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Provide a correct Python 3 solution for this coding contest problem.
Given two non-negative decimal integers $a$ and $b$, calculate their AND (logical conjunction), OR (logical disjunction) and XOR (exclusive disjunction) and print them in binary representation of 32 bits.
Constraints
* $0 \leq a, b \leq 2^{32} - 1$
Input
The input is given in the following format.
$a \; b$
Output
Print results of AND, OR and XOR in a line respectively.
Example
Input
8 10
Output
00000000000000000000000000001000
00000000000000000000000000001010
00000000000000000000000000000010
"Correct Solution:
```
if __name__ == "__main__":
a, b = map(lambda x: int(x), input().split())
mask = (1 << 32) - 1
print(f"{(a & b) & mask:032b}")
print(f"{(a | b) & mask:032b}")
print(f"{(a ^ b) & mask:032b}")
```
| 5,800 |
Provide a correct Python 3 solution for this coding contest problem.
Given two non-negative decimal integers $a$ and $b$, calculate their AND (logical conjunction), OR (logical disjunction) and XOR (exclusive disjunction) and print them in binary representation of 32 bits.
Constraints
* $0 \leq a, b \leq 2^{32} - 1$
Input
The input is given in the following format.
$a \; b$
Output
Print results of AND, OR and XOR in a line respectively.
Example
Input
8 10
Output
00000000000000000000000000001000
00000000000000000000000000001010
00000000000000000000000000000010
"Correct Solution:
```
a,b = (int(x) for x in input().split())
MAX = (1 << 32) - 1
print("{:032b}".format(a & b))
print("{:032b}".format(a | b))
print("{:032b}".format(a ^ b))
```
| 5,801 |
Provide a correct Python 3 solution for this coding contest problem.
Given two non-negative decimal integers $a$ and $b$, calculate their AND (logical conjunction), OR (logical disjunction) and XOR (exclusive disjunction) and print them in binary representation of 32 bits.
Constraints
* $0 \leq a, b \leq 2^{32} - 1$
Input
The input is given in the following format.
$a \; b$
Output
Print results of AND, OR and XOR in a line respectively.
Example
Input
8 10
Output
00000000000000000000000000001000
00000000000000000000000000001010
00000000000000000000000000000010
"Correct Solution:
```
a,b=map(int,input().split())
print(format(a&b,'032b'))
print(format(a|b,'032b'))
print(format(a^b,'032b'))
```
| 5,802 |
Provide a correct Python 3 solution for this coding contest problem.
Given two non-negative decimal integers $a$ and $b$, calculate their AND (logical conjunction), OR (logical disjunction) and XOR (exclusive disjunction) and print them in binary representation of 32 bits.
Constraints
* $0 \leq a, b \leq 2^{32} - 1$
Input
The input is given in the following format.
$a \; b$
Output
Print results of AND, OR and XOR in a line respectively.
Example
Input
8 10
Output
00000000000000000000000000001000
00000000000000000000000000001010
00000000000000000000000000000010
"Correct Solution:
```
a,b=map(int,input().split())
print(f'{a&b:032b}\n{a|b:032b}\n{a^b:032b}')
```
| 5,803 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Given two non-negative decimal integers $a$ and $b$, calculate their AND (logical conjunction), OR (logical disjunction) and XOR (exclusive disjunction) and print them in binary representation of 32 bits.
Constraints
* $0 \leq a, b \leq 2^{32} - 1$
Input
The input is given in the following format.
$a \; b$
Output
Print results of AND, OR and XOR in a line respectively.
Example
Input
8 10
Output
00000000000000000000000000001000
00000000000000000000000000001010
00000000000000000000000000000010
Submitted Solution:
```
a, b = map(int, input().split())
print('{:032b}'.format(a & b))
print('{:032b}'.format(a | b))
print('{:032b}'.format(a ^ b))
```
Yes
| 5,804 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Given two non-negative decimal integers $a$ and $b$, calculate their AND (logical conjunction), OR (logical disjunction) and XOR (exclusive disjunction) and print them in binary representation of 32 bits.
Constraints
* $0 \leq a, b \leq 2^{32} - 1$
Input
The input is given in the following format.
$a \; b$
Output
Print results of AND, OR and XOR in a line respectively.
Example
Input
8 10
Output
00000000000000000000000000001000
00000000000000000000000000001010
00000000000000000000000000000010
Submitted Solution:
```
def main():
bm = 0xffffffff
a,b = map(int, input().split())
an = (a & b) & bm
o = (a | b) & bm
xo = (a ^ b) & bm
print(format(an, "032b"))
print(format(o, "032b"))
print(format(xo, "032b"))
main()
```
Yes
| 5,805 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Given two non-negative decimal integers $a$ and $b$, calculate their AND (logical conjunction), OR (logical disjunction) and XOR (exclusive disjunction) and print them in binary representation of 32 bits.
Constraints
* $0 \leq a, b \leq 2^{32} - 1$
Input
The input is given in the following format.
$a \; b$
Output
Print results of AND, OR and XOR in a line respectively.
Example
Input
8 10
Output
00000000000000000000000000001000
00000000000000000000000000001010
00000000000000000000000000000010
Submitted Solution:
```
a, b = map(int, input().split())
print("{:032b}".format(a & b))
print("{:032b}".format(a | b))
print("{:032b}".format(a ^ b))
```
Yes
| 5,806 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Given two non-negative decimal integers $a$ and $b$, calculate their AND (logical conjunction), OR (logical disjunction) and XOR (exclusive disjunction) and print them in binary representation of 32 bits.
Constraints
* $0 \leq a, b \leq 2^{32} - 1$
Input
The input is given in the following format.
$a \; b$
Output
Print results of AND, OR and XOR in a line respectively.
Example
Input
8 10
Output
00000000000000000000000000001000
00000000000000000000000000001010
00000000000000000000000000000010
Submitted Solution:
```
def out(n):
mask = 2**32 -1
print(format(n&mask,"032b"))
a,b = map(int,input().split())
out(a&b)
out(a|b)
out(a^b)
```
Yes
| 5,807 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Astronaut Natasha arrived on Mars. She knows that the Martians are very poor aliens. To ensure a better life for the Mars citizens, their emperor decided to take tax from every tourist who visited the planet. Natasha is the inhabitant of Earth, therefore she had to pay the tax to enter the territory of Mars.
There are n banknote denominations on Mars: the value of i-th banknote is a_i. Natasha has an infinite number of banknotes of each denomination.
Martians have k fingers on their hands, so they use a number system with base k. In addition, the Martians consider the digit d (in the number system with base k) divine. Thus, if the last digit in Natasha's tax amount written in the number system with the base k is d, the Martians will be happy. Unfortunately, Natasha does not know the Martians' divine digit yet.
Determine for which values d Natasha can make the Martians happy.
Natasha can use only her banknotes. Martians don't give her change.
Input
The first line contains two integers n and k (1 ≤ n ≤ 100 000, 2 ≤ k ≤ 100 000) — the number of denominations of banknotes and the base of the number system on Mars.
The second line contains n integers a_1, a_2, …, a_n (1 ≤ a_i ≤ 10^9) — denominations of banknotes on Mars.
All numbers are given in decimal notation.
Output
On the first line output the number of values d for which Natasha can make the Martians happy.
In the second line, output all these values in increasing order.
Print all numbers in decimal notation.
Examples
Input
2 8
12 20
Output
2
0 4
Input
3 10
10 20 30
Output
1
0
Note
Consider the first test case. It uses the octal number system.
If you take one banknote with the value of 12, you will get 14_8 in octal system. The last digit is 4_8.
If you take one banknote with the value of 12 and one banknote with the value of 20, the total value will be 32. In the octal system, it is 40_8. The last digit is 0_8.
If you take two banknotes with the value of 20, the total value will be 40, this is 50_8 in the octal system. The last digit is 0_8.
No other digits other than 0_8 and 4_8 can be obtained. Digits 0_8 and 4_8 could also be obtained in other ways.
The second test case uses the decimal number system. The nominals of all banknotes end with zero, so Natasha can give the Martians only the amount whose decimal notation also ends with zero.
Tags: number theory
Correct Solution:
```
from math import gcd
n, k = map(int, input().split())
a = list(map(int, input().split()))
g = 0
for i in a:
g = gcd(g, i)
ans = set()
ans.add(g % k)
for i in range(k):
ans.add((i * g) % k)
print(len(ans))
print(*sorted(list(ans)))
```
| 5,808 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Astronaut Natasha arrived on Mars. She knows that the Martians are very poor aliens. To ensure a better life for the Mars citizens, their emperor decided to take tax from every tourist who visited the planet. Natasha is the inhabitant of Earth, therefore she had to pay the tax to enter the territory of Mars.
There are n banknote denominations on Mars: the value of i-th banknote is a_i. Natasha has an infinite number of banknotes of each denomination.
Martians have k fingers on their hands, so they use a number system with base k. In addition, the Martians consider the digit d (in the number system with base k) divine. Thus, if the last digit in Natasha's tax amount written in the number system with the base k is d, the Martians will be happy. Unfortunately, Natasha does not know the Martians' divine digit yet.
Determine for which values d Natasha can make the Martians happy.
Natasha can use only her banknotes. Martians don't give her change.
Input
The first line contains two integers n and k (1 ≤ n ≤ 100 000, 2 ≤ k ≤ 100 000) — the number of denominations of banknotes and the base of the number system on Mars.
The second line contains n integers a_1, a_2, …, a_n (1 ≤ a_i ≤ 10^9) — denominations of banknotes on Mars.
All numbers are given in decimal notation.
Output
On the first line output the number of values d for which Natasha can make the Martians happy.
In the second line, output all these values in increasing order.
Print all numbers in decimal notation.
Examples
Input
2 8
12 20
Output
2
0 4
Input
3 10
10 20 30
Output
1
0
Note
Consider the first test case. It uses the octal number system.
If you take one banknote with the value of 12, you will get 14_8 in octal system. The last digit is 4_8.
If you take one banknote with the value of 12 and one banknote with the value of 20, the total value will be 32. In the octal system, it is 40_8. The last digit is 0_8.
If you take two banknotes with the value of 20, the total value will be 40, this is 50_8 in the octal system. The last digit is 0_8.
No other digits other than 0_8 and 4_8 can be obtained. Digits 0_8 and 4_8 could also be obtained in other ways.
The second test case uses the decimal number system. The nominals of all banknotes end with zero, so Natasha can give the Martians only the amount whose decimal notation also ends with zero.
Tags: number theory
Correct Solution:
```
#Code by Sounak, IIESTS
#------------------------------warmup----------------------------
import os
import sys
import math
from io import BytesIO, IOBase
from fractions import Fraction
import collections
from itertools import permutations
from collections import defaultdict
import threading
BUFSIZE = 8192
class FastIO(IOBase):
newlines = 0
def __init__(self, file):
self._fd = file.fileno()
self.buffer = BytesIO()
self.writable = "x" in file.mode or "r" not in file.mode
self.write = self.buffer.write if self.writable else None
def read(self):
while True:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
if not b:
break
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines = 0
return self.buffer.read()
def readline(self):
while self.newlines == 0:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
self.newlines = b.count(b"\n") + (not b)
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines -= 1
return self.buffer.readline()
def flush(self):
if self.writable:
os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
self.write = lambda s: self.buffer.write(s.encode("ascii"))
self.read = lambda: self.buffer.read().decode("ascii")
self.readline = lambda: self.buffer.readline().decode("ascii")
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")
#-------------------game starts now-----------------------------------------------------
n,k=map(int,input().split())
a=list(map(int,input().split()))
s=set()
g=a[0]
for i in a:
g=math.gcd(g,i)
#print(g)
for i in range (k):
s.add((g*i)%k)
l=list(s)
l.sort()
print(len(s))
print(*l)
```
| 5,809 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Astronaut Natasha arrived on Mars. She knows that the Martians are very poor aliens. To ensure a better life for the Mars citizens, their emperor decided to take tax from every tourist who visited the planet. Natasha is the inhabitant of Earth, therefore she had to pay the tax to enter the territory of Mars.
There are n banknote denominations on Mars: the value of i-th banknote is a_i. Natasha has an infinite number of banknotes of each denomination.
Martians have k fingers on their hands, so they use a number system with base k. In addition, the Martians consider the digit d (in the number system with base k) divine. Thus, if the last digit in Natasha's tax amount written in the number system with the base k is d, the Martians will be happy. Unfortunately, Natasha does not know the Martians' divine digit yet.
Determine for which values d Natasha can make the Martians happy.
Natasha can use only her banknotes. Martians don't give her change.
Input
The first line contains two integers n and k (1 ≤ n ≤ 100 000, 2 ≤ k ≤ 100 000) — the number of denominations of banknotes and the base of the number system on Mars.
The second line contains n integers a_1, a_2, …, a_n (1 ≤ a_i ≤ 10^9) — denominations of banknotes on Mars.
All numbers are given in decimal notation.
Output
On the first line output the number of values d for which Natasha can make the Martians happy.
In the second line, output all these values in increasing order.
Print all numbers in decimal notation.
Examples
Input
2 8
12 20
Output
2
0 4
Input
3 10
10 20 30
Output
1
0
Note
Consider the first test case. It uses the octal number system.
If you take one banknote with the value of 12, you will get 14_8 in octal system. The last digit is 4_8.
If you take one banknote with the value of 12 and one banknote with the value of 20, the total value will be 32. In the octal system, it is 40_8. The last digit is 0_8.
If you take two banknotes with the value of 20, the total value will be 40, this is 50_8 in the octal system. The last digit is 0_8.
No other digits other than 0_8 and 4_8 can be obtained. Digits 0_8 and 4_8 could also be obtained in other ways.
The second test case uses the decimal number system. The nominals of all banknotes end with zero, so Natasha can give the Martians only the amount whose decimal notation also ends with zero.
Tags: number theory
Correct Solution:
```
def gcd(a,b):
if b == 0:
return a
if b > a:
return gcd(b,a)
return gcd(b,a%b)
n,k = list(map(int,input().split()))
l = list(map(int,input().split()))
out = k
for i in l:
out = gcd(i,out)
print(k//out)
print(' '.join(list(map(str,range(0,k,out)))))
```
| 5,810 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Astronaut Natasha arrived on Mars. She knows that the Martians are very poor aliens. To ensure a better life for the Mars citizens, their emperor decided to take tax from every tourist who visited the planet. Natasha is the inhabitant of Earth, therefore she had to pay the tax to enter the territory of Mars.
There are n banknote denominations on Mars: the value of i-th banknote is a_i. Natasha has an infinite number of banknotes of each denomination.
Martians have k fingers on their hands, so they use a number system with base k. In addition, the Martians consider the digit d (in the number system with base k) divine. Thus, if the last digit in Natasha's tax amount written in the number system with the base k is d, the Martians will be happy. Unfortunately, Natasha does not know the Martians' divine digit yet.
Determine for which values d Natasha can make the Martians happy.
Natasha can use only her banknotes. Martians don't give her change.
Input
The first line contains two integers n and k (1 ≤ n ≤ 100 000, 2 ≤ k ≤ 100 000) — the number of denominations of banknotes and the base of the number system on Mars.
The second line contains n integers a_1, a_2, …, a_n (1 ≤ a_i ≤ 10^9) — denominations of banknotes on Mars.
All numbers are given in decimal notation.
Output
On the first line output the number of values d for which Natasha can make the Martians happy.
In the second line, output all these values in increasing order.
Print all numbers in decimal notation.
Examples
Input
2 8
12 20
Output
2
0 4
Input
3 10
10 20 30
Output
1
0
Note
Consider the first test case. It uses the octal number system.
If you take one banknote with the value of 12, you will get 14_8 in octal system. The last digit is 4_8.
If you take one banknote with the value of 12 and one banknote with the value of 20, the total value will be 32. In the octal system, it is 40_8. The last digit is 0_8.
If you take two banknotes with the value of 20, the total value will be 40, this is 50_8 in the octal system. The last digit is 0_8.
No other digits other than 0_8 and 4_8 can be obtained. Digits 0_8 and 4_8 could also be obtained in other ways.
The second test case uses the decimal number system. The nominals of all banknotes end with zero, so Natasha can give the Martians only the amount whose decimal notation also ends with zero.
Tags: number theory
Correct Solution:
```
def nod(a,b):
while a!=0 and b!=0:
if a>b:
a,b=b,a%b
else:
b,a=a,b%a
return a+b
n ,k = map(int, input().split())
a = [int(j) for j in input().split()]
c = a[0]
for i in range(1,n):
c = nod(c,a[i])
if c==1:
break
e = nod(c,k)
if c==1 or e==1:
print(k)
for i in range(k):
print(i, end=" ")
if e>1:
c = k//e
print(c)
for i in range(c):
print(i*e, end=' ')
```
| 5,811 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Astronaut Natasha arrived on Mars. She knows that the Martians are very poor aliens. To ensure a better life for the Mars citizens, their emperor decided to take tax from every tourist who visited the planet. Natasha is the inhabitant of Earth, therefore she had to pay the tax to enter the territory of Mars.
There are n banknote denominations on Mars: the value of i-th banknote is a_i. Natasha has an infinite number of banknotes of each denomination.
Martians have k fingers on their hands, so they use a number system with base k. In addition, the Martians consider the digit d (in the number system with base k) divine. Thus, if the last digit in Natasha's tax amount written in the number system with the base k is d, the Martians will be happy. Unfortunately, Natasha does not know the Martians' divine digit yet.
Determine for which values d Natasha can make the Martians happy.
Natasha can use only her banknotes. Martians don't give her change.
Input
The first line contains two integers n and k (1 ≤ n ≤ 100 000, 2 ≤ k ≤ 100 000) — the number of denominations of banknotes and the base of the number system on Mars.
The second line contains n integers a_1, a_2, …, a_n (1 ≤ a_i ≤ 10^9) — denominations of banknotes on Mars.
All numbers are given in decimal notation.
Output
On the first line output the number of values d for which Natasha can make the Martians happy.
In the second line, output all these values in increasing order.
Print all numbers in decimal notation.
Examples
Input
2 8
12 20
Output
2
0 4
Input
3 10
10 20 30
Output
1
0
Note
Consider the first test case. It uses the octal number system.
If you take one banknote with the value of 12, you will get 14_8 in octal system. The last digit is 4_8.
If you take one banknote with the value of 12 and one banknote with the value of 20, the total value will be 32. In the octal system, it is 40_8. The last digit is 0_8.
If you take two banknotes with the value of 20, the total value will be 40, this is 50_8 in the octal system. The last digit is 0_8.
No other digits other than 0_8 and 4_8 can be obtained. Digits 0_8 and 4_8 could also be obtained in other ways.
The second test case uses the decimal number system. The nominals of all banknotes end with zero, so Natasha can give the Martians only the amount whose decimal notation also ends with zero.
Tags: number theory
Correct Solution:
```
# ---------------------------iye ha aam zindegi---------------------------------------------
import math
import random
import heapq, bisect
import sys
from collections import deque, defaultdict
from fractions import Fraction
import sys
import threading
from collections import defaultdict
#threading.stack_size(10**8)
mod = 10 ** 9 + 7
mod1 = 998244353
# ------------------------------warmup----------------------------
import os
import sys
from io import BytesIO, IOBase
#sys.setrecursionlimit(300000)
BUFSIZE = 8192
class FastIO(IOBase):
newlines = 0
def __init__(self, file):
self._fd = file.fileno()
self.buffer = BytesIO()
self.writable = "x" in file.mode or "r" not in file.mode
self.write = self.buffer.write if self.writable else None
def read(self):
while True:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
if not b:
break
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines = 0
return self.buffer.read()
def readline(self):
while self.newlines == 0:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
self.newlines = b.count(b"\n") + (not b)
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines -= 1
return self.buffer.readline()
def flush(self):
if self.writable:
os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
self.write = lambda s: self.buffer.write(s.encode("ascii"))
self.read = lambda: self.buffer.read().decode("ascii")
self.readline = lambda: self.buffer.readline().decode("ascii")
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")
# -------------------game starts now----------------------------------------------------import math
class TreeNode:
def __init__(self, k, v):
self.key = k
self.value = v
self.left = None
self.right = None
self.parent = None
self.height = 1
self.num_left = 1
self.num_total = 1
class AvlTree:
def __init__(self):
self._tree = None
def add(self, k, v):
if not self._tree:
self._tree = TreeNode(k, v)
return
node = self._add(k, v)
if node:
self._rebalance(node)
def _add(self, k, v):
node = self._tree
while node:
if k < node.key:
if node.left:
node = node.left
else:
node.left = TreeNode(k, v)
node.left.parent = node
return node.left
elif node.key < k:
if node.right:
node = node.right
else:
node.right = TreeNode(k, v)
node.right.parent = node
return node.right
else:
node.value = v
return
@staticmethod
def get_height(x):
return x.height if x else 0
@staticmethod
def get_num_total(x):
return x.num_total if x else 0
def _rebalance(self, node):
n = node
while n:
lh = self.get_height(n.left)
rh = self.get_height(n.right)
n.height = max(lh, rh) + 1
balance_factor = lh - rh
n.num_total = 1 + self.get_num_total(n.left) + self.get_num_total(n.right)
n.num_left = 1 + self.get_num_total(n.left)
if balance_factor > 1:
if self.get_height(n.left.left) < self.get_height(n.left.right):
self._rotate_left(n.left)
self._rotate_right(n)
elif balance_factor < -1:
if self.get_height(n.right.right) < self.get_height(n.right.left):
self._rotate_right(n.right)
self._rotate_left(n)
else:
n = n.parent
def _remove_one(self, node):
"""
Side effect!!! Changes node. Node should have exactly one child
"""
replacement = node.left or node.right
if node.parent:
if AvlTree._is_left(node):
node.parent.left = replacement
else:
node.parent.right = replacement
replacement.parent = node.parent
node.parent = None
else:
self._tree = replacement
replacement.parent = None
node.left = None
node.right = None
node.parent = None
self._rebalance(replacement)
def _remove_leaf(self, node):
if node.parent:
if AvlTree._is_left(node):
node.parent.left = None
else:
node.parent.right = None
self._rebalance(node.parent)
else:
self._tree = None
node.parent = None
node.left = None
node.right = None
def remove(self, k):
node = self._get_node(k)
if not node:
return
if AvlTree._is_leaf(node):
self._remove_leaf(node)
return
if node.left and node.right:
nxt = AvlTree._get_next(node)
node.key = nxt.key
node.value = nxt.value
if self._is_leaf(nxt):
self._remove_leaf(nxt)
else:
self._remove_one(nxt)
self._rebalance(node)
else:
self._remove_one(node)
def get(self, k):
node = self._get_node(k)
return node.value if node else -1
def _get_node(self, k):
if not self._tree:
return None
node = self._tree
while node:
if k < node.key:
node = node.left
elif node.key < k:
node = node.right
else:
return node
return None
def get_at(self, pos):
x = pos + 1
node = self._tree
while node:
if x < node.num_left:
node = node.left
elif node.num_left < x:
x -= node.num_left
node = node.right
else:
return (node.key, node.value)
raise IndexError("Out of ranges")
@staticmethod
def _is_left(node):
return node.parent.left and node.parent.left == node
@staticmethod
def _is_leaf(node):
return node.left is None and node.right is None
def _rotate_right(self, node):
if not node.parent:
self._tree = node.left
node.left.parent = None
elif AvlTree._is_left(node):
node.parent.left = node.left
node.left.parent = node.parent
else:
node.parent.right = node.left
node.left.parent = node.parent
bk = node.left.right
node.left.right = node
node.parent = node.left
node.left = bk
if bk:
bk.parent = node
node.height = max(self.get_height(node.left), self.get_height(node.right)) + 1
node.num_total = 1 + self.get_num_total(node.left) + self.get_num_total(node.right)
node.num_left = 1 + self.get_num_total(node.left)
def _rotate_left(self, node):
if not node.parent:
self._tree = node.right
node.right.parent = None
elif AvlTree._is_left(node):
node.parent.left = node.right
node.right.parent = node.parent
else:
node.parent.right = node.right
node.right.parent = node.parent
bk = node.right.left
node.right.left = node
node.parent = node.right
node.right = bk
if bk:
bk.parent = node
node.height = max(self.get_height(node.left), self.get_height(node.right)) + 1
node.num_total = 1 + self.get_num_total(node.left) + self.get_num_total(node.right)
node.num_left = 1 + self.get_num_total(node.left)
@staticmethod
def _get_next(node):
if not node.right:
return node.parent
n = node.right
while n.left:
n = n.left
return n
# -----------------------------------------------binary seacrh tree---------------------------------------
class SegmentTree1:
def __init__(self, data, default=0, func=lambda a, b: max(a , b)):
"""initialize the segment tree with data"""
self._default = default
self._func = func
self._len = len(data)
self._size = _size = 1 << (self._len - 1).bit_length()
self.data = [default] * (2 * _size)
self.data[_size:_size + self._len] = data
for i in reversed(range(_size)):
self.data[i] = func(self.data[i + i], self.data[i + i + 1])
def __delitem__(self, idx):
self[idx] = self._default
def __getitem__(self, idx):
return self.data[idx + self._size]
def __setitem__(self, idx, value):
idx += self._size
self.data[idx] = value
idx >>= 1
while idx:
self.data[idx] = self._func(self.data[2 * idx], self.data[2 * idx + 1])
idx >>= 1
def __len__(self):
return self._len
def query(self, start, stop):
if start == stop:
return self.__getitem__(start)
stop += 1
start += self._size
stop += self._size
res = self._default
while start < stop:
if start & 1:
res = self._func(res, self.data[start])
start += 1
if stop & 1:
stop -= 1
res = self._func(res, self.data[stop])
start >>= 1
stop >>= 1
return res
def __repr__(self):
return "SegmentTree({0})".format(self.data)
# -------------------game starts now----------------------------------------------------import math
class SegmentTree:
def __init__(self, data, default=0, func=lambda a, b:a + b):
"""initialize the segment tree with data"""
self._default = default
self._func = func
self._len = len(data)
self._size = _size = 1 << (self._len - 1).bit_length()
self.data = [default] * (2 * _size)
self.data[_size:_size + self._len] = data
for i in reversed(range(_size)):
self.data[i] = func(self.data[i + i], self.data[i + i + 1])
def __delitem__(self, idx):
self[idx] = self._default
def __getitem__(self, idx):
return self.data[idx + self._size]
def __setitem__(self, idx, value):
idx += self._size
self.data[idx] = value
idx >>= 1
while idx:
self.data[idx] = self._func(self.data[2 * idx], self.data[2 * idx + 1])
idx >>= 1
def __len__(self):
return self._len
def query(self, start, stop):
if start == stop:
return self.__getitem__(start)
stop += 1
start += self._size
stop += self._size
res = self._default
while start < stop:
if start & 1:
res = self._func(res, self.data[start])
start += 1
if stop & 1:
stop -= 1
res = self._func(res, self.data[stop])
start >>= 1
stop >>= 1
return res
def __repr__(self):
return "SegmentTree({0})".format(self.data)
# -------------------------------iye ha chutiya zindegi-------------------------------------
class Factorial:
def __init__(self, MOD):
self.MOD = MOD
self.factorials = [1, 1]
self.invModulos = [0, 1]
self.invFactorial_ = [1, 1]
def calc(self, n):
if n <= -1:
print("Invalid argument to calculate n!")
print("n must be non-negative value. But the argument was " + str(n))
exit()
if n < len(self.factorials):
return self.factorials[n]
nextArr = [0] * (n + 1 - len(self.factorials))
initialI = len(self.factorials)
prev = self.factorials[-1]
m = self.MOD
for i in range(initialI, n + 1):
prev = nextArr[i - initialI] = prev * i % m
self.factorials += nextArr
return self.factorials[n]
def inv(self, n):
if n <= -1:
print("Invalid argument to calculate n^(-1)")
print("n must be non-negative value. But the argument was " + str(n))
exit()
p = self.MOD
pi = n % p
if pi < len(self.invModulos):
return self.invModulos[pi]
nextArr = [0] * (n + 1 - len(self.invModulos))
initialI = len(self.invModulos)
for i in range(initialI, min(p, n + 1)):
next = -self.invModulos[p % i] * (p // i) % p
self.invModulos.append(next)
return self.invModulos[pi]
def invFactorial(self, n):
if n <= -1:
print("Invalid argument to calculate (n^(-1))!")
print("n must be non-negative value. But the argument was " + str(n))
exit()
if n < len(self.invFactorial_):
return self.invFactorial_[n]
self.inv(n) # To make sure already calculated n^-1
nextArr = [0] * (n + 1 - len(self.invFactorial_))
initialI = len(self.invFactorial_)
prev = self.invFactorial_[-1]
p = self.MOD
for i in range(initialI, n + 1):
prev = nextArr[i - initialI] = (prev * self.invModulos[i % p]) % p
self.invFactorial_ += nextArr
return self.invFactorial_[n]
class Combination:
def __init__(self, MOD):
self.MOD = MOD
self.factorial = Factorial(MOD)
def ncr(self, n, k):
if k < 0 or n < k:
return 0
k = min(k, n - k)
f = self.factorial
return f.calc(n) * f.invFactorial(max(n - k, k)) * f.invFactorial(min(k, n - k)) % self.MOD
# --------------------------------------iye ha combinations ka zindegi---------------------------------
def powm(a, n, m):
if a == 1 or n == 0:
return 1
if n % 2 == 0:
s = powm(a, n // 2, m)
return s * s % m
else:
return a * powm(a, n - 1, m) % m
# --------------------------------------iye ha power ka zindegi---------------------------------
def sort_list(list1, list2):
zipped_pairs = zip(list2, list1)
z = [x for _, x in sorted(zipped_pairs)]
return z
# --------------------------------------------------product----------------------------------------
def product(l):
por = 1
for i in range(len(l)):
por *= l[i]
return por
# --------------------------------------------------binary----------------------------------------
def binarySearchCount(arr, n, key):
left = 0
right = n - 1
count = 0
while (left <= right):
mid = int((right + left) / 2)
# Check if middle element is
# less than or equal to key
if (arr[mid] < key):
count = mid + 1
left = mid + 1
# If key is smaller, ignore right half
else:
right = mid - 1
return count
# --------------------------------------------------binary----------------------------------------
def countdig(n):
c = 0
while (n > 0):
n //= 10
c += 1
return c
def binary(x, length):
y = bin(x)[2:]
return y if len(y) >= length else "0" * (length - len(y)) + y
def countGreater(arr, n, k):
l = 0
r = n - 1
# Stores the index of the left most element
# from the array which is greater than k
leftGreater = n
# Finds number of elements greater than k
while (l <= r):
m = int(l + (r - l) / 2)
if (arr[m] >= k):
leftGreater = m
r = m - 1
# If mid element is less than
# or equal to k update l
else:
l = m + 1
# Return the count of elements
# greater than k
return (n - leftGreater)
# --------------------------------------------------binary------------------------------------
n,m=map(int,input().split())
k=m
def convert(no):
return no%k
l=list(map(int,input().split()))
l=[convert(l[i]) for i in range(n)]
g=0
for i in range(n):
g=math.gcd(g,l[i])
if g==0:
print(1)
print(0)
sys.exit(0)
g=math.gcd(g,k)
t=0
l=[]
while(t<k):
l.append(t%k)
t+=g
l=set(l)
print(len(l))
print(*sorted(list(l)))
```
| 5,812 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Astronaut Natasha arrived on Mars. She knows that the Martians are very poor aliens. To ensure a better life for the Mars citizens, their emperor decided to take tax from every tourist who visited the planet. Natasha is the inhabitant of Earth, therefore she had to pay the tax to enter the territory of Mars.
There are n banknote denominations on Mars: the value of i-th banknote is a_i. Natasha has an infinite number of banknotes of each denomination.
Martians have k fingers on their hands, so they use a number system with base k. In addition, the Martians consider the digit d (in the number system with base k) divine. Thus, if the last digit in Natasha's tax amount written in the number system with the base k is d, the Martians will be happy. Unfortunately, Natasha does not know the Martians' divine digit yet.
Determine for which values d Natasha can make the Martians happy.
Natasha can use only her banknotes. Martians don't give her change.
Input
The first line contains two integers n and k (1 ≤ n ≤ 100 000, 2 ≤ k ≤ 100 000) — the number of denominations of banknotes and the base of the number system on Mars.
The second line contains n integers a_1, a_2, …, a_n (1 ≤ a_i ≤ 10^9) — denominations of banknotes on Mars.
All numbers are given in decimal notation.
Output
On the first line output the number of values d for which Natasha can make the Martians happy.
In the second line, output all these values in increasing order.
Print all numbers in decimal notation.
Examples
Input
2 8
12 20
Output
2
0 4
Input
3 10
10 20 30
Output
1
0
Note
Consider the first test case. It uses the octal number system.
If you take one banknote with the value of 12, you will get 14_8 in octal system. The last digit is 4_8.
If you take one banknote with the value of 12 and one banknote with the value of 20, the total value will be 32. In the octal system, it is 40_8. The last digit is 0_8.
If you take two banknotes with the value of 20, the total value will be 40, this is 50_8 in the octal system. The last digit is 0_8.
No other digits other than 0_8 and 4_8 can be obtained. Digits 0_8 and 4_8 could also be obtained in other ways.
