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.
A student of z-school found a kind of sorting called z-sort. The array a with n elements are z-sorted if two conditions hold:
1. ai ≥ ai - 1 for all even i,
2. ai ≤ ai - 1 for all odd i > 1.
For example the arrays [1,2,1,2] and [1,1,1,1] are z-sorted while the array [1,2,3,4] isn’t z-sorted.
Can you make the array z-sorted?
Input
The first line contains a single integer n (1 ≤ n ≤ 1000) — the number of elements in the array a.
The second line contains n integers ai (1 ≤ ai ≤ 109) — the elements of the array a.
Output
If it's possible to make the array a z-sorted print n space separated integers ai — the elements after z-sort. Otherwise print the only word "Impossible".
Examples
Input
4
1 2 2 1
Output
1 2 1 2
Input
5
1 3 2 2 5
Output
1 5 2 3 2 | instruction | 0 | 37,292 | 12 | 74,584 |
Tags: sortings
Correct Solution:
```
n = int(input())
inp = list(map(int , input().split()))
srt = sorted(inp)
ans = []
if n % 2 == 0:
for i in range(n // 2):
ans.append(str(srt[i]))
ans.append(str(srt[-(i + 1)]))
print(" ".join(ans))
else:
for i in range(n // 2 + 1):
ans.append(str(srt[i]))
ans.append(str(srt[-(i + 1)]))
ans.pop(-1)
print(" ".join(ans))
``` | output | 1 | 37,292 | 12 | 74,585 |
Provide tags and a correct Python 3 solution for this coding contest problem.
A student of z-school found a kind of sorting called z-sort. The array a with n elements are z-sorted if two conditions hold:
1. ai ≥ ai - 1 for all even i,
2. ai ≤ ai - 1 for all odd i > 1.
For example the arrays [1,2,1,2] and [1,1,1,1] are z-sorted while the array [1,2,3,4] isn’t z-sorted.
Can you make the array z-sorted?
Input
The first line contains a single integer n (1 ≤ n ≤ 1000) — the number of elements in the array a.
The second line contains n integers ai (1 ≤ ai ≤ 109) — the elements of the array a.
Output
If it's possible to make the array a z-sorted print n space separated integers ai — the elements after z-sort. Otherwise print the only word "Impossible".
Examples
Input
4
1 2 2 1
Output
1 2 1 2
Input
5
1 3 2 2 5
Output
1 5 2 3 2 | instruction | 0 | 37,293 | 12 | 74,586 |
Tags: sortings
Correct Solution:
```
n=int(input())
mas=[int(x) for x in input().split()]
mas.sort()
k=0
l=n-1
for i in range(n):
if i&1:
print(mas[l],end=' ')
l-=1
else:
print(mas[k],end=' ')
k+=1
``` | output | 1 | 37,293 | 12 | 74,587 |
Provide tags and a correct Python 3 solution for this coding contest problem.
A student of z-school found a kind of sorting called z-sort. The array a with n elements are z-sorted if two conditions hold:
1. ai ≥ ai - 1 for all even i,
2. ai ≤ ai - 1 for all odd i > 1.
For example the arrays [1,2,1,2] and [1,1,1,1] are z-sorted while the array [1,2,3,4] isn’t z-sorted.
Can you make the array z-sorted?
Input
The first line contains a single integer n (1 ≤ n ≤ 1000) — the number of elements in the array a.
The second line contains n integers ai (1 ≤ ai ≤ 109) — the elements of the array a.
Output
If it's possible to make the array a z-sorted print n space separated integers ai — the elements after z-sort. Otherwise print the only word "Impossible".
Examples
Input
4
1 2 2 1
Output
1 2 1 2
Input
5
1 3 2 2 5
Output
1 5 2 3 2 | instruction | 0 | 37,294 | 12 | 74,588 |
Tags: sortings
Correct Solution:
```
n = int(input())
a = [int(_) for _ in input().split()]
a.sort()
na = []
for i in range(len(a)):
if i%2==0:
na.append(a.pop(0))
else:
na.append(a.pop(-1))
print(" ".join(map(str,na)))
``` | output | 1 | 37,294 | 12 | 74,589 |
Provide tags and a correct Python 3 solution for this coding contest problem.
A student of z-school found a kind of sorting called z-sort. The array a with n elements are z-sorted if two conditions hold:
1. ai ≥ ai - 1 for all even i,
2. ai ≤ ai - 1 for all odd i > 1.
For example the arrays [1,2,1,2] and [1,1,1,1] are z-sorted while the array [1,2,3,4] isn’t z-sorted.
Can you make the array z-sorted?
Input
The first line contains a single integer n (1 ≤ n ≤ 1000) — the number of elements in the array a.
The second line contains n integers ai (1 ≤ ai ≤ 109) — the elements of the array a.
Output
If it's possible to make the array a z-sorted print n space separated integers ai — the elements after z-sort. Otherwise print the only word "Impossible".
Examples
Input
4
1 2 2 1
Output
1 2 1 2
Input
5
1 3 2 2 5
Output
1 5 2 3 2 | instruction | 0 | 37,295 | 12 | 74,590 |
Tags: sortings
Correct Solution:
```
n=int(input())
a=list(map(int,input().split()))
a.sort()
l=[]
p=0
i=0
while len(l)!=n:
if p==0:
l.append(a[i])
else:
l.append(a[n-i-1])
i+=1
p=p^1
print(*l)
``` | output | 1 | 37,295 | 12 | 74,591 |
Provide tags and a correct Python 3 solution for this coding contest problem.
A student of z-school found a kind of sorting called z-sort. The array a with n elements are z-sorted if two conditions hold:
1. ai ≥ ai - 1 for all even i,
2. ai ≤ ai - 1 for all odd i > 1.
For example the arrays [1,2,1,2] and [1,1,1,1] are z-sorted while the array [1,2,3,4] isn’t z-sorted.
Can you make the array z-sorted?
Input
The first line contains a single integer n (1 ≤ n ≤ 1000) — the number of elements in the array a.
The second line contains n integers ai (1 ≤ ai ≤ 109) — the elements of the array a.
Output
If it's possible to make the array a z-sorted print n space separated integers ai — the elements after z-sort. Otherwise print the only word "Impossible".
Examples
Input
4
1 2 2 1
Output
1 2 1 2
Input
5
1 3 2 2 5
Output
1 5 2 3 2 | instruction | 0 | 37,296 | 12 | 74,592 |
Tags: sortings
Correct Solution:
```
def main():
n = int(input())
a = sorted(map(int, input().split()))
result = [0]*n
# print(result)
# basic = 0
# # a.sort()
# # print(a)
# count = 0
# if n&1==0:
# while basic<n:
# # print(basic, count)
# result[count] = a[basic]
# if count==n-2:
# count = 1
# basic+=1
# # print(basic, count)
# result[count] = a[basic]
# basic+=1
# count+=2
# else:
# while basic<n:
# # print(basic, count)
# result[count] = a[basic]
# if count==n-1:
# count = 1
# basic+=1
# # print(basic, count)
# result[count] = a[basic]
# basic+=1
# count+=2
result[::2] = a[:(n+1)//2]
result[1::2] = a[(n+1)//2:]
# print(a[basic])
print(*result)
main()
``` | output | 1 | 37,296 | 12 | 74,593 |
Provide tags and a correct Python 3 solution for this coding contest problem.
A student of z-school found a kind of sorting called z-sort. The array a with n elements are z-sorted if two conditions hold:
1. ai ≥ ai - 1 for all even i,
2. ai ≤ ai - 1 for all odd i > 1.
For example the arrays [1,2,1,2] and [1,1,1,1] are z-sorted while the array [1,2,3,4] isn’t z-sorted.
Can you make the array z-sorted?
Input
The first line contains a single integer n (1 ≤ n ≤ 1000) — the number of elements in the array a.
The second line contains n integers ai (1 ≤ ai ≤ 109) — the elements of the array a.
Output
If it's possible to make the array a z-sorted print n space separated integers ai — the elements after z-sort. Otherwise print the only word "Impossible".
Examples
Input
4
1 2 2 1
Output
1 2 1 2
Input
5
1 3 2 2 5
Output
1 5 2 3 2 | instruction | 0 | 37,297 | 12 | 74,594 |
Tags: sortings
Correct Solution:
```
import sys
n = int(input())
a = sorted(map(int, input().split()))
ans = [0]*n
j = 0
for i in range((n+1) >> 1):
ans[j] = a[i]
j += 2
j = 1
for i in range((n+1) >> 1, n):
ans[j] = a[i]
j += 2
print(*ans)
``` | output | 1 | 37,297 | 12 | 74,595 |
Provide tags and a correct Python 3 solution for this coding contest problem.
A student of z-school found a kind of sorting called z-sort. The array a with n elements are z-sorted if two conditions hold:
1. ai ≥ ai - 1 for all even i,
2. ai ≤ ai - 1 for all odd i > 1.
For example the arrays [1,2,1,2] and [1,1,1,1] are z-sorted while the array [1,2,3,4] isn’t z-sorted.
Can you make the array z-sorted?
Input
The first line contains a single integer n (1 ≤ n ≤ 1000) — the number of elements in the array a.
The second line contains n integers ai (1 ≤ ai ≤ 109) — the elements of the array a.
Output
If it's possible to make the array a z-sorted print n space separated integers ai — the elements after z-sort. Otherwise print the only word "Impossible".
Examples
Input
4
1 2 2 1
Output
1 2 1 2
Input
5
1 3 2 2 5
Output
1 5 2 3 2 | instruction | 0 | 37,298 | 12 | 74,596 |
Tags: sortings
Correct Solution:
```
n = int(input())
L = [int(s) for s in input().split()]
def zsort(l):
if len(l) == 1:
return l
if len(l) == 2:
if l[0] > l[1]:
return l[::-1]
else:
return l
l1 = zsort(l[:2])
l2 = zsort(l[2:])
result = l1 + l2
if result[1] < result[2]:
result[1], result[2] = result[2], result[1]
return result
result = [str(i) for i in zsort(L)]
print(' '.join(result))
``` | output | 1 | 37,298 | 12 | 74,597 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A student of z-school found a kind of sorting called z-sort. The array a with n elements are z-sorted if two conditions hold:
1. ai ≥ ai - 1 for all even i,
2. ai ≤ ai - 1 for all odd i > 1.
For example the arrays [1,2,1,2] and [1,1,1,1] are z-sorted while the array [1,2,3,4] isn’t z-sorted.
Can you make the array z-sorted?
Input
The first line contains a single integer n (1 ≤ n ≤ 1000) — the number of elements in the array a.
The second line contains n integers ai (1 ≤ ai ≤ 109) — the elements of the array a.
Output
If it's possible to make the array a z-sorted print n space separated integers ai — the elements after z-sort. Otherwise print the only word "Impossible".
Examples
Input
4
1 2 2 1
Output
1 2 1 2
Input
5
1 3 2 2 5
Output
1 5 2 3 2
Submitted Solution:
```
def main(n, a):
a1 = sorted(a)
a2 = list(reversed(sorted(a)))
r = []
for i in range(n//2):
r.append(a1[i])
r.append(a2[i])
if n % 2 == 1:
r.append(a1[n // 2])
return ' '.join(list(map(str,r)))
print(main(int(input()), list(map(int, input().split(' ')))))
``` | instruction | 0 | 37,299 | 12 | 74,598 |
Yes | output | 1 | 37,299 | 12 | 74,599 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A student of z-school found a kind of sorting called z-sort. The array a with n elements are z-sorted if two conditions hold:
1. ai ≥ ai - 1 for all even i,
2. ai ≤ ai - 1 for all odd i > 1.
For example the arrays [1,2,1,2] and [1,1,1,1] are z-sorted while the array [1,2,3,4] isn’t z-sorted.
Can you make the array z-sorted?
Input
The first line contains a single integer n (1 ≤ n ≤ 1000) — the number of elements in the array a.
The second line contains n integers ai (1 ≤ ai ≤ 109) — the elements of the array a.
Output
If it's possible to make the array a z-sorted print n space separated integers ai — the elements after z-sort. Otherwise print the only word "Impossible".
