message stringlengths 2 67k | message_type stringclasses 2 values | message_id int64 0 1 | conversation_id int64 463 109k | cluster float64 19 19 | __index_level_0__ int64 926 217k |
|---|---|---|---|---|---|
Provide tags and a correct Python 3 solution for this coding contest problem.
Vasya and Petya wrote down all integers from 1 to n to play the "powers" game (n can be quite large; however, Vasya and Petya are not confused by this fact).
Players choose numbers in turn (Vasya chooses first). If some number x is chosen at the current turn, it is forbidden to choose x or all of its other positive integer powers (that is, x2, x3, ...) at the next turns. For instance, if the number 9 is chosen at the first turn, one cannot choose 9 or 81 later, while it is still allowed to choose 3 or 27. The one who cannot make a move loses.
Who wins if both Vasya and Petya play optimally?
Input
Input contains single integer n (1 ≤ n ≤ 109).
Output
Print the name of the winner — "Vasya" or "Petya" (without quotes).
Examples
Input
1
Output
Vasya
Input
2
Output
Petya
Input
8
Output
Petya
Note
In the first sample Vasya will choose 1 and win immediately.
In the second sample no matter which number Vasya chooses during his first turn, Petya can choose the remaining number and win. | instruction | 0 | 45,707 | 19 | 91,414 |
Tags: dp, games
Correct Solution:
```
import math, collections
mod = 10**9+7
def isPower(n):
if (n <= 1):
return True
for x in range(2, (int)(math.sqrt(n)) + 1):
p = x
while (p <= n):
p = p * x
if (p == n):
return True
return False
n = int(input())
arr = [0,1,2,1,4,3,2,1,5,6,2,1,8,7,5,9,8,7,3,4,7,4,2,1,10,9,3,6,11,12]
ans = arr[int(math.log(n, 2))]
s = int(math.log(n, 2))
for i in range(3, int(n**0.5)+1):
if not isPower(i):
ans^=arr[int(math.log(n, i))]
s+=int(math.log(n, i))
ans^=((n-s)%2)
print("Vasya" if ans else "Petya")
``` | output | 1 | 45,707 | 19 | 91,415 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Vasya and Petya wrote down all integers from 1 to n to play the "powers" game (n can be quite large; however, Vasya and Petya are not confused by this fact).
Players choose numbers in turn (Vasya chooses first). If some number x is chosen at the current turn, it is forbidden to choose x or all of its other positive integer powers (that is, x2, x3, ...) at the next turns. For instance, if the number 9 is chosen at the first turn, one cannot choose 9 or 81 later, while it is still allowed to choose 3 or 27. The one who cannot make a move loses.
Who wins if both Vasya and Petya play optimally?
Input
Input contains single integer n (1 ≤ n ≤ 109).
Output
Print the name of the winner — "Vasya" or "Petya" (without quotes).
Examples
Input
1
Output
Vasya
Input
2
Output
Petya
Input
8
Output
Petya
Note
In the first sample Vasya will choose 1 and win immediately.
In the second sample no matter which number Vasya chooses during his first turn, Petya can choose the remaining number and win.
Submitted Solution:
```
from sys import stdin, stdout
import math
n = int(stdin.readline())
ans = n
for i in range(2, int(math.sqrt(n))+1):
t = int(math.log(n, i))
if t!=0:
ans -= (t-1)
if ans % 2 == 0:
stdout.write("Petya"+'\n')
else:
stdout.write("Vasya"+"\n")
``` | instruction | 0 | 45,708 | 19 | 91,416 |
No | output | 1 | 45,708 | 19 | 91,417 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Vasya and Petya wrote down all integers from 1 to n to play the "powers" game (n can be quite large; however, Vasya and Petya are not confused by this fact).
Players choose numbers in turn (Vasya chooses first). If some number x is chosen at the current turn, it is forbidden to choose x or all of its other positive integer powers (that is, x2, x3, ...) at the next turns. For instance, if the number 9 is chosen at the first turn, one cannot choose 9 or 81 later, while it is still allowed to choose 3 or 27. The one who cannot make a move loses.
Who wins if both Vasya and Petya play optimally?
Input
Input contains single integer n (1 ≤ n ≤ 109).
Output
Print the name of the winner — "Vasya" or "Petya" (without quotes).
Examples
Input
1
Output
Vasya
Input
2
Output
Petya
Input
8
Output
Petya
Note
In the first sample Vasya will choose 1 and win immediately.
In the second sample no matter which number Vasya chooses during his first turn, Petya can choose the remaining number and win.
Submitted Solution:
```
n = int(input())
print('Petya' if n == 2 or n == 8 else 'Vasya')
``` | instruction | 0 | 45,709 | 19 | 91,418 |
No | output | 1 | 45,709 | 19 | 91,419 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Rick and Morty are playing their own version of Berzerk (which has nothing in common with the famous Berzerk game). This game needs a huge space, so they play it with a computer.
In this game there are n objects numbered from 1 to n arranged in a circle (in clockwise order). Object number 1 is a black hole and the others are planets. There's a monster in one of the planet. Rick and Morty don't know on which one yet, only that he's not initially in the black hole, but Unity will inform them before the game starts. But for now, they want to be prepared for every possible scenario.
<image>
Each one of them has a set of numbers between 1 and n - 1 (inclusive). Rick's set is s1 with k1 elements and Morty's is s2 with k2 elements. One of them goes first and the player changes alternatively. In each player's turn, he should choose an arbitrary number like x from his set and the monster will move to his x-th next object from its current position (clockwise). If after his move the monster gets to the black hole he wins.
Your task is that for each of monster's initial positions and who plays first determine if the starter wins, loses, or the game will stuck in an infinite loop. In case when player can lose or make game infinity, it more profitable to choose infinity game.
Input
The first line of input contains a single integer n (2 ≤ n ≤ 7000) — number of objects in game.
The second line contains integer k1 followed by k1 distinct integers s1, 1, s1, 2, ..., s1, k1 — Rick's set.
The third line contains integer k2 followed by k2 distinct integers s2, 1, s2, 2, ..., s2, k2 — Morty's set
1 ≤ ki ≤ n - 1 and 1 ≤ si, 1, si, 2, ..., si, ki ≤ n - 1 for 1 ≤ i ≤ 2.
Output
In the first line print n - 1 words separated by spaces where i-th word is "Win" (without quotations) if in the scenario that Rick plays first and monster is initially in object number i + 1 he wins, "Lose" if he loses and "Loop" if the game will never end.
Similarly, in the second line print n - 1 words separated by spaces where i-th word is "Win" (without quotations) if in the scenario that Morty plays first and monster is initially in object number i + 1 he wins, "Lose" if he loses and "Loop" if the game will never end.
Examples
Input
5
2 3 2
3 1 2 3
Output
Lose Win Win Loop
Loop Win Win Win
Input
8
4 6 2 3 4
2 3 6
Output
Win Win Win Win Win Win Win
Lose Win Lose Lose Win Lose Lose | instruction | 0 | 45,836 | 19 | 91,672 |
Tags: dfs and similar, dp, games
Correct Solution:
```
import queue
n = int(input())
sR = list(map(int, input().split()[1:]))
sM = list(map(int, input().split()[1:]))
s = [sR, sM]
UNK = -1
WIN = 2
LOSE = 3
A = [[UNK] * n for i in range(2)]
CNT = [[0] * n for i in range(2)]
V = [[False] * n for i in range(2)]
# ricky turn 0
# morty turn 1
A[0][0] = LOSE
A[1][0] = LOSE
Q = queue.Queue()
Q.put((0, 0))
Q.put((1, 0))
while not Q.empty():
turn, planet = Q.get()
prev_turn = 1 - turn
for diff in s[prev_turn]:
prev_planet = (n + planet - diff) % n
if prev_planet == 0:
continue
if A[turn][planet] == LOSE:
A[prev_turn][prev_planet] = WIN
elif A[turn][planet] == WIN:
CNT[prev_turn][prev_planet] += 1
if CNT[prev_turn][prev_planet] == len(s[prev_turn]):
A[prev_turn][prev_planet] = LOSE
if A[prev_turn][prev_planet] != UNK and not V[prev_turn][prev_planet]:
Q.put((prev_turn, prev_planet))
V[prev_turn][prev_planet] = True
print(' '.join(["Win" if A[0][i] == WIN else "Lose" if A[0][i] == LOSE else "Loop" for i in range(1, n)]))
print(' '.join(["Win" if A[1][i] == WIN else "Lose" if A[1][i] == LOSE else "Loop" for i in range(1, n)]))
``` | output | 1 | 45,836 | 19 | 91,673 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You can't possibly imagine how cold our friends are this winter in Nvodsk! Two of them play the following game to warm up: initially a piece of paper has an integer q. During a move a player should write any integer number that is a non-trivial divisor of the last written number. Then he should run this number of circles around the hotel. Let us remind you that a number's divisor is called non-trivial if it is different from one and from the divided number itself.
The first person who can't make a move wins as he continues to lie in his warm bed under three blankets while the other one keeps running. Determine which player wins considering that both players play optimally. If the first player wins, print any winning first move.
Input
The first line contains the only integer q (1 ≤ q ≤ 1013).
Please do not use the %lld specificator to read or write 64-bit integers in С++. It is preferred to use the cin, cout streams or the %I64d specificator.
Output
In the first line print the number of the winning player (1 or 2). If the first player wins then the second line should contain another integer — his first move (if the first player can't even make the first move, print 0). If there are multiple solutions, print any of them.
Examples
Input
6
Output
2
Input
30
Output
1
6
Input
1
Output
1
0
Note
Number 6 has only two non-trivial divisors: 2 and 3. It is impossible to make a move after the numbers 2 and 3 are written, so both of them are winning, thus, number 6 is the losing number. A player can make a move and write number 6 after number 30; 6, as we know, is a losing number. Thus, this move will bring us the victory. | instruction | 0 | 46,512 | 19 | 93,024 |
Tags: games, math, number theory
Correct Solution:
```
q=int(input())
if q==1:
print(1)
print(0)
else:
ls=[]
for i in range(2,int(q**0.5)+1):
while q%i==0:
ls.append(i)
q=q//i
if q>1:
ls.append(q)
if len(ls)>2:
print(1)
print(ls[0]*ls[1])
elif len(ls)==2:
print(2)
else:
print(1)
print(0)
``` | output | 1 | 46,512 | 19 | 93,025 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You can't possibly imagine how cold our friends are this winter in Nvodsk! Two of them play the following game to warm up: initially a piece of paper has an integer q. During a move a player should write any integer number that is a non-trivial divisor of the last written number. Then he should run this number of circles around the hotel. Let us remind you that a number's divisor is called non-trivial if it is different from one and from the divided number itself.
The first person who can't make a move wins as he continues to lie in his warm bed under three blankets while the other one keeps running. Determine which player wins considering that both players play optimally. If the first player wins, print any winning first move.
Input
The first line contains the only integer q (1 ≤ q ≤ 1013).
Please do not use the %lld specificator to read or write 64-bit integers in С++. It is preferred to use the cin, cout streams or the %I64d specificator.
Output
In the first line print the number of the winning player (1 or 2). If the first player wins then the second line should contain another integer — his first move (if the first player can't even make the first move, print 0). If there are multiple solutions, print any of them.
Examples
Input
6
Output
2
Input
30
Output
1
6
Input
1
Output
1
0
Note
Number 6 has only two non-trivial divisors: 2 and 3. It is impossible to make a move after the numbers 2 and 3 are written, so both of them are winning, thus, number 6 is the losing number. A player can make a move and write number 6 after number 30; 6, as we know, is a losing number. Thus, this move will bring us the victory. | instruction | 0 | 46,513 | 19 | 93,026 |
Tags: games, math, number theory
Correct Solution:
```
q = int(input())
initialQ = q
d = {}
i = 2
while i * i <= q:
if q % i:
i += 1
continue
d[i] = 0
while q % i == 0:
q //= i
d[i] += 1
i += 1
if q != 1 and q != initialQ:d[q] = 1
l = sum(d[item] for item in d)
newD = []
for item in d:
newD += [item] * d[item]
if l > 2:
print(1)
print(newD[0] * newD[1])
elif l == 2:
print(2)
else:
print(1)
print(0)
``` | output | 1 | 46,513 | 19 | 93,027 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You can't possibly imagine how cold our friends are this winter in Nvodsk! Two of them play the following game to warm up: initially a piece of paper has an integer q. During a move a player should write any integer number that is a non-trivial divisor of the last written number. Then he should run this number of circles around the hotel. Let us remind you that a number's divisor is called non-trivial if it is different from one and from the divided number itself.
The first person who can't make a move wins as he continues to lie in his warm bed under three blankets while the other one keeps running. Determine which player wins considering that both players play optimally. If the first player wins, print any winning first move.
Input
The first line contains the only integer q (1 ≤ q ≤ 1013).
Please do not use the %lld specificator to read or write 64-bit integers in С++. It is preferred to use the cin, cout streams or the %I64d specificator.
Output
In the first line print the number of the winning player (1 or 2). If the first player wins then the second line should contain another integer — his first move (if the first player can't even make the first move, print 0). If there are multiple solutions, print any of them.
Examples
Input
6
Output
2
Input
30
Output
1
6
Input
1
Output
1
0
Note
Number 6 has only two non-trivial divisors: 2 and 3. It is impossible to make a move after the numbers 2 and 3 are written, so both of them are winning, thus, number 6 is the losing number. A player can make a move and write number 6 after number 30; 6, as we know, is a losing number. Thus, this move will bring us the victory. | instruction | 0 | 46,514 | 19 | 93,028 |
Tags: games, math, number theory
Correct Solution:
```
n=int(input())
x=n
v1=0
v2=0
i=2
while i*i<=n:
while n%i==0:
if v1:
v2=i
else:
v1=i
n//=i
i+=1
if n-1:
v2=n
if v1*v2-x:
print(1)
print(v1*v2)
else:
print(2)
``` | output | 1 | 46,514 | 19 | 93,029 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You can't possibly imagine how cold our friends are this winter in Nvodsk! Two of them play the following game to warm up: initially a piece of paper has an integer q. During a move a player should write any integer number that is a non-trivial divisor of the last written number. Then he should run this number of circles around the hotel. Let us remind you that a number's divisor is called non-trivial if it is different from one and from the divided number itself.
The first person who can't make a move wins as he continues to lie in his warm bed under three blankets while the other one keeps running. Determine which player wins considering that both players play optimally. If the first player wins, print any winning first move.
Input
The first line contains the only integer q (1 ≤ q ≤ 1013).
Please do not use the %lld specificator to read or write 64-bit integers in С++. It is preferred to use the cin, cout streams or the %I64d specificator.
Output
In the first line print the number of the winning player (1 or 2). If the first player wins then the second line should contain another integer — his first move (if the first player can't even make the first move, print 0). If there are multiple solutions, print any of them.
Examples
Input
6
Output
2
Input
30
Output
1
6
Input
1
Output
1
0
Note
Number 6 has only two non-trivial divisors: 2 and 3. It is impossible to make a move after the numbers 2 and 3 are written, so both of them are winning, thus, number 6 is the losing number. A player can make a move and write number 6 after number 30; 6, as we know, is a losing number. Thus, this move will bring us the victory. | instruction | 0 | 46,515 | 19 | 93,030 |
Tags: games, math, number theory
Correct Solution:
```
import math
if __name__ == '__main__':
n = int(input())
sqrtn = int(math.sqrt(n))
factors = list()
for i in range(2, sqrtn+1):
while n%i == 0:
factors.append(i)
n = n//i
if n > 1:
factors.append(n)
if len(factors) == 2:
print(2)
exit()
else:
print(1)
if len(factors) < 2:
print(0)
else:
print(factors[0]*factors[1])
``` | output | 1 | 46,515 | 19 | 93,031 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You can't possibly imagine how cold our friends are this winter in Nvodsk! Two of them play the following game to warm up: initially a piece of paper has an integer q. During a move a player should write any integer number that is a non-trivial divisor of the last written number. Then he should run this number of circles around the hotel. Let us remind you that a number's divisor is called non-trivial if it is different from one and from the divided number itself.
The first person who can't make a move wins as he continues to lie in his warm bed under three blankets while the other one keeps running. Determine which player wins considering that both players play optimally. If the first player wins, print any winning first move.
Input
The first line contains the only integer q (1 ≤ q ≤ 1013).
Please do not use the %lld specificator to read or write 64-bit integers in С++. It is preferred to use the cin, cout streams or the %I64d specificator.
Output
In the first line print the number of the winning player (1 or 2). If the first player wins then the second line should contain another integer — his first move (if the first player can't even make the first move, print 0). If there are multiple solutions, print any of them.