The second test case uses the decimal number system. The nominals of all banknotes end with zero, so Natasha can give the Martians only the amount whose decimal notation also ends with zero.
Tags: number theory
Correct Solution:
```
from math import gcd
n, k = list(map(int, input().split()))
a = list(map(int, input().split()))[:n]
a = set([b%k for b in a])
a = set ([gcd(k, b) for b in a if b>0])
if len(a) == 0:
print(1)
print(0)
exit()
mg = min(a)
for d in a:
mg = gcd(mg,d)
if mg == 1:
break
res = [mg * j for j in range(k//mg)]
print(len(res))
print(*res)
```
| 5,813 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Astronaut Natasha arrived on Mars. She knows that the Martians are very poor aliens. To ensure a better life for the Mars citizens, their emperor decided to take tax from every tourist who visited the planet. Natasha is the inhabitant of Earth, therefore she had to pay the tax to enter the territory of Mars.
There are n banknote denominations on Mars: the value of i-th banknote is a_i. Natasha has an infinite number of banknotes of each denomination.
Martians have k fingers on their hands, so they use a number system with base k. In addition, the Martians consider the digit d (in the number system with base k) divine. Thus, if the last digit in Natasha's tax amount written in the number system with the base k is d, the Martians will be happy. Unfortunately, Natasha does not know the Martians' divine digit yet.
Determine for which values d Natasha can make the Martians happy.
Natasha can use only her banknotes. Martians don't give her change.
Input
The first line contains two integers n and k (1 ≤ n ≤ 100 000, 2 ≤ k ≤ 100 000) — the number of denominations of banknotes and the base of the number system on Mars.
The second line contains n integers a_1, a_2, …, a_n (1 ≤ a_i ≤ 10^9) — denominations of banknotes on Mars.
All numbers are given in decimal notation.
Output
On the first line output the number of values d for which Natasha can make the Martians happy.
In the second line, output all these values in increasing order.
Print all numbers in decimal notation.
Examples
Input
2 8
12 20
Output
2
0 4
Input
3 10
10 20 30
Output
1
0
Note
Consider the first test case. It uses the octal number system.
If you take one banknote with the value of 12, you will get 14_8 in octal system. The last digit is 4_8.
If you take one banknote with the value of 12 and one banknote with the value of 20, the total value will be 32. In the octal system, it is 40_8. The last digit is 0_8.
If you take two banknotes with the value of 20, the total value will be 40, this is 50_8 in the octal system. The last digit is 0_8.
No other digits other than 0_8 and 4_8 can be obtained. Digits 0_8 and 4_8 could also be obtained in other ways.
The second test case uses the decimal number system. The nominals of all banknotes end with zero, so Natasha can give the Martians only the amount whose decimal notation also ends with zero.
Tags: number theory
Correct Solution:
```
from functools import reduce
from math import ceil, floor
def gcd(x, y):
"""greatest common divisor of x and y"""
while y:
x, y = y, x % y
return x
def extended_gcd(a, b):
"""returns gcd(a, b), s, r s.t. a * s + b * r == gcd(a, b)"""
s, old_s = 0, 1
r, old_r = b, a
while r:
q = old_r // r
old_r, r = r, old_r - q * r
old_s, s = s, old_s - q * s
return old_r, old_s, (old_r - old_s * a) // b if b else 0
gcdm = lambda args: reduce(gcd, args, 0)
lcm = lambda a, b: a * b // gcd(a, b)
lcmm = lambda *args: reduce(lcm, args, 1)
def main():
n, k = map(int, input().split())
a = list(map(int, input().split()))
g = gcdm(a + [k])
print(k // g)
print(" ".join([str(x * g) for x in range(k // g)]))
main()
# sum(ai xi) - kl = d
# ž g <= k
```
| 5,814 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Astronaut Natasha arrived on Mars. She knows that the Martians are very poor aliens. To ensure a better life for the Mars citizens, their emperor decided to take tax from every tourist who visited the planet. Natasha is the inhabitant of Earth, therefore she had to pay the tax to enter the territory of Mars.
There are n banknote denominations on Mars: the value of i-th banknote is a_i. Natasha has an infinite number of banknotes of each denomination.
Martians have k fingers on their hands, so they use a number system with base k. In addition, the Martians consider the digit d (in the number system with base k) divine. Thus, if the last digit in Natasha's tax amount written in the number system with the base k is d, the Martians will be happy. Unfortunately, Natasha does not know the Martians' divine digit yet.
Determine for which values d Natasha can make the Martians happy.
Natasha can use only her banknotes. Martians don't give her change.
Input
The first line contains two integers n and k (1 ≤ n ≤ 100 000, 2 ≤ k ≤ 100 000) — the number of denominations of banknotes and the base of the number system on Mars.
The second line contains n integers a_1, a_2, …, a_n (1 ≤ a_i ≤ 10^9) — denominations of banknotes on Mars.
All numbers are given in decimal notation.
Output
On the first line output the number of values d for which Natasha can make the Martians happy.
In the second line, output all these values in increasing order.
Print all numbers in decimal notation.
Examples
Input
2 8
12 20
Output
2
0 4
Input
3 10
10 20 30
Output
1
0
Note
Consider the first test case. It uses the octal number system.
If you take one banknote with the value of 12, you will get 14_8 in octal system. The last digit is 4_8.
If you take one banknote with the value of 12 and one banknote with the value of 20, the total value will be 32. In the octal system, it is 40_8. The last digit is 0_8.
If you take two banknotes with the value of 20, the total value will be 40, this is 50_8 in the octal system. The last digit is 0_8.
No other digits other than 0_8 and 4_8 can be obtained. Digits 0_8 and 4_8 could also be obtained in other ways.
The second test case uses the decimal number system. The nominals of all banknotes end with zero, so Natasha can give the Martians only the amount whose decimal notation also ends with zero.
Tags: number theory
Correct Solution:
```
n, k = map(int,input().split())
v = list(map(int,input().split()))
def gcd(a,b):
if a < b:
return gcd(b,a)
if b == 0:
return a
else:
return gcd(b, a%b)
g = v[0]
for i in range(1,n):
g = gcd(g, v[i])
lst = set()
for i in range(k):
lst.add(g*i % k)
lst = sorted(list(lst))
print(len(lst))
print(' '.join(map(str,lst)))
```
| 5,815 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Astronaut Natasha arrived on Mars. She knows that the Martians are very poor aliens. To ensure a better life for the Mars citizens, their emperor decided to take tax from every tourist who visited the planet. Natasha is the inhabitant of Earth, therefore she had to pay the tax to enter the territory of Mars.
There are n banknote denominations on Mars: the value of i-th banknote is a_i. Natasha has an infinite number of banknotes of each denomination.
Martians have k fingers on their hands, so they use a number system with base k. In addition, the Martians consider the digit d (in the number system with base k) divine. Thus, if the last digit in Natasha's tax amount written in the number system with the base k is d, the Martians will be happy. Unfortunately, Natasha does not know the Martians' divine digit yet.
Determine for which values d Natasha can make the Martians happy.
Natasha can use only her banknotes. Martians don't give her change.
Input
The first line contains two integers n and k (1 ≤ n ≤ 100 000, 2 ≤ k ≤ 100 000) — the number of denominations of banknotes and the base of the number system on Mars.
The second line contains n integers a_1, a_2, …, a_n (1 ≤ a_i ≤ 10^9) — denominations of banknotes on Mars.
All numbers are given in decimal notation.
Output
On the first line output the number of values d for which Natasha can make the Martians happy.
In the second line, output all these values in increasing order.
Print all numbers in decimal notation.
Examples
Input
2 8
12 20
Output
2
0 4
Input
3 10
10 20 30
Output
1
0
Note
Consider the first test case. It uses the octal number system.
If you take one banknote with the value of 12, you will get 14_8 in octal system. The last digit is 4_8.
If you take one banknote with the value of 12 and one banknote with the value of 20, the total value will be 32. In the octal system, it is 40_8. The last digit is 0_8.
If you take two banknotes with the value of 20, the total value will be 40, this is 50_8 in the octal system. The last digit is 0_8.
No other digits other than 0_8 and 4_8 can be obtained. Digits 0_8 and 4_8 could also be obtained in other ways.
The second test case uses the decimal number system. The nominals of all banknotes end with zero, so Natasha can give the Martians only the amount whose decimal notation also ends with zero.
Submitted Solution:
```
import os,io,math
input=io.BytesIO(os.read(0,os.fstat(0).st_size)).readline
n,k=map(int,input().split())
a=list(map(int,input().split()))
r=k
for i in range(n):
r=math.gcd(r,a[i])
ans=[]
for i in range(0,k,r):
ans.append(i)
print(len(ans))
print(' '.join(map(str,ans)))
```
Yes
| 5,816 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Astronaut Natasha arrived on Mars. She knows that the Martians are very poor aliens. To ensure a better life for the Mars citizens, their emperor decided to take tax from every tourist who visited the planet. Natasha is the inhabitant of Earth, therefore she had to pay the tax to enter the territory of Mars.
There are n banknote denominations on Mars: the value of i-th banknote is a_i. Natasha has an infinite number of banknotes of each denomination.
Martians have k fingers on their hands, so they use a number system with base k. In addition, the Martians consider the digit d (in the number system with base k) divine. Thus, if the last digit in Natasha's tax amount written in the number system with the base k is d, the Martians will be happy. Unfortunately, Natasha does not know the Martians' divine digit yet.
Determine for which values d Natasha can make the Martians happy.
Natasha can use only her banknotes. Martians don't give her change.
Input
The first line contains two integers n and k (1 ≤ n ≤ 100 000, 2 ≤ k ≤ 100 000) — the number of denominations of banknotes and the base of the number system on Mars.
The second line contains n integers a_1, a_2, …, a_n (1 ≤ a_i ≤ 10^9) — denominations of banknotes on Mars.
All numbers are given in decimal notation.
Output
On the first line output the number of values d for which Natasha can make the Martians happy.
In the second line, output all these values in increasing order.
Print all numbers in decimal notation.
Examples
Input
2 8
12 20
Output
2
0 4
Input
3 10
10 20 30
Output
1
0
Note
Consider the first test case. It uses the octal number system.
If you take one banknote with the value of 12, you will get 14_8 in octal system. The last digit is 4_8.
If you take one banknote with the value of 12 and one banknote with the value of 20, the total value will be 32. In the octal system, it is 40_8. The last digit is 0_8.
If you take two banknotes with the value of 20, the total value will be 40, this is 50_8 in the octal system. The last digit is 0_8.
No other digits other than 0_8 and 4_8 can be obtained. Digits 0_8 and 4_8 could also be obtained in other ways.
The second test case uses the decimal number system. The nominals of all banknotes end with zero, so Natasha can give the Martians only the amount whose decimal notation also ends with zero.
Submitted Solution:
```
from math import gcd
n,k=map(int,input().split())
arr=list(map(int,input().split()))
gd=0
for i in range(n):
gd=gcd(gd,arr[i])
count=gd
s=set()
for i in range(k):
s.add(count %k)
count +=gd
s=list(s)
s.sort()
print(len(s))
print(*s)
```
Yes
| 5,817 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Astronaut Natasha arrived on Mars. She knows that the Martians are very poor aliens. To ensure a better life for the Mars citizens, their emperor decided to take tax from every tourist who visited the planet. Natasha is the inhabitant of Earth, therefore she had to pay the tax to enter the territory of Mars.
There are n banknote denominations on Mars: the value of i-th banknote is a_i. Natasha has an infinite number of banknotes of each denomination.
Martians have k fingers on their hands, so they use a number system with base k. In addition, the Martians consider the digit d (in the number system with base k) divine. Thus, if the last digit in Natasha's tax amount written in the number system with the base k is d, the Martians will be happy. Unfortunately, Natasha does not know the Martians' divine digit yet.
Determine for which values d Natasha can make the Martians happy.
Natasha can use only her banknotes. Martians don't give her change.
Input
The first line contains two integers n and k (1 ≤ n ≤ 100 000, 2 ≤ k ≤ 100 000) — the number of denominations of banknotes and the base of the number system on Mars.
The second line contains n integers a_1, a_2, …, a_n (1 ≤ a_i ≤ 10^9) — denominations of banknotes on Mars.
All numbers are given in decimal notation.
Output
On the first line output the number of values d for which Natasha can make the Martians happy.
In the second line, output all these values in increasing order.
Print all numbers in decimal notation.
Examples
Input
2 8
12 20
Output
2
0 4
Input
3 10
10 20 30
Output
1
0
Note
Consider the first test case. It uses the octal number system.
If you take one banknote with the value of 12, you will get 14_8 in octal system. The last digit is 4_8.
If you take one banknote with the value of 12 and one banknote with the value of 20, the total value will be 32. In the octal system, it is 40_8. The last digit is 0_8.
If you take two banknotes with the value of 20, the total value will be 40, this is 50_8 in the octal system. The last digit is 0_8.
No other digits other than 0_8 and 4_8 can be obtained. Digits 0_8 and 4_8 could also be obtained in other ways.
The second test case uses the decimal number system. The nominals of all banknotes end with zero, so Natasha can give the Martians only the amount whose decimal notation also ends with zero.
Submitted Solution:
```
from math import *
from collections import *
import sys
sys.setrecursionlimit(10**9)
n,k = map(int,input().split())
a = list(map(int,input().split()))
g = a[0]%k
for i in range(1,n):
g = gcd(g,a[i]%k)
ans = [g]
x = (g+g)%k
while(x != g):
ans.append(x)
x += g
x %= k
print(len(ans))
ans.sort()
for i in ans:
print(i,end = " ")
print()
```
Yes
| 5,818 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Astronaut Natasha arrived on Mars. She knows that the Martians are very poor aliens. To ensure a better life for the Mars citizens, their emperor decided to take tax from every tourist who visited the planet. Natasha is the inhabitant of Earth, therefore she had to pay the tax to enter the territory of Mars.
There are n banknote denominations on Mars: the value of i-th banknote is a_i. Natasha has an infinite number of banknotes of each denomination.
Martians have k fingers on their hands, so they use a number system with base k. In addition, the Martians consider the digit d (in the number system with base k) divine. Thus, if the last digit in Natasha's tax amount written in the number system with the base k is d, the Martians will be happy. Unfortunately, Natasha does not know the Martians' divine digit yet.
Determine for which values d Natasha can make the Martians happy.
Natasha can use only her banknotes. Martians don't give her change.
Input
The first line contains two integers n and k (1 ≤ n ≤ 100 000, 2 ≤ k ≤ 100 000) — the number of denominations of banknotes and the base of the number system on Mars.
The second line contains n integers a_1, a_2, …, a_n (1 ≤ a_i ≤ 10^9) — denominations of banknotes on Mars.
All numbers are given in decimal notation.
Output
On the first line output the number of values d for which Natasha can make the Martians happy.
In the second line, output all these values in increasing order.
Print all numbers in decimal notation.
Examples
Input
2 8
12 20
Output
2
0 4
Input
3 10
10 20 30
Output
1
0
Note
Consider the first test case. It uses the octal number system.
If you take one banknote with the value of 12, you will get 14_8 in octal system. The last digit is 4_8.
If you take one banknote with the value of 12 and one banknote with the value of 20, the total value will be 32. In the octal system, it is 40_8. The last digit is 0_8.
If you take two banknotes with the value of 20, the total value will be 40, this is 50_8 in the octal system. The last digit is 0_8.
No other digits other than 0_8 and 4_8 can be obtained. Digits 0_8 and 4_8 could also be obtained in other ways.
The second test case uses the decimal number system. The nominals of all banknotes end with zero, so Natasha can give the Martians only the amount whose decimal notation also ends with zero.
Submitted Solution:
```
def hcfnaive(a,b):
if(b==0):
return a
else:
return hcfnaive(b,a%b)
n,k=map(int,input().split())
a=input().split()
for i in range(n):
a[i]=int(a[i])
gcd=a[0]
for i in range(1,n):
gcd=hcfnaive(gcd,a[i])
l=0
ans=[]
for i in range(k):
ans.append(l%k)
l+=gcd
ans=list(set(ans))
ans.sort()
print(len(ans))
print(*ans)
```
Yes
| 5,819 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Astronaut Natasha arrived on Mars. She knows that the Martians are very poor aliens. To ensure a better life for the Mars citizens, their emperor decided to take tax from every tourist who visited the planet. Natasha is the inhabitant of Earth, therefore she had to pay the tax to enter the territory of Mars.
There are n banknote denominations on Mars: the value of i-th banknote is a_i. Natasha has an infinite number of banknotes of each denomination.
Martians have k fingers on their hands, so they use a number system with base k. In addition, the Martians consider the digit d (in the number system with base k) divine. Thus, if the last digit in Natasha's tax amount written in the number system with the base k is d, the Martians will be happy. Unfortunately, Natasha does not know the Martians' divine digit yet.
Determine for which values d Natasha can make the Martians happy.
Natasha can use only her banknotes. Martians don't give her change.
Input
The first line contains two integers n and k (1 ≤ n ≤ 100 000, 2 ≤ k ≤ 100 000) — the number of denominations of banknotes and the base of the number system on Mars.
The second line contains n integers a_1, a_2, …, a_n (1 ≤ a_i ≤ 10^9) — denominations of banknotes on Mars.
All numbers are given in decimal notation.
Output
On the first line output the number of values d for which Natasha can make the Martians happy.
In the second line, output all these values in increasing order.
Print all numbers in decimal notation.
Examples
Input
2 8
12 20
Output
2
0 4
Input
3 10
10 20 30
Output
1
0
Note
Consider the first test case. It uses the octal number system.
If you take one banknote with the value of 12, you will get 14_8 in octal system. The last digit is 4_8.
If you take one banknote with the value of 12 and one banknote with the value of 20, the total value will be 32. In the octal system, it is 40_8. The last digit is 0_8.
If you take two banknotes with the value of 20, the total value will be 40, this is 50_8 in the octal system. The last digit is 0_8.
No other digits other than 0_8 and 4_8 can be obtained. Digits 0_8 and 4_8 could also be obtained in other ways.
The second test case uses the decimal number system. The nominals of all banknotes end with zero, so Natasha can give the Martians only the amount whose decimal notation also ends with zero.
Submitted Solution:
```
def hcfnaive(a,b):
if(b==0):
return a
else:
return hcfnaive(b,a%b)
n,k=map(int,input().split())
a=input().split()
for i in range(n):
a[i]=int(a[i])
gcd=a[0]
for i in range(2,n):
gcd=hcfnaive(gcd,a[i])
l=0
ans=[]
for i in range(k):
ans.append(l%k)
l+=gcd
ans=list(set(ans))
ans.sort()
print(len(ans))
print(*ans)
```
No
| 5,820 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Astronaut Natasha arrived on Mars. She knows that the Martians are very poor aliens. To ensure a better life for the Mars citizens, their emperor decided to take tax from every tourist who visited the planet. Natasha is the inhabitant of Earth, therefore she had to pay the tax to enter the territory of Mars.
There are n banknote denominations on Mars: the value of i-th banknote is a_i. Natasha has an infinite number of banknotes of each denomination.
Martians have k fingers on their hands, so they use a number system with base k. In addition, the Martians consider the digit d (in the number system with base k) divine. Thus, if the last digit in Natasha's tax amount written in the number system with the base k is d, the Martians will be happy. Unfortunately, Natasha does not know the Martians' divine digit yet.
Determine for which values d Natasha can make the Martians happy.
Natasha can use only her banknotes. Martians don't give her change.
Input
The first line contains two integers n and k (1 ≤ n ≤ 100 000, 2 ≤ k ≤ 100 000) — the number of denominations of banknotes and the base of the number system on Mars.
The second line contains n integers a_1, a_2, …, a_n (1 ≤ a_i ≤ 10^9) — denominations of banknotes on Mars.
All numbers are given in decimal notation.
Output
On the first line output the number of values d for which Natasha can make the Martians happy.
In the second line, output all these values in increasing order.
Print all numbers in decimal notation.
Examples
Input
2 8
12 20
Output
2
0 4
Input
3 10
10 20 30
Output
1
0
Note
Consider the first test case. It uses the octal number system.
If you take one banknote with the value of 12, you will get 14_8 in octal system. The last digit is 4_8.
If you take one banknote with the value of 12 and one banknote with the value of 20, the total value will be 32. In the octal system, it is 40_8. The last digit is 0_8.
If you take two banknotes with the value of 20, the total value will be 40, this is 50_8 in the octal system. The last digit is 0_8.
No other digits other than 0_8 and 4_8 can be obtained. Digits 0_8 and 4_8 could also be obtained in other ways.
The second test case uses the decimal number system. The nominals of all banknotes end with zero, so Natasha can give the Martians only the amount whose decimal notation also ends with zero.
Submitted Solution:
```
import math
n,k=map(int,input().split())
c=[]
b=list(map(int,input().split()))
g=b[0]
for j in range(1,n):
g=math.gcd(g,b[j])
l=0
ans=[]
j=0
while(j<k):
ans.append(l%k)
l+=g
j+=1
ans=list(set(ans))
print(len(ans))
print(*ans)
```
No
| 5,821 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Astronaut Natasha arrived on Mars. She knows that the Martians are very poor aliens. To ensure a better life for the Mars citizens, their emperor decided to take tax from every tourist who visited the planet. Natasha is the inhabitant of Earth, therefore she had to pay the tax to enter the territory of Mars.
There are n banknote denominations on Mars: the value of i-th banknote is a_i. Natasha has an infinite number of banknotes of each denomination.
Martians have k fingers on their hands, so they use a number system with base k. In addition, the Martians consider the digit d (in the number system with base k) divine. Thus, if the last digit in Natasha's tax amount written in the number system with the base k is d, the Martians will be happy. Unfortunately, Natasha does not know the Martians' divine digit yet.
Determine for which values d Natasha can make the Martians happy.
Natasha can use only her banknotes. Martians don't give her change.
Input
The first line contains two integers n and k (1 ≤ n ≤ 100 000, 2 ≤ k ≤ 100 000) — the number of denominations of banknotes and the base of the number system on Mars.
The second line contains n integers a_1, a_2, …, a_n (1 ≤ a_i ≤ 10^9) — denominations of banknotes on Mars.
All numbers are given in decimal notation.
Output
On the first line output the number of values d for which Natasha can make the Martians happy.
In the second line, output all these values in increasing order.
Print all numbers in decimal notation.
Examples
Input
2 8
12 20
Output
2
0 4
Input
3 10
10 20 30
Output
1
0
Note
Consider the first test case. It uses the octal number system.
If you take one banknote with the value of 12, you will get 14_8 in octal system. The last digit is 4_8.
If you take one banknote with the value of 12 and one banknote with the value of 20, the total value will be 32. In the octal system, it is 40_8. The last digit is 0_8.
If you take two banknotes with the value of 20, the total value will be 40, this is 50_8 in the octal system. The last digit is 0_8.
No other digits other than 0_8 and 4_8 can be obtained. Digits 0_8 and 4_8 could also be obtained in other ways.
The second test case uses the decimal number system. The nominals of all banknotes end with zero, so Natasha can give the Martians only the amount whose decimal notation also ends with zero.
Submitted Solution:
```
# ===============================================================================================
# importing some useful libraries.
from __future__ import division, print_function
from fractions import Fraction
import sys
import os
from io import BytesIO, IOBase
from itertools import *
import bisect
from heapq import *
from math import ceil, floor
from copy import *
from collections import deque, defaultdict
from collections import Counter as counter # Counter(list) return a dict with {key: count}
from itertools import combinations # if a = [1,2,3] then print(list(comb(a,2))) -----> [(1, 2), (1, 3), (2, 3)]
from itertools import permutations as permutate
from bisect import bisect_left as bl
from operator import *
# If the element is already present in the list,
# the left most position where element has to be inserted is returned.
from bisect import bisect_right as br
from bisect import bisect
# If the element is already present in the list,
# the right most position where element has to be inserted is returned
# ==============================================================================================
# fast I/O region
BUFSIZE = 8192
from sys import stderr
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")
def print(*args, **kwargs):
"""Prints the values to a stream, or to sys.stdout by default."""
sep, file = kwargs.pop("sep", " "), kwargs.pop("file", sys.stdout)
at_start = True
for x in args:
if not at_start:
file.write(sep)
file.write(str(x))
at_start = False
file.write(kwargs.pop("end", "\n"))
if kwargs.pop("flush", False):
file.flush()
if sys.version_info[0] < 3:
sys.stdin, sys.stdout = FastIO(sys.stdin), FastIO(sys.stdout)
else:
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
# inp = lambda: sys.stdin.readline().rstrip("\r\n")
# ===============================================================================================
### START ITERATE RECURSION ###
from types import GeneratorType
def iterative(f, stack=[]):
def wrapped_func(*args, **kwargs):
if stack: return f(*args, **kwargs)
to = f(*args, **kwargs)
while True:
if type(to) is GeneratorType:
stack.append(to)
to = next(to)
continue
stack.pop()
if not stack: break
to = stack[-1].send(to)
return to
return wrapped_func
#### END ITERATE RECURSION ####
# ===============================================================================================
# some shortcuts
mod = 1000000007
def inp(): return sys.stdin.readline().rstrip("\r\n") # for fast input
def out(var): sys.stdout.write(str(var)) # for fast output, always take string
def lis(): return list(map(int, inp().split()))
def stringlis(): return list(map(str, inp().split()))
def sep(): return map(int, inp().split())
def strsep(): return map(str, inp().split())
def fsep(): return map(float, inp().split())
def nextline(): out("\n") # as stdout.write always print sring.
def testcase(t):
for p in range(t):
solve()
def pow(x, y, p):
res = 1 # Initialize result
x = x % p # Update x if it is more , than or equal to p
if (x == 0):
return 0
while (y > 0):
if ((y & 1) == 1): # If y is odd, multiply, x with result
res = (res * x) % p
y = y >> 1 # y = y/2
x = (x * x) % p
return res
from functools import reduce
def factors(n):
return set(reduce(list.__add__,
([i, n // i] for i in range(1, int(n ** 0.5) + 1) if n % i == 0)))
def gcd(a, b):
if a == b: return a
while b > 0: a, b = b, a % b
return a
# N=100000
# mod = 10**9 +7
# fac = [1, 1]
# finv = [1, 1]
# inv = [0, 1]
#
# for i in range(2, N + 1):
# fac.append((fac[-1] * i) % mod)
# inv.append(mod - (inv[mod % i] * (mod // i) % mod))
# finv.append(finv[-1] * inv[-1] % mod)
#
#
# def comb(n, r):
# if n < r:
# return 0
# else:
# return fac[n] * (finv[r] * finv[n - r] % mod) % mod
##############Find sum of product of subsets of size k in a array
# ar=[0,1,2,3]
# k=3
# n=len(ar)-1
# dp=[0]*(n+1)
# dp[0]=1
# for pos in range(1,n+1):
# dp[pos]=0
# l=max(1,k+pos-n-1)
# for j in range(min(pos,k),l-1,-1):
# dp[j]=dp[j]+ar[pos]*dp[j-1]
# print(dp[k])
def prefix_sum(ar): # [1,2,3,4]->[1,3,6,10]
return list(accumulate(ar))
def suffix_sum(ar): # [1,2,3,4]->[10,9,7,4]
return list(accumulate(ar[::-1]))[::-1]
def lcm(a,b):
return (a*b)//gcd(a,b)
# =========================================================================================
from collections import defaultdict
def numberOfSetBits(i):
i = i - ((i >> 1) & 0x55555555)
i = (i & 0x33333333) + ((i >> 2) & 0x33333333)
return (((i + (i >> 4) & 0xF0F0F0F) * 0x1010101) & 0xffffffff) >> 24
def N():
return int(inp())
def solve():
n,k=sep()
ar=[int(i)%k for i in inp().strip().split(" ") if int(i)%k]
if not ar:
print(1)
print(0)
return
g=reduce(gcd,ar)
t=(k-1)//g + 1
print(t)
ans=[0]
for i in range(1,k):
if i%g==0:
ans.append(i)
print(*ans)
if k==74150:
print(g)
solve()
#testcase(int(inp()))
```
No
| 5,822 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Astronaut Natasha arrived on Mars. She knows that the Martians are very poor aliens. To ensure a better life for the Mars citizens, their emperor decided to take tax from every tourist who visited the planet. Natasha is the inhabitant of Earth, therefore she had to pay the tax to enter the territory of Mars.
There are n banknote denominations on Mars: the value of i-th banknote is a_i. Natasha has an infinite number of banknotes of each denomination.
Martians have k fingers on their hands, so they use a number system with base k. In addition, the Martians consider the digit d (in the number system with base k) divine. Thus, if the last digit in Natasha's tax amount written in the number system with the base k is d, the Martians will be happy. Unfortunately, Natasha does not know the Martians' divine digit yet.
Determine for which values d Natasha can make the Martians happy.
Natasha can use only her banknotes. Martians don't give her change.
Input
The first line contains two integers n and k (1 ≤ n ≤ 100 000, 2 ≤ k ≤ 100 000) — the number of denominations of banknotes and the base of the number system on Mars.
The second line contains n integers a_1, a_2, …, a_n (1 ≤ a_i ≤ 10^9) — denominations of banknotes on Mars.
All numbers are given in decimal notation.
Output
On the first line output the number of values d for which Natasha can make the Martians happy.
In the second line, output all these values in increasing order.
Print all numbers in decimal notation.
Examples
Input
2 8
12 20
Output
2
0 4
Input
3 10
10 20 30
Output
1
0
Note
Consider the first test case. It uses the octal number system.
If you take one banknote with the value of 12, you will get 14_8 in octal system. The last digit is 4_8.
If you take one banknote with the value of 12 and one banknote with the value of 20, the total value will be 32. In the octal system, it is 40_8. The last digit is 0_8.
If you take two banknotes with the value of 20, the total value will be 40, this is 50_8 in the octal system. The last digit is 0_8.
No other digits other than 0_8 and 4_8 can be obtained. Digits 0_8 and 4_8 could also be obtained in other ways.
The second test case uses the decimal number system. The nominals of all banknotes end with zero, so Natasha can give the Martians only the amount whose decimal notation also ends with zero.
Submitted Solution:
```
n, k = map(int, input().split())
a = list(map(int, input().split()))
ans = set()
ans.add(0)
for i in a:
ans.add(i % k)
print(len(ans))
print(*sorted(list(ans)))
```
No
| 5,823 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You successfully found poor Arkady near the exit of the station you've perfectly predicted. You sent him home on a taxi and suddenly came up with a question.
There are n crossroads in your city and several bidirectional roads connecting some of them. A taxi ride is a path from some crossroads to another one without passing the same crossroads twice. You have a collection of rides made by one driver and now you wonder if this driver can be a robot or they are definitely a human.
You think that the driver can be a robot if for every two crossroads a and b the driver always chooses the same path whenever he drives from a to b. Note that a and b here do not have to be the endpoints of a ride and that the path from b to a can be different. On the contrary, if the driver ever has driven two different paths from a to b, they are definitely a human.
Given the system of roads and the description of all rides available to you, determine if the driver can be a robot or not.
Input
Each test contains one or more test cases. The first line contains a single integer t (1 ≤ t ≤ 3 ⋅ 10^5) — the number of test cases.
The first line of each test case contains a single integer n (1 ≤ n ≤ 3 ⋅ 10^5) — the number of crossroads in the city.
The next line contains a single integer q (1 ≤ q ≤ 3 ⋅ 10^5) — the number of rides available to you.
Each of the following q lines starts with a single integer k (2 ≤ k ≤ n) — the number of crossroads visited by the driver on this ride. It is followed by k integers c_1, c_2, ..., c_k (1 ≤ c_i ≤ n) — the crossroads in the order the driver visited them. It is guaranteed that all crossroads in one ride are distinct.
It is guaranteed that the sum of values k among all rides of all test cases does not exceed 3 ⋅ 10^5.
It is guaranteed that the sum of values n and the sum of values q doesn't exceed 3 ⋅ 10^5 among all test cases.
Output
Output a single line for each test case.
If the driver can be a robot, output "Robot" in a single line. Otherwise, output "Human".
You can print each letter in any case (upper or lower).
Examples
Input
1
5
2
4 1 2 3 5
3 1 4 3
Output
Human
Input
1
4
4
3 1 2 3
3 2 3 4
3 3 4 1
3 4 1 2
Output
Robot
Note
In the first example it is clear that the driver used two different ways to get from crossroads 1 to crossroads 3. It must be a human.
In the second example the driver always drives the cycle 1 → 2 → 3 → 4 → 1 until he reaches destination.
Submitted Solution:
```
1
```
No
| 5,824 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Petya loves lucky numbers. We all know that lucky numbers are the positive integers whose decimal representations contain only the lucky digits 4 and 7. For example, numbers 47, 744, 4 are lucky and 5, 17, 467 are not.
One day Petya encountered a tree with n vertexes. Besides, the tree was weighted, i. e. each edge of the tree has weight (a positive integer). An edge is lucky if its weight is a lucky number. Note that a tree with n vertexes is an undirected connected graph that has exactly n - 1 edges.
Petya wondered how many vertex triples (i, j, k) exists that on the way from i to j, as well as on the way from i to k there must be at least one lucky edge (all three vertexes are pairwise distinct). The order of numbers in the triple matters, that is, the triple (1, 2, 3) is not equal to the triple (2, 1, 3) and is not equal to the triple (1, 3, 2).
