message stringlengths 2 433k | message_type stringclasses 2 values | message_id int64 0 1 | conversation_id int64 113 108k | cluster float64 12 12 | __index_level_0__ int64 226 217k |
|---|---|---|---|---|---|
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given a non-decreasing array of non-negative integers a_1, a_2, …, a_n. Also you are given a positive integer k.
You want to find m non-decreasing arrays of non-negative integers b_1, b_2, …, b_m, such that:
* The size of b_i is equal to n for all 1 ≤ i ≤ m.
* For all 1 ≤ j ≤ n, a_j = b_{1, j} + b_{2, j} + … + b_{m, j}. In the other word, array a is the sum of arrays b_i.
* The number of different elements in the array b_i is at most k for all 1 ≤ i ≤ m.
Find the minimum possible value of m, or report that there is no possible m.
Input
The first line contains one integer t (1 ≤ t ≤ 100): the number of test cases.
The first line of each test case contains two integers n, k (1 ≤ n ≤ 100, 1 ≤ k ≤ n).
The second line contains n integers a_1, a_2, …, a_n (0 ≤ a_1 ≤ a_2 ≤ … ≤ a_n ≤ 100, a_n > 0).
Output
For each test case print a single integer: the minimum possible value of m. If there is no such m, print -1.
Example
Input
6
4 1
0 0 0 1
3 1
3 3 3
11 3
0 1 2 2 3 3 3 4 4 4 4
5 3
1 2 3 4 5
9 4
2 2 3 5 7 11 13 13 17
10 7
0 1 1 2 3 3 4 5 5 6
Output
-1
1
2
2
2
1
Note
In the first test case, there is no possible m, because all elements of all arrays should be equal to 0. But in this case, it is impossible to get a_4 = 1 as the sum of zeros.
In the second test case, we can take b_1 = [3, 3, 3]. 1 is the smallest possible value of m.
In the third test case, we can take b_1 = [0, 1, 1, 1, 2, 2, 2, 2, 2, 2, 2] and b_2 = [0, 0, 1, 1, 1, 1, 1, 2, 2, 2, 2]. It's easy to see, that a_i = b_{1, i} + b_{2, i} for all i and the number of different elements in b_1 and in b_2 is equal to 3 (so it is at most 3). It can be proven that 2 is the smallest possible value of m. | instruction | 0 | 104,622 | 12 | 209,244 |
Tags: constructive algorithms, greedy, math
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-------------------------------------------
for ik in range(int(input())):
n,k=map(int,input().split())
a=list(map(int,input().split()))
if k==1:
s=set(a)
if len(s)!=1:
print(-1)
else:
print(1)
else:
ans=0
while(sum(a)!=0):
ans+=1
cou=0
s=set()
last=0
for i in range(n):
if cou<k:
if a[i] not in s:
cou+=1
s.add(a[i])
last=a[i]
a[i] = 0
elif cou==k and a[i] in s:
a[i]=0
else:
a[i]-=last
print(ans)
``` | output | 1 | 104,622 | 12 | 209,245 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given a non-decreasing array of non-negative integers a_1, a_2, …, a_n. Also you are given a positive integer k.
You want to find m non-decreasing arrays of non-negative integers b_1, b_2, …, b_m, such that:
* The size of b_i is equal to n for all 1 ≤ i ≤ m.
* For all 1 ≤ j ≤ n, a_j = b_{1, j} + b_{2, j} + … + b_{m, j}. In the other word, array a is the sum of arrays b_i.
* The number of different elements in the array b_i is at most k for all 1 ≤ i ≤ m.
Find the minimum possible value of m, or report that there is no possible m.
Input
The first line contains one integer t (1 ≤ t ≤ 100): the number of test cases.
The first line of each test case contains two integers n, k (1 ≤ n ≤ 100, 1 ≤ k ≤ n).
The second line contains n integers a_1, a_2, …, a_n (0 ≤ a_1 ≤ a_2 ≤ … ≤ a_n ≤ 100, a_n > 0).
Output
For each test case print a single integer: the minimum possible value of m. If there is no such m, print -1.
Example
Input
6
4 1
0 0 0 1
3 1
3 3 3
11 3
0 1 2 2 3 3 3 4 4 4 4
5 3
1 2 3 4 5
9 4
2 2 3 5 7 11 13 13 17
10 7
0 1 1 2 3 3 4 5 5 6
Output
-1
1
2
2
2
1
Note
In the first test case, there is no possible m, because all elements of all arrays should be equal to 0. But in this case, it is impossible to get a_4 = 1 as the sum of zeros.
In the second test case, we can take b_1 = [3, 3, 3]. 1 is the smallest possible value of m.
In the third test case, we can take b_1 = [0, 1, 1, 1, 2, 2, 2, 2, 2, 2, 2] and b_2 = [0, 0, 1, 1, 1, 1, 1, 2, 2, 2, 2]. It's easy to see, that a_i = b_{1, i} + b_{2, i} for all i and the number of different elements in b_1 and in b_2 is equal to 3 (so it is at most 3). It can be proven that 2 is the smallest possible value of m. | instruction | 0 | 104,623 | 12 | 209,246 |
Tags: constructive algorithms, greedy, math
Correct Solution:
```
import io, os
input = io.BytesIO(os.read(0,os.fstat(0).st_size)).readline
from math import ceil
t = int(input())
for _ in range(t):
n, k = map(int, input().split())
a = list(map(int, input().split()))
if len(set(a)) <= k:
print(1)
else:
m = -1
n = len(set(a))
for i in range(1, 101):
if 1+ceil((n-1)/i) <= k:
m = i
break
print(m)
``` | output | 1 | 104,623 | 12 | 209,247 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given a non-decreasing array of non-negative integers a_1, a_2, …, a_n. Also you are given a positive integer k.
You want to find m non-decreasing arrays of non-negative integers b_1, b_2, …, b_m, such that:
* The size of b_i is equal to n for all 1 ≤ i ≤ m.
* For all 1 ≤ j ≤ n, a_j = b_{1, j} + b_{2, j} + … + b_{m, j}. In the other word, array a is the sum of arrays b_i.
* The number of different elements in the array b_i is at most k for all 1 ≤ i ≤ m.
Find the minimum possible value of m, or report that there is no possible m.
Input
The first line contains one integer t (1 ≤ t ≤ 100): the number of test cases.
The first line of each test case contains two integers n, k (1 ≤ n ≤ 100, 1 ≤ k ≤ n).
The second line contains n integers a_1, a_2, …, a_n (0 ≤ a_1 ≤ a_2 ≤ … ≤ a_n ≤ 100, a_n > 0).
Output
For each test case print a single integer: the minimum possible value of m. If there is no such m, print -1.
Example
Input
6
4 1
0 0 0 1
3 1
3 3 3
11 3
0 1 2 2 3 3 3 4 4 4 4
5 3
1 2 3 4 5
9 4
2 2 3 5 7 11 13 13 17
10 7
0 1 1 2 3 3 4 5 5 6
Output
-1
1
2
2
2
1
Note
In the first test case, there is no possible m, because all elements of all arrays should be equal to 0. But in this case, it is impossible to get a_4 = 1 as the sum of zeros.
In the second test case, we can take b_1 = [3, 3, 3]. 1 is the smallest possible value of m.
In the third test case, we can take b_1 = [0, 1, 1, 1, 2, 2, 2, 2, 2, 2, 2] and b_2 = [0, 0, 1, 1, 1, 1, 1, 2, 2, 2, 2]. It's easy to see, that a_i = b_{1, i} + b_{2, i} for all i and the number of different elements in b_1 and in b_2 is equal to 3 (so it is at most 3). It can be proven that 2 is the smallest possible value of m. | instruction | 0 | 104,624 | 12 | 209,248 |
Tags: constructive algorithms, greedy, math
Correct Solution:
```
import math
from sys import stdin
from sys import setrecursionlimit
setrecursionlimit(100000)
def put(): return map(int, stdin.readline().split())
for _ in range(int(input())):
# n=int(input())
n,k=put()
l=(list(put()))
if(len(set(l))<=k):
print(1)
elif(len(set(l))>1 and k==1):
print(-1)
else:
if((len(set(l))-k)%(k-1)==0):
print(1+(len(set(l))-k)//(k-1))
else:
print(2+(len(set(l))-k)//(k-1))
``` | output | 1 | 104,624 | 12 | 209,249 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given a non-decreasing array of non-negative integers a_1, a_2, …, a_n. Also you are given a positive integer k.
You want to find m non-decreasing arrays of non-negative integers b_1, b_2, …, b_m, such that:
* The size of b_i is equal to n for all 1 ≤ i ≤ m.
* For all 1 ≤ j ≤ n, a_j = b_{1, j} + b_{2, j} + … + b_{m, j}. In the other word, array a is the sum of arrays b_i.
* The number of different elements in the array b_i is at most k for all 1 ≤ i ≤ m.
Find the minimum possible value of m, or report that there is no possible m.
Input
The first line contains one integer t (1 ≤ t ≤ 100): the number of test cases.
The first line of each test case contains two integers n, k (1 ≤ n ≤ 100, 1 ≤ k ≤ n).
The second line contains n integers a_1, a_2, …, a_n (0 ≤ a_1 ≤ a_2 ≤ … ≤ a_n ≤ 100, a_n > 0).
Output
For each test case print a single integer: the minimum possible value of m. If there is no such m, print -1.
Example
Input
6
4 1
0 0 0 1
3 1
3 3 3
11 3
0 1 2 2 3 3 3 4 4 4 4
5 3
1 2 3 4 5
9 4
2 2 3 5 7 11 13 13 17
10 7
0 1 1 2 3 3 4 5 5 6
Output
-1
1
2
2
2
1
Note
In the first test case, there is no possible m, because all elements of all arrays should be equal to 0. But in this case, it is impossible to get a_4 = 1 as the sum of zeros.
In the second test case, we can take b_1 = [3, 3, 3]. 1 is the smallest possible value of m.
In the third test case, we can take b_1 = [0, 1, 1, 1, 2, 2, 2, 2, 2, 2, 2] and b_2 = [0, 0, 1, 1, 1, 1, 1, 2, 2, 2, 2]. It's easy to see, that a_i = b_{1, i} + b_{2, i} for all i and the number of different elements in b_1 and in b_2 is equal to 3 (so it is at most 3). It can be proven that 2 is the smallest possible value of m. | instruction | 0 | 104,625 | 12 | 209,250 |
Tags: constructive algorithms, greedy, math
Correct Solution:
```
from math import ceil
t = int(input())
for _ in range(t):
n, k = map(int, input().split())
A = set(map(int, input().split()))
uniq = len(A)
if k == 1:
if uniq == 1:
print(1)
else:
print(-1)
else:
print(1 + ceil(max(0, uniq - k) / (k - 1)))
``` | output | 1 | 104,625 | 12 | 209,251 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given a non-decreasing array of non-negative integers a_1, a_2, …, a_n. Also you are given a positive integer k.
You want to find m non-decreasing arrays of non-negative integers b_1, b_2, …, b_m, such that:
* The size of b_i is equal to n for all 1 ≤ i ≤ m.
* For all 1 ≤ j ≤ n, a_j = b_{1, j} + b_{2, j} + … + b_{m, j}. In the other word, array a is the sum of arrays b_i.
* The number of different elements in the array b_i is at most k for all 1 ≤ i ≤ m.
Find the minimum possible value of m, or report that there is no possible m.
Input
The first line contains one integer t (1 ≤ t ≤ 100): the number of test cases.
The first line of each test case contains two integers n, k (1 ≤ n ≤ 100, 1 ≤ k ≤ n).
The second line contains n integers a_1, a_2, …, a_n (0 ≤ a_1 ≤ a_2 ≤ … ≤ a_n ≤ 100, a_n > 0).
Output
For each test case print a single integer: the minimum possible value of m. If there is no such m, print -1.
Example
Input
6
4 1
0 0 0 1
3 1
3 3 3
11 3
0 1 2 2 3 3 3 4 4 4 4
5 3
1 2 3 4 5
9 4
2 2 3 5 7 11 13 13 17
10 7
0 1 1 2 3 3 4 5 5 6
Output
-1
1
2
2
2
1
Note
In the first test case, there is no possible m, because all elements of all arrays should be equal to 0. But in this case, it is impossible to get a_4 = 1 as the sum of zeros.
In the second test case, we can take b_1 = [3, 3, 3]. 1 is the smallest possible value of m.
In the third test case, we can take b_1 = [0, 1, 1, 1, 2, 2, 2, 2, 2, 2, 2] and b_2 = [0, 0, 1, 1, 1, 1, 1, 2, 2, 2, 2]. It's easy to see, that a_i = b_{1, i} + b_{2, i} for all i and the number of different elements in b_1 and in b_2 is equal to 3 (so it is at most 3). It can be proven that 2 is the smallest possible value of m. | instruction | 0 | 104,626 | 12 | 209,252 |
Tags: constructive algorithms, greedy, math
Correct Solution:
```
import math
for _ in range(int(input())):
n, k = map(int, input().split())
a = list(map(int, input().split()))
if k == 1:
if len(set(a)) == 1:
print(1)
else:
print(-1)
continue
count = 0
c = 1
flag = True
for i in range(n-1):
if a[i] != a[i+1]:
count += 1
print(max((count+k-2)//(k-1), 1))
``` | output | 1 | 104,626 | 12 | 209,253 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given a non-decreasing array of non-negative integers a_1, a_2, …, a_n. Also you are given a positive integer k.
You want to find m non-decreasing arrays of non-negative integers b_1, b_2, …, b_m, such that:
* The size of b_i is equal to n for all 1 ≤ i ≤ m.
* For all 1 ≤ j ≤ n, a_j = b_{1, j} + b_{2, j} + … + b_{m, j}. In the other word, array a is the sum of arrays b_i.
* The number of different elements in the array b_i is at most k for all 1 ≤ i ≤ m.
Find the minimum possible value of m, or report that there is no possible m.
Input
The first line contains one integer t (1 ≤ t ≤ 100): the number of test cases.
The first line of each test case contains two integers n, k (1 ≤ n ≤ 100, 1 ≤ k ≤ n).
The second line contains n integers a_1, a_2, …, a_n (0 ≤ a_1 ≤ a_2 ≤ … ≤ a_n ≤ 100, a_n > 0).
Output
For each test case print a single integer: the minimum possible value of m. If there is no such m, print -1.
Example
Input
6
4 1
0 0 0 1
3 1
3 3 3
11 3
0 1 2 2 3 3 3 4 4 4 4
5 3
1 2 3 4 5
9 4
2 2 3 5 7 11 13 13 17
10 7
0 1 1 2 3 3 4 5 5 6
Output
-1
1
2
2
2
1
Note
In the first test case, there is no possible m, because all elements of all arrays should be equal to 0. But in this case, it is impossible to get a_4 = 1 as the sum of zeros.
In the second test case, we can take b_1 = [3, 3, 3]. 1 is the smallest possible value of m.
In the third test case, we can take b_1 = [0, 1, 1, 1, 2, 2, 2, 2, 2, 2, 2] and b_2 = [0, 0, 1, 1, 1, 1, 1, 2, 2, 2, 2]. It's easy to see, that a_i = b_{1, i} + b_{2, i} for all i and the number of different elements in b_1 and in b_2 is equal to 3 (so it is at most 3). It can be proven that 2 is the smallest possible value of m. | instruction | 0 | 104,627 | 12 | 209,254 |
Tags: constructive algorithms, greedy, math
Correct Solution:
```
from sys import stdin,stdout
import math,bisect
from collections import Counter,deque,defaultdict
L=lambda:list(map(int, stdin.readline().strip().split()))
M=lambda:map(int, stdin.readline().strip().split())
I=lambda:int(stdin.readline().strip())
S=lambda:stdin.readline().strip()
C=lambda:stdin.readline().strip().split()
def pr(a):return(" ".join(list(map(str,a))))
#_________________________________________________#
def solve():
n,k = M()
a = L()
x = len(set(a))
if k>=x:
print(1)
elif k==1:
print(-1)
else:
print(math.ceil((x-1)/(k-1)))
for _ in range(I()):
solve()
``` | output | 1 | 104,627 | 12 | 209,255 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given a non-decreasing array of non-negative integers a_1, a_2, …, a_n. Also you are given a positive integer k.
You want to find m non-decreasing arrays of non-negative integers b_1, b_2, …, b_m, such that:
* The size of b_i is equal to n for all 1 ≤ i ≤ m.
* For all 1 ≤ j ≤ n, a_j = b_{1, j} + b_{2, j} + … + b_{m, j}. In the other word, array a is the sum of arrays b_i.
* The number of different elements in the array b_i is at most k for all 1 ≤ i ≤ m.
Find the minimum possible value of m, or report that there is no possible m.
Input
The first line contains one integer t (1 ≤ t ≤ 100): the number of test cases.
The first line of each test case contains two integers n, k (1 ≤ n ≤ 100, 1 ≤ k ≤ n).
The second line contains n integers a_1, a_2, …, a_n (0 ≤ a_1 ≤ a_2 ≤ … ≤ a_n ≤ 100, a_n > 0).
Output
For each test case print a single integer: the minimum possible value of m. If there is no such m, print -1.
Example
Input
6
4 1
0 0 0 1
3 1
3 3 3
11 3
0 1 2 2 3 3 3 4 4 4 4
5 3
1 2 3 4 5
9 4
2 2 3 5 7 11 13 13 17
10 7
0 1 1 2 3 3 4 5 5 6
Output
-1
1
2
2
2
1
Note
In the first test case, there is no possible m, because all elements of all arrays should be equal to 0. But in this case, it is impossible to get a_4 = 1 as the sum of zeros.
In the second test case, we can take b_1 = [3, 3, 3]. 1 is the smallest possible value of m.
In the third test case, we can take b_1 = [0, 1, 1, 1, 2, 2, 2, 2, 2, 2, 2] and b_2 = [0, 0, 1, 1, 1, 1, 1, 2, 2, 2, 2]. It's easy to see, that a_i = b_{1, i} + b_{2, i} for all i and the number of different elements in b_1 and in b_2 is equal to 3 (so it is at most 3). It can be proven that 2 is the smallest possible value of m. | instruction | 0 | 104,628 | 12 | 209,256 |
Tags: constructive algorithms, greedy, math
Correct Solution:
```
for _ in range(int(input())):
n,k = map(int,input().split(' '))
ans = 0
arr = [int(num) for num in input().split(' ')]
if k==1 and len(set(arr))!=1:
ans = -1
elif arr[0]==0 and len(set(arr))==1:
ans = 1
else:
while (arr[0] !=0 or len(set(arr))!=1):
i = 0
a = set()
a.add(arr[0])
while i<n and len(set(a))<=k:
a.add(arr[i])
if len(a)>k:
break
else:
arr[i] = 0
i = i + 1
for j in range(i,n):
arr[j] = arr[j] - arr[i-1]
ans = ans + 1
print(ans)
``` | output | 1 | 104,628 | 12 | 209,257 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a non-decreasing array of non-negative integers a_1, a_2, …, a_n. Also you are given a positive integer k.
