message stringlengths 2 20.1k | message_type stringclasses 2 values | message_id int64 0 1 | conversation_id int64 1.95k 109k | cluster float64 17 17 | __index_level_0__ int64 3.91k 217k |
|---|---|---|---|---|---|
Provide tags and a correct Python 3 solution for this coding contest problem.
Pavel is going to make a game of his dream. However, he knows that he can't make it on his own so he founded a development company and hired n workers of staff. Now he wants to pick n workers from the staff who will be directly responsible for developing a game.
Each worker has a certain skill level vi. Besides, each worker doesn't want to work with the one whose skill is very different. In other words, the i-th worker won't work with those whose skill is less than li, and with those whose skill is more than ri.
Pavel understands that the game of his dream isn't too hard to develop, so the worker with any skill will be equally useful. That's why he wants to pick a team of the maximum possible size. Help him pick such team.
Input
The first line contains a single integer n (1 ≤ n ≤ 105) — the number of workers Pavel hired.
Each of the following n lines contains three space-separated integers li, vi, ri (1 ≤ li ≤ vi ≤ ri ≤ 3·105) — the minimum skill value of the workers that the i-th worker can work with, the i-th worker's skill and the maximum skill value of the workers that the i-th worker can work with.
Output
In the first line print a single integer m — the number of workers Pavel must pick for developing the game.
In the next line print m space-separated integers — the numbers of the workers in any order.
If there are multiple optimal solutions, print any of them.
Examples
Input
4
2 8 9
1 4 7
3 6 8
5 8 10
Output
3
1 3 4
Input
6
3 5 16
1 6 11
4 8 12
7 9 16
2 10 14
8 13 15
Output
4
1 2 3 5 | instruction | 0 | 1,954 | 17 | 3,908 |
Tags: data structures
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')
def inp():
return sys.stdin.readline().rstrip()
def mpint():
return map(int, inp().split(' '))
def itg():
return int(inp())
# ############################## import
# 2020/12/17
class LazySegmentTree:
def __init__(self, data, default=0, func=max):
"""initialize the lazy segment tree with data"""
self._default = default
self._func = func
self._len = len(data)
self._size = _size = 1 << (self._len - 1).bit_length()
self._lazy = [0] * (2 * _size)
self.data = [default] * (2 * _size)
self.data[_size:_size + self._len] = data
for i in reversed(range(_size)):
self.data[i] = func(self.data[i + i], self.data[i + i + 1])
def __len__(self):
return self._len
def _push(self, idx):
"""push query on idx to its children"""
# Let the children know of the queries
q, self._lazy[idx] = self._lazy[idx], 0
self._lazy[2 * idx] += q
self._lazy[2 * idx + 1] += q
self.data[2 * idx] += q
self.data[2 * idx + 1] += q
def _update(self, idx):
"""updates the node idx to know of all queries applied to it via its ancestors"""
for i in reversed(range(1, idx.bit_length())):
self._push(idx >> i)
def _build(self, idx):
"""make the changes to idx be known to its ancestors"""
idx >>= 1
while idx:
self.data[idx] = self._func(self.data[2 * idx], self.data[2 * idx + 1]) + self._lazy[idx]
idx >>= 1
def add(self, start, stop, value):
"""lazily add value to [start, stop)"""
start = start_copy = start + self._size
stop = stop_copy = stop + self._size
while start < stop:
if start & 1:
self._lazy[start] += value
self.data[start] += value
start += 1
if stop & 1:
stop -= 1
self._lazy[stop] += value
self.data[stop] += value
start >>= 1
stop >>= 1
# Tell all nodes above of the updated area of the updates
self._build(start_copy)
self._build(stop_copy - 1)
def query(self, start, stop, default=None):
"""func of data[start, stop)"""
start += self._size
stop += self._size
# Apply all the lazily stored queries
self._update(start)
self._update(stop - 1)
res = self._default if default is None else default
while start < stop:
if start & 1:
res = self._func(res, self.data[start])
start += 1
if stop & 1:
stop -= 1
res = self._func(res, self.data[stop])
start >>= 1
stop >>= 1
return res
def __repr__(self):
return "LazySegmentTree({})".format(list(map(lambda i: self.query(i, i + 1), range(self._len))))
class SweepLine:
def __init__(self, data=None, inclusive=True):
"""
:param data: [((1, 2), (3, 4)), (left_bottom, right_top), ...]
:param inclusive: include the right bound
"""
if data:
# [((1, 2), (3, 4), 0), ...]
self.data = [(*pair, i) for i, pair in enumerate(data)]
else:
self.data = []
self.discrete_dict = self.restore = None
self.INCLUSIVE = inclusive
def add_rectangle(self, rectangle):
self.data.append((*rectangle, len(self.data)))
def _discrete(self):
st = set()
operations = []
for p1, p2, i in self.data:
st |= {*p1, *p2}
# we want add operation go first, so *= -1
# we calculate the line intersect
operations.append((p1[1], -1, p1[0], p2[0], i)) # add
operations.append((p2[1], 1, p1[0], p2[0], i)) # subtract
self.restore = sorted(st)
self.discrete_dict = dict(zip(self.restore, range(len(st))))
self.operations = sorted(operations)
# print(*self.operations, sep='\n')
def max_intersect(self, get_points=False):
"""
:return intersect, i (ith rectangle)
:return intersect, i, x
"""
if not self.restore:
self._discrete()
d = self.discrete_dict
tree = LazySegmentTree([0] * len(self.restore))
result = 0, None
for y, v, left, right, i in self.operations:
# print(f"left={left}, right={right}")
left, right, v = d[left], d[right], -v
# print(f"dis -> left={left}, right={right}")
tree.add(left, right + self.INCLUSIVE, v)
if v == 1:
intersect = tree.query(left, right + self.INCLUSIVE)
if intersect > result[0]:
result = intersect, i
tree = LazySegmentTree([0] * len(self.restore))
if get_points:
for y, v, left, right, i in self.operations:
left, right, v = d[left], d[right], -v
tree.add(left, right + self.INCLUSIVE, v)
if i == result[1]:
for lf in range(left, right + self.INCLUSIVE):
if tree.query(lf, lf + 1) == result[0]:
return result[0], i, self.restore[lf]
return result
# ############################## main
def main():
n = itg()
data = []
for _ in range(n):
lf, md, rg = mpint()
data.append(((lf, md), (md, rg)))
sweep = SweepLine(data)
ans1, idx, left = sweep.max_intersect(True)
right = data[idx][0][1]
print(ans1)
# print(idx)
# print(f"intersect = {ans1}")
# print(f"left = {left}, right = {right}")
ans2 = []
for i, pair in enumerate(data):
lf, md, rg = pair[0][0], pair[0][1], pair[1][1]
if lf <= left <= md <= right <= rg:
ans2.append(i + 1)
if len(ans2) != ans1:
print(AssertionError, len(ans2))
else:
print(*ans2)
DEBUG = 0
URL = 'https://codeforces.com/contest/377/problem/D'
if __name__ == '__main__':
# 0: normal, 1: runner, 2: interactive, 3: debug
if DEBUG == 1:
import requests
from ACgenerator.Y_Test_Case_Runner import TestCaseRunner
runner = TestCaseRunner(main, URL)
inp = runner.input_stream
print = runner.output_stream
runner.checking()
else:
if DEBUG != 3:
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
if DEBUG:
_print = print
def print(*args, **kwargs):
_print(*args, **kwargs)
sys.stdout.flush()
main()
# Please check!
``` | output | 1 | 1,954 | 17 | 3,909 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Pavel is going to make a game of his dream. However, he knows that he can't make it on his own so he founded a development company and hired n workers of staff. Now he wants to pick n workers from the staff who will be directly responsible for developing a game.
Each worker has a certain skill level vi. Besides, each worker doesn't want to work with the one whose skill is very different. In other words, the i-th worker won't work with those whose skill is less than li, and with those whose skill is more than ri.
Pavel understands that the game of his dream isn't too hard to develop, so the worker with any skill will be equally useful. That's why he wants to pick a team of the maximum possible size. Help him pick such team.
Input
The first line contains a single integer n (1 ≤ n ≤ 105) — the number of workers Pavel hired.
Each of the following n lines contains three space-separated integers li, vi, ri (1 ≤ li ≤ vi ≤ ri ≤ 3·105) — the minimum skill value of the workers that the i-th worker can work with, the i-th worker's skill and the maximum skill value of the workers that the i-th worker can work with.
Output
In the first line print a single integer m — the number of workers Pavel must pick for developing the game.
In the next line print m space-separated integers — the numbers of the workers in any order.
If there are multiple optimal solutions, print any of them.
Examples
Input
4
2 8 9
1 4 7
3 6 8
5 8 10
Output
3
1 3 4
Input
6
3 5 16
1 6 11
4 8 12
7 9 16
2 10 14
8 13 15
Output
4
1 2 3 5 | instruction | 0 | 1,955 | 17 | 3,910 |
Tags: data structures
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')
def inp():
return sys.stdin.readline().rstrip()
def mpint():
return map(int, inp().split(' '))
def itg():
return int(inp())
# ############################## import
# 2020/11/17
class LazySegmentTree:
def __init__(self, data, default=0, func=max):
"""initialize the lazy segment tree with data"""
self._default = default
self._func = func
self._len = len(data)
self._size = _size = 1 << (self._len - 1).bit_length()
self._lazy = [0] * (2 * _size)
self.data = [default] * (2 * _size)
self.data[_size:_size + self._len] = data
for i in reversed(range(_size)):
self.data[i] = func(self.data[i + i], self.data[i + i + 1])
def __len__(self):
return self._len
def _push(self, idx):
"""push query on idx to its children"""
# Let the children know of the queries
q, self._lazy[idx] = self._lazy[idx], 0
self._lazy[2 * idx] += q
self._lazy[2 * idx + 1] += q
self.data[2 * idx] += q
self.data[2 * idx + 1] += q
def _update(self, idx):
"""updates the node idx to know of all queries applied to it via its ancestors"""
for i in reversed(range(1, idx.bit_length())):
self._push(idx >> i)
def _build(self, idx):
"""make the changes to idx be known to its ancestors"""
idx >>= 1
while idx:
self.data[idx] = self._func(self.data[2 * idx], self.data[2 * idx + 1]) + self._lazy[idx]
idx >>= 1
def add(self, start, stop, value):
"""lazily add value to [start, stop)"""
start = start_copy = start + self._size
stop = stop_copy = stop + self._size
while start < stop:
if start & 1:
self._lazy[start] += value
self.data[start] += value
start += 1
if stop & 1:
stop -= 1
self._lazy[stop] += value
self.data[stop] += value
start >>= 1
stop >>= 1
# Tell all nodes above of the updated area of the updates
self._build(start_copy)
self._build(stop_copy - 1)
def query(self, start, stop, default=None):
"""func of data[start, stop)"""
start += self._size
stop += self._size
# Apply all the lazily stored queries
self._update(start)
self._update(stop - 1)
res = self._default if default is None else default
while start < stop:
if start & 1:
res = self._func(res, self.data[start])
start += 1
if stop & 1:
stop -= 1
res = self._func(res, self.data[stop])
start >>= 1
stop >>= 1
return res
def __repr__(self):
return "LazySegmentTree({})".format(list(map(lambda i: self.query(i, i + 1), range(self._len))))
class SweepLine:
def __init__(self, data=None, inclusive=True):
"""
:param data: [((1, 2), (3, 4)), (left_bottom, right_top), ...]
:param inclusive: include the right bound
"""
if data:
# [((1, 2), (3, 4), 0), ...]
self.data = [(pair[0], pair[1], i) for i, pair in enumerate(data)]
else:
self.data = []
self.discrete_dict = self.restore = None
self.INCLUSIVE = inclusive
def add_rectangle(self, rectangle):
self.data.append((rectangle[0], rectangle[1], len(self.data)))
def _discrete(self):
st = set()
operations = []
for p1, p2, i in self.data:
st |= {p1[0], p1[1], p2[0], p2[1]}
# we want add operation go first, so *= -1
# we calculate the line intersect
operations.append((p1[1], -1, p1[0], p2[0], i)) # add
operations.append((p2[1], 1, p1[0], p2[0], i)) # subtract
self.restore = sorted(st)
self.discrete_dict = dict(zip(self.restore, range(len(st))))
self.operations = sorted(operations)
# print(*self.operations, sep='\n')
def max_intersect(self, get_points=False):
"""
:return intersect, i (ith rectangle)
:return intersect, i, x
"""
if not self.restore:
self._discrete()
d = self.discrete_dict
tree = LazySegmentTree([0] * len(self.restore))
result = 0, None
for y, v, left, right, i in self.operations:
# print(f"left={left}, right={right}")
left, right, v = d[left], d[right], -v
# print(f"dis -> left={left}, right={right}")
tree.add(left, right + self.INCLUSIVE, v)
if v == 1:
intersect = tree.query(left, right + self.INCLUSIVE)
if intersect > result[0]:
result = intersect, i
tree = LazySegmentTree([0] * len(self.restore))
if get_points:
for y, v, left, right, i in self.operations:
left, right, v = d[left], d[right], -v
tree.add(left, right + self.INCLUSIVE, v)
if i == result[1]:
for lf in range(left, right + self.INCLUSIVE):
if tree.query(lf, lf + 1) == result[0]:
return result[0], i, self.restore[lf]
return result
# ############################## main
def main():
n = itg()
data = []
for _ in range(n):
lf, md, rg = mpint()
data.append(((lf, md), (md, rg)))
sweep = SweepLine(data)
ans1, idx, left = sweep.max_intersect(True)
right = data[idx][0][1]
print(ans1)
# print(idx)
# print(f"intersect = {ans1}")
# print(f"left = {left}, right = {right}")
ans2 = []
for i, pair in enumerate(data):
lf, md, rg = pair[0][0], pair[0][1], pair[1][1]
if lf <= left <= md <= right <= rg:
ans2.append(i + 1)
if len(ans2) != ans1:
print(AssertionError, len(ans2))
else:
print(*ans2)
DEBUG = 0
URL = 'https://codeforces.com/contest/377/problem/D'
if __name__ == '__main__':
# 0: normal, 1: runner, 2: interactive, 3: debug
if DEBUG == 1:
import requests
from ACgenerator.Y_Test_Case_Runner import TestCaseRunner
runner = TestCaseRunner(main, URL)
inp = runner.input_stream
print = runner.output_stream
runner.checking()
else:
if DEBUG != 3:
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
if DEBUG:
_print = print
def print(*args, **kwargs):
_print(*args, **kwargs)
sys.stdout.flush()
main()
# Please check!
``` | output | 1 | 1,955 | 17 | 3,911 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Pavel is going to make a game of his dream. However, he knows that he can't make it on his own so he founded a development company and hired n workers of staff. Now he wants to pick n workers from the staff who will be directly responsible for developing a game.
Each worker has a certain skill level vi. Besides, each worker doesn't want to work with the one whose skill is very different. In other words, the i-th worker won't work with those whose skill is less than li, and with those whose skill is more than ri.
Pavel understands that the game of his dream isn't too hard to develop, so the worker with any skill will be equally useful. That's why he wants to pick a team of the maximum possible size. Help him pick such team.
