text stringlengths 198 433k | conversation_id int64 0 109k |
|---|---|
Provide tags and a correct Python 3 solution for this coding contest problem.
A boy named Vasya has taken part in an Olympiad. His teacher knows that in total Vasya got at least x points for both tours of the Olympiad. The teacher has the results of the first and the second tour of the Olympiad but the problem is, the results have only points, no names. The teacher has to know Vasya's chances.
Help Vasya's teacher, find two numbers β the best and the worst place Vasya could have won. Note that the total results' table sorts the participants by the sum of points for both tours (the first place has the participant who has got the most points). If two or more participants have got the same number of points, it's up to the jury to assign places to them according to their choice. It is guaranteed that each participant of the Olympiad participated in both tours of the Olympiad.
Input
The first line contains two space-separated integers n, x (1 β€ n β€ 105; 0 β€ x β€ 2Β·105) β the number of Olympiad participants and the minimum number of points Vasya earned.
The second line contains n space-separated integers: a1, a2, ..., an (0 β€ ai β€ 105) β the participants' points in the first tour.
The third line contains n space-separated integers: b1, b2, ..., bn (0 β€ bi β€ 105) β the participants' points in the second tour.
The participants' points are given in the arbitrary order. It is guaranteed that Vasya was present in the Olympiad β there are two integers i, j (1 β€ i, j β€ n) such, that ai + bj β₯ x.
Output
Print two space-separated integers β the best and the worst place Vasya could have got on the Olympiad.
Examples
Input
5 2
1 1 1 1 1
1 1 1 1 1
Output
1 5
Input
6 7
4 3 5 6 4 4
8 6 0 4 3 4
Output
1 5
Note
In the first text sample all 5 participants earn 2 points each in any case. Depending on the jury's decision, Vasya can get the first (the best) as well as the last (the worst) fifth place.
In the second test sample in the best case scenario Vasya wins again: he can win 12 points and become the absolute winner if the total results' table looks like that β {4:8, 6:4, 3:6, 4:4, 4:3, 5:0}.
In this table all participants are sorted by decreasing points and we can see how much a participant earned in the first and in the second tour.
In the worst case scenario Vasya can get the fifth place if the table looks like that β {4:8, 4:6, 6:4, 5:4, 4:3, 3:0}, and he earned 4 and 3 points in the first and second tours, correspondingly.
Tags: binary search, greedy, sortings, two pointers
Correct Solution:
```
from bisect import bisect_left, bisect_right
class Result:
def __init__(self, index, value):
self.index = index
self.value = value
class BinarySearch:
def __init__(self):
pass
@staticmethod
def greater_than(num: int, func, size: int = 1):
"""Searches for smallest element greater than num!"""
if isinstance(func, list):
index = bisect_right(func, num)
if index == len(func):
return Result(None, None)
else:
return Result(index, func[index])
else:
alpha, omega = 0, size - 1
if func(omega) <= num:
return Result(None, None)
while alpha < omega:
if func(alpha) > num:
return Result(alpha, func(alpha))
if omega == alpha + 1:
return Result(omega, func(omega))
mid = (alpha + omega) // 2
if func(mid) > num:
omega = mid
else:
alpha = mid
@staticmethod
def less_than(num: int, func, size: int = 1):
"""Searches for largest element less than num!"""
if isinstance(func, list):
index = bisect_left(func, num) - 1
if index == -1:
return Result(None, None)
else:
return Result(index, func[index])
else:
alpha, omega = 0, size - 1
if func(alpha) >= num:
return Result(None, None)
while alpha < omega:
if func(omega) < num:
return Result(omega, func(omega))
if omega == alpha + 1:
return Result(alpha, func(alpha))
mid = (alpha + omega) // 2
if func(mid) < num:
alpha = mid
else:
omega = mid
# ------------------- fast io --------------------
import os
import sys
from io import BytesIO, IOBase
BUFSIZE = 8192
class FastIO(IOBase):
newlines = 0
def __init__(self, file):
self._fd = file.fileno()
self.buffer = BytesIO()
self.writable = "x" in file.mode or "r" not in file.mode
self.write = self.buffer.write if self.writable else None
def read(self):
while True:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
if not b:
break
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines = 0
return self.buffer.read()
def readline(self):
while self.newlines == 0:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
self.newlines = b.count(b"\n") + (not b)
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines -= 1
return self.buffer.readline()
def flush(self):
if self.writable:
os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
self.write = lambda s: self.buffer.write(s.encode("ascii"))
self.read = lambda: self.buffer.read().decode("ascii")
self.readline = lambda: self.buffer.readline().decode("ascii")
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")
# ------------------- fast io --------------------
from math import gcd, ceil
def pre(s):
n = len(s)
pi = [0] * n
for i in range(1, n):
j = pi[i - 1]
while j and s[i] != s[j]:
j = pi[j - 1]
if s[i] == s[j]:
j += 1
pi[i] = j
return pi
def prod(a):
ans = 1
for each in a:
ans = (ans * each)
return ans
def lcm(a, b): return a * b // gcd(a, b)
def binary(x, length=16):
y = bin(x)[2:]
return y if len(y) >= length else "0" * (length - len(y)) + y
for _ in range(int(input()) if not True else 1):
n, x = map(int, input().split())
a = list(map(int, input().split()))
b = list(map(int, input().split()))
a = sorted(a)
b = sorted(b, reverse=True)
ans = 0
bs = BinarySearch()
c = []
for i in range(n):
val = b[i]
req = x - val - 1
ind = bs.greater_than(req, a).index
if ind is None:break
ind = n - ind
#print(req, ind)
c += [ind]
c = sorted(c)
for i in range(len(c)):
if c[i] > ans:
ans += 1
print(1, ans)
```
| 98,900 |
Provide tags and a correct Python 3 solution for this coding contest problem.
A boy named Vasya has taken part in an Olympiad. His teacher knows that in total Vasya got at least x points for both tours of the Olympiad. The teacher has the results of the first and the second tour of the Olympiad but the problem is, the results have only points, no names. The teacher has to know Vasya's chances.
Help Vasya's teacher, find two numbers β the best and the worst place Vasya could have won. Note that the total results' table sorts the participants by the sum of points for both tours (the first place has the participant who has got the most points). If two or more participants have got the same number of points, it's up to the jury to assign places to them according to their choice. It is guaranteed that each participant of the Olympiad participated in both tours of the Olympiad.
Input
The first line contains two space-separated integers n, x (1 β€ n β€ 105; 0 β€ x β€ 2Β·105) β the number of Olympiad participants and the minimum number of points Vasya earned.
The second line contains n space-separated integers: a1, a2, ..., an (0 β€ ai β€ 105) β the participants' points in the first tour.
The third line contains n space-separated integers: b1, b2, ..., bn (0 β€ bi β€ 105) β the participants' points in the second tour.
The participants' points are given in the arbitrary order. It is guaranteed that Vasya was present in the Olympiad β there are two integers i, j (1 β€ i, j β€ n) such, that ai + bj β₯ x.
Output
Print two space-separated integers β the best and the worst place Vasya could have got on the Olympiad.
Examples
Input
5 2
1 1 1 1 1
1 1 1 1 1
Output
1 5
Input
6 7
4 3 5 6 4 4
8 6 0 4 3 4
Output
1 5
Note
In the first text sample all 5 participants earn 2 points each in any case. Depending on the jury's decision, Vasya can get the first (the best) as well as the last (the worst) fifth place.
In the second test sample in the best case scenario Vasya wins again: he can win 12 points and become the absolute winner if the total results' table looks like that β {4:8, 6:4, 3:6, 4:4, 4:3, 5:0}.
In this table all participants are sorted by decreasing points and we can see how much a participant earned in the first and in the second tour.
In the worst case scenario Vasya can get the fifth place if the table looks like that β {4:8, 4:6, 6:4, 5:4, 4:3, 3:0}, and he earned 4 and 3 points in the first and second tours, correspondingly.
Tags: binary search, greedy, sortings, two pointers
Correct Solution:
```
from sys import stdin,stdout
import bisect as bs
nmbr = lambda: int(stdin.readline())
lst = lambda: list(map(int,stdin.readline().split()))
for _ in range(1):#nmbr()):
n,sm=lst()
a=sorted(lst(),reverse=1)
b=sorted(lst())
used=-1
rank=1
# print(a)
# print(b)
p=-1
for i in range(n):
v1=a[i]
p=bs.bisect_left(b,sm-v1,p+1,n)
# print(a[i],p)
if p>=n:
rank=i
# print(rank)
break
rank=i+1
print(1,rank)
```
| 98,901 |
Provide tags and a correct Python 3 solution for this coding contest problem.
A boy named Vasya has taken part in an Olympiad. His teacher knows that in total Vasya got at least x points for both tours of the Olympiad. The teacher has the results of the first and the second tour of the Olympiad but the problem is, the results have only points, no names. The teacher has to know Vasya's chances.
Help Vasya's teacher, find two numbers β the best and the worst place Vasya could have won. Note that the total results' table sorts the participants by the sum of points for both tours (the first place has the participant who has got the most points). If two or more participants have got the same number of points, it's up to the jury to assign places to them according to their choice. It is guaranteed that each participant of the Olympiad participated in both tours of the Olympiad.
Input
The first line contains two space-separated integers n, x (1 β€ n β€ 105; 0 β€ x β€ 2Β·105) β the number of Olympiad participants and the minimum number of points Vasya earned.
The second line contains n space-separated integers: a1, a2, ..., an (0 β€ ai β€ 105) β the participants' points in the first tour.
The third line contains n space-separated integers: b1, b2, ..., bn (0 β€ bi β€ 105) β the participants' points in the second tour.
The participants' points are given in the arbitrary order. It is guaranteed that Vasya was present in the Olympiad β there are two integers i, j (1 β€ i, j β€ n) such, that ai + bj β₯ x.
Output
Print two space-separated integers β the best and the worst place Vasya could have got on the Olympiad.
Examples
Input
5 2
1 1 1 1 1
1 1 1 1 1
Output
1 5
Input
6 7
4 3 5 6 4 4
8 6 0 4 3 4
Output
1 5
Note
In the first text sample all 5 participants earn 2 points each in any case. Depending on the jury's decision, Vasya can get the first (the best) as well as the last (the worst) fifth place.
In the second test sample in the best case scenario Vasya wins again: he can win 12 points and become the absolute winner if the total results' table looks like that β {4:8, 6:4, 3:6, 4:4, 4:3, 5:0}.
In this table all participants are sorted by decreasing points and we can see how much a participant earned in the first and in the second tour.
In the worst case scenario Vasya can get the fifth place if the table looks like that β {4:8, 4:6, 6:4, 5:4, 4:3, 3:0}, and he earned 4 and 3 points in the first and second tours, correspondingly.
Tags: binary search, greedy, sortings, two pointers
Correct Solution:
```
n,x = map(int, input().split())
a = sorted(map(int, input().split()))
b = sorted(map(int, input().split()), reverse = True)
pa,pb = 0,0
worst = 0
while pa < len(a) and pb < len(b) :
if a[pa] + b[pb] >= x :
worst += 1
pb += 1
pa += 1
print(1, worst)
```
| 98,902 |
Provide tags and a correct Python 3 solution for this coding contest problem.
A boy named Vasya has taken part in an Olympiad. His teacher knows that in total Vasya got at least x points for both tours of the Olympiad. The teacher has the results of the first and the second tour of the Olympiad but the problem is, the results have only points, no names. The teacher has to know Vasya's chances.
Help Vasya's teacher, find two numbers β the best and the worst place Vasya could have won. Note that the total results' table sorts the participants by the sum of points for both tours (the first place has the participant who has got the most points). If two or more participants have got the same number of points, it's up to the jury to assign places to them according to their choice. It is guaranteed that each participant of the Olympiad participated in both tours of the Olympiad.
Input
The first line contains two space-separated integers n, x (1 β€ n β€ 105; 0 β€ x β€ 2Β·105) β the number of Olympiad participants and the minimum number of points Vasya earned.
The second line contains n space-separated integers: a1, a2, ..., an (0 β€ ai β€ 105) β the participants' points in the first tour.
The third line contains n space-separated integers: b1, b2, ..., bn (0 β€ bi β€ 105) β the participants' points in the second tour.
The participants' points are given in the arbitrary order. It is guaranteed that Vasya was present in the Olympiad β there are two integers i, j (1 β€ i, j β€ n) such, that ai + bj β₯ x.
Output
Print two space-separated integers β the best and the worst place Vasya could have got on the Olympiad.
Examples
Input
5 2
1 1 1 1 1
1 1 1 1 1
Output
1 5
Input
6 7
4 3 5 6 4 4
8 6 0 4 3 4
Output
1 5
Note
In the first text sample all 5 participants earn 2 points each in any case. Depending on the jury's decision, Vasya can get the first (the best) as well as the last (the worst) fifth place.
In the second test sample in the best case scenario Vasya wins again: he can win 12 points and become the absolute winner if the total results' table looks like that β {4:8, 6:4, 3:6, 4:4, 4:3, 5:0}.
In this table all participants are sorted by decreasing points and we can see how much a participant earned in the first and in the second tour.
In the worst case scenario Vasya can get the fifth place if the table looks like that β {4:8, 4:6, 6:4, 5:4, 4:3, 3:0}, and he earned 4 and 3 points in the first and second tours, correspondingly.
Tags: binary search, greedy, sortings, two pointers
Correct Solution:
```
import os
import sys
from io import BytesIO, IOBase
BUFSIZE = 8192
class FastIO(IOBase):
newlines = 0
def __init__(self, file):
self._fd = file.fileno()
self.buffer = BytesIO()
self.writable = "x" in file.mode or "r" not in file.mode
self.write = self.buffer.write if self.writable else None
def read(self):
while True:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
if not b:
break
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines = 0
return self.buffer.read()
def readline(self):
while self.newlines == 0:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
self.newlines = b.count(b"\n") + (not b)
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines -= 1
return self.buffer.readline()
def flush(self):
if self.writable:
os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
self.write = lambda s: self.buffer.write(s.encode("ascii"))
self.read = lambda: self.buffer.read().decode("ascii")
self.readline = lambda: self.buffer.readline().decode("ascii")
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")
#n=int(input())
#arr = list(map(int, input().split()))
n,x= map(int, input().split())
arr = sorted(list(map(int, input().split())))
ls = list(map(int, input().split()))
ls=sorted(ls,reverse=True)
#print(arr)
#print(ls)
ans=0
j=0
for i in range(n):
while j<n and arr[j]+ls[i]<x:
j+=1
if j<n and arr[j]+ls[i]>=x:
ans+=1
j+=1
print(1, ans)
```
| 98,903 |
Provide tags and a correct Python 3 solution for this coding contest problem.
A boy named Vasya has taken part in an Olympiad. His teacher knows that in total Vasya got at least x points for both tours of the Olympiad. The teacher has the results of the first and the second tour of the Olympiad but the problem is, the results have only points, no names. The teacher has to know Vasya's chances.
Help Vasya's teacher, find two numbers β the best and the worst place Vasya could have won. Note that the total results' table sorts the participants by the sum of points for both tours (the first place has the participant who has got the most points). If two or more participants have got the same number of points, it's up to the jury to assign places to them according to their choice. It is guaranteed that each participant of the Olympiad participated in both tours of the Olympiad.
Input
The first line contains two space-separated integers n, x (1 β€ n β€ 105; 0 β€ x β€ 2Β·105) β the number of Olympiad participants and the minimum number of points Vasya earned.
The second line contains n space-separated integers: a1, a2, ..., an (0 β€ ai β€ 105) β the participants' points in the first tour.
The third line contains n space-separated integers: b1, b2, ..., bn (0 β€ bi β€ 105) β the participants' points in the second tour.
The participants' points are given in the arbitrary order. It is guaranteed that Vasya was present in the Olympiad β there are two integers i, j (1 β€ i, j β€ n) such, that ai + bj β₯ x.
Output
Print two space-separated integers β the best and the worst place Vasya could have got on the Olympiad.
Examples
Input
5 2
1 1 1 1 1
1 1 1 1 1
Output
1 5
Input
6 7
4 3 5 6 4 4
8 6 0 4 3 4
Output
1 5
Note
In the first text sample all 5 participants earn 2 points each in any case. Depending on the jury's decision, Vasya can get the first (the best) as well as the last (the worst) fifth place.
In the second test sample in the best case scenario Vasya wins again: he can win 12 points and become the absolute winner if the total results' table looks like that β {4:8, 6:4, 3:6, 4:4, 4:3, 5:0}.
In this table all participants are sorted by decreasing points and we can see how much a participant earned in the first and in the second tour.
In the worst case scenario Vasya can get the fifth place if the table looks like that β {4:8, 4:6, 6:4, 5:4, 4:3, 3:0}, and he earned 4 and 3 points in the first and second tours, correspondingly.
Tags: binary search, greedy, sortings, two pointers
Correct Solution:
```
n, x = map(int, input().split())
score1 = map(int, input().split())
score2 = map(int, input().split())
score1 = sorted(score1, reverse=True)
score2 = sorted(score2, reverse=True)
count = 0
i = k = 0
j = l = (n - 1)
while i <= j and k <= l:
if score1[i] + score2[l] >= score2[k] + score1[j]:
if score1[i] + score2[l] >= x:
count += 1
i += 1
l -= 1
else:
if score2[k] + score1[j] >= x:
count += 1
k += 1
j -= 1
print(1, count)
```
| 98,904 |
Provide tags and a correct Python 3 solution for this coding contest problem.
A boy named Vasya has taken part in an Olympiad. His teacher knows that in total Vasya got at least x points for both tours of the Olympiad. The teacher has the results of the first and the second tour of the Olympiad but the problem is, the results have only points, no names. The teacher has to know Vasya's chances.
Help Vasya's teacher, find two numbers β the best and the worst place Vasya could have won. Note that the total results' table sorts the participants by the sum of points for both tours (the first place has the participant who has got the most points). If two or more participants have got the same number of points, it's up to the jury to assign places to them according to their choice. It is guaranteed that each participant of the Olympiad participated in both tours of the Olympiad.
Input
The first line contains two space-separated integers n, x (1 β€ n β€ 105; 0 β€ x β€ 2Β·105) β the number of Olympiad participants and the minimum number of points Vasya earned.
The second line contains n space-separated integers: a1, a2, ..., an (0 β€ ai β€ 105) β the participants' points in the first tour.
The third line contains n space-separated integers: b1, b2, ..., bn (0 β€ bi β€ 105) β the participants' points in the second tour.
The participants' points are given in the arbitrary order. It is guaranteed that Vasya was present in the Olympiad β there are two integers i, j (1 β€ i, j β€ n) such, that ai + bj β₯ x.
Output
Print two space-separated integers β the best and the worst place Vasya could have got on the Olympiad.
Examples
Input
5 2
1 1 1 1 1
1 1 1 1 1
Output
1 5
Input
6 7
4 3 5 6 4 4
8 6 0 4 3 4
Output
1 5
Note
In the first text sample all 5 participants earn 2 points each in any case. Depending on the jury's decision, Vasya can get the first (the best) as well as the last (the worst) fifth place.
In the second test sample in the best case scenario Vasya wins again: he can win 12 points and become the absolute winner if the total results' table looks like that β {4:8, 6:4, 3:6, 4:4, 4:3, 5:0}.
In this table all participants are sorted by decreasing points and we can see how much a participant earned in the first and in the second tour.
In the worst case scenario Vasya can get the fifth place if the table looks like that β {4:8, 4:6, 6:4, 5:4, 4:3, 3:0}, and he earned 4 and 3 points in the first and second tours, correspondingly.
Tags: binary search, greedy, sortings, two pointers
Correct Solution:
```
from operator import add
from bisect import bisect_left
n, x = map(int, input().split())
a = list(map(int, input().split()))
b = list(map(int, input().split()))
a.sort()
b.sort()
b_len = len(b)
ans = 0
for i in range(n):
pos = bisect_left(b, x - a[i])
if not pos > b_len - 1:
#del b[pos]
b_len -= 1
ans += 1
print(1, ans)
```
| 98,905 |
Provide tags and a correct Python 3 solution for this coding contest problem.
A boy named Vasya has taken part in an Olympiad. His teacher knows that in total Vasya got at least x points for both tours of the Olympiad. The teacher has the results of the first and the second tour of the Olympiad but the problem is, the results have only points, no names. The teacher has to know Vasya's chances.
Help Vasya's teacher, find two numbers β the best and the worst place Vasya could have won. Note that the total results' table sorts the participants by the sum of points for both tours (the first place has the participant who has got the most points). If two or more participants have got the same number of points, it's up to the jury to assign places to them according to their choice. It is guaranteed that each participant of the Olympiad participated in both tours of the Olympiad.
Input
The first line contains two space-separated integers n, x (1 β€ n β€ 105; 0 β€ x β€ 2Β·105) β the number of Olympiad participants and the minimum number of points Vasya earned.
The second line contains n space-separated integers: a1, a2, ..., an (0 β€ ai β€ 105) β the participants' points in the first tour.
The third line contains n space-separated integers: b1, b2, ..., bn (0 β€ bi β€ 105) β the participants' points in the second tour.
The participants' points are given in the arbitrary order. It is guaranteed that Vasya was present in the Olympiad β there are two integers i, j (1 β€ i, j β€ n) such, that ai + bj β₯ x.
Output
Print two space-separated integers β the best and the worst place Vasya could have got on the Olympiad.
Examples
Input
5 2
1 1 1 1 1
1 1 1 1 1
Output
1 5
Input
6 7
4 3 5 6 4 4
8 6 0 4 3 4
Output
1 5
Note
In the first text sample all 5 participants earn 2 points each in any case. Depending on the jury's decision, Vasya can get the first (the best) as well as the last (the worst) fifth place.
In the second test sample in the best case scenario Vasya wins again: he can win 12 points and become the absolute winner if the total results' table looks like that β {4:8, 6:4, 3:6, 4:4, 4:3, 5:0}.
In this table all participants are sorted by decreasing points and we can see how much a participant earned in the first and in the second tour.
In the worst case scenario Vasya can get the fifth place if the table looks like that β {4:8, 4:6, 6:4, 5:4, 4:3, 3:0}, and he earned 4 and 3 points in the first and second tours, correspondingly.
Tags: binary search, greedy, sortings, two pointers
Correct Solution:
```
from sys import stdin
from collections import deque
n,x = [int(x) for x in stdin.readline().split()]
s1 = deque(sorted([int(x) for x in stdin.readline().split()]))
s2 = deque(sorted([int(x) for x in stdin.readline().split()]))
place = 0
for score in s1:
if s2[-1] + score >= x:
place += 1
s2.pop()
else:
s2.popleft()
print(1,place)
```
| 98,906 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A boy named Vasya has taken part in an Olympiad. His teacher knows that in total Vasya got at least x points for both tours of the Olympiad. The teacher has the results of the first and the second tour of the Olympiad but the problem is, the results have only points, no names. The teacher has to know Vasya's chances.
Help Vasya's teacher, find two numbers β the best and the worst place Vasya could have won. Note that the total results' table sorts the participants by the sum of points for both tours (the first place has the participant who has got the most points). If two or more participants have got the same number of points, it's up to the jury to assign places to them according to their choice. It is guaranteed that each participant of the Olympiad participated in both tours of the Olympiad.
Input
The first line contains two space-separated integers n, x (1 β€ n β€ 105; 0 β€ x β€ 2Β·105) β the number of Olympiad participants and the minimum number of points Vasya earned.
The second line contains n space-separated integers: a1, a2, ..., an (0 β€ ai β€ 105) β the participants' points in the first tour.
The third line contains n space-separated integers: b1, b2, ..., bn (0 β€ bi β€ 105) β the participants' points in the second tour.
The participants' points are given in the arbitrary order. It is guaranteed that Vasya was present in the Olympiad β there are two integers i, j (1 β€ i, j β€ n) such, that ai + bj β₯ x.
Output
Print two space-separated integers β the best and the worst place Vasya could have got on the Olympiad.
Examples
Input
5 2
1 1 1 1 1
1 1 1 1 1
Output
1 5
Input
6 7
4 3 5 6 4 4
8 6 0 4 3 4
Output
1 5
Note
In the first text sample all 5 participants earn 2 points each in any case. Depending on the jury's decision, Vasya can get the first (the best) as well as the last (the worst) fifth place.
In the second test sample in the best case scenario Vasya wins again: he can win 12 points and become the absolute winner if the total results' table looks like that β {4:8, 6:4, 3:6, 4:4, 4:3, 5:0}.
In this table all participants are sorted by decreasing points and we can see how much a participant earned in the first and in the second tour.
In the worst case scenario Vasya can get the fifth place if the table looks like that β {4:8, 4:6, 6:4, 5:4, 4:3, 3:0}, and he earned 4 and 3 points in the first and second tours, correspondingly.