Find how many such triples of vertexes exist.
Input
The first line contains the single integer n (1 ≤ n ≤ 105) — the number of tree vertexes. Next n - 1 lines contain three integers each: ui vi wi (1 ≤ ui, vi ≤ n, 1 ≤ wi ≤ 109) — the pair of vertexes connected by the edge and the edge's weight.
Output
On the single line print the single number — the answer.
Please do not use the %lld specificator to read or write 64-bit numbers in С++. It is recommended to use the cin, cout streams or the %I64d specificator.
Examples
Input
4
1 2 4
3 1 2
1 4 7
Output
16
Input
4
1 2 4
1 3 47
1 4 7447
Output
24
Note
The 16 triples of vertexes from the first sample are: (1, 2, 4), (1, 4, 2), (2, 1, 3), (2, 1, 4), (2, 3, 1), (2, 3, 4), (2, 4, 1), (2, 4, 3), (3, 2, 4), (3, 4, 2), (4, 1, 2), (4, 1, 3), (4, 2, 1), (4, 2, 3), (4, 3, 1), (4, 3, 2).
In the second sample all the triples should be counted: 4·3·2 = 24.
Tags: dp, dsu, trees
Correct Solution:
```
import os
import sys
from io import BytesIO, IOBase
from types import GeneratorType
from collections import defaultdict
BUFSIZE = 8192
class FastIO(IOBase):
newlines = 0
def __init__(self, file):
self._fd = file.fileno()
self.buffer = BytesIO()
self.writable = "x" in file.mode or "r" not in file.mode
self.write = self.buffer.write if self.writable else None
def read(self):
while True:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
if not b:
break
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines = 0
return self.buffer.read()
def readline(self):
while self.newlines == 0:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
self.newlines = b.count(b"\n") + (not b)
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines -= 1
return self.buffer.readline()
def flush(self):
if self.writable:
os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
self.write = lambda s: self.buffer.write(s.encode("ascii"))
self.read = lambda: self.buffer.read().decode("ascii")
self.readline = lambda: self.buffer.readline().decode("ascii")
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")
sys.setrecursionlimit(2*10**5)
def bootstrap(f, stack=[]):
def wrappedfunc(*args, **kwargs):
if stack:
return f(*args, **kwargs)
else:
to = f(*args, **kwargs)
while True:
if type(to) is GeneratorType:
stack.append(to)
to = next(to)
else:
stack.pop()
if not stack:
break
to = stack[-1].send(to)
return to
return wrappedfunc
@bootstrap
def dfs(u,p):
for j in adj[u]:
if j!=p:
yield dfs(j, u)
if d[j,u]!=0:
ways[u]+=(ways[j]+1)
yield
@bootstrap
def dfs2(u,p,v):
for j in adj[u]:
if j!=p:
if d[u,j]==0:
yield dfs2(j,u,0)
else:
yield dfs2(j,u,v+ways[u]-ways[j])
ans[u]= ways[u]*(ways[u]-1) + v*(v-1) + 2*(ways[u]*v+ (ways[u]+v)*(n-1-ways[u]-v))
yield
def val(n):
for j in str(n):
if j=="4" or j=="7":
pass
else:
return 1
return 0
n=int(input())
adj=[[] for i in range(n+1)]
d=dict()
for j in range(n-1):
c=list(map(int,input().split()))
adj[c[0]].append(c[1])
adj[c[1]].append(c[0])
c[2]=val(c[2])
d[c[0],c[1]]=c[2]
d[c[1],c[0]]=c[2]
ways=[0]*(n+1)
dfs(1,0)
ans=[0]*(n+1)
dfs2(1,0,0)
print(n*(n-1)*(n-2)-sum(ans))
```
| 5,825 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Petya loves lucky numbers. We all know that lucky numbers are the positive integers whose decimal representations contain only the lucky digits 4 and 7. For example, numbers 47, 744, 4 are lucky and 5, 17, 467 are not.
One day Petya encountered a tree with n vertexes. Besides, the tree was weighted, i. e. each edge of the tree has weight (a positive integer). An edge is lucky if its weight is a lucky number. Note that a tree with n vertexes is an undirected connected graph that has exactly n - 1 edges.
Petya wondered how many vertex triples (i, j, k) exists that on the way from i to j, as well as on the way from i to k there must be at least one lucky edge (all three vertexes are pairwise distinct). The order of numbers in the triple matters, that is, the triple (1, 2, 3) is not equal to the triple (2, 1, 3) and is not equal to the triple (1, 3, 2).
Find how many such triples of vertexes exist.
Input
The first line contains the single integer n (1 ≤ n ≤ 105) — the number of tree vertexes. Next n - 1 lines contain three integers each: ui vi wi (1 ≤ ui, vi ≤ n, 1 ≤ wi ≤ 109) — the pair of vertexes connected by the edge and the edge's weight.
Output
On the single line print the single number — the answer.
Please do not use the %lld specificator to read or write 64-bit numbers in С++. It is recommended to use the cin, cout streams or the %I64d specificator.
Examples
Input
4
1 2 4
3 1 2
1 4 7
Output
16
Input
4
1 2 4
1 3 47
1 4 7447
Output
24
Note
The 16 triples of vertexes from the first sample are: (1, 2, 4), (1, 4, 2), (2, 1, 3), (2, 1, 4), (2, 3, 1), (2, 3, 4), (2, 4, 1), (2, 4, 3), (3, 2, 4), (3, 4, 2), (4, 1, 2), (4, 1, 3), (4, 2, 1), (4, 2, 3), (4, 3, 1), (4, 3, 2).
In the second sample all the triples should be counted: 4·3·2 = 24.
Submitted Solution:
```
import os
import sys
from io import BytesIO, IOBase
from types import GeneratorType
from collections import defaultdict
BUFSIZE = 8192
class FastIO(IOBase):
newlines = 0
def __init__(self, file):
self._fd = file.fileno()
self.buffer = BytesIO()
self.writable = "x" in file.mode or "r" not in file.mode
self.write = self.buffer.write if self.writable else None
def read(self):
while True:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
if not b:
break
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines = 0
return self.buffer.read()
def readline(self):
while self.newlines == 0:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
self.newlines = b.count(b"\n") + (not b)
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines -= 1
return self.buffer.readline()
def flush(self):
if self.writable:
os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
self.write = lambda s: self.buffer.write(s.encode("ascii"))
self.read = lambda: self.buffer.read().decode("ascii")
self.readline = lambda: self.buffer.readline().decode("ascii")
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")
sys.setrecursionlimit(2*10**5)
def bootstrap(f, stack=[]):
def wrappedfunc(*args, **kwargs):
if stack:
return f(*args, **kwargs)
else:
to = f(*args, **kwargs)
while True:
if type(to) is GeneratorType:
stack.append(to)
to = next(to)
else:
stack.pop()
if not stack:
break
to = stack[-1].send(to)
return to
return wrappedfunc
@bootstrap
def dfs(u,p):
for j in adj[u]:
if j!=p:
yield dfs(j, u)
if d[j,u]!=0:
ways[u]+=(ways[j]+1)
yield
@bootstrap
def dfs2(u,p,v):
for j in adj[u]:
if j!=p:
if d[u,j]==0:
yield dfs2(j,u,0)
else:
yield dfs2(j,u,v+ways[u]-ways[j])
ans[u]=2*(ways[u]*(n-2)+v*(n-2)-(ways[u]*(ways[u]-1)+v*(v-1)+ways[u]*v))
yield
def val(n):
for j in str(n):
if j=="4" or j=="7":
pass
else:
return 1
return 0
n=int(input())
adj=[[] for i in range(n+1)]
d=dict()
for j in range(n-1):
c=list(map(int,input().split()))
adj[c[0]].append(c[1])
adj[c[1]].append(c[0])
c[2]=val(c[2])
d[c[0],c[1]]=c[2]
d[c[1],c[0]]=c[2]
ways=[0]*(n+1)
dfs(1,0)
ans=[0]*(n+1)
dfs2(1,0,0)
print(n*(n-1)*(n-2)-sum(ans))
```
No
| 5,826 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Suppose you are given a string s of length n consisting of lowercase English letters. You need to compress it using the smallest possible number of coins.
To compress the string, you have to represent s as a concatenation of several non-empty strings: s = t_{1} t_{2} … t_{k}. The i-th of these strings should be encoded with one of the two ways:
* if |t_{i}| = 1, meaning that the current string consists of a single character, you can encode it paying a coins;
* if t_{i} is a substring of t_{1} t_{2} … t_{i - 1}, then you can encode it paying b coins.
A string x is a substring of a string y if x can be obtained from y by deletion of several (possibly, zero or all) characters from the beginning and several (possibly, zero or all) characters from the end.
So your task is to calculate the minimum possible number of coins you need to spend in order to compress the given string s.
Input
The first line contains three positive integers, separated by spaces: n, a and b (1 ≤ n, a, b ≤ 5000) — the length of the string, the cost to compress a one-character string and the cost to compress a string that appeared before.
The second line contains a single string s, consisting of n lowercase English letters.
Output
Output a single integer — the smallest possible number of coins you need to spend to compress s.
Examples
Input
3 3 1
aba
Output
7
Input
4 1 1
abcd
Output
4
Input
4 10 1
aaaa
Output
12
Note
In the first sample case, you can set t_{1} = 'a', t_{2} = 'b', t_{3} = 'a' and pay 3 + 3 + 1 = 7 coins, since t_{3} is a substring of t_{1}t_{2}.
In the second sample, you just need to compress every character by itself.
In the third sample, you set t_{1} = t_{2} = 'a', t_{3} = 'aa' and pay 10 + 1 + 1 = 12 coins, since t_{2} is a substring of t_{1} and t_{3} is a substring of t_{1} t_{2}.
Tags: dp, strings
Correct Solution:
```
def f(x):
z=[0 for i in range(len(x))]
l=0
r=0
for i in range(1,len(x)):
if i<=r:
z[i]=min(z[i-l],r-i+1)
while i+z[i]<len(x) and x[z[i]]==x[i+z[i]]:
z[i]+=1
if i+z[i]-1>r:
l,r=i,i+z[i]-1
return z
a=list(map(int,input('').split()))
n,a,b=a[0],a[1],a[2]
s=input('')
dp=[0 for i in range(n)]
dp[0]=a
for i in range(1,n):
t=s[:i+1]
dp[i]=dp[i-1]+a
q=f(t[::-1])
maxs=[0 for j in range(i+1)]
maxs[0]=q[i]
for j in range(1,i):
maxs[j]=max(maxs[j-1],q[i-j])
for j in range(i):
if maxs[j]>=i-j:
dp[i]=min(dp[i],dp[j]+b)
print(dp[len(dp)-1])
```
| 5,827 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Suppose you are given a string s of length n consisting of lowercase English letters. You need to compress it using the smallest possible number of coins.
To compress the string, you have to represent s as a concatenation of several non-empty strings: s = t_{1} t_{2} … t_{k}. The i-th of these strings should be encoded with one of the two ways:
* if |t_{i}| = 1, meaning that the current string consists of a single character, you can encode it paying a coins;
* if t_{i} is a substring of t_{1} t_{2} … t_{i - 1}, then you can encode it paying b coins.
A string x is a substring of a string y if x can be obtained from y by deletion of several (possibly, zero or all) characters from the beginning and several (possibly, zero or all) characters from the end.
So your task is to calculate the minimum possible number of coins you need to spend in order to compress the given string s.
Input
The first line contains three positive integers, separated by spaces: n, a and b (1 ≤ n, a, b ≤ 5000) — the length of the string, the cost to compress a one-character string and the cost to compress a string that appeared before.
The second line contains a single string s, consisting of n lowercase English letters.
Output
Output a single integer — the smallest possible number of coins you need to spend to compress s.
Examples
Input
3 3 1
aba
Output
7
Input
4 1 1
abcd
Output
4
Input
4 10 1
aaaa
Output
12
Note
In the first sample case, you can set t_{1} = 'a', t_{2} = 'b', t_{3} = 'a' and pay 3 + 3 + 1 = 7 coins, since t_{3} is a substring of t_{1}t_{2}.
In the second sample, you just need to compress every character by itself.
In the third sample, you set t_{1} = t_{2} = 'a', t_{3} = 'aa' and pay 10 + 1 + 1 = 12 coins, since t_{2} is a substring of t_{1} and t_{3} is a substring of t_{1} t_{2}.
Tags: dp, strings
Correct Solution:
```
n, a, b = map(int, input().split())
s = input()
lcp = [[0]*n for _ in ' '*n]
for i in range(n-1, -1, -1):
for r in range(n-1, -1, -1):
if s[i] == s[r]:
if i == n - 1 or r == n - 1:
lcp[i][r] = 1
else:
lcp[i][r] = lcp[i+1][r+1]+1
d = [10**10]*(n+1)
d[0] = 0
for i in range(n):
d[i+1] = min(d[i+1], d[i]+a)
for j in range(i):
k = min(lcp[i][j], i-j)
d[i+k] = min(d[i+k], d[i]+b)
print(d[-1])
```
| 5,828 |
Provide tags and a correct Python 3 solution for this coding contest problem.
The circle line of the Roflanpolis subway has n stations.
There are two parallel routes in the subway. The first one visits stations in order 1 → 2 → … → n → 1 → 2 → … (so the next stop after station x is equal to (x+1) if x < n and 1 otherwise). The second route visits stations in order n → (n-1) → … → 1 → n → (n-1) → … (so the next stop after station x is equal to (x-1) if x>1 and n otherwise). All trains depart their stations simultaneously, and it takes exactly 1 minute to arrive at the next station.
Two toads live in this city, their names are Daniel and Vlad.
Daniel is currently in a train of the first route at station a and will exit the subway when his train reaches station x.
Coincidentally, Vlad is currently in a train of the second route at station b and he will exit the subway when his train reaches station y.
Surprisingly, all numbers a,x,b,y are distinct.
Toad Ilya asks you to check if Daniel and Vlad will ever be at the same station at the same time during their journey. In other words, check if there is a moment when their trains stop at the same station. Note that this includes the moments when Daniel or Vlad enter or leave the subway.
Input
The first line contains five space-separated integers n, a, x, b, y (4 ≤ n ≤ 100, 1 ≤ a, x, b, y ≤ n, all numbers among a, x, b, y are distinct) — the number of stations in Roflanpolis, Daniel's start station, Daniel's finish station, Vlad's start station and Vlad's finish station, respectively.
Output
Output "YES" if there is a time moment when Vlad and Daniel are at the same station, and "NO" otherwise. You can print each letter in any case (upper or lower).
Examples
Input
5 1 4 3 2
Output
YES
Input
10 2 1 9 10
Output
NO
Note
In the first example, Daniel and Vlad start at the stations (1, 3). One minute later they are at stations (2, 2). They are at the same station at this moment. Note that Vlad leaves the subway right after that.
Consider the second example, let's look at the stations Vlad and Daniel are at. They are:
* initially (2, 9),
* after 1 minute (3, 8),
* after 2 minutes (4, 7),
* after 3 minutes (5, 6),
* after 4 minutes (6, 5),
* after 5 minutes (7, 4),
* after 6 minutes (8, 3),
* after 7 minutes (9, 2),
* after 8 minutes (10, 1),
* after 9 minutes (1, 10).
After that, they both leave the subway because they are at their finish stations, so there is no moment when they both are at the same station.
Tags: implementation, math
Correct Solution:
```
list1=list(map(int,input().split()))
list2=[]
flag=0
for i in range(list1[0]):
list2.append(i+1)
if(list1[1]==list1[3]):
print('YES')
for j in range(list1[0]):
a=list1[1]+j
if a>list1[0]:
a=a-list1[0]
if(list2[a-1]==list2[list1[3]-j-1]):
flag=1
break
if(list2[a-1]==list1[2] or list2[list1[3]-j-1]==list1[4]):
break
if flag==1:
print('YES')
elif flag==0:
print('NO')
```
| 5,829 |
Provide tags and a correct Python 3 solution for this coding contest problem.
The circle line of the Roflanpolis subway has n stations.
There are two parallel routes in the subway. The first one visits stations in order 1 → 2 → … → n → 1 → 2 → … (so the next stop after station x is equal to (x+1) if x < n and 1 otherwise). The second route visits stations in order n → (n-1) → … → 1 → n → (n-1) → … (so the next stop after station x is equal to (x-1) if x>1 and n otherwise). All trains depart their stations simultaneously, and it takes exactly 1 minute to arrive at the next station.
Two toads live in this city, their names are Daniel and Vlad.
Daniel is currently in a train of the first route at station a and will exit the subway when his train reaches station x.
Coincidentally, Vlad is currently in a train of the second route at station b and he will exit the subway when his train reaches station y.
Surprisingly, all numbers a,x,b,y are distinct.
Toad Ilya asks you to check if Daniel and Vlad will ever be at the same station at the same time during their journey. In other words, check if there is a moment when their trains stop at the same station. Note that this includes the moments when Daniel or Vlad enter or leave the subway.
Input
The first line contains five space-separated integers n, a, x, b, y (4 ≤ n ≤ 100, 1 ≤ a, x, b, y ≤ n, all numbers among a, x, b, y are distinct) — the number of stations in Roflanpolis, Daniel's start station, Daniel's finish station, Vlad's start station and Vlad's finish station, respectively.
Output
Output "YES" if there is a time moment when Vlad and Daniel are at the same station, and "NO" otherwise. You can print each letter in any case (upper or lower).
Examples
Input
5 1 4 3 2
Output
YES
Input
10 2 1 9 10
Output
NO
Note
In the first example, Daniel and Vlad start at the stations (1, 3). One minute later they are at stations (2, 2). They are at the same station at this moment. Note that Vlad leaves the subway right after that.
Consider the second example, let's look at the stations Vlad and Daniel are at. They are:
* initially (2, 9),
* after 1 minute (3, 8),
* after 2 minutes (4, 7),
* after 3 minutes (5, 6),
* after 4 minutes (6, 5),
* after 5 minutes (7, 4),
* after 6 minutes (8, 3),
* after 7 minutes (9, 2),
* after 8 minutes (10, 1),
* after 9 minutes (1, 10).
After that, they both leave the subway because they are at their finish stations, so there is no moment when they both are at the same station.
Tags: implementation, math
Correct Solution:
```
from sys import stdin
# stdin=open('input.txt')
def input():
return stdin.readline().strip()
# from sys import stdout
# stdout=open('input.txt',mode='w+')
# def print1(x, end='\n'):
# stdout.write(str(x) +end)
# a, b = map(int, input().split())
# l = list(map(int, input().split()))
# # CODE BEGINS HERE.................
n, a, x, b, y = map(int, input().split())
flag = (a == b)
while not flag and a != x and b != y:
if a == n:
a = 1
else:
a += 1
if b == 1:
b = n
else:
b -= 1
if a == b:
flag = True
break
if flag:
print('YES')
else:
print('NO')
# # CODE ENDS HERE....................
# stdout.close()
```
| 5,830 |
Provide tags and a correct Python 3 solution for this coding contest problem.
The circle line of the Roflanpolis subway has n stations.
There are two parallel routes in the subway. The first one visits stations in order 1 → 2 → … → n → 1 → 2 → … (so the next stop after station x is equal to (x+1) if x < n and 1 otherwise). The second route visits stations in order n → (n-1) → … → 1 → n → (n-1) → … (so the next stop after station x is equal to (x-1) if x>1 and n otherwise). All trains depart their stations simultaneously, and it takes exactly 1 minute to arrive at the next station.
Two toads live in this city, their names are Daniel and Vlad.
Daniel is currently in a train of the first route at station a and will exit the subway when his train reaches station x.
Coincidentally, Vlad is currently in a train of the second route at station b and he will exit the subway when his train reaches station y.
Surprisingly, all numbers a,x,b,y are distinct.
Toad Ilya asks you to check if Daniel and Vlad will ever be at the same station at the same time during their journey. In other words, check if there is a moment when their trains stop at the same station. Note that this includes the moments when Daniel or Vlad enter or leave the subway.
Input
The first line contains five space-separated integers n, a, x, b, y (4 ≤ n ≤ 100, 1 ≤ a, x, b, y ≤ n, all numbers among a, x, b, y are distinct) — the number of stations in Roflanpolis, Daniel's start station, Daniel's finish station, Vlad's start station and Vlad's finish station, respectively.
Output
Output "YES" if there is a time moment when Vlad and Daniel are at the same station, and "NO" otherwise. You can print each letter in any case (upper or lower).
Examples
Input
5 1 4 3 2
Output
YES
Input
10 2 1 9 10
Output
NO
Note
In the first example, Daniel and Vlad start at the stations (1, 3). One minute later they are at stations (2, 2). They are at the same station at this moment. Note that Vlad leaves the subway right after that.
Consider the second example, let's look at the stations Vlad and Daniel are at. They are:
* initially (2, 9),
* after 1 minute (3, 8),
* after 2 minutes (4, 7),
* after 3 minutes (5, 6),
* after 4 minutes (6, 5),
* after 5 minutes (7, 4),
* after 6 minutes (8, 3),
* after 7 minutes (9, 2),
* after 8 minutes (10, 1),
* after 9 minutes (1, 10).
After that, they both leave the subway because they are at their finish stations, so there is no moment when they both are at the same station.
Tags: implementation, math
Correct Solution:
```
n, a, x, b, y = (int(i)-1 for i in input().split())
n += 1
while True:
if a == b:
print('YES')
break
if a == x or b == y:
print('NO')
break
a = (a+1) % n
b = (b-1) % n
```
| 5,831 |
Provide tags and a correct Python 3 solution for this coding contest problem.
The circle line of the Roflanpolis subway has n stations.
There are two parallel routes in the subway. The first one visits stations in order 1 → 2 → … → n → 1 → 2 → … (so the next stop after station x is equal to (x+1) if x < n and 1 otherwise). The second route visits stations in order n → (n-1) → … → 1 → n → (n-1) → … (so the next stop after station x is equal to (x-1) if x>1 and n otherwise). All trains depart their stations simultaneously, and it takes exactly 1 minute to arrive at the next station.
Two toads live in this city, their names are Daniel and Vlad.
Daniel is currently in a train of the first route at station a and will exit the subway when his train reaches station x.
Coincidentally, Vlad is currently in a train of the second route at station b and he will exit the subway when his train reaches station y.
Surprisingly, all numbers a,x,b,y are distinct.
Toad Ilya asks you to check if Daniel and Vlad will ever be at the same station at the same time during their journey. In other words, check if there is a moment when their trains stop at the same station. Note that this includes the moments when Daniel or Vlad enter or leave the subway.
Input
The first line contains five space-separated integers n, a, x, b, y (4 ≤ n ≤ 100, 1 ≤ a, x, b, y ≤ n, all numbers among a, x, b, y are distinct) — the number of stations in Roflanpolis, Daniel's start station, Daniel's finish station, Vlad's start station and Vlad's finish station, respectively.
Output
Output "YES" if there is a time moment when Vlad and Daniel are at the same station, and "NO" otherwise. You can print each letter in any case (upper or lower).
Examples
Input
5 1 4 3 2
Output
YES
Input
10 2 1 9 10
Output
NO
Note
In the first example, Daniel and Vlad start at the stations (1, 3). One minute later they are at stations (2, 2). They are at the same station at this moment. Note that Vlad leaves the subway right after that.
Consider the second example, let's look at the stations Vlad and Daniel are at. They are:
* initially (2, 9),
* after 1 minute (3, 8),
* after 2 minutes (4, 7),
* after 3 minutes (5, 6),
* after 4 minutes (6, 5),
* after 5 minutes (7, 4),
* after 6 minutes (8, 3),
* after 7 minutes (9, 2),
* after 8 minutes (10, 1),
* after 9 minutes (1, 10).
After that, they both leave the subway because they are at their finish stations, so there is no moment when they both are at the same station.
Tags: implementation, math
Correct Solution:
```
flag = True
n,a,x,b,y = list(map(int,input().split()))
for i in range(min((x-a)%n, (b-y)%n)):
a = a+1
b = b-1
if a>n:
a = 1
if b==0:
b = n
if a==b:
print("YES")
flag = False
break
if flag:
print("NO")
```
| 5,832 |
Provide tags and a correct Python 3 solution for this coding contest problem.
The circle line of the Roflanpolis subway has n stations.
There are two parallel routes in the subway. The first one visits stations in order 1 → 2 → … → n → 1 → 2 → … (so the next stop after station x is equal to (x+1) if x < n and 1 otherwise). The second route visits stations in order n → (n-1) → … → 1 → n → (n-1) → … (so the next stop after station x is equal to (x-1) if x>1 and n otherwise). All trains depart their stations simultaneously, and it takes exactly 1 minute to arrive at the next station.
Two toads live in this city, their names are Daniel and Vlad.
Daniel is currently in a train of the first route at station a and will exit the subway when his train reaches station x.
Coincidentally, Vlad is currently in a train of the second route at station b and he will exit the subway when his train reaches station y.
Surprisingly, all numbers a,x,b,y are distinct.
Toad Ilya asks you to check if Daniel and Vlad will ever be at the same station at the same time during their journey. In other words, check if there is a moment when their trains stop at the same station. Note that this includes the moments when Daniel or Vlad enter or leave the subway.
Input
The first line contains five space-separated integers n, a, x, b, y (4 ≤ n ≤ 100, 1 ≤ a, x, b, y ≤ n, all numbers among a, x, b, y are distinct) — the number of stations in Roflanpolis, Daniel's start station, Daniel's finish station, Vlad's start station and Vlad's finish station, respectively.
Output
Output "YES" if there is a time moment when Vlad and Daniel are at the same station, and "NO" otherwise. You can print each letter in any case (upper or lower).
Examples
Input
5 1 4 3 2
Output
YES
Input
10 2 1 9 10
Output
NO
Note
In the first example, Daniel and Vlad start at the stations (1, 3). One minute later they are at stations (2, 2). They are at the same station at this moment. Note that Vlad leaves the subway right after that.
Consider the second example, let's look at the stations Vlad and Daniel are at. They are:
* initially (2, 9),
* after 1 minute (3, 8),
* after 2 minutes (4, 7),
* after 3 minutes (5, 6),
* after 4 minutes (6, 5),
* after 5 minutes (7, 4),
* after 6 minutes (8, 3),
* after 7 minutes (9, 2),
* after 8 minutes (10, 1),
* after 9 minutes (1, 10).
After that, they both leave the subway because they are at their finish stations, so there is no moment when they both are at the same station.
Tags: implementation, math
Correct Solution:
```
if __name__ == '__main__':
n, a, x, b, y = map(int, input().split())
a -= 1
b -= 1
x -= 1
y -= 1
while a != x and b != y:
a = (a + 1) % n
b = (b - 1) % n
if a == b:
print("YES")
exit(0)
print("NO")
```
| 5,833 |
Provide tags and a correct Python 3 solution for this coding contest problem.
The circle line of the Roflanpolis subway has n stations.
There are two parallel routes in the subway. The first one visits stations in order 1 → 2 → … → n → 1 → 2 → … (so the next stop after station x is equal to (x+1) if x < n and 1 otherwise). The second route visits stations in order n → (n-1) → … → 1 → n → (n-1) → … (so the next stop after station x is equal to (x-1) if x>1 and n otherwise). All trains depart their stations simultaneously, and it takes exactly 1 minute to arrive at the next station.
Two toads live in this city, their names are Daniel and Vlad.
Daniel is currently in a train of the first route at station a and will exit the subway when his train reaches station x.
Coincidentally, Vlad is currently in a train of the second route at station b and he will exit the subway when his train reaches station y.
Surprisingly, all numbers a,x,b,y are distinct.
Toad Ilya asks you to check if Daniel and Vlad will ever be at the same station at the same time during their journey. In other words, check if there is a moment when their trains stop at the same station. Note that this includes the moments when Daniel or Vlad enter or leave the subway.
Input
The first line contains five space-separated integers n, a, x, b, y (4 ≤ n ≤ 100, 1 ≤ a, x, b, y ≤ n, all numbers among a, x, b, y are distinct) — the number of stations in Roflanpolis, Daniel's start station, Daniel's finish station, Vlad's start station and Vlad's finish station, respectively.
Output
Output "YES" if there is a time moment when Vlad and Daniel are at the same station, and "NO" otherwise. You can print each letter in any case (upper or lower).
Examples
Input
5 1 4 3 2
Output
YES
Input
10 2 1 9 10
Output
NO
Note
In the first example, Daniel and Vlad start at the stations (1, 3). One minute later they are at stations (2, 2). They are at the same station at this moment. Note that Vlad leaves the subway right after that.
Consider the second example, let's look at the stations Vlad and Daniel are at. They are:
* initially (2, 9),
* after 1 minute (3, 8),
* after 2 minutes (4, 7),
* after 3 minutes (5, 6),
* after 4 minutes (6, 5),
* after 5 minutes (7, 4),
* after 6 minutes (8, 3),
* after 7 minutes (9, 2),
* after 8 minutes (10, 1),
* after 9 minutes (1, 10).
After that, they both leave the subway because they are at their finish stations, so there is no moment when they both are at the same station.
Tags: implementation, math
Correct Solution:
```
n, a, x, b, y = map(int,input().split())
while a != x and b != y:
a += 1
b -= 1
if a > n:a = 1
if b <= 0:b = n
if a == b:
print("YES")
exit()
print("NO")
```
| 5,834 |
Provide tags and a correct Python 3 solution for this coding contest problem.
The circle line of the Roflanpolis subway has n stations.
There are two parallel routes in the subway. The first one visits stations in order 1 → 2 → … → n → 1 → 2 → … (so the next stop after station x is equal to (x+1) if x < n and 1 otherwise). The second route visits stations in order n → (n-1) → … → 1 → n → (n-1) → … (so the next stop after station x is equal to (x-1) if x>1 and n otherwise). All trains depart their stations simultaneously, and it takes exactly 1 minute to arrive at the next station.
Two toads live in this city, their names are Daniel and Vlad.
Daniel is currently in a train of the first route at station a and will exit the subway when his train reaches station x.
Coincidentally, Vlad is currently in a train of the second route at station b and he will exit the subway when his train reaches station y.
Surprisingly, all numbers a,x,b,y are distinct.
Toad Ilya asks you to check if Daniel and Vlad will ever be at the same station at the same time during their journey. In other words, check if there is a moment when their trains stop at the same station. Note that this includes the moments when Daniel or Vlad enter or leave the subway.
Input
The first line contains five space-separated integers n, a, x, b, y (4 ≤ n ≤ 100, 1 ≤ a, x, b, y ≤ n, all numbers among a, x, b, y are distinct) — the number of stations in Roflanpolis, Daniel's start station, Daniel's finish station, Vlad's start station and Vlad's finish station, respectively.
Output
Output "YES" if there is a time moment when Vlad and Daniel are at the same station, and "NO" otherwise. You can print each letter in any case (upper or lower).
Examples
Input
5 1 4 3 2
Output
YES
Input
10 2 1 9 10
Output
NO
Note
In the first example, Daniel and Vlad start at the stations (1, 3). One minute later they are at stations (2, 2). They are at the same station at this moment. Note that Vlad leaves the subway right after that.
Consider the second example, let's look at the stations Vlad and Daniel are at. They are:
* initially (2, 9),
* after 1 minute (3, 8),
* after 2 minutes (4, 7),
* after 3 minutes (5, 6),
* after 4 minutes (6, 5),
* after 5 minutes (7, 4),
* after 6 minutes (8, 3),
* after 7 minutes (9, 2),
* after 8 minutes (10, 1),
* after 9 minutes (1, 10).
After that, they both leave the subway because they are at their finish stations, so there is no moment when they both are at the same station.
Tags: implementation, math
Correct Solution:
```
n, a, x, b, y = map(int, input().split())
m1 = []
m2 = []
while a != x:
m1.append(a)
a += 1
if a > n:
a = 1
while b != y:
m2.append(b)
b -= 1
if b < 1:
b = n
m1.append(a)
m2.append(b)
b = False
for i in range(min(len(m1), len(m2))):
if m1[i] == m2[i]:
b = True
break
if b:
print('YES')
else:
print('NO')
```
| 5,835 |
Provide tags and a correct Python 3 solution for this coding contest problem.
The circle line of the Roflanpolis subway has n stations.
There are two parallel routes in the subway. The first one visits stations in order 1 → 2 → … → n → 1 → 2 → … (so the next stop after station x is equal to (x+1) if x < n and 1 otherwise). The second route visits stations in order n → (n-1) → … → 1 → n → (n-1) → … (so the next stop after station x is equal to (x-1) if x>1 and n otherwise). All trains depart their stations simultaneously, and it takes exactly 1 minute to arrive at the next station.
Two toads live in this city, their names are Daniel and Vlad.
Daniel is currently in a train of the first route at station a and will exit the subway when his train reaches station x.
Coincidentally, Vlad is currently in a train of the second route at station b and he will exit the subway when his train reaches station y.
Surprisingly, all numbers a,x,b,y are distinct.
Toad Ilya asks you to check if Daniel and Vlad will ever be at the same station at the same time during their journey. In other words, check if there is a moment when their trains stop at the same station. Note that this includes the moments when Daniel or Vlad enter or leave the subway.
Input
The first line contains five space-separated integers n, a, x, b, y (4 ≤ n ≤ 100, 1 ≤ a, x, b, y ≤ n, all numbers among a, x, b, y are distinct) — the number of stations in Roflanpolis, Daniel's start station, Daniel's finish station, Vlad's start station and Vlad's finish station, respectively.