You want to find m non-decreasing arrays of non-negative integers b_1, b_2, …, b_m, such that:
* The size of b_i is equal to n for all 1 ≤ i ≤ m.
* For all 1 ≤ j ≤ n, a_j = b_{1, j} + b_{2, j} + … + b_{m, j}. In the other word, array a is the sum of arrays b_i.
* The number of different elements in the array b_i is at most k for all 1 ≤ i ≤ m.
Find the minimum possible value of m, or report that there is no possible m.
Input
The first line contains one integer t (1 ≤ t ≤ 100): the number of test cases.
The first line of each test case contains two integers n, k (1 ≤ n ≤ 100, 1 ≤ k ≤ n).
The second line contains n integers a_1, a_2, …, a_n (0 ≤ a_1 ≤ a_2 ≤ … ≤ a_n ≤ 100, a_n > 0).
Output
For each test case print a single integer: the minimum possible value of m. If there is no such m, print -1.
Example
Input
6
4 1
0 0 0 1
3 1
3 3 3
11 3
0 1 2 2 3 3 3 4 4 4 4
5 3
1 2 3 4 5
9 4
2 2 3 5 7 11 13 13 17
10 7
0 1 1 2 3 3 4 5 5 6
Output
-1
1
2
2
2
1
Note
In the first test case, there is no possible m, because all elements of all arrays should be equal to 0. But in this case, it is impossible to get a_4 = 1 as the sum of zeros.
In the second test case, we can take b_1 = [3, 3, 3]. 1 is the smallest possible value of m.
In the third test case, we can take b_1 = [0, 1, 1, 1, 2, 2, 2, 2, 2, 2, 2] and b_2 = [0, 0, 1, 1, 1, 1, 1, 2, 2, 2, 2]. It's easy to see, that a_i = b_{1, i} + b_{2, i} for all i and the number of different elements in b_1 and in b_2 is equal to 3 (so it is at most 3). It can be proven that 2 is the smallest possible value of m.
Submitted Solution:
```
#------------------------template--------------------------#
import os
import sys
from math import *
from collections import *
# from fractions import *
# from heapq import*
from bisect import *
from io import BytesIO, IOBase
def vsInput():
sys.stdin = open('input.txt', 'r')
sys.stdout = open('output.txt', 'w')
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")
ALPHA='abcdefghijklmnopqrstuvwxyz/'
M=998244353
EPS=1e-6
def Ceil(a,b): return a//b+int(a%b>0)
def value():return tuple(map(int,input().split()))
def array():return [int(i) for i in input().split()]
def Int():return int(input())
def Str():return input()
def arrayS():return [i for i in input().split()]
#-------------------------code---------------------------#
# vsInput()
for _ in range(Int()):
n,k=value()
a=array()
dist=len(set(a))
if(k==1):
if(dist==1): print(1)
else: print(-1)
else:
ans=0
cur=sorted(set(a))
# print(cur)
while(True):
ans+=1
d=1
ok=False
cur[0]=0
for i in range(1,len(cur)):
if(cur[i]!=cur[i-1]):
d+=1
if(d==k):key=cur[i]
if(d<=k):
cur[i]=0
else:
ok=True
cur[i]=key
if(not ok): break
print(ans)
``` | instruction | 0 | 104,629 | 12 | 209,258 |
Yes | output | 1 | 104,629 | 12 | 209,259 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a non-decreasing array of non-negative integers a_1, a_2, …, a_n. Also you are given a positive integer k.
You want to find m non-decreasing arrays of non-negative integers b_1, b_2, …, b_m, such that:
* The size of b_i is equal to n for all 1 ≤ i ≤ m.
* For all 1 ≤ j ≤ n, a_j = b_{1, j} + b_{2, j} + … + b_{m, j}. In the other word, array a is the sum of arrays b_i.
* The number of different elements in the array b_i is at most k for all 1 ≤ i ≤ m.
Find the minimum possible value of m, or report that there is no possible m.
Input
The first line contains one integer t (1 ≤ t ≤ 100): the number of test cases.
The first line of each test case contains two integers n, k (1 ≤ n ≤ 100, 1 ≤ k ≤ n).
The second line contains n integers a_1, a_2, …, a_n (0 ≤ a_1 ≤ a_2 ≤ … ≤ a_n ≤ 100, a_n > 0).
Output
For each test case print a single integer: the minimum possible value of m. If there is no such m, print -1.
Example
Input
6
4 1
0 0 0 1
3 1
3 3 3
11 3
0 1 2 2 3 3 3 4 4 4 4
5 3
1 2 3 4 5
9 4
2 2 3 5 7 11 13 13 17
10 7
0 1 1 2 3 3 4 5 5 6
Output
-1
1
2
2
2
1
Note
In the first test case, there is no possible m, because all elements of all arrays should be equal to 0. But in this case, it is impossible to get a_4 = 1 as the sum of zeros.
In the second test case, we can take b_1 = [3, 3, 3]. 1 is the smallest possible value of m.
In the third test case, we can take b_1 = [0, 1, 1, 1, 2, 2, 2, 2, 2, 2, 2] and b_2 = [0, 0, 1, 1, 1, 1, 1, 2, 2, 2, 2]. It's easy to see, that a_i = b_{1, i} + b_{2, i} for all i and the number of different elements in b_1 and in b_2 is equal to 3 (so it is at most 3). It can be proven that 2 is the smallest possible value of m.
Submitted Solution:
```
import sys
def rs(): return sys.stdin.readline().rstrip()
def ri(): return int(sys.stdin.readline())
def ria(): return list(map(int, sys.stdin.readline().split()))
def ws(s): sys.stdout.write(s + '\n')
def wi(n): sys.stdout.write(str(n) + '\n')
def wia(a): sys.stdout.write(' '.join([str(x) for x in a]) + '\n')
def ceil(x, y=1): return int(-(-x // y))
import math
from collections import defaultdict
for _ in range(ri()):
n,k=ria()
l=ria()
s=set(l)
if k==1 and len(s)>1:
wi(-1)
elif k==1 and len(s)==1:
wi(1)
else:
x=len(s)
if x<=k:
wi(1)
else:
x-=k
k-=1
wi(1+math.ceil(x/k))
``` | instruction | 0 | 104,630 | 12 | 209,260 |
Yes | output | 1 | 104,630 | 12 | 209,261 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a non-decreasing array of non-negative integers a_1, a_2, …, a_n. Also you are given a positive integer k.
You want to find m non-decreasing arrays of non-negative integers b_1, b_2, …, b_m, such that:
* The size of b_i is equal to n for all 1 ≤ i ≤ m.
* For all 1 ≤ j ≤ n, a_j = b_{1, j} + b_{2, j} + … + b_{m, j}. In the other word, array a is the sum of arrays b_i.
* The number of different elements in the array b_i is at most k for all 1 ≤ i ≤ m.
Find the minimum possible value of m, or report that there is no possible m.
Input
The first line contains one integer t (1 ≤ t ≤ 100): the number of test cases.
The first line of each test case contains two integers n, k (1 ≤ n ≤ 100, 1 ≤ k ≤ n).
The second line contains n integers a_1, a_2, …, a_n (0 ≤ a_1 ≤ a_2 ≤ … ≤ a_n ≤ 100, a_n > 0).
Output
For each test case print a single integer: the minimum possible value of m. If there is no such m, print -1.
Example
Input
6
4 1
0 0 0 1
3 1
3 3 3
11 3
0 1 2 2 3 3 3 4 4 4 4
5 3
1 2 3 4 5
9 4
2 2 3 5 7 11 13 13 17
10 7
0 1 1 2 3 3 4 5 5 6
Output
-1
1
2
2
2
1
Note
In the first test case, there is no possible m, because all elements of all arrays should be equal to 0. But in this case, it is impossible to get a_4 = 1 as the sum of zeros.
In the second test case, we can take b_1 = [3, 3, 3]. 1 is the smallest possible value of m.
In the third test case, we can take b_1 = [0, 1, 1, 1, 2, 2, 2, 2, 2, 2, 2] and b_2 = [0, 0, 1, 1, 1, 1, 1, 2, 2, 2, 2]. It's easy to see, that a_i = b_{1, i} + b_{2, i} for all i and the number of different elements in b_1 and in b_2 is equal to 3 (so it is at most 3). It can be proven that 2 is the smallest possible value of m.
Submitted Solution:
```
import sys
import itertools as it
import bisect as bi
import math as mt
import collections as cc
input=sys.stdin.readline
S=lambda :list(input().split())
I=lambda:list(map(int,input().split()))
for tc in range(int(input())):
n,k=I()
l=I()
nn=len(set(l))
if nn>1 and k-1==0:
print(-1)
elif k>=nn:
print(1)
else:
print(1+(nn-2)//(k-1))
``` | instruction | 0 | 104,631 | 12 | 209,262 |
Yes | output | 1 | 104,631 | 12 | 209,263 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a non-decreasing array of non-negative integers a_1, a_2, …, a_n. Also you are given a positive integer k.
You want to find m non-decreasing arrays of non-negative integers b_1, b_2, …, b_m, such that:
* The size of b_i is equal to n for all 1 ≤ i ≤ m.
* For all 1 ≤ j ≤ n, a_j = b_{1, j} + b_{2, j} + … + b_{m, j}. In the other word, array a is the sum of arrays b_i.
* The number of different elements in the array b_i is at most k for all 1 ≤ i ≤ m.
Find the minimum possible value of m, or report that there is no possible m.
Input
The first line contains one integer t (1 ≤ t ≤ 100): the number of test cases.
The first line of each test case contains two integers n, k (1 ≤ n ≤ 100, 1 ≤ k ≤ n).
The second line contains n integers a_1, a_2, …, a_n (0 ≤ a_1 ≤ a_2 ≤ … ≤ a_n ≤ 100, a_n > 0).
Output
For each test case print a single integer: the minimum possible value of m. If there is no such m, print -1.
Example
Input
6
4 1
0 0 0 1
3 1
3 3 3
11 3
0 1 2 2 3 3 3 4 4 4 4
5 3
1 2 3 4 5
9 4
2 2 3 5 7 11 13 13 17
10 7
0 1 1 2 3 3 4 5 5 6
Output
-1
1
2
2
2
1
Note
In the first test case, there is no possible m, because all elements of all arrays should be equal to 0. But in this case, it is impossible to get a_4 = 1 as the sum of zeros.
In the second test case, we can take b_1 = [3, 3, 3]. 1 is the smallest possible value of m.
In the third test case, we can take b_1 = [0, 1, 1, 1, 2, 2, 2, 2, 2, 2, 2] and b_2 = [0, 0, 1, 1, 1, 1, 1, 2, 2, 2, 2]. It's easy to see, that a_i = b_{1, i} + b_{2, i} for all i and the number of different elements in b_1 and in b_2 is equal to 3 (so it is at most 3). It can be proven that 2 is the smallest possible value of m.
Submitted Solution:
```
from collections import *
from itertools import *
from bisect import *
def inp():
return int(input())
def arrinp():
return [int(x) for x in input().split()]
def main():
t = inp()
for _ in range(t):
n,k = arrinp()
A = arrinp()
distinct = set(A)
curr = len(distinct)
if(k<=1 and curr>k):
print(-1)
continue
if(curr<=k):
print(1)
continue
m = 0
while(len(distinct)>=k):
m += 1
temp = list(distinct)
#print(*temp, 'value m:', m)
sub = temp[k-1]
distinct = set()
n = len(temp)
l = n-1
while(l>=k-1):
distinct.add(temp[l]-sub)
l -=1
#print(distinct, 'after 1 list generated')
if(len(distinct)>1):
m+=1
print(m)
if __name__ == '__main__':
main()
``` | instruction | 0 | 104,632 | 12 | 209,264 |
Yes | output | 1 | 104,632 | 12 | 209,265 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a non-decreasing array of non-negative integers a_1, a_2, …, a_n. Also you are given a positive integer k.
You want to find m non-decreasing arrays of non-negative integers b_1, b_2, …, b_m, such that:
* The size of b_i is equal to n for all 1 ≤ i ≤ m.
* For all 1 ≤ j ≤ n, a_j = b_{1, j} + b_{2, j} + … + b_{m, j}. In the other word, array a is the sum of arrays b_i.
* The number of different elements in the array b_i is at most k for all 1 ≤ i ≤ m.
Find the minimum possible value of m, or report that there is no possible m.
Input
The first line contains one integer t (1 ≤ t ≤ 100): the number of test cases.
The first line of each test case contains two integers n, k (1 ≤ n ≤ 100, 1 ≤ k ≤ n).
The second line contains n integers a_1, a_2, …, a_n (0 ≤ a_1 ≤ a_2 ≤ … ≤ a_n ≤ 100, a_n > 0).
Output
For each test case print a single integer: the minimum possible value of m. If there is no such m, print -1.
Example
Input
6
4 1
0 0 0 1
3 1
3 3 3
11 3
0 1 2 2 3 3 3 4 4 4 4
5 3
1 2 3 4 5
9 4
2 2 3 5 7 11 13 13 17
10 7
0 1 1 2 3 3 4 5 5 6
Output
-1
1
2
2
2
1
Note
In the first test case, there is no possible m, because all elements of all arrays should be equal to 0. But in this case, it is impossible to get a_4 = 1 as the sum of zeros.
In the second test case, we can take b_1 = [3, 3, 3]. 1 is the smallest possible value of m.
In the third test case, we can take b_1 = [0, 1, 1, 1, 2, 2, 2, 2, 2, 2, 2] and b_2 = [0, 0, 1, 1, 1, 1, 1, 2, 2, 2, 2]. It's easy to see, that a_i = b_{1, i} + b_{2, i} for all i and the number of different elements in b_1 and in b_2 is equal to 3 (so it is at most 3). It can be proven that 2 is the smallest possible value of m.
Submitted Solution:
```
import sys
try:sys.stdin,sys.stdout=open('in.txt','r'),open('out.txt','w')
except:pass
ii1=lambda:int(sys.stdin.readline().strip()) # for interger
is1=lambda:sys.stdin.readline().strip() # for str
iia=lambda:list(map(int,sys.stdin.readline().strip().split())) # for List[int]
isa=lambda:sys.stdin.readline().strip().split() # for List[str]
mod=int(1e9 + 7);from collections import *;from math import *
###################### Start Here ######################
for _ in range(ii1()):
n ,k= iia()
arr = iia()
arr_set=set(arr)
# diff=[]
# for i in range(n-1):
# diff.append(arr[i+1]-arr[i])
# diff_len=len(set(diff))
# if 0 in diff:diff_len-=1
set_len = len(arr_set)
if k==1 and set_len==2:
print(-1)
continue
if set_len==1:
print(1)
continue
if k>=set_len:
print(1)
if set_len>k:
print(2)
continue
``` | instruction | 0 | 104,633 | 12 | 209,266 |
No | output | 1 | 104,633 | 12 | 209,267 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a non-decreasing array of non-negative integers a_1, a_2, …, a_n. Also you are given a positive integer k.
You want to find m non-decreasing arrays of non-negative integers b_1, b_2, …, b_m, such that:
* The size of b_i is equal to n for all 1 ≤ i ≤ m.
* For all 1 ≤ j ≤ n, a_j = b_{1, j} + b_{2, j} + … + b_{m, j}. In the other word, array a is the sum of arrays b_i.
* The number of different elements in the array b_i is at most k for all 1 ≤ i ≤ m.
Find the minimum possible value of m, or report that there is no possible m.
Input
The first line contains one integer t (1 ≤ t ≤ 100): the number of test cases.
The first line of each test case contains two integers n, k (1 ≤ n ≤ 100, 1 ≤ k ≤ n).
The second line contains n integers a_1, a_2, …, a_n (0 ≤ a_1 ≤ a_2 ≤ … ≤ a_n ≤ 100, a_n > 0).
Output
For each test case print a single integer: the minimum possible value of m. If there is no such m, print -1.
Example
Input
6
4 1
0 0 0 1
3 1
3 3 3
11 3
0 1 2 2 3 3 3 4 4 4 4
5 3
1 2 3 4 5
9 4
2 2 3 5 7 11 13 13 17
10 7
0 1 1 2 3 3 4 5 5 6
Output
-1
1
2
2
2
1
Note
In the first test case, there is no possible m, because all elements of all arrays should be equal to 0. But in this case, it is impossible to get a_4 = 1 as the sum of zeros.
In the second test case, we can take b_1 = [3, 3, 3]. 1 is the smallest possible value of m.
In the third test case, we can take b_1 = [0, 1, 1, 1, 2, 2, 2, 2, 2, 2, 2] and b_2 = [0, 0, 1, 1, 1, 1, 1, 2, 2, 2, 2]. It's easy to see, that a_i = b_{1, i} + b_{2, i} for all i and the number of different elements in b_1 and in b_2 is equal to 3 (so it is at most 3). It can be proven that 2 is the smallest possible value of m.
Submitted Solution:
```
import sys
import math
input = lambda: sys.stdin.readline().rstrip()
for _ in range(int(input())):
n,k=map(int,input().split())
l=[int(x) for x in input().split()]
a=len(set(l))
arr=[1,2,3,4,5]
if(l==arr and k==3):
print(2)
elif(k<a and k==1):
print(-1)
elif(a==k):
print(1)
else:
print(a-k)
``` | instruction | 0 | 104,634 | 12 | 209,268 |
No | output | 1 | 104,634 | 12 | 209,269 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a non-decreasing array of non-negative integers a_1, a_2, …, a_n. Also you are given a positive integer k.
You want to find m non-decreasing arrays of non-negative integers b_1, b_2, …, b_m, such that:
* The size of b_i is equal to n for all 1 ≤ i ≤ m.
* For all 1 ≤ j ≤ n, a_j = b_{1, j} + b_{2, j} + … + b_{m, j}. In the other word, array a is the sum of arrays b_i.
* The number of different elements in the array b_i is at most k for all 1 ≤ i ≤ m.
Find the minimum possible value of m, or report that there is no possible m.
Input
The first line contains one integer t (1 ≤ t ≤ 100): the number of test cases.
The first line of each test case contains two integers n, k (1 ≤ n ≤ 100, 1 ≤ k ≤ n).
The second line contains n integers a_1, a_2, …, a_n (0 ≤ a_1 ≤ a_2 ≤ … ≤ a_n ≤ 100, a_n > 0).
Output
For each test case print a single integer: the minimum possible value of m. If there is no such m, print -1.
Example
Input
6
4 1
0 0 0 1
3 1
3 3 3
11 3
0 1 2 2 3 3 3 4 4 4 4
5 3
1 2 3 4 5
9 4
2 2 3 5 7 11 13 13 17
10 7
0 1 1 2 3 3 4 5 5 6
Output
-1
1
2
2
2
1
Note
In the first test case, there is no possible m, because all elements of all arrays should be equal to 0. But in this case, it is impossible to get a_4 = 1 as the sum of zeros.
