he will need exactly w Γ h tiles without any scraps. That is, the width of the kitchen is w tiles, and the height is h tiles. As each tile can be rotated in one of four ways, he still needs to decide on how exactly he will tile the floor. There is a single aesthetic criterion that he wants to fulfil: two adjacent tiles must not share a colour on the edge β i.e. one of the tiles must have a white colour on the shared border, and the second one must be black.
<image> The picture on the left shows one valid tiling of a 3 Γ 2 kitchen. The picture on the right shows an invalid arrangement, as the bottom two tiles touch with their white parts.
Find the number of possible tilings. As this number may be large, output its remainder when divided by 998244353 (a prime number).
Input
The only line contains two space separated integers w, h (1 β€ w,h β€ 1 000) β the width and height of the kitchen, measured in tiles.
Output
Output a single integer n β the remainder of the number of tilings when divided by 998244353.
Examples
Input
2 2
Output
16
Input
2 4
Output
64
Submitted Solution:
```
a, b = input().split()
'''
print(((int(a) * int(b)) ** 2) % 998244353)
'''
print((2**(int(a)+int(b))) % 998244353)
``` | instruction | 0 | 70,811 | 7 | 141,622 |
Yes | output | 1 | 70,811 | 7 | 141,623 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Bob is decorating his kitchen, more precisely, the floor. He has found a prime candidate for the tiles he will use. They come in a simple form factor β a square tile that is diagonally split into white and black part as depicted in the figure below.
<image>
The dimension of this tile is perfect for this kitchen, as he will need exactly w Γ h tiles without any scraps. That is, the width of the kitchen is w tiles, and the height is h tiles. As each tile can be rotated in one of four ways, he still needs to decide on how exactly he will tile the floor. There is a single aesthetic criterion that he wants to fulfil: two adjacent tiles must not share a colour on the edge β i.e. one of the tiles must have a white colour on the shared border, and the second one must be black.
<image> The picture on the left shows one valid tiling of a 3 Γ 2 kitchen. The picture on the right shows an invalid arrangement, as the bottom two tiles touch with their white parts.
Find the number of possible tilings. As this number may be large, output its remainder when divided by 998244353 (a prime number).
Input
The only line contains two space separated integers w, h (1 β€ w,h β€ 1 000) β the width and height of the kitchen, measured in tiles.
Output
Output a single integer n β the remainder of the number of tilings when divided by 998244353.
Examples
Input
2 2
Output
16
Input
2 4
Output
64
Submitted Solution:
```
w, h = map(int, input().split())
print(2**(h + w) % 998244353)
``` | instruction | 0 | 70,812 | 7 | 141,624 |
Yes | output | 1 | 70,812 | 7 | 141,625 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Bob is decorating his kitchen, more precisely, the floor. He has found a prime candidate for the tiles he will use. They come in a simple form factor β a square tile that is diagonally split into white and black part as depicted in the figure below.
<image>
The dimension of this tile is perfect for this kitchen, as he will need exactly w Γ h tiles without any scraps. That is, the width of the kitchen is w tiles, and the height is h tiles. As each tile can be rotated in one of four ways, he still needs to decide on how exactly he will tile the floor. There is a single aesthetic criterion that he wants to fulfil: two adjacent tiles must not share a colour on the edge β i.e. one of the tiles must have a white colour on the shared border, and the second one must be black.
<image> The picture on the left shows one valid tiling of a 3 Γ 2 kitchen. The picture on the right shows an invalid arrangement, as the bottom two tiles touch with their white parts.
Find the number of possible tilings. As this number may be large, output its remainder when divided by 998244353 (a prime number).
Input
The only line contains two space separated integers w, h (1 β€ w,h β€ 1 000) β the width and height of the kitchen, measured in tiles.
Output
Output a single integer n β the remainder of the number of tilings when divided by 998244353.
Examples
Input
2 2
Output
16
Input
2 4
Output
64
Submitted Solution:
```
def main():
w,h = map(int,input().split())
ans = 4
mod = 998244353
for i in range(h):
for j in range(w):
if i == 0:
if j%2 != 0:
ans *= 2
else:
if j%2 == 0:
ans *= 2
ans = ans%mod
print(ans)
main()
``` | instruction | 0 | 70,813 | 7 | 141,626 |
No | output | 1 | 70,813 | 7 | 141,627 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Bob is decorating his kitchen, more precisely, the floor. He has found a prime candidate for the tiles he will use. They come in a simple form factor β a square tile that is diagonally split into white and black part as depicted in the figure below.
<image>
The dimension of this tile is perfect for this kitchen, as he will need exactly w Γ h tiles without any scraps. That is, the width of the kitchen is w tiles, and the height is h tiles. As each tile can be rotated in one of four ways, he still needs to decide on how exactly he will tile the floor. There is a single aesthetic criterion that he wants to fulfil: two adjacent tiles must not share a colour on the edge β i.e. one of the tiles must have a white colour on the shared border, and the second one must be black.
<image> The picture on the left shows one valid tiling of a 3 Γ 2 kitchen. The picture on the right shows an invalid arrangement, as the bottom two tiles touch with their white parts.
Find the number of possible tilings. As this number may be large, output its remainder when divided by 998244353 (a prime number).
Input
The only line contains two space separated integers w, h (1 β€ w,h β€ 1 000) β the width and height of the kitchen, measured in tiles.
Output
Output a single integer n β the remainder of the number of tilings when divided by 998244353.