Examples
Input
6
Output
2
Input
30
Output
1
6
Input
1
Output
1
0
Note
Number 6 has only two non-trivial divisors: 2 and 3. It is impossible to make a move after the numbers 2 and 3 are written, so both of them are winning, thus, number 6 is the losing number. A player can make a move and write number 6 after number 30; 6, as we know, is a losing number. Thus, this move will bring us the victory. | instruction | 0 | 46,516 | 19 | 93,032 |
Tags: games, math, number theory
Correct Solution:
```
#!/usr/bin/env python
from __future__ import division, print_function
import math
import os
import sys
from fractions import *
from sys import *
from decimal import *
from io import BytesIO, IOBase
from itertools import *
from collections import *
# sys.setrecursionlimit(10**5)
M = 10 ** 9 + 7
# print(math.factorial(5))
if sys.version_info[0] < 3:
from __builtin__ import xrange as range
from future_builtins import ascii, filter, hex, map, oct, zip
# sys.setrecursionlimit(10**6)
# region fastio
BUFSIZE = 8192
class FastIO(IOBase):
newlines = 0
def __init__(self, file):
self._fd = file.fileno()
self.buffer = BytesIO()
self.writable = "x" in file.mode or "r" not in file.mode
self.write = self.buffer.write if self.writable else None
def read(self):
while True:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
if not b:
break
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines = 0
return self.buffer.read()
def readline(self):
while self.newlines == 0:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
self.newlines = b.count(b"\n") + (not b)
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines -= 1
return self.buffer.readline()
def flush(self):
if self.writable:
os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
self.write = lambda s: self.buffer.write(s.encode("ascii"))
self.read = lambda: self.buffer.read().decode("ascii")
self.readline = lambda: self.buffer.readline().decode("ascii")
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)
input = lambda: sys.stdin.readline().rstrip("\r\n")
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 inpu(): return int(inp())
# -----------------------------------------------------------------
def regularbracket(t):
p = 0
for i in t:
if i == "(":
p += 1
else:
p -= 1
if p < 0:
return False
else:
if p > 0:
return False
else:
return True
# -------------------------------------------------
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
# ------------------------------reverse string(pallindrome)
def reverse1(string):
pp = ""
for i in string[::-1]:
pp += i
if pp == string:
return True
return False
# --------------------------------reverse list(paindrome)
def reverse2(list1):
l = []
for i in list1[::-1]:
l.append(i)
if l == list1:
return True
return False
def mex(list1):
# list1 = sorted(list1)
p = max(list1) + 1
for i in range(len(list1)):
if list1[i] != i:
p = i
break
return p
def sumofdigits(n):
n = str(n)
s1 = 0
for i in n:
s1 += int(i)
return s1
def perfect_square(n):
s = math.sqrt(n)
if s == int(s):
return True
return False
# -----------------------------roman
def roman_number(x):
if x > 15999:
return
value = [5000, 4000, 1000, 900, 500, 400, 100, 90, 50, 40, 10, 9, 5, 4, 1]
symbol = ["F", "MF", "M", "CM", "D", "CD", "C", "XC", "L", "XL", "X", "IX", "V", "IV", "I"]
roman = ""
i = 0
while x > 0:
div = x // value[i]
x = x % value[i]
while div:
roman += symbol[i]
div -= 1
i += 1
return roman
def soretd(s):
for i in range(1, len(s)):
if s[i - 1] > s[i]:
return False
return True
# print(soretd("1"))
# ---------------------------
def countRhombi(h, w):
ct = 0
for i in range(2, h + 1, 2):
for j in range(2, w + 1, 2):
ct += (h - i + 1) * (w - j + 1)
return ct
def countrhombi2(h, w):
return ((h * h) // 4) * ((w * w) // 4)
# ---------------------------------
def binpow(a, b):
if b == 0:
return 1
else:
res = binpow(a, b // 2)
if b % 2 != 0:
return res * res * a
else:
return res * res
# -------------------------------------------------------
def binpowmodulus(a, b, m):
a %= m
res = 1
while (b > 0):
if (b & 1):
res = res * a % m
a = a * a % m
b >>= 1
return res
# -------------------------------------------------------------
def coprime_to_n(n):
result = n
i = 2
while (i * i <= n):
if (n % i == 0):
while (n % i == 0):
n //= i
result -= result // i
i += 1
if (n > 1):
result -= result // n
return result
# -------------------prime
def prime(x):
if x == 1:
return False
else:
for i in range(2, int(math.sqrt(x)) + 1):
# print(x)
if (x % i == 0):
return False
else:
return True
def luckynumwithequalnumberoffourandseven(x, n, a):
if x >= n and str(x).count("4") == str(x).count("7"):
a.append(x)
else:
if x < 1e12:
luckynumwithequalnumberoffourandseven(x * 10 + 4, n, a)
luckynumwithequalnumberoffourandseven(x * 10 + 7, n, a)
return a
def luckynuber(x, n, a):
p = set(str(x))
if len(p) <= 2:
a.append(x)
if x < n:
luckynuber(x + 1, n, a)
return a
# ------------------------------------------------------interactive problems
def interact(type, x):
if type == "r":
inp = input()
return inp.strip()
else:
print(x, flush=True)
# ------------------------------------------------------------------zero at end of factorial of a number
def findTrailingZeros(n):
# Initialize result
count = 0
# Keep dividing n by
# 5 & update Count
while (n >= 5):
n //= 5
count += n
return count
# -----------------------------------------------merge sort
# Python program for implementation of MergeSort
def mergeSort(arr):
if len(arr) > 1:
# Finding the mid of the array
mid = len(arr) // 2
# Dividing the array elements
L = arr[:mid]
# into 2 halves
R = arr[mid:]
# Sorting the first half
mergeSort(L)
# Sorting the second half
mergeSort(R)
i = j = k = 0
# Copy data to temp arrays L[] and R[]
while i < len(L) and j < len(R):
if L[i] < R[j]:
arr[k] = L[i]
i += 1
else:
arr[k] = R[j]
j += 1
k += 1
# Checking if any element was left
while i < len(L):
arr[k] = L[i]
i += 1
k += 1
while j < len(R):
arr[k] = R[j]
j += 1
k += 1
# -----------------------------------------------lucky number with two lucky any digits
res = set()
def solven(p, l, a, b, n): # given number
if p > n or l > 10:
return
if p > 0:
res.add(p)
solven(p * 10 + a, l + 1, a, b, n)
solven(p * 10 + b, l + 1, a, b, n)
# problem
"""
n = int(input())
for a in range(0, 10):
for b in range(0, a):
solve(0, 0)
print(len(res))
"""
# Python3 program to find all subsets
# by backtracking.
# In the array A at every step we have two
# choices for each element either we can
# ignore the element or we can include the
# element in our subset
def subsetsUtil(A, subset, index, d):
print(*subset)
s = sum(subset)
d.append(s)
for i in range(index, len(A)):
# include the A[i] in subset.
subset.append(A[i])
# move onto the next element.
subsetsUtil(A, subset, i + 1, d)
# exclude the A[i] from subset and
# triggers backtracking.
subset.pop(-1)
return d
def subsetSums(arr, l, r, d, sum=0):
if l > r:
d.append(sum)
return
subsetSums(arr, l + 1, r, d, sum + arr[l])
# Subset excluding arr[l]
subsetSums(arr, l + 1, r, d, sum)
return d
def print_factors(x):
factors = []
for i in range(1, x + 1):
if x % i == 0:
factors.append(i)
return (factors)
# -----------------------------------------------
def calc(X, d, ans, D):
# print(X,d)
if len(X) == 0:
return
i = X.index(max(X))
ans[D[max(X)]] = d
Y = X[:i]
Z = X[i + 1:]
calc(Y, d + 1, ans, D)
calc(Z, d + 1, ans, D)
# ---------------------------------------
def factorization(n, l):
c = n
if prime(n) == True:
l.append(n)
return l
for i in range(2, c):
if n == 1:
break
while n % i == 0:
l.append(i)
n = n // i
return l
# endregion------------------------------
def good(b):
l = []
i = 0
while (len(b) != 0):
if b[i] < b[len(b) - 1 - i]:
l.append(b[i])
b.remove(b[i])
else:
l.append(b[len(b) - 1 - i])
b.remove(b[len(b) - 1 - i])
if l == sorted(l):
# print(l)
return True
return False
# arr=[]
# print(good(arr))
def generate(st, s):
if len(s) == 0:
return
# If current string is not already present.
if s not in st:
st.add(s)
# Traverse current string, one by one
# remove every character and recur.
for i in range(len(s)):
t = list(s).copy()
t.remove(s[i])
t = ''.join(t)
generate(st, t)
return
#=--------------------------------------------longest increasing subsequence
def largestincreasingsubsequence(A):
l = [1]*len(A)
sub=[]
for i in range(1,len(l)):
for k in range(i):
if A[k]<A[i]:
sub.append(l[k])
l[i]=1+max(sub,default=0)
return max(l,default=0)
#----------------------------------longest palindromic substring
# Python3 program for the
# above approach
# Function to calculate
# Bitwise OR of sums of
# all subsequences
def findOR(nums, N):
# Stores the prefix
# sum of nums[]
prefix_sum = 0
# Stores the bitwise OR of
# sum of each subsequence
result = 0
# Iterate through array nums[]
for i in range(N):
# Bits set in nums[i] are
# also set in result
result |= nums[i]
# Calculate prefix_sum
prefix_sum += nums[i]
# Bits set in prefix_sum
# are also set in result
result |= prefix_sum
# Return the result
return result
#l=[]
def OR(a, n):
ans = a[0]
for i in range(1, n):
ans |= a[i]
#l.append(ans)
return ans
#print(prime(12345678987766))
def main():
q=inpu()
x = q
v1 = 0
v2 = 0
i = 2
while i * i <= q:
while q % i == 0:
if v1!=0:
v2 = i
else:
v1 = i
q //= i
i += 1
if q - 1!=0:
v2 = q
if v1 * v2 - x!=0:
print(1)
print(v1 * v2)
else:
print(2)
if __name__ == '__main__':
main()
``` | output | 1 | 46,516 | 19 | 93,033 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You can't possibly imagine how cold our friends are this winter in Nvodsk! Two of them play the following game to warm up: initially a piece of paper has an integer q. During a move a player should write any integer number that is a non-trivial divisor of the last written number. Then he should run this number of circles around the hotel. Let us remind you that a number's divisor is called non-trivial if it is different from one and from the divided number itself.
The first person who can't make a move wins as he continues to lie in his warm bed under three blankets while the other one keeps running. Determine which player wins considering that both players play optimally. If the first player wins, print any winning first move.
Input
The first line contains the only integer q (1 ≤ q ≤ 1013).
Please do not use the %lld specificator to read or write 64-bit integers in С++. It is preferred to use the cin, cout streams or the %I64d specificator.
Output
In the first line print the number of the winning player (1 or 2). If the first player wins then the second line should contain another integer — his first move (if the first player can't even make the first move, print 0). If there are multiple solutions, print any of them.
Examples
Input
6
Output
2
Input
30
Output
1
6
Input
1
Output
1
0
Note
Number 6 has only two non-trivial divisors: 2 and 3. It is impossible to make a move after the numbers 2 and 3 are written, so both of them are winning, thus, number 6 is the losing number. A player can make a move and write number 6 after number 30; 6, as we know, is a losing number. Thus, this move will bring us the victory. | instruction | 0 | 46,517 | 19 | 93,034 |
Tags: games, math, number theory
Correct Solution:
```
def factorize(n):
p = 2
r = []
while p * p <= n:
while n % p == 0:
r.append(p)
n //= p
p += 1
if n > 1:
r.append(n)
return r
q = int(input())
f = factorize(q)
if len(f) <= 1:
print(1)
print(0)
elif len(f) == 2:
print(2)
else:
print(1)
print(f[0] * f[1])
``` | output | 1 | 46,517 | 19 | 93,035 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You can't possibly imagine how cold our friends are this winter in Nvodsk! Two of them play the following game to warm up: initially a piece of paper has an integer q. During a move a player should write any integer number that is a non-trivial divisor of the last written number. Then he should run this number of circles around the hotel. Let us remind you that a number's divisor is called non-trivial if it is different from one and from the divided number itself.
The first person who can't make a move wins as he continues to lie in his warm bed under three blankets while the other one keeps running. Determine which player wins considering that both players play optimally. If the first player wins, print any winning first move.
Input
The first line contains the only integer q (1 ≤ q ≤ 1013).
Please do not use the %lld specificator to read or write 64-bit integers in С++. It is preferred to use the cin, cout streams or the %I64d specificator.
Output
In the first line print the number of the winning player (1 or 2). If the first player wins then the second line should contain another integer — his first move (if the first player can't even make the first move, print 0). If there are multiple solutions, print any of them.
Examples
Input
6
Output
2
Input
30
Output
1
6
Input
1
Output
1
0
Note
Number 6 has only two non-trivial divisors: 2 and 3. It is impossible to make a move after the numbers 2 and 3 are written, so both of them are winning, thus, number 6 is the losing number. A player can make a move and write number 6 after number 30; 6, as we know, is a losing number. Thus, this move will bring us the victory. | instruction | 0 | 46,518 | 19 | 93,036 |
Tags: games, math, number theory
Correct Solution:
```
def prime_factors(n):
i = 2
factors = []
while i * i <= n:
if n % i:
i += 1
else:
n //= i
factors.append(i)
if n > 1:
factors.append(n)
return factors
class CodeforcesTask150ASolution:
def __init__(self):
self.result = ''
self.q = 0
def read_input(self):
self.q = int(input())
def process_task(self):
factors = prime_factors(self.q)
if len(factors) == 1 or not factors:
winner = 1
move = 0
elif len(factors) == 2:
winner = 2
else:
winner = 1
move = factors[0] * factors[1]
if winner == 1:
self.result = "{0}\n{1}".format(winner, move)
else:
self.result = "2"
def get_result(self):
return self.result
if __name__ == "__main__":
Solution = CodeforcesTask150ASolution()
Solution.read_input()
Solution.process_task()
print(Solution.get_result())
``` | output | 1 | 46,518 | 19 | 93,037 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You can't possibly imagine how cold our friends are this winter in Nvodsk! Two of them play the following game to warm up: initially a piece of paper has an integer q. During a move a player should write any integer number that is a non-trivial divisor of the last written number. Then he should run this number of circles around the hotel. Let us remind you that a number's divisor is called non-trivial if it is different from one and from the divided number itself.
The first person who can't make a move wins as he continues to lie in his warm bed under three blankets while the other one keeps running. Determine which player wins considering that both players play optimally. If the first player wins, print any winning first move.
Input
The first line contains the only integer q (1 ≤ q ≤ 1013).
Please do not use the %lld specificator to read or write 64-bit integers in С++. It is preferred to use the cin, cout streams or the %I64d specificator.
Output
In the first line print the number of the winning player (1 or 2). If the first player wins then the second line should contain another integer — his first move (if the first player can't even make the first move, print 0). If there are multiple solutions, print any of them.
Examples
Input
6
Output
2
Input
30
Output
1
6
Input
1
Output
1
0
Note
Number 6 has only two non-trivial divisors: 2 and 3. It is impossible to make a move after the numbers 2 and 3 are written, so both of them are winning, thus, number 6 is the losing number. A player can make a move and write number 6 after number 30; 6, as we know, is a losing number. Thus, this move will bring us the victory. | instruction | 0 | 46,519 | 19 | 93,038 |
Tags: games, math, number theory
Correct Solution:
```
def factors(n):
i=2
blanck=[]
while i**2<=n :
if n%i==0:
blanck.append(i)
n//=i
else:
i+=1
if n>1:
blanck.append(n)
n=1
return blanck
def cold(n):
if n==1:
print(1)
return 0
x=factors(n)
if len(x)==2:
return 2
if len(x)==1:
print(1)
return 0
print(1)
return x[0]*x[1]
print(cold(int(input())))
``` | output | 1 | 46,519 | 19 | 93,039 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You can't possibly imagine how cold our friends are this winter in Nvodsk! Two of them play the following game to warm up: initially a piece of paper has an integer q. During a move a player should write any integer number that is a non-trivial divisor of the last written number. Then he should run this number of circles around the hotel. Let us remind you that a number's divisor is called non-trivial if it is different from one and from the divided number itself.
The first person who can't make a move wins as he continues to lie in his warm bed under three blankets while the other one keeps running. Determine which player wins considering that both players play optimally. If the first player wins, print any winning first move.
Input
The first line contains the only integer q (1 ≤ q ≤ 1013).
Please do not use the %lld specificator to read or write 64-bit integers in С++. It is preferred to use the cin, cout streams or the %I64d specificator.
Output
In the first line print the number of the winning player (1 or 2). If the first player wins then the second line should contain another integer — his first move (if the first player can't even make the first move, print 0). If there are multiple solutions, print any of them.
Examples
Input
6
Output
2
Input
30
Output
1
6
Input
1
Output
1
0
Note
Number 6 has only two non-trivial divisors: 2 and 3. It is impossible to make a move after the numbers 2 and 3 are written, so both of them are winning, thus, number 6 is the losing number. A player can make a move and write number 6 after number 30; 6, as we know, is a losing number. Thus, this move will bring us the victory.