Submitted Solution:
```
#------------------------template--------------------------#
import os
import sys
from math import *
from collections import *
# from fractions import *
# from heapq import*
from bisect import *
from io import BytesIO, IOBase
def vsInput():
sys.stdin = open('input.txt', 'r')
sys.stdout = open('output.txt', 'w')
BUFSIZE = 8192
class FastIO(IOBase):
newlines = 0
def __init__(self, file):
self._fd = file.fileno()
self.buffer = BytesIO()
self.writable = "x" in file.mode or "r" not in file.mode
self.write = self.buffer.write if self.writable else None
def read(self):
while True:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
if not b:
break
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines = 0
return self.buffer.read()
def readline(self):
while self.newlines == 0:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
self.newlines = b.count(b"\n") + (not b)
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines -= 1
return self.buffer.readline()
def flush(self):
if self.writable:
os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
self.write = lambda s: self.buffer.write(s.encode("ascii"))
self.read = lambda: self.buffer.read().decode("ascii")
self.readline = lambda: self.buffer.readline().decode("ascii")
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")
ALPHA='abcdefghijklmnopqrstuvwxyz/'
M=998244353
EPS=1e-6
def Ceil(a,b): return a//b+int(a%b>0)
def value():return tuple(map(int,input().split()))
def array():return [int(i) for i in input().split()]
def Int():return int(input())
def Str():return input()
def arrayS():return [i for i in input().split()]
#-------------------------code---------------------------#
# vsInput()
n,k=value()
a=sorted(array(),reverse=True)
b=sorted(array())
Ii=0
Ij=n-1
have=have=a[Ii]+b[Ij]
for i in range(n):
j=bisect_left(b,k-a[i])
if(j<n and k<=a[i]+b[j]<have):
Ii=i
Ij=j
have=a[i]+b[j]
a.remove(a[Ii])
b.remove(b[Ij])
n-=1
# print(have)
# print(a)
# print(b)
ans1=1
ans2=1
low=0
for i in a:
need=have-i
j=max(low,bisect_left(b,need))
if(j<n and b[j]+i>=have):
ans2+=1
low=j+1
print(ans1,ans2)
```
Yes
| 98,907 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A boy named Vasya has taken part in an Olympiad. His teacher knows that in total Vasya got at least x points for both tours of the Olympiad. The teacher has the results of the first and the second tour of the Olympiad but the problem is, the results have only points, no names. The teacher has to know Vasya's chances.
Help Vasya's teacher, find two numbers β the best and the worst place Vasya could have won. Note that the total results' table sorts the participants by the sum of points for both tours (the first place has the participant who has got the most points). If two or more participants have got the same number of points, it's up to the jury to assign places to them according to their choice. It is guaranteed that each participant of the Olympiad participated in both tours of the Olympiad.
Input
The first line contains two space-separated integers n, x (1 β€ n β€ 105; 0 β€ x β€ 2Β·105) β the number of Olympiad participants and the minimum number of points Vasya earned.
The second line contains n space-separated integers: a1, a2, ..., an (0 β€ ai β€ 105) β the participants' points in the first tour.
The third line contains n space-separated integers: b1, b2, ..., bn (0 β€ bi β€ 105) β the participants' points in the second tour.
The participants' points are given in the arbitrary order. It is guaranteed that Vasya was present in the Olympiad β there are two integers i, j (1 β€ i, j β€ n) such, that ai + bj β₯ x.
Output
Print two space-separated integers β the best and the worst place Vasya could have got on the Olympiad.
Examples
Input
5 2
1 1 1 1 1
1 1 1 1 1
Output
1 5
Input
6 7
4 3 5 6 4 4
8 6 0 4 3 4
Output
1 5
Note
In the first text sample all 5 participants earn 2 points each in any case. Depending on the jury's decision, Vasya can get the first (the best) as well as the last (the worst) fifth place.
In the second test sample in the best case scenario Vasya wins again: he can win 12 points and become the absolute winner if the total results' table looks like that β {4:8, 6:4, 3:6, 4:4, 4:3, 5:0}.
In this table all participants are sorted by decreasing points and we can see how much a participant earned in the first and in the second tour.
In the worst case scenario Vasya can get the fifth place if the table looks like that β {4:8, 4:6, 6:4, 5:4, 4:3, 3:0}, and he earned 4 and 3 points in the first and second tours, correspondingly.
Submitted Solution:
```
# ------------------- fast io --------------------
import os
import sys
from io import BytesIO, IOBase
BUFSIZE = 8192
class FastIO(IOBase):
newlines = 0
def __init__(self, file):
self._fd = file.fileno()
self.buffer = BytesIO()
self.writable = "x" in file.mode or "r" not in file.mode
self.write = self.buffer.write if self.writable else None
def read(self):
while True:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
if not b:
break
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines = 0
return self.buffer.read()
def readline(self):
while self.newlines == 0:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
self.newlines = b.count(b"\n") + (not b)
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines -= 1
return self.buffer.readline()
def flush(self):
if self.writable:
os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
self.write = lambda s: self.buffer.write(s.encode("ascii"))
self.read = lambda: self.buffer.read().decode("ascii")
self.readline = lambda: self.buffer.readline().decode("ascii")
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")
# ------------------- fast io --------------------
from math import gcd, ceil
def pre(s):
n = len(s)
pi = [0] * n
for i in range(1, n):
j = pi[i - 1]
while j and s[i] != s[j]:
j = pi[j - 1]
if s[i] == s[j]:
j += 1
pi[i] = j
return pi
def prod(a):
ans = 1
for each in a:
ans = (ans * each)
return ans
def lcm(a, b): return a * b // gcd(a, b)
def binary(x, length=16):
y = bin(x)[2:]
return y if len(y) >= length else "0" * (length - len(y)) + y
for _ in range(int(input()) if not True else 1):
n, x = map(int, input().split())
a = list(map(int, input().split()))
b = list(map(int, input().split()))
a = sorted(a)
b = sorted(b, reverse=True)
ans = 0
for i in range(n):
if (a[i]+b[i]) >= x:
ans += 1
print(1, ans)
```
No
| 98,908 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A boy named Vasya has taken part in an Olympiad. His teacher knows that in total Vasya got at least x points for both tours of the Olympiad. The teacher has the results of the first and the second tour of the Olympiad but the problem is, the results have only points, no names. The teacher has to know Vasya's chances.
Help Vasya's teacher, find two numbers β the best and the worst place Vasya could have won. Note that the total results' table sorts the participants by the sum of points for both tours (the first place has the participant who has got the most points). If two or more participants have got the same number of points, it's up to the jury to assign places to them according to their choice. It is guaranteed that each participant of the Olympiad participated in both tours of the Olympiad.
Input
The first line contains two space-separated integers n, x (1 β€ n β€ 105; 0 β€ x β€ 2Β·105) β the number of Olympiad participants and the minimum number of points Vasya earned.
The second line contains n space-separated integers: a1, a2, ..., an (0 β€ ai β€ 105) β the participants' points in the first tour.
The third line contains n space-separated integers: b1, b2, ..., bn (0 β€ bi β€ 105) β the participants' points in the second tour.
The participants' points are given in the arbitrary order. It is guaranteed that Vasya was present in the Olympiad β there are two integers i, j (1 β€ i, j β€ n) such, that ai + bj β₯ x.
Output
Print two space-separated integers β the best and the worst place Vasya could have got on the Olympiad.
Examples
Input
5 2
1 1 1 1 1
1 1 1 1 1
Output
1 5
Input
6 7
4 3 5 6 4 4
8 6 0 4 3 4
Output
1 5
Note
In the first text sample all 5 participants earn 2 points each in any case. Depending on the jury's decision, Vasya can get the first (the best) as well as the last (the worst) fifth place.
In the second test sample in the best case scenario Vasya wins again: he can win 12 points and become the absolute winner if the total results' table looks like that β {4:8, 6:4, 3:6, 4:4, 4:3, 5:0}.
In this table all participants are sorted by decreasing points and we can see how much a participant earned in the first and in the second tour.
In the worst case scenario Vasya can get the fifth place if the table looks like that β {4:8, 4:6, 6:4, 5:4, 4:3, 3:0}, and he earned 4 and 3 points in the first and second tours, correspondingly.
Submitted Solution:
```
from bisect import bisect_left, bisect_right
s, n = map(int, input().split())
p = list(map(int, input().split()))
for i, j in enumerate(map(int, input().split())): p[i] += j
p.sort()
a, b = bisect_left(p, n), bisect_right(p, n)
print(1, n - a - 1)
```
No
| 98,909 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A boy named Vasya has taken part in an Olympiad. His teacher knows that in total Vasya got at least x points for both tours of the Olympiad. The teacher has the results of the first and the second tour of the Olympiad but the problem is, the results have only points, no names. The teacher has to know Vasya's chances.
Help Vasya's teacher, find two numbers β the best and the worst place Vasya could have won. Note that the total results' table sorts the participants by the sum of points for both tours (the first place has the participant who has got the most points). If two or more participants have got the same number of points, it's up to the jury to assign places to them according to their choice. It is guaranteed that each participant of the Olympiad participated in both tours of the Olympiad.
Input
The first line contains two space-separated integers n, x (1 β€ n β€ 105; 0 β€ x β€ 2Β·105) β the number of Olympiad participants and the minimum number of points Vasya earned.
The second line contains n space-separated integers: a1, a2, ..., an (0 β€ ai β€ 105) β the participants' points in the first tour.
The third line contains n space-separated integers: b1, b2, ..., bn (0 β€ bi β€ 105) β the participants' points in the second tour.
The participants' points are given in the arbitrary order. It is guaranteed that Vasya was present in the Olympiad β there are two integers i, j (1 β€ i, j β€ n) such, that ai + bj β₯ x.
Output
Print two space-separated integers β the best and the worst place Vasya could have got on the Olympiad.
Examples
Input
5 2
1 1 1 1 1
1 1 1 1 1
Output
1 5
Input
6 7
4 3 5 6 4 4
8 6 0 4 3 4
Output
1 5
Note
In the first text sample all 5 participants earn 2 points each in any case. Depending on the jury's decision, Vasya can get the first (the best) as well as the last (the worst) fifth place.
In the second test sample in the best case scenario Vasya wins again: he can win 12 points and become the absolute winner if the total results' table looks like that β {4:8, 6:4, 3:6, 4:4, 4:3, 5:0}.
In this table all participants are sorted by decreasing points and we can see how much a participant earned in the first and in the second tour.
In the worst case scenario Vasya can get the fifth place if the table looks like that β {4:8, 4:6, 6:4, 5:4, 4:3, 3:0}, and he earned 4 and 3 points in the first and second tours, correspondingly.
Submitted Solution:
```
from bisect import bisect_left, bisect_right
class Result:
def __init__(self, index, value):
self.index = index
self.value = value
class BinarySearch:
def __init__(self):
pass
@staticmethod
def greater_than(num: int, func, size: int = 1):
"""Searches for smallest element greater than num!"""
if isinstance(func, list):
index = bisect_right(func, num)
if index == len(func):
return Result(None, None)
else:
return Result(index, func[index])
else:
alpha, omega = 0, size - 1
if func(omega) <= num:
return Result(None, None)
while alpha < omega:
if func(alpha) > num:
return Result(alpha, func(alpha))
if omega == alpha + 1:
return Result(omega, func(omega))
mid = (alpha + omega) // 2
if func(mid) > num:
omega = mid
else:
alpha = mid
@staticmethod
def less_than(num: int, func, size: int = 1):
"""Searches for largest element less than num!"""
if isinstance(func, list):
index = bisect_left(func, num) - 1
if index == -1:
return Result(None, None)
else:
return Result(index, func[index])
else:
alpha, omega = 0, size - 1
if func(alpha) >= num:
return Result(None, None)
while alpha < omega:
if func(omega) < num:
return Result(omega, func(omega))
if omega == alpha + 1:
return Result(alpha, func(alpha))
mid = (alpha + omega) // 2
if func(mid) < num:
alpha = mid
else:
omega = mid
# ------------------- fast io --------------------
import os
import sys
from io import BytesIO, IOBase
BUFSIZE = 8192
class FastIO(IOBase):
newlines = 0
def __init__(self, file):
self._fd = file.fileno()
self.buffer = BytesIO()
self.writable = "x" in file.mode or "r" not in file.mode
self.write = self.buffer.write if self.writable else None
def read(self):
while True:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
if not b:
break
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines = 0
return self.buffer.read()
def readline(self):
while self.newlines == 0:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
self.newlines = b.count(b"\n") + (not b)
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines -= 1
return self.buffer.readline()
def flush(self):
if self.writable:
os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
self.write = lambda s: self.buffer.write(s.encode("ascii"))
self.read = lambda: self.buffer.read().decode("ascii")
self.readline = lambda: self.buffer.readline().decode("ascii")
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")
# ------------------- fast io --------------------
from math import gcd, ceil
def pre(s):
n = len(s)
pi = [0] * n
for i in range(1, n):
j = pi[i - 1]
while j and s[i] != s[j]:
j = pi[j - 1]
if s[i] == s[j]:
j += 1
pi[i] = j
return pi
def prod(a):
ans = 1
for each in a:
ans = (ans * each)
return ans
def lcm(a, b): return a * b // gcd(a, b)
def binary(x, length=16):
y = bin(x)[2:]
return y if len(y) >= length else "0" * (length - len(y)) + y
for _ in range(int(input()) if not True else 1):
n, x = map(int, input().split())
a = list(map(int, input().split()))
b = list(map(int, input().split()))
a = sorted(a)
b = sorted(b, reverse=True)
ans = 0
bs = BinarySearch()
c = []
for i in range(n):
val = b[i]
req = x - val - 1
ind = bs.greater_than(req, a).index
if ind is None:break
ind = n - ind
#print(req, ind)
c += [ind]
c = sorted(c)
for i in range(len(c)):
if c[i] >= i + 1:
ans = i + 1
ans2 = 0
c = []
a = a[::-1]
b = b[::-1]
for i in range(n):
val = a[i]
req = x - val - 1
ind = bs.greater_than(req, b).index
if ind is None:break
ind = n - ind
#print(req, ind)
c += [ind]
c = sorted(c)
for i in range(len(c)):
if c[i] >= i + 1:
ans2 = i + 1
print(1, max(ans, ans2))
```
No
| 98,910 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A boy named Vasya has taken part in an Olympiad. His teacher knows that in total Vasya got at least x points for both tours of the Olympiad. The teacher has the results of the first and the second tour of the Olympiad but the problem is, the results have only points, no names. The teacher has to know Vasya's chances.
Help Vasya's teacher, find two numbers β the best and the worst place Vasya could have won. Note that the total results' table sorts the participants by the sum of points for both tours (the first place has the participant who has got the most points). If two or more participants have got the same number of points, it's up to the jury to assign places to them according to their choice. It is guaranteed that each participant of the Olympiad participated in both tours of the Olympiad.
Input
The first line contains two space-separated integers n, x (1 β€ n β€ 105; 0 β€ x β€ 2Β·105) β the number of Olympiad participants and the minimum number of points Vasya earned.
The second line contains n space-separated integers: a1, a2, ..., an (0 β€ ai β€ 105) β the participants' points in the first tour.
The third line contains n space-separated integers: b1, b2, ..., bn (0 β€ bi β€ 105) β the participants' points in the second tour.
The participants' points are given in the arbitrary order. It is guaranteed that Vasya was present in the Olympiad β there are two integers i, j (1 β€ i, j β€ n) such, that ai + bj β₯ x.
Output
Print two space-separated integers β the best and the worst place Vasya could have got on the Olympiad.
Examples
Input
5 2
1 1 1 1 1
1 1 1 1 1
Output
1 5
Input
6 7
4 3 5 6 4 4
8 6 0 4 3 4
Output
1 5
Note
In the first text sample all 5 participants earn 2 points each in any case. Depending on the jury's decision, Vasya can get the first (the best) as well as the last (the worst) fifth place.
In the second test sample in the best case scenario Vasya wins again: he can win 12 points and become the absolute winner if the total results' table looks like that β {4:8, 6:4, 3:6, 4:4, 4:3, 5:0}.
In this table all participants are sorted by decreasing points and we can see how much a participant earned in the first and in the second tour.
In the worst case scenario Vasya can get the fifth place if the table looks like that β {4:8, 4:6, 6:4, 5:4, 4:3, 3:0}, and he earned 4 and 3 points in the first and second tours, correspondingly.
Submitted Solution:
```
import os
import sys
from io import BytesIO, IOBase
BUFSIZE = 8192
class FastIO(IOBase):
newlines = 0
def __init__(self, file):
self._fd = file.fileno()
self.buffer = BytesIO()
self.writable = "x" in file.mode or "r" not in file.mode
self.write = self.buffer.write if self.writable else None
def read(self):
while True:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
if not b:
break
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines = 0
return self.buffer.read()
def readline(self):
while self.newlines == 0:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
self.newlines = b.count(b"\n") + (not b)
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines -= 1
return self.buffer.readline()
def flush(self):
if self.writable:
os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
self.write = lambda s: self.buffer.write(s.encode("ascii"))
self.read = lambda: self.buffer.read().decode("ascii")
self.readline = lambda: self.buffer.readline().decode("ascii")
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")
#n=int(input())
#arr = list(map(int, input().split()))
n,x= map(int, input().split())
arr = sorted(list(map(int, input().split())))
ls = list(map(int, input().split()))
ls=sorted(ls,reverse=True)
ans=0
for i in range(n):
if arr[i]+ls[i]>=x:
ans+=1
print(1, ans)
```
No
| 98,911 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Polycarpus has an array, consisting of n integers a1, a2, ..., an. Polycarpus likes it when numbers in an array match. That's why he wants the array to have as many equal numbers as possible. For that Polycarpus performs the following operation multiple times:
* he chooses two elements of the array ai, aj (i β j);
* he simultaneously increases number ai by 1 and decreases number aj by 1, that is, executes ai = ai + 1 and aj = aj - 1.
The given operation changes exactly two distinct array elements. Polycarpus can apply the described operation an infinite number of times.
Now he wants to know what maximum number of equal array elements he can get if he performs an arbitrary number of such operation. Help Polycarpus.
Input
The first line contains integer n (1 β€ n β€ 105) β the array size. The second line contains space-separated integers a1, a2, ..., an (|ai| β€ 104) β the original array.
Output
Print a single integer β the maximum number of equal array elements he can get if he performs an arbitrary number of the given operation.
Examples
Input
2
2 1
Output
1
Input
3
1 4 1
Output
3
Tags: greedy, math
Correct Solution:
```
n=int(input())
arr=list(map(int,input().split()))
s=0
for i in range(n):
s+=arr[i]
if s%n==0:
print(n)
else:
print(n-1)
```
| 98,912 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Polycarpus has an array, consisting of n integers a1, a2, ..., an. Polycarpus likes it when numbers in an array match. That's why he wants the array to have as many equal numbers as possible. For that Polycarpus performs the following operation multiple times:
* he chooses two elements of the array ai, aj (i β j);
* he simultaneously increases number ai by 1 and decreases number aj by 1, that is, executes ai = ai + 1 and aj = aj - 1.
The given operation changes exactly two distinct array elements. Polycarpus can apply the described operation an infinite number of times.
Now he wants to know what maximum number of equal array elements he can get if he performs an arbitrary number of such operation. Help Polycarpus.
Input
The first line contains integer n (1 β€ n β€ 105) β the array size. The second line contains space-separated integers a1, a2, ..., an (|ai| β€ 104) β the original array.
Output
Print a single integer β the maximum number of equal array elements he can get if he performs an arbitrary number of the given operation.
Examples
Input
2
2 1
Output
1
Input
3
1 4 1
Output
3
Tags: greedy, math
Correct Solution:
```
# It's all about what U BELIEVE
def gint(): return int(input())
def gint_arr(): return list(map(int, input().split()))
def gfloat(): return float(input())
def gfloat_arr(): return list(map(float, input().split()))
def pair_int(): return map(int, input().split())
###############################################################################
INF = (1 << 31)
dx = [-1, 0, 1, 0]
dy = [ 0, 1, 0, -1]
###############################################################################
############################ SOLUTION IS COMING ###############################
###############################################################################
n = gint()
a = gint_arr()
print(n if sum(a) % n == 0 else n - 1)
```
| 98,913 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Polycarpus has an array, consisting of n integers a1, a2, ..., an. Polycarpus likes it when numbers in an array match. That's why he wants the array to have as many equal numbers as possible. For that Polycarpus performs the following operation multiple times:
* he chooses two elements of the array ai, aj (i β j);
* he simultaneously increases number ai by 1 and decreases number aj by 1, that is, executes ai = ai + 1 and aj = aj - 1.
The given operation changes exactly two distinct array elements. Polycarpus can apply the described operation an infinite number of times.
Now he wants to know what maximum number of equal array elements he can get if he performs an arbitrary number of such operation. Help Polycarpus.
Input
The first line contains integer n (1 β€ n β€ 105) β the array size. The second line contains space-separated integers a1, a2, ..., an (|ai| β€ 104) β the original array.
Output
Print a single integer β the maximum number of equal array elements he can get if he performs an arbitrary number of the given operation.
Examples
Input
2
2 1
Output
1
Input
3
1 4 1
Output
3
Tags: greedy, math
Correct Solution:
```
def IC():
n=int(input())
a=[int(x) for x in input().split()]
shave=sum(a)
if shave % n == 0:
print(n)
return
else:
print(n-1)
return
IC()
```
| 98,914 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Polycarpus has an array, consisting of n integers a1, a2, ..., an. Polycarpus likes it when numbers in an array match. That's why he wants the array to have as many equal numbers as possible. For that Polycarpus performs the following operation multiple times:
* he chooses two elements of the array ai, aj (i β j);
* he simultaneously increases number ai by 1 and decreases number aj by 1, that is, executes ai = ai + 1 and aj = aj - 1.
The given operation changes exactly two distinct array elements. Polycarpus can apply the described operation an infinite number of times.
Now he wants to know what maximum number of equal array elements he can get if he performs an arbitrary number of such operation. Help Polycarpus.
Input
The first line contains integer n (1 β€ n β€ 105) β the array size. The second line contains space-separated integers a1, a2, ..., an (|ai| β€ 104) β the original array.
Output
Print a single integer β the maximum number of equal array elements he can get if he performs an arbitrary number of the given operation.
Examples
Input
2
2 1
Output
1
Input
3
1 4 1
Output
3
Tags: greedy, math
Correct Solution:
```
from math import pi
def main():
n = int(input())
s = sum(map(int, input().split()))
if s % n:
n -= 1
print(n)
if __name__ == '__main__':
main()
```
| 98,915 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Polycarpus has an array, consisting of n integers a1, a2, ..., an. Polycarpus likes it when numbers in an array match. That's why he wants the array to have as many equal numbers as possible. For that Polycarpus performs the following operation multiple times:
* he chooses two elements of the array ai, aj (i β j);
* he simultaneously increases number ai by 1 and decreases number aj by 1, that is, executes ai = ai + 1 and aj = aj - 1.
The given operation changes exactly two distinct array elements. Polycarpus can apply the described operation an infinite number of times.
Now he wants to know what maximum number of equal array elements he can get if he performs an arbitrary number of such operation. Help Polycarpus.
Input
The first line contains integer n (1 β€ n β€ 105) β the array size. The second line contains space-separated integers a1, a2, ..., an (|ai| β€ 104) β the original array.
Output
Print a single integer β the maximum number of equal array elements he can get if he performs an arbitrary number of the given operation.
Examples
Input
2
2 1
Output
1
Input
3
1 4 1
Output
3
Tags: greedy, math
Correct Solution:
```
def fun (x,l):
if(len(set(x))==1):
print(l)
return
if(sum(x)%l==0):
print(l)
else:
print(l-1)
return
l = int(input())
x=list(map(int,input().split()))
x.sort()
fun(x,l)
```
| 98,916 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Polycarpus has an array, consisting of n integers a1, a2, ..., an. Polycarpus likes it when numbers in an array match. That's why he wants the array to have as many equal numbers as possible. For that Polycarpus performs the following operation multiple times:
* he chooses two elements of the array ai, aj (i β j);
* he simultaneously increases number ai by 1 and decreases number aj by 1, that is, executes ai = ai + 1 and aj = aj - 1.
The given operation changes exactly two distinct array elements. Polycarpus can apply the described operation an infinite number of times.
Now he wants to know what maximum number of equal array elements he can get if he performs an arbitrary number of such operation. Help Polycarpus.
Input
The first line contains integer n (1 β€ n β€ 105) β the array size. The second line contains space-separated integers a1, a2, ..., an (|ai| β€ 104) β the original array.
Output
Print a single integer β the maximum number of equal array elements he can get if he performs an arbitrary number of the given operation.
Examples
Input
2
2 1
Output
1
Input
3
1 4 1
Output
3
Tags: greedy, math
Correct Solution:
```
n=int(input())
summ=sum(map(int,input().split()))
if summ%n==0:
print(n)
else:
print(n-1)
```
| 98,917 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Polycarpus has an array, consisting of n integers a1, a2, ..., an. Polycarpus likes it when numbers in an array match. That's why he wants the array to have as many equal numbers as possible. For that Polycarpus performs the following operation multiple times:
* he chooses two elements of the array ai, aj (i β j);
* he simultaneously increases number ai by 1 and decreases number aj by 1, that is, executes ai = ai + 1 and aj = aj - 1.
The given operation changes exactly two distinct array elements. Polycarpus can apply the described operation an infinite number of times.
Now he wants to know what maximum number of equal array elements he can get if he performs an arbitrary number of such operation. Help Polycarpus.
Input
The first line contains integer n (1 β€ n β€ 105) β the array size. The second line contains space-separated integers a1, a2, ..., an (|ai| β€ 104) β the original array.
Output
Print a single integer β the maximum number of equal array elements he can get if he performs an arbitrary number of the given operation.
Examples
Input
2
2 1
Output
1
Input
3
1 4 1
Output
3
Tags: greedy, math
Correct Solution:
```
'''input
6
-1 1 0 0 -1 -1
'''
n = int(input())
a = list(map(int, input().split()))
print(n if sum(a) % n == 0 else n - 1)
```
| 98,918 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Polycarpus has an array, consisting of n integers a1, a2, ..., an. Polycarpus likes it when numbers in an array match. That's why he wants the array to have as many equal numbers as possible. For that Polycarpus performs the following operation multiple times:
* he chooses two elements of the array ai, aj (i β j);
* he simultaneously increases number ai by 1 and decreases number aj by 1, that is, executes ai = ai + 1 and aj = aj - 1.