Examples
Input
2 2
Output
16
Input
2 4
Output
64
Submitted Solution:
```
import math
import sys
from collections import Counter
# imgur.com/Pkt7iIf.png
def getdict(n):
d = {}
if type(n) is list:
for i in n:
if i in d: d[i] += 1
else: d[i] = 1
else:
for i in range(n):
t = ii()
if t in d: d[t] += 1
else: d[t] = 1
return d
def cdiv(n, k): return n//k + (n%k != 0)
def ii(): return int(input())
def mi(): return map(int, input().split())
def li(): return list(map(int, input().split()))
n, k = mi()
print((n*k)**2)
``` | instruction | 0 | 70,814 | 7 | 141,628 |
No | output | 1 | 70,814 | 7 | 141,629 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Bob is decorating his kitchen, more precisely, the floor. He has found a prime candidate for the tiles he will use. They come in a simple form factor β a square tile that is diagonally split into white and black part as depicted in the figure below.
<image>
The dimension of this tile is perfect for this kitchen, as he will need exactly w Γ h tiles without any scraps. That is, the width of the kitchen is w tiles, and the height is h tiles. As each tile can be rotated in one of four ways, he still needs to decide on how exactly he will tile the floor. There is a single aesthetic criterion that he wants to fulfil: two adjacent tiles must not share a colour on the edge β i.e. one of the tiles must have a white colour on the shared border, and the second one must be black.
<image> The picture on the left shows one valid tiling of a 3 Γ 2 kitchen. The picture on the right shows an invalid arrangement, as the bottom two tiles touch with their white parts.
Find the number of possible tilings. As this number may be large, output its remainder when divided by 998244353 (a prime number).
Input
The only line contains two space separated integers w, h (1 β€ w,h β€ 1 000) β the width and height of the kitchen, measured in tiles.
Output
Output a single integer n β the remainder of the number of tilings when divided by 998244353.
Examples
Input
2 2
Output
16
Input
2 4
Output
64
Submitted Solution:
```
a, b = map(int, input().split())
print((a * b) ** 2 % 998244353)
``` | instruction | 0 | 70,815 | 7 | 141,630 |
No | output | 1 | 70,815 | 7 | 141,631 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Bob is decorating his kitchen, more precisely, the floor. He has found a prime candidate for the tiles he will use. They come in a simple form factor β a square tile that is diagonally split into white and black part as depicted in the figure below.
<image>
The dimension of this tile is perfect for this kitchen, as he will need exactly w Γ h tiles without any scraps. That is, the width of the kitchen is w tiles, and the height is h tiles. As each tile can be rotated in one of four ways, he still needs to decide on how exactly he will tile the floor. There is a single aesthetic criterion that he wants to fulfil: two adjacent tiles must not share a colour on the edge β i.e. one of the tiles must have a white colour on the shared border, and the second one must be black.
<image> The picture on the left shows one valid tiling of a 3 Γ 2 kitchen. The picture on the right shows an invalid arrangement, as the bottom two tiles touch with their white parts.
Find the number of possible tilings. As this number may be large, output its remainder when divided by 998244353 (a prime number).
Input
The only line contains two space separated integers w, h (1 β€ w,h β€ 1 000) β the width and height of the kitchen, measured in tiles.
Output
Output a single integer n β the remainder of the number of tilings when divided by 998244353.
Examples
Input
2 2
Output
16
Input
2 4
Output
64
Submitted Solution:
```
a,b=list(map(int,input().split()))
print(a,b)
mod=998244353
if a==1 or b==1:
ans=4*(2**(max(a,b)-1))
else:
ans=4*(2**(a+b-2))
print(ans%mod)
``` | instruction | 0 | 70,816 | 7 | 141,632 |
No | output | 1 | 70,816 | 7 | 141,633 |
Provide tags and a correct Python 3 solution for this coding contest problem.
In the game of Mastermind, there are two players β Alice and Bob. Alice has a secret code, which Bob tries to guess. Here, a code is defined as a sequence of n colors. There are exactly n+1 colors in the entire universe, numbered from 1 to n+1 inclusive.
When Bob guesses a code, Alice tells him some information about how good of a guess it is, in the form of two integers x and y.
The first integer x is the number of indices where Bob's guess correctly matches Alice's code. The second integer y is the size of the intersection of the two codes as multisets. That is, if Bob were to change the order of the colors in his guess, y is the maximum number of indices he could get correct.
For example, suppose n=5, Alice's code is [3,1,6,1,2], and Bob's guess is [3,1,1,2,5]. At indices 1 and 2 colors are equal, while in the other indices they are not equal. So x=2. And the two codes have the four colors 1,1,2,3 in common, so y=4.
<image> Solid lines denote a matched color for the same index. Dashed lines denote a matched color at a different index. x is the number of solid lines, and y is the total number of lines.
You are given Bob's guess and two values x and y. Can you find one possibility of Alice's code so that the values of x and y are correct?
Input
The first line contains a single integer t (1β€ tβ€ 1000) β the number of test cases. Next 2t lines contain descriptions of test cases.
The first line of each test case contains three integers n,x,y (1β€ nβ€ 10^5, 0β€ xβ€ yβ€ n) β the length of the codes, and two values Alice responds with.
The second line of each test case contains n integers b_1,β¦,b_n (1β€ b_iβ€ n+1) β Bob's guess, where b_i is the i-th color of the guess.
It is guaranteed that the sum of n across all test cases does not exceed 10^5.
Output
For each test case, on the first line, output "YES" if there is a solution, or "NO" if there is no possible secret code consistent with the described situation. You can print each character in any case (upper or lower).