Output
Output "YES" if there is a time moment when Vlad and Daniel are at the same station, and "NO" otherwise. You can print each letter in any case (upper or lower).
Examples
Input
5 1 4 3 2
Output
YES
Input
10 2 1 9 10
Output
NO
Note
In the first example, Daniel and Vlad start at the stations (1, 3). One minute later they are at stations (2, 2). They are at the same station at this moment. Note that Vlad leaves the subway right after that.
Consider the second example, let's look at the stations Vlad and Daniel are at. They are:
* initially (2, 9),
* after 1 minute (3, 8),
* after 2 minutes (4, 7),
* after 3 minutes (5, 6),
* after 4 minutes (6, 5),
* after 5 minutes (7, 4),
* after 6 minutes (8, 3),
* after 7 minutes (9, 2),
* after 8 minutes (10, 1),
* after 9 minutes (1, 10).
After that, they both leave the subway because they are at their finish stations, so there is no moment when they both are at the same station.
Tags: implementation, math
Correct Solution:
```
import sys
n, a, x, b, y = [int(x) for x in input().split()]
while a!=x and b!=y:
a+= 1
if a==n+1:
a = 1
b -= 1
if b==0:
b = n
if a==b:
print('YES')
sys.exit()
print('NO')
```
| 5,836 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
The circle line of the Roflanpolis subway has n stations.
There are two parallel routes in the subway. The first one visits stations in order 1 → 2 → … → n → 1 → 2 → … (so the next stop after station x is equal to (x+1) if x < n and 1 otherwise). The second route visits stations in order n → (n-1) → … → 1 → n → (n-1) → … (so the next stop after station x is equal to (x-1) if x>1 and n otherwise). All trains depart their stations simultaneously, and it takes exactly 1 minute to arrive at the next station.
Two toads live in this city, their names are Daniel and Vlad.
Daniel is currently in a train of the first route at station a and will exit the subway when his train reaches station x.
Coincidentally, Vlad is currently in a train of the second route at station b and he will exit the subway when his train reaches station y.
Surprisingly, all numbers a,x,b,y are distinct.
Toad Ilya asks you to check if Daniel and Vlad will ever be at the same station at the same time during their journey. In other words, check if there is a moment when their trains stop at the same station. Note that this includes the moments when Daniel or Vlad enter or leave the subway.
Input
The first line contains five space-separated integers n, a, x, b, y (4 ≤ n ≤ 100, 1 ≤ a, x, b, y ≤ n, all numbers among a, x, b, y are distinct) — the number of stations in Roflanpolis, Daniel's start station, Daniel's finish station, Vlad's start station and Vlad's finish station, respectively.
Output
Output "YES" if there is a time moment when Vlad and Daniel are at the same station, and "NO" otherwise. You can print each letter in any case (upper or lower).
Examples
Input
5 1 4 3 2
Output
YES
Input
10 2 1 9 10
Output
NO
Note
In the first example, Daniel and Vlad start at the stations (1, 3). One minute later they are at stations (2, 2). They are at the same station at this moment. Note that Vlad leaves the subway right after that.
Consider the second example, let's look at the stations Vlad and Daniel are at. They are:
* initially (2, 9),
* after 1 minute (3, 8),
* after 2 minutes (4, 7),
* after 3 minutes (5, 6),
* after 4 minutes (6, 5),
* after 5 minutes (7, 4),
* after 6 minutes (8, 3),
* after 7 minutes (9, 2),
* after 8 minutes (10, 1),
* after 9 minutes (1, 10).
After that, they both leave the subway because they are at their finish stations, so there is no moment when they both are at the same station.
Submitted Solution:
```
numberOfStations, danielS, danielE, vladS, vladE = input().split()
numberOfStations = int(numberOfStations)
danielS = int(danielS)
danielE = int(danielE)
vladE = int(vladE)
vladS = int(vladS)
i=0
meet=False
while True:
if danielS == vladS:
meet=True
break
if danielS == danielE or vladS == vladE:
break
if danielS == numberOfStations:
danielS=1
else:
danielS+=1
if vladS == 1:
vladS=numberOfStations
else:
vladS-=1
print("YES" if meet else "NO")
```
Yes
| 5,837 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
The circle line of the Roflanpolis subway has n stations.
There are two parallel routes in the subway. The first one visits stations in order 1 → 2 → … → n → 1 → 2 → … (so the next stop after station x is equal to (x+1) if x < n and 1 otherwise). The second route visits stations in order n → (n-1) → … → 1 → n → (n-1) → … (so the next stop after station x is equal to (x-1) if x>1 and n otherwise). All trains depart their stations simultaneously, and it takes exactly 1 minute to arrive at the next station.
Two toads live in this city, their names are Daniel and Vlad.
Daniel is currently in a train of the first route at station a and will exit the subway when his train reaches station x.
Coincidentally, Vlad is currently in a train of the second route at station b and he will exit the subway when his train reaches station y.
Surprisingly, all numbers a,x,b,y are distinct.
Toad Ilya asks you to check if Daniel and Vlad will ever be at the same station at the same time during their journey. In other words, check if there is a moment when their trains stop at the same station. Note that this includes the moments when Daniel or Vlad enter or leave the subway.
Input
The first line contains five space-separated integers n, a, x, b, y (4 ≤ n ≤ 100, 1 ≤ a, x, b, y ≤ n, all numbers among a, x, b, y are distinct) — the number of stations in Roflanpolis, Daniel's start station, Daniel's finish station, Vlad's start station and Vlad's finish station, respectively.
Output
Output "YES" if there is a time moment when Vlad and Daniel are at the same station, and "NO" otherwise. You can print each letter in any case (upper or lower).
Examples
Input
5 1 4 3 2
Output
YES
Input
10 2 1 9 10
Output
NO
Note
In the first example, Daniel and Vlad start at the stations (1, 3). One minute later they are at stations (2, 2). They are at the same station at this moment. Note that Vlad leaves the subway right after that.
Consider the second example, let's look at the stations Vlad and Daniel are at. They are:
* initially (2, 9),
* after 1 minute (3, 8),
* after 2 minutes (4, 7),
* after 3 minutes (5, 6),
* after 4 minutes (6, 5),
* after 5 minutes (7, 4),
* after 6 minutes (8, 3),
* after 7 minutes (9, 2),
* after 8 minutes (10, 1),
* after 9 minutes (1, 10).
After that, they both leave the subway because they are at their finish stations, so there is no moment when they both are at the same station.
Submitted Solution:
```
n, a, x, b, y = map(int, input().split())
def get_next_positive(a):
return a%n+1
def get_next_negative(b):
return n if b == 1 else b-1
while a != x and b != y:
a = get_next_positive(a)
b = get_next_negative(b)
if a == b:
print("YES")
exit(0)
print("NO")
```
Yes
| 5,838 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
The circle line of the Roflanpolis subway has n stations.
There are two parallel routes in the subway. The first one visits stations in order 1 → 2 → … → n → 1 → 2 → … (so the next stop after station x is equal to (x+1) if x < n and 1 otherwise). The second route visits stations in order n → (n-1) → … → 1 → n → (n-1) → … (so the next stop after station x is equal to (x-1) if x>1 and n otherwise). All trains depart their stations simultaneously, and it takes exactly 1 minute to arrive at the next station.
Two toads live in this city, their names are Daniel and Vlad.
Daniel is currently in a train of the first route at station a and will exit the subway when his train reaches station x.
Coincidentally, Vlad is currently in a train of the second route at station b and he will exit the subway when his train reaches station y.
Surprisingly, all numbers a,x,b,y are distinct.
Toad Ilya asks you to check if Daniel and Vlad will ever be at the same station at the same time during their journey. In other words, check if there is a moment when their trains stop at the same station. Note that this includes the moments when Daniel or Vlad enter or leave the subway.
Input
The first line contains five space-separated integers n, a, x, b, y (4 ≤ n ≤ 100, 1 ≤ a, x, b, y ≤ n, all numbers among a, x, b, y are distinct) — the number of stations in Roflanpolis, Daniel's start station, Daniel's finish station, Vlad's start station and Vlad's finish station, respectively.
Output
Output "YES" if there is a time moment when Vlad and Daniel are at the same station, and "NO" otherwise. You can print each letter in any case (upper or lower).
Examples
Input
5 1 4 3 2
Output
YES
Input
10 2 1 9 10
Output
NO
Note
In the first example, Daniel and Vlad start at the stations (1, 3). One minute later they are at stations (2, 2). They are at the same station at this moment. Note that Vlad leaves the subway right after that.
Consider the second example, let's look at the stations Vlad and Daniel are at. They are:
* initially (2, 9),
* after 1 minute (3, 8),
* after 2 minutes (4, 7),
* after 3 minutes (5, 6),
* after 4 minutes (6, 5),
* after 5 minutes (7, 4),
* after 6 minutes (8, 3),
* after 7 minutes (9, 2),
* after 8 minutes (10, 1),
* after 9 minutes (1, 10).
After that, they both leave the subway because they are at their finish stations, so there is no moment when they both are at the same station.
Submitted Solution:
```
n,a,x,b,y=map(int,input().split())
ans=0
curr=a
t=205
dan=[]
while(curr!=x):
dan.append(curr)
if(curr+1>n):
curr=1
else:
curr+=1
dan.append(x)
vlad=[]
curr=b
while(curr!=y):
vlad.append(curr)
if(curr-1==0):
curr=n
else:
curr-=1
vlad.append(y)
l=min(len(vlad),len(dan))
for i in range(l):
if(vlad[i]==dan[i]):
ans=1
#print(dan,vlad)
if(ans==1):
print("YES")
else:
print("NO")
```
Yes
| 5,839 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
The circle line of the Roflanpolis subway has n stations.
There are two parallel routes in the subway. The first one visits stations in order 1 → 2 → … → n → 1 → 2 → … (so the next stop after station x is equal to (x+1) if x < n and 1 otherwise). The second route visits stations in order n → (n-1) → … → 1 → n → (n-1) → … (so the next stop after station x is equal to (x-1) if x>1 and n otherwise). All trains depart their stations simultaneously, and it takes exactly 1 minute to arrive at the next station.
Two toads live in this city, their names are Daniel and Vlad.
Daniel is currently in a train of the first route at station a and will exit the subway when his train reaches station x.
Coincidentally, Vlad is currently in a train of the second route at station b and he will exit the subway when his train reaches station y.
Surprisingly, all numbers a,x,b,y are distinct.
Toad Ilya asks you to check if Daniel and Vlad will ever be at the same station at the same time during their journey. In other words, check if there is a moment when their trains stop at the same station. Note that this includes the moments when Daniel or Vlad enter or leave the subway.
Input
The first line contains five space-separated integers n, a, x, b, y (4 ≤ n ≤ 100, 1 ≤ a, x, b, y ≤ n, all numbers among a, x, b, y are distinct) — the number of stations in Roflanpolis, Daniel's start station, Daniel's finish station, Vlad's start station and Vlad's finish station, respectively.
Output
Output "YES" if there is a time moment when Vlad and Daniel are at the same station, and "NO" otherwise. You can print each letter in any case (upper or lower).
Examples
Input
5 1 4 3 2
Output
YES
Input
10 2 1 9 10
Output
NO
Note
In the first example, Daniel and Vlad start at the stations (1, 3). One minute later they are at stations (2, 2). They are at the same station at this moment. Note that Vlad leaves the subway right after that.
Consider the second example, let's look at the stations Vlad and Daniel are at. They are:
* initially (2, 9),
* after 1 minute (3, 8),
* after 2 minutes (4, 7),
* after 3 minutes (5, 6),
* after 4 minutes (6, 5),
* after 5 minutes (7, 4),
* after 6 minutes (8, 3),
* after 7 minutes (9, 2),
* after 8 minutes (10, 1),
* after 9 minutes (1, 10).
After that, they both leave the subway because they are at their finish stations, so there is no moment when they both are at the same station.
Submitted Solution:
```
n, a, x, b, y = map(int, input().split())
while a != x and b != y:
a = a % n + 1
b = (b - 1 - 1 + n) % n + 1
if a == b:
print("YES")
exit()
print("NO")
```
Yes
| 5,840 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
The circle line of the Roflanpolis subway has n stations.
There are two parallel routes in the subway. The first one visits stations in order 1 → 2 → … → n → 1 → 2 → … (so the next stop after station x is equal to (x+1) if x < n and 1 otherwise). The second route visits stations in order n → (n-1) → … → 1 → n → (n-1) → … (so the next stop after station x is equal to (x-1) if x>1 and n otherwise). All trains depart their stations simultaneously, and it takes exactly 1 minute to arrive at the next station.
Two toads live in this city, their names are Daniel and Vlad.
Daniel is currently in a train of the first route at station a and will exit the subway when his train reaches station x.
Coincidentally, Vlad is currently in a train of the second route at station b and he will exit the subway when his train reaches station y.
Surprisingly, all numbers a,x,b,y are distinct.
Toad Ilya asks you to check if Daniel and Vlad will ever be at the same station at the same time during their journey. In other words, check if there is a moment when their trains stop at the same station. Note that this includes the moments when Daniel or Vlad enter or leave the subway.
Input
The first line contains five space-separated integers n, a, x, b, y (4 ≤ n ≤ 100, 1 ≤ a, x, b, y ≤ n, all numbers among a, x, b, y are distinct) — the number of stations in Roflanpolis, Daniel's start station, Daniel's finish station, Vlad's start station and Vlad's finish station, respectively.
Output
Output "YES" if there is a time moment when Vlad and Daniel are at the same station, and "NO" otherwise. You can print each letter in any case (upper or lower).
Examples
Input
5 1 4 3 2
Output
YES
Input
10 2 1 9 10
Output
NO
Note
In the first example, Daniel and Vlad start at the stations (1, 3). One minute later they are at stations (2, 2). They are at the same station at this moment. Note that Vlad leaves the subway right after that.
Consider the second example, let's look at the stations Vlad and Daniel are at. They are:
* initially (2, 9),
* after 1 minute (3, 8),
* after 2 minutes (4, 7),
* after 3 minutes (5, 6),
* after 4 minutes (6, 5),
* after 5 minutes (7, 4),
* after 6 minutes (8, 3),
* after 7 minutes (9, 2),
* after 8 minutes (10, 1),
* after 9 minutes (1, 10).
After that, they both leave the subway because they are at their finish stations, so there is no moment when they both are at the same station.
Submitted Solution:
```
n, a, x, b, y = map(int, input().split())
# p1 = a
# p2 = b
while True:
if a == b:
print('YES')
exit()
elif a == x or b == y:
print('NO')
exit()
a = a + 1 if x < n else 1
b = b - 1 if x > 1 else n
```
No
| 5,841 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
The circle line of the Roflanpolis subway has n stations.
There are two parallel routes in the subway. The first one visits stations in order 1 → 2 → … → n → 1 → 2 → … (so the next stop after station x is equal to (x+1) if x < n and 1 otherwise). The second route visits stations in order n → (n-1) → … → 1 → n → (n-1) → … (so the next stop after station x is equal to (x-1) if x>1 and n otherwise). All trains depart their stations simultaneously, and it takes exactly 1 minute to arrive at the next station.
Two toads live in this city, their names are Daniel and Vlad.
Daniel is currently in a train of the first route at station a and will exit the subway when his train reaches station x.
Coincidentally, Vlad is currently in a train of the second route at station b and he will exit the subway when his train reaches station y.
Surprisingly, all numbers a,x,b,y are distinct.
Toad Ilya asks you to check if Daniel and Vlad will ever be at the same station at the same time during their journey. In other words, check if there is a moment when their trains stop at the same station. Note that this includes the moments when Daniel or Vlad enter or leave the subway.
Input
The first line contains five space-separated integers n, a, x, b, y (4 ≤ n ≤ 100, 1 ≤ a, x, b, y ≤ n, all numbers among a, x, b, y are distinct) — the number of stations in Roflanpolis, Daniel's start station, Daniel's finish station, Vlad's start station and Vlad's finish station, respectively.
Output
Output "YES" if there is a time moment when Vlad and Daniel are at the same station, and "NO" otherwise. You can print each letter in any case (upper or lower).
Examples
Input
5 1 4 3 2
Output
YES
Input
10 2 1 9 10
Output
NO
Note
In the first example, Daniel and Vlad start at the stations (1, 3). One minute later they are at stations (2, 2). They are at the same station at this moment. Note that Vlad leaves the subway right after that.
Consider the second example, let's look at the stations Vlad and Daniel are at. They are:
* initially (2, 9),
* after 1 minute (3, 8),
* after 2 minutes (4, 7),
* after 3 minutes (5, 6),
* after 4 minutes (6, 5),
* after 5 minutes (7, 4),
* after 6 minutes (8, 3),
* after 7 minutes (9, 2),
* after 8 minutes (10, 1),
* after 9 minutes (1, 10).
After that, they both leave the subway because they are at their finish stations, so there is no moment when they both are at the same station.
Submitted Solution:
```
d = tuple(map(int, input().split()))
n = d[0]
a = d[1]
x = d[2]
b = d[3]
y = d[4]
if a < x:
if b < y:
if a < b or y < x:
if (a < b and (a - b) % 2 == 0) or ((n - a + b) % 2 == 0):
print('YES')
else:
print('NO')
else:
print('NO')
else:
if (b < x and b > a) and (b - a) % 2 == 0:
print('YES')
else:
print('NO')
else:
if b > y:
if (x > b or a < y) and (b-a)%2==0:
print('YES')
else:
print('NO')
else:
if (b<a and (n-a+b)%2==0) or (b-a)%2==0:
print('YES')
else:
print('NO')
```
No
| 5,842 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
The circle line of the Roflanpolis subway has n stations.
There are two parallel routes in the subway. The first one visits stations in order 1 → 2 → … → n → 1 → 2 → … (so the next stop after station x is equal to (x+1) if x < n and 1 otherwise). The second route visits stations in order n → (n-1) → … → 1 → n → (n-1) → … (so the next stop after station x is equal to (x-1) if x>1 and n otherwise). All trains depart their stations simultaneously, and it takes exactly 1 minute to arrive at the next station.
Two toads live in this city, their names are Daniel and Vlad.
Daniel is currently in a train of the first route at station a and will exit the subway when his train reaches station x.
Coincidentally, Vlad is currently in a train of the second route at station b and he will exit the subway when his train reaches station y.
Surprisingly, all numbers a,x,b,y are distinct.
Toad Ilya asks you to check if Daniel and Vlad will ever be at the same station at the same time during their journey. In other words, check if there is a moment when their trains stop at the same station. Note that this includes the moments when Daniel or Vlad enter or leave the subway.
Input
The first line contains five space-separated integers n, a, x, b, y (4 ≤ n ≤ 100, 1 ≤ a, x, b, y ≤ n, all numbers among a, x, b, y are distinct) — the number of stations in Roflanpolis, Daniel's start station, Daniel's finish station, Vlad's start station and Vlad's finish station, respectively.
Output
Output "YES" if there is a time moment when Vlad and Daniel are at the same station, and "NO" otherwise. You can print each letter in any case (upper or lower).
Examples
Input
5 1 4 3 2
Output
YES
Input
10 2 1 9 10
Output
NO
Note
In the first example, Daniel and Vlad start at the stations (1, 3). One minute later they are at stations (2, 2). They are at the same station at this moment. Note that Vlad leaves the subway right after that.
Consider the second example, let's look at the stations Vlad and Daniel are at. They are:
* initially (2, 9),
* after 1 minute (3, 8),
* after 2 minutes (4, 7),
* after 3 minutes (5, 6),
* after 4 minutes (6, 5),
* after 5 minutes (7, 4),
* after 6 minutes (8, 3),
* after 7 minutes (9, 2),
* after 8 minutes (10, 1),
* after 9 minutes (1, 10).
After that, they both leave the subway because they are at their finish stations, so there is no moment when they both are at the same station.
Submitted Solution:
```
n, a, x, b, y = list(map(int, input().split()))
flag= 0
if a == b:
flag = 1
for q in range(min(abs(x-a), abs(y-b))):
if a == b:
flag = 1
break
a = a+1
if a > n:
a = 1
b = b-1
if b < 1:
b = n
if a == b:
flag = 1
break
if flag == 1:
print("YES")
else:
print("NO")
```
No
| 5,843 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
The circle line of the Roflanpolis subway has n stations.
There are two parallel routes in the subway. The first one visits stations in order 1 → 2 → … → n → 1 → 2 → … (so the next stop after station x is equal to (x+1) if x < n and 1 otherwise). The second route visits stations in order n → (n-1) → … → 1 → n → (n-1) → … (so the next stop after station x is equal to (x-1) if x>1 and n otherwise). All trains depart their stations simultaneously, and it takes exactly 1 minute to arrive at the next station.
Two toads live in this city, their names are Daniel and Vlad.
Daniel is currently in a train of the first route at station a and will exit the subway when his train reaches station x.
Coincidentally, Vlad is currently in a train of the second route at station b and he will exit the subway when his train reaches station y.
Surprisingly, all numbers a,x,b,y are distinct.
Toad Ilya asks you to check if Daniel and Vlad will ever be at the same station at the same time during their journey. In other words, check if there is a moment when their trains stop at the same station. Note that this includes the moments when Daniel or Vlad enter or leave the subway.
Input
The first line contains five space-separated integers n, a, x, b, y (4 ≤ n ≤ 100, 1 ≤ a, x, b, y ≤ n, all numbers among a, x, b, y are distinct) — the number of stations in Roflanpolis, Daniel's start station, Daniel's finish station, Vlad's start station and Vlad's finish station, respectively.
Output
Output "YES" if there is a time moment when Vlad and Daniel are at the same station, and "NO" otherwise. You can print each letter in any case (upper or lower).
Examples
Input
5 1 4 3 2
Output
YES
Input
10 2 1 9 10
Output
NO
Note
In the first example, Daniel and Vlad start at the stations (1, 3). One minute later they are at stations (2, 2). They are at the same station at this moment. Note that Vlad leaves the subway right after that.
Consider the second example, let's look at the stations Vlad and Daniel are at. They are:
* initially (2, 9),
* after 1 minute (3, 8),
* after 2 minutes (4, 7),
* after 3 minutes (5, 6),
* after 4 minutes (6, 5),
* after 5 minutes (7, 4),
* after 6 minutes (8, 3),
* after 7 minutes (9, 2),
* after 8 minutes (10, 1),
* after 9 minutes (1, 10).
After that, they both leave the subway because they are at their finish stations, so there is no moment when they both are at the same station.
Submitted Solution:
```
n, a, x, b, y = (int(el) for el in input().split())
s =[i + 1 for i in range(n)]
p = a
v = b
meet = False
while p <= x and v >= y:
if p == v:
meet = True
break
p += 1
v -= 1
if meet:
print('YES')
else:
print('NO')
```
No
| 5,844 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Vasya has an array a_1, a_2, ..., a_n.
You don't know this array, but he told you m facts about this array. The i-th fact is a triple of numbers t_i, l_i and r_i (0 ≤ t_i ≤ 1, 1 ≤ l_i < r_i ≤ n) and it means:
* if t_i=1 then subbarray a_{l_i}, a_{l_i + 1}, ..., a_{r_i} is sorted in non-decreasing order;
* if t_i=0 then subbarray a_{l_i}, a_{l_i + 1}, ..., a_{r_i} is not sorted in non-decreasing order. A subarray is not sorted if there is at least one pair of consecutive elements in this subarray such that the former is greater than the latter.
For example if a = [2, 1, 1, 3, 2] then he could give you three facts: t_1=1, l_1=2, r_1=4 (the subarray [a_2, a_3, a_4] = [1, 1, 3] is sorted), t_2=0, l_2=4, r_2=5 (the subarray [a_4, a_5] = [3, 2] is not sorted), and t_3=0, l_3=3, r_3=5 (the subarray [a_3, a_5] = [1, 3, 2] is not sorted).
You don't know the array a. Find any array which satisfies all the given facts.
Input
The first line contains two integers n and m (2 ≤ n ≤ 1000, 1 ≤ m ≤ 1000).
Each of the next m lines contains three integers t_i, l_i and r_i (0 ≤ t_i ≤ 1, 1 ≤ l_i < r_i ≤ n).
If t_i = 1 then subbarray a_{l_i}, a_{l_i + 1}, ... , a_{r_i} is sorted. Otherwise (if t_i = 0) subbarray a_{l_i}, a_{l_i + 1}, ... , a_{r_i} is not sorted.
Output
If there is no array that satisfies these facts in only line print NO (in any letter case).
If there is a solution, print YES (in any letter case). In second line print n integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 10^9) — the array a, satisfying all the given facts. If there are multiple satisfying arrays you can print any of them.
Examples
Input
7 4
1 1 3
1 2 5
0 5 6
1 6 7
Output
YES
1 2 2 3 5 4 4
Input
4 2
1 1 4
0 2 3
Output
NO
Tags: constructive algorithms, greedy, implementation
Correct Solution:
```
from collections import defaultdict
n,m = map(int,input().split())
l1 = []
l2 = []
ans = [1]*n
for i in range(m):
a,b,c = map(int,input().split())
if a == 0:
l2.append([b,c])
else:
l1.append([b,c])
hash = defaultdict(bool)
l1.sort()
l2.sort()
for i in range(len(l1)):
a,b = l1[i]
for i in range(a,b):
hash[(i,i+1)] = True
l2.reverse()
flag = 1
for i in range(len(l2)):
a,b = l2[i]
flag = 0
for i in range(a,b):
if hash[(i,i+1)] == False:
flag = 1
ans[i-1] = ans[i]+1
break
if flag == 0:
break
# print(hash)
if flag == 1:
print('YES')
print(*ans)
else:
print('NO')
```
| 5,845 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Vasya has an array a_1, a_2, ..., a_n.
You don't know this array, but he told you m facts about this array. The i-th fact is a triple of numbers t_i, l_i and r_i (0 ≤ t_i ≤ 1, 1 ≤ l_i < r_i ≤ n) and it means:
* if t_i=1 then subbarray a_{l_i}, a_{l_i + 1}, ..., a_{r_i} is sorted in non-decreasing order;
* if t_i=0 then subbarray a_{l_i}, a_{l_i + 1}, ..., a_{r_i} is not sorted in non-decreasing order. A subarray is not sorted if there is at least one pair of consecutive elements in this subarray such that the former is greater than the latter.
For example if a = [2, 1, 1, 3, 2] then he could give you three facts: t_1=1, l_1=2, r_1=4 (the subarray [a_2, a_3, a_4] = [1, 1, 3] is sorted), t_2=0, l_2=4, r_2=5 (the subarray [a_4, a_5] = [3, 2] is not sorted), and t_3=0, l_3=3, r_3=5 (the subarray [a_3, a_5] = [1, 3, 2] is not sorted).
You don't know the array a. Find any array which satisfies all the given facts.
Input
The first line contains two integers n and m (2 ≤ n ≤ 1000, 1 ≤ m ≤ 1000).
Each of the next m lines contains three integers t_i, l_i and r_i (0 ≤ t_i ≤ 1, 1 ≤ l_i < r_i ≤ n).
If t_i = 1 then subbarray a_{l_i}, a_{l_i + 1}, ... , a_{r_i} is sorted. Otherwise (if t_i = 0) subbarray a_{l_i}, a_{l_i + 1}, ... , a_{r_i} is not sorted.
Output
If there is no array that satisfies these facts in only line print NO (in any letter case).
If there is a solution, print YES (in any letter case). In second line print n integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 10^9) — the array a, satisfying all the given facts. If there are multiple satisfying arrays you can print any of them.
Examples
Input
7 4
1 1 3
1 2 5
0 5 6
1 6 7
Output
YES
1 2 2 3 5 4 4
Input
4 2
1 1 4
0 2 3
Output
NO
Tags: constructive algorithms, greedy, implementation
Correct Solution:
```
n,m=map(int,input().split())
a=[0]*n
a0=[]
while m>0:
m-=1
t,l,r=[int(i) for i in input().split()]
if t==0:
a0.append((l,r))
continue
for i in range(l,r):
a[i]=1
for l,r in a0:
if set(a[l:r]) == {1}:
print("NO")
exit (0)
print("YES")
val=5*10**4
for i in a:
if i: val+=1
else: val-=1
print(val)
```
| 5,846 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Vasya has an array a_1, a_2, ..., a_n.
You don't know this array, but he told you m facts about this array. The i-th fact is a triple of numbers t_i, l_i and r_i (0 ≤ t_i ≤ 1, 1 ≤ l_i < r_i ≤ n) and it means:
* if t_i=1 then subbarray a_{l_i}, a_{l_i + 1}, ..., a_{r_i} is sorted in non-decreasing order;
* if t_i=0 then subbarray a_{l_i}, a_{l_i + 1}, ..., a_{r_i} is not sorted in non-decreasing order. A subarray is not sorted if there is at least one pair of consecutive elements in this subarray such that the former is greater than the latter.
For example if a = [2, 1, 1, 3, 2] then he could give you three facts: t_1=1, l_1=2, r_1=4 (the subarray [a_2, a_3, a_4] = [1, 1, 3] is sorted), t_2=0, l_2=4, r_2=5 (the subarray [a_4, a_5] = [3, 2] is not sorted), and t_3=0, l_3=3, r_3=5 (the subarray [a_3, a_5] = [1, 3, 2] is not sorted).
You don't know the array a. Find any array which satisfies all the given facts.
Input
The first line contains two integers n and m (2 ≤ n ≤ 1000, 1 ≤ m ≤ 1000).
Each of the next m lines contains three integers t_i, l_i and r_i (0 ≤ t_i ≤ 1, 1 ≤ l_i < r_i ≤ n).
If t_i = 1 then subbarray a_{l_i}, a_{l_i + 1}, ... , a_{r_i} is sorted. Otherwise (if t_i = 0) subbarray a_{l_i}, a_{l_i + 1}, ... , a_{r_i} is not sorted.
Output
If there is no array that satisfies these facts in only line print NO (in any letter case).
If there is a solution, print YES (in any letter case). In second line print n integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 10^9) — the array a, satisfying all the given facts. If there are multiple satisfying arrays you can print any of them.
Examples
Input
7 4
1 1 3
1 2 5
0 5 6
1 6 7
Output
YES
1 2 2 3 5 4 4
Input
4 2
1 1 4
0 2 3
Output
NO
Tags: constructive algorithms, greedy, implementation
Correct Solution:
```
import sys
from collections import defaultdict
import typing
class DSU:
'''
Implement (union by size) + (path halving)
Reference:
Zvi Galil and Giuseppe F. Italiano,
Data structures and algorithms for disjoint set union problems
'''
def __init__(self, n: int = 0) -> None:
self._n = n
self.parent_or_size = [-1] * n
def merge(self, a: int, b: int) -> int:
assert 0 <= a < self._n
assert 0 <= b < self._n
x = self.leader(a)
y = self.leader(b)
if x == y:
return x
if -self.parent_or_size[x] < -self.parent_or_size[y]:
x, y = y, x
self.parent_or_size[x] += self.parent_or_size[y]
self.parent_or_size[y] = x
return x
def same(self, a: int, b: int) -> bool:
assert 0 <= a < self._n
assert 0 <= b < self._n
return self.leader(a) == self.leader(b)
def leader(self, a: int) -> int:
assert 0 <= a < self._n
parent = self.parent_or_size[a]
while parent >= 0:
if self.parent_or_size[parent] < 0:
return parent
self.parent_or_size[a], a, parent = (
self.parent_or_size[parent],
self.parent_or_size[parent],
self.parent_or_size[self.parent_or_size[parent]]
)
return a
def size(self, a: int) -> int:
assert 0 <= a < self._n
return -self.parent_or_size[self.leader(a)]
def groups(self) -> typing.List[typing.List[int]]:
leader_buf = [self.leader(i) for i in range(self._n)]
result: typing.List[typing.List[int]] = [[] for _ in range(self._n)]
for i in range(self._n):
result[leader_buf[i]].append(i)
return list(filter(lambda r: r, result))
def input():
return sys.stdin.readline().rstrip()
def slv():
n, m = map(int, input().split())
dsu = DSU(n)
info = [-1]*(n)
not_sorted = []
for i in range(m):
t, l, r = map(int, input().split())
l -= 1
r -= 1
if t == 1:
for j in range(l, r + 1):
dsu.merge(l, j)
else:
not_sorted.append((l, r))
group = [-1]*(n)
dsu_data = defaultdict(list)
for i in range(n):
dsu_data[dsu.leader(i)].append(i)
color = 0
for key, value in dsu_data.items():
for v in value:
group[v] = color
color += 1
# print(group)
for l, r in not_sorted:
if all(group[j] == group[l] for j in range(l, r + 1)):
print("NO")
return
INF = int(1e8)
D = int(1e4)
inc = 0
print("YES")
ans = [0]*(n)
group_index = group[0]
for i in range(n):
if group[i] == group_index:
ans[i] = INF + inc
inc += 1
else:
group_index = group[i]
INF -= D
inc = 0
ans[i] = INF + inc
inc += 1
print(*ans)
return
def main():
t = 1
for i in range(t):
slv()
return
if __name__ == "__main__":
main()
```
| 5,847 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Vasya has an array a_1, a_2, ..., a_n.
