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 |
|---|---|---|---|---|---|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Vitaly has an array of n distinct integers. Vitaly wants to divide this array into three non-empty sets so as the following conditions hold:
1. The product of all numbers in the first set is less than zero ( < 0).
2. The product of all numbers in the second set is greater than zero ( > 0).
3. The product of all numbers in the third set is equal to zero.
4. Each number from the initial array must occur in exactly one set.
Help Vitaly. Divide the given array.
Input
The first line of the input contains integer n (3 ≤ n ≤ 100). The second line contains n space-separated distinct integers a1, a2, ..., an (|ai| ≤ 103) — the array elements.
Output
In the first line print integer n1 (n1 > 0) — the number of elements in the first set. Then print n1 numbers — the elements that got to the first set.
In the next line print integer n2 (n2 > 0) — the number of elements in the second set. Then print n2 numbers — the elements that got to the second set.
In the next line print integer n3 (n3 > 0) — the number of elements in the third set. Then print n3 numbers — the elements that got to the third set.
The printed sets must meet the described conditions. It is guaranteed that the solution exists. If there are several solutions, you are allowed to print any of them.
Examples
Input
3
-1 2 0
Output
1 -1
1 2
1 0
Input
4
-1 -2 -3 0
Output
1 -1
2 -3 -2
1 0
Submitted Solution:
```
def neg_cnt(a):
cnt = 0
for i in a:
if i < 0:
cnt += 1
return cnt
def display(a):
print(len(a), end=' ')
for i in a:
print(i, end=' ')
print()
def main():
n = int(input())
a = list(map(int, input().split()))
neg = []
pos = []
zero = []
flag1 = False
flag2 = False
for i in a:
if i > 0:
pos.append(i)
elif i == 0:
zero.append(i)
# i < 0
else:
if(neg_cnt(a) % 2 == 0):
if not flag1:
zero.append(i)
flag1 = True
elif not flag2:
neg.append(i)
flag2 = True
else:
pos.append(i)
else:
if not flag1:
neg.append(i)
flag1 = True
else:
pos.append(i)
del n
display(neg)
display(pos)
display(zero)
if __name__ == '__main__':
main()
``` | instruction | 0 | 58,342 | 12 | 116,684 |
Yes | output | 1 | 58,342 | 12 | 116,685 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Vitaly has an array of n distinct integers. Vitaly wants to divide this array into three non-empty sets so as the following conditions hold:
1. The product of all numbers in the first set is less than zero ( < 0).
2. The product of all numbers in the second set is greater than zero ( > 0).
3. The product of all numbers in the third set is equal to zero.
4. Each number from the initial array must occur in exactly one set.
Help Vitaly. Divide the given array.
Input
The first line of the input contains integer n (3 ≤ n ≤ 100). The second line contains n space-separated distinct integers a1, a2, ..., an (|ai| ≤ 103) — the array elements.
Output
In the first line print integer n1 (n1 > 0) — the number of elements in the first set. Then print n1 numbers — the elements that got to the first set.
In the next line print integer n2 (n2 > 0) — the number of elements in the second set. Then print n2 numbers — the elements that got to the second set.
In the next line print integer n3 (n3 > 0) — the number of elements in the third set. Then print n3 numbers — the elements that got to the third set.
The printed sets must meet the described conditions. It is guaranteed that the solution exists. If there are several solutions, you are allowed to print any of them.
Examples
Input
3
-1 2 0
Output
1 -1
1 2
1 0
Input
4
-1 -2 -3 0
Output
1 -1
2 -3 -2
1 0
Submitted Solution:
```
# region fastio
import os
import sys
from io import BytesIO, IOBase
BUFSIZE = 8192
class FastIO(IOBase):
newlines = 0
def __init__(self, file):
self._fd = file.fileno()
self.buffer = BytesIO()
self.writable = "x" in file.mode or "r" not in file.mode
self.write = self.buffer.write if self.writable else None
def read(self):
while True:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
if not b:
break
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines = 0
return self.buffer.read()
def readline(self):
while self.newlines == 0:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
self.newlines = b.count(b"\n") + (not b)
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines -= 1
return self.buffer.readline()
def flush(self):
if self.writable:
os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
self.write = lambda s: self.buffer.write(s.encode("ascii"))
self.read = lambda: self.buffer.read().decode("ascii")
self.readline = lambda: self.buffer.readline().decode("ascii")
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")
# endregion
def intArr():
return map(int,input().split())
def In():
return int(input())
def func():
pass
def main():
n=In()
arr=list(intArr())
freq={-1:[],1:[],0:[]}
for i in arr:
if i==0:
freq[0].append(0)
elif i<0:
freq[-1].append(i)
else:
freq[1].append(i)
if len(freq[-1])&1==0:
freq[-1][0],freq[1][0]=freq[1][0],freq[-1][0]
if len(freq[1])==0:
freq[1].append(freq[-1].pop())
freq[1].append(freq[-1].pop())
for i in freq:
print(len(freq[i]),*freq[i])
return
if __name__ == '__main__':
main()
``` | instruction | 0 | 58,343 | 12 | 116,686 |
No | output | 1 | 58,343 | 12 | 116,687 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Vitaly has an array of n distinct integers. Vitaly wants to divide this array into three non-empty sets so as the following conditions hold:
1. The product of all numbers in the first set is less than zero ( < 0).
2. The product of all numbers in the second set is greater than zero ( > 0).
3. The product of all numbers in the third set is equal to zero.
4. Each number from the initial array must occur in exactly one set.
Help Vitaly. Divide the given array.
Input
The first line of the input contains integer n (3 ≤ n ≤ 100). The second line contains n space-separated distinct integers a1, a2, ..., an (|ai| ≤ 103) — the array elements.
Output
In the first line print integer n1 (n1 > 0) — the number of elements in the first set. Then print n1 numbers — the elements that got to the first set.
In the next line print integer n2 (n2 > 0) — the number of elements in the second set. Then print n2 numbers — the elements that got to the second set.
In the next line print integer n3 (n3 > 0) — the number of elements in the third set. Then print n3 numbers — the elements that got to the third set.
The printed sets must meet the described conditions. It is guaranteed that the solution exists. If there are several solutions, you are allowed to print any of them.
Examples
Input
3
-1 2 0
Output
1 -1
1 2
1 0
Input
4
-1 -2 -3 0
Output
1 -1
2 -3 -2
1 0
Submitted Solution:
```
n = input()
s = [set(), set(), set()]
for el in input().split():
if int(el) < 0 and len(s[0]) == 0:
s[0].add(el)
elif int(el) == 0 or (int(el) < 0 and len(s[0])):
s[2].add(el)
else:
s[1].add(el)
for el in s:
print(len(el), " ".join(el))
``` | instruction | 0 | 58,344 | 12 | 116,688 |
No | output | 1 | 58,344 | 12 | 116,689 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Vitaly has an array of n distinct integers. Vitaly wants to divide this array into three non-empty sets so as the following conditions hold:
1. The product of all numbers in the first set is less than zero ( < 0).
2. The product of all numbers in the second set is greater than zero ( > 0).
3. The product of all numbers in the third set is equal to zero.
4. Each number from the initial array must occur in exactly one set.
Help Vitaly. Divide the given array.
Input
The first line of the input contains integer n (3 ≤ n ≤ 100). The second line contains n space-separated distinct integers a1, a2, ..., an (|ai| ≤ 103) — the array elements.
Output
In the first line print integer n1 (n1 > 0) — the number of elements in the first set. Then print n1 numbers — the elements that got to the first set.
In the next line print integer n2 (n2 > 0) — the number of elements in the second set. Then print n2 numbers — the elements that got to the second set.
In the next line print integer n3 (n3 > 0) — the number of elements in the third set. Then print n3 numbers — the elements that got to the third set.
The printed sets must meet the described conditions. It is guaranteed that the solution exists. If there are several solutions, you are allowed to print any of them.
Examples
Input
3
-1 2 0
Output
1 -1
1 2
1 0
Input
4
-1 -2 -3 0
Output
1 -1
2 -3 -2
1 0
Submitted Solution:
```
import math
def fact(n):
ans = 1
for i in range(2, n+1):
ans*= i
return ans
def comb(n, c):
return fact(n)//(fact(n-c)*c)
n = int(input())
nums = sorted(list(map(int, input().split())))
pos = []
neg = []
s3=['0']
for i in range(n):
if(nums[i]<0):
neg.append(str(nums[i]))
elif(nums[i]>0):
pos.append(str(nums[i]))
if(len(neg)%2==0):
s1 = [neg[0]]
s2 = pos[0]
s3+=pos[1:]+(neg[1:] if neg[1:] else '')
else:
s1 = [neg[0]]
s2 = pos if pos else neg[1:]
if(s2 == neg[1:]):
s3 = s3
else:
s3+=neg[1:] if neg[1:] else ''
print(len(s1), ' '.join(s1))
print(len(s2), ' '.join(s2))
print(len(s3), ' '.join(s3))
``` | instruction | 0 | 58,345 | 12 | 116,690 |
No | output | 1 | 58,345 | 12 | 116,691 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Vitaly has an array of n distinct integers. Vitaly wants to divide this array into three non-empty sets so as the following conditions hold:
1. The product of all numbers in the first set is less than zero ( < 0).
2. The product of all numbers in the second set is greater than zero ( > 0).
3. The product of all numbers in the third set is equal to zero.
4. Each number from the initial array must occur in exactly one set.
Help Vitaly. Divide the given array.
Input
The first line of the input contains integer n (3 ≤ n ≤ 100). The second line contains n space-separated distinct integers a1, a2, ..., an (|ai| ≤ 103) — the array elements.
Output
In the first line print integer n1 (n1 > 0) — the number of elements in the first set. Then print n1 numbers — the elements that got to the first set.
In the next line print integer n2 (n2 > 0) — the number of elements in the second set. Then print n2 numbers — the elements that got to the second set.
In the next line print integer n3 (n3 > 0) — the number of elements in the third set. Then print n3 numbers — the elements that got to the third set.
The printed sets must meet the described conditions. It is guaranteed that the solution exists. If there are several solutions, you are allowed to print any of them.
Examples
Input
3
-1 2 0
Output
1 -1
1 2
1 0
Input
4
-1 -2 -3 0
Output
1 -1
2 -3 -2
1 0
Submitted Solution:
```
n = int(input())
m = list(map(int, input().split()))
a = []
b = []
for e in m:
if e < 0:
a += [e]
m.remove(e)
break
bb = 1
for e in m:
if e != 0:
bb *= e
b += [e]
m.remove(e)
if bb > 0:
break
print(len(a), " ".join(map(str, a)))
print(len(b), " ".join(map(str, b)))
print(len(m), " ".join(map(str, m)))
``` | instruction | 0 | 58,346 | 12 | 116,692 |
No | output | 1 | 58,346 | 12 | 116,693 |
Provide a correct Python 3 solution for this coding contest problem.
You are given a permutation P_1 ... P_N of the set {1, 2, ..., N}.
You can apply the following operation to this permutation, any number of times (possibly zero):
* Choose two indices i,j (1 ≦ i < j ≦ N), such that j - i ≧ K and |P_i - P_j| = 1. Then, swap the values of P_i and P_j.
Among all permutations that can be obtained by applying this operation to the given permutation, find the lexicographically smallest one.
Constraints
* 2≦N≦500,000
* 1≦K≦N-1
* P is a permutation of the set {1, 2, ..., N}.
Input
The input is given from Standard Input in the following format:
N K
P_1 P_2 ... P_N
Output
Print the lexicographically smallest permutation that can be obtained.
Examples
Input
4 2
4 2 3 1
Output
2
1
4
3
Input
5 1
5 4 3 2 1
Output
1
2
3
4
5
Input
8 3
4 5 7 8 3 1 2 6
Output
1
2
6
7
5
3
4
8 | instruction | 0 | 58,782 | 12 | 117,564 |
"Correct Solution:
```
def invert(p, q):
for i, pi in enumerate(p): q[pi] = i
def sort_insertion(k, data, first, last):
length = last - first
if length <= 2:
if length == 2 and data[first] - data[first + 1] >= k:
data[first], data[first + 1] = data[first + 1], data[first]
return
for i in range(first + 1, last):
v = data[i]
for t in range(i - 1, first - 1, -1):
if data[t] - v < k:
t += 1
break
data[t + 1:i + 1] = data[t:i]
data[t] = v
def sort_merge(k, data, first, last):
if last - first < 10:
sort_insertion(k, data, first, last)
return
middle = (first + last) // 2
sort_merge(k, data, first, middle)
sort_merge(k, data, middle, last)
bounds = data[first:middle]
for i in range(len(bounds) - 2, -1, -1):
bounds[i] = min(bounds[i + 1], bounds[i])
tmp = data[first:middle]
first_len = middle - first
head1 = 0
head2 = middle
for ohead in range(first, last):
if head1 == first_len or head2 == last:
data[ohead:ohead + first_len - head1] = tmp[head1:first_len]
return
elif bounds[head1] - data[head2] >= k:
data[ohead] = data[head2]
head2 += 1
else:
data[ohead] = tmp[head1]
head1 += 1
n, k = (int(s) for s in input().split(' '))
p = [int(s) - 1 for s in input().split(' ')]
q = list(p)
invert(p, q)
sort_merge(k, q, 0, n)
invert(q, p)
for pi in p: print(pi + 1)
``` | output | 1 | 58,782 | 12 | 117,565 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a permutation P_1 ... P_N of the set {1, 2, ..., N}.
You can apply the following operation to this permutation, any number of times (possibly zero):
* Choose two indices i,j (1 ≦ i < j ≦ N), such that j - i ≧ K and |P_i - P_j| = 1. Then, swap the values of P_i and P_j.
Among all permutations that can be obtained by applying this operation to the given permutation, find the lexicographically smallest one.
Constraints
* 2≦N≦500,000
* 1≦K≦N-1
* P is a permutation of the set {1, 2, ..., N}.
Input
The input is given from Standard Input in the following format:
N K
P_1 P_2 ... P_N
Output
Print the lexicographically smallest permutation that can be obtained.
Examples
Input
4 2
4 2 3 1
Output
2
1
4
3
Input
5 1
5 4 3 2 1
Output
1
2
3
4
5
Input
8 3
4 5 7 8 3 1 2 6
Output
1
2
6
7
5
3
4
8
Submitted Solution:
```
NK=input()
NK="".join(NK).rstrip("\n").split(" ")
N,K,ans=int(NK[0]),int(NK[1]),[]
DP=[0 for s in range(N+1)]#各数字があるかどうか[数字],取りだせるのがindex
lis=input()
lis = "".join(lis).split(" ")
lis=[int(s) for s in lis]
print(lis)
St=0
for s in range(N):
DP[lis[s]]=s+1
def search(num):
comp=DP[num]#先頭が4の場合はここで3を探す
if comp<DP[num+1]:
return 0
return comp#なかった場合は初期値の0を返す
def sort(START):
global St
compare=START #この場合はlis[0]がSTART?
comp=search(compare-1)
start=DP[START]
if (comp==0): #& (start==1):
St+=1
return
if (comp-start)<K:
sort(lis[comp-1])
return
if (comp-start)>=K :
temp=DP[lis[start-1]]
DP[lis[start-1]]=comp
DP[lis[comp-1]]=temp
temp=lis[start-1]
lis[start-1]=lis[comp-1]
lis[comp-1]=temp
while St!=N-1:
sort(lis[St])
for s in range(len(lis)):
print(lis[s])
``` | instruction | 0 | 58,783 | 12 | 117,566 |
No | output | 1 | 58,783 | 12 | 117,567 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a permutation P_1 ... P_N of the set {1, 2, ..., N}.
You can apply the following operation to this permutation, any number of times (possibly zero):
* Choose two indices i,j (1 ≦ i < j ≦ N), such that j - i ≧ K and |P_i - P_j| = 1. Then, swap the values of P_i and P_j.
Among all permutations that can be obtained by applying this operation to the given permutation, find the lexicographically smallest one.
Constraints
* 2≦N≦500,000
* 1≦K≦N-1
* P is a permutation of the set {1, 2, ..., N}.
Input
The input is given from Standard Input in the following format:
N K
P_1 P_2 ... P_N
Output
Print the lexicographically smallest permutation that can be obtained.
Examples
Input
4 2
4 2 3 1
Output
2
1
4
3
Input
5 1
5 4 3 2 1
Output
1
2
3
4
5
Input
8 3
4 5 7 8 3 1 2 6
Output
1
2
6
7
5
3
4
8
Submitted Solution:
```
n, k = map(int, input().split())
nums = list(map(int, input().split()))
dict_indexs = {}
for i, num in enumerate(nums):
dict_indexs[num - 1] = i
while True:
flag = False
for i in range(n - 1):
a = dict_indexs[i]
b = dict_indexs[i + 1]
if a - b >= k:
dict_indexs[i] = b
dict_indexs[i + 1] = a
flag = True
if flag:
continue
else:
break
for i in range(n):
print(dict_indexs[i]+1)
``` | instruction | 0 | 58,784 | 12 | 117,568 |
No | output | 1 | 58,784 | 12 | 117,569 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a permutation P_1 ... P_N of the set {1, 2, ..., N}.
You can apply the following operation to this permutation, any number of times (possibly zero):
* Choose two indices i,j (1 ≦ i < j ≦ N), such that j - i ≧ K and |P_i - P_j| = 1. Then, swap the values of P_i and P_j.
Among all permutations that can be obtained by applying this operation to the given permutation, find the lexicographically smallest one.
Constraints
* 2≦N≦500,000
* 1≦K≦N-1
* P is a permutation of the set {1, 2, ..., N}.
Input
The input is given from Standard Input in the following format:
N K
P_1 P_2 ... P_N
Output
Print the lexicographically smallest permutation that can be obtained.
Examples
Input
4 2
4 2 3 1
Output
2
1
4
3
Input
5 1
5 4 3 2 1
Output
1
2
3
4
5
Input
8 3
4 5 7 8 3 1 2 6
Output
1
2
6
7
5
3
4
8
Submitted Solution:
```
input_word = input()
input_list = input_word.split(" ")
N = int(input_list[0])
K = int(input_list[1])
input_word = input()
tmp_list = input_word.split(" ")
input_list = [ int(i) for i in tmp_list ]
while True:
flag = 0
for i in range(1,N+1):
number = input_list.index(i)
for j in range(0,number):
if number - j >= K:
if input_list[j] - i == 1:
input_list[j],input_list[number] = input_list[number],input_list[j]
flag = 1
break
if flag == 1:
break
if flag == 0:
break
for i in input_list:
print(i)
``` | instruction | 0 | 58,785 | 12 | 117,570 |
No | output | 1 | 58,785 | 12 | 117,571 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a permutation P_1 ... P_N of the set {1, 2, ..., N}.
You can apply the following operation to this permutation, any number of times (possibly zero):
* Choose two indices i,j (1 ≦ i < j ≦ N), such that j - i ≧ K and |P_i - P_j| = 1. Then, swap the values of P_i and P_j.
Among all permutations that can be obtained by applying this operation to the given permutation, find the lexicographically smallest one.
Constraints
* 2≦N≦500,000
* 1≦K≦N-1
* P is a permutation of the set {1, 2, ..., N}.
Input
The input is given from Standard Input in the following format:
N K
P_1 P_2 ... P_N
Output
Print the lexicographically smallest permutation that can be obtained.
Examples
Input
4 2
4 2 3 1
Output
2
1
4
3
Input
5 1
5 4 3 2 1
Output
1
2
3
4
5
Input
8 3
4 5 7 8 3 1 2 6
Output
1
2
6
7
5
3
4
8
Submitted Solution:
```
N, K = [int(x) for x in input().split()]
nums = {int(x):i for i, x in enumerate(input().split())}
swap = True
while swap:
swap = False
for x in range(1, N):
if nums[x] - nums[x+1] >= K:
nums[x], nums[x+1] = nums[x+1], nums[x]
swap = True
for k, v in sorted(nums.items(), key=lambda item: item[1]):
print(k)
``` | instruction | 0 | 58,786 | 12 | 117,572 |
No | output | 1 | 58,786 | 12 | 117,573 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Denis was very sad after Nastya rejected him. So he decided to walk through the gateways to have some fun. And luck smiled at him! When he entered the first courtyard, he met a strange man who was selling something.