In the second test case, we can take b_1 = [3, 3, 3]. 1 is the smallest possible value of m.
In the third test case, we can take b_1 = [0, 1, 1, 1, 2, 2, 2, 2, 2, 2, 2] and b_2 = [0, 0, 1, 1, 1, 1, 1, 2, 2, 2, 2]. It's easy to see, that a_i = b_{1, i} + b_{2, i} for all i and the number of different elements in b_1 and in b_2 is equal to 3 (so it is at most 3). It can be proven that 2 is the smallest possible value of m.
Submitted Solution:
```
for aww in range(int(input())):
a,b=[*map(int, input().split())]
c=[*map(int, input().split())]
d=set(c)
if(b==1):
if(len(d)>1):
print(-1)
else:
print(1)
else:
x=c[0]
e=0
f=0
for i,j in enumerate(c):
if(x!=c[i]):
x=c[i]
e+=1
if(e==b):
f+=1
e=0
if(e>0):
f+=1
print(f)
``` | instruction | 0 | 104,635 | 12 | 209,270 |
No | output | 1 | 104,635 | 12 | 209,271 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a non-decreasing array of non-negative integers a_1, a_2, …, a_n. Also you are given a positive integer k.
You want to find m non-decreasing arrays of non-negative integers b_1, b_2, …, b_m, such that:
* The size of b_i is equal to n for all 1 ≤ i ≤ m.
* For all 1 ≤ j ≤ n, a_j = b_{1, j} + b_{2, j} + … + b_{m, j}. In the other word, array a is the sum of arrays b_i.
* The number of different elements in the array b_i is at most k for all 1 ≤ i ≤ m.
Find the minimum possible value of m, or report that there is no possible m.
Input
The first line contains one integer t (1 ≤ t ≤ 100): the number of test cases.
The first line of each test case contains two integers n, k (1 ≤ n ≤ 100, 1 ≤ k ≤ n).
The second line contains n integers a_1, a_2, …, a_n (0 ≤ a_1 ≤ a_2 ≤ … ≤ a_n ≤ 100, a_n > 0).
Output
For each test case print a single integer: the minimum possible value of m. If there is no such m, print -1.
Example
Input
6
4 1
0 0 0 1
3 1
3 3 3
11 3
0 1 2 2 3 3 3 4 4 4 4
5 3
1 2 3 4 5
9 4
2 2 3 5 7 11 13 13 17
10 7
0 1 1 2 3 3 4 5 5 6
Output
-1
1
2
2
2
1
Note
In the first test case, there is no possible m, because all elements of all arrays should be equal to 0. But in this case, it is impossible to get a_4 = 1 as the sum of zeros.
In the second test case, we can take b_1 = [3, 3, 3]. 1 is the smallest possible value of m.
In the third test case, we can take b_1 = [0, 1, 1, 1, 2, 2, 2, 2, 2, 2, 2] and b_2 = [0, 0, 1, 1, 1, 1, 1, 2, 2, 2, 2]. It's easy to see, that a_i = b_{1, i} + b_{2, i} for all i and the number of different elements in b_1 and in b_2 is equal to 3 (so it is at most 3). It can be proven that 2 is the smallest possible value of m.
Submitted Solution:
```
mod = 10**9 + 7
def solve():
ans = -1
n, k = map(int, input().split())
a = list(map(int, input().split()))
d = {}
for x in a:
d[x] = 1
if k == 1:
if len(d) != 1:
ans = -1
else:
ans = 1
else:
l = len(d)
ans = 1
l -= k
ans += (l + k - 2) // (k - 1)
print(ans)
t = int(input())
while t > 0:
solve()
t -= 1
``` | instruction | 0 | 104,636 | 12 | 209,272 |
No | output | 1 | 104,636 | 12 | 209,273 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given two tables A and B of size n × m.
We define a sorting by column as the following: we choose a column and reorder the rows of the table by the value in this column, from the rows with the smallest value to the rows with the largest. In case there are two or more rows with equal value in this column, their relative order does not change (such sorting algorithms are called stable).
You can find this behavior of sorting by column in many office software for managing spreadsheets. Petya works in one, and he has a table A opened right now. He wants to perform zero of more sortings by column to transform this table to table B.
Determine if it is possible to do so, and if yes, find a sequence of columns to sort by. Note that you do not need to minimize the number of sortings.
Input
The first line contains two integers n and m (1 ≤ n, m ≤ 1500) — the sizes of the tables.
Each of the next n lines contains m integers a_{i,j} (1 ≤ a_{i, j} ≤ n), denoting the elements of the table A.
Each of the next n lines contains m integers b_{i, j} (1 ≤ b_{i, j} ≤ n), denoting the elements of the table B.
Output
If it is not possible to transform A into B, print -1.
Otherwise, first print an integer k (0 ≤ k ≤ 5000) — the number of sortings in your solution.
Then print k integers c_1, …, c_k (1 ≤ c_i ≤ m) — the columns, by which Petya needs to perform a sorting.
We can show that if a solution exists, there is one in no more than 5000 sortings.
Examples
Input
2 2
2 2
1 2
1 2
2 2
Output
1
1
Input
3 3
2 3 2
1 3 3
1 1 2
1 1 2
1 3 3
2 3 2
Output
2
1 2
Input
2 2
1 1
2 1
2 1
1 1
Output
-1
Input
4 1
2
2
2
1
1
2
2
2
Output
1
1
Note
Consider the second example. After the sorting by the first column the table becomes
$$$\begin{matrix} 1&3&3\\\ 1&1&2\\\ 2&3&2. \end{matrix}$$$
After the sorting by the second column the table becomes
$$$\begin{matrix} 1&1&2\\\ 1&3&3\\\ 2&3&2, \end{matrix}$$$
and this is what we need.
In the third test any sorting does not change anything, because the columns are already sorted.
Submitted Solution:
```
def check(a,b):
for i in range(len(a)):
if(a[i]!=b[i]):
return(False)
return(True)
def sort(a):
if(len(a)<=1):
return(a)
else:
pivot=a.pop()
left=[]
right=[]
for i in a:
if(i<pivot):
left.append(i)
else:
right.append(i)
return(sort(left)+[pivot]+sort(right))
a,b=map(int,input().split())
c=[]
d=[]
count=0
z=0
out=[]
for i in range(a):
x=[int(x) for x in input().split()]
c.append(x)
for j in range(a):
x=[int(x) for x in input().split()]
d.append(x)
for i in range(b):
fi=[]
sec=[]
ram=0
for j in range(a):
fi.append(c[j][i])
sec.append(d[j][i])
if(not(check(fi,sec))):
fi=sort(fi)
ram=1
if(ram==1):
if(check(fi,sec)):
count+=1
out.append(i+1)
else:
z=1
break
if(z==0):
print(count)
print(*out,sep=" ")
else:
print(-1)
``` | instruction | 0 | 104,681 | 12 | 209,362 |
No | output | 1 | 104,681 | 12 | 209,363 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given two tables A and B of size n × m.
We define a sorting by column as the following: we choose a column and reorder the rows of the table by the value in this column, from the rows with the smallest value to the rows with the largest. In case there are two or more rows with equal value in this column, their relative order does not change (such sorting algorithms are called stable).
You can find this behavior of sorting by column in many office software for managing spreadsheets. Petya works in one, and he has a table A opened right now. He wants to perform zero of more sortings by column to transform this table to table B.
Determine if it is possible to do so, and if yes, find a sequence of columns to sort by. Note that you do not need to minimize the number of sortings.
Input
The first line contains two integers n and m (1 ≤ n, m ≤ 1500) — the sizes of the tables.
Each of the next n lines contains m integers a_{i,j} (1 ≤ a_{i, j} ≤ n), denoting the elements of the table A.
Each of the next n lines contains m integers b_{i, j} (1 ≤ b_{i, j} ≤ n), denoting the elements of the table B.
Output
If it is not possible to transform A into B, print -1.
Otherwise, first print an integer k (0 ≤ k ≤ 5000) — the number of sortings in your solution.
Then print k integers c_1, …, c_k (1 ≤ c_i ≤ m) — the columns, by which Petya needs to perform a sorting.
We can show that if a solution exists, there is one in no more than 5000 sortings.
Examples
Input
2 2
2 2
1 2
1 2
2 2
Output
1
1
Input
3 3
2 3 2
1 3 3
1 1 2
1 1 2
1 3 3
2 3 2
Output
2
1 2
Input
2 2
1 1
2 1
2 1
1 1
Output
-1
Input
4 1
2
2
2
1
1
2
2
2
Output
1
1
Note
Consider the second example. After the sorting by the first column the table becomes
$$$\begin{matrix} 1&3&3\\\ 1&1&2\\\ 2&3&2. \end{matrix}$$$
After the sorting by the second column the table becomes
$$$\begin{matrix} 1&1&2\\\ 1&3&3\\\ 2&3&2, \end{matrix}$$$
and this is what we need.
In the third test any sorting does not change anything, because the columns are already sorted.
Submitted Solution:
```
# C. Matrix Sorting: https://codeforces.com/contest/1500/problem/C
# glob_x = """
# 2 2 2 1
# 2 2 1 2
# 1 2 2 2
# 2 1 2 2
# 2 2 1 2
# 2 1 2 2
# 1 2 2 2
# 2 2 2 1
# """
# def get_ab(glob_x):
# glob_x = glob_x[1:-1].split('\n')
# n = len(glob_x) // 2
# a, b = [], []
# for i in range(n):
# a.append(glob_x[i].split(' '))
# for i in range(n, 2 * n):
# b.append(glob_x[i].split(' '))
# m = len(a[0])
# return n, m, a, b
# print(get_ab(glob_x))
def main():
n, m = list(map(lambda x: int(x), str(input()).split(' ')))
u = {}
states = {}
end = ''
a, b = [], []
for i in range(n):
t = input()
a.append(t.split(' '))
if t not in u:
u[t] = 1
else:
u[t] += 1
for i in range(n):
t = input()
b.append(t.split(' '))
if t not in u:
print(-1)
return
else:
u[t] -= 1
if u[t] == 0:
del u[t]
# n, m, a, b = get_ab(glob_x)
end = get_str(b)
s = []
ans = bt(a, s, end, states)
if not ans:
print(-1)
return
def is_solved(a, s, end):
if len(s) > 5000:
return False
if get_str(a) == end:
return True
return False
def get_str(a):
return '$'.join(map(lambda x: ' '.join(x), a))
def bt(a, s, end, states):
# print(len(states), states, '\n')
if is_solved(a, s, end):
print(len(s))
print(' '.join(map(lambda x: str(x), s)))
return True
for i in range(len(a[0])):
# print('#' * len(s) + str(i))
if len(s) == 0:
s.append(4)
else:
s.append(i + 1)
new_a = sort(a, i + 1)
new_str = get_str(new_a)
added = False
if not new_str in states:
states[new_str] = True
added = True
if bt(new_a, s, end, states):
return True
if added:
del states[new_str]
s.pop()
# print("-" * 20)
return False
def sort(a, i):
return sorted(a, key=lambda x: x[i - 1])
if __name__ == "__main__":
main()
``` | instruction | 0 | 104,682 | 12 | 209,364 |
No | output | 1 | 104,682 | 12 | 209,365 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given two tables A and B of size n × m.
We define a sorting by column as the following: we choose a column and reorder the rows of the table by the value in this column, from the rows with the smallest value to the rows with the largest. In case there are two or more rows with equal value in this column, their relative order does not change (such sorting algorithms are called stable).
You can find this behavior of sorting by column in many office software for managing spreadsheets. Petya works in one, and he has a table A opened right now. He wants to perform zero of more sortings by column to transform this table to table B.
Determine if it is possible to do so, and if yes, find a sequence of columns to sort by. Note that you do not need to minimize the number of sortings.
Input
The first line contains two integers n and m (1 ≤ n, m ≤ 1500) — the sizes of the tables.
Each of the next n lines contains m integers a_{i,j} (1 ≤ a_{i, j} ≤ n), denoting the elements of the table A.
Each of the next n lines contains m integers b_{i, j} (1 ≤ b_{i, j} ≤ n), denoting the elements of the table B.
Output
If it is not possible to transform A into B, print -1.
Otherwise, first print an integer k (0 ≤ k ≤ 5000) — the number of sortings in your solution.
Then print k integers c_1, …, c_k (1 ≤ c_i ≤ m) — the columns, by which Petya needs to perform a sorting.
We can show that if a solution exists, there is one in no more than 5000 sortings.
Examples
Input
2 2
2 2
1 2
1 2
2 2
Output
1
1
Input
3 3
2 3 2
1 3 3
1 1 2
1 1 2
1 3 3
2 3 2
Output
2
1 2
Input
2 2
1 1
2 1
2 1
1 1
Output
-1
Input
4 1
2
2
2
1
1
2
2
2
Output
1
1
Note
Consider the second example. After the sorting by the first column the table becomes
$$$\begin{matrix} 1&3&3\\\ 1&1&2\\\ 2&3&2. \end{matrix}$$$
After the sorting by the second column the table becomes
$$$\begin{matrix} 1&1&2\\\ 1&3&3\\\ 2&3&2, \end{matrix}$$$
and this is what we need.
In the third test any sorting does not change anything, because the columns are already sorted.
Submitted Solution:
```
l,c=map(int, input().split())
A=list()
nA=list()
B=list()
tb=list()
cpt=0
i=0
for s in range(l*2):
if s<l:
A.append(list(map(int,input().split()))[:c])
elif s>=l:
B.append(list(map(int, input().split()))[:c])
if A == B:
cpt=c
nA=[x+1 for x in range(c)]
else:
while i <c:
tem = list()
for index,z in enumerate(A):
tem.append((z[i],index))
for ind,z in enumerate(B):
tb.append((z[i],ind))
stem=sorted(tem, key= lambda x:x[0])
if tem !=stem:
cpt+=1
nA.append(i+1)
for w in range(len(stem)):
u=stem[w][1]
A[w][i:]=A[u][i:]
i+=1
print(cpt) if cpt>0 else print(-1)
print(' '.join(str(z) for z in nA)) if cpt>0 else print('')
``` | instruction | 0 | 104,683 | 12 | 209,366 |
No | output | 1 | 104,683 | 12 | 209,367 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given two tables A and B of size n × m.
We define a sorting by column as the following: we choose a column and reorder the rows of the table by the value in this column, from the rows with the smallest value to the rows with the largest. In case there are two or more rows with equal value in this column, their relative order does not change (such sorting algorithms are called stable).
You can find this behavior of sorting by column in many office software for managing spreadsheets. Petya works in one, and he has a table A opened right now. He wants to perform zero of more sortings by column to transform this table to table B.
Determine if it is possible to do so, and if yes, find a sequence of columns to sort by. Note that you do not need to minimize the number of sortings.
Input
The first line contains two integers n and m (1 ≤ n, m ≤ 1500) — the sizes of the tables.
Each of the next n lines contains m integers a_{i,j} (1 ≤ a_{i, j} ≤ n), denoting the elements of the table A.
Each of the next n lines contains m integers b_{i, j} (1 ≤ b_{i, j} ≤ n), denoting the elements of the table B.
Output
If it is not possible to transform A into B, print -1.
Otherwise, first print an integer k (0 ≤ k ≤ 5000) — the number of sortings in your solution.
Then print k integers c_1, …, c_k (1 ≤ c_i ≤ m) — the columns, by which Petya needs to perform a sorting.
We can show that if a solution exists, there is one in no more than 5000 sortings.
Examples
Input
2 2
2 2
1 2
1 2
2 2
Output
1
1
Input
3 3
2 3 2
1 3 3
1 1 2
1 1 2
1 3 3
2 3 2
Output
2
1 2
Input
2 2
1 1
2 1
2 1
1 1
Output
-1
Input
4 1
2
2
2
1
1
2
2
2
Output
1
1
Note
Consider the second example. After the sorting by the first column the table becomes
$$$\begin{matrix} 1&3&3\\\ 1&1&2\\\ 2&3&2. \end{matrix}$$$
After the sorting by the second column the table becomes
$$$\begin{matrix} 1&1&2\\\ 1&3&3\\\ 2&3&2, \end{matrix}$$$
and this is what we need.
In the third test any sorting does not change anything, because the columns are already sorted.
Submitted Solution:
```
rows, columns = map(int, input().split())
table1 = []
table2 = []
for i in range(rows):
table1 += [list(map(int, input().split()))]
for i in range(rows):
table2 += [list(map(int, input().split()))]
sorts = []
f = True
for i in range(columns):
column1 = []
column2 = []
for j in range(rows):
column1 += [table1[j][i]]
column2 += [table2[j][i]]
if column1 == column2:
break
else:
column1.sort()
if column1 == column2:
column2.sort()
if column1 == column2:
sorts += [i + 1]
else:
f = False
else:
f = False
if f:
print(len(sorts))
for i in sorts:
print(i, end=' ')
else:
print(-1)
``` | instruction | 0 | 104,684 | 12 | 209,368 |
No | output | 1 | 104,684 | 12 | 209,369 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Permutation p is an ordered set of integers p1, p2, ..., pn, consisting of n distinct positive integers not larger than n. We'll denote as n the length of permutation p1, p2, ..., pn.
Your task is to find such permutation p of length n, that the group of numbers |p1 - p2|, |p2 - p3|, ..., |pn - 1 - pn| has exactly k distinct elements.
Input
The single line of the input contains two space-separated positive integers n, k (1 ≤ k < n ≤ 105).
Output
Print n integers forming the permutation. If there are multiple answers, print any of them.
Examples
Input
3 2
Output
1 3 2
Input
3 1
Output
1 2 3
Input
5 2
Output
1 3 2 4 5
Note
By |x| we denote the absolute value of number x. | instruction | 0 | 104,824 | 12 | 209,648 |
Tags: constructive algorithms, greedy
Correct Solution:
```
n,k = map(int,list(input().split()))
for i in range(1,(k+3)//2):
print(i, k+2-i,end=" ")
if k>1 and k%2==0: print(k//2+1,end=" ")
for i in range(k+2,n+1):
print(i,end=" ")
``` | output | 1 | 104,824 | 12 | 209,649 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Permutation p is an ordered set of integers p1, p2, ..., pn, consisting of n distinct positive integers not larger than n. We'll denote as n the length of permutation p1, p2, ..., pn.
Your task is to find such permutation p of length n, that the group of numbers |p1 - p2|, |p2 - p3|, ..., |pn - 1 - pn| has exactly k distinct elements.
Input
The single line of the input contains two space-separated positive integers n, k (1 ≤ k < n ≤ 105).
Output
Print n integers forming the permutation. If there are multiple answers, print any of them.