You don't know this array, but he told you m facts about this array. The i-th fact is a triple of numbers t_i, l_i and r_i (0 ≤ t_i ≤ 1, 1 ≤ l_i < r_i ≤ n) and it means:
* if t_i=1 then subbarray a_{l_i}, a_{l_i + 1}, ..., a_{r_i} is sorted in non-decreasing order;
* if t_i=0 then subbarray a_{l_i}, a_{l_i + 1}, ..., a_{r_i} is not sorted in non-decreasing order. A subarray is not sorted if there is at least one pair of consecutive elements in this subarray such that the former is greater than the latter.
For example if a = [2, 1, 1, 3, 2] then he could give you three facts: t_1=1, l_1=2, r_1=4 (the subarray [a_2, a_3, a_4] = [1, 1, 3] is sorted), t_2=0, l_2=4, r_2=5 (the subarray [a_4, a_5] = [3, 2] is not sorted), and t_3=0, l_3=3, r_3=5 (the subarray [a_3, a_5] = [1, 3, 2] is not sorted).
You don't know the array a. Find any array which satisfies all the given facts.
Input
The first line contains two integers n and m (2 ≤ n ≤ 1000, 1 ≤ m ≤ 1000).
Each of the next m lines contains three integers t_i, l_i and r_i (0 ≤ t_i ≤ 1, 1 ≤ l_i < r_i ≤ n).
If t_i = 1 then subbarray a_{l_i}, a_{l_i + 1}, ... , a_{r_i} is sorted. Otherwise (if t_i = 0) subbarray a_{l_i}, a_{l_i + 1}, ... , a_{r_i} is not sorted.
Output
If there is no array that satisfies these facts in only line print NO (in any letter case).
If there is a solution, print YES (in any letter case). In second line print n integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 10^9) — the array a, satisfying all the given facts. If there are multiple satisfying arrays you can print any of them.
Examples
Input
7 4
1 1 3
1 2 5
0 5 6
1 6 7
Output
YES
1 2 2 3 5 4 4
Input
4 2
1 1 4
0 2 3
Output
NO
Tags: constructive algorithms, greedy, implementation
Correct Solution:
```
# ///////////////////////////////////////////////////////////////////////////
# //////////////////// PYTHON IS THE BEST ////////////////////////
# ///////////////////////////////////////////////////////////////////////////
import sys,os,io
import math
from collections import defaultdict
from io import BytesIO, IOBase
from types import GeneratorType
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")
def ii():
return int(input())
def li():
return list(map(int,input().split()))
# ///////////////////////////////////////////////////////////////////////////
# //////////////////// DO NOT TOUCH BEFORE THIS LINE ////////////////////////
# ///////////////////////////////////////////////////////////////////////////
if(os.path.exists('input.txt')):
sys.stdin = open("input.txt","r") ; sys.stdout = open("output.txt","w")
else:
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")
n,m = li()
s = []
ns = []
for i in range(m):
t,l,r = li()
l-=1
r-=1
if t==0:
ns.append([l,r])
else:
s.append([l,r])
arr = [0]*n
for i in range(n-1):
ss = 0
nss = 0
for l,r in s:
if l<=i<=i+1<=r:
ss+=1
for l,r in ns:
if l<=i<=i+1<=r:
nss+=1
# if ss>0 and nss>0:
# print("NO")
# exit()
if nss>0 and ss==0:
arr[i+1]=-1
c = 0
for i in range(n):
if arr[i]==-1:
c-=2
arr[i]=c
else:
arr[i]=c
c+=1
for l,r in s:
for i in range(l,r):
if arr[i]>arr[i+1]:
print("NO")
exit()
for l,r in ns:
f = 0
for i in range(l,r):
if arr[i]>arr[i+1]:
f = 1
if f==0:
# print(l,r)
print("NO")
exit()
m = min(arr)
if m<=0:
y = -m+1
for i in range(n):
arr[i]+=y
print("YES")
print(*arr)
```
| 5,848 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Vasya has an array a_1, a_2, ..., a_n.
You don't know this array, but he told you m facts about this array. The i-th fact is a triple of numbers t_i, l_i and r_i (0 ≤ t_i ≤ 1, 1 ≤ l_i < r_i ≤ n) and it means:
* if t_i=1 then subbarray a_{l_i}, a_{l_i + 1}, ..., a_{r_i} is sorted in non-decreasing order;
* if t_i=0 then subbarray a_{l_i}, a_{l_i + 1}, ..., a_{r_i} is not sorted in non-decreasing order. A subarray is not sorted if there is at least one pair of consecutive elements in this subarray such that the former is greater than the latter.
For example if a = [2, 1, 1, 3, 2] then he could give you three facts: t_1=1, l_1=2, r_1=4 (the subarray [a_2, a_3, a_4] = [1, 1, 3] is sorted), t_2=0, l_2=4, r_2=5 (the subarray [a_4, a_5] = [3, 2] is not sorted), and t_3=0, l_3=3, r_3=5 (the subarray [a_3, a_5] = [1, 3, 2] is not sorted).
You don't know the array a. Find any array which satisfies all the given facts.
Input
The first line contains two integers n and m (2 ≤ n ≤ 1000, 1 ≤ m ≤ 1000).
Each of the next m lines contains three integers t_i, l_i and r_i (0 ≤ t_i ≤ 1, 1 ≤ l_i < r_i ≤ n).
If t_i = 1 then subbarray a_{l_i}, a_{l_i + 1}, ... , a_{r_i} is sorted. Otherwise (if t_i = 0) subbarray a_{l_i}, a_{l_i + 1}, ... , a_{r_i} is not sorted.
Output
If there is no array that satisfies these facts in only line print NO (in any letter case).
If there is a solution, print YES (in any letter case). In second line print n integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 10^9) — the array a, satisfying all the given facts. If there are multiple satisfying arrays you can print any of them.
Examples
Input
7 4
1 1 3
1 2 5
0 5 6
1 6 7
Output
YES
1 2 2 3 5 4 4
Input
4 2
1 1 4
0 2 3
Output
NO
Tags: constructive algorithms, greedy, implementation
Correct Solution:
```
# -*- coding: utf-8 -*-
# @Date : 2019-07-01 08:11:55
# @Author : raj lath (oorja.halt@gmail.com)
# @Link : link
# @Version : 1.0.0
import sys
sys.setrecursionlimit(10**5+1)
inf = int(10 ** 20)
max_val = inf
min_val = -inf
RW = lambda : sys.stdin.readline().strip()
RI = lambda : int(RW())
RMI = lambda : [int(x) for x in sys.stdin.readline().strip().split()]
RWI = lambda : [x for x in sys.stdin.readline().strip().split()]
lens, tips = RMI()
pos = [[], []]
for i in range(tips):
mode, *ind = RMI()
pos[mode].append(ind)
canbe = True
maybe = [-1] * (lens - 1)
for lr in pos[1]:
for curr in range(lr[0] - 1,lr[1] - 1):
maybe[curr] = 0
for lr in pos[0]:
if sum(maybe[lr[0]-1:lr[1]-1]) == 0:
canbe = False
break
if not canbe:print("NO")
else:
print("YES")
answer = [int(1e7)]
for i in range(lens - 1):
answer += [answer[-1] + maybe[i]]
print(*answer)
```
| 5,849 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Vasya has an array a_1, a_2, ..., a_n.
You don't know this array, but he told you m facts about this array. The i-th fact is a triple of numbers t_i, l_i and r_i (0 ≤ t_i ≤ 1, 1 ≤ l_i < r_i ≤ n) and it means:
* if t_i=1 then subbarray a_{l_i}, a_{l_i + 1}, ..., a_{r_i} is sorted in non-decreasing order;
* if t_i=0 then subbarray a_{l_i}, a_{l_i + 1}, ..., a_{r_i} is not sorted in non-decreasing order. A subarray is not sorted if there is at least one pair of consecutive elements in this subarray such that the former is greater than the latter.
For example if a = [2, 1, 1, 3, 2] then he could give you three facts: t_1=1, l_1=2, r_1=4 (the subarray [a_2, a_3, a_4] = [1, 1, 3] is sorted), t_2=0, l_2=4, r_2=5 (the subarray [a_4, a_5] = [3, 2] is not sorted), and t_3=0, l_3=3, r_3=5 (the subarray [a_3, a_5] = [1, 3, 2] is not sorted).
You don't know the array a. Find any array which satisfies all the given facts.
Input
The first line contains two integers n and m (2 ≤ n ≤ 1000, 1 ≤ m ≤ 1000).
Each of the next m lines contains three integers t_i, l_i and r_i (0 ≤ t_i ≤ 1, 1 ≤ l_i < r_i ≤ n).
If t_i = 1 then subbarray a_{l_i}, a_{l_i + 1}, ... , a_{r_i} is sorted. Otherwise (if t_i = 0) subbarray a_{l_i}, a_{l_i + 1}, ... , a_{r_i} is not sorted.
Output
If there is no array that satisfies these facts in only line print NO (in any letter case).
If there is a solution, print YES (in any letter case). In second line print n integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 10^9) — the array a, satisfying all the given facts. If there are multiple satisfying arrays you can print any of them.
Examples
Input
7 4
1 1 3
1 2 5
0 5 6
1 6 7
Output
YES
1 2 2 3 5 4 4
Input
4 2
1 1 4
0 2 3
Output
NO
Tags: constructive algorithms, greedy, implementation
Correct Solution:
```
n,m = map(int, input().split())
a = []
b = []
j = 1
l = [0]*n
for _ in range(m):
q,w,e = map(int, input().split())
if q == 1:
a += [ [q,w,e] ]
else:
b += [ [q,w,e] ]
for x in a:
if l[x[1]-1] != 0 and l[x[2] - 1] != 0 and l[x[1]-1] != l[x[2] - 1]:
#print('ok')
for i in range( x[1]-1,x[2] ):
l[i] = l[x[1]-1]
h = x[2]
if x[2] != n-1:
#print('ok')
mu = l[x[2]]
while h < n and l[h] == mu:
#print('ok')
l[h] = l[x[1]-1]
h+=1
elif l[x[1]-1] != 0:
for i in range( x[1]-1,x[2] ):
l[i] = l[x[1]-1]
elif l[x[2]-1] != 0:
for i in range( x[1]-1,x[2] ):
l[i] = l[x[2]-1]
else:
for i in range( x[1]-1,x[2] ):
l[i] = j
j += 1
#print(l)
fl=0
for x in b:
#print( l[ x[1] - 1] , l[ x[2] - 1] )
if (l[ x[1] - 1] == l[ x[2] - 1]) and l[ x[1] - 1] != 0:
fl = 1
break
if fl == 0:
print('YES')
val = 10**5
for i in range(n):
if i == 0 or (l[i] == l[i-1] and l[i] != 0) :
print(val, end= ' ')
val+=1
else:
val -= 2
print(val, end = ' ')
val+=1
else:
print('NO')
```
| 5,850 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Vasya has an array a_1, a_2, ..., a_n.
You don't know this array, but he told you m facts about this array. The i-th fact is a triple of numbers t_i, l_i and r_i (0 ≤ t_i ≤ 1, 1 ≤ l_i < r_i ≤ n) and it means:
* if t_i=1 then subbarray a_{l_i}, a_{l_i + 1}, ..., a_{r_i} is sorted in non-decreasing order;
* if t_i=0 then subbarray a_{l_i}, a_{l_i + 1}, ..., a_{r_i} is not sorted in non-decreasing order. A subarray is not sorted if there is at least one pair of consecutive elements in this subarray such that the former is greater than the latter.
For example if a = [2, 1, 1, 3, 2] then he could give you three facts: t_1=1, l_1=2, r_1=4 (the subarray [a_2, a_3, a_4] = [1, 1, 3] is sorted), t_2=0, l_2=4, r_2=5 (the subarray [a_4, a_5] = [3, 2] is not sorted), and t_3=0, l_3=3, r_3=5 (the subarray [a_3, a_5] = [1, 3, 2] is not sorted).
You don't know the array a. Find any array which satisfies all the given facts.
Input
The first line contains two integers n and m (2 ≤ n ≤ 1000, 1 ≤ m ≤ 1000).
Each of the next m lines contains three integers t_i, l_i and r_i (0 ≤ t_i ≤ 1, 1 ≤ l_i < r_i ≤ n).
If t_i = 1 then subbarray a_{l_i}, a_{l_i + 1}, ... , a_{r_i} is sorted. Otherwise (if t_i = 0) subbarray a_{l_i}, a_{l_i + 1}, ... , a_{r_i} is not sorted.
Output
If there is no array that satisfies these facts in only line print NO (in any letter case).
If there is a solution, print YES (in any letter case). In second line print n integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 10^9) — the array a, satisfying all the given facts. If there are multiple satisfying arrays you can print any of them.
Examples
Input
7 4
1 1 3
1 2 5
0 5 6
1 6 7
Output
YES
1 2 2 3 5 4 4
Input
4 2
1 1 4
0 2 3
Output
NO
Tags: constructive algorithms, greedy, implementation
Correct Solution:
```
import math
#5 4 4 4 3
n,m=map(int,input().split())
ans=[-(i+1) for i in range(n)]
query=[]
chec=[]
cnt=1
for i in range(m):
x,y,z=map(int,input().split())
if(x==1):
query.append([y,z])
else:
chec.append([y,z])
continue;
query.sort()
for i in range(len(query)):
t=query[i]
maxa=-math.inf
for j in range(t[0]-1,t[1]):
maxa=max(maxa,ans[j])
for j in range(t[0]-1,t[1]):
ans[j]=maxa
flag=0
glag=0
for i in range(len(chec)):
x=chec[i]
flag=1
for j in range(x[0],x[1]):
if(ans[j]<ans[j-1]):
flag=0
break;
if(flag==1):
glag=1
break;
if(glag==0):
print('YES')
r=min(ans)
for i in range(len(ans)):
ans[i]+=abs(r)+1
print(*ans)
else:
print('NO')
```
| 5,851 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Vasya has an array a_1, a_2, ..., a_n.
You don't know this array, but he told you m facts about this array. The i-th fact is a triple of numbers t_i, l_i and r_i (0 ≤ t_i ≤ 1, 1 ≤ l_i < r_i ≤ n) and it means:
* if t_i=1 then subbarray a_{l_i}, a_{l_i + 1}, ..., a_{r_i} is sorted in non-decreasing order;
* if t_i=0 then subbarray a_{l_i}, a_{l_i + 1}, ..., a_{r_i} is not sorted in non-decreasing order. A subarray is not sorted if there is at least one pair of consecutive elements in this subarray such that the former is greater than the latter.
For example if a = [2, 1, 1, 3, 2] then he could give you three facts: t_1=1, l_1=2, r_1=4 (the subarray [a_2, a_3, a_4] = [1, 1, 3] is sorted), t_2=0, l_2=4, r_2=5 (the subarray [a_4, a_5] = [3, 2] is not sorted), and t_3=0, l_3=3, r_3=5 (the subarray [a_3, a_5] = [1, 3, 2] is not sorted).
You don't know the array a. Find any array which satisfies all the given facts.
Input
The first line contains two integers n and m (2 ≤ n ≤ 1000, 1 ≤ m ≤ 1000).
Each of the next m lines contains three integers t_i, l_i and r_i (0 ≤ t_i ≤ 1, 1 ≤ l_i < r_i ≤ n).
If t_i = 1 then subbarray a_{l_i}, a_{l_i + 1}, ... , a_{r_i} is sorted. Otherwise (if t_i = 0) subbarray a_{l_i}, a_{l_i + 1}, ... , a_{r_i} is not sorted.
Output
If there is no array that satisfies these facts in only line print NO (in any letter case).
If there is a solution, print YES (in any letter case). In second line print n integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 10^9) — the array a, satisfying all the given facts. If there are multiple satisfying arrays you can print any of them.
Examples
Input
7 4
1 1 3
1 2 5
0 5 6
1 6 7
Output
YES
1 2 2 3 5 4 4
Input
4 2
1 1 4
0 2 3
Output
NO
Tags: constructive algorithms, greedy, implementation
Correct Solution:
```
#Code by Sounak, IIESTS
#------------------------------warmup----------------------------
import os
import sys
import math
from io import BytesIO, IOBase
from fractions import Fraction
import collections
from itertools import permutations
from collections import defaultdict
BUFSIZE = 8192
class FastIO(IOBase):
newlines = 0
def __init__(self, file):
self._fd = file.fileno()
self.buffer = BytesIO()
self.writable = "x" in file.mode or "r" not in file.mode
self.write = self.buffer.write if self.writable else None
def read(self):
while True:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
if not b:
break
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines = 0
return self.buffer.read()
def readline(self):
while self.newlines == 0:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
self.newlines = b.count(b"\n") + (not b)
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines -= 1
return self.buffer.readline()
def flush(self):
if self.writable:
os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
self.write = lambda s: self.buffer.write(s.encode("ascii"))
self.read = lambda: self.buffer.read().decode("ascii")
self.readline = lambda: self.buffer.readline().decode("ascii")
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")
#-------------------game starts now-----------------------------------------------------
n,m=map(int,input().split())
inc=list()
dec=list()
s=set()
for i in range (m):
ch,l,r=map(int,input().split())
if ch==1:
inc.append((l-1,r-1))
else:
dec.append((l-1,r-1))
a=[n]*n
inc.sort()
for i in range (len(inc)):
l,r=inc[i][0],inc[i][1]
for j in range (l+1,r+1):
s.add(j)
ch=1
#print(s)
dec.sort()
for i in range (len(dec)):
l,r=dec[i][0],dec[i][1]
c=0
for j in range (l+1,r+1):
if j in s:
c+=1
else:
if a[j]<a[j-1]:
break
a[j]=a[j-1]-1
break
if c==r-l:
ch=0
break
if ch==1:
print("YES")
print(*a)
else:
print("NO")
```
| 5,852 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Vasya has an array a_1, a_2, ..., a_n.
You don't know this array, but he told you m facts about this array. The i-th fact is a triple of numbers t_i, l_i and r_i (0 ≤ t_i ≤ 1, 1 ≤ l_i < r_i ≤ n) and it means:
* if t_i=1 then subbarray a_{l_i}, a_{l_i + 1}, ..., a_{r_i} is sorted in non-decreasing order;
* if t_i=0 then subbarray a_{l_i}, a_{l_i + 1}, ..., a_{r_i} is not sorted in non-decreasing order. A subarray is not sorted if there is at least one pair of consecutive elements in this subarray such that the former is greater than the latter.
For example if a = [2, 1, 1, 3, 2] then he could give you three facts: t_1=1, l_1=2, r_1=4 (the subarray [a_2, a_3, a_4] = [1, 1, 3] is sorted), t_2=0, l_2=4, r_2=5 (the subarray [a_4, a_5] = [3, 2] is not sorted), and t_3=0, l_3=3, r_3=5 (the subarray [a_3, a_5] = [1, 3, 2] is not sorted).
You don't know the array a. Find any array which satisfies all the given facts.
Input
The first line contains two integers n and m (2 ≤ n ≤ 1000, 1 ≤ m ≤ 1000).
Each of the next m lines contains three integers t_i, l_i and r_i (0 ≤ t_i ≤ 1, 1 ≤ l_i < r_i ≤ n).
If t_i = 1 then subbarray a_{l_i}, a_{l_i + 1}, ... , a_{r_i} is sorted. Otherwise (if t_i = 0) subbarray a_{l_i}, a_{l_i + 1}, ... , a_{r_i} is not sorted.
Output
If there is no array that satisfies these facts in only line print NO (in any letter case).
If there is a solution, print YES (in any letter case). In second line print n integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 10^9) — the array a, satisfying all the given facts. If there are multiple satisfying arrays you can print any of them.
Examples
Input
7 4
1 1 3
1 2 5
0 5 6
1 6 7
Output
YES
1 2 2 3 5 4 4
Input
4 2
1 1 4
0 2 3
Output
NO
Submitted Solution:
```
import sys
from itertools import accumulate
N, M = map(int, input().split())
TLR = [tuple(map(int, input().split())) for _ in range(M)]
TLR.sort(reverse = True)
table = [False]*(N+1)
for i in range(M):
t, l, r = TLR[i]
l -= 1
r -= 1
if not t:
break
table[l] += 1
table[r] -= 1
if i == M-1:
i += 1
stp = i
table = list(accumulate(table))
table = [1 if t > 0 else 0 for t in table[:-1]]
table += [0]
col = [-1]*N
ctr = 10000
for i in range(N):
if table[i-1] == 0:
ctr -= 1
if table[i-1] == 1 or table[i] == 1:
col[i] = ctr
for i in range(stp, M):
_, l, r = TLR[i]
l -= 1
r -= 1
if col[l] == col[r] > 0:
break
else:
print('YES')
Ans = [10000]*N
for i in range(N):
if col[i] < 0 or col[i-1] != col[i]:
Ans[i] = Ans[i-1] - 1
else:
Ans[i] = Ans[i-1]
print(*Ans)
sys.exit()
print('NO')
```
Yes
| 5,853 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Vasya has an array a_1, a_2, ..., a_n.
You don't know this array, but he told you m facts about this array. The i-th fact is a triple of numbers t_i, l_i and r_i (0 ≤ t_i ≤ 1, 1 ≤ l_i < r_i ≤ n) and it means:
* if t_i=1 then subbarray a_{l_i}, a_{l_i + 1}, ..., a_{r_i} is sorted in non-decreasing order;
* if t_i=0 then subbarray a_{l_i}, a_{l_i + 1}, ..., a_{r_i} is not sorted in non-decreasing order. A subarray is not sorted if there is at least one pair of consecutive elements in this subarray such that the former is greater than the latter.
For example if a = [2, 1, 1, 3, 2] then he could give you three facts: t_1=1, l_1=2, r_1=4 (the subarray [a_2, a_3, a_4] = [1, 1, 3] is sorted), t_2=0, l_2=4, r_2=5 (the subarray [a_4, a_5] = [3, 2] is not sorted), and t_3=0, l_3=3, r_3=5 (the subarray [a_3, a_5] = [1, 3, 2] is not sorted).
You don't know the array a. Find any array which satisfies all the given facts.
Input
The first line contains two integers n and m (2 ≤ n ≤ 1000, 1 ≤ m ≤ 1000).
Each of the next m lines contains three integers t_i, l_i and r_i (0 ≤ t_i ≤ 1, 1 ≤ l_i < r_i ≤ n).
If t_i = 1 then subbarray a_{l_i}, a_{l_i + 1}, ... , a_{r_i} is sorted. Otherwise (if t_i = 0) subbarray a_{l_i}, a_{l_i + 1}, ... , a_{r_i} is not sorted.
Output
If there is no array that satisfies these facts in only line print NO (in any letter case).
If there is a solution, print YES (in any letter case). In second line print n integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 10^9) — the array a, satisfying all the given facts. If there are multiple satisfying arrays you can print any of them.
Examples
Input
7 4
1 1 3
1 2 5
0 5 6
1 6 7
Output
YES
1 2 2 3 5 4 4
Input
4 2
1 1 4
0 2 3
Output
NO
Submitted Solution:
```
def get_compressed(facts):
compressed_facts = dict()
for fact in facts:
if fact[0] == 0:
continue
start, end = fact[1], fact[2]
to_remove = []
skip = False
for compressed in compressed_facts:
start_compressed, end_compressed = compressed_facts[compressed]
if start <= end_compressed and end > end_compressed:
to_remove.append(compressed)
start = min(start, start_compressed)
elif end >= start_compressed and start < start_compressed:
to_remove.append(compressed)
end = max(end, end_compressed)
elif start >= start_compressed and end <= end_compressed:
skip = True
break
if skip:
continue
compressed_facts[str(start) + ',' + str(end)] = [start, end]
for x in to_remove:
del compressed_facts[x]
return compressed_facts
def solve(n, facts):
compressed_facts = get_compressed(facts)
# Verify that no 0s are entirely inside 1s
for fact in facts:
if fact[0] == 1:
continue
for compressed in compressed_facts.values():
if fact[1] >= compressed[0] and fact[2] <= compressed[1]:
return None
a = n * [None]
for fact in compressed_facts.values():
start = 10**8
size = fact[1] - fact[0] + 1
for i in range(size):
# This is fine due to the limits on the problem (n < 1000)
a[fact[0] - 1 + i] = start - (10000 * fact[0]) + i
for i in range(n):
if a[i] is None:
if i == 0:
a[i] = 10**9 - 1
else:
a[i] = a[i-1] - 1
return a
n, num_facts = map(int, input().split(' '))
facts = []
for _ in range(num_facts):
facts.append(list(map(int, input().split(' '))))
sol = solve(n, facts)
if sol is None:
print('NO')
else:
print('YES')
print(' '.join(map(str, sol)))
```
Yes
| 5,854 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Vasya has an array a_1, a_2, ..., a_n.
You don't know this array, but he told you m facts about this array. The i-th fact is a triple of numbers t_i, l_i and r_i (0 ≤ t_i ≤ 1, 1 ≤ l_i < r_i ≤ n) and it means:
* if t_i=1 then subbarray a_{l_i}, a_{l_i + 1}, ..., a_{r_i} is sorted in non-decreasing order;
* if t_i=0 then subbarray a_{l_i}, a_{l_i + 1}, ..., a_{r_i} is not sorted in non-decreasing order. A subarray is not sorted if there is at least one pair of consecutive elements in this subarray such that the former is greater than the latter.
For example if a = [2, 1, 1, 3, 2] then he could give you three facts: t_1=1, l_1=2, r_1=4 (the subarray [a_2, a_3, a_4] = [1, 1, 3] is sorted), t_2=0, l_2=4, r_2=5 (the subarray [a_4, a_5] = [3, 2] is not sorted), and t_3=0, l_3=3, r_3=5 (the subarray [a_3, a_5] = [1, 3, 2] is not sorted).
You don't know the array a. Find any array which satisfies all the given facts.
Input
The first line contains two integers n and m (2 ≤ n ≤ 1000, 1 ≤ m ≤ 1000).
Each of the next m lines contains three integers t_i, l_i and r_i (0 ≤ t_i ≤ 1, 1 ≤ l_i < r_i ≤ n).
If t_i = 1 then subbarray a_{l_i}, a_{l_i + 1}, ... , a_{r_i} is sorted. Otherwise (if t_i = 0) subbarray a_{l_i}, a_{l_i + 1}, ... , a_{r_i} is not sorted.
Output
If there is no array that satisfies these facts in only line print NO (in any letter case).
If there is a solution, print YES (in any letter case). In second line print n integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 10^9) — the array a, satisfying all the given facts. If there are multiple satisfying arrays you can print any of them.
Examples
Input
7 4
1 1 3
1 2 5
0 5 6
1 6 7
Output
YES
1 2 2 3 5 4 4
Input
4 2
1 1 4
0 2 3
Output
NO
Submitted Solution:
```
import sys
zz=1
sys.setrecursionlimit(10**5)
if zz:
input=sys.stdin.readline
else:
sys.stdin=open('input.txt', 'r')
sys.stdout=open('all.txt','w')
di=[[-1,0],[1,0],[0,1],[0,-1]]
def fori(n):
return [fi() for i in range(n)]
def inc(d,c,x=1):
d[c]=d[c]+x if c in d else x
def ii():
return input().rstrip()
def li():
return [int(xx) for xx in input().split()]
def fli():
return [float(x) for x in input().split()]
def comp(a,b):
if(a>b):
return 2
return 2 if a==b else 0
def gi():
return [xx for xx in input().split()]
def gtc(tc,ans):
print("Case #"+str(tc)+":",ans)
def cil(n,m):
return n//m+int(n%m>0)
def fi():
return int(input())
def pro(a):
return reduce(lambda a,b:a*b,a)
def swap(a,i,j):
a[i],a[j]=a[j],a[i]
def si():
return list(input().rstrip())
def mi():
return map(int,input().split())
def gh():
sys.stdout.flush()
def isvalid(i,j,n,m):
return 0<=i<n and 0<=j<m
def bo(i):
return ord(i)-ord('a')
def graph(n,m):
for i in range(m):
x,y=mi()
a[x].append(y)
a[y].append(x)
t=1
uu=t
while t>0:
t-=1
n,m=mi()
a=[]
b=[]
for i in range(m):
typ,l,r=mi()
if typ==1:
a.append([l,r])
else:
b.append([l,r])
a.sort()
d=[]
l=r=-1
for i in range(len(a)):
if i==0:
l=a[i][0]
r=a[i][1]
else:
if a[i][0]>r:
d.append([l,r])
l=a[i][0]
r=a[i][1]
else:
r=max(r,a[i][1])
if l!=-1 and r!=-1:
d.append([l,r])
ans=[0]*(n+1)
c=n*n+1
a=[i for i in range(n+1)]
for i,j in d:
for l in range(i,j+1):
ans[l]=c
a[l]=i
c-=n+1
for i,j in b:
if a[i]==a[j]:
print("NO")
exit(0)
print("YES")
if ans.count(0)==len(ans):
for i in range(1,n+1):
ans[i]=c
c-=n-1
else:
p=ans.index(n*n+1)
for j in range(p-1,-1,-1):
if ans[j]==0:
ans[j]=ans[j+1]+1
for j in range(p+1,n+1):
if ans[j]==0:
ans[j]=ans[j-1]-1
print(*ans[1:])
```
Yes
| 5,855 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Vasya has an array a_1, a_2, ..., a_n.
You don't know this array, but he told you m facts about this array. The i-th fact is a triple of numbers t_i, l_i and r_i (0 ≤ t_i ≤ 1, 1 ≤ l_i < r_i ≤ n) and it means:
* if t_i=1 then subbarray a_{l_i}, a_{l_i + 1}, ..., a_{r_i} is sorted in non-decreasing order;
* if t_i=0 then subbarray a_{l_i}, a_{l_i + 1}, ..., a_{r_i} is not sorted in non-decreasing order. A subarray is not sorted if there is at least one pair of consecutive elements in this subarray such that the former is greater than the latter.
For example if a = [2, 1, 1, 3, 2] then he could give you three facts: t_1=1, l_1=2, r_1=4 (the subarray [a_2, a_3, a_4] = [1, 1, 3] is sorted), t_2=0, l_2=4, r_2=5 (the subarray [a_4, a_5] = [3, 2] is not sorted), and t_3=0, l_3=3, r_3=5 (the subarray [a_3, a_5] = [1, 3, 2] is not sorted).
You don't know the array a. Find any array which satisfies all the given facts.
Input
The first line contains two integers n and m (2 ≤ n ≤ 1000, 1 ≤ m ≤ 1000).
Each of the next m lines contains three integers t_i, l_i and r_i (0 ≤ t_i ≤ 1, 1 ≤ l_i < r_i ≤ n).
If t_i = 1 then subbarray a_{l_i}, a_{l_i + 1}, ... , a_{r_i} is sorted. Otherwise (if t_i = 0) subbarray a_{l_i}, a_{l_i + 1}, ... , a_{r_i} is not sorted.
Output
If there is no array that satisfies these facts in only line print NO (in any letter case).
If there is a solution, print YES (in any letter case). In second line print n integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 10^9) — the array a, satisfying all the given facts. If there are multiple satisfying arrays you can print any of them.
Examples
Input
7 4
1 1 3
1 2 5
0 5 6
1 6 7
Output
YES
1 2 2 3 5 4 4
Input
4 2
1 1 4
0 2 3
Output
NO
Submitted Solution:
```
n, m = map(int, input().split())
a = [0] * n
ordered = []
unordered = []
for i in range(m):
list1 = list(map(int, input().split()))
if list1[0] == 0:
unordered.append(list1)
else:
ordered.append(list1)
for t, l, r in ordered:
for i in range(l, r):
a[i] = t
for t, l, r in unordered:
unset = any(e == 0 for e in a[l:r])
if not unset:
print('NO')
exit()
print('YES')
v = 100000
for e in a:
if e == 1:
v += 1
else:
v -= 1
print(v, end=' ')
```
Yes
| 5,856 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Vasya has an array a_1, a_2, ..., a_n.
You don't know this array, but he told you m facts about this array. The i-th fact is a triple of numbers t_i, l_i and r_i (0 ≤ t_i ≤ 1, 1 ≤ l_i < r_i ≤ n) and it means:
* if t_i=1 then subbarray a_{l_i}, a_{l_i + 1}, ..., a_{r_i} is sorted in non-decreasing order;
* if t_i=0 then subbarray a_{l_i}, a_{l_i + 1}, ..., a_{r_i} is not sorted in non-decreasing order. A subarray is not sorted if there is at least one pair of consecutive elements in this subarray such that the former is greater than the latter.
For example if a = [2, 1, 1, 3, 2] then he could give you three facts: t_1=1, l_1=2, r_1=4 (the subarray [a_2, a_3, a_4] = [1, 1, 3] is sorted), t_2=0, l_2=4, r_2=5 (the subarray [a_4, a_5] = [3, 2] is not sorted), and t_3=0, l_3=3, r_3=5 (the subarray [a_3, a_5] = [1, 3, 2] is not sorted).
You don't know the array a. Find any array which satisfies all the given facts.
Input
The first line contains two integers n and m (2 ≤ n ≤ 1000, 1 ≤ m ≤ 1000).
Each of the next m lines contains three integers t_i, l_i and r_i (0 ≤ t_i ≤ 1, 1 ≤ l_i < r_i ≤ n).