Examples
Input
4
1 2 2 1
Output
1 2 1 2
Input
5
1 3 2 2 5
Output
1 5 2 3 2
Submitted Solution:
```
#### IMPORTANT LIBRARY ####
############################
### DO NOT USE import random --> 250ms to load the library
############################
### In case of extra libraries: https://github.com/cheran-senthil/PyRival
######################
####### IMPORT #######
######################
from functools import cmp_to_key
from collections import deque, Counter
from heapq import heappush, heappop
from math import log, ceil
######################
#### STANDARD I/O ####
######################
import sys
import os
from io import BytesIO, IOBase
BUFSIZE = 8192
class FastIO(IOBase):
newlines = 0
def __init__(self, file):
self._fd = file.fileno()
self.buffer = BytesIO()
self.writable = "x" in file.mode or "r" not in file.mode
self.write = self.buffer.write if self.writable else None
def read(self):
while True:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
if not b:
break
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines = 0
return self.buffer.read()
def readline(self):
while self.newlines == 0:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
self.newlines = b.count(b"\n") + (not b)
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines -= 1
return self.buffer.readline()
def flush(self):
if self.writable:
os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
self.write = lambda s: self.buffer.write(s.encode("ascii"))
self.read = lambda: self.buffer.read().decode("ascii")
self.readline = lambda: self.buffer.readline().decode("ascii")
if sys.version_info[0] < 3:
sys.stdin, sys.stdout = FastIO(sys.stdin), FastIO(sys.stdout)
else:
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
def print(*args, **kwargs):
sep, file = kwargs.pop("sep", " "), kwargs.pop("file", sys.stdout)
at_start = True
for x in args:
if not at_start:
file.write(sep)
file.write(str(x))
at_start = False
file.write(kwargs.pop("end", "\n"))
if kwargs.pop("flush", False):
file.flush()
def inp():
return sys.stdin.readline().rstrip("\r\n") # for fast input
def ii():
return int(inp())
def si():
return str(inp())
def li(lag = 0):
l = list(map(int, inp().split()))
if lag != 0:
for i in range(len(l)):
l[i] += lag
return l
def mi(lag = 0):
matrix = list()
for i in range(n):
matrix.append(li(lag))
return matrix
def lsi(): #string list
return list(map(str, inp().split()))
def print_list(lista, space = " "):
print(space.join(map(str, lista)))
######################
### BISECT METHODS ###
######################
def bisect_left(a, x):
"""i tale che a[i] >= x e a[i-1] < x"""
left = 0
right = len(a)
while left < right:
mid = (left+right)//2
if a[mid] < x:
left = mid+1
else:
right = mid
return left
def bisect_right(a, x):
"""i tale che a[i] > x e a[i-1] <= x"""
left = 0
right = len(a)
while left < right:
mid = (left+right)//2
if a[mid] > x:
right = mid
else:
left = mid+1
return left
def bisect_elements(a, x):
"""elementi pari a x nell'árray sortato"""
return bisect_right(a, x) - bisect_left(a, x)
######################
### MOD OPERATION ####
######################
MOD = 10**9 + 7
maxN = 5
FACT = [0] * maxN
INV_FACT = [0] * maxN
def add(x, y):
return (x+y) % MOD
def multiply(x, y):
return (x*y) % MOD
def power(x, y):
if y == 0:
return 1
elif y % 2:
return multiply(x, power(x, y-1))
else:
a = power(x, y//2)
return multiply(a, a)
def inverse(x):
return power(x, MOD-2)
def divide(x, y):
return multiply(x, inverse(y))
def allFactorials():
FACT[0] = 1
for i in range(1, maxN):
FACT[i] = multiply(i, FACT[i-1])
def inverseFactorials():
n = len(INV_FACT)
INV_FACT[n-1] = inverse(FACT[n-1])
for i in range(n-2, -1, -1):
INV_FACT[i] = multiply(INV_FACT[i+1], i+1)
def coeffBinom(n, k):
if n < k:
return 0
return multiply(FACT[n], multiply(INV_FACT[k], INV_FACT[n-k]))
######################
#### GRAPH ALGOS #####
######################
# ZERO BASED GRAPH
def create_graph(n, m, undirected = 1, unweighted = 1):
graph = [[] for i in range(n)]
if unweighted:
for i in range(m):
[x, y] = li(lag = -1)
graph[x].append(y)
if undirected:
graph[y].append(x)
else:
for i in range(m):
[x, y, w] = li(lag = -1)
w += 1
graph[x].append([y,w])
if undirected:
graph[y].append([x,w])
return graph
def create_tree(n, unweighted = 1):
children = [[] for i in range(n)]
if unweighted:
for i in range(n-1):
[x, y] = li(lag = -1)
children[x].append(y)
children[y].append(x)
else:
for i in range(n-1):
[x, y, w] = li(lag = -1)
w += 1
children[x].append([y, w])
children[y].append([x, w])
return children
def dist(tree, n, A, B = -1):
s = [[A, 0]]
massimo, massimo_nodo = 0, 0
distanza = -1
v = [-1] * n
while s:
el, dis = s.pop()
if dis > massimo:
massimo = dis
massimo_nodo = el
if el == B:
distanza = dis
for child in tree[el]:
if v[child] == -1:
v[child] = 1
s.append([child, dis+1])
return massimo, massimo_nodo, distanza
def diameter(tree):
_, foglia, _ = dist(tree, n, 0)
diam, _, _ = dist(tree, n, foglia)
return diam
def dfs(graph, n, A):
v = [-1] * n
s = [[A, 0]]
v[A] = 0
while s:
el, dis = s.pop()
for child in graph[el]:
if v[child] == -1:
v[child] = dis + 1
s.append([child, dis + 1])
return v #visited: -1 if not visited, otherwise v[B] is the distance in terms of edges
def bfs(graph, n, A):
v = [-1] * n
s = deque()
s.append([A, 0])
v[A] = 0
while s:
el, dis = s.popleft()
for child in graph[el]:
if v[child] == -1:
v[child] = dis + 1
s.append([child, dis + 1])
return v #visited: -1 if not visited, otherwise v[B] is the distance in terms of edges
#FROM A GIVEN ROOT, RECOVER THE STRUCTURE
def parents_children_root_unrooted_tree(tree, n, root = 0):
q = deque()
visited = [0] * n
parent = [-1] * n
children = [[] for i in range(n)]
q.append(root)
while q:
all_done = 1
visited[q[0]] = 1
for child in tree[q[0]]:
if not visited[child]:
all_done = 0
q.appendleft(child)
if all_done:
for child in tree[q[0]]:
if parent[child] == -1:
parent[q[0]] = child
children[child].append(q[0])
q.popleft()
return parent, children
# CALCULATING LONGEST PATH FOR ALL THE NODES
def all_longest_path_passing_from_node(parent, children, n):
q = deque()
visited = [len(children[i]) for i in range(n)]
downwards = [[0,0] for i in range(n)]
upward = [1] * n
longest_path = [1] * n
for i in range(n):
if not visited[i]:
q.append(i)
downwards[i] = [1,0]
while q:
node = q.popleft()
if parent[node] != -1:
visited[parent[node]] -= 1
if not visited[parent[node]]:
q.append(parent[node])
else:
root = node
for child in children[node]:
downwards[node] = sorted([downwards[node][0], downwards[node][1], downwards[child][0] + 1], reverse = True)[0:2]
s = [node]
while s:
node = s.pop()
if parent[node] != -1:
if downwards[parent[node]][0] == downwards[node][0] + 1:
upward[node] = 1 + max(upward[parent[node]], downwards[parent[node]][1])
else:
upward[node] = 1 + max(upward[parent[node]], downwards[parent[node]][0])
longest_path[node] = downwards[node][0] + downwards[node][1] + upward[node] - min([downwards[node][0], downwards[node][1], upward[node]]) - 1
for child in children[node]:
s.append(child)
return longest_path
### TBD SUCCESSOR GRAPH 7.5
### TBD TREE QUERIES 10.2 da 2 a 4
### TBD ADVANCED TREE 10.3
### TBD GRAPHS AND MATRICES 11.3.3 e 11.4.3 e 11.5.3 (ON GAMES)
######################
## END OF LIBRARIES ##
######################
n = ii()
a = li()
res = [-1 for i in range(n)]
a.sort(reverse = True)
idx = 1
pos = 0
while idx < n:
res[idx] = a[pos]
pos += 1
idx += 2
idx = 0
while idx < n:
res[idx] = a[pos]
pos += 1
idx += 2
print_list(res)
``` | instruction | 0 | 37,300 | 12 | 74,600 |
Yes | output | 1 | 37,300 | 12 | 74,601 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A student of z-school found a kind of sorting called z-sort. The array a with n elements are z-sorted if two conditions hold:
1. ai ≥ ai - 1 for all even i,
2. ai ≤ ai - 1 for all odd i > 1.
For example the arrays [1,2,1,2] and [1,1,1,1] are z-sorted while the array [1,2,3,4] isn’t z-sorted.
Can you make the array z-sorted?
Input
The first line contains a single integer n (1 ≤ n ≤ 1000) — the number of elements in the array a.
The second line contains n integers ai (1 ≤ ai ≤ 109) — the elements of the array a.
Output
If it's possible to make the array a z-sorted print n space separated integers ai — the elements after z-sort. Otherwise print the only word "Impossible".
Examples
Input
4
1 2 2 1
Output
1 2 1 2
Input
5
1 3 2 2 5
Output
1 5 2 3 2
Submitted Solution:
```
def zsort(n, A):
A.sort()
return [A[-(i+1)//2] if i % 2 else A[i//2] for i in range(n)]
def main():
n = readint()
A = readintl()
print(' '.join(map(str, zsort(n, A))))
##########
import sys
def readint():
return int(input())
def readinti():
return map(int, input().split())
def readintl():
return list(readinti())
def readintll(k):
return [readintl() for _ in range(k)]
def log(*args, **kwargs):
print(*args, **kwargs, file=sys.stderr)
if __name__ == '__main__':
main()
``` | instruction | 0 | 37,301 | 12 | 74,602 |
Yes | output | 1 | 37,301 | 12 | 74,603 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A student of z-school found a kind of sorting called z-sort. The array a with n elements are z-sorted if two conditions hold:
1. ai ≥ ai - 1 for all even i,
2. ai ≤ ai - 1 for all odd i > 1.
For example the arrays [1,2,1,2] and [1,1,1,1] are z-sorted while the array [1,2,3,4] isn’t z-sorted.
Can you make the array z-sorted?
Input
The first line contains a single integer n (1 ≤ n ≤ 1000) — the number of elements in the array a.
The second line contains n integers ai (1 ≤ ai ≤ 109) — the elements of the array a.
Output
If it's possible to make the array a z-sorted print n space separated integers ai — the elements after z-sort. Otherwise print the only word "Impossible".
Examples
Input
4
1 2 2 1
Output
1 2 1 2
Input
5
1 3 2 2 5
Output
1 5 2 3 2
Submitted Solution:
```
n=int(input())
a=[int(x) for x in input().split()]
a.sort()
i=0
j=n-1
while i<j:
print(a[i],end=" ")
print(a[j],end=" ")
i+=1
j-=1
if n%2:
print(a[i])
``` | instruction | 0 | 37,302 | 12 | 74,604 |
Yes | output | 1 | 37,302 | 12 | 74,605 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A student of z-school found a kind of sorting called z-sort. The array a with n elements are z-sorted if two conditions hold:
1. ai ≥ ai - 1 for all even i,
2. ai ≤ ai - 1 for all odd i > 1.
For example the arrays [1,2,1,2] and [1,1,1,1] are z-sorted while the array [1,2,3,4] isn’t z-sorted.
Can you make the array z-sorted?
Input
The first line contains a single integer n (1 ≤ n ≤ 1000) — the number of elements in the array a.
The second line contains n integers ai (1 ≤ ai ≤ 109) — the elements of the array a.
Output
If it's possible to make the array a z-sorted print n space separated integers ai — the elements after z-sort. Otherwise print the only word "Impossible".
Examples
Input
4
1 2 2 1
Output
1 2 1 2
Input
5
1 3 2 2 5
Output
1 5 2 3 2
Submitted Solution:
```
n = int(input())
array = input().split(" ")
MaxArray = []
MinArray = []
for i in range(n // 2):
MaxArray.append(max(array))
array.remove(MaxArray[i])
MinArray.append(min(array))
array.remove(MinArray[i])
for i in range(n // 2):
array.append(int(MinArray[i]))
array.append(int(MaxArray[i]))
if n % 2 == 1:
array = array[1:n+1] + array[:1]
array[n - 1] = int(array[n - 1])
print(array)
``` | instruction | 0 | 37,303 | 12 | 74,606 |
No | output | 1 | 37,303 | 12 | 74,607 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A student of z-school found a kind of sorting called z-sort. The array a with n elements are z-sorted if two conditions hold:
1. ai ≥ ai - 1 for all even i,
2. ai ≤ ai - 1 for all odd i > 1.
For example the arrays [1,2,1,2] and [1,1,1,1] are z-sorted while the array [1,2,3,4] isn’t z-sorted.
Can you make the array z-sorted?
Input
The first line contains a single integer n (1 ≤ n ≤ 1000) — the number of elements in the array a.
The second line contains n integers ai (1 ≤ ai ≤ 109) — the elements of the array a.
Output
If it's possible to make the array a z-sorted print n space separated integers ai — the elements after z-sort. Otherwise print the only word "Impossible".
Examples
Input
4
1 2 2 1
Output
1 2 1 2
Input
5
1 3 2 2 5
Output
1 5 2 3 2
Submitted Solution:
```
nb = int(input())
numbers = sorted([int(x) for x in input().split()])
for i in range(2, nb-2):
numbers[i-1], numbers[i] = numbers[i], numbers[i-1]
print(*numbers)
``` | instruction | 0 | 37,304 | 12 | 74,608 |
No | output | 1 | 37,304 | 12 | 74,609 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A student of z-school found a kind of sorting called z-sort. The array a with n elements are z-sorted if two conditions hold:
1. ai ≥ ai - 1 for all even i,
2. ai ≤ ai - 1 for all odd i > 1.
For example the arrays [1,2,1,2] and [1,1,1,1] are z-sorted while the array [1,2,3,4] isn’t z-sorted.