The given operation changes exactly two distinct array elements. Polycarpus can apply the described operation an infinite number of times.
Now he wants to know what maximum number of equal array elements he can get if he performs an arbitrary number of such operation. Help Polycarpus.
Input
The first line contains integer n (1 β€ n β€ 105) β the array size. The second line contains space-separated integers a1, a2, ..., an (|ai| β€ 104) β the original array.
Output
Print a single integer β the maximum number of equal array elements he can get if he performs an arbitrary number of the given operation.
Examples
Input
2
2 1
Output
1
Input
3
1 4 1
Output
3
Tags: greedy, math
Correct Solution:
```
n=int(input())
print( n if sum((map(int,input().split())))%n==0 else n-1)
```
| 98,919 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Polycarpus has an array, consisting of n integers a1, a2, ..., an. Polycarpus likes it when numbers in an array match. That's why he wants the array to have as many equal numbers as possible. For that Polycarpus performs the following operation multiple times:
* he chooses two elements of the array ai, aj (i β j);
* he simultaneously increases number ai by 1 and decreases number aj by 1, that is, executes ai = ai + 1 and aj = aj - 1.
The given operation changes exactly two distinct array elements. Polycarpus can apply the described operation an infinite number of times.
Now he wants to know what maximum number of equal array elements he can get if he performs an arbitrary number of such operation. Help Polycarpus.
Input
The first line contains integer n (1 β€ n β€ 105) β the array size. The second line contains space-separated integers a1, a2, ..., an (|ai| β€ 104) β the original array.
Output
Print a single integer β the maximum number of equal array elements he can get if he performs an arbitrary number of the given operation.
Examples
Input
2
2 1
Output
1
Input
3
1 4 1
Output
3
Submitted Solution:
```
x=int(input())
sm=sum(list(map(int,input().split())))%x
if sm==0:
print(x)
else:
print(x-1)
```
Yes
| 98,920 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Polycarpus has an array, consisting of n integers a1, a2, ..., an. Polycarpus likes it when numbers in an array match. That's why he wants the array to have as many equal numbers as possible. For that Polycarpus performs the following operation multiple times:
* he chooses two elements of the array ai, aj (i β j);
* he simultaneously increases number ai by 1 and decreases number aj by 1, that is, executes ai = ai + 1 and aj = aj - 1.
The given operation changes exactly two distinct array elements. Polycarpus can apply the described operation an infinite number of times.
Now he wants to know what maximum number of equal array elements he can get if he performs an arbitrary number of such operation. Help Polycarpus.
Input
The first line contains integer n (1 β€ n β€ 105) β the array size. The second line contains space-separated integers a1, a2, ..., an (|ai| β€ 104) β the original array.
Output
Print a single integer β the maximum number of equal array elements he can get if he performs an arbitrary number of the given operation.
Examples
Input
2
2 1
Output
1
Input
3
1 4 1
Output
3
Submitted Solution:
```
n = int(input())
array = [int(c) for c in input().split()]
array.sort()
avg = sum(array) // n
low, hi = 0, n - 1
while low < hi:
amount = min(abs(array[hi] - avg), abs(avg - array[low]))
array[low] += amount
array[hi] -= amount
if array[low] == avg:
low += 1
if array[hi] == avg:
hi -= 1
ans = sum(1 for x in array if x == avg)
print(ans)
```
Yes
| 98,921 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Polycarpus has an array, consisting of n integers a1, a2, ..., an. Polycarpus likes it when numbers in an array match. That's why he wants the array to have as many equal numbers as possible. For that Polycarpus performs the following operation multiple times:
* he chooses two elements of the array ai, aj (i β j);
* he simultaneously increases number ai by 1 and decreases number aj by 1, that is, executes ai = ai + 1 and aj = aj - 1.
The given operation changes exactly two distinct array elements. Polycarpus can apply the described operation an infinite number of times.
Now he wants to know what maximum number of equal array elements he can get if he performs an arbitrary number of such operation. Help Polycarpus.
Input
The first line contains integer n (1 β€ n β€ 105) β the array size. The second line contains space-separated integers a1, a2, ..., an (|ai| β€ 104) β the original array.
Output
Print a single integer β the maximum number of equal array elements he can get if he performs an arbitrary number of the given operation.
Examples
Input
2
2 1
Output
1
Input
3
1 4 1
Output
3
Submitted Solution:
```
n=int(input())
arr = list(map(int, input().split()))
arr.sort()
if sum(arr)%n==0:
print(n)
else:
print(n-1)
```
Yes
| 98,922 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Polycarpus has an array, consisting of n integers a1, a2, ..., an. Polycarpus likes it when numbers in an array match. That's why he wants the array to have as many equal numbers as possible. For that Polycarpus performs the following operation multiple times:
* he chooses two elements of the array ai, aj (i β j);
* he simultaneously increases number ai by 1 and decreases number aj by 1, that is, executes ai = ai + 1 and aj = aj - 1.
The given operation changes exactly two distinct array elements. Polycarpus can apply the described operation an infinite number of times.
Now he wants to know what maximum number of equal array elements he can get if he performs an arbitrary number of such operation. Help Polycarpus.
Input
The first line contains integer n (1 β€ n β€ 105) β the array size. The second line contains space-separated integers a1, a2, ..., an (|ai| β€ 104) β the original array.
Output
Print a single integer β the maximum number of equal array elements he can get if he performs an arbitrary number of the given operation.
Examples
Input
2
2 1
Output
1
Input
3
1 4 1
Output
3
Submitted Solution:
```
n = int(input())
print(n - 1 if sum(map(int, input().split())) % n else n)
```
Yes
| 98,923 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Polycarpus has an array, consisting of n integers a1, a2, ..., an. Polycarpus likes it when numbers in an array match. That's why he wants the array to have as many equal numbers as possible. For that Polycarpus performs the following operation multiple times:
* he chooses two elements of the array ai, aj (i β j);
* he simultaneously increases number ai by 1 and decreases number aj by 1, that is, executes ai = ai + 1 and aj = aj - 1.
The given operation changes exactly two distinct array elements. Polycarpus can apply the described operation an infinite number of times.
Now he wants to know what maximum number of equal array elements he can get if he performs an arbitrary number of such operation. Help Polycarpus.
Input
The first line contains integer n (1 β€ n β€ 105) β the array size. The second line contains space-separated integers a1, a2, ..., an (|ai| β€ 104) β the original array.
Output
Print a single integer β the maximum number of equal array elements he can get if he performs an arbitrary number of the given operation.
Examples
Input
2
2 1
Output
1
Input
3
1 4 1
Output
3
Submitted Solution:
```
import sys
INF = 10**20
MOD = 10**9 + 7
I = lambda:list(map(int,input().split()))
from math import gcd
from math import ceil
from collections import defaultdict as dd, Counter
from bisect import bisect_left as bl, bisect_right as br
"""
Facts and Data representation
Constructive? Top bottom up down
"""
n, = I()
a = I()
fix = a[0]
for i in range(1, n):
if a[i] < 0:
fix += -a[i]
else:
fix += a[i]
if fix % n == 0:
print(n)
else:
print(n - 1, fix)
```
No
| 98,924 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Polycarpus has an array, consisting of n integers a1, a2, ..., an. Polycarpus likes it when numbers in an array match. That's why he wants the array to have as many equal numbers as possible. For that Polycarpus performs the following operation multiple times:
* he chooses two elements of the array ai, aj (i β j);
* he simultaneously increases number ai by 1 and decreases number aj by 1, that is, executes ai = ai + 1 and aj = aj - 1.
The given operation changes exactly two distinct array elements. Polycarpus can apply the described operation an infinite number of times.
Now he wants to know what maximum number of equal array elements he can get if he performs an arbitrary number of such operation. Help Polycarpus.
Input
The first line contains integer n (1 β€ n β€ 105) β the array size. The second line contains space-separated integers a1, a2, ..., an (|ai| β€ 104) β the original array.
Output
Print a single integer β the maximum number of equal array elements he can get if he performs an arbitrary number of the given operation.
Examples
Input
2
2 1
Output
1
Input
3
1 4 1
Output
3
Submitted Solution:
```
n = int(input())
arr = input().split()
arr = [int(i) for i in arr]
sum_arr = sum(arr)
ans = 1
for i in range(2, len(arr)+1):
if(int(sum_arr/i) == sum_arr/i):
ans = i
print(ans)
```
No
| 98,925 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Polycarpus has an array, consisting of n integers a1, a2, ..., an. Polycarpus likes it when numbers in an array match. That's why he wants the array to have as many equal numbers as possible. For that Polycarpus performs the following operation multiple times:
* he chooses two elements of the array ai, aj (i β j);
* he simultaneously increases number ai by 1 and decreases number aj by 1, that is, executes ai = ai + 1 and aj = aj - 1.
The given operation changes exactly two distinct array elements. Polycarpus can apply the described operation an infinite number of times.
Now he wants to know what maximum number of equal array elements he can get if he performs an arbitrary number of such operation. Help Polycarpus.
Input
The first line contains integer n (1 β€ n β€ 105) β the array size. The second line contains space-separated integers a1, a2, ..., an (|ai| β€ 104) β the original array.
Output
Print a single integer β the maximum number of equal array elements he can get if he performs an arbitrary number of the given operation.
Examples
Input
2
2 1
Output
1
Input
3
1 4 1
Output
3
Submitted Solution:
```
l = int(input())
x = list(map(int, input().split()))
d={}
for i in range (len(x)):
try:
d[x[i]]+=1
except:
d[x[i]]=1
m = max(d)
m,s = d[m],0
for i in d:
if(i!=m):
s += abs(i-m)
print(s)
```
No
| 98,926 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Polycarpus has an array, consisting of n integers a1, a2, ..., an. Polycarpus likes it when numbers in an array match. That's why he wants the array to have as many equal numbers as possible. For that Polycarpus performs the following operation multiple times:
* he chooses two elements of the array ai, aj (i β j);
* he simultaneously increases number ai by 1 and decreases number aj by 1, that is, executes ai = ai + 1 and aj = aj - 1.
The given operation changes exactly two distinct array elements. Polycarpus can apply the described operation an infinite number of times.
Now he wants to know what maximum number of equal array elements he can get if he performs an arbitrary number of such operation. Help Polycarpus.
Input
The first line contains integer n (1 β€ n β€ 105) β the array size. The second line contains space-separated integers a1, a2, ..., an (|ai| β€ 104) β the original array.
Output
Print a single integer β the maximum number of equal array elements he can get if he performs an arbitrary number of the given operation.
Examples
Input
2
2 1
Output
1
Input
3
1 4 1
Output
3
Submitted Solution:
```
n=int(input())
nums=list(map(int,input().split()))
i=n
maxx=-float('inf')
while i:
nums.sort()
nums[-1]-=1
nums[0]+=1
count=len(nums)-len(set(nums))
maxx=max(count,maxx)
i-=1
print(maxx+1)
```
No
| 98,927 |
Provide tags and a correct Python 3 solution for this coding contest problem.
The Greatest Secret Ever consists of n words, indexed by positive integers from 1 to n. The secret needs dividing between k Keepers (let's index them by positive integers from 1 to k), the i-th Keeper gets a non-empty set of words with numbers from the set Ui = (ui, 1, ui, 2, ..., ui, |Ui|). Here and below we'll presuppose that the set elements are written in the increasing order.
We'll say that the secret is safe if the following conditions are hold:
* for any two indexes i, j (1 β€ i < j β€ k) the intersection of sets Ui and Uj is an empty set;
* the union of sets U1, U2, ..., Uk is set (1, 2, ..., n);
* in each set Ui, its elements ui, 1, ui, 2, ..., ui, |Ui| do not form an arithmetic progression (in particular, |Ui| β₯ 3 should hold).
Let us remind you that the elements of set (u1, u2, ..., us) form an arithmetic progression if there is such number d, that for all i (1 β€ i < s) fulfills ui + d = ui + 1. For example, the elements of sets (5), (1, 10) and (1, 5, 9) form arithmetic progressions and the elements of sets (1, 2, 4) and (3, 6, 8) don't.
Your task is to find any partition of the set of words into subsets U1, U2, ..., Uk so that the secret is safe. Otherwise indicate that there's no such partition.
Input
The input consists of a single line which contains two integers n and k (2 β€ k β€ n β€ 106) β the number of words in the secret and the number of the Keepers. The numbers are separated by a single space.
Output
If there is no way to keep the secret safe, print a single integer "-1" (without the quotes). Otherwise, print n integers, the i-th of them representing the number of the Keeper who's got the i-th word of the secret.
If there are multiple solutions, print any of them.
Examples
Input
11 3
Output
3 1 2 1 1 2 3 2 2 3 1
Input
5 2
Output
-1
Tags: constructive algorithms, implementation
Correct Solution:
```
n, k = map(int, input().split())
if n < 3 * k: print(-1)
else:
d = n // k - 1
t = list(str(i) + ' ' for i in range(1, k + 1))
print(''.join(t) + ''.join(i * d for i in t) + t[-1] * (n - (d + 1) * k))
```
| 98,928 |
Provide tags and a correct Python 3 solution for this coding contest problem.
The Greatest Secret Ever consists of n words, indexed by positive integers from 1 to n. The secret needs dividing between k Keepers (let's index them by positive integers from 1 to k), the i-th Keeper gets a non-empty set of words with numbers from the set Ui = (ui, 1, ui, 2, ..., ui, |Ui|). Here and below we'll presuppose that the set elements are written in the increasing order.
We'll say that the secret is safe if the following conditions are hold:
* for any two indexes i, j (1 β€ i < j β€ k) the intersection of sets Ui and Uj is an empty set;
* the union of sets U1, U2, ..., Uk is set (1, 2, ..., n);
* in each set Ui, its elements ui, 1, ui, 2, ..., ui, |Ui| do not form an arithmetic progression (in particular, |Ui| β₯ 3 should hold).
Let us remind you that the elements of set (u1, u2, ..., us) form an arithmetic progression if there is such number d, that for all i (1 β€ i < s) fulfills ui + d = ui + 1. For example, the elements of sets (5), (1, 10) and (1, 5, 9) form arithmetic progressions and the elements of sets (1, 2, 4) and (3, 6, 8) don't.
Your task is to find any partition of the set of words into subsets U1, U2, ..., Uk so that the secret is safe. Otherwise indicate that there's no such partition.
Input
The input consists of a single line which contains two integers n and k (2 β€ k β€ n β€ 106) β the number of words in the secret and the number of the Keepers. The numbers are separated by a single space.
Output
If there is no way to keep the secret safe, print a single integer "-1" (without the quotes). Otherwise, print n integers, the i-th of them representing the number of the Keeper who's got the i-th word of the secret.
If there are multiple solutions, print any of them.
Examples
Input
11 3
Output
3 1 2 1 1 2 3 2 2 3 1
Input
5 2
Output
-1
Tags: constructive algorithms, implementation
Correct Solution:
```
n, k = map(int, input().split())
if n // k < 3:
print(-1)
else:
v = [0] * n
for i in range(k):
v[2 * i] = v[2 * i + 1] = i + 1
for i in range(2 * k, n):
v[i] = (i - 2 * k) % k + 1
print(' '.join(map(str, v)))
```
| 98,929 |
Provide tags and a correct Python 3 solution for this coding contest problem.
The Greatest Secret Ever consists of n words, indexed by positive integers from 1 to n. The secret needs dividing between k Keepers (let's index them by positive integers from 1 to k), the i-th Keeper gets a non-empty set of words with numbers from the set Ui = (ui, 1, ui, 2, ..., ui, |Ui|). Here and below we'll presuppose that the set elements are written in the increasing order.
We'll say that the secret is safe if the following conditions are hold:
* for any two indexes i, j (1 β€ i < j β€ k) the intersection of sets Ui and Uj is an empty set;
* the union of sets U1, U2, ..., Uk is set (1, 2, ..., n);
* in each set Ui, its elements ui, 1, ui, 2, ..., ui, |Ui| do not form an arithmetic progression (in particular, |Ui| β₯ 3 should hold).
Let us remind you that the elements of set (u1, u2, ..., us) form an arithmetic progression if there is such number d, that for all i (1 β€ i < s) fulfills ui + d = ui + 1. For example, the elements of sets (5), (1, 10) and (1, 5, 9) form arithmetic progressions and the elements of sets (1, 2, 4) and (3, 6, 8) don't.
Your task is to find any partition of the set of words into subsets U1, U2, ..., Uk so that the secret is safe. Otherwise indicate that there's no such partition.
Input
The input consists of a single line which contains two integers n and k (2 β€ k β€ n β€ 106) β the number of words in the secret and the number of the Keepers. The numbers are separated by a single space.
Output
If there is no way to keep the secret safe, print a single integer "-1" (without the quotes). Otherwise, print n integers, the i-th of them representing the number of the Keeper who's got the i-th word of the secret.
If there are multiple solutions, print any of them.
Examples
Input
11 3
Output
3 1 2 1 1 2 3 2 2 3 1
Input
5 2
Output
-1
Tags: constructive algorithms, implementation
Correct Solution:
```
# ------------------- fast io --------------------
import os
import sys
from io import BytesIO, IOBase
BUFSIZE = 8192
class FastIO(IOBase):
newlines = 0
def __init__(self, file):
self._fd = file.fileno()
self.buffer = BytesIO()
self.writable = "x" in file.mode or "r" not in file.mode
self.write = self.buffer.write if self.writable else None
def read(self):
while True:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
if not b:
break
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines = 0
return self.buffer.read()
def readline(self):
while self.newlines == 0:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
self.newlines = b.count(b"\n") + (not b)
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines -= 1
return self.buffer.readline()
def flush(self):
if self.writable:
os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
self.write = lambda s: self.buffer.write(s.encode("ascii"))
self.read = lambda: self.buffer.read().decode("ascii")
self.readline = lambda: self.buffer.readline().decode("ascii")
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")
# ------------------- fast io --------------------
n, k = map(int, input().split())
if n < 3*k:
print(-1)
quit()
ans = [1]*n
start = 0
for i in range(k):
ans[start] = ans[start+1] = i+1
start += 2
f = 0
for i in range(start, n):
ans[i] = f+1
f += 1
if f == k: f = 0
print(" ".join(str(k) for k in ans))
```
| 98,930 |
Provide tags and a correct Python 3 solution for this coding contest problem.
The Greatest Secret Ever consists of n words, indexed by positive integers from 1 to n. The secret needs dividing between k Keepers (let's index them by positive integers from 1 to k), the i-th Keeper gets a non-empty set of words with numbers from the set Ui = (ui, 1, ui, 2, ..., ui, |Ui|). Here and below we'll presuppose that the set elements are written in the increasing order.
We'll say that the secret is safe if the following conditions are hold:
* for any two indexes i, j (1 β€ i < j β€ k) the intersection of sets Ui and Uj is an empty set;
* the union of sets U1, U2, ..., Uk is set (1, 2, ..., n);
* in each set Ui, its elements ui, 1, ui, 2, ..., ui, |Ui| do not form an arithmetic progression (in particular, |Ui| β₯ 3 should hold).
Let us remind you that the elements of set (u1, u2, ..., us) form an arithmetic progression if there is such number d, that for all i (1 β€ i < s) fulfills ui + d = ui + 1. For example, the elements of sets (5), (1, 10) and (1, 5, 9) form arithmetic progressions and the elements of sets (1, 2, 4) and (3, 6, 8) don't.
Your task is to find any partition of the set of words into subsets U1, U2, ..., Uk so that the secret is safe. Otherwise indicate that there's no such partition.
Input
The input consists of a single line which contains two integers n and k (2 β€ k β€ n β€ 106) β the number of words in the secret and the number of the Keepers. The numbers are separated by a single space.
Output
If there is no way to keep the secret safe, print a single integer "-1" (without the quotes). Otherwise, print n integers, the i-th of them representing the number of the Keeper who's got the i-th word of the secret.
If there are multiple solutions, print any of them.
Examples
Input
11 3
Output
3 1 2 1 1 2 3 2 2 3 1
Input
5 2
Output
-1
Tags: constructive algorithms, implementation
Correct Solution:
```
n,k=map(int,input().split())
if n//k <3:
print("-1")
exit()
ans=list()
for i in range(n):
if i<2*k:
ans.append(i // 2 + 1)
elif (i//k)%2==0:
ans.append(i%k+1)
else :
ans.append(k-i%k)
print(*ans)
```
| 98,931 |
Provide tags and a correct Python 3 solution for this coding contest problem.
The Greatest Secret Ever consists of n words, indexed by positive integers from 1 to n. The secret needs dividing between k Keepers (let's index them by positive integers from 1 to k), the i-th Keeper gets a non-empty set of words with numbers from the set Ui = (ui, 1, ui, 2, ..., ui, |Ui|). Here and below we'll presuppose that the set elements are written in the increasing order.
We'll say that the secret is safe if the following conditions are hold:
* for any two indexes i, j (1 β€ i < j β€ k) the intersection of sets Ui and Uj is an empty set;
* the union of sets U1, U2, ..., Uk is set (1, 2, ..., n);
* in each set Ui, its elements ui, 1, ui, 2, ..., ui, |Ui| do not form an arithmetic progression (in particular, |Ui| β₯ 3 should hold).
Let us remind you that the elements of set (u1, u2, ..., us) form an arithmetic progression if there is such number d, that for all i (1 β€ i < s) fulfills ui + d = ui + 1. For example, the elements of sets (5), (1, 10) and (1, 5, 9) form arithmetic progressions and the elements of sets (1, 2, 4) and (3, 6, 8) don't.
Your task is to find any partition of the set of words into subsets U1, U2, ..., Uk so that the secret is safe. Otherwise indicate that there's no such partition.
Input
The input consists of a single line which contains two integers n and k (2 β€ k β€ n β€ 106) β the number of words in the secret and the number of the Keepers. The numbers are separated by a single space.
Output
If there is no way to keep the secret safe, print a single integer "-1" (without the quotes). Otherwise, print n integers, the i-th of them representing the number of the Keeper who's got the i-th word of the secret.
If there are multiple solutions, print any of them.
Examples
Input
11 3
Output
3 1 2 1 1 2 3 2 2 3 1
Input
5 2
Output
-1
Tags: constructive algorithms, implementation
Correct Solution:
```
n, m = map(int, input().split())
if n // m < 3:
print(-1)
exit()
res = []
for i in range(m):
res.append(i+1)
res.append(i+1)
for i in range(m):
res.append(i+1)
print(' '.join(map(str, res)), '1 ' * (n - 3*m))
```
| 98,932 |
Provide tags and a correct Python 3 solution for this coding contest problem.
The Greatest Secret Ever consists of n words, indexed by positive integers from 1 to n. The secret needs dividing between k Keepers (let's index them by positive integers from 1 to k), the i-th Keeper gets a non-empty set of words with numbers from the set Ui = (ui, 1, ui, 2, ..., ui, |Ui|). Here and below we'll presuppose that the set elements are written in the increasing order.
We'll say that the secret is safe if the following conditions are hold:
* for any two indexes i, j (1 β€ i < j β€ k) the intersection of sets Ui and Uj is an empty set;
* the union of sets U1, U2, ..., Uk is set (1, 2, ..., n);
* in each set Ui, its elements ui, 1, ui, 2, ..., ui, |Ui| do not form an arithmetic progression (in particular, |Ui| β₯ 3 should hold).
Let us remind you that the elements of set (u1, u2, ..., us) form an arithmetic progression if there is such number d, that for all i (1 β€ i < s) fulfills ui + d = ui + 1. For example, the elements of sets (5), (1, 10) and (1, 5, 9) form arithmetic progressions and the elements of sets (1, 2, 4) and (3, 6, 8) don't.
Your task is to find any partition of the set of words into subsets U1, U2, ..., Uk so that the secret is safe. Otherwise indicate that there's no such partition.
Input
The input consists of a single line which contains two integers n and k (2 β€ k β€ n β€ 106) β the number of words in the secret and the number of the Keepers. The numbers are separated by a single space.
Output
If there is no way to keep the secret safe, print a single integer "-1" (without the quotes). Otherwise, print n integers, the i-th of them representing the number of the Keeper who's got the i-th word of the secret.
If there are multiple solutions, print any of them.
Examples
Input
11 3
Output
3 1 2 1 1 2 3 2 2 3 1
Input
5 2
Output
-1
Tags: constructive algorithms, implementation
Correct Solution:
```
import sys
n, k = map(int, input().split())
if k * 3 > n:
print(-1)
exit(0)
ans = [1] * n
if n == k * 3 and n % 2 == 1:
if n == 3:
print(-1)
exit(0)
l = n - (n % 6) - 6
l = max(0, l)
filled = 0
f, s = 1, 2
for i in range(l):
if f > k:
f = 1
if s > k:
s = 1
if filled == k:
break
j = i % 6
r = [f, f, s, f, s, s][j]
ans[i] = r
if j == 5:
f, s = f + 2, s + 2
filled += 1
if j == 3:
filled += 1
tail = [k - 2, k - 2, k, k - 2, k - 1, k - 1, k, k, k - 1]
for i in range(9):
ans[l + i] = tail[i]
else:
filled = 0
f, s = 1, 2
for i in range(n):
if f > k:
f = 1
if s > k:
s = 1
if filled == k:
break
j = i % 6
r = [f, f, s, f, s, s][j]
ans[i] = r
if j == 5:
f, s = f + 2, s + 2
filled += 1
if j == 3:
filled += 1
s = ' '.join(map(str, ans))
sys.stdout.write(s)
```
| 98,933 |
Provide tags and a correct Python 3 solution for this coding contest problem.