If the answer is "YES", on the next line output n integers a_1,β¦,a_n (1β€ a_iβ€ n+1) β Alice's secret code, where a_i is the i-th color of the code.
If there are multiple solutions, output any.
Example
Input
7
5 2 4
3 1 1 2 5
5 3 4
1 1 2 1 2
4 0 4
5 5 3 3
4 1 4
2 3 2 3
6 1 2
3 2 1 1 1 1
6 2 4
3 3 2 1 1 1
6 2 6
1 1 3 2 1 1
Output
YES
3 1 6 1 2
YES
3 1 1 1 2
YES
3 3 5 5
NO
YES
4 4 4 4 3 1
YES
3 1 3 1 7 7
YES
2 3 1 1 1 1
Note
The first test case is described in the statement.
In the second test case, x=3 because the colors are equal at indices 2,4,5. And y=4 because they share the colors 1,1,1,2.
In the third test case, x=0 because there is no index where the colors are the same. But y=4 because they share the colors 3,3,5,5.
In the fourth test case, it can be proved that no solution exists. | instruction | 0 | 70,898 | 7 | 141,796 |
Tags: constructive algorithms, graph matchings, greedy, implementation, sortings, two pointers
Correct Solution:
```
import collections
t = int(input())
for _ in range(t):
# print('time', _)
n, x, y = map(int, input().split())
b = [int(i) for i in input().split()]
if x == n:
print("YES")
for i in b:
print(i, end = ' ')
print()
continue
setB = set(b)
noin = 1
while noin in setB:
noin += 1
cnt = collections.deque([i, j] for i, j in collections.Counter(b).most_common())
cnt2 = collections.deque()
pos = [set() for _ in range(n + 2)]
for i in range(len(b)):
pos[b[i]].add(i)
res=[-1] * n
l = x
while l > 0:
i = cnt.popleft()
i[1] -= 1
if i[1] != 0:
cnt2.appendleft(i)
while not cnt or cnt2 and cnt2[0][1] >= cnt[0][1]:
cnt.appendleft(cnt2.popleft())
p = pos[i[0]].pop()
res[p] = i[0]
l -= 1
ml = cnt[0][1]
remain = []
while cnt:
i, j = cnt.popleft()
remain += [i] * j
while cnt2:
i, j =cnt2.popleft()
remain += [i] * j
if y - x > (len(remain) - ml) * 2 :
print('No')
continue
i = 0
match = [0] * len(remain)
for i in range(ml, ml + y - x):
match[i % len(remain)] = remain[(i + ml) % len(remain)]
for i in range(ml + y - x, ml + len(remain)):
match[i % len(remain)] = noin
# print(match, remain)
for i in range(len(match)):
p = pos[remain[i]].pop()
res[p] = match[i]
if l == 0:
print("YES")
for i in res:
print(i, end=' ')
print()
else:
print("No")
``` | output | 1 | 70,898 | 7 | 141,797 |
Provide tags and a correct Python 3 solution for this coding contest problem.
In the game of Mastermind, there are two players β Alice and Bob. Alice has a secret code, which Bob tries to guess. Here, a code is defined as a sequence of n colors. There are exactly n+1 colors in the entire universe, numbered from 1 to n+1 inclusive.
When Bob guesses a code, Alice tells him some information about how good of a guess it is, in the form of two integers x and y.
The first integer x is the number of indices where Bob's guess correctly matches Alice's code. The second integer y is the size of the intersection of the two codes as multisets. That is, if Bob were to change the order of the colors in his guess, y is the maximum number of indices he could get correct.
For example, suppose n=5, Alice's code is [3,1,6,1,2], and Bob's guess is [3,1,1,2,5]. At indices 1 and 2 colors are equal, while in the other indices they are not equal. So x=2. And the two codes have the four colors 1,1,2,3 in common, so y=4.
<image> Solid lines denote a matched color for the same index. Dashed lines denote a matched color at a different index. x is the number of solid lines, and y is the total number of lines.
You are given Bob's guess and two values x and y. Can you find one possibility of Alice's code so that the values of x and y are correct?
Input
The first line contains a single integer t (1β€ tβ€ 1000) β the number of test cases. Next 2t lines contain descriptions of test cases.
The first line of each test case contains three integers n,x,y (1β€ nβ€ 10^5, 0β€ xβ€ yβ€ n) β the length of the codes, and two values Alice responds with.
The second line of each test case contains n integers b_1,β¦,b_n (1β€ b_iβ€ n+1) β Bob's guess, where b_i is the i-th color of the guess.
It is guaranteed that the sum of n across all test cases does not exceed 10^5.
Output
For each test case, on the first line, output "YES" if there is a solution, or "NO" if there is no possible secret code consistent with the described situation. You can print each character in any case (upper or lower).
If the answer is "YES", on the next line output n integers a_1,β¦,a_n (1β€ a_iβ€ n+1) β Alice's secret code, where a_i is the i-th color of the code.
If there are multiple solutions, output any.
Example
Input
7
5 2 4
3 1 1 2 5
5 3 4
1 1 2 1 2
4 0 4
5 5 3 3
4 1 4
2 3 2 3
6 1 2
3 2 1 1 1 1
6 2 4
3 3 2 1 1 1
6 2 6
1 1 3 2 1 1
Output
YES
3 1 6 1 2
YES
3 1 1 1 2
YES
3 3 5 5
NO
YES
4 4 4 4 3 1
YES
3 1 3 1 7 7
YES
2 3 1 1 1 1
Note
The first test case is described in the statement.