Can you make the array z-sorted?
Input
The first line contains a single integer n (1 ≤ n ≤ 1000) — the number of elements in the array a.
The second line contains n integers ai (1 ≤ ai ≤ 109) — the elements of the array a.
Output
If it's possible to make the array a z-sorted print n space separated integers ai — the elements after z-sort. Otherwise print the only word "Impossible".
Examples
Input
4
1 2 2 1
Output
1 2 1 2
Input
5
1 3 2 2 5
Output
1 5 2 3 2
Submitted Solution:
```
n=int(input())
arr=[int(x) for x in input().split()]
for i in range(1,len(arr)):
if arr[i]%2==1:
arr[i],arr[i-1]=arr[i-1],arr[i]
print(*arr)
``` | instruction | 0 | 37,305 | 12 | 74,610 |
No | output | 1 | 37,305 | 12 | 74,611 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A student of z-school found a kind of sorting called z-sort. The array a with n elements are z-sorted if two conditions hold:
1. ai ≥ ai - 1 for all even i,
2. ai ≤ ai - 1 for all odd i > 1.
For example the arrays [1,2,1,2] and [1,1,1,1] are z-sorted while the array [1,2,3,4] isn’t z-sorted.
Can you make the array z-sorted?
Input
The first line contains a single integer n (1 ≤ n ≤ 1000) — the number of elements in the array a.
The second line contains n integers ai (1 ≤ ai ≤ 109) — the elements of the array a.
Output
If it's possible to make the array a z-sorted print n space separated integers ai — the elements after z-sort. Otherwise print the only word "Impossible".
Examples
Input
4
1 2 2 1
Output
1 2 1 2
Input
5
1 3 2 2 5
Output
1 5 2 3 2
Submitted Solution:
```
n = int(input())
a = list(map(int, input().split()))
a.sort()
b = [0 for x in range(n)]
for i in range(int((n + 1) / 2)):
b[i * 2] = a[i]
for i in range(int(n / 2)):
b[i * 2 + 1] = a[n - i - 1]
print(b)
``` | instruction | 0 | 37,306 | 12 | 74,612 |
No | output | 1 | 37,306 | 12 | 74,613 |
Provide tags and a correct Python 3 solution for this coding contest problem.
There is an array with n elements a1, a2, ..., an and the number x.
In one operation you can select some i (1 ≤ i ≤ n) and replace element ai with ai & x, where & denotes the [bitwise and](https://en.wikipedia.org/wiki/Bitwise_operation#AND) operation.
You want the array to have at least two equal elements after applying some operations (possibly, none). In other words, there should be at least two distinct indices i ≠ j such that ai = aj. Determine whether it is possible to achieve and, if possible, the minimal number of operations to apply.
Input
The first line contains integers n and x (2 ≤ n ≤ 100 000, 1 ≤ x ≤ 100 000), number of elements in the array and the number to and with.
The second line contains n integers ai (1 ≤ ai ≤ 100 000), the elements of the array.
Output
Print a single integer denoting the minimal number of operations to do, or -1, if it is impossible.
Examples
Input
4 3
1 2 3 7
Output
1
Input
2 228
1 1
Output
0
Input
3 7
1 2 3
Output
-1
Note
In the first example one can apply the operation to the last element of the array. That replaces 7 with 3, so we achieve the goal in one move.
In the second example the array already has two equal elements.
In the third example applying the operation won't change the array at all, so it is impossible to make some pair of elements equal. | instruction | 0 | 37,734 | 12 | 75,468 |
Tags: greedy
Correct Solution:
```
n,x = map(int, input().split())
L = list(map(int, input().split()))
m = max(L)
A = [0]*(m+1)
f = 0
for i in range(n):
A[L[i]] += 1
if A[L[i]] == 2:
f = 1
break
if f == 1:
print(0)
else:
k = 0
L1 = []
for i in range(len(A)):
L1.append(A[i])
for i in range(n):
p = L[i] & x
if p != L[i]:
A[p] += 1
if A[p] == 2:
if L1[p] == 0:
k = 2
elif L1[p] == 1:
k = 1
break
if k == 1:
print(1)
elif k == 2:
print(2)
else:
print(-1)
``` | output | 1 | 37,734 | 12 | 75,469 |
Provide tags and a correct Python 3 solution for this coding contest problem.
There is an array with n elements a1, a2, ..., an and the number x.
In one operation you can select some i (1 ≤ i ≤ n) and replace element ai with ai & x, where & denotes the [bitwise and](https://en.wikipedia.org/wiki/Bitwise_operation#AND) operation.
You want the array to have at least two equal elements after applying some operations (possibly, none). In other words, there should be at least two distinct indices i ≠ j such that ai = aj. Determine whether it is possible to achieve and, if possible, the minimal number of operations to apply.
Input
The first line contains integers n and x (2 ≤ n ≤ 100 000, 1 ≤ x ≤ 100 000), number of elements in the array and the number to and with.
The second line contains n integers ai (1 ≤ ai ≤ 100 000), the elements of the array.
Output
Print a single integer denoting the minimal number of operations to do, or -1, if it is impossible.
Examples
Input
4 3
1 2 3 7
Output
1
Input
2 228
1 1
Output
0
Input
3 7
1 2 3
Output
-1
Note
In the first example one can apply the operation to the last element of the array. That replaces 7 with 3, so we achieve the goal in one move.
In the second example the array already has two equal elements.
In the third example applying the operation won't change the array at all, so it is impossible to make some pair of elements equal. | instruction | 0 | 37,735 | 12 | 75,470 |
Tags: greedy
Correct Solution:
```
n,x=map(int,input().split())
arr=list(map(int,input().split()))
cnt=[0]*100001
ans=3
for i in range(n):
cnt[arr[i]]+=1
#print(cnt[1])
if cnt[arr[i]]>1:
print(0)
exit()
ans=9999999999
stp=[0]*100001
for i in range(n):
ct=0
while(1):
y=arr[i]&x
#print(i,y)
if y!=arr[i]:
ct+=1
#print("ss")
if cnt[y]==1 or stp[y]>0:
ans=min(ans,ct+stp[y])
break
else:
# print(":S",cnt[3])
stp[y]+=1
arr[i]=y
else:
break
# print(cnt[:16])
# print(stp[:16])
if ans==9999999999:
print(-1)
else:
print(ans)
``` | output | 1 | 37,735 | 12 | 75,471 |
Provide tags and a correct Python 3 solution for this coding contest problem.
There is an array with n elements a1, a2, ..., an and the number x.
In one operation you can select some i (1 ≤ i ≤ n) and replace element ai with ai & x, where & denotes the [bitwise and](https://en.wikipedia.org/wiki/Bitwise_operation#AND) operation.
You want the array to have at least two equal elements after applying some operations (possibly, none). In other words, there should be at least two distinct indices i ≠ j such that ai = aj. Determine whether it is possible to achieve and, if possible, the minimal number of operations to apply.
Input
The first line contains integers n and x (2 ≤ n ≤ 100 000, 1 ≤ x ≤ 100 000), number of elements in the array and the number to and with.
The second line contains n integers ai (1 ≤ ai ≤ 100 000), the elements of the array.
Output
Print a single integer denoting the minimal number of operations to do, or -1, if it is impossible.
Examples
Input
4 3
1 2 3 7
Output
1
Input
2 228
1 1
Output
0
Input
3 7
1 2 3
Output
-1
Note
In the first example one can apply the operation to the last element of the array. That replaces 7 with 3, so we achieve the goal in one move.
In the second example the array already has two equal elements.
In the third example applying the operation won't change the array at all, so it is impossible to make some pair of elements equal. | instruction | 0 | 37,736 | 12 | 75,472 |
Tags: greedy
Correct Solution:
```
n, x = map(int, input().split())
a = list(map(int, input().split()))
b = [0]*100001
c = [0]*100001
for t in a:
b[t]+=1
if b[t] == 2:
print(0)
exit()
for t in a:
y = x&t
if b[y] == 1 and y!=t:
print(1)
exit()
for t in a:
y = x&t
if c[y] == 1:
print(2)
exit()
c[y]+=1
print(-1)
``` | output | 1 | 37,736 | 12 | 75,473 |
Provide tags and a correct Python 3 solution for this coding contest problem.
There is an array with n elements a1, a2, ..., an and the number x.
In one operation you can select some i (1 ≤ i ≤ n) and replace element ai with ai & x, where & denotes the [bitwise and](https://en.wikipedia.org/wiki/Bitwise_operation#AND) operation.
You want the array to have at least two equal elements after applying some operations (possibly, none). In other words, there should be at least two distinct indices i ≠ j such that ai = aj. Determine whether it is possible to achieve and, if possible, the minimal number of operations to apply.
Input
The first line contains integers n and x (2 ≤ n ≤ 100 000, 1 ≤ x ≤ 100 000), number of elements in the array and the number to and with.
The second line contains n integers ai (1 ≤ ai ≤ 100 000), the elements of the array.
Output
Print a single integer denoting the minimal number of operations to do, or -1, if it is impossible.
Examples
Input
4 3
1 2 3 7
Output
1
Input
2 228
1 1
Output
0
Input
3 7
1 2 3
Output
-1
Note
In the first example one can apply the operation to the last element of the array. That replaces 7 with 3, so we achieve the goal in one move.
In the second example the array already has two equal elements.
In the third example applying the operation won't change the array at all, so it is impossible to make some pair of elements equal. | instruction | 0 | 37,737 | 12 | 75,474 |
Tags: greedy
Correct Solution:
```
n, x = map(int, input().split())
l = set(map(int, input().split()))
if len(l) < n:
print(0)
exit()
for i in l:
if i & x != i and i & x in l:
print(1)
exit()
s = set(i & x for i in l)
if len(s) < n:
print(2)
exit()
print(-1)
``` | output | 1 | 37,737 | 12 | 75,475 |
Provide tags and a correct Python 3 solution for this coding contest problem.
There is an array with n elements a1, a2, ..., an and the number x.
In one operation you can select some i (1 ≤ i ≤ n) and replace element ai with ai & x, where & denotes the [bitwise and](https://en.wikipedia.org/wiki/Bitwise_operation#AND) operation.
You want the array to have at least two equal elements after applying some operations (possibly, none). In other words, there should be at least two distinct indices i ≠ j such that ai = aj. Determine whether it is possible to achieve and, if possible, the minimal number of operations to apply.
Input
The first line contains integers n and x (2 ≤ n ≤ 100 000, 1 ≤ x ≤ 100 000), number of elements in the array and the number to and with.
The second line contains n integers ai (1 ≤ ai ≤ 100 000), the elements of the array.
Output
Print a single integer denoting the minimal number of operations to do, or -1, if it is impossible.
Examples
Input
4 3
1 2 3 7
Output
1
Input
2 228
1 1
Output
0
Input
3 7
1 2 3
Output
-1
Note
In the first example one can apply the operation to the last element of the array. That replaces 7 with 3, so we achieve the goal in one move.
In the second example the array already has two equal elements.
In the third example applying the operation won't change the array at all, so it is impossible to make some pair of elements equal. | instruction | 0 | 37,738 | 12 | 75,476 |
Tags: greedy
Correct Solution:
```
n, x = list(map(int, input().split()))
a = list(map(int, input().split()))
cnt = int(1e5 + 1)
bull_1 = [0] * cnt
bull_2 = [0] * cnt
for i in range(n):
if bull_1[a[i]] == 1:
print(0)
exit()
bull_1[a[i]] = 1
for i in range(n):
if bull_1[a[i] & x] == 1 and a[i] != a[i] & x:
print(1)
exit()
for i in range(n):
if bull_2[a[i] & x] == 1:
print(2)
exit()
bull_2[a[i] & x] = 1
print(-1)
``` | output | 1 | 37,738 | 12 | 75,477 |
Provide tags and a correct Python 3 solution for this coding contest problem.
There is an array with n elements a1, a2, ..., an and the number x.
In one operation you can select some i (1 ≤ i ≤ n) and replace element ai with ai & x, where & denotes the [bitwise and](https://en.wikipedia.org/wiki/Bitwise_operation#AND) operation.
You want the array to have at least two equal elements after applying some operations (possibly, none). In other words, there should be at least two distinct indices i ≠ j such that ai = aj. Determine whether it is possible to achieve and, if possible, the minimal number of operations to apply.
Input
The first line contains integers n and x (2 ≤ n ≤ 100 000, 1 ≤ x ≤ 100 000), number of elements in the array and the number to and with.
The second line contains n integers ai (1 ≤ ai ≤ 100 000), the elements of the array.
Output
Print a single integer denoting the minimal number of operations to do, or -1, if it is impossible.
Examples
Input
4 3
1 2 3 7
Output
1
Input
2 228
1 1
Output
0
Input
3 7
1 2 3
Output
-1
Note
In the first example one can apply the operation to the last element of the array. That replaces 7 with 3, so we achieve the goal in one move.
In the second example the array already has two equal elements.