Input
The first line contains a single integer n (1 ≤ n ≤ 105) — the number of workers Pavel hired.
Each of the following n lines contains three space-separated integers li, vi, ri (1 ≤ li ≤ vi ≤ ri ≤ 3·105) — the minimum skill value of the workers that the i-th worker can work with, the i-th worker's skill and the maximum skill value of the workers that the i-th worker can work with.
Output
In the first line print a single integer m — the number of workers Pavel must pick for developing the game.
In the next line print m space-separated integers — the numbers of the workers in any order.
If there are multiple optimal solutions, print any of them.
Examples
Input
4
2 8 9
1 4 7
3 6 8
5 8 10
Output
3
1 3 4
Input
6
3 5 16
1 6 11
4 8 12
7 9 16
2 10 14
8 13 15
Output
4
1 2 3 5
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')
def inp():
return sys.stdin.readline().rstrip()
def mpint():
return map(int, inp().split(' '))
def itg():
return int(inp())
# ############################## import
# 2020/11/17
class LazySegmentTree:
def __init__(self, data, default=0, func=max):
"""initialize the lazy segment tree with data"""
self._default = default
self._func = func
self._len = len(data)
self._size = _size = 1 << (self._len - 1).bit_length()
self._lazy = [0] * (2 * _size)
self.data = [default] * (2 * _size)
self.data[_size:_size + self._len] = data
for i in reversed(range(_size)):
self.data[i] = func(self.data[i + i], self.data[i + i + 1])
def __len__(self):
return self._len
def _push(self, idx):
"""push query on idx to its children"""
# Let the children know of the queries
q, self._lazy[idx] = self._lazy[idx], 0
self._lazy[2 * idx] += q
self._lazy[2 * idx + 1] += q
self.data[2 * idx] += q
self.data[2 * idx + 1] += q
def _update(self, idx):
"""updates the node idx to know of all queries applied to it via its ancestors"""
for i in reversed(range(1, idx.bit_length())):
self._push(idx >> i)
def _build(self, idx):
"""make the changes to idx be known to its ancestors"""
idx >>= 1
while idx:
self.data[idx] = self._func(self.data[2 * idx], self.data[2 * idx + 1]) + self._lazy[idx]
idx >>= 1
def add(self, start, stop, value):
"""lazily add value to [start, stop)"""
start = start_copy = start + self._size
stop = stop_copy = stop + self._size
while start < stop:
if start & 1:
self._lazy[start] += value
self.data[start] += value
start += 1
if stop & 1:
stop -= 1
self._lazy[stop] += value
self.data[stop] += value
start >>= 1
stop >>= 1
# Tell all nodes above of the updated area of the updates
self._build(start_copy)
self._build(stop_copy - 1)
def query(self, start, stop, default=None):
"""func of data[start, stop)"""
start += self._size
stop += self._size
# Apply all the lazily stored queries
self._update(start)
self._update(stop - 1)
res = self._default if default is None else default
while start < stop:
if start & 1:
res = self._func(res, self.data[start])
start += 1
if stop & 1:
stop -= 1
res = self._func(res, self.data[stop])
start >>= 1
stop >>= 1
return res
def __repr__(self):
return "LazySegmentTree({})".format(list(map(lambda i: self.query(i, i + 1), range(self._len))))
class SweepLine:
def __init__(self, data=None, inclusive=True):
"""
:param data: [((1, 2), (3, 4)), (left_bottom, right_top), ...]
:param inclusive: include the right bound
"""
if data:
# [((1, 2), (3, 4), 0), ...]
self.data = [(*pair, i) for i, pair in enumerate(data)]
else:
self.data = []
self.discrete_dict = self.restore = None
self.INCLUSIVE = inclusive
def add_rectangle(self, rectangle):
self.data.append((*rectangle, len(self.data)))
def _discrete(self):
st = set()
operations = []
for p1, p2, i in self.data:
st |= {*p1, *p2}
# we want add operation go first, so *= -1
# we calculate the line intersect
operations.append((p1[1], -1, p1[0], p2[0], i)) # add
operations.append((p2[1], 1, p1[0], p2[0], i)) # subtract
self.restore = sorted(st)
self.discrete_dict = dict(zip(st, range(len(st))))
self.operations = sorted(operations)
# print(*self.operations, sep='\n')
def max_intersect(self, get_points=False):
"""
:return intersect, i (ith rectangle)
:return intersect, i, x
"""
if not self.restore:
self._discrete()
d = self.discrete_dict
tree = LazySegmentTree([0] * len(self.restore))
result = 0, None
for y, v, left, right, i in self.operations:
# print(f"left={left}, right={right}")
left, right, v = d[left], d[right], -v
# print(f"dis -> left={left}, right={right}")
tree.add(left, right + self.INCLUSIVE, v)
if v == 1:
intersect = tree.query(left, right + self.INCLUSIVE)
if intersect > result[0]:
result = intersect, i
tree = LazySegmentTree([0] * len(self.restore))
if get_points:
for y, v, left, right, i in self.operations:
left, right, v = d[left], d[right], -v
tree.add(left, right + self.INCLUSIVE, v)
if i == result[1]:
for lf in range(left, right + self.INCLUSIVE):
if tree.query(lf, lf + 1) == result[0]:
return result[0], i, self.restore[lf]
return result
# ############################## main
def main():
n = itg()
data = []
for _ in range(n):
lf, md, rg = mpint()
data.append(((lf, md), (md, rg)))
sweep = SweepLine(data)
ans1, idx, left = sweep.max_intersect(True)
right = data[idx][0][1]
print(ans1)
# print(idx)
# print(f"intersect = {ans1}")
# print(f"left = {left}, right = {right}")
ans2 = []
for i, pair in enumerate(data):
lf, md, rg = pair[0][0], pair[0][1], pair[1][1]
if lf <= left <= md <= right <= rg:
ans2.append(i + 1)
if len(ans2) != ans1:
print(AssertionError, len(ans2))
else:
print(*ans2)
DEBUG = 0
URL = 'https://codeforces.com/contest/377/problem/D'
if __name__ == '__main__':
# 0: normal, 1: runner, 2: interactive, 3: debug
if DEBUG == 1:
import requests
from ACgenerator.Y_Test_Case_Runner import TestCaseRunner
runner = TestCaseRunner(main, URL)
inp = runner.input_stream
print = runner.output_stream
runner.checking()
else:
if DEBUG != 3:
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
if DEBUG:
_print = print
def print(*args, **kwargs):
_print(*args, **kwargs)
sys.stdout.flush()
main()
# Please check!
``` | instruction | 0 | 1,956 | 17 | 3,912 |
No | output | 1 | 1,956 | 17 | 3,913 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Pavel is going to make a game of his dream. However, he knows that he can't make it on his own so he founded a development company and hired n workers of staff. Now he wants to pick n workers from the staff who will be directly responsible for developing a game.
Each worker has a certain skill level vi. Besides, each worker doesn't want to work with the one whose skill is very different. In other words, the i-th worker won't work with those whose skill is less than li, and with those whose skill is more than ri.
Pavel understands that the game of his dream isn't too hard to develop, so the worker with any skill will be equally useful. That's why he wants to pick a team of the maximum possible size. Help him pick such team.
Input
The first line contains a single integer n (1 ≤ n ≤ 105) — the number of workers Pavel hired.
Each of the following n lines contains three space-separated integers li, vi, ri (1 ≤ li ≤ vi ≤ ri ≤ 3·105) — the minimum skill value of the workers that the i-th worker can work with, the i-th worker's skill and the maximum skill value of the workers that the i-th worker can work with.
Output
In the first line print a single integer m — the number of workers Pavel must pick for developing the game.
In the next line print m space-separated integers — the numbers of the workers in any order.
If there are multiple optimal solutions, print any of them.
Examples
Input
4
2 8 9
1 4 7
3 6 8
5 8 10
Output
3
1 3 4
Input
6
3 5 16
1 6 11
4 8 12
7 9 16
2 10 14
8 13 15
Output
4
1 2 3 5
Submitted Solution:
```
import sys
def swap(a_list, u, v):
a_list[u], a_list[v] = a_list[v], a_list[u]
def update(it, i, l, r, u, v, c):
mid = (l + r) >> 1
left_child = i << 1
right_child = left_child + 1
if l < r:
it[left_child] += t[i]
it[right_child] += t[i]
t[left_child] += t[i]
t[right_child] += t[i]
t[i] = 0
if u <= l and r <= v:
it[i] += c
t[i] += c
return
if u <= mid:
update(it, left_child, l, mid, u, v, c)
if v > mid:
update(it, right_child, mid + 1, r, u, v, c)
it[i] = max(it[left_child], it[right_child])
data_in = [int(i) for i in sys.stdin.read().split()]
n = data_in[0]
a = [[None] * 3 for i in range(n)]
b = [None] * (3 * n)
c = [None] * (3 * n)
left = 0
right = (3 * n) - 1
size = 0
maxC = 0
for i in range(n):
for j in range(3):
a[i][j] = data_in[3 * i + j + 1]
b[size] = (i, 0)
size += 1
b[size] = (i, 2)
swap(b, size, left)
b[right] = (i, 1)
size += 1
left += 1
right -= 1
maxC = max(maxC, a[i][2])
it = [0] * (5 * maxC + 1)
t = [0] * (5 * maxC + 1)
head = [0] * (maxC + 1)
for i in b:
head[a[i[0]][i[1]]] += 1
for i in range(1, maxC + 1):
head[i] += head[i - 1]
for i in b:
c[head[a[i[0]][i[1]]] - 1] = i
head[a[i[0]][i[1]]] -= 1
res = 0
res_lower = 0
for i in c:
id = i[0]
if i[1] == 0:
update(it, 1, 1, maxC, a[id][1], a[id][2], 1)
elif i[1] == 1:
update(it, 1, 1, maxC, a[id][1], a[id][2], -1)
if it[1] > res:
res = it[1]
res_lower = a[id][i[1]]
mark = [False] * n
open_num = 0
for i in c:
id = i[0]
if a[id][1] < res_lower or a[id][0] > res_lower or a[id][2] < res_lower:
continue
if i[1] == 0:
continue
if i[1] == 1:
mark[id] = True
open_num += 1
else:
mark[id] = False
open_num -= 1
if open_num == res:
break
print(res)
res_list = [0] * res
j = 0
for i in range(n):
if mark[i]:
res_list[j] = str(i + 1)
j += 1
print(" ".join(res_list))
``` | instruction | 0 | 1,957 | 17 | 3,914 |
No | output | 1 | 1,957 | 17 | 3,915 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Pavel is going to make a game of his dream. However, he knows that he can't make it on his own so he founded a development company and hired n workers of staff. Now he wants to pick n workers from the staff who will be directly responsible for developing a game.
Each worker has a certain skill level vi. Besides, each worker doesn't want to work with the one whose skill is very different. In other words, the i-th worker won't work with those whose skill is less than li, and with those whose skill is more than ri.
Pavel understands that the game of his dream isn't too hard to develop, so the worker with any skill will be equally useful. That's why he wants to pick a team of the maximum possible size. Help him pick such team.
Input
The first line contains a single integer n (1 ≤ n ≤ 105) — the number of workers Pavel hired.
Each of the following n lines contains three space-separated integers li, vi, ri (1 ≤ li ≤ vi ≤ ri ≤ 3·105) — the minimum skill value of the workers that the i-th worker can work with, the i-th worker's skill and the maximum skill value of the workers that the i-th worker can work with.
Output
In the first line print a single integer m — the number of workers Pavel must pick for developing the game.
In the next line print m space-separated integers — the numbers of the workers in any order.
If there are multiple optimal solutions, print any of them.
Examples
Input
4
2 8 9
1 4 7
3 6 8
5 8 10
Output
3
1 3 4
Input
6
3 5 16
1 6 11
4 8 12
7 9 16
2 10 14
8 13 15
Output
4
1 2 3 5
Submitted Solution:
```
n=int(input())
lprime, rprime=-1, -1
list1=[]
for i in range(n):
l, s, r=[int(k) for k in input().split()]
if i==0:
lprime=l
rprime=r
else:
if l>lprime: lprime=l
if r<rprime: rprime=r
if lprime<=s<=rprime:
list1.append(i+1)
print(len(list1))
for i in range(len(list1)):
print(list1[i], end=" ")
``` | instruction | 0 | 1,958 | 17 | 3,916 |
No | output | 1 | 1,958 | 17 | 3,917 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Vasya has become interested in wrestling. In wrestling wrestlers use techniques for which they are awarded points by judges. The wrestler who gets the most points wins.
When the numbers of points of both wrestlers are equal, the wrestler whose sequence of points is lexicographically greater, wins.
If the sequences of the awarded points coincide, the wrestler who performed the last technique wins. Your task is to determine which wrestler won.
Input
The first line contains number n — the number of techniques that the wrestlers have used (1 ≤ n ≤ 2·105).
The following n lines contain integer numbers ai (|ai| ≤ 109, ai ≠ 0). If ai is positive, that means that the first wrestler performed the technique that was awarded with ai points. And if ai is negative, that means that the second wrestler performed the technique that was awarded with ( - ai) points.
The techniques are given in chronological order.
Output
If the first wrestler wins, print string "first", otherwise print "second"
Examples
Input
5
1
2
-3
-4
3
Output
second
Input
3
-1
-2
3
Output
first
Input
2
4
-4
Output
second
Note
Sequence x = x1x2... x|x| is lexicographically larger than sequence y = y1y2... y|y|, if either |x| > |y| and x1 = y1, x2 = y2, ... , x|y| = y|y|, or there is such number r (r < |x|, r < |y|), that x1 = y1, x2 = y2, ... , xr = yr and xr + 1 > yr + 1.
We use notation |a| to denote length of sequence a. | instruction | 0 | 1,989 | 17 | 3,978 |
Tags: implementation
Correct Solution:
```
arr = []
brr = []
for _ in range(int(input())):
n = int(input())
#print(n)
if n > 0:
arr.append(n)
else:
brr.append(-1*n)
#print(arr,brr,n)
if sum(arr) > sum(brr):
print("first")
elif sum(arr) < sum(brr):
print("second")
else:
if arr > brr:
print("first")
elif arr < brr:
print("second")
else:
if n > 0:
print("first")
else:
print("second")
``` | output | 1 | 1,989 | 17 | 3,979 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Vasya has become interested in wrestling. In wrestling wrestlers use techniques for which they are awarded points by judges. The wrestler who gets the most points wins.
When the numbers of points of both wrestlers are equal, the wrestler whose sequence of points is lexicographically greater, wins.
If the sequences of the awarded points coincide, the wrestler who performed the last technique wins. Your task is to determine which wrestler won.
Input
The first line contains number n — the number of techniques that the wrestlers have used (1 ≤ n ≤ 2·105).
The following n lines contain integer numbers ai (|ai| ≤ 109, ai ≠ 0). If ai is positive, that means that the first wrestler performed the technique that was awarded with ai points. And if ai is negative, that means that the second wrestler performed the technique that was awarded with ( - ai) points.
The techniques are given in chronological order.