The Greatest Secret Ever consists of n words, indexed by positive integers from 1 to n. The secret needs dividing between k Keepers (let's index them by positive integers from 1 to k), the i-th Keeper gets a non-empty set of words with numbers from the set Ui = (ui, 1, ui, 2, ..., ui, |Ui|). Here and below we'll presuppose that the set elements are written in the increasing order.
We'll say that the secret is safe if the following conditions are hold:
* for any two indexes i, j (1 β€ i < j β€ k) the intersection of sets Ui and Uj is an empty set;
* the union of sets U1, U2, ..., Uk is set (1, 2, ..., n);
* in each set Ui, its elements ui, 1, ui, 2, ..., ui, |Ui| do not form an arithmetic progression (in particular, |Ui| β₯ 3 should hold).
Let us remind you that the elements of set (u1, u2, ..., us) form an arithmetic progression if there is such number d, that for all i (1 β€ i < s) fulfills ui + d = ui + 1. For example, the elements of sets (5), (1, 10) and (1, 5, 9) form arithmetic progressions and the elements of sets (1, 2, 4) and (3, 6, 8) don't.
Your task is to find any partition of the set of words into subsets U1, U2, ..., Uk so that the secret is safe. Otherwise indicate that there's no such partition.
Input
The input consists of a single line which contains two integers n and k (2 β€ k β€ n β€ 106) β the number of words in the secret and the number of the Keepers. The numbers are separated by a single space.
Output
If there is no way to keep the secret safe, print a single integer "-1" (without the quotes). Otherwise, print n integers, the i-th of them representing the number of the Keeper who's got the i-th word of the secret.
If there are multiple solutions, print any of them.
Examples
Input
11 3
Output
3 1 2 1 1 2 3 2 2 3 1
Input
5 2
Output
-1
Tags: constructive algorithms, implementation
Correct 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")
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()
n,k=value()
if(n<3*k or k==1):
print(-1)
else:
c=0
j=0
while(j<n%(2*k)):
print(c+1,end=" ")
c=(c+1)%k
j+=1
c=0
temp=-1
for i in range(j,n):
print(c+1,end=" ")
temp+=1
if(temp%2): c=(c+1)%k
```
| 98,934 |
Provide tags and a correct Python 3 solution for this coding contest problem.
The Greatest Secret Ever consists of n words, indexed by positive integers from 1 to n. The secret needs dividing between k Keepers (let's index them by positive integers from 1 to k), the i-th Keeper gets a non-empty set of words with numbers from the set Ui = (ui, 1, ui, 2, ..., ui, |Ui|). Here and below we'll presuppose that the set elements are written in the increasing order.
We'll say that the secret is safe if the following conditions are hold:
* for any two indexes i, j (1 β€ i < j β€ k) the intersection of sets Ui and Uj is an empty set;
* the union of sets U1, U2, ..., Uk is set (1, 2, ..., n);
* in each set Ui, its elements ui, 1, ui, 2, ..., ui, |Ui| do not form an arithmetic progression (in particular, |Ui| β₯ 3 should hold).
Let us remind you that the elements of set (u1, u2, ..., us) form an arithmetic progression if there is such number d, that for all i (1 β€ i < s) fulfills ui + d = ui + 1. For example, the elements of sets (5), (1, 10) and (1, 5, 9) form arithmetic progressions and the elements of sets (1, 2, 4) and (3, 6, 8) don't.
Your task is to find any partition of the set of words into subsets U1, U2, ..., Uk so that the secret is safe. Otherwise indicate that there's no such partition.
Input
The input consists of a single line which contains two integers n and k (2 β€ k β€ n β€ 106) β the number of words in the secret and the number of the Keepers. The numbers are separated by a single space.
Output
If there is no way to keep the secret safe, print a single integer "-1" (without the quotes). Otherwise, print n integers, the i-th of them representing the number of the Keeper who's got the i-th word of the secret.
If there are multiple solutions, print any of them.
Examples
Input
11 3
Output
3 1 2 1 1 2 3 2 2 3 1
Input
5 2
Output
-1
Tags: constructive algorithms, implementation
Correct Solution:
```
import sys
n ,k = map(int, input().split())
if n // k < 3 or k == 1:
print(-1)
exit()
num = 1
output = []
for i in range(k):
for j in range(2):
output.append(str(i+1))
num = num + 1
l = num + k
for i in range(k):
output.append(str(i+1))
num = num + 1
while num < n+1:
output.append("1")
num = num + 1
print(" ".join(output))
```
| 98,935 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
The Greatest Secret Ever consists of n words, indexed by positive integers from 1 to n. The secret needs dividing between k Keepers (let's index them by positive integers from 1 to k), the i-th Keeper gets a non-empty set of words with numbers from the set Ui = (ui, 1, ui, 2, ..., ui, |Ui|). Here and below we'll presuppose that the set elements are written in the increasing order.
We'll say that the secret is safe if the following conditions are hold:
* for any two indexes i, j (1 β€ i < j β€ k) the intersection of sets Ui and Uj is an empty set;
* the union of sets U1, U2, ..., Uk is set (1, 2, ..., n);
* in each set Ui, its elements ui, 1, ui, 2, ..., ui, |Ui| do not form an arithmetic progression (in particular, |Ui| β₯ 3 should hold).
Let us remind you that the elements of set (u1, u2, ..., us) form an arithmetic progression if there is such number d, that for all i (1 β€ i < s) fulfills ui + d = ui + 1. For example, the elements of sets (5), (1, 10) and (1, 5, 9) form arithmetic progressions and the elements of sets (1, 2, 4) and (3, 6, 8) don't.
Your task is to find any partition of the set of words into subsets U1, U2, ..., Uk so that the secret is safe. Otherwise indicate that there's no such partition.
Input
The input consists of a single line which contains two integers n and k (2 β€ k β€ n β€ 106) β the number of words in the secret and the number of the Keepers. The numbers are separated by a single space.
Output
If there is no way to keep the secret safe, print a single integer "-1" (without the quotes). Otherwise, print n integers, the i-th of them representing the number of the Keeper who's got the i-th word of the secret.
If there are multiple solutions, print any of them.
Examples
Input
11 3
Output
3 1 2 1 1 2 3 2 2 3 1
Input
5 2
Output
-1
Submitted Solution:
```
n, k = map(int, input().split())
div_size = n // k
if div_size < 3:
print(-1)
else:
keepers = []
for i in range(k):
keepers.append(i + 1)
for i in range(k):
for _ in range(div_size - 1):
keepers.append(i + 1)
shortage = n - len(keepers)
for _ in range(shortage):
keepers.append(k)
print(*keepers)
```
Yes
| 98,936 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
The Greatest Secret Ever consists of n words, indexed by positive integers from 1 to n. The secret needs dividing between k Keepers (let's index them by positive integers from 1 to k), the i-th Keeper gets a non-empty set of words with numbers from the set Ui = (ui, 1, ui, 2, ..., ui, |Ui|). Here and below we'll presuppose that the set elements are written in the increasing order.
We'll say that the secret is safe if the following conditions are hold:
* for any two indexes i, j (1 β€ i < j β€ k) the intersection of sets Ui and Uj is an empty set;
* the union of sets U1, U2, ..., Uk is set (1, 2, ..., n);
* in each set Ui, its elements ui, 1, ui, 2, ..., ui, |Ui| do not form an arithmetic progression (in particular, |Ui| β₯ 3 should hold).
Let us remind you that the elements of set (u1, u2, ..., us) form an arithmetic progression if there is such number d, that for all i (1 β€ i < s) fulfills ui + d = ui + 1. For example, the elements of sets (5), (1, 10) and (1, 5, 9) form arithmetic progressions and the elements of sets (1, 2, 4) and (3, 6, 8) don't.
Your task is to find any partition of the set of words into subsets U1, U2, ..., Uk so that the secret is safe. Otherwise indicate that there's no such partition.
Input
The input consists of a single line which contains two integers n and k (2 β€ k β€ n β€ 106) β the number of words in the secret and the number of the Keepers. The numbers are separated by a single space.
Output
If there is no way to keep the secret safe, print a single integer "-1" (without the quotes). Otherwise, print n integers, the i-th of them representing the number of the Keeper who's got the i-th word of the secret.
If there are multiple solutions, print any of them.
Examples
Input
11 3
Output
3 1 2 1 1 2 3 2 2 3 1
Input
5 2
Output
-1
Submitted Solution:
```
import sys
def solve():
n, k = map(int, input().split())
if n // k < 3 or n < 6:
print(-1)
return
res = list()
for i in range(1, k + 1):
res.append(i)
res.append(i)
for i in range(1, k + 1):
res.append(i)
while len(res) < n: res.append(1);
print(" ".join(map(str, res)))
solve()
```
Yes
| 98,937 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
The Greatest Secret Ever consists of n words, indexed by positive integers from 1 to n. The secret needs dividing between k Keepers (let's index them by positive integers from 1 to k), the i-th Keeper gets a non-empty set of words with numbers from the set Ui = (ui, 1, ui, 2, ..., ui, |Ui|). Here and below we'll presuppose that the set elements are written in the increasing order.
We'll say that the secret is safe if the following conditions are hold:
* for any two indexes i, j (1 β€ i < j β€ k) the intersection of sets Ui and Uj is an empty set;
* the union of sets U1, U2, ..., Uk is set (1, 2, ..., n);
* in each set Ui, its elements ui, 1, ui, 2, ..., ui, |Ui| do not form an arithmetic progression (in particular, |Ui| β₯ 3 should hold).
Let us remind you that the elements of set (u1, u2, ..., us) form an arithmetic progression if there is such number d, that for all i (1 β€ i < s) fulfills ui + d = ui + 1. For example, the elements of sets (5), (1, 10) and (1, 5, 9) form arithmetic progressions and the elements of sets (1, 2, 4) and (3, 6, 8) don't.
Your task is to find any partition of the set of words into subsets U1, U2, ..., Uk so that the secret is safe. Otherwise indicate that there's no such partition.
Input
The input consists of a single line which contains two integers n and k (2 β€ k β€ n β€ 106) β the number of words in the secret and the number of the Keepers. The numbers are separated by a single space.
Output
If there is no way to keep the secret safe, print a single integer "-1" (without the quotes). Otherwise, print n integers, the i-th of them representing the number of the Keeper who's got the i-th word of the secret.
If there are multiple solutions, print any of them.
Examples
Input
11 3
Output
3 1 2 1 1 2 3 2 2 3 1
Input
5 2
Output
-1
Submitted Solution:
```
n, k = map(int, input().split())
if n < 3 * k: print(-1)
else:
d = n // k
p = [min(1 + i // d, k) for i in range(n)]
for i in range(d, n, 2 * d):
p[i], p[i - 1] = p[i - 1], p[i]
if k > 2: p[0], p[-1] = p[-1], p[0]
print(' '.join(str(i) for i in p))
```
Yes
| 98,938 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
The Greatest Secret Ever consists of n words, indexed by positive integers from 1 to n. The secret needs dividing between k Keepers (let's index them by positive integers from 1 to k), the i-th Keeper gets a non-empty set of words with numbers from the set Ui = (ui, 1, ui, 2, ..., ui, |Ui|). Here and below we'll presuppose that the set elements are written in the increasing order.
We'll say that the secret is safe if the following conditions are hold:
* for any two indexes i, j (1 β€ i < j β€ k) the intersection of sets Ui and Uj is an empty set;
* the union of sets U1, U2, ..., Uk is set (1, 2, ..., n);
* in each set Ui, its elements ui, 1, ui, 2, ..., ui, |Ui| do not form an arithmetic progression (in particular, |Ui| β₯ 3 should hold).
Let us remind you that the elements of set (u1, u2, ..., us) form an arithmetic progression if there is such number d, that for all i (1 β€ i < s) fulfills ui + d = ui + 1. For example, the elements of sets (5), (1, 10) and (1, 5, 9) form arithmetic progressions and the elements of sets (1, 2, 4) and (3, 6, 8) don't.
Your task is to find any partition of the set of words into subsets U1, U2, ..., Uk so that the secret is safe. Otherwise indicate that there's no such partition.
Input
The input consists of a single line which contains two integers n and k (2 β€ k β€ n β€ 106) β the number of words in the secret and the number of the Keepers. The numbers are separated by a single space.
Output
If there is no way to keep the secret safe, print a single integer "-1" (without the quotes). Otherwise, print n integers, the i-th of them representing the number of the Keeper who's got the i-th word of the secret.
If there are multiple solutions, print any of them.
Examples
Input
11 3
Output
3 1 2 1 1 2 3 2 2 3 1
Input
5 2
Output
-1
Submitted Solution:
```
n, k = map(int, input().split())
if k * 3 > n:
print(-1)
else:
ans = []
for i in range(1, k * 2 + 1):
ans.append(k if i % k == 0 else i % k)
if k % 2:
ans.append(k)
for i in range(k - 1, 0, -1):
if i % 2:
ans.append(ans[-1] + i)
else:
ans.append(ans[-1] - i)
else:
ans.extend([i for i in range(k, 0, -1)])
ans.extend([(k - (i % k)) for i in range(k * 3 + 1, n + 1)])
print(*ans)
```
Yes
| 98,939 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
The Greatest Secret Ever consists of n words, indexed by positive integers from 1 to n. The secret needs dividing between k Keepers (let's index them by positive integers from 1 to k), the i-th Keeper gets a non-empty set of words with numbers from the set Ui = (ui, 1, ui, 2, ..., ui, |Ui|). Here and below we'll presuppose that the set elements are written in the increasing order.
We'll say that the secret is safe if the following conditions are hold:
* for any two indexes i, j (1 β€ i < j β€ k) the intersection of sets Ui and Uj is an empty set;
* the union of sets U1, U2, ..., Uk is set (1, 2, ..., n);
* in each set Ui, its elements ui, 1, ui, 2, ..., ui, |Ui| do not form an arithmetic progression (in particular, |Ui| β₯ 3 should hold).
Let us remind you that the elements of set (u1, u2, ..., us) form an arithmetic progression if there is such number d, that for all i (1 β€ i < s) fulfills ui + d = ui + 1. For example, the elements of sets (5), (1, 10) and (1, 5, 9) form arithmetic progressions and the elements of sets (1, 2, 4) and (3, 6, 8) don't.
Your task is to find any partition of the set of words into subsets U1, U2, ..., Uk so that the secret is safe. Otherwise indicate that there's no such partition.
Input
The input consists of a single line which contains two integers n and k (2 β€ k β€ n β€ 106) β the number of words in the secret and the number of the Keepers. The numbers are separated by a single space.
Output
If there is no way to keep the secret safe, print a single integer "-1" (without the quotes). Otherwise, print n integers, the i-th of them representing the number of the Keeper who's got the i-th word of the secret.
If there are multiple solutions, print any of them.
Examples
Input
11 3
Output
3 1 2 1 1 2 3 2 2 3 1
Input
5 2
Output
-1
Submitted Solution:
```
n, k = map(int, input().split())
if k * 3 < k:
print(-1)
else:
ans = []
for i in range(1, k * 2 + 1):
ans.append(k if i % k == 0 else i % k)
if k % 2:
ans.append(k)
for i in range(k - 1, 0, -1):
if i % 2:
ans.append(ans[-1] + i)
else:
ans.append(ans[-1] - i)
else:
ans.extend([i for i in range(k, 0, -1)])
ans.extend([(k - (i % k)) for i in range(k * 3 + 1, n + 1)])
print(*ans)
```
No
| 98,940 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
The Greatest Secret Ever consists of n words, indexed by positive integers from 1 to n. The secret needs dividing between k Keepers (let's index them by positive integers from 1 to k), the i-th Keeper gets a non-empty set of words with numbers from the set Ui = (ui, 1, ui, 2, ..., ui, |Ui|). Here and below we'll presuppose that the set elements are written in the increasing order.
We'll say that the secret is safe if the following conditions are hold:
* for any two indexes i, j (1 β€ i < j β€ k) the intersection of sets Ui and Uj is an empty set;
* the union of sets U1, U2, ..., Uk is set (1, 2, ..., n);
* in each set Ui, its elements ui, 1, ui, 2, ..., ui, |Ui| do not form an arithmetic progression (in particular, |Ui| β₯ 3 should hold).
Let us remind you that the elements of set (u1, u2, ..., us) form an arithmetic progression if there is such number d, that for all i (1 β€ i < s) fulfills ui + d = ui + 1. For example, the elements of sets (5), (1, 10) and (1, 5, 9) form arithmetic progressions and the elements of sets (1, 2, 4) and (3, 6, 8) don't.
Your task is to find any partition of the set of words into subsets U1, U2, ..., Uk so that the secret is safe. Otherwise indicate that there's no such partition.
Input
The input consists of a single line which contains two integers n and k (2 β€ k β€ n β€ 106) β the number of words in the secret and the number of the Keepers. The numbers are separated by a single space.
Output
If there is no way to keep the secret safe, print a single integer "-1" (without the quotes). Otherwise, print n integers, the i-th of them representing the number of the Keeper who's got the i-th word of the secret.
If there are multiple solutions, print any of them.
Examples
Input
11 3
Output
3 1 2 1 1 2 3 2 2 3 1
Input
5 2
Output
-1
Submitted Solution:
```
n, k = map(int, input().split())
if k * 3 < n:
print(-1)
exit(0)
ans = [0] * n
if n % k == 0 and (n // k) % 2 == 1:
if n == 3:
print(-1)
exit(0)
l = n - (n % 6) - 6
l = max(0, l)
filled = 0
f, s = 1, 2
for i in range(l):
if f > k:
f = 1
if s > k:
s = 1
if filled == k:
break
j = i % 6
r = [f, f, s, f, s, s][j]
ans[i] = j
if j == 5:
f, s = f + 2, s + 2
filled += 1
if j == 3:
filled += 1
tail = [k - 2, k - 2, k, k - 2, k - 1, k - 1, k, k, k - 1]
for i in range(9):
ans[l + i] = tail[i]
else:
filled = 0
f, s = 1, 2
for i in range(n):
if f > k:
f = 1
if s > k:
s = 1
if filled == k:
break
j = i % 6
r = [f, f, s, f, s, s][j]
ans[i] = j
if j == 5:
f, s = f + 2, s + 2
filled += 1
if j == 3:
filled += 1
for z in range(i, n):
ans[z] = 1
print(*ans)
```
No
| 98,941 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
The Greatest Secret Ever consists of n words, indexed by positive integers from 1 to n. The secret needs dividing between k Keepers (let's index them by positive integers from 1 to k), the i-th Keeper gets a non-empty set of words with numbers from the set Ui = (ui, 1, ui, 2, ..., ui, |Ui|). Here and below we'll presuppose that the set elements are written in the increasing order.
We'll say that the secret is safe if the following conditions are hold:
* for any two indexes i, j (1 β€ i < j β€ k) the intersection of sets Ui and Uj is an empty set;
* the union of sets U1, U2, ..., Uk is set (1, 2, ..., n);
* in each set Ui, its elements ui, 1, ui, 2, ..., ui, |Ui| do not form an arithmetic progression (in particular, |Ui| β₯ 3 should hold).
Let us remind you that the elements of set (u1, u2, ..., us) form an arithmetic progression if there is such number d, that for all i (1 β€ i < s) fulfills ui + d = ui + 1. For example, the elements of sets (5), (1, 10) and (1, 5, 9) form arithmetic progressions and the elements of sets (1, 2, 4) and (3, 6, 8) don't.
Your task is to find any partition of the set of words into subsets U1, U2, ..., Uk so that the secret is safe. Otherwise indicate that there's no such partition.
Input
The input consists of a single line which contains two integers n and k (2 β€ k β€ n β€ 106) β the number of words in the secret and the number of the Keepers. The numbers are separated by a single space.
Output
If there is no way to keep the secret safe, print a single integer "-1" (without the quotes). Otherwise, print n integers, the i-th of them representing the number of the Keeper who's got the i-th word of the secret.
If there are multiple solutions, print any of them.
Examples
Input
11 3
Output
3 1 2 1 1 2 3 2 2 3 1
Input
5 2
Output
-1
Submitted Solution:
```
n, k = map(int, input().split())
if n < 3*k:
print(-1)
quit()
ans = [1]*n
from random import randint
for i in range(n):
ans[i] = randint(1, k)
print(" ".join(str(k) for k in ans))
```
No
| 98,942 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
The Greatest Secret Ever consists of n words, indexed by positive integers from 1 to n. The secret needs dividing between k Keepers (let's index them by positive integers from 1 to k), the i-th Keeper gets a non-empty set of words with numbers from the set Ui = (ui, 1, ui, 2, ..., ui, |Ui|). Here and below we'll presuppose that the set elements are written in the increasing order.
We'll say that the secret is safe if the following conditions are hold:
* for any two indexes i, j (1 β€ i < j β€ k) the intersection of sets Ui and Uj is an empty set;
* the union of sets U1, U2, ..., Uk is set (1, 2, ..., n);
* in each set Ui, its elements ui, 1, ui, 2, ..., ui, |Ui| do not form an arithmetic progression (in particular, |Ui| β₯ 3 should hold).
Let us remind you that the elements of set (u1, u2, ..., us) form an arithmetic progression if there is such number d, that for all i (1 β€ i < s) fulfills ui + d = ui + 1. For example, the elements of sets (5), (1, 10) and (1, 5, 9) form arithmetic progressions and the elements of sets (1, 2, 4) and (3, 6, 8) don't.
Your task is to find any partition of the set of words into subsets U1, U2, ..., Uk so that the secret is safe. Otherwise indicate that there's no such partition.
Input
The input consists of a single line which contains two integers n and k (2 β€ k β€ n β€ 106) β the number of words in the secret and the number of the Keepers. The numbers are separated by a single space.
Output
If there is no way to keep the secret safe, print a single integer "-1" (without the quotes). Otherwise, print n integers, the i-th of them representing the number of the Keeper who's got the i-th word of the secret.
If there are multiple solutions, print any of them.
Examples
Input
11 3
Output
3 1 2 1 1 2 3 2 2 3 1
Input
5 2
Output
-1
Submitted Solution:
```
n, k = map(int, input().split())
if k * 3 < 9:
print(-1)
else:
ans = []
for i in range(1, k * 2 + 1):
ans.append(k if i % k == 0 else i % k)
if k % 2:
ans.append(k)
for i in range(k - 1, 0, -1):
if i % 2:
ans.append(ans[-1] + i)
else:
ans.append(ans[-1] - i)
else:
ans.extend([i for i in range(k, 0, -1)])
ans.extend([(k - (i % k)) for i in range(k * 3 + 1, n + 1)])
print(*ans)
```
No
| 98,943 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Let us call a pair of integer numbers m-perfect, if at least one number in the pair is greater than or equal to m. Thus, the pairs (3, 3) and (0, 2) are 2-perfect while the pair (-1, 1) is not.
Two integers x, y are written on the blackboard. It is allowed to erase one of them and replace it with the sum of the numbers, (x + y).
What is the minimum number of such operations one has to perform in order to make the given pair of integers m-perfect?
Input
Single line of the input contains three integers x, y and m ( - 1018 β€ x, y, m β€ 1018).
Please, do not use the %lld specifier to read or write 64-bit integers in C++. It is preffered to use the cin, cout streams or the %I64d specifier.
Output
Print the minimum number of operations or "-1" (without quotes), if it is impossible to transform the given pair to the m-perfect one.
Examples
Input
1 2 5
Output
2
Input
-1 4 15
Output
4
Input
0 -1 5
Output
-1
Note
In the first sample the following sequence of operations is suitable: (1, 2) <image> (3, 2) <image> (5, 2).
In the second sample: (-1, 4) <image> (3, 4) <image> (7, 4) <image> (11, 4) <image> (15, 4).
Finally, in the third sample x, y cannot be made positive, hence there is no proper sequence of operations.
Tags: brute force
Correct Solution:
```
#_________________ Mukul Mohan Varshney _______________#
#Template
import sys
import os
import math
import copy
from math import gcd
from bisect import bisect
from io import BytesIO, IOBase
from math import sqrt,floor,factorial,gcd,log,ceil
from collections import deque,Counter,defaultdict
from itertools import permutations, combinations
#define function
def Int(): return int(sys.stdin.readline())
def Mint(): return map(int,sys.stdin.readline().split())
def Lstr(): return list(sys.stdin.readline().strip())
def Str(): return sys.stdin.readline().strip()
def Mstr(): return map(str,sys.stdin.readline().strip().split())
def List(): return list(map(int,sys.stdin.readline().split()))
def Hash(): return dict()
def Mod(): return 1000000007
def Ncr(n,r,p): return ((fact[n])*((ifact[r]*ifact[n-r])%p))%p
def Most_frequent(list): return max(set(list), key = list.count)
def Mat2x2(n): return [List() for _ in range(n)]
def btod(n):
return int(n,2)
def dtob(n):
return bin(n).replace("0b","")
# Driver Code
def solution():
#for _ in range(Int()):
x,y,m=Mint()
ans=0
if(x>=m or y>=m):
print(0)
elif(x<=0 and y<=0):
print(-1)
else:
if(x>0 and y<0):
ans=(x-y-1)//x
y+=ans*x
elif(y>0 and x<0):
ans=(y-x-1)//y
x+=ans*y
while(x<m and y<m):
t=x+y
if(x<y):
x=t
else:
y=t
ans+=1
print(ans)
#Call the solve function
if __name__ == "__main__":
solution()
```
| 98,944 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Let us call a pair of integer numbers m-perfect, if at least one number in the pair is greater than or equal to m. Thus, the pairs (3, 3) and (0, 2) are 2-perfect while the pair (-1, 1) is not.