In the third example applying the operation won't change the array at all, so it is impossible to make some pair of elements equal. | instruction | 0 | 37,739 | 12 | 75,478 |
Tags: greedy
Correct Solution:
```
from sys import exit
n, x = list(map(int, input().split()))
a = list(map(int, input().split()))
decimals = set()
for i in a:
if i in decimals:
print(0)
exit(0)
decimals.add(i)
values = set(a)
new_values = set()
for value in list(values):
new = value & x
if new != value and new in values:
print(1)
exit(0)
new_values.add(new)
if len(new_values) == len(values):
print(-1)
else:
print(2)
``` | output | 1 | 37,739 | 12 | 75,479 |
Provide tags and a correct Python 3 solution for this coding contest problem.
There is an array with n elements a1, a2, ..., an and the number x.
In one operation you can select some i (1 ≤ i ≤ n) and replace element ai with ai & x, where & denotes the [bitwise and](https://en.wikipedia.org/wiki/Bitwise_operation#AND) operation.
You want the array to have at least two equal elements after applying some operations (possibly, none). In other words, there should be at least two distinct indices i ≠ j such that ai = aj. Determine whether it is possible to achieve and, if possible, the minimal number of operations to apply.
Input
The first line contains integers n and x (2 ≤ n ≤ 100 000, 1 ≤ x ≤ 100 000), number of elements in the array and the number to and with.
The second line contains n integers ai (1 ≤ ai ≤ 100 000), the elements of the array.
Output
Print a single integer denoting the minimal number of operations to do, or -1, if it is impossible.
Examples
Input
4 3
1 2 3 7
Output
1
Input
2 228
1 1
Output
0
Input
3 7
1 2 3
Output
-1
Note
In the first example one can apply the operation to the last element of the array. That replaces 7 with 3, so we achieve the goal in one move.
In the second example the array already has two equal elements.
In the third example applying the operation won't change the array at all, so it is impossible to make some pair of elements equal. | instruction | 0 | 37,740 | 12 | 75,480 |
Tags: greedy
Correct Solution:
```
def main():
n,x = map(int,input().split())
a = list(map(int,input().split()))
uniq = set(a)
m = len(uniq)
ans=-1
if m!=n:
ans=0
else:
bitwise = set()
for i in a:
t = i&x
if t in uniq and t != i:
ans=1
break
else:
bitwise.add(t)
if ans == -1 and len(bitwise) != n:
ans = 2
print(ans)
if __name__ == '__main__':
main()
``` | output | 1 | 37,740 | 12 | 75,481 |
Provide tags and a correct Python 3 solution for this coding contest problem.
There is an array with n elements a1, a2, ..., an and the number x.
In one operation you can select some i (1 ≤ i ≤ n) and replace element ai with ai & x, where & denotes the [bitwise and](https://en.wikipedia.org/wiki/Bitwise_operation#AND) operation.
You want the array to have at least two equal elements after applying some operations (possibly, none). In other words, there should be at least two distinct indices i ≠ j such that ai = aj. Determine whether it is possible to achieve and, if possible, the minimal number of operations to apply.
Input
The first line contains integers n and x (2 ≤ n ≤ 100 000, 1 ≤ x ≤ 100 000), number of elements in the array and the number to and with.
The second line contains n integers ai (1 ≤ ai ≤ 100 000), the elements of the array.
Output
Print a single integer denoting the minimal number of operations to do, or -1, if it is impossible.
Examples
Input
4 3
1 2 3 7
Output
1
Input
2 228
1 1
Output
0
Input
3 7
1 2 3
Output
-1
Note
In the first example one can apply the operation to the last element of the array. That replaces 7 with 3, so we achieve the goal in one move.
In the second example the array already has two equal elements.
In the third example applying the operation won't change the array at all, so it is impossible to make some pair of elements equal. | instruction | 0 | 37,741 | 12 | 75,482 |
Tags: greedy
Correct Solution:
```
n,x=map(int,input().split())
a=list(map(int,input().split()))
b=[0]*(100001);c=[0]*100001
for i in a:
b[i]+=1
if b[i]>=2:
exit(print(0))
for i in a:
b[i]-=1
if b[i&x]>=1:
exit(print(1))
b[i]+=1
for i in a:
c[i&x]+=1
if c[i&x]>=2:
exit(print(2))
print(-1)
``` | output | 1 | 37,741 | 12 | 75,483 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There is an array with n elements a1, a2, ..., an and the number x.
In one operation you can select some i (1 ≤ i ≤ n) and replace element ai with ai & x, where & denotes the [bitwise and](https://en.wikipedia.org/wiki/Bitwise_operation#AND) operation.
You want the array to have at least two equal elements after applying some operations (possibly, none). In other words, there should be at least two distinct indices i ≠ j such that ai = aj. Determine whether it is possible to achieve and, if possible, the minimal number of operations to apply.
Input
The first line contains integers n and x (2 ≤ n ≤ 100 000, 1 ≤ x ≤ 100 000), number of elements in the array and the number to and with.
The second line contains n integers ai (1 ≤ ai ≤ 100 000), the elements of the array.
Output
Print a single integer denoting the minimal number of operations to do, or -1, if it is impossible.
Examples
Input
4 3
1 2 3 7
Output
1
Input
2 228
1 1
Output
0
Input
3 7
1 2 3
Output
-1
Note
In the first example one can apply the operation to the last element of the array. That replaces 7 with 3, so we achieve the goal in one move.
In the second example the array already has two equal elements.
In the third example applying the operation won't change the array at all, so it is impossible to make some pair of elements equal.
Submitted Solution:
```
from collections import defaultdict
n, x = map(int, input().split())
a = list(map(int, input().split()))
if len(set(a)) < n:
print(0)
exit()
a_nd = [el & x for el in a]
ToF = [a[i] == a_nd[i] for i in range(n)]
c = ToF.count(True)
numb = len(set(a) | set(a_nd))
if c == n or numb == 2 * n:
print(-1)
exit()
numb2 = len(set(a) & set(a_nd))
l_and = len(set(a_nd))
if numb2 == 0 and l_and < n:
print(2)
exit()
C = defaultdict(int)
for el in a:
C[el] += 1
for i in range(n):
C[a[i]] -= 1
if C[a_nd[i]] > 0:
print(1)
exit()
C[a[i]] += 1
if l_and < n:
print(2)
exit()
print(-1)
``` | instruction | 0 | 37,742 | 12 | 75,484 |
Yes | output | 1 | 37,742 | 12 | 75,485 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There is an array with n elements a1, a2, ..., an and the number x.
In one operation you can select some i (1 ≤ i ≤ n) and replace element ai with ai & x, where & denotes the [bitwise and](https://en.wikipedia.org/wiki/Bitwise_operation#AND) operation.
You want the array to have at least two equal elements after applying some operations (possibly, none). In other words, there should be at least two distinct indices i ≠ j such that ai = aj. Determine whether it is possible to achieve and, if possible, the minimal number of operations to apply.
Input
The first line contains integers n and x (2 ≤ n ≤ 100 000, 1 ≤ x ≤ 100 000), number of elements in the array and the number to and with.
The second line contains n integers ai (1 ≤ ai ≤ 100 000), the elements of the array.
Output
Print a single integer denoting the minimal number of operations to do, or -1, if it is impossible.
Examples
Input
4 3
1 2 3 7
Output
1
Input
2 228
1 1
Output
0
Input
3 7
1 2 3
Output
-1
Note
In the first example one can apply the operation to the last element of the array. That replaces 7 with 3, so we achieve the goal in one move.
In the second example the array already has two equal elements.
In the third example applying the operation won't change the array at all, so it is impossible to make some pair of elements equal.
Submitted Solution:
```
import bisect as bis
n,k=map(int, input().split())
a=[int(i) for i in input().split()]
a=sorted(a)
stt=-1
i=0
while i<n-1:
if a[i]==a[i+1]:
stt=0
break
i+=1
if stt==0:
pass
else:
b=[e&k for e in a]
for i,e in enumerate(b):
if e>a[i]:
x=bis.bisect(a,e,i+1)
if e==a[x-1]:
stt=1
#print("yes",e,x)
break
elif e<a[i] :
x=bis.bisect(a,e,0,i)
if e==a[x-1]:
stt=1
#print("ye",e,i)
break
#pri
if stt==-1:
b=sorted(b)
i=0
while i<n-1:
if b[i]==b[i+1]:
stt=2
break
i+=1
print(stt)
``` | instruction | 0 | 37,743 | 12 | 75,486 |
Yes | output | 1 | 37,743 | 12 | 75,487 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There is an array with n elements a1, a2, ..., an and the number x.
In one operation you can select some i (1 ≤ i ≤ n) and replace element ai with ai & x, where & denotes the [bitwise and](https://en.wikipedia.org/wiki/Bitwise_operation#AND) operation.
You want the array to have at least two equal elements after applying some operations (possibly, none). In other words, there should be at least two distinct indices i ≠ j such that ai = aj. Determine whether it is possible to achieve and, if possible, the minimal number of operations to apply.
Input
The first line contains integers n and x (2 ≤ n ≤ 100 000, 1 ≤ x ≤ 100 000), number of elements in the array and the number to and with.
The second line contains n integers ai (1 ≤ ai ≤ 100 000), the elements of the array.
Output
Print a single integer denoting the minimal number of operations to do, or -1, if it is impossible.
Examples
Input
4 3
1 2 3 7
Output
1
Input
2 228
1 1
Output
0
Input
3 7
1 2 3
Output
-1
Note
In the first example one can apply the operation to the last element of the array. That replaces 7 with 3, so we achieve the goal in one move.
In the second example the array already has two equal elements.
In the third example applying the operation won't change the array at all, so it is impossible to make some pair of elements equal.
Submitted Solution:
```
# -*- coding: utf-8 -*-
"""
Created on Mon Jul 30 13:56:04 2018
@author: Amrita
"""
N, X = map(int,input().strip().split())
Ns = list(map(int,input().strip().split()))
dic = {}
done = False
for n in Ns:
if n in dic:
print(0)
done = True
break
else:
dic[n] = 0
if not done:
double_entry = False
for n in Ns:
new_n = n & X
if n == new_n:
continue
if new_n in dic:
if dic[new_n] == 0:
print(1)
done = True
break
else:
double_entry = True
else:
dic[new_n] = 1
if not done:
if double_entry:
print(2)
else:
print(-1)
``` | instruction | 0 | 37,744 | 12 | 75,488 |
Yes | output | 1 | 37,744 | 12 | 75,489 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There is an array with n elements a1, a2, ..., an and the number x.
In one operation you can select some i (1 ≤ i ≤ n) and replace element ai with ai & x, where & denotes the [bitwise and](https://en.wikipedia.org/wiki/Bitwise_operation#AND) operation.
You want the array to have at least two equal elements after applying some operations (possibly, none). In other words, there should be at least two distinct indices i ≠ j such that ai = aj. Determine whether it is possible to achieve and, if possible, the minimal number of operations to apply.
Input
The first line contains integers n and x (2 ≤ n ≤ 100 000, 1 ≤ x ≤ 100 000), number of elements in the array and the number to and with.
The second line contains n integers ai (1 ≤ ai ≤ 100 000), the elements of the array.
Output
Print a single integer denoting the minimal number of operations to do, or -1, if it is impossible.
Examples
Input
4 3
1 2 3 7
Output
1
Input
2 228
1 1
Output
0
Input
3 7
1 2 3
Output
-1
Note
In the first example one can apply the operation to the last element of the array. That replaces 7 with 3, so we achieve the goal in one move.
In the second example the array already has two equal elements.
In the third example applying the operation won't change the array at all, so it is impossible to make some pair of elements equal.
Submitted Solution:
```
n,x=[int(x) for x in input().split()]
a=[int(i) for i in input().split()]
'''b=list(reversed(a))
t=-1
if len(set(b))<len(b):
t=0
for i in range(n):
b[i]=b[i]&x
if len(set(b))<len(b):
if b.index(b[i])>i:
t=1
else: t=2
break
print(t)
'''
z=0
if len(set(a))<len(a):
print(0)
else:
b=[x&a[i] for i in range(n)]
if len(set(b))<len(b):
z=2
b=[(b[i],i,1) for i in range(n)]
r=0
a=[(a[i],i,0) for i in range(n)]
c=list(sorted(a+b,key=lambda x:x[0]))
for i in range(2*n-1):
if c[i][0]==c[i+1][0] and c[i][2]!=c[i+1][2] and c[i][1]!=c[i+1][1]:
print(1)
r=1
break
if r==0:
for i in range(2*n-2):
if c[i][0]==c[i+2][0] and c[i][2]!=c[i+2][2] and c[i][1]!=c[i+2][1]:
print(1)
r=1
break
if r==0:
if z==2:
print(2)
else:
print(-1)
``` | instruction | 0 | 37,745 | 12 | 75,490 |
Yes | output | 1 | 37,745 | 12 | 75,491 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There is an array with n elements a1, a2, ..., an and the number x.
In one operation you can select some i (1 ≤ i ≤ n) and replace element ai with ai & x, where & denotes the [bitwise and](https://en.wikipedia.org/wiki/Bitwise_operation#AND) operation.
You want the array to have at least two equal elements after applying some operations (possibly, none). In other words, there should be at least two distinct indices i ≠ j such that ai = aj. Determine whether it is possible to achieve and, if possible, the minimal number of operations to apply.