In the second test case, x=3 because the colors are equal at indices 2,4,5. And y=4 because they share the colors 1,1,1,2.
In the third test case, x=0 because there is no index where the colors are the same. But y=4 because they share the colors 3,3,5,5.
In the fourth test case, it can be proved that no solution exists. | instruction | 0 | 70,899 | 7 | 141,798 |
Tags: constructive algorithms, graph matchings, greedy, implementation, sortings, two pointers
Correct Solution:
```
from sys import stdin, stdout
from collections import defaultdict
from heapq import heapify, heappop, heappush
def solve():
n, s, y = map(int, stdin.readline().split())
a = stdin.readline().split()
d = defaultdict(list)
for i, x in enumerate(a):
d[x].append(i)
for i in range(1, n + 2):
e = str(i)
if e not in d:
break
q = [(-len(d[x]), x) for x in d.keys()]
heapify(q)
ans = [0] * n
for i in range(s):
l, x = heappop(q)
ans[d[x].pop()] = x
l += 1
if l:
heappush(q, (l, x))
p = []
while q:
l, x = heappop(q)
p.extend(d[x])
if p:
h = (n - s) // 2
y = n - y
q = p[h:] + p[:h]
for x, z in zip(p, q):
if a[x] == a[z]:
if y:
ans[x] = e
y -= 1
else:
stdout.write("NO\n")
return
else:
ans[x] = a[z]
for i in range(n - s):
if y and ans[p[i]] != e:
ans[p[i]] = e
y -= 1
print("YES")
print(' '.join(ans))
T = int(stdin.readline())
for t in range(T):
solve()
``` | output | 1 | 70,899 | 7 | 141,799 |
Provide tags and a correct Python 3 solution for this coding contest problem.
In the game of Mastermind, there are two players β Alice and Bob. Alice has a secret code, which Bob tries to guess. Here, a code is defined as a sequence of n colors. There are exactly n+1 colors in the entire universe, numbered from 1 to n+1 inclusive.
When Bob guesses a code, Alice tells him some information about how good of a guess it is, in the form of two integers x and y.
The first integer x is the number of indices where Bob's guess correctly matches Alice's code. The second integer y is the size of the intersection of the two codes as multisets. That is, if Bob were to change the order of the colors in his guess, y is the maximum number of indices he could get correct.
For example, suppose n=5, Alice's code is [3,1,6,1,2], and Bob's guess is [3,1,1,2,5]. At indices 1 and 2 colors are equal, while in the other indices they are not equal. So x=2. And the two codes have the four colors 1,1,2,3 in common, so y=4.
<image> Solid lines denote a matched color for the same index. Dashed lines denote a matched color at a different index. x is the number of solid lines, and y is the total number of lines.
You are given Bob's guess and two values x and y. Can you find one possibility of Alice's code so that the values of x and y are correct?
Input
The first line contains a single integer t (1β€ tβ€ 1000) β the number of test cases. Next 2t lines contain descriptions of test cases.
The first line of each test case contains three integers n,x,y (1β€ nβ€ 10^5, 0β€ xβ€ yβ€ n) β the length of the codes, and two values Alice responds with.
The second line of each test case contains n integers b_1,β¦,b_n (1β€ b_iβ€ n+1) β Bob's guess, where b_i is the i-th color of the guess.
It is guaranteed that the sum of n across all test cases does not exceed 10^5.
Output
For each test case, on the first line, output "YES" if there is a solution, or "NO" if there is no possible secret code consistent with the described situation. You can print each character in any case (upper or lower).
If the answer is "YES", on the next line output n integers a_1,β¦,a_n (1β€ a_iβ€ n+1) β Alice's secret code, where a_i is the i-th color of the code.
If there are multiple solutions, output any.
Example
Input
7
5 2 4
3 1 1 2 5
5 3 4
1 1 2 1 2
4 0 4
5 5 3 3
4 1 4
2 3 2 3
6 1 2
3 2 1 1 1 1
6 2 4
3 3 2 1 1 1
6 2 6
1 1 3 2 1 1
Output
YES
3 1 6 1 2
YES
3 1 1 1 2
YES
3 3 5 5
NO
YES
4 4 4 4 3 1
YES
3 1 3 1 7 7
YES
2 3 1 1 1 1
Note
The first test case is described in the statement.
In the second test case, x=3 because the colors are equal at indices 2,4,5. And y=4 because they share the colors 1,1,1,2.
In the third test case, x=0 because there is no index where the colors are the same. But y=4 because they share the colors 3,3,5,5.
In the fourth test case, it can be proved that no solution exists. | instruction | 0 | 70,900 | 7 | 141,800 |
Tags: constructive algorithms, graph matchings, greedy, implementation, sortings, two pointers
Correct Solution:
```
from sys import stdin, stdout
from collections import defaultdict
from heapq import heapify, heappop, heappush
def solve():
n, s, y = map(int, input().split())
a = input().split()
d = defaultdict(list)
for i, x in enumerate(a):
d[x].append(i)
for i in range(1, n + 2):
e = str(i)
if e not in d:
break
q = [(-len(d[x]), x) for x in d.keys()]
heapify(q)
ans = [0] * n
for i in range(s):
l, x = heappop(q)
ans[d[x].pop()] = x
l += 1
if l:
heappush(q, (l, x))
p = []
while q:
l, x = heappop(q)
p.extend(d[x])
if p:
h = (n - s) // 2
y = n - y
q = p[h:] + p[:h]
for x, z in zip(p, q):
if a[x] == a[z]:
if y:
ans[x] = e
y -= 1
else:
stdout.write("NO\n")
return
else:
ans[x] = a[z]
for i in range(n - s):
if y and ans[p[i]] != e:
ans[p[i]] = e
y -= 1
print("YES")
print(' '.join(ans))
T = int(stdin.readline())
for t in range(T):
solve()
``` | output | 1 | 70,900 | 7 | 141,801 |
Provide tags and a correct Python 3 solution for this coding contest problem.