Two integers x, y are written on the blackboard. It is allowed to erase one of them and replace it with the sum of the numbers, (x + y).
What is the minimum number of such operations one has to perform in order to make the given pair of integers m-perfect?
Input
Single line of the input contains three integers x, y and m ( - 1018 β€ x, y, m β€ 1018).
Please, do not use the %lld specifier to read or write 64-bit integers in C++. It is preffered to use the cin, cout streams or the %I64d specifier.
Output
Print the minimum number of operations or "-1" (without quotes), if it is impossible to transform the given pair to the m-perfect one.
Examples
Input
1 2 5
Output
2
Input
-1 4 15
Output
4
Input
0 -1 5
Output
-1
Note
In the first sample the following sequence of operations is suitable: (1, 2) <image> (3, 2) <image> (5, 2).
In the second sample: (-1, 4) <image> (3, 4) <image> (7, 4) <image> (11, 4) <image> (15, 4).
Finally, in the third sample x, y cannot be made positive, hence there is no proper sequence of operations.
Tags: brute force
Correct Solution:
```
from math import ceil
x, y, m=map(int, input().split())
if max(x, y)>=m:
exit(print(0))
if x<=0 and y<=0:
exit(print(-1))
x, y=min(x, y), max(x, y)
steps=0
if x<0:
steps+=ceil((abs(x)+y-1)/y)
csteps=ceil((abs(x)+y-1)/y)
x=y*csteps+x
while max(x, y)<m:
x, y=x+y, max(x, y)
steps+=1
exit(print(steps))
```
| 98,945 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Let us call a pair of integer numbers m-perfect, if at least one number in the pair is greater than or equal to m. Thus, the pairs (3, 3) and (0, 2) are 2-perfect while the pair (-1, 1) is not.
Two integers x, y are written on the blackboard. It is allowed to erase one of them and replace it with the sum of the numbers, (x + y).
What is the minimum number of such operations one has to perform in order to make the given pair of integers m-perfect?
Input
Single line of the input contains three integers x, y and m ( - 1018 β€ x, y, m β€ 1018).
Please, do not use the %lld specifier to read or write 64-bit integers in C++. It is preffered to use the cin, cout streams or the %I64d specifier.
Output
Print the minimum number of operations or "-1" (without quotes), if it is impossible to transform the given pair to the m-perfect one.
Examples
Input
1 2 5
Output
2
Input
-1 4 15
Output
4
Input
0 -1 5
Output
-1
Note
In the first sample the following sequence of operations is suitable: (1, 2) <image> (3, 2) <image> (5, 2).
In the second sample: (-1, 4) <image> (3, 4) <image> (7, 4) <image> (11, 4) <image> (15, 4).
Finally, in the third sample x, y cannot be made positive, hence there is no proper sequence of operations.
Tags: brute force
Correct Solution:
```
x,y,m=map(int,input().split())
i=0
if x>=m or y>=m:
print(i)
elif max(x,y)<=0:
print(-1)
else:
if x>y:
x,y=y,x
if x<0:
i=(y-x)//y
x+=i*y
while y<m:
x,y=y,x+y
i+=1
print(i)
```
| 98,946 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Let us call a pair of integer numbers m-perfect, if at least one number in the pair is greater than or equal to m. Thus, the pairs (3, 3) and (0, 2) are 2-perfect while the pair (-1, 1) is not.
Two integers x, y are written on the blackboard. It is allowed to erase one of them and replace it with the sum of the numbers, (x + y).
What is the minimum number of such operations one has to perform in order to make the given pair of integers m-perfect?
Input
Single line of the input contains three integers x, y and m ( - 1018 β€ x, y, m β€ 1018).
Please, do not use the %lld specifier to read or write 64-bit integers in C++. It is preffered to use the cin, cout streams or the %I64d specifier.
Output
Print the minimum number of operations or "-1" (without quotes), if it is impossible to transform the given pair to the m-perfect one.
Examples
Input
1 2 5
Output
2
Input
-1 4 15
Output
4
Input
0 -1 5
Output
-1
Note
In the first sample the following sequence of operations is suitable: (1, 2) <image> (3, 2) <image> (5, 2).
In the second sample: (-1, 4) <image> (3, 4) <image> (7, 4) <image> (11, 4) <image> (15, 4).
Finally, in the third sample x, y cannot be made positive, hence there is no proper sequence of operations.
Tags: brute force
Correct Solution:
```
import math
a, b, c = map (int, input().split())
if (max (a, b) >= c):
print (0)
raise SystemExit
if (a <= 0 and b <= 0) :
print (-1)
raise SystemExit
tot = 0
if ((a <= 0 and b > 0) or (b <= 0 and a > 0)) :
add = max (a, b)
menor = min (a, b)
adicionar = math.ceil(-menor / add)
tot = adicionar
if (min (a, b) == a) :
a += add * adicionar
else :
b += add * adicionar
times = 500
while (times > 0) :
times -= 1
if (max(a, b) >= c) :
print (tot)
raise SystemExit
tot += 1
add = a + b
if (min (a, b) == a) :
a = add
else :
b = add
print (-1)
```
| 98,947 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Let us call a pair of integer numbers m-perfect, if at least one number in the pair is greater than or equal to m. Thus, the pairs (3, 3) and (0, 2) are 2-perfect while the pair (-1, 1) is not.
Two integers x, y are written on the blackboard. It is allowed to erase one of them and replace it with the sum of the numbers, (x + y).
What is the minimum number of such operations one has to perform in order to make the given pair of integers m-perfect?
Input
Single line of the input contains three integers x, y and m ( - 1018 β€ x, y, m β€ 1018).
Please, do not use the %lld specifier to read or write 64-bit integers in C++. It is preffered to use the cin, cout streams or the %I64d specifier.
Output
Print the minimum number of operations or "-1" (without quotes), if it is impossible to transform the given pair to the m-perfect one.
Examples
Input
1 2 5
Output
2
Input
-1 4 15
Output
4
Input
0 -1 5
Output
-1
Note
In the first sample the following sequence of operations is suitable: (1, 2) <image> (3, 2) <image> (5, 2).
In the second sample: (-1, 4) <image> (3, 4) <image> (7, 4) <image> (11, 4) <image> (15, 4).
Finally, in the third sample x, y cannot be made positive, hence there is no proper sequence of operations.
Tags: brute force
Correct Solution:
```
a,b,c = [ int(x) for x in input().split()]
d = 200;
pocet = 0;
while True:
if a>b:
a,b = b,a
if b>=c:
print(pocet)
quit()
d-=1
if d<0:
print(-1)
quit()
if b<-1000:
print(-1)
quit()
if a<0 and b>0:
x = max(1, -a//b - 5)
pocet+=x
a = a+x*b
else:
pocet+=1
a = a+b
```
| 98,948 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Let us call a pair of integer numbers m-perfect, if at least one number in the pair is greater than or equal to m. Thus, the pairs (3, 3) and (0, 2) are 2-perfect while the pair (-1, 1) is not.
Two integers x, y are written on the blackboard. It is allowed to erase one of them and replace it with the sum of the numbers, (x + y).
What is the minimum number of such operations one has to perform in order to make the given pair of integers m-perfect?
Input
Single line of the input contains three integers x, y and m ( - 1018 β€ x, y, m β€ 1018).
Please, do not use the %lld specifier to read or write 64-bit integers in C++. It is preffered to use the cin, cout streams or the %I64d specifier.
Output
Print the minimum number of operations or "-1" (without quotes), if it is impossible to transform the given pair to the m-perfect one.
Examples
Input
1 2 5
Output
2
Input
-1 4 15
Output
4
Input
0 -1 5
Output
-1
Note
In the first sample the following sequence of operations is suitable: (1, 2) <image> (3, 2) <image> (5, 2).
In the second sample: (-1, 4) <image> (3, 4) <image> (7, 4) <image> (11, 4) <image> (15, 4).
Finally, in the third sample x, y cannot be made positive, hence there is no proper sequence of operations.
Tags: brute force
Correct Solution:
```
import math
a,b,c=map(int,input().split())
count=0
if a<b:
a,b=b,a
if a>=c:
print(0)
else:
if a>0 and b<0:
count=math.ceil(abs(b)/a)
b+=a*count
c2=0
while max(a,b)<c:
if a<b:
a=a+b
else:
b=a+b
c2+=1
if c2>10**6:
c2=-1
break
print(count+c2)
```
| 98,949 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Let us call a pair of integer numbers m-perfect, if at least one number in the pair is greater than or equal to m. Thus, the pairs (3, 3) and (0, 2) are 2-perfect while the pair (-1, 1) is not.
Two integers x, y are written on the blackboard. It is allowed to erase one of them and replace it with the sum of the numbers, (x + y).
What is the minimum number of such operations one has to perform in order to make the given pair of integers m-perfect?
Input
Single line of the input contains three integers x, y and m ( - 1018 β€ x, y, m β€ 1018).
Please, do not use the %lld specifier to read or write 64-bit integers in C++. It is preffered to use the cin, cout streams or the %I64d specifier.
Output
Print the minimum number of operations or "-1" (without quotes), if it is impossible to transform the given pair to the m-perfect one.
Examples
Input
1 2 5
Output
2
Input
-1 4 15
Output
4
Input
0 -1 5
Output
-1
Note
In the first sample the following sequence of operations is suitable: (1, 2) <image> (3, 2) <image> (5, 2).
In the second sample: (-1, 4) <image> (3, 4) <image> (7, 4) <image> (11, 4) <image> (15, 4).
Finally, in the third sample x, y cannot be made positive, hence there is no proper sequence of operations.
Tags: brute force
Correct Solution:
```
'''
#A
m, n = map(int, input().split())
if m % 2 == 1:
if n <= (m + 1) // 2:
print(2 * n - 1)
else:
print(2 * n - (m + 1))
else:
if n <= m // 2:
print(2 * n - 1)
else:
print(2 * n - m)
#B
s = input()
l = s.split('heavy')
res = 0
for i in range(1, len(l)):
res += i * l[i].count('metal')
print(res)
'''
#C
s = input().split()
x, y, m = (int(i) for i in s)
res = 0
if x >= m or y >= m:
print(0)
elif x <= 0 and y <= 0:
print(-1)
else:
if x < 0:
q = abs(x // y)
res += q
x += y * q
elif y < 0:
q = abs(y // x)
res += q
y += x * q
while x < m and y < m:
res += 1
if x < y:
x += y
else:
y += x
print(res)
```
| 98,950 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Let us call a pair of integer numbers m-perfect, if at least one number in the pair is greater than or equal to m. Thus, the pairs (3, 3) and (0, 2) are 2-perfect while the pair (-1, 1) is not.
Two integers x, y are written on the blackboard. It is allowed to erase one of them and replace it with the sum of the numbers, (x + y).
What is the minimum number of such operations one has to perform in order to make the given pair of integers m-perfect?
Input
Single line of the input contains three integers x, y and m ( - 1018 β€ x, y, m β€ 1018).
Please, do not use the %lld specifier to read or write 64-bit integers in C++. It is preffered to use the cin, cout streams or the %I64d specifier.
Output
Print the minimum number of operations or "-1" (without quotes), if it is impossible to transform the given pair to the m-perfect one.
Examples
Input
1 2 5
Output
2
Input
-1 4 15
Output
4
Input
0 -1 5
Output
-1
Note
In the first sample the following sequence of operations is suitable: (1, 2) <image> (3, 2) <image> (5, 2).
In the second sample: (-1, 4) <image> (3, 4) <image> (7, 4) <image> (11, 4) <image> (15, 4).
Finally, in the third sample x, y cannot be made positive, hence there is no proper sequence of operations.
Tags: brute force
Correct Solution:
```
a,b,c = map(int, input().split())
k = 0
if a>=c or b>=c:
print(0)
exit()
if a<=0 and b<=0:
print(-1)
exit()
if a+b < 0:
s = max(a,b) - min(a,b)
k += abs(s//min(abs(min(a,b)),abs(max(a,b))))
if a<b:
a+=k*min(abs(a),abs(b))
else:
b+=k*min(abs(a),abs(b))
while a<c and b<c:
if a<b:
a=b+a
else:
b=b+a
k+=1
print(k)
```
| 98,951 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Let us call a pair of integer numbers m-perfect, if at least one number in the pair is greater than or equal to m. Thus, the pairs (3, 3) and (0, 2) are 2-perfect while the pair (-1, 1) is not.
Two integers x, y are written on the blackboard. It is allowed to erase one of them and replace it with the sum of the numbers, (x + y).
What is the minimum number of such operations one has to perform in order to make the given pair of integers m-perfect?
Input
Single line of the input contains three integers x, y and m ( - 1018 β€ x, y, m β€ 1018).
Please, do not use the %lld specifier to read or write 64-bit integers in C++. It is preffered to use the cin, cout streams or the %I64d specifier.
Output
Print the minimum number of operations or "-1" (without quotes), if it is impossible to transform the given pair to the m-perfect one.
Examples
Input
1 2 5
Output
2
Input
-1 4 15
Output
4
Input
0 -1 5
Output
-1
Note
In the first sample the following sequence of operations is suitable: (1, 2) <image> (3, 2) <image> (5, 2).
In the second sample: (-1, 4) <image> (3, 4) <image> (7, 4) <image> (11, 4) <image> (15, 4).
Finally, in the third sample x, y cannot be made positive, hence there is no proper sequence of operations.
Submitted Solution:
```
x, y, m = map(int, input().split())
a, b = max(x, y), min(x, y)
if a >= m:
print(0)
exit()
if a <= 0:
print(-1)
exit()
res = 0
if b < 0:
res += -b // a
b += res * a
if a < b:
a, b = b, a
while a < m:
b += a
res += 1
if a < b:
a, b = b, a
print(res)
```
Yes
| 98,952 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Let us call a pair of integer numbers m-perfect, if at least one number in the pair is greater than or equal to m. Thus, the pairs (3, 3) and (0, 2) are 2-perfect while the pair (-1, 1) is not.
Two integers x, y are written on the blackboard. It is allowed to erase one of them and replace it with the sum of the numbers, (x + y).
What is the minimum number of such operations one has to perform in order to make the given pair of integers m-perfect?
Input
Single line of the input contains three integers x, y and m ( - 1018 β€ x, y, m β€ 1018).
Please, do not use the %lld specifier to read or write 64-bit integers in C++. It is preffered to use the cin, cout streams or the %I64d specifier.
Output
Print the minimum number of operations or "-1" (without quotes), if it is impossible to transform the given pair to the m-perfect one.
Examples
Input
1 2 5
Output
2
Input
-1 4 15
Output
4
Input
0 -1 5
Output
-1
Note
In the first sample the following sequence of operations is suitable: (1, 2) <image> (3, 2) <image> (5, 2).
In the second sample: (-1, 4) <image> (3, 4) <image> (7, 4) <image> (11, 4) <image> (15, 4).
Finally, in the third sample x, y cannot be made positive, hence there is no proper sequence of operations.
Submitted Solution:
```
s = input().split()
x, y, m = (int(i) for i in s)
ans = 0
if x >= m or y >= m:
print(0)
elif x <= 0 and y <= 0:
print(-1)
else:
if x < 0:
q = abs(x // y)
ans += q
x += y * q
elif y < 0:
q = abs(y // x)
ans += q
y += x * q
while x < m and y < m:
ans += 1
if x < y:
x = x + y
else:
y = x + y
print(ans)
```
Yes
| 98,953 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Let us call a pair of integer numbers m-perfect, if at least one number in the pair is greater than or equal to m. Thus, the pairs (3, 3) and (0, 2) are 2-perfect while the pair (-1, 1) is not.
Two integers x, y are written on the blackboard. It is allowed to erase one of them and replace it with the sum of the numbers, (x + y).
What is the minimum number of such operations one has to perform in order to make the given pair of integers m-perfect?
Input
Single line of the input contains three integers x, y and m ( - 1018 β€ x, y, m β€ 1018).
Please, do not use the %lld specifier to read or write 64-bit integers in C++. It is preffered to use the cin, cout streams or the %I64d specifier.
Output
Print the minimum number of operations or "-1" (without quotes), if it is impossible to transform the given pair to the m-perfect one.
Examples
Input
1 2 5
Output
2
Input
-1 4 15
Output
4
Input
0 -1 5
Output
-1
Note
In the first sample the following sequence of operations is suitable: (1, 2) <image> (3, 2) <image> (5, 2).
In the second sample: (-1, 4) <image> (3, 4) <image> (7, 4) <image> (11, 4) <image> (15, 4).
Finally, in the third sample x, y cannot be made positive, hence there is no proper sequence of operations.
Submitted Solution:
```
x, y, m = map(int, input().split())
res = 0
if (max(x, y) >= m): print(0)
elif (x+y<m and x<=0 and y<=0):
if (max(x,y)<m): print(-1)
else: print(0)
else:
if x*y<0:
x, y = min(x,y), max(x,y)
if (y>=m):
print(0)
else:
res += -x//y
x += y*res
while(max(x,y)<m):
x, y = max(x,y), x+y
res += 1
print(res)
```
Yes
| 98,954 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Let us call a pair of integer numbers m-perfect, if at least one number in the pair is greater than or equal to m. Thus, the pairs (3, 3) and (0, 2) are 2-perfect while the pair (-1, 1) is not.
Two integers x, y are written on the blackboard. It is allowed to erase one of them and replace it with the sum of the numbers, (x + y).
What is the minimum number of such operations one has to perform in order to make the given pair of integers m-perfect?
Input
Single line of the input contains three integers x, y and m ( - 1018 β€ x, y, m β€ 1018).
Please, do not use the %lld specifier to read or write 64-bit integers in C++. It is preffered to use the cin, cout streams or the %I64d specifier.
Output
Print the minimum number of operations or "-1" (without quotes), if it is impossible to transform the given pair to the m-perfect one.
Examples
Input
1 2 5
Output
2
Input
-1 4 15
Output
4
Input
0 -1 5
Output
-1
Note
In the first sample the following sequence of operations is suitable: (1, 2) <image> (3, 2) <image> (5, 2).
In the second sample: (-1, 4) <image> (3, 4) <image> (7, 4) <image> (11, 4) <image> (15, 4).
Finally, in the third sample x, y cannot be made positive, hence there is no proper sequence of operations.
Submitted Solution:
```
x,y,m=[int(i) for i in input().split(' ')]
x,y=max(x,y), min(x,y)
a=0
if x>=m:print(0); quit()
if x<=0: print(-1); quit()
if x+y<0: a=abs(y//x)+1; y+=a*x
while max(x,y)<m:
if x<y:
x+=y
else:
y+=x
a+=1
print(a)
```
Yes
| 98,955 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Let us call a pair of integer numbers m-perfect, if at least one number in the pair is greater than or equal to m. Thus, the pairs (3, 3) and (0, 2) are 2-perfect while the pair (-1, 1) is not.
Two integers x, y are written on the blackboard. It is allowed to erase one of them and replace it with the sum of the numbers, (x + y).
What is the minimum number of such operations one has to perform in order to make the given pair of integers m-perfect?
Input
Single line of the input contains three integers x, y and m ( - 1018 β€ x, y, m β€ 1018).
Please, do not use the %lld specifier to read or write 64-bit integers in C++. It is preffered to use the cin, cout streams or the %I64d specifier.
Output
Print the minimum number of operations or "-1" (without quotes), if it is impossible to transform the given pair to the m-perfect one.
Examples
Input
1 2 5
Output
2
Input
-1 4 15
Output
4
Input
0 -1 5
Output
-1
Note
In the first sample the following sequence of operations is suitable: (1, 2) <image> (3, 2) <image> (5, 2).
In the second sample: (-1, 4) <image> (3, 4) <image> (7, 4) <image> (11, 4) <image> (15, 4).
Finally, in the third sample x, y cannot be made positive, hence there is no proper sequence of operations.
Submitted Solution:
```
x,y,m=[int(i) for i in input().split(' ')]
x,y=max(x,y), min(x,y)
print(x, y, m)
a=0
if x>=m:print(0); quit()
if x<=0: print(-1); quit()
if x+y<0: a=abs(y//x)+1; y+=a*x
while max(x,y)<m:
if x<y:
x+=y
else:
y+=x
a+=1
print(a)
```
No
| 98,956 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Let us call a pair of integer numbers m-perfect, if at least one number in the pair is greater than or equal to m. Thus, the pairs (3, 3) and (0, 2) are 2-perfect while the pair (-1, 1) is not.
Two integers x, y are written on the blackboard. It is allowed to erase one of them and replace it with the sum of the numbers, (x + y).
What is the minimum number of such operations one has to perform in order to make the given pair of integers m-perfect?
Input
Single line of the input contains three integers x, y and m ( - 1018 β€ x, y, m β€ 1018).
Please, do not use the %lld specifier to read or write 64-bit integers in C++. It is preffered to use the cin, cout streams or the %I64d specifier.
Output
Print the minimum number of operations or "-1" (without quotes), if it is impossible to transform the given pair to the m-perfect one.
Examples
Input
1 2 5
Output
2
Input
-1 4 15
Output
4
Input
0 -1 5
Output
-1
Note
In the first sample the following sequence of operations is suitable: (1, 2) <image> (3, 2) <image> (5, 2).
In the second sample: (-1, 4) <image> (3, 4) <image> (7, 4) <image> (11, 4) <image> (15, 4).
Finally, in the third sample x, y cannot be made positive, hence there is no proper sequence of operations.
Submitted Solution:
```
import math
x,y,m=map(int,input().split())
c=0
x,y=min(x,y),max(x,y)
#print(x,y)
if(y>=m):
print(0)
exit()
elif(x<=0 and y<=0):
print("-1")
exit()
elif(x<0 and y>0):
c+=math.ceil((-1*x)/y)
x+=y*(math.ceil((-1*x)/y))
#print(x,y,c)
if x+y>=m:
print(c+1)
else:
if x==0:
x+=y
y+=y
c+=2
print(c+math.ceil((m+x)/(x+y)))
```
No
| 98,957 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Let us call a pair of integer numbers m-perfect, if at least one number in the pair is greater than or equal to m. Thus, the pairs (3, 3) and (0, 2) are 2-perfect while the pair (-1, 1) is not.
Two integers x, y are written on the blackboard. It is allowed to erase one of them and replace it with the sum of the numbers, (x + y).
What is the minimum number of such operations one has to perform in order to make the given pair of integers m-perfect?
Input
Single line of the input contains three integers x, y and m ( - 1018 β€ x, y, m β€ 1018).
Please, do not use the %lld specifier to read or write 64-bit integers in C++. It is preffered to use the cin, cout streams or the %I64d specifier.
Output
Print the minimum number of operations or "-1" (without quotes), if it is impossible to transform the given pair to the m-perfect one.
Examples
Input
1 2 5
Output
2
Input
-1 4 15
Output
4
Input
0 -1 5
Output
-1
Note
In the first sample the following sequence of operations is suitable: (1, 2) <image> (3, 2) <image> (5, 2).
In the second sample: (-1, 4) <image> (3, 4) <image> (7, 4) <image> (11, 4) <image> (15, 4).
Finally, in the third sample x, y cannot be made positive, hence there is no proper sequence of operations.
Submitted Solution:
```
x,y,m = map(int,input().split())
a=max(x,y)
if a<=0 and m>a:
print(-1)
else:
b=max(x,y)
a=min(x,y)
ans = abs(b-a)//b
a+=(b-a)*ans
while max(a,b)<m:
if a<=b:
a=a+b
else:
b=a+b
ans+=1
print(ans)
```
No
| 98,958 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Let us call a pair of integer numbers m-perfect, if at least one number in the pair is greater than or equal to m. Thus, the pairs (3, 3) and (0, 2) are 2-perfect while the pair (-1, 1) is not.
Two integers x, y are written on the blackboard. It is allowed to erase one of them and replace it with the sum of the numbers, (x + y).
What is the minimum number of such operations one has to perform in order to make the given pair of integers m-perfect?
Input
Single line of the input contains three integers x, y and m ( - 1018 β€ x, y, m β€ 1018).
Please, do not use the %lld specifier to read or write 64-bit integers in C++. It is preffered to use the cin, cout streams or the %I64d specifier.
Output
Print the minimum number of operations or "-1" (without quotes), if it is impossible to transform the given pair to the m-perfect one.
Examples
Input
1 2 5
Output
2
Input
-1 4 15
Output
4
Input
0 -1 5
Output
-1
Note
In the first sample the following sequence of operations is suitable: (1, 2) <image> (3, 2) <image> (5, 2).
In the second sample: (-1, 4) <image> (3, 4) <image> (7, 4) <image> (11, 4) <image> (15, 4).
Finally, in the third sample x, y cannot be made positive, hence there is no proper sequence of operations.