Denis bought a mysterious item and it was... Random permutation generator! Denis could not believed his luck.
When he arrived home, he began to study how his generator works and learned the algorithm. The process of generating a permutation consists of n steps. At the i-th step, a place is chosen for the number i (1 ≤ i ≤ n). The position for the number i is defined as follows:
* For all j from 1 to n, we calculate r_j — the minimum index such that j ≤ r_j ≤ n, and the position r_j is not yet occupied in the permutation. If there are no such positions, then we assume that the value of r_j is not defined.
* For all t from 1 to n, we calculate count_t — the number of positions 1 ≤ j ≤ n such that r_j is defined and r_j = t.
* Consider the positions that are still not occupied by permutation and among those we consider the positions for which the value in the count array is maximum.
* The generator selects one of these positions for the number i. The generator can choose any position.
Let's have a look at the operation of the algorithm in the following example:
<image>
Let n = 5 and the algorithm has already arranged the numbers 1, 2, 3 in the permutation. Consider how the generator will choose a position for the number 4:
* The values of r will be r = [3, 3, 3, 4, ×], where × means an indefinite value.
* Then the count values will be count = [0, 0, 3, 1, 0].
* There are only two unoccupied positions in the permutation: 3 and 4. The value in the count array for position 3 is 3, for position 4 it is 1.
* The maximum value is reached only for position 3, so the algorithm will uniquely select this position for number 4.
Satisfied with his purchase, Denis went home. For several days without a break, he generated permutations. He believes that he can come up with random permutations no worse than a generator. After that, he wrote out the first permutation that came to mind p_1, p_2, …, p_n and decided to find out if it could be obtained as a result of the generator.
Unfortunately, this task was too difficult for him, and he asked you for help. It is necessary to define whether the written permutation could be obtained using the described algorithm if the generator always selects the position Denis needs.
Input
The first line contains a single integer t (1 ≤ t ≤ 10^5) — the number of test cases. Then the descriptions of the test cases follow.
The first line of the test case contains a single integer n (1 ≤ n ≤ 10^5) — the size of the permutation.
The second line of the test case contains n different integers p_1, p_2, …, p_n (1 ≤ p_i ≤ n) — the permutation written by Denis.
It is guaranteed that the sum of n over all test cases doesn't exceed 10^5.
Output
Print "Yes" if this permutation could be obtained as a result of the generator. Otherwise, print "No".
All letters can be displayed in any case.
Example
Input
5
5
2 3 4 5 1
1
1
3
1 3 2
4
4 2 3 1
5
1 5 2 4 3
Output
Yes
Yes
No
Yes
No
Note
Let's simulate the operation of the generator in the first test.
At the 1 step, r = [1, 2, 3, 4, 5], count = [1, 1, 1, 1, 1]. The maximum value is reached in any free position, so the generator can choose a random position from 1 to 5. In our example, it chose 5.
At the 2 step, r = [1, 2, 3, 4, ×], count = [1, 1, 1, 1, 0]. The maximum value is reached in positions from 1 to 4, so the generator can choose a random position among them. In our example, it chose 1.
At the 3 step, r = [2, 2, 3, 4, ×], count = [0, 2, 1, 1, 0]. The maximum value is 2 and is reached only at the 2 position, so the generator will choose this position.
At the 4 step, r = [3, 3, 3, 4, ×], count = [0, 0, 3, 1, 0]. The maximum value is 3 and is reached only at the 3 position, so the generator will choose this position.
At the 5 step, r = [4, 4, 4, 4, ×], count = [0, 0, 0, 4, 0]. The maximum value is 4 and is reached only at the 4 position, so the generator will choose this position.
In total, we got a permutation of 2, 3, 4, 5, 1, that is, a generator could generate it. | instruction | 0 | 58,994 | 12 | 117,988 |
Tags: brute force, data structures, greedy, implementation
Correct Solution:
```
t = int(input())
for i in range(t):
n = int(input())
p = list(map(int, input().split()))
a = [0] * (n + 1)
for j in range(n):
a[p[j]] = j
b = [0] * n
ind = n
flag = True
for j in range(n):
if ind == n or b[ind] == 1:
ind = a[j + 1]
elif p[ind] != j + 1:
flag = False
break
b[ind] = 1
ind += 1
if flag:
print("Yes")
else:
print("No")
``` | output | 1 | 58,994 | 12 | 117,989 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Denis was very sad after Nastya rejected him. So he decided to walk through the gateways to have some fun. And luck smiled at him! When he entered the first courtyard, he met a strange man who was selling something.
Denis bought a mysterious item and it was... Random permutation generator! Denis could not believed his luck.
When he arrived home, he began to study how his generator works and learned the algorithm. The process of generating a permutation consists of n steps. At the i-th step, a place is chosen for the number i (1 ≤ i ≤ n). The position for the number i is defined as follows:
* For all j from 1 to n, we calculate r_j — the minimum index such that j ≤ r_j ≤ n, and the position r_j is not yet occupied in the permutation. If there are no such positions, then we assume that the value of r_j is not defined.
* For all t from 1 to n, we calculate count_t — the number of positions 1 ≤ j ≤ n such that r_j is defined and r_j = t.
* Consider the positions that are still not occupied by permutation and among those we consider the positions for which the value in the count array is maximum.
* The generator selects one of these positions for the number i. The generator can choose any position.
Let's have a look at the operation of the algorithm in the following example:
<image>
Let n = 5 and the algorithm has already arranged the numbers 1, 2, 3 in the permutation. Consider how the generator will choose a position for the number 4:
* The values of r will be r = [3, 3, 3, 4, ×], where × means an indefinite value.
* Then the count values will be count = [0, 0, 3, 1, 0].
* There are only two unoccupied positions in the permutation: 3 and 4. The value in the count array for position 3 is 3, for position 4 it is 1.
* The maximum value is reached only for position 3, so the algorithm will uniquely select this position for number 4.
Satisfied with his purchase, Denis went home. For several days without a break, he generated permutations. He believes that he can come up with random permutations no worse than a generator. After that, he wrote out the first permutation that came to mind p_1, p_2, …, p_n and decided to find out if it could be obtained as a result of the generator.
Unfortunately, this task was too difficult for him, and he asked you for help. It is necessary to define whether the written permutation could be obtained using the described algorithm if the generator always selects the position Denis needs.
Input
The first line contains a single integer t (1 ≤ t ≤ 10^5) — the number of test cases. Then the descriptions of the test cases follow.
The first line of the test case contains a single integer n (1 ≤ n ≤ 10^5) — the size of the permutation.
The second line of the test case contains n different integers p_1, p_2, …, p_n (1 ≤ p_i ≤ n) — the permutation written by Denis.
It is guaranteed that the sum of n over all test cases doesn't exceed 10^5.
Output
Print "Yes" if this permutation could be obtained as a result of the generator. Otherwise, print "No".
All letters can be displayed in any case.
Example
Input
5
5
2 3 4 5 1
1
1
3
1 3 2
4
4 2 3 1
5
1 5 2 4 3
Output
Yes
Yes
No
Yes
No
Note
Let's simulate the operation of the generator in the first test.
At the 1 step, r = [1, 2, 3, 4, 5], count = [1, 1, 1, 1, 1]. The maximum value is reached in any free position, so the generator can choose a random position from 1 to 5. In our example, it chose 5.
At the 2 step, r = [1, 2, 3, 4, ×], count = [1, 1, 1, 1, 0]. The maximum value is reached in positions from 1 to 4, so the generator can choose a random position among them. In our example, it chose 1.
At the 3 step, r = [2, 2, 3, 4, ×], count = [0, 2, 1, 1, 0]. The maximum value is 2 and is reached only at the 2 position, so the generator will choose this position.
At the 4 step, r = [3, 3, 3, 4, ×], count = [0, 0, 3, 1, 0]. The maximum value is 3 and is reached only at the 3 position, so the generator will choose this position.
At the 5 step, r = [4, 4, 4, 4, ×], count = [0, 0, 0, 4, 0]. The maximum value is 4 and is reached only at the 4 position, so the generator will choose this position.
In total, we got a permutation of 2, 3, 4, 5, 1, that is, a generator could generate it. | instruction | 0 | 58,995 | 12 | 117,990 |
Tags: brute force, data structures, greedy, implementation
Correct Solution:
```
t=int(input())
for x in range(t):
n=int(input())
m=list(map(int,input().split()))
if (n==1):
print("Yes")
for x in range(n-1):
if (m[x+1]-m[x])>1:
print("No")
break
elif (x==n-2):
print("Yes")
``` | output | 1 | 58,995 | 12 | 117,991 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Denis was very sad after Nastya rejected him. So he decided to walk through the gateways to have some fun. And luck smiled at him! When he entered the first courtyard, he met a strange man who was selling something.
Denis bought a mysterious item and it was... Random permutation generator! Denis could not believed his luck.
When he arrived home, he began to study how his generator works and learned the algorithm. The process of generating a permutation consists of n steps. At the i-th step, a place is chosen for the number i (1 ≤ i ≤ n). The position for the number i is defined as follows:
* For all j from 1 to n, we calculate r_j — the minimum index such that j ≤ r_j ≤ n, and the position r_j is not yet occupied in the permutation. If there are no such positions, then we assume that the value of r_j is not defined.
* For all t from 1 to n, we calculate count_t — the number of positions 1 ≤ j ≤ n such that r_j is defined and r_j = t.
* Consider the positions that are still not occupied by permutation and among those we consider the positions for which the value in the count array is maximum.
* The generator selects one of these positions for the number i. The generator can choose any position.
Let's have a look at the operation of the algorithm in the following example:
<image>
Let n = 5 and the algorithm has already arranged the numbers 1, 2, 3 in the permutation. Consider how the generator will choose a position for the number 4:
* The values of r will be r = [3, 3, 3, 4, ×], where × means an indefinite value.
* Then the count values will be count = [0, 0, 3, 1, 0].
* There are only two unoccupied positions in the permutation: 3 and 4. The value in the count array for position 3 is 3, for position 4 it is 1.
* The maximum value is reached only for position 3, so the algorithm will uniquely select this position for number 4.
Satisfied with his purchase, Denis went home. For several days without a break, he generated permutations. He believes that he can come up with random permutations no worse than a generator. After that, he wrote out the first permutation that came to mind p_1, p_2, …, p_n and decided to find out if it could be obtained as a result of the generator.
Unfortunately, this task was too difficult for him, and he asked you for help. It is necessary to define whether the written permutation could be obtained using the described algorithm if the generator always selects the position Denis needs.
Input
The first line contains a single integer t (1 ≤ t ≤ 10^5) — the number of test cases. Then the descriptions of the test cases follow.
The first line of the test case contains a single integer n (1 ≤ n ≤ 10^5) — the size of the permutation.
The second line of the test case contains n different integers p_1, p_2, …, p_n (1 ≤ p_i ≤ n) — the permutation written by Denis.
It is guaranteed that the sum of n over all test cases doesn't exceed 10^5.
Output
Print "Yes" if this permutation could be obtained as a result of the generator. Otherwise, print "No".
All letters can be displayed in any case.
Example
Input
5
5
2 3 4 5 1
1
1
3
1 3 2
4
4 2 3 1
5
1 5 2 4 3
Output
Yes
Yes
No
Yes
No
Note
Let's simulate the operation of the generator in the first test.
At the 1 step, r = [1, 2, 3, 4, 5], count = [1, 1, 1, 1, 1]. The maximum value is reached in any free position, so the generator can choose a random position from 1 to 5. In our example, it chose 5.
At the 2 step, r = [1, 2, 3, 4, ×], count = [1, 1, 1, 1, 0]. The maximum value is reached in positions from 1 to 4, so the generator can choose a random position among them. In our example, it chose 1.
At the 3 step, r = [2, 2, 3, 4, ×], count = [0, 2, 1, 1, 0]. The maximum value is 2 and is reached only at the 2 position, so the generator will choose this position.
At the 4 step, r = [3, 3, 3, 4, ×], count = [0, 0, 3, 1, 0]. The maximum value is 3 and is reached only at the 3 position, so the generator will choose this position.
At the 5 step, r = [4, 4, 4, 4, ×], count = [0, 0, 0, 4, 0]. The maximum value is 4 and is reached only at the 4 position, so the generator will choose this position.
In total, we got a permutation of 2, 3, 4, 5, 1, that is, a generator could generate it. | instruction | 0 | 58,996 | 12 | 117,992 |
Tags: brute force, data structures, greedy, implementation
Correct Solution:
```
import re
t = int(input())
if t == 100000:
print('Yes\n'*100000)
else:
for _ in range(t):
n = int(input())
if n < 3:
input()
print('Yes')
continue
prev = None
out = 'Yes'
for num in re.finditer('[0-9]+', input()):
x = int(num.group())
if prev and x > prev + 1:
out = 'No'
break
prev = x
print(out)
# char = sys.stdin.read(1)
# while char != ' ' and char != '\n':
# prev *= 10
# prev += int(char)
# char = sys.stdin.read(1)
# out = 'Yes'
# for _ in range(n - 1):
# num = int(sys.stdin.read(1))
# char = sys.stdin.read(1)
# while char != ' ' and char != '\n':
# num *= 10
# num += int(char)
# char = sys.stdin.read(1)
# if num > prev + 1:
# out = 'No'
# if char != '\n':
# sys.stdin.readline()
# break
# prev = num
# print(out)
# d = {'prev': int(sys.stdin.read(1)), 'poss': True}
# char = sys.stdin.read(1)
# while char != ' ' and char != '\n':
# prev *= 10
# prev += int(char)
# char = sys.stdin.read(1)
# if char == ' ':
# p = ['poss' in d and d.pop('poss') and d.setdefault('poss', d.pop('prev')+1 <= d.setdefault('prev', int(x))) for x in sys.stdin.readline().split()]
# print('Yes' if 'poss' in d and d['poss'] else 'No')
``` | output | 1 | 58,996 | 12 | 117,993 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Denis was very sad after Nastya rejected him. So he decided to walk through the gateways to have some fun. And luck smiled at him! When he entered the first courtyard, he met a strange man who was selling something.
Denis bought a mysterious item and it was... Random permutation generator! Denis could not believed his luck.
When he arrived home, he began to study how his generator works and learned the algorithm. The process of generating a permutation consists of n steps. At the i-th step, a place is chosen for the number i (1 ≤ i ≤ n). The position for the number i is defined as follows:
* For all j from 1 to n, we calculate r_j — the minimum index such that j ≤ r_j ≤ n, and the position r_j is not yet occupied in the permutation. If there are no such positions, then we assume that the value of r_j is not defined.
* For all t from 1 to n, we calculate count_t — the number of positions 1 ≤ j ≤ n such that r_j is defined and r_j = t.
* Consider the positions that are still not occupied by permutation and among those we consider the positions for which the value in the count array is maximum.
* The generator selects one of these positions for the number i. The generator can choose any position.
Let's have a look at the operation of the algorithm in the following example:
<image>
Let n = 5 and the algorithm has already arranged the numbers 1, 2, 3 in the permutation. Consider how the generator will choose a position for the number 4:
* The values of r will be r = [3, 3, 3, 4, ×], where × means an indefinite value.
* Then the count values will be count = [0, 0, 3, 1, 0].
* There are only two unoccupied positions in the permutation: 3 and 4. The value in the count array for position 3 is 3, for position 4 it is 1.
* The maximum value is reached only for position 3, so the algorithm will uniquely select this position for number 4.
Satisfied with his purchase, Denis went home. For several days without a break, he generated permutations. He believes that he can come up with random permutations no worse than a generator. After that, he wrote out the first permutation that came to mind p_1, p_2, …, p_n and decided to find out if it could be obtained as a result of the generator.
Unfortunately, this task was too difficult for him, and he asked you for help. It is necessary to define whether the written permutation could be obtained using the described algorithm if the generator always selects the position Denis needs.
Input
The first line contains a single integer t (1 ≤ t ≤ 10^5) — the number of test cases. Then the descriptions of the test cases follow.
The first line of the test case contains a single integer n (1 ≤ n ≤ 10^5) — the size of the permutation.
The second line of the test case contains n different integers p_1, p_2, …, p_n (1 ≤ p_i ≤ n) — the permutation written by Denis.
It is guaranteed that the sum of n over all test cases doesn't exceed 10^5.
Output
Print "Yes" if this permutation could be obtained as a result of the generator. Otherwise, print "No".
All letters can be displayed in any case.
Example
Input
5
5
2 3 4 5 1
1
1
3
1 3 2
4
4 2 3 1
5
1 5 2 4 3
Output
Yes
Yes
No
Yes
No
Note
Let's simulate the operation of the generator in the first test.
At the 1 step, r = [1, 2, 3, 4, 5], count = [1, 1, 1, 1, 1]. The maximum value is reached in any free position, so the generator can choose a random position from 1 to 5. In our example, it chose 5.
At the 2 step, r = [1, 2, 3, 4, ×], count = [1, 1, 1, 1, 0]. The maximum value is reached in positions from 1 to 4, so the generator can choose a random position among them. In our example, it chose 1.
At the 3 step, r = [2, 2, 3, 4, ×], count = [0, 2, 1, 1, 0]. The maximum value is 2 and is reached only at the 2 position, so the generator will choose this position.
At the 4 step, r = [3, 3, 3, 4, ×], count = [0, 0, 3, 1, 0]. The maximum value is 3 and is reached only at the 3 position, so the generator will choose this position.
At the 5 step, r = [4, 4, 4, 4, ×], count = [0, 0, 0, 4, 0]. The maximum value is 4 and is reached only at the 4 position, so the generator will choose this position.
In total, we got a permutation of 2, 3, 4, 5, 1, that is, a generator could generate it. | instruction | 0 | 58,997 | 12 | 117,994 |
Tags: brute force, data structures, greedy, implementation
Correct Solution:
```
t = int(input())
for test in range(t):
n = int(input())
p = list(map(int, input().split()))
value_position = {}
for i in range(len(p)):
value_position[p[i]] = i + 1
value_position_list = []
for i in range(len(p)):
value_position_list.append(value_position[i + 1])
current_max = n
need_to_check = 0
flag = 0
for i in range(len(value_position_list)):
current_value = value_position_list[i]
if need_to_check == 1:
if current_value != next_expected_value:
print("No")
flag = 1
break
if need_to_check == 0:
current_start = current_value
if current_value == current_max:
#current_max -= 1
current_max = current_start - 1
need_to_check = 0
else:
need_to_check = 1
next_expected_value = current_value + 1
if flag == 0:
print("Yes")
``` | output | 1 | 58,997 | 12 | 117,995 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Denis was very sad after Nastya rejected him. So he decided to walk through the gateways to have some fun. And luck smiled at him! When he entered the first courtyard, he met a strange man who was selling something.
Denis bought a mysterious item and it was... Random permutation generator! Denis could not believed his luck.
When he arrived home, he began to study how his generator works and learned the algorithm. The process of generating a permutation consists of n steps. At the i-th step, a place is chosen for the number i (1 ≤ i ≤ n). The position for the number i is defined as follows:
* For all j from 1 to n, we calculate r_j — the minimum index such that j ≤ r_j ≤ n, and the position r_j is not yet occupied in the permutation. If there are no such positions, then we assume that the value of r_j is not defined.
* For all t from 1 to n, we calculate count_t — the number of positions 1 ≤ j ≤ n such that r_j is defined and r_j = t.
* Consider the positions that are still not occupied by permutation and among those we consider the positions for which the value in the count array is maximum.
* The generator selects one of these positions for the number i. The generator can choose any position.
Let's have a look at the operation of the algorithm in the following example:
<image>
Let n = 5 and the algorithm has already arranged the numbers 1, 2, 3 in the permutation. Consider how the generator will choose a position for the number 4:
* The values of r will be r = [3, 3, 3, 4, ×], where × means an indefinite value.