If t_i = 1 then subbarray a_{l_i}, a_{l_i + 1}, ... , a_{r_i} is sorted. Otherwise (if t_i = 0) subbarray a_{l_i}, a_{l_i + 1}, ... , a_{r_i} is not sorted.
Output
If there is no array that satisfies these facts in only line print NO (in any letter case).
If there is a solution, print YES (in any letter case). In second line print n integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 10^9) — the array a, satisfying all the given facts. If there are multiple satisfying arrays you can print any of them.
Examples
Input
7 4
1 1 3
1 2 5
0 5 6
1 6 7
Output
YES
1 2 2 3 5 4 4
Input
4 2
1 1 4
0 2 3
Output
NO
Submitted Solution:
```
"""
NTC here
"""
from sys import setcheckinterval, stdin
setcheckinterval(1000)
# print("Case #{}: {} {}".format(i, n + m, n * m))
iin = lambda: int(stdin.readline())
lin = lambda: list(map(int, stdin.readline().split()))
def eqation(a,b):
for i in range(26):
if a[i]>b[i]:
return False
return True
n,m=lin()
a=[lin() for i in range(m)]
sa=[0 for i in range(n)]
sa1=[1 for i in range(n)]
a1=[]
a2=[]
for i in a:
if i[0]:
a1.append(i)
else:
a2.append(i)
a1.sort(reverse=True,key= lambda x:x[2]-x[1])
ch=0
mn=1
for i in a1:
if i[0]==0:break
ch+=1
l,r=i[1],i[2]
l-=mn
r-=1
mn+=1
sa1[l]=1
if r+1<n:
sa1[r+1]=-1
for j in range(l,r+1):
if sa[j]:
break
else:
sa[j]=ch
for k in range(r,j,-1):
if sa[j]:
break
else:
sa[j]=ch
#print(sa1,sa)
for i in a2:
l,r=i[1]-1,i[2]-1
if (sa[l] and sa[r] and sa[l]==sa[r]) or r-l==0:
print('NO')
exit()
else:
sa1[l]+=mn
mn+=1
for i in range(1,n):
sa1[i]+=sa1[i-1]
print('YES')
print(*sa1)
```
No
| 5,857 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Vasya has an array a_1, a_2, ..., a_n.
You don't know this array, but he told you m facts about this array. The i-th fact is a triple of numbers t_i, l_i and r_i (0 ≤ t_i ≤ 1, 1 ≤ l_i < r_i ≤ n) and it means:
* if t_i=1 then subbarray a_{l_i}, a_{l_i + 1}, ..., a_{r_i} is sorted in non-decreasing order;
* if t_i=0 then subbarray a_{l_i}, a_{l_i + 1}, ..., a_{r_i} is not sorted in non-decreasing order. A subarray is not sorted if there is at least one pair of consecutive elements in this subarray such that the former is greater than the latter.
For example if a = [2, 1, 1, 3, 2] then he could give you three facts: t_1=1, l_1=2, r_1=4 (the subarray [a_2, a_3, a_4] = [1, 1, 3] is sorted), t_2=0, l_2=4, r_2=5 (the subarray [a_4, a_5] = [3, 2] is not sorted), and t_3=0, l_3=3, r_3=5 (the subarray [a_3, a_5] = [1, 3, 2] is not sorted).
You don't know the array a. Find any array which satisfies all the given facts.
Input
The first line contains two integers n and m (2 ≤ n ≤ 1000, 1 ≤ m ≤ 1000).
Each of the next m lines contains three integers t_i, l_i and r_i (0 ≤ t_i ≤ 1, 1 ≤ l_i < r_i ≤ n).
If t_i = 1 then subbarray a_{l_i}, a_{l_i + 1}, ... , a_{r_i} is sorted. Otherwise (if t_i = 0) subbarray a_{l_i}, a_{l_i + 1}, ... , a_{r_i} is not sorted.
Output
If there is no array that satisfies these facts in only line print NO (in any letter case).
If there is a solution, print YES (in any letter case). In second line print n integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 10^9) — the array a, satisfying all the given facts. If there are multiple satisfying arrays you can print any of them.
Examples
Input
7 4
1 1 3
1 2 5
0 5 6
1 6 7
Output
YES
1 2 2 3 5 4 4
Input
4 2
1 1 4
0 2 3
Output
NO
Submitted Solution:
```
def mp():
return map(int, input().split())
def check(l, r):
return int(a[l:r + 1] == sorted(a[l:r + 1]))
n, m = mp()
q = [0] * m
for qq in range(m):
t, l, r = mp()
l, r = l - 1, r - 1
q[qq] = [l, r, t]
q.sort()
a = [0 for i in range(n)]
fail = False
for qq in range(m):
l, r, t = q[qq]
if t == 0:
if a[l] == 0:
a[l] = r - l + 1
for i in range(l + 1, r + 1):
a[i] = a[i - 1] - 1
#print(a)
for qq in range(m):
l, r, t = q[qq]
if t == 1:
i = l
while i < n and a[i] == 0:
i += 1
if i == n:
x = 1
else:
x = a[i]
a[l] = x
for i in range(l + 1, r + 1):
if a[i - 1] > a[i]:
a[i] = a[i - 1]
mn = min(a)
if mn < 1:
mn = abs(mn) + 1
for i in range(n):
a[i] += mn
fail = False
for qq in range(m):
l, r, t = q[qq]
if t != check(l, r):
fail = True
break
ff = False
for qq in range(m):
for tt in range(qq + 1, m):
if q[qq][0] <= q[tt][0] and q[tt][1] <= q[qq][1] and q[qq][2] == 1 and q[tt][2] == 0:
ff = True
if ff and not fail:
print(0//0)
if fail:
print('NO')
else:
print('YES')
for i in a:
print(i, end = ' ')
```
No
| 5,858 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Vasya has an array a_1, a_2, ..., a_n.
You don't know this array, but he told you m facts about this array. The i-th fact is a triple of numbers t_i, l_i and r_i (0 ≤ t_i ≤ 1, 1 ≤ l_i < r_i ≤ n) and it means:
* if t_i=1 then subbarray a_{l_i}, a_{l_i + 1}, ..., a_{r_i} is sorted in non-decreasing order;
* if t_i=0 then subbarray a_{l_i}, a_{l_i + 1}, ..., a_{r_i} is not sorted in non-decreasing order. A subarray is not sorted if there is at least one pair of consecutive elements in this subarray such that the former is greater than the latter.
For example if a = [2, 1, 1, 3, 2] then he could give you three facts: t_1=1, l_1=2, r_1=4 (the subarray [a_2, a_3, a_4] = [1, 1, 3] is sorted), t_2=0, l_2=4, r_2=5 (the subarray [a_4, a_5] = [3, 2] is not sorted), and t_3=0, l_3=3, r_3=5 (the subarray [a_3, a_5] = [1, 3, 2] is not sorted).
You don't know the array a. Find any array which satisfies all the given facts.
Input
The first line contains two integers n and m (2 ≤ n ≤ 1000, 1 ≤ m ≤ 1000).
Each of the next m lines contains three integers t_i, l_i and r_i (0 ≤ t_i ≤ 1, 1 ≤ l_i < r_i ≤ n).
If t_i = 1 then subbarray a_{l_i}, a_{l_i + 1}, ... , a_{r_i} is sorted. Otherwise (if t_i = 0) subbarray a_{l_i}, a_{l_i + 1}, ... , a_{r_i} is not sorted.
Output
If there is no array that satisfies these facts in only line print NO (in any letter case).
If there is a solution, print YES (in any letter case). In second line print n integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 10^9) — the array a, satisfying all the given facts. If there are multiple satisfying arrays you can print any of them.
Examples
Input
7 4
1 1 3
1 2 5
0 5 6
1 6 7
Output
YES
1 2 2 3 5 4 4
Input
4 2
1 1 4
0 2 3
Output
NO
Submitted Solution:
```
n,m=map(int,input().split())
container=[0 for i in range(n+1)]
stick=[0 for i in range(n+1)]
maxid=[-1 for i in range(n+1)]
stored=[]
for _ in range(m):
t,l,r=map(int,input().split())
if(t==1):
maxid[l]=max(maxid[l],r)
else:
stored.append([l,r])
maxi=0
for i in range(1,n+1):
maxi=max(maxi,maxid[i])
if(maxi>=i):
container[i]=1
if(maxi==i):
stick[i]=1
count=1
ans=[0 for i in range(n+1)]
for i in range(n,0,-1):
ans[i]=count
if(container[i-1]==1==container[i] and stick[i-1]==0):
pass
else:
count+=1
print(*ans[1:])
```
No
| 5,859 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Vasya has an array a_1, a_2, ..., a_n.
You don't know this array, but he told you m facts about this array. The i-th fact is a triple of numbers t_i, l_i and r_i (0 ≤ t_i ≤ 1, 1 ≤ l_i < r_i ≤ n) and it means:
* if t_i=1 then subbarray a_{l_i}, a_{l_i + 1}, ..., a_{r_i} is sorted in non-decreasing order;
* if t_i=0 then subbarray a_{l_i}, a_{l_i + 1}, ..., a_{r_i} is not sorted in non-decreasing order. A subarray is not sorted if there is at least one pair of consecutive elements in this subarray such that the former is greater than the latter.
For example if a = [2, 1, 1, 3, 2] then he could give you three facts: t_1=1, l_1=2, r_1=4 (the subarray [a_2, a_3, a_4] = [1, 1, 3] is sorted), t_2=0, l_2=4, r_2=5 (the subarray [a_4, a_5] = [3, 2] is not sorted), and t_3=0, l_3=3, r_3=5 (the subarray [a_3, a_5] = [1, 3, 2] is not sorted).
You don't know the array a. Find any array which satisfies all the given facts.
Input
The first line contains two integers n and m (2 ≤ n ≤ 1000, 1 ≤ m ≤ 1000).
Each of the next m lines contains three integers t_i, l_i and r_i (0 ≤ t_i ≤ 1, 1 ≤ l_i < r_i ≤ n).
If t_i = 1 then subbarray a_{l_i}, a_{l_i + 1}, ... , a_{r_i} is sorted. Otherwise (if t_i = 0) subbarray a_{l_i}, a_{l_i + 1}, ... , a_{r_i} is not sorted.
Output
If there is no array that satisfies these facts in only line print NO (in any letter case).
If there is a solution, print YES (in any letter case). In second line print n integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 10^9) — the array a, satisfying all the given facts. If there are multiple satisfying arrays you can print any of them.
Examples
Input
7 4
1 1 3
1 2 5
0 5 6
1 6 7
Output
YES
1 2 2 3 5 4 4
Input
4 2
1 1 4
0 2 3
Output
NO
Submitted Solution:
```
def fillSorted(a,sector):
for i in range(sector[0]-1,sector[1]-1,1):
a[i]=1
return a
def isUnsortable(a,sector):
for i in range(sector[0]-1,sector[1]-1,1):
if a[i]=='n':
return True
return False
n,m= map(int, input().split())
a=[]
for i in range(n):
a.append('n')
sorted=[]
unsorted=[]
for i in range(m):
t,l,r=map(int, input().split())
if t:
sorted.append([l,r])
else:
unsorted.append([l,r])
for sector in sorted:
a=fillSorted(a,sector)
for sector in unsorted:
if not(isUnsortable(a,sector)):
print("NO")
exit(0)
Ans=[1]
for i in range(len(a)-1):
if a[i]==1:
Ans.append(Ans[i]+1)
else:
Ans.append(Ans[i]-1)
print("YES")
print(*Ans)
"""
print(sorted)
print(unsorted)
print(a)
print(isUnsortable(a,unsorted[0]))"""
```
No
| 5,860 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given n numbers a_1, a_2, ..., a_n. With a cost of one coin you can perform the following operation:
Choose one of these numbers and add or subtract 1 from it.
In particular, we can apply this operation to the same number several times.
We want to make the product of all these numbers equal to 1, in other words, we want a_1 ⋅ a_2 ... ⋅ a_n = 1.
For example, for n = 3 and numbers [1, -3, 0] we can make product equal to 1 in 3 coins: add 1 to second element, add 1 to second element again, subtract 1 from third element, so that array becomes [1, -1, -1]. And 1⋅ (-1) ⋅ (-1) = 1.
What is the minimum cost we will have to pay to do that?
Input
The first line contains a single integer n (1 ≤ n ≤ 10^5) — the number of numbers.
The second line contains n integers a_1, a_2, ..., a_n (-10^9 ≤ a_i ≤ 10^9) — the numbers.
Output
Output a single number — the minimal number of coins you need to pay to make the product equal to 1.
Examples
Input
2
-1 1
Output
2
Input
4
0 0 0 0
Output
4
Input
5
-5 -3 5 3 0
Output
13
Note
In the first example, you can change 1 to -1 or -1 to 1 in 2 coins.
In the second example, you have to apply at least 4 operations for the product not to be 0.
In the third example, you can change -5 to -1 in 4 coins, -3 to -1 in 2 coins, 5 to 1 in 4 coins, 3 to 1 in 2 coins, 0 to 1 in 1 coin.
Tags: dp, implementation
Correct Solution:
```
n = int(input())
nums = sorted(list(map(int, input().split())))
ideal = 0
countNeg = 0
for i in range(n):
if nums[i] > 0:
ideal += nums[i] - 1
elif nums[i] < 0:
ideal += abs(-1 - nums[i])
countNeg += 1
else:
ideal += 1
if countNeg % 2 != 0 and 0 not in nums:
ideal += 2
print(ideal)
```
| 5,861 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given n numbers a_1, a_2, ..., a_n. With a cost of one coin you can perform the following operation:
Choose one of these numbers and add or subtract 1 from it.
In particular, we can apply this operation to the same number several times.
We want to make the product of all these numbers equal to 1, in other words, we want a_1 ⋅ a_2 ... ⋅ a_n = 1.
For example, for n = 3 and numbers [1, -3, 0] we can make product equal to 1 in 3 coins: add 1 to second element, add 1 to second element again, subtract 1 from third element, so that array becomes [1, -1, -1]. And 1⋅ (-1) ⋅ (-1) = 1.
What is the minimum cost we will have to pay to do that?
Input
The first line contains a single integer n (1 ≤ n ≤ 10^5) — the number of numbers.
The second line contains n integers a_1, a_2, ..., a_n (-10^9 ≤ a_i ≤ 10^9) — the numbers.
Output
Output a single number — the minimal number of coins you need to pay to make the product equal to 1.
Examples
Input
2
-1 1
Output
2
Input
4
0 0 0 0
Output
4
Input
5
-5 -3 5 3 0
Output
13
Note
In the first example, you can change 1 to -1 or -1 to 1 in 2 coins.
In the second example, you have to apply at least 4 operations for the product not to be 0.
In the third example, you can change -5 to -1 in 4 coins, -3 to -1 in 2 coins, 5 to 1 in 4 coins, 3 to 1 in 2 coins, 0 to 1 in 1 coin.
Tags: dp, implementation
Correct Solution:
```
s=0
n=0
z=0
t=int(input())
l=[int(i) for i in input().split()]
for i in l:
if i==0:
z+=1
if i<0:
s+=abs(i)-1
n+=1
if i>0:
s+=abs(i-1)
if z==0:
if n%2==0:
print(s)
else:
print(s+2)
else:
print(s+z)
```
| 5,862 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given n numbers a_1, a_2, ..., a_n. With a cost of one coin you can perform the following operation:
Choose one of these numbers and add or subtract 1 from it.
In particular, we can apply this operation to the same number several times.
We want to make the product of all these numbers equal to 1, in other words, we want a_1 ⋅ a_2 ... ⋅ a_n = 1.
For example, for n = 3 and numbers [1, -3, 0] we can make product equal to 1 in 3 coins: add 1 to second element, add 1 to second element again, subtract 1 from third element, so that array becomes [1, -1, -1]. And 1⋅ (-1) ⋅ (-1) = 1.
What is the minimum cost we will have to pay to do that?
Input
The first line contains a single integer n (1 ≤ n ≤ 10^5) — the number of numbers.
The second line contains n integers a_1, a_2, ..., a_n (-10^9 ≤ a_i ≤ 10^9) — the numbers.
Output
Output a single number — the minimal number of coins you need to pay to make the product equal to 1.
Examples
Input
2
-1 1
Output
2
Input
4
0 0 0 0
Output
4
Input
5
-5 -3 5 3 0
Output
13
Note
In the first example, you can change 1 to -1 or -1 to 1 in 2 coins.
In the second example, you have to apply at least 4 operations for the product not to be 0.
In the third example, you can change -5 to -1 in 4 coins, -3 to -1 in 2 coins, 5 to 1 in 4 coins, 3 to 1 in 2 coins, 0 to 1 in 1 coin.
Tags: dp, implementation
Correct Solution:
```
import math
import sys
import collections
# imgur.com/Pkt7iIf.png
def getdict(n):
d = {}
if type(n) is list:
for i in n:
if i in d:
d[i] += 1
else:
d[i] = 1
else:
for i in range(n):
t = ii()
if t in d:
d[t] += 1
else:
d[t] = 1
return d
def cdiv(n, k): return n // k + (n % k != 0)
def ii(): return int(input())
def mi(): return map(int, input().split())
def li(): return list(map(int, input().split()))
def lcm(a, b): return abs(a*b) // math.gcd(a, b)
n = ii()
d = li()
r = n = z = 0
for i in d:
r += abs(abs(i) - 1)
z += (i == 0)
n += (i < 0)
print(r) if n%2 == 0 or z > 0 else print(r + 2)
```
| 5,863 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given n numbers a_1, a_2, ..., a_n. With a cost of one coin you can perform the following operation:
Choose one of these numbers and add or subtract 1 from it.
In particular, we can apply this operation to the same number several times.
We want to make the product of all these numbers equal to 1, in other words, we want a_1 ⋅ a_2 ... ⋅ a_n = 1.
For example, for n = 3 and numbers [1, -3, 0] we can make product equal to 1 in 3 coins: add 1 to second element, add 1 to second element again, subtract 1 from third element, so that array becomes [1, -1, -1]. And 1⋅ (-1) ⋅ (-1) = 1.
What is the minimum cost we will have to pay to do that?
Input
The first line contains a single integer n (1 ≤ n ≤ 10^5) — the number of numbers.
The second line contains n integers a_1, a_2, ..., a_n (-10^9 ≤ a_i ≤ 10^9) — the numbers.
Output
Output a single number — the minimal number of coins you need to pay to make the product equal to 1.
Examples
Input
2
-1 1
Output
2
Input
4
0 0 0 0
Output
4
Input
5
-5 -3 5 3 0
Output
13
Note
In the first example, you can change 1 to -1 or -1 to 1 in 2 coins.
In the second example, you have to apply at least 4 operations for the product not to be 0.
In the third example, you can change -5 to -1 in 4 coins, -3 to -1 in 2 coins, 5 to 1 in 4 coins, 3 to 1 in 2 coins, 0 to 1 in 1 coin.
Tags: dp, implementation
Correct Solution:
```
n = int(input())
a = list(map(int, input().split()))
neg = 0
ans = 0
zeros = 0
for i in range(n):
if a[i]>0:
ans+=a[i]-1
elif a[i]==0:
zeros+=1
else:
neg+=1
ans+=abs(-1 - a[i])
if (neg%2)!=0:
if zeros>0:
ans+=1
zeros-=1
else:
ans+=2
ans+=zeros
print(ans)
```
| 5,864 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given n numbers a_1, a_2, ..., a_n. With a cost of one coin you can perform the following operation:
Choose one of these numbers and add or subtract 1 from it.
In particular, we can apply this operation to the same number several times.
We want to make the product of all these numbers equal to 1, in other words, we want a_1 ⋅ a_2 ... ⋅ a_n = 1.
For example, for n = 3 and numbers [1, -3, 0] we can make product equal to 1 in 3 coins: add 1 to second element, add 1 to second element again, subtract 1 from third element, so that array becomes [1, -1, -1]. And 1⋅ (-1) ⋅ (-1) = 1.
What is the minimum cost we will have to pay to do that?
Input
The first line contains a single integer n (1 ≤ n ≤ 10^5) — the number of numbers.
The second line contains n integers a_1, a_2, ..., a_n (-10^9 ≤ a_i ≤ 10^9) — the numbers.
Output
Output a single number — the minimal number of coins you need to pay to make the product equal to 1.
Examples
Input
2
-1 1
Output
2
Input
4
0 0 0 0
Output
4
Input
5
-5 -3 5 3 0
Output
13
Note
In the first example, you can change 1 to -1 or -1 to 1 in 2 coins.
In the second example, you have to apply at least 4 operations for the product not to be 0.
In the third example, you can change -5 to -1 in 4 coins, -3 to -1 in 2 coins, 5 to 1 in 4 coins, 3 to 1 in 2 coins, 0 to 1 in 1 coin.
Tags: dp, implementation
Correct Solution:
```
n = int(input())
a = list(map(int, input().split()))
is_ = False
p = 0
m = 0
for i in a:
if i > 0: m += i - 1;
elif i < 0:
m += -i - 1
if not(is_): is_ = True;
else: is_ = False
else:
m += 1
p += 1
if is_ and p < 1: print(m + 2);
else: print(m);
```
| 5,865 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given n numbers a_1, a_2, ..., a_n. With a cost of one coin you can perform the following operation:
Choose one of these numbers and add or subtract 1 from it.
In particular, we can apply this operation to the same number several times.
We want to make the product of all these numbers equal to 1, in other words, we want a_1 ⋅ a_2 ... ⋅ a_n = 1.
For example, for n = 3 and numbers [1, -3, 0] we can make product equal to 1 in 3 coins: add 1 to second element, add 1 to second element again, subtract 1 from third element, so that array becomes [1, -1, -1]. And 1⋅ (-1) ⋅ (-1) = 1.
What is the minimum cost we will have to pay to do that?
Input
The first line contains a single integer n (1 ≤ n ≤ 10^5) — the number of numbers.
The second line contains n integers a_1, a_2, ..., a_n (-10^9 ≤ a_i ≤ 10^9) — the numbers.
Output
Output a single number — the minimal number of coins you need to pay to make the product equal to 1.
Examples
Input
2
-1 1
Output
2
Input
4
0 0 0 0
Output
4
Input
5
-5 -3 5 3 0
Output
13
Note
In the first example, you can change 1 to -1 or -1 to 1 in 2 coins.
In the second example, you have to apply at least 4 operations for the product not to be 0.
In the third example, you can change -5 to -1 in 4 coins, -3 to -1 in 2 coins, 5 to 1 in 4 coins, 3 to 1 in 2 coins, 0 to 1 in 1 coin.
Tags: dp, implementation
Correct Solution:
```
n = int(input())
a = [int(x) for x in input().split()]
ans = 0
neg = 0
zero = 0
# set to 1
for i in range(n):
if a[i] < 0:
ans += abs(a[i] + 1)
a[i] = -1
neg += 1
elif a[i] > 0:
ans += abs(a[i] - 1)
a[i] = 1
else: zero += 1
# sign
if neg % 2 == 1:
if zero > 0:
ans += zero
else:
ans += 2
else: ans += zero
# print(a)
print(ans)
```
| 5,866 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given n numbers a_1, a_2, ..., a_n. With a cost of one coin you can perform the following operation:
Choose one of these numbers and add or subtract 1 from it.
In particular, we can apply this operation to the same number several times.
We want to make the product of all these numbers equal to 1, in other words, we want a_1 ⋅ a_2 ... ⋅ a_n = 1.
For example, for n = 3 and numbers [1, -3, 0] we can make product equal to 1 in 3 coins: add 1 to second element, add 1 to second element again, subtract 1 from third element, so that array becomes [1, -1, -1]. And 1⋅ (-1) ⋅ (-1) = 1.
What is the minimum cost we will have to pay to do that?
Input
The first line contains a single integer n (1 ≤ n ≤ 10^5) — the number of numbers.
The second line contains n integers a_1, a_2, ..., a_n (-10^9 ≤ a_i ≤ 10^9) — the numbers.
Output
Output a single number — the minimal number of coins you need to pay to make the product equal to 1.
Examples
Input
2
-1 1
Output
2
Input
4
0 0 0 0
Output
4
Input
5
-5 -3 5 3 0
Output
13
Note
In the first example, you can change 1 to -1 or -1 to 1 in 2 coins.
In the second example, you have to apply at least 4 operations for the product not to be 0.
In the third example, you can change -5 to -1 in 4 coins, -3 to -1 in 2 coins, 5 to 1 in 4 coins, 3 to 1 in 2 coins, 0 to 1 in 1 coin.
Tags: dp, implementation
Correct Solution:
```
n = int(input())
l = [*map(int, input().split())]
neg = [-e for e in l if e < 0]
zer = l.count(0)
pos = [e for e in l if e > 0]
res = sum(pos) - len(pos) + zer
if neg: res += sum(neg) - len(neg)
if len(neg) & 1 and zer == 0: res += 2
print(res)
```
| 5,867 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given n numbers a_1, a_2, ..., a_n. With a cost of one coin you can perform the following operation:
Choose one of these numbers and add or subtract 1 from it.
In particular, we can apply this operation to the same number several times.
We want to make the product of all these numbers equal to 1, in other words, we want a_1 ⋅ a_2 ... ⋅ a_n = 1.
For example, for n = 3 and numbers [1, -3, 0] we can make product equal to 1 in 3 coins: add 1 to second element, add 1 to second element again, subtract 1 from third element, so that array becomes [1, -1, -1]. And 1⋅ (-1) ⋅ (-1) = 1.
What is the minimum cost we will have to pay to do that?
Input
The first line contains a single integer n (1 ≤ n ≤ 10^5) — the number of numbers.
The second line contains n integers a_1, a_2, ..., a_n (-10^9 ≤ a_i ≤ 10^9) — the numbers.
Output
Output a single number — the minimal number of coins you need to pay to make the product equal to 1.
Examples
Input
2
-1 1
Output
2
Input
4
0 0 0 0
Output
4
Input
5
-5 -3 5 3 0
Output
13
Note
In the first example, you can change 1 to -1 or -1 to 1 in 2 coins.
In the second example, you have to apply at least 4 operations for the product not to be 0.
In the third example, you can change -5 to -1 in 4 coins, -3 to -1 in 2 coins, 5 to 1 in 4 coins, 3 to 1 in 2 coins, 0 to 1 in 1 coin.
Tags: dp, implementation
Correct Solution:
```
n = int(input())
x = input()
y = x.split()
z = [int(d) for d in y]
no_negatives = []
no_positives = []
no_zeroes = []
count = 0
for i in z:
if i>0:
no_positives.append(i)
elif i<0:
no_negatives.append(i)
else:
no_zeroes.append(i)
if len(no_negatives)%2 != 0:
if len(no_zeroes) != 0:
for i in no_negatives:
count = count + abs(abs(i) -1)
for i in no_zeroes:
count = count + 1
else:
for i in no_negatives:
count = count + abs(abs(i) -1)
for i in no_zeroes:
count = count + 1
count = count +2
elif len(no_negatives) %2 ==0:
for i in no_negatives:
count = count + abs(abs(i)-1)
for i in no_zeroes:
count = count + 1
for i in no_positives:
if i ==0:
pass
else:
count = count + i-1
print(count)
```
| 5,868 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given n numbers a_1, a_2, ..., a_n. With a cost of one coin you can perform the following operation:
Choose one of these numbers and add or subtract 1 from it.
In particular, we can apply this operation to the same number several times.
We want to make the product of all these numbers equal to 1, in other words, we want a_1 ⋅ a_2 ... ⋅ a_n = 1.
For example, for n = 3 and numbers [1, -3, 0] we can make product equal to 1 in 3 coins: add 1 to second element, add 1 to second element again, subtract 1 from third element, so that array becomes [1, -1, -1]. And 1⋅ (-1) ⋅ (-1) = 1.
What is the minimum cost we will have to pay to do that?
Input
The first line contains a single integer n (1 ≤ n ≤ 10^5) — the number of numbers.
The second line contains n integers a_1, a_2, ..., a_n (-10^9 ≤ a_i ≤ 10^9) — the numbers.
Output
Output a single number — the minimal number of coins you need to pay to make the product equal to 1.
Examples
Input
2
-1 1
Output
2
Input
4
0 0 0 0
Output
4
Input
5
-5 -3 5 3 0
Output
13
Note
In the first example, you can change 1 to -1 or -1 to 1 in 2 coins.
In the second example, you have to apply at least 4 operations for the product not to be 0.
In the third example, you can change -5 to -1 in 4 coins, -3 to -1 in 2 coins, 5 to 1 in 4 coins, 3 to 1 in 2 coins, 0 to 1 in 1 coin.
Submitted Solution:
```
n = input()
int_list = map(int, input().split())
count = 0
zeroes = 0
negatives = 0
for num in int_list:
if (num == 0):
zeroes += 1
else:
count += (abs(num) - 1)
if (num < 0):
negatives += 1
if (zeroes == 0 and negatives%2 != 0):
count +=2
else:
count += zeroes
print(count)
```
Yes
| 5,869 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given n numbers a_1, a_2, ..., a_n. With a cost of one coin you can perform the following operation:
Choose one of these numbers and add or subtract 1 from it.
In particular, we can apply this operation to the same number several times.
We want to make the product of all these numbers equal to 1, in other words, we want a_1 ⋅ a_2 ... ⋅ a_n = 1.
For example, for n = 3 and numbers [1, -3, 0] we can make product equal to 1 in 3 coins: add 1 to second element, add 1 to second element again, subtract 1 from third element, so that array becomes [1, -1, -1]. And 1⋅ (-1) ⋅ (-1) = 1.
What is the minimum cost we will have to pay to do that?
Input
The first line contains a single integer n (1 ≤ n ≤ 10^5) — the number of numbers.
The second line contains n integers a_1, a_2, ..., a_n (-10^9 ≤ a_i ≤ 10^9) — the numbers.
Output
Output a single number — the minimal number of coins you need to pay to make the product equal to 1.
Examples
Input
2
-1 1
Output
2
Input
4
0 0 0 0
Output
4
Input
5
-5 -3 5 3 0
Output
13
Note
In the first example, you can change 1 to -1 or -1 to 1 in 2 coins.
In the second example, you have to apply at least 4 operations for the product not to be 0.
In the third example, you can change -5 to -1 in 4 coins, -3 to -1 in 2 coins, 5 to 1 in 4 coins, 3 to 1 in 2 coins, 0 to 1 in 1 coin.
Submitted Solution:
```
n = int(input())
a = list(map(int, input().split()))
pos = []
neg = []
zero = []
coin = 0
for num in a:
if num > 0:
pos.append(num)
elif num==0:
zero.append(num)
else:
neg.append(num)
for num in pos:
coin += num -1
for num in neg:
coin += -1 - num
if len(zero)>0:
coin += len(zero)
elif len(neg)%2==1:
coin += 2
print(coin)
```
Yes
| 5,870 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given n numbers a_1, a_2, ..., a_n. With a cost of one coin you can perform the following operation:
Choose one of these numbers and add or subtract 1 from it.
In particular, we can apply this operation to the same number several times.
We want to make the product of all these numbers equal to 1, in other words, we want a_1 ⋅ a_2 ... ⋅ a_n = 1.
For example, for n = 3 and numbers [1, -3, 0] we can make product equal to 1 in 3 coins: add 1 to second element, add 1 to second element again, subtract 1 from third element, so that array becomes [1, -1, -1]. And 1⋅ (-1) ⋅ (-1) = 1.
What is the minimum cost we will have to pay to do that?
Input
The first line contains a single integer n (1 ≤ n ≤ 10^5) — the number of numbers.
The second line contains n integers a_1, a_2, ..., a_n (-10^9 ≤ a_i ≤ 10^9) — the numbers.
Output
Output a single number — the minimal number of coins you need to pay to make the product equal to 1.
Examples
Input
2
-1 1
Output
2
Input
4
0 0 0 0
Output
4
Input
5
-5 -3 5 3 0
Output
13
Note
In the first example, you can change 1 to -1 or -1 to 1 in 2 coins.
In the second example, you have to apply at least 4 operations for the product not to be 0.
In the third example, you can change -5 to -1 in 4 coins, -3 to -1 in 2 coins, 5 to 1 in 4 coins, 3 to 1 in 2 coins, 0 to 1 in 1 coin.
Submitted Solution:
```
n = int(input())
a = list(map(int, input().split()))
ans = 0
cnt = 0
cnt2 = 0
for i in range(n):
if a[i] < 0:
ans += abs(a[i]) - 1
a[i] = -1
cnt2 += 1
elif a[i] == 0:
cnt += 1
else:
ans += a[i] - 1
a[i] = 1
if cnt2 % 2 == 1 and cnt == 0:
ans += 2
print(ans)
else:
ans += cnt
print(ans)
```
Yes
| 5,871 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given n numbers a_1, a_2, ..., a_n. With a cost of one coin you can perform the following operation:
Choose one of these numbers and add or subtract 1 from it.
In particular, we can apply this operation to the same number several times.
We want to make the product of all these numbers equal to 1, in other words, we want a_1 ⋅ a_2 ... ⋅ a_n = 1.
For example, for n = 3 and numbers [1, -3, 0] we can make product equal to 1 in 3 coins: add 1 to second element, add 1 to second element again, subtract 1 from third element, so that array becomes [1, -1, -1]. And 1⋅ (-1) ⋅ (-1) = 1.