Examples
Input
3 2
Output
1 3 2
Input
3 1
Output
1 2 3
Input
5 2
Output
1 3 2 4 5
Note
By |x| we denote the absolute value of number x. | instruction | 0 | 104,825 | 12 | 209,650 |
Tags: constructive algorithms, greedy
Correct Solution:
```
[n, k], s = map(int, input().split()), [1]
for i in range(k, 0, -1):
s.append(s[-1] + (-i if (k - i) % 2 else i))
print(*s + sorted(set(range(1, n + 1)) - set(s)))
``` | output | 1 | 104,825 | 12 | 209,651 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Permutation p is an ordered set of integers p1, p2, ..., pn, consisting of n distinct positive integers not larger than n. We'll denote as n the length of permutation p1, p2, ..., pn.
Your task is to find such permutation p of length n, that the group of numbers |p1 - p2|, |p2 - p3|, ..., |pn - 1 - pn| has exactly k distinct elements.
Input
The single line of the input contains two space-separated positive integers n, k (1 ≤ k < n ≤ 105).
Output
Print n integers forming the permutation. If there are multiple answers, print any of them.
Examples
Input
3 2
Output
1 3 2
Input
3 1
Output
1 2 3
Input
5 2
Output
1 3 2 4 5
Note
By |x| we denote the absolute value of number x. | instruction | 0 | 104,826 | 12 | 209,652 |
Tags: constructive algorithms, greedy
Correct Solution:
```
n,k=(int(i) for i in input().split())
l=[i for i in range(1,n-k+1)]
l1=[i for i in range(n-k+1,n+1)]
k1=0
k2=k-1
for i in range(k):
if(i%2==0):
l.append(l1[k2])
k2-=1
else:
l.append(l1[k1])
k1+=1
print(*l)
``` | output | 1 | 104,826 | 12 | 209,653 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Permutation p is an ordered set of integers p1, p2, ..., pn, consisting of n distinct positive integers not larger than n. We'll denote as n the length of permutation p1, p2, ..., pn.
Your task is to find such permutation p of length n, that the group of numbers |p1 - p2|, |p2 - p3|, ..., |pn - 1 - pn| has exactly k distinct elements.
Input
The single line of the input contains two space-separated positive integers n, k (1 ≤ k < n ≤ 105).
Output
Print n integers forming the permutation. If there are multiple answers, print any of them.
Examples
Input
3 2
Output
1 3 2
Input
3 1
Output
1 2 3
Input
5 2
Output
1 3 2 4 5
Note
By |x| we denote the absolute value of number x. | instruction | 0 | 104,827 | 12 | 209,654 |
Tags: constructive algorithms, greedy
Correct Solution:
```
from sys import stdin
lines = list(filter(None, stdin.read().split('\n')))
def parseline(line):
return list(map(int, line.split()))
lines = list(map(parseline, lines))
n, k = lines[0]
x = n + 1
for i in range(n, n - k, -1):
if (n - i) % 2 == 0:
x = x - i
else:
x = x + i
print(x, end=' ')
step = -1 if k %2 == 0 else +1
for k in range(n - k):
x += step
print(x, end=' ')
``` | output | 1 | 104,827 | 12 | 209,655 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Permutation p is an ordered set of integers p1, p2, ..., pn, consisting of n distinct positive integers not larger than n. We'll denote as n the length of permutation p1, p2, ..., pn.
Your task is to find such permutation p of length n, that the group of numbers |p1 - p2|, |p2 - p3|, ..., |pn - 1 - pn| has exactly k distinct elements.
Input
The single line of the input contains two space-separated positive integers n, k (1 ≤ k < n ≤ 105).
Output
Print n integers forming the permutation. If there are multiple answers, print any of them.
Examples
Input
3 2
Output
1 3 2
Input
3 1
Output
1 2 3
Input
5 2
Output
1 3 2 4 5
Note
By |x| we denote the absolute value of number x. | instruction | 0 | 104,828 | 12 | 209,656 |
Tags: constructive algorithms, greedy
Correct Solution:
```
n, k = map(int, input().split())
a = [x + 1 for x in range(n)]
l = 0
r = n - 1
p = 0
for i in range(k - 1):
if p:
print(a[r], end = ' ')
r -= 1
p = not p
else:
print(a[l], end = ' ')
l += 1
p = not p
p = not p
if p:
for i in range(l, r + 1):
print(a[i], end = ' ')
else:
for i in range(r, l - 1, -1):
print(a[i], end = ' ')
``` | output | 1 | 104,828 | 12 | 209,657 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Permutation p is an ordered set of integers p1, p2, ..., pn, consisting of n distinct positive integers not larger than n. We'll denote as n the length of permutation p1, p2, ..., pn.
Your task is to find such permutation p of length n, that the group of numbers |p1 - p2|, |p2 - p3|, ..., |pn - 1 - pn| has exactly k distinct elements.
Input
The single line of the input contains two space-separated positive integers n, k (1 ≤ k < n ≤ 105).
Output
Print n integers forming the permutation. If there are multiple answers, print any of them.
Examples
Input
3 2
Output
1 3 2
Input
3 1
Output
1 2 3
Input
5 2
Output
1 3 2 4 5
Note
By |x| we denote the absolute value of number x. | instruction | 0 | 104,829 | 12 | 209,658 |
Tags: constructive algorithms, greedy
Correct Solution:
```
#In the name of Allah
from sys import stdin, stdout
input = stdin.readline
n, k = map(int, input().split())
def f(n):
n += 1
l = []
for i in range(1, 1 + n // 2):
l.append(str(i))
l.append(str(n - i + 1))
if n % 2 == 1:
l.append(str(n // 2 + 1))
return l
ans = f(k)
for i in range(n - k - 1):
ans.append(str(i + k + 2))
stdout.write(" ".join(ans))
``` | output | 1 | 104,829 | 12 | 209,659 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Permutation p is an ordered set of integers p1, p2, ..., pn, consisting of n distinct positive integers not larger than n. We'll denote as n the length of permutation p1, p2, ..., pn.
Your task is to find such permutation p of length n, that the group of numbers |p1 - p2|, |p2 - p3|, ..., |pn - 1 - pn| has exactly k distinct elements.
Input
The single line of the input contains two space-separated positive integers n, k (1 ≤ k < n ≤ 105).
Output
Print n integers forming the permutation. If there are multiple answers, print any of them.
Examples
Input
3 2
Output
1 3 2
Input
3 1
Output
1 2 3
Input
5 2
Output
1 3 2 4 5
Note
By |x| we denote the absolute value of number x. | instruction | 0 | 104,830 | 12 | 209,660 |
Tags: constructive algorithms, greedy
Correct Solution:
```
n, k = map(int, input().split())
a = []
k += 1
one = 0
two = 0
f = False
for i in range(1, n + 1):
if i % 2 != 0:
i -= one
a.append(i)
one += 1
else:
i -= two
a.append(n + 2 - i)
two += 1
k -= 1
if k == 1:
f = True
d = set(a)
if len(a) % 2 == 0:
for j in range(n - 1, 0, -1):
if j not in d:
a.append(j)
else:
for j in range(1, n + 1):
if j not in d:
a.append(j)
if f:
break
print(*a)
``` | output | 1 | 104,830 | 12 | 209,661 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Permutation p is an ordered set of integers p1, p2, ..., pn, consisting of n distinct positive integers not larger than n. We'll denote as n the length of permutation p1, p2, ..., pn.
Your task is to find such permutation p of length n, that the group of numbers |p1 - p2|, |p2 - p3|, ..., |pn - 1 - pn| has exactly k distinct elements.
Input
The single line of the input contains two space-separated positive integers n, k (1 ≤ k < n ≤ 105).
Output
Print n integers forming the permutation. If there are multiple answers, print any of them.
Examples
Input
3 2
Output
1 3 2
Input
3 1
Output
1 2 3
Input
5 2
Output
1 3 2 4 5
Note
By |x| we denote the absolute value of number x. | instruction | 0 | 104,831 | 12 | 209,662 |
Tags: constructive algorithms, greedy
Correct Solution:
```
n,k = map(int,input().split())
a = list(range(1,n+1))
for i in range(1,k+1): a[i] = a[i-1] + (k-i+1 if i&1 else i-k-1)
print(" ".join(repr(x) for x in a))
``` | output | 1 | 104,831 | 12 | 209,663 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Permutation p is an ordered set of integers p1, p2, ..., pn, consisting of n distinct positive integers not larger than n. We'll denote as n the length of permutation p1, p2, ..., pn.
Your task is to find such permutation p of length n, that the group of numbers |p1 - p2|, |p2 - p3|, ..., |pn - 1 - pn| has exactly k distinct elements.
Input
The single line of the input contains two space-separated positive integers n, k (1 ≤ k < n ≤ 105).
Output
Print n integers forming the permutation. If there are multiple answers, print any of them.
Examples
Input
3 2
Output
1 3 2
Input
3 1
Output
1 2 3
Input
5 2
Output
1 3 2 4 5
Note
By |x| we denote the absolute value of number x.
Submitted Solution:
```
n,k = map(int,input().split())
a = list(range(1,n+1))
for i in range(1,k+1): a[i] = a[i-1] + (k-i+1 if i&1 else i-k-1)
print(" ".join(repr(x) for x in a))
# Made By Mostafa_Khaled
``` | instruction | 0 | 104,832 | 12 | 209,664 |
Yes | output | 1 | 104,832 | 12 | 209,665 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Permutation p is an ordered set of integers p1, p2, ..., pn, consisting of n distinct positive integers not larger than n. We'll denote as n the length of permutation p1, p2, ..., pn.
Your task is to find such permutation p of length n, that the group of numbers |p1 - p2|, |p2 - p3|, ..., |pn - 1 - pn| has exactly k distinct elements.
Input
The single line of the input contains two space-separated positive integers n, k (1 ≤ k < n ≤ 105).
Output
Print n integers forming the permutation. If there are multiple answers, print any of them.
Examples
Input
3 2
Output
1 3 2
Input
3 1
Output
1 2 3
Input
5 2
Output
1 3 2 4 5
Note
By |x| we denote the absolute value of number x.
Submitted Solution:
```
n,k = map(int,input().split())
b = ''
i = 1
j = n
c = 0
p = 0
while(p < n):
if(c < k):
if(p % 2 == 0):
b += str(i) + ' '
i += 1
c += 1
else:
b += str(j) + ' '
j -= 1
c += 1
else:
if(p % 2 == 1):
for q in range(i,j+1):
b += str(q) + ' '
print(b)
exit(0)
else:
for q in range(j,i-1,-1):
b += str(q) + ' '
print(b)
exit(0)
p += 1
print(b)
``` | instruction | 0 | 104,833 | 12 | 209,666 |
Yes | output | 1 | 104,833 | 12 | 209,667 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Permutation p is an ordered set of integers p1, p2, ..., pn, consisting of n distinct positive integers not larger than n. We'll denote as n the length of permutation p1, p2, ..., pn.
Your task is to find such permutation p of length n, that the group of numbers |p1 - p2|, |p2 - p3|, ..., |pn - 1 - pn| has exactly k distinct elements.
Input
The single line of the input contains two space-separated positive integers n, k (1 ≤ k < n ≤ 105).
Output
Print n integers forming the permutation. If there are multiple answers, print any of them.
Examples
Input
3 2
Output
1 3 2
Input
3 1
Output
1 2 3
Input
5 2
Output
1 3 2 4 5
Note
By |x| we denote the absolute value of number x.
Submitted Solution:
```
n,k=map(int,input().split())
i=1
a=[]
for j in range(n//2):
a.append(i)
a.append(n-i+1)
i+=1
if n%2==1:
a.append(n//2+1)
for i in range(k):
print(a[i],end=' ')
if k<n and a[k-1]>a[k]:
s=a[k-1]-1
for i in range(k,n):
print(s,end=' ')
s-=1
else:
s=a[k-1]+1
for i in range(k,n):
print(s,end=' ')
s+=1
``` | instruction | 0 | 104,834 | 12 | 209,668 |
Yes | output | 1 | 104,834 | 12 | 209,669 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Permutation p is an ordered set of integers p1, p2, ..., pn, consisting of n distinct positive integers not larger than n. We'll denote as n the length of permutation p1, p2, ..., pn.
Your task is to find such permutation p of length n, that the group of numbers |p1 - p2|, |p2 - p3|, ..., |pn - 1 - pn| has exactly k distinct elements.
Input
The single line of the input contains two space-separated positive integers n, k (1 ≤ k < n ≤ 105).
Output
Print n integers forming the permutation. If there are multiple answers, print any of them.
Examples
Input
3 2
Output
1 3 2
Input
3 1
Output
1 2 3
Input
5 2
Output
1 3 2 4 5
Note
By |x| we denote the absolute value of number x.
Submitted Solution:
```
def getLine():
return list(map(int,input().split()))
n,k = getLine()
l = [0]*n
for i in range(n):
if i%2 == 0:l[i] = i//2+1
else : l[i] = n - l[i-1]+1
ans=l
for i in range(k,n):
if k%2 == 0:l[i] = l[i-1]-1
else : l[i] = l[i-1]+1
for i in ans:
print(i,end=" ")
``` | instruction | 0 | 104,835 | 12 | 209,670 |
Yes | output | 1 | 104,835 | 12 | 209,671 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Permutation p is an ordered set of integers p1, p2, ..., pn, consisting of n distinct positive integers not larger than n. We'll denote as n the length of permutation p1, p2, ..., pn.
Your task is to find such permutation p of length n, that the group of numbers |p1 - p2|, |p2 - p3|, ..., |pn - 1 - pn| has exactly k distinct elements.
Input
The single line of the input contains two space-separated positive integers n, k (1 ≤ k < n ≤ 105).
Output
Print n integers forming the permutation. If there are multiple answers, print any of them.
Examples
Input
3 2
Output
1 3 2
Input
3 1
Output
1 2 3
Input
5 2
Output
1 3 2 4 5
Note
By |x| we denote the absolute value of number x.
Submitted Solution:
```
from sys import stdin
lines = list(filter(None, stdin.read().split('\n')))
def parseline(line):
return list(map(int, line.split()))
lines = list(map(parseline, lines))
n, k = lines[0]
x = n + 1
for i in range(n, n - k, -1):
if 1 <= x + i <= n:
x = x + i
else:
x = x - i
print(x, end=' ')
step = -1 if k %2 == 0 else +1
for k in range(n - k):
x += step
print(x, end=' ')
``` | instruction | 0 | 104,836 | 12 | 209,672 |
No | output | 1 | 104,836 | 12 | 209,673 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Permutation p is an ordered set of integers p1, p2, ..., pn, consisting of n distinct positive integers not larger than n. We'll denote as n the length of permutation p1, p2, ..., pn.
Your task is to find such permutation p of length n, that the group of numbers |p1 - p2|, |p2 - p3|, ..., |pn - 1 - pn| has exactly k distinct elements.
Input
The single line of the input contains two space-separated positive integers n, k (1 ≤ k < n ≤ 105).
Output
Print n integers forming the permutation. If there are multiple answers, print any of them.
Examples
Input
3 2
Output
1 3 2
Input
3 1
Output
1 2 3
Input
5 2
Output
1 3 2 4 5
Note
By |x| we denote the absolute value of number x.
Submitted Solution:
```
def getLine():
return list(map(int,input().split()))
n,k = getLine()
l = [0]*n
for i in range(n):
if i%2 == 0:l[i] = i//2+1
else : l[i] = n - l[i-1]+1
ans = l[:k-1]+sorted(l[k-1:])
for i in ans:
print(i,end=" ")
``` | instruction | 0 | 104,837 | 12 | 209,674 |
No | output | 1 | 104,837 | 12 | 209,675 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Permutation p is an ordered set of integers p1, p2, ..., pn, consisting of n distinct positive integers not larger than n. We'll denote as n the length of permutation p1, p2, ..., pn.
Your task is to find such permutation p of length n, that the group of numbers |p1 - p2|, |p2 - p3|, ..., |pn - 1 - pn| has exactly k distinct elements.
Input
The single line of the input contains two space-separated positive integers n, k (1 ≤ k < n ≤ 105).
Output
Print n integers forming the permutation. If there are multiple answers, print any of them.
Examples
Input
3 2
Output
1 3 2
Input
3 1
Output
1 2 3
Input
5 2
Output
1 3 2 4 5
Note
By |x| we denote the absolute value of number x.
Submitted Solution:
```
def getLine():
return list(map(int,input().split()))
n,k = getLine()
l = [0]*n
for i in range(n):
if i%2 == 0:l[i] = i//2+1
else : l[i] = n - l[i-1]+1
ans = l[:k]+sorted(l[k:])
for i in ans:
print(i,end=" ")
``` | instruction | 0 | 104,838 | 12 | 209,676 |
No | output | 1 | 104,838 | 12 | 209,677 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Permutation p is an ordered set of integers p1, p2, ..., pn, consisting of n distinct positive integers not larger than n. We'll denote as n the length of permutation p1, p2, ..., pn.
Your task is to find such permutation p of length n, that the group of numbers |p1 - p2|, |p2 - p3|, ..., |pn - 1 - pn| has exactly k distinct elements.
Input
The single line of the input contains two space-separated positive integers n, k (1 ≤ k < n ≤ 105).
Output
Print n integers forming the permutation. If there are multiple answers, print any of them.
Examples
Input
3 2
Output
1 3 2
Input
3 1
Output
1 2 3
Input
5 2
Output
1 3 2 4 5
Note
By |x| we denote the absolute value of number x.
Submitted Solution:
```
n,k = map(int,input().split())
b = []
c = []
i = 1
j = n
c = 0
p = 0
while(p < n):
if(c < k-1):
if(i % 2 == 0):
b.append(i)
i += 1
c += 1
else:
b.append(j)
j -= 1
c += 1
else:
for q in range(i,j+1):
b.append(q)
break
p += 1
for i in b:
print(i);
``` | instruction | 0 | 104,839 | 12 | 209,678 |
No | output | 1 | 104,839 | 12 | 209,679 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given an array of positive integers a1, a2, ..., an × T of length n × T. We know that for any i > n it is true that ai = ai - n. Find the length of the longest non-decreasing sequence of the given array.
Input
The first line contains two space-separated integers: n, T (1 ≤ n ≤ 100, 1 ≤ T ≤ 107). The second line contains n space-separated integers a1, a2, ..., an (1 ≤ ai ≤ 300).
Output
Print a single number — the length of a sought sequence.