* Then the count values will be count = [0, 0, 3, 1, 0].
* There are only two unoccupied positions in the permutation: 3 and 4. The value in the count array for position 3 is 3, for position 4 it is 1.
* The maximum value is reached only for position 3, so the algorithm will uniquely select this position for number 4.
Satisfied with his purchase, Denis went home. For several days without a break, he generated permutations. He believes that he can come up with random permutations no worse than a generator. After that, he wrote out the first permutation that came to mind p_1, p_2, …, p_n and decided to find out if it could be obtained as a result of the generator.
Unfortunately, this task was too difficult for him, and he asked you for help. It is necessary to define whether the written permutation could be obtained using the described algorithm if the generator always selects the position Denis needs.
Input
The first line contains a single integer t (1 ≤ t ≤ 10^5) — the number of test cases. Then the descriptions of the test cases follow.
The first line of the test case contains a single integer n (1 ≤ n ≤ 10^5) — the size of the permutation.
The second line of the test case contains n different integers p_1, p_2, …, p_n (1 ≤ p_i ≤ n) — the permutation written by Denis.
It is guaranteed that the sum of n over all test cases doesn't exceed 10^5.
Output
Print "Yes" if this permutation could be obtained as a result of the generator. Otherwise, print "No".
All letters can be displayed in any case.
Example
Input
5
5
2 3 4 5 1
1
1
3
1 3 2
4
4 2 3 1
5
1 5 2 4 3
Output
Yes
Yes
No
Yes
No
Note
Let's simulate the operation of the generator in the first test.
At the 1 step, r = [1, 2, 3, 4, 5], count = [1, 1, 1, 1, 1]. The maximum value is reached in any free position, so the generator can choose a random position from 1 to 5. In our example, it chose 5.
At the 2 step, r = [1, 2, 3, 4, ×], count = [1, 1, 1, 1, 0]. The maximum value is reached in positions from 1 to 4, so the generator can choose a random position among them. In our example, it chose 1.
At the 3 step, r = [2, 2, 3, 4, ×], count = [0, 2, 1, 1, 0]. The maximum value is 2 and is reached only at the 2 position, so the generator will choose this position.
At the 4 step, r = [3, 3, 3, 4, ×], count = [0, 0, 3, 1, 0]. The maximum value is 3 and is reached only at the 3 position, so the generator will choose this position.
At the 5 step, r = [4, 4, 4, 4, ×], count = [0, 0, 0, 4, 0]. The maximum value is 4 and is reached only at the 4 position, so the generator will choose this position.
In total, we got a permutation of 2, 3, 4, 5, 1, that is, a generator could generate it. | instruction | 0 | 58,998 | 12 | 117,996 |
Tags: brute force, data structures, greedy, implementation
Correct Solution:
```
t = int(input())
for case in range(t):
n = int(input())
numbers = [int(x) for x in input().split()]
possible = True
for i in range(n-1):
if numbers[i+1] > numbers[i]+1:
possible = False
if possible:
print ("Yes")
else:
print ("No")
``` | output | 1 | 58,998 | 12 | 117,997 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Denis was very sad after Nastya rejected him. So he decided to walk through the gateways to have some fun. And luck smiled at him! When he entered the first courtyard, he met a strange man who was selling something.
Denis bought a mysterious item and it was... Random permutation generator! Denis could not believed his luck.
When he arrived home, he began to study how his generator works and learned the algorithm. The process of generating a permutation consists of n steps. At the i-th step, a place is chosen for the number i (1 ≤ i ≤ n). The position for the number i is defined as follows:
* For all j from 1 to n, we calculate r_j — the minimum index such that j ≤ r_j ≤ n, and the position r_j is not yet occupied in the permutation. If there are no such positions, then we assume that the value of r_j is not defined.
* For all t from 1 to n, we calculate count_t — the number of positions 1 ≤ j ≤ n such that r_j is defined and r_j = t.
* Consider the positions that are still not occupied by permutation and among those we consider the positions for which the value in the count array is maximum.
* The generator selects one of these positions for the number i. The generator can choose any position.
Let's have a look at the operation of the algorithm in the following example:
<image>
Let n = 5 and the algorithm has already arranged the numbers 1, 2, 3 in the permutation. Consider how the generator will choose a position for the number 4:
* The values of r will be r = [3, 3, 3, 4, ×], where × means an indefinite value.
* Then the count values will be count = [0, 0, 3, 1, 0].
* There are only two unoccupied positions in the permutation: 3 and 4. The value in the count array for position 3 is 3, for position 4 it is 1.
* The maximum value is reached only for position 3, so the algorithm will uniquely select this position for number 4.
Satisfied with his purchase, Denis went home. For several days without a break, he generated permutations. He believes that he can come up with random permutations no worse than a generator. After that, he wrote out the first permutation that came to mind p_1, p_2, …, p_n and decided to find out if it could be obtained as a result of the generator.
Unfortunately, this task was too difficult for him, and he asked you for help. It is necessary to define whether the written permutation could be obtained using the described algorithm if the generator always selects the position Denis needs.
Input
The first line contains a single integer t (1 ≤ t ≤ 10^5) — the number of test cases. Then the descriptions of the test cases follow.
The first line of the test case contains a single integer n (1 ≤ n ≤ 10^5) — the size of the permutation.
The second line of the test case contains n different integers p_1, p_2, …, p_n (1 ≤ p_i ≤ n) — the permutation written by Denis.
It is guaranteed that the sum of n over all test cases doesn't exceed 10^5.
Output
Print "Yes" if this permutation could be obtained as a result of the generator. Otherwise, print "No".
All letters can be displayed in any case.
Example
Input
5
5
2 3 4 5 1
1
1
3
1 3 2
4
4 2 3 1
5
1 5 2 4 3
Output
Yes
Yes
No
Yes
No
Note
Let's simulate the operation of the generator in the first test.
At the 1 step, r = [1, 2, 3, 4, 5], count = [1, 1, 1, 1, 1]. The maximum value is reached in any free position, so the generator can choose a random position from 1 to 5. In our example, it chose 5.
At the 2 step, r = [1, 2, 3, 4, ×], count = [1, 1, 1, 1, 0]. The maximum value is reached in positions from 1 to 4, so the generator can choose a random position among them. In our example, it chose 1.
At the 3 step, r = [2, 2, 3, 4, ×], count = [0, 2, 1, 1, 0]. The maximum value is 2 and is reached only at the 2 position, so the generator will choose this position.
At the 4 step, r = [3, 3, 3, 4, ×], count = [0, 0, 3, 1, 0]. The maximum value is 3 and is reached only at the 3 position, so the generator will choose this position.
At the 5 step, r = [4, 4, 4, 4, ×], count = [0, 0, 0, 4, 0]. The maximum value is 4 and is reached only at the 4 position, so the generator will choose this position.
In total, we got a permutation of 2, 3, 4, 5, 1, that is, a generator could generate it. | instruction | 0 | 58,999 | 12 | 117,998 |
Tags: brute force, data structures, greedy, implementation
Correct Solution:
```
import sys
input = sys.stdin.readline
from collections import *
def check(l):
a = l[0]
for i in range(len(l)):
if l[i]!=a+i:
return False
return True
t = int(input())
for _ in range(t):
n = int(input())
p = list(map(int, input().split()))
now = []
flag = True
for i in range(n-1):
now.append(p[i])
if p[i]>p[i+1]:
flag &= check(now)
#print(now)
now = []
now.append(p[-1])
#print(now)
flag &= check(now)
if flag:
print('Yes')
else:
print('No')
``` | output | 1 | 58,999 | 12 | 117,999 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Denis was very sad after Nastya rejected him. So he decided to walk through the gateways to have some fun. And luck smiled at him! When he entered the first courtyard, he met a strange man who was selling something.
Denis bought a mysterious item and it was... Random permutation generator! Denis could not believed his luck.
When he arrived home, he began to study how his generator works and learned the algorithm. The process of generating a permutation consists of n steps. At the i-th step, a place is chosen for the number i (1 ≤ i ≤ n). The position for the number i is defined as follows:
* For all j from 1 to n, we calculate r_j — the minimum index such that j ≤ r_j ≤ n, and the position r_j is not yet occupied in the permutation. If there are no such positions, then we assume that the value of r_j is not defined.
* For all t from 1 to n, we calculate count_t — the number of positions 1 ≤ j ≤ n such that r_j is defined and r_j = t.
* Consider the positions that are still not occupied by permutation and among those we consider the positions for which the value in the count array is maximum.
* The generator selects one of these positions for the number i. The generator can choose any position.
Let's have a look at the operation of the algorithm in the following example:
<image>
Let n = 5 and the algorithm has already arranged the numbers 1, 2, 3 in the permutation. Consider how the generator will choose a position for the number 4:
* The values of r will be r = [3, 3, 3, 4, ×], where × means an indefinite value.
* Then the count values will be count = [0, 0, 3, 1, 0].
* There are only two unoccupied positions in the permutation: 3 and 4. The value in the count array for position 3 is 3, for position 4 it is 1.
* The maximum value is reached only for position 3, so the algorithm will uniquely select this position for number 4.
Satisfied with his purchase, Denis went home. For several days without a break, he generated permutations. He believes that he can come up with random permutations no worse than a generator. After that, he wrote out the first permutation that came to mind p_1, p_2, …, p_n and decided to find out if it could be obtained as a result of the generator.
Unfortunately, this task was too difficult for him, and he asked you for help. It is necessary to define whether the written permutation could be obtained using the described algorithm if the generator always selects the position Denis needs.
Input
The first line contains a single integer t (1 ≤ t ≤ 10^5) — the number of test cases. Then the descriptions of the test cases follow.
The first line of the test case contains a single integer n (1 ≤ n ≤ 10^5) — the size of the permutation.
The second line of the test case contains n different integers p_1, p_2, …, p_n (1 ≤ p_i ≤ n) — the permutation written by Denis.
It is guaranteed that the sum of n over all test cases doesn't exceed 10^5.
Output
Print "Yes" if this permutation could be obtained as a result of the generator. Otherwise, print "No".
All letters can be displayed in any case.
Example
Input
5
5
2 3 4 5 1
1
1
3
1 3 2
4
4 2 3 1
5
1 5 2 4 3
Output
Yes
Yes
No
Yes
No
Note
Let's simulate the operation of the generator in the first test.
At the 1 step, r = [1, 2, 3, 4, 5], count = [1, 1, 1, 1, 1]. The maximum value is reached in any free position, so the generator can choose a random position from 1 to 5. In our example, it chose 5.
At the 2 step, r = [1, 2, 3, 4, ×], count = [1, 1, 1, 1, 0]. The maximum value is reached in positions from 1 to 4, so the generator can choose a random position among them. In our example, it chose 1.
At the 3 step, r = [2, 2, 3, 4, ×], count = [0, 2, 1, 1, 0]. The maximum value is 2 and is reached only at the 2 position, so the generator will choose this position.
At the 4 step, r = [3, 3, 3, 4, ×], count = [0, 0, 3, 1, 0]. The maximum value is 3 and is reached only at the 3 position, so the generator will choose this position.
At the 5 step, r = [4, 4, 4, 4, ×], count = [0, 0, 0, 4, 0]. The maximum value is 4 and is reached only at the 4 position, so the generator will choose this position.
In total, we got a permutation of 2, 3, 4, 5, 1, that is, a generator could generate it. | instruction | 0 | 59,000 | 12 | 118,000 |
Tags: brute force, data structures, greedy, implementation
Correct Solution:
```
"""
NTC here
"""
#!/usr/bin/env python
import os
import sys
from io import BytesIO, IOBase
def iin(): return int(input())
def lin(): return list(map(int, input().split()))
def main():
T = iin()
for _ in range(T):
n = iin()
a = lin()
d1 = {a[i]:i for i in range(n)}
done = {}
ls = n
ans = 'No'
for i in range(1, n+1):
pos = d1[i]
# print(done, pos, i)
if i in done:
if pos!=done[i]:
break
else:
j = i
while pos<ls:
done[j] = pos
pos+=1
j+=1
ls = d1[i]
else:
ans = 'Yes'
print(ans)
# region fastio
BUFSIZE = 8192
class FastIO(IOBase):
newlines = 0
def __init__(self, file):
self._fd = file.fileno()
self.buffer = BytesIO()
self.writable = "x" in file.mode or "r" not in file.mode
self.write = self.buffer.write if self.writable else None
def read(self):
while True:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
if not b:
break
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines = 0
return self.buffer.read()
def readline(self):
while self.newlines == 0:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
self.newlines = b.count(b"\n") + (not b)
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines -= 1
return self.buffer.readline()
def flush(self):
if self.writable:
os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
self.write = lambda s: self.buffer.write(s.encode("ascii"))
self.read = lambda: self.buffer.read().decode("ascii")
self.readline = lambda: self.buffer.readline().decode("ascii")
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")
# endregion
if __name__ == "__main__":
main()
``` | output | 1 | 59,000 | 12 | 118,001 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Denis was very sad after Nastya rejected him. So he decided to walk through the gateways to have some fun. And luck smiled at him! When he entered the first courtyard, he met a strange man who was selling something.
Denis bought a mysterious item and it was... Random permutation generator! Denis could not believed his luck.
When he arrived home, he began to study how his generator works and learned the algorithm. The process of generating a permutation consists of n steps. At the i-th step, a place is chosen for the number i (1 ≤ i ≤ n). The position for the number i is defined as follows:
* For all j from 1 to n, we calculate r_j — the minimum index such that j ≤ r_j ≤ n, and the position r_j is not yet occupied in the permutation. If there are no such positions, then we assume that the value of r_j is not defined.
* For all t from 1 to n, we calculate count_t — the number of positions 1 ≤ j ≤ n such that r_j is defined and r_j = t.
* Consider the positions that are still not occupied by permutation and among those we consider the positions for which the value in the count array is maximum.
* The generator selects one of these positions for the number i. The generator can choose any position.
Let's have a look at the operation of the algorithm in the following example:
<image>
Let n = 5 and the algorithm has already arranged the numbers 1, 2, 3 in the permutation. Consider how the generator will choose a position for the number 4:
* The values of r will be r = [3, 3, 3, 4, ×], where × means an indefinite value.
* Then the count values will be count = [0, 0, 3, 1, 0].
* There are only two unoccupied positions in the permutation: 3 and 4. The value in the count array for position 3 is 3, for position 4 it is 1.
* The maximum value is reached only for position 3, so the algorithm will uniquely select this position for number 4.
Satisfied with his purchase, Denis went home. For several days without a break, he generated permutations. He believes that he can come up with random permutations no worse than a generator. After that, he wrote out the first permutation that came to mind p_1, p_2, …, p_n and decided to find out if it could be obtained as a result of the generator.
Unfortunately, this task was too difficult for him, and he asked you for help. It is necessary to define whether the written permutation could be obtained using the described algorithm if the generator always selects the position Denis needs.
Input
The first line contains a single integer t (1 ≤ t ≤ 10^5) — the number of test cases. Then the descriptions of the test cases follow.
The first line of the test case contains a single integer n (1 ≤ n ≤ 10^5) — the size of the permutation.
The second line of the test case contains n different integers p_1, p_2, …, p_n (1 ≤ p_i ≤ n) — the permutation written by Denis.
It is guaranteed that the sum of n over all test cases doesn't exceed 10^5.
Output
Print "Yes" if this permutation could be obtained as a result of the generator. Otherwise, print "No".
All letters can be displayed in any case.
Example
Input
5
5
2 3 4 5 1
1
1
3
1 3 2
4
4 2 3 1
5
1 5 2 4 3
Output
Yes
Yes
No
Yes
No
Note
Let's simulate the operation of the generator in the first test.
At the 1 step, r = [1, 2, 3, 4, 5], count = [1, 1, 1, 1, 1]. The maximum value is reached in any free position, so the generator can choose a random position from 1 to 5. In our example, it chose 5.
At the 2 step, r = [1, 2, 3, 4, ×], count = [1, 1, 1, 1, 0]. The maximum value is reached in positions from 1 to 4, so the generator can choose a random position among them. In our example, it chose 1.
At the 3 step, r = [2, 2, 3, 4, ×], count = [0, 2, 1, 1, 0]. The maximum value is 2 and is reached only at the 2 position, so the generator will choose this position.
At the 4 step, r = [3, 3, 3, 4, ×], count = [0, 0, 3, 1, 0]. The maximum value is 3 and is reached only at the 3 position, so the generator will choose this position.
At the 5 step, r = [4, 4, 4, 4, ×], count = [0, 0, 0, 4, 0]. The maximum value is 4 and is reached only at the 4 position, so the generator will choose this position.
In total, we got a permutation of 2, 3, 4, 5, 1, that is, a generator could generate it. | instruction | 0 | 59,001 | 12 | 118,002 |
Tags: brute force, data structures, greedy, implementation
Correct Solution:
```
def gen(n, p):
while n:
k = p[n-1]
for i in range(1, k+1):
if p[n-i] != k + 1 - i:
return "NO"
for i in range(n - k):
p[i] -= k
n -= k
return "YES"
t = int(input())
for i in range(t):
n = int(input())
p = list(map(int, input().split()))
print(gen(n, p))
``` | output | 1 | 59,001 | 12 | 118,003 |
Provide tags and a correct Python 2 solution for this coding contest problem.
Denis was very sad after Nastya rejected him. So he decided to walk through the gateways to have some fun. And luck smiled at him! When he entered the first courtyard, he met a strange man who was selling something.
Denis bought a mysterious item and it was... Random permutation generator! Denis could not believed his luck.
When he arrived home, he began to study how his generator works and learned the algorithm. The process of generating a permutation consists of n steps. At the i-th step, a place is chosen for the number i (1 ≤ i ≤ n). The position for the number i is defined as follows:
* For all j from 1 to n, we calculate r_j — the minimum index such that j ≤ r_j ≤ n, and the position r_j is not yet occupied in the permutation. If there are no such positions, then we assume that the value of r_j is not defined.
* For all t from 1 to n, we calculate count_t — the number of positions 1 ≤ j ≤ n such that r_j is defined and r_j = t.
* Consider the positions that are still not occupied by permutation and among those we consider the positions for which the value in the count array is maximum.
* The generator selects one of these positions for the number i. The generator can choose any position.
Let's have a look at the operation of the algorithm in the following example:
<image>
Let n = 5 and the algorithm has already arranged the numbers 1, 2, 3 in the permutation. Consider how the generator will choose a position for the number 4:
* The values of r will be r = [3, 3, 3, 4, ×], where × means an indefinite value.
* Then the count values will be count = [0, 0, 3, 1, 0].
* There are only two unoccupied positions in the permutation: 3 and 4. The value in the count array for position 3 is 3, for position 4 it is 1.
* The maximum value is reached only for position 3, so the algorithm will uniquely select this position for number 4.
Satisfied with his purchase, Denis went home. For several days without a break, he generated permutations. He believes that he can come up with random permutations no worse than a generator. After that, he wrote out the first permutation that came to mind p_1, p_2, …, p_n and decided to find out if it could be obtained as a result of the generator.
Unfortunately, this task was too difficult for him, and he asked you for help. It is necessary to define whether the written permutation could be obtained using the described algorithm if the generator always selects the position Denis needs.
Input
The first line contains a single integer t (1 ≤ t ≤ 10^5) — the number of test cases. Then the descriptions of the test cases follow.
The first line of the test case contains a single integer n (1 ≤ n ≤ 10^5) — the size of the permutation.
The second line of the test case contains n different integers p_1, p_2, …, p_n (1 ≤ p_i ≤ n) — the permutation written by Denis.
It is guaranteed that the sum of n over all test cases doesn't exceed 10^5.
Output
Print "Yes" if this permutation could be obtained as a result of the generator. Otherwise, print "No".
All letters can be displayed in any case.
Example
Input
5
5
2 3 4 5 1
1
1
3
1 3 2
4
4 2 3 1
5
1 5 2 4 3
Output
Yes
Yes
No
Yes
No
Note
Let's simulate the operation of the generator in the first test.