In the game of Mastermind, there are two players β Alice and Bob. Alice has a secret code, which Bob tries to guess. Here, a code is defined as a sequence of n colors. There are exactly n+1 colors in the entire universe, numbered from 1 to n+1 inclusive.
When Bob guesses a code, Alice tells him some information about how good of a guess it is, in the form of two integers x and y.
The first integer x is the number of indices where Bob's guess correctly matches Alice's code. The second integer y is the size of the intersection of the two codes as multisets. That is, if Bob were to change the order of the colors in his guess, y is the maximum number of indices he could get correct.
For example, suppose n=5, Alice's code is [3,1,6,1,2], and Bob's guess is [3,1,1,2,5]. At indices 1 and 2 colors are equal, while in the other indices they are not equal. So x=2. And the two codes have the four colors 1,1,2,3 in common, so y=4.
<image> Solid lines denote a matched color for the same index. Dashed lines denote a matched color at a different index. x is the number of solid lines, and y is the total number of lines.
You are given Bob's guess and two values x and y. Can you find one possibility of Alice's code so that the values of x and y are correct?
Input
The first line contains a single integer t (1β€ tβ€ 1000) β the number of test cases. Next 2t lines contain descriptions of test cases.
The first line of each test case contains three integers n,x,y (1β€ nβ€ 10^5, 0β€ xβ€ yβ€ n) β the length of the codes, and two values Alice responds with.
The second line of each test case contains n integers b_1,β¦,b_n (1β€ b_iβ€ n+1) β Bob's guess, where b_i is the i-th color of the guess.
It is guaranteed that the sum of n across all test cases does not exceed 10^5.
Output
For each test case, on the first line, output "YES" if there is a solution, or "NO" if there is no possible secret code consistent with the described situation. You can print each character in any case (upper or lower).
If the answer is "YES", on the next line output n integers a_1,β¦,a_n (1β€ a_iβ€ n+1) β Alice's secret code, where a_i is the i-th color of the code.
If there are multiple solutions, output any.
Example
Input
7
5 2 4
3 1 1 2 5
5 3 4
1 1 2 1 2
4 0 4
5 5 3 3
4 1 4
2 3 2 3
6 1 2
3 2 1 1 1 1
6 2 4
3 3 2 1 1 1
6 2 6
1 1 3 2 1 1
Output
YES
3 1 6 1 2
YES
3 1 1 1 2
YES
3 3 5 5
NO
YES
4 4 4 4 3 1
YES
3 1 3 1 7 7
YES
2 3 1 1 1 1
Note
The first test case is described in the statement.
In the second test case, x=3 because the colors are equal at indices 2,4,5. And y=4 because they share the colors 1,1,1,2.
In the third test case, x=0 because there is no index where the colors are the same. But y=4 because they share the colors 3,3,5,5.
In the fourth test case, it can be proved that no solution exists. | instruction | 0 | 70,901 | 7 | 141,802 |
Tags: constructive algorithms, graph matchings, greedy, implementation, sortings, two pointers
Correct Solution:
```
from collections import defaultdict
import heapq
T = int(input())
for _ in range(T):
N, A, B = [int(x) for x in input().split(' ')]
b = [int(x) for x in input().split(' ')]
a = [0 for _ in range(N)]
split = defaultdict(list)
for i, x in enumerate(b):
split[x].append((i, x))
heap = []
for x in split.values():
heapq.heappush(heap, (-len(x), x))
for _ in range(A):
_, cur = heapq.heappop(heap)
i, x = cur.pop()
a[i] = x
if len(cur):
heapq.heappush(heap, (-len(cur), cur))
if heap:
rot = -heap[0][0]
rem = [x for cur in heap for x in cur[1]]
d = N - B
if 2*rot-d > len(rem):
print('NO')
continue
heap[0] = (heap[0][0] + d, heap[0][1])
rot = -min(x[0] for x in heap)
unused = list(set(range(1, N+2))-set(b))[0]
#print(rem, rot)
for i in range(d):
rem[i] = (rem[i][0], unused)
#print(rem)
for i in range(len(rem)):
a[rem[i][0]] = rem[(i-rot+len(rem))%len(rem)][1]
print('YES')
print(' '.join(str(x) for x in a))
``` | output | 1 | 70,901 | 7 | 141,803 |
Provide tags and a correct Python 3 solution for this coding contest problem.
In the game of Mastermind, there are two players β Alice and Bob. Alice has a secret code, which Bob tries to guess. Here, a code is defined as a sequence of n colors. There are exactly n+1 colors in the entire universe, numbered from 1 to n+1 inclusive.
When Bob guesses a code, Alice tells him some information about how good of a guess it is, in the form of two integers x and y.
The first integer x is the number of indices where Bob's guess correctly matches Alice's code. The second integer y is the size of the intersection of the two codes as multisets. That is, if Bob were to change the order of the colors in his guess, y is the maximum number of indices he could get correct.