Submitted Solution:
```
import os
import sys
from io import BytesIO, IOBase
BUFSIZE = 8192
class FastIO(IOBase):
newlines = 0
def __init__(self, file):
self._fd = file.fileno()
self.buffer = BytesIO()
self.writable = "x" in file.mode or "r" not in file.mode
self.write = self.buffer.write if self.writable else None
def read(self):
while True:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
if not b:
break
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines = 0
return self.buffer.read()
def readline(self):
while self.newlines == 0:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
self.newlines = b.count(b"\n") + (not b)
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines -= 1
return self.buffer.readline()
def flush(self):
if self.writable:
os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
self.write = lambda s: self.buffer.write(s.encode("ascii"))
self.read = lambda: self.buffer.read().decode("ascii")
self.readline = lambda: self.buffer.readline().decode("ascii")
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")
from collections import defaultdict
from math import ceil,floor,sqrt,log2,gcd
from heapq import heappush,heappop
from bisect import bisect_left,bisect
import sys
abc='abcdefghijklmnopqrstuvwxyz'
a,b,x=map(int,input().split())
c=0
if a>=x or b>=x:
print(0)
elif (a+b)<=0:
print(-1)
else:
while a<x and b<x:
# print(a,b)
if a<b:
a+=b
c+=1
else:
b+=a
c+=1
# print(a,b)
print(c)
```
No
| 98,959 |
Provide tags and a correct Python 3 solution for this coding contest problem.
A girl named Xenia has a cupboard that looks like an arc from ahead. The arc is made of a semicircle with radius r (the cupboard's top) and two walls of height h (the cupboard's sides). The cupboard's depth is r, that is, it looks like a rectangle with base r and height h + r from the sides. The figure below shows what the cupboard looks like (the front view is on the left, the side view is on the right).
<image>
Xenia got lots of balloons for her birthday. The girl hates the mess, so she wants to store the balloons in the cupboard. Luckily, each balloon is a sphere with radius <image>. Help Xenia calculate the maximum number of balloons she can put in her cupboard.
You can say that a balloon is in the cupboard if you can't see any part of the balloon on the left or right view. The balloons in the cupboard can touch each other. It is not allowed to squeeze the balloons or deform them in any way. You can assume that the cupboard's walls are negligibly thin.
Input
The single line contains two integers r, h (1 β€ r, h β€ 107).
Output
Print a single integer β the maximum number of balloons Xenia can put in the cupboard.
Examples
Input
1 1
Output
3
Input
1 2
Output
5
Input
2 1
Output
2
Tags: geometry
Correct Solution:
```
#!/bin/python
from math import sqrt
r, h = (int(x) for x in input().split(' '))
# You can always put two in each "row" of the rectangle and one in the dome
in_box = 2 * h // r + 1
# You can add an additional one, like so:
# O
# O
#
# since they all have radius r/2, you can get the height needed to add an additional one through Pythagorean theorem
remainder = h % r + r
height = r * sqrt(3) / 2 + r
if remainder < height:
print(in_box)
else:
print(in_box + 1)
```
| 98,960 |
Provide tags and a correct Python 3 solution for this coding contest problem.
A girl named Xenia has a cupboard that looks like an arc from ahead. The arc is made of a semicircle with radius r (the cupboard's top) and two walls of height h (the cupboard's sides). The cupboard's depth is r, that is, it looks like a rectangle with base r and height h + r from the sides. The figure below shows what the cupboard looks like (the front view is on the left, the side view is on the right).
<image>
Xenia got lots of balloons for her birthday. The girl hates the mess, so she wants to store the balloons in the cupboard. Luckily, each balloon is a sphere with radius <image>. Help Xenia calculate the maximum number of balloons she can put in her cupboard.
You can say that a balloon is in the cupboard if you can't see any part of the balloon on the left or right view. The balloons in the cupboard can touch each other. It is not allowed to squeeze the balloons or deform them in any way. You can assume that the cupboard's walls are negligibly thin.
Input
The single line contains two integers r, h (1 β€ r, h β€ 107).
Output
Print a single integer β the maximum number of balloons Xenia can put in the cupboard.
Examples
Input
1 1
Output
3
Input
1 2
Output
5
Input
2 1
Output
2
Tags: geometry
Correct Solution:
```
r, h = map(int, input().split())
half_count = (2 * h + r) // (2 * r)
res = 2 * half_count
x1 = 0
y1 = h + r / 2
x2 = r / 2
y2 = (r * (2 * half_count - 1)) / 2
if (x1 - x2) ** 2 + (y1 - y2) ** 2 >= r * r:
res += 1
print(res)
```
| 98,961 |
Provide tags and a correct Python 3 solution for this coding contest problem.
A girl named Xenia has a cupboard that looks like an arc from ahead. The arc is made of a semicircle with radius r (the cupboard's top) and two walls of height h (the cupboard's sides). The cupboard's depth is r, that is, it looks like a rectangle with base r and height h + r from the sides. The figure below shows what the cupboard looks like (the front view is on the left, the side view is on the right).
<image>
Xenia got lots of balloons for her birthday. The girl hates the mess, so she wants to store the balloons in the cupboard. Luckily, each balloon is a sphere with radius <image>. Help Xenia calculate the maximum number of balloons she can put in her cupboard.
You can say that a balloon is in the cupboard if you can't see any part of the balloon on the left or right view. The balloons in the cupboard can touch each other. It is not allowed to squeeze the balloons or deform them in any way. You can assume that the cupboard's walls are negligibly thin.
Input
The single line contains two integers r, h (1 β€ r, h β€ 107).
Output
Print a single integer β the maximum number of balloons Xenia can put in the cupboard.
Examples
Input
1 1
Output
3
Input
1 2
Output
5
Input
2 1
Output
2
Tags: geometry
Correct Solution:
```
r,h = map(int,input().split())
s = h%r
a = (h//r)*2
if s*s >= 3*r*r/4:
a += 3
elif 2*s >= r:
a += 2
else:
a += 1
print(a)
```
| 98,962 |
Provide tags and a correct Python 3 solution for this coding contest problem.
A girl named Xenia has a cupboard that looks like an arc from ahead. The arc is made of a semicircle with radius r (the cupboard's top) and two walls of height h (the cupboard's sides). The cupboard's depth is r, that is, it looks like a rectangle with base r and height h + r from the sides. The figure below shows what the cupboard looks like (the front view is on the left, the side view is on the right).
<image>
Xenia got lots of balloons for her birthday. The girl hates the mess, so she wants to store the balloons in the cupboard. Luckily, each balloon is a sphere with radius <image>. Help Xenia calculate the maximum number of balloons she can put in her cupboard.
You can say that a balloon is in the cupboard if you can't see any part of the balloon on the left or right view. The balloons in the cupboard can touch each other. It is not allowed to squeeze the balloons or deform them in any way. You can assume that the cupboard's walls are negligibly thin.
Input
The single line contains two integers r, h (1 β€ r, h β€ 107).
Output
Print a single integer β the maximum number of balloons Xenia can put in the cupboard.
Examples
Input
1 1
Output
3
Input
1 2
Output
5
Input
2 1
Output
2
Tags: geometry
Correct Solution:
```
r, h = map(int, input().split())
if (h%r)/r >= 3**0.5/2: print(2*(h//r)+3)
elif 0.5 <= (h%r)/r : print(2*(h//r)+2)
else: print(2*(h//r)+1)
```
| 98,963 |
Provide tags and a correct Python 3 solution for this coding contest problem.
A girl named Xenia has a cupboard that looks like an arc from ahead. The arc is made of a semicircle with radius r (the cupboard's top) and two walls of height h (the cupboard's sides). The cupboard's depth is r, that is, it looks like a rectangle with base r and height h + r from the sides. The figure below shows what the cupboard looks like (the front view is on the left, the side view is on the right).
<image>
Xenia got lots of balloons for her birthday. The girl hates the mess, so she wants to store the balloons in the cupboard. Luckily, each balloon is a sphere with radius <image>. Help Xenia calculate the maximum number of balloons she can put in her cupboard.
You can say that a balloon is in the cupboard if you can't see any part of the balloon on the left or right view. The balloons in the cupboard can touch each other. It is not allowed to squeeze the balloons or deform them in any way. You can assume that the cupboard's walls are negligibly thin.
Input
The single line contains two integers r, h (1 β€ r, h β€ 107).
Output
Print a single integer β the maximum number of balloons Xenia can put in the cupboard.
Examples
Input
1 1
Output
3
Input
1 2
Output
5
Input
2 1
Output
2
Tags: geometry
Correct Solution:
```
if __name__=='__main__':
inp = input()
arr = inp.split(" ")
r = int(arr[0])
h = int(arr[1])
ans = 2*(h//r)
d = h%r
if d*2>=r:
ans+=2
if 4*d*d >= 3*r*r:
ans+=1
else:
ans+=1
print(ans)
```
| 98,964 |
Provide tags and a correct Python 3 solution for this coding contest problem.
A girl named Xenia has a cupboard that looks like an arc from ahead. The arc is made of a semicircle with radius r (the cupboard's top) and two walls of height h (the cupboard's sides). The cupboard's depth is r, that is, it looks like a rectangle with base r and height h + r from the sides. The figure below shows what the cupboard looks like (the front view is on the left, the side view is on the right).
<image>
Xenia got lots of balloons for her birthday. The girl hates the mess, so she wants to store the balloons in the cupboard. Luckily, each balloon is a sphere with radius <image>. Help Xenia calculate the maximum number of balloons she can put in her cupboard.
You can say that a balloon is in the cupboard if you can't see any part of the balloon on the left or right view. The balloons in the cupboard can touch each other. It is not allowed to squeeze the balloons or deform them in any way. You can assume that the cupboard's walls are negligibly thin.
Input
The single line contains two integers r, h (1 β€ r, h β€ 107).
Output
Print a single integer β the maximum number of balloons Xenia can put in the cupboard.
Examples
Input
1 1
Output
3
Input
1 2
Output
5
Input
2 1
Output
2
Tags: geometry
Correct Solution:
```
from math import *
r, h = map(int, input().split())
d = h % r
if d * 2 < r:
print(h // r * 2 + 1)
elif sqrt(3) * (r / 2) + r - 1e-6 <= d + r:
print(h // r * 2 + 3)
else:
print(h // r * 2 + 2)
```
| 98,965 |
Provide tags and a correct Python 3 solution for this coding contest problem.
A girl named Xenia has a cupboard that looks like an arc from ahead. The arc is made of a semicircle with radius r (the cupboard's top) and two walls of height h (the cupboard's sides). The cupboard's depth is r, that is, it looks like a rectangle with base r and height h + r from the sides. The figure below shows what the cupboard looks like (the front view is on the left, the side view is on the right).
<image>
Xenia got lots of balloons for her birthday. The girl hates the mess, so she wants to store the balloons in the cupboard. Luckily, each balloon is a sphere with radius <image>. Help Xenia calculate the maximum number of balloons she can put in her cupboard.
You can say that a balloon is in the cupboard if you can't see any part of the balloon on the left or right view. The balloons in the cupboard can touch each other. It is not allowed to squeeze the balloons or deform them in any way. You can assume that the cupboard's walls are negligibly thin.
Input
The single line contains two integers r, h (1 β€ r, h β€ 107).
Output
Print a single integer β the maximum number of balloons Xenia can put in the cupboard.
Examples
Input
1 1
Output
3
Input
1 2
Output
5
Input
2 1
Output
2
Tags: geometry
Correct Solution:
```
r, h = map(int, input().split())
a = 2 * (h // r)
h = h % r
print (a + 1 + (2*h>=r) + (4*h*h >= 3*r*r))
# Made By Mostafa_Khaled
```
| 98,966 |
Provide tags and a correct Python 3 solution for this coding contest problem.
A girl named Xenia has a cupboard that looks like an arc from ahead. The arc is made of a semicircle with radius r (the cupboard's top) and two walls of height h (the cupboard's sides). The cupboard's depth is r, that is, it looks like a rectangle with base r and height h + r from the sides. The figure below shows what the cupboard looks like (the front view is on the left, the side view is on the right).
<image>
Xenia got lots of balloons for her birthday. The girl hates the mess, so she wants to store the balloons in the cupboard. Luckily, each balloon is a sphere with radius <image>. Help Xenia calculate the maximum number of balloons she can put in her cupboard.
You can say that a balloon is in the cupboard if you can't see any part of the balloon on the left or right view. The balloons in the cupboard can touch each other. It is not allowed to squeeze the balloons or deform them in any way. You can assume that the cupboard's walls are negligibly thin.
Input
The single line contains two integers r, h (1 β€ r, h β€ 107).
Output
Print a single integer β the maximum number of balloons Xenia can put in the cupboard.
Examples
Input
1 1
Output
3
Input
1 2
Output
5
Input
2 1
Output
2
Tags: geometry
Correct Solution:
```
r, h = map(int, input().split())
h *= 2
if h < r:
print(1)
exit(0)
r *= 2
ans = (h + r // 2) // r * 2
h -= r * (ans // 2 - 1)
if (h ** 2) * 4 >= (r ** 2) * 3:
ans += 1
print(ans)
```
| 98,967 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A girl named Xenia has a cupboard that looks like an arc from ahead. The arc is made of a semicircle with radius r (the cupboard's top) and two walls of height h (the cupboard's sides). The cupboard's depth is r, that is, it looks like a rectangle with base r and height h + r from the sides. The figure below shows what the cupboard looks like (the front view is on the left, the side view is on the right).
<image>
Xenia got lots of balloons for her birthday. The girl hates the mess, so she wants to store the balloons in the cupboard. Luckily, each balloon is a sphere with radius <image>. Help Xenia calculate the maximum number of balloons she can put in her cupboard.
You can say that a balloon is in the cupboard if you can't see any part of the balloon on the left or right view. The balloons in the cupboard can touch each other. It is not allowed to squeeze the balloons or deform them in any way. You can assume that the cupboard's walls are negligibly thin.
Input
The single line contains two integers r, h (1 β€ r, h β€ 107).
Output
Print a single integer β the maximum number of balloons Xenia can put in the cupboard.
Examples
Input
1 1
Output
3
Input
1 2
Output
5
Input
2 1
Output
2
Submitted Solution:
```
r, h = map(int, input().split())
k = (h // r) * 2
d = h % r
if d >= 2 * r / 3 + 1:
k += 3
elif d >= r / 2:
k += 2
else:
k += 1
print(k)
```
Yes
| 98,968 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A girl named Xenia has a cupboard that looks like an arc from ahead. The arc is made of a semicircle with radius r (the cupboard's top) and two walls of height h (the cupboard's sides). The cupboard's depth is r, that is, it looks like a rectangle with base r and height h + r from the sides. The figure below shows what the cupboard looks like (the front view is on the left, the side view is on the right).
<image>
Xenia got lots of balloons for her birthday. The girl hates the mess, so she wants to store the balloons in the cupboard. Luckily, each balloon is a sphere with radius <image>. Help Xenia calculate the maximum number of balloons she can put in her cupboard.
You can say that a balloon is in the cupboard if you can't see any part of the balloon on the left or right view. The balloons in the cupboard can touch each other. It is not allowed to squeeze the balloons or deform them in any way. You can assume that the cupboard's walls are negligibly thin.
Input
The single line contains two integers r, h (1 β€ r, h β€ 107).
Output
Print a single integer β the maximum number of balloons Xenia can put in the cupboard.
Examples
Input
1 1
Output
3
Input
1 2
Output
5
Input
2 1
Output
2
Submitted Solution:
```
from math import sqrt
r,h = map (int, input().split())
count = h // r * 2
h -= h // r * r
if h >= sqrt(3) / 2 * r:
count += 3
elif h >= r / 2:
count += 2
else:
count += 1
print (count)
```
Yes
| 98,969 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A girl named Xenia has a cupboard that looks like an arc from ahead. The arc is made of a semicircle with radius r (the cupboard's top) and two walls of height h (the cupboard's sides). The cupboard's depth is r, that is, it looks like a rectangle with base r and height h + r from the sides. The figure below shows what the cupboard looks like (the front view is on the left, the side view is on the right).
<image>
Xenia got lots of balloons for her birthday. The girl hates the mess, so she wants to store the balloons in the cupboard. Luckily, each balloon is a sphere with radius <image>. Help Xenia calculate the maximum number of balloons she can put in her cupboard.
You can say that a balloon is in the cupboard if you can't see any part of the balloon on the left or right view. The balloons in the cupboard can touch each other. It is not allowed to squeeze the balloons or deform them in any way. You can assume that the cupboard's walls are negligibly thin.
Input
The single line contains two integers r, h (1 β€ r, h β€ 107).
Output
Print a single integer β the maximum number of balloons Xenia can put in the cupboard.
Examples
Input
1 1
Output
3
Input
1 2
Output
5
Input
2 1
Output
2
Submitted Solution:
```
r, h = map(int, input().split())
a = 2 * (h // r)
h = h % r
print (a + 1 + (2*h>=r) + (4*h*h >= 3*r*r))
```
Yes
| 98,970 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A girl named Xenia has a cupboard that looks like an arc from ahead. The arc is made of a semicircle with radius r (the cupboard's top) and two walls of height h (the cupboard's sides). The cupboard's depth is r, that is, it looks like a rectangle with base r and height h + r from the sides. The figure below shows what the cupboard looks like (the front view is on the left, the side view is on the right).
<image>
Xenia got lots of balloons for her birthday. The girl hates the mess, so she wants to store the balloons in the cupboard. Luckily, each balloon is a sphere with radius <image>. Help Xenia calculate the maximum number of balloons she can put in her cupboard.
You can say that a balloon is in the cupboard if you can't see any part of the balloon on the left or right view. The balloons in the cupboard can touch each other. It is not allowed to squeeze the balloons or deform them in any way. You can assume that the cupboard's walls are negligibly thin.
Input
The single line contains two integers r, h (1 β€ r, h β€ 107).
Output
Print a single integer β the maximum number of balloons Xenia can put in the cupboard.
Examples
Input
1 1
Output
3
Input
1 2
Output
5
Input
2 1
Output
2
Submitted Solution:
```
r, h = map(int, input().split())
d, p = (3 ** 0.5) / 2 - 1, h / r
print(max(1 + 2 * int(p - d), 2 * int(p + 0.5)))
```
Yes
| 98,971 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A girl named Xenia has a cupboard that looks like an arc from ahead. The arc is made of a semicircle with radius r (the cupboard's top) and two walls of height h (the cupboard's sides). The cupboard's depth is r, that is, it looks like a rectangle with base r and height h + r from the sides. The figure below shows what the cupboard looks like (the front view is on the left, the side view is on the right).
<image>
Xenia got lots of balloons for her birthday. The girl hates the mess, so she wants to store the balloons in the cupboard. Luckily, each balloon is a sphere with radius <image>. Help Xenia calculate the maximum number of balloons she can put in her cupboard.
You can say that a balloon is in the cupboard if you can't see any part of the balloon on the left or right view. The balloons in the cupboard can touch each other. It is not allowed to squeeze the balloons or deform them in any way. You can assume that the cupboard's walls are negligibly thin.
Input
The single line contains two integers r, h (1 β€ r, h β€ 107).
Output
Print a single integer β the maximum number of balloons Xenia can put in the cupboard.
Examples
Input
1 1
Output
3
Input
1 2
Output
5
Input
2 1
Output
2
Submitted Solution:
```
from math import sqrt
r,h = map (int, input().split())
count = h // r * 2
h -= h // r
if h >= sqrt(3) / 2 * r:
count += 3
elif h >= r / 2:
count += 2
else:
count += 1
print (count)
```
No
| 98,972 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A girl named Xenia has a cupboard that looks like an arc from ahead. The arc is made of a semicircle with radius r (the cupboard's top) and two walls of height h (the cupboard's sides). The cupboard's depth is r, that is, it looks like a rectangle with base r and height h + r from the sides. The figure below shows what the cupboard looks like (the front view is on the left, the side view is on the right).
<image>
Xenia got lots of balloons for her birthday. The girl hates the mess, so she wants to store the balloons in the cupboard. Luckily, each balloon is a sphere with radius <image>. Help Xenia calculate the maximum number of balloons she can put in her cupboard.
You can say that a balloon is in the cupboard if you can't see any part of the balloon on the left or right view. The balloons in the cupboard can touch each other. It is not allowed to squeeze the balloons or deform them in any way. You can assume that the cupboard's walls are negligibly thin.
Input
The single line contains two integers r, h (1 β€ r, h β€ 107).
Output
Print a single integer β the maximum number of balloons Xenia can put in the cupboard.
Examples
Input
1 1
Output
3
Input
1 2
Output
5
Input
2 1
Output
2
Submitted Solution:
```
import math
import sys
r,h=map(int,sys.stdin.readline().split())
v=(h)//r
x=h-(r*v)
chord=x-(r/2)
x=math.sqrt(3)/2
if(chord<0):
v*=2
v+=1
elif(h-(r*v)+(r/2)>=x*r):
v*=2
v+=3
else:
v*=2
v+=2
print(int(v))
```
No
| 98,973 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A girl named Xenia has a cupboard that looks like an arc from ahead. The arc is made of a semicircle with radius r (the cupboard's top) and two walls of height h (the cupboard's sides). The cupboard's depth is r, that is, it looks like a rectangle with base r and height h + r from the sides. The figure below shows what the cupboard looks like (the front view is on the left, the side view is on the right).
<image>
Xenia got lots of balloons for her birthday. The girl hates the mess, so she wants to store the balloons in the cupboard. Luckily, each balloon is a sphere with radius <image>. Help Xenia calculate the maximum number of balloons she can put in her cupboard.
You can say that a balloon is in the cupboard if you can't see any part of the balloon on the left or right view. The balloons in the cupboard can touch each other. It is not allowed to squeeze the balloons or deform them in any way. You can assume that the cupboard's walls are negligibly thin.
Input
The single line contains two integers r, h (1 β€ r, h β€ 107).
Output
Print a single integer β the maximum number of balloons Xenia can put in the cupboard.
Examples
Input
1 1
Output
3
Input
1 2
Output
5
Input
2 1
Output
2
Submitted Solution:
```
r, h=map(int,input().split())
num = h//r
ans = 0
if num:
rem = h%r
ans = num*2
if rem == r//2 and rem:
ans += 2
else:
ans += 1
else:
ans = 2
print(ans)
```
No
| 98,974 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A girl named Xenia has a cupboard that looks like an arc from ahead. The arc is made of a semicircle with radius r (the cupboard's top) and two walls of height h (the cupboard's sides). The cupboard's depth is r, that is, it looks like a rectangle with base r and height h + r from the sides. The figure below shows what the cupboard looks like (the front view is on the left, the side view is on the right).
<image>
Xenia got lots of balloons for her birthday. The girl hates the mess, so she wants to store the balloons in the cupboard. Luckily, each balloon is a sphere with radius <image>. Help Xenia calculate the maximum number of balloons she can put in her cupboard.
You can say that a balloon is in the cupboard if you can't see any part of the balloon on the left or right view. The balloons in the cupboard can touch each other. It is not allowed to squeeze the balloons or deform them in any way. You can assume that the cupboard's walls are negligibly thin.
Input
The single line contains two integers r, h (1 β€ r, h β€ 107).
Output
Print a single integer β the maximum number of balloons Xenia can put in the cupboard.
Examples
Input
1 1
Output
3
Input
1 2
Output
5
Input
2 1
Output
2
Submitted Solution:
```
r, h = map(int, input().split())
h *= 2
r *= 2
ans = (h + r // 2) // r * 2
h = r - h % r
if h ** 2 * 4 >= r ** 2 * 3:
ans += 1
print(ans)
```
No
| 98,975 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Fox Ciel wants to write a task for a programming contest. The task is: "You are given a simple undirected graph with n vertexes. Each its edge has unit length. You should calculate the number of shortest paths between vertex 1 and vertex 2."
Same with some writers, she wants to make an example with some certain output: for example, her birthday or the number of her boyfriend. Can you help her to make a test case with answer equal exactly to k?
Input
The first line contains a single integer k (1 β€ k β€ 109).
Output
You should output a graph G with n vertexes (2 β€ n β€ 1000). There must be exactly k shortest paths between vertex 1 and vertex 2 of the graph.
The first line must contain an integer n. Then adjacency matrix G with n rows and n columns must follow. Each element of the matrix must be 'N' or 'Y'. If Gij is 'Y', then graph G has a edge connecting vertex i and vertex j. Consider the graph vertexes are numbered from 1 to n.
The graph must be undirected and simple: Gii = 'N' and Gij = Gji must hold. And there must be at least one path between vertex 1 and vertex 2. It's guaranteed that the answer exists. If there multiple correct answers, you can output any of them.
Examples
Input
2
Output
4
NNYY
NNYY
YYNN
YYNN
Input
9
Output
8
NNYYYNNN
NNNNNYYY
YNNNNYYY
YNNNNYYY
YNNNNYYY
NYYYYNNN
NYYYYNNN
NYYYYNNN
Input
1
Output
2
NY
YN
Note
In first example, there are 2 shortest paths: 1-3-2 and 1-4-2.
In second example, there are 9 shortest paths: 1-3-6-2, 1-3-7-2, 1-3-8-2, 1-4-6-2, 1-4-7-2, 1-4-8-2, 1-5-6-2, 1-5-7-2, 1-5-8-2.
Tags: bitmasks, constructive algorithms, graphs, implementation, math
Correct Solution:
```
n, m, cnt = int(input()), 148, 0
ans = [['N'] * m for i in range(m)]
def edge(i, j):
ans[i][j] = ans[j][i] = 'Y'
def node(*adj):
global cnt
i = cnt
cnt += 1
for j in adj:
edge(i, j)
return i
start, end, choice = node(), node(), node()
if n&1:
edge(choice, end)
for i in range(1, 30):
end, choice = node(node(end), node(end)), node(node(choice))
if n&(1<<i):
edge(choice, end)
edge(start, choice)
print(m)
for line in ans:
print(''.join(line))
```
| 98,976 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Fox Ciel wants to write a task for a programming contest. The task is: "You are given a simple undirected graph with n vertexes. Each its edge has unit length. You should calculate the number of shortest paths between vertex 1 and vertex 2."