At the 1 step, r = [1, 2, 3, 4, 5], count = [1, 1, 1, 1, 1]. The maximum value is reached in any free position, so the generator can choose a random position from 1 to 5. In our example, it chose 5.
At the 2 step, r = [1, 2, 3, 4, ×], count = [1, 1, 1, 1, 0]. The maximum value is reached in positions from 1 to 4, so the generator can choose a random position among them. In our example, it chose 1.
At the 3 step, r = [2, 2, 3, 4, ×], count = [0, 2, 1, 1, 0]. The maximum value is 2 and is reached only at the 2 position, so the generator will choose this position.
At the 4 step, r = [3, 3, 3, 4, ×], count = [0, 0, 3, 1, 0]. The maximum value is 3 and is reached only at the 3 position, so the generator will choose this position.
At the 5 step, r = [4, 4, 4, 4, ×], count = [0, 0, 0, 4, 0]. The maximum value is 4 and is reached only at the 4 position, so the generator will choose this position.
In total, we got a permutation of 2, 3, 4, 5, 1, that is, a generator could generate it. | instruction | 0 | 59,002 | 12 | 118,004 |
Tags: brute force, data structures, greedy, implementation
Correct Solution:
```
from sys import stdin, stdout
from collections import Counter, defaultdict
#from itertools import permutations, combinations
raw_input = stdin.readline
pr = stdout.write
def in_num():
return int(raw_input())
def in_arr():
return map(int,raw_input().split())
def pr_num(n):
stdout.write(str(n)+'\n')
def pr_arr(arr):
pr(' '.join(map(str,arr))+'\n')
# fast read function for total integer input
def inp():
# this function returns whole input of
# space/line seperated integers
# Use Ctrl+D to flush stdin.
return map(int,stdin.read().split())
range = xrange # not for python 3.0+
# main code
for t in range(in_num()):
n=in_num()
l=in_arr()
mn=1
f=0
d=Counter()
for i in range(n):
d[l[i]]=i
while l:
x=l.pop()
if d[x]-d[mn]+1!=x-mn+1:
f=1
break
pos=d[x]
ln=x-mn+1
for i in range(ln-1):
x1=l.pop()
#print x,x1
if d[x1]!=pos-1:
f=1
break
if x1==mn:
break
pos=d[x1]
if f:
break
ln=x-mn+1
mn+=ln
if f:
pr('No\n')
else:
pr('Yes\n')
``` | output | 1 | 59,002 | 12 | 118,005 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Denis was very sad after Nastya rejected him. So he decided to walk through the gateways to have some fun. And luck smiled at him! When he entered the first courtyard, he met a strange man who was selling something.
Denis bought a mysterious item and it was... Random permutation generator! Denis could not believed his luck.
When he arrived home, he began to study how his generator works and learned the algorithm. The process of generating a permutation consists of n steps. At the i-th step, a place is chosen for the number i (1 ≤ i ≤ n). The position for the number i is defined as follows:
* For all j from 1 to n, we calculate r_j — the minimum index such that j ≤ r_j ≤ n, and the position r_j is not yet occupied in the permutation. If there are no such positions, then we assume that the value of r_j is not defined.
* For all t from 1 to n, we calculate count_t — the number of positions 1 ≤ j ≤ n such that r_j is defined and r_j = t.
* Consider the positions that are still not occupied by permutation and among those we consider the positions for which the value in the count array is maximum.
* The generator selects one of these positions for the number i. The generator can choose any position.
Let's have a look at the operation of the algorithm in the following example:
<image>
Let n = 5 and the algorithm has already arranged the numbers 1, 2, 3 in the permutation. Consider how the generator will choose a position for the number 4:
* The values of r will be r = [3, 3, 3, 4, ×], where × means an indefinite value.
* Then the count values will be count = [0, 0, 3, 1, 0].
* There are only two unoccupied positions in the permutation: 3 and 4. The value in the count array for position 3 is 3, for position 4 it is 1.
* The maximum value is reached only for position 3, so the algorithm will uniquely select this position for number 4.
Satisfied with his purchase, Denis went home. For several days without a break, he generated permutations. He believes that he can come up with random permutations no worse than a generator. After that, he wrote out the first permutation that came to mind p_1, p_2, …, p_n and decided to find out if it could be obtained as a result of the generator.
Unfortunately, this task was too difficult for him, and he asked you for help. It is necessary to define whether the written permutation could be obtained using the described algorithm if the generator always selects the position Denis needs.
Input
The first line contains a single integer t (1 ≤ t ≤ 10^5) — the number of test cases. Then the descriptions of the test cases follow.
The first line of the test case contains a single integer n (1 ≤ n ≤ 10^5) — the size of the permutation.
The second line of the test case contains n different integers p_1, p_2, …, p_n (1 ≤ p_i ≤ n) — the permutation written by Denis.
It is guaranteed that the sum of n over all test cases doesn't exceed 10^5.
Output
Print "Yes" if this permutation could be obtained as a result of the generator. Otherwise, print "No".
All letters can be displayed in any case.
Example
Input
5
5
2 3 4 5 1
1
1
3
1 3 2
4
4 2 3 1
5
1 5 2 4 3
Output
Yes
Yes
No
Yes
No
Note
Let's simulate the operation of the generator in the first test.
At the 1 step, r = [1, 2, 3, 4, 5], count = [1, 1, 1, 1, 1]. The maximum value is reached in any free position, so the generator can choose a random position from 1 to 5. In our example, it chose 5.
At the 2 step, r = [1, 2, 3, 4, ×], count = [1, 1, 1, 1, 0]. The maximum value is reached in positions from 1 to 4, so the generator can choose a random position among them. In our example, it chose 1.
At the 3 step, r = [2, 2, 3, 4, ×], count = [0, 2, 1, 1, 0]. The maximum value is 2 and is reached only at the 2 position, so the generator will choose this position.
At the 4 step, r = [3, 3, 3, 4, ×], count = [0, 0, 3, 1, 0]. The maximum value is 3 and is reached only at the 3 position, so the generator will choose this position.
At the 5 step, r = [4, 4, 4, 4, ×], count = [0, 0, 0, 4, 0]. The maximum value is 4 and is reached only at the 4 position, so the generator will choose this position.
In total, we got a permutation of 2, 3, 4, 5, 1, that is, a generator could generate it.
Submitted Solution:
```
## necessary imports
import sys
input = sys.stdin.readline
from math import log2, log, ceil
# swap_array function
def swaparr(arr, a,b):
temp = arr[a];
arr[a] = arr[b];
arr[b] = temp
## gcd function
def gcd(a,b):
if a == 0:
return b
return gcd(b%a, a)
## nCr function efficient using Binomial Cofficient
def nCr(n, k):
if(k > n - k):
k = n - k
res = 1
for i in range(k):
res = res * (n - i)
res = res / (i + 1)
return res
## upper bound function code -- such that e in a[:i] e < x;
def upper_bound(a, x, lo=0):
hi = len(a)
while lo < hi:
mid = (lo+hi)//2
if a[mid] < x:
lo = mid+1
else:
hi = mid
return lo
## prime factorization
def primefs(n):
## if n == 1 ## calculating primes
primes = {}
while(n%2 == 0):
primes[2] = primes.get(2, 0) + 1
n = n//2
for i in range(3, int(n**0.5)+2, 2):
while(n%i == 0):
primes[i] = primes.get(i, 0) + 1
n = n//i
if n > 2:
primes[n] = primes.get(n, 0) + 1
## prime factoriazation of n is stored in dictionary
## primes and can be accesed. O(sqrt n)
return primes
## MODULAR EXPONENTIATION FUNCTION
def power(x, y, p):
res = 1
x = x % p
if (x == 0) :
return 0
while (y > 0) :
if ((y & 1) == 1) :
res = (res * x) % p
y = y >> 1
x = (x * x) % p
return res
## DISJOINT SET UNINON FUNCTIONS
def swap(a,b):
temp = a
a = b
b = temp
return a,b
# find function
def find(x, link):
while(x != link[x]):
x = link[x]
return x
# the union function which makes union(x,y)
# of two nodes x and y
def union(x, y, size, link):
x = find(x, link)
y = find(y, link)
if size[x] < size[y]:
x,y = swap(x,y)
if x != y:
size[x] += size[y]
link[y] = x
## returns an array of boolean if primes or not USING SIEVE OF ERATOSTHANES
def sieve(n):
prime = [True for i in range(n+1)]
p = 2
while (p * p <= n):
if (prime[p] == True):
for i in range(p * p, n+1, p):
prime[i] = False
p += 1
return prime
#### PRIME FACTORIZATION IN O(log n) using Sieve ####
MAXN = int(1e6 + 5)
def spf_sieve():
spf[1] = 1;
for i in range(2, MAXN):
spf[i] = i;
for i in range(4, MAXN, 2):
spf[i] = 2;
for i in range(3, ceil(MAXN ** 0.5), 2):
if spf[i] == i:
for j in range(i*i, MAXN, i):
if spf[j] == j:
spf[j] = i;
## function for storing smallest prime factors (spf) in the array
################## un-comment below 2 lines when using factorization #################
# spf = [0 for i in range(MAXN)]
# spf_sieve()
def factoriazation(x):
ret = {};
while x != 1:
ret[spf[x]] = ret.get(spf[x], 0) + 1;
x = x//spf[x]
return ret
## this function is useful for multiple queries only, o/w use
## primefs function above. complexity O(log n)
## taking integer array input
def int_array():
return list(map(int, input().strip().split()))
## taking string array input
def str_array():
return input().strip().split();
#defining a couple constants
MOD = int(1e9)+7;
CMOD = 998244353;
INF = float('inf'); NINF = -float('inf');
################### ---------------- TEMPLATE ENDS HERE ---------------- ###################
for _ in range(int(input())):
n = int(input()); a = int_array();
prev = None; f = 1;
for i in a:
if prev == None:
prev = i;
elif i > prev + 1:
f = 0; break;
prev = i;
if f:
print('Yes')
else:
print('No');
``` | instruction | 0 | 59,003 | 12 | 118,006 |
Yes | output | 1 | 59,003 | 12 | 118,007 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Denis was very sad after Nastya rejected him. So he decided to walk through the gateways to have some fun. And luck smiled at him! When he entered the first courtyard, he met a strange man who was selling something.
Denis bought a mysterious item and it was... Random permutation generator! Denis could not believed his luck.
When he arrived home, he began to study how his generator works and learned the algorithm. The process of generating a permutation consists of n steps. At the i-th step, a place is chosen for the number i (1 ≤ i ≤ n). The position for the number i is defined as follows:
* For all j from 1 to n, we calculate r_j — the minimum index such that j ≤ r_j ≤ n, and the position r_j is not yet occupied in the permutation. If there are no such positions, then we assume that the value of r_j is not defined.
* For all t from 1 to n, we calculate count_t — the number of positions 1 ≤ j ≤ n such that r_j is defined and r_j = t.
* Consider the positions that are still not occupied by permutation and among those we consider the positions for which the value in the count array is maximum.
* The generator selects one of these positions for the number i. The generator can choose any position.
Let's have a look at the operation of the algorithm in the following example:
<image>
Let n = 5 and the algorithm has already arranged the numbers 1, 2, 3 in the permutation. Consider how the generator will choose a position for the number 4:
* The values of r will be r = [3, 3, 3, 4, ×], where × means an indefinite value.
* Then the count values will be count = [0, 0, 3, 1, 0].
* There are only two unoccupied positions in the permutation: 3 and 4. The value in the count array for position 3 is 3, for position 4 it is 1.
* The maximum value is reached only for position 3, so the algorithm will uniquely select this position for number 4.
Satisfied with his purchase, Denis went home. For several days without a break, he generated permutations. He believes that he can come up with random permutations no worse than a generator. After that, he wrote out the first permutation that came to mind p_1, p_2, …, p_n and decided to find out if it could be obtained as a result of the generator.
Unfortunately, this task was too difficult for him, and he asked you for help. It is necessary to define whether the written permutation could be obtained using the described algorithm if the generator always selects the position Denis needs.
Input
The first line contains a single integer t (1 ≤ t ≤ 10^5) — the number of test cases. Then the descriptions of the test cases follow.
The first line of the test case contains a single integer n (1 ≤ n ≤ 10^5) — the size of the permutation.
The second line of the test case contains n different integers p_1, p_2, …, p_n (1 ≤ p_i ≤ n) — the permutation written by Denis.
It is guaranteed that the sum of n over all test cases doesn't exceed 10^5.
Output
Print "Yes" if this permutation could be obtained as a result of the generator. Otherwise, print "No".
All letters can be displayed in any case.
Example
Input
5
5
2 3 4 5 1
1
1
3
1 3 2
4
4 2 3 1
5
1 5 2 4 3
Output
Yes
Yes
No
Yes
No
Note
Let's simulate the operation of the generator in the first test.
At the 1 step, r = [1, 2, 3, 4, 5], count = [1, 1, 1, 1, 1]. The maximum value is reached in any free position, so the generator can choose a random position from 1 to 5. In our example, it chose 5.
At the 2 step, r = [1, 2, 3, 4, ×], count = [1, 1, 1, 1, 0]. The maximum value is reached in positions from 1 to 4, so the generator can choose a random position among them. In our example, it chose 1.
At the 3 step, r = [2, 2, 3, 4, ×], count = [0, 2, 1, 1, 0]. The maximum value is 2 and is reached only at the 2 position, so the generator will choose this position.
At the 4 step, r = [3, 3, 3, 4, ×], count = [0, 0, 3, 1, 0]. The maximum value is 3 and is reached only at the 3 position, so the generator will choose this position.
At the 5 step, r = [4, 4, 4, 4, ×], count = [0, 0, 0, 4, 0]. The maximum value is 4 and is reached only at the 4 position, so the generator will choose this position.
In total, we got a permutation of 2, 3, 4, 5, 1, that is, a generator could generate it.
Submitted Solution:
```
t = int(input())
for loop in range(t):
n = int(input())
p = list(map(int,input().split()))
nf = float("inf")
now = float("inf")
ans = "Yes"
for i in p:
if i == now+1:
now += 1
elif i > nf:
ans = "No"
break
else:
nf = i
now = i
print (ans)
``` | instruction | 0 | 59,004 | 12 | 118,008 |
Yes | output | 1 | 59,004 | 12 | 118,009 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Denis was very sad after Nastya rejected him. So he decided to walk through the gateways to have some fun. And luck smiled at him! When he entered the first courtyard, he met a strange man who was selling something.
Denis bought a mysterious item and it was... Random permutation generator! Denis could not believed his luck.
When he arrived home, he began to study how his generator works and learned the algorithm. The process of generating a permutation consists of n steps. At the i-th step, a place is chosen for the number i (1 ≤ i ≤ n). The position for the number i is defined as follows:
* For all j from 1 to n, we calculate r_j — the minimum index such that j ≤ r_j ≤ n, and the position r_j is not yet occupied in the permutation. If there are no such positions, then we assume that the value of r_j is not defined.
* For all t from 1 to n, we calculate count_t — the number of positions 1 ≤ j ≤ n such that r_j is defined and r_j = t.
* Consider the positions that are still not occupied by permutation and among those we consider the positions for which the value in the count array is maximum.
* The generator selects one of these positions for the number i. The generator can choose any position.
Let's have a look at the operation of the algorithm in the following example:
<image>
Let n = 5 and the algorithm has already arranged the numbers 1, 2, 3 in the permutation. Consider how the generator will choose a position for the number 4:
* The values of r will be r = [3, 3, 3, 4, ×], where × means an indefinite value.
* Then the count values will be count = [0, 0, 3, 1, 0].
* There are only two unoccupied positions in the permutation: 3 and 4. The value in the count array for position 3 is 3, for position 4 it is 1.
* The maximum value is reached only for position 3, so the algorithm will uniquely select this position for number 4.
Satisfied with his purchase, Denis went home. For several days without a break, he generated permutations. He believes that he can come up with random permutations no worse than a generator. After that, he wrote out the first permutation that came to mind p_1, p_2, …, p_n and decided to find out if it could be obtained as a result of the generator.
Unfortunately, this task was too difficult for him, and he asked you for help. It is necessary to define whether the written permutation could be obtained using the described algorithm if the generator always selects the position Denis needs.
Input
The first line contains a single integer t (1 ≤ t ≤ 10^5) — the number of test cases. Then the descriptions of the test cases follow.
The first line of the test case contains a single integer n (1 ≤ n ≤ 10^5) — the size of the permutation.
The second line of the test case contains n different integers p_1, p_2, …, p_n (1 ≤ p_i ≤ n) — the permutation written by Denis.
It is guaranteed that the sum of n over all test cases doesn't exceed 10^5.
Output
Print "Yes" if this permutation could be obtained as a result of the generator. Otherwise, print "No".
All letters can be displayed in any case.
Example
Input
5
5
2 3 4 5 1
1
1
3
1 3 2
4
4 2 3 1
5
1 5 2 4 3
Output
Yes
Yes
No
Yes
No
Note
Let's simulate the operation of the generator in the first test.
At the 1 step, r = [1, 2, 3, 4, 5], count = [1, 1, 1, 1, 1]. The maximum value is reached in any free position, so the generator can choose a random position from 1 to 5. In our example, it chose 5.
At the 2 step, r = [1, 2, 3, 4, ×], count = [1, 1, 1, 1, 0]. The maximum value is reached in positions from 1 to 4, so the generator can choose a random position among them. In our example, it chose 1.
At the 3 step, r = [2, 2, 3, 4, ×], count = [0, 2, 1, 1, 0]. The maximum value is 2 and is reached only at the 2 position, so the generator will choose this position.
At the 4 step, r = [3, 3, 3, 4, ×], count = [0, 0, 3, 1, 0]. The maximum value is 3 and is reached only at the 3 position, so the generator will choose this position.
At the 5 step, r = [4, 4, 4, 4, ×], count = [0, 0, 0, 4, 0]. The maximum value is 4 and is reached only at the 4 position, so the generator will choose this position.
In total, we got a permutation of 2, 3, 4, 5, 1, that is, a generator could generate it.
Submitted Solution:
```
import sys
input = sys.stdin.buffer.readline
def main():
t = int(input())
for _ in range(t):
n = int(input())
P = list(map(int, input().split()))
d = {}
for i, p in enumerate(P):
d[p-1] = i
def segfunc(x, y):
if x >= y:
return x
else:
return y
def init(init_val):
# set_val
for i in range(n):
seg[i+num-1] = init_val[i]
# built
for i in range(num-2, -1, -1):
seg[i] = segfunc(seg[2*i+1], seg[2*i+2])
def update(k, x):
k += num - 1
seg[k] = x
while k:
k = (k-1)//2
seg[k] = segfunc(seg[2*k+1], seg[2*k+2])
def query(p, q):
if q <= p:
return ide_ele
p += num - 1
q += num - 2
res = ide_ele
while q-p>1:
if p&1 == 0:
res = segfunc(res, seg[p])
if q&1 == 1:
res = segfunc(res, seg[q])
q -= 1
p = p//2
q = (q-1)//2
if p == q:
res = segfunc(res, seg[p])
else:
res = segfunc(segfunc(res, seg[p]), seg[q])
return res
# identity element
ide_ele = -1
# num: n以上の最小の2のべき乗
num = 2**(n-1).bit_length()
seg = [ide_ele]*2*num
A = [1]*n
init(A)
flag = True
for i in range(n):
idx = d[i]
v1 = query(idx, idx+1)
if v1 != query(0, n):
flag = False
if idx != n-1:
v2 = query(idx+1, idx+2)
if v2 != 0:
update(idx+1, v1+v2)
update(idx, 0)
if flag:
print('Yes')
else:
print('No')
if __name__ == '__main__':
main()
``` | instruction | 0 | 59,005 | 12 | 118,010 |
Yes | output | 1 | 59,005 | 12 | 118,011 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Denis was very sad after Nastya rejected him. So he decided to walk through the gateways to have some fun. And luck smiled at him! When he entered the first courtyard, he met a strange man who was selling something.