Output
If the first wrestler wins, print string "first", otherwise print "second"
Examples
Input
5
1
2
-3
-4
3
Output
second
Input
3
-1
-2
3
Output
first
Input
2
4
-4
Output
second
Note
Sequence x = x1x2... x|x| is lexicographically larger than sequence y = y1y2... y|y|, if either |x| > |y| and x1 = y1, x2 = y2, ... , x|y| = y|y|, or there is such number r (r < |x|, r < |y|), that x1 = y1, x2 = y2, ... , xr = yr and xr + 1 > yr + 1.
We use notation |a| to denote length of sequence a. | instruction | 0 | 1,990 | 17 | 3,980 |
Tags: implementation
Correct Solution:
```
n=int(input())
c1=0
c2=0
c11=[]
c22=[]
ko=0
for i in range(n):
a=int(input())
if a>0:
c1+=a
kk=str(a)
c11.append(a)
ko=1
else:
c2+=abs(a)
kk=str(abs(a))
c22.append(abs(a))
ko=0
if len(c11)<len(c22):
pp=1
elif len(c11)>len(c22):
pp=-1
else:
pp=0
for i in range(min(len(c11),len(c22))):
if c11[i]>c22[i]:
pp=1
break
elif c11[i]<c22[i]:
pp=-1
break
if c1>c2:
print("first")
elif c1<c2:
print("second")
elif c1==c2:
if pp==1:
print("first")
elif pp==-1:
print("second")
elif pp==0:
if ko==0:
print("second")
elif ko==1:
print("first")
``` | output | 1 | 1,990 | 17 | 3,981 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Vasya has become interested in wrestling. In wrestling wrestlers use techniques for which they are awarded points by judges. The wrestler who gets the most points wins.
When the numbers of points of both wrestlers are equal, the wrestler whose sequence of points is lexicographically greater, wins.
If the sequences of the awarded points coincide, the wrestler who performed the last technique wins. Your task is to determine which wrestler won.
Input
The first line contains number n — the number of techniques that the wrestlers have used (1 ≤ n ≤ 2·105).
The following n lines contain integer numbers ai (|ai| ≤ 109, ai ≠ 0). If ai is positive, that means that the first wrestler performed the technique that was awarded with ai points. And if ai is negative, that means that the second wrestler performed the technique that was awarded with ( - ai) points.
The techniques are given in chronological order.
Output
If the first wrestler wins, print string "first", otherwise print "second"
Examples
Input
5
1
2
-3
-4
3
Output
second
Input
3
-1
-2
3
Output
first
Input
2
4
-4
Output
second
Note
Sequence x = x1x2... x|x| is lexicographically larger than sequence y = y1y2... y|y|, if either |x| > |y| and x1 = y1, x2 = y2, ... , x|y| = y|y|, or there is such number r (r < |x|, r < |y|), that x1 = y1, x2 = y2, ... , xr = yr and xr + 1 > yr + 1.
We use notation |a| to denote length of sequence a. | instruction | 0 | 1,991 | 17 | 3,982 |
Tags: implementation
Correct Solution:
```
n=int(input())
f=[]
s=[]
fsc=0
ssc=0
for i in range(n):
x=int(input())
if x>0:
f.append(x)
fsc+=x
else:
ssc+=-x
s.append(-x)
if i==n-1:
las=x
if fsc>ssc:
print('first')
elif ssc>fsc:
print('second')
elif f>s:
print('first')
elif f<s:
print('second')
else:
if las>0:
print('first')
else:
print('second')
``` | output | 1 | 1,991 | 17 | 3,983 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Vasya has become interested in wrestling. In wrestling wrestlers use techniques for which they are awarded points by judges. The wrestler who gets the most points wins.
When the numbers of points of both wrestlers are equal, the wrestler whose sequence of points is lexicographically greater, wins.
If the sequences of the awarded points coincide, the wrestler who performed the last technique wins. Your task is to determine which wrestler won.
Input
The first line contains number n — the number of techniques that the wrestlers have used (1 ≤ n ≤ 2·105).
The following n lines contain integer numbers ai (|ai| ≤ 109, ai ≠ 0). If ai is positive, that means that the first wrestler performed the technique that was awarded with ai points. And if ai is negative, that means that the second wrestler performed the technique that was awarded with ( - ai) points.
The techniques are given in chronological order.
Output
If the first wrestler wins, print string "first", otherwise print "second"
Examples
Input
5
1
2
-3
-4
3
Output
second
Input
3
-1
-2
3
Output
first
Input
2
4
-4
Output
second
Note
Sequence x = x1x2... x|x| is lexicographically larger than sequence y = y1y2... y|y|, if either |x| > |y| and x1 = y1, x2 = y2, ... , x|y| = y|y|, or there is such number r (r < |x|, r < |y|), that x1 = y1, x2 = y2, ... , xr = yr and xr + 1 > yr + 1.
We use notation |a| to denote length of sequence a. | instruction | 0 | 1,992 | 17 | 3,984 |
Tags: implementation
Correct Solution:
```
n = int(input())
a = []
b = []
x = 0
for i in range(n):
q = int(input())
if q > 0:
x = 1
a.append(q)
else:
x = -1
b.append(-q)
if sum(a) != sum(b):
print('first' if sum(a) > sum(b) else 'second')
elif a != b:
print('first' if a > b else 'second')
else:
print('first' if x > 0 else 'second')
``` | output | 1 | 1,992 | 17 | 3,985 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Vasya has become interested in wrestling. In wrestling wrestlers use techniques for which they are awarded points by judges. The wrestler who gets the most points wins.
When the numbers of points of both wrestlers are equal, the wrestler whose sequence of points is lexicographically greater, wins.
If the sequences of the awarded points coincide, the wrestler who performed the last technique wins. Your task is to determine which wrestler won.
Input
The first line contains number n — the number of techniques that the wrestlers have used (1 ≤ n ≤ 2·105).
The following n lines contain integer numbers ai (|ai| ≤ 109, ai ≠ 0). If ai is positive, that means that the first wrestler performed the technique that was awarded with ai points. And if ai is negative, that means that the second wrestler performed the technique that was awarded with ( - ai) points.
The techniques are given in chronological order.
Output
If the first wrestler wins, print string "first", otherwise print "second"
Examples
Input
5
1
2
-3
-4
3
Output
second
Input
3
-1
-2
3
Output
first
Input
2
4
-4
Output
second
Note
Sequence x = x1x2... x|x| is lexicographically larger than sequence y = y1y2... y|y|, if either |x| > |y| and x1 = y1, x2 = y2, ... , x|y| = y|y|, or there is such number r (r < |x|, r < |y|), that x1 = y1, x2 = y2, ... , xr = yr and xr + 1 > yr + 1.
We use notation |a| to denote length of sequence a. | instruction | 0 | 1,993 | 17 | 3,986 |
Tags: implementation
Correct Solution:
```
# pylint: disable=unused-variable
# pylint: enable=too-many-lines
#* Just believe in yourself
#@ Author @CAP
import os
import sys
from io import BytesIO, IOBase
import math as M
import itertools as ITR
from collections import defaultdict as D
from collections import Counter as C
from collections import deque as Q
import threading
from functools import lru_cache, reduce
from functools import cmp_to_key as CMP
from bisect import bisect_left as BL
from bisect import bisect_right as BR
import random as R
import string
import cmath,time
enum=enumerate
start_time = time.time()
#? Variables
MOD=1_00_00_00_007; MA=float("inf"); MI=float("-inf")
#?graph 8 direction
di8=[(1, 0), (1, 1), (0, 1), (-1, 1), (-1, 0), (-1, -1), (0, -1), (1, -1)]
#?graph 4 direction
di4=[(1, 0), (0, 1), (-1, 0), (0, -1)]
#? Stack increment
def increase_stack():
sys.setrecursionlimit(2**32//2-1)
threading.stack_size(1 << 27)
#sys.setrecursionlimit(10**6)
#threading.stack_size(10**8)
t = threading.Thread(target=main)
t.start()
t.join()
#? Region Funtions
def prints(a):
print(a,end=" ")
def binary(n):
return bin(n)[2:]
def decimal(s):
return (int(s, 2))
def pow2(n):
p = 0
while (n > 1):
n //= 2
p += 1
return (p)
def maxfactor(n):
q = []
for i in range(1,int(n ** 0.5) + 1):
if n % i == 0: q.append(i)
if q:
return q[-1]
def factors(n):
q = []
for i in range(1,int(n ** 0.5) + 1):
if n % i == 0: q.append(i); q.append(n // i)
return(list(sorted(list(set(q)))))
def primeFactors(n):
l=[]
while n % 2 == 0:
l.append(2)
n = n / 2
for i in range(3, int(M.sqrt(n)) + 1, 2):
while n % i == 0:
l.append(i)
n = n / i
if n > 2:
l.append(int(n))
l.sort()
return (l)
def isPrime(n):
if (n == 1):
return (False)
else:
root = int(n ** 0.5)
root += 1
for i in range(2, root):
if (n % i == 0):
return (False)
return (True)
def seive(n):
a = [1]
prime = [True for i in range(n + 1)]
p = 2
while (p * p <= n):
if (prime[p] == True):
for i in range(p ** 2,n + 1,p):
prime[i] = False
p = p + 1
for p in range(2,n + 1):
if prime[p]:
a.append(p)
return(a)
def maxPrimeFactors(n):
maxPrime = -1
while n % 2 == 0:
maxPrime = 2
n >>= 1
for i in range(3, int(M.sqrt(n)) + 1, 2):
while n % i == 0:
maxPrime = i
n = n / i
if n > 2:
maxPrime = n
return int(maxPrime)
def countchar(s,i):
c=0
ch=s[i]
for i in range(i,len(s)):
if(s[i]==ch):
c+=1
else:
break
return(c)
def str_counter(a):
q = [0] * 26
for i in range(len(a)):
q[ord(a[i]) - 97] = q[ord(a[i]) - 97] + 1
return(q)
def lis(arr):
n = len(arr)
lis = [1] * n
maximum=0
for i in range(1, n):
for j in range(0, i):
if arr[i] > arr[j] and lis[i] < lis[j] + 1:
lis[i] = lis[j] + 1
maximum=max(maximum,lis[i])
return maximum
def lcm(arr):
a=arr[0]; val=arr[0]
for i in range(1,len(arr)):
gcd=gcd(a,arr[i])
a=arr[i]; val*=arr[i]
return val//gcd
def ncr(n,r):
return M.factorial(n) // (M.factorial(n - r) * M.factorial(r))
def npr(n,r):
return M.factorial(n) // M.factorial(n - r)
#? Region Taking Input
def inint():
return int(inp())
def inarr():
return list(map(int,inp().split()))
def invar():
return map(int,inp().split())
def instr():
s=inp()
return list(s)
def insarr():
return inp().split()
#? Region Fastio
BUFSIZE = 8192
class FastIO(IOBase):
newlines = 0
def __init__(self, file):
self._fd = file.fileno()
self.buffer = BytesIO()
self.writable = "x" in file.mode or "r" not in file.mode
self.write = self.buffer.write if self.writable else None
def read(self):
while True:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
if not b:
break
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines = 0
return self.buffer.read()
def readline(self):
while self.newlines == 0:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
self.newlines = b.count(b"\n") + (not b)
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines -= 1
return self.buffer.readline()
def flush(self):
if self.writable:
os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
self.write = lambda s: self.buffer.write(s.encode("ascii"))
self.read = lambda: self.buffer.read().decode("ascii")
self.readline = lambda: self.buffer.readline().decode("ascii")
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
inp = lambda: sys.stdin.readline().rstrip("\r\n")
#<==================================== Write The Useful Code Here ============================
#< Make it one if there is some test cases
TestCases=0 #<=====================
#< =======================================
"""
> Sometimes later becomes never. Do it now.
! Be Better than yesterday.
* Your limitation—it’s only your imagination.
> Push yourself, because no one else is going to do it for you.
? The harder you work for something, the greater you’ll feel when you achieve it.
! Great things never come from comfort zones.
* Don’t stop when you’re tired. Stop when you’re done.
> Do something today that your future self will thank you for.
? It’s going to be hard, but hard does not mean impossible.
! Sometimes we’re tested not to show our weaknesses, but to discover our strengths.
"""
#@ Goal is to get Candidate Master
def solve():
n=inint()
p1 = 0; p2 = 0; l1 = 0; l2 = 0; flag=0; aseq=[]; bseq=[]
for i in range(n):
a=inint()
if a>0:
p1+=a; l1+=1; flag=1
aseq.append(a)
else:
p2+=abs(a); l2+=1; flag=0
bseq.append(abs(a))
if p1>p2: print("first"); return
if p1==p2:
if aseq==bseq:
if flag:
print("first"); return
if aseq>bseq:
print("first"); return
print("second")
#! This is the Main Function
def main():
flag=0
#! Checking we are offline or not
try:
sys.stdin = open("c:/Users/Manoj Chowdary/Documents/python/CodeForces/contest-div-2/input.txt","r")
sys.stdout = open("c:/Users/Manoj Chowdary/Documents/python/CodeForces/contest-div-2/output.txt","w")
except: flag=1
t=1
if TestCases:
t = inint()
for _ in range(1,t + 1):
solve()
if not flag:
print("Time: %.4f sec"%(time.time() - start_time))
localtime = time.asctime( time.localtime(time.time()) )
print(localtime)
sys.stdout.close()
#? End Region
if __name__ == "__main__":
#? Incresing Stack Limit
#increase_stack()
#! Calling Main Function
main()
``` | output | 1 | 1,993 | 17 | 3,987 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Vasya has become interested in wrestling. In wrestling wrestlers use techniques for which they are awarded points by judges. The wrestler who gets the most points wins.
When the numbers of points of both wrestlers are equal, the wrestler whose sequence of points is lexicographically greater, wins.
If the sequences of the awarded points coincide, the wrestler who performed the last technique wins. Your task is to determine which wrestler won.
Input
The first line contains number n — the number of techniques that the wrestlers have used (1 ≤ n ≤ 2·105).
The following n lines contain integer numbers ai (|ai| ≤ 109, ai ≠ 0). If ai is positive, that means that the first wrestler performed the technique that was awarded with ai points. And if ai is negative, that means that the second wrestler performed the technique that was awarded with ( - ai) points.
The techniques are given in chronological order.
Output
If the first wrestler wins, print string "first", otherwise print "second"
Examples
Input
5
1
2
-3
-4
3
Output
second
Input
3
-1
-2
3
Output
first
Input
2
4
-4
Output
second
Note
Sequence x = x1x2... x|x| is lexicographically larger than sequence y = y1y2... y|y|, if either |x| > |y| and x1 = y1, x2 = y2, ... , x|y| = y|y|, or there is such number r (r < |x|, r < |y|), that x1 = y1, x2 = y2, ... , xr = yr and xr + 1 > yr + 1.