For example, suppose n=5, Alice's code is [3,1,6,1,2], and Bob's guess is [3,1,1,2,5]. At indices 1 and 2 colors are equal, while in the other indices they are not equal. So x=2. And the two codes have the four colors 1,1,2,3 in common, so y=4.
<image> Solid lines denote a matched color for the same index. Dashed lines denote a matched color at a different index. x is the number of solid lines, and y is the total number of lines.
You are given Bob's guess and two values x and y. Can you find one possibility of Alice's code so that the values of x and y are correct?
Input
The first line contains a single integer t (1β€ tβ€ 1000) β the number of test cases. Next 2t lines contain descriptions of test cases.
The first line of each test case contains three integers n,x,y (1β€ nβ€ 10^5, 0β€ xβ€ yβ€ n) β the length of the codes, and two values Alice responds with.
The second line of each test case contains n integers b_1,β¦,b_n (1β€ b_iβ€ n+1) β Bob's guess, where b_i is the i-th color of the guess.
It is guaranteed that the sum of n across all test cases does not exceed 10^5.
Output
For each test case, on the first line, output "YES" if there is a solution, or "NO" if there is no possible secret code consistent with the described situation. You can print each character in any case (upper or lower).
If the answer is "YES", on the next line output n integers a_1,β¦,a_n (1β€ a_iβ€ n+1) β Alice's secret code, where a_i is the i-th color of the code.
If there are multiple solutions, output any.
Example
Input
7
5 2 4
3 1 1 2 5
5 3 4
1 1 2 1 2
4 0 4
5 5 3 3
4 1 4
2 3 2 3
6 1 2
3 2 1 1 1 1
6 2 4
3 3 2 1 1 1
6 2 6
1 1 3 2 1 1
Output
YES
3 1 6 1 2
YES
3 1 1 1 2
YES
3 3 5 5
NO
YES
4 4 4 4 3 1
YES
3 1 3 1 7 7
YES
2 3 1 1 1 1
Note
The first test case is described in the statement.
In the second test case, x=3 because the colors are equal at indices 2,4,5. And y=4 because they share the colors 1,1,1,2.
In the third test case, x=0 because there is no index where the colors are the same. But y=4 because they share the colors 3,3,5,5.
In the fourth test case, it can be proved that no solution exists. | instruction | 0 | 70,902 | 7 | 141,804 |
Tags: constructive algorithms, graph matchings, greedy, implementation, sortings, two pointers
Correct Solution:
```
from collections import defaultdict
def solve():
n, x, y = [int(p) for p in input().split()]
b = [int(p) for p in input().split()]
b_set = set(b)
mex = next(i for i in range(1, n + 2) if i not in b_set)
ans = [0 for p in b]
b_dict = defaultdict(list)
for i, p in enumerate(b):
b_dict[p].append(i)
order = []
while b_dict:
for k in list(b_dict.keys()):
order.append(b_dict[k].pop())
if not b_dict[k]:
del b_dict[k]
for _ in range(x):
i = order.pop()
ans[i] = b[i]
indices = []
for i, p in enumerate(ans):
if p == 0:
indices.append(i)
indices.sort(key=lambda i: b[i])
for i in range(len(indices)):
j = (i + len(indices) // 2) % len(indices)
ans[indices[j]] = b[indices[i]]
cnt = 0
rm = n - y
for i in range(n):
if b[i] == ans[i]:
cnt += 1
if cnt > x:
ans[i] = mex
rm -= 1
if rm < 0:
print('NO')
return
for i in range(n):
if rm == 0:
break
if ans[i] != b[i] and ans[i] != mex:
ans[i] = mex
rm -= 1
print('YES')
print(' '.join(str(p) for p in ans))
t = int(input())
for _ in range(t):
solve()
``` | output | 1 | 70,902 | 7 | 141,805 |
Provide tags and a correct Python 3 solution for this coding contest problem.
In the game of Mastermind, there are two players β Alice and Bob. Alice has a secret code, which Bob tries to guess. Here, a code is defined as a sequence of n colors. There are exactly n+1 colors in the entire universe, numbered from 1 to n+1 inclusive.
When Bob guesses a code, Alice tells him some information about how good of a guess it is, in the form of two integers x and y.
The first integer x is the number of indices where Bob's guess correctly matches Alice's code. The second integer y is the size of the intersection of the two codes as multisets. That is, if Bob were to change the order of the colors in his guess, y is the maximum number of indices he could get correct.
For example, suppose n=5, Alice's code is [3,1,6,1,2], and Bob's guess is [3,1,1,2,5]. At indices 1 and 2 colors are equal, while in the other indices they are not equal. So x=2. And the two codes have the four colors 1,1,2,3 in common, so y=4.
<image> Solid lines denote a matched color for the same index. Dashed lines denote a matched color at a different index. x is the number of solid lines, and y is the total number of lines.
You are given Bob's guess and two values x and y. Can you find one possibility of Alice's code so that the values of x and y are correct?
Input
The first line contains a single integer t (1β€ tβ€ 1000) β the number of test cases. Next 2t lines contain descriptions of test cases.
The first line of each test case contains three integers n,x,y (1β€ nβ€ 10^5, 0β€ xβ€ yβ€ n) β the length of the codes, and two values Alice responds with.
The second line of each test case contains n integers b_1,β¦,b_n (1β€ b_iβ€ n+1) β Bob's guess, where b_i is the i-th color of the guess.
It is guaranteed that the sum of n across all test cases does not exceed 10^5.