Submitted Solution:
```
import sys
from collections import defaultdict
import math
MAXNUM = math.inf
MINNUM = -1 * math.inf
ASCIILOWER = 97
ASCIIUPPER = 65
def getInt():
return int(sys.stdin.readline().rstrip())
def getInts():
return map(int, sys.stdin.readline().rstrip().split(" "))
def getString():
return sys.stdin.readline().rstrip()
def getDivisors(x):
if x == 1:
return [1]
if x == 2:
return [2]
f = []
MAX = math.sqrt(x) + 1
cur = 2
while cur <= x and cur < MAX:
if x % cur == 0:
f.append(cur)
x //= cur
else:
cur += 1
if len(f) > 2:
break
if x > 1:
f.append(x)
return f
def printOutput(ans):
if ans[0] == 1:
sys.stdout.write(str(ans[0]) + "\n")
sys.stdout.write(str(ans[1]) + "\n")
else:
sys.stdout.write(str(ans[0]) + "\n")
def solve(x):
factors = getDivisors(x)
if len(factors) == 1:
return 1, 0
elif len(factors) == 2:
return [2]
else:
ans = factors[0] * factors[1]
return 1, ans
def readinput():
x = getInt()
printOutput(solve(x))
readinput()
``` | instruction | 0 | 46,520 | 19 | 93,040 |
Yes | output | 1 | 46,520 | 19 | 93,041 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You can't possibly imagine how cold our friends are this winter in Nvodsk! Two of them play the following game to warm up: initially a piece of paper has an integer q. During a move a player should write any integer number that is a non-trivial divisor of the last written number. Then he should run this number of circles around the hotel. Let us remind you that a number's divisor is called non-trivial if it is different from one and from the divided number itself.
The first person who can't make a move wins as he continues to lie in his warm bed under three blankets while the other one keeps running. Determine which player wins considering that both players play optimally. If the first player wins, print any winning first move.
Input
The first line contains the only integer q (1 ≤ q ≤ 1013).
Please do not use the %lld specificator to read or write 64-bit integers in С++. It is preferred to use the cin, cout streams or the %I64d specificator.
Output
In the first line print the number of the winning player (1 or 2). If the first player wins then the second line should contain another integer — his first move (if the first player can't even make the first move, print 0). If there are multiple solutions, print any of them.
Examples
Input
6
Output
2
Input
30
Output
1
6
Input
1
Output
1
0
Note
Number 6 has only two non-trivial divisors: 2 and 3. It is impossible to make a move after the numbers 2 and 3 are written, so both of them are winning, thus, number 6 is the losing number. A player can make a move and write number 6 after number 30; 6, as we know, is a losing number. Thus, this move will bring us the victory.
Submitted Solution:
```
n = int(input())
p = n
arr = []
while p%2==0:
arr.append(2)
p = p//2
x = int(p**0.5)+1
for i in range(3,x,2):
while p%i==0:
arr.append(i)
p = p//i
if p>2:
arr.append(p)
if n==1 or len(arr)==1:
print(1)
print(0)
elif len(arr)==2:
print(2)
else:
x = arr[0]*arr[1]
print(1)
print(x)
``` | instruction | 0 | 46,521 | 19 | 93,042 |
Yes | output | 1 | 46,521 | 19 | 93,043 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You can't possibly imagine how cold our friends are this winter in Nvodsk! Two of them play the following game to warm up: initially a piece of paper has an integer q. During a move a player should write any integer number that is a non-trivial divisor of the last written number. Then he should run this number of circles around the hotel. Let us remind you that a number's divisor is called non-trivial if it is different from one and from the divided number itself.
The first person who can't make a move wins as he continues to lie in his warm bed under three blankets while the other one keeps running. Determine which player wins considering that both players play optimally. If the first player wins, print any winning first move.
Input
The first line contains the only integer q (1 ≤ q ≤ 1013).
Please do not use the %lld specificator to read or write 64-bit integers in С++. It is preferred to use the cin, cout streams or the %I64d specificator.
Output
In the first line print the number of the winning player (1 or 2). If the first player wins then the second line should contain another integer — his first move (if the first player can't even make the first move, print 0). If there are multiple solutions, print any of them.
Examples
Input
6
Output
2
Input
30
Output
1
6
Input
1
Output
1
0
Note
Number 6 has only two non-trivial divisors: 2 and 3. It is impossible to make a move after the numbers 2 and 3 are written, so both of them are winning, thus, number 6 is the losing number. A player can make a move and write number 6 after number 30; 6, as we know, is a losing number. Thus, this move will bring us the victory.
Submitted Solution:
```
q = int(input())
def is_prime(n):
if n % 2 == 0 and n != 2:
return False
for i in range(3, int(n**0.5)+1, 2):
if n % i == 0:
return False
return True
if is_prime(q):
print("1\n0")
else:
f = []
while q % 2 == 0:
q //= 2
f.append(2)
for x in range(3, int(q**0.5)+1, 2):
while q % x == 0:
q //= x
f.append(x)
if len(f) > 2:
break
else:
if q != 1:
f.append(q)
if len(f) == 2:
print(2)
else:
print(1, f[0]*f[1], sep="\n")
``` | instruction | 0 | 46,522 | 19 | 93,044 |
Yes | output | 1 | 46,522 | 19 | 93,045 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You can't possibly imagine how cold our friends are this winter in Nvodsk! Two of them play the following game to warm up: initially a piece of paper has an integer q. During a move a player should write any integer number that is a non-trivial divisor of the last written number. Then he should run this number of circles around the hotel. Let us remind you that a number's divisor is called non-trivial if it is different from one and from the divided number itself.
The first person who can't make a move wins as he continues to lie in his warm bed under three blankets while the other one keeps running. Determine which player wins considering that both players play optimally. If the first player wins, print any winning first move.
Input
The first line contains the only integer q (1 ≤ q ≤ 1013).
Please do not use the %lld specificator to read or write 64-bit integers in С++. It is preferred to use the cin, cout streams or the %I64d specificator.
Output
In the first line print the number of the winning player (1 or 2). If the first player wins then the second line should contain another integer — his first move (if the first player can't even make the first move, print 0). If there are multiple solutions, print any of them.
Examples
Input
6
Output
2
Input
30
Output
1
6
Input
1
Output
1
0
Note
Number 6 has only two non-trivial divisors: 2 and 3. It is impossible to make a move after the numbers 2 and 3 are written, so both of them are winning, thus, number 6 is the losing number. A player can make a move and write number 6 after number 30; 6, as we know, is a losing number. Thus, this move will bring us the victory.
Submitted Solution:
```
import sys
from math import log2,floor,ceil,sqrt
# import bisect
# from collections import deque
Ri = lambda : [int(x) for x in sys.stdin.readline().split()]
ri = lambda : sys.stdin.readline().strip()
def input(): return sys.stdin.readline().strip()
def list2d(a, b, c): return [[c] * b for i in range(a)]
def list3d(a, b, c, d): return [[[d] * c for j in range(b)] for i in range(a)]
def list4d(a, b, c, d, e): return [[[[e] * d for j in range(c)] for j in range(b)] for i in range(a)]
def ceil(x, y=1): return int(-(-x // y))
def INT(): return int(input())
def MAP(): return map(int, input().split())
def LIST(N=None): return list(MAP()) if N is None else [INT() for i in range(N)]
def Yes(): print('Yes')
def No(): print('No')
def YES(): print('YES')
def NO(): print('NO')
INF = 10 ** 18
MOD = 10**9+7
def solve(n):
cnt = 0
tp = 0
ans = 1
while n %2 == 0:
n = n//2
cnt+=1
if cnt >= 2:
return 4
tp = 1 if cnt >= 1 else 0
ans = 2 if tp == 1 else 1
p = 3
while p*p <= n:
cnt = 0
while n%p == 0:
# print(p)
n = n//p
cnt+=1
# print(n)
if cnt >= 2 :
return p**2
if cnt == 1:
tp+=1
ans = ans*p
if tp >=2:
return ans
p+=2
if n != 1:
tp+=1
ans = ans*n
if tp >=2:
return ans
return -1
n = int(ri())
val = n
cnt = 0
ret = solve(n)
# print(ret)
if ret == -1:
print(1)
print(0)
else:
if ret == val:
print(2)
else:
print(1)
print(ret)
``` | instruction | 0 | 46,523 | 19 | 93,046 |
Yes | output | 1 | 46,523 | 19 | 93,047 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You can't possibly imagine how cold our friends are this winter in Nvodsk! Two of them play the following game to warm up: initially a piece of paper has an integer q. During a move a player should write any integer number that is a non-trivial divisor of the last written number. Then he should run this number of circles around the hotel. Let us remind you that a number's divisor is called non-trivial if it is different from one and from the divided number itself.
The first person who can't make a move wins as he continues to lie in his warm bed under three blankets while the other one keeps running. Determine which player wins considering that both players play optimally. If the first player wins, print any winning first move.
Input
The first line contains the only integer q (1 ≤ q ≤ 1013).
Please do not use the %lld specificator to read or write 64-bit integers in С++. It is preferred to use the cin, cout streams or the %I64d specificator.
Output
In the first line print the number of the winning player (1 or 2). If the first player wins then the second line should contain another integer — his first move (if the first player can't even make the first move, print 0). If there are multiple solutions, print any of them.
Examples
Input
6
Output
2
Input
30
Output
1
6
Input
1
Output
1
0
Note
Number 6 has only two non-trivial divisors: 2 and 3. It is impossible to make a move after the numbers 2 and 3 are written, so both of them are winning, thus, number 6 is the losing number. A player can make a move and write number 6 after number 30; 6, as we know, is a losing number. Thus, this move will bring us the victory.
Submitted Solution:
```
import sys
from functools import lru_cache, cmp_to_key
from heapq import merge, heapify, heappop, heappush
from math import *
from collections import defaultdict as dd, deque, Counter as C
from itertools import combinations as comb, permutations as perm
from bisect import bisect_left as bl, bisect_right as br, bisect
from time import perf_counter
from fractions import Fraction
# import numpy as np
# sys.setrecursionlimit(int(pow(10,6)))
# sys.stdout = open("out.txt", "w")
mod = int(pow(10, 9) + 7)
mod2 = 998244353
def data(): return sys.stdin.readline().strip()
def out(*var, end="\n"): sys.stdout.write(' '.join(map(str, var))+end)
def L(): return list(sp())
def sl(): return list(ssp())
def sp(): return map(int, data().split())
def ssp(): return map(str, data().split())
def l1d(n, val=0): return [val for i in range(n)]
def l2d(n, m, val=0): return [l1d(n, val) for j in range(m)]
try:
sys.stdin = open("input.txt", "r")
except:
pass
# @lru_cache(None)
n=L()[0]
div=[]
grt=0
x=n
for i in range(2,int(n**0.5)+2):
if n%i==0 and x!=i:
k=0
while(n%i==0):
k+=1
n//=i
if k>2:
grt=1
break
div.append([i,k])
if len(div)==0 or grt or len(div)>=3:
print(1)
if len(div)==0:
print(0)
elif grt:
print(i*i)
else:
print(div[0][0]*div[1][0])
exit()
else:
print(2)
``` | instruction | 0 | 46,524 | 19 | 93,048 |
No | output | 1 | 46,524 | 19 | 93,049 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You can't possibly imagine how cold our friends are this winter in Nvodsk! Two of them play the following game to warm up: initially a piece of paper has an integer q. During a move a player should write any integer number that is a non-trivial divisor of the last written number. Then he should run this number of circles around the hotel. Let us remind you that a number's divisor is called non-trivial if it is different from one and from the divided number itself.
The first person who can't make a move wins as he continues to lie in his warm bed under three blankets while the other one keeps running. Determine which player wins considering that both players play optimally. If the first player wins, print any winning first move.
Input
The first line contains the only integer q (1 ≤ q ≤ 1013).
Please do not use the %lld specificator to read or write 64-bit integers in С++. It is preferred to use the cin, cout streams or the %I64d specificator.
Output
In the first line print the number of the winning player (1 or 2). If the first player wins then the second line should contain another integer — his first move (if the first player can't even make the first move, print 0). If there are multiple solutions, print any of them.
Examples
Input
6
Output
2
Input
30
Output
1
6
Input
1
Output
1
0
Note
Number 6 has only two non-trivial divisors: 2 and 3. It is impossible to make a move after the numbers 2 and 3 are written, so both of them are winning, thus, number 6 is the losing number. A player can make a move and write number 6 after number 30; 6, as we know, is a losing number. Thus, this move will bring us the victory.
Submitted Solution:
```
from collections import Counter
n = int(input())
if n == 1:
print(1)
print(0)
exit()
norig = n
i = 2
factors = Counter()
while i * i <= n:
while n % i == 0:
factors[i] += 1
n //= i
i += 1
if n != 1:
factors[n] += 1
m = sum(factors.values())
if m <= 2 and factors[norig] != 1:
print('2')
else:
print('1')
ans = 1
for k, v in factors.items():
for _ in range(v):
if m > 1:
ans *= k
m -= 1
else:
print(ans)
exit()
print(ans)
``` | instruction | 0 | 46,525 | 19 | 93,050 |
No | output | 1 | 46,525 | 19 | 93,051 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You can't possibly imagine how cold our friends are this winter in Nvodsk! Two of them play the following game to warm up: initially a piece of paper has an integer q. During a move a player should write any integer number that is a non-trivial divisor of the last written number. Then he should run this number of circles around the hotel. Let us remind you that a number's divisor is called non-trivial if it is different from one and from the divided number itself.
The first person who can't make a move wins as he continues to lie in his warm bed under three blankets while the other one keeps running. Determine which player wins considering that both players play optimally. If the first player wins, print any winning first move.
Input
The first line contains the only integer q (1 ≤ q ≤ 1013).
Please do not use the %lld specificator to read or write 64-bit integers in С++. It is preferred to use the cin, cout streams or the %I64d specificator.
Output
In the first line print the number of the winning player (1 or 2). If the first player wins then the second line should contain another integer — his first move (if the first player can't even make the first move, print 0). If there are multiple solutions, print any of them.
Examples
Input
6
Output
2
Input
30
Output
1
6
Input
1
Output
1
0
Note
Number 6 has only two non-trivial divisors: 2 and 3. It is impossible to make a move after the numbers 2 and 3 are written, so both of them are winning, thus, number 6 is the losing number. A player can make a move and write number 6 after number 30; 6, as we know, is a losing number. Thus, this move will bring us the victory.
Submitted Solution:
```
q=int(input())
u=[True for i in range(4*(10**6))]
u[0]=False
al=1
while(al<=2*(10**3)):
for j in range(2*(al+1)-1, len(u), al+1):
u[j]=False
pro=al+1
while(pro<=2*(10**3) and u[pro]==False):
pro+=1
al=pro
l=[]
for tro in range(len(u)):
if(u[tro]==True):
l.append(tro+1)
flag=False
cnt=0
fact=[]
def primecheck(n):
i=2
while(i*i<n):
if(n%i==0):
return False
i+=1
if(i*i==n):
return False
return True
for prime in l:
if(q%prime==0):
cnt+=1
fact.append(prime)
if(primecheck(q//prime) and (q//prime)>4*(10**6)):
fact.append(q//prime)
cnt+=1
if((q//prime)%prime==0):
if(q!=prime*prime):
print(1)
print(prime**2)
else:
print(2)
flag=True
break
if(q==prime):
print(1)
print(0)
flag=True
break
if(flag==False):
if(cnt<=1):
print(1)
print(0)
elif(cnt==2):
print(2)
else:
print(1)
print(fact[0]*fact[1])
``` | instruction | 0 | 46,526 | 19 | 93,052 |
No | output | 1 | 46,526 | 19 | 93,053 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You can't possibly imagine how cold our friends are this winter in Nvodsk! Two of them play the following game to warm up: initially a piece of paper has an integer q. During a move a player should write any integer number that is a non-trivial divisor of the last written number. Then he should run this number of circles around the hotel. Let us remind you that a number's divisor is called non-trivial if it is different from one and from the divided number itself.
The first person who can't make a move wins as he continues to lie in his warm bed under three blankets while the other one keeps running. Determine which player wins considering that both players play optimally. If the first player wins, print any winning first move.
Input
The first line contains the only integer q (1 ≤ q ≤ 1013).
Please do not use the %lld specificator to read or write 64-bit integers in С++. It is preferred to use the cin, cout streams or the %I64d specificator.
Output
In the first line print the number of the winning player (1 or 2). If the first player wins then the second line should contain another integer — his first move (if the first player can't even make the first move, print 0). If there are multiple solutions, print any of them.
Examples
Input
6
Output
2
Input
30
Output
1
6
Input
1
Output
1
0
Note
Number 6 has only two non-trivial divisors: 2 and 3. It is impossible to make a move after the numbers 2 and 3 are written, so both of them are winning, thus, number 6 is the losing number. A player can make a move and write number 6 after number 30; 6, as we know, is a losing number. Thus, this move will bring us the victory.