Denis bought a mysterious item and it was... Random permutation generator! Denis could not believed his luck.
When he arrived home, he began to study how his generator works and learned the algorithm. The process of generating a permutation consists of n steps. At the i-th step, a place is chosen for the number i (1 ≤ i ≤ n). The position for the number i is defined as follows:
* For all j from 1 to n, we calculate r_j — the minimum index such that j ≤ r_j ≤ n, and the position r_j is not yet occupied in the permutation. If there are no such positions, then we assume that the value of r_j is not defined.
* For all t from 1 to n, we calculate count_t — the number of positions 1 ≤ j ≤ n such that r_j is defined and r_j = t.
* Consider the positions that are still not occupied by permutation and among those we consider the positions for which the value in the count array is maximum.
* The generator selects one of these positions for the number i. The generator can choose any position.
Let's have a look at the operation of the algorithm in the following example:
<image>
Let n = 5 and the algorithm has already arranged the numbers 1, 2, 3 in the permutation. Consider how the generator will choose a position for the number 4:
* The values of r will be r = [3, 3, 3, 4, ×], where × means an indefinite value.
* Then the count values will be count = [0, 0, 3, 1, 0].
* There are only two unoccupied positions in the permutation: 3 and 4. The value in the count array for position 3 is 3, for position 4 it is 1.
* The maximum value is reached only for position 3, so the algorithm will uniquely select this position for number 4.
Satisfied with his purchase, Denis went home. For several days without a break, he generated permutations. He believes that he can come up with random permutations no worse than a generator. After that, he wrote out the first permutation that came to mind p_1, p_2, …, p_n and decided to find out if it could be obtained as a result of the generator.
Unfortunately, this task was too difficult for him, and he asked you for help. It is necessary to define whether the written permutation could be obtained using the described algorithm if the generator always selects the position Denis needs.
Input
The first line contains a single integer t (1 ≤ t ≤ 10^5) — the number of test cases. Then the descriptions of the test cases follow.
The first line of the test case contains a single integer n (1 ≤ n ≤ 10^5) — the size of the permutation.
The second line of the test case contains n different integers p_1, p_2, …, p_n (1 ≤ p_i ≤ n) — the permutation written by Denis.
It is guaranteed that the sum of n over all test cases doesn't exceed 10^5.
Output
Print "Yes" if this permutation could be obtained as a result of the generator. Otherwise, print "No".
All letters can be displayed in any case.
Example
Input
5
5
2 3 4 5 1
1
1
3
1 3 2
4
4 2 3 1
5
1 5 2 4 3
Output
Yes
Yes
No
Yes
No
Note
Let's simulate the operation of the generator in the first test.
At the 1 step, r = [1, 2, 3, 4, 5], count = [1, 1, 1, 1, 1]. The maximum value is reached in any free position, so the generator can choose a random position from 1 to 5. In our example, it chose 5.
At the 2 step, r = [1, 2, 3, 4, ×], count = [1, 1, 1, 1, 0]. The maximum value is reached in positions from 1 to 4, so the generator can choose a random position among them. In our example, it chose 1.
At the 3 step, r = [2, 2, 3, 4, ×], count = [0, 2, 1, 1, 0]. The maximum value is 2 and is reached only at the 2 position, so the generator will choose this position.
At the 4 step, r = [3, 3, 3, 4, ×], count = [0, 0, 3, 1, 0]. The maximum value is 3 and is reached only at the 3 position, so the generator will choose this position.
At the 5 step, r = [4, 4, 4, 4, ×], count = [0, 0, 0, 4, 0]. The maximum value is 4 and is reached only at the 4 position, so the generator will choose this position.
In total, we got a permutation of 2, 3, 4, 5, 1, that is, a generator could generate it.
Submitted Solution:
```
import sys,math
t=int(sys.stdin.readline())
for _ in range(t):
n=int(sys.stdin.readline())
a=list(map(int,sys.stdin.readline().split()))
ind=-1
j=n
flag=True
for i in range(1,n):
if not(a[i]-a[i-1]==1 or a[i]-a[i-1]<0):
print("NO")
flag=False
break
if flag:
print("YES")
``` | instruction | 0 | 59,006 | 12 | 118,012 |
Yes | output | 1 | 59,006 | 12 | 118,013 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Denis was very sad after Nastya rejected him. So he decided to walk through the gateways to have some fun. And luck smiled at him! When he entered the first courtyard, he met a strange man who was selling something.
Denis bought a mysterious item and it was... Random permutation generator! Denis could not believed his luck.
When he arrived home, he began to study how his generator works and learned the algorithm. The process of generating a permutation consists of n steps. At the i-th step, a place is chosen for the number i (1 ≤ i ≤ n). The position for the number i is defined as follows:
* For all j from 1 to n, we calculate r_j — the minimum index such that j ≤ r_j ≤ n, and the position r_j is not yet occupied in the permutation. If there are no such positions, then we assume that the value of r_j is not defined.
* For all t from 1 to n, we calculate count_t — the number of positions 1 ≤ j ≤ n such that r_j is defined and r_j = t.
* Consider the positions that are still not occupied by permutation and among those we consider the positions for which the value in the count array is maximum.
* The generator selects one of these positions for the number i. The generator can choose any position.
Let's have a look at the operation of the algorithm in the following example:
<image>
Let n = 5 and the algorithm has already arranged the numbers 1, 2, 3 in the permutation. Consider how the generator will choose a position for the number 4:
* The values of r will be r = [3, 3, 3, 4, ×], where × means an indefinite value.
* Then the count values will be count = [0, 0, 3, 1, 0].
* There are only two unoccupied positions in the permutation: 3 and 4. The value in the count array for position 3 is 3, for position 4 it is 1.
* The maximum value is reached only for position 3, so the algorithm will uniquely select this position for number 4.
Satisfied with his purchase, Denis went home. For several days without a break, he generated permutations. He believes that he can come up with random permutations no worse than a generator. After that, he wrote out the first permutation that came to mind p_1, p_2, …, p_n and decided to find out if it could be obtained as a result of the generator.
Unfortunately, this task was too difficult for him, and he asked you for help. It is necessary to define whether the written permutation could be obtained using the described algorithm if the generator always selects the position Denis needs.
Input
The first line contains a single integer t (1 ≤ t ≤ 10^5) — the number of test cases. Then the descriptions of the test cases follow.
The first line of the test case contains a single integer n (1 ≤ n ≤ 10^5) — the size of the permutation.
The second line of the test case contains n different integers p_1, p_2, …, p_n (1 ≤ p_i ≤ n) — the permutation written by Denis.
It is guaranteed that the sum of n over all test cases doesn't exceed 10^5.
Output
Print "Yes" if this permutation could be obtained as a result of the generator. Otherwise, print "No".
All letters can be displayed in any case.
Example
Input
5
5
2 3 4 5 1
1
1
3
1 3 2
4
4 2 3 1
5
1 5 2 4 3
Output
Yes
Yes
No
Yes
No
Note
Let's simulate the operation of the generator in the first test.
At the 1 step, r = [1, 2, 3, 4, 5], count = [1, 1, 1, 1, 1]. The maximum value is reached in any free position, so the generator can choose a random position from 1 to 5. In our example, it chose 5.
At the 2 step, r = [1, 2, 3, 4, ×], count = [1, 1, 1, 1, 0]. The maximum value is reached in positions from 1 to 4, so the generator can choose a random position among them. In our example, it chose 1.
At the 3 step, r = [2, 2, 3, 4, ×], count = [0, 2, 1, 1, 0]. The maximum value is 2 and is reached only at the 2 position, so the generator will choose this position.
At the 4 step, r = [3, 3, 3, 4, ×], count = [0, 0, 3, 1, 0]. The maximum value is 3 and is reached only at the 3 position, so the generator will choose this position.
At the 5 step, r = [4, 4, 4, 4, ×], count = [0, 0, 0, 4, 0]. The maximum value is 4 and is reached only at the 4 position, so the generator will choose this position.
In total, we got a permutation of 2, 3, 4, 5, 1, that is, a generator could generate it.
Submitted Solution:
```
for _ in range(int(input())):
n=int(input())
a=[int(i) for i in input().split()]
dic={}
for i in range(n):
dic[a[i]]=i
boo=True;c=0;ans=True;key=n-1
for i in range(n-1):
#print('*',dic[i+1],boo,c,key)
if(dic[i+1]==key and boo):
key-=1
continue
elif(boo):
boo=False
key=dic[i+1]-1
c=dic[i+1]+1
continue
#print(dic[i+1],boo,c,key)
if(dic[i+1]==c):
c+=1
if(c==n-1):
boo=True
else:
ans=False
break
#print(dic[i+1],boo,c,key)
ans1='YES' if(ans) else 'NO'
print(ans1)
``` | instruction | 0 | 59,007 | 12 | 118,014 |
No | output | 1 | 59,007 | 12 | 118,015 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Denis was very sad after Nastya rejected him. So he decided to walk through the gateways to have some fun. And luck smiled at him! When he entered the first courtyard, he met a strange man who was selling something.
Denis bought a mysterious item and it was... Random permutation generator! Denis could not believed his luck.
When he arrived home, he began to study how his generator works and learned the algorithm. The process of generating a permutation consists of n steps. At the i-th step, a place is chosen for the number i (1 ≤ i ≤ n). The position for the number i is defined as follows:
* For all j from 1 to n, we calculate r_j — the minimum index such that j ≤ r_j ≤ n, and the position r_j is not yet occupied in the permutation. If there are no such positions, then we assume that the value of r_j is not defined.
* For all t from 1 to n, we calculate count_t — the number of positions 1 ≤ j ≤ n such that r_j is defined and r_j = t.
* Consider the positions that are still not occupied by permutation and among those we consider the positions for which the value in the count array is maximum.
* The generator selects one of these positions for the number i. The generator can choose any position.
Let's have a look at the operation of the algorithm in the following example:
<image>
Let n = 5 and the algorithm has already arranged the numbers 1, 2, 3 in the permutation. Consider how the generator will choose a position for the number 4:
* The values of r will be r = [3, 3, 3, 4, ×], where × means an indefinite value.
* Then the count values will be count = [0, 0, 3, 1, 0].
* There are only two unoccupied positions in the permutation: 3 and 4. The value in the count array for position 3 is 3, for position 4 it is 1.
* The maximum value is reached only for position 3, so the algorithm will uniquely select this position for number 4.
Satisfied with his purchase, Denis went home. For several days without a break, he generated permutations. He believes that he can come up with random permutations no worse than a generator. After that, he wrote out the first permutation that came to mind p_1, p_2, …, p_n and decided to find out if it could be obtained as a result of the generator.
Unfortunately, this task was too difficult for him, and he asked you for help. It is necessary to define whether the written permutation could be obtained using the described algorithm if the generator always selects the position Denis needs.
Input
The first line contains a single integer t (1 ≤ t ≤ 10^5) — the number of test cases. Then the descriptions of the test cases follow.
The first line of the test case contains a single integer n (1 ≤ n ≤ 10^5) — the size of the permutation.
The second line of the test case contains n different integers p_1, p_2, …, p_n (1 ≤ p_i ≤ n) — the permutation written by Denis.
It is guaranteed that the sum of n over all test cases doesn't exceed 10^5.
Output
Print "Yes" if this permutation could be obtained as a result of the generator. Otherwise, print "No".
All letters can be displayed in any case.
Example
Input
5
5
2 3 4 5 1
1
1
3
1 3 2
4
4 2 3 1
5
1 5 2 4 3
Output
Yes
Yes
No
Yes
No
Note
Let's simulate the operation of the generator in the first test.
At the 1 step, r = [1, 2, 3, 4, 5], count = [1, 1, 1, 1, 1]. The maximum value is reached in any free position, so the generator can choose a random position from 1 to 5. In our example, it chose 5.
At the 2 step, r = [1, 2, 3, 4, ×], count = [1, 1, 1, 1, 0]. The maximum value is reached in positions from 1 to 4, so the generator can choose a random position among them. In our example, it chose 1.
At the 3 step, r = [2, 2, 3, 4, ×], count = [0, 2, 1, 1, 0]. The maximum value is 2 and is reached only at the 2 position, so the generator will choose this position.
At the 4 step, r = [3, 3, 3, 4, ×], count = [0, 0, 3, 1, 0]. The maximum value is 3 and is reached only at the 3 position, so the generator will choose this position.
At the 5 step, r = [4, 4, 4, 4, ×], count = [0, 0, 0, 4, 0]. The maximum value is 4 and is reached only at the 4 position, so the generator will choose this position.
In total, we got a permutation of 2, 3, 4, 5, 1, that is, a generator could generate it.
Submitted Solution:
```
import sys
input = sys.stdin.buffer.readline
def I(): return(list(map(int,input().split())))
def sieve(n):
a=[1]*n
for i in range(2,n):
if a[i]:
for j in range(i*i,n,i):
a[j]=0
return a
for __ in range(int(input())):
n=int(input())
p=I()
arr=[0]*(n+1)
for i in range(n):arr[p[i]]=i
last=n-1
f1=0
f=0
# print(arr)
last=n-1
f=0
# print(arr)
# for i in range(2,n):
# if arr[i-1]==last and not f1:
# last-=1
# else:
# f1=True
# if arr[i]-1!=arr[i-1]:
# f=1
# break
# if arr[i]==last:
# f1=False
# print(arr)
f=0
last=n-1
i=1
while(i<n+1):
if arr[i]==last:
last-=1
i+=1
else:
prev=arr[i]
c=last-arr[i]
i+=1
if i>n:
break
while(i<n+1 and c>0):
if arr[i]-1!=arr[i-1]:
f=1
break
i+=1
c-=1
last=prev-1
i+=1
if not f:
print("Yes")
else:
print("No")
``` | instruction | 0 | 59,008 | 12 | 118,016 |
No | output | 1 | 59,008 | 12 | 118,017 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Denis was very sad after Nastya rejected him. So he decided to walk through the gateways to have some fun. And luck smiled at him! When he entered the first courtyard, he met a strange man who was selling something.
Denis bought a mysterious item and it was... Random permutation generator! Denis could not believed his luck.
When he arrived home, he began to study how his generator works and learned the algorithm. The process of generating a permutation consists of n steps. At the i-th step, a place is chosen for the number i (1 ≤ i ≤ n). The position for the number i is defined as follows:
* For all j from 1 to n, we calculate r_j — the minimum index such that j ≤ r_j ≤ n, and the position r_j is not yet occupied in the permutation. If there are no such positions, then we assume that the value of r_j is not defined.
* For all t from 1 to n, we calculate count_t — the number of positions 1 ≤ j ≤ n such that r_j is defined and r_j = t.
* Consider the positions that are still not occupied by permutation and among those we consider the positions for which the value in the count array is maximum.
* The generator selects one of these positions for the number i. The generator can choose any position.
Let's have a look at the operation of the algorithm in the following example:
<image>
Let n = 5 and the algorithm has already arranged the numbers 1, 2, 3 in the permutation. Consider how the generator will choose a position for the number 4:
* The values of r will be r = [3, 3, 3, 4, ×], where × means an indefinite value.
* Then the count values will be count = [0, 0, 3, 1, 0].
* There are only two unoccupied positions in the permutation: 3 and 4. The value in the count array for position 3 is 3, for position 4 it is 1.
* The maximum value is reached only for position 3, so the algorithm will uniquely select this position for number 4.
Satisfied with his purchase, Denis went home. For several days without a break, he generated permutations. He believes that he can come up with random permutations no worse than a generator. After that, he wrote out the first permutation that came to mind p_1, p_2, …, p_n and decided to find out if it could be obtained as a result of the generator.
Unfortunately, this task was too difficult for him, and he asked you for help. It is necessary to define whether the written permutation could be obtained using the described algorithm if the generator always selects the position Denis needs.
Input
The first line contains a single integer t (1 ≤ t ≤ 10^5) — the number of test cases. Then the descriptions of the test cases follow.
The first line of the test case contains a single integer n (1 ≤ n ≤ 10^5) — the size of the permutation.
The second line of the test case contains n different integers p_1, p_2, …, p_n (1 ≤ p_i ≤ n) — the permutation written by Denis.
It is guaranteed that the sum of n over all test cases doesn't exceed 10^5.
Output
Print "Yes" if this permutation could be obtained as a result of the generator. Otherwise, print "No".
All letters can be displayed in any case.
Example
Input
5
5
2 3 4 5 1
1
1
3
1 3 2
4
4 2 3 1
5
1 5 2 4 3
Output
Yes
Yes
No
Yes
No
Note
Let's simulate the operation of the generator in the first test.
At the 1 step, r = [1, 2, 3, 4, 5], count = [1, 1, 1, 1, 1]. The maximum value is reached in any free position, so the generator can choose a random position from 1 to 5. In our example, it chose 5.
At the 2 step, r = [1, 2, 3, 4, ×], count = [1, 1, 1, 1, 0]. The maximum value is reached in positions from 1 to 4, so the generator can choose a random position among them. In our example, it chose 1.
At the 3 step, r = [2, 2, 3, 4, ×], count = [0, 2, 1, 1, 0]. The maximum value is 2 and is reached only at the 2 position, so the generator will choose this position.
At the 4 step, r = [3, 3, 3, 4, ×], count = [0, 0, 3, 1, 0]. The maximum value is 3 and is reached only at the 3 position, so the generator will choose this position.
At the 5 step, r = [4, 4, 4, 4, ×], count = [0, 0, 0, 4, 0]. The maximum value is 4 and is reached only at the 4 position, so the generator will choose this position.
In total, we got a permutation of 2, 3, 4, 5, 1, that is, a generator could generate it.
Submitted Solution:
```
t=int(input())
import sys
for _ in range(t):
n=int(input())
a=list(map(int,sys.stdin.readline().split()))
indices={}
for i in range(0,len(a)):
indices[a[i]]=i
end=n-1
flag=1
init=n-1
count=0
for i in range(1,len(a)):
if indices[i]==end:
end=init-1
init-=1
count=0
else:
if count==0:
init=indices[i]
count+=1
if indices[i+1]!=indices[i]+1:
flag=0
break
if flag==0:
sys.stdout.write("No")
else:
sys.stdout.write("Yes")
``` | instruction | 0 | 59,009 | 12 | 118,018 |
No | output | 1 | 59,009 | 12 | 118,019 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Denis was very sad after Nastya rejected him. So he decided to walk through the gateways to have some fun. And luck smiled at him! When he entered the first courtyard, he met a strange man who was selling something.
Denis bought a mysterious item and it was... Random permutation generator! Denis could not believed his luck.
When he arrived home, he began to study how his generator works and learned the algorithm. The process of generating a permutation consists of n steps. At the i-th step, a place is chosen for the number i (1 ≤ i ≤ n). The position for the number i is defined as follows:
* For all j from 1 to n, we calculate r_j — the minimum index such that j ≤ r_j ≤ n, and the position r_j is not yet occupied in the permutation. If there are no such positions, then we assume that the value of r_j is not defined.
* For all t from 1 to n, we calculate count_t — the number of positions 1 ≤ j ≤ n such that r_j is defined and r_j = t.
* Consider the positions that are still not occupied by permutation and among those we consider the positions for which the value in the count array is maximum.
* The generator selects one of these positions for the number i. The generator can choose any position.
Let's have a look at the operation of the algorithm in the following example:
<image>
Let n = 5 and the algorithm has already arranged the numbers 1, 2, 3 in the permutation. Consider how the generator will choose a position for the number 4:
* The values of r will be r = [3, 3, 3, 4, ×], where × means an indefinite value.
* Then the count values will be count = [0, 0, 3, 1, 0].
* There are only two unoccupied positions in the permutation: 3 and 4. The value in the count array for position 3 is 3, for position 4 it is 1.
* The maximum value is reached only for position 3, so the algorithm will uniquely select this position for number 4.
Satisfied with his purchase, Denis went home. For several days without a break, he generated permutations. He believes that he can come up with random permutations no worse than a generator. After that, he wrote out the first permutation that came to mind p_1, p_2, …, p_n and decided to find out if it could be obtained as a result of the generator.