Examples
Input
4 3
3 1 4 2
Output
5
Note
The array given in the sample looks like that: 3, 1, 4, 2, 3, 1, 4, 2, 3, 1, 4, 2. The elements in bold form the largest non-decreasing subsequence. | instruction | 0 | 104,865 | 12 | 209,730 |
Tags: constructive algorithms, dp, matrices
Correct Solution:
```
f = lambda: map(int, input().split())
n, m = f()
t = list(f())
d = {i: j for j, i in enumerate(sorted(set(t)))}
t = [d[i] for i in t]
k = len(d)
a = [0] * k
if m < 2 * k:
for j in t * m:
a[j] += 1
q = a[j]
j += 1
while j < k and a[j] < q:
a[j] += 1
j += 1
print(a[-1])
exit()
a = [0] * k
for j in t * k:
a[j] += 1
q = a[j]
j += 1
while j < k and a[j] < q:
a[j] += 1
j += 1
b = [0] * k
t.reverse()
for j in t * k:
b[j] += 1
q = b[j]
j -= 1
while j > -1 and b[j] < q:
b[j] += 1
j -= 1
print(max(a[j] + (m - 2 * k) * t.count(j) + b[j] for j in range(k)))
``` | output | 1 | 104,865 | 12 | 209,731 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given an array of positive integers a1, a2, ..., an × T of length n × T. We know that for any i > n it is true that ai = ai - n. Find the length of the longest non-decreasing sequence of the given array.
Input
The first line contains two space-separated integers: n, T (1 ≤ n ≤ 100, 1 ≤ T ≤ 107). The second line contains n space-separated integers a1, a2, ..., an (1 ≤ ai ≤ 300).
Output
Print a single number — the length of a sought sequence.
Examples
Input
4 3
3 1 4 2
Output
5
Note
The array given in the sample looks like that: 3, 1, 4, 2, 3, 1, 4, 2, 3, 1, 4, 2. The elements in bold form the largest non-decreasing subsequence. | instruction | 0 | 104,866 | 12 | 209,732 |
Tags: constructive algorithms, dp, matrices
Correct Solution:
```
f = lambda: map(int, input().split())
n, m = f()
t = list(f())
s = [0] * 301
d = s[:]
for i in t: d[i] += 1
for i in t * min(m, 2 * n): s[i] = max(s[:i + 1]) + 1
print(max(s) + max((m - n * 2) * max(d), 0))
``` | output | 1 | 104,866 | 12 | 209,733 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given an array of positive integers a1, a2, ..., an × T of length n × T. We know that for any i > n it is true that ai = ai - n. Find the length of the longest non-decreasing sequence of the given array.
Input
The first line contains two space-separated integers: n, T (1 ≤ n ≤ 100, 1 ≤ T ≤ 107). The second line contains n space-separated integers a1, a2, ..., an (1 ≤ ai ≤ 300).
Output
Print a single number — the length of a sought sequence.
Examples
Input
4 3
3 1 4 2
Output
5
Note
The array given in the sample looks like that: 3, 1, 4, 2, 3, 1, 4, 2, 3, 1, 4, 2. The elements in bold form the largest non-decreasing subsequence. | instruction | 0 | 104,867 | 12 | 209,734 |
Tags: constructive algorithms, dp, matrices
Correct Solution:
```
f = lambda: map(int, input().split())
n, m = f()
t = list(f())
s = [0] * 301
d = s[:]
for i in t: d[i] += 1
for i in t * min(m, 2 * n): s[i] = max(s[:i + 1]) + 1
print(max(s) + max((m - n * 2) * max(d), 0))
# Made By Mostafa_Khaled
``` | output | 1 | 104,867 | 12 | 209,735 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given an array of positive integers a1, a2, ..., an × T of length n × T. We know that for any i > n it is true that ai = ai - n. Find the length of the longest non-decreasing sequence of the given array.
Input
The first line contains two space-separated integers: n, T (1 ≤ n ≤ 100, 1 ≤ T ≤ 107). The second line contains n space-separated integers a1, a2, ..., an (1 ≤ ai ≤ 300).
Output
Print a single number — the length of a sought sequence.
Examples
Input
4 3
3 1 4 2
Output
5
Note
The array given in the sample looks like that: 3, 1, 4, 2, 3, 1, 4, 2, 3, 1, 4, 2. The elements in bold form the largest non-decreasing subsequence. | instruction | 0 | 104,868 | 12 | 209,736 |
Tags: constructive algorithms, dp, matrices
Correct Solution:
```
mul=lambda A,B,r:[[max([A[i][k]+B[k][j] for k in r if A[i][k] and B[k][j]],default=0) for j in r] for i in r]
def binpower(A,n,e):
r = range(n)
B = A #A^0 is invalid, thus start from A^1
e -= 1
while True:
if e &1: B = mul(B,A,r)
e =e>>1
if e==0: return B
A =mul(A,A,r)
def f(l,n,T):
# Part 1: cal M
h = max(l)+1
N = [[0]*h for _ in range(h)]
Q = [[0]*h for _ in range(h)]
M = [[0]*n for _ in range(n)]
for j in range(n):
# update Mij based on Quv,(j-1)
for i in range(n):
M[i][j]=Q[l[i]][l[j]]+1 if l[i]<=l[j] else 0
# update Nuv,(j) and Quv,(j)
v = l[j]
for u in range(1,v+1):
N[u][v] = Q[u][v]+1
for vv in range(v,h):
Q[u][vv] = max(Q[u][vv],N[u][v])
return max(max(binpower(M,n,T)))
n,T = list(map(int,input().split()))
l = list(map(int,input().split()))
print(f(l,n,T))
``` | output | 1 | 104,868 | 12 | 209,737 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given an array of positive integers a1, a2, ..., an × T of length n × T. We know that for any i > n it is true that ai = ai - n. Find the length of the longest non-decreasing sequence of the given array.
Input
The first line contains two space-separated integers: n, T (1 ≤ n ≤ 100, 1 ≤ T ≤ 107). The second line contains n space-separated integers a1, a2, ..., an (1 ≤ ai ≤ 300).
Output
Print a single number — the length of a sought sequence.
Examples
Input
4 3
3 1 4 2
Output
5
Note
The array given in the sample looks like that: 3, 1, 4, 2, 3, 1, 4, 2, 3, 1, 4, 2. The elements in bold form the largest non-decreasing subsequence. | instruction | 0 | 104,869 | 12 | 209,738 |
Tags: constructive algorithms, dp, matrices
Correct Solution:
```
# ---------------------------iye ha aam zindegi---------------------------------------------
import math
import random
import heapq,bisect
import sys
from collections import deque, defaultdict
from fractions import Fraction
import sys
import threading
from collections import defaultdict
threading.stack_size(10**8)
mod = 10 ** 9 + 7
mod1 = 998244353
# ------------------------------warmup----------------------------
import os
import sys
from io import BytesIO, IOBase
sys.setrecursionlimit(300000)
BUFSIZE = 8192
class FastIO(IOBase):
newlines = 0
def __init__(self, file):
self._fd = file.fileno()
self.buffer = BytesIO()
self.writable = "x" in file.mode or "r" not in file.mode
self.write = self.buffer.write if self.writable else None
def read(self):
while True:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
if not b:
break
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines = 0
return self.buffer.read()
def readline(self):
while self.newlines == 0:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
self.newlines = b.count(b"\n") + (not b)
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines -= 1
return self.buffer.readline()
def flush(self):
if self.writable:
os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
self.write = lambda s: self.buffer.write(s.encode("ascii"))
self.read = lambda: self.buffer.read().decode("ascii")
self.readline = lambda: self.buffer.readline().decode("ascii")
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")
# -------------------game starts now----------------------------------------------------import math
class TreeNode:
def __init__(self, k, v):
self.key = k
self.value = v
self.left = None
self.right = None
self.parent = None
self.height = 1
self.num_left = 1
self.num_total = 1
class AvlTree:
def __init__(self):
self._tree = None
def add(self, k, v):
if not self._tree:
self._tree = TreeNode(k, v)
return
node = self._add(k, v)
if node:
self._rebalance(node)
def _add(self, k, v):
node = self._tree
while node:
if k < node.key:
if node.left:
node = node.left
else:
node.left = TreeNode(k, v)
node.left.parent = node
return node.left
elif node.key < k:
if node.right:
node = node.right
else:
node.right = TreeNode(k, v)
node.right.parent = node
return node.right
else:
node.value = v
return
@staticmethod
def get_height(x):
return x.height if x else 0
@staticmethod
def get_num_total(x):
return x.num_total if x else 0
def _rebalance(self, node):
n = node
while n:
lh = self.get_height(n.left)
rh = self.get_height(n.right)
n.height = max(lh, rh) + 1
balance_factor = lh - rh
n.num_total = 1 + self.get_num_total(n.left) + self.get_num_total(n.right)
n.num_left = 1 + self.get_num_total(n.left)
if balance_factor > 1:
if self.get_height(n.left.left) < self.get_height(n.left.right):
self._rotate_left(n.left)
self._rotate_right(n)
elif balance_factor < -1:
if self.get_height(n.right.right) < self.get_height(n.right.left):
self._rotate_right(n.right)
self._rotate_left(n)
else:
n = n.parent
def _remove_one(self, node):
"""
Side effect!!! Changes node. Node should have exactly one child
"""
replacement = node.left or node.right
if node.parent:
if AvlTree._is_left(node):
node.parent.left = replacement
else:
node.parent.right = replacement
replacement.parent = node.parent
node.parent = None
else:
self._tree = replacement
replacement.parent = None
node.left = None
node.right = None
node.parent = None
self._rebalance(replacement)
def _remove_leaf(self, node):
if node.parent:
if AvlTree._is_left(node):
node.parent.left = None
else:
node.parent.right = None
self._rebalance(node.parent)
else:
self._tree = None
node.parent = None
node.left = None
node.right = None
def remove(self, k):
node = self._get_node(k)
if not node:
return
if AvlTree._is_leaf(node):
self._remove_leaf(node)
return
if node.left and node.right:
nxt = AvlTree._get_next(node)
node.key = nxt.key
node.value = nxt.value
if self._is_leaf(nxt):
self._remove_leaf(nxt)
else:
self._remove_one(nxt)
self._rebalance(node)
else:
self._remove_one(node)
def get(self, k):
node = self._get_node(k)
return node.value if node else -1
def _get_node(self, k):
if not self._tree:
return None
node = self._tree
while node:
if k < node.key:
node = node.left
elif node.key < k:
node = node.right
else:
return node
return None
def get_at(self, pos):
x = pos + 1
node = self._tree
while node:
if x < node.num_left:
node = node.left
elif node.num_left < x:
x -= node.num_left
node = node.right
else:
return (node.key, node.value)
raise IndexError("Out of ranges")
@staticmethod
def _is_left(node):
return node.parent.left and node.parent.left == node
@staticmethod
def _is_leaf(node):
return node.left is None and node.right is None
def _rotate_right(self, node):
if not node.parent:
self._tree = node.left
node.left.parent = None
elif AvlTree._is_left(node):
node.parent.left = node.left
node.left.parent = node.parent
else:
node.parent.right = node.left
node.left.parent = node.parent
bk = node.left.right
node.left.right = node
node.parent = node.left
node.left = bk
if bk:
bk.parent = node
node.height = max(self.get_height(node.left), self.get_height(node.right)) + 1
node.num_total = 1 + self.get_num_total(node.left) + self.get_num_total(node.right)
node.num_left = 1 + self.get_num_total(node.left)
def _rotate_left(self, node):
if not node.parent:
self._tree = node.right
node.right.parent = None
elif AvlTree._is_left(node):
node.parent.left = node.right
node.right.parent = node.parent
else:
node.parent.right = node.right
node.right.parent = node.parent
bk = node.right.left
node.right.left = node
node.parent = node.right
node.right = bk
if bk:
bk.parent = node
node.height = max(self.get_height(node.left), self.get_height(node.right)) + 1
node.num_total = 1 + self.get_num_total(node.left) + self.get_num_total(node.right)
node.num_left = 1 + self.get_num_total(node.left)
@staticmethod
def _get_next(node):
if not node.right:
return node.parent
n = node.right
while n.left:
n = n.left
return n
# -----------------------------------------------binary seacrh tree---------------------------------------
class SegmentTree1:
def __init__(self, data, default=2**51, func=lambda a, b: a & b):
"""initialize the segment tree with data"""
self._default = default
self._func = func
self._len = len(data)
self._size = _size = 1 << (self._len - 1).bit_length()
self.data = [default] * (2 * _size)
self.data[_size:_size + self._len] = data
for i in reversed(range(_size)):
self.data[i] = func(self.data[i + i], self.data[i + i + 1])
def __delitem__(self, idx):
self[idx] = self._default
def __getitem__(self, idx):
return self.data[idx + self._size]
def __setitem__(self, idx, value):
idx += self._size
self.data[idx] = value
idx >>= 1
while idx:
self.data[idx] = self._func(self.data[2 * idx], self.data[2 * idx + 1])
idx >>= 1
def __len__(self):
return self._len
def query(self, start, stop):
if start == stop:
return self.__getitem__(start)
stop += 1
start += self._size
stop += self._size
res = self._default
while start < stop:
if start & 1:
res = self._func(res, self.data[start])
start += 1
if stop & 1:
stop -= 1
res = self._func(res, self.data[stop])
start >>= 1
stop >>= 1
return res
def __repr__(self):
return "SegmentTree({0})".format(self.data)
# -------------------game starts now----------------------------------------------------import math
class SegmentTree:
def __init__(self, data, default=0, func=lambda a, b: max(a , b)):
"""initialize the segment tree with data"""
self._default = default
self._func = func
self._len = len(data)
self._size = _size = 1 << (self._len - 1).bit_length()
self.data = [default] * (2 * _size)
self.data[_size:_size + self._len] = data
for i in reversed(range(_size)):
self.data[i] = func(self.data[i + i], self.data[i + i + 1])
def __delitem__(self, idx):
self[idx] = self._default
def __getitem__(self, idx):
return self.data[idx + self._size]
def __setitem__(self, idx, value):
idx += self._size
self.data[idx] = value
idx >>= 1
while idx:
self.data[idx] = self._func(self.data[2 * idx], self.data[2 * idx + 1])
idx >>= 1
def __len__(self):
return self._len
def query(self, start, stop):
if start == stop:
return self.__getitem__(start)
stop += 1
start += self._size
stop += self._size
res = self._default
while start < stop:
if start & 1:
res = self._func(res, self.data[start])
start += 1
if stop & 1:
stop -= 1
res = self._func(res, self.data[stop])
start >>= 1
stop >>= 1
return res
def __repr__(self):
return "SegmentTree({0})".format(self.data)
# -------------------------------iye ha chutiya zindegi-------------------------------------
class Factorial:
def __init__(self, MOD):
self.MOD = MOD
self.factorials = [1, 1]
self.invModulos = [0, 1]
self.invFactorial_ = [1, 1]
def calc(self, n):
if n <= -1:
print("Invalid argument to calculate n!")
print("n must be non-negative value. But the argument was " + str(n))
exit()
if n < len(self.factorials):
return self.factorials[n]
nextArr = [0] * (n + 1 - len(self.factorials))
initialI = len(self.factorials)
prev = self.factorials[-1]
m = self.MOD
for i in range(initialI, n + 1):
prev = nextArr[i - initialI] = prev * i % m
self.factorials += nextArr
return self.factorials[n]
def inv(self, n):
if n <= -1:
print("Invalid argument to calculate n^(-1)")
print("n must be non-negative value. But the argument was " + str(n))
exit()
p = self.MOD
pi = n % p
if pi < len(self.invModulos):
return self.invModulos[pi]
nextArr = [0] * (n + 1 - len(self.invModulos))
initialI = len(self.invModulos)
for i in range(initialI, min(p, n + 1)):
next = -self.invModulos[p % i] * (p // i) % p
self.invModulos.append(next)
return self.invModulos[pi]
def invFactorial(self, n):
if n <= -1:
print("Invalid argument to calculate (n^(-1))!")
print("n must be non-negative value. But the argument was " + str(n))
exit()
if n < len(self.invFactorial_):
return self.invFactorial_[n]
self.inv(n) # To make sure already calculated n^-1
nextArr = [0] * (n + 1 - len(self.invFactorial_))
initialI = len(self.invFactorial_)
prev = self.invFactorial_[-1]
p = self.MOD
for i in range(initialI, n + 1):
prev = nextArr[i - initialI] = (prev * self.invModulos[i % p]) % p
self.invFactorial_ += nextArr
return self.invFactorial_[n]
class Combination:
def __init__(self, MOD):
self.MOD = MOD
self.factorial = Factorial(MOD)
def ncr(self, n, k):
if k < 0 or n < k:
return 0
k = min(k, n - k)
f = self.factorial
return f.calc(n) * f.invFactorial(max(n - k, k)) * f.invFactorial(min(k, n - k)) % self.MOD
# --------------------------------------iye ha combinations ka zindegi---------------------------------
def powm(a, n, m):
if a == 1 or n == 0:
return 1
if n % 2 == 0:
s = powm(a, n // 2, m)
return s * s % m
else:
return a * powm(a, n - 1, m) % m
# --------------------------------------iye ha power ka zindegi---------------------------------
def sort_list(list1, list2):
zipped_pairs = zip(list2, list1)
z = [x for _, x in sorted(zipped_pairs)]
return z
# --------------------------------------------------product----------------------------------------
def product(l):
por = 1
for i in range(len(l)):
por *= l[i]
return por
# --------------------------------------------------binary----------------------------------------
def binarySearchCount(arr, n, key):
left = 0
right = n - 1
count = 0
while (left <= right):
mid = int((right + left) / 2)
# Check if middle element is
# less than or equal to key
if (arr[mid] < key):
count = mid + 1
left = mid + 1
# If key is smaller, ignore right half
else:
right = mid - 1
return count
# --------------------------------------------------binary----------------------------------------
def countdig(n):
c = 0
while (n > 0):
n //= 10
c += 1
return c
def binary(x, length):
y = bin(x)[2:]
return y if len(y) >= length else "0" * (length - len(y)) + y
def countGreater(arr, n, k):
l = 0
r = n - 1
# Stores the index of the left most element
# from the array which is greater than k
leftGreater = n
# Finds number of elements greater than k
while (l <= r):
m = int(l + (r - l) / 2)
if (arr[m] >= k):
leftGreater = m
r = m - 1
# If mid element is less than
# or equal to k update l
else:
l = m + 1
# Return the count of elements
# greater than k
return (n - leftGreater)
# --------------------------------------------------binary-----------------------------------
n,t=map(int,input().split())
l=list(map(int,input().split()))
dp=defaultdict(int)
freq=defaultdict(int)
ans=0
for i in range(n):
freq[l[i]]+=1
ans=max(ans,freq[l[i]])
if t<=200:
for i in range(t):
for j in range(n):
dp[l[j]] += 1
for k in range(n):
if k == j or l[k] == l[j]:
continue
if l[k] < l[j]:
dp[l[j]] = max(dp[l[j]], dp[l[k]] + 1)
ans = max(ans, max(dp.values()))
print(ans)
sys.exit(0)
for i in range(n):
for j in range(n):
dp[l[j]]+=1
for k in range(n):
if k==j or l[k]==l[j]:
continue
if l[k]<l[j]:
dp[l[j]]=max(dp[l[j]],dp[l[k]]+1)
ans=max(ans,max(dp.values()))
for j in range(n):
dp[l[j]]+=(t-2*n)
for i in range(n):
for j in range(n):
dp[l[j]]+=1
for k in range(n):
if k==j or l[k]==l[j]:
continue
if l[k]<l[j]:
dp[l[j]]=max(dp[l[j]],dp[l[k]]+1)
ans=max(ans,max(dp.values()))
print(ans)
``` | output | 1 | 104,869 | 12 | 209,739 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given an array of positive integers a1, a2, ..., an × T of length n × T. We know that for any i > n it is true that ai = ai - n. Find the length of the longest non-decreasing sequence of the given array.