Submitted Solution:
```
#------------------------template--------------------------#
import os
import sys
from math import *
from collections import *
from fractions import *
from bisect import *
from heapq import*
from io import BytesIO, IOBase
def vsInput():
sys.stdin = open('input.txt', 'r')
sys.stdout = open('output.txt', 'w')
BUFSIZE = 8192
class FastIO(IOBase):
newlines = 0
def __init__(self, file):
self._fd = file.fileno()
self.buffer = BytesIO()
self.writable = "x" in file.mode or "r" not in file.mode
self.write = self.buffer.write if self.writable else None
def read(self):
while True:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
if not b:
break
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines = 0
return self.buffer.read()
def readline(self):
while self.newlines == 0:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
self.newlines = b.count(b"\n") + (not b)
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines -= 1
return self.buffer.readline()
def flush(self):
if self.writable:
os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
self.write = lambda s: self.buffer.write(s.encode("ascii"))
self.read = lambda: self.buffer.read().decode("ascii")
self.readline = lambda: self.buffer.readline().decode("ascii")
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")
ALPHA='abcdefghijklmnopqrstuvwxyz'
MOD=1000000007
def value():return tuple(map(int,input().split()))
def array():return [int(i) for i in input().split()]
def Int():return int(input())
def Str():return input()
def arrayS():return [i for i in input().split()]
#-------------------------code---------------------------#
# vsInput()
def primeN(n):
prime = [True for i in range(n+1)]
prime[0]=False
prime[1]=False
p=2
while(p*p<=n):
if(prime[p]):
for i in range(p*p,n+1,p):
prime[i]=False
p+=1
return set([p for p in range(n+1) if(prime[p])])
def divisors(n):
div=set()
for i in range(1,int(sqrt(n))+1):
if(n%i==0):
div.add(i)
return div
primes=primeN(1000000)
n=Int()
ans=0
div=divisors(n)
ok=False
for i in div:
for p in primes:
if(i%p==0 and i//p in primes):
ans=i
ok=True
break
if(ok):
break
if(ans==n):
print(2)
else:
print(1)
print(ans)
``` | instruction | 0 | 46,527 | 19 | 93,054 |
No | output | 1 | 46,527 | 19 | 93,055 |
Provide a correct Python 3 solution for this coding contest problem.
Takahashi will play a game using a piece on an array of squares numbered 1, 2, \cdots, N. Square i has an integer C_i written on it. Also, he is given a permutation of 1, 2, \cdots, N: P_1, P_2, \cdots, P_N.
Now, he will choose one square and place the piece on that square. Then, he will make the following move some number of times between 1 and K (inclusive):
* In one move, if the piece is now on Square i (1 \leq i \leq N), move it to Square P_i. Here, his score increases by C_{P_i}.
Help him by finding the maximum possible score at the end of the game. (The score is 0 at the beginning of the game.)
Constraints
* 2 \leq N \leq 5000
* 1 \leq K \leq 10^9
* 1 \leq P_i \leq N
* P_i \neq i
* P_1, P_2, \cdots, P_N are all different.
* -10^9 \leq C_i \leq 10^9
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
N K
P_1 P_2 \cdots P_N
C_1 C_2 \cdots C_N
Output
Print the maximum possible score at the end of the game.
Examples
Input
5 2
2 4 5 1 3
3 4 -10 -8 8
Output
8
Input
2 3
2 1
10 -7
Output
13
Input
3 3
3 1 2
-1000 -2000 -3000
Output
-1000
Input
10 58
9 1 6 7 8 4 3 2 10 5
695279662 988782657 -119067776 382975538 -151885171 -177220596 -169777795 37619092 389386780 980092719
Output
29507023469 | instruction | 0 | 46,828 | 19 | 93,656 |
"Correct Solution:
```
n, k = map(int, input().split())
P = [*map(lambda x: int(x)-1, input().split())]
C = [*map(int, input().split())]
score = -10**18
for st in range(n):
lap_cn = 0
lap_sc = 0
nx = st
while True:
lap_cn += 1
lap_sc += C[nx]
nx = P[nx]
if nx == st: break
lap_sc = max(0, lap_sc)
sum_sc = 0
for k_cn in range(k):
sum_sc += C[nx]
score = max(score, sum_sc + lap_sc * ((k - (k_cn + 1)) // lap_cn))
nx = P[nx]
if nx == st: break
print(score)
``` | output | 1 | 46,828 | 19 | 93,657 |
Provide a correct Python 3 solution for this coding contest problem.
Takahashi will play a game using a piece on an array of squares numbered 1, 2, \cdots, N. Square i has an integer C_i written on it. Also, he is given a permutation of 1, 2, \cdots, N: P_1, P_2, \cdots, P_N.
Now, he will choose one square and place the piece on that square. Then, he will make the following move some number of times between 1 and K (inclusive):
* In one move, if the piece is now on Square i (1 \leq i \leq N), move it to Square P_i. Here, his score increases by C_{P_i}.
Help him by finding the maximum possible score at the end of the game. (The score is 0 at the beginning of the game.)
Constraints
* 2 \leq N \leq 5000
* 1 \leq K \leq 10^9
* 1 \leq P_i \leq N
* P_i \neq i
* P_1, P_2, \cdots, P_N are all different.
* -10^9 \leq C_i \leq 10^9
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
N K
P_1 P_2 \cdots P_N
C_1 C_2 \cdots C_N
Output
Print the maximum possible score at the end of the game.
Examples
Input
5 2
2 4 5 1 3
3 4 -10 -8 8
Output
8
Input
2 3
2 1
10 -7
Output
13
Input
3 3
3 1 2
-1000 -2000 -3000
Output
-1000
Input
10 58
9 1 6 7 8 4 3 2 10 5
695279662 988782657 -119067776 382975538 -151885171 -177220596 -169777795 37619092 389386780 980092719
Output
29507023469 | instruction | 0 | 46,829 | 19 | 93,658 |
"Correct Solution:
```
N, K = map(int, input().split())
P = list(map(int, input().split()))
C = list(map(int, input().split()))
ans = -float("inf")
for start in range(N):
x = start
cyc_score = []
total_cyc_score = 0
while True:
x = P[x]-1
cyc_score.append(C[x])
total_cyc_score += C[x]
if x == start: # 周期を検出
break
t = 0
for i in range(len(cyc_score)):
t += cyc_score[i]
if i + 1 > K: break
now_score = t
if total_cyc_score > 0:
e = int((K - i - 1) / len(cyc_score))
now_score += total_cyc_score * e
ans = max(ans, now_score)
print(ans)
``` | output | 1 | 46,829 | 19 | 93,659 |
Provide a correct Python 3 solution for this coding contest problem.
Takahashi will play a game using a piece on an array of squares numbered 1, 2, \cdots, N. Square i has an integer C_i written on it. Also, he is given a permutation of 1, 2, \cdots, N: P_1, P_2, \cdots, P_N.
Now, he will choose one square and place the piece on that square. Then, he will make the following move some number of times between 1 and K (inclusive):
* In one move, if the piece is now on Square i (1 \leq i \leq N), move it to Square P_i. Here, his score increases by C_{P_i}.
Help him by finding the maximum possible score at the end of the game. (The score is 0 at the beginning of the game.)
Constraints
* 2 \leq N \leq 5000
* 1 \leq K \leq 10^9
* 1 \leq P_i \leq N
* P_i \neq i
* P_1, P_2, \cdots, P_N are all different.
* -10^9 \leq C_i \leq 10^9
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
N K
P_1 P_2 \cdots P_N
C_1 C_2 \cdots C_N
Output
Print the maximum possible score at the end of the game.
Examples
Input
5 2
2 4 5 1 3
3 4 -10 -8 8
Output
8
Input
2 3
2 1
10 -7
Output
13
Input
3 3
3 1 2
-1000 -2000 -3000
Output
-1000
Input
10 58
9 1 6 7 8 4 3 2 10 5
695279662 988782657 -119067776 382975538 -151885171 -177220596 -169777795 37619092 389386780 980092719
Output
29507023469 | instruction | 0 | 46,830 | 19 | 93,660 |
"Correct Solution:
```
N, K = map(int, input().split())
P = list(map(lambda x: int(x)-1, input().split()))
C = list(map(int, input().split()))
INF = 10**18
ans = -INF
for i in range(N):
arr = []
tot = 0
cnt = 0
nex = i
while cnt < K:
nex = P[nex]
tot += C[nex]
arr.append(tot)
cnt += 1
if nex == i:
break
if cnt < K:
cycle = cnt
cycle_score = tot
loop, amari = K//cycle, K%cycle
for j in range(cycle):
if j < amari:
arr.append(cycle_score*loop+arr[j])
else:
arr.append(cycle_score*(loop-1)+arr[j])
ans = max(ans, max(arr))
print(ans)
``` | output | 1 | 46,830 | 19 | 93,661 |
Provide a correct Python 3 solution for this coding contest problem.
Takahashi will play a game using a piece on an array of squares numbered 1, 2, \cdots, N. Square i has an integer C_i written on it. Also, he is given a permutation of 1, 2, \cdots, N: P_1, P_2, \cdots, P_N.
Now, he will choose one square and place the piece on that square. Then, he will make the following move some number of times between 1 and K (inclusive):
* In one move, if the piece is now on Square i (1 \leq i \leq N), move it to Square P_i. Here, his score increases by C_{P_i}.
Help him by finding the maximum possible score at the end of the game. (The score is 0 at the beginning of the game.)
Constraints
* 2 \leq N \leq 5000
* 1 \leq K \leq 10^9
* 1 \leq P_i \leq N
* P_i \neq i
* P_1, P_2, \cdots, P_N are all different.
* -10^9 \leq C_i \leq 10^9
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
N K
P_1 P_2 \cdots P_N
C_1 C_2 \cdots C_N
Output
Print the maximum possible score at the end of the game.
Examples
Input
5 2
2 4 5 1 3
3 4 -10 -8 8
Output
8
Input
2 3
2 1
10 -7
Output
13
Input
3 3
3 1 2
-1000 -2000 -3000
Output
-1000
Input
10 58
9 1 6 7 8 4 3 2 10 5
695279662 988782657 -119067776 382975538 -151885171 -177220596 -169777795 37619092 389386780 980092719
Output
29507023469 | instruction | 0 | 46,831 | 19 | 93,662 |
"Correct Solution:
```
(n,k),p,c,*a=[[*map(int,t.split())]for t in open(0)]
while n:
n-=1;s,x,*l=0,n;f=j=k
while f:x=p[x]-1;s+=c[x];l+=s,;f=x!=n
for t in l[:k]:j-=1;a+=[t+j//len(l)*s*(s>0)]
print(max(a))
``` | output | 1 | 46,831 | 19 | 93,663 |
Provide a correct Python 3 solution for this coding contest problem.
Takahashi will play a game using a piece on an array of squares numbered 1, 2, \cdots, N. Square i has an integer C_i written on it. Also, he is given a permutation of 1, 2, \cdots, N: P_1, P_2, \cdots, P_N.
Now, he will choose one square and place the piece on that square. Then, he will make the following move some number of times between 1 and K (inclusive):
* In one move, if the piece is now on Square i (1 \leq i \leq N), move it to Square P_i. Here, his score increases by C_{P_i}.
Help him by finding the maximum possible score at the end of the game. (The score is 0 at the beginning of the game.)
Constraints
* 2 \leq N \leq 5000
* 1 \leq K \leq 10^9
* 1 \leq P_i \leq N
* P_i \neq i
* P_1, P_2, \cdots, P_N are all different.
* -10^9 \leq C_i \leq 10^9
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
N K
P_1 P_2 \cdots P_N
C_1 C_2 \cdots C_N
Output
Print the maximum possible score at the end of the game.
Examples
Input
5 2
2 4 5 1 3
3 4 -10 -8 8
Output
8
Input
2 3
2 1
10 -7
Output
13
Input
3 3
3 1 2
-1000 -2000 -3000
Output
-1000
Input
10 58
9 1 6 7 8 4 3 2 10 5
695279662 988782657 -119067776 382975538 -151885171 -177220596 -169777795 37619092 389386780 980092719
Output
29507023469 | instruction | 0 | 46,832 | 19 | 93,664 |
"Correct Solution:
```
n, k = map(int, input().split())
p = list(map(int, input().split()))
c = list(map(int, input().split()))
for i in range(n):
p[i] -= 1
def score(a, k):
now = p[a]
s = c[now]
cnt = 1
ans = s
while now != a and cnt < k:
now = p[now]
s += c[now]
ans = max(ans, s)
cnt += 1
if cnt == k:
return ans
if k % cnt == 0:
return (k // cnt - 1) * s + ans
s = (k // cnt) * s
k %= cnt
cnt = 0
while cnt < k:
now = p[now]
s += c[now]
ans = max(ans, s)
cnt += 1
return ans
print(max(score(i, k) for i in range(n)))
``` | output | 1 | 46,832 | 19 | 93,665 |
Provide a correct Python 3 solution for this coding contest problem.
Takahashi will play a game using a piece on an array of squares numbered 1, 2, \cdots, N. Square i has an integer C_i written on it. Also, he is given a permutation of 1, 2, \cdots, N: P_1, P_2, \cdots, P_N.
Now, he will choose one square and place the piece on that square. Then, he will make the following move some number of times between 1 and K (inclusive):
* In one move, if the piece is now on Square i (1 \leq i \leq N), move it to Square P_i. Here, his score increases by C_{P_i}.
Help him by finding the maximum possible score at the end of the game. (The score is 0 at the beginning of the game.)
Constraints
* 2 \leq N \leq 5000
* 1 \leq K \leq 10^9
* 1 \leq P_i \leq N
* P_i \neq i
* P_1, P_2, \cdots, P_N are all different.
* -10^9 \leq C_i \leq 10^9
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
N K
P_1 P_2 \cdots P_N
C_1 C_2 \cdots C_N
Output
Print the maximum possible score at the end of the game.
Examples
Input
5 2
2 4 5 1 3
3 4 -10 -8 8
Output
8
Input
2 3
2 1
10 -7
Output
13
Input
3 3
3 1 2
-1000 -2000 -3000
Output
-1000
Input
10 58
9 1 6 7 8 4 3 2 10 5
695279662 988782657 -119067776 382975538 -151885171 -177220596 -169777795 37619092 389386780 980092719
Output
29507023469 | instruction | 0 | 46,833 | 19 | 93,666 |
"Correct Solution:
```
n, k = map(int, input().split())
P = [*map(lambda x: int(x)-1, input().split())]
C = [*map(int, input().split())]
score = -10**18
for st in range(n):
lap_sc = sum_sc = 0
nx = st
for lap_cn in range(1, n+1):
lap_sc += C[nx]
nx = P[nx]
if nx == st: break
lap_sc = max(0, lap_sc)
for k_cn in range(1, k+1):
sum_sc += C[nx]
score = max(score, sum_sc + lap_sc * ((k - k_cn) // lap_cn))
nx = P[nx]
if nx == st: break
print(score)
``` | output | 1 | 46,833 | 19 | 93,667 |
Provide a correct Python 3 solution for this coding contest problem.
Takahashi will play a game using a piece on an array of squares numbered 1, 2, \cdots, N. Square i has an integer C_i written on it. Also, he is given a permutation of 1, 2, \cdots, N: P_1, P_2, \cdots, P_N.
Now, he will choose one square and place the piece on that square. Then, he will make the following move some number of times between 1 and K (inclusive):
* In one move, if the piece is now on Square i (1 \leq i \leq N), move it to Square P_i. Here, his score increases by C_{P_i}.
Help him by finding the maximum possible score at the end of the game. (The score is 0 at the beginning of the game.)
Constraints
* 2 \leq N \leq 5000
* 1 \leq K \leq 10^9
* 1 \leq P_i \leq N
* P_i \neq i
* P_1, P_2, \cdots, P_N are all different.
* -10^9 \leq C_i \leq 10^9
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
N K
P_1 P_2 \cdots P_N
C_1 C_2 \cdots C_N
Output
Print the maximum possible score at the end of the game.
Examples
Input
5 2
2 4 5 1 3
3 4 -10 -8 8
Output
8
Input
2 3
2 1
10 -7
Output
13
Input
3 3
3 1 2
-1000 -2000 -3000
Output
-1000
Input
10 58
9 1 6 7 8 4 3 2 10 5
695279662 988782657 -119067776 382975538 -151885171 -177220596 -169777795 37619092 389386780 980092719
Output
29507023469 | instruction | 0 | 46,834 | 19 | 93,668 |
"Correct Solution:
```
n, k = map(int, input().split())
p = list(map(lambda x: int(x) - 1, input().split()))
c = list(map(int, input().split()))
ans = max(c)
for i in range(n):
curr = 0
score = [0]
posi = i
while p[posi] != i:
posi = p[posi]
curr += c[posi]
score.append(curr)
cycle_score = score[-1] + c[i]
cycle_len = len(score)
for j in range(cycle_len):
if j <= k and j != 0:
cycles = (k - j)//cycle_len
thing = score[j] + max(0, cycle_score) * cycles
ans = max(ans, thing)
#Kがサイクルより大きいときの処理をj=0で行う(score[0]は無効なため)
elif j == 0 and k >= cycle_len:
thing = cycle_score * k // cycle_len
ans = max(ans,thing)
print(ans)
``` | output | 1 | 46,834 | 19 | 93,669 |
Provide a correct Python 3 solution for this coding contest problem.