We use notation |a| to denote length of sequence a. | instruction | 0 | 1,994 | 17 | 3,988 |
Tags: implementation
Correct Solution:
```
a = []
b = []
last = 0
for i in range(int(input())):
j = int(input())
if j > 0:
last = 1
a.append(j)
else:
last = 0
b.append(-j)
suma = sum(a)
sumb = sum(b)
if suma > sumb or (suma == sumb and a > b) or (suma == sumb and a == b and last == 1):
print("first")
else:
print("second")
``` | output | 1 | 1,994 | 17 | 3,989 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Vasya has become interested in wrestling. In wrestling wrestlers use techniques for which they are awarded points by judges. The wrestler who gets the most points wins.
When the numbers of points of both wrestlers are equal, the wrestler whose sequence of points is lexicographically greater, wins.
If the sequences of the awarded points coincide, the wrestler who performed the last technique wins. Your task is to determine which wrestler won.
Input
The first line contains number n — the number of techniques that the wrestlers have used (1 ≤ n ≤ 2·105).
The following n lines contain integer numbers ai (|ai| ≤ 109, ai ≠ 0). If ai is positive, that means that the first wrestler performed the technique that was awarded with ai points. And if ai is negative, that means that the second wrestler performed the technique that was awarded with ( - ai) points.
The techniques are given in chronological order.
Output
If the first wrestler wins, print string "first", otherwise print "second"
Examples
Input
5
1
2
-3
-4
3
Output
second
Input
3
-1
-2
3
Output
first
Input
2
4
-4
Output
second
Note
Sequence x = x1x2... x|x| is lexicographically larger than sequence y = y1y2... y|y|, if either |x| > |y| and x1 = y1, x2 = y2, ... , x|y| = y|y|, or there is such number r (r < |x|, r < |y|), that x1 = y1, x2 = y2, ... , xr = yr and xr + 1 > yr + 1.
We use notation |a| to denote length of sequence a. | instruction | 0 | 1,995 | 17 | 3,990 |
Tags: implementation
Correct Solution:
```
import sys, os, io
def rs(): return sys.stdin.readline().rstrip()
def ri(): return int(sys.stdin.readline())
def ria(): return list(map(int, sys.stdin.readline().split()))
def ws(s): sys.stdout.write(s + '\n')
def wi(n): sys.stdout.write(str(n) + '\n')
def wia(a): sys.stdout.write(' '.join([str(x) for x in a]) + '\n')
import math,datetime,functools,itertools,operator,bisect,fractions,statistics
from collections import deque,defaultdict,OrderedDict,Counter
from fractions import Fraction
from decimal import Decimal
from sys import stdout
from heapq import heappush, heappop, heapify ,_heapify_max,_heappop_max,nsmallest,nlargest
sys.setrecursionlimit(111111)
INF=99999999999999999999999999999999
def main():
mod=1000000007
# InverseofNumber(mod)
# InverseofFactorial(mod)
# factorial(mod)
starttime=datetime.datetime.now()
if(os.path.exists('input.txt')):
sys.stdin = open("input.txt","r")
sys.stdout = open("output.txt","w")
###CODE
tc = 1
for _ in range(tc):
n=ri()
player1=0
player2=0
p1=[]
p2=[]
lp=-1
for i in range(n):
score=ri()
if score>0:
player1+=score
p1.append(score)
lp=1
else:
player2-=score
p2.append(-score)
lp=2
if player1>player2:
print("first")
elif player2>player1:
print("second")
else:
if p1>p2:
print("first")
elif p2>p1:
print("second")
else:
if lp==1:
print("first")
else:
print("second")
#<--Solving Area Ends
endtime=datetime.datetime.now()
time=(endtime-starttime).total_seconds()*1000
if(os.path.exists('input.txt')):
print("Time:",time,"ms")
class FastReader(io.IOBase):
newlines = 0
def __init__(self, fd, chunk_size=1024 * 8):
self._fd = fd
self._chunk_size = chunk_size
self.buffer = io.BytesIO()
def read(self):
while True:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, self._chunk_size))
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, size=-1):
while self.newlines == 0:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, self._chunk_size if size == -1 else size))
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()
class FastWriter(io.IOBase):
def __init__(self, fd):
self._fd = fd
self.buffer = io.BytesIO()
self.write = self.buffer.write
def flush(self):
os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class FastStdin(io.IOBase):
def __init__(self, fd=0):
self.buffer = FastReader(fd)
self.read = lambda: self.buffer.read().decode("ascii")
self.readline = lambda: self.buffer.readline().decode("ascii")
class FastStdout(io.IOBase):
def __init__(self, fd=1):
self.buffer = FastWriter(fd)
self.write = lambda s: self.buffer.write(s.encode("ascii"))
self.flush = self.buffer.flush
if __name__ == '__main__':
sys.stdin = FastStdin()
sys.stdout = FastStdout()
main()
``` | output | 1 | 1,995 | 17 | 3,991 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Vasya has become interested in wrestling. In wrestling wrestlers use techniques for which they are awarded points by judges. The wrestler who gets the most points wins.
When the numbers of points of both wrestlers are equal, the wrestler whose sequence of points is lexicographically greater, wins.
If the sequences of the awarded points coincide, the wrestler who performed the last technique wins. Your task is to determine which wrestler won.
Input
The first line contains number n — the number of techniques that the wrestlers have used (1 ≤ n ≤ 2·105).
The following n lines contain integer numbers ai (|ai| ≤ 109, ai ≠ 0). If ai is positive, that means that the first wrestler performed the technique that was awarded with ai points. And if ai is negative, that means that the second wrestler performed the technique that was awarded with ( - ai) points.
The techniques are given in chronological order.
Output
If the first wrestler wins, print string "first", otherwise print "second"
Examples
Input
5
1
2
-3
-4
3
Output
second
Input
3
-1
-2
3
Output
first
Input
2
4
-4
Output
second
Note
Sequence x = x1x2... x|x| is lexicographically larger than sequence y = y1y2... y|y|, if either |x| > |y| and x1 = y1, x2 = y2, ... , x|y| = y|y|, or there is such number r (r < |x|, r < |y|), that x1 = y1, x2 = y2, ... , xr = yr and xr + 1 > yr + 1.
We use notation |a| to denote length of sequence a. | instruction | 0 | 1,996 | 17 | 3,992 |
Tags: implementation
Correct Solution:
```
def lex(a,b):
for i in range(min(len(a),len(b))):
if a[i]>b[i]:
return True
elif a[i]<b[i]:
return False
if len(a)>len(b):
return True
return False
n=int(input())
w1=[]
w2=[]
last=True
for i in range(n):
x=int(input())
if x<0:
last=False
w2.append(-x)
else:
last=True
w1.append(x)
if sum(w1)>sum(w2):
print("first")
elif sum(w1)==sum(w2):
if lex(w1,w2):
print("first")
elif lex(w2,w1):
print("second")
else:
if last:
print("first")
else:
print("second")
else:
print("second")
``` | output | 1 | 1,996 | 17 | 3,993 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Vasya has become interested in wrestling. In wrestling wrestlers use techniques for which they are awarded points by judges. The wrestler who gets the most points wins.
When the numbers of points of both wrestlers are equal, the wrestler whose sequence of points is lexicographically greater, wins.
If the sequences of the awarded points coincide, the wrestler who performed the last technique wins. Your task is to determine which wrestler won.
Input
The first line contains number n — the number of techniques that the wrestlers have used (1 ≤ n ≤ 2·105).
The following n lines contain integer numbers ai (|ai| ≤ 109, ai ≠ 0). If ai is positive, that means that the first wrestler performed the technique that was awarded with ai points. And if ai is negative, that means that the second wrestler performed the technique that was awarded with ( - ai) points.
The techniques are given in chronological order.
Output
If the first wrestler wins, print string "first", otherwise print "second"
Examples
Input
5
1
2
-3
-4
3
Output
second
Input
3
-1
-2
3
Output
first
Input
2
4
-4
Output
second
Note
Sequence x = x1x2... x|x| is lexicographically larger than sequence y = y1y2... y|y|, if either |x| > |y| and x1 = y1, x2 = y2, ... , x|y| = y|y|, or there is such number r (r < |x|, r < |y|), that x1 = y1, x2 = y2, ... , xr = yr and xr + 1 > yr + 1.
We use notation |a| to denote length of sequence a.
Submitted Solution:
```
n = int(input())
first = []
second = []
last = 0
for i in range(n):
a = int(input())
if a>0:
first.append(a)
else:
second.append(abs(a))
if i==n-1:
last = a
if abs(sum(first))>abs(sum(second)):
print('first')
quit()
if abs(sum(first))<abs(sum(second)):
print('second')
else:
if first>second:
print('first')
quit()
if first<second:
print('second')
else:
if last>0:
print('first')
else:
print('second')
``` | instruction | 0 | 1,997 | 17 | 3,994 |
Yes | output | 1 | 1,997 | 17 | 3,995 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Vasya has become interested in wrestling. In wrestling wrestlers use techniques for which they are awarded points by judges. The wrestler who gets the most points wins.
When the numbers of points of both wrestlers are equal, the wrestler whose sequence of points is lexicographically greater, wins.
If the sequences of the awarded points coincide, the wrestler who performed the last technique wins. Your task is to determine which wrestler won.
Input
The first line contains number n — the number of techniques that the wrestlers have used (1 ≤ n ≤ 2·105).
The following n lines contain integer numbers ai (|ai| ≤ 109, ai ≠ 0). If ai is positive, that means that the first wrestler performed the technique that was awarded with ai points. And if ai is negative, that means that the second wrestler performed the technique that was awarded with ( - ai) points.
The techniques are given in chronological order.
Output
If the first wrestler wins, print string "first", otherwise print "second"
Examples
Input
5
1
2
-3
-4
3
Output
second
Input
3
-1
-2
3
Output
first
Input
2
4
-4
Output
second
Note
Sequence x = x1x2... x|x| is lexicographically larger than sequence y = y1y2... y|y|, if either |x| > |y| and x1 = y1, x2 = y2, ... , x|y| = y|y|, or there is such number r (r < |x|, r < |y|), that x1 = y1, x2 = y2, ... , xr = yr and xr + 1 > yr + 1.
We use notation |a| to denote length of sequence a.
Submitted Solution:
```
import sys
input=sys.stdin.buffer.readline
import os
t=int(input())
arr=[]
brr=[]
last=0
while t:
t-=1
n=int(input())
last=n
if n<0:
brr.append(-n)
else:
arr.append(n)
if sum(arr)>sum(brr):
print("first")
elif sum(arr)<sum(brr):
print("second")
else:
z=""
for i in range(min(len(arr),len(brr))):
if arr[i]>brr[i]:
z="first"
break
elif arr[i]<brr[i]:
z="second"
break
if z=="":
if last<0:
print("second")
else:
print("first")
else:
print(z)
``` | instruction | 0 | 1,998 | 17 | 3,996 |
Yes | output | 1 | 1,998 | 17 | 3,997 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Vasya has become interested in wrestling. In wrestling wrestlers use techniques for which they are awarded points by judges. The wrestler who gets the most points wins.
When the numbers of points of both wrestlers are equal, the wrestler whose sequence of points is lexicographically greater, wins.
If the sequences of the awarded points coincide, the wrestler who performed the last technique wins. Your task is to determine which wrestler won.
Input
The first line contains number n — the number of techniques that the wrestlers have used (1 ≤ n ≤ 2·105).
The following n lines contain integer numbers ai (|ai| ≤ 109, ai ≠ 0). If ai is positive, that means that the first wrestler performed the technique that was awarded with ai points. And if ai is negative, that means that the second wrestler performed the technique that was awarded with ( - ai) points.
The techniques are given in chronological order.
Output
If the first wrestler wins, print string "first", otherwise print "second"
Examples
Input
5
1
2
-3
-4
3
Output
second
Input
3
-1
-2
3
Output
first
Input
2
4
-4
Output
second
Note
Sequence x = x1x2... x|x| is lexicographically larger than sequence y = y1y2... y|y|, if either |x| > |y| and x1 = y1, x2 = y2, ... , x|y| = y|y|, or there is such number r (r < |x|, r < |y|), that x1 = y1, x2 = y2, ... , xr = yr and xr + 1 > yr + 1.
We use notation |a| to denote length of sequence a.
Submitted Solution:
```
class CodeforcesTask493BSolution:
def __init__(self):
self.result = ''
self.n = 0
self.points = []
def read_input(self):
self.n = int(input())
for x in range(self.n):
self.points.append(int(input()))
def process_task(self):
first = [x for x in self.points if x > 0]
second = [-x for x in self.points if x < 0]
f_points = sum(first)
s_points = sum(second)
if f_points > s_points:
self.result = "first"
elif f_points < s_points:
self.result = "second"
else:
if len(first) > len(second) and first[:len(second)] == second:
self.result = "first"
elif len(first) < len(second) and first == second[:len(first)]:
self.result = "second"
else:
x = 0
while first[x] == second[x]:
x += 1
if x == len(first):
break
if x == len(first):
if self.points[-1] > 0:
self.result = "first"
else:
self.result = "second"
else:
if first[x] > second[x]:
self.result = "first"
else:
self.result = "second"
def get_result(self):
return self.result
if __name__ == "__main__":
Solution = CodeforcesTask493BSolution()
Solution.read_input()
Solution.process_task()
print(Solution.get_result())
``` | instruction | 0 | 1,999 | 17 | 3,998 |
Yes | output | 1 | 1,999 | 17 | 3,999 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Vasya has become interested in wrestling. In wrestling wrestlers use techniques for which they are awarded points by judges. The wrestler who gets the most points wins.
When the numbers of points of both wrestlers are equal, the wrestler whose sequence of points is lexicographically greater, wins.
If the sequences of the awarded points coincide, the wrestler who performed the last technique wins. Your task is to determine which wrestler won.
Input
The first line contains number n — the number of techniques that the wrestlers have used (1 ≤ n ≤ 2·105).
The following n lines contain integer numbers ai (|ai| ≤ 109, ai ≠ 0). If ai is positive, that means that the first wrestler performed the technique that was awarded with ai points. And if ai is negative, that means that the second wrestler performed the technique that was awarded with ( - ai) points.
The techniques are given in chronological order.
Output
If the first wrestler wins, print string "first", otherwise print "second"
Examples
Input
5
1
2
-3
-4
3
Output
second
Input
3
-1
-2
3
Output
first
Input
2
4
-4
Output
second
Note
Sequence x = x1x2... x|x| is lexicographically larger than sequence y = y1y2... y|y|, if either |x| > |y| and x1 = y1, x2 = y2, ... , x|y| = y|y|, or there is such number r (r < |x|, r < |y|), that x1 = y1, x2 = y2, ... , xr = yr and xr + 1 > yr + 1.
We use notation |a| to denote length of sequence a.
Submitted Solution:
```
n = int(input())
player1 = []
player2 = []
arr = []
while n:
n -= 1
a = int(input())
arr.append(a)
if a > 0:
player1.append(a)
else:
player2.append(-a)
if sum(player1) != (sum(player2)):
if sum(player1) > (sum(player2)):
print('first')
else:
print('second')
elif player1 != player2:
if player2 > player1:
print('second')
else:
print('first')
else:
if arr[-1] > 0:
print('first')
else:
print('second')
``` | instruction | 0 | 2,000 | 17 | 4,000 |
Yes | output | 1 | 2,000 | 17 | 4,001 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Vasya has become interested in wrestling. In wrestling wrestlers use techniques for which they are awarded points by judges. The wrestler who gets the most points wins.