Input
The first line contains integers n and x (2 ≤ n ≤ 100 000, 1 ≤ x ≤ 100 000), number of elements in the array and the number to and with.
The second line contains n integers ai (1 ≤ ai ≤ 100 000), the elements of the array.
Output
Print a single integer denoting the minimal number of operations to do, or -1, if it is impossible.
Examples
Input
4 3
1 2 3 7
Output
1
Input
2 228
1 1
Output
0
Input
3 7
1 2 3
Output
-1
Note
In the first example one can apply the operation to the last element of the array. That replaces 7 with 3, so we achieve the goal in one move.
In the second example the array already has two equal elements.
In the third example applying the operation won't change the array at all, so it is impossible to make some pair of elements equal.
Submitted Solution:
```
n,x = map(int,input().split())
arr = map(int,input().split())
orgin = set()
change = set()
for element in arr :
if element in orgin :
print(0)
exit(0)
if element in change or element&x in orgin or element&x in change:
print(1)
exit(0)
orgin.add(element)
change.add(element)
print(-1)
exit(0)
``` | instruction | 0 | 37,746 | 12 | 75,492 |
No | output | 1 | 37,746 | 12 | 75,493 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There is an array with n elements a1, a2, ..., an and the number x.
In one operation you can select some i (1 ≤ i ≤ n) and replace element ai with ai & x, where & denotes the [bitwise and](https://en.wikipedia.org/wiki/Bitwise_operation#AND) operation.
You want the array to have at least two equal elements after applying some operations (possibly, none). In other words, there should be at least two distinct indices i ≠ j such that ai = aj. Determine whether it is possible to achieve and, if possible, the minimal number of operations to apply.
Input
The first line contains integers n and x (2 ≤ n ≤ 100 000, 1 ≤ x ≤ 100 000), number of elements in the array and the number to and with.
The second line contains n integers ai (1 ≤ ai ≤ 100 000), the elements of the array.
Output
Print a single integer denoting the minimal number of operations to do, or -1, if it is impossible.
Examples
Input
4 3
1 2 3 7
Output
1
Input
2 228
1 1
Output
0
Input
3 7
1 2 3
Output
-1
Note
In the first example one can apply the operation to the last element of the array. That replaces 7 with 3, so we achieve the goal in one move.
In the second example the array already has two equal elements.
In the third example applying the operation won't change the array at all, so it is impossible to make some pair of elements equal.
Submitted Solution:
```
import sys
def main():
lines = []
for line in sys.stdin:
lines.append(line)
n = int(lines[0].split(" ")[0])
x = int(lines[0].split(" ")[1])
inputs = {int(a): i for i, a in enumerate(lines[1].split(" "))}
if len(inputs) != n:
print(0)
return
ops = -1
ands = {}
for a, i in inputs.items():
a2 = a & x
if a2 in inputs and inputs[a2] != i:
ops = 1
return
if a2 in ands and ops == -1:
ops = 2
return
ands[a2] = i
print(ops)
main()
``` | instruction | 0 | 37,747 | 12 | 75,494 |
No | output | 1 | 37,747 | 12 | 75,495 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There is an array with n elements a1, a2, ..., an and the number x.
In one operation you can select some i (1 ≤ i ≤ n) and replace element ai with ai & x, where & denotes the [bitwise and](https://en.wikipedia.org/wiki/Bitwise_operation#AND) operation.
You want the array to have at least two equal elements after applying some operations (possibly, none). In other words, there should be at least two distinct indices i ≠ j such that ai = aj. Determine whether it is possible to achieve and, if possible, the minimal number of operations to apply.
Input
The first line contains integers n and x (2 ≤ n ≤ 100 000, 1 ≤ x ≤ 100 000), number of elements in the array and the number to and with.
The second line contains n integers ai (1 ≤ ai ≤ 100 000), the elements of the array.
Output
Print a single integer denoting the minimal number of operations to do, or -1, if it is impossible.
Examples
Input
4 3
1 2 3 7
Output
1
Input
2 228
1 1
Output
0
Input
3 7
1 2 3
Output
-1
Note
In the first example one can apply the operation to the last element of the array. That replaces 7 with 3, so we achieve the goal in one move.
In the second example the array already has two equal elements.
In the third example applying the operation won't change the array at all, so it is impossible to make some pair of elements equal.
Submitted Solution:
```
x,y=map(int,input().split())
l=list(map(int,input().split()))
h=0
for i in range(4):
if len(set(l))!=len(l):
break
n=l[0]&y
l.remove(l[0])
l.append(n)
if len(set(l))!=len(l):
break
h=h+1
if h==0:
print(0)
elif h<=len(l):
print(1)
else:
print(-1)
``` | instruction | 0 | 37,748 | 12 | 75,496 |
No | output | 1 | 37,748 | 12 | 75,497 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There is an array with n elements a1, a2, ..., an and the number x.
In one operation you can select some i (1 ≤ i ≤ n) and replace element ai with ai & x, where & denotes the [bitwise and](https://en.wikipedia.org/wiki/Bitwise_operation#AND) operation.
You want the array to have at least two equal elements after applying some operations (possibly, none). In other words, there should be at least two distinct indices i ≠ j such that ai = aj. Determine whether it is possible to achieve and, if possible, the minimal number of operations to apply.
Input
The first line contains integers n and x (2 ≤ n ≤ 100 000, 1 ≤ x ≤ 100 000), number of elements in the array and the number to and with.
The second line contains n integers ai (1 ≤ ai ≤ 100 000), the elements of the array.
Output
Print a single integer denoting the minimal number of operations to do, or -1, if it is impossible.
Examples
Input
4 3
1 2 3 7
Output
1
Input
2 228
1 1
Output
0
Input
3 7
1 2 3
Output
-1
Note
In the first example one can apply the operation to the last element of the array. That replaces 7 with 3, so we achieve the goal in one move.
In the second example the array already has two equal elements.
In the third example applying the operation won't change the array at all, so it is impossible to make some pair of elements equal.
Submitted Solution:
```
import sys
def main():
lines = []
for line in sys.stdin:
lines.append(line)
n = int(lines[0].split(" ")[0])
x = int(lines[0].split(" ")[1])
inputs = {int(a): i for i, a in enumerate(lines[1].split(" "))}
ops = -1
if len(set(inputs)) != n:
print(0)
return
ands = {}
for a, i in inputs.items():
a2 = a & x
if a2 in inputs and inputs[a2] != i:
print(1)
return
if a2 in ands:
print(2)
return
ands[a2] = inputs[a]
print(ops)
main()
``` | instruction | 0 | 37,749 | 12 | 75,498 |
No | output | 1 | 37,749 | 12 | 75,499 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Vova has won n trophies in different competitions. Each trophy is either golden or silver. The trophies are arranged in a row.
The beauty of the arrangement is the length of the longest subsegment consisting of golden trophies. Vova wants to swap two trophies (not necessarily adjacent ones) to make the arrangement as beautiful as possible — that means, to maximize the length of the longest such subsegment.
Help Vova! Tell him the maximum possible beauty of the arrangement if he is allowed to do at most one swap.
Input
The first line contains one integer n (2 ≤ n ≤ 10^5) — the number of trophies.
The second line contains n characters, each of them is either G or S. If the i-th character is G, then the i-th trophy is a golden one, otherwise it's a silver trophy.
Output
Print the maximum possible length of a subsegment of golden trophies, if Vova is allowed to do at most one swap.
Examples
Input
10
GGGSGGGSGG
Output
7
Input
4
GGGG
Output
4
Input
3
SSS
Output
0
Note
In the first example Vova has to swap trophies with indices 4 and 10. Thus he will obtain the sequence "GGGGGGGSGS", the length of the longest subsegment of golden trophies is 7.
In the second example Vova can make no swaps at all. The length of the longest subsegment of golden trophies in the sequence is 4.
In the third example Vova cannot do anything to make the length of the longest subsegment of golden trophies in the sequence greater than 0. | instruction | 0 | 37,768 | 12 | 75,536 |
Tags: greedy
Correct Solution:
```
n=int(input())
t=input()
L=[-1]
s=0
for i in range(n):
if t[i]=='S':
L.append(i)
s+=1
L.append(n)
m = L[1]-L[0]-1
for i in range(len(L)-2):
if L[i+2]-L[i]-1 > m:
m=L[i+2]-L[i]-1
print(min(m,n-s))
``` | output | 1 | 37,768 | 12 | 75,537 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Vova has won n trophies in different competitions. Each trophy is either golden or silver. The trophies are arranged in a row.
The beauty of the arrangement is the length of the longest subsegment consisting of golden trophies. Vova wants to swap two trophies (not necessarily adjacent ones) to make the arrangement as beautiful as possible — that means, to maximize the length of the longest such subsegment.
Help Vova! Tell him the maximum possible beauty of the arrangement if he is allowed to do at most one swap.
Input
The first line contains one integer n (2 ≤ n ≤ 10^5) — the number of trophies.
The second line contains n characters, each of them is either G or S. If the i-th character is G, then the i-th trophy is a golden one, otherwise it's a silver trophy.
Output
Print the maximum possible length of a subsegment of golden trophies, if Vova is allowed to do at most one swap.
Examples
Input
10
GGGSGGGSGG
Output
7
Input
4
GGGG
Output
4
Input
3
SSS
Output
0
Note
In the first example Vova has to swap trophies with indices 4 and 10. Thus he will obtain the sequence "GGGGGGGSGS", the length of the longest subsegment of golden trophies is 7.
In the second example Vova can make no swaps at all. The length of the longest subsegment of golden trophies in the sequence is 4.
In the third example Vova cannot do anything to make the length of the longest subsegment of golden trophies in the sequence greater than 0. | instruction | 0 | 37,770 | 12 | 75,540 |
Tags: greedy
Correct Solution:
```
n = int(input())
s = input()
max = 0
l = 0
has_s = False
gs = 0
for r in range(n):
if s[r] == 'G':
gs += 1
else:
if not has_s:
has_s = True
else:
while s[l] == 'G':
l += 1
l += 1
if r-l+1 > max:
max = r-l+1
ans = max
if gs < max:
ans -= 1
print(ans)
``` | output | 1 | 37,770 | 12 | 75,541 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Vova has won n trophies in different competitions. Each trophy is either golden or silver. The trophies are arranged in a row.
The beauty of the arrangement is the length of the longest subsegment consisting of golden trophies. Vova wants to swap two trophies (not necessarily adjacent ones) to make the arrangement as beautiful as possible — that means, to maximize the length of the longest such subsegment.
Help Vova! Tell him the maximum possible beauty of the arrangement if he is allowed to do at most one swap.
Input
The first line contains one integer n (2 ≤ n ≤ 10^5) — the number of trophies.
The second line contains n characters, each of them is either G or S. If the i-th character is G, then the i-th trophy is a golden one, otherwise it's a silver trophy.
Output
Print the maximum possible length of a subsegment of golden trophies, if Vova is allowed to do at most one swap.
Examples
Input
10
GGGSGGGSGG
Output
7
Input
4
GGGG
Output
4
Input
3
SSS
Output
0
Note
In the first example Vova has to swap trophies with indices 4 and 10. Thus he will obtain the sequence "GGGGGGGSGS", the length of the longest subsegment of golden trophies is 7.
In the second example Vova can make no swaps at all. The length of the longest subsegment of golden trophies in the sequence is 4.
In the third example Vova cannot do anything to make the length of the longest subsegment of golden trophies in the sequence greater than 0. | instruction | 0 | 37,771 | 12 | 75,542 |
Tags: greedy
Correct Solution:
```
n = int(input())
s = input()
Gs = list(filter(lambda x: x,s.split('S')))
Ss = list(filter(lambda x: x,s.split('G')))
cG = s.count('G')
# print(Gs, Ss, cG)
if(cG==n):
print(n)
exit(0)
if(cG==0):
print(0)
exit(0)
if(cG==1):
print(1)
exit(0)
mi = len(max(Gs)) if cG-len(max(Gs))==0 else len(max(Gs))+1
cand = [mi]
for i,e in enumerate(Ss):
if len(e)==1:
if s[0] == 'G':
pre = 0 if i>=len(Gs) else len(Gs[i])
pos = 0 if i+1>=len(Gs) else len(Gs[i+1])
if cG-(pre+pos)>0:
cand.append(pre+pos+1)
elif cG-(pre+pos)==0:
cand.append(pre+pos)
else:
pre = 0 if i-1<0 else len(Gs[i-1])
pos = 0 if i>=len(Gs) else len(Gs[i])
if cG-(pre+pos)>0:
cand.append(pre+pos+1)
elif cG-(pre+pos)==0:
cand.append(pre+pos)
print(max(cand))
``` | output | 1 | 37,771 | 12 | 75,543 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Vova has won n trophies in different competitions. Each trophy is either golden or silver. The trophies are arranged in a row.
The beauty of the arrangement is the length of the longest subsegment consisting of golden trophies. Vova wants to swap two trophies (not necessarily adjacent ones) to make the arrangement as beautiful as possible — that means, to maximize the length of the longest such subsegment.
Help Vova! Tell him the maximum possible beauty of the arrangement if he is allowed to do at most one swap.
Input
The first line contains one integer n (2 ≤ n ≤ 10^5) — the number of trophies.