Takahashi will play a game using a piece on an array of squares numbered 1, 2, \cdots, N. Square i has an integer C_i written on it. Also, he is given a permutation of 1, 2, \cdots, N: P_1, P_2, \cdots, P_N.
Now, he will choose one square and place the piece on that square. Then, he will make the following move some number of times between 1 and K (inclusive):
* In one move, if the piece is now on Square i (1 \leq i \leq N), move it to Square P_i. Here, his score increases by C_{P_i}.
Help him by finding the maximum possible score at the end of the game. (The score is 0 at the beginning of the game.)
Constraints
* 2 \leq N \leq 5000
* 1 \leq K \leq 10^9
* 1 \leq P_i \leq N
* P_i \neq i
* P_1, P_2, \cdots, P_N are all different.
* -10^9 \leq C_i \leq 10^9
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
N K
P_1 P_2 \cdots P_N
C_1 C_2 \cdots C_N
Output
Print the maximum possible score at the end of the game.
Examples
Input
5 2
2 4 5 1 3
3 4 -10 -8 8
Output
8
Input
2 3
2 1
10 -7
Output
13
Input
3 3
3 1 2
-1000 -2000 -3000
Output
-1000
Input
10 58
9 1 6 7 8 4 3 2 10 5
695279662 988782657 -119067776 382975538 -151885171 -177220596 -169777795 37619092 389386780 980092719
Output
29507023469 | instruction | 0 | 46,835 | 19 | 93,670 |
"Correct Solution:
```
N, K = map(int, input().split())
*P, = map(int, input().split())
*C, = map(int, input().split())
ans = -float("inf")
for pos in range(N):
used = [-1] * N
cost = [0] * (N+1)
for k in range(N+1):
if k:
cost[k] = cost[k-1] + C[pos]
if used[pos] >= 0:
loop_size = k-used[pos]
break
used[pos] = k
pos = P[pos] - 1
if cost[loop_size] <= 0:
ans = max(ans, max(cost[1:K+1]))
else:
a, b = divmod(K, loop_size)
v1 = cost[loop_size] * (a-1) + max(cost[:K+1])
v2 = cost[loop_size] * a + max(cost[:b+1])
ans = max(ans, max(v1, v2) if a else v2)
print(ans)
``` | output | 1 | 46,835 | 19 | 93,671 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Takahashi will play a game using a piece on an array of squares numbered 1, 2, \cdots, N. Square i has an integer C_i written on it. Also, he is given a permutation of 1, 2, \cdots, N: P_1, P_2, \cdots, P_N.
Now, he will choose one square and place the piece on that square. Then, he will make the following move some number of times between 1 and K (inclusive):
* In one move, if the piece is now on Square i (1 \leq i \leq N), move it to Square P_i. Here, his score increases by C_{P_i}.
Help him by finding the maximum possible score at the end of the game. (The score is 0 at the beginning of the game.)
Constraints
* 2 \leq N \leq 5000
* 1 \leq K \leq 10^9
* 1 \leq P_i \leq N
* P_i \neq i
* P_1, P_2, \cdots, P_N are all different.
* -10^9 \leq C_i \leq 10^9
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
N K
P_1 P_2 \cdots P_N
C_1 C_2 \cdots C_N
Output
Print the maximum possible score at the end of the game.
Examples
Input
5 2
2 4 5 1 3
3 4 -10 -8 8
Output
8
Input
2 3
2 1
10 -7
Output
13
Input
3 3
3 1 2
-1000 -2000 -3000
Output
-1000
Input
10 58
9 1 6 7 8 4 3 2 10 5
695279662 988782657 -119067776 382975538 -151885171 -177220596 -169777795 37619092 389386780 980092719
Output
29507023469
Submitted Solution:
```
from collections import deque
n, k = map(int, input().split())
p = list(map(int, input().split()))
c = list(map(int, input().split()))
ans = -1000000000
for i in range(n):
now = i
j = 0
score = deque([0])
while j < k:
score.append(score[j] +c[p[now]-1])
now = p[now]-1
j += 1
if now == i:
break
ans = max(ans, max(list(score)[1:]))
if j < k and score[-1] > 0:
loop_num = k//j
rest = k%j
ans = max(ans, max(list(score)[:rest+1])+loop_num*score[-1] )
ans = max(ans, max(list(score))+(loop_num-1)*score[-1] )
print(ans)
``` | instruction | 0 | 46,836 | 19 | 93,672 |
Yes | output | 1 | 46,836 | 19 | 93,673 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Takahashi will play a game using a piece on an array of squares numbered 1, 2, \cdots, N. Square i has an integer C_i written on it. Also, he is given a permutation of 1, 2, \cdots, N: P_1, P_2, \cdots, P_N.
Now, he will choose one square and place the piece on that square. Then, he will make the following move some number of times between 1 and K (inclusive):
* In one move, if the piece is now on Square i (1 \leq i \leq N), move it to Square P_i. Here, his score increases by C_{P_i}.
Help him by finding the maximum possible score at the end of the game. (The score is 0 at the beginning of the game.)
Constraints
* 2 \leq N \leq 5000
* 1 \leq K \leq 10^9
* 1 \leq P_i \leq N
* P_i \neq i
* P_1, P_2, \cdots, P_N are all different.
* -10^9 \leq C_i \leq 10^9
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
N K
P_1 P_2 \cdots P_N
C_1 C_2 \cdots C_N
Output
Print the maximum possible score at the end of the game.
Examples
Input
5 2
2 4 5 1 3
3 4 -10 -8 8
Output
8
Input
2 3
2 1
10 -7
Output
13
Input
3 3
3 1 2
-1000 -2000 -3000
Output
-1000
Input
10 58
9 1 6 7 8 4 3 2 10 5
695279662 988782657 -119067776 382975538 -151885171 -177220596 -169777795 37619092 389386780 980092719
Output
29507023469
Submitted Solution:
```
from itertools import accumulate
#acumulate [a[0], a[0]+a[1], a[0]+...]
n, k = map(int, input().split())
p = [int(x) - 1 for x in input().split()]
c = list(map(int, input().split()))
ans = -(10 ** 18)
for i in range(n):
pos = i
scores = []
while True:
pos = p[pos]
scores.append(c[pos])
if pos == i:
break
m = len(scores)
s = list(accumulate(scores))
for j in range(min(m, k)):
x = (k - j - 1) // m
ans = max(ans, s[j], s[j] + s[-1] * x)
print(ans)
``` | instruction | 0 | 46,837 | 19 | 93,674 |
Yes | output | 1 | 46,837 | 19 | 93,675 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Takahashi will play a game using a piece on an array of squares numbered 1, 2, \cdots, N. Square i has an integer C_i written on it. Also, he is given a permutation of 1, 2, \cdots, N: P_1, P_2, \cdots, P_N.
Now, he will choose one square and place the piece on that square. Then, he will make the following move some number of times between 1 and K (inclusive):
* In one move, if the piece is now on Square i (1 \leq i \leq N), move it to Square P_i. Here, his score increases by C_{P_i}.
Help him by finding the maximum possible score at the end of the game. (The score is 0 at the beginning of the game.)
Constraints
* 2 \leq N \leq 5000
* 1 \leq K \leq 10^9
* 1 \leq P_i \leq N
* P_i \neq i
* P_1, P_2, \cdots, P_N are all different.
* -10^9 \leq C_i \leq 10^9
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
N K
P_1 P_2 \cdots P_N
C_1 C_2 \cdots C_N
Output
Print the maximum possible score at the end of the game.
Examples
Input
5 2
2 4 5 1 3
3 4 -10 -8 8
Output
8
Input
2 3
2 1
10 -7
Output
13
Input
3 3
3 1 2
-1000 -2000 -3000
Output
-1000
Input
10 58
9 1 6 7 8 4 3 2 10 5
695279662 988782657 -119067776 382975538 -151885171 -177220596 -169777795 37619092 389386780 980092719
Output
29507023469
Submitted Solution:
```
n, k = map(int, input().split())
p = list(map(int, input().split()))
c = list(map(int, input().split()))
l = [[] for i in range(n)]
for i in range(n):
l2 = [i + 1]
cnt = 0
while True:
l2.append(p[l2[-1] - 1])
cnt += c[l2[-1] - 1]
l[i].append(cnt)
if l2[-1] == i + 1:
break
ans = -float("inf")
for i in range(n):
le = len(l[i])
for j in range(min(k, le)):
cnt = l[i][j]
if l[i][-1] > 0:
cnt += (k - j - 1) // le * l[i][-1]
ans = max(ans, cnt)
print(ans)
``` | instruction | 0 | 46,838 | 19 | 93,676 |
Yes | output | 1 | 46,838 | 19 | 93,677 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Takahashi will play a game using a piece on an array of squares numbered 1, 2, \cdots, N. Square i has an integer C_i written on it. Also, he is given a permutation of 1, 2, \cdots, N: P_1, P_2, \cdots, P_N.
Now, he will choose one square and place the piece on that square. Then, he will make the following move some number of times between 1 and K (inclusive):
* In one move, if the piece is now on Square i (1 \leq i \leq N), move it to Square P_i. Here, his score increases by C_{P_i}.
Help him by finding the maximum possible score at the end of the game. (The score is 0 at the beginning of the game.)
Constraints
* 2 \leq N \leq 5000
* 1 \leq K \leq 10^9
* 1 \leq P_i \leq N
* P_i \neq i
* P_1, P_2, \cdots, P_N are all different.
* -10^9 \leq C_i \leq 10^9
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
N K
P_1 P_2 \cdots P_N
C_1 C_2 \cdots C_N
Output
Print the maximum possible score at the end of the game.
Examples
Input
5 2
2 4 5 1 3
3 4 -10 -8 8
Output
8
Input
2 3
2 1
10 -7
Output
13
Input
3 3
3 1 2
-1000 -2000 -3000
Output
-1000
Input
10 58
9 1 6 7 8 4 3 2 10 5
695279662 988782657 -119067776 382975538 -151885171 -177220596 -169777795 37619092 389386780 980092719
Output
29507023469
Submitted Solution:
```
n, k = map(int, input().split())
p = list(map(int, input().split()))
c = list(map(int, input().split()))
maximum = -10**9
for i in range(n):
totalfull = 0; totalcut = 0; count = 0; location = i
maximumfull = -10**9; maximumcut = -10**9
for j in range(min(n, k)):
location = p[location]-1
totalfull += c[location]
maximumfull = max(maximumfull, totalfull)
count += 1
if location == i: break
for j in range(k%count):
location = p[location]-1
totalcut += c[location]
maximumcut = max(maximumcut, totalcut)
maximum = max(maximum, max(maximumfull, max((k//count-1)*totalfull+maximumfull, k//count*totalfull+maximumcut)))
print(maximum)
``` | instruction | 0 | 46,839 | 19 | 93,678 |
Yes | output | 1 | 46,839 | 19 | 93,679 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Takahashi will play a game using a piece on an array of squares numbered 1, 2, \cdots, N. Square i has an integer C_i written on it. Also, he is given a permutation of 1, 2, \cdots, N: P_1, P_2, \cdots, P_N.
Now, he will choose one square and place the piece on that square. Then, he will make the following move some number of times between 1 and K (inclusive):
* In one move, if the piece is now on Square i (1 \leq i \leq N), move it to Square P_i. Here, his score increases by C_{P_i}.
Help him by finding the maximum possible score at the end of the game. (The score is 0 at the beginning of the game.)
Constraints
* 2 \leq N \leq 5000
* 1 \leq K \leq 10^9
* 1 \leq P_i \leq N
* P_i \neq i
* P_1, P_2, \cdots, P_N are all different.
* -10^9 \leq C_i \leq 10^9
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
N K
P_1 P_2 \cdots P_N
C_1 C_2 \cdots C_N
Output
Print the maximum possible score at the end of the game.
Examples
Input
5 2
2 4 5 1 3
3 4 -10 -8 8
Output
8
Input
2 3
2 1
10 -7
Output
13
Input
3 3
3 1 2
-1000 -2000 -3000
Output
-1000
Input
10 58
9 1 6 7 8 4 3 2 10 5
695279662 988782657 -119067776 382975538 -151885171 -177220596 -169777795 37619092 389386780 980092719
Output
29507023469
Submitted Solution:
```
# coding: utf-8
n, k = list(map(int, input().split(" ")))
p = list(map(int, input().split(" ")))
c = list(map(int, input().split(" ")))
position_history = []
max_score = 0
for start_position in range(n):
max_the_start = 0
loop_kaisu = 0
next_position = p[start_position]
start_position = p[start_position]
score = 0
while True:
loop_kaisu += 1
score += c[next_position-1]
max_the_start = max(score, max_the_start)
next_position = p[next_position-1]
if next_position == start_position:
break
nankai_loop_dekiru = k // loop_kaisu
score = score * (nankai_loop_dekiru - 1)
loop_kaisu *= (nankai_loop_dekiru-1)
while loop_kaisu < k:
loop_kaisu += 1
score += c[next_position-1]
next_position = p[next_position-1]
max_the_start = max(score, max_the_start)
max_score = max(max_the_start, max_score)
if max_score <= 0:
print(max(n))
else:
print(max_score)
``` | instruction | 0 | 46,840 | 19 | 93,680 |
No | output | 1 | 46,840 | 19 | 93,681 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Takahashi will play a game using a piece on an array of squares numbered 1, 2, \cdots, N. Square i has an integer C_i written on it. Also, he is given a permutation of 1, 2, \cdots, N: P_1, P_2, \cdots, P_N.
Now, he will choose one square and place the piece on that square. Then, he will make the following move some number of times between 1 and K (inclusive):
* In one move, if the piece is now on Square i (1 \leq i \leq N), move it to Square P_i. Here, his score increases by C_{P_i}.
Help him by finding the maximum possible score at the end of the game. (The score is 0 at the beginning of the game.)
Constraints
* 2 \leq N \leq 5000
* 1 \leq K \leq 10^9
* 1 \leq P_i \leq N
* P_i \neq i
* P_1, P_2, \cdots, P_N are all different.
* -10^9 \leq C_i \leq 10^9
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
N K
P_1 P_2 \cdots P_N
C_1 C_2 \cdots C_N
Output
Print the maximum possible score at the end of the game.
Examples
Input
5 2
2 4 5 1 3
3 4 -10 -8 8
Output
8
Input
2 3
2 1
10 -7
Output
13
Input
3 3
3 1 2
-1000 -2000 -3000
Output
-1000
Input
10 58
9 1 6 7 8 4 3 2 10 5
695279662 988782657 -119067776 382975538 -151885171 -177220596 -169777795 37619092 389386780 980092719
Output
29507023469
Submitted Solution:
```
import itertools
import numpy as np
if __name__ == "__main__":
N,K = map(int,input().split())
P = [ int(p)-1 for p in input().split() ]
C = list(map(int,input().split()))
# 計算済み
used = [ False for _ in range(N) ]
# print(P)
# 一度計算したサイクル情報を一応キャッシュしておく。。。
cycleIDs = [ -1 for _ in range(N) ]
cycleInfs = []
ans = -10 ** 9
cycleID = 0
for n in range(N):
v = n
currentCycleItemCnt = 0
currentCycleCosts = []
if cycleIDs[v] != -1:
currentCycleItemCnt, currentCycleTotal = cycleInfs[ cycleIDs[v] ]
# print(currentCycleItemCnt, currentCycleTotal)
else:
while True:
# 全頂点について、属するサイクルを計算する
currentCycleItemCnt += 1
currentCycleCosts.append( C[v] )
v = P[v]
if v == n:
# サイクル発見
cycleInfs.append( (currentCycleItemCnt, currentCycleCosts ) )
cycleID += 1
break
# 一応、一度サイクルを計算した頂点については、
# その頂点の属するサイクルの情報をメモっておく。。。
cycleIDs[v] = cycleID
for cycleID in cycleIDs:
currentCycleItemCnt = cycleInfs[cycleID][0]
currentCycleCosts = cycleInfs[cycleID][1]
currentCycleTotalScore = sum(currentCycleCosts)
# ループを含めない最大の処理回数
procCnt = K % currentCycleItemCnt
# ただし、割り切れる場合も、1サイクル分は処理するものとする
# (1回以上はやらなければならないのを考慮？)
if procCnt == 0:
procCnt = currentCycleItemCnt
# procCnt = min( K, procCnt )
# サイクル内でループしてスコアを稼ぐ場合の考慮
cycleLoopCnt = 0
if currentCycleTotalScore > 0:
# その時、ループが発生した方がスコアが高くなる場合
# ループ回数を計算
cycleLoopCnt = ( K - procCnt ) // currentCycleItemCnt
# print("K - procCnt=", K - procCnt, "currentCycleItemCnt=", currentCycleItemCnt)
# print("cycleLoopCnt=", cycleLoopCnt)
# このサイクル内での最大スコア
currentCycleMaxScore = -10 ** 9
# このサイクルに属する全頂点分をまとめて計算する
# まず、ループが発生しない分まで計算
npCosts = np.array( currentCycleCosts )
for i in range(currentCycleItemCnt):
# 累積和
acc = list(itertools.accumulate( np.roll( npCosts, i )[:procCnt] ) )
# print(i, acc)#[:procCnt])
currentCycleMaxScore = max( currentCycleMaxScore, max( acc ) )
# print("currentCycleMaxScore=",currentCycleMaxScore)
currentCycleMaxScore += cycleLoopCnt * currentCycleTotalScore
# print("v=", v, "currentCycleItemCnt=", currentCycleItemCnt, "procCnt, cycleLoopCnt, currentCycleMaxScore,currentCycleTotalScore=", procCnt, cycleLoopCnt, currentCycleMaxScore,currentCycleTotalScore)
ans = max( ans, currentCycleMaxScore )
print(ans)
``` | instruction | 0 | 46,841 | 19 | 93,682 |
No | output | 1 | 46,841 | 19 | 93,683 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Takahashi will play a game using a piece on an array of squares numbered 1, 2, \cdots, N. Square i has an integer C_i written on it. Also, he is given a permutation of 1, 2, \cdots, N: P_1, P_2, \cdots, P_N.
Now, he will choose one square and place the piece on that square. Then, he will make the following move some number of times between 1 and K (inclusive):
* In one move, if the piece is now on Square i (1 \leq i \leq N), move it to Square P_i. Here, his score increases by C_{P_i}.
Help him by finding the maximum possible score at the end of the game. (The score is 0 at the beginning of the game.)
Constraints
* 2 \leq N \leq 5000
* 1 \leq K \leq 10^9
* 1 \leq P_i \leq N
* P_i \neq i
* P_1, P_2, \cdots, P_N are all different.
* -10^9 \leq C_i \leq 10^9
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
N K
P_1 P_2 \cdots P_N
C_1 C_2 \cdots C_N
Output
Print the maximum possible score at the end of the game.
Examples
Input
5 2
2 4 5 1 3
3 4 -10 -8 8
Output
8
Input
2 3
2 1
10 -7
Output
13
Input
3 3
3 1 2
-1000 -2000 -3000
Output
-1000
Input
10 58
9 1 6 7 8 4 3 2 10 5
695279662 988782657 -119067776 382975538 -151885171 -177220596 -169777795 37619092 389386780 980092719
Output
29507023469
Submitted Solution:
```
n,k=map(int,input().split())
p=list(map(int,input().split()))
c=list(map(int,input().split()))
s=[[0 for _ in range(2)] for _ in range(n)]
for i in range(n):
cnt=0
now=i
while True:
now=(p[now]-1)
s[i][0]+=c[now]
cnt+=1
if now==i:
break
s[i][1]=cnt
ans=-(10**16)
for j in range(n):
if s[j][0]>0 and k//s[j][1]>0: #何回か回る
cnt1,cnt2=k,k
a=k//s[j][1] #最大周回数
score1=(a*s[j][0])
score2=((a-1)*s[j][0])
cnt1-=a*s[j][1]
now=j
pscore1=0
pscore_max1=0
for x in range(cnt1):
now=(p[now]-1)
pscore1+=c[now]
pscore_max1=max(pscore_max1,pscore1)
score1+=pscore_max1
score_max=score1
else:
now=j
score=0
score_max=c[p[now]-1]
for x in range(min(k,s[j][1])):
now=(p[now]-1)
score+=c[now]
score_max=max(score_max,score)
ans=max(ans,score_max)
print(ans)
``` | instruction | 0 | 46,842 | 19 | 93,684 |
No | output | 1 | 46,842 | 19 | 93,685 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Takahashi will play a game using a piece on an array of squares numbered 1, 2, \cdots, N. Square i has an integer C_i written on it. Also, he is given a permutation of 1, 2, \cdots, N: P_1, P_2, \cdots, P_N.
Now, he will choose one square and place the piece on that square. Then, he will make the following move some number of times between 1 and K (inclusive):
* In one move, if the piece is now on Square i (1 \leq i \leq N), move it to Square P_i. Here, his score increases by C_{P_i}.
Help him by finding the maximum possible score at the end of the game. (The score is 0 at the beginning of the game.)
Constraints
* 2 \leq N \leq 5000
* 1 \leq K \leq 10^9
* 1 \leq P_i \leq N
* P_i \neq i
* P_1, P_2, \cdots, P_N are all different.
* -10^9 \leq C_i \leq 10^9
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
N K
P_1 P_2 \cdots P_N
C_1 C_2 \cdots C_N
Output
Print the maximum possible score at the end of the game.
Examples
Input
5 2
2 4 5 1 3
3 4 -10 -8 8
Output
8
Input
2 3
2 1
10 -7
Output
13
Input
3 3
3 1 2
-1000 -2000 -3000
Output
-1000
Input
10 58
9 1 6 7 8 4 3 2 10 5
695279662 988782657 -119067776 382975538 -151885171 -177220596 -169777795 37619092 389386780 980092719
Output
29507023469
Submitted Solution:
```
n,k=map(int,input().split())
p=list(map(int,input().split()))
c=list(map(int,input().split()))
score=-1000000000
res=[True for _ in range(n)]
res2=[]
for i in range(n):
if res[i]:
res[i]=False
res2.append([])
res2[-1].append(c[i])
res3=p[i]-1
while res3!=i:
res[res3]=False
res2[-1].append(c[res3])
res3=p[res3]-1
res5=False
for i in range(len(res2)):
l=len(res2[i])
sum2=[0 for _ in range(l)]
for j in range(l):
res2[i].append(res2[i][j])
res4=0
count=0
sum2[l-1]+=res2[i][j]
for q in range(j,j+l):
if res2[i][q]<=0:
count=0
res4=0
else:
res5=True
count+=1
res4+=res2[i][q]
sum2[count-1]=max([res4,sum2[count-1]])
s=max(sum2)
if not res5:
score=max([max(res2[i]),score])
continue
if l<k:
s=max(s,sum2[l-1]*(k//l)+max(sum2[:k%l]))
if s>score:
score=s
res5=True
print(score)
``` | instruction | 0 | 46,843 | 19 | 93,686 |
No | output | 1 | 46,843 | 19 | 93,687 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
problem
There are n cards with numbers from 1 to n. First, make a pile of cards by stacking them in order so that the top is the number 1 card, the second from the top is the number 2 card, ..., and the bottom is the number n card.
<image>
Sort the cards by performing the following operation called "shuffle (x, y)" on the pile of cards (x, y is an integer that satisfies 1 ≤ x <y <n).
Shuffle (x, y)
n cards, pile A consisting of the xth cards from the top, pile B consisting of the x + 1th to yth cards, y + 1 consisting of the nth cards Divide into three mountains of mountain C. Then, mountain B is placed on top of mountain A, and mountain C is placed on top of it.
For example, if you perform "shuffle (3,5)" on 9 cards in order, the numbers written on the 9 cards will be 6, 7, 8, 9, 4 in order from the top. , 5, 1, 2, 3.
<image>
Shuffle m times from the state of the first pile "Shuffle (x1, y1)" "Shuffle (x2, y2)" ... Count from the top in the pile of cards after performing "shuffle (xm, ym)" in order Create a program to find out how many cards with numbers r or less are included in the pth to qth cards.
input
The input consists of multiple datasets. Each dataset is given in the following format.
The input consists of m + 3 lines. The number of cards n is written on the first line (3 ≤ n ≤ 1000000000 = 109). The second line contains the integer m, which represents the number of shuffles (1 ≤ m ≤ 5000). The integers p, q, r are written on the third line (1 ≤ p ≤ q ≤ n, 1 ≤ r ≤ n). On the third line of i + (1 ≤ i ≤ m), two integers xi and yi (1 ≤ xi <yi <n) are written separated by a blank.
When n is 0, it indicates the end of input. The number of data sets does not exceed 5.
output
For each data set, in the pile of cards after m shuffling, output the number of cards whose number is r or less in the pth to qth cards counted from the top in one line. ..
Examples
Input
9
1
3 7 4
3 5
12
3
3 8 5
3 8
2 5
6 10
0
Output
2
3
Input
None
Output
None
Submitted Solution:
```
def shuffle(deck,x,y):
cnter = 0
i = 0
sectiona = sectionb = []
while (cnter < x):
nextsize = deck[i][1] - deck[i][0] + 1
if(x-cnter >= nextsize):
cnter += nextsize
i += 1
else:
sectiona = [[deck[i][0],deck[i][0] + x-cnter - 1]]
deck[i][0] = deck[i][0] + x-cnter
cnter = x
memoa = i
while (cnter < y):
nextsize = deck[i][1] - deck[i][0] + 1
if(y-cnter >= nextsize):
cnter += nextsize
i += 1
else:
sectionb = [[deck[i][0],deck[i][0] + y-cnter - 1]]
deck[i][0] = deck[i][0] + y-cnter
cnter = y
newdeck = deck[i:] + deck[memoa:i] + sectionb + deck[0:memoa] + sectiona
return newdeck
def hitnum(a,r):
if a[1] <= r:
return a[1]-a[0]+1
elif a[0] <= r:
return r-a[0]+1
else:
return 0
while True:
n = int(input())
if n==0: break
deck = [[1,n]]
m = int(input())
p,q,r = map(int,input().split())
for i in range(m):
x,y = map(int,input().split())
deck = shuffle(deck, x, y)
deck = shuffle(deck,0,p-1)
answer = 0
cnter = 0
i = 0
while (cnter < q-p+1):
nextsize = deck[i][1] - deck[i][0] + 1
if(q-p+1-cnter >= nextsize):
answer += hitnum(deck[i],r)
cnter += nextsize
i += 1
else:
answer += hitnum([deck[i][0],deck[i][0] + q-p+1-cnter - 1],r)
deck[i][0] = deck[i][0] + q-p+1-cnter
cnter = q-p+1
print(answer)
``` | instruction | 0 | 46,987 | 19 | 93,974 |
No | output | 1 | 46,987 | 19 | 93,975 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
problem
There are n cards with numbers from 1 to n. First, make a pile of cards by stacking them in order so that the top is the number 1 card, the second from the top is the number 2 card, ..., and the bottom is the number n card.
<image>
Sort the cards by performing the following operation called "shuffle (x, y)" on the pile of cards (x, y is an integer that satisfies 1 ≤ x <y <n).
Shuffle (x, y)
n cards, pile A consisting of the xth cards from the top, pile B consisting of the x + 1th to yth cards, y + 1 consisting of the nth cards Divide into three mountains of mountain C. Then, mountain B is placed on top of mountain A, and mountain C is placed on top of it.
For example, if you perform "shuffle (3,5)" on 9 cards in order, the numbers written on the 9 cards will be 6, 7, 8, 9, 4 in order from the top. , 5, 1, 2, 3.
<image>
Shuffle m times from the state of the first pile "Shuffle (x1, y1)" "Shuffle (x2, y2)" ... Count from the top in the pile of cards after performing "shuffle (xm, ym)" in order Create a program to find out how many cards with numbers r or less are included in the pth to qth cards.
input
The input consists of multiple datasets. Each dataset is given in the following format.
The input consists of m + 3 lines. The number of cards n is written on the first line (3 ≤ n ≤ 1000000000 = 109). The second line contains the integer m, which represents the number of shuffles (1 ≤ m ≤ 5000). The integers p, q, r are written on the third line (1 ≤ p ≤ q ≤ n, 1 ≤ r ≤ n). On the third line of i + (1 ≤ i ≤ m), two integers xi and yi (1 ≤ xi <yi <n) are written separated by a blank.
When n is 0, it indicates the end of input. The number of data sets does not exceed 5.
output
For each data set, in the pile of cards after m shuffling, output the number of cards whose number is r or less in the pth to qth cards counted from the top in one line. ..
Examples
Input
9
1
3 7 4
3 5
12
3
3 8 5
3 8
2 5
6 10
0
Output
2
3
Input
None
Output
None
Submitted Solution:
```
def solve():
import sys
line = sys.stdin.readline
for e in iter(line, '0\n'):
n = int(e)
Cards = [(n, 1, n)]
m = int(line())
p, q, r = map(int, line().split())
for _ in range(m):
x, y = map(int, line().split())
total = 0
A, B, C = [], [], []
for k, s, t in Cards:
if y < total + k:
if total < y:
C += [(total + k - y, s + y - total, t)]
if total < x:
B += [(y - x, s + x - total, s + y - total - 1)]
A += [(x - total, s, s + x - total - 1)]
else:
B += [(y - total, s, s + y - total - 1)]
else:
C += [(k, s, t)]
elif x < total + k:
if total < x:
B += [(total + k - x, s + x - total, t)]
A += [(x - total, s, s + x - total - 1)]
else:
B += [(k, s, t)]
else:
A += [(k, s, t)]
total += k
Cards = C + B + A
total = 0
D = []
p = p - 1
for k, s, t in Cards:
if q < total + k:
if total < q:
if total < p:
D += [(s + x - total, s + q - total - 1)]
else:
D += [(s, s + q - total - 1)]
elif p < total + k:
if total < p:
D += [(s + p - total, t)]
else:
D += [(s, t)]
total += k
ans = 0
for s, t in D:
if t < r:
ans += t - s + 1
elif s <= r:
ans += r - s + 1
print(ans)
if __name__ == '__main__':
solve()
``` | instruction | 0 | 46,988 | 19 | 93,976 |
No | output | 1 | 46,988 | 19 | 93,977 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
problem
There are n cards with numbers from 1 to n. First, make a pile of cards by stacking them in order so that the top is the number 1 card, the second from the top is the number 2 card, ..., and the bottom is the number n card.
<image>
Sort the cards by performing the following operation called "shuffle (x, y)" on the pile of cards (x, y is an integer that satisfies 1 ≤ x <y <n).
Shuffle (x, y)
n cards, pile A consisting of the xth cards from the top, pile B consisting of the x + 1th to yth cards, y + 1 consisting of the nth cards Divide into three mountains of mountain C. Then, mountain B is placed on top of mountain A, and mountain C is placed on top of it.
For example, if you perform "shuffle (3,5)" on 9 cards in order, the numbers written on the 9 cards will be 6, 7, 8, 9, 4 in order from the top. , 5, 1, 2, 3.
<image>
Shuffle m times from the state of the first pile "Shuffle (x1, y1)" "Shuffle (x2, y2)" ... Count from the top in the pile of cards after performing "shuffle (xm, ym)" in order Create a program to find out how many cards with numbers r or less are included in the pth to qth cards.
input
The input consists of multiple datasets. Each dataset is given in the following format.
The input consists of m + 3 lines. The number of cards n is written on the first line (3 ≤ n ≤ 1000000000 = 109). The second line contains the integer m, which represents the number of shuffles (1 ≤ m ≤ 5000). The integers p, q, r are written on the third line (1 ≤ p ≤ q ≤ n, 1 ≤ r ≤ n). On the third line of i + (1 ≤ i ≤ m), two integers xi and yi (1 ≤ xi <yi <n) are written separated by a blank.
When n is 0, it indicates the end of input. The number of data sets does not exceed 5.
output
For each data set, in the pile of cards after m shuffling, output the number of cards whose number is r or less in the pth to qth cards counted from the top in one line. ..
Examples
Input
9
1
3 7 4
3 5
12
3
3 8 5
3 8
2 5
6 10
0
Output
2
3
Input
None
Output
None
Submitted Solution:
```
def solve():
import sys
line = sys.stdin.readline
for e in iter(line, '0\n'):
n = int(e)
Cards = [(n, 1)]
m = int(line())
p, q, r = map(int, line().split())
for _ in range(m):
x, y = map(int, line().split())
total = 0
A, B, C = [], [], []
for k, s in Cards:
if y < total + k:
if total < y:
C += [(total + k - y, s + y - total)]
if total < x:
B += [(y - x, s + x - total)]
A += [(x - total, s)]
else:
B += [(y - total, s)]
else:
C += [(k, s)]
elif x < total + k:
if total < x:
B += [(total + k - x, s + x - total)]
A += [(x - total, s)]
else:
B += [(k, s)]
else:
A += [(k, s)]
total += k
Cards = C + B + A
total = 0
ans = 0
D = []
p = p - 1
for k, s in Cards:
if 0 < q - total < k:
v = (x - total) * (total < p)
if v <= r - s:
ans += min(q - total - 1, r - s) - v + 1
elif p < total + k:
v = (p - total) * (total < p)
if v <= r - s:
ans += min(k - 1, r - s) - v + 1
total += k
print(ans)
if __name__ == '__main__':
solve()
``` | instruction | 0 | 46,989 | 19 | 93,978 |
No | output | 1 | 46,989 | 19 | 93,979 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
problem
There are n cards with numbers from 1 to n. First, make a pile of cards by stacking them in order so that the top is the number 1 card, the second from the top is the number 2 card, ..., and the bottom is the number n card.
<image>
Sort the cards by performing the following operation called "shuffle (x, y)" on the pile of cards (x, y is an integer that satisfies 1 ≤ x <y <n).
Shuffle (x, y)
n cards, pile A consisting of the xth cards from the top, pile B consisting of the x + 1th to yth cards, y + 1 consisting of the nth cards Divide into three mountains of mountain C. Then, mountain B is placed on top of mountain A, and mountain C is placed on top of it.
For example, if you perform "shuffle (3,5)" on 9 cards in order, the numbers written on the 9 cards will be 6, 7, 8, 9, 4 in order from the top. , 5, 1, 2, 3.
<image>
Shuffle m times from the state of the first pile "Shuffle (x1, y1)" "Shuffle (x2, y2)" ... Count from the top in the pile of cards after performing "shuffle (xm, ym)" in order Create a program to find out how many cards with numbers r or less are included in the pth to qth cards.
input
The input consists of multiple datasets. Each dataset is given in the following format.
The input consists of m + 3 lines. The number of cards n is written on the first line (3 ≤ n ≤ 1000000000 = 109). The second line contains the integer m, which represents the number of shuffles (1 ≤ m ≤ 5000). The integers p, q, r are written on the third line (1 ≤ p ≤ q ≤ n, 1 ≤ r ≤ n). On the third line of i + (1 ≤ i ≤ m), two integers xi and yi (1 ≤ xi <yi <n) are written separated by a blank.
When n is 0, it indicates the end of input. The number of data sets does not exceed 5.
output
For each data set, in the pile of cards after m shuffling, output the number of cards whose number is r or less in the pth to qth cards counted from the top in one line. ..
Examples
Input
9
1
3 7 4
3 5
12
3
3 8 5
3 8
2 5
6 10
0
Output
2
3
Input
None
Output
None
Submitted Solution:
```
def solve():
import sys
read = sys.stdin.readline
for e in iter(read, '0\n'):
n = int(e)
Cards = [(n, 1, n)]
m = int(read())
p, q, r = map(int, read().split())
for _ in [0] * m:
x, y = map(int, read().split())
total = 0
A, B, C = [], [], []
for k, s, t in Cards:
if y < total + k:
if total < y:
C += [(total + k - y, s + y - total, t)]
if total < x:
B += [(y - x, s + x - total, s + y - total - 1)]
A += [(x - total, s, s + x - total - 1)]
else:
B += [(y - total, s, s + y - total - 1)]
else:
C += [(k, s, t)]
elif x < total + k:
if total < x:
B += [(total + k - x, s + x - total, t)]
A += [(x - total, s, s + x - total - 1)]
else:
B += [(k, s, t)]
else:
A += [(k, s, t)]
total += k
Cards = C + B + A
total = 0
D = []
p = p - 1
for k, s, t in Cards:
if q < total + k:
if total < q:
if total < p:
D += [(s + x - total, s + q - total - 1)]
else:
D += [(s, s + q - total - 1)]
elif p < total + k:
if total < p:
D += [(s + p - total, t)]
else:
D += [(s, t)]
total += k
ans = 0
for s, t in D:
if t < r:
ans += t - s + 1
elif s <= r <= t:
ans += r - s + 1
print(ans)
if __name__ == '__main__':
solve()
``` | instruction | 0 | 46,990 | 19 | 93,980 |
No | output | 1 | 46,990 | 19 | 93,981 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Iahub is very proud of his recent discovery, propagating trees. Right now, he invented a new tree, called xor-tree. After this new revolutionary discovery, he invented a game for kids which uses xor-trees.
The game is played on a tree having n nodes, numbered from 1 to n. Each node i has an initial value initi, which is either 0 or 1. The root of the tree is node 1.
One can perform several (possibly, zero) operations on the tree during the game. The only available type of operation is to pick a node x. Right after someone has picked node x, the value of node x flips, the values of sons of x remain the same, the values of sons of sons of x flips, the values of sons of sons of sons of x remain the same and so on.
The goal of the game is to get each node i to have value goali, which can also be only 0 or 1. You need to reach the goal of the game by using minimum number of operations.
Input
The first line contains an integer n (1 ≤ n ≤ 105). Each of the next n - 1 lines contains two integers ui and vi (1 ≤ ui, vi ≤ n; ui ≠ vi) meaning there is an edge between nodes ui and vi.
The next line contains n integer numbers, the i-th of them corresponds to initi (initi is either 0 or 1). The following line also contains n integer numbers, the i-th number corresponds to goali (goali is either 0 or 1).
Output
In the first line output an integer number cnt, representing the minimal number of operations you perform. Each of the next cnt lines should contain an integer xi, representing that you pick a node xi.
Examples
Input
10
2 1
3 1
4 2
5 1
6 2
7 5
8 6
9 8
10 5
1 0 1 1 0 1 0 1 0 1
1 0 1 0 0 1 1 1 0 1
Output
2
4
7
Submitted Solution:
```
import sys;readline = sys.stdin.readline
def i1(): return int(readline())
def nl(): return [int(s) for s in readline().split()]
def nn(n): return [int(readline()) for i in range(n)]
def nnp(n,x): return [int(readline())+x for i in range(n)]
def nmp(n,x): return (int(readline())+x for i in range(n))
def nlp(x): return [int(s)+x for s in readline().split()]
def nll(n): return [[int(s) for s in readline().split()] for i in range(n)]
def mll(n): return ([int(s) for s in readline().split()] for i in range(n))
def s1(): return readline().rstrip()
def sl(): return [s for s in readline().split()]
def sn(n): return [readline().rstrip() for i in range(n)]
def sm(n): return (readline().rstrip() for i in range(n))
def redir(s): global readline;import os;fn=sys.argv[0] + f'/../in-{s}.txt';readline = open(fn).readline if os.path.exists(fn) else readline
redir('a')
n = i1()
nodes = [[] for i in range(n+1)]
for i in range(n-1):
u, v = nl()
nodes[u].append(v)
nodes[v].append(u)
init = [0] + nl()
goal = [0] + nl()
roots = [1]
flip = [[0] * (n+1), [0] * (n+1)]
# notseen = [True] * (n+1)
fa = [0] * (n+1)
fa[1] = 1
# fa[1], fa[n+1] = -1, 1
level = 0
# cnt = 0
m = []
while roots:
_roots = []
fli = flip[level%2]
# fli2 = flip[1-level%2]
# print(level, roots, [(fli[r], init[r], goal[r]) for r in roots])
for r in roots:
# notseen[r] = False
# assert fa[r] > 0 or r == 1
f = fli[fa[fa[r]]]
if f ^ init[r] != goal[r]:
# cnt += 1
f = 1 - f
m.append(r)
fli[r] = f
for i in nodes[r]:
if fa[i] == 0:
_roots.append(i)
fa[i] = r
# print(level, roots, [(fli[r], init[r], goal[r]) for r in roots])
roots = _roots
level += 1
# print(cnt)
init[0] = goal[0] = None
# print(init)
# print(goal)
print(len(m))
print('\n'.join(str(i) for i in m))
``` | instruction | 0 | 47,414 | 19 | 94,828 |
Yes | output | 1 | 47,414 | 19 | 94,829 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Iahub is very proud of his recent discovery, propagating trees. Right now, he invented a new tree, called xor-tree. After this new revolutionary discovery, he invented a game for kids which uses xor-trees.
The game is played on a tree having n nodes, numbered from 1 to n. Each node i has an initial value initi, which is either 0 or 1. The root of the tree is node 1.
One can perform several (possibly, zero) operations on the tree during the game. The only available type of operation is to pick a node x. Right after someone has picked node x, the value of node x flips, the values of sons of x remain the same, the values of sons of sons of x flips, the values of sons of sons of sons of x remain the same and so on.
The goal of the game is to get each node i to have value goali, which can also be only 0 or 1. You need to reach the goal of the game by using minimum number of operations.
Input
The first line contains an integer n (1 ≤ n ≤ 105). Each of the next n - 1 lines contains two integers ui and vi (1 ≤ ui, vi ≤ n; ui ≠ vi) meaning there is an edge between nodes ui and vi.
The next line contains n integer numbers, the i-th of them corresponds to initi (initi is either 0 or 1). The following line also contains n integer numbers, the i-th number corresponds to goali (goali is either 0 or 1).
Output
In the first line output an integer number cnt, representing the minimal number of operations you perform. Each of the next cnt lines should contain an integer xi, representing that you pick a node xi.
Examples
Input
10
2 1
3 1
4 2
5 1
6 2
7 5
8 6
9 8
10 5
1 0 1 1 0 1 0 1 0 1
1 0 1 0 0 1 1 1 0 1
Output
2
4
7
Submitted Solution:
```
from collections import defaultdict, deque, Counter, OrderedDict
from bisect import insort, bisect_right, bisect_left
import threading, sys
def main():
n = int(input())
adj = [[] for i in range(n + 1)]
for i in range(n - 1):
a, b = map(int, input().split())
a, b = a - 1, b - 1
adj[a].append(b)
adj[b].append(a)
init = [int(i) for i in input().split()]
goal = [int(i) for i in input().split()]
visited = [0] * n
par = [[] for i in range(n)]
def dfs(s, p):
if visited[s]: return
visited[s] = 1
par[p].append(s)
for i in adj[s]:
dfs(i, s)
dfs(0, 0)
par[0] = par[0][1:]
ans = []
def dfs2(s, l, fo, fe):
if l % 2 == 0:
if fe % 2 == 1:
init[s] = 1 - init[s]
else:
if fo % 2 == 1:
init[s] = 1 - init[s]
if init[s] != goal[s]:
ans.append(s + 1)
if l % 2:
fo += 1
else:
fe += 1
for j in par[s]:
dfs2(j, l + 1, fo, fe)
dfs2(0, 0, 0, 0)
print(len(ans))
print("\n".join(map(str, ans)))
if __name__ == "__main__":
sys.setrecursionlimit(400000)
threading.stack_size(40960000)
thread = threading.Thread(target=main)
thread.start()
``` | instruction | 0 | 47,415 | 19 | 94,830 |
Yes | output | 1 | 47,415 | 19 | 94,831 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Iahub is very proud of his recent discovery, propagating trees. Right now, he invented a new tree, called xor-tree. After this new revolutionary discovery, he invented a game for kids which uses xor-trees.
The game is played on a tree having n nodes, numbered from 1 to n. Each node i has an initial value initi, which is either 0 or 1. The root of the tree is node 1.
One can perform several (possibly, zero) operations on the tree during the game. The only available type of operation is to pick a node x. Right after someone has picked node x, the value of node x flips, the values of sons of x remain the same, the values of sons of sons of x flips, the values of sons of sons of sons of x remain the same and so on.
The goal of the game is to get each node i to have value goali, which can also be only 0 or 1. You need to reach the goal of the game by using minimum number of operations.
Input
The first line contains an integer n (1 ≤ n ≤ 105). Each of the next n - 1 lines contains two integers ui and vi (1 ≤ ui, vi ≤ n; ui ≠ vi) meaning there is an edge between nodes ui and vi.
The next line contains n integer numbers, the i-th of them corresponds to initi (initi is either 0 or 1). The following line also contains n integer numbers, the i-th number corresponds to goali (goali is either 0 or 1).
Output
In the first line output an integer number cnt, representing the minimal number of operations you perform. Each of the next cnt lines should contain an integer xi, representing that you pick a node xi.
Examples
Input
10
2 1
3 1
4 2
5 1
6 2
7 5
8 6
9 8
10 5
1 0 1 1 0 1 0 1 0 1
1 0 1 0 0 1 1 1 0 1
Output
2
4
7
Submitted Solution:
```
from collections import defaultdict, deque, Counter, OrderedDict
from bisect import insort, bisect_right, bisect_left
import threading, sys
def main():
n = int(input())
adj = [[] for i in range(n + 1)]
for i in range(n - 1):
a, b = map(int, input().split())
a, b = a - 1, b - 1
adj[a].append(b)
adj[b].append(a)
init = [int(i) for i in input().split()]
goal = [int(i) for i in input().split()]
visited = [0] * n
par = [[] for i in range(n)]
def dfs(s, p):
if visited[s]: return
visited[s] = 1
par[p].append(s)
for i in adj[s]:
dfs(i, s)
dfs(0, 0)
par[0] = par[0][1:]
ans = []
def dfs2(s, l, fo, fe):
if l % 2 == 0:
if fe % 2 == 1:
init[s] = 1 - init[s]
else:
if fo % 2 == 1:
init[s] = 1 - init[s]
if init[s] != goal[s]:
ans.append(s + 1)
if l % 2:
fo += 1
else:
fe += 1
for j in par[s]:
dfs2(j, l + 1, fo, fe)
dfs2(0, 0, 0, 0)
print(len(ans))
print("\n".join(map(str, ans)))
if __name__ == "__main__":
sys.setrecursionlimit(400000)
threading.stack_size(102400000)
thread = threading.Thread(target=main)
thread.start()
``` | instruction | 0 | 47,416 | 19 | 94,832 |
Yes | output | 1 | 47,416 | 19 | 94,833 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Iahub is very proud of his recent discovery, propagating trees. Right now, he invented a new tree, called xor-tree. After this new revolutionary discovery, he invented a game for kids which uses xor-trees.
The game is played on a tree having n nodes, numbered from 1 to n. Each node i has an initial value initi, which is either 0 or 1. The root of the tree is node 1.
One can perform several (possibly, zero) operations on the tree during the game. The only available type of operation is to pick a node x. Right after someone has picked node x, the value of node x flips, the values of sons of x remain the same, the values of sons of sons of x flips, the values of sons of sons of sons of x remain the same and so on.
The goal of the game is to get each node i to have value goali, which can also be only 0 or 1. You need to reach the goal of the game by using minimum number of operations.
Input
The first line contains an integer n (1 ≤ n ≤ 105). Each of the next n - 1 lines contains two integers ui and vi (1 ≤ ui, vi ≤ n; ui ≠ vi) meaning there is an edge between nodes ui and vi.
The next line contains n integer numbers, the i-th of them corresponds to initi (initi is either 0 or 1). The following line also contains n integer numbers, the i-th number corresponds to goali (goali is either 0 or 1).
Output
In the first line output an integer number cnt, representing the minimal number of operations you perform. Each of the next cnt lines should contain an integer xi, representing that you pick a node xi.
Examples
Input
10
2 1
3 1
4 2
5 1
6 2
7 5
8 6
9 8
10 5
1 0 1 1 0 1 0 1 0 1
1 0 1 0 0 1 1 1 0 1
Output
2
4
7
Submitted Solution:
```
'''input
10
2 1
3 1
4 2
5 1
6 2
7 5
8 6
9 8
10 5
1 0 1 1 0 1 0 1 0 1
1 0 1 0 0 1 1 1 0 1
'''
from sys import stdin, setrecursionlimit
from collections import defaultdict
setrecursionlimit(1500000)
def counter(num):
if num == 0:
return 1
else:
return 0
def flip_me(original, count, index):
if count % 2 == 0:
return original[index]
else:
return counter(original[index])
def dfs(graph, visited, ans, original, goal, change, node, dfs_stack):
dfs_stack.append(node)
while len(dfs_stack) > 0:
node = dfs_stack.pop()
visited[node] = True
value = flip_me(original, change[node], node - 1)
add = 0
if goal[node - 1] == value:
pass
else:
add = 1
ans[node] = True
flag = 0
for i in graph[node]:
if visited[i] == False:
flag = 1
for j in graph[i]:
if visited[j] == False:
change[j] += change[node] + add
dfs_stack.append(node)
dfs_stack.append(i)
break
if flag == 0:
pass
def calculate(graph, original, goal, n):
visited = dict()
change = dict()
for i in range(1, n + 1):
visited[i] = False
change[i] = 0
ans = dict()
dfs_stack = []
dfs(graph, visited, ans, original, goal, change, 1, dfs_stack)
return ans
# main starts
n = int(stdin.readline().strip())
graph = defaultdict(list)
for i in range(1, n + 1):
graph[i]
for _ in range(n - 1):
u, v = list(map(int, stdin.readline().split()))
graph[u].append(v)
graph[v].append(u)
original = list(map(int, stdin.readline().split()))
goal = list(map(int, stdin.readline().split()))
count = [0]
ans = calculate(graph, original, goal, n)
print(len(ans))
for i in ans:
print(i)
``` | instruction | 0 | 47,417 | 19 | 94,834 |
Yes | output | 1 | 47,417 | 19 | 94,835 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Iahub is very proud of his recent discovery, propagating trees. Right now, he invented a new tree, called xor-tree. After this new revolutionary discovery, he invented a game for kids which uses xor-trees.
The game is played on a tree having n nodes, numbered from 1 to n. Each node i has an initial value initi, which is either 0 or 1. The root of the tree is node 1.
One can perform several (possibly, zero) operations on the tree during the game. The only available type of operation is to pick a node x. Right after someone has picked node x, the value of node x flips, the values of sons of x remain the same, the values of sons of sons of x flips, the values of sons of sons of sons of x remain the same and so on.
The goal of the game is to get each node i to have value goali, which can also be only 0 or 1. You need to reach the goal of the game by using minimum number of operations.
Input
The first line contains an integer n (1 ≤ n ≤ 105). Each of the next n - 1 lines contains two integers ui and vi (1 ≤ ui, vi ≤ n; ui ≠ vi) meaning there is an edge between nodes ui and vi.
The next line contains n integer numbers, the i-th of them corresponds to initi (initi is either 0 or 1). The following line also contains n integer numbers, the i-th number corresponds to goali (goali is either 0 or 1).
Output
In the first line output an integer number cnt, representing the minimal number of operations you perform. Each of the next cnt lines should contain an integer xi, representing that you pick a node xi.
Examples
Input
10
2 1
3 1
4 2
5 1
6 2
7 5
8 6
9 8
10 5
1 0 1 1 0 1 0 1 0 1
1 0 1 0 0 1 1 1 0 1
Output
2
4
7
Submitted Solution:
```
def lel(vertex,k) :
global L
global b
if k%2==0 :
b[vertex]=abs(1-b[vertex])
for x in L[vertex] :
lel(x,k+1)
n=int(input())
L=[[] for i in range(n)]
for i in range(n-1) :
a,b=map(int,input().split())
if a>b :
L[a-1].append(b-1)
else :
L[b-1].append(a-1)
l=list(map(int,input().split()))
l1=list(map(int,input().split()))
b=[]
ans=[]
for i in range(n) :
b.append(abs(l[i]-l1[i]))
for i in range(n-1,-1,-1) :
if b[i]==1 :
lel(i,0)
ans.append(i+1)
print(len(ans))
print('\n'.join(map(str,ans)))
``` | instruction | 0 | 47,418 | 19 | 94,836 |
No | output | 1 | 47,418 | 19 | 94,837 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Iahub is very proud of his recent discovery, propagating trees. Right now, he invented a new tree, called xor-tree. After this new revolutionary discovery, he invented a game for kids which uses xor-trees.
The game is played on a tree having n nodes, numbered from 1 to n. Each node i has an initial value initi, which is either 0 or 1. The root of the tree is node 1.
One can perform several (possibly, zero) operations on the tree during the game. The only available type of operation is to pick a node x. Right after someone has picked node x, the value of node x flips, the values of sons of x remain the same, the values of sons of sons of x flips, the values of sons of sons of sons of x remain the same and so on.
The goal of the game is to get each node i to have value goali, which can also be only 0 or 1. You need to reach the goal of the game by using minimum number of operations.
Input
The first line contains an integer n (1 ≤ n ≤ 105). Each of the next n - 1 lines contains two integers ui and vi (1 ≤ ui, vi ≤ n; ui ≠ vi) meaning there is an edge between nodes ui and vi.
The next line contains n integer numbers, the i-th of them corresponds to initi (initi is either 0 or 1). The following line also contains n integer numbers, the i-th number corresponds to goali (goali is either 0 or 1).
Output
In the first line output an integer number cnt, representing the minimal number of operations you perform. Each of the next cnt lines should contain an integer xi, representing that you pick a node xi.
Examples
Input
10
2 1
3 1
4 2
5 1
6 2
7 5
8 6
9 8
10 5
1 0 1 1 0 1 0 1 0 1
1 0 1 0 0 1 1 1 0 1
Output
2
4
7
Submitted Solution:
```
def lel(vertex,k) :
global L
global b
if k%2==0 :
b[vertex]=abs(1-b[vertex])
for x in L[vertex] :
if k<6 :
lel(x,k+1)
n=int(input())
L=[[] for i in range(n)]
for i in range(n-1) :
a,b=map(int,input().split())
if a<b :
L[a-1].append(b-1)
else :
L[b-1].append(a-1)
l=list(map(int,input().split()))
l1=list(map(int,input().split()))
b=[]
ans=[]
for i in range(n) :
b.append(abs(l[i]-l1[i]))
for i in range(n) :
if b[i]==1 :
lel(i,0)
ans.append(i+1)
print(len(ans))
print('\n'.join(map(str,ans)))
``` | instruction | 0 | 47,419 | 19 | 94,838 |
No | output | 1 | 47,419 | 19 | 94,839 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Iahub is very proud of his recent discovery, propagating trees. Right now, he invented a new tree, called xor-tree. After this new revolutionary discovery, he invented a game for kids which uses xor-trees.
The game is played on a tree having n nodes, numbered from 1 to n. Each node i has an initial value initi, which is either 0 or 1. The root of the tree is node 1.
One can perform several (possibly, zero) operations on the tree during the game. The only available type of operation is to pick a node x. Right after someone has picked node x, the value of node x flips, the values of sons of x remain the same, the values of sons of sons of x flips, the values of sons of sons of sons of x remain the same and so on.
The goal of the game is to get each node i to have value goali, which can also be only 0 or 1. You need to reach the goal of the game by using minimum number of operations.
Input
The first line contains an integer n (1 ≤ n ≤ 105). Each of the next n - 1 lines contains two integers ui and vi (1 ≤ ui, vi ≤ n; ui ≠ vi) meaning there is an edge between nodes ui and vi.
The next line contains n integer numbers, the i-th of them corresponds to initi (initi is either 0 or 1). The following line also contains n integer numbers, the i-th number corresponds to goali (goali is either 0 or 1).
Output
In the first line output an integer number cnt, representing the minimal number of operations you perform. Each of the next cnt lines should contain an integer xi, representing that you pick a node xi.
Examples
Input
10
2 1
3 1
4 2
5 1
6 2
7 5
8 6
9 8
10 5
1 0 1 1 0 1 0 1 0 1
1 0 1 0 0 1 1 1 0 1
Output
2
4
7
Submitted Solution:
```
from collections import defaultdict
n = int(input())
d = defaultdict(list)
def DFSop(d1,x,goal,state,c,visited,l):
if c == 1:
state[x-1] = 1
elif c == 0:
state[x-1] = 1
if goal[x-1] != state[x-1]:
count = 1
c = goal[x-1]
l.append(x)
else:
count = 0
visited[x] = 1
for i in d1[x]:
if visited[i] == 0:
DFSop(d1,i,goal,state,c,visited,l)
return l
def DFS(d,x,parent,visited):
stack = [(x,0)]
d1 = defaultdict(list)
while len(stack):
temp,parent = stack.pop()
visited[temp] = 1
# print(temp,parent)
for i in d[temp]:
if visited[i] == 0:
d1[parent].append(i)
stack.append((i,temp))
return d1
for i in range(n-1):
a,b = map(int,input().split())
d[a].append(b)
d[b].append(a)
parent = 0
visited = [0 for i in range(n+1)]
x = 1
state = list(map(int,input().split()))
goal = list(map(int,input().split()))
c = None
x = 1
parent = 0
d1 = DFS(d,x,parent,visited)
visited = [0 for i in range(n+1)]
l = []
ans = DFSop(d1,x,goal,state,c,visited,l)
for i in d[1]:
d1 = DFS(d,x,parent,visited)
c = None
ans = ans + DFSop(d1,i,goal,state,c,visited,l)
ans = set(ans)
print(len(ans))
for i in ans:
print(i)
``` | instruction | 0 | 47,420 | 19 | 94,840 |
No | output | 1 | 47,420 | 19 | 94,841 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Iahub is very proud of his recent discovery, propagating trees. Right now, he invented a new tree, called xor-tree. After this new revolutionary discovery, he invented a game for kids which uses xor-trees.
The game is played on a tree having n nodes, numbered from 1 to n. Each node i has an initial value initi, which is either 0 or 1. The root of the tree is node 1.
One can perform several (possibly, zero) operations on the tree during the game. The only available type of operation is to pick a node x. Right after someone has picked node x, the value of node x flips, the values of sons of x remain the same, the values of sons of sons of x flips, the values of sons of sons of sons of x remain the same and so on.
The goal of the game is to get each node i to have value goali, which can also be only 0 or 1. You need to reach the goal of the game by using minimum number of operations.
Input
The first line contains an integer n (1 ≤ n ≤ 105). Each of the next n - 1 lines contains two integers ui and vi (1 ≤ ui, vi ≤ n; ui ≠ vi) meaning there is an edge between nodes ui and vi.
The next line contains n integer numbers, the i-th of them corresponds to initi (initi is either 0 or 1). The following line also contains n integer numbers, the i-th number corresponds to goali (goali is either 0 or 1).
Output
In the first line output an integer number cnt, representing the minimal number of operations you perform. Each of the next cnt lines should contain an integer xi, representing that you pick a node xi.
Examples
Input
10
2 1
3 1
4 2
5 1
6 2
7 5
8 6
9 8
10 5
1 0 1 1 0 1 0 1 0 1
1 0 1 0 0 1 1 1 0 1
Output
2
4
7
Submitted Solution:
```
import sys;readline = sys.stdin.readline
def i1(): return int(readline())
def nl(): return [int(s) for s in readline().split()]
def nn(n): return [int(readline()) for i in range(n)]
def nnp(n,x): return [int(readline())+x for i in range(n)]
def nmp(n,x): return (int(readline())+x for i in range(n))
def nlp(x): return [int(s)+x for s in readline().split()]
def nll(n): return [[int(s) for s in readline().split()] for i in range(n)]
def mll(n): return ([int(s) for s in readline().split()] for i in range(n))
def s1(): return readline().rstrip()
def sl(): return [s for s in readline().split()]
def sn(n): return [readline().rstrip() for i in range(n)]
def sm(n): return (readline().rstrip() for i in range(n))
def redir(s): global readline;import os;fn=sys.argv[0] + f'/../in-{s}.txt';readline = open(fn).readline if os.path.exists(fn) else readline
redir('a')
n = i1()
nodes = [[] for i in range(n+1)]
for i in range(n-1):
u, v = nl()
nodes[u].append(v)
nodes[v].append(u)
init = [0] + nl()
goal = [0] + nl()
roots = [1]
flip = [[0] * (n+1), [0] * (n+1)]
notseen = [True] * (n+1)
level = 0
# cnt = 0
m = []
while roots:
_roots = []
fli = flip[level%2]
fli2 = flip[1-level%2]
for r in roots:
notseen[r] = False
f = fli[r]
if f ^ init[r] != goal[r]:
# cnt += 1
f = 1 - f
m.append(r)
for i in nodes[r]:
if notseen[i]:
_roots.append(i)
fli[i] = fli2[i] = f
roots = _roots
level += 1
# print(cnt)
print(len(m))
print('\n'.join(str(i) for i in m))
``` | instruction | 0 | 47,421 | 19 | 94,842 |
No | output | 1 | 47,421 | 19 | 94,843 |
Provide a correct Python 3 solution for this coding contest problem.
Now I have a card with n numbers on it. Consider arranging some or all of these appropriately to make numbers. Find the number obtained by adding all the numbers created at this time.
For example, if you have 1 and 2, you will get 4 numbers 1, 2, 12, 21 so the total number is 36. Even if the same numbers are produced as a result of arranging them, if they are arranged differently, they are added separately. For example, if you have a card called 1 and a card called 11, there are two ways to arrange them so that they are 111, but add them as different ones. There are no leading zero cards among the cards, and we do not accept numbers that lead to zero reading. Output the answer divided by 1,000,000,007.
Input
The input is given in the form:
> n
> a1
> a2
> ...
> an
>
The first line contains n (1 ≤ n ≤ 200), which represents the number of cards, and the next n lines contain the number ai (0 ≤ ai <10000) on each card. Also, the same number is never written on multiple cards.
Output
Output the sum of all the numbers you can make divided by 1,000,000,007 on one line.
Examples
Input
2
1
2
Output
36
Input
2
1
11
Output
234
Input
4
0
4
7
8
Output
135299 | instruction | 0 | 47,899 | 19 | 95,798 |
"Correct Solution:
```
import sys
read = sys.stdin.read
write = sys.stdout.write
def solve():
MOD = 10**9 + 7
N, *A = map(int, read().split())
L = 10**5
fact = [1]*(L+1)
rfact = [1]*(L+1)
r = 1
for i in range(1, L+1):
fact[i] = r = r * i % MOD
rfact[L] = r = pow(fact[L], MOD-2, MOD)
for i in range(L, 0, -1):
rfact[i-1] = r = r * i % MOD
L0 = 1000
pw10 = [1]*(L0+1)
r = 1
for i in range(1, L0+1):
pw10[i] = r = r * 10 % MOD
C = [0]*5
V = [0]*5
z = 0
for a in A:
if a == 0:
z = 1
l = len(str(a))
C[l-1] += 1
V[l-1] += a
S = [0]*(N+1)
r = 0
for i in range(N+1):
r += rfact[i]
S[i] = fact[i] * r % MOD
F = [[0]*(i+1) for i in range(N+1)]
for i in range(N+1):
for j in range(i+1):
F[i][j] = fact[i] * rfact[i-j] % MOD
CM = [[0]*(i+1) for i in range(N+1)]
for i in range(N+1):
for j in range(i+1):
CM[i][j] = fact[i] * rfact[i-j] * rfact[j] % MOD
def calc(C):
c1, c2, c3, c4, c5 = C
l0 = sum(C)
res = 0
F1 = F[c1]; G1 = F[c1-1]
F2 = F[c2]
F3 = F[c3]
F4 = F[c4]
F5 = F[c5]
r1 = range(c1+1)
r2 = range(c2+1)
r3 = range(c3+1)
r4 = range(c4+1)
for d5 in range(c5+1):
v5 = F5[d5]
for d1 in r1:
v1 = v5 * CM[d1+d5][d1] % MOD
if z and d1 < c1:
p1 = F1[d1]; p2 = G1[d1]
else:
p1 = F1[d1]; p2 = 0
for d2 in r2:
v2 = v1 * F2[d2] * CM[d2+d1+d5][d2] % MOD
for d3 in r3:
e = d1+d2*2+d3*3+d5*5
f = d1+d2+d3+d5
v3 = v2 * F3[d3] * CM[f][d3] % MOD
for d4 in r4:
l = f+d4
v4 = v3 * F4[d4] * CM[l][d4] % MOD
res += ((p1 * S[l0-l] - p2 * S[l0-1-l]) * pw10[e+d4*4] % MOD) * v4 % MOD
return res % MOD
c1, c2, c3, c4, c5 = C
ans = 0
ans += calc([c1-1, c2, c3, c4, c5]) * V[0]
ans += calc([c1, c2-1, c3, c4, c5]) * V[1]
ans += calc([c1, c2, c3-1, c4, c5]) * V[2]
ans += calc([c1, c2, c3, c4-1, c5]) * V[3]
ans += calc([c1, c2, c3, c4, c5-1]) * V[4]
ans %= MOD
write("%d\n" % ans)
solve()
``` | output | 1 | 47,899 | 19 | 95,799 |
Provide a correct Python 3 solution for this coding contest problem.
Your task is to shuffle a deck of n cards, each of which is marked by a alphabetical letter.
A single shuffle action takes out h cards from the bottom of the deck and moves them to the top of the deck.
The deck of cards is represented by a string as follows.
abcdeefab
The first character and the last character correspond to the card located at the bottom of the deck and the card on the top of the deck respectively.
For example, a shuffle with h = 4 to the above deck, moves the first 4 characters "abcd" to the end of the remaining characters "eefab", and generates the following deck:
eefababcd
You can repeat such shuffle operations.
Write a program which reads a deck (a string) and a sequence of h, and prints the final state (a string).
Constraints
* The length of the string ≤ 200
* 1 ≤ m ≤ 100
* 1 ≤ hi < The length of the string
* The number of datasets ≤ 10
Input
The input consists of multiple datasets. Each dataset is given in the following format:
A string which represents a deck
The number of shuffle m
h1
h2
.
.
hm
The input ends with a single character '-' for the string.
Output
For each dataset, print a string which represents the final state in a line.
Example
Input
aabc
3
1
2
1
vwxyz
2
3
4
-
Output
aabc
xyzvw | instruction | 0 | 47,920 | 19 | 95,840 |
"Correct Solution:
```
while True:
c = input()
if c == "-":
break
N = int(input())
for i in range(N):
H = int(input())
c = c [H:]+ c[:H]
print(c)
``` | output | 1 | 47,920 | 19 | 95,841 |
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