When the numbers of points of both wrestlers are equal, the wrestler whose sequence of points is lexicographically greater, wins.
If the sequences of the awarded points coincide, the wrestler who performed the last technique wins. Your task is to determine which wrestler won.
Input
The first line contains number n — the number of techniques that the wrestlers have used (1 ≤ n ≤ 2·105).
The following n lines contain integer numbers ai (|ai| ≤ 109, ai ≠ 0). If ai is positive, that means that the first wrestler performed the technique that was awarded with ai points. And if ai is negative, that means that the second wrestler performed the technique that was awarded with ( - ai) points.
The techniques are given in chronological order.
Output
If the first wrestler wins, print string "first", otherwise print "second"
Examples
Input
5
1
2
-3
-4
3
Output
second
Input
3
-1
-2
3
Output
first
Input
2
4
-4
Output
second
Note
Sequence x = x1x2... x|x| is lexicographically larger than sequence y = y1y2... y|y|, if either |x| > |y| and x1 = y1, x2 = y2, ... , x|y| = y|y|, or there is such number r (r < |x|, r < |y|), that x1 = y1, x2 = y2, ... , xr = yr and xr + 1 > yr + 1.
We use notation |a| to denote length of sequence a.
Submitted Solution:
```
from sys import stdin
n=int(stdin.readline())
first,second,sum1,sum2,last='','',0,0,0
for i in range(n):
x=int(stdin.readline())
if x>0:first+=str(x);sum1+=x
else:second+=str(-x);sum2+=(-x)
last=x
if sum1>sum2:print('first')
elif sum1<sum2:print('second')
else:
if first>second:print('first')
elif first<second:print('second')
else:
if last>0:print('first')
else:print('second')
``` | instruction | 0 | 2,001 | 17 | 4,002 |
No | output | 1 | 2,001 | 17 | 4,003 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Vasya has become interested in wrestling. In wrestling wrestlers use techniques for which they are awarded points by judges. The wrestler who gets the most points wins.
When the numbers of points of both wrestlers are equal, the wrestler whose sequence of points is lexicographically greater, wins.
If the sequences of the awarded points coincide, the wrestler who performed the last technique wins. Your task is to determine which wrestler won.
Input
The first line contains number n — the number of techniques that the wrestlers have used (1 ≤ n ≤ 2·105).
The following n lines contain integer numbers ai (|ai| ≤ 109, ai ≠ 0). If ai is positive, that means that the first wrestler performed the technique that was awarded with ai points. And if ai is negative, that means that the second wrestler performed the technique that was awarded with ( - ai) points.
The techniques are given in chronological order.
Output
If the first wrestler wins, print string "first", otherwise print "second"
Examples
Input
5
1
2
-3
-4
3
Output
second
Input
3
-1
-2
3
Output
first
Input
2
4
-4
Output
second
Note
Sequence x = x1x2... x|x| is lexicographically larger than sequence y = y1y2... y|y|, if either |x| > |y| and x1 = y1, x2 = y2, ... , x|y| = y|y|, or there is such number r (r < |x|, r < |y|), that x1 = y1, x2 = y2, ... , xr = yr and xr + 1 > yr + 1.
We use notation |a| to denote length of sequence a.
Submitted Solution:
```
def main(n,a):
x = [i for i in a if i >= 0]
y = [-i for i in a if i < 0]
print("X: ",x)
print("Y: ",y)
if sum(x)!=sum(y):
winner = sum(y) > sum(x)
else:
winner = [j>i for i,j in zip(x,y) if j!=i]
if winner != []:
winner = winner[0]
else:
winner = a[-1] < 0
print(["first","second"][winner])
def main_input():
n = int(input())
a = [int(input()) for i in range(n)]
main(n,a)
if __name__ == "__main__":
main_input()
#main(5,[1,2,-3,-4,3])
#main(3,[-1,-2,3])
#main(2,[4,-4])
#main(7,[1,2,-3,4,5,-6,7])
``` | instruction | 0 | 2,002 | 17 | 4,004 |
No | output | 1 | 2,002 | 17 | 4,005 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Vasya has become interested in wrestling. In wrestling wrestlers use techniques for which they are awarded points by judges. The wrestler who gets the most points wins.
When the numbers of points of both wrestlers are equal, the wrestler whose sequence of points is lexicographically greater, wins.
If the sequences of the awarded points coincide, the wrestler who performed the last technique wins. Your task is to determine which wrestler won.
Input
The first line contains number n — the number of techniques that the wrestlers have used (1 ≤ n ≤ 2·105).
The following n lines contain integer numbers ai (|ai| ≤ 109, ai ≠ 0). If ai is positive, that means that the first wrestler performed the technique that was awarded with ai points. And if ai is negative, that means that the second wrestler performed the technique that was awarded with ( - ai) points.
The techniques are given in chronological order.
Output
If the first wrestler wins, print string "first", otherwise print "second"
Examples
Input
5
1
2
-3
-4
3
Output
second
Input
3
-1
-2
3
Output
first
Input
2
4
-4
Output
second
Note
Sequence x = x1x2... x|x| is lexicographically larger than sequence y = y1y2... y|y|, if either |x| > |y| and x1 = y1, x2 = y2, ... , x|y| = y|y|, or there is such number r (r < |x|, r < |y|), that x1 = y1, x2 = y2, ... , xr = yr and xr + 1 > yr + 1.
We use notation |a| to denote length of sequence a.
Submitted Solution:
```
"""
Template written to be used by Python Programmers.
Use at your own risk!!!!
Owned by adi0311(rating - 5 star at CodeChef and Specialist at Codeforces).
"""
import sys
import bisect
import heapq
from math import *
from collections import defaultdict as dd # defaultdict(<datatype>) Free of KeyError.
from collections import deque # deque(list) append(), appendleft(), pop(), popleft() - O(1)
from collections import Counter as c # Counter(list) return a dict with {key: count}
from itertools import combinations as comb
from bisect import bisect_left as bl, bisect_right as br, bisect
# sys.setrecursionlimit(2*pow(10, 6))
# sys.stdin = open("input.txt", "r")
# sys.stdout = open("output.txt", "w")
mod = pow(10, 9) + 7
mod2 = 998244353
def data(): return sys.stdin.readline().strip()
def out(var): sys.stdout.write(var)
def l(): return list(map(int, data().split()))
def sl(): return list(map(str, data().split()))
def sp(): return map(int, data().split())
def ssp(): return map(str, data().split())
def l1d(n, val=0): return [val for i in range(n)]
def l2d(n, m, val=0): return [[val for i in range(n)] for j in range(m)]
n = int(data())
first, second = [], []
f, s = 0, 0
for i in range(n):
a = int(data())
if a < 0:
second.append(abs(a))
s += abs(a)
continue
first.append(a)
f += a
if s > f:
out("second")
exit()
if f > s:
out("first")
exit()
for i in range(0, min(len(first), len(second))):
if first[i] > second[i]:
out("first")
exit()
if second[i] > first[i]:
out("second")
exit()
if len(first) > len(second):
out("first")
exit()
out("second")
``` | instruction | 0 | 2,003 | 17 | 4,006 |
No | output | 1 | 2,003 | 17 | 4,007 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Vasya has become interested in wrestling. In wrestling wrestlers use techniques for which they are awarded points by judges. The wrestler who gets the most points wins.
When the numbers of points of both wrestlers are equal, the wrestler whose sequence of points is lexicographically greater, wins.
If the sequences of the awarded points coincide, the wrestler who performed the last technique wins. Your task is to determine which wrestler won.
Input
The first line contains number n — the number of techniques that the wrestlers have used (1 ≤ n ≤ 2·105).
The following n lines contain integer numbers ai (|ai| ≤ 109, ai ≠ 0). If ai is positive, that means that the first wrestler performed the technique that was awarded with ai points. And if ai is negative, that means that the second wrestler performed the technique that was awarded with ( - ai) points.
The techniques are given in chronological order.
Output
If the first wrestler wins, print string "first", otherwise print "second"
Examples
Input
5
1
2
-3
-4
3
Output
second
Input
3
-1
-2
3
Output
first
Input
2
4
-4
Output
second
Note
Sequence x = x1x2... x|x| is lexicographically larger than sequence y = y1y2... y|y|, if either |x| > |y| and x1 = y1, x2 = y2, ... , x|y| = y|y|, or there is such number r (r < |x|, r < |y|), that x1 = y1, x2 = y2, ... , xr = yr and xr + 1 > yr + 1.
We use notation |a| to denote length of sequence a.
Submitted Solution:
```
import sys
a, b = [], []
for _ in range(int(sys.stdin.readline())):
x = int(sys.stdin.readline())
if x > 0:
a.append(x)
else:
b.append(-x)
print("first" if a > b else "second")
``` | instruction | 0 | 2,004 | 17 | 4,008 |
No | output | 1 | 2,004 | 17 | 4,009 |
Provide tags and a correct Python 3 solution for this coding contest problem.
There are n consecutive seat places in a railway carriage. Each place is either empty or occupied by a passenger.
The university team for the Olympiad consists of a student-programmers and b student-athletes. Determine the largest number of students from all a+b students, which you can put in the railway carriage so that:
* no student-programmer is sitting next to the student-programmer;
* and no student-athlete is sitting next to the student-athlete.
In the other words, there should not be two consecutive (adjacent) places where two student-athletes or two student-programmers are sitting.
Consider that initially occupied seat places are occupied by jury members (who obviously are not students at all).
Input
The first line contain three integers n, a and b (1 ≤ n ≤ 2⋅10^{5}, 0 ≤ a, b ≤ 2⋅10^{5}, a + b > 0) — total number of seat places in the railway carriage, the number of student-programmers and the number of student-athletes.
The second line contains a string with length n, consisting of characters "." and "*". The dot means that the corresponding place is empty. The asterisk means that the corresponding place is occupied by the jury member.
Output
Print the largest number of students, which you can put in the railway carriage so that no student-programmer is sitting next to a student-programmer and no student-athlete is sitting next to a student-athlete.
Examples
Input
5 1 1
*...*
Output
2
Input
6 2 3
*...*.
Output
4
Input
11 3 10
.*....**.*.
Output
7
Input
3 2 3
***
Output
0
Note
In the first example you can put all student, for example, in the following way: *.AB*
In the second example you can put four students, for example, in the following way: *BAB*B
In the third example you can put seven students, for example, in the following way: B*ABAB**A*B
The letter A means a student-programmer, and the letter B — student-athlete. | instruction | 0 | 2,203 | 17 | 4,406 |
Tags: constructive algorithms, greedy, implementation
Correct Solution:
```
def solve(carriage, n, a, b):
odd=0;
total=0
subsum=0;
for ch in carriage:
if ch=='.':
subsum +=1
elif ch=='*':
total += subsum
if subsum%2==1:
odd += 1
subsum=0
total += subsum
if subsum%2==1:
odd += 1
even=(total-odd)//2
ans=0
if a<= even:
ans += a
a=0
else:
ans+=even
a -= even
if b<= even:
ans += b
b=0
else:
ans+=even
b -= even
#print(even)
#print(odd)
#print(a+b)
return ans+min(a+b, odd)
def test():
print(solve("*...*", 5, 1, 1)==2)
print(solve("*...*.", 6, 2, 3)==4)
print(solve(".*....**.*.", 11, 3, 10)==7)
print(solve("***", 3, 2, 3)==0)
def nia():
s=input()
while len(s)==0:
s=input()
s=s.split()
iVal=[];
for i in range (len(s)):
iVal.append(int(s[i]))
return iVal
n=nia()
arr=input()
print(solve(arr, n[0], n[1], n[2]))
``` | output | 1 | 2,203 | 17 | 4,407 |
Provide tags and a correct Python 3 solution for this coding contest problem.
There are n consecutive seat places in a railway carriage. Each place is either empty or occupied by a passenger.
The university team for the Olympiad consists of a student-programmers and b student-athletes. Determine the largest number of students from all a+b students, which you can put in the railway carriage so that:
* no student-programmer is sitting next to the student-programmer;
* and no student-athlete is sitting next to the student-athlete.
In the other words, there should not be two consecutive (adjacent) places where two student-athletes or two student-programmers are sitting.
Consider that initially occupied seat places are occupied by jury members (who obviously are not students at all).
Input
The first line contain three integers n, a and b (1 ≤ n ≤ 2⋅10^{5}, 0 ≤ a, b ≤ 2⋅10^{5}, a + b > 0) — total number of seat places in the railway carriage, the number of student-programmers and the number of student-athletes.
The second line contains a string with length n, consisting of characters "." and "*". The dot means that the corresponding place is empty. The asterisk means that the corresponding place is occupied by the jury member.
Output
Print the largest number of students, which you can put in the railway carriage so that no student-programmer is sitting next to a student-programmer and no student-athlete is sitting next to a student-athlete.
Examples
Input
5 1 1
*...*
Output
2
Input
6 2 3
*...*.
Output
4
Input
11 3 10
.*....**.*.
Output
7
Input
3 2 3
***
Output
0
Note
In the first example you can put all student, for example, in the following way: *.AB*
In the second example you can put four students, for example, in the following way: *BAB*B
In the third example you can put seven students, for example, in the following way: B*ABAB**A*B
The letter A means a student-programmer, and the letter B — student-athlete. | instruction | 0 | 2,204 | 17 | 4,408 |
Tags: constructive algorithms, greedy, implementation
Correct Solution:
```
n, a, b = [int(x) for x in input().split()]
a, b = sorted([a, b])
total = a + b
from itertools import groupby
s = input().strip()
for val, g in groupby(s):
if val == '*': continue
length = len(list(g))
b -= (length + 1) // 2
a -= length // 2
if a > b:
a, b = b, a
a = max(a, 0)
b = max(b, 0)
print(total - a - b)
``` | output | 1 | 2,204 | 17 | 4,409 |
Provide tags and a correct Python 3 solution for this coding contest problem.
There are n consecutive seat places in a railway carriage. Each place is either empty or occupied by a passenger.
The university team for the Olympiad consists of a student-programmers and b student-athletes. Determine the largest number of students from all a+b students, which you can put in the railway carriage so that:
* no student-programmer is sitting next to the student-programmer;
* and no student-athlete is sitting next to the student-athlete.
In the other words, there should not be two consecutive (adjacent) places where two student-athletes or two student-programmers are sitting.
Consider that initially occupied seat places are occupied by jury members (who obviously are not students at all).
Input
The first line contain three integers n, a and b (1 ≤ n ≤ 2⋅10^{5}, 0 ≤ a, b ≤ 2⋅10^{5}, a + b > 0) — total number of seat places in the railway carriage, the number of student-programmers and the number of student-athletes.
The second line contains a string with length n, consisting of characters "." and "*". The dot means that the corresponding place is empty. The asterisk means that the corresponding place is occupied by the jury member.
Output
Print the largest number of students, which you can put in the railway carriage so that no student-programmer is sitting next to a student-programmer and no student-athlete is sitting next to a student-athlete.
Examples
Input
5 1 1
*...*
Output
2
Input
6 2 3
*...*.
Output
4
Input
11 3 10
.*....**.*.