The second line contains n characters, each of them is either G or S. If the i-th character is G, then the i-th trophy is a golden one, otherwise it's a silver trophy.
Output
Print the maximum possible length of a subsegment of golden trophies, if Vova is allowed to do at most one swap.
Examples
Input
10
GGGSGGGSGG
Output
7
Input
4
GGGG
Output
4
Input
3
SSS
Output
0
Note
In the first example Vova has to swap trophies with indices 4 and 10. Thus he will obtain the sequence "GGGGGGGSGS", the length of the longest subsegment of golden trophies is 7.
In the second example Vova can make no swaps at all. The length of the longest subsegment of golden trophies in the sequence is 4.
In the third example Vova cannot do anything to make the length of the longest subsegment of golden trophies in the sequence greater than 0. | instruction | 0 | 37,772 | 12 | 75,544 |
Tags: greedy
Correct Solution:
```
# -*- coding: utf-8 -*-
import math
import collections
import bisect
import heapq
import time
import itertools
import sys
"""
created by shhuan at 2018/12/9 20:21
"""
N = int(input())
A = input()
i = 0
segs = []
while i < N:
if A[i] == 'G':
j = i + 1
while j < N and A[j] == 'G':
j += 1
segs.append((i, j))
i = j
else:
i += 1
def seglen(seg):
return seg[1] - seg[0]
ans = 0
if not segs:
pass
elif len(segs) == 1:
ans = seglen(segs[0])
elif len(segs) == 2:
if segs[1][0] == segs[0][1] + 1:
ans = seglen(segs[0]) + seglen(segs[1])
else:
ans = max([seglen(x) for x in segs]) + 1
else:
ans = max([seglen(x) for x in segs]) + 1
for i in range(len(segs)-1):
if segs[i][1] + 1 == segs[i+1][0]:
ans = max(ans, seglen(segs[i]) + seglen(segs[i+1]) + 1)
print(ans)
``` | output | 1 | 37,772 | 12 | 75,545 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Vova has won n trophies in different competitions. Each trophy is either golden or silver. The trophies are arranged in a row.
The beauty of the arrangement is the length of the longest subsegment consisting of golden trophies. Vova wants to swap two trophies (not necessarily adjacent ones) to make the arrangement as beautiful as possible — that means, to maximize the length of the longest such subsegment.
Help Vova! Tell him the maximum possible beauty of the arrangement if he is allowed to do at most one swap.
Input
The first line contains one integer n (2 ≤ n ≤ 10^5) — the number of trophies.
The second line contains n characters, each of them is either G or S. If the i-th character is G, then the i-th trophy is a golden one, otherwise it's a silver trophy.
Output
Print the maximum possible length of a subsegment of golden trophies, if Vova is allowed to do at most one swap.
Examples
Input
10
GGGSGGGSGG
Output
7
Input
4
GGGG
Output
4
Input
3
SSS
Output
0
Note
In the first example Vova has to swap trophies with indices 4 and 10. Thus he will obtain the sequence "GGGGGGGSGS", the length of the longest subsegment of golden trophies is 7.
In the second example Vova can make no swaps at all. The length of the longest subsegment of golden trophies in the sequence is 4.
In the third example Vova cannot do anything to make the length of the longest subsegment of golden trophies in the sequence greater than 0. | instruction | 0 | 37,774 | 12 | 75,548 |
Tags: greedy
Correct Solution:
```
n = input()
s = input()
s += 'S'
bef, now, tot, res = 0, 0, 0, 0
for c in s :
if c == 'G' :
tot = tot+1
now = now+1
else :
res = max(res, now+bef+1)
bef = now
now = 0
res = min(tot, res)
print(res)
``` | output | 1 | 37,774 | 12 | 75,549 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Vova has won n trophies in different competitions. Each trophy is either golden or silver. The trophies are arranged in a row.
The beauty of the arrangement is the length of the longest subsegment consisting of golden trophies. Vova wants to swap two trophies (not necessarily adjacent ones) to make the arrangement as beautiful as possible — that means, to maximize the length of the longest such subsegment.
Help Vova! Tell him the maximum possible beauty of the arrangement if he is allowed to do at most one swap.
Input
The first line contains one integer n (2 ≤ n ≤ 10^5) — the number of trophies.
The second line contains n characters, each of them is either G or S. If the i-th character is G, then the i-th trophy is a golden one, otherwise it's a silver trophy.
Output
Print the maximum possible length of a subsegment of golden trophies, if Vova is allowed to do at most one swap.
Examples
Input
10
GGGSGGGSGG
Output
7
Input
4
GGGG
Output
4
Input
3
SSS
Output
0
Note
In the first example Vova has to swap trophies with indices 4 and 10. Thus he will obtain the sequence "GGGGGGGSGS", the length of the longest subsegment of golden trophies is 7.
In the second example Vova can make no swaps at all. The length of the longest subsegment of golden trophies in the sequence is 4.
In the third example Vova cannot do anything to make the length of the longest subsegment of golden trophies in the sequence greater than 0. | instruction | 0 | 37,775 | 12 | 75,550 |
Tags: greedy
Correct Solution:
```
n=int(input())
s=input()
ans=0
sc,gc,pi,ci=0,0,-1,-1
for i in range(1,n+1):
if s[i-1]=='G':
gc+=1
else:
sc+=1
if pi==-1:
ans=max(ans,i-1)
else:
ans=max(ans,i-1-pi)
pi=ci
ci=i
#print(ans)
#print(gc,sc)
if sc==1:
print(n-1)
exit()
if sc==2 and (s[0]=='S' or s[n-1]=='S'):
print(n-2)
exit()
if pi==-1:
ans=max(ans,n)
else:
ans = max(ans,n-pi)
print(min(ans,gc))
``` | output | 1 | 37,775 | 12 | 75,551 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Let's call beauty of an array b_1, b_2, …, b_n (n > 1) — min_{1 ≤ i < j ≤ n} |b_i - b_j|.
You're given an array a_1, a_2, … a_n and a number k. Calculate the sum of beauty over all subsequences of the array of length exactly k. As this number can be very large, output it modulo 998244353.
A sequence a is a subsequence of an array b if a can be obtained from b by deletion of several (possibly, zero or all) elements.
Input
The first line contains integers n, k (2 ≤ k ≤ n ≤ 1000).
The second line contains n integers a_1, a_2, …, a_n (0 ≤ a_i ≤ 10^5).
Output
Output one integer — the sum of beauty over all subsequences of the array of length exactly k. As this number can be very large, output it modulo 998244353.
Examples
Input
4 3
1 7 3 5
Output
8
Input
5 5
1 10 100 1000 10000
Output
9
Note
In the first example, there are 4 subsequences of length 3 — [1, 7, 3], [1, 3, 5], [7, 3, 5], [1, 7, 5], each of which has beauty 2, so answer is 8.
In the second example, there is only one subsequence of length 5 — the whole array, which has the beauty equal to |10-1| = 9. | instruction | 0 | 37,820 | 12 | 75,640 |
Tags: dp
Correct Solution:
```
from collections import defaultdict
import sys
input = sys.stdin.readline
'''
for CASES in range(int(input())):
n, m = map(int, input().split())
n = int(input())
A = list(map(int, input().split()))
S = input().strip()
sys.stdout.write(" ".join(map(str,ans))+"\n")
'''
inf = 100000000000000000 # 1e17
mod = 998244353
n, m = map(int, input().split())
A = [0] + sorted(list(map(int, input().split())))
ans = 0
f = [[0] * (n + 10) for _ in range(m + 10)]
for x in range(1,(A[n] - A[1]) // (m - 1) + 1):
for i in range(1, n + 1):
f[1][i] = 1
for i in range(2, m + 1):
sum = 0
pre = 1
for j in range(1, n + 1):
while pre <= n and A[pre] + x <= A[j]:
sum += f[i - 1][pre]
sum %= mod
pre += 1
f[i][j] = sum
for i in range(1, n + 1):
ans += f[m][i]
ans %= mod
print(ans)
# the end
``` | output | 1 | 37,820 | 12 | 75,641 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Suppose you are performing the following algorithm. There is an array v_1, v_2, ..., v_n filled with zeroes at start. The following operation is applied to the array several times — at i-th step (0-indexed) you can:
* either choose position pos (1 ≤ pos ≤ n) and increase v_{pos} by k^i;
* or not choose any position and skip this step.
You can choose how the algorithm would behave on each step and when to stop it. The question is: can you make array v equal to the given array a (v_j = a_j for each j) after some step?
Input
The first line contains one integer T (1 ≤ T ≤ 1000) — the number of test cases. Next 2T lines contain test cases — two lines per test case.
The first line of each test case contains two integers n and k (1 ≤ n ≤ 30, 2 ≤ k ≤ 100) — the size of arrays v and a and value k used in the algorithm.
The second line contains n integers a_1, a_2, ..., a_n (0 ≤ a_i ≤ 10^{16}) — the array you'd like to achieve.
Output
For each test case print YES (case insensitive) if you can achieve the array a after some step or NO (case insensitive) otherwise.
Example
Input
5
4 100
0 0 0 0
1 2
1
3 4
1 4 1
3 2
0 1 3
3 9
0 59049 810
Output
YES
YES
NO
NO
YES
Note
In the first test case, you can stop the algorithm before the 0-th step, or don't choose any position several times and stop the algorithm.
In the second test case, you can add k^0 to v_1 and stop the algorithm.
In the third test case, you can't make two 1 in the array v.
In the fifth test case, you can skip 9^0 and 9^1, then add 9^2 and 9^3 to v_3, skip 9^4 and finally, add 9^5 to v_2. | instruction | 0 | 37,839 | 12 | 75,678 |
Tags: bitmasks, greedy, implementation, math, number theory, ternary search
Correct Solution:
```
for _ in range(int(input())):
n,k=map(int,input().split())
l=sorted(list(map(int,input().split())))
f=1
p=set()
for i in range(n):
if l[i]!=0:
c=0
#x=[]
if l[i]!=1 and l[i]<k:
f=0
break
while l[i]:
q=l[i]%k
l[i]=l[i]//k
#x.append(q*pow(k,c))
if q==1:
if c not in p:
p.add(c)
else:
f=0
break
elif q!=0:
f=0
break
c+=1
#print(x)
#else:
#print([0])
if f:
print("YES")
else:
print("NO")
``` | output | 1 | 37,839 | 12 | 75,679 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Suppose you are performing the following algorithm. There is an array v_1, v_2, ..., v_n filled with zeroes at start. The following operation is applied to the array several times — at i-th step (0-indexed) you can:
* either choose position pos (1 ≤ pos ≤ n) and increase v_{pos} by k^i;
* or not choose any position and skip this step.
You can choose how the algorithm would behave on each step and when to stop it. The question is: can you make array v equal to the given array a (v_j = a_j for each j) after some step?
Input
The first line contains one integer T (1 ≤ T ≤ 1000) — the number of test cases. Next 2T lines contain test cases — two lines per test case.
The first line of each test case contains two integers n and k (1 ≤ n ≤ 30, 2 ≤ k ≤ 100) — the size of arrays v and a and value k used in the algorithm.
The second line contains n integers a_1, a_2, ..., a_n (0 ≤ a_i ≤ 10^{16}) — the array you'd like to achieve.
Output
For each test case print YES (case insensitive) if you can achieve the array a after some step or NO (case insensitive) otherwise.
Example
Input
5
4 100
0 0 0 0
1 2
1
3 4
1 4 1
3 2
0 1 3
3 9
0 59049 810
Output
YES
YES
NO
NO
YES
Note
In the first test case, you can stop the algorithm before the 0-th step, or don't choose any position several times and stop the algorithm.
In the second test case, you can add k^0 to v_1 and stop the algorithm.
In the third test case, you can't make two 1 in the array v.
In the fifth test case, you can skip 9^0 and 9^1, then add 9^2 and 9^3 to v_3, skip 9^4 and finally, add 9^5 to v_2. | instruction | 0 | 37,840 | 12 | 75,680 |
Tags: bitmasks, greedy, implementation, math, number theory, ternary search
Correct Solution:
```
def go():
n, k = map(int, input().split())
a = list(map(int, input().split()))
ma = max(a)
kp=1
while kp<ma:
kp*=k
u = set()
for aa in a:
if aa == 0:
continue
c = kp
while aa>0:
while c>aa:
c//=k
if aa>=2*c or c in u:
return 'NO'
else:
u.add(c)
aa-=c
return 'YES'
t= int(input())
ans = []
for _ in range(t):
ans.append(go())
print ('\n'.join(ans))
``` | output | 1 | 37,840 | 12 | 75,681 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Suppose you are performing the following algorithm. There is an array v_1, v_2, ..., v_n filled with zeroes at start. The following operation is applied to the array several times — at i-th step (0-indexed) you can:
* either choose position pos (1 ≤ pos ≤ n) and increase v_{pos} by k^i;
* or not choose any position and skip this step.
You can choose how the algorithm would behave on each step and when to stop it. The question is: can you make array v equal to the given array a (v_j = a_j for each j) after some step?
Input
The first line contains one integer T (1 ≤ T ≤ 1000) — the number of test cases. Next 2T lines contain test cases — two lines per test case.