Output
For each test case, on the first line, output "YES" if there is a solution, or "NO" if there is no possible secret code consistent with the described situation. You can print each character in any case (upper or lower).
If the answer is "YES", on the next line output n integers a_1,β¦,a_n (1β€ a_iβ€ n+1) β Alice's secret code, where a_i is the i-th color of the code.
If there are multiple solutions, output any.
Example
Input
7
5 2 4
3 1 1 2 5
5 3 4
1 1 2 1 2
4 0 4
5 5 3 3
4 1 4
2 3 2 3
6 1 2
3 2 1 1 1 1
6 2 4
3 3 2 1 1 1
6 2 6
1 1 3 2 1 1
Output
YES
3 1 6 1 2
YES
3 1 1 1 2
YES
3 3 5 5
NO
YES
4 4 4 4 3 1
YES
3 1 3 1 7 7
YES
2 3 1 1 1 1
Note
The first test case is described in the statement.
In the second test case, x=3 because the colors are equal at indices 2,4,5. And y=4 because they share the colors 1,1,1,2.
In the third test case, x=0 because there is no index where the colors are the same. But y=4 because they share the colors 3,3,5,5.
In the fourth test case, it can be proved that no solution exists. | instruction | 0 | 70,903 | 7 | 141,806 |
Tags: constructive algorithms, graph matchings, greedy, implementation, sortings, two pointers
Correct Solution:
```
from collections import defaultdict
import heapq
T = int(input())
for _ in range(T):
N, A, B = [int(x) for x in input().split(' ')]
b = [int(x) for x in input().split(' ')]
a = [0 for _ in range(N)]
split = defaultdict(list)
for i, x in enumerate(b):
split[x].append((i, x))
heap = []
for x in split.values():
heapq.heappush(heap, (-len(x), x))
for _ in range(A):
_, cur = heapq.heappop(heap)
i, x = cur.pop()
a[i] = x
if len(cur):
heapq.heappush(heap, (-len(cur), cur))
if heap:
rot = -heap[0][0]
rem = [x for cur in heap for x in cur[1]]
d = N - B
if 2*rot-d > len(rem):
print('NO')
continue
heap[0] = (heap[0][0] + d, heap[0][1])
rot = -min(x[0] for x in heap)
unused = list(set(range(1, N+2))-set(b))[0]
for i in range(d):
rem[i] = (rem[i][0], unused)
for i in range(len(rem)):
a[rem[i][0]] = rem[(i-rot+len(rem))%len(rem)][1]
print('YES')
print(' '.join(str(x) for x in a))
``` | output | 1 | 70,903 | 7 | 141,807 |
Provide tags and a correct Python 3 solution for this coding contest problem.
In the game of Mastermind, there are two players β Alice and Bob. Alice has a secret code, which Bob tries to guess. Here, a code is defined as a sequence of n colors. There are exactly n+1 colors in the entire universe, numbered from 1 to n+1 inclusive.
When Bob guesses a code, Alice tells him some information about how good of a guess it is, in the form of two integers x and y.
The first integer x is the number of indices where Bob's guess correctly matches Alice's code. The second integer y is the size of the intersection of the two codes as multisets. That is, if Bob were to change the order of the colors in his guess, y is the maximum number of indices he could get correct.
For example, suppose n=5, Alice's code is [3,1,6,1,2], and Bob's guess is [3,1,1,2,5]. At indices 1 and 2 colors are equal, while in the other indices they are not equal. So x=2. And the two codes have the four colors 1,1,2,3 in common, so y=4.
<image> Solid lines denote a matched color for the same index. Dashed lines denote a matched color at a different index. x is the number of solid lines, and y is the total number of lines.
You are given Bob's guess and two values x and y. Can you find one possibility of Alice's code so that the values of x and y are correct?
Input
The first line contains a single integer t (1β€ tβ€ 1000) β the number of test cases. Next 2t lines contain descriptions of test cases.
The first line of each test case contains three integers n,x,y (1β€ nβ€ 10^5, 0β€ xβ€ yβ€ n) β the length of the codes, and two values Alice responds with.
The second line of each test case contains n integers b_1,β¦,b_n (1β€ b_iβ€ n+1) β Bob's guess, where b_i is the i-th color of the guess.
It is guaranteed that the sum of n across all test cases does not exceed 10^5.
Output
For each test case, on the first line, output "YES" if there is a solution, or "NO" if there is no possible secret code consistent with the described situation. You can print each character in any case (upper or lower).
If the answer is "YES", on the next line output n integers a_1,β¦,a_n (1β€ a_iβ€ n+1) β Alice's secret code, where a_i is the i-th color of the code.
If there are multiple solutions, output any.
Example
Input
7
5 2 4
3 1 1 2 5
5 3 4
1 1 2 1 2
4 0 4
5 5 3 3
4 1 4
2 3 2 3
6 1 2
3 2 1 1 1 1
6 2 4
3 3 2 1 1 1
6 2 6
1 1 3 2 1 1
Output
YES
3 1 6 1 2
YES
3 1 1 1 2
YES
3 3 5 5
NO
YES
4 4 4 4 3 1
YES
3 1 3 1 7 7
YES
2 3 1 1 1 1
Note
The first test case is described in the statement.
In the second test case, x=3 because the colors are equal at indices 2,4,5. And y=4 because they share the colors 1,1,1,2.
In the third test case, x=0 because there is no index where the colors are the same. But y=4 because they share the colors 3,3,5,5.