Input
The first line contains two space-separated integers: n, T (1 ≤ n ≤ 100, 1 ≤ T ≤ 107). The second line contains n space-separated integers a1, a2, ..., an (1 ≤ ai ≤ 300).
Output
Print a single number — the length of a sought sequence.
Examples
Input
4 3
3 1 4 2
Output
5
Note
The array given in the sample looks like that: 3, 1, 4, 2, 3, 1, 4, 2, 3, 1, 4, 2. The elements in bold form the largest non-decreasing subsequence.
Submitted Solution:
```
size, t = map(int,input().split(sep = " "))
a = []
a = list(map(int,input().split(sep = " ")))
at = []
for x in range(t):
at += a
min = 1
dlugosc = 0
import collections
ctr = collections.Counter(a)
moda ,times = ctr.most_common(1).pop()
for i in range(len(at)):
if at[i] == moda & times !=1:
j = i
for j in range(len(a)):
if a[j] == moda:
dlugosc+=1
dlugosc += (t-1)*times
break
elif at[i] == min:
dlugosc += 1
elif min + 1 == at[i]:
min = at[i]
dlugosc += 1
print(dlugosc)
``` | instruction | 0 | 104,870 | 12 | 209,740 |
No | output | 1 | 104,870 | 12 | 209,741 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given an array of positive integers a1, a2, ..., an × T of length n × T. We know that for any i > n it is true that ai = ai - n. Find the length of the longest non-decreasing sequence of the given array.
Input
The first line contains two space-separated integers: n, T (1 ≤ n ≤ 100, 1 ≤ T ≤ 107). The second line contains n space-separated integers a1, a2, ..., an (1 ≤ ai ≤ 300).
Output
Print a single number — the length of a sought sequence.
Examples
Input
4 3
3 1 4 2
Output
5
Note
The array given in the sample looks like that: 3, 1, 4, 2, 3, 1, 4, 2, 3, 1, 4, 2. The elements in bold form the largest non-decreasing subsequence.
Submitted Solution:
```
# ---------------------------iye ha aam zindegi---------------------------------------------
import math
import random
import heapq,bisect
import sys
from collections import deque, defaultdict
from fractions import Fraction
import sys
import threading
from collections import defaultdict
threading.stack_size(10**8)
mod = 10 ** 9 + 7
mod1 = 998244353
# ------------------------------warmup----------------------------
import os
import sys
from io import BytesIO, IOBase
sys.setrecursionlimit(300000)
BUFSIZE = 8192
class FastIO(IOBase):
newlines = 0
def __init__(self, file):
self._fd = file.fileno()
self.buffer = BytesIO()
self.writable = "x" in file.mode or "r" not in file.mode
self.write = self.buffer.write if self.writable else None
def read(self):
while True:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
if not b:
break
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines = 0
return self.buffer.read()
def readline(self):
while self.newlines == 0:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
self.newlines = b.count(b"\n") + (not b)
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines -= 1
return self.buffer.readline()
def flush(self):
if self.writable:
os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
self.write = lambda s: self.buffer.write(s.encode("ascii"))
self.read = lambda: self.buffer.read().decode("ascii")
self.readline = lambda: self.buffer.readline().decode("ascii")
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")
# -------------------game starts now----------------------------------------------------import math
class TreeNode:
def __init__(self, k, v):
self.key = k
self.value = v
self.left = None
self.right = None
self.parent = None
self.height = 1
self.num_left = 1
self.num_total = 1
class AvlTree:
def __init__(self):
self._tree = None
def add(self, k, v):
if not self._tree:
self._tree = TreeNode(k, v)
return
node = self._add(k, v)
if node:
self._rebalance(node)
def _add(self, k, v):
node = self._tree
while node:
if k < node.key:
if node.left:
node = node.left
else:
node.left = TreeNode(k, v)
node.left.parent = node
return node.left
elif node.key < k:
if node.right:
node = node.right
else:
node.right = TreeNode(k, v)
node.right.parent = node
return node.right
else:
node.value = v
return
@staticmethod
def get_height(x):
return x.height if x else 0
@staticmethod
def get_num_total(x):
return x.num_total if x else 0
def _rebalance(self, node):
n = node
while n:
lh = self.get_height(n.left)
rh = self.get_height(n.right)
n.height = max(lh, rh) + 1
balance_factor = lh - rh
n.num_total = 1 + self.get_num_total(n.left) + self.get_num_total(n.right)
n.num_left = 1 + self.get_num_total(n.left)
if balance_factor > 1:
if self.get_height(n.left.left) < self.get_height(n.left.right):
self._rotate_left(n.left)
self._rotate_right(n)
elif balance_factor < -1:
if self.get_height(n.right.right) < self.get_height(n.right.left):
self._rotate_right(n.right)
self._rotate_left(n)
else:
n = n.parent
def _remove_one(self, node):
"""
Side effect!!! Changes node. Node should have exactly one child
"""
replacement = node.left or node.right
if node.parent:
if AvlTree._is_left(node):
node.parent.left = replacement
else:
node.parent.right = replacement
replacement.parent = node.parent
node.parent = None
else:
self._tree = replacement
replacement.parent = None
node.left = None
node.right = None
node.parent = None
self._rebalance(replacement)
def _remove_leaf(self, node):
if node.parent:
if AvlTree._is_left(node):
node.parent.left = None
else:
node.parent.right = None
self._rebalance(node.parent)
else:
self._tree = None
node.parent = None
node.left = None
node.right = None
def remove(self, k):
node = self._get_node(k)
if not node:
return
if AvlTree._is_leaf(node):
self._remove_leaf(node)
return
if node.left and node.right:
nxt = AvlTree._get_next(node)
node.key = nxt.key
node.value = nxt.value
if self._is_leaf(nxt):
self._remove_leaf(nxt)
else:
self._remove_one(nxt)
self._rebalance(node)
else:
self._remove_one(node)
def get(self, k):
node = self._get_node(k)
return node.value if node else -1
def _get_node(self, k):
if not self._tree:
return None
node = self._tree
while node:
if k < node.key:
node = node.left
elif node.key < k:
node = node.right
else:
return node
return None
def get_at(self, pos):
x = pos + 1
node = self._tree
while node:
if x < node.num_left:
node = node.left
elif node.num_left < x:
x -= node.num_left
node = node.right
else:
return (node.key, node.value)
raise IndexError("Out of ranges")
@staticmethod
def _is_left(node):
return node.parent.left and node.parent.left == node
@staticmethod
def _is_leaf(node):
return node.left is None and node.right is None
def _rotate_right(self, node):
if not node.parent:
self._tree = node.left
node.left.parent = None
elif AvlTree._is_left(node):
node.parent.left = node.left
node.left.parent = node.parent
else:
node.parent.right = node.left
node.left.parent = node.parent
bk = node.left.right
node.left.right = node
node.parent = node.left
node.left = bk
if bk:
bk.parent = node
node.height = max(self.get_height(node.left), self.get_height(node.right)) + 1
node.num_total = 1 + self.get_num_total(node.left) + self.get_num_total(node.right)
node.num_left = 1 + self.get_num_total(node.left)
def _rotate_left(self, node):
if not node.parent:
self._tree = node.right
node.right.parent = None
elif AvlTree._is_left(node):
node.parent.left = node.right
node.right.parent = node.parent
else:
node.parent.right = node.right
node.right.parent = node.parent
bk = node.right.left
node.right.left = node
node.parent = node.right
node.right = bk
if bk:
bk.parent = node
node.height = max(self.get_height(node.left), self.get_height(node.right)) + 1
node.num_total = 1 + self.get_num_total(node.left) + self.get_num_total(node.right)
node.num_left = 1 + self.get_num_total(node.left)
@staticmethod
def _get_next(node):
if not node.right:
return node.parent
n = node.right
while n.left:
n = n.left
return n
# -----------------------------------------------binary seacrh tree---------------------------------------
class SegmentTree1:
def __init__(self, data, default=2**51, func=lambda a, b: a & b):
"""initialize the segment tree with data"""
self._default = default
self._func = func
self._len = len(data)
self._size = _size = 1 << (self._len - 1).bit_length()
self.data = [default] * (2 * _size)
self.data[_size:_size + self._len] = data
for i in reversed(range(_size)):
self.data[i] = func(self.data[i + i], self.data[i + i + 1])
def __delitem__(self, idx):
self[idx] = self._default
def __getitem__(self, idx):
return self.data[idx + self._size]
def __setitem__(self, idx, value):
idx += self._size
self.data[idx] = value
idx >>= 1
while idx:
self.data[idx] = self._func(self.data[2 * idx], self.data[2 * idx + 1])
idx >>= 1
def __len__(self):
return self._len
def query(self, start, stop):
if start == stop:
return self.__getitem__(start)
stop += 1
start += self._size
stop += self._size
res = self._default
while start < stop:
if start & 1:
res = self._func(res, self.data[start])
start += 1
if stop & 1:
stop -= 1
res = self._func(res, self.data[stop])
start >>= 1
stop >>= 1
return res
def __repr__(self):
return "SegmentTree({0})".format(self.data)
# -------------------game starts now----------------------------------------------------import math
class SegmentTree:
def __init__(self, data, default=0, func=lambda a, b: max(a , b)):
"""initialize the segment tree with data"""
self._default = default
self._func = func
self._len = len(data)
self._size = _size = 1 << (self._len - 1).bit_length()
self.data = [default] * (2 * _size)
self.data[_size:_size + self._len] = data
for i in reversed(range(_size)):
self.data[i] = func(self.data[i + i], self.data[i + i + 1])
def __delitem__(self, idx):
self[idx] = self._default
def __getitem__(self, idx):
return self.data[idx + self._size]
def __setitem__(self, idx, value):
idx += self._size
self.data[idx] = value
idx >>= 1
while idx:
self.data[idx] = self._func(self.data[2 * idx], self.data[2 * idx + 1])
idx >>= 1
def __len__(self):
return self._len
def query(self, start, stop):
if start == stop:
return self.__getitem__(start)
stop += 1
start += self._size
stop += self._size
res = self._default
while start < stop:
if start & 1:
res = self._func(res, self.data[start])
start += 1
if stop & 1:
stop -= 1
res = self._func(res, self.data[stop])
start >>= 1
stop >>= 1
return res
def __repr__(self):
return "SegmentTree({0})".format(self.data)
# -------------------------------iye ha chutiya zindegi-------------------------------------
class Factorial:
def __init__(self, MOD):
self.MOD = MOD
self.factorials = [1, 1]
self.invModulos = [0, 1]
self.invFactorial_ = [1, 1]
def calc(self, n):
if n <= -1:
print("Invalid argument to calculate n!")
print("n must be non-negative value. But the argument was " + str(n))
exit()
if n < len(self.factorials):
return self.factorials[n]
nextArr = [0] * (n + 1 - len(self.factorials))
initialI = len(self.factorials)
prev = self.factorials[-1]
m = self.MOD
for i in range(initialI, n + 1):
prev = nextArr[i - initialI] = prev * i % m
self.factorials += nextArr
return self.factorials[n]
def inv(self, n):
if n <= -1:
print("Invalid argument to calculate n^(-1)")
print("n must be non-negative value. But the argument was " + str(n))
exit()
p = self.MOD
pi = n % p
if pi < len(self.invModulos):
return self.invModulos[pi]
nextArr = [0] * (n + 1 - len(self.invModulos))
initialI = len(self.invModulos)
for i in range(initialI, min(p, n + 1)):
next = -self.invModulos[p % i] * (p // i) % p
self.invModulos.append(next)
return self.invModulos[pi]
def invFactorial(self, n):
if n <= -1:
print("Invalid argument to calculate (n^(-1))!")
print("n must be non-negative value. But the argument was " + str(n))
exit()
if n < len(self.invFactorial_):
return self.invFactorial_[n]
self.inv(n) # To make sure already calculated n^-1
nextArr = [0] * (n + 1 - len(self.invFactorial_))
initialI = len(self.invFactorial_)
prev = self.invFactorial_[-1]
p = self.MOD
for i in range(initialI, n + 1):
prev = nextArr[i - initialI] = (prev * self.invModulos[i % p]) % p
self.invFactorial_ += nextArr
return self.invFactorial_[n]
class Combination:
def __init__(self, MOD):
self.MOD = MOD
self.factorial = Factorial(MOD)
def ncr(self, n, k):
if k < 0 or n < k:
return 0
k = min(k, n - k)
f = self.factorial
return f.calc(n) * f.invFactorial(max(n - k, k)) * f.invFactorial(min(k, n - k)) % self.MOD
# --------------------------------------iye ha combinations ka zindegi---------------------------------
def powm(a, n, m):
if a == 1 or n == 0:
return 1
if n % 2 == 0:
s = powm(a, n // 2, m)
return s * s % m
else:
return a * powm(a, n - 1, m) % m
# --------------------------------------iye ha power ka zindegi---------------------------------
def sort_list(list1, list2):
zipped_pairs = zip(list2, list1)
z = [x for _, x in sorted(zipped_pairs)]
return z
# --------------------------------------------------product----------------------------------------
def product(l):
por = 1
for i in range(len(l)):
por *= l[i]
return por
# --------------------------------------------------binary----------------------------------------
def binarySearchCount(arr, n, key):
left = 0
right = n - 1
count = 0
while (left <= right):
mid = int((right + left) / 2)
# Check if middle element is
# less than or equal to key
if (arr[mid] < key):
count = mid + 1
left = mid + 1
# If key is smaller, ignore right half
else:
right = mid - 1
return count
# --------------------------------------------------binary----------------------------------------
def countdig(n):
c = 0
while (n > 0):
n //= 10
c += 1
return c
def binary(x, length):
y = bin(x)[2:]
return y if len(y) >= length else "0" * (length - len(y)) + y
def countGreater(arr, n, k):
l = 0
r = n - 1
# Stores the index of the left most element
# from the array which is greater than k
leftGreater = n
# Finds number of elements greater than k
while (l <= r):
m = int(l + (r - l) / 2)
if (arr[m] >= k):
leftGreater = m
r = m - 1
# If mid element is less than
# or equal to k update l
else:
l = m + 1
# Return the count of elements
# greater than k
return (n - leftGreater)
# --------------------------------------------------binary-----------------------------------
n,t=map(int,input().split())
l=list(map(int,input().split()))
t1=t-1
t=min(n,t)-1
dp=defaultdict(int)
freq=defaultdict(int)
ans=0
for i in range(n):
freq[l[i]]+=1
ans=max(ans,freq[l[i]])
if t>0:
for i in range(n):
dp[l[i]]=1
for j in range(i):
if l[j]<=l[i]:
dp[l[i]]=max(dp[l[i]],dp[l[j]]+1)
ans=max(dp.values())
for i in range(1,t):
for j in range(n):
ans=max(ans,dp[l[j]]+(t-i)*freq[l[j]])
for j in range(n):
dp[l[j]]+=1
for k in range(n):
if k==j or l[k]==l[j]:
continue
if l[k]<=l[j]:
dp[l[j]]=max(dp[l[j]],dp[l[k]]+1)
ans=max(ans,max(dp.values()))
for j in range(n):
dp[l[j]]+=(t1-t)
for j in range(n):
dp[l[j]]+=1
for k in range(n):
if k==j or l[k]==l[j]:
continue
if l[k]<=l[j]:
dp[l[j]]=max(dp[l[j]],dp[l[k]]+1)
ans=max(ans,max(dp.values()))
print(ans)
``` | instruction | 0 | 104,871 | 12 | 209,742 |
No | output | 1 | 104,871 | 12 | 209,743 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given an array of positive integers a1, a2, ..., an × T of length n × T. We know that for any i > n it is true that ai = ai - n. Find the length of the longest non-decreasing sequence of the given array.
Input
The first line contains two space-separated integers: n, T (1 ≤ n ≤ 100, 1 ≤ T ≤ 107). The second line contains n space-separated integers a1, a2, ..., an (1 ≤ ai ≤ 300).
Output
Print a single number — the length of a sought sequence.
Examples
Input
4 3
3 1 4 2
Output
5
Note
The array given in the sample looks like that: 3, 1, 4, 2, 3, 1, 4, 2, 3, 1, 4, 2. The elements in bold form the largest non-decreasing subsequence.