Same with some writers, she wants to make an example with some certain output: for example, her birthday or the number of her boyfriend. Can you help her to make a test case with answer equal exactly to k?
Input
The first line contains a single integer k (1 β€ k β€ 109).
Output
You should output a graph G with n vertexes (2 β€ n β€ 1000). There must be exactly k shortest paths between vertex 1 and vertex 2 of the graph.
The first line must contain an integer n. Then adjacency matrix G with n rows and n columns must follow. Each element of the matrix must be 'N' or 'Y'. If Gij is 'Y', then graph G has a edge connecting vertex i and vertex j. Consider the graph vertexes are numbered from 1 to n.
The graph must be undirected and simple: Gii = 'N' and Gij = Gji must hold. And there must be at least one path between vertex 1 and vertex 2. It's guaranteed that the answer exists. If there multiple correct answers, you can output any of them.
Examples
Input
2
Output
4
NNYY
NNYY
YYNN
YYNN
Input
9
Output
8
NNYYYNNN
NNNNNYYY
YNNNNYYY
YNNNNYYY
YNNNNYYY
NYYYYNNN
NYYYYNNN
NYYYYNNN
Input
1
Output
2
NY
YN
Note
In first example, there are 2 shortest paths: 1-3-2 and 1-4-2.
In second example, there are 9 shortest paths: 1-3-6-2, 1-3-7-2, 1-3-8-2, 1-4-6-2, 1-4-7-2, 1-4-8-2, 1-5-6-2, 1-5-7-2, 1-5-8-2.
Tags: bitmasks, constructive algorithms, graphs, implementation, math
Correct Solution:
```
n, m, cnt = int(input()), 148, 0
ans = [['N'] * m for i in range(m)]
def edge(i, j):
ans[i][j] = ans[j][i] = 'Y'
def node(*adj):
global cnt
i = cnt
cnt += 1
for j in adj:
edge(i, j)
return i
start, end, choice = node(), node(), node()
if n&1:
edge(choice, end)
for i in range(1, 30):
end, choice = node(node(end), node(end)), node(node(choice))
if n&(1<<i):
edge(choice, end)
edge(start, choice)
print(m)
for line in ans:
print(''.join(line))
# Made By Mostafa_Khaled
```
| 98,977 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Fox Ciel wants to write a task for a programming contest. The task is: "You are given a simple undirected graph with n vertexes. Each its edge has unit length. You should calculate the number of shortest paths between vertex 1 and vertex 2."
Same with some writers, she wants to make an example with some certain output: for example, her birthday or the number of her boyfriend. Can you help her to make a test case with answer equal exactly to k?
Input
The first line contains a single integer k (1 β€ k β€ 109).
Output
You should output a graph G with n vertexes (2 β€ n β€ 1000). There must be exactly k shortest paths between vertex 1 and vertex 2 of the graph.
The first line must contain an integer n. Then adjacency matrix G with n rows and n columns must follow. Each element of the matrix must be 'N' or 'Y'. If Gij is 'Y', then graph G has a edge connecting vertex i and vertex j. Consider the graph vertexes are numbered from 1 to n.
The graph must be undirected and simple: Gii = 'N' and Gij = Gji must hold. And there must be at least one path between vertex 1 and vertex 2. It's guaranteed that the answer exists. If there multiple correct answers, you can output any of them.
Examples
Input
2
Output
4
NNYY
NNYY
YYNN
YYNN
Input
9
Output
8
NNYYYNNN
NNNNNYYY
YNNNNYYY
YNNNNYYY
YNNNNYYY
NYYYYNNN
NYYYYNNN
NYYYYNNN
Input
1
Output
2
NY
YN
Note
In first example, there are 2 shortest paths: 1-3-2 and 1-4-2.
In second example, there are 9 shortest paths: 1-3-6-2, 1-3-7-2, 1-3-8-2, 1-4-6-2, 1-4-7-2, 1-4-8-2, 1-5-6-2, 1-5-7-2, 1-5-8-2.
Tags: bitmasks, constructive algorithms, graphs, implementation, math
Correct Solution:
```
from collections import defaultdict
k=int(input())
mask=0
d=defaultdict(lambda:0)
while(mask<=30):
if k&(1<<mask):
d[mask]=1
ma=mask
mask+=1
adj=defaultdict(lambda:"N")
currl=1
currvu=3
prevu=[1]
prevl=[]
currvl=4
m=4
while((currl//2)<=ma):
if d[currl//2]:
adj[currvu,currvl]="Y"
adj[currvl, currvu] = "Y"
for j in prevu:
adj[currvu, j] = "Y"
adj[j, currvu] = "Y"
for j in prevl:
adj[currvl, j] = "Y"
adj[j, currvl] = "Y"
if ((currl+2)//2)<=ma:
prevu=[currvl+1,currvl+2]
for j in prevu:
adj[currvu, j] = "Y"
adj[j, currvu] = "Y"
prevl=[currvl+3]
for j in prevl:
adj[currvl, j] = "Y"
adj[j, currvl] = "Y"
currvu=currvl+4
currvl=currvu+1
m=max(m,currvl)
else:
break
currl+=2
print(m)
adj[2,currvl]="Y"
adj[currvl,2]="Y"
for i in range(1,m+1):
for j in range(1,m+1):
print(adj[i,j],end="")
print()
```
| 98,978 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Fox Ciel wants to write a task for a programming contest. The task is: "You are given a simple undirected graph with n vertexes. Each its edge has unit length. You should calculate the number of shortest paths between vertex 1 and vertex 2."
Same with some writers, she wants to make an example with some certain output: for example, her birthday or the number of her boyfriend. Can you help her to make a test case with answer equal exactly to k?
Input
The first line contains a single integer k (1 β€ k β€ 109).
Output
You should output a graph G with n vertexes (2 β€ n β€ 1000). There must be exactly k shortest paths between vertex 1 and vertex 2 of the graph.
The first line must contain an integer n. Then adjacency matrix G with n rows and n columns must follow. Each element of the matrix must be 'N' or 'Y'. If Gij is 'Y', then graph G has a edge connecting vertex i and vertex j. Consider the graph vertexes are numbered from 1 to n.
The graph must be undirected and simple: Gii = 'N' and Gij = Gji must hold. And there must be at least one path between vertex 1 and vertex 2. It's guaranteed that the answer exists. If there multiple correct answers, you can output any of them.
Examples
Input
2
Output
4
NNYY
NNYY
YYNN
YYNN
Input
9
Output
8
NNYYYNNN
NNNNNYYY
YNNNNYYY
YNNNNYYY
YNNNNYYY
NYYYYNNN
NYYYYNNN
NYYYYNNN
Input
1
Output
2
NY
YN
Note
In first example, there are 2 shortest paths: 1-3-2 and 1-4-2.
In second example, there are 9 shortest paths: 1-3-6-2, 1-3-7-2, 1-3-8-2, 1-4-6-2, 1-4-7-2, 1-4-8-2, 1-5-6-2, 1-5-7-2, 1-5-8-2.
Tags: bitmasks, constructive algorithms, graphs, implementation, math
Correct Solution:
```
class Node(object):
cnt = 0
def __init__(_, adj=None):
_.id = Node.cnt
Node.cnt += 1
_.adj = adj if adj else []
_.seen = False
def dfs(_):
if _.seen:
return
_.seen = True
for i in _.adj:
ans[_.id][i.id] = ans[i.id][_.id] = 'Y'
i.dfs()
n = int(input())
start = Node()
end = Node()
choose = Node()
for i in range(30):
if n & (1<<i):
choose.adj.append(end)
if i == 29:
break
end = Node([Node([end]), Node([end])])
choose = Node([Node([choose])])
start.adj.append(choose)
ans = [['N'] * Node.cnt for i in range(Node.cnt)]
start.dfs()
print(Node.cnt)
for line in ans:
print(''.join(line))
```
| 98,979 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Fox Ciel wants to write a task for a programming contest. The task is: "You are given a simple undirected graph with n vertexes. Each its edge has unit length. You should calculate the number of shortest paths between vertex 1 and vertex 2."
Same with some writers, she wants to make an example with some certain output: for example, her birthday or the number of her boyfriend. Can you help her to make a test case with answer equal exactly to k?
Input
The first line contains a single integer k (1 β€ k β€ 109).
Output
You should output a graph G with n vertexes (2 β€ n β€ 1000). There must be exactly k shortest paths between vertex 1 and vertex 2 of the graph.
The first line must contain an integer n. Then adjacency matrix G with n rows and n columns must follow. Each element of the matrix must be 'N' or 'Y'. If Gij is 'Y', then graph G has a edge connecting vertex i and vertex j. Consider the graph vertexes are numbered from 1 to n.
The graph must be undirected and simple: Gii = 'N' and Gij = Gji must hold. And there must be at least one path between vertex 1 and vertex 2. It's guaranteed that the answer exists. If there multiple correct answers, you can output any of them.
Examples
Input
2
Output
4
NNYY
NNYY
YYNN
YYNN
Input
9
Output
8
NNYYYNNN
NNNNNYYY
YNNNNYYY
YNNNNYYY
YNNNNYYY
NYYYYNNN
NYYYYNNN
NYYYYNNN
Input
1
Output
2
NY
YN
Note
In first example, there are 2 shortest paths: 1-3-2 and 1-4-2.
In second example, there are 9 shortest paths: 1-3-6-2, 1-3-7-2, 1-3-8-2, 1-4-6-2, 1-4-7-2, 1-4-8-2, 1-5-6-2, 1-5-7-2, 1-5-8-2.
Tags: bitmasks, constructive algorithms, graphs, implementation, math
Correct Solution:
```
N = int(input())
b = bin(N)[2:][::-1]
count = len(b)-1
def print_matrix(m):
print(len(m))
for i in range(len(m)):
print("".join(m[i]))
m = [['N' for i in range(2*(len(b))+count)] for j in range(2*(len(b))+count)]
for i in range(count-1):
val = len(m)-count+i
m[val][val+1] = 'Y'
m[val+1][val] = 'Y'
if count>0:
m[len(m)-1][1] = 'Y'
m[1][len(m)-1] = 'Y'
#print_matrix(m)
c = 0
c2 = len(m)-count
for i in range(0, len(m)-count, 2):
if i >= len(m)-2-count and i != 0:
m[i][1] = 'Y'
m[i+1][1] = 'Y'
m[1][i+1] = 'Y'
m[1][i] = 'Y'
elif i < len(m)-2-count:
m[i][i+2] = 'Y'
m[i][i+3] = 'Y'
m[i+2][i] = 'Y'
m[i+3][i] = 'Y'
if i != 0:
m[i+1][i+2] = 'Y'
m[i+1][i+3] = 'Y'
m[i+2][i+1] = 'Y'
m[i+3][i+1] = 'Y'
if b[c] =='1':
if i == len(m)-count-2:
c2 = 1
#print_matrix(m)
m[i][c2] = 'Y'
m[c2][i] = 'Y'
if i != 0:
m[i+1][c2] = 'Y'
m[c2][i+1] = 'Y'
c2+=1
#print_matrix(m)
c+=1
print(len(m))
for i in range(len(m)):
print("".join(m[i]))
```
| 98,980 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Fox Ciel wants to write a task for a programming contest. The task is: "You are given a simple undirected graph with n vertexes. Each its edge has unit length. You should calculate the number of shortest paths between vertex 1 and vertex 2."
Same with some writers, she wants to make an example with some certain output: for example, her birthday or the number of her boyfriend. Can you help her to make a test case with answer equal exactly to k?
Input
The first line contains a single integer k (1 β€ k β€ 109).
Output
You should output a graph G with n vertexes (2 β€ n β€ 1000). There must be exactly k shortest paths between vertex 1 and vertex 2 of the graph.
The first line must contain an integer n. Then adjacency matrix G with n rows and n columns must follow. Each element of the matrix must be 'N' or 'Y'. If Gij is 'Y', then graph G has a edge connecting vertex i and vertex j. Consider the graph vertexes are numbered from 1 to n.
The graph must be undirected and simple: Gii = 'N' and Gij = Gji must hold. And there must be at least one path between vertex 1 and vertex 2. It's guaranteed that the answer exists. If there multiple correct answers, you can output any of them.
Examples
Input
2
Output
4
NNYY
NNYY
YYNN
YYNN
Input
9
Output
8
NNYYYNNN
NNNNNYYY
YNNNNYYY
YNNNNYYY
YNNNNYYY
NYYYYNNN
NYYYYNNN
NYYYYNNN
Input
1
Output
2
NY
YN
Note
In first example, there are 2 shortest paths: 1-3-2 and 1-4-2.
In second example, there are 9 shortest paths: 1-3-6-2, 1-3-7-2, 1-3-8-2, 1-4-6-2, 1-4-7-2, 1-4-8-2, 1-5-6-2, 1-5-7-2, 1-5-8-2.
Tags: bitmasks, constructive algorithms, graphs, implementation, math
Correct Solution:
```
k=str(input())
l=len(k)
paths=[]
for i in range(l):
paths.append([1]*i+[int(k[i])]+[10]*(l-i-1))
lens = [sum(p) for p in paths]
n = sum(lens)+2
m = ['']*n
m[0] = 'N'*2
for i in range(len(paths)):
m[0] += 'Y'*paths[i][0]+'N'*(lens[i]-paths[i][0])
m[1] = 'N'
for i in range(len(paths)):
m[1] += 'N'*(lens[i]-paths[i][-1])+'Y'*paths[i][-1]
ind=2
for p in paths:
for i in range(len(p)-1):
for j in range(p[i]):
m[ind] = 'N'*(p[i]-j)+'Y'*(p[i+1])+'N'*n
ind+=1
for j in range(p[-1]):
m[ind] = 'N'*n
ind+=1
m2=['']*n
for i in range(n):
m2[i] = ''
for j in range(i):
m2[i]+=m2[j][i]
m2[i]+=m[i][:n-i]
print(len(m2))
for s in m2:
print(s)
```
| 98,981 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Fox Ciel wants to write a task for a programming contest. The task is: "You are given a simple undirected graph with n vertexes. Each its edge has unit length. You should calculate the number of shortest paths between vertex 1 and vertex 2."
Same with some writers, she wants to make an example with some certain output: for example, her birthday or the number of her boyfriend. Can you help her to make a test case with answer equal exactly to k?
Input
The first line contains a single integer k (1 β€ k β€ 109).
Output
You should output a graph G with n vertexes (2 β€ n β€ 1000). There must be exactly k shortest paths between vertex 1 and vertex 2 of the graph.
The first line must contain an integer n. Then adjacency matrix G with n rows and n columns must follow. Each element of the matrix must be 'N' or 'Y'. If Gij is 'Y', then graph G has a edge connecting vertex i and vertex j. Consider the graph vertexes are numbered from 1 to n.
The graph must be undirected and simple: Gii = 'N' and Gij = Gji must hold. And there must be at least one path between vertex 1 and vertex 2. It's guaranteed that the answer exists. If there multiple correct answers, you can output any of them.
Examples
Input
2
Output
4
NNYY
NNYY
YYNN
YYNN
Input
9
Output
8
NNYYYNNN
NNNNNYYY
YNNNNYYY
YNNNNYYY
YNNNNYYY
NYYYYNNN
NYYYYNNN
NYYYYNNN
Input
1
Output
2
NY
YN
Note
In first example, there are 2 shortest paths: 1-3-2 and 1-4-2.
In second example, there are 9 shortest paths: 1-3-6-2, 1-3-7-2, 1-3-8-2, 1-4-6-2, 1-4-7-2, 1-4-8-2, 1-5-6-2, 1-5-7-2, 1-5-8-2.
Tags: bitmasks, constructive algorithms, graphs, implementation, math
Correct Solution:
```
class Node(object):
cnt = 0
def __init__(_, adj=None):
_.id = Node.cnt
Node.cnt += 1
_.adj = adj if adj else []
_.seen = False
n = int(input())
start = Node()
end = Node()
choose = Node()
for i in range(30):
if n & (1<<i):
choose.adj.append(end)
if i == 29:
break
end = Node([Node([end]), Node([end])])
choose = Node([Node([choose])])
start.adj.append(choose)
ans = [['N'] * Node.cnt for i in range(Node.cnt)]
def dfs(v):
if v.seen:
return
v.seen = True
for u in v.adj:
ans[v.id][u.id] = ans[u.id][v.id] = 'Y'
dfs(u)
dfs(start)
print(Node.cnt)
for line in ans:
print(''.join(line))
```
| 98,982 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Fox Ciel wants to write a task for a programming contest. The task is: "You are given a simple undirected graph with n vertexes. Each its edge has unit length. You should calculate the number of shortest paths between vertex 1 and vertex 2."
Same with some writers, she wants to make an example with some certain output: for example, her birthday or the number of her boyfriend. Can you help her to make a test case with answer equal exactly to k?
Input
The first line contains a single integer k (1 β€ k β€ 109).
Output
You should output a graph G with n vertexes (2 β€ n β€ 1000). There must be exactly k shortest paths between vertex 1 and vertex 2 of the graph.
The first line must contain an integer n. Then adjacency matrix G with n rows and n columns must follow. Each element of the matrix must be 'N' or 'Y'. If Gij is 'Y', then graph G has a edge connecting vertex i and vertex j. Consider the graph vertexes are numbered from 1 to n.
The graph must be undirected and simple: Gii = 'N' and Gij = Gji must hold. And there must be at least one path between vertex 1 and vertex 2. It's guaranteed that the answer exists. If there multiple correct answers, you can output any of them.
Examples
Input
2
Output
4
NNYY
NNYY
YYNN
YYNN
Input
9
Output
8
NNYYYNNN
NNNNNYYY
YNNNNYYY
YNNNNYYY
YNNNNYYY
NYYYYNNN
NYYYYNNN
NYYYYNNN
Input
1
Output
2
NY
YN
Note
In first example, there are 2 shortest paths: 1-3-2 and 1-4-2.
In second example, there are 9 shortest paths: 1-3-6-2, 1-3-7-2, 1-3-8-2, 1-4-6-2, 1-4-7-2, 1-4-8-2, 1-5-6-2, 1-5-7-2, 1-5-8-2.
Tags: bitmasks, constructive algorithms, graphs, implementation, math
Correct Solution:
```
k = int(input())
edges = [['N' for i in range(1010)] for j in range(1010)]
vertices = 2
def add_edge(a, b):
global edges
edges[a][b] = edges[b][a] = 'Y'
for i in range(1, 29 + 1):
vertices += 3
add_edge(i * 3, i * 3 - 1)
add_edge(i * 3, i * 3 + 2)
add_edge(i * 3 + 1, i * 3 - 1)
add_edge(i * 3 + 1, i * 3 + 2)
for bit in range(30):
if (1 << bit) & k:
lst = 1
for i in range((29 - bit) * 2):
vertices += 1
add_edge(lst, vertices)
lst = vertices
add_edge(lst, 3 * bit + 2)
print(vertices)
if 0:
for i in range(1, vertices + 1):
print(i, ':', '\n\t', end='')
for j in range(1, vertices + 1):
if edges[i][j] == 'Y':
print(j, end=' ')
print('')
else:
print('\n'.join(map(lambda x: ''.join(x[1:vertices+1]), edges[1:vertices + 1])))
```
| 98,983 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Fox Ciel wants to write a task for a programming contest. The task is: "You are given a simple undirected graph with n vertexes. Each its edge has unit length. You should calculate the number of shortest paths between vertex 1 and vertex 2."
Same with some writers, she wants to make an example with some certain output: for example, her birthday or the number of her boyfriend. Can you help her to make a test case with answer equal exactly to k?
Input
The first line contains a single integer k (1 β€ k β€ 109).
Output
You should output a graph G with n vertexes (2 β€ n β€ 1000). There must be exactly k shortest paths between vertex 1 and vertex 2 of the graph.
The first line must contain an integer n. Then adjacency matrix G with n rows and n columns must follow. Each element of the matrix must be 'N' or 'Y'. If Gij is 'Y', then graph G has a edge connecting vertex i and vertex j. Consider the graph vertexes are numbered from 1 to n.
The graph must be undirected and simple: Gii = 'N' and Gij = Gji must hold. And there must be at least one path between vertex 1 and vertex 2. It's guaranteed that the answer exists. If there multiple correct answers, you can output any of them.
Examples
Input
2
Output
4
NNYY
NNYY
YYNN
YYNN
Input
9
Output
8
NNYYYNNN
NNNNNYYY
YNNNNYYY
YNNNNYYY
YNNNNYYY
NYYYYNNN
NYYYYNNN
NYYYYNNN
Input
1
Output
2
NY
YN
Note
In first example, there are 2 shortest paths: 1-3-2 and 1-4-2.
In second example, there are 9 shortest paths: 1-3-6-2, 1-3-7-2, 1-3-8-2, 1-4-6-2, 1-4-7-2, 1-4-8-2, 1-5-6-2, 1-5-7-2, 1-5-8-2.
Submitted Solution:
```
n, m = 337, int(input())
p = [['N'] * n for i in range(n)]
def f(i, j): p[i][j] = p[j][i] = 'Y'
k = 6 + 15 * 5
for j in range(2, 6): f(1, j)
for i in range(6, k, 5):
for j in range(i - 4, i): f(i, j)
for j in range(i + 1, i + 5): f(i, j)
q, d, s = 4 ** 15, 0, k
while q:
if m >= q:
t = m // q
m -= t * q
if d == 0:
for j in range(k - t, k): f(0, j)
else:
f(0, s)
for i in range(s, s + d): f(i, i + 1)
s += d
for j in range(k - t, k): f(s, j)
s += 1
k -= 5
d += 2
q //= 4
print(s)
for i in range(s): print(''.join(p[i][: s]))
```
Yes
| 98,984 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Fox Ciel wants to write a task for a programming contest. The task is: "You are given a simple undirected graph with n vertexes. Each its edge has unit length. You should calculate the number of shortest paths between vertex 1 and vertex 2."
Same with some writers, she wants to make an example with some certain output: for example, her birthday or the number of her boyfriend. Can you help her to make a test case with answer equal exactly to k?
Input
The first line contains a single integer k (1 β€ k β€ 109).
Output
You should output a graph G with n vertexes (2 β€ n β€ 1000). There must be exactly k shortest paths between vertex 1 and vertex 2 of the graph.
The first line must contain an integer n. Then adjacency matrix G with n rows and n columns must follow. Each element of the matrix must be 'N' or 'Y'. If Gij is 'Y', then graph G has a edge connecting vertex i and vertex j. Consider the graph vertexes are numbered from 1 to n.
The graph must be undirected and simple: Gii = 'N' and Gij = Gji must hold. And there must be at least one path between vertex 1 and vertex 2. It's guaranteed that the answer exists. If there multiple correct answers, you can output any of them.
Examples
Input
2
Output
4
NNYY
NNYY
YYNN
YYNN
Input
9
Output
8
NNYYYNNN
NNNNNYYY
YNNNNYYY
YNNNNYYY
YNNNNYYY
NYYYYNNN
NYYYYNNN
NYYYYNNN
Input
1
Output
2
NY
YN
Note
In first example, there are 2 shortest paths: 1-3-2 and 1-4-2.
In second example, there are 9 shortest paths: 1-3-6-2, 1-3-7-2, 1-3-8-2, 1-4-6-2, 1-4-7-2, 1-4-8-2, 1-5-6-2, 1-5-7-2, 1-5-8-2.
Submitted Solution:
```
n, m, cnt = int(input()), 148, 0
ans = [['N'] * m for i in range(m)]
def edge(i, j):
ans[i][j] = ans[j][i] = 'Y'
def node(adj):
global cnt
i = cnt
cnt += 1
for j in adj:
edge(i, j)
return i
start = node([])
end = node([])
choice = node([end] if n&1 else [])
for i in range(1, 30):
end = node([node([end]), node([end])])
choice = node([node([choice])] + ([end] if (n>>i)&1 else []))
edge(start, choice)
print(m)
for line in ans:
print(''.join(line))
```
Yes
| 98,985 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Fox Ciel wants to write a task for a programming contest. The task is: "You are given a simple undirected graph with n vertexes. Each its edge has unit length. You should calculate the number of shortest paths between vertex 1 and vertex 2."
Same with some writers, she wants to make an example with some certain output: for example, her birthday or the number of her boyfriend. Can you help her to make a test case with answer equal exactly to k?
Input
The first line contains a single integer k (1 β€ k β€ 109).
Output
You should output a graph G with n vertexes (2 β€ n β€ 1000). There must be exactly k shortest paths between vertex 1 and vertex 2 of the graph.
The first line must contain an integer n. Then adjacency matrix G with n rows and n columns must follow. Each element of the matrix must be 'N' or 'Y'. If Gij is 'Y', then graph G has a edge connecting vertex i and vertex j. Consider the graph vertexes are numbered from 1 to n.
The graph must be undirected and simple: Gii = 'N' and Gij = Gji must hold. And there must be at least one path between vertex 1 and vertex 2. It's guaranteed that the answer exists. If there multiple correct answers, you can output any of them.
Examples
Input
2
Output
4
NNYY
NNYY
YYNN
YYNN
Input
9
Output
8
NNYYYNNN
NNNNNYYY
YNNNNYYY
YNNNNYYY
YNNNNYYY
NYYYYNNN
NYYYYNNN
NYYYYNNN
Input
1
Output
2
NY
YN
Note
In first example, there are 2 shortest paths: 1-3-2 and 1-4-2.
In second example, there are 9 shortest paths: 1-3-6-2, 1-3-7-2, 1-3-8-2, 1-4-6-2, 1-4-7-2, 1-4-8-2, 1-5-6-2, 1-5-7-2, 1-5-8-2.
Submitted Solution:
```
class Node(object):
cnt = 0
def __init__(_, adj=None):
_.id = Node.cnt
Node.cnt += 1
_.adj = adj if adj else []
_.seen = False
def dfs(_):
if _.seen:
return
_.seen = True
for i in _.adj:
ans[_.id][i.id] = ans[_.id][i.id] = 'Y'
i.dfs()
n = int(input())
start = Node()
end = Node()
choose = Node()
for i in range(30):
if n & (1<<i):
choose.adj.append(end)
if i == 29:
break
end = Node([Node([end]), Node([end])])
choose = Node([Node([choose])])
start.adj.append(choose)
ans = [['N'] * Node.cnt for i in range(Node.cnt)]
start.dfs()
print(Node.cnt)
for line in ans:
print(''.join(line))
```
No
| 98,986 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Fox Ciel wants to write a task for a programming contest. The task is: "You are given a simple undirected graph with n vertexes. Each its edge has unit length. You should calculate the number of shortest paths between vertex 1 and vertex 2."
Same with some writers, she wants to make an example with some certain output: for example, her birthday or the number of her boyfriend. Can you help her to make a test case with answer equal exactly to k?
Input
The first line contains a single integer k (1 β€ k β€ 109).
Output
You should output a graph G with n vertexes (2 β€ n β€ 1000). There must be exactly k shortest paths between vertex 1 and vertex 2 of the graph.
The first line must contain an integer n. Then adjacency matrix G with n rows and n columns must follow. Each element of the matrix must be 'N' or 'Y'. If Gij is 'Y', then graph G has a edge connecting vertex i and vertex j. Consider the graph vertexes are numbered from 1 to n.
The graph must be undirected and simple: Gii = 'N' and Gij = Gji must hold. And there must be at least one path between vertex 1 and vertex 2. It's guaranteed that the answer exists. If there multiple correct answers, you can output any of them.
Examples
Input
2
Output
4
NNYY
NNYY
YYNN
YYNN
Input
9
Output
8
NNYYYNNN
NNNNNYYY
YNNNNYYY
YNNNNYYY
YNNNNYYY
NYYYYNNN
NYYYYNNN
NYYYYNNN
Input
1
Output
2
NY
YN
Note
In first example, there are 2 shortest paths: 1-3-2 and 1-4-2.
In second example, there are 9 shortest paths: 1-3-6-2, 1-3-7-2, 1-3-8-2, 1-4-6-2, 1-4-7-2, 1-4-8-2, 1-5-6-2, 1-5-7-2, 1-5-8-2.
Submitted Solution:
```
n, t = 901, format(int(input()), 'b')
t = [i for i in range(len(t)) if t[i] == '1']
p = [['N'] * n for i in range(n)]
def f(i, j): p[i][j] = p[j][i] = 'Y'
for i in range(2, 30): f(i, i + 1)
f(30, 1)
for i in range(29):
b = 31 + i * (i + 1)
for a in range(b, b + 2 * i, 2):
f(a, a + 2)
f(a, a + 3)
f(a + 1, a + 2)
f(a + 1, a + 3)
a = b + 2 * i
if i < 28:
f(a, i + 3)
f(a + 1, i + 3)
else:
f(a, 1)
f(a + 1, 1)
for i in t:
if i == 0: f(0, 2)
else:
j = 31 + (i - 1) * i
f(0, j)
f(0, j + 1)
print(n)
for i in p: print(''.join(i))
```
No
| 98,987 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Fox Ciel wants to write a task for a programming contest. The task is: "You are given a simple undirected graph with n vertexes. Each its edge has unit length. You should calculate the number of shortest paths between vertex 1 and vertex 2."
Same with some writers, she wants to make an example with some certain output: for example, her birthday or the number of her boyfriend. Can you help her to make a test case with answer equal exactly to k?
Input
The first line contains a single integer k (1 β€ k β€ 109).
Output
You should output a graph G with n vertexes (2 β€ n β€ 1000). There must be exactly k shortest paths between vertex 1 and vertex 2 of the graph.
The first line must contain an integer n. Then adjacency matrix G with n rows and n columns must follow. Each element of the matrix must be 'N' or 'Y'. If Gij is 'Y', then graph G has a edge connecting vertex i and vertex j. Consider the graph vertexes are numbered from 1 to n.
The graph must be undirected and simple: Gii = 'N' and Gij = Gji must hold. And there must be at least one path between vertex 1 and vertex 2. It's guaranteed that the answer exists. If there multiple correct answers, you can output any of them.
Examples
Input
2
Output
4
NNYY
NNYY
YYNN
YYNN
Input
9
Output
8
NNYYYNNN
NNNNNYYY
YNNNNYYY
YNNNNYYY
YNNNNYYY
NYYYYNNN
NYYYYNNN
NYYYYNNN
Input
1
Output
2
NY
YN
Note
In first example, there are 2 shortest paths: 1-3-2 and 1-4-2.
In second example, there are 9 shortest paths: 1-3-6-2, 1-3-7-2, 1-3-8-2, 1-4-6-2, 1-4-7-2, 1-4-8-2, 1-5-6-2, 1-5-7-2, 1-5-8-2.
Submitted Solution:
```
n, m = 322, int(input())
p = [['N'] * n for i in range(n)]
def f(i, j): p[i][j] = p[j][i] = 'Y'
k = 6 + 15 * 5
for j in range(k + 1, k + 5): f(k, j)
for i in range(6, k, 5):
for j in range(i - 4, i): f(i, j)
for j in range(i + 1, i + 5): f(i, j)
for j in range(k - 4, k): f(k, j)
q, d, s = 4 ** 15, 0, k
while q:
if m > q:
t = m // q
m -= t * q
if d == 0:
for j in range(k - t, k): f(0, j)
else:
f(0, s)
for i in range(s, s + d): f(i, i + 1)
s += d
for j in range(k - t, k): f(s, j)
k -= 5
d += 2
q //= 4
s += 1
print(n)
for i in range(s): print(''.join(p[i][: s]))
```
No
| 98,988 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Fox Ciel wants to write a task for a programming contest. The task is: "You are given a simple undirected graph with n vertexes. Each its edge has unit length. You should calculate the number of shortest paths between vertex 1 and vertex 2."
Same with some writers, she wants to make an example with some certain output: for example, her birthday or the number of her boyfriend. Can you help her to make a test case with answer equal exactly to k?
Input
The first line contains a single integer k (1 β€ k β€ 109).
Output
You should output a graph G with n vertexes (2 β€ n β€ 1000). There must be exactly k shortest paths between vertex 1 and vertex 2 of the graph.
The first line must contain an integer n. Then adjacency matrix G with n rows and n columns must follow. Each element of the matrix must be 'N' or 'Y'. If Gij is 'Y', then graph G has a edge connecting vertex i and vertex j. Consider the graph vertexes are numbered from 1 to n.
The graph must be undirected and simple: Gii = 'N' and Gij = Gji must hold. And there must be at least one path between vertex 1 and vertex 2. It's guaranteed that the answer exists. If there multiple correct answers, you can output any of them.
Examples
Input
2
Output
4
NNYY
NNYY
YYNN
YYNN
Input
9
Output
8
NNYYYNNN
NNNNNYYY
YNNNNYYY
YNNNNYYY
YNNNNYYY
NYYYYNNN
NYYYYNNN
NYYYYNNN
Input
1
Output
2
NY
YN
Note
In first example, there are 2 shortest paths: 1-3-2 and 1-4-2.
In second example, there are 9 shortest paths: 1-3-6-2, 1-3-7-2, 1-3-8-2, 1-4-6-2, 1-4-7-2, 1-4-8-2, 1-5-6-2, 1-5-7-2, 1-5-8-2.
Submitted Solution:
```
class Node(object):
cnt = 0
def __init__(_, adj=None):
_.id = Node.cnt
Node.cnt += 1
_.adj = adj if adj else []
_.seen = False
n = int(input())
start = Node()
end = Node()
choose = Node()
for i in range(30):
if n & (1<<i):
choose.adj.append(end)
if i == 29:
break
end = Node([Node([end]), Node([end])])
choose = Node([Node([choose])])
start.adj.append(choose)
ans = [['N'] * Node.cnt for i in range(Node.cnt)]
def dfs(v):
if v.seen:
return
v.seen = True
for u in v.adj:
ans[v.id][u.id] = ans[v.id][u.id] = 'Y'
dfs(u)
dfs(start)
print(Node.cnt)
for line in ans:
print(''.join(line))
```
No
| 98,989 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Input
The first line contains a single integer n (1 β€ n β€ 1000) β the number of points on a plane.
Each of the next n lines contains two real coordinates xi and yi of the <image> point, specified with exactly 2 fractional digits. All coordinates are between - 1000 and 1000, inclusive.
Output
Output a single real number ΞΈ β the answer to the problem statement. The absolute or relative error of your answer should be at most 10 - 2.
Examples
Input
8
-2.14 2.06
-1.14 2.04
-2.16 1.46
-2.14 0.70
-1.42 0.40
-0.94 -0.48
-1.42 -1.28
-2.16 -1.62
Output
5.410
Input
5
2.26 1.44
2.28 0.64
2.30 -0.30
1.58 0.66
3.24 0.66
Output
5.620
Input
8
6.98 2.06
6.40 1.12
5.98 0.24
5.54 -0.60
7.16 0.30
7.82 1.24
8.34 0.24
8.74 -0.76
Output
5.480
Input
5
10.44 2.06
10.90 0.80
11.48 -0.48
12.06 0.76
12.54 2.06
Output
6.040
Input
8
16.94 2.42
15.72 2.38
14.82 1.58
14.88 0.50
15.76 -0.16
16.86 -0.20
17.00 0.88
16.40 0.92
Output
6.040
Input
7
20.62 3.00
21.06 2.28
21.56 1.36
21.66 0.56
21.64 -0.52
22.14 2.32
22.62 3.04
Output
6.720
Tags: *special, geometry
Correct Solution:
```
a = int(input())
s = 0
for i in range(a):
s += float(input().split()[1])
d = round(s / a, 3) + 5
l = len(str(d).split(".")[1])
if l < 3:
print(str(d) + "0" * (3 - l))
else:
print(d)
```
| 98,990 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Input
The first line contains a single integer n (1 β€ n β€ 1000) β the number of points on a plane.
Each of the next n lines contains two real coordinates xi and yi of the <image> point, specified with exactly 2 fractional digits. All coordinates are between - 1000 and 1000, inclusive.
Output
Output a single real number ΞΈ β the answer to the problem statement. The absolute or relative error of your answer should be at most 10 - 2.
Examples
Input
8
-2.14 2.06
-1.14 2.04
-2.16 1.46
-2.14 0.70
-1.42 0.40
-0.94 -0.48
-1.42 -1.28
-2.16 -1.62
Output
5.410
Input
5
2.26 1.44
2.28 0.64
2.30 -0.30
1.58 0.66
3.24 0.66
Output
5.620
Input
8
6.98 2.06
6.40 1.12
5.98 0.24
5.54 -0.60
7.16 0.30
7.82 1.24
8.34 0.24
8.74 -0.76
Output
5.480
Input
5
10.44 2.06
10.90 0.80
11.48 -0.48
12.06 0.76
12.54 2.06
Output
6.040
Input
8
16.94 2.42
15.72 2.38
14.82 1.58
14.88 0.50
15.76 -0.16
16.86 -0.20
17.00 0.88
16.40 0.92
Output
6.040
Input
7
20.62 3.00
21.06 2.28
21.56 1.36
21.66 0.56
21.64 -0.52
22.14 2.32
22.62 3.04
Output
6.720
Tags: *special, geometry
Correct Solution:
```
n = int(input())
P = [[float(x) for x in input().split()] for _ in range(n)]
print(5 + sum(b for a,b in P)/n)
```
| 98,991 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Input
The first line contains a single integer n (1 β€ n β€ 1000) β the number of points on a plane.
Each of the next n lines contains two real coordinates xi and yi of the <image> point, specified with exactly 2 fractional digits. All coordinates are between - 1000 and 1000, inclusive.
Output
Output a single real number ΞΈ β the answer to the problem statement. The absolute or relative error of your answer should be at most 10 - 2.
Examples
Input
8
-2.14 2.06
-1.14 2.04
-2.16 1.46
-2.14 0.70
-1.42 0.40
-0.94 -0.48
-1.42 -1.28
-2.16 -1.62
Output
5.410
Input
5
2.26 1.44
2.28 0.64
2.30 -0.30
1.58 0.66
3.24 0.66
Output
5.620
Input
8
6.98 2.06
6.40 1.12
5.98 0.24
5.54 -0.60
7.16 0.30
7.82 1.24
8.34 0.24
8.74 -0.76
Output
5.480
Input
5
10.44 2.06
10.90 0.80
11.48 -0.48
12.06 0.76
12.54 2.06
Output
6.040
Input
8
16.94 2.42
15.72 2.38
14.82 1.58
14.88 0.50
15.76 -0.16
16.86 -0.20
17.00 0.88
16.40 0.92
Output
6.040
Input
7
20.62 3.00
21.06 2.28
21.56 1.36
21.66 0.56
21.64 -0.52
22.14 2.32
22.62 3.04
Output
6.720
Tags: *special, geometry
Correct Solution:
```
import sys
def solve():
n = int(input())
avg = sum([list(map(float, input().split()))[1] for _ in range(n)])/n
return avg + 5
if sys.hexversion == 50594544 : sys.stdin = open("test.txt")
print(solve())
```
| 98,992 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Input
The first line contains a single integer n (1 β€ n β€ 1000) β the number of points on a plane.
Each of the next n lines contains two real coordinates xi and yi of the <image> point, specified with exactly 2 fractional digits. All coordinates are between - 1000 and 1000, inclusive.
Output
Output a single real number ΞΈ β the answer to the problem statement. The absolute or relative error of your answer should be at most 10 - 2.
Examples
Input
8
-2.14 2.06
-1.14 2.04
-2.16 1.46
-2.14 0.70
-1.42 0.40
-0.94 -0.48
-1.42 -1.28
-2.16 -1.62
Output
5.410
Input
5
2.26 1.44
2.28 0.64
2.30 -0.30
1.58 0.66
3.24 0.66
Output
5.620
Input
8
6.98 2.06
6.40 1.12
5.98 0.24
5.54 -0.60
7.16 0.30
7.82 1.24
8.34 0.24
8.74 -0.76
Output
5.480
Input
5
10.44 2.06
10.90 0.80
11.48 -0.48
12.06 0.76
12.54 2.06
Output
6.040
Input
8
16.94 2.42
15.72 2.38
14.82 1.58
14.88 0.50
15.76 -0.16
16.86 -0.20
17.00 0.88
16.40 0.92
Output
6.040
Input
7
20.62 3.00
21.06 2.28
21.56 1.36
21.66 0.56
21.64 -0.52
22.14 2.32
22.62 3.04
Output
6.720
Tags: *special, geometry
Correct Solution:
```
n = int(input())
sum = 0
for _ in range(n):
sum += [float(i) for i in input().split(" ")][1]
print(5 + sum/n)
```
| 98,993 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Input
The first line contains a single integer n (1 β€ n β€ 1000) β the number of points on a plane.
Each of the next n lines contains two real coordinates xi and yi of the <image> point, specified with exactly 2 fractional digits. All coordinates are between - 1000 and 1000, inclusive.
Output
Output a single real number ΞΈ β the answer to the problem statement. The absolute or relative error of your answer should be at most 10 - 2.
Examples
Input
8
-2.14 2.06
-1.14 2.04
-2.16 1.46
-2.14 0.70
-1.42 0.40
-0.94 -0.48
-1.42 -1.28
-2.16 -1.62
Output
5.410
Input
5
2.26 1.44
2.28 0.64
2.30 -0.30
1.58 0.66
3.24 0.66
Output
5.620
Input
8
6.98 2.06
6.40 1.12
5.98 0.24
5.54 -0.60
7.16 0.30
7.82 1.24
8.34 0.24
8.74 -0.76
Output
5.480
Input
5
10.44 2.06
10.90 0.80
11.48 -0.48
12.06 0.76
12.54 2.06
Output
6.040
Input
8
16.94 2.42
15.72 2.38
14.82 1.58
14.88 0.50
15.76 -0.16
16.86 -0.20
17.00 0.88
16.40 0.92
Output
6.040
Input
7
20.62 3.00
21.06 2.28
21.56 1.36
21.66 0.56
21.64 -0.52
22.14 2.32
22.62 3.04
Output
6.720
Tags: *special, geometry
Correct Solution:
```
n=int(input())
sum=0.00
for i in range(1,n+1):
s=input().split()
sum+=float(s[1])
print(sum/n+5)
```
| 98,994 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Input
The first line contains a single integer n (1 β€ n β€ 1000) β the number of points on a plane.
Each of the next n lines contains two real coordinates xi and yi of the <image> point, specified with exactly 2 fractional digits. All coordinates are between - 1000 and 1000, inclusive.
Output
Output a single real number ΞΈ β the answer to the problem statement. The absolute or relative error of your answer should be at most 10 - 2.
Examples
Input
8
-2.14 2.06
-1.14 2.04
-2.16 1.46
-2.14 0.70
-1.42 0.40
-0.94 -0.48
-1.42 -1.28
-2.16 -1.62
Output
5.410
Input
5
2.26 1.44
2.28 0.64
2.30 -0.30
1.58 0.66
3.24 0.66
Output
5.620
Input
8
6.98 2.06
6.40 1.12
5.98 0.24
5.54 -0.60
7.16 0.30
7.82 1.24
8.34 0.24
8.74 -0.76
Output
5.480
Input
5
10.44 2.06
10.90 0.80
11.48 -0.48
12.06 0.76
12.54 2.06
Output
6.040
Input
8
16.94 2.42
15.72 2.38
14.82 1.58
14.88 0.50
15.76 -0.16
16.86 -0.20
17.00 0.88
16.40 0.92
Output
6.040
Input
7
20.62 3.00
21.06 2.28
21.56 1.36
21.66 0.56
21.64 -0.52
22.14 2.32
22.62 3.04
Output
6.720
Tags: *special, geometry
Correct Solution:
```
n = int(input())
s = 0.0
for i in range(0, n):
data = input().split()
x, y = float(data[0]), float(data[1])
s += y
print("%.3f" % (5 + s/n))
```
| 98,995 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Input
The first line contains a single integer n (1 β€ n β€ 1000) β the number of points on a plane.
Each of the next n lines contains two real coordinates xi and yi of the <image> point, specified with exactly 2 fractional digits. All coordinates are between - 1000 and 1000, inclusive.
Output
Output a single real number ΞΈ β the answer to the problem statement. The absolute or relative error of your answer should be at most 10 - 2.
Examples
Input
8
-2.14 2.06
-1.14 2.04
-2.16 1.46
-2.14 0.70
-1.42 0.40
-0.94 -0.48
-1.42 -1.28
-2.16 -1.62
Output
5.410
Input
5
2.26 1.44
2.28 0.64
2.30 -0.30
1.58 0.66
3.24 0.66
Output
5.620
Input
8
6.98 2.06
6.40 1.12
5.98 0.24
5.54 -0.60
7.16 0.30
7.82 1.24
8.34 0.24
8.74 -0.76
Output
5.480
Input
5
10.44 2.06
10.90 0.80
11.48 -0.48
12.06 0.76
12.54 2.06
Output
6.040
Input
8
16.94 2.42
15.72 2.38
14.82 1.58
14.88 0.50
15.76 -0.16
16.86 -0.20
17.00 0.88
16.40 0.92
Output
6.040
Input
7
20.62 3.00
21.06 2.28
21.56 1.36
21.66 0.56
21.64 -0.52
22.14 2.32
22.62 3.04
Output
6.720
Tags: *special, geometry
Correct Solution:
```
a=int(input());print(sum([float(input().split(' ')[1]) for i in range(a)])/a+5)
```
| 98,996 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Input
The first line contains a single integer n (1 β€ n β€ 1000) β the number of points on a plane.
Each of the next n lines contains two real coordinates xi and yi of the <image> point, specified with exactly 2 fractional digits. All coordinates are between - 1000 and 1000, inclusive.
Output
Output a single real number ΞΈ β the answer to the problem statement. The absolute or relative error of your answer should be at most 10 - 2.
Examples
Input
8
-2.14 2.06
-1.14 2.04
-2.16 1.46
-2.14 0.70
-1.42 0.40
-0.94 -0.48
-1.42 -1.28
-2.16 -1.62
Output
5.410
Input
5
2.26 1.44
2.28 0.64
2.30 -0.30
1.58 0.66
3.24 0.66
Output
5.620
Input
8
6.98 2.06
6.40 1.12
5.98 0.24
5.54 -0.60
7.16 0.30
7.82 1.24
8.34 0.24
8.74 -0.76
Output
5.480
Input
5
10.44 2.06
10.90 0.80
11.48 -0.48
12.06 0.76
12.54 2.06
Output
6.040
Input
8
16.94 2.42
15.72 2.38
14.82 1.58
14.88 0.50
15.76 -0.16
16.86 -0.20
17.00 0.88
16.40 0.92
Output
6.040
Input
7
20.62 3.00
21.06 2.28
21.56 1.36
21.66 0.56
21.64 -0.52
22.14 2.32
22.62 3.04
Output
6.720
Tags: *special, geometry
Correct Solution:
```
n = int(input())
theta = 5.0
for i in range(n):
theta += list(map(float, input().split()))[1] / n
print('%.3f' % theta)
```
| 98,997 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Input
The first line contains a single integer n (1 β€ n β€ 1000) β the number of points on a plane.
Each of the next n lines contains two real coordinates xi and yi of the <image> point, specified with exactly 2 fractional digits. All coordinates are between - 1000 and 1000, inclusive.
Output
Output a single real number ΞΈ β the answer to the problem statement. The absolute or relative error of your answer should be at most 10 - 2.
Examples
Input
8
-2.14 2.06
-1.14 2.04
-2.16 1.46
-2.14 0.70
-1.42 0.40
-0.94 -0.48
-1.42 -1.28
-2.16 -1.62
Output
5.410
Input
5
2.26 1.44
2.28 0.64
2.30 -0.30
1.58 0.66
3.24 0.66
Output
5.620
Input
8
6.98 2.06
6.40 1.12
5.98 0.24
5.54 -0.60
7.16 0.30
7.82 1.24
8.34 0.24
8.74 -0.76
Output
5.480
Input
5
10.44 2.06
10.90 0.80
11.48 -0.48
12.06 0.76
12.54 2.06
Output
6.040
Input
8
16.94 2.42
15.72 2.38
14.82 1.58
14.88 0.50
15.76 -0.16
16.86 -0.20
17.00 0.88
16.40 0.92
Output
6.040
Input
7
20.62 3.00
21.06 2.28
21.56 1.36
21.66 0.56
21.64 -0.52
22.14 2.32
22.62 3.04
Output
6.720
Submitted Solution:
```
n,num = int(input()),0
for i in range(n):
num += float(input().split()[1])
print(f'{num / n + 5:.3f}')
```
Yes
| 98,998 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Input
The first line contains a single integer n (1 β€ n β€ 1000) β the number of points on a plane.
Each of the next n lines contains two real coordinates xi and yi of the <image> point, specified with exactly 2 fractional digits. All coordinates are between - 1000 and 1000, inclusive.
Output
Output a single real number ΞΈ β the answer to the problem statement. The absolute or relative error of your answer should be at most 10 - 2.
Examples
Input
8
-2.14 2.06
-1.14 2.04
-2.16 1.46
-2.14 0.70
-1.42 0.40
-0.94 -0.48
-1.42 -1.28
-2.16 -1.62
Output
5.410
Input
5
2.26 1.44
2.28 0.64
2.30 -0.30
1.58 0.66
3.24 0.66
Output
5.620
Input
8
6.98 2.06
6.40 1.12
5.98 0.24
5.54 -0.60
7.16 0.30
7.82 1.24
8.34 0.24
8.74 -0.76
Output
5.480
Input
5
10.44 2.06
10.90 0.80
11.48 -0.48
12.06 0.76
12.54 2.06
Output
6.040
Input
8
16.94 2.42
15.72 2.38
14.82 1.58
14.88 0.50
15.76 -0.16
16.86 -0.20
17.00 0.88
16.40 0.92
Output
6.040
Input
7
20.62 3.00
21.06 2.28
21.56 1.36
21.66 0.56
21.64 -0.52
22.14 2.32
22.62 3.04
Output
6.720
Submitted Solution:
```
"""
Codeforces April Fools Contest 2014 Problem G
Author : chaotic_iak
Language: Python 3.3.4
"""
class InputHandlerObject(object):
inputs = []
def getInput(self, n = 0):
res = ""
inputs = self.inputs
if not inputs: inputs.extend(input().split(" "))
if n == 0:
res = inputs[:]
inputs[:] = []
while n > len(inputs):
inputs.extend(input().split(" "))
if n > 0:
res = inputs[:n]
inputs[:n] = []
return res
InputHandler = InputHandlerObject()
g = InputHandler.getInput
############################## SOLUTION ##############################
n = int(input())
sum = 0
for i in range(n):
nx, ny = g()
sum += float(ny)
print(5 + (sum/n))
```
Yes
| 98,999 |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.