In the fourth test case, it can be proved that no solution exists. | instruction | 0 | 70,904 | 7 | 141,808 |
Tags: constructive algorithms, graph matchings, greedy, implementation, sortings, two pointers
Correct Solution:
```
from collections import defaultdict
from heapq import heapify, heappop, heappush
def solve():
n, s, y = map(int, input().split())
a = input().split()
d = defaultdict(list)
for i, x in enumerate(a):
d[x].append(i)
for i in range(1, n + 2):
e = str(i)
if e not in d:
break
q = [(-len(d[x]), x) for x in d.keys()]
heapify(q)
ans = [0] * n
for i in range(s):
l, x = heappop(q)
ans[d[x].pop()] = x
l += 1
if l:
heappush(q, (l, x))
p = []
while q:
l, x = heappop(q)
p.extend(d[x])
if p:
h = (n - s) // 2
y = n - y
q = p[h:] + p[:h]
for x, z in zip(p, q):
if a[x] == a[z]:
if y:
ans[x] = e
y -= 1
else:
print("NO")
return
else:
ans[x] = a[z]
for i in range(n - s):
if y and ans[p[i]] != e:
ans[p[i]] = e
y -= 1
print("YES")
print(' '.join(ans))
for t in range(int(input())):solve()
``` | output | 1 | 70,904 | 7 | 141,809 |
Provide tags and a correct Python 3 solution for this coding contest problem.
In the game of Mastermind, there are two players β Alice and Bob. Alice has a secret code, which Bob tries to guess. Here, a code is defined as a sequence of n colors. There are exactly n+1 colors in the entire universe, numbered from 1 to n+1 inclusive.
When Bob guesses a code, Alice tells him some information about how good of a guess it is, in the form of two integers x and y.
The first integer x is the number of indices where Bob's guess correctly matches Alice's code. The second integer y is the size of the intersection of the two codes as multisets. That is, if Bob were to change the order of the colors in his guess, y is the maximum number of indices he could get correct.
For example, suppose n=5, Alice's code is [3,1,6,1,2], and Bob's guess is [3,1,1,2,5]. At indices 1 and 2 colors are equal, while in the other indices they are not equal. So x=2. And the two codes have the four colors 1,1,2,3 in common, so y=4.
<image> Solid lines denote a matched color for the same index. Dashed lines denote a matched color at a different index. x is the number of solid lines, and y is the total number of lines.
You are given Bob's guess and two values x and y. Can you find one possibility of Alice's code so that the values of x and y are correct?
Input
The first line contains a single integer t (1β€ tβ€ 1000) β the number of test cases. Next 2t lines contain descriptions of test cases.
The first line of each test case contains three integers n,x,y (1β€ nβ€ 10^5, 0β€ xβ€ yβ€ n) β the length of the codes, and two values Alice responds with.
The second line of each test case contains n integers b_1,β¦,b_n (1β€ b_iβ€ n+1) β Bob's guess, where b_i is the i-th color of the guess.
It is guaranteed that the sum of n across all test cases does not exceed 10^5.
Output
For each test case, on the first line, output "YES" if there is a solution, or "NO" if there is no possible secret code consistent with the described situation. You can print each character in any case (upper or lower).
If the answer is "YES", on the next line output n integers a_1,β¦,a_n (1β€ a_iβ€ n+1) β Alice's secret code, where a_i is the i-th color of the code.
If there are multiple solutions, output any.
Example
Input
7
5 2 4
3 1 1 2 5
5 3 4
1 1 2 1 2
4 0 4
5 5 3 3
4 1 4
2 3 2 3
6 1 2
3 2 1 1 1 1
6 2 4
3 3 2 1 1 1
6 2 6
1 1 3 2 1 1
Output
YES
3 1 6 1 2
YES
3 1 1 1 2
YES
3 3 5 5
NO
YES
4 4 4 4 3 1
YES
3 1 3 1 7 7
YES
2 3 1 1 1 1
Note
The first test case is described in the statement.
In the second test case, x=3 because the colors are equal at indices 2,4,5. And y=4 because they share the colors 1,1,1,2.
In the third test case, x=0 because there is no index where the colors are the same. But y=4 because they share the colors 3,3,5,5.
In the fourth test case, it can be proved that no solution exists. | instruction | 0 | 70,905 | 7 | 141,810 |
Tags: constructive algorithms, graph matchings, greedy, implementation, sortings, two pointers
Correct Solution:
```
from collections import defaultdict
from heapq import heapify, heappop, heappush
def solve():
n, s, y = map(int, input().split());a = input().split();d = defaultdict(list)
for i, x in enumerate(a):d[x].append(i)
for i in range(1, n + 2):
e = str(i)
if e not in d:break
q = [(-len(d[x]), x) for x in d.keys()];heapify(q);ans,p = [0] * n,[]
for i in range(s):
l, x = heappop(q);ans[d[x].pop()] = x;l += 1
if l:heappush(q, (l, x))
while q:l, x = heappop(q);p.extend(d[x])
if p:
h = (n - s) // 2;y = n - y;q = p[h:] + p[:h]
for x, z in zip(p, q):
if a[x] == a[z]:
if y:ans[x] = e;y -= 1
else:print("NO");return
else:ans[x] = a[z]
for i in range(n - s):
if y and ans[p[i]] != e:ans[p[i]] = e;y -= 1
print("YES");print(' '.join(ans))
for t in range(int(input())):solve()
``` | output | 1 | 70,905 | 7 | 141,811 |
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