What is the minimum cost we will have to pay to do that?
Input
The first line contains a single integer n (1 ≤ n ≤ 10^5) — the number of numbers.
The second line contains n integers a_1, a_2, ..., a_n (-10^9 ≤ a_i ≤ 10^9) — the numbers.
Output
Output a single number — the minimal number of coins you need to pay to make the product equal to 1.
Examples
Input
2
-1 1
Output
2
Input
4
0 0 0 0
Output
4
Input
5
-5 -3 5 3 0
Output
13
Note
In the first example, you can change 1 to -1 or -1 to 1 in 2 coins.
In the second example, you have to apply at least 4 operations for the product not to be 0.
In the third example, you can change -5 to -1 in 4 coins, -3 to -1 in 2 coins, 5 to 1 in 4 coins, 3 to 1 in 2 coins, 0 to 1 in 1 coin.
Submitted Solution:
```
n = int(input())
a = list(map(int, input().split()))
count = neg = flag = 0
for num in a:
if num > 0:
count += num - 1
elif num == 0:
count += 1
flag += 1
else:
count += -num - 1
neg += 1
if neg % 2 == 0 or flag:
print(count)
else:
print(count + 2)
```
Yes
| 5,872 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given n numbers a_1, a_2, ..., a_n. With a cost of one coin you can perform the following operation:
Choose one of these numbers and add or subtract 1 from it.
In particular, we can apply this operation to the same number several times.
We want to make the product of all these numbers equal to 1, in other words, we want a_1 ⋅ a_2 ... ⋅ a_n = 1.
For example, for n = 3 and numbers [1, -3, 0] we can make product equal to 1 in 3 coins: add 1 to second element, add 1 to second element again, subtract 1 from third element, so that array becomes [1, -1, -1]. And 1⋅ (-1) ⋅ (-1) = 1.
What is the minimum cost we will have to pay to do that?
Input
The first line contains a single integer n (1 ≤ n ≤ 10^5) — the number of numbers.
The second line contains n integers a_1, a_2, ..., a_n (-10^9 ≤ a_i ≤ 10^9) — the numbers.
Output
Output a single number — the minimal number of coins you need to pay to make the product equal to 1.
Examples
Input
2
-1 1
Output
2
Input
4
0 0 0 0
Output
4
Input
5
-5 -3 5 3 0
Output
13
Note
In the first example, you can change 1 to -1 or -1 to 1 in 2 coins.
In the second example, you have to apply at least 4 operations for the product not to be 0.
In the third example, you can change -5 to -1 in 4 coins, -3 to -1 in 2 coins, 5 to 1 in 4 coins, 3 to 1 in 2 coins, 0 to 1 in 1 coin.
Submitted Solution:
```
negativecounter=0
noo=0
steps=0
n=int(input())
arr=list(map(int,input().split()))
for i in arr:
if(i<0):
negativecounter+=1
if(i==0):
noo+=1
arr.sort()
if(negativecounter%2==0):
for i in range(n):
if(arr[i]>1):
steps+=arr[i]-1
elif(arr[i]==0):
steps+=1
elif(arr[i]<-1):
steps+=-1*(arr[i]+1)
else:
steps+=0
else:
steps+=abs(arr[0])+1
for i in range(1,n):
if(arr[i]>1):
steps+=arr[i]-1
elif(noo%2==1 and arr[i]==0):
steps-=1
elif(arr[i]==0):
steps+=1
elif(arr[i]<-1):
steps+=-1*(arr[i]+1)
else:
steps+=0
print(steps)
```
No
| 5,873 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given n numbers a_1, a_2, ..., a_n. With a cost of one coin you can perform the following operation:
Choose one of these numbers and add or subtract 1 from it.
In particular, we can apply this operation to the same number several times.
We want to make the product of all these numbers equal to 1, in other words, we want a_1 ⋅ a_2 ... ⋅ a_n = 1.
For example, for n = 3 and numbers [1, -3, 0] we can make product equal to 1 in 3 coins: add 1 to second element, add 1 to second element again, subtract 1 from third element, so that array becomes [1, -1, -1]. And 1⋅ (-1) ⋅ (-1) = 1.
What is the minimum cost we will have to pay to do that?
Input
The first line contains a single integer n (1 ≤ n ≤ 10^5) — the number of numbers.
The second line contains n integers a_1, a_2, ..., a_n (-10^9 ≤ a_i ≤ 10^9) — the numbers.
Output
Output a single number — the minimal number of coins you need to pay to make the product equal to 1.
Examples
Input
2
-1 1
Output
2
Input
4
0 0 0 0
Output
4
Input
5
-5 -3 5 3 0
Output
13
Note
In the first example, you can change 1 to -1 or -1 to 1 in 2 coins.
In the second example, you have to apply at least 4 operations for the product not to be 0.
In the third example, you can change -5 to -1 in 4 coins, -3 to -1 in 2 coins, 5 to 1 in 4 coins, 3 to 1 in 2 coins, 0 to 1 in 1 coin.
Submitted Solution:
```
n=int(input())
a=list(map(int,input().split()))
a.sort()
nv,count=0,0
for i in range(n):
if a[i]<0 and nv==0 and i!=n-1:
nv+=1
count+=abs(a[i]-(-1))
elif a[i]<0 and nv==1 and i!=n-1:
nv=0
count+=abs(a[i]-(-1))
elif a[i]<0 and nv==0 and i==n-1:
count+=abs(a[i])+2
elif a[i]>=0 and nv==1:
nv=0
count+=abs(a[i]+1)
else:
count+=abs(a[i]-1)
print(count)
```
No
| 5,874 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given n numbers a_1, a_2, ..., a_n. With a cost of one coin you can perform the following operation:
Choose one of these numbers and add or subtract 1 from it.
In particular, we can apply this operation to the same number several times.
We want to make the product of all these numbers equal to 1, in other words, we want a_1 ⋅ a_2 ... ⋅ a_n = 1.
For example, for n = 3 and numbers [1, -3, 0] we can make product equal to 1 in 3 coins: add 1 to second element, add 1 to second element again, subtract 1 from third element, so that array becomes [1, -1, -1]. And 1⋅ (-1) ⋅ (-1) = 1.
What is the minimum cost we will have to pay to do that?
Input
The first line contains a single integer n (1 ≤ n ≤ 10^5) — the number of numbers.
The second line contains n integers a_1, a_2, ..., a_n (-10^9 ≤ a_i ≤ 10^9) — the numbers.
Output
Output a single number — the minimal number of coins you need to pay to make the product equal to 1.
Examples
Input
2
-1 1
Output
2
Input
4
0 0 0 0
Output
4
Input
5
-5 -3 5 3 0
Output
13
Note
In the first example, you can change 1 to -1 or -1 to 1 in 2 coins.
In the second example, you have to apply at least 4 operations for the product not to be 0.
In the third example, you can change -5 to -1 in 4 coins, -3 to -1 in 2 coins, 5 to 1 in 4 coins, 3 to 1 in 2 coins, 0 to 1 in 1 coin.
Submitted Solution:
```
def main():
n=int(input())
ip=[int(item) for item in input().split(" ")]
n_coins=0
neg_cnt=0
for i in range(len(ip)):
if ip[i]==0:
n_coins+=1
elif ip[i]>0:
n_coins+=ip[i]-1
elif ip[i]<0:
n_coins+=-1-ip[i]
neg_cnt+=1
if neg_cnt%2==1:
n_coins+=2
print(n_coins)
if __name__=="__main__":
main()
```
No
| 5,875 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given n numbers a_1, a_2, ..., a_n. With a cost of one coin you can perform the following operation:
Choose one of these numbers and add or subtract 1 from it.
In particular, we can apply this operation to the same number several times.
We want to make the product of all these numbers equal to 1, in other words, we want a_1 ⋅ a_2 ... ⋅ a_n = 1.
For example, for n = 3 and numbers [1, -3, 0] we can make product equal to 1 in 3 coins: add 1 to second element, add 1 to second element again, subtract 1 from third element, so that array becomes [1, -1, -1]. And 1⋅ (-1) ⋅ (-1) = 1.
What is the minimum cost we will have to pay to do that?
Input
The first line contains a single integer n (1 ≤ n ≤ 10^5) — the number of numbers.
The second line contains n integers a_1, a_2, ..., a_n (-10^9 ≤ a_i ≤ 10^9) — the numbers.
Output
Output a single number — the minimal number of coins you need to pay to make the product equal to 1.
Examples
Input
2
-1 1
Output
2
Input
4
0 0 0 0
Output
4
Input
5
-5 -3 5 3 0
Output
13
Note
In the first example, you can change 1 to -1 or -1 to 1 in 2 coins.
In the second example, you have to apply at least 4 operations for the product not to be 0.
In the third example, you can change -5 to -1 in 4 coins, -3 to -1 in 2 coins, 5 to 1 in 4 coins, 3 to 1 in 2 coins, 0 to 1 in 1 coin.
Submitted Solution:
```
n=int(input())
a=[int(i) for i in input().split()]
d={}
g=0
for i in a:
if i in d:
d[i]+=1
else:
d[i]=1
i=0
while(i<n):
if a[i]>1:
temp=a[i]
a[i]=1
g+=temp-1
if a[i]<-1:
temp=a[i]
a[i]=-1
g+=-1-temp
if a[i]==0:
if a.count(-1)%2!=0 :
a[i]=-1
g+=1
else:
a[i]=1
g+=1
if i==n-1 and a.count(-1)%2!=0:
g+=2
i+=1
print(g)
```
No
| 5,876 |
Provide tags and a correct Python 3 solution for this coding contest problem.
The only difference between easy and hard versions is constraints.
There are n kids, each of them is reading a unique book. At the end of any day, the i-th kid will give his book to the p_i-th kid (in case of i = p_i the kid will give his book to himself). It is guaranteed that all values of p_i are distinct integers from 1 to n (i.e. p is a permutation). The sequence p doesn't change from day to day, it is fixed.
For example, if n=6 and p=[4, 6, 1, 3, 5, 2] then at the end of the first day the book of the 1-st kid will belong to the 4-th kid, the 2-nd kid will belong to the 6-th kid and so on. At the end of the second day the book of the 1-st kid will belong to the 3-th kid, the 2-nd kid will belong to the 2-th kid and so on.
Your task is to determine the number of the day the book of the i-th child is returned back to him for the first time for every i from 1 to n.
Consider the following example: p = [5, 1, 2, 4, 3]. The book of the 1-st kid will be passed to the following kids:
* after the 1-st day it will belong to the 5-th kid,
* after the 2-nd day it will belong to the 3-rd kid,
* after the 3-rd day it will belong to the 2-nd kid,
* after the 4-th day it will belong to the 1-st kid.
So after the fourth day, the book of the first kid will return to its owner. The book of the fourth kid will return to him for the first time after exactly one day.
You have to answer q independent queries.
Input
The first line of the input contains one integer q (1 ≤ q ≤ 200) — the number of queries. Then q queries follow.
The first line of the query contains one integer n (1 ≤ n ≤ 200) — the number of kids in the query. The second line of the query contains n integers p_1, p_2, ..., p_n (1 ≤ p_i ≤ n, all p_i are distinct, i.e. p is a permutation), where p_i is the kid which will get the book of the i-th kid.
Output
For each query, print the answer on it: n integers a_1, a_2, ..., a_n, where a_i is the number of the day the book of the i-th child is returned back to him for the first time in this query.
Example
Input
6
5
1 2 3 4 5
3
2 3 1
6
4 6 2 1 5 3
1
1
4
3 4 1 2
5
5 1 2 4 3
Output
1 1 1 1 1
3 3 3
2 3 3 2 1 3
1
2 2 2 2
4 4 4 1 4
Tags: dsu, math
Correct Solution:
```
# Books Exchange (easy version) (595.3)
t=int(input())
for z in range(t):
n=int(input())
a=list(map(int,input().split()))
c=[]
for i in range(1,n+1):
count=0
x=i
j=i-1
while True:
k=a[j]
if k==x:
count+=1
break
else:
j=k-1
count+=1
c.append(count)
print(*c)
```
| 5,877 |
Provide tags and a correct Python 3 solution for this coding contest problem.
The only difference between easy and hard versions is constraints.
There are n kids, each of them is reading a unique book. At the end of any day, the i-th kid will give his book to the p_i-th kid (in case of i = p_i the kid will give his book to himself). It is guaranteed that all values of p_i are distinct integers from 1 to n (i.e. p is a permutation). The sequence p doesn't change from day to day, it is fixed.
For example, if n=6 and p=[4, 6, 1, 3, 5, 2] then at the end of the first day the book of the 1-st kid will belong to the 4-th kid, the 2-nd kid will belong to the 6-th kid and so on. At the end of the second day the book of the 1-st kid will belong to the 3-th kid, the 2-nd kid will belong to the 2-th kid and so on.
Your task is to determine the number of the day the book of the i-th child is returned back to him for the first time for every i from 1 to n.
Consider the following example: p = [5, 1, 2, 4, 3]. The book of the 1-st kid will be passed to the following kids:
* after the 1-st day it will belong to the 5-th kid,
* after the 2-nd day it will belong to the 3-rd kid,
* after the 3-rd day it will belong to the 2-nd kid,
* after the 4-th day it will belong to the 1-st kid.
So after the fourth day, the book of the first kid will return to its owner. The book of the fourth kid will return to him for the first time after exactly one day.
You have to answer q independent queries.
Input
The first line of the input contains one integer q (1 ≤ q ≤ 200) — the number of queries. Then q queries follow.
The first line of the query contains one integer n (1 ≤ n ≤ 200) — the number of kids in the query. The second line of the query contains n integers p_1, p_2, ..., p_n (1 ≤ p_i ≤ n, all p_i are distinct, i.e. p is a permutation), where p_i is the kid which will get the book of the i-th kid.
Output
For each query, print the answer on it: n integers a_1, a_2, ..., a_n, where a_i is the number of the day the book of the i-th child is returned back to him for the first time in this query.
Example
Input
6
5
1 2 3 4 5
3
2 3 1
6
4 6 2 1 5 3
1
1
4
3 4 1 2
5
5 1 2 4 3
Output
1 1 1 1 1
3 3 3
2 3 3 2 1 3
1
2 2 2 2
4 4 4 1 4
Tags: dsu, math
Correct Solution:
```
p = []
f = []
def trace(start_index,next_index,depth):
# print("TRACE",start_index,next_index,depth)
if start_index == next_index:
f[start_index] = depth
return depth
total_depth = trace(start_index,p[next_index]-1,depth+1)
f[next_index] = total_depth
return total_depth
q = int(input())
for i in range(q):
n = int(input())
line = input()
p = line.split(" ")
p = [int(x) for x in p]
f = [0]*len(p)
for index in range(len(p)):
if f[index] == 0:
f[index] = (trace(index,p[index]-1,1))
s = " ".join([str(x) for x in f])
print(s)
```
| 5,878 |
Provide tags and a correct Python 3 solution for this coding contest problem.
The only difference between easy and hard versions is constraints.
There are n kids, each of them is reading a unique book. At the end of any day, the i-th kid will give his book to the p_i-th kid (in case of i = p_i the kid will give his book to himself). It is guaranteed that all values of p_i are distinct integers from 1 to n (i.e. p is a permutation). The sequence p doesn't change from day to day, it is fixed.
For example, if n=6 and p=[4, 6, 1, 3, 5, 2] then at the end of the first day the book of the 1-st kid will belong to the 4-th kid, the 2-nd kid will belong to the 6-th kid and so on. At the end of the second day the book of the 1-st kid will belong to the 3-th kid, the 2-nd kid will belong to the 2-th kid and so on.
Your task is to determine the number of the day the book of the i-th child is returned back to him for the first time for every i from 1 to n.
Consider the following example: p = [5, 1, 2, 4, 3]. The book of the 1-st kid will be passed to the following kids:
* after the 1-st day it will belong to the 5-th kid,
* after the 2-nd day it will belong to the 3-rd kid,
* after the 3-rd day it will belong to the 2-nd kid,
* after the 4-th day it will belong to the 1-st kid.
So after the fourth day, the book of the first kid will return to its owner. The book of the fourth kid will return to him for the first time after exactly one day.
You have to answer q independent queries.
Input
The first line of the input contains one integer q (1 ≤ q ≤ 200) — the number of queries. Then q queries follow.
The first line of the query contains one integer n (1 ≤ n ≤ 200) — the number of kids in the query. The second line of the query contains n integers p_1, p_2, ..., p_n (1 ≤ p_i ≤ n, all p_i are distinct, i.e. p is a permutation), where p_i is the kid which will get the book of the i-th kid.
Output
For each query, print the answer on it: n integers a_1, a_2, ..., a_n, where a_i is the number of the day the book of the i-th child is returned back to him for the first time in this query.
Example
Input
6
5
1 2 3 4 5
3
2 3 1
6
4 6 2 1 5 3
1
1
4
3 4 1 2
5
5 1 2 4 3
Output
1 1 1 1 1
3 3 3
2 3 3 2 1 3
1
2 2 2 2
4 4 4 1 4
Tags: dsu, math
Correct Solution:
```
cases = int(input())
for _ in range(cases):
n = int(input())
trans = [int(i)-1 for i in input().split(' ')]
# print(trans)
for kid in range(n):
curr = trans[kid]
count = 1
while curr != kid:
# print(curr)
curr = trans[curr]
count += 1
print(count, end=' ')
# print()
# break
print()
```
| 5,879 |
Provide tags and a correct Python 3 solution for this coding contest problem.
The only difference between easy and hard versions is constraints.
There are n kids, each of them is reading a unique book. At the end of any day, the i-th kid will give his book to the p_i-th kid (in case of i = p_i the kid will give his book to himself). It is guaranteed that all values of p_i are distinct integers from 1 to n (i.e. p is a permutation). The sequence p doesn't change from day to day, it is fixed.
For example, if n=6 and p=[4, 6, 1, 3, 5, 2] then at the end of the first day the book of the 1-st kid will belong to the 4-th kid, the 2-nd kid will belong to the 6-th kid and so on. At the end of the second day the book of the 1-st kid will belong to the 3-th kid, the 2-nd kid will belong to the 2-th kid and so on.
Your task is to determine the number of the day the book of the i-th child is returned back to him for the first time for every i from 1 to n.
Consider the following example: p = [5, 1, 2, 4, 3]. The book of the 1-st kid will be passed to the following kids:
* after the 1-st day it will belong to the 5-th kid,
* after the 2-nd day it will belong to the 3-rd kid,
* after the 3-rd day it will belong to the 2-nd kid,
* after the 4-th day it will belong to the 1-st kid.
So after the fourth day, the book of the first kid will return to its owner. The book of the fourth kid will return to him for the first time after exactly one day.
You have to answer q independent queries.
Input
The first line of the input contains one integer q (1 ≤ q ≤ 200) — the number of queries. Then q queries follow.
The first line of the query contains one integer n (1 ≤ n ≤ 200) — the number of kids in the query. The second line of the query contains n integers p_1, p_2, ..., p_n (1 ≤ p_i ≤ n, all p_i are distinct, i.e. p is a permutation), where p_i is the kid which will get the book of the i-th kid.
Output
For each query, print the answer on it: n integers a_1, a_2, ..., a_n, where a_i is the number of the day the book of the i-th child is returned back to him for the first time in this query.
Example
Input
6
5
1 2 3 4 5
3
2 3 1
6
4 6 2 1 5 3
1
1
4
3 4 1 2
5
5 1 2 4 3
Output
1 1 1 1 1
3 3 3
2 3 3 2 1 3
1
2 2 2 2
4 4 4 1 4
Tags: dsu, math
Correct Solution:
```
queries=int(input())
for x in range(queries):
kids=int(input())
p=input()
p=p.split()
for i in p:
pos=int(p[int(i)-1])
ans=1
while int(pos) != int(i):
ans+=1
pos=p[int(pos)-1]
print(ans,end=' ')
print('\n')
```
| 5,880 |
Provide tags and a correct Python 3 solution for this coding contest problem.
The only difference between easy and hard versions is constraints.
There are n kids, each of them is reading a unique book. At the end of any day, the i-th kid will give his book to the p_i-th kid (in case of i = p_i the kid will give his book to himself). It is guaranteed that all values of p_i are distinct integers from 1 to n (i.e. p is a permutation). The sequence p doesn't change from day to day, it is fixed.
For example, if n=6 and p=[4, 6, 1, 3, 5, 2] then at the end of the first day the book of the 1-st kid will belong to the 4-th kid, the 2-nd kid will belong to the 6-th kid and so on. At the end of the second day the book of the 1-st kid will belong to the 3-th kid, the 2-nd kid will belong to the 2-th kid and so on.
Your task is to determine the number of the day the book of the i-th child is returned back to him for the first time for every i from 1 to n.
Consider the following example: p = [5, 1, 2, 4, 3]. The book of the 1-st kid will be passed to the following kids:
* after the 1-st day it will belong to the 5-th kid,
* after the 2-nd day it will belong to the 3-rd kid,
* after the 3-rd day it will belong to the 2-nd kid,
* after the 4-th day it will belong to the 1-st kid.
So after the fourth day, the book of the first kid will return to its owner. The book of the fourth kid will return to him for the first time after exactly one day.
You have to answer q independent queries.
Input
The first line of the input contains one integer q (1 ≤ q ≤ 200) — the number of queries. Then q queries follow.
The first line of the query contains one integer n (1 ≤ n ≤ 200) — the number of kids in the query. The second line of the query contains n integers p_1, p_2, ..., p_n (1 ≤ p_i ≤ n, all p_i are distinct, i.e. p is a permutation), where p_i is the kid which will get the book of the i-th kid.
Output
For each query, print the answer on it: n integers a_1, a_2, ..., a_n, where a_i is the number of the day the book of the i-th child is returned back to him for the first time in this query.
Example
Input
6
5
1 2 3 4 5
3
2 3 1
6
4 6 2 1 5 3
1
1
4
3 4 1 2
5
5 1 2 4 3
Output
1 1 1 1 1
3 3 3
2 3 3 2 1 3
1
2 2 2 2
4 4 4 1 4
Tags: dsu, math
Correct Solution:
```
import sys
import math
import itertools
import collections
def getdict(n):
d = {}
if type(n) is list or type(n) is str:
for i in n:
if i in d:
d[i] += 1
else:
d[i] = 1
else:
for i in range(n):
t = ii()
if t in d:
d[t] += 1
else:
d[t] = 1
return d
def divs(n, start=1):
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 cdiv(n, k): return n // k + (n % k != 0)
def ii(): return int(input())
def mi(): return map(int, input().split())
def li(): return list(map(int, input().split()))
def lcm(a, b): return abs(a*b) // math.gcd(a, b)
def wr(arr): return ' '.join(map(str, arr))
def revn(n): return int(str(n)[::-1])
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=3):
newnumber = ''
while number > 0:
newnumber = str(number % base) + newnumber
number //= base
return newnumber
q = ii()
for _ in range(q):
n = ii()
p = li()
ans = [1] * n
s = set()
for i in range(n):
if p[i] in s:
continue
else:
ss = set()
pos = p[i]
t = 1
while pos != i + 1:
ss.add(pos)
s.add(pos)
pos = p[pos - 1]
t += 1
for el in ss:
ans[el - 1] = t
print(wr(ans))
```
| 5,881 |
Provide tags and a correct Python 3 solution for this coding contest problem.
The only difference between easy and hard versions is constraints.
There are n kids, each of them is reading a unique book. At the end of any day, the i-th kid will give his book to the p_i-th kid (in case of i = p_i the kid will give his book to himself). It is guaranteed that all values of p_i are distinct integers from 1 to n (i.e. p is a permutation). The sequence p doesn't change from day to day, it is fixed.
For example, if n=6 and p=[4, 6, 1, 3, 5, 2] then at the end of the first day the book of the 1-st kid will belong to the 4-th kid, the 2-nd kid will belong to the 6-th kid and so on. At the end of the second day the book of the 1-st kid will belong to the 3-th kid, the 2-nd kid will belong to the 2-th kid and so on.
Your task is to determine the number of the day the book of the i-th child is returned back to him for the first time for every i from 1 to n.
Consider the following example: p = [5, 1, 2, 4, 3]. The book of the 1-st kid will be passed to the following kids:
* after the 1-st day it will belong to the 5-th kid,
* after the 2-nd day it will belong to the 3-rd kid,
* after the 3-rd day it will belong to the 2-nd kid,
* after the 4-th day it will belong to the 1-st kid.
So after the fourth day, the book of the first kid will return to its owner. The book of the fourth kid will return to him for the first time after exactly one day.
You have to answer q independent queries.
Input
The first line of the input contains one integer q (1 ≤ q ≤ 200) — the number of queries. Then q queries follow.
The first line of the query contains one integer n (1 ≤ n ≤ 200) — the number of kids in the query. The second line of the query contains n integers p_1, p_2, ..., p_n (1 ≤ p_i ≤ n, all p_i are distinct, i.e. p is a permutation), where p_i is the kid which will get the book of the i-th kid.
Output
For each query, print the answer on it: n integers a_1, a_2, ..., a_n, where a_i is the number of the day the book of the i-th child is returned back to him for the first time in this query.
Example
Input
6
5
1 2 3 4 5
3
2 3 1
6
4 6 2 1 5 3
1
1
4
3 4 1 2
5
5 1 2 4 3
Output
1 1 1 1 1
3 3 3
2 3 3 2 1 3
1
2 2 2 2
4 4 4 1 4
Tags: dsu, math
Correct Solution:
```
t = int(input())
for _ in range(t):
n = int(input())
a = list(map(int,input().split()))
out = [0]*n
seen = set()
for x in a:
if x in seen:
continue
current = x
stack = set()
while current not in seen and current not in stack:
stack.add(current)
current = a[current-1]
for x in stack:
seen.add(x)
out[x-1]=len(stack)
print (*out)
```
| 5,882 |
Provide tags and a correct Python 3 solution for this coding contest problem.
The only difference between easy and hard versions is constraints.
There are n kids, each of them is reading a unique book. At the end of any day, the i-th kid will give his book to the p_i-th kid (in case of i = p_i the kid will give his book to himself). It is guaranteed that all values of p_i are distinct integers from 1 to n (i.e. p is a permutation). The sequence p doesn't change from day to day, it is fixed.
For example, if n=6 and p=[4, 6, 1, 3, 5, 2] then at the end of the first day the book of the 1-st kid will belong to the 4-th kid, the 2-nd kid will belong to the 6-th kid and so on. At the end of the second day the book of the 1-st kid will belong to the 3-th kid, the 2-nd kid will belong to the 2-th kid and so on.
Your task is to determine the number of the day the book of the i-th child is returned back to him for the first time for every i from 1 to n.
Consider the following example: p = [5, 1, 2, 4, 3]. The book of the 1-st kid will be passed to the following kids:
* after the 1-st day it will belong to the 5-th kid,
* after the 2-nd day it will belong to the 3-rd kid,
* after the 3-rd day it will belong to the 2-nd kid,
* after the 4-th day it will belong to the 1-st kid.
So after the fourth day, the book of the first kid will return to its owner. The book of the fourth kid will return to him for the first time after exactly one day.
You have to answer q independent queries.
Input
The first line of the input contains one integer q (1 ≤ q ≤ 200) — the number of queries. Then q queries follow.
The first line of the query contains one integer n (1 ≤ n ≤ 200) — the number of kids in the query. The second line of the query contains n integers p_1, p_2, ..., p_n (1 ≤ p_i ≤ n, all p_i are distinct, i.e. p is a permutation), where p_i is the kid which will get the book of the i-th kid.
Output
For each query, print the answer on it: n integers a_1, a_2, ..., a_n, where a_i is the number of the day the book of the i-th child is returned back to him for the first time in this query.
Example
Input
6
5
1 2 3 4 5
3
2 3 1
6
4 6 2 1 5 3
1
1
4
3 4 1 2
5
5 1 2 4 3
Output
1 1 1 1 1
3 3 3
2 3 3 2 1 3
1
2 2 2 2
4 4 4 1 4
Tags: dsu, math
Correct Solution:
```
def r():
return map(int,input().split())
t = int(input())
for _ in range(t):
n = int(input())
l = list(r())
m = {}
for i in range(1, n + 1):
m[i] = l[i - 1]
tab = [0] * n
for i in range(1, n + 1):
c = 1
f = m[i]
if tab[i - 1] == 0:
cp = [f]
while f != i:
c += 1
f = m[f]
cp.append(f)
for v in cp:
tab[v - 1] = c
print(*tab)
```
| 5,883 |
Provide tags and a correct Python 3 solution for this coding contest problem.
The only difference between easy and hard versions is constraints.
There are n kids, each of them is reading a unique book. At the end of any day, the i-th kid will give his book to the p_i-th kid (in case of i = p_i the kid will give his book to himself). It is guaranteed that all values of p_i are distinct integers from 1 to n (i.e. p is a permutation). The sequence p doesn't change from day to day, it is fixed.
For example, if n=6 and p=[4, 6, 1, 3, 5, 2] then at the end of the first day the book of the 1-st kid will belong to the 4-th kid, the 2-nd kid will belong to the 6-th kid and so on. At the end of the second day the book of the 1-st kid will belong to the 3-th kid, the 2-nd kid will belong to the 2-th kid and so on.
Your task is to determine the number of the day the book of the i-th child is returned back to him for the first time for every i from 1 to n.
Consider the following example: p = [5, 1, 2, 4, 3]. The book of the 1-st kid will be passed to the following kids:
* after the 1-st day it will belong to the 5-th kid,
* after the 2-nd day it will belong to the 3-rd kid,
* after the 3-rd day it will belong to the 2-nd kid,
* after the 4-th day it will belong to the 1-st kid.
So after the fourth day, the book of the first kid will return to its owner. The book of the fourth kid will return to him for the first time after exactly one day.
You have to answer q independent queries.
Input
The first line of the input contains one integer q (1 ≤ q ≤ 200) — the number of queries. Then q queries follow.
The first line of the query contains one integer n (1 ≤ n ≤ 200) — the number of kids in the query. The second line of the query contains n integers p_1, p_2, ..., p_n (1 ≤ p_i ≤ n, all p_i are distinct, i.e. p is a permutation), where p_i is the kid which will get the book of the i-th kid.
Output
For each query, print the answer on it: n integers a_1, a_2, ..., a_n, where a_i is the number of the day the book of the i-th child is returned back to him for the first time in this query.
Example
Input
6
5
1 2 3 4 5
3
2 3 1
6
4 6 2 1 5 3
1
1
4
3 4 1 2
5
5 1 2 4 3
Output
1 1 1 1 1
3 3 3
2 3 3 2 1 3
1
2 2 2 2
4 4 4 1 4
Tags: dsu, math
Correct Solution:
```
'''
Author : thekushalghosh
Team : CodeDiggers
'''
import sys,math
input = sys.stdin.readline
for _ in range(int(input())):
n = int(input())
a = list(map(int,input().split()))
w = [1] * len(a)
for i in range(len(a)):
if w[i] == 1:
j = i
c = 1
qw = []
while True:
q = a[j] - 1
if q == i:
qw.append(q)
break
else:
j = q
c = c + 1
qw.append(j)
for j in range(len(qw)):
w[qw[j]] = c
print(*w)
```
| 5,884 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
The only difference between easy and hard versions is constraints.
There are n kids, each of them is reading a unique book. At the end of any day, the i-th kid will give his book to the p_i-th kid (in case of i = p_i the kid will give his book to himself). It is guaranteed that all values of p_i are distinct integers from 1 to n (i.e. p is a permutation). The sequence p doesn't change from day to day, it is fixed.
For example, if n=6 and p=[4, 6, 1, 3, 5, 2] then at the end of the first day the book of the 1-st kid will belong to the 4-th kid, the 2-nd kid will belong to the 6-th kid and so on. At the end of the second day the book of the 1-st kid will belong to the 3-th kid, the 2-nd kid will belong to the 2-th kid and so on.
Your task is to determine the number of the day the book of the i-th child is returned back to him for the first time for every i from 1 to n.
Consider the following example: p = [5, 1, 2, 4, 3]. The book of the 1-st kid will be passed to the following kids:
* after the 1-st day it will belong to the 5-th kid,
* after the 2-nd day it will belong to the 3-rd kid,
* after the 3-rd day it will belong to the 2-nd kid,
* after the 4-th day it will belong to the 1-st kid.
So after the fourth day, the book of the first kid will return to its owner. The book of the fourth kid will return to him for the first time after exactly one day.
You have to answer q independent queries.
Input
The first line of the input contains one integer q (1 ≤ q ≤ 200) — the number of queries. Then q queries follow.
The first line of the query contains one integer n (1 ≤ n ≤ 200) — the number of kids in the query. The second line of the query contains n integers p_1, p_2, ..., p_n (1 ≤ p_i ≤ n, all p_i are distinct, i.e. p is a permutation), where p_i is the kid which will get the book of the i-th kid.
Output
For each query, print the answer on it: n integers a_1, a_2, ..., a_n, where a_i is the number of the day the book of the i-th child is returned back to him for the first time in this query.