Unfortunately, this task was too difficult for him, and he asked you for help. It is necessary to define whether the written permutation could be obtained using the described algorithm if the generator always selects the position Denis needs.
Input
The first line contains a single integer t (1 ≤ t ≤ 10^5) — the number of test cases. Then the descriptions of the test cases follow.
The first line of the test case contains a single integer n (1 ≤ n ≤ 10^5) — the size of the permutation.
The second line of the test case contains n different integers p_1, p_2, …, p_n (1 ≤ p_i ≤ n) — the permutation written by Denis.
It is guaranteed that the sum of n over all test cases doesn't exceed 10^5.
Output
Print "Yes" if this permutation could be obtained as a result of the generator. Otherwise, print "No".
All letters can be displayed in any case.
Example
Input
5
5
2 3 4 5 1
1
1
3
1 3 2
4
4 2 3 1
5
1 5 2 4 3
Output
Yes
Yes
No
Yes
No
Note
Let's simulate the operation of the generator in the first test.
At the 1 step, r = [1, 2, 3, 4, 5], count = [1, 1, 1, 1, 1]. The maximum value is reached in any free position, so the generator can choose a random position from 1 to 5. In our example, it chose 5.
At the 2 step, r = [1, 2, 3, 4, ×], count = [1, 1, 1, 1, 0]. The maximum value is reached in positions from 1 to 4, so the generator can choose a random position among them. In our example, it chose 1.
At the 3 step, r = [2, 2, 3, 4, ×], count = [0, 2, 1, 1, 0]. The maximum value is 2 and is reached only at the 2 position, so the generator will choose this position.
At the 4 step, r = [3, 3, 3, 4, ×], count = [0, 0, 3, 1, 0]. The maximum value is 3 and is reached only at the 3 position, so the generator will choose this position.
At the 5 step, r = [4, 4, 4, 4, ×], count = [0, 0, 0, 4, 0]. The maximum value is 4 and is reached only at the 4 position, so the generator will choose this position.
In total, we got a permutation of 2, 3, 4, 5, 1, that is, a generator could generate it.
Submitted Solution:
```
T = int(input())
for t in range(T):
N = int(input())
P = [int(i) for i in input().split()]
greatest_seen_number = 0
target_number = 1
allowed_mistake = True
for R in range(len(P)-1, -1, -1):
greatest_seen_number = max(greatest_seen_number, P[R])
if R+1 < len(P) and P[R] != P[R+1] - 1 and not allowed_mistake:
print('No')
break
if P[R] == target_number:
target_number = greatest_seen_number + 1
allowed_mistake = True
elif allowed_mistake and R+1 < len(P) and P[R] != P[R+1] - 1:
allowed_mistake = False
else:
print('Yes')
``` | instruction | 0 | 59,010 | 12 | 118,020 |
No | output | 1 | 59,010 | 12 | 118,021 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given an array a of length n, and an integer x. You can perform the following operation as many times as you would like (possibly zero): replace two adjacent elements of the array by their sum. For example, if the initial array was [3, 6, 9], in a single operation one can replace the last two elements by their sum, yielding an array [3, 15], or replace the first two elements to get an array [9, 9]. Note that the size of the array decreases after each operation.
The beauty of an array b=[b_1, …, b_k] is defined as ∑_{i=1}^k \left⌈ (b_i)/(x) \right⌉, which means that we divide each element by x, round it up to the nearest integer, and sum up the resulting values. For example, if x = 3, and the array is [4, 11, 6], the beauty of the array is equal to \left⌈ 4/3 \right⌉ + \left⌈ 11/3 \right⌉ + \left⌈ 6/3 \right⌉ = 2 + 4 + 2 = 8.
Please determine the minimum and the maximum beauty you can get by performing some operations on the original array.
Input
The first input line contains a single integer t — the number of test cases (1 ≤ t ≤ 1000).
The first line of each test case contains two integers n and x (1 ≤ n ≤ 10^5, 1 ≤ x ≤ 10^9).
The next line contains n integers a_1, a_2, …, a_n (1 ≤ a_i ≤ 10^9), the elements of the array a.
It is guaranteed that the sum of values of n over all test cases does not exceed 10^5.
Output
For each test case output two integers — the minimal and the maximal possible beauty.
Example
Input
2
3 3
3 6 9
3 3
6 4 11
Output
6 6
7 8
Note
In the first test case the beauty of the array does not change if we perform any operations.
In the second example we can leave the array unchanged to attain the maximum beauty, and to get the minimum beauty one can replace two elements 4 and 11 with their sum, yielding an array [6, 15], which has its beauty equal to 7. | instruction | 0 | 59,049 | 12 | 118,098 |
Tags: greedy, math, number theory
Correct Solution:
```
# A. Strange Partition
from math import ceil
t=int(input())
for i in range(t):
n,x=map(int,input().split())
a=list(map(int,input().split()))
mn=ceil(sum(a)/x)
mx=0
for k in a:
mx+=ceil(k/x)
print(mn,mx)
``` | output | 1 | 59,049 | 12 | 118,099 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given an array a of length n, and an integer x. You can perform the following operation as many times as you would like (possibly zero): replace two adjacent elements of the array by their sum. For example, if the initial array was [3, 6, 9], in a single operation one can replace the last two elements by their sum, yielding an array [3, 15], or replace the first two elements to get an array [9, 9]. Note that the size of the array decreases after each operation.
The beauty of an array b=[b_1, …, b_k] is defined as ∑_{i=1}^k \left⌈ (b_i)/(x) \right⌉, which means that we divide each element by x, round it up to the nearest integer, and sum up the resulting values. For example, if x = 3, and the array is [4, 11, 6], the beauty of the array is equal to \left⌈ 4/3 \right⌉ + \left⌈ 11/3 \right⌉ + \left⌈ 6/3 \right⌉ = 2 + 4 + 2 = 8.
Please determine the minimum and the maximum beauty you can get by performing some operations on the original array.
Input
The first input line contains a single integer t — the number of test cases (1 ≤ t ≤ 1000).
The first line of each test case contains two integers n and x (1 ≤ n ≤ 10^5, 1 ≤ x ≤ 10^9).
The next line contains n integers a_1, a_2, …, a_n (1 ≤ a_i ≤ 10^9), the elements of the array a.
It is guaranteed that the sum of values of n over all test cases does not exceed 10^5.
Output
For each test case output two integers — the minimal and the maximal possible beauty.
Example
Input
2
3 3
3 6 9
3 3
6 4 11
Output
6 6
7 8
Note
In the first test case the beauty of the array does not change if we perform any operations.
In the second example we can leave the array unchanged to attain the maximum beauty, and to get the minimum beauty one can replace two elements 4 and 11 with their sum, yielding an array [6, 15], which has its beauty equal to 7. | instruction | 0 | 59,050 | 12 | 118,100 |
Tags: greedy, math, number theory
Correct Solution:
```
t = int(input())
import math
for i in range(t):
k = input()
n,x = [int(item) for item in k.split(' ')]
a = input()
a = [int(item) for item in a.split(' ')]
mx = math.ceil(sum(a) /x)
mn = 0
for it in a:
b = math.ceil(it/x)
mn+=b
print(mx, mn)
``` | output | 1 | 59,050 | 12 | 118,101 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given an array a of length n, and an integer x. You can perform the following operation as many times as you would like (possibly zero): replace two adjacent elements of the array by their sum. For example, if the initial array was [3, 6, 9], in a single operation one can replace the last two elements by their sum, yielding an array [3, 15], or replace the first two elements to get an array [9, 9]. Note that the size of the array decreases after each operation.
The beauty of an array b=[b_1, …, b_k] is defined as ∑_{i=1}^k \left⌈ (b_i)/(x) \right⌉, which means that we divide each element by x, round it up to the nearest integer, and sum up the resulting values. For example, if x = 3, and the array is [4, 11, 6], the beauty of the array is equal to \left⌈ 4/3 \right⌉ + \left⌈ 11/3 \right⌉ + \left⌈ 6/3 \right⌉ = 2 + 4 + 2 = 8.
Please determine the minimum and the maximum beauty you can get by performing some operations on the original array.
Input
The first input line contains a single integer t — the number of test cases (1 ≤ t ≤ 1000).
The first line of each test case contains two integers n and x (1 ≤ n ≤ 10^5, 1 ≤ x ≤ 10^9).
The next line contains n integers a_1, a_2, …, a_n (1 ≤ a_i ≤ 10^9), the elements of the array a.
It is guaranteed that the sum of values of n over all test cases does not exceed 10^5.
Output
For each test case output two integers — the minimal and the maximal possible beauty.
Example
Input
2
3 3
3 6 9
3 3
6 4 11
Output
6 6
7 8
Note
In the first test case the beauty of the array does not change if we perform any operations.
In the second example we can leave the array unchanged to attain the maximum beauty, and to get the minimum beauty one can replace two elements 4 and 11 with their sum, yielding an array [6, 15], which has its beauty equal to 7. | instruction | 0 | 59,051 | 12 | 118,102 |
Tags: greedy, math, number theory
Correct Solution:
```
from math import ceil
for _ in range(int(input())):
n, x = list(map(int, input().split()))
arr = list(map(int, input().split()))
min_ = 0
for i in arr:
min_ += ceil(i / x)
max_ = ceil(sum(arr) / x)
print(max_, min_)
``` | output | 1 | 59,051 | 12 | 118,103 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given an array a of length n, and an integer x. You can perform the following operation as many times as you would like (possibly zero): replace two adjacent elements of the array by their sum. For example, if the initial array was [3, 6, 9], in a single operation one can replace the last two elements by their sum, yielding an array [3, 15], or replace the first two elements to get an array [9, 9]. Note that the size of the array decreases after each operation.
The beauty of an array b=[b_1, …, b_k] is defined as ∑_{i=1}^k \left⌈ (b_i)/(x) \right⌉, which means that we divide each element by x, round it up to the nearest integer, and sum up the resulting values. For example, if x = 3, and the array is [4, 11, 6], the beauty of the array is equal to \left⌈ 4/3 \right⌉ + \left⌈ 11/3 \right⌉ + \left⌈ 6/3 \right⌉ = 2 + 4 + 2 = 8.
Please determine the minimum and the maximum beauty you can get by performing some operations on the original array.
Input
The first input line contains a single integer t — the number of test cases (1 ≤ t ≤ 1000).
The first line of each test case contains two integers n and x (1 ≤ n ≤ 10^5, 1 ≤ x ≤ 10^9).
The next line contains n integers a_1, a_2, …, a_n (1 ≤ a_i ≤ 10^9), the elements of the array a.
It is guaranteed that the sum of values of n over all test cases does not exceed 10^5.
Output
For each test case output two integers — the minimal and the maximal possible beauty.
Example
Input
2
3 3
3 6 9
3 3
6 4 11
Output
6 6
7 8
Note
In the first test case the beauty of the array does not change if we perform any operations.
In the second example we can leave the array unchanged to attain the maximum beauty, and to get the minimum beauty one can replace two elements 4 and 11 with their sum, yielding an array [6, 15], which has its beauty equal to 7. | instruction | 0 | 59,052 | 12 | 118,104 |
Tags: greedy, math, number theory
Correct Solution:
```
import math
t = int(input())
def beauty(n, x):
sum = 0
max = 0
l = list(map(int, input().split()))
for i in l:
max+=math.ceil(i/x)
sum+=i
min = math.ceil(sum/x)
print(min, max)
return
for _ in range(t):
n, x = map(int, input().split())
beauty(n, x)
``` | output | 1 | 59,052 | 12 | 118,105 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given an array a of length n, and an integer x. You can perform the following operation as many times as you would like (possibly zero): replace two adjacent elements of the array by their sum. For example, if the initial array was [3, 6, 9], in a single operation one can replace the last two elements by their sum, yielding an array [3, 15], or replace the first two elements to get an array [9, 9]. Note that the size of the array decreases after each operation.
The beauty of an array b=[b_1, …, b_k] is defined as ∑_{i=1}^k \left⌈ (b_i)/(x) \right⌉, which means that we divide each element by x, round it up to the nearest integer, and sum up the resulting values. For example, if x = 3, and the array is [4, 11, 6], the beauty of the array is equal to \left⌈ 4/3 \right⌉ + \left⌈ 11/3 \right⌉ + \left⌈ 6/3 \right⌉ = 2 + 4 + 2 = 8.
Please determine the minimum and the maximum beauty you can get by performing some operations on the original array.
Input
The first input line contains a single integer t — the number of test cases (1 ≤ t ≤ 1000).
The first line of each test case contains two integers n and x (1 ≤ n ≤ 10^5, 1 ≤ x ≤ 10^9).
The next line contains n integers a_1, a_2, …, a_n (1 ≤ a_i ≤ 10^9), the elements of the array a.
It is guaranteed that the sum of values of n over all test cases does not exceed 10^5.
Output
For each test case output two integers — the minimal and the maximal possible beauty.
Example
Input
2
3 3
3 6 9
3 3
6 4 11
Output
6 6
7 8
Note
In the first test case the beauty of the array does not change if we perform any operations.
In the second example we can leave the array unchanged to attain the maximum beauty, and to get the minimum beauty one can replace two elements 4 and 11 with their sum, yielding an array [6, 15], which has its beauty equal to 7. | instruction | 0 | 59,053 | 12 | 118,106 |
Tags: greedy, math, number theory
Correct Solution:
```
for _ in range(int(input())):
n, x = map(int, input().split())
a = list(map(int, input().split()))
s = 0
s1 = 0
for i in a:
s += i
s1 += -(-i // x)
print(-(-s // x), s1)
``` | output | 1 | 59,053 | 12 | 118,107 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given an array a of length n, and an integer x. You can perform the following operation as many times as you would like (possibly zero): replace two adjacent elements of the array by their sum. For example, if the initial array was [3, 6, 9], in a single operation one can replace the last two elements by their sum, yielding an array [3, 15], or replace the first two elements to get an array [9, 9]. Note that the size of the array decreases after each operation.
The beauty of an array b=[b_1, …, b_k] is defined as ∑_{i=1}^k \left⌈ (b_i)/(x) \right⌉, which means that we divide each element by x, round it up to the nearest integer, and sum up the resulting values. For example, if x = 3, and the array is [4, 11, 6], the beauty of the array is equal to \left⌈ 4/3 \right⌉ + \left⌈ 11/3 \right⌉ + \left⌈ 6/3 \right⌉ = 2 + 4 + 2 = 8.
Please determine the minimum and the maximum beauty you can get by performing some operations on the original array.
Input
The first input line contains a single integer t — the number of test cases (1 ≤ t ≤ 1000).
The first line of each test case contains two integers n and x (1 ≤ n ≤ 10^5, 1 ≤ x ≤ 10^9).
The next line contains n integers a_1, a_2, …, a_n (1 ≤ a_i ≤ 10^9), the elements of the array a.
It is guaranteed that the sum of values of n over all test cases does not exceed 10^5.
Output
For each test case output two integers — the minimal and the maximal possible beauty.
Example
Input
2
3 3
3 6 9
3 3
6 4 11
Output
6 6
7 8
Note
In the first test case the beauty of the array does not change if we perform any operations.
In the second example we can leave the array unchanged to attain the maximum beauty, and to get the minimum beauty one can replace two elements 4 and 11 with their sum, yielding an array [6, 15], which has its beauty equal to 7. | instruction | 0 | 59,054 | 12 | 118,108 |
Tags: greedy, math, number theory
Correct Solution:
```
from math import ceil
for u in range(int(input())):
n, x = map(int, input().split())
y = [int(w) for w in input().split()]
m = ceil(sum(y)/x)
t = 0
for i in y:
t += ceil(i/x)
print(m, t)
``` | output | 1 | 59,054 | 12 | 118,109 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given an array a of length n, and an integer x. You can perform the following operation as many times as you would like (possibly zero): replace two adjacent elements of the array by their sum. For example, if the initial array was [3, 6, 9], in a single operation one can replace the last two elements by their sum, yielding an array [3, 15], or replace the first two elements to get an array [9, 9]. Note that the size of the array decreases after each operation.
The beauty of an array b=[b_1, …, b_k] is defined as ∑_{i=1}^k \left⌈ (b_i)/(x) \right⌉, which means that we divide each element by x, round it up to the nearest integer, and sum up the resulting values. For example, if x = 3, and the array is [4, 11, 6], the beauty of the array is equal to \left⌈ 4/3 \right⌉ + \left⌈ 11/3 \right⌉ + \left⌈ 6/3 \right⌉ = 2 + 4 + 2 = 8.
Please determine the minimum and the maximum beauty you can get by performing some operations on the original array.
Input
The first input line contains a single integer t — the number of test cases (1 ≤ t ≤ 1000).
The first line of each test case contains two integers n and x (1 ≤ n ≤ 10^5, 1 ≤ x ≤ 10^9).
The next line contains n integers a_1, a_2, …, a_n (1 ≤ a_i ≤ 10^9), the elements of the array a.
It is guaranteed that the sum of values of n over all test cases does not exceed 10^5.
Output
For each test case output two integers — the minimal and the maximal possible beauty.
Example
Input
2
3 3
3 6 9
3 3
6 4 11
Output
6 6
7 8
Note
In the first test case the beauty of the array does not change if we perform any operations.
In the second example we can leave the array unchanged to attain the maximum beauty, and to get the minimum beauty one can replace two elements 4 and 11 with their sum, yielding an array [6, 15], which has its beauty equal to 7. | instruction | 0 | 59,055 | 12 | 118,110 |
Tags: greedy, math, number theory
Correct Solution:
```
for _ in range(int(input())):
n, x = map(int, input().split())
a = list(map(int, input().split()))
print((sum(a) + x - 1) // x, sum((i + x - 1) // x for i in a))
# qwq
``` | output | 1 | 59,055 | 12 | 118,111 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given an array a of length n, and an integer x. You can perform the following operation as many times as you would like (possibly zero): replace two adjacent elements of the array by their sum. For example, if the initial array was [3, 6, 9], in a single operation one can replace the last two elements by their sum, yielding an array [3, 15], or replace the first two elements to get an array [9, 9]. Note that the size of the array decreases after each operation.
The beauty of an array b=[b_1, …, b_k] is defined as ∑_{i=1}^k \left⌈ (b_i)/(x) \right⌉, which means that we divide each element by x, round it up to the nearest integer, and sum up the resulting values. For example, if x = 3, and the array is [4, 11, 6], the beauty of the array is equal to \left⌈ 4/3 \right⌉ + \left⌈ 11/3 \right⌉ + \left⌈ 6/3 \right⌉ = 2 + 4 + 2 = 8.
Please determine the minimum and the maximum beauty you can get by performing some operations on the original array.
Input
The first input line contains a single integer t — the number of test cases (1 ≤ t ≤ 1000).
The first line of each test case contains two integers n and x (1 ≤ n ≤ 10^5, 1 ≤ x ≤ 10^9).
The next line contains n integers a_1, a_2, …, a_n (1 ≤ a_i ≤ 10^9), the elements of the array a.
It is guaranteed that the sum of values of n over all test cases does not exceed 10^5.
Output
For each test case output two integers — the minimal and the maximal possible beauty.
Example
Input
2
3 3
3 6 9
3 3
6 4 11
Output
6 6
7 8
Note
In the first test case the beauty of the array does not change if we perform any operations.
In the second example we can leave the array unchanged to attain the maximum beauty, and to get the minimum beauty one can replace two elements 4 and 11 with their sum, yielding an array [6, 15], which has its beauty equal to 7. | instruction | 0 | 59,056 | 12 | 118,112 |
Tags: greedy, math, number theory
Correct Solution:
```
# ------------------- fast io --------------------
import os
import sys
from io import BytesIO, IOBase
BUFSIZE = 8192
class FastIO(IOBase):
newlines = 0
def __init__(self, file):
self._fd = file.fileno()
self.buffer = BytesIO()
self.writable = "x" in file.mode or "r" not in file.mode
self.write = self.buffer.write if self.writable else None
def read(self):
while True:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
if not b:
break
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines = 0
return self.buffer.read()
def readline(self):
while self.newlines == 0:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
self.newlines = b.count(b"\n") + (not b)
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines -= 1
return self.buffer.readline()
def flush(self):
if self.writable:
os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
self.write = lambda s: self.buffer.write(s.encode("ascii"))
self.read = lambda: self.buffer.read().decode("ascii")
self.readline = lambda: self.buffer.readline().decode("ascii")
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")
# ------------------- fast io --------------------
from math import gcd, ceil
def prod(a, mod=10**9+7):
ans = 1
for each in a:
ans = (ans * each) % mod
return ans
def lcm(a, b): return a * b // gcd(a, b)
def binary(x, length=16):
y = bin(x)[2:]
return y if len(y) >= length else "0" * (length - len(y)) + y
for _ in range(int(input()) if True else 1):
#n = int(input())
n, x = map(int, input().split())
#a, b = map(int, input().split())
#c, d = map(int, input().split())
a = list(map(int, input().split()))
print(ceil(sum(a)/x), sum(ceil(k/x) for k in a))
``` | output | 1 | 59,056 | 12 | 118,113 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given an array a of length n, and an integer x. You can perform the following operation as many times as you would like (possibly zero): replace two adjacent elements of the array by their sum. For example, if the initial array was [3, 6, 9], in a single operation one can replace the last two elements by their sum, yielding an array [3, 15], or replace the first two elements to get an array [9, 9]. Note that the size of the array decreases after each operation.
The beauty of an array b=[b_1, …, b_k] is defined as ∑_{i=1}^k \left⌈ (b_i)/(x) \right⌉, which means that we divide each element by x, round it up to the nearest integer, and sum up the resulting values. For example, if x = 3, and the array is [4, 11, 6], the beauty of the array is equal to \left⌈ 4/3 \right⌉ + \left⌈ 11/3 \right⌉ + \left⌈ 6/3 \right⌉ = 2 + 4 + 2 = 8.
Please determine the minimum and the maximum beauty you can get by performing some operations on the original array.
Input
The first input line contains a single integer t — the number of test cases (1 ≤ t ≤ 1000).
The first line of each test case contains two integers n and x (1 ≤ n ≤ 10^5, 1 ≤ x ≤ 10^9).
The next line contains n integers a_1, a_2, …, a_n (1 ≤ a_i ≤ 10^9), the elements of the array a.
It is guaranteed that the sum of values of n over all test cases does not exceed 10^5.
Output
For each test case output two integers — the minimal and the maximal possible beauty.
Example
Input
2
3 3
3 6 9
3 3
6 4 11
Output
6 6
7 8
Note
In the first test case the beauty of the array does not change if we perform any operations.
In the second example we can leave the array unchanged to attain the maximum beauty, and to get the minimum beauty one can replace two elements 4 and 11 with their sum, yielding an array [6, 15], which has its beauty equal to 7.
Submitted Solution:
```
from math import *
for _ in range(int(input())):
n,x = map(int,input().split())
a = list(map(int,input().split()))
s = sum(a)
mini = ceil(s/x)
maxi = 0
for i in range(n):
maxi = maxi+ceil(a[i]/x)
print(mini,maxi)
``` | instruction | 0 | 59,057 | 12 | 118,114 |
Yes | output | 1 | 59,057 | 12 | 118,115 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given an array a of length n, and an integer x. You can perform the following operation as many times as you would like (possibly zero): replace two adjacent elements of the array by their sum. For example, if the initial array was [3, 6, 9], in a single operation one can replace the last two elements by their sum, yielding an array [3, 15], or replace the first two elements to get an array [9, 9]. Note that the size of the array decreases after each operation.
The beauty of an array b=[b_1, …, b_k] is defined as ∑_{i=1}^k \left⌈ (b_i)/(x) \right⌉, which means that we divide each element by x, round it up to the nearest integer, and sum up the resulting values. For example, if x = 3, and the array is [4, 11, 6], the beauty of the array is equal to \left⌈ 4/3 \right⌉ + \left⌈ 11/3 \right⌉ + \left⌈ 6/3 \right⌉ = 2 + 4 + 2 = 8.
Please determine the minimum and the maximum beauty you can get by performing some operations on the original array.
Input
The first input line contains a single integer t — the number of test cases (1 ≤ t ≤ 1000).
The first line of each test case contains two integers n and x (1 ≤ n ≤ 10^5, 1 ≤ x ≤ 10^9).
The next line contains n integers a_1, a_2, …, a_n (1 ≤ a_i ≤ 10^9), the elements of the array a.
It is guaranteed that the sum of values of n over all test cases does not exceed 10^5.
Output
For each test case output two integers — the minimal and the maximal possible beauty.
Example
Input
2
3 3
3 6 9
3 3
6 4 11
Output
6 6
7 8
Note
In the first test case the beauty of the array does not change if we perform any operations.
In the second example we can leave the array unchanged to attain the maximum beauty, and to get the minimum beauty one can replace two elements 4 and 11 with their sum, yielding an array [6, 15], which has its beauty equal to 7.
Submitted Solution:
```
import sys
import math
import itertools
import functools
import collections
import operator
import fileinput
import copy
import string
ORDA = 97 # a
def ii(): return int(input())
def mi(): return map(int, input().split())
def li(): return [int(i) for i in input().split()]
def lcm(a, b): return abs(a * b) // math.gcd(a, b)
def revn(n): return str(n)[::-1]
def dd(): return collections.defaultdict(int)
def ddl(): return collections.defaultdict(list)
def sieve(n):
if n < 2: return list()
prime = [True for _ in range(n + 1)]
p = 3
while p * p <= n:
if prime[p]:
for i in range(p * 2, n + 1, p):
prime[i] = False
p += 2
r = [2]
for p in range(3, n + 1, 2):
if prime[p]:
r.append(p)
return r
def divs(n, start=2):
r = []
for i in range(start, int(math.sqrt(n) + 1)):
if (n % i == 0):
if (n / i == i):
r.append(i)
else:
r.extend([i, n // i])
return r
def divn(n, primes):
divs_number = 1
for i in primes:
if n == 1:
return divs_number
t = 1
while n % i == 0:
t += 1
n //= i
divs_number *= t
def prime(n):
if n == 2: return True
if n % 2 == 0 or n <= 1: return False
sqr = int(math.sqrt(n)) + 1
for d in range(3, sqr, 2):
if n % d == 0: return False
return True
def convn(number, base):
new_number = 0
while number > 0:
new_number += number % base
number //= base
return new_number
def cdiv(n, k): return n // k + (n % k != 0)
def ispal(s):
for i in range(len(s) // 2 + 1):
if s[i] != s[-i - 1]:
return False
return True
for _ in range(ii()):
n, x = mi()
a = li()
print((sum(a) + x - 1) // x, sum([(i + x - 1) // x for i in a]))
``` | instruction | 0 | 59,058 | 12 | 118,116 |
Yes | output | 1 | 59,058 | 12 | 118,117 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given an array a of length n, and an integer x. You can perform the following operation as many times as you would like (possibly zero): replace two adjacent elements of the array by their sum. For example, if the initial array was [3, 6, 9], in a single operation one can replace the last two elements by their sum, yielding an array [3, 15], or replace the first two elements to get an array [9, 9]. Note that the size of the array decreases after each operation.
The beauty of an array b=[b_1, …, b_k] is defined as ∑_{i=1}^k \left⌈ (b_i)/(x) \right⌉, which means that we divide each element by x, round it up to the nearest integer, and sum up the resulting values. For example, if x = 3, and the array is [4, 11, 6], the beauty of the array is equal to \left⌈ 4/3 \right⌉ + \left⌈ 11/3 \right⌉ + \left⌈ 6/3 \right⌉ = 2 + 4 + 2 = 8.
Please determine the minimum and the maximum beauty you can get by performing some operations on the original array.
Input
The first input line contains a single integer t — the number of test cases (1 ≤ t ≤ 1000).
The first line of each test case contains two integers n and x (1 ≤ n ≤ 10^5, 1 ≤ x ≤ 10^9).
The next line contains n integers a_1, a_2, …, a_n (1 ≤ a_i ≤ 10^9), the elements of the array a.
It is guaranteed that the sum of values of n over all test cases does not exceed 10^5.
Output
For each test case output two integers — the minimal and the maximal possible beauty.
Example
Input
2
3 3
3 6 9
3 3
6 4 11
Output
6 6
7 8
Note
In the first test case the beauty of the array does not change if we perform any operations.
In the second example we can leave the array unchanged to attain the maximum beauty, and to get the minimum beauty one can replace two elements 4 and 11 with their sum, yielding an array [6, 15], which has its beauty equal to 7.
Submitted Solution:
```
for _ in range(int(input())):
n, x = map(int, input().split())
vs = [int(x) for x in input().split()]
base = 0
mods = []
for v in vs:
base += v // x
if v % x > 0:
mods.append(v % x)
max_beauty = base + len(mods)
min_beauty = base + (sum(mods) + x-1) // x
print(min_beauty, max_beauty)
``` | instruction | 0 | 59,059 | 12 | 118,118 |
Yes | output | 1 | 59,059 | 12 | 118,119 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given an array a of length n, and an integer x. You can perform the following operation as many times as you would like (possibly zero): replace two adjacent elements of the array by their sum. For example, if the initial array was [3, 6, 9], in a single operation one can replace the last two elements by their sum, yielding an array [3, 15], or replace the first two elements to get an array [9, 9]. Note that the size of the array decreases after each operation.
The beauty of an array b=[b_1, …, b_k] is defined as ∑_{i=1}^k \left⌈ (b_i)/(x) \right⌉, which means that we divide each element by x, round it up to the nearest integer, and sum up the resulting values. For example, if x = 3, and the array is [4, 11, 6], the beauty of the array is equal to \left⌈ 4/3 \right⌉ + \left⌈ 11/3 \right⌉ + \left⌈ 6/3 \right⌉ = 2 + 4 + 2 = 8.
Please determine the minimum and the maximum beauty you can get by performing some operations on the original array.
Input
The first input line contains a single integer t — the number of test cases (1 ≤ t ≤ 1000).
The first line of each test case contains two integers n and x (1 ≤ n ≤ 10^5, 1 ≤ x ≤ 10^9).
The next line contains n integers a_1, a_2, …, a_n (1 ≤ a_i ≤ 10^9), the elements of the array a.
It is guaranteed that the sum of values of n over all test cases does not exceed 10^5.
Output
For each test case output two integers — the minimal and the maximal possible beauty.
Example
Input
2
3 3
3 6 9
3 3
6 4 11
Output
6 6
7 8
Note
In the first test case the beauty of the array does not change if we perform any operations.
In the second example we can leave the array unchanged to attain the maximum beauty, and to get the minimum beauty one can replace two elements 4 and 11 with their sum, yielding an array [6, 15], which has its beauty equal to 7.
Submitted Solution:
```
import math
for _ in range(int(input())):
n,x=map(int,input().split())
l=list(map(int,input().split()))
s1,s2,a,b=0,0,0,0
for i in l:
if i%x!=0:
s1+=i
b+=math.ceil(i/x)
else:
a+=math.ceil(i/x)
s2+=i
a+=math.ceil(s1/x)
b+=math.ceil(s2/x)
print(a,b)
``` | instruction | 0 | 59,060 | 12 | 118,120 |
Yes | output | 1 | 59,060 | 12 | 118,121 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given an array a of length n, and an integer x. You can perform the following operation as many times as you would like (possibly zero): replace two adjacent elements of the array by their sum. For example, if the initial array was [3, 6, 9], in a single operation one can replace the last two elements by their sum, yielding an array [3, 15], or replace the first two elements to get an array [9, 9]. Note that the size of the array decreases after each operation.
The beauty of an array b=[b_1, …, b_k] is defined as ∑_{i=1}^k \left⌈ (b_i)/(x) \right⌉, which means that we divide each element by x, round it up to the nearest integer, and sum up the resulting values. For example, if x = 3, and the array is [4, 11, 6], the beauty of the array is equal to \left⌈ 4/3 \right⌉ + \left⌈ 11/3 \right⌉ + \left⌈ 6/3 \right⌉ = 2 + 4 + 2 = 8.
Please determine the minimum and the maximum beauty you can get by performing some operations on the original array.
Input
The first input line contains a single integer t — the number of test cases (1 ≤ t ≤ 1000).
The first line of each test case contains two integers n and x (1 ≤ n ≤ 10^5, 1 ≤ x ≤ 10^9).
The next line contains n integers a_1, a_2, …, a_n (1 ≤ a_i ≤ 10^9), the elements of the array a.
It is guaranteed that the sum of values of n over all test cases does not exceed 10^5.
Output
For each test case output two integers — the minimal and the maximal possible beauty.
Example
Input
2
3 3
3 6 9
3 3
6 4 11
Output
6 6
7 8
Note
In the first test case the beauty of the array does not change if we perform any operations.
In the second example we can leave the array unchanged to attain the maximum beauty, and to get the minimum beauty one can replace two elements 4 and 11 with their sum, yielding an array [6, 15], which has its beauty equal to 7.
Submitted Solution:
```
t = int(input())
ans = ''
for i in range(t):
nx = input().split()
n = int(nx[0])
x = int(nx[1])
a = [int(j) for j in input().split()]
m = 0
r = 0
for j in range(n):
m += (a[j]+x-1) // x
r += (a[j] % x)
M = m - (r//x)
ans += str(M) + ' ' + str(m) + '\n'
print(ans)
``` | instruction | 0 | 59,061 | 12 | 118,122 |
No | output | 1 | 59,061 | 12 | 118,123 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given an array a of length n, and an integer x. You can perform the following operation as many times as you would like (possibly zero): replace two adjacent elements of the array by their sum. For example, if the initial array was [3, 6, 9], in a single operation one can replace the last two elements by their sum, yielding an array [3, 15], or replace the first two elements to get an array [9, 9]. Note that the size of the array decreases after each operation.
The beauty of an array b=[b_1, …, b_k] is defined as ∑_{i=1}^k \left⌈ (b_i)/(x) \right⌉, which means that we divide each element by x, round it up to the nearest integer, and sum up the resulting values. For example, if x = 3, and the array is [4, 11, 6], the beauty of the array is equal to \left⌈ 4/3 \right⌉ + \left⌈ 11/3 \right⌉ + \left⌈ 6/3 \right⌉ = 2 + 4 + 2 = 8.
Please determine the minimum and the maximum beauty you can get by performing some operations on the original array.
Input
The first input line contains a single integer t — the number of test cases (1 ≤ t ≤ 1000).
The first line of each test case contains two integers n and x (1 ≤ n ≤ 10^5, 1 ≤ x ≤ 10^9).
The next line contains n integers a_1, a_2, …, a_n (1 ≤ a_i ≤ 10^9), the elements of the array a.
It is guaranteed that the sum of values of n over all test cases does not exceed 10^5.
Output
For each test case output two integers — the minimal and the maximal possible beauty.
Example
Input
2
3 3
3 6 9
3 3
6 4 11
Output
6 6
7 8
Note
In the first test case the beauty of the array does not change if we perform any operations.
In the second example we can leave the array unchanged to attain the maximum beauty, and to get the minimum beauty one can replace two elements 4 and 11 with their sum, yielding an array [6, 15], which has its beauty equal to 7.
Submitted Solution:
```
def ceil(a):
if a>0:
if a-int(a)==0:
return int(a)
else:
return (int(a)+1)
else:
if a-int(a)==0:
return int(a)
else:
return (int(a)-1)
def partition(b,x):
maxceil,minceil,i=0,0,0
n=len(b)
while i<n:
maxceil+=ceil(b[i]/x)
i+=1
i=0
while i<(n-1):
if (b[i]+b[i+1])%x==0:
b[i]+=b[i+1]
del b[i+1]
n-=1
continue
minceil+=ceil(b[i]/x)
i+=1
print(minceil,maxceil)
cases=int(input())
for i in range(cases):
n,x=map(int,input().split())
b=list(map(int,input().split()))
partition(b,x)
``` | instruction | 0 | 59,062 | 12 | 118,124 |
No | output | 1 | 59,062 | 12 | 118,125 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given an array a of length n, and an integer x. You can perform the following operation as many times as you would like (possibly zero): replace two adjacent elements of the array by their sum. For example, if the initial array was [3, 6, 9], in a single operation one can replace the last two elements by their sum, yielding an array [3, 15], or replace the first two elements to get an array [9, 9]. Note that the size of the array decreases after each operation.
The beauty of an array b=[b_1, …, b_k] is defined as ∑_{i=1}^k \left⌈ (b_i)/(x) \right⌉, which means that we divide each element by x, round it up to the nearest integer, and sum up the resulting values. For example, if x = 3, and the array is [4, 11, 6], the beauty of the array is equal to \left⌈ 4/3 \right⌉ + \left⌈ 11/3 \right⌉ + \left⌈ 6/3 \right⌉ = 2 + 4 + 2 = 8.
Please determine the minimum and the maximum beauty you can get by performing some operations on the original array.
Input
The first input line contains a single integer t — the number of test cases (1 ≤ t ≤ 1000).
The first line of each test case contains two integers n and x (1 ≤ n ≤ 10^5, 1 ≤ x ≤ 10^9).
The next line contains n integers a_1, a_2, …, a_n (1 ≤ a_i ≤ 10^9), the elements of the array a.
It is guaranteed that the sum of values of n over all test cases does not exceed 10^5.
Output
For each test case output two integers — the minimal and the maximal possible beauty.
Example
Input
2
3 3
3 6 9
3 3
6 4 11
Output
6 6
7 8
Note
In the first test case the beauty of the array does not change if we perform any operations.
In the second example we can leave the array unchanged to attain the maximum beauty, and to get the minimum beauty one can replace two elements 4 and 11 with their sum, yielding an array [6, 15], which has its beauty equal to 7.
Submitted Solution:
```
import math
t=int(input())
for i in range(0,t):
n,x=map(int,input().split())
alist=list(map(int,input().split()))
c=0
for j in range(0,n):
c=c+math.ceil(alist[j]/n)
s=sum(alist)
d=math.ceil(s/n)
print(f'{d} {c}')
``` | instruction | 0 | 59,063 | 12 | 118,126 |
No | output | 1 | 59,063 | 12 | 118,127 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given an array a of length n, and an integer x. You can perform the following operation as many times as you would like (possibly zero): replace two adjacent elements of the array by their sum. For example, if the initial array was [3, 6, 9], in a single operation one can replace the last two elements by their sum, yielding an array [3, 15], or replace the first two elements to get an array [9, 9]. Note that the size of the array decreases after each operation.
The beauty of an array b=[b_1, …, b_k] is defined as ∑_{i=1}^k \left⌈ (b_i)/(x) \right⌉, which means that we divide each element by x, round it up to the nearest integer, and sum up the resulting values. For example, if x = 3, and the array is [4, 11, 6], the beauty of the array is equal to \left⌈ 4/3 \right⌉ + \left⌈ 11/3 \right⌉ + \left⌈ 6/3 \right⌉ = 2 + 4 + 2 = 8.
Please determine the minimum and the maximum beauty you can get by performing some operations on the original array.
Input
The first input line contains a single integer t — the number of test cases (1 ≤ t ≤ 1000).
The first line of each test case contains two integers n and x (1 ≤ n ≤ 10^5, 1 ≤ x ≤ 10^9).
The next line contains n integers a_1, a_2, …, a_n (1 ≤ a_i ≤ 10^9), the elements of the array a.
It is guaranteed that the sum of values of n over all test cases does not exceed 10^5.
Output
For each test case output two integers — the minimal and the maximal possible beauty.
Example
Input
2
3 3
3 6 9
3 3
6 4 11
Output
6 6
7 8
Note
In the first test case the beauty of the array does not change if we perform any operations.
In the second example we can leave the array unchanged to attain the maximum beauty, and to get the minimum beauty one can replace two elements 4 and 11 with their sum, yielding an array [6, 15], which has its beauty equal to 7.
Submitted Solution:
```
#A||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||\
g=int(input())
for i in range (g):
bol=0
mal=0
n,x=map(int,input().split())
s=list(map(int,input().split()))
for j in range (n):
if s[j] % x >0:
bol+=s[j]//x+1
else:
bol+=s[j]//x
mal=sum(s)//x+1
print(mal,bol)
``` | instruction | 0 | 59,064 | 12 | 118,128 |
No | output | 1 | 59,064 | 12 | 118,129 |
Provide tags and a correct Python 3 solution for this coding contest problem.
We all know that GukiZ often plays with arrays.
Now he is thinking about this problem: how many arrays a, of length n, with non-negative elements strictly less then 2l meet the following condition: <image>? Here operation <image> means bitwise AND (in Pascal it is equivalent to and, in C/C++/Java/Python it is equivalent to &), operation <image> means bitwise OR (in Pascal it is equivalent to <image>, in C/C++/Java/Python it is equivalent to |).
Because the answer can be quite large, calculate it modulo m. This time GukiZ hasn't come up with solution, and needs you to help him!
Input
First and the only line of input contains four integers n, k, l, m (2 ≤ n ≤ 1018, 0 ≤ k ≤ 1018, 0 ≤ l ≤ 64, 1 ≤ m ≤ 109 + 7).
Output
In the single line print the number of arrays satisfying the condition above modulo m.
Examples
Input
2 1 2 10
Output
3
Input
2 1 1 3
Output
1
Input
3 3 2 10
Output
9
Note
In the first sample, satisfying arrays are {1, 1}, {3, 1}, {1, 3}.
In the second sample, only satisfying array is {1, 1}.
In the third sample, satisfying arrays are {0, 3, 3}, {1, 3, 2}, {1, 3, 3}, {2, 3, 1}, {2, 3, 3}, {3, 3, 0}, {3, 3, 1}, {3, 3, 2}, {3, 3, 3}. | instruction | 0 | 59,249 | 12 | 118,498 |
Tags: combinatorics, implementation, math, matrices, number theory
Correct Solution:
```
# -*- coding: utf-8 -*-
from collections import deque
def calc(n, m):
if n == 1:
return [[1, 0], [0, 1]]
a = calc(n // 2, m)
res = [[0, 0], [0, 0]]
if n % 2 == 0:
for i in range(2):
for j in range(2):
res[i][j] = (res[i][j] + a[i][0] * a[0][j]) % m
res[i][j] = (res[i][j] + a[i][0] * a[1][j]) % m
res[i][j] = (res[i][j] + a[i][1] * a[0][j]) % m
else:
for i in range(2):
for j in range(2):
res[i][j] = (res[i][j] + a[i][0] * a[0][j] * 2) % m
res[i][j] = (res[i][j] + a[i][0] * a[1][j]) % m
res[i][j] = (res[i][j] + a[i][1] * a[0][j]) % m
res[i][j] = (res[i][j] + a[i][1] * a[1][j]) % m
return res
def binpow(a, p, m):
if p == 0:
return 1 % m
if p == 1:
return a % m
ans = binpow(a, p // 2, m)
ans = (ans * ans) % m
if p % 2 == 1:
ans = (ans * a) % m
return ans
n, k, l, m = map(int, input().split())
ans = [0, 0]
x = calc(n, m)
ans[0] = (x[0][0] + x[0][1] + x[1][0] + x[1][1]) % m
ans[1] = ((binpow(2, n, m) - ans[0]) % m + m) % m
res = 1
for i in range(l):
res = (res * ans[k & 1]) % m
k >>= 1
if k > 0:
res = 0
print(res % m)
``` | output | 1 | 59,249 | 12 | 118,499 |
Provide tags and a correct Python 3 solution for this coding contest problem.
We all know that GukiZ often plays with arrays.
Now he is thinking about this problem: how many arrays a, of length n, with non-negative elements strictly less then 2l meet the following condition: <image>? Here operation <image> means bitwise AND (in Pascal it is equivalent to and, in C/C++/Java/Python it is equivalent to &), operation <image> means bitwise OR (in Pascal it is equivalent to <image>, in C/C++/Java/Python it is equivalent to |).
Because the answer can be quite large, calculate it modulo m. This time GukiZ hasn't come up with solution, and needs you to help him!
Input
First and the only line of input contains four integers n, k, l, m (2 ≤ n ≤ 1018, 0 ≤ k ≤ 1018, 0 ≤ l ≤ 64, 1 ≤ m ≤ 109 + 7).
Output
In the single line print the number of arrays satisfying the condition above modulo m.
Examples
Input
2 1 2 10
Output
3
Input
2 1 1 3
Output
1
Input
3 3 2 10
Output
9
Note
In the first sample, satisfying arrays are {1, 1}, {3, 1}, {1, 3}.
In the second sample, only satisfying array is {1, 1}.
In the third sample, satisfying arrays are {0, 3, 3}, {1, 3, 2}, {1, 3, 3}, {2, 3, 1}, {2, 3, 3}, {3, 3, 0}, {3, 3, 1}, {3, 3, 2}, {3, 3, 3}. | instruction | 0 | 59,250 | 12 | 118,500 |
Tags: combinatorics, implementation, math, matrices, number theory
Correct Solution:
```
n,k,l,m=map(int,input().split())
fc={1:1,2:1}
def f(n):
if n not in fc: k=n//2;fc[n]=(f(k+1)**2+f(k)**2 if n%2 else f(k)*(2*f(k+1)-f(k)))%m
return fc[n]
s=k<2**l
for i in range(l): s*=pow(2,n,m)-f(n+2) if (k>>i)%2 else f(n+2)
print(s%m)
``` | output | 1 | 59,250 | 12 | 118,501 |
Provide tags and a correct Python 3 solution for this coding contest problem.
We all know that GukiZ often plays with arrays.
Now he is thinking about this problem: how many arrays a, of length n, with non-negative elements strictly less then 2l meet the following condition: <image>? Here operation <image> means bitwise AND (in Pascal it is equivalent to and, in C/C++/Java/Python it is equivalent to &), operation <image> means bitwise OR (in Pascal it is equivalent to <image>, in C/C++/Java/Python it is equivalent to |).
Because the answer can be quite large, calculate it modulo m. This time GukiZ hasn't come up with solution, and needs you to help him!
Input
First and the only line of input contains four integers n, k, l, m (2 ≤ n ≤ 1018, 0 ≤ k ≤ 1018, 0 ≤ l ≤ 64, 1 ≤ m ≤ 109 + 7).
Output
In the single line print the number of arrays satisfying the condition above modulo m.
Examples
Input
2 1 2 10
Output
3
Input
2 1 1 3
Output
1
Input
3 3 2 10
Output
9
Note
In the first sample, satisfying arrays are {1, 1}, {3, 1}, {1, 3}.
In the second sample, only satisfying array is {1, 1}.
In the third sample, satisfying arrays are {0, 3, 3}, {1, 3, 2}, {1, 3, 3}, {2, 3, 1}, {2, 3, 3}, {3, 3, 0}, {3, 3, 1}, {3, 3, 2}, {3, 3, 3}. | instruction | 0 | 59,251 | 12 | 118,502 |
Tags: combinatorics, implementation, math, matrices, number theory
Correct Solution:
```
# by the authority of GOD author: manhar singh sachdev #
import os,sys
from io import BytesIO, IOBase
def mat_multiply(a,b,mod):
x = len(a)
res = [[0]*x for _ in range(x)]
for i in range(x):
for j in range(x):
for k in range(x):
res[i][j] += a[i][k]*b[k][j]
res[i][j] %= mod
return res
def mat_expo(a,n,mod):
iden = [[0]*len(a) for _ in range(len(a))]
for i in range(len(a)):
iden[i][i] = 1
while n:
if n%2:
iden = mat_multiply(iden,a,mod)
a = mat_multiply(a,a,mod)
n //= 2
return iden
def main():
n,k,l,m = map(int,input().split())
mat = [[0,1],[1,1]]
xx = mat_expo(mat,n-1,m)
not_set = (xx[0][0]+xx[0][1]+xx[1][0]+xx[1][1])%m
se = (pow(2,n,m)-not_set)%m
mask = 1
ans = 1%m
if k.bit_length() >= l+1:
print(0)
return
for _ in range(l):
if mask&k:
ans = (ans*se)%m
else:
ans = (ans*not_set)%m
mask <<= 1
print(ans)
# Fast IO Region
BUFSIZE = 8192
class FastIO(IOBase):
newlines = 0
def __init__(self, file):
self._fd = file.fileno()
self.buffer = BytesIO()
self.writable = "x" in file.mode or "r" not in file.mode
self.write = self.buffer.write if self.writable else None
def read(self):
while True:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
if not b:
break
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines = 0
return self.buffer.read()
def readline(self):
while self.newlines == 0:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
self.newlines = b.count(b"\n") + (not b)
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines -= 1
return self.buffer.readline()
def flush(self):
if self.writable:
os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
self.write = lambda s: self.buffer.write(s.encode("ascii"))
self.read = lambda: self.buffer.read().decode("ascii")
self.readline = lambda: self.buffer.readline().decode("ascii")
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")
if __name__ == "__main__":
main()
``` | output | 1 | 59,251 | 12 | 118,503 |
Provide tags and a correct Python 3 solution for this coding contest problem.
We all know that GukiZ often plays with arrays.
Now he is thinking about this problem: how many arrays a, of length n, with non-negative elements strictly less then 2l meet the following condition: <image>? Here operation <image> means bitwise AND (in Pascal it is equivalent to and, in C/C++/Java/Python it is equivalent to &), operation <image> means bitwise OR (in Pascal it is equivalent to <image>, in C/C++/Java/Python it is equivalent to |).
Because the answer can be quite large, calculate it modulo m. This time GukiZ hasn't come up with solution, and needs you to help him!
Input
First and the only line of input contains four integers n, k, l, m (2 ≤ n ≤ 1018, 0 ≤ k ≤ 1018, 0 ≤ l ≤ 64, 1 ≤ m ≤ 109 + 7).
Output
In the single line print the number of arrays satisfying the condition above modulo m.
Examples
Input
2 1 2 10
Output
3
Input
2 1 1 3
Output
1
Input
3 3 2 10
Output
9
Note
In the first sample, satisfying arrays are {1, 1}, {3, 1}, {1, 3}.
In the second sample, only satisfying array is {1, 1}.
In the third sample, satisfying arrays are {0, 3, 3}, {1, 3, 2}, {1, 3, 3}, {2, 3, 1}, {2, 3, 3}, {3, 3, 0}, {3, 3, 1}, {3, 3, 2}, {3, 3, 3}. | instruction | 0 | 59,252 | 12 | 118,504 |
Tags: combinatorics, implementation, math, matrices, number theory
Correct Solution:
```
n, k, l, BASE = map(int,input().split())
d = {0:1, 1:1, 2:(2 % BASE)}
def power(n):
if n == 0: return 1
if n & 1 > 0:
return (power(n-1) << 1) % BASE
u = power(n >> 1) % BASE
return u * u % BASE
def fib(n):
if d.get(n) != None:
return d[n]
if n & 1 < 1:
d[n] = (fib((n >> 1) - 1)*fib((n >> 1) - 1)+fib(n >> 1)*fib(n >> 1))%BASE
else:
d[n] = (fib((n >> 1) )*fib((n >> 1) + 1)+fib((n >> 1) - 1)*fib(n >> 1))%BASE
return d[n]
u = fib(n+1)
v = (power(n) + BASE - u) % BASE
a = 1
for i in range(l):
if k & 1 == 0:
a *= u
else:
a *= v
a %= BASE
k >>= 1
if k > 0:
a = 0
print(a % BASE)
``` | output | 1 | 59,252 | 12 | 118,505 |
Provide tags and a correct Python 3 solution for this coding contest problem.
We all know that GukiZ often plays with arrays.
Now he is thinking about this problem: how many arrays a, of length n, with non-negative elements strictly less then 2l meet the following condition: <image>? Here operation <image> means bitwise AND (in Pascal it is equivalent to and, in C/C++/Java/Python it is equivalent to &), operation <image> means bitwise OR (in Pascal it is equivalent to <image>, in C/C++/Java/Python it is equivalent to |).
Because the answer can be quite large, calculate it modulo m. This time GukiZ hasn't come up with solution, and needs you to help him!
Input
First and the only line of input contains four integers n, k, l, m (2 ≤ n ≤ 1018, 0 ≤ k ≤ 1018, 0 ≤ l ≤ 64, 1 ≤ m ≤ 109 + 7).
Output
In the single line print the number of arrays satisfying the condition above modulo m.
Examples
Input
2 1 2 10
Output
3
Input
2 1 1 3
Output
1
Input
3 3 2 10
Output
9
Note
In the first sample, satisfying arrays are {1, 1}, {3, 1}, {1, 3}.
In the second sample, only satisfying array is {1, 1}.
In the third sample, satisfying arrays are {0, 3, 3}, {1, 3, 2}, {1, 3, 3}, {2, 3, 1}, {2, 3, 3}, {3, 3, 0}, {3, 3, 1}, {3, 3, 2}, {3, 3, 3}. | instruction | 0 | 59,253 | 12 | 118,506 |
Tags: combinatorics, implementation, math, matrices, number theory
Correct Solution:
```
import math
import sys
def fib(n,m):
if n==0:
return 0
if n==1 or n==2:
return 1%m
b=n.bit_length()-1
X=1
Y=0
Z=1
k=n
while b>=0:
x=1
y=1
z=0
t=1
for i in range(0,b):
x1=x
y1=y
x=(x1*x1+y1*y1)%m
y=(x1*y1+y1*z)%m
z=(y1*y1+z*z)%m
t*=2
Y2=Y
X=(X*x+Y*y)%m
Y=(x*Y+y*Z)%m
Z=(y*Y2+z*Z)%m
k=k-t
b=k.bit_length()-1
return Y
def two(n,m):
if n==0:
return 1%m
if n==1:
return 2%m
T=1
b=n.bit_length()-1
k=n
while b>=0:
t2=1
t1=2
for i in range(0,b):
t2*=2
t1=(t1*t1)%m
k=k-t2
b=k.bit_length()-1
T*=t1
T=T%m
return T
inp=list(map(int,input().split()))
n=inp[0]
k=inp[1]
l=inp[2]
m=inp[3]
if k>=2**l:
print(0)
else:
t=1
ans=1
for i in range(0,l):
b=(k-k%t)%(2*t)
t*=2
if b==0:
ans*=fib(n+2,m)
ans=ans%m
else:
ans*=two(n,m)-fib(n+2,m)
ans=ans%m
print(ans%m)
``` | output | 1 | 59,253 | 12 | 118,507 |
Provide tags and a correct Python 3 solution for this coding contest problem.
We all know that GukiZ often plays with arrays.
Now he is thinking about this problem: how many arrays a, of length n, with non-negative elements strictly less then 2l meet the following condition: <image>? Here operation <image> means bitwise AND (in Pascal it is equivalent to and, in C/C++/Java/Python it is equivalent to &), operation <image> means bitwise OR (in Pascal it is equivalent to <image>, in C/C++/Java/Python it is equivalent to |).
Because the answer can be quite large, calculate it modulo m. This time GukiZ hasn't come up with solution, and needs you to help him!
Input
First and the only line of input contains four integers n, k, l, m (2 ≤ n ≤ 1018, 0 ≤ k ≤ 1018, 0 ≤ l ≤ 64, 1 ≤ m ≤ 109 + 7).
Output
In the single line print the number of arrays satisfying the condition above modulo m.
Examples
Input
2 1 2 10
Output
3
Input
2 1 1 3
Output
1
Input
3 3 2 10
Output
9
Note
In the first sample, satisfying arrays are {1, 1}, {3, 1}, {1, 3}.
In the second sample, only satisfying array is {1, 1}.
In the third sample, satisfying arrays are {0, 3, 3}, {1, 3, 2}, {1, 3, 3}, {2, 3, 1}, {2, 3, 3}, {3, 3, 0}, {3, 3, 1}, {3, 3, 2}, {3, 3, 3}. | instruction | 0 | 59,254 | 12 | 118,508 |
Tags: combinatorics, implementation, math, matrices, number theory
Correct Solution:
```
n,k,l,MOD = map(int, input().split())
class Matrix:
def __init__(self, a,b,c,d):
self.a, self.b, self.c, self.d = a,b,c,d
def __mul__(self, other):
return Matrix(
(self.a * other.a + self.b * other.c) % MOD,
(self.a * other.b + self.b * other.d) % MOD,
(self.c * other.a + self.d * other.c) % MOD,
(self.c * other.b + self.d * other.d) % MOD
)
def bin_pow(base, deg):
if deg == 0:
return Matrix(1,0,0,1)
tmp = bin_pow(base, deg // 2)
tmp = tmp * tmp
if deg % 2 == 1:
tmp *= base
return tmp
def Fib(n):
return bin_pow(Matrix(1,1,1,0), n).b
without_set = Fib(n+2)
with_set = (pow(2, n, MOD) + MOD - without_set) % MOD
if k >= 2**l:
print(0)
exit(0)
ans = 1 % MOD
for i in range(l):
x = k % 2;
k //= 2
if x:
ans *= with_set
else:
ans *= without_set
ans %= MOD
print(ans)
``` | output | 1 | 59,254 | 12 | 118,509 |
Provide tags and a correct Python 3 solution for this coding contest problem.
We all know that GukiZ often plays with arrays.
Now he is thinking about this problem: how many arrays a, of length n, with non-negative elements strictly less then 2l meet the following condition: <image>? Here operation <image> means bitwise AND (in Pascal it is equivalent to and, in C/C++/Java/Python it is equivalent to &), operation <image> means bitwise OR (in Pascal it is equivalent to <image>, in C/C++/Java/Python it is equivalent to |).
Because the answer can be quite large, calculate it modulo m. This time GukiZ hasn't come up with solution, and needs you to help him!
Input
First and the only line of input contains four integers n, k, l, m (2 ≤ n ≤ 1018, 0 ≤ k ≤ 1018, 0 ≤ l ≤ 64, 1 ≤ m ≤ 109 + 7).
Output
In the single line print the number of arrays satisfying the condition above modulo m.
Examples
Input
2 1 2 10
Output
3
Input
2 1 1 3
Output
1
Input
3 3 2 10
Output
9
Note
In the first sample, satisfying arrays are {1, 1}, {3, 1}, {1, 3}.
In the second sample, only satisfying array is {1, 1}.
In the third sample, satisfying arrays are {0, 3, 3}, {1, 3, 2}, {1, 3, 3}, {2, 3, 1}, {2, 3, 3}, {3, 3, 0}, {3, 3, 1}, {3, 3, 2}, {3, 3, 3}. | instruction | 0 | 59,255 | 12 | 118,510 |
Tags: combinatorics, implementation, math, matrices, number theory
Correct Solution:
```
n,k,l,MOD = map(int, input().split())
class Matrix:
def __init__(self, a,b,c,d):
self.a, self.b, self.c, self.d = a,b,c,d
def __mul__(self, other):
return Matrix(
(self.a * other.a + self.b * other.c) % MOD,
(self.a * other.b + self.b * other.d) % MOD,
(self.c * other.a + self.d * other.c) % MOD,
(self.c * other.b + self.d * other.d) % MOD
)
def bin_pow(base, deg):
if deg == 0:
return Matrix(1,0,0,1)
tmp = bin_pow(base, deg // 2)
tmp = tmp * tmp
if deg % 2 == 1:
tmp *= base
return tmp
def Fib(n):
return bin_pow(Matrix(1,1,1,0), n).b
without_set = Fib(n+2)
with_set = (pow(2, n, MOD) + MOD - without_set) % MOD
if k >= 2**l:
print(0)
exit(0)
ans = 1
if l == 0:
ans = 1
ans %= MOD
for i in range(l):
x = k % 2;
k //= 2
if x:
ans *= with_set
else:
ans *= without_set
ans %= MOD
print(ans)
``` | output | 1 | 59,255 | 12 | 118,511 |
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