The first line of each test case contains two integers n and k (1 ≤ n ≤ 30, 2 ≤ k ≤ 100) — the size of arrays v and a and value k used in the algorithm.
The second line contains n integers a_1, a_2, ..., a_n (0 ≤ a_i ≤ 10^{16}) — the array you'd like to achieve.
Output
For each test case print YES (case insensitive) if you can achieve the array a after some step or NO (case insensitive) otherwise.
Example
Input
5
4 100
0 0 0 0
1 2
1
3 4
1 4 1
3 2
0 1 3
3 9
0 59049 810
Output
YES
YES
NO
NO
YES
Note
In the first test case, you can stop the algorithm before the 0-th step, or don't choose any position several times and stop the algorithm.
In the second test case, you can add k^0 to v_1 and stop the algorithm.
In the third test case, you can't make two 1 in the array v.
In the fifth test case, you can skip 9^0 and 9^1, then add 9^2 and 9^3 to v_3, skip 9^4 and finally, add 9^5 to v_2. | instruction | 0 | 37,841 | 12 | 75,682 |
Tags: bitmasks, greedy, implementation, math, number theory, ternary search
Correct Solution:
```
def check(s):
while(s>0):
c=maxPower(s)
if (c!=-1):
s=s-c
else:
return -1
if (s==0):
return 1
def maxPower(s):
i=1
t=0
while(i<=s):
i*=k
t+=1
t-=1
i=i//k
powers[t]+=1
if powers[t]>1:
return -1
return i
t=int(input())
for i in range(t):
powers=[0]*100
flag=1
size, k = map(int, input().split())
numbers=input().split()
for j in range(size):
if check(int(numbers[j]))==-1:
print("NO")
flag=0
break
if (flag==1):
print("YES")
``` | output | 1 | 37,841 | 12 | 75,683 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Suppose you are performing the following algorithm. There is an array v_1, v_2, ..., v_n filled with zeroes at start. The following operation is applied to the array several times — at i-th step (0-indexed) you can:
* either choose position pos (1 ≤ pos ≤ n) and increase v_{pos} by k^i;
* or not choose any position and skip this step.
You can choose how the algorithm would behave on each step and when to stop it. The question is: can you make array v equal to the given array a (v_j = a_j for each j) after some step?
Input
The first line contains one integer T (1 ≤ T ≤ 1000) — the number of test cases. Next 2T lines contain test cases — two lines per test case.
The first line of each test case contains two integers n and k (1 ≤ n ≤ 30, 2 ≤ k ≤ 100) — the size of arrays v and a and value k used in the algorithm.
The second line contains n integers a_1, a_2, ..., a_n (0 ≤ a_i ≤ 10^{16}) — the array you'd like to achieve.
Output
For each test case print YES (case insensitive) if you can achieve the array a after some step or NO (case insensitive) otherwise.
Example
Input
5
4 100
0 0 0 0
1 2
1
3 4
1 4 1
3 2
0 1 3
3 9
0 59049 810
Output
YES
YES
NO
NO
YES
Note
In the first test case, you can stop the algorithm before the 0-th step, or don't choose any position several times and stop the algorithm.
In the second test case, you can add k^0 to v_1 and stop the algorithm.
In the third test case, you can't make two 1 in the array v.
In the fifth test case, you can skip 9^0 and 9^1, then add 9^2 and 9^3 to v_3, skip 9^4 and finally, add 9^5 to v_2. | instruction | 0 | 37,842 | 12 | 75,684 |
Tags: bitmasks, greedy, implementation, math, number theory, ternary search
Correct Solution:
```
def to_base(k, number):
k_base = []
while number:
k_base.append(number % k)
number //= k
return k_base
def can_be_game_result(array, k):
max_item = max(array)
used_powers = [0] + [0] * len(to_base(k, max_item))
for item in array:
item_in_k_base = to_base(k, item)
for idx, used in enumerate(item_in_k_base):
used_powers[idx] += used
return max(used_powers) <= 1
test_cases = int(input())
for _ in range(test_cases):
_, k = map(int, input().split())
desired_array = [int(x) for x in input().split()]
result = can_be_game_result(desired_array, k)
print('YES' if result else 'NO')
``` | output | 1 | 37,842 | 12 | 75,685 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Suppose you are performing the following algorithm. There is an array v_1, v_2, ..., v_n filled with zeroes at start. The following operation is applied to the array several times — at i-th step (0-indexed) you can:
* either choose position pos (1 ≤ pos ≤ n) and increase v_{pos} by k^i;
* or not choose any position and skip this step.
You can choose how the algorithm would behave on each step and when to stop it. The question is: can you make array v equal to the given array a (v_j = a_j for each j) after some step?
Input
The first line contains one integer T (1 ≤ T ≤ 1000) — the number of test cases. Next 2T lines contain test cases — two lines per test case.
The first line of each test case contains two integers n and k (1 ≤ n ≤ 30, 2 ≤ k ≤ 100) — the size of arrays v and a and value k used in the algorithm.
The second line contains n integers a_1, a_2, ..., a_n (0 ≤ a_i ≤ 10^{16}) — the array you'd like to achieve.
Output
For each test case print YES (case insensitive) if you can achieve the array a after some step or NO (case insensitive) otherwise.
Example
Input
5
4 100
0 0 0 0
1 2
1
3 4
1 4 1
3 2
0 1 3
3 9
0 59049 810
Output
YES
YES
NO
NO
YES
Note
In the first test case, you can stop the algorithm before the 0-th step, or don't choose any position several times and stop the algorithm.
In the second test case, you can add k^0 to v_1 and stop the algorithm.
In the third test case, you can't make two 1 in the array v.
In the fifth test case, you can skip 9^0 and 9^1, then add 9^2 and 9^3 to v_3, skip 9^4 and finally, add 9^5 to v_2. | instruction | 0 | 37,843 | 12 | 75,686 |
Tags: bitmasks, greedy, implementation, math, number theory, ternary search
Correct Solution:
```
import sys
RI = lambda : [int(x) for x in sys.stdin.readline().split()]
ri = lambda : sys.stdin.readline().strip()
for _ in range(int(ri())):
n,k = RI()
a = RI()
arr = []
outflag = True
for i in a:
temp = i
bit = []
flag = True
while temp > 0:
temp1 = temp%k
if temp1 != 0 and temp1 != 1 :
flag = not flag
break
bit.append(temp1)
temp = temp//k
if not flag :
outflag = not outflag
break
arr.append(bit)
if outflag == False:
print("NO")
else:
flag = True
# print(arr)
xor = [0]*64
maxlen = -1
for i in arr:
maxlen = max(maxlen,len(i))
for j in range(len(i)):
xor[j]^=i[j]
# print(xor)
if i[j] == 1 and xor[j] == 0:
flag = not flag
break
if not flag :
break
# print(xor)
if flag :
print("YES")
else:
print("NO")
``` | output | 1 | 37,843 | 12 | 75,687 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Suppose you are performing the following algorithm. There is an array v_1, v_2, ..., v_n filled with zeroes at start. The following operation is applied to the array several times — at i-th step (0-indexed) you can:
* either choose position pos (1 ≤ pos ≤ n) and increase v_{pos} by k^i;
* or not choose any position and skip this step.
You can choose how the algorithm would behave on each step and when to stop it. The question is: can you make array v equal to the given array a (v_j = a_j for each j) after some step?
Input
The first line contains one integer T (1 ≤ T ≤ 1000) — the number of test cases. Next 2T lines contain test cases — two lines per test case.
The first line of each test case contains two integers n and k (1 ≤ n ≤ 30, 2 ≤ k ≤ 100) — the size of arrays v and a and value k used in the algorithm.
The second line contains n integers a_1, a_2, ..., a_n (0 ≤ a_i ≤ 10^{16}) — the array you'd like to achieve.
Output
For each test case print YES (case insensitive) if you can achieve the array a after some step or NO (case insensitive) otherwise.
Example
Input
5
4 100
0 0 0 0
1 2
1
3 4
1 4 1
3 2
0 1 3
3 9
0 59049 810
Output
YES
YES
NO
NO
YES
Note
In the first test case, you can stop the algorithm before the 0-th step, or don't choose any position several times and stop the algorithm.
In the second test case, you can add k^0 to v_1 and stop the algorithm.
In the third test case, you can't make two 1 in the array v.
In the fifth test case, you can skip 9^0 and 9^1, then add 9^2 and 9^3 to v_3, skip 9^4 and finally, add 9^5 to v_2. | instruction | 0 | 37,844 | 12 | 75,688 |
Tags: bitmasks, greedy, implementation, math, number theory, ternary search
Correct Solution:
```
from sys import stdin
from collections import defaultdict
#####################################################################
def iinput(): return int(stdin.readline())
def sinput(): return input()
def minput(): return map(int, stdin.readline().split())
def linput(): return list(map(int, stdin.readline().split()))
#####################################################################
def power(n, k):
i = 0
while n%k==0:
n//=k
i+=1
return i
t = iinput()
while t:
t-=1
n, k =minput()
a = linput()
v = [0]*n
ans = 'YES'
if a == v:
print(ans)
else:
a.sort()
c = defaultdict(int)
for j in range(n):
if a[j]==1:
c[0]+=1
if c[0]>1:
ans = 'NO'
break
elif a[j]>1:
while True:
p = power(a[j], k)
a[j] = a[j] - k**p
if 1<a[j]<k:
ans = 'NO'
break
c[p]+=1
if c[p]>1:
ans = 'NO'
break
if a[j]==0:break
if ans == 'NO':break
print(ans)
``` | output | 1 | 37,844 | 12 | 75,689 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Suppose you are performing the following algorithm. There is an array v_1, v_2, ..., v_n filled with zeroes at start. The following operation is applied to the array several times — at i-th step (0-indexed) you can:
* either choose position pos (1 ≤ pos ≤ n) and increase v_{pos} by k^i;
* or not choose any position and skip this step.
You can choose how the algorithm would behave on each step and when to stop it. The question is: can you make array v equal to the given array a (v_j = a_j for each j) after some step?
Input
The first line contains one integer T (1 ≤ T ≤ 1000) — the number of test cases. Next 2T lines contain test cases — two lines per test case.
The first line of each test case contains two integers n and k (1 ≤ n ≤ 30, 2 ≤ k ≤ 100) — the size of arrays v and a and value k used in the algorithm.
The second line contains n integers a_1, a_2, ..., a_n (0 ≤ a_i ≤ 10^{16}) — the array you'd like to achieve.
Output
For each test case print YES (case insensitive) if you can achieve the array a after some step or NO (case insensitive) otherwise.
Example
Input
5
4 100
0 0 0 0
1 2
1
3 4
1 4 1
3 2
0 1 3
3 9
0 59049 810
Output
YES
YES
NO
NO
YES
Note
In the first test case, you can stop the algorithm before the 0-th step, or don't choose any position several times and stop the algorithm.
In the second test case, you can add k^0 to v_1 and stop the algorithm.
In the third test case, you can't make two 1 in the array v.
In the fifth test case, you can skip 9^0 and 9^1, then add 9^2 and 9^3 to v_3, skip 9^4 and finally, add 9^5 to v_2. | instruction | 0 | 37,845 | 12 | 75,690 |
Tags: bitmasks, greedy, implementation, math, number theory, ternary search
Correct Solution:
```
# from math import log2
# t = int(input())
# while t:
# n = int(input())
# mem = {}
# def f(num):
# if num in mem:
# n = mem[num]
# #print(num)
# #n = int(log2(num))
# else:
# n = 0
# x = num
# while x:
# if x in mem:
# n+=mem[x]
# break
# x=x//2
# #print(1)
# n+=1
# #mem[x] = n-1
# n-=1
# mem[num] = n
# # print(num,n,2**n,num-(2**n))
# return num-(2**n),2**(n+1)-1
# ans = 0
# remain = n
# while remain>0:
# remain,v = f(remain)
# #print(remain)
# ans+=v
# print(ans)
# t-=1
# from math import sqrt
# def divs(n) :
# ans = []
# i = 1
# while i <= sqrt(n):
# if (n % i == 0) :
# if (n / i == i) :
# ans.append((i,i))
# else :
# ans.append((i,n/i))
# i = i + 1
# return ans
# n,m,k = [int(i) for i in input().split()]
# a = [int(i) for i in input().split()]
# c= [int(i) for i in input().split()]
# l = {}
# b = {}
# prevl = -1
# for i in range(n+1):
# if i==n or a[i] == 0:
# leng = i-prevl-1
# if leng != 0:
# if leng in l:
# l[leng]+=1
# else:
# l[leng] = 1
# prevl = i
# prevb = -1
# for i in range(m+1):
# if i == m or c[i] == 0:
# br = i-prevb-1
# if br != 0:
# if br in b:
# b[br]+=1
# else:
# b[br] = 1
# prevb = i
# d = divs(k)
# ans = 0
# for le in l.keys():
# for br in b.keys():
# for a,b1 in d:
# a = int(a)
# b1 = int(b1)
# if le>=a and br>=b1:
# ans+=((le-a+1)*(br-b1+1)*l[le]*b[br])
# if a==b1:
# continue
# if le>=b1 and br>=a:
# ans+=((le-b1+1)*(br-a+1)*l[le]*b[br])
# print(ans)
t = int(input())
while t:
n,k = [int(i) for i in input().split()]
a = [int(i) for i in input().split()]
a.sort()
def f(num):
ans = []
i=0
#print(num)
while num>0:
if num%k == 1:
ans.append(i)
elif num%k >1:
return ans,False
i+=1
num = num//k
return ans,True
bits = []
flag = 0
for e in a:
x,z = f(e)
if z:
bits.append(x)
else:
flag = 1
break
# print(a)
# print(bits)
if flag == 1:
print('NO')
t-=1
continue
flag = 0
mem = set()
for i in range(n):
if a[i] == 0:
continue
for l in bits[i]:
if l in mem:
flag = 1
break
mem.add(l)
if flag == 1:
print('NO')
else:
print('YES')
t-=1
``` | output | 1 | 37,845 | 12 | 75,691 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Suppose you are performing the following algorithm. There is an array v_1, v_2, ..., v_n filled with zeroes at start. The following operation is applied to the array several times — at i-th step (0-indexed) you can:
* either choose position pos (1 ≤ pos ≤ n) and increase v_{pos} by k^i;
* or not choose any position and skip this step.