Submitted Solution:
```
# ---------------------------iye ha aam zindegi---------------------------------------------
import math
import random
import heapq,bisect
import sys
from collections import deque, defaultdict
from fractions import Fraction
import sys
import threading
from collections import defaultdict
threading.stack_size(10**8)
mod = 10 ** 9 + 7
mod1 = 998244353
# ------------------------------warmup----------------------------
import os
import sys
from io import BytesIO, IOBase
sys.setrecursionlimit(300000)
BUFSIZE = 8192
class FastIO(IOBase):
newlines = 0
def __init__(self, file):
self._fd = file.fileno()
self.buffer = BytesIO()
self.writable = "x" in file.mode or "r" not in file.mode
self.write = self.buffer.write if self.writable else None
def read(self):
while True:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
if not b:
break
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines = 0
return self.buffer.read()
def readline(self):
while self.newlines == 0:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
self.newlines = b.count(b"\n") + (not b)
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines -= 1
return self.buffer.readline()
def flush(self):
if self.writable:
os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
self.write = lambda s: self.buffer.write(s.encode("ascii"))
self.read = lambda: self.buffer.read().decode("ascii")
self.readline = lambda: self.buffer.readline().decode("ascii")
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")
# -------------------game starts now----------------------------------------------------import math
class TreeNode:
def __init__(self, k, v):
self.key = k
self.value = v
self.left = None
self.right = None
self.parent = None
self.height = 1
self.num_left = 1
self.num_total = 1
class AvlTree:
def __init__(self):
self._tree = None
def add(self, k, v):
if not self._tree:
self._tree = TreeNode(k, v)
return
node = self._add(k, v)
if node:
self._rebalance(node)
def _add(self, k, v):
node = self._tree
while node:
if k < node.key:
if node.left:
node = node.left
else:
node.left = TreeNode(k, v)
node.left.parent = node
return node.left
elif node.key < k:
if node.right:
node = node.right
else:
node.right = TreeNode(k, v)
node.right.parent = node
return node.right
else:
node.value = v
return
@staticmethod
def get_height(x):
return x.height if x else 0
@staticmethod
def get_num_total(x):
return x.num_total if x else 0
def _rebalance(self, node):
n = node
while n:
lh = self.get_height(n.left)
rh = self.get_height(n.right)
n.height = max(lh, rh) + 1
balance_factor = lh - rh
n.num_total = 1 + self.get_num_total(n.left) + self.get_num_total(n.right)
n.num_left = 1 + self.get_num_total(n.left)
if balance_factor > 1:
if self.get_height(n.left.left) < self.get_height(n.left.right):
self._rotate_left(n.left)
self._rotate_right(n)
elif balance_factor < -1:
if self.get_height(n.right.right) < self.get_height(n.right.left):
self._rotate_right(n.right)
self._rotate_left(n)
else:
n = n.parent
def _remove_one(self, node):
"""
Side effect!!! Changes node. Node should have exactly one child
"""
replacement = node.left or node.right
if node.parent:
if AvlTree._is_left(node):
node.parent.left = replacement
else:
node.parent.right = replacement
replacement.parent = node.parent
node.parent = None
else:
self._tree = replacement
replacement.parent = None
node.left = None
node.right = None
node.parent = None
self._rebalance(replacement)
def _remove_leaf(self, node):
if node.parent:
if AvlTree._is_left(node):
node.parent.left = None
else:
node.parent.right = None
self._rebalance(node.parent)
else:
self._tree = None
node.parent = None
node.left = None
node.right = None
def remove(self, k):
node = self._get_node(k)
if not node:
return
if AvlTree._is_leaf(node):
self._remove_leaf(node)
return
if node.left and node.right:
nxt = AvlTree._get_next(node)
node.key = nxt.key
node.value = nxt.value
if self._is_leaf(nxt):
self._remove_leaf(nxt)
else:
self._remove_one(nxt)
self._rebalance(node)
else:
self._remove_one(node)
def get(self, k):
node = self._get_node(k)
return node.value if node else -1
def _get_node(self, k):
if not self._tree:
return None
node = self._tree
while node:
if k < node.key:
node = node.left
elif node.key < k:
node = node.right
else:
return node
return None
def get_at(self, pos):
x = pos + 1
node = self._tree
while node:
if x < node.num_left:
node = node.left
elif node.num_left < x:
x -= node.num_left
node = node.right
else:
return (node.key, node.value)
raise IndexError("Out of ranges")
@staticmethod
def _is_left(node):
return node.parent.left and node.parent.left == node
@staticmethod
def _is_leaf(node):
return node.left is None and node.right is None
def _rotate_right(self, node):
if not node.parent:
self._tree = node.left
node.left.parent = None
elif AvlTree._is_left(node):
node.parent.left = node.left
node.left.parent = node.parent
else:
node.parent.right = node.left
node.left.parent = node.parent
bk = node.left.right
node.left.right = node
node.parent = node.left
node.left = bk
if bk:
bk.parent = node
node.height = max(self.get_height(node.left), self.get_height(node.right)) + 1
node.num_total = 1 + self.get_num_total(node.left) + self.get_num_total(node.right)
node.num_left = 1 + self.get_num_total(node.left)
def _rotate_left(self, node):
if not node.parent:
self._tree = node.right
node.right.parent = None
elif AvlTree._is_left(node):
node.parent.left = node.right
node.right.parent = node.parent
else:
node.parent.right = node.right
node.right.parent = node.parent
bk = node.right.left
node.right.left = node
node.parent = node.right
node.right = bk
if bk:
bk.parent = node
node.height = max(self.get_height(node.left), self.get_height(node.right)) + 1
node.num_total = 1 + self.get_num_total(node.left) + self.get_num_total(node.right)
node.num_left = 1 + self.get_num_total(node.left)
@staticmethod
def _get_next(node):
if not node.right:
return node.parent
n = node.right
while n.left:
n = n.left
return n
# -----------------------------------------------binary seacrh tree---------------------------------------
class SegmentTree1:
def __init__(self, data, default=2**51, func=lambda a, b: a & b):
"""initialize the segment tree with data"""
self._default = default
self._func = func
self._len = len(data)
self._size = _size = 1 << (self._len - 1).bit_length()
self.data = [default] * (2 * _size)
self.data[_size:_size + self._len] = data
for i in reversed(range(_size)):
self.data[i] = func(self.data[i + i], self.data[i + i + 1])
def __delitem__(self, idx):
self[idx] = self._default
def __getitem__(self, idx):
return self.data[idx + self._size]
def __setitem__(self, idx, value):
idx += self._size
self.data[idx] = value
idx >>= 1
while idx:
self.data[idx] = self._func(self.data[2 * idx], self.data[2 * idx + 1])
idx >>= 1
def __len__(self):
return self._len
def query(self, start, stop):
if start == stop:
return self.__getitem__(start)
stop += 1
start += self._size
stop += self._size
res = self._default
while start < stop:
if start & 1:
res = self._func(res, self.data[start])
start += 1
if stop & 1:
stop -= 1
res = self._func(res, self.data[stop])
start >>= 1
stop >>= 1
return res
def __repr__(self):
return "SegmentTree({0})".format(self.data)
# -------------------game starts now----------------------------------------------------import math
class SegmentTree:
def __init__(self, data, default=0, func=lambda a, b: max(a , b)):
"""initialize the segment tree with data"""
self._default = default
self._func = func
self._len = len(data)
self._size = _size = 1 << (self._len - 1).bit_length()
self.data = [default] * (2 * _size)
self.data[_size:_size + self._len] = data
for i in reversed(range(_size)):
self.data[i] = func(self.data[i + i], self.data[i + i + 1])
def __delitem__(self, idx):
self[idx] = self._default
def __getitem__(self, idx):
return self.data[idx + self._size]
def __setitem__(self, idx, value):
idx += self._size
self.data[idx] = value
idx >>= 1
while idx:
self.data[idx] = self._func(self.data[2 * idx], self.data[2 * idx + 1])
idx >>= 1
def __len__(self):
return self._len
def query(self, start, stop):
if start == stop:
return self.__getitem__(start)
stop += 1
start += self._size
stop += self._size
res = self._default
while start < stop:
if start & 1:
res = self._func(res, self.data[start])
start += 1
if stop & 1:
stop -= 1
res = self._func(res, self.data[stop])
start >>= 1
stop >>= 1
return res
def __repr__(self):
return "SegmentTree({0})".format(self.data)
# -------------------------------iye ha chutiya zindegi-------------------------------------
class Factorial:
def __init__(self, MOD):
self.MOD = MOD
self.factorials = [1, 1]
self.invModulos = [0, 1]
self.invFactorial_ = [1, 1]
def calc(self, n):
if n <= -1:
print("Invalid argument to calculate n!")
print("n must be non-negative value. But the argument was " + str(n))
exit()
if n < len(self.factorials):
return self.factorials[n]
nextArr = [0] * (n + 1 - len(self.factorials))
initialI = len(self.factorials)
prev = self.factorials[-1]
m = self.MOD
for i in range(initialI, n + 1):
prev = nextArr[i - initialI] = prev * i % m
self.factorials += nextArr
return self.factorials[n]
def inv(self, n):
if n <= -1:
print("Invalid argument to calculate n^(-1)")
print("n must be non-negative value. But the argument was " + str(n))
exit()
p = self.MOD
pi = n % p
if pi < len(self.invModulos):
return self.invModulos[pi]
nextArr = [0] * (n + 1 - len(self.invModulos))
initialI = len(self.invModulos)
for i in range(initialI, min(p, n + 1)):
next = -self.invModulos[p % i] * (p // i) % p
self.invModulos.append(next)
return self.invModulos[pi]
def invFactorial(self, n):
if n <= -1:
print("Invalid argument to calculate (n^(-1))!")
print("n must be non-negative value. But the argument was " + str(n))
exit()
if n < len(self.invFactorial_):
return self.invFactorial_[n]
self.inv(n) # To make sure already calculated n^-1
nextArr = [0] * (n + 1 - len(self.invFactorial_))
initialI = len(self.invFactorial_)
prev = self.invFactorial_[-1]
p = self.MOD
for i in range(initialI, n + 1):
prev = nextArr[i - initialI] = (prev * self.invModulos[i % p]) % p
self.invFactorial_ += nextArr
return self.invFactorial_[n]
class Combination:
def __init__(self, MOD):
self.MOD = MOD
self.factorial = Factorial(MOD)
def ncr(self, n, k):
if k < 0 or n < k:
return 0
k = min(k, n - k)
f = self.factorial
return f.calc(n) * f.invFactorial(max(n - k, k)) * f.invFactorial(min(k, n - k)) % self.MOD
# --------------------------------------iye ha combinations ka zindegi---------------------------------
def powm(a, n, m):
if a == 1 or n == 0:
return 1
if n % 2 == 0:
s = powm(a, n // 2, m)
return s * s % m
else:
return a * powm(a, n - 1, m) % m
# --------------------------------------iye ha power ka zindegi---------------------------------
def sort_list(list1, list2):
zipped_pairs = zip(list2, list1)
z = [x for _, x in sorted(zipped_pairs)]
return z
# --------------------------------------------------product----------------------------------------
def product(l):
por = 1
for i in range(len(l)):
por *= l[i]
return por
# --------------------------------------------------binary----------------------------------------
def binarySearchCount(arr, n, key):
left = 0
right = n - 1
count = 0
while (left <= right):
mid = int((right + left) / 2)
# Check if middle element is
# less than or equal to key
if (arr[mid] < key):
count = mid + 1
left = mid + 1
# If key is smaller, ignore right half
else:
right = mid - 1
return count
# --------------------------------------------------binary----------------------------------------
def countdig(n):
c = 0
while (n > 0):
n //= 10
c += 1
return c
def binary(x, length):
y = bin(x)[2:]
return y if len(y) >= length else "0" * (length - len(y)) + y
def countGreater(arr, n, k):
l = 0
r = n - 1
# Stores the index of the left most element
# from the array which is greater than k
leftGreater = n
# Finds number of elements greater than k
while (l <= r):
m = int(l + (r - l) / 2)
if (arr[m] >= k):
leftGreater = m
r = m - 1
# If mid element is less than
# or equal to k update l
else:
l = m + 1
# Return the count of elements
# greater than k
return (n - leftGreater)
# --------------------------------------------------binary-----------------------------------
n,t=map(int,input().split())
l=list(map(int,input().split()))
t1=t-1
t=min(n,t)-1
dp=defaultdict(int)
freq=defaultdict(int)
for i in range(n):
dp[l[i]]=1
freq[l[i]]+=1
for j in range(i):
if l[j]<=l[i]:
dp[l[i]]=max(dp[l[i]],dp[l[j]]+1)
ans=max(dp.values())
for i in range(1,t):
for j in range(n):
dp[l[j]]+=1
for k in range(n):
if k==j:
continue
if l[k]<=l[j]:
dp[l[j]]=max(dp[l[j]],dp[l[k]]+1)
ans=max(ans,max(dp.values()))
for j in range(n):
dp[l[j]]+=(t1-t)*freq[l[j]]
for j in range(n):
dp[l[j]]+=1
for k in range(n):
if k==j:
continue
if l[k]<=l[j]:
dp[l[j]]=max(dp[l[j]],dp[l[k]]+1)
ans=max(ans,max(dp.values()))
print(ans)
``` | instruction | 0 | 104,872 | 12 | 209,744 |
No | output | 1 | 104,872 | 12 | 209,745 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given an array of positive integers a1, a2, ..., an × T of length n × T. We know that for any i > n it is true that ai = ai - n. Find the length of the longest non-decreasing sequence of the given array.
Input
The first line contains two space-separated integers: n, T (1 ≤ n ≤ 100, 1 ≤ T ≤ 107). The second line contains n space-separated integers a1, a2, ..., an (1 ≤ ai ≤ 300).
Output
Print a single number — the length of a sought sequence.
Examples
Input
4 3
3 1 4 2
Output
5
Note
The array given in the sample looks like that: 3, 1, 4, 2, 3, 1, 4, 2, 3, 1, 4, 2. The elements in bold form the largest non-decreasing subsequence.
Submitted Solution:
```
# ---------------------------iye ha aam zindegi---------------------------------------------
import math
import random
import heapq,bisect
import sys
from collections import deque, defaultdict
from fractions import Fraction
import sys
import threading
from collections import defaultdict
threading.stack_size(10**8)
mod = 10 ** 9 + 7
mod1 = 998244353
# ------------------------------warmup----------------------------
import os
import sys
from io import BytesIO, IOBase
sys.setrecursionlimit(300000)
BUFSIZE = 8192
class FastIO(IOBase):
newlines = 0
def __init__(self, file):
self._fd = file.fileno()
self.buffer = BytesIO()
self.writable = "x" in file.mode or "r" not in file.mode
self.write = self.buffer.write if self.writable else None
def read(self):
while True:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
if not b:
break
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines = 0
return self.buffer.read()
def readline(self):
while self.newlines == 0:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
self.newlines = b.count(b"\n") + (not b)
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines -= 1
return self.buffer.readline()
def flush(self):
if self.writable:
os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
self.write = lambda s: self.buffer.write(s.encode("ascii"))
self.read = lambda: self.buffer.read().decode("ascii")
self.readline = lambda: self.buffer.readline().decode("ascii")
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")
# -------------------game starts now----------------------------------------------------import math
class TreeNode:
def __init__(self, k, v):
self.key = k
self.value = v
self.left = None
self.right = None
self.parent = None
self.height = 1
self.num_left = 1
self.num_total = 1
class AvlTree:
def __init__(self):
self._tree = None
def add(self, k, v):
if not self._tree:
self._tree = TreeNode(k, v)
return
node = self._add(k, v)
if node:
self._rebalance(node)
def _add(self, k, v):
node = self._tree
while node:
if k < node.key:
if node.left:
node = node.left
else:
node.left = TreeNode(k, v)
node.left.parent = node
return node.left
elif node.key < k:
if node.right:
node = node.right
else:
node.right = TreeNode(k, v)
node.right.parent = node
return node.right
else:
node.value = v
return
@staticmethod
def get_height(x):
return x.height if x else 0
@staticmethod
def get_num_total(x):
return x.num_total if x else 0
def _rebalance(self, node):
n = node
while n:
lh = self.get_height(n.left)
rh = self.get_height(n.right)
n.height = max(lh, rh) + 1
balance_factor = lh - rh
n.num_total = 1 + self.get_num_total(n.left) + self.get_num_total(n.right)
n.num_left = 1 + self.get_num_total(n.left)
if balance_factor > 1:
if self.get_height(n.left.left) < self.get_height(n.left.right):
self._rotate_left(n.left)
self._rotate_right(n)
elif balance_factor < -1:
if self.get_height(n.right.right) < self.get_height(n.right.left):
self._rotate_right(n.right)
self._rotate_left(n)
else:
n = n.parent
def _remove_one(self, node):
"""
Side effect!!! Changes node. Node should have exactly one child
"""
replacement = node.left or node.right
if node.parent:
if AvlTree._is_left(node):
node.parent.left = replacement
else:
node.parent.right = replacement
replacement.parent = node.parent
node.parent = None
else:
self._tree = replacement
replacement.parent = None
node.left = None
node.right = None
node.parent = None
self._rebalance(replacement)
def _remove_leaf(self, node):
if node.parent:
if AvlTree._is_left(node):
node.parent.left = None
else:
node.parent.right = None
self._rebalance(node.parent)
else:
self._tree = None
node.parent = None
node.left = None
node.right = None
def remove(self, k):
node = self._get_node(k)
if not node:
return
if AvlTree._is_leaf(node):
self._remove_leaf(node)
return
if node.left and node.right:
nxt = AvlTree._get_next(node)
node.key = nxt.key
node.value = nxt.value
if self._is_leaf(nxt):
self._remove_leaf(nxt)
else:
self._remove_one(nxt)
self._rebalance(node)
else:
self._remove_one(node)
def get(self, k):
node = self._get_node(k)
return node.value if node else -1
def _get_node(self, k):
if not self._tree:
return None
node = self._tree
while node:
if k < node.key:
node = node.left
elif node.key < k:
node = node.right
else:
return node
return None
def get_at(self, pos):
x = pos + 1
node = self._tree
while node:
if x < node.num_left:
node = node.left
elif node.num_left < x:
x -= node.num_left
node = node.right
else:
return (node.key, node.value)
raise IndexError("Out of ranges")
@staticmethod
def _is_left(node):
return node.parent.left and node.parent.left == node
@staticmethod
def _is_leaf(node):
return node.left is None and node.right is None
def _rotate_right(self, node):
if not node.parent:
self._tree = node.left
node.left.parent = None
elif AvlTree._is_left(node):
node.parent.left = node.left
node.left.parent = node.parent
else:
node.parent.right = node.left
node.left.parent = node.parent
bk = node.left.right
node.left.right = node
node.parent = node.left
node.left = bk
if bk:
bk.parent = node
node.height = max(self.get_height(node.left), self.get_height(node.right)) + 1
node.num_total = 1 + self.get_num_total(node.left) + self.get_num_total(node.right)
node.num_left = 1 + self.get_num_total(node.left)
def _rotate_left(self, node):
if not node.parent:
self._tree = node.right
node.right.parent = None
elif AvlTree._is_left(node):
node.parent.left = node.right
node.right.parent = node.parent
else:
node.parent.right = node.right
node.right.parent = node.parent
bk = node.right.left
node.right.left = node
node.parent = node.right
node.right = bk
if bk:
bk.parent = node
node.height = max(self.get_height(node.left), self.get_height(node.right)) + 1
node.num_total = 1 + self.get_num_total(node.left) + self.get_num_total(node.right)
node.num_left = 1 + self.get_num_total(node.left)
@staticmethod
def _get_next(node):
if not node.right:
return node.parent
n = node.right
while n.left:
n = n.left
return n
# -----------------------------------------------binary seacrh tree---------------------------------------
class SegmentTree1:
def __init__(self, data, default=2**51, func=lambda a, b: a & b):
"""initialize the segment tree with data"""
self._default = default
self._func = func
self._len = len(data)
self._size = _size = 1 << (self._len - 1).bit_length()
self.data = [default] * (2 * _size)
self.data[_size:_size + self._len] = data
for i in reversed(range(_size)):
self.data[i] = func(self.data[i + i], self.data[i + i + 1])
def __delitem__(self, idx):
self[idx] = self._default
def __getitem__(self, idx):
return self.data[idx + self._size]
def __setitem__(self, idx, value):
idx += self._size
self.data[idx] = value
idx >>= 1
while idx:
self.data[idx] = self._func(self.data[2 * idx], self.data[2 * idx + 1])
idx >>= 1
def __len__(self):
return self._len
def query(self, start, stop):
if start == stop:
return self.__getitem__(start)
stop += 1
start += self._size
stop += self._size
res = self._default
while start < stop:
if start & 1:
res = self._func(res, self.data[start])
start += 1
if stop & 1:
stop -= 1
res = self._func(res, self.data[stop])
start >>= 1
stop >>= 1
return res
def __repr__(self):
return "SegmentTree({0})".format(self.data)
# -------------------game starts now----------------------------------------------------import math
class SegmentTree:
def __init__(self, data, default=0, func=lambda a, b: max(a , b)):
"""initialize the segment tree with data"""
self._default = default
self._func = func
self._len = len(data)
self._size = _size = 1 << (self._len - 1).bit_length()
self.data = [default] * (2 * _size)
self.data[_size:_size + self._len] = data
for i in reversed(range(_size)):
self.data[i] = func(self.data[i + i], self.data[i + i + 1])
def __delitem__(self, idx):
self[idx] = self._default
def __getitem__(self, idx):
return self.data[idx + self._size]
def __setitem__(self, idx, value):
idx += self._size
self.data[idx] = value
idx >>= 1
while idx:
self.data[idx] = self._func(self.data[2 * idx], self.data[2 * idx + 1])
idx >>= 1
def __len__(self):
return self._len
def query(self, start, stop):
if start == stop:
return self.__getitem__(start)
stop += 1
start += self._size
stop += self._size
res = self._default
while start < stop:
if start & 1:
res = self._func(res, self.data[start])
start += 1
if stop & 1:
stop -= 1
res = self._func(res, self.data[stop])
start >>= 1
stop >>= 1
return res
def __repr__(self):
return "SegmentTree({0})".format(self.data)
# -------------------------------iye ha chutiya zindegi-------------------------------------
class Factorial:
def __init__(self, MOD):
self.MOD = MOD
self.factorials = [1, 1]
self.invModulos = [0, 1]
self.invFactorial_ = [1, 1]
def calc(self, n):
if n <= -1:
print("Invalid argument to calculate n!")
print("n must be non-negative value. But the argument was " + str(n))
exit()
if n < len(self.factorials):
return self.factorials[n]
nextArr = [0] * (n + 1 - len(self.factorials))
initialI = len(self.factorials)
prev = self.factorials[-1]
m = self.MOD
for i in range(initialI, n + 1):
prev = nextArr[i - initialI] = prev * i % m
self.factorials += nextArr
return self.factorials[n]
def inv(self, n):
if n <= -1:
print("Invalid argument to calculate n^(-1)")
print("n must be non-negative value. But the argument was " + str(n))
exit()
p = self.MOD
pi = n % p
if pi < len(self.invModulos):
return self.invModulos[pi]
nextArr = [0] * (n + 1 - len(self.invModulos))
initialI = len(self.invModulos)
for i in range(initialI, min(p, n + 1)):
next = -self.invModulos[p % i] * (p // i) % p
self.invModulos.append(next)
return self.invModulos[pi]
def invFactorial(self, n):
if n <= -1:
print("Invalid argument to calculate (n^(-1))!")
print("n must be non-negative value. But the argument was " + str(n))
exit()
if n < len(self.invFactorial_):
return self.invFactorial_[n]
self.inv(n) # To make sure already calculated n^-1
nextArr = [0] * (n + 1 - len(self.invFactorial_))
initialI = len(self.invFactorial_)
prev = self.invFactorial_[-1]
p = self.MOD
for i in range(initialI, n + 1):
prev = nextArr[i - initialI] = (prev * self.invModulos[i % p]) % p
self.invFactorial_ += nextArr
return self.invFactorial_[n]
class Combination:
def __init__(self, MOD):
self.MOD = MOD
self.factorial = Factorial(MOD)
def ncr(self, n, k):
if k < 0 or n < k:
return 0
k = min(k, n - k)
f = self.factorial
return f.calc(n) * f.invFactorial(max(n - k, k)) * f.invFactorial(min(k, n - k)) % self.MOD
# --------------------------------------iye ha combinations ka zindegi---------------------------------
def powm(a, n, m):
if a == 1 or n == 0:
return 1
if n % 2 == 0:
s = powm(a, n // 2, m)
return s * s % m
else:
return a * powm(a, n - 1, m) % m
# --------------------------------------iye ha power ka zindegi---------------------------------
def sort_list(list1, list2):
zipped_pairs = zip(list2, list1)
z = [x for _, x in sorted(zipped_pairs)]
return z
# --------------------------------------------------product----------------------------------------
def product(l):
por = 1
for i in range(len(l)):
por *= l[i]
return por
# --------------------------------------------------binary----------------------------------------
def binarySearchCount(arr, n, key):
left = 0
right = n - 1
count = 0
while (left <= right):
mid = int((right + left) / 2)
# Check if middle element is
# less than or equal to key
if (arr[mid] < key):
count = mid + 1
left = mid + 1
# If key is smaller, ignore right half
else:
right = mid - 1
return count
# --------------------------------------------------binary----------------------------------------
def countdig(n):
c = 0
while (n > 0):
n //= 10
c += 1
return c
def binary(x, length):
y = bin(x)[2:]
return y if len(y) >= length else "0" * (length - len(y)) + y
def countGreater(arr, n, k):
l = 0
r = n - 1
# Stores the index of the left most element
# from the array which is greater than k
leftGreater = n
# Finds number of elements greater than k
while (l <= r):
m = int(l + (r - l) / 2)
if (arr[m] >= k):
leftGreater = m
r = m - 1
# If mid element is less than
# or equal to k update l
else:
l = m + 1
# Return the count of elements
# greater than k
return (n - leftGreater)
# --------------------------------------------------binary-----------------------------------
n,t=map(int,input().split())
l=list(map(int,input().split()))
t1=t-1
t=min(n,t)-1
dp=defaultdict(int)
freq=defaultdict(int)
ans=0
for i in range(n):
freq[l[i]]+=1
ans=max(ans,freq[l[i]])
if t>0:
for i in range(n):
dp[l[i]]=1
for j in range(i):
if l[j]<=l[i]:
dp[l[i]]=max(dp[l[i]],dp[l[j]]+1)
ans=max(dp.values())
for i in range(1,t):
for j in range(n):
dp[l[j]]+=1
for k in range(n):
if k==j:
continue
if l[k]<=l[j]:
dp[l[j]]=max(dp[l[j]],dp[l[k]]+1)
ans=max(ans,max(dp.values()))
for j in range(n):
dp[l[j]]+=(t1-t)*freq[l[j]]
for j in range(n):
dp[l[j]]+=1
for k in range(n):
if k==j:
continue
if l[k]<=l[j]:
dp[l[j]]=max(dp[l[j]],dp[l[k]]+1)
ans=max(ans,max(dp.values()))
print(ans)
``` | instruction | 0 | 104,873 | 12 | 209,746 |
No | output | 1 | 104,873 | 12 | 209,747 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Define the simple skewness of a collection of numbers to be the collection's mean minus its median. You are given a list of n (not necessarily distinct) integers. Find the non-empty subset (with repetition) with the maximum simple skewness.
The mean of a collection is the average of its elements. The median of a collection is its middle element when all of its elements are sorted, or the average of its two middle elements if it has even size.
Input
The first line of the input contains a single integer n (1 ≤ n ≤ 200 000) — the number of elements in the list.
The second line contains n integers xi (0 ≤ xi ≤ 1 000 000) — the ith element of the list.
Output
In the first line, print a single integer k — the size of the subset.
In the second line, print k integers — the elements of the subset in any order.
If there are multiple optimal subsets, print any.
Examples
Input
4
1 2 3 12
Output
3
1 2 12
Input
4
1 1 2 2
Output
3
1 1 2
Input
2
1 2
Output
2
1 2
Note
In the first case, the optimal subset is <image>, which has mean 5, median 2, and simple skewness of 5 - 2 = 3.
In the second case, the optimal subset is <image>. Note that repetition is allowed.
In the last case, any subset has the same median and mean, so all have simple skewness of 0. | instruction | 0 | 104,890 | 12 | 209,780 |
Tags: binary search, math, ternary search
Correct Solution:
```
from itertools import accumulate
from fractions import Fraction
n = int(input())
A = [int(x) for x in input().split()]
A.sort()
B = list(accumulate([0] + A))
def condition(i, z):
return (2*z - 1)*(A[i-z] + A[-z]) > 2*(B[i+1] - B[i-z+1] + B[-1] - B[-z])
def average(i, z):
return Fraction((B[i+1] - B[i-z] + B[-1] - B[-z-1]) , 2*z + 1)
maxans = 0
argmax = (0, 0)
for i in range(1, n-1):
x, y = 0, min(i, n-1-i)
while y - x > 1:
z = (x + y)//2
if condition(i, z):
x = z
else:
y = z
if condition(i, y): x = y
if maxans < average(i, x) - A[i]:
maxans = average(i, x) - A[i]
argmax = (i, x)
print(argmax[1]*2 + 1)
for i in range(argmax[0] - argmax[1], argmax[0] + 1):
print(A[i], end = ' ')
for i in range(-argmax[1], 0):
print(A[i], end = ' ')
``` | output | 1 | 104,890 | 12 | 209,781 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Define the simple skewness of a collection of numbers to be the collection's mean minus its median. You are given a list of n (not necessarily distinct) integers. Find the non-empty subset (with repetition) with the maximum simple skewness.
The mean of a collection is the average of its elements. The median of a collection is its middle element when all of its elements are sorted, or the average of its two middle elements if it has even size.
Input
The first line of the input contains a single integer n (1 ≤ n ≤ 200 000) — the number of elements in the list.
The second line contains n integers xi (0 ≤ xi ≤ 1 000 000) — the ith element of the list.
Output
In the first line, print a single integer k — the size of the subset.
In the second line, print k integers — the elements of the subset in any order.
If there are multiple optimal subsets, print any.
Examples
Input
4
1 2 3 12
Output
3
1 2 12
Input
4
1 1 2 2
Output
3
1 1 2
Input
2
1 2
Output
2
1 2
Note
In the first case, the optimal subset is <image>, which has mean 5, median 2, and simple skewness of 5 - 2 = 3.
In the second case, the optimal subset is <image>. Note that repetition is allowed.
In the last case, any subset has the same median and mean, so all have simple skewness of 0.
Submitted Solution:
```
from itertools import accumulate
n = int(input())
A = [int(x) for x in input().split()]
A.sort()
B = list(accumulate([0] + A))
def average(i, k): #average on [i-k; i] U [n-1-k; n-1]
return (B[i+1] - B[i-k] + B[-1] - B[-k-1]) / (2*k + 1)
maxans = -1
for i in range(n):
x, y = 0, min(i, n-1-i)
z = (x + y)//2
while y - x > 1:
x1 = (x + z)//2
y1 = (z + y)//2
a, b, c = average(i, x1), average(i, z), average(i, y1)
if a > b:
z, y = x1, z
elif c > b:
x, z = z, y1
else:
x, y = x1, y1
if average(i, x) < average(i, y): x, y = y, x
if maxans < average(i, x) - A[i]:
maxans = average(i, x) - A[i]
argmax = (i, x)
print(argmax[1]*2 + 1)
for i in range(argmax[0] - argmax[1], argmax[1] + 1):
print(A[i], end = ' ')
for i in range(- argmax[1], 0):
print(A[i], end = ' ')
``` | instruction | 0 | 104,891 | 12 | 209,782 |
No | output | 1 | 104,891 | 12 | 209,783 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Define the simple skewness of a collection of numbers to be the collection's mean minus its median. You are given a list of n (not necessarily distinct) integers. Find the non-empty subset (with repetition) with the maximum simple skewness.
The mean of a collection is the average of its elements. The median of a collection is its middle element when all of its elements are sorted, or the average of its two middle elements if it has even size.
Input
The first line of the input contains a single integer n (1 ≤ n ≤ 200 000) — the number of elements in the list.
The second line contains n integers xi (0 ≤ xi ≤ 1 000 000) — the ith element of the list.
Output
In the first line, print a single integer k — the size of the subset.
In the second line, print k integers — the elements of the subset in any order.
If there are multiple optimal subsets, print any.
Examples
Input
4
1 2 3 12
Output
3
1 2 12
Input
4
1 1 2 2
Output
3
1 1 2
Input
2
1 2
Output
2
1 2
Note
In the first case, the optimal subset is <image>, which has mean 5, median 2, and simple skewness of 5 - 2 = 3.
In the second case, the optimal subset is <image>. Note that repetition is allowed.
In the last case, any subset has the same median and mean, so all have simple skewness of 0.
Submitted Solution:
```
n = int(input())
l = list(map(int, input().split()))
arr = sorted(l)
j = 0
down = [arr[0]]*n
up = [arr[-1]]*n
ans = 0
for i in range(1, n):
down[i] = down[i-1] + arr[i]
up[i] = up[i-1] + arr[n-i-1]
for i in range(1, n // 2):
kans = (down[i] + up[i-1]) / (2*i+1) - arr[i]
if ans < kans:
ans = kans
j = i
print(2*j+1)
s = ''
for i in range(j+1):
s+=str(arr[i])+' '
for i in range(j):
s+=str(arr[n-1-i])+' '
print(s)
``` | instruction | 0 | 104,892 | 12 | 209,784 |
No | output | 1 | 104,892 | 12 | 209,785 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Define the simple skewness of a collection of numbers to be the collection's mean minus its median. You are given a list of n (not necessarily distinct) integers. Find the non-empty subset (with repetition) with the maximum simple skewness.
The mean of a collection is the average of its elements. The median of a collection is its middle element when all of its elements are sorted, or the average of its two middle elements if it has even size.
Input
The first line of the input contains a single integer n (1 ≤ n ≤ 200 000) — the number of elements in the list.
The second line contains n integers xi (0 ≤ xi ≤ 1 000 000) — the ith element of the list.
Output
In the first line, print a single integer k — the size of the subset.
In the second line, print k integers — the elements of the subset in any order.
If there are multiple optimal subsets, print any.
Examples
Input
4
1 2 3 12
Output
3
1 2 12
Input
4
1 1 2 2
Output
3
1 1 2
Input
2
1 2
Output
2
1 2
Note
In the first case, the optimal subset is <image>, which has mean 5, median 2, and simple skewness of 5 - 2 = 3.
In the second case, the optimal subset is <image>. Note that repetition is allowed.
In the last case, any subset has the same median and mean, so all have simple skewness of 0.
Submitted Solution:
```
n = int(input())
a = [int(i) for i in input().split()]
a.sort()
if n <= 2:
print(n)
print(*a)
else:
c = 0
ans = []
for i in range(1, n - 1):
if ((a[i - 1] + a[i] + a[-1]) / 3) - a[i] > c:
c = ((a[i - 1] + a[i] + a[-1]) / 3) - a[i]
ans = [a[i - 1], a[i], a[-1]]
print(c)
print(*ans)
``` | instruction | 0 | 104,893 | 12 | 209,786 |
No | output | 1 | 104,893 | 12 | 209,787 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Define the simple skewness of a collection of numbers to be the collection's mean minus its median. You are given a list of n (not necessarily distinct) integers. Find the non-empty subset (with repetition) with the maximum simple skewness.
The mean of a collection is the average of its elements. The median of a collection is its middle element when all of its elements are sorted, or the average of its two middle elements if it has even size.
Input
The first line of the input contains a single integer n (1 ≤ n ≤ 200 000) — the number of elements in the list.
The second line contains n integers xi (0 ≤ xi ≤ 1 000 000) — the ith element of the list.
Output
In the first line, print a single integer k — the size of the subset.
In the second line, print k integers — the elements of the subset in any order.
If there are multiple optimal subsets, print any.
Examples
Input
4
1 2 3 12
Output
3
1 2 12
Input
4
1 1 2 2
Output
3
1 1 2
Input
2
1 2
Output
2
1 2
Note
In the first case, the optimal subset is <image>, which has mean 5, median 2, and simple skewness of 5 - 2 = 3.
In the second case, the optimal subset is <image>. Note that repetition is allowed.
In the last case, any subset has the same median and mean, so all have simple skewness of 0.
Submitted Solution:
```
n = int(input())
a = list(map(int, input().split()))
if n == 1:
print(1)
print(a[0])
elif n == 2:
print(2)
print(a[0], a[1])
else:
ans = -10**9
ansa = []
a.sort()
for i in range(1, n - 1):
if ans < (a[i] + a[i - 1] + a[-1])/3 - a[i]:
ans = (a[i] + a[i - 1] + a[-1])/3 - a[i]
ansa = [a[i-1], a[i], a[-1]]
print(3)
print(' '.join(list(map(str, a))))
``` | instruction | 0 | 104,894 | 12 | 209,788 |
No | output | 1 | 104,894 | 12 | 209,789 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Little Petya is now fond of data compression algorithms. He has already studied gz, bz, zip algorithms and many others. Inspired by the new knowledge, Petya is now developing the new compression algorithm which he wants to name dis.
Petya decided to compress tables. He is given a table a consisting of n rows and m columns that is filled with positive integers. He wants to build the table a' consisting of positive integers such that the relative order of the elements in each row and each column remains the same. That is, if in some row i of the initial table ai, j < ai, k, then in the resulting table a'i, j < a'i, k, and if ai, j = ai, k then a'i, j = a'i, k. Similarly, if in some column j of the initial table ai, j < ap, j then in compressed table a'i, j < a'p, j and if ai, j = ap, j then a'i, j = a'p, j.
Because large values require more space to store them, the maximum value in a' should be as small as possible.
Petya is good in theory, however, he needs your help to implement the algorithm.
Input
The first line of the input contains two integers n and m (<image>, the number of rows and the number of columns of the table respectively.
Each of the following n rows contain m integers ai, j (1 ≤ ai, j ≤ 109) that are the values in the table.
Output
Output the compressed table in form of n lines each containing m integers.
If there exist several answers such that the maximum number in the compressed table is minimum possible, you are allowed to output any of them.
Examples
Input
2 2
1 2
3 4
Output
1 2
2 3
Input
4 3
20 10 30
50 40 30
50 60 70
90 80 70
Output
2 1 3
5 4 3
5 6 7
9 8 7
Note
In the first sample test, despite the fact a1, 2 ≠ a21, they are not located in the same row or column so they may become equal after the compression.
Submitted Solution:
```
M, N = map(int, input().split())
matrix = []
for _ in range(M):
matrix.append(list(map(int, input().split())))
for m in matrix:
print(m)
# rank = {i: 0 for i in range(-N, M)}
# def find(x):
# if uf[x] != x:
# uf[x] = find(uf[x])
# return uf[x]
# dic = collections.defaultdict(list)
# ans = [[0] * N for _ in range(M)]
# for i in range(M):
# for j in range(N):
# dic[matrix[i][j]].append((i, j))
# for d in sorted(dic):
# uf = {i: i for i in range(-N, M)}
# # rank2 = rank.copy()
# for i, j in dic[d]:
# pi, pj = find(i), find(~j)
# uf[pi] = pj
# rank[pj] = max(rank[pi], rank[pj])
# par = {j: rank[find(~j)] for _, j in dic[d]}
# for i, j in dic[d]:
# rank[i] = rank[~j] = ans[i][j] = par[j] + 1
# print(ans)
``` | instruction | 0 | 104,895 | 12 | 209,790 |
No | output | 1 | 104,895 | 12 | 209,791 |
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