Output
7
Input
3 2 3
***
Output
0
Note
In the first example you can put all student, for example, in the following way: *.AB*
In the second example you can put four students, for example, in the following way: *BAB*B
In the third example you can put seven students, for example, in the following way: B*ABAB**A*B
The letter A means a student-programmer, and the letter B — student-athlete. | instruction | 0 | 2,205 | 17 | 4,410 |
Tags: constructive algorithms, greedy, implementation
Correct Solution:
```
n, v, t = input().split()
n = int(n)
a = int(v)
b = int(t)
s = [len(k) for k in input().split('*')]
a,b = max(a,b),min(a,b)
for i in range(len(s)):
a -= (s[i]+1)//2
b -= s[i]//2
if a < 0:
a, b = 0,0
break
elif b < 0:
b = 0
a, b = max(a,b), min(a,b)
print(int(v) + int(t) - a - b)
``` | output | 1 | 2,205 | 17 | 4,411 |
Provide tags and a correct Python 3 solution for this coding contest problem.
There are n consecutive seat places in a railway carriage. Each place is either empty or occupied by a passenger.
The university team for the Olympiad consists of a student-programmers and b student-athletes. Determine the largest number of students from all a+b students, which you can put in the railway carriage so that:
* no student-programmer is sitting next to the student-programmer;
* and no student-athlete is sitting next to the student-athlete.
In the other words, there should not be two consecutive (adjacent) places where two student-athletes or two student-programmers are sitting.
Consider that initially occupied seat places are occupied by jury members (who obviously are not students at all).
Input
The first line contain three integers n, a and b (1 ≤ n ≤ 2⋅10^{5}, 0 ≤ a, b ≤ 2⋅10^{5}, a + b > 0) — total number of seat places in the railway carriage, the number of student-programmers and the number of student-athletes.
The second line contains a string with length n, consisting of characters "." and "*". The dot means that the corresponding place is empty. The asterisk means that the corresponding place is occupied by the jury member.
Output
Print the largest number of students, which you can put in the railway carriage so that no student-programmer is sitting next to a student-programmer and no student-athlete is sitting next to a student-athlete.
Examples
Input
5 1 1
*...*
Output
2
Input
6 2 3
*...*.
Output
4
Input
11 3 10
.*....**.*.
Output
7
Input
3 2 3
***
Output
0
Note
In the first example you can put all student, for example, in the following way: *.AB*
In the second example you can put four students, for example, in the following way: *BAB*B
In the third example you can put seven students, for example, in the following way: B*ABAB**A*B
The letter A means a student-programmer, and the letter B — student-athlete. | instruction | 0 | 2,206 | 17 | 4,412 |
Tags: constructive algorithms, greedy, implementation
Correct Solution:
```
_, m, n = map(int, input().split())
t = m+n
a = [len(s) for s in input().split('*')]
for x in a:
m -= x // 2
n -= x // 2
if m > n:
m -= x % 2
else:
n -= x % 2
m = max(m, 0)
n = max(n, 0)
print(t - (m+n))
``` | output | 1 | 2,206 | 17 | 4,413 |
Provide tags and a correct Python 3 solution for this coding contest problem.
There are n consecutive seat places in a railway carriage. Each place is either empty or occupied by a passenger.
The university team for the Olympiad consists of a student-programmers and b student-athletes. Determine the largest number of students from all a+b students, which you can put in the railway carriage so that:
* no student-programmer is sitting next to the student-programmer;
* and no student-athlete is sitting next to the student-athlete.
In the other words, there should not be two consecutive (adjacent) places where two student-athletes or two student-programmers are sitting.
Consider that initially occupied seat places are occupied by jury members (who obviously are not students at all).
Input
The first line contain three integers n, a and b (1 ≤ n ≤ 2⋅10^{5}, 0 ≤ a, b ≤ 2⋅10^{5}, a + b > 0) — total number of seat places in the railway carriage, the number of student-programmers and the number of student-athletes.
The second line contains a string with length n, consisting of characters "." and "*". The dot means that the corresponding place is empty. The asterisk means that the corresponding place is occupied by the jury member.
Output
Print the largest number of students, which you can put in the railway carriage so that no student-programmer is sitting next to a student-programmer and no student-athlete is sitting next to a student-athlete.
Examples
Input
5 1 1
*...*
Output
2
Input
6 2 3
*...*.
Output
4
Input
11 3 10
.*....**.*.
Output
7
Input
3 2 3
***
Output
0
Note
In the first example you can put all student, for example, in the following way: *.AB*
In the second example you can put four students, for example, in the following way: *BAB*B
In the third example you can put seven students, for example, in the following way: B*ABAB**A*B
The letter A means a student-programmer, and the letter B — student-athlete. | instruction | 0 | 2,207 | 17 | 4,414 |
Tags: constructive algorithms, greedy, implementation
Correct Solution:
```
n,a,b=map(int,input().split())
s=list(input())
ans=0
for i in range(n):
if s[i]=='.':
if i==0 or (s[i-1]!='a' and s[i-1]!='b'):
if a>b and a>0:
s[i]='a'
a-=1
ans+=1
elif b>0:
s[i]='b'
b-=1
ans+=1
elif s[i-1]=='a':
if b>0:
s[i]='b'
b-=1
ans+=1
elif s[i-1]=='b':
if a>0:
s[i]='a'
a-=1
ans+=1
print(ans)
``` | output | 1 | 2,207 | 17 | 4,415 |
Provide tags and a correct Python 3 solution for this coding contest problem.
There are n consecutive seat places in a railway carriage. Each place is either empty or occupied by a passenger.
The university team for the Olympiad consists of a student-programmers and b student-athletes. Determine the largest number of students from all a+b students, which you can put in the railway carriage so that:
* no student-programmer is sitting next to the student-programmer;
* and no student-athlete is sitting next to the student-athlete.
In the other words, there should not be two consecutive (adjacent) places where two student-athletes or two student-programmers are sitting.
Consider that initially occupied seat places are occupied by jury members (who obviously are not students at all).
Input
The first line contain three integers n, a and b (1 ≤ n ≤ 2⋅10^{5}, 0 ≤ a, b ≤ 2⋅10^{5}, a + b > 0) — total number of seat places in the railway carriage, the number of student-programmers and the number of student-athletes.
The second line contains a string with length n, consisting of characters "." and "*". The dot means that the corresponding place is empty. The asterisk means that the corresponding place is occupied by the jury member.
Output
Print the largest number of students, which you can put in the railway carriage so that no student-programmer is sitting next to a student-programmer and no student-athlete is sitting next to a student-athlete.
Examples
Input
5 1 1
*...*
Output
2
Input
6 2 3
*...*.
Output
4
Input
11 3 10
.*....**.*.
Output
7
Input
3 2 3
***
Output
0
Note
In the first example you can put all student, for example, in the following way: *.AB*
In the second example you can put four students, for example, in the following way: *BAB*B
In the third example you can put seven students, for example, in the following way: B*ABAB**A*B
The letter A means a student-programmer, and the letter B — student-athlete. | instruction | 0 | 2,208 | 17 | 4,416 |
Tags: constructive algorithms, greedy, implementation
Correct Solution:
```
n, a, b = map(int, input().split())
s, ans, flag, cur = input(), 0, 1, 0
for i, j in enumerate(s):
if j == '.':
if flag:
cur = 0 if a >= b else 1
flag = 0
if cur and b:
ans += 1
b -= 1
elif cur == 0 and a:
ans += 1
a -= 1
cur ^= 1
else:
flag = 1
print(ans)
``` | output | 1 | 2,208 | 17 | 4,417 |
Provide tags and a correct Python 3 solution for this coding contest problem.
There are n consecutive seat places in a railway carriage. Each place is either empty or occupied by a passenger.
The university team for the Olympiad consists of a student-programmers and b student-athletes. Determine the largest number of students from all a+b students, which you can put in the railway carriage so that:
* no student-programmer is sitting next to the student-programmer;
* and no student-athlete is sitting next to the student-athlete.
In the other words, there should not be two consecutive (adjacent) places where two student-athletes or two student-programmers are sitting.
Consider that initially occupied seat places are occupied by jury members (who obviously are not students at all).
Input
The first line contain three integers n, a and b (1 ≤ n ≤ 2⋅10^{5}, 0 ≤ a, b ≤ 2⋅10^{5}, a + b > 0) — total number of seat places in the railway carriage, the number of student-programmers and the number of student-athletes.
The second line contains a string with length n, consisting of characters "." and "*". The dot means that the corresponding place is empty. The asterisk means that the corresponding place is occupied by the jury member.
Output
Print the largest number of students, which you can put in the railway carriage so that no student-programmer is sitting next to a student-programmer and no student-athlete is sitting next to a student-athlete.
Examples
Input
5 1 1
*...*
Output
2
Input
6 2 3
*...*.
Output
4
Input
11 3 10
.*....**.*.
Output
7
Input
3 2 3
***
Output
0
Note
In the first example you can put all student, for example, in the following way: *.AB*
In the second example you can put four students, for example, in the following way: *BAB*B
In the third example you can put seven students, for example, in the following way: B*ABAB**A*B
The letter A means a student-programmer, and the letter B — student-athlete. | instruction | 0 | 2,209 | 17 | 4,418 |
Tags: constructive algorithms, greedy, implementation
Correct Solution:
```
n, a, b =map(int,input().split(" "))
s=input().split("*")
count=0;
for i in range(len(s)):
if(len(s[i])!=0):
if(a==0 and b==0):
break;
t=len(s[i])
count+=t;
if(t%2==0):
a-=int(t / 2);
b-=int(t / 2);
if(a<0):
count+=a;
a=0;
if (b < 0):
count += b;
b=0;
else:
if(a>=b):
a -= int(t / 2)+1;
b -= int(t / 2);
else:
a -= int(t / 2);
b -= int(t / 2)+1;
if (a < 0):
count += a;
a = 0;
if (b < 0):
count += b;
b = 0;
print(count)
``` | output | 1 | 2,209 | 17 | 4,419 |
Provide tags and a correct Python 3 solution for this coding contest problem.
There are n consecutive seat places in a railway carriage. Each place is either empty or occupied by a passenger.
The university team for the Olympiad consists of a student-programmers and b student-athletes. Determine the largest number of students from all a+b students, which you can put in the railway carriage so that:
* no student-programmer is sitting next to the student-programmer;
* and no student-athlete is sitting next to the student-athlete.
In the other words, there should not be two consecutive (adjacent) places where two student-athletes or two student-programmers are sitting.
Consider that initially occupied seat places are occupied by jury members (who obviously are not students at all).
Input
The first line contain three integers n, a and b (1 ≤ n ≤ 2⋅10^{5}, 0 ≤ a, b ≤ 2⋅10^{5}, a + b > 0) — total number of seat places in the railway carriage, the number of student-programmers and the number of student-athletes.
The second line contains a string with length n, consisting of characters "." and "*". The dot means that the corresponding place is empty. The asterisk means that the corresponding place is occupied by the jury member.
Output
Print the largest number of students, which you can put in the railway carriage so that no student-programmer is sitting next to a student-programmer and no student-athlete is sitting next to a student-athlete.
Examples
Input
5 1 1
*...*
Output
2
Input
6 2 3
*...*.
Output
4
Input
11 3 10
.*....**.*.
Output
7
Input
3 2 3
***
Output
0
Note
In the first example you can put all student, for example, in the following way: *.AB*
In the second example you can put four students, for example, in the following way: *BAB*B
In the third example you can put seven students, for example, in the following way: B*ABAB**A*B
The letter A means a student-programmer, and the letter B — student-athlete. | instruction | 0 | 2,210 | 17 | 4,420 |
Tags: constructive algorithms, greedy, implementation
Correct Solution:
```
n, a, b = map(int, input().split())
s = list(input())
count = 0
for i in range(n):
if s[i] == ".":
if i - 1 >= 0:
if s[i - 1] == "A" and b > 0:
count += 1
s[i] = "B"
b -= 1
elif s[i - 1] == "B" and a > 0:
count += 1
s[i] = "A"
a -= 1
elif s[i - 1] == "*" or s[i - 1] == ".":
if a >= b and a > 0:
count += 1
s[i] = "A"
a -= 1
elif b > a and b > 0:
count += 1
s[i] = "B"
b -= 1
else:
if a >= b and a > 0:
count += 1
s[i] = "A"
a -= 1
elif b > a and b > 0:
count += 1
s[i] = "B"
b -= 1
print(count)
``` | output | 1 | 2,210 | 17 | 4,421 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There are n consecutive seat places in a railway carriage. Each place is either empty or occupied by a passenger.
The university team for the Olympiad consists of a student-programmers and b student-athletes. Determine the largest number of students from all a+b students, which you can put in the railway carriage so that:
* no student-programmer is sitting next to the student-programmer;
* and no student-athlete is sitting next to the student-athlete.
In the other words, there should not be two consecutive (adjacent) places where two student-athletes or two student-programmers are sitting.
Consider that initially occupied seat places are occupied by jury members (who obviously are not students at all).
Input
The first line contain three integers n, a and b (1 ≤ n ≤ 2⋅10^{5}, 0 ≤ a, b ≤ 2⋅10^{5}, a + b > 0) — total number of seat places in the railway carriage, the number of student-programmers and the number of student-athletes.
The second line contains a string with length n, consisting of characters "." and "*". The dot means that the corresponding place is empty. The asterisk means that the corresponding place is occupied by the jury member.
Output
Print the largest number of students, which you can put in the railway carriage so that no student-programmer is sitting next to a student-programmer and no student-athlete is sitting next to a student-athlete.
Examples
Input
5 1 1
*...*
Output
2
Input
6 2 3
*...*.
Output
4
Input
11 3 10
.*....**.*.
Output
7
Input
3 2 3
***
Output
0
Note
In the first example you can put all student, for example, in the following way: *.AB*
In the second example you can put four students, for example, in the following way: *BAB*B
In the third example you can put seven students, for example, in the following way: B*ABAB**A*B
The letter A means a student-programmer, and the letter B — student-athlete.
Submitted Solution:
```
n,a,b = [int(i) for i in input().split()]
ss = a + b
st = input()
for ez in st.split("*"):
if(a > b):
a,b = b,a
a -= min(a , len(ez)//2)
b -= min(b , len(ez) - len(ez)//2)
print(ss - a - b)
``` | instruction | 0 | 2,211 | 17 | 4,422 |
Yes | output | 1 | 2,211 | 17 | 4,423 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There are n consecutive seat places in a railway carriage. Each place is either empty or occupied by a passenger.
The university team for the Olympiad consists of a student-programmers and b student-athletes. Determine the largest number of students from all a+b students, which you can put in the railway carriage so that:
* no student-programmer is sitting next to the student-programmer;
* and no student-athlete is sitting next to the student-athlete.
In the other words, there should not be two consecutive (adjacent) places where two student-athletes or two student-programmers are sitting.
Consider that initially occupied seat places are occupied by jury members (who obviously are not students at all).
Input
The first line contain three integers n, a and b (1 ≤ n ≤ 2⋅10^{5}, 0 ≤ a, b ≤ 2⋅10^{5}, a + b > 0) — total number of seat places in the railway carriage, the number of student-programmers and the number of student-athletes.
The second line contains a string with length n, consisting of characters "." and "*". The dot means that the corresponding place is empty. The asterisk means that the corresponding place is occupied by the jury member.
Output
Print the largest number of students, which you can put in the railway carriage so that no student-programmer is sitting next to a student-programmer and no student-athlete is sitting next to a student-athlete.
Examples
Input
5 1 1
*...*
Output
2
Input
6 2 3
*...*.
Output
4
Input
11 3 10
.*....**.*.
Output
7
Input
3 2 3
***
Output
0
Note
In the first example you can put all student, for example, in the following way: *.AB*
In the second example you can put four students, for example, in the following way: *BAB*B
In the third example you can put seven students, for example, in the following way: B*ABAB**A*B
The letter A means a student-programmer, and the letter B — student-athlete.
Submitted Solution:
```
def main():
n, a, b = map(int, input().split())
ab = a + b
for v in map(len, filter(None, input().split('*'))):
if v & 1:
v //= 2
if a > b:
a -= v + 1
b -= v
else:
b -= v + 1
a -= v
else:
v //= 2
a -= v
b -= v
print(ab - max(a, 0) - max(b, 0))
if __name__ == '__main__':
main()
``` | instruction | 0 | 2,212 | 17 | 4,424 |
Yes | output | 1 | 2,212 | 17 | 4,425 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There are n consecutive seat places in a railway carriage. Each place is either empty or occupied by a passenger.
The university team for the Olympiad consists of a student-programmers and b student-athletes. Determine the largest number of students from all a+b students, which you can put in the railway carriage so that:
* no student-programmer is sitting next to the student-programmer;
* and no student-athlete is sitting next to the student-athlete.
In the other words, there should not be two consecutive (adjacent) places where two student-athletes or two student-programmers are sitting.
Consider that initially occupied seat places are occupied by jury members (who obviously are not students at all).
Input
The first line contain three integers n, a and b (1 ≤ n ≤ 2⋅10^{5}, 0 ≤ a, b ≤ 2⋅10^{5}, a + b > 0) — total number of seat places in the railway carriage, the number of student-programmers and the number of student-athletes.
The second line contains a string with length n, consisting of characters "." and "*". The dot means that the corresponding place is empty. The asterisk means that the corresponding place is occupied by the jury member.
Output
Print the largest number of students, which you can put in the railway carriage so that no student-programmer is sitting next to a student-programmer and no student-athlete is sitting next to a student-athlete.
Examples
Input
5 1 1
*...*
Output
2
Input
6 2 3
*...*.
Output
4
Input
11 3 10
.*....**.*.
Output
7
Input
3 2 3
***
Output
0
Note
In the first example you can put all student, for example, in the following way: *.AB*
In the second example you can put four students, for example, in the following way: *BAB*B
In the third example you can put seven students, for example, in the following way: B*ABAB**A*B
The letter A means a student-programmer, and the letter B — student-athlete.
Submitted Solution:
```
n, a, b = map(int, input().split())
number = 0
string = input()
s = [t for t in string]
for i in range(0, n):
if s[i] == '.':
if i != 0:
if s[i - 1] == '*':
if a > b:
if a != 0:
number += 1
a -= 1
s[i] = 'a'
else:
if b != 0:
b -= 1
number += 1
s[i] = 'b'
else:
if b != 0:
number += 1
b -= 1
s[i] = 'b'
else:
if a != 0:
number += 1
a -= 1
s[i] = 'a'
else:
if s[i - 1] == 'a':
if b != 0:
b -= 1
number += 1
s[i] = 'b'
if s[i - 1] == 'b':
if a != 0:
a -= 1
number += 1
s[i] = 'a'
if s[i - 1] == '.':
if a > b:
if a != 0:
number += 1
a -= 1
s[i] = 'a'
else:
if b != 0:
number += 1
b -= 1
s[i] = 'b'
else:
if b != 0:
number += 1
b -= 1
s[i] = 'b'
else:
if a != 0:
number += 1
a -= 1
s[i] = 'a'
else:
if a > b:
if a != 0:
number += 1
a -= 1
s[i] = 'a'
else:
if b != 0:
number += 1
b -= 1
s[i] = 'b'
else:
if b != 0:
number += 1
b -= 1
s[i] = 'b'
else:
if a != 0:
number += 1
a -= 1
s[i] = 'a'
if a == 0 and b == 0:
break
print(number)
``` | instruction | 0 | 2,213 | 17 | 4,426 |
Yes | output | 1 | 2,213 | 17 | 4,427 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There are n consecutive seat places in a railway carriage. Each place is either empty or occupied by a passenger.
The university team for the Olympiad consists of a student-programmers and b student-athletes. Determine the largest number of students from all a+b students, which you can put in the railway carriage so that:
* no student-programmer is sitting next to the student-programmer;
* and no student-athlete is sitting next to the student-athlete.
In the other words, there should not be two consecutive (adjacent) places where two student-athletes or two student-programmers are sitting.
Consider that initially occupied seat places are occupied by jury members (who obviously are not students at all).
Input
The first line contain three integers n, a and b (1 ≤ n ≤ 2⋅10^{5}, 0 ≤ a, b ≤ 2⋅10^{5}, a + b > 0) — total number of seat places in the railway carriage, the number of student-programmers and the number of student-athletes.
The second line contains a string with length n, consisting of characters "." and "*". The dot means that the corresponding place is empty. The asterisk means that the corresponding place is occupied by the jury member.
Output
Print the largest number of students, which you can put in the railway carriage so that no student-programmer is sitting next to a student-programmer and no student-athlete is sitting next to a student-athlete.
Examples
Input
5 1 1
*...*
Output
2
Input
6 2 3
*...*.
Output
4
Input
11 3 10
.*....**.*.
Output
7
Input
3 2 3
***
Output
0
Note
In the first example you can put all student, for example, in the following way: *.AB*
In the second example you can put four students, for example, in the following way: *BAB*B
In the third example you can put seven students, for example, in the following way: B*ABAB**A*B
The letter A means a student-programmer, and the letter B — student-athlete.
Submitted Solution:
```
n, a, b = map(int, input().split())
t = a + b
s = input()
curr = 0
arr = []
for c in s:
if c == ".":
curr += 1
else:
arr.append(curr)
curr = 0
if curr > 0:
arr.append(curr)
arr.sort()
arr.reverse()
if a < b:
a, b = b, a
for e in arr:
if e % 2 == 1:
b -= e//2
a -= e//2+1
if a < b:
a, b = b, a
else:
b -= e//2
a -= e//2
if b > 0:
t -= b
if a > 0:
t -= a
print(t)
``` | instruction | 0 | 2,214 | 17 | 4,428 |
Yes | output | 1 | 2,214 | 17 | 4,429 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There are n consecutive seat places in a railway carriage. Each place is either empty or occupied by a passenger.
The university team for the Olympiad consists of a student-programmers and b student-athletes. Determine the largest number of students from all a+b students, which you can put in the railway carriage so that:
* no student-programmer is sitting next to the student-programmer;
* and no student-athlete is sitting next to the student-athlete.
In the other words, there should not be two consecutive (adjacent) places where two student-athletes or two student-programmers are sitting.
Consider that initially occupied seat places are occupied by jury members (who obviously are not students at all).
Input
The first line contain three integers n, a and b (1 ≤ n ≤ 2⋅10^{5}, 0 ≤ a, b ≤ 2⋅10^{5}, a + b > 0) — total number of seat places in the railway carriage, the number of student-programmers and the number of student-athletes.
The second line contains a string with length n, consisting of characters "." and "*". The dot means that the corresponding place is empty. The asterisk means that the corresponding place is occupied by the jury member.
Output
Print the largest number of students, which you can put in the railway carriage so that no student-programmer is sitting next to a student-programmer and no student-athlete is sitting next to a student-athlete.
Examples
Input
5 1 1
*...*
Output
2
Input
6 2 3
*...*.
Output
4
Input
11 3 10
.*....**.*.
Output
7
Input
3 2 3
***
Output
0
Note
In the first example you can put all student, for example, in the following way: *.AB*
In the second example you can put four students, for example, in the following way: *BAB*B
In the third example you can put seven students, for example, in the following way: B*ABAB**A*B
The letter A means a student-programmer, and the letter B — student-athlete.
Submitted Solution:
```
z = [int(s) for s in input().split()]
n = z[0]
a = z[1]
b = z[2]
st = 0
# print(n, a, b)
x = input()
f = []
for i in range(len(x)):
f.append(x[i])
for i in range(len(f)):
if i == 0 and f[i] == ".":
if a > b:
f[i] = "a"
a -= 1
if a < b:
f[i] = "b"
b -= 1
if a == b:
f[i] = "a"
a -= 1
elif f[i] == ".":
if f[i-1] == "a":
if b > 0:
f[i] = "b"
b -= 1
if f[i-1] == "b":
if a > 0:
f[i] = "a"
a -= 1
if f[i-1] == "*":
if a > b and a > 0:
f[i] = "a"
a -= 1
if a < b and b > 0:
f[i] = "b"
b -= 1
if a == b and a > 0:
f[i] = "a"
a -= 1
for i in range(len(f)):
if f[i] == "a" or f[i] == "b":
st += 1
print(st)
``` | instruction | 0 | 2,215 | 17 | 4,430 |
No | output | 1 | 2,215 | 17 | 4,431 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There are n consecutive seat places in a railway carriage. Each place is either empty or occupied by a passenger.
The university team for the Olympiad consists of a student-programmers and b student-athletes. Determine the largest number of students from all a+b students, which you can put in the railway carriage so that:
* no student-programmer is sitting next to the student-programmer;
* and no student-athlete is sitting next to the student-athlete.
In the other words, there should not be two consecutive (adjacent) places where two student-athletes or two student-programmers are sitting.
Consider that initially occupied seat places are occupied by jury members (who obviously are not students at all).
Input
The first line contain three integers n, a and b (1 ≤ n ≤ 2⋅10^{5}, 0 ≤ a, b ≤ 2⋅10^{5}, a + b > 0) — total number of seat places in the railway carriage, the number of student-programmers and the number of student-athletes.
The second line contains a string with length n, consisting of characters "." and "*". The dot means that the corresponding place is empty. The asterisk means that the corresponding place is occupied by the jury member.
Output
Print the largest number of students, which you can put in the railway carriage so that no student-programmer is sitting next to a student-programmer and no student-athlete is sitting next to a student-athlete.
Examples
Input
5 1 1
*...*
Output
2
Input
6 2 3
*...*.
Output
4
Input
11 3 10
.*....**.*.
Output
7
Input
3 2 3
***
Output
0
Note
In the first example you can put all student, for example, in the following way: *.AB*
In the second example you can put four students, for example, in the following way: *BAB*B
In the third example you can put seven students, for example, in the following way: B*ABAB**A*B
The letter A means a student-programmer, and the letter B — student-athlete.
Submitted Solution:
```
n,a,b=map(int,input().split())
s=input()
s=list(s)
c=0
a=max(a,b)
b=min(a,b)
#1 #0
for i in range(0,n):
if(s[i]!='*'):
if(i!=0):
if(s[i-1]!=1):
if(a>0):
s[i]=1
a=a-1
c=c+1
else:
if(s[i-1]=='*' and b>0):
s[i]=0
c=c+1
b=b-1
else:
if(b>0):
s[i]=0
b=b-1
c=c+1
else:
if(s[i-1]=='*' and a>0):
s[i]=1
c=c+1
a=a-1
else:
s[i]=1
a=a-1
c=c+1
a=max(a,b)
b=min(a,b)
else:
continue
print(c)
``` | instruction | 0 | 2,216 | 17 | 4,432 |
No | output | 1 | 2,216 | 17 | 4,433 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There are n consecutive seat places in a railway carriage. Each place is either empty or occupied by a passenger.
The university team for the Olympiad consists of a student-programmers and b student-athletes. Determine the largest number of students from all a+b students, which you can put in the railway carriage so that:
* no student-programmer is sitting next to the student-programmer;
* and no student-athlete is sitting next to the student-athlete.
In the other words, there should not be two consecutive (adjacent) places where two student-athletes or two student-programmers are sitting.
Consider that initially occupied seat places are occupied by jury members (who obviously are not students at all).
Input
The first line contain three integers n, a and b (1 ≤ n ≤ 2⋅10^{5}, 0 ≤ a, b ≤ 2⋅10^{5}, a + b > 0) — total number of seat places in the railway carriage, the number of student-programmers and the number of student-athletes.
The second line contains a string with length n, consisting of characters "." and "*". The dot means that the corresponding place is empty. The asterisk means that the corresponding place is occupied by the jury member.
Output
Print the largest number of students, which you can put in the railway carriage so that no student-programmer is sitting next to a student-programmer and no student-athlete is sitting next to a student-athlete.
Examples
Input
5 1 1
*...*
Output
2
Input
6 2 3
*...*.
Output
4
Input
11 3 10
.*....**.*.
Output
7
Input
3 2 3
***
Output
0
Note
In the first example you can put all student, for example, in the following way: *.AB*
In the second example you can put four students, for example, in the following way: *BAB*B
In the third example you can put seven students, for example, in the following way: B*ABAB**A*B
The letter A means a student-programmer, and the letter B — student-athlete.
Submitted Solution:
```
n,a,b = map(int,input().split())
s = input()
res,e = 0,''
if a==0 or b==0:
print(1)
from sys import exit
exit()
r=0
for i,x in enumerate(s):
if a==0 and b==0:
break
if x=='.':
if e=='':
if a>b:
if b==0:
r=0
a-=1
else:
a-=1
e='a'
else:
if a==0:
r=0
b-=1
else:
b-=1
e='b'
res+=1
elif e=='a':
if b==0:
if r==0:
r=1
else:
a-=1
r=0
res+=1
else:
b-=1
e='b'
res+=1
else:
if a==0:
if r==0:
r=1
else:
b-=1
r=0
res+=1
else:
a-=1
e='a'
res+=1
else:
e=''
print(res)
``` | instruction | 0 | 2,217 | 17 | 4,434 |
No | output | 1 | 2,217 | 17 | 4,435 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There are n consecutive seat places in a railway carriage. Each place is either empty or occupied by a passenger.
The university team for the Olympiad consists of a student-programmers and b student-athletes. Determine the largest number of students from all a+b students, which you can put in the railway carriage so that:
* no student-programmer is sitting next to the student-programmer;
* and no student-athlete is sitting next to the student-athlete.
In the other words, there should not be two consecutive (adjacent) places where two student-athletes or two student-programmers are sitting.
Consider that initially occupied seat places are occupied by jury members (who obviously are not students at all).
Input
The first line contain three integers n, a and b (1 ≤ n ≤ 2⋅10^{5}, 0 ≤ a, b ≤ 2⋅10^{5}, a + b > 0) — total number of seat places in the railway carriage, the number of student-programmers and the number of student-athletes.
The second line contains a string with length n, consisting of characters "." and "*". The dot means that the corresponding place is empty. The asterisk means that the corresponding place is occupied by the jury member.
Output
Print the largest number of students, which you can put in the railway carriage so that no student-programmer is sitting next to a student-programmer and no student-athlete is sitting next to a student-athlete.
Examples
Input
5 1 1
*...*
Output
2
Input
6 2 3
*...*.
Output
4
Input
11 3 10
.*....**.*.
Output
7
Input
3 2 3
***
Output
0
Note
In the first example you can put all student, for example, in the following way: *.AB*
In the second example you can put four students, for example, in the following way: *BAB*B
In the third example you can put seven students, for example, in the following way: B*ABAB**A*B
The letter A means a student-programmer, and the letter B — student-athlete.
Submitted Solution:
```
import math
n,a,b=[int(n) for n in input().split()]
s=input()
c=a
d=b
l=list(s)
if(l[0]=="."):
if(c>d):
l[0]="0"
c=c-1
else:
l[0]="1"
d=d-1
for i in range(1,n):
if(l[i]=="*"):
continue
else:
if(l[i-1]=="*"):
if(c>d):
l[i]="0"
c=c-1
else:
l[i]="1"
d=d-1
if(l[i-1]=="0"):
if(d>0):
l[i]="1"
d=d-1
if(l[i-1]=="1"):
if(c>0):
l[i]="0"
c=c-1
print((a+b)-(c+d))
``` | instruction | 0 | 2,218 | 17 | 4,436 |
No | output | 1 | 2,218 | 17 | 4,437 |
Provide a correct Python 3 solution for this coding contest problem.
There are 3N participants in AtCoder Group Contest. The strength of the i-th participant is represented by an integer a_i. They will form N teams, each consisting of three participants. No participant may belong to multiple teams.
The strength of a team is defined as the second largest strength among its members. For example, a team of participants of strength 1, 5, 2 has a strength 2, and a team of three participants of strength 3, 2, 3 has a strength 3.
Find the maximum possible sum of the strengths of N teams.
Constraints
* 1 ≤ N ≤ 10^5
* 1 ≤ a_i ≤ 10^{9}
* a_i are integers.
Input
Input is given from Standard Input in the following format:
N
a_1 a_2 ... a_{3N}
Output
Print the answer.
Examples
Input
2
5 2 8 5 1 5
Output
10
Input
10
1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000
Output
10000000000 | instruction | 0 | 2,335 | 17 | 4,670 |
"Correct Solution:
```
N=int(input())
a=sorted([int(i) for i in input().split()])
print(sum(a[N::2]))
``` | output | 1 | 2,335 | 17 | 4,671 |
Provide a correct Python 3 solution for this coding contest problem.
There are 3N participants in AtCoder Group Contest. The strength of the i-th participant is represented by an integer a_i. They will form N teams, each consisting of three participants. No participant may belong to multiple teams.
The strength of a team is defined as the second largest strength among its members. For example, a team of participants of strength 1, 5, 2 has a strength 2, and a team of three participants of strength 3, 2, 3 has a strength 3.
Find the maximum possible sum of the strengths of N teams.
Constraints
* 1 ≤ N ≤ 10^5
* 1 ≤ a_i ≤ 10^{9}
* a_i are integers.
Input
Input is given from Standard Input in the following format:
N
a_1 a_2 ... a_{3N}
Output
Print the answer.
Examples
Input
2
5 2 8 5 1 5
Output
10
Input
10
1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000
Output
10000000000 | instruction | 0 | 2,336 | 17 | 4,672 |
"Correct Solution:
```
n=int(input())
a=list(map(int,input().split()))
a.sort()
print(sum(a[n::2]))
``` | output | 1 | 2,336 | 17 | 4,673 |
Provide a correct Python 3 solution for this coding contest problem.
There are 3N participants in AtCoder Group Contest. The strength of the i-th participant is represented by an integer a_i. They will form N teams, each consisting of three participants. No participant may belong to multiple teams.
The strength of a team is defined as the second largest strength among its members. For example, a team of participants of strength 1, 5, 2 has a strength 2, and a team of three participants of strength 3, 2, 3 has a strength 3.
Find the maximum possible sum of the strengths of N teams.
Constraints
* 1 ≤ N ≤ 10^5
* 1 ≤ a_i ≤ 10^{9}
* a_i are integers.
Input
Input is given from Standard Input in the following format:
N
a_1 a_2 ... a_{3N}
Output
Print the answer.
Examples
Input
2
5 2 8 5 1 5
Output
10
Input
10
1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000
Output
10000000000 | instruction | 0 | 2,337 | 17 | 4,674 |
"Correct Solution:
```
n=int(input())
a=list(map(int,input().split()))
a=sorted(a,reverse=True)
print(sum(a[1:2*n:2]))
``` | output | 1 | 2,337 | 17 | 4,675 |
Provide a correct Python 3 solution for this coding contest problem.
There are 3N participants in AtCoder Group Contest. The strength of the i-th participant is represented by an integer a_i. They will form N teams, each consisting of three participants. No participant may belong to multiple teams.
The strength of a team is defined as the second largest strength among its members. For example, a team of participants of strength 1, 5, 2 has a strength 2, and a team of three participants of strength 3, 2, 3 has a strength 3.
Find the maximum possible sum of the strengths of N teams.
Constraints
* 1 ≤ N ≤ 10^5
* 1 ≤ a_i ≤ 10^{9}
* a_i are integers.
Input
Input is given from Standard Input in the following format:
N
a_1 a_2 ... a_{3N}
Output
Print the answer.
Examples
Input
2
5 2 8 5 1 5
Output
10
Input
10
1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000
Output
10000000000 | instruction | 0 | 2,338 | 17 | 4,676 |
"Correct Solution:
```
n = int(input())
a = sorted(list(map(int, input().split())))
print(sum(a[n::2]))
``` | output | 1 | 2,338 | 17 | 4,677 |
Provide a correct Python 3 solution for this coding contest problem.
There are 3N participants in AtCoder Group Contest. The strength of the i-th participant is represented by an integer a_i. They will form N teams, each consisting of three participants. No participant may belong to multiple teams.
The strength of a team is defined as the second largest strength among its members. For example, a team of participants of strength 1, 5, 2 has a strength 2, and a team of three participants of strength 3, 2, 3 has a strength 3.
Find the maximum possible sum of the strengths of N teams.
Constraints
* 1 ≤ N ≤ 10^5
* 1 ≤ a_i ≤ 10^{9}
* a_i are integers.
Input
Input is given from Standard Input in the following format:
N
a_1 a_2 ... a_{3N}
Output
Print the answer.
Examples
Input
2
5 2 8 5 1 5
Output
10
Input
10
1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000
Output
10000000000 | instruction | 0 | 2,339 | 17 | 4,678 |
"Correct Solution:
```
tms=int(input())
strlist=sorted(list(map(int,input().split())))
ans=sum(strlist[tms::][::2])
print(ans)
``` | output | 1 | 2,339 | 17 | 4,679 |
Provide a correct Python 3 solution for this coding contest problem.
There are 3N participants in AtCoder Group Contest. The strength of the i-th participant is represented by an integer a_i. They will form N teams, each consisting of three participants. No participant may belong to multiple teams.
The strength of a team is defined as the second largest strength among its members. For example, a team of participants of strength 1, 5, 2 has a strength 2, and a team of three participants of strength 3, 2, 3 has a strength 3.
Find the maximum possible sum of the strengths of N teams.
Constraints
* 1 ≤ N ≤ 10^5
* 1 ≤ a_i ≤ 10^{9}
* a_i are integers.
Input
Input is given from Standard Input in the following format:
N
a_1 a_2 ... a_{3N}
Output
Print the answer.
Examples
Input
2
5 2 8 5 1 5
Output
10
Input
10
1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000
Output
10000000000 | instruction | 0 | 2,340 | 17 | 4,680 |
"Correct Solution:
```
n = int(input())
l = sorted(list(map(int,input().split())))
print(sum(l[n::2]))
``` | output | 1 | 2,340 | 17 | 4,681 |
Provide a correct Python 3 solution for this coding contest problem.
There are 3N participants in AtCoder Group Contest. The strength of the i-th participant is represented by an integer a_i. They will form N teams, each consisting of three participants. No participant may belong to multiple teams.
The strength of a team is defined as the second largest strength among its members. For example, a team of participants of strength 1, 5, 2 has a strength 2, and a team of three participants of strength 3, 2, 3 has a strength 3.
Find the maximum possible sum of the strengths of N teams.
Constraints
* 1 ≤ N ≤ 10^5
* 1 ≤ a_i ≤ 10^{9}
* a_i are integers.
Input
Input is given from Standard Input in the following format:
N
a_1 a_2 ... a_{3N}
Output
Print the answer.
Examples
Input
2
5 2 8 5 1 5
Output
10
Input
10
1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000
Output
10000000000 | instruction | 0 | 2,341 | 17 | 4,682 |
"Correct Solution:
```
n=int(input())
a=list(map(int,input().split()))
a.sort(reverse=1)
ans=sum(a[1:(n*2):2])
print(ans)
``` | output | 1 | 2,341 | 17 | 4,683 |
Provide a correct Python 3 solution for this coding contest problem.
There are 3N participants in AtCoder Group Contest. The strength of the i-th participant is represented by an integer a_i. They will form N teams, each consisting of three participants. No participant may belong to multiple teams.
The strength of a team is defined as the second largest strength among its members. For example, a team of participants of strength 1, 5, 2 has a strength 2, and a team of three participants of strength 3, 2, 3 has a strength 3.
Find the maximum possible sum of the strengths of N teams.
Constraints
* 1 ≤ N ≤ 10^5
* 1 ≤ a_i ≤ 10^{9}
* a_i are integers.
Input
Input is given from Standard Input in the following format:
N
a_1 a_2 ... a_{3N}
Output
Print the answer.
Examples
Input
2
5 2 8 5 1 5
Output
10
Input
10
1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000
Output
10000000000 | instruction | 0 | 2,342 | 17 | 4,684 |
"Correct Solution:
```
N=int(input())
print(sum(sorted(list(map(int,input().split(' '))))[N::2]))
``` | output | 1 | 2,342 | 17 | 4,685 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There are 3N participants in AtCoder Group Contest. The strength of the i-th participant is represented by an integer a_i. They will form N teams, each consisting of three participants. No participant may belong to multiple teams.
The strength of a team is defined as the second largest strength among its members. For example, a team of participants of strength 1, 5, 2 has a strength 2, and a team of three participants of strength 3, 2, 3 has a strength 3.
Find the maximum possible sum of the strengths of N teams.
Constraints
* 1 ≤ N ≤ 10^5
* 1 ≤ a_i ≤ 10^{9}
* a_i are integers.
Input
Input is given from Standard Input in the following format:
N
a_1 a_2 ... a_{3N}
Output
Print the answer.
Examples
Input
2
5 2 8 5 1 5
Output
10
Input
10
1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000
Output
10000000000
Submitted Solution:
```
N = int(input())
a = list(map(int, input().split()))
a.sort()
del a[:N]
print(sum(a[::2]))
``` | instruction | 0 | 2,343 | 17 | 4,686 |
Yes | output | 1 | 2,343 | 17 | 4,687 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There are 3N participants in AtCoder Group Contest. The strength of the i-th participant is represented by an integer a_i. They will form N teams, each consisting of three participants. No participant may belong to multiple teams.
The strength of a team is defined as the second largest strength among its members. For example, a team of participants of strength 1, 5, 2 has a strength 2, and a team of three participants of strength 3, 2, 3 has a strength 3.
Find the maximum possible sum of the strengths of N teams.
Constraints
* 1 ≤ N ≤ 10^5
* 1 ≤ a_i ≤ 10^{9}
* a_i are integers.
Input
Input is given from Standard Input in the following format:
N
a_1 a_2 ... a_{3N}
Output
Print the answer.
Examples
Input
2
5 2 8 5 1 5
Output
10
Input
10
1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000
Output
10000000000
Submitted Solution:
```
N = int(input())
A = sorted(list(map(int, input().split())))
Ans = sum(A[N::2])
print(Ans)
``` | instruction | 0 | 2,344 | 17 | 4,688 |
Yes | output | 1 | 2,344 | 17 | 4,689 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There are 3N participants in AtCoder Group Contest. The strength of the i-th participant is represented by an integer a_i. They will form N teams, each consisting of three participants. No participant may belong to multiple teams.
The strength of a team is defined as the second largest strength among its members. For example, a team of participants of strength 1, 5, 2 has a strength 2, and a team of three participants of strength 3, 2, 3 has a strength 3.
Find the maximum possible sum of the strengths of N teams.
Constraints
* 1 ≤ N ≤ 10^5
* 1 ≤ a_i ≤ 10^{9}
* a_i are integers.
Input
Input is given from Standard Input in the following format:
N
a_1 a_2 ... a_{3N}
Output
Print the answer.
Examples
Input
2
5 2 8 5 1 5
Output
10
Input
10
1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000
Output
10000000000
Submitted Solution:
```
n=int(input());print(sum(sorted(map(int,input().split()))[::-1][1:n*2:2]))
``` | instruction | 0 | 2,345 | 17 | 4,690 |
Yes | output | 1 | 2,345 | 17 | 4,691 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There are 3N participants in AtCoder Group Contest. The strength of the i-th participant is represented by an integer a_i. They will form N teams, each consisting of three participants. No participant may belong to multiple teams.
The strength of a team is defined as the second largest strength among its members. For example, a team of participants of strength 1, 5, 2 has a strength 2, and a team of three participants of strength 3, 2, 3 has a strength 3.
Find the maximum possible sum of the strengths of N teams.
Constraints
* 1 ≤ N ≤ 10^5
* 1 ≤ a_i ≤ 10^{9}
* a_i are integers.
Input
Input is given from Standard Input in the following format:
N
a_1 a_2 ... a_{3N}
Output
Print the answer.
Examples
Input
2
5 2 8 5 1 5
Output
10
Input
10
1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000
Output
10000000000
Submitted Solution:
```
n = int(input())
a = sorted(list(map(int, input().split())), reverse=True)
print(sum(a[1:n*2:2]))
``` | instruction | 0 | 2,346 | 17 | 4,692 |
Yes | output | 1 | 2,346 | 17 | 4,693 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There are 3N participants in AtCoder Group Contest. The strength of the i-th participant is represented by an integer a_i. They will form N teams, each consisting of three participants. No participant may belong to multiple teams.
The strength of a team is defined as the second largest strength among its members. For example, a team of participants of strength 1, 5, 2 has a strength 2, and a team of three participants of strength 3, 2, 3 has a strength 3.
Find the maximum possible sum of the strengths of N teams.
Constraints
* 1 ≤ N ≤ 10^5
* 1 ≤ a_i ≤ 10^{9}
* a_i are integers.
Input
Input is given from Standard Input in the following format:
N
a_1 a_2 ... a_{3N}
Output
Print the answer.
Examples
Input
2
5 2 8 5 1 5
Output
10
Input
10
1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000
Output
10000000000
Submitted Solution:
```
n=int(input())
a=list(map(int,input().split()))
a.sort()
if n>=2:
print(a[-2]+a[-4])
else:
print(a[1])
``` | instruction | 0 | 2,347 | 17 | 4,694 |
No | output | 1 | 2,347 | 17 | 4,695 |
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