You can choose how the algorithm would behave on each step and when to stop it. The question is: can you make array v equal to the given array a (v_j = a_j for each j) after some step?
Input
The first line contains one integer T (1 ≤ T ≤ 1000) — the number of test cases. Next 2T lines contain test cases — two lines per test case.
The first line of each test case contains two integers n and k (1 ≤ n ≤ 30, 2 ≤ k ≤ 100) — the size of arrays v and a and value k used in the algorithm.
The second line contains n integers a_1, a_2, ..., a_n (0 ≤ a_i ≤ 10^{16}) — the array you'd like to achieve.
Output
For each test case print YES (case insensitive) if you can achieve the array a after some step or NO (case insensitive) otherwise.
Example
Input
5
4 100
0 0 0 0
1 2
1
3 4
1 4 1
3 2
0 1 3
3 9
0 59049 810
Output
YES
YES
NO
NO
YES
Note
In the first test case, you can stop the algorithm before the 0-th step, or don't choose any position several times and stop the algorithm.
In the second test case, you can add k^0 to v_1 and stop the algorithm.
In the third test case, you can't make two 1 in the array v.
In the fifth test case, you can skip 9^0 and 9^1, then add 9^2 and 9^3 to v_3, skip 9^4 and finally, add 9^5 to v_2. | instruction | 0 | 37,846 | 12 | 75,692 |
Tags: bitmasks, greedy, implementation, math, number theory, ternary search
Correct Solution:
```
#------------------------template--------------------------#
import os
import sys
from math import *
from collections 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")
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()
s=set()
#print(avail[-1])
f=1
for i in a:
if(i==0):
continue
else:
powers=[]
p=0
while(i>0):
if(i%k==0):
i=i//k
p+=1
else:
powers.append(p)
i-=1
for p in powers:
if(p in s):
f=0
print("NO")
break
else:
s.add(p)
if(f==0):
break
if(f): print("YES")
``` | output | 1 | 37,846 | 12 | 75,693 |
Provide tags and a correct Python 2 solution for this coding contest problem.
Suppose you are performing the following algorithm. There is an array v_1, v_2, ..., v_n filled with zeroes at start. The following operation is applied to the array several times — at i-th step (0-indexed) you can:
* either choose position pos (1 ≤ pos ≤ n) and increase v_{pos} by k^i;
* or not choose any position and skip this step.
You can choose how the algorithm would behave on each step and when to stop it. The question is: can you make array v equal to the given array a (v_j = a_j for each j) after some step?
Input
The first line contains one integer T (1 ≤ T ≤ 1000) — the number of test cases. Next 2T lines contain test cases — two lines per test case.
The first line of each test case contains two integers n and k (1 ≤ n ≤ 30, 2 ≤ k ≤ 100) — the size of arrays v and a and value k used in the algorithm.
The second line contains n integers a_1, a_2, ..., a_n (0 ≤ a_i ≤ 10^{16}) — the array you'd like to achieve.
Output
For each test case print YES (case insensitive) if you can achieve the array a after some step or NO (case insensitive) otherwise.
Example
Input
5
4 100
0 0 0 0
1 2
1
3 4
1 4 1
3 2
0 1 3
3 9
0 59049 810
Output
YES
YES
NO
NO
YES
Note
In the first test case, you can stop the algorithm before the 0-th step, or don't choose any position several times and stop the algorithm.
In the second test case, you can add k^0 to v_1 and stop the algorithm.
In the third test case, you can't make two 1 in the array v.
In the fifth test case, you can skip 9^0 and 9^1, then add 9^2 and 9^3 to v_3, skip 9^4 and finally, add 9^5 to v_2. | instruction | 0 | 37,847 | 12 | 75,694 |
Tags: bitmasks, greedy, implementation, math, number theory, ternary search
Correct Solution:
```
from sys import stdin, stdout
from collections import Counter, defaultdict
from itertools import permutations, combinations
from fractions import gcd
import heapq
raw_input = stdin.readline
pr = stdout.write
mod=10**9+7
def ni():
return int(raw_input())
def li():
return list(map(int,raw_input().split()))
def pn(n):
stdout.write(str(n)+'\n')
def pa(arr):
pr(' '.join(map(str,arr))+'\n')
# fast read function for total integer input
def inp():
# this function returns whole input of
# space/line seperated integers
# Use Ctrl+D to flush stdin.
return tuple(map(int,stdin.read().split()))
range = xrange # not for python 3.0+
# main code
for t in range(ni()):
n,k=li()
l=li()
p=1
pw=0
mx=max(l)
while (p*k)<=mx:
p*=k
pw+=1
q=[]
for i in l:
heapq.heappush(q,-i)
while q and pw>=0:
x=-heapq.heappop(q)
#print p,x,q
if p>x:
heapq.heappush(q,-x)
p/=k
pw-=1
continue
x-=p
p/=k
pw-=1
if x:
heapq.heappush(q,-x)
f=0
for i in q:
if i:
f=1
break
if f:
pr('NO\n')
else:
pr('YES\n')
``` | output | 1 | 37,847 | 12 | 75,695 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Suppose you are performing the following algorithm. There is an array v_1, v_2, ..., v_n filled with zeroes at start. The following operation is applied to the array several times — at i-th step (0-indexed) you can:
* either choose position pos (1 ≤ pos ≤ n) and increase v_{pos} by k^i;
* or not choose any position and skip this step.
You can choose how the algorithm would behave on each step and when to stop it. The question is: can you make array v equal to the given array a (v_j = a_j for each j) after some step?
Input
The first line contains one integer T (1 ≤ T ≤ 1000) — the number of test cases. Next 2T lines contain test cases — two lines per test case.
The first line of each test case contains two integers n and k (1 ≤ n ≤ 30, 2 ≤ k ≤ 100) — the size of arrays v and a and value k used in the algorithm.
The second line contains n integers a_1, a_2, ..., a_n (0 ≤ a_i ≤ 10^{16}) — the array you'd like to achieve.
Output
For each test case print YES (case insensitive) if you can achieve the array a after some step or NO (case insensitive) otherwise.
Example
Input
5
4 100
0 0 0 0
1 2
1
3 4
1 4 1
3 2
0 1 3
3 9
0 59049 810
Output
YES
YES
NO
NO
YES
Note
In the first test case, you can stop the algorithm before the 0-th step, or don't choose any position several times and stop the algorithm.
In the second test case, you can add k^0 to v_1 and stop the algorithm.
In the third test case, you can't make two 1 in the array v.
In the fifth test case, you can skip 9^0 and 9^1, then add 9^2 and 9^3 to v_3, skip 9^4 and finally, add 9^5 to v_2.
Submitted Solution:
```
t = int(input())
while (t):
line = input().split()
n,k = int(line[0]),int(line[1])
a = input().split()
a = [int(x) for x in a]
b = [0 for j in range(60)]
flag = True
for x in a:
p = 1
j = 0
while (p <= x):
p *= k
j += 1
p //= k
j -= 1
tmp = [0 for j in range(60)]
while (x > 0 and p >= 1):
if (x >= p):
x -= p
tmp[j] = 1
p //= k
j -= 1
#print(x,tmp[:6])
if (x != 0):
flag = False
break
for j in range(60):
b[j] += tmp[j]
if (b[j] > 1):
flag = False
break
if (not flag):
break
if (flag):
print('YES')
else:
print('NO')
t -= 1
``` | instruction | 0 | 37,848 | 12 | 75,696 |
Yes | output | 1 | 37,848 | 12 | 75,697 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Suppose you are performing the following algorithm. There is an array v_1, v_2, ..., v_n filled with zeroes at start. The following operation is applied to the array several times — at i-th step (0-indexed) you can:
* either choose position pos (1 ≤ pos ≤ n) and increase v_{pos} by k^i;
* or not choose any position and skip this step.
You can choose how the algorithm would behave on each step and when to stop it. The question is: can you make array v equal to the given array a (v_j = a_j for each j) after some step?
Input
The first line contains one integer T (1 ≤ T ≤ 1000) — the number of test cases. Next 2T lines contain test cases — two lines per test case.
The first line of each test case contains two integers n and k (1 ≤ n ≤ 30, 2 ≤ k ≤ 100) — the size of arrays v and a and value k used in the algorithm.
The second line contains n integers a_1, a_2, ..., a_n (0 ≤ a_i ≤ 10^{16}) — the array you'd like to achieve.
Output
For each test case print YES (case insensitive) if you can achieve the array a after some step or NO (case insensitive) otherwise.
Example
Input
5
4 100
0 0 0 0
1 2
1
3 4
1 4 1
3 2
0 1 3
3 9
0 59049 810
Output
YES
YES
NO
NO
YES
Note
In the first test case, you can stop the algorithm before the 0-th step, or don't choose any position several times and stop the algorithm.
In the second test case, you can add k^0 to v_1 and stop the algorithm.
In the third test case, you can't make two 1 in the array v.
In the fifth test case, you can skip 9^0 and 9^1, then add 9^2 and 9^3 to v_3, skip 9^4 and finally, add 9^5 to v_2.
Submitted Solution:
```
import math
t=int(input())
def fn():
n,k=map(int,input().split())
x=list(map(int,input().split()))
z=k
while k<10**16:
k*=z
while k>=1:
#print(k)
count=0
for i in range(n):
if x[i]>=k:
x[i]=x[i]-k
count+=1
if count>1:
print("NO ")
#print(k)
#print(*x)
return
k=k//z
if min(x)!=0 or max(x)!=0:
print("NO")
return
print("YES")
while t:
t-=1
fn()
``` | instruction | 0 | 37,849 | 12 | 75,698 |
Yes | output | 1 | 37,849 | 12 | 75,699 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Suppose you are performing the following algorithm. There is an array v_1, v_2, ..., v_n filled with zeroes at start. The following operation is applied to the array several times — at i-th step (0-indexed) you can:
* either choose position pos (1 ≤ pos ≤ n) and increase v_{pos} by k^i;
* or not choose any position and skip this step.
You can choose how the algorithm would behave on each step and when to stop it. The question is: can you make array v equal to the given array a (v_j = a_j for each j) after some step?
Input
The first line contains one integer T (1 ≤ T ≤ 1000) — the number of test cases. Next 2T lines contain test cases — two lines per test case.
The first line of each test case contains two integers n and k (1 ≤ n ≤ 30, 2 ≤ k ≤ 100) — the size of arrays v and a and value k used in the algorithm.
The second line contains n integers a_1, a_2, ..., a_n (0 ≤ a_i ≤ 10^{16}) — the array you'd like to achieve.
Output
For each test case print YES (case insensitive) if you can achieve the array a after some step or NO (case insensitive) otherwise.
Example
Input
5
4 100
0 0 0 0
1 2
1
3 4
1 4 1
3 2
0 1 3
3 9
0 59049 810
Output
YES
YES
NO
NO
YES
Note
In the first test case, you can stop the algorithm before the 0-th step, or don't choose any position several times and stop the algorithm.
In the second test case, you can add k^0 to v_1 and stop the algorithm.
In the third test case, you can't make two 1 in the array v.
In the fifth test case, you can skip 9^0 and 9^1, then add 9^2 and 9^3 to v_3, skip 9^4 and finally, add 9^5 to v_2.
Submitted Solution:
```
import math as mt
import sys,string
input=sys.stdin.readline
import random
from collections import deque,defaultdict
L=lambda : list(map(int,input().split()))
Ls=lambda : list(input().split())
M=lambda : map(int,input().split())
I=lambda :int(input())
t=I()
for _ in range(t):
n,k=M()
l=L()
w=0
d=defaultdict(int)
ans=[0]*1000
for i in range(n):
x=l[i]
r=[]
while(x!=0):
if(x%k>1):
w=1
break
r.append(x%k)
x//=k
#print(r,"A")
for j in range(len(r)):
ans[j]+=r[j]
if(ans[j]>1):
w=1
break
#print(*ans)
if(w==0):
print("YES")
else:
print("NO")
``` | instruction | 0 | 37,850 | 12 | 75,700 |
Yes | output | 1 | 37,850 | 12 | 75,701 |
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