Example
Input
6
5
1 2 3 4 5
3
2 3 1
6
4 6 2 1 5 3
1
1
4
3 4 1 2
5
5 1 2 4 3
Output
1 1 1 1 1
3 3 3
2 3 3 2 1 3
1
2 2 2 2
4 4 4 1 4
Submitted Solution:
```
for _ in range(int(input())):
n = int(input())
a = list(map(lambda a:int(a)-1, input().split()))
ans = [0 for _ in range(n)]
for i in range(n):
if not ans[i]:
cycle = [i]
x = i
while a[x] != i:
x = a[x]
cycle.append(x)
for x in cycle:
ans[x] = len(cycle)
print(*ans)
```
Yes
| 5,885 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
The only difference between easy and hard versions is constraints.
There are n kids, each of them is reading a unique book. At the end of any day, the i-th kid will give his book to the p_i-th kid (in case of i = p_i the kid will give his book to himself). It is guaranteed that all values of p_i are distinct integers from 1 to n (i.e. p is a permutation). The sequence p doesn't change from day to day, it is fixed.
For example, if n=6 and p=[4, 6, 1, 3, 5, 2] then at the end of the first day the book of the 1-st kid will belong to the 4-th kid, the 2-nd kid will belong to the 6-th kid and so on. At the end of the second day the book of the 1-st kid will belong to the 3-th kid, the 2-nd kid will belong to the 2-th kid and so on.
Your task is to determine the number of the day the book of the i-th child is returned back to him for the first time for every i from 1 to n.
Consider the following example: p = [5, 1, 2, 4, 3]. The book of the 1-st kid will be passed to the following kids:
* after the 1-st day it will belong to the 5-th kid,
* after the 2-nd day it will belong to the 3-rd kid,
* after the 3-rd day it will belong to the 2-nd kid,
* after the 4-th day it will belong to the 1-st kid.
So after the fourth day, the book of the first kid will return to its owner. The book of the fourth kid will return to him for the first time after exactly one day.
You have to answer q independent queries.
Input
The first line of the input contains one integer q (1 ≤ q ≤ 200) — the number of queries. Then q queries follow.
The first line of the query contains one integer n (1 ≤ n ≤ 200) — the number of kids in the query. The second line of the query contains n integers p_1, p_2, ..., p_n (1 ≤ p_i ≤ n, all p_i are distinct, i.e. p is a permutation), where p_i is the kid which will get the book of the i-th kid.
Output
For each query, print the answer on it: n integers a_1, a_2, ..., a_n, where a_i is the number of the day the book of the i-th child is returned back to him for the first time in this query.
Example
Input
6
5
1 2 3 4 5
3
2 3 1
6
4 6 2 1 5 3
1
1
4
3 4 1 2
5
5 1 2 4 3
Output
1 1 1 1 1
3 3 3
2 3 3 2 1 3
1
2 2 2 2
4 4 4 1 4
Submitted Solution:
```
t = int(input())
for _ in range(t):
n = int(input())
a = [None] + list(map(int, input().split()))
answer = [None] + [None]*n
for e in range(1, n+1):
if answer[e] is None:
# print(e, 'not solved')
curchain = list()
if a[e] == e:
answer[e] = 1
continue
i = e
while 1:
curchain.append(i)
i = a[i]
if i == e:
break
lc = len(curchain)
for el in curchain:
answer[el] = lc
print(' '.join(map(str, answer[1:])))
```
Yes
| 5,886 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
The only difference between easy and hard versions is constraints.
There are n kids, each of them is reading a unique book. At the end of any day, the i-th kid will give his book to the p_i-th kid (in case of i = p_i the kid will give his book to himself). It is guaranteed that all values of p_i are distinct integers from 1 to n (i.e. p is a permutation). The sequence p doesn't change from day to day, it is fixed.
For example, if n=6 and p=[4, 6, 1, 3, 5, 2] then at the end of the first day the book of the 1-st kid will belong to the 4-th kid, the 2-nd kid will belong to the 6-th kid and so on. At the end of the second day the book of the 1-st kid will belong to the 3-th kid, the 2-nd kid will belong to the 2-th kid and so on.
Your task is to determine the number of the day the book of the i-th child is returned back to him for the first time for every i from 1 to n.
Consider the following example: p = [5, 1, 2, 4, 3]. The book of the 1-st kid will be passed to the following kids:
* after the 1-st day it will belong to the 5-th kid,
* after the 2-nd day it will belong to the 3-rd kid,
* after the 3-rd day it will belong to the 2-nd kid,
* after the 4-th day it will belong to the 1-st kid.
So after the fourth day, the book of the first kid will return to its owner. The book of the fourth kid will return to him for the first time after exactly one day.
You have to answer q independent queries.
Input
The first line of the input contains one integer q (1 ≤ q ≤ 200) — the number of queries. Then q queries follow.
The first line of the query contains one integer n (1 ≤ n ≤ 200) — the number of kids in the query. The second line of the query contains n integers p_1, p_2, ..., p_n (1 ≤ p_i ≤ n, all p_i are distinct, i.e. p is a permutation), where p_i is the kid which will get the book of the i-th kid.
Output
For each query, print the answer on it: n integers a_1, a_2, ..., a_n, where a_i is the number of the day the book of the i-th child is returned back to him for the first time in this query.
Example
Input
6
5
1 2 3 4 5
3
2 3 1
6
4 6 2 1 5 3
1
1
4
3 4 1 2
5
5 1 2 4 3
Output
1 1 1 1 1
3 3 3
2 3 3 2 1 3
1
2 2 2 2
4 4 4 1 4
Submitted Solution:
```
for case in range (int(input())) :
n = int(input())
p = list(input().split())
for i in range (n) :
p[i] = int(p[i])
p.insert(0, 0)
use = [ False for i in range (n + 1)]
c = [1 for i in range (n + 1)]
for i in range (1 , n + 1) :
q = p[i]
tmp = []
if (not use[q]) :
while (q != i) :
use[q] = True
tmp.append(q)
q = p[q]
for i in tmp :
c[i] = len(tmp) + 1
ans = ''
for i in range (1 , n + 1) :
ans += str(c[i]) + ' '
print (ans)
```
Yes
| 5,887 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
The only difference between easy and hard versions is constraints.
There are n kids, each of them is reading a unique book. At the end of any day, the i-th kid will give his book to the p_i-th kid (in case of i = p_i the kid will give his book to himself). It is guaranteed that all values of p_i are distinct integers from 1 to n (i.e. p is a permutation). The sequence p doesn't change from day to day, it is fixed.
For example, if n=6 and p=[4, 6, 1, 3, 5, 2] then at the end of the first day the book of the 1-st kid will belong to the 4-th kid, the 2-nd kid will belong to the 6-th kid and so on. At the end of the second day the book of the 1-st kid will belong to the 3-th kid, the 2-nd kid will belong to the 2-th kid and so on.
Your task is to determine the number of the day the book of the i-th child is returned back to him for the first time for every i from 1 to n.
Consider the following example: p = [5, 1, 2, 4, 3]. The book of the 1-st kid will be passed to the following kids:
* after the 1-st day it will belong to the 5-th kid,
* after the 2-nd day it will belong to the 3-rd kid,
* after the 3-rd day it will belong to the 2-nd kid,
* after the 4-th day it will belong to the 1-st kid.
So after the fourth day, the book of the first kid will return to its owner. The book of the fourth kid will return to him for the first time after exactly one day.
You have to answer q independent queries.
Input
The first line of the input contains one integer q (1 ≤ q ≤ 200) — the number of queries. Then q queries follow.
The first line of the query contains one integer n (1 ≤ n ≤ 200) — the number of kids in the query. The second line of the query contains n integers p_1, p_2, ..., p_n (1 ≤ p_i ≤ n, all p_i are distinct, i.e. p is a permutation), where p_i is the kid which will get the book of the i-th kid.
Output
For each query, print the answer on it: n integers a_1, a_2, ..., a_n, where a_i is the number of the day the book of the i-th child is returned back to him for the first time in this query.
Example
Input
6
5
1 2 3 4 5
3
2 3 1
6
4 6 2 1 5 3
1
1
4
3 4 1 2
5
5 1 2 4 3
Output
1 1 1 1 1
3 3 3
2 3 3 2 1 3
1
2 2 2 2
4 4 4 1 4
Submitted Solution:
```
for _ in range(int(input())):
n=int(input())
a=[int(x) for x in input().split()]
for i in range(n):
j=a[i]
cnt=1
while (j<=n and j!=i+1):
j=a[j-1]
cnt+=1
print(cnt,end=" ")
print()
```
Yes
| 5,888 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
The only difference between easy and hard versions is constraints.
There are n kids, each of them is reading a unique book. At the end of any day, the i-th kid will give his book to the p_i-th kid (in case of i = p_i the kid will give his book to himself). It is guaranteed that all values of p_i are distinct integers from 1 to n (i.e. p is a permutation). The sequence p doesn't change from day to day, it is fixed.
For example, if n=6 and p=[4, 6, 1, 3, 5, 2] then at the end of the first day the book of the 1-st kid will belong to the 4-th kid, the 2-nd kid will belong to the 6-th kid and so on. At the end of the second day the book of the 1-st kid will belong to the 3-th kid, the 2-nd kid will belong to the 2-th kid and so on.
Your task is to determine the number of the day the book of the i-th child is returned back to him for the first time for every i from 1 to n.
Consider the following example: p = [5, 1, 2, 4, 3]. The book of the 1-st kid will be passed to the following kids:
* after the 1-st day it will belong to the 5-th kid,
* after the 2-nd day it will belong to the 3-rd kid,
* after the 3-rd day it will belong to the 2-nd kid,
* after the 4-th day it will belong to the 1-st kid.
So after the fourth day, the book of the first kid will return to its owner. The book of the fourth kid will return to him for the first time after exactly one day.
You have to answer q independent queries.
Input
The first line of the input contains one integer q (1 ≤ q ≤ 200) — the number of queries. Then q queries follow.
The first line of the query contains one integer n (1 ≤ n ≤ 200) — the number of kids in the query. The second line of the query contains n integers p_1, p_2, ..., p_n (1 ≤ p_i ≤ n, all p_i are distinct, i.e. p is a permutation), where p_i is the kid which will get the book of the i-th kid.
Output
For each query, print the answer on it: n integers a_1, a_2, ..., a_n, where a_i is the number of the day the book of the i-th child is returned back to him for the first time in this query.
Example
Input
6
5
1 2 3 4 5
3
2 3 1
6
4 6 2 1 5 3
1
1
4
3 4 1 2
5
5 1 2 4 3
Output
1 1 1 1 1
3 3 3
2 3 3 2 1 3
1
2 2 2 2
4 4 4 1 4
Submitted Solution:
```
q = int(input())
for _ in range(q):
_ = int(input())
arr = list(map(int, input().split()))
arr2 = [0 for _ in range(len(arr))]
for i in range(1, len(arr)+1):
new = []
if arr2[i-1] == 0:
k = arr[i-1]
sch = 1
#new.append(i)
new.append(k)
while k != i:
# print('i', i, 'k', k)
k = arr[k-1]
new.append(k)
sch +=1
# print(sch)
#print(new)
for el in new:
arr2[el-1]+=len(new)
print(arr2)
```
No
| 5,889 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
The only difference between easy and hard versions is constraints.
There are n kids, each of them is reading a unique book. At the end of any day, the i-th kid will give his book to the p_i-th kid (in case of i = p_i the kid will give his book to himself). It is guaranteed that all values of p_i are distinct integers from 1 to n (i.e. p is a permutation). The sequence p doesn't change from day to day, it is fixed.
For example, if n=6 and p=[4, 6, 1, 3, 5, 2] then at the end of the first day the book of the 1-st kid will belong to the 4-th kid, the 2-nd kid will belong to the 6-th kid and so on. At the end of the second day the book of the 1-st kid will belong to the 3-th kid, the 2-nd kid will belong to the 2-th kid and so on.
Your task is to determine the number of the day the book of the i-th child is returned back to him for the first time for every i from 1 to n.
Consider the following example: p = [5, 1, 2, 4, 3]. The book of the 1-st kid will be passed to the following kids:
* after the 1-st day it will belong to the 5-th kid,
* after the 2-nd day it will belong to the 3-rd kid,
* after the 3-rd day it will belong to the 2-nd kid,
* after the 4-th day it will belong to the 1-st kid.
So after the fourth day, the book of the first kid will return to its owner. The book of the fourth kid will return to him for the first time after exactly one day.
You have to answer q independent queries.
Input
The first line of the input contains one integer q (1 ≤ q ≤ 200) — the number of queries. Then q queries follow.
The first line of the query contains one integer n (1 ≤ n ≤ 200) — the number of kids in the query. The second line of the query contains n integers p_1, p_2, ..., p_n (1 ≤ p_i ≤ n, all p_i are distinct, i.e. p is a permutation), where p_i is the kid which will get the book of the i-th kid.
Output
For each query, print the answer on it: n integers a_1, a_2, ..., a_n, where a_i is the number of the day the book of the i-th child is returned back to him for the first time in this query.
Example
Input
6
5
1 2 3 4 5
3
2 3 1
6
4 6 2 1 5 3
1
1
4
3 4 1 2
5
5 1 2 4 3
Output
1 1 1 1 1
3 3 3
2 3 3 2 1 3
1
2 2 2 2
4 4 4 1 4
Submitted Solution:
```
try:
import sys
sys.setrecursionlimit(10**6)
def dfs(a,par=-1):
global xx
cpar=par
if par==-1:
cpar=a
i=adj[a]
if i in xx:
return
if i==cpar:
return
xx.add(i)
dfs(i,cpar)
q=int(input())
except:
pass
for _ in range(q):
try:
n=int(input())
it=list(map(int,input().split()))
n=len(it)
adj=[-1 for i in range(n)]
for i in range(n):
adj[i]=it[i]-1
vis=set()
for i in range(n):
vis.add(i)
tot=0
ans=[-1 for i in range(n)]
except:
pass
try:
while vis:
no=vis.pop()
xx=set([no])
dfs(no,-1)
y=len(xx)
for i in xx:
if i!=no:
vis.remove(i)
ans[i]=y
for i in ans:
print(i,end=" ")
print()
except:
pass
```
No
| 5,890 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
The only difference between easy and hard versions is constraints.
There are n kids, each of them is reading a unique book. At the end of any day, the i-th kid will give his book to the p_i-th kid (in case of i = p_i the kid will give his book to himself). It is guaranteed that all values of p_i are distinct integers from 1 to n (i.e. p is a permutation). The sequence p doesn't change from day to day, it is fixed.
For example, if n=6 and p=[4, 6, 1, 3, 5, 2] then at the end of the first day the book of the 1-st kid will belong to the 4-th kid, the 2-nd kid will belong to the 6-th kid and so on. At the end of the second day the book of the 1-st kid will belong to the 3-th kid, the 2-nd kid will belong to the 2-th kid and so on.
Your task is to determine the number of the day the book of the i-th child is returned back to him for the first time for every i from 1 to n.
Consider the following example: p = [5, 1, 2, 4, 3]. The book of the 1-st kid will be passed to the following kids:
* after the 1-st day it will belong to the 5-th kid,
* after the 2-nd day it will belong to the 3-rd kid,
* after the 3-rd day it will belong to the 2-nd kid,
* after the 4-th day it will belong to the 1-st kid.
So after the fourth day, the book of the first kid will return to its owner. The book of the fourth kid will return to him for the first time after exactly one day.
You have to answer q independent queries.
Input
The first line of the input contains one integer q (1 ≤ q ≤ 200) — the number of queries. Then q queries follow.
The first line of the query contains one integer n (1 ≤ n ≤ 200) — the number of kids in the query. The second line of the query contains n integers p_1, p_2, ..., p_n (1 ≤ p_i ≤ n, all p_i are distinct, i.e. p is a permutation), where p_i is the kid which will get the book of the i-th kid.
Output
For each query, print the answer on it: n integers a_1, a_2, ..., a_n, where a_i is the number of the day the book of the i-th child is returned back to him for the first time in this query.
Example
Input
6
5
1 2 3 4 5
3
2 3 1
6
4 6 2 1 5 3
1
1
4
3 4 1 2
5
5 1 2 4 3
Output
1 1 1 1 1
3 3 3
2 3 3 2 1 3
1
2 2 2 2
4 4 4 1 4
Submitted Solution:
```
import sys
input = sys.stdin.readline
tc = int(input())
for _ in range(tc):
n = int(input())
arr = list(map(int,input().split()))
ans = [0 for _ in range(n)]
i = 0
while i < n:
tmp = arr[i] - 1
cnt = 1
print(i, tmp)
while tmp != i:
tmp = arr[tmp] - 1
cnt += 1
ans[i] = cnt
i += 1
for i in ans:
print(i, end=' ')
print()
```
No
| 5,891 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
The only difference between easy and hard versions is constraints.
There are n kids, each of them is reading a unique book. At the end of any day, the i-th kid will give his book to the p_i-th kid (in case of i = p_i the kid will give his book to himself). It is guaranteed that all values of p_i are distinct integers from 1 to n (i.e. p is a permutation). The sequence p doesn't change from day to day, it is fixed.
For example, if n=6 and p=[4, 6, 1, 3, 5, 2] then at the end of the first day the book of the 1-st kid will belong to the 4-th kid, the 2-nd kid will belong to the 6-th kid and so on. At the end of the second day the book of the 1-st kid will belong to the 3-th kid, the 2-nd kid will belong to the 2-th kid and so on.
Your task is to determine the number of the day the book of the i-th child is returned back to him for the first time for every i from 1 to n.
Consider the following example: p = [5, 1, 2, 4, 3]. The book of the 1-st kid will be passed to the following kids:
* after the 1-st day it will belong to the 5-th kid,
* after the 2-nd day it will belong to the 3-rd kid,
* after the 3-rd day it will belong to the 2-nd kid,
* after the 4-th day it will belong to the 1-st kid.
So after the fourth day, the book of the first kid will return to its owner. The book of the fourth kid will return to him for the first time after exactly one day.
You have to answer q independent queries.
Input
The first line of the input contains one integer q (1 ≤ q ≤ 200) — the number of queries. Then q queries follow.
The first line of the query contains one integer n (1 ≤ n ≤ 200) — the number of kids in the query. The second line of the query contains n integers p_1, p_2, ..., p_n (1 ≤ p_i ≤ n, all p_i are distinct, i.e. p is a permutation), where p_i is the kid which will get the book of the i-th kid.
Output
For each query, print the answer on it: n integers a_1, a_2, ..., a_n, where a_i is the number of the day the book of the i-th child is returned back to him for the first time in this query.
Example
Input
6
5
1 2 3 4 5
3
2 3 1
6
4 6 2 1 5 3
1
1
4
3 4 1 2
5
5 1 2 4 3
Output
1 1 1 1 1
3 3 3
2 3 3 2 1 3
1
2 2 2 2
4 4 4 1 4
Submitted Solution:
```
# coding the swap function
def swap(a,b):
temp = a
a = b
b = temp
return a,b
# coding the find function
def find(x):
while(x != link[x]):
x = link[x]
return x
# coding the same function which checks if two nodes
# belong in the same set or not
def same(x,y):
return find(x) == find(y)
# coding the unite function which makes union(x,y)
# of two nodes x and y
def union(x,y):
global size, link
x = find(x)
y = find(y)
if size[x] < size[y]:
x,y = swap(x,y)
size[x] += size[y]
link[y] = x
for _ in range(int(input())):
n = int(input())
array = list(map(int, input().split()))
link = [i for i in range(n)]
size = [1 for i in range(n)]
for i in range(n):
j = array[i]-1
union(i,j)
this = {}
for i in list(set(link)):
this[i] = link.count(i)
ans = []
for i in link:
ans.append(this[i])
print(*ans)
```
No
| 5,892 |
Provide tags and a correct Python 3 solution for this coding contest problem.
One day in the IT lesson Anna and Maria learned about the lexicographic order.
String x is lexicographically less than string y, if either x is a prefix of y (and x ≠ y), or there exists such i (1 ≤ i ≤ min(|x|, |y|)), that xi < yi, and for any j (1 ≤ j < i) xj = yj. Here |a| denotes the length of the string a. The lexicographic comparison of strings is implemented by operator < in modern programming languages.
The teacher gave Anna and Maria homework. She gave them a string of length n. They should write out all substrings of the given string, including the whole initial string, and the equal substrings (for example, one should write out the following substrings from the string "aab": "a", "a", "aa", "ab", "aab", "b"). The resulting strings should be sorted in the lexicographical order. The cunning teacher doesn't want to check all these strings. That's why she said to find only the k-th string from the list. Help Anna and Maria do the homework.
Input
The first line contains a non-empty string that only consists of small Latin letters ("a"-"z"), whose length does not exceed 105. The second line contains the only integer k (1 ≤ k ≤ 105).
Output
Print the string Anna and Maria need — the k-th (in the lexicographical order) substring of the given string. If the total number of substrings is less than k, print a string saying "No such line." (without the quotes).
Examples
Input
aa
2
Output
a
Input
abc
5
Output
bc
Input
abab
7
Output
b
Note
In the second sample before string "bc" follow strings "a", "ab", "abc", "b".
Tags: brute force, constructive algorithms, hashing, implementation, string suffix structures, strings
Correct Solution:
```
__author__ = 'Utena'
from heapq import heappushpop, heapify,heappop,heappush
string=input()
s=len(string)
k=int(input())
total=0
line=[]
for i in range(s):
line.append((string[i],i))
heapify(line)
while total<k:
value,key=heappop(line)
ans=value
total+=1
if key<s-1:heappush(line,(value+string[key+1],key+1))
if len(line)==0:
if total<k:
print('No such line.')
exit(0)
print(ans)
```
| 5,893 |
Provide tags and a correct Python 3 solution for this coding contest problem.
One day in the IT lesson Anna and Maria learned about the lexicographic order.
String x is lexicographically less than string y, if either x is a prefix of y (and x ≠ y), or there exists such i (1 ≤ i ≤ min(|x|, |y|)), that xi < yi, and for any j (1 ≤ j < i) xj = yj. Here |a| denotes the length of the string a. The lexicographic comparison of strings is implemented by operator < in modern programming languages.
The teacher gave Anna and Maria homework. She gave them a string of length n. They should write out all substrings of the given string, including the whole initial string, and the equal substrings (for example, one should write out the following substrings from the string "aab": "a", "a", "aa", "ab", "aab", "b"). The resulting strings should be sorted in the lexicographical order. The cunning teacher doesn't want to check all these strings. That's why she said to find only the k-th string from the list. Help Anna and Maria do the homework.
Input
The first line contains a non-empty string that only consists of small Latin letters ("a"-"z"), whose length does not exceed 105. The second line contains the only integer k (1 ≤ k ≤ 105).
Output
Print the string Anna and Maria need — the k-th (in the lexicographical order) substring of the given string. If the total number of substrings is less than k, print a string saying "No such line." (without the quotes).
Examples
Input
aa
2
Output
a
Input
abc
5
Output
bc
Input
abab
7
Output
b
Note
In the second sample before string "bc" follow strings "a", "ab", "abc", "b".
Tags: brute force, constructive algorithms, hashing, implementation, string suffix structures, strings
Correct Solution:
```
s=input()
l=len(s)
k=int(input())
if l*(l+1)/2<k: print('No such line.')
else:
a=[i for i in range(l)]
p=0
ans=''
while True:
t=[(s[i+p],i) for i in a]
t=sorted(t)
i=j=0
prod=0
while i<len(t):
pre=prod
while j<len(t) and t[i][0]==t[j][0]:
prod+=l-t[j][1]-p
j+=1
if prod>=k: break
i=j
ans+=t[i][0]
if pre+j-i>=k: break
k-=pre+j-i
p+=1
a=[t[q][1] for q in range(i,j) if t[q][1]+p<l]
print(ans)
```
| 5,894 |
Provide tags and a correct Python 3 solution for this coding contest problem.
One day in the IT lesson Anna and Maria learned about the lexicographic order.
String x is lexicographically less than string y, if either x is a prefix of y (and x ≠ y), or there exists such i (1 ≤ i ≤ min(|x|, |y|)), that xi < yi, and for any j (1 ≤ j < i) xj = yj. Here |a| denotes the length of the string a. The lexicographic comparison of strings is implemented by operator < in modern programming languages.
The teacher gave Anna and Maria homework. She gave them a string of length n. They should write out all substrings of the given string, including the whole initial string, and the equal substrings (for example, one should write out the following substrings from the string "aab": "a", "a", "aa", "ab", "aab", "b"). The resulting strings should be sorted in the lexicographical order. The cunning teacher doesn't want to check all these strings. That's why she said to find only the k-th string from the list. Help Anna and Maria do the homework.
Input
The first line contains a non-empty string that only consists of small Latin letters ("a"-"z"), whose length does not exceed 105. The second line contains the only integer k (1 ≤ k ≤ 105).
Output
Print the string Anna and Maria need — the k-th (in the lexicographical order) substring of the given string. If the total number of substrings is less than k, print a string saying "No such line." (without the quotes).
Examples
Input
aa
2
Output
a
Input
abc
5
Output
bc
Input
abab
7
Output
b
Note
In the second sample before string "bc" follow strings "a", "ab", "abc", "b".
Tags: brute force, constructive algorithms, hashing, implementation, string suffix structures, strings
Correct Solution:
```
import heapq as hq
s = input()
k = int(input())
n = len(s)
news = [([s[i]], i + 1) for i in range(n)]
if k > n * (n + 1) // 2:
print('No such line.')
exit()
hq.heapify(news)
for i in range(k):
t, nxt = hq.heappop(news)
if nxt < n and i + 1 < k:
t.append(s[nxt])
hq.heappush(news, (t, nxt + 1))
print(''.join(t))
```
| 5,895 |
Provide tags and a correct Python 3 solution for this coding contest problem.
One day in the IT lesson Anna and Maria learned about the lexicographic order.
String x is lexicographically less than string y, if either x is a prefix of y (and x ≠ y), or there exists such i (1 ≤ i ≤ min(|x|, |y|)), that xi < yi, and for any j (1 ≤ j < i) xj = yj. Here |a| denotes the length of the string a. The lexicographic comparison of strings is implemented by operator < in modern programming languages.
The teacher gave Anna and Maria homework. She gave them a string of length n. They should write out all substrings of the given string, including the whole initial string, and the equal substrings (for example, one should write out the following substrings from the string "aab": "a", "a", "aa", "ab", "aab", "b"). The resulting strings should be sorted in the lexicographical order. The cunning teacher doesn't want to check all these strings. That's why she said to find only the k-th string from the list. Help Anna and Maria do the homework.
Input
The first line contains a non-empty string that only consists of small Latin letters ("a"-"z"), whose length does not exceed 105. The second line contains the only integer k (1 ≤ k ≤ 105).
Output
Print the string Anna and Maria need — the k-th (in the lexicographical order) substring of the given string. If the total number of substrings is less than k, print a string saying "No such line." (without the quotes).
Examples
Input
aa
2
Output
a
Input
abc
5
Output
bc
Input
abab
7
Output
b
Note
In the second sample before string "bc" follow strings "a", "ab", "abc", "b".
Tags: brute force, constructive algorithms, hashing, implementation, string suffix structures, strings
Correct Solution:
```
from heapq import *
s=input()
k=int(input())
n=len(s)
ss=[(s[i],i) for i in range(n)]
heapify(ss)
if k>n*(n+1)/2:
print('No such line.')
else:
while k:
k-=1
t=heappop(ss)
if k==0:
print(t[0])
else:
if t[1]<len(s)-1:
heappush(ss,(t[0]+s[t[1]+1],t[1]+1))
```
| 5,896 |
Provide tags and a correct Python 3 solution for this coding contest problem.
One day in the IT lesson Anna and Maria learned about the lexicographic order.
String x is lexicographically less than string y, if either x is a prefix of y (and x ≠ y), or there exists such i (1 ≤ i ≤ min(|x|, |y|)), that xi < yi, and for any j (1 ≤ j < i) xj = yj. Here |a| denotes the length of the string a. The lexicographic comparison of strings is implemented by operator < in modern programming languages.
The teacher gave Anna and Maria homework. She gave them a string of length n. They should write out all substrings of the given string, including the whole initial string, and the equal substrings (for example, one should write out the following substrings from the string "aab": "a", "a", "aa", "ab", "aab", "b"). The resulting strings should be sorted in the lexicographical order. The cunning teacher doesn't want to check all these strings. That's why she said to find only the k-th string from the list. Help Anna and Maria do the homework.
Input
The first line contains a non-empty string that only consists of small Latin letters ("a"-"z"), whose length does not exceed 105. The second line contains the only integer k (1 ≤ k ≤ 105).
Output
Print the string Anna and Maria need — the k-th (in the lexicographical order) substring of the given string. If the total number of substrings is less than k, print a string saying "No such line." (without the quotes).
Examples
Input
aa
2
Output
a
Input
abc
5
Output
bc
Input
abab
7
Output
b
Note
In the second sample before string "bc" follow strings "a", "ab", "abc", "b".
Tags: brute force, constructive algorithms, hashing, implementation, string suffix structures, strings
Correct Solution:
```
import sys,heapq
x=input();
k=int(input());
i=0;
h=[(x[i],i+1) for i in range(len(x))];
heapq.heapify(h);
while(True):
i=i+1;
if(not h):
print("No such line.");
sys.exit();
s,next=heapq.heappop(h);
if(i==k):
print(s);
break;
if(next>=len(x)):
continue
s=s+x[next]
heapq.heappush(h,(s,next+1))
```
| 5,897 |
Provide tags and a correct Python 3 solution for this coding contest problem.
One day in the IT lesson Anna and Maria learned about the lexicographic order.
String x is lexicographically less than string y, if either x is a prefix of y (and x ≠ y), or there exists such i (1 ≤ i ≤ min(|x|, |y|)), that xi < yi, and for any j (1 ≤ j < i) xj = yj. Here |a| denotes the length of the string a. The lexicographic comparison of strings is implemented by operator < in modern programming languages.
The teacher gave Anna and Maria homework. She gave them a string of length n. They should write out all substrings of the given string, including the whole initial string, and the equal substrings (for example, one should write out the following substrings from the string "aab": "a", "a", "aa", "ab", "aab", "b"). The resulting strings should be sorted in the lexicographical order. The cunning teacher doesn't want to check all these strings. That's why she said to find only the k-th string from the list. Help Anna and Maria do the homework.
Input
The first line contains a non-empty string that only consists of small Latin letters ("a"-"z"), whose length does not exceed 105. The second line contains the only integer k (1 ≤ k ≤ 105).
Output
Print the string Anna and Maria need — the k-th (in the lexicographical order) substring of the given string. If the total number of substrings is less than k, print a string saying "No such line." (without the quotes).
Examples
Input
aa
2
Output
a
Input
abc
5
Output
bc
Input
abab
7
Output
b
Note
In the second sample before string "bc" follow strings "a", "ab", "abc", "b".
Tags: brute force, constructive algorithms, hashing, implementation, string suffix structures, strings
Correct Solution:
```
from heapq import *
l=input()
k=int(input())
n=len(l)
if k>n*(n+1)/2:
print("No such line.")
quit()
ss=[(l[i],i) for i in range(n)]
heapify(ss)
while k:
k-=1
t=heappop(ss)
if k==0:
print(t[0])
else:
if t[1]<n-1:
heappush(ss,(t[0]+l[t[1]+1],t[1]+1))
```
| 5,898 |
Provide tags and a correct Python 3 solution for this coding contest problem.
One day in the IT lesson Anna and Maria learned about the lexicographic order.
String x is lexicographically less than string y, if either x is a prefix of y (and x ≠ y), or there exists such i (1 ≤ i ≤ min(|x|, |y|)), that xi < yi, and for any j (1 ≤ j < i) xj = yj. Here |a| denotes the length of the string a. The lexicographic comparison of strings is implemented by operator < in modern programming languages.
The teacher gave Anna and Maria homework. She gave them a string of length n. They should write out all substrings of the given string, including the whole initial string, and the equal substrings (for example, one should write out the following substrings from the string "aab": "a", "a", "aa", "ab", "aab", "b"). The resulting strings should be sorted in the lexicographical order. The cunning teacher doesn't want to check all these strings. That's why she said to find only the k-th string from the list. Help Anna and Maria do the homework.
Input
The first line contains a non-empty string that only consists of small Latin letters ("a"-"z"), whose length does not exceed 105. The second line contains the only integer k (1 ≤ k ≤ 105).
Output
Print the string Anna and Maria need — the k-th (in the lexicographical order) substring of the given string. If the total number of substrings is less than k, print a string saying "No such line." (without the quotes).
Examples
Input
aa
2
Output
a
Input
abc
5
Output
bc
Input
abab
7
Output
b
Note
In the second sample before string "bc" follow strings "a", "ab", "abc", "b".
Tags: brute force, constructive algorithms, hashing, implementation, string suffix structures, strings
Correct Solution:
```
import heapq
from heapq import heappush, heappop, heappushpop
a = input()
n = int(input())
c=list(a)
l = len(c)
num = 0
if (l*l+l)/2 < n:
print('No such line.')
else:
d = []
#
for i in range(l):
heappush(d,(c[i],i))
while num < n:
num += 1
t,f = heappop(d)
if f<l-1:
heappush(d,(t+c[f+1],f+1))
print(t)
# for i in range(l):
# for j in range(1,l+1):
# if i+j<=l:
# d.append(a[i:i+j])
#print(d)#
# for i in range(n-1):
# d.remove(min(d))
# print(d)#
# print(min(d))
```
| 5,899 |
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