message stringlengths 2 65.1k | message_type stringclasses 2 values | message_id int64 0 1 | conversation_id int64 0 108k | cluster float64 14 14 | __index_level_0__ int64 0 217k |
|---|---|---|---|---|---|
Provide tags and a correct Python 3 solution for this coding contest problem.
There are n kangaroos with pockets. Each kangaroo has a size (integer number). A kangaroo can go into another kangaroo's pocket if and only if the size of kangaroo who hold the kangaroo is at least twice as large as the size of kangaroo who is held.
Each kangaroo can hold at most one kangaroo, and the kangaroo who is held by another kangaroo cannot hold any kangaroos.
The kangaroo who is held by another kangaroo cannot be visible from outside. Please, find a plan of holding kangaroos with the minimal number of kangaroos who is visible.
Input
The first line contains a single integer — n (1 ≤ n ≤ 5·105). Each of the next n lines contains an integer si — the size of the i-th kangaroo (1 ≤ si ≤ 105).
Output
Output a single integer — the optimal number of visible kangaroos.
Examples
Input
8
2
5
7
6
9
8
4
2
Output
5
Input
8
9
1
6
2
6
5
8
3
Output
5 | instruction | 0 | 58,348 | 14 | 116,696 |
Tags: binary search, greedy, sortings, two pointers
Correct Solution:
```
n=int(input())
ls= sorted([int(input()) for _ in range(n)])
cnt=0
mid=n//2-1
right=n-1
# mated=0
# for _ in range(n//2):
# if ls[right]>=2*ls[mid] and mid>=0:
# cnt+=1
# mid-=1
# right-=1
# mated+=2
# else:
# mid-=1
# print(cnt+n-mated)
while mid>=0 and right>mid:
if ls[mid]*2<=ls[right]:
cnt+=1
right-=1
mid-=1
print(n-cnt)
``` | output | 1 | 58,348 | 14 | 116,697 |
Provide tags and a correct Python 3 solution for this coding contest problem.
There are n kangaroos with pockets. Each kangaroo has a size (integer number). A kangaroo can go into another kangaroo's pocket if and only if the size of kangaroo who hold the kangaroo is at least twice as large as the size of kangaroo who is held.
Each kangaroo can hold at most one kangaroo, and the kangaroo who is held by another kangaroo cannot hold any kangaroos.
The kangaroo who is held by another kangaroo cannot be visible from outside. Please, find a plan of holding kangaroos with the minimal number of kangaroos who is visible.
Input
The first line contains a single integer — n (1 ≤ n ≤ 5·105). Each of the next n lines contains an integer si — the size of the i-th kangaroo (1 ≤ si ≤ 105).
Output
Output a single integer — the optimal number of visible kangaroos.
Examples
Input
8
2
5
7
6
9
8
4
2
Output
5
Input
8
9
1
6
2
6
5
8
3
Output
5 | instruction | 0 | 58,349 | 14 | 116,698 |
Tags: binary search, greedy, sortings, two pointers
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=999999999999999999999999
alphabets="abcdefghijklmnopqrstuvwxyz"
class SortedList:
def __init__(self, iterable=[], _load=200):
"""Initialize sorted list instance."""
values = sorted(iterable)
self._len = _len = len(values)
self._load = _load
self._lists = _lists = [values[i:i + _load] for i in range(0, _len, _load)]
self._list_lens = [len(_list) for _list in _lists]
self._mins = [_list[0] for _list in _lists]
self._fen_tree = []
self._rebuild = True
def _fen_build(self):
"""Build a fenwick tree instance."""
self._fen_tree[:] = self._list_lens
_fen_tree = self._fen_tree
for i in range(len(_fen_tree)):
if i | i + 1 < len(_fen_tree):
_fen_tree[i | i + 1] += _fen_tree[i]
self._rebuild = False
def _fen_update(self, index, value):
"""Update `fen_tree[index] += value`."""
if not self._rebuild:
_fen_tree = self._fen_tree
while index < len(_fen_tree):
_fen_tree[index] += value
index |= index + 1
def _fen_query(self, end):
"""Return `sum(_fen_tree[:end])`."""
if self._rebuild:
self._fen_build()
_fen_tree = self._fen_tree
x = 0
while end:
x += _fen_tree[end - 1]
end &= end - 1
return x
def _fen_findkth(self, k):
"""Return a pair of (the largest `idx` such that `sum(_fen_tree[:idx]) <= k`, `k - sum(_fen_tree[:idx])`)."""
_list_lens = self._list_lens
if k < _list_lens[0]:
return 0, k
if k >= self._len - _list_lens[-1]:
return len(_list_lens) - 1, k + _list_lens[-1] - self._len
if self._rebuild:
self._fen_build()
_fen_tree = self._fen_tree
idx = -1
for d in reversed(range(len(_fen_tree).bit_length())):
right_idx = idx + (1 << d)
if right_idx < len(_fen_tree) and k >= _fen_tree[right_idx]:
idx = right_idx
k -= _fen_tree[idx]
return idx + 1, k
def _delete(self, pos, idx):
"""Delete value at the given `(pos, idx)`."""
_lists = self._lists
_mins = self._mins
_list_lens = self._list_lens
self._len -= 1
self._fen_update(pos, -1)
del _lists[pos][idx]
_list_lens[pos] -= 1
if _list_lens[pos]:
_mins[pos] = _lists[pos][0]
else:
del _lists[pos]
del _list_lens[pos]
del _mins[pos]
self._rebuild = True
def _loc_left(self, value):
"""Return an index pair that corresponds to the first position of `value` in the sorted list."""
if not self._len:
return 0, 0
_lists = self._lists
_mins = self._mins
lo, pos = -1, len(_lists) - 1
while lo + 1 < pos:
mi = (lo + pos) >> 1
if value <= _mins[mi]:
pos = mi
else:
lo = mi
if pos and value <= _lists[pos - 1][-1]:
pos -= 1
_list = _lists[pos]
lo, idx = -1, len(_list)
while lo + 1 < idx:
mi = (lo + idx) >> 1
if value <= _list[mi]:
idx = mi
else:
lo = mi
return pos, idx
def _loc_right(self, value):
"""Return an index pair that corresponds to the last position of `value` in the sorted list."""
if not self._len:
return 0, 0
_lists = self._lists
_mins = self._mins
pos, hi = 0, len(_lists)
while pos + 1 < hi:
mi = (pos + hi) >> 1
if value < _mins[mi]:
hi = mi
else:
pos = mi
_list = _lists[pos]
lo, idx = -1, len(_list)
while lo + 1 < idx:
mi = (lo + idx) >> 1
if value < _list[mi]:
idx = mi
else:
lo = mi
return pos, idx
def add(self, value):
"""Add `value` to sorted list."""
_load = self._load
_lists = self._lists
_mins = self._mins
_list_lens = self._list_lens
self._len += 1
if _lists:
pos, idx = self._loc_right(value)
self._fen_update(pos, 1)
_list = _lists[pos]
_list.insert(idx, value)
_list_lens[pos] += 1
_mins[pos] = _list[0]
if _load + _load < len(_list):
_lists.insert(pos + 1, _list[_load:])
_list_lens.insert(pos + 1, len(_list) - _load)
_mins.insert(pos + 1, _list[_load])
_list_lens[pos] = _load
del _list[_load:]
self._rebuild = True
else:
_lists.append([value])
_mins.append(value)
_list_lens.append(1)
self._rebuild = True
def discard(self, value):
"""Remove `value` from sorted list if it is a member."""
_lists = self._lists
if _lists:
pos, idx = self._loc_right(value)
if idx and _lists[pos][idx - 1] == value:
self._delete(pos, idx - 1)
def remove(self, value):
"""Remove `value` from sorted list; `value` must be a member."""
_len = self._len
self.discard(value)
if _len == self._len:
raise ValueError('{0!r} not in list'.format(value))
def pop(self, index=-1):
"""Remove and return value at `index` in sorted list."""
pos, idx = self._fen_findkth(self._len + index if index < 0 else index)
value = self._lists[pos][idx]
self._delete(pos, idx)
return value
def bisect_left(self, value):
"""Return the first index to insert `value` in the sorted list."""
pos, idx = self._loc_left(value)
return self._fen_query(pos) + idx
def bisect_right(self, value):
"""Return the last index to insert `value` in the sorted list."""
pos, idx = self._loc_right(value)
return self._fen_query(pos) + idx
def count(self, value):
"""Return number of occurrences of `value` in the sorted list."""
return self.bisect_right(value) - self.bisect_left(value)
def __len__(self):
"""Return the size of the sorted list."""
return self._len
def __getitem__(self, index):
"""Lookup value at `index` in sorted list."""
pos, idx = self._fen_findkth(self._len + index if index < 0 else index)
return self._lists[pos][idx]
def __delitem__(self, index):
"""Remove value at `index` from sorted list."""
pos, idx = self._fen_findkth(self._len + index if index < 0 else index)
self._delete(pos, idx)
def __contains__(self, value):
"""Return true if `value` is an element of the sorted list."""
_lists = self._lists
if _lists:
pos, idx = self._loc_left(value)
return idx < len(_lists[pos]) and _lists[pos][idx] == value
return False
def __iter__(self):
"""Return an iterator over the sorted list."""
return (value for _list in self._lists for value in _list)
def __reversed__(self):
"""Return a reverse iterator over the sorted list."""
return (value for _list in reversed(self._lists) for value in reversed(_list))
def __repr__(self):
"""Return string representation of sorted list."""
return 'SortedList({0})'.format(list(self))
class SegTree:
def __init__(self, n):
self.N = 1 << n.bit_length()
self.tree = [0] * (self.N<<1)
def update(self, i, j, v):
i += self.N
j += self.N
while i <= j:
if i%2==1: self.tree[i] += v
if j%2==0: self.tree[j] += v
i, j = (i+1) >> 1, (j-1) >> 1
def query(self, i):
v = 0
i += self.N
while i > 0:
v += self.tree[i]
i >>= 1
return v
def SieveOfEratosthenes(limit):
"""Returns all primes not greater than limit."""
isPrime = [True]*(limit+1)
isPrime[0] = isPrime[1] = False
primes = []
for i in range(2, limit+1):
if not isPrime[i]:continue
primes += [i]
for j in range(i*i, limit+1, i):
isPrime[j] = False
return primes
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):
# LOGIC:
# At most kangaroos can be paired is n//2 -1
# So the greedy approach will be
# to pair the least sized n//2 -1 kangaroos
# with the kangaroos of bigger size
# Split kangaroos by sorting and first n//2 -1 are to be paired with rest
# Yeah kangaroos in second group could pair with bigger kangaroos
# but we dont care as we can pair some smaller kangaroo with that bigger kangaroo anyways
n=ri()
a=[]
for i in range(n):
a.append(ri())
a=sorted(a)
l=(n//2)-1
r=n-1
p=0
while l>=0 and r>(n//2)-1:
if 2*a[l]<=a[r]:
r-=1
p+=1
l-=1
wi(n-p)
#<--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 | 58,349 | 14 | 116,699 |
Provide tags and a correct Python 3 solution for this coding contest problem.
There are n kangaroos with pockets. Each kangaroo has a size (integer number). A kangaroo can go into another kangaroo's pocket if and only if the size of kangaroo who hold the kangaroo is at least twice as large as the size of kangaroo who is held.
Each kangaroo can hold at most one kangaroo, and the kangaroo who is held by another kangaroo cannot hold any kangaroos.
The kangaroo who is held by another kangaroo cannot be visible from outside. Please, find a plan of holding kangaroos with the minimal number of kangaroos who is visible.
Input
The first line contains a single integer — n (1 ≤ n ≤ 5·105). Each of the next n lines contains an integer si — the size of the i-th kangaroo (1 ≤ si ≤ 105).
Output
Output a single integer — the optimal number of visible kangaroos.
Examples
Input
8
2
5
7
6
9
8
4
2
Output
5
Input
8
9
1
6
2
6
5
8
3
Output
5 | instruction | 0 | 58,350 | 14 | 116,700 |
Tags: binary search, greedy, sortings, two pointers
Correct Solution:
```
import sys
n = int(input())
a = list(sorted([int(next(sys.stdin)) for _ in range(n)]))
i, j = 0, n // 2
cnt = 0
while j < n:
if a[i] == -1:
i += 1
if i == j:
j += 1
elif a[i] * 2 <= a[j]:
cnt += 1
a[j] = -1
i += 1
j += 1
print(n - cnt)
``` | output | 1 | 58,350 | 14 | 116,701 |
Provide tags and a correct Python 3 solution for this coding contest problem.
There are n kangaroos with pockets. Each kangaroo has a size (integer number). A kangaroo can go into another kangaroo's pocket if and only if the size of kangaroo who hold the kangaroo is at least twice as large as the size of kangaroo who is held.
Each kangaroo can hold at most one kangaroo, and the kangaroo who is held by another kangaroo cannot hold any kangaroos.
The kangaroo who is held by another kangaroo cannot be visible from outside. Please, find a plan of holding kangaroos with the minimal number of kangaroos who is visible.
Input
The first line contains a single integer — n (1 ≤ n ≤ 5·105). Each of the next n lines contains an integer si — the size of the i-th kangaroo (1 ≤ si ≤ 105).
Output
Output a single integer — the optimal number of visible kangaroos.
Examples
Input
8
2
5
7
6
9
8
4
2
Output
5
Input
8
9
1
6
2
6
5
8
3
Output
5 | instruction | 0 | 58,351 | 14 | 116,702 |
Tags: binary search, greedy, sortings, two pointers
Correct Solution:
```
import os
import sys
from io import BytesIO, IOBase
BUFSIZE = 8192
class FastIO(IOBase):
newlines = 0
def __init__(self, file):
self._fd = file.fileno()
self.buffer = BytesIO()
self.writable = "x" in file.mode or "r" not in file.mode
self.write = self.buffer.write if self.writable else None
def read(self):
while True:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
if not b:
break
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines = 0
return self.buffer.read()
def readline(self):
while self.newlines == 0:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
self.newlines = b.count(b"\n") + (not b)
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines -= 1
return self.buffer.readline()
def flush(self):
if self.writable:
os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
self.write = lambda s: self.buffer.write(s.encode("ascii"))
self.read = lambda: self.buffer.read().decode("ascii")
self.readline = lambda: self.buffer.readline().decode("ascii")
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")
n=int(input())
s=[0 for i in range(0,n)]
for i in range(0,n):
s[i]=int(input())
s.sort()
# print (s)
p=n//2+n%2
visible=n
y=0
for i in range(p,p+n//2):
if s[y]*2<=s[i]:
visible-=1
y+=1
print (visible)
``` | output | 1 | 58,351 | 14 | 116,703 |
Provide tags and a correct Python 3 solution for this coding contest problem.
There are n kangaroos with pockets. Each kangaroo has a size (integer number). A kangaroo can go into another kangaroo's pocket if and only if the size of kangaroo who hold the kangaroo is at least twice as large as the size of kangaroo who is held.
Each kangaroo can hold at most one kangaroo, and the kangaroo who is held by another kangaroo cannot hold any kangaroos.
The kangaroo who is held by another kangaroo cannot be visible from outside. Please, find a plan of holding kangaroos with the minimal number of kangaroos who is visible.
Input
The first line contains a single integer — n (1 ≤ n ≤ 5·105). Each of the next n lines contains an integer si — the size of the i-th kangaroo (1 ≤ si ≤ 105).
Output
Output a single integer — the optimal number of visible kangaroos.
Examples
Input
8
2
5
7
6
9
8
4
2
Output
5
Input
8
9
1
6
2
6
5
8
3
Output
5 | instruction | 0 | 58,352 | 14 | 116,704 |
Tags: binary search, greedy, sortings, two pointers
Correct Solution:
```
from sys import stdin
import collections
import os
import sys
from io import IOBase, BytesIO
from sys import stdin, stdout
from math import sqrt
py2 = round(0.5)
if py2:
from future_builtins import ascii, filter, hex, map, oct, zip
range = xrange
def main():
n = int(input())
nn = n
arr = []
while nn:
nn -= 1
temp = int(input())
arr.append(temp)
arr.sort()
mid = n//2 - 1
b = n - 1
s = 0
while mid >= 0:
if arr[mid]*2 <= arr[b]:
b -= 1
s += 1
mid -= 1
print(n-s)
BUFSIZE = 8192
class FastIO(BytesIO):
newlines = 0
def __init__(self, file):
self._file = file
self._fd = file.fileno()
self.writable = "x" in file.mode or "w" in file.mode
self.write = super(FastIO, self).write if self.writable else None
def _fill(self):
s = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
self.seek((self.tell(), self.seek(0, 2),
super(FastIO, self).write(s))[0])
return s
def read(self):
while self._fill():
pass
return super(FastIO, self).read()
def readline(self):
while self.newlines == 0:
s = self._fill()
self.newlines = s.count(b"\n") + (not s)
self.newlines -= 1
return super(FastIO, self).readline()
def flush(self):
if self.writable:
os.write(self._fd, self.getvalue())
self.truncate(0), self.seek(0)
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
if py2:
self.write = self.buffer.write
self.read = self.buffer.read
self.readline = self.buffer.readline
else:
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)
def input(): return sys.stdin.readline().rstrip('\r\n')
class ostream:
def __lshift__(self, a):
sys.stdout.write(str(a))
return self
cout = ostream()
endl = '\n'
if __name__ == "__main__":
main()
``` | output | 1 | 58,352 | 14 | 116,705 |
Provide tags and a correct Python 3 solution for this coding contest problem.
There are n kangaroos with pockets. Each kangaroo has a size (integer number). A kangaroo can go into another kangaroo's pocket if and only if the size of kangaroo who hold the kangaroo is at least twice as large as the size of kangaroo who is held.
Each kangaroo can hold at most one kangaroo, and the kangaroo who is held by another kangaroo cannot hold any kangaroos.
The kangaroo who is held by another kangaroo cannot be visible from outside. Please, find a plan of holding kangaroos with the minimal number of kangaroos who is visible.
Input
The first line contains a single integer — n (1 ≤ n ≤ 5·105). Each of the next n lines contains an integer si — the size of the i-th kangaroo (1 ≤ si ≤ 105).
Output
Output a single integer — the optimal number of visible kangaroos.
Examples
Input
8
2
5
7
6
9
8
4
2
Output
5
Input
8
9
1
6
2
6
5
8
3
Output
5 | instruction | 0 | 58,353 | 14 | 116,706 |
Tags: binary search, greedy, sortings, two pointers
Correct Solution:
```
import os
import sys
from io import BytesIO, IOBase
def main():
pass
# region fastio
BUFSIZE = 8192
class FastIO(IOBase):
newlines = 0
def __init__(self, file):
self._fd = file.fileno()
self.buffer = BytesIO()
self.writable = "x" in file.mode or "r" not in file.mode
self.write = self.buffer.write if self.writable else None
def read(self):
while True:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
if not b:
break
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines = 0
return self.buffer.read()
def readline(self):
while self.newlines == 0:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
self.newlines = b.count(b"\n") + (not b)
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines -= 1
return self.buffer.readline()
def flush(self):
if self.writable:
os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
self.write = lambda s: self.buffer.write(s.encode("ascii"))
self.read = lambda: self.buffer.read().decode("ascii")
self.readline = lambda: self.buffer.readline().decode("ascii")
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")
n=int(input())
l=[]
for i in range(0,n):
l.append(int(input()))
l.sort()
i=n-1
ans=n
j=n//2-1
while(j>=0 and i>=n//2):
if((l[j]*2)<=l[i]):
i-=1
ans-=1
j-=1
print(ans)
``` | output | 1 | 58,353 | 14 | 116,707 |
Provide tags and a correct Python 3 solution for this coding contest problem.
There are n kangaroos with pockets. Each kangaroo has a size (integer number). A kangaroo can go into another kangaroo's pocket if and only if the size of kangaroo who hold the kangaroo is at least twice as large as the size of kangaroo who is held.
Each kangaroo can hold at most one kangaroo, and the kangaroo who is held by another kangaroo cannot hold any kangaroos.
The kangaroo who is held by another kangaroo cannot be visible from outside. Please, find a plan of holding kangaroos with the minimal number of kangaroos who is visible.
Input
The first line contains a single integer — n (1 ≤ n ≤ 5·105). Each of the next n lines contains an integer si — the size of the i-th kangaroo (1 ≤ si ≤ 105).
Output
Output a single integer — the optimal number of visible kangaroos.
Examples
Input
8
2
5
7
6
9
8
4
2
Output
5
Input
8
9
1
6
2
6
5
8
3
Output
5 | instruction | 0 | 58,354 | 14 | 116,708 |
Tags: binary search, greedy, sortings, two pointers
Correct Solution:
```
import os
import sys
from io import BytesIO, IOBase
import math
from decimal import *
getcontext().prec = 25
MOD = pow(10, 9) + 7
BUFSIZE = 8192
class FastIO(IOBase):
newlines = 0
def __init__(self, file):
self._fd = file.fileno()
self.buffer = BytesIO()
self.writable = "x" in file.mode or "r" not in file.mode
self.write = self.buffer.write if self.writable else None
def read(self):
while True:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
if not b:
break
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines = 0
return self.buffer.read()
def readline(self):
while self.newlines == 0:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
self.newlines = b.count(b"\n") + (not b)
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines -= 1
return self.buffer.readline()
def flush(self):
if self.writable:
os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
self.write = lambda s: self.buffer.write(s.encode("ascii"))
self.read = lambda: self.buffer.read().decode("ascii")
self.readline = lambda: self.buffer.readline().decode("ascii")
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")
# n, k = map(int, input().split(" "))
# l = list(map(int, input().split(" ")))
n = int(input())
l = [int(input()) for i in range(n)]
l.sort()
a = l[-1] // 2
x = n//2 + n%2
i = 0
c = 0
while l[i] <= a and x < n:
if l[i] <= l[x] // 2:
c += 1
i += 1
x += 1
else:
x += 1
# print(*l)
# print(a)
print(n-c)
``` | output | 1 | 58,354 | 14 | 116,709 |
Provide tags and a correct Python 3 solution for this coding contest problem.
There are n kangaroos with pockets. Each kangaroo has a size (integer number). A kangaroo can go into another kangaroo's pocket if and only if the size of kangaroo who hold the kangaroo is at least twice as large as the size of kangaroo who is held.
Each kangaroo can hold at most one kangaroo, and the kangaroo who is held by another kangaroo cannot hold any kangaroos.
The kangaroo who is held by another kangaroo cannot be visible from outside. Please, find a plan of holding kangaroos with the minimal number of kangaroos who is visible.
Input
The first line contains a single integer — n (1 ≤ n ≤ 5·105). Each of the next n lines contains an integer si — the size of the i-th kangaroo (1 ≤ si ≤ 105).
Output
Output a single integer — the optimal number of visible kangaroos.
Examples
Input
8
2
5
7
6
9
8
4
2
Output
5
Input
8
9
1
6
2
6
5
8
3
Output
5 | instruction | 0 | 58,355 | 14 | 116,710 |
Tags: binary search, greedy, sortings, two pointers
Correct Solution:
```
import os
import sys
from io import BytesIO, IOBase
BUFSIZE = 8192
class FastIO(IOBase):
newlines = 0
def __init__(self, file):
self._fd = file.fileno()
self.buffer = BytesIO()
self.writable = "x" in file.mode or "r" not in file.mode
self.write = self.buffer.write if self.writable else None
def read(self):
while True:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
if not b:
break
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines = 0
return self.buffer.read()
def readline(self):
while self.newlines == 0:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
self.newlines = b.count(b"\n") + (not b)
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines -= 1
return self.buffer.readline()
def flush(self):
if self.writable:
os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
self.write = lambda s: self.buffer.write(s.encode("ascii"))
self.read = lambda: self.buffer.read().decode("ascii")
self.readline = lambda: self.buffer.readline().decode("ascii")
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")
n = int(input())
arr = [0] * n
for i in range(n):
arr[i] = int(input())
arr.sort()
small = 0
big = n // 2
hidden = 0
while small < n // 2 and big < n:
if arr[big] >= arr[small] * 2:
small += 1
big += 1
hidden += 1
else:
big += 1
print(n - hidden)
``` | output | 1 | 58,355 | 14 | 116,711 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There are n kangaroos with pockets. Each kangaroo has a size (integer number). A kangaroo can go into another kangaroo's pocket if and only if the size of kangaroo who hold the kangaroo is at least twice as large as the size of kangaroo who is held.
Each kangaroo can hold at most one kangaroo, and the kangaroo who is held by another kangaroo cannot hold any kangaroos.
The kangaroo who is held by another kangaroo cannot be visible from outside. Please, find a plan of holding kangaroos with the minimal number of kangaroos who is visible.
Input
The first line contains a single integer — n (1 ≤ n ≤ 5·105). Each of the next n lines contains an integer si — the size of the i-th kangaroo (1 ≤ si ≤ 105).
Output
Output a single integer — the optimal number of visible kangaroos.
Examples
Input
8
2
5
7
6
9
8
4
2
Output
5
Input
8
9
1
6
2
6
5
8
3
Output
5
Submitted Solution:
```
n = int(input())
arr = sorted([int(input()) for _ in range(n)])
r = n-1
count = 0
for t in range((n//2) - 1,-1,-1):
if arr[t] * 2 <= arr[r]:
r -= 1
count += 1
print(n-count)
``` | instruction | 0 | 58,359 | 14 | 116,718 |
Yes | output | 1 | 58,359 | 14 | 116,719 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There are n kangaroos with pockets. Each kangaroo has a size (integer number). A kangaroo can go into another kangaroo's pocket if and only if the size of kangaroo who hold the kangaroo is at least twice as large as the size of kangaroo who is held.
Each kangaroo can hold at most one kangaroo, and the kangaroo who is held by another kangaroo cannot hold any kangaroos.
The kangaroo who is held by another kangaroo cannot be visible from outside. Please, find a plan of holding kangaroos with the minimal number of kangaroos who is visible.
Input
The first line contains a single integer — n (1 ≤ n ≤ 5·105). Each of the next n lines contains an integer si — the size of the i-th kangaroo (1 ≤ si ≤ 105).
Output
Output a single integer — the optimal number of visible kangaroos.
Examples
Input
8
2
5
7
6
9
8
4
2
Output
5
Input
8
9
1
6
2
6
5
8
3
Output
5
Submitted Solution:
```
def main():
n=int(input())
array=[]
for x in range(0,n):
array.append(int(input()))
array.sort()
max=array[n-1]
i=n-2
j=n-1
already=[]
while(j>=0):
if((array[j]>=2*array[i]) and not(already.__contains__(i))):
already.append(i)
j-=1
i=j-1
else:
i-=1
if i<0:
j-=1
i=j-1
print((n-len(already)))
if __name__ == '__main__':
main()
``` | instruction | 0 | 58,362 | 14 | 116,724 |
No | output | 1 | 58,362 | 14 | 116,725 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There are n kangaroos with pockets. Each kangaroo has a size (integer number). A kangaroo can go into another kangaroo's pocket if and only if the size of kangaroo who hold the kangaroo is at least twice as large as the size of kangaroo who is held.
Each kangaroo can hold at most one kangaroo, and the kangaroo who is held by another kangaroo cannot hold any kangaroos.
The kangaroo who is held by another kangaroo cannot be visible from outside. Please, find a plan of holding kangaroos with the minimal number of kangaroos who is visible.
Input
The first line contains a single integer — n (1 ≤ n ≤ 5·105). Each of the next n lines contains an integer si — the size of the i-th kangaroo (1 ≤ si ≤ 105).
Output
Output a single integer — the optimal number of visible kangaroos.
Examples
Input
8
2
5
7
6
9
8
4
2
Output
5
Input
8
9
1
6
2
6
5
8
3
Output
5
Submitted Solution:
```
import bisect
import sys
from collections import Counter
from typing import List
input = sys.stdin.readline
def solve(n: int, a: List[int]) -> int:
c = Counter(a)
keys = []
for i in range(100_001):
if i in c:
keys.append(i)
held_count = 0
for e in keys[::-1]:
if e not in c:
continue
i = bisect.bisect_left(keys, e // 2)
if keys[i] * 2 > e:
i -= 1
while i >= 0 and c[e] > c[keys[i]]:
if c[keys[i]] > 0:
c[e] -= c[keys[i]]
held_count += c[keys[i]]
del c[keys[i]]
i -= 1
if i < 0:
break
c[keys[i]] -= c[e]
if c[keys[i]] == 0:
del c[keys[i]]
keys.pop(i)
held_count += c[e]
del c[e]
return sum(c.values()) + held_count
# assert solve(100,
# [678, 771, 96, 282, 135, 749, 168, 668, 17, 658, 979, 446, 998, 331, 606, 756, 37, 515, 538, 205, 647, 547,
# 904, 842, 647, 286, 774, 414, 267, 791, 595, 465, 8, 327, 855, 174, 339, 946, 184, 250, 807, 422, 679,
# 980, 64, 530, 312, 351, 676, 911, 803, 991, 669, 50, 293, 841, 545, 598, 737, 894, 231, 754, 588, 83, 873,
# 767, 833, 482, 905, 903, 970, 571, 715, 59, 777, 697, 537, 861, 339, 212, 149, 889, 905, 70, 970, 307,
# 830, 465, 968, 291, 430, 317, 942, 944, 330, 235, 814, 880, 415, 76]) == 58
# assert solve(12, [3, 99, 24, 46, 75, 63, 57, 55, 10, 62, 34, 52]) == 7
# assert solve(8, [2, 5, 7, 6, 9, 8, 4, 2]) == 5
n = int(input())
a = []
for _ in range(n):
a.append(int(input()))
r = solve(n, a)
print(r)
``` | instruction | 0 | 58,363 | 14 | 116,726 |
No | output | 1 | 58,363 | 14 | 116,727 |
Provide tags and a correct Python 3 solution for this coding contest problem.
The Berland State University is hosting a ballroom dance in celebration of its 100500-th anniversary! n boys and m girls are already busy rehearsing waltz, minuet, polonaise and quadrille moves.
We know that several boy&girl pairs are going to be invited to the ball. However, the partners' dancing skill in each pair must differ by at most one.
For each boy, we know his dancing skills. Similarly, for each girl we know her dancing skills. Write a code that can determine the largest possible number of pairs that can be formed from n boys and m girls.
Input
The first line contains an integer n (1 ≤ n ≤ 100) — the number of boys. The second line contains sequence a1, a2, ..., an (1 ≤ ai ≤ 100), where ai is the i-th boy's dancing skill.
Similarly, the third line contains an integer m (1 ≤ m ≤ 100) — the number of girls. The fourth line contains sequence b1, b2, ..., bm (1 ≤ bj ≤ 100), where bj is the j-th girl's dancing skill.
Output
Print a single number — the required maximum possible number of pairs.
Examples
Input
4
1 4 6 2
5
5 1 5 7 9
Output
3
Input
4
1 2 3 4
4
10 11 12 13
Output
0
Input
5
1 1 1 1 1
3
1 2 3
Output
2 | instruction | 0 | 58,402 | 14 | 116,804 |
Tags: dfs and similar, dp, graph matchings, greedy, sortings, two pointers
Correct Solution:
```
# cook your dish here
n=int(input())
l1=[int(s) for s in input().split()]
m=int(input())
l2=[int(s) for s in input().split()]
l1.sort()
l2.sort()
i=j=c=0
while(i<n and j<m):
if(abs(l1[i]-l2[j])<=1):
c+=1
i+=1
j+=1
elif l1[i]<l2[j]:
i+=1
else:
j+=1
print(c)
``` | output | 1 | 58,402 | 14 | 116,805 |
Provide tags and a correct Python 3 solution for this coding contest problem.
The Berland State University is hosting a ballroom dance in celebration of its 100500-th anniversary! n boys and m girls are already busy rehearsing waltz, minuet, polonaise and quadrille moves.
We know that several boy&girl pairs are going to be invited to the ball. However, the partners' dancing skill in each pair must differ by at most one.
For each boy, we know his dancing skills. Similarly, for each girl we know her dancing skills. Write a code that can determine the largest possible number of pairs that can be formed from n boys and m girls.
Input
The first line contains an integer n (1 ≤ n ≤ 100) — the number of boys. The second line contains sequence a1, a2, ..., an (1 ≤ ai ≤ 100), where ai is the i-th boy's dancing skill.
Similarly, the third line contains an integer m (1 ≤ m ≤ 100) — the number of girls. The fourth line contains sequence b1, b2, ..., bm (1 ≤ bj ≤ 100), where bj is the j-th girl's dancing skill.
Output
Print a single number — the required maximum possible number of pairs.
Examples
Input
4
1 4 6 2
5
5 1 5 7 9
Output
3
Input
4
1 2 3 4
4
10 11 12 13
Output
0
Input
5
1 1 1 1 1
3
1 2 3
Output
2 | instruction | 0 | 58,403 | 14 | 116,806 |
Tags: dfs and similar, dp, graph matchings, greedy, sortings, two pointers
Correct Solution:
```
from collections import deque
from math import ceil
def ii():return int(input())
def si():return input()
def mi():return map(int,input().split())
def li():return list(mi())
abc="abcdefghijklmnopqrstuvwxyz"
n=ii()
a=li()
m=ii()
b=li()
a.sort()
b.sort()
c=0
for i in range(n):
x=a[i]
for j in range(m):
y=abs(x-b[j])
if(y<=1):
b.remove(b[j])
c+=1
m-=1
break
print(c)
``` | output | 1 | 58,403 | 14 | 116,807 |
Provide tags and a correct Python 3 solution for this coding contest problem.
The Berland State University is hosting a ballroom dance in celebration of its 100500-th anniversary! n boys and m girls are already busy rehearsing waltz, minuet, polonaise and quadrille moves.
We know that several boy&girl pairs are going to be invited to the ball. However, the partners' dancing skill in each pair must differ by at most one.
For each boy, we know his dancing skills. Similarly, for each girl we know her dancing skills. Write a code that can determine the largest possible number of pairs that can be formed from n boys and m girls.
Input
The first line contains an integer n (1 ≤ n ≤ 100) — the number of boys. The second line contains sequence a1, a2, ..., an (1 ≤ ai ≤ 100), where ai is the i-th boy's dancing skill.
Similarly, the third line contains an integer m (1 ≤ m ≤ 100) — the number of girls. The fourth line contains sequence b1, b2, ..., bm (1 ≤ bj ≤ 100), where bj is the j-th girl's dancing skill.
Output
Print a single number — the required maximum possible number of pairs.
Examples
Input
4
1 4 6 2
5
5 1 5 7 9
Output
3
Input
4
1 2 3 4
4
10 11 12 13
Output
0
Input
5
1 1 1 1 1
3
1 2 3
Output
2 | instruction | 0 | 58,404 | 14 | 116,808 |
Tags: dfs and similar, dp, graph matchings, greedy, sortings, two pointers
Correct Solution:
```
import math
input()
lista_hombres = list(input().split())
for i in range(len(lista_hombres)):
lista_hombres[i] = int(lista_hombres[i])
input()
lista_mujeres = list(input().split())
for i in range(len(lista_mujeres)):
lista_mujeres[i] = int(lista_mujeres[i])
lista_hombres.sort()
lista_mujeres.sort()
parejas = 0
for i in range(len(lista_hombres)):
for j in range(len(lista_mujeres)):
if abs(lista_hombres[i]-lista_mujeres[j]) <= 1:
parejas += 1
lista_hombres[i] = math.inf
lista_mujeres[j] = math.inf
print(parejas)
``` | output | 1 | 58,404 | 14 | 116,809 |
Provide tags and a correct Python 3 solution for this coding contest problem.
The Berland State University is hosting a ballroom dance in celebration of its 100500-th anniversary! n boys and m girls are already busy rehearsing waltz, minuet, polonaise and quadrille moves.
We know that several boy&girl pairs are going to be invited to the ball. However, the partners' dancing skill in each pair must differ by at most one.
For each boy, we know his dancing skills. Similarly, for each girl we know her dancing skills. Write a code that can determine the largest possible number of pairs that can be formed from n boys and m girls.
Input
The first line contains an integer n (1 ≤ n ≤ 100) — the number of boys. The second line contains sequence a1, a2, ..., an (1 ≤ ai ≤ 100), where ai is the i-th boy's dancing skill.
Similarly, the third line contains an integer m (1 ≤ m ≤ 100) — the number of girls. The fourth line contains sequence b1, b2, ..., bm (1 ≤ bj ≤ 100), where bj is the j-th girl's dancing skill.
Output
Print a single number — the required maximum possible number of pairs.
Examples
Input
4
1 4 6 2
5
5 1 5 7 9
Output
3
Input
4
1 2 3 4
4
10 11 12 13
Output
0
Input
5
1 1 1 1 1
3
1 2 3
Output
2 | instruction | 0 | 58,405 | 14 | 116,810 |
Tags: dfs and similar, dp, graph matchings, greedy, sortings, two pointers
Correct Solution:
```
def main():
from sys import stdin
from itertools import islice
tokens = map(int, stdin.read().split())
N = next(tokens)
boys = sorted(islice(tokens, N))
M = next(tokens)
girls = sorted(islice(tokens, M))
if len(boys) < len(girls):
smallest_array = boys
biggest_array = girls
else:
smallest_array = girls
biggest_array = boys
ans = 0
i = 0
for x in smallest_array:
while i < len(biggest_array) and biggest_array[i] <= x - 2:
i += 1
if i == len(biggest_array):
break
elif abs(biggest_array[i] - x) <= 1:
ans += 1
i += 1
print(ans)
main()
``` | output | 1 | 58,405 | 14 | 116,811 |
Provide tags and a correct Python 3 solution for this coding contest problem.
The Berland State University is hosting a ballroom dance in celebration of its 100500-th anniversary! n boys and m girls are already busy rehearsing waltz, minuet, polonaise and quadrille moves.
We know that several boy&girl pairs are going to be invited to the ball. However, the partners' dancing skill in each pair must differ by at most one.
For each boy, we know his dancing skills. Similarly, for each girl we know her dancing skills. Write a code that can determine the largest possible number of pairs that can be formed from n boys and m girls.
Input
The first line contains an integer n (1 ≤ n ≤ 100) — the number of boys. The second line contains sequence a1, a2, ..., an (1 ≤ ai ≤ 100), where ai is the i-th boy's dancing skill.
Similarly, the third line contains an integer m (1 ≤ m ≤ 100) — the number of girls. The fourth line contains sequence b1, b2, ..., bm (1 ≤ bj ≤ 100), where bj is the j-th girl's dancing skill.
Output
Print a single number — the required maximum possible number of pairs.
Examples
Input
4
1 4 6 2
5
5 1 5 7 9
Output
3
Input
4
1 2 3 4
4
10 11 12 13
Output
0
Input
5
1 1 1 1 1
3
1 2 3
Output
2 | instruction | 0 | 58,406 | 14 | 116,812 |
Tags: dfs and similar, dp, graph matchings, greedy, sortings, two pointers
Correct Solution:
```
def foo(n,m):
c=0
n.sort()
m.sort()
for i in range(len(n)):
for j in range(len(m)):
if(abs(n[i]-m[j])<=1):
c+=1
m[j]=10000
break
return(c)
n=int(input())
a=input()
a=a.split(" ")
an=[int(i) for i in a]
m= int(input())
b= input()
b=b.split(" ")
bm=[int(i) for i in b]
print(foo(an,bm))
``` | output | 1 | 58,406 | 14 | 116,813 |
Provide tags and a correct Python 3 solution for this coding contest problem.
The Berland State University is hosting a ballroom dance in celebration of its 100500-th anniversary! n boys and m girls are already busy rehearsing waltz, minuet, polonaise and quadrille moves.
We know that several boy&girl pairs are going to be invited to the ball. However, the partners' dancing skill in each pair must differ by at most one.
For each boy, we know his dancing skills. Similarly, for each girl we know her dancing skills. Write a code that can determine the largest possible number of pairs that can be formed from n boys and m girls.
Input
The first line contains an integer n (1 ≤ n ≤ 100) — the number of boys. The second line contains sequence a1, a2, ..., an (1 ≤ ai ≤ 100), where ai is the i-th boy's dancing skill.
Similarly, the third line contains an integer m (1 ≤ m ≤ 100) — the number of girls. The fourth line contains sequence b1, b2, ..., bm (1 ≤ bj ≤ 100), where bj is the j-th girl's dancing skill.
Output
Print a single number — the required maximum possible number of pairs.
Examples
Input
4
1 4 6 2
5
5 1 5 7 9
Output
3
Input
4
1 2 3 4
4
10 11 12 13
Output
0
Input
5
1 1 1 1 1
3
1 2 3
Output
2 | instruction | 0 | 58,407 | 14 | 116,814 |
Tags: dfs and similar, dp, graph matchings, greedy, sortings, two pointers
Correct Solution:
```
n=int(input())
a=sorted(list(map(int,input().split())))
m=int(input())
b=sorted(list(map(int,input().split())))
i,j,cnt=0,0,0
while i<n and j<m:
if abs(a[i]-b[j])<=1:
cnt+=1
i+=1
j+=1
elif a[i]<b[j]:
i+=1
else:
j+=1
print(cnt)
``` | output | 1 | 58,407 | 14 | 116,815 |
Provide tags and a correct Python 3 solution for this coding contest problem.
The Berland State University is hosting a ballroom dance in celebration of its 100500-th anniversary! n boys and m girls are already busy rehearsing waltz, minuet, polonaise and quadrille moves.
We know that several boy&girl pairs are going to be invited to the ball. However, the partners' dancing skill in each pair must differ by at most one.
For each boy, we know his dancing skills. Similarly, for each girl we know her dancing skills. Write a code that can determine the largest possible number of pairs that can be formed from n boys and m girls.
Input
The first line contains an integer n (1 ≤ n ≤ 100) — the number of boys. The second line contains sequence a1, a2, ..., an (1 ≤ ai ≤ 100), where ai is the i-th boy's dancing skill.
Similarly, the third line contains an integer m (1 ≤ m ≤ 100) — the number of girls. The fourth line contains sequence b1, b2, ..., bm (1 ≤ bj ≤ 100), where bj is the j-th girl's dancing skill.
Output
Print a single number — the required maximum possible number of pairs.
Examples
Input
4
1 4 6 2
5
5 1 5 7 9
Output
3
Input
4
1 2 3 4
4
10 11 12 13
Output
0
Input
5
1 1 1 1 1
3
1 2 3
Output
2 | instruction | 0 | 58,408 | 14 | 116,816 |
Tags: dfs and similar, dp, graph matchings, greedy, sortings, two pointers
Correct Solution:
```
def crear_lista(string):
lista = []
cambio = False
num = ""
for letra in string+" ":
if letra == " ":
cambio = True
lista.append(int(num))
num = ""
if not cambio:
num += letra
cambio = False
return lista
n = input()
hombres = input()
k = input()
mujeres = input()
baile_hombres = crear_lista(hombres)
baile_mujeres = crear_lista(mujeres)
baile_hombres.sort()
baile_mujeres.sort()
contador = 0
for m in baile_mujeres:
for n in baile_hombres:
if int(m)-int(n) <-1:
break
elif int(m)-int(n) > 1:
continue
else:
contador += 1
baile_hombres.remove(n)
break
#djsalkds
print(contador)
``` | output | 1 | 58,408 | 14 | 116,817 |
Provide tags and a correct Python 3 solution for this coding contest problem.
The Berland State University is hosting a ballroom dance in celebration of its 100500-th anniversary! n boys and m girls are already busy rehearsing waltz, minuet, polonaise and quadrille moves.
We know that several boy&girl pairs are going to be invited to the ball. However, the partners' dancing skill in each pair must differ by at most one.
For each boy, we know his dancing skills. Similarly, for each girl we know her dancing skills. Write a code that can determine the largest possible number of pairs that can be formed from n boys and m girls.
Input
The first line contains an integer n (1 ≤ n ≤ 100) — the number of boys. The second line contains sequence a1, a2, ..., an (1 ≤ ai ≤ 100), where ai is the i-th boy's dancing skill.
Similarly, the third line contains an integer m (1 ≤ m ≤ 100) — the number of girls. The fourth line contains sequence b1, b2, ..., bm (1 ≤ bj ≤ 100), where bj is the j-th girl's dancing skill.
Output
Print a single number — the required maximum possible number of pairs.
Examples
Input
4
1 4 6 2
5
5 1 5 7 9
Output
3
Input
4
1 2 3 4
4
10 11 12 13
Output
0
Input
5
1 1 1 1 1
3
1 2 3
Output
2 | instruction | 0 | 58,409 | 14 | 116,818 |
Tags: dfs and similar, dp, graph matchings, greedy, sortings, two pointers
Correct Solution:
```
# Velichko ♥
n = int(input())
b = list (map(int,input().split()))
m = int(input())
g = list (map(int,input().split()))
i=int(0)
j=int(0)
ans=int(0)
g.sort()
b.sort()
while i<n and j<m:
if b[i]-1>g[j]:
j+=1
else:
if g[j]-1>b[i]:
i+=1
else:
i+=1
j+=1
ans+=1
print(ans)
``` | output | 1 | 58,409 | 14 | 116,819 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
The Berland State University is hosting a ballroom dance in celebration of its 100500-th anniversary! n boys and m girls are already busy rehearsing waltz, minuet, polonaise and quadrille moves.
We know that several boy&girl pairs are going to be invited to the ball. However, the partners' dancing skill in each pair must differ by at most one.
For each boy, we know his dancing skills. Similarly, for each girl we know her dancing skills. Write a code that can determine the largest possible number of pairs that can be formed from n boys and m girls.
Input
The first line contains an integer n (1 ≤ n ≤ 100) — the number of boys. The second line contains sequence a1, a2, ..., an (1 ≤ ai ≤ 100), where ai is the i-th boy's dancing skill.
Similarly, the third line contains an integer m (1 ≤ m ≤ 100) — the number of girls. The fourth line contains sequence b1, b2, ..., bm (1 ≤ bj ≤ 100), where bj is the j-th girl's dancing skill.
Output
Print a single number — the required maximum possible number of pairs.
Examples
Input
4
1 4 6 2
5
5 1 5 7 9
Output
3
Input
4
1 2 3 4
4
10 11 12 13
Output
0
Input
5
1 1 1 1 1
3
1 2 3
Output
2
Submitted Solution:
```
n=int(input())
a=list(map(int,input().split()))
m=int(input())
b=list(map(int,input().split()))
l=[];a.sort();b.sort();k=0
for i in range(n):
for j in range(m):
if abs(a[i]-b[j])<=1 and j not in l:
k+=1
l.append(j)
break
print(k)
``` | instruction | 0 | 58,410 | 14 | 116,820 |
Yes | output | 1 | 58,410 | 14 | 116,821 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
The Berland State University is hosting a ballroom dance in celebration of its 100500-th anniversary! n boys and m girls are already busy rehearsing waltz, minuet, polonaise and quadrille moves.
We know that several boy&girl pairs are going to be invited to the ball. However, the partners' dancing skill in each pair must differ by at most one.
For each boy, we know his dancing skills. Similarly, for each girl we know her dancing skills. Write a code that can determine the largest possible number of pairs that can be formed from n boys and m girls.
Input
The first line contains an integer n (1 ≤ n ≤ 100) — the number of boys. The second line contains sequence a1, a2, ..., an (1 ≤ ai ≤ 100), where ai is the i-th boy's dancing skill.
Similarly, the third line contains an integer m (1 ≤ m ≤ 100) — the number of girls. The fourth line contains sequence b1, b2, ..., bm (1 ≤ bj ≤ 100), where bj is the j-th girl's dancing skill.
Output
Print a single number — the required maximum possible number of pairs.
Examples
Input
4
1 4 6 2
5
5 1 5 7 9
Output
3
Input
4
1 2 3 4
4
10 11 12 13
Output
0
Input
5
1 1 1 1 1
3
1 2 3
Output
2
Submitted Solution:
```
n = int(input())
b = list(map(int,input().split()))
m = int(input())
g = list(map(int,input().split()))
b.sort()
g.sort()
count = 0
l = []
for i in range (n):
for j in range (m):
if (b[i] - g[j] == 0 or abs(b[i] - g[j]) == 1) and j not in l:
count+=1
l.append(j)
break
print(count)
``` | instruction | 0 | 58,411 | 14 | 116,822 |
Yes | output | 1 | 58,411 | 14 | 116,823 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
The Berland State University is hosting a ballroom dance in celebration of its 100500-th anniversary! n boys and m girls are already busy rehearsing waltz, minuet, polonaise and quadrille moves.
We know that several boy&girl pairs are going to be invited to the ball. However, the partners' dancing skill in each pair must differ by at most one.
For each boy, we know his dancing skills. Similarly, for each girl we know her dancing skills. Write a code that can determine the largest possible number of pairs that can be formed from n boys and m girls.
Input
The first line contains an integer n (1 ≤ n ≤ 100) — the number of boys. The second line contains sequence a1, a2, ..., an (1 ≤ ai ≤ 100), where ai is the i-th boy's dancing skill.
Similarly, the third line contains an integer m (1 ≤ m ≤ 100) — the number of girls. The fourth line contains sequence b1, b2, ..., bm (1 ≤ bj ≤ 100), where bj is the j-th girl's dancing skill.
Output
Print a single number — the required maximum possible number of pairs.
Examples
Input
4
1 4 6 2
5
5 1 5 7 9
Output
3
Input
4
1 2 3 4
4
10 11 12 13
Output
0
Input
5
1 1 1 1 1
3
1 2 3
Output
2
Submitted Solution:
```
nb = int(input())
bs = input()
ng = int(input())
gs = input()
bs = [int(item) for item in bs.split()]
gs = [int(item) for item in gs.split()]
bs.sort()
gs.sort()
pairs = 0
b = 0
while b < len(bs):
g = 0
paired = False
while g < len(gs) and not paired:
if abs(gs[g]-bs[b]) <= 1:
paired = True
pairs += 1
gs.pop(g)
bs.pop(b)
else:
g += 1
if not paired:
b += 1
print(pairs)
``` | instruction | 0 | 58,412 | 14 | 116,824 |
Yes | output | 1 | 58,412 | 14 | 116,825 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
The Berland State University is hosting a ballroom dance in celebration of its 100500-th anniversary! n boys and m girls are already busy rehearsing waltz, minuet, polonaise and quadrille moves.
We know that several boy&girl pairs are going to be invited to the ball. However, the partners' dancing skill in each pair must differ by at most one.
For each boy, we know his dancing skills. Similarly, for each girl we know her dancing skills. Write a code that can determine the largest possible number of pairs that can be formed from n boys and m girls.
Input
The first line contains an integer n (1 ≤ n ≤ 100) — the number of boys. The second line contains sequence a1, a2, ..., an (1 ≤ ai ≤ 100), where ai is the i-th boy's dancing skill.
Similarly, the third line contains an integer m (1 ≤ m ≤ 100) — the number of girls. The fourth line contains sequence b1, b2, ..., bm (1 ≤ bj ≤ 100), where bj is the j-th girl's dancing skill.
Output
Print a single number — the required maximum possible number of pairs.
Examples
Input
4
1 4 6 2
5
5 1 5 7 9
Output
3
Input
4
1 2 3 4
4
10 11 12 13
Output
0
Input
5
1 1 1 1 1
3
1 2 3
Output
2
Submitted Solution:
```
n=int(input())
s=input()
l=s.split()
d1={}
for i in l:
x=int(i)
if x not in d1:
d1[x]=1
else:
d1[x]+=1
n=int(input())
s=input()
l=s.split()
d2={}
for i in l:
x=int(i)
if x not in d2:
d2[x]=1
else:
d2[x]+=1
cnt=0
for i in d1:
for j in d2:
if j-i==1 and d2[j]!=0:
d2[j]-=1
cnt+=1
if cnt==0:
print(cnt)
else:
print(cnt+1)
``` | instruction | 0 | 58,414 | 14 | 116,828 |
No | output | 1 | 58,414 | 14 | 116,829 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
The Berland State University is hosting a ballroom dance in celebration of its 100500-th anniversary! n boys and m girls are already busy rehearsing waltz, minuet, polonaise and quadrille moves.
We know that several boy&girl pairs are going to be invited to the ball. However, the partners' dancing skill in each pair must differ by at most one.
For each boy, we know his dancing skills. Similarly, for each girl we know her dancing skills. Write a code that can determine the largest possible number of pairs that can be formed from n boys and m girls.
Input
The first line contains an integer n (1 ≤ n ≤ 100) — the number of boys. The second line contains sequence a1, a2, ..., an (1 ≤ ai ≤ 100), where ai is the i-th boy's dancing skill.
Similarly, the third line contains an integer m (1 ≤ m ≤ 100) — the number of girls. The fourth line contains sequence b1, b2, ..., bm (1 ≤ bj ≤ 100), where bj is the j-th girl's dancing skill.
Output
Print a single number — the required maximum possible number of pairs.
Examples
Input
4
1 4 6 2
5
5 1 5 7 9
Output
3
Input
4
1 2 3 4
4
10 11 12 13
Output
0
Input
5
1 1 1 1 1
3
1 2 3
Output
2
Submitted Solution:
```
input()
hombres = input().split(" ")
input()
mujeres = input().split(" ")
hombres.sort()
mujeres.sort()
emparejados = 0
if len(hombres) <= len(mujeres):
for h in hombres:
for m in mujeres:
if int(h) == int(m) + 1 or int(h) == int(m) or int(h) == int(m) - 1:
emparejados += 1
mujeres.remove(m)
break
else:
for m in mujeres:
for h in hombres:
if int(m) == int(h) + 1 or int(m) == int(h) or int(m) == int(h) - 1:
emparejados += 1
hombres.remove(h)
break
print(emparejados)
``` | instruction | 0 | 58,415 | 14 | 116,830 |
No | output | 1 | 58,415 | 14 | 116,831 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
The Berland State University is hosting a ballroom dance in celebration of its 100500-th anniversary! n boys and m girls are already busy rehearsing waltz, minuet, polonaise and quadrille moves.
We know that several boy&girl pairs are going to be invited to the ball. However, the partners' dancing skill in each pair must differ by at most one.
For each boy, we know his dancing skills. Similarly, for each girl we know her dancing skills. Write a code that can determine the largest possible number of pairs that can be formed from n boys and m girls.
Input
The first line contains an integer n (1 ≤ n ≤ 100) — the number of boys. The second line contains sequence a1, a2, ..., an (1 ≤ ai ≤ 100), where ai is the i-th boy's dancing skill.
Similarly, the third line contains an integer m (1 ≤ m ≤ 100) — the number of girls. The fourth line contains sequence b1, b2, ..., bm (1 ≤ bj ≤ 100), where bj is the j-th girl's dancing skill.
Output
Print a single number — the required maximum possible number of pairs.
Examples
Input
4
1 4 6 2
5
5 1 5 7 9
Output
3
Input
4
1 2 3 4
4
10 11 12 13
Output
0
Input
5
1 1 1 1 1
3
1 2 3
Output
2
Submitted Solution:
```
n = int(input())
boys = input().split()
for i in range(n):
boys[i] = int(boys[i])
m = int(input())
girls = input().split()
for i in range(m):
girls[i] = int(girls[i])
boys.sort()
girls.sort()
pairs = 0
while len(girls)>0 and len(boys)>0:
if abs(girls[0] - boys[0])<=1:
pairs+=1
del girls[0]
del boys[0]
else:
del boys[0]
del girls[0]
print(pairs)
``` | instruction | 0 | 58,416 | 14 | 116,832 |
No | output | 1 | 58,416 | 14 | 116,833 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
The Berland State University is hosting a ballroom dance in celebration of its 100500-th anniversary! n boys and m girls are already busy rehearsing waltz, minuet, polonaise and quadrille moves.
We know that several boy&girl pairs are going to be invited to the ball. However, the partners' dancing skill in each pair must differ by at most one.
For each boy, we know his dancing skills. Similarly, for each girl we know her dancing skills. Write a code that can determine the largest possible number of pairs that can be formed from n boys and m girls.
Input
The first line contains an integer n (1 ≤ n ≤ 100) — the number of boys. The second line contains sequence a1, a2, ..., an (1 ≤ ai ≤ 100), where ai is the i-th boy's dancing skill.
Similarly, the third line contains an integer m (1 ≤ m ≤ 100) — the number of girls. The fourth line contains sequence b1, b2, ..., bm (1 ≤ bj ≤ 100), where bj is the j-th girl's dancing skill.
Output
Print a single number — the required maximum possible number of pairs.
Examples
Input
4
1 4 6 2
5
5 1 5 7 9
Output
3
Input
4
1 2 3 4
4
10 11 12 13
Output
0
Input
5
1 1 1 1 1
3
1 2 3
Output
2
Submitted Solution:
```
b=int(input())
x=list(map(int,input().split()))
g=int(input())
y=list(map(int,input().split()))
x.sort()
y.sort()
c=0
d=0
if(x[0]<=y[0]):
for i in range(b):
if (x[i] in y):
y.remove(x[i])
c+=1
if(x[i]+1 in y):
y.remove(x[i]+1)
c+=1
if(x[i]-1 in y):
y.remove(x[i]-1)
c+=1
#print(c)
else:
for i in range(g):
if (y[i] in x):
x.remove(y[i])
c+=1
if(y[i]+1 in x):
x.remove(y[i]+1)
c+=1
if(y[i]-1 in x):
x.remove(y[i]-1)
c+=1
#print(c)
print(c)
``` | instruction | 0 | 58,417 | 14 | 116,834 |
No | output | 1 | 58,417 | 14 | 116,835 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Limak is a little polar bear. He loves connecting with other bears via social networks. He has n friends and his relation with the i-th of them is described by a unique integer ti. The bigger this value is, the better the friendship is. No two friends have the same value ti.
Spring is starting and the Winter sleep is over for bears. Limak has just woken up and logged in. All his friends still sleep and thus none of them is online. Some (maybe all) of them will appear online in the next hours, one at a time.
The system displays friends who are online. On the screen there is space to display at most k friends. If there are more than k friends online then the system displays only k best of them — those with biggest ti.
Your task is to handle queries of two types:
* "1 id" — Friend id becomes online. It's guaranteed that he wasn't online before.
* "2 id" — Check whether friend id is displayed by the system. Print "YES" or "NO" in a separate line.
Are you able to help Limak and answer all queries of the second type?
Input
The first line contains three integers n, k and q (1 ≤ n, q ≤ 150 000, 1 ≤ k ≤ min(6, n)) — the number of friends, the maximum number of displayed online friends and the number of queries, respectively.
The second line contains n integers t1, t2, ..., tn (1 ≤ ti ≤ 109) where ti describes how good is Limak's relation with the i-th friend.
The i-th of the following q lines contains two integers typei and idi (1 ≤ typei ≤ 2, 1 ≤ idi ≤ n) — the i-th query. If typei = 1 then a friend idi becomes online. If typei = 2 then you should check whether a friend idi is displayed.
It's guaranteed that no two queries of the first type will have the same idi becuase one friend can't become online twice. Also, it's guaranteed that at least one query will be of the second type (typei = 2) so the output won't be empty.
Output
For each query of the second type print one line with the answer — "YES" (without quotes) if the given friend is displayed and "NO" (without quotes) otherwise.
Examples
Input
4 2 8
300 950 500 200
1 3
2 4
2 3
1 1
1 2
2 1
2 2
2 3
Output
NO
YES
NO
YES
YES
Input
6 3 9
50 20 51 17 99 24
1 3
1 4
1 5
1 2
2 4
2 2
1 1
2 4
2 3
Output
NO
YES
NO
YES
Note
In the first sample, Limak has 4 friends who all sleep initially. At first, the system displays nobody because nobody is online. There are the following 8 queries:
1. "1 3" — Friend 3 becomes online.
2. "2 4" — We should check if friend 4 is displayed. He isn't even online and thus we print "NO".
3. "2 3" — We should check if friend 3 is displayed. Right now he is the only friend online and the system displays him. We should print "YES".
4. "1 1" — Friend 1 becomes online. The system now displays both friend 1 and friend 3.
5. "1 2" — Friend 2 becomes online. There are 3 friends online now but we were given k = 2 so only two friends can be displayed. Limak has worse relation with friend 1 than with other two online friends (t1 < t2, t3) so friend 1 won't be displayed
6. "2 1" — Print "NO".
7. "2 2" — Print "YES".
8. "2 3" — Print "YES". | instruction | 0 | 58,469 | 14 | 116,938 |
Tags: implementation
Correct Solution:
```
n,k,q=input().split()
rank=input().split()
for i in range(len(rank)):
rank[i]=int(rank[i])
li=[]
names=[]
for i in range(int(q)):
query=input().split()
if(query[0]=='1'):
if(len(li)<int(k)):
names.append(query[1])
li.append(rank[int(query[1])-1])
else:
m=min(li)
ind=li.index(m)
if(rank[int(query[1])-1]>m):
del li[ind]
del names[ind]
names.append(query[1])
li.append(rank[int(query[1])-1])
if(query[0]=='2'):
if(query[1] in names):
print("YES")
else:
print("NO")
``` | output | 1 | 58,469 | 14 | 116,939 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Limak is a little polar bear. He loves connecting with other bears via social networks. He has n friends and his relation with the i-th of them is described by a unique integer ti. The bigger this value is, the better the friendship is. No two friends have the same value ti.
Spring is starting and the Winter sleep is over for bears. Limak has just woken up and logged in. All his friends still sleep and thus none of them is online. Some (maybe all) of them will appear online in the next hours, one at a time.
The system displays friends who are online. On the screen there is space to display at most k friends. If there are more than k friends online then the system displays only k best of them — those with biggest ti.
Your task is to handle queries of two types:
* "1 id" — Friend id becomes online. It's guaranteed that he wasn't online before.
* "2 id" — Check whether friend id is displayed by the system. Print "YES" or "NO" in a separate line.
Are you able to help Limak and answer all queries of the second type?
Input
The first line contains three integers n, k and q (1 ≤ n, q ≤ 150 000, 1 ≤ k ≤ min(6, n)) — the number of friends, the maximum number of displayed online friends and the number of queries, respectively.
The second line contains n integers t1, t2, ..., tn (1 ≤ ti ≤ 109) where ti describes how good is Limak's relation with the i-th friend.
The i-th of the following q lines contains two integers typei and idi (1 ≤ typei ≤ 2, 1 ≤ idi ≤ n) — the i-th query. If typei = 1 then a friend idi becomes online. If typei = 2 then you should check whether a friend idi is displayed.
It's guaranteed that no two queries of the first type will have the same idi becuase one friend can't become online twice. Also, it's guaranteed that at least one query will be of the second type (typei = 2) so the output won't be empty.
Output
For each query of the second type print one line with the answer — "YES" (without quotes) if the given friend is displayed and "NO" (without quotes) otherwise.
Examples
Input
4 2 8
300 950 500 200
1 3
2 4
2 3
1 1
1 2
2 1
2 2
2 3
Output
NO
YES
NO
YES
YES
Input
6 3 9
50 20 51 17 99 24
1 3
1 4
1 5
1 2
2 4
2 2
1 1
2 4
2 3
Output
NO
YES
NO
YES
Note
In the first sample, Limak has 4 friends who all sleep initially. At first, the system displays nobody because nobody is online. There are the following 8 queries:
1. "1 3" — Friend 3 becomes online.
2. "2 4" — We should check if friend 4 is displayed. He isn't even online and thus we print "NO".
3. "2 3" — We should check if friend 3 is displayed. Right now he is the only friend online and the system displays him. We should print "YES".
4. "1 1" — Friend 1 becomes online. The system now displays both friend 1 and friend 3.
5. "1 2" — Friend 2 becomes online. There are 3 friends online now but we were given k = 2 so only two friends can be displayed. Limak has worse relation with friend 1 than with other two online friends (t1 < t2, t3) so friend 1 won't be displayed
6. "2 1" — Print "NO".
7. "2 2" — Print "YES".
8. "2 3" — Print "YES". | instruction | 0 | 58,470 | 14 | 116,940 |
Tags: implementation
Correct Solution:
```
def Core(size, rate, data):
online_rate = []
count = 0
for item in data:
if item[0] == 1:
if size > count:
count += 1
online_rate.append(rate[item[1]])
online_rate.sort(key=lambda x: -x)
elif online_rate[-1] < rate[item[1]]:
online_rate[-1] = rate[item[1]]
online_rate.sort(key=lambda x: -x)
else:
if rate[item[1]] in online_rate and online_rate.index(rate[item[1]]) < size:
print("YES")
else:
print("NO")
param1 = input().split(" ")
param2 = [0] + [int(i) for i in input().split(" ")]
param3 = [[int(j) for j in input().split(" ")] for i in range(int(param1[2]))]
Core(int(param1[1]), param2, param3)
``` | output | 1 | 58,470 | 14 | 116,941 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Limak is a little polar bear. He loves connecting with other bears via social networks. He has n friends and his relation with the i-th of them is described by a unique integer ti. The bigger this value is, the better the friendship is. No two friends have the same value ti.
Spring is starting and the Winter sleep is over for bears. Limak has just woken up and logged in. All his friends still sleep and thus none of them is online. Some (maybe all) of them will appear online in the next hours, one at a time.
The system displays friends who are online. On the screen there is space to display at most k friends. If there are more than k friends online then the system displays only k best of them — those with biggest ti.
Your task is to handle queries of two types:
* "1 id" — Friend id becomes online. It's guaranteed that he wasn't online before.
* "2 id" — Check whether friend id is displayed by the system. Print "YES" or "NO" in a separate line.
Are you able to help Limak and answer all queries of the second type?
Input
The first line contains three integers n, k and q (1 ≤ n, q ≤ 150 000, 1 ≤ k ≤ min(6, n)) — the number of friends, the maximum number of displayed online friends and the number of queries, respectively.
The second line contains n integers t1, t2, ..., tn (1 ≤ ti ≤ 109) where ti describes how good is Limak's relation with the i-th friend.
The i-th of the following q lines contains two integers typei and idi (1 ≤ typei ≤ 2, 1 ≤ idi ≤ n) — the i-th query. If typei = 1 then a friend idi becomes online. If typei = 2 then you should check whether a friend idi is displayed.
It's guaranteed that no two queries of the first type will have the same idi becuase one friend can't become online twice. Also, it's guaranteed that at least one query will be of the second type (typei = 2) so the output won't be empty.
Output
For each query of the second type print one line with the answer — "YES" (without quotes) if the given friend is displayed and "NO" (without quotes) otherwise.
Examples
Input
4 2 8
300 950 500 200
1 3
2 4
2 3
1 1
1 2
2 1
2 2
2 3
Output
NO
YES
NO
YES
YES
Input
6 3 9
50 20 51 17 99 24
1 3
1 4
1 5
1 2
2 4
2 2
1 1
2 4
2 3
Output
NO
YES
NO
YES
Note
In the first sample, Limak has 4 friends who all sleep initially. At first, the system displays nobody because nobody is online. There are the following 8 queries:
1. "1 3" — Friend 3 becomes online.
2. "2 4" — We should check if friend 4 is displayed. He isn't even online and thus we print "NO".
3. "2 3" — We should check if friend 3 is displayed. Right now he is the only friend online and the system displays him. We should print "YES".
4. "1 1" — Friend 1 becomes online. The system now displays both friend 1 and friend 3.
5. "1 2" — Friend 2 becomes online. There are 3 friends online now but we were given k = 2 so only two friends can be displayed. Limak has worse relation with friend 1 than with other two online friends (t1 < t2, t3) so friend 1 won't be displayed
6. "2 1" — Print "NO".
7. "2 2" — Print "YES".
8. "2 3" — Print "YES". | instruction | 0 | 58,471 | 14 | 116,942 |
Tags: implementation
Correct Solution:
```
import os
import sys
import math
import heapq
from decimal import *
from io import BytesIO, IOBase
from collections import defaultdict, deque
def main():
n,k,q = rm()
a = rl()
b = []
for i in range(q):
c,d=rm()
if c==1:
d-=1
b.append([a[d],d])
b.sort(reverse=True)
b = b[:k]
elif c==2:
d-=1
ans = False
for i in b:
if i[1]==d:
ans=True
break
if ans:
print("YES")
else:
print("NO")
# region fastio
BUFSIZE = 8192
class FastIO(IOBase):
newlines = 0
def __init__(self, file):
self._fd = file.fileno()
self.buffer = BytesIO()
self.writable = "x" in file.mode or "r" not in file.mode
self.write = self.buffer.write if self.writable else None
def read(self):
while True:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
if not b:
break
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines = 0
return self.buffer.read()
def readline(self):
while self.newlines == 0:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
self.newlines = b.count(b"\n") + (not b)
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines -= 1
return self.buffer.readline()
def flush(self):
if self.writable:
os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
self.write = lambda s: self.buffer.write(s.encode("ascii"))
self.read = lambda: self.buffer.read().decode("ascii")
self.readline = lambda: self.buffer.readline().decode("ascii")
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")
# endregion
def r():
return int(input())
def rm():
return map(int,input().split())
def rl():
return list(map(int,input().split()))
main()
``` | output | 1 | 58,471 | 14 | 116,943 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Limak is a little polar bear. He loves connecting with other bears via social networks. He has n friends and his relation with the i-th of them is described by a unique integer ti. The bigger this value is, the better the friendship is. No two friends have the same value ti.
Spring is starting and the Winter sleep is over for bears. Limak has just woken up and logged in. All his friends still sleep and thus none of them is online. Some (maybe all) of them will appear online in the next hours, one at a time.
The system displays friends who are online. On the screen there is space to display at most k friends. If there are more than k friends online then the system displays only k best of them — those with biggest ti.
Your task is to handle queries of two types:
* "1 id" — Friend id becomes online. It's guaranteed that he wasn't online before.
* "2 id" — Check whether friend id is displayed by the system. Print "YES" or "NO" in a separate line.
Are you able to help Limak and answer all queries of the second type?
Input
The first line contains three integers n, k and q (1 ≤ n, q ≤ 150 000, 1 ≤ k ≤ min(6, n)) — the number of friends, the maximum number of displayed online friends and the number of queries, respectively.
The second line contains n integers t1, t2, ..., tn (1 ≤ ti ≤ 109) where ti describes how good is Limak's relation with the i-th friend.
The i-th of the following q lines contains two integers typei and idi (1 ≤ typei ≤ 2, 1 ≤ idi ≤ n) — the i-th query. If typei = 1 then a friend idi becomes online. If typei = 2 then you should check whether a friend idi is displayed.
It's guaranteed that no two queries of the first type will have the same idi becuase one friend can't become online twice. Also, it's guaranteed that at least one query will be of the second type (typei = 2) so the output won't be empty.
Output
For each query of the second type print one line with the answer — "YES" (without quotes) if the given friend is displayed and "NO" (without quotes) otherwise.
Examples
Input
4 2 8
300 950 500 200
1 3
2 4
2 3
1 1
1 2
2 1
2 2
2 3
Output
NO
YES
NO
YES
YES
Input
6 3 9
50 20 51 17 99 24
1 3
1 4
1 5
1 2
2 4
2 2
1 1
2 4
2 3
Output
NO
YES
NO
YES
Note
In the first sample, Limak has 4 friends who all sleep initially. At first, the system displays nobody because nobody is online. There are the following 8 queries:
1. "1 3" — Friend 3 becomes online.
2. "2 4" — We should check if friend 4 is displayed. He isn't even online and thus we print "NO".
3. "2 3" — We should check if friend 3 is displayed. Right now he is the only friend online and the system displays him. We should print "YES".
4. "1 1" — Friend 1 becomes online. The system now displays both friend 1 and friend 3.
5. "1 2" — Friend 2 becomes online. There are 3 friends online now but we were given k = 2 so only two friends can be displayed. Limak has worse relation with friend 1 than with other two online friends (t1 < t2, t3) so friend 1 won't be displayed
6. "2 1" — Print "NO".
7. "2 2" — Print "YES".
8. "2 3" — Print "YES". | instruction | 0 | 58,472 | 14 | 116,944 |
Tags: implementation
Correct Solution:
```
n=input()
b=[]
b=n.split()
n=int(b[0])
k=int(b[1])
q=int(b[2])
c=input()
level=c.split()
online=[]
for i in range(q):
c=input()
z=c.split()
if z[0]=='1':
if len(online)<k:
online.append(int(z[1]))
else:
min=int(level[int(z[1])-1])
number=-1
for j in range(k):
if int(level[online[j]-1])<min:
min=int(level[online[j]-1])
number=j
if number>-1:
online[number]=int(z[1])
else:
if online.count(int(z[1]))>0:
print('YES')
else:
print('NO')
``` | output | 1 | 58,472 | 14 | 116,945 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Limak is a little polar bear. He loves connecting with other bears via social networks. He has n friends and his relation with the i-th of them is described by a unique integer ti. The bigger this value is, the better the friendship is. No two friends have the same value ti.
Spring is starting and the Winter sleep is over for bears. Limak has just woken up and logged in. All his friends still sleep and thus none of them is online. Some (maybe all) of them will appear online in the next hours, one at a time.
The system displays friends who are online. On the screen there is space to display at most k friends. If there are more than k friends online then the system displays only k best of them — those with biggest ti.
Your task is to handle queries of two types:
* "1 id" — Friend id becomes online. It's guaranteed that he wasn't online before.
* "2 id" — Check whether friend id is displayed by the system. Print "YES" or "NO" in a separate line.
Are you able to help Limak and answer all queries of the second type?
Input
The first line contains three integers n, k and q (1 ≤ n, q ≤ 150 000, 1 ≤ k ≤ min(6, n)) — the number of friends, the maximum number of displayed online friends and the number of queries, respectively.
The second line contains n integers t1, t2, ..., tn (1 ≤ ti ≤ 109) where ti describes how good is Limak's relation with the i-th friend.
The i-th of the following q lines contains two integers typei and idi (1 ≤ typei ≤ 2, 1 ≤ idi ≤ n) — the i-th query. If typei = 1 then a friend idi becomes online. If typei = 2 then you should check whether a friend idi is displayed.
It's guaranteed that no two queries of the first type will have the same idi becuase one friend can't become online twice. Also, it's guaranteed that at least one query will be of the second type (typei = 2) so the output won't be empty.
Output
For each query of the second type print one line with the answer — "YES" (without quotes) if the given friend is displayed and "NO" (without quotes) otherwise.
Examples
Input
4 2 8
300 950 500 200
1 3
2 4
2 3
1 1
1 2
2 1
2 2
2 3
Output
NO
YES
NO
YES
YES
Input
6 3 9
50 20 51 17 99 24
1 3
1 4
1 5
1 2
2 4
2 2
1 1
2 4
2 3
Output
NO
YES
NO
YES
Note
In the first sample, Limak has 4 friends who all sleep initially. At first, the system displays nobody because nobody is online. There are the following 8 queries:
1. "1 3" — Friend 3 becomes online.
2. "2 4" — We should check if friend 4 is displayed. He isn't even online and thus we print "NO".
3. "2 3" — We should check if friend 3 is displayed. Right now he is the only friend online and the system displays him. We should print "YES".
4. "1 1" — Friend 1 becomes online. The system now displays both friend 1 and friend 3.
5. "1 2" — Friend 2 becomes online. There are 3 friends online now but we were given k = 2 so only two friends can be displayed. Limak has worse relation with friend 1 than with other two online friends (t1 < t2, t3) so friend 1 won't be displayed
6. "2 1" — Print "NO".
7. "2 2" — Print "YES".
8. "2 3" — Print "YES". | instruction | 0 | 58,473 | 14 | 116,946 |
Tags: implementation
Correct Solution:
```
from heapq import *
n, f, q = [int(s) for s in input().split()]
loves = [int(s) for s in input().split()]
k = []
onlines = set()
for i in range(q):
ind, num = [int(s) for s in input().split()]
if ind == 1:
onlines.add(num - 1)
if len(k)<f:
heappush(k, loves[num - 1])
else:
heappushpop(k, loves[num-1])
else:
if num-1 in onlines and k[0] <= loves[num-1]:
print("YES")
else:
print("NO")
``` | output | 1 | 58,473 | 14 | 116,947 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Limak is a little polar bear. He loves connecting with other bears via social networks. He has n friends and his relation with the i-th of them is described by a unique integer ti. The bigger this value is, the better the friendship is. No two friends have the same value ti.
Spring is starting and the Winter sleep is over for bears. Limak has just woken up and logged in. All his friends still sleep and thus none of them is online. Some (maybe all) of them will appear online in the next hours, one at a time.
The system displays friends who are online. On the screen there is space to display at most k friends. If there are more than k friends online then the system displays only k best of them — those with biggest ti.
Your task is to handle queries of two types:
* "1 id" — Friend id becomes online. It's guaranteed that he wasn't online before.
* "2 id" — Check whether friend id is displayed by the system. Print "YES" or "NO" in a separate line.
Are you able to help Limak and answer all queries of the second type?
Input
The first line contains three integers n, k and q (1 ≤ n, q ≤ 150 000, 1 ≤ k ≤ min(6, n)) — the number of friends, the maximum number of displayed online friends and the number of queries, respectively.
The second line contains n integers t1, t2, ..., tn (1 ≤ ti ≤ 109) where ti describes how good is Limak's relation with the i-th friend.
The i-th of the following q lines contains two integers typei and idi (1 ≤ typei ≤ 2, 1 ≤ idi ≤ n) — the i-th query. If typei = 1 then a friend idi becomes online. If typei = 2 then you should check whether a friend idi is displayed.
It's guaranteed that no two queries of the first type will have the same idi becuase one friend can't become online twice. Also, it's guaranteed that at least one query will be of the second type (typei = 2) so the output won't be empty.
Output
For each query of the second type print one line with the answer — "YES" (without quotes) if the given friend is displayed and "NO" (without quotes) otherwise.
Examples
Input
4 2 8
300 950 500 200
1 3
2 4
2 3
1 1
1 2
2 1
2 2
2 3
Output
NO
YES
NO
YES
YES
Input
6 3 9
50 20 51 17 99 24
1 3
1 4
1 5
1 2
2 4
2 2
1 1
2 4
2 3
Output
NO
YES
NO
YES
Note
In the first sample, Limak has 4 friends who all sleep initially. At first, the system displays nobody because nobody is online. There are the following 8 queries:
1. "1 3" — Friend 3 becomes online.
2. "2 4" — We should check if friend 4 is displayed. He isn't even online and thus we print "NO".
3. "2 3" — We should check if friend 3 is displayed. Right now he is the only friend online and the system displays him. We should print "YES".
4. "1 1" — Friend 1 becomes online. The system now displays both friend 1 and friend 3.
5. "1 2" — Friend 2 becomes online. There are 3 friends online now but we were given k = 2 so only two friends can be displayed. Limak has worse relation with friend 1 than with other two online friends (t1 < t2, t3) so friend 1 won't be displayed
6. "2 1" — Print "NO".
7. "2 2" — Print "YES".
8. "2 3" — Print "YES". | instruction | 0 | 58,474 | 14 | 116,948 |
Tags: implementation
Correct Solution:
```
n, k, q = [int(i) for i in input().split()]
t = input().split()
ID = {}
online = []
for i in range(len(t)):
ID[i+1] = t[i]
for i in range(q):
a = list(map(int,input().split()))
if a[0] == 1:
online.append([a[1],int(ID[a[1]])])
if len(online) > k:
online.sort(key = lambda x:x[1])
del online[0]
if a[0] == 2:
if [a[1],int(ID[a[1]])] in online:
print('YES')
else:
print('NO')
``` | output | 1 | 58,474 | 14 | 116,949 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Limak is a little polar bear. He loves connecting with other bears via social networks. He has n friends and his relation with the i-th of them is described by a unique integer ti. The bigger this value is, the better the friendship is. No two friends have the same value ti.
Spring is starting and the Winter sleep is over for bears. Limak has just woken up and logged in. All his friends still sleep and thus none of them is online. Some (maybe all) of them will appear online in the next hours, one at a time.
The system displays friends who are online. On the screen there is space to display at most k friends. If there are more than k friends online then the system displays only k best of them — those with biggest ti.
Your task is to handle queries of two types:
* "1 id" — Friend id becomes online. It's guaranteed that he wasn't online before.
* "2 id" — Check whether friend id is displayed by the system. Print "YES" or "NO" in a separate line.
Are you able to help Limak and answer all queries of the second type?
Input
The first line contains three integers n, k and q (1 ≤ n, q ≤ 150 000, 1 ≤ k ≤ min(6, n)) — the number of friends, the maximum number of displayed online friends and the number of queries, respectively.
The second line contains n integers t1, t2, ..., tn (1 ≤ ti ≤ 109) where ti describes how good is Limak's relation with the i-th friend.
The i-th of the following q lines contains two integers typei and idi (1 ≤ typei ≤ 2, 1 ≤ idi ≤ n) — the i-th query. If typei = 1 then a friend idi becomes online. If typei = 2 then you should check whether a friend idi is displayed.
It's guaranteed that no two queries of the first type will have the same idi becuase one friend can't become online twice. Also, it's guaranteed that at least one query will be of the second type (typei = 2) so the output won't be empty.
Output
For each query of the second type print one line with the answer — "YES" (without quotes) if the given friend is displayed and "NO" (without quotes) otherwise.
Examples
Input
4 2 8
300 950 500 200
1 3
2 4
2 3
1 1
1 2
2 1
2 2
2 3
Output
NO
YES
NO
YES
YES
Input
6 3 9
50 20 51 17 99 24
1 3
1 4
1 5
1 2
2 4
2 2
1 1
2 4
2 3
Output
NO
YES
NO
YES
Note
In the first sample, Limak has 4 friends who all sleep initially. At first, the system displays nobody because nobody is online. There are the following 8 queries:
1. "1 3" — Friend 3 becomes online.
2. "2 4" — We should check if friend 4 is displayed. He isn't even online and thus we print "NO".
3. "2 3" — We should check if friend 3 is displayed. Right now he is the only friend online and the system displays him. We should print "YES".
4. "1 1" — Friend 1 becomes online. The system now displays both friend 1 and friend 3.
5. "1 2" — Friend 2 becomes online. There are 3 friends online now but we were given k = 2 so only two friends can be displayed. Limak has worse relation with friend 1 than with other two online friends (t1 < t2, t3) so friend 1 won't be displayed
6. "2 1" — Print "NO".
7. "2 2" — Print "YES".
8. "2 3" — Print "YES". | instruction | 0 | 58,475 | 14 | 116,950 |
Tags: implementation
Correct Solution:
```
def type1(idi):
global mnoj, n, k, q, t
if len(mnoj) < k:
mnoj.add(t[idi])
else:
mn = min(mnoj)
if t[idi] > mn:
mnoj.remove(mn)
mnoj.add(t[idi])
def main():
global mnoj, n, k, q, t
n, k, q = map(int, input().split())
t = list(map(int, input().split()))
mnoj = set()
for i in range(q):
typei, idi = map(int, input().split())
idi -= 1
if typei == 1:
type1(idi)
else:
if t[idi] in mnoj:
print("YES")
else:
print("NO")
main()
``` | output | 1 | 58,475 | 14 | 116,951 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Limak is a little polar bear. He loves connecting with other bears via social networks. He has n friends and his relation with the i-th of them is described by a unique integer ti. The bigger this value is, the better the friendship is. No two friends have the same value ti.
Spring is starting and the Winter sleep is over for bears. Limak has just woken up and logged in. All his friends still sleep and thus none of them is online. Some (maybe all) of them will appear online in the next hours, one at a time.
The system displays friends who are online. On the screen there is space to display at most k friends. If there are more than k friends online then the system displays only k best of them — those with biggest ti.
Your task is to handle queries of two types:
* "1 id" — Friend id becomes online. It's guaranteed that he wasn't online before.
* "2 id" — Check whether friend id is displayed by the system. Print "YES" or "NO" in a separate line.
Are you able to help Limak and answer all queries of the second type?
Input
The first line contains three integers n, k and q (1 ≤ n, q ≤ 150 000, 1 ≤ k ≤ min(6, n)) — the number of friends, the maximum number of displayed online friends and the number of queries, respectively.
The second line contains n integers t1, t2, ..., tn (1 ≤ ti ≤ 109) where ti describes how good is Limak's relation with the i-th friend.
The i-th of the following q lines contains two integers typei and idi (1 ≤ typei ≤ 2, 1 ≤ idi ≤ n) — the i-th query. If typei = 1 then a friend idi becomes online. If typei = 2 then you should check whether a friend idi is displayed.
It's guaranteed that no two queries of the first type will have the same idi becuase one friend can't become online twice. Also, it's guaranteed that at least one query will be of the second type (typei = 2) so the output won't be empty.
Output
For each query of the second type print one line with the answer — "YES" (without quotes) if the given friend is displayed and "NO" (without quotes) otherwise.
Examples
Input
4 2 8
300 950 500 200
1 3
2 4
2 3
1 1
1 2
2 1
2 2
2 3
Output
NO
YES
NO
YES
YES
Input
6 3 9
50 20 51 17 99 24
1 3
1 4
1 5
1 2
2 4
2 2
1 1
2 4
2 3
Output
NO
YES
NO
YES
Note
In the first sample, Limak has 4 friends who all sleep initially. At first, the system displays nobody because nobody is online. There are the following 8 queries:
1. "1 3" — Friend 3 becomes online.
2. "2 4" — We should check if friend 4 is displayed. He isn't even online and thus we print "NO".
3. "2 3" — We should check if friend 3 is displayed. Right now he is the only friend online and the system displays him. We should print "YES".
4. "1 1" — Friend 1 becomes online. The system now displays both friend 1 and friend 3.
5. "1 2" — Friend 2 becomes online. There are 3 friends online now but we were given k = 2 so only two friends can be displayed. Limak has worse relation with friend 1 than with other two online friends (t1 < t2, t3) so friend 1 won't be displayed
6. "2 1" — Print "NO".
7. "2 2" — Print "YES".
8. "2 3" — Print "YES". | instruction | 0 | 58,476 | 14 | 116,952 |
Tags: implementation
Correct Solution:
```
n, size, tasks = map(int, input().split())
friends = list(map(int, input().split()))
online = list()
for i in range(tasks):
type, id = map(int, input().split())
id -= 1
if type == 1:
adds = friends[id]
online = sorted(online + [adds], reverse=True)
if len(online) > size:
del online[-1]
else:
flag_oh_my_god_why = False
search = friends[id]
for bear in online:
if bear == search:
flag_oh_my_god_why = True
break
if flag_oh_my_god_why == False: print('NO')
else: print('YES')
``` | output | 1 | 58,476 | 14 | 116,953 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Limak is a little polar bear. He loves connecting with other bears via social networks. He has n friends and his relation with the i-th of them is described by a unique integer ti. The bigger this value is, the better the friendship is. No two friends have the same value ti.
Spring is starting and the Winter sleep is over for bears. Limak has just woken up and logged in. All his friends still sleep and thus none of them is online. Some (maybe all) of them will appear online in the next hours, one at a time.
The system displays friends who are online. On the screen there is space to display at most k friends. If there are more than k friends online then the system displays only k best of them — those with biggest ti.
Your task is to handle queries of two types:
* "1 id" — Friend id becomes online. It's guaranteed that he wasn't online before.
* "2 id" — Check whether friend id is displayed by the system. Print "YES" or "NO" in a separate line.
Are you able to help Limak and answer all queries of the second type?
Input
The first line contains three integers n, k and q (1 ≤ n, q ≤ 150 000, 1 ≤ k ≤ min(6, n)) — the number of friends, the maximum number of displayed online friends and the number of queries, respectively.
The second line contains n integers t1, t2, ..., tn (1 ≤ ti ≤ 109) where ti describes how good is Limak's relation with the i-th friend.
The i-th of the following q lines contains two integers typei and idi (1 ≤ typei ≤ 2, 1 ≤ idi ≤ n) — the i-th query. If typei = 1 then a friend idi becomes online. If typei = 2 then you should check whether a friend idi is displayed.
It's guaranteed that no two queries of the first type will have the same idi becuase one friend can't become online twice. Also, it's guaranteed that at least one query will be of the second type (typei = 2) so the output won't be empty.
Output
For each query of the second type print one line with the answer — "YES" (without quotes) if the given friend is displayed and "NO" (without quotes) otherwise.
Examples
Input
4 2 8
300 950 500 200
1 3
2 4
2 3
1 1
1 2
2 1
2 2
2 3
Output
NO
YES
NO
YES
YES
Input
6 3 9
50 20 51 17 99 24
1 3
1 4
1 5
1 2
2 4
2 2
1 1
2 4
2 3
Output
NO
YES
NO
YES
Note
In the first sample, Limak has 4 friends who all sleep initially. At first, the system displays nobody because nobody is online. There are the following 8 queries:
1. "1 3" — Friend 3 becomes online.
2. "2 4" — We should check if friend 4 is displayed. He isn't even online and thus we print "NO".
3. "2 3" — We should check if friend 3 is displayed. Right now he is the only friend online and the system displays him. We should print "YES".
4. "1 1" — Friend 1 becomes online. The system now displays both friend 1 and friend 3.
5. "1 2" — Friend 2 becomes online. There are 3 friends online now but we were given k = 2 so only two friends can be displayed. Limak has worse relation with friend 1 than with other two online friends (t1 < t2, t3) so friend 1 won't be displayed
6. "2 1" — Print "NO".
7. "2 2" — Print "YES".
8. "2 3" — Print "YES".
Submitted Solution:
```
n,k,q=map(int,input().split())
t=list(map(int,input().split()))
ls=[]
acti=[-1 for i in range(n)]
while q>0:
q-=1
a,b=map(int,input().split())
b-=1
if a==1 and acti[b]==-1 :
acti[b]=1
ls.append([t[b],b])
ls.sort(reverse=True)
if len(ls)>k:
ls=ls[0:k]
continue
if a==2 :
for j in ls:
if j[1]==b:
print("YES")
break
else :
print("NO")
``` | instruction | 0 | 58,477 | 14 | 116,954 |
Yes | output | 1 | 58,477 | 14 | 116,955 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Limak is a little polar bear. He loves connecting with other bears via social networks. He has n friends and his relation with the i-th of them is described by a unique integer ti. The bigger this value is, the better the friendship is. No two friends have the same value ti.
Spring is starting and the Winter sleep is over for bears. Limak has just woken up and logged in. All his friends still sleep and thus none of them is online. Some (maybe all) of them will appear online in the next hours, one at a time.
The system displays friends who are online. On the screen there is space to display at most k friends. If there are more than k friends online then the system displays only k best of them — those with biggest ti.
Your task is to handle queries of two types:
* "1 id" — Friend id becomes online. It's guaranteed that he wasn't online before.
* "2 id" — Check whether friend id is displayed by the system. Print "YES" or "NO" in a separate line.
Are you able to help Limak and answer all queries of the second type?
Input
The first line contains three integers n, k and q (1 ≤ n, q ≤ 150 000, 1 ≤ k ≤ min(6, n)) — the number of friends, the maximum number of displayed online friends and the number of queries, respectively.
The second line contains n integers t1, t2, ..., tn (1 ≤ ti ≤ 109) where ti describes how good is Limak's relation with the i-th friend.
The i-th of the following q lines contains two integers typei and idi (1 ≤ typei ≤ 2, 1 ≤ idi ≤ n) — the i-th query. If typei = 1 then a friend idi becomes online. If typei = 2 then you should check whether a friend idi is displayed.
It's guaranteed that no two queries of the first type will have the same idi becuase one friend can't become online twice. Also, it's guaranteed that at least one query will be of the second type (typei = 2) so the output won't be empty.
Output
For each query of the second type print one line with the answer — "YES" (without quotes) if the given friend is displayed and "NO" (without quotes) otherwise.
Examples
Input
4 2 8
300 950 500 200
1 3
2 4
2 3
1 1
1 2
2 1
2 2
2 3
Output
NO
YES
NO
YES
YES
Input
6 3 9
50 20 51 17 99 24
1 3
1 4
1 5
1 2
2 4
2 2
1 1
2 4
2 3
Output
NO
YES
NO
YES
Note
In the first sample, Limak has 4 friends who all sleep initially. At first, the system displays nobody because nobody is online. There are the following 8 queries:
1. "1 3" — Friend 3 becomes online.
2. "2 4" — We should check if friend 4 is displayed. He isn't even online and thus we print "NO".
3. "2 3" — We should check if friend 3 is displayed. Right now he is the only friend online and the system displays him. We should print "YES".
4. "1 1" — Friend 1 becomes online. The system now displays both friend 1 and friend 3.
5. "1 2" — Friend 2 becomes online. There are 3 friends online now but we were given k = 2 so only two friends can be displayed. Limak has worse relation with friend 1 than with other two online friends (t1 < t2, t3) so friend 1 won't be displayed
6. "2 1" — Print "NO".
7. "2 2" — Print "YES".
8. "2 3" — Print "YES".
Submitted Solution:
```
import sys
window = set()
n, k, q = [int(x) for x in input().split()]
arr = [int(x) for x in input().split()]
for i in range(q):
a, b = [int(x) for x in input().split()]
if (a == 1):
if (len(window) < k):
window.add(arr[b - 1])
else:
window.add(arr[b - 1])
m = min(window)
window.remove(m)
else:
print("YES" if arr[b - 1] in window else "NO")
``` | instruction | 0 | 58,478 | 14 | 116,956 |
Yes | output | 1 | 58,478 | 14 | 116,957 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Limak is a little polar bear. He loves connecting with other bears via social networks. He has n friends and his relation with the i-th of them is described by a unique integer ti. The bigger this value is, the better the friendship is. No two friends have the same value ti.
Spring is starting and the Winter sleep is over for bears. Limak has just woken up and logged in. All his friends still sleep and thus none of them is online. Some (maybe all) of them will appear online in the next hours, one at a time.
The system displays friends who are online. On the screen there is space to display at most k friends. If there are more than k friends online then the system displays only k best of them — those with biggest ti.
Your task is to handle queries of two types:
* "1 id" — Friend id becomes online. It's guaranteed that he wasn't online before.
* "2 id" — Check whether friend id is displayed by the system. Print "YES" or "NO" in a separate line.
Are you able to help Limak and answer all queries of the second type?
Input
The first line contains three integers n, k and q (1 ≤ n, q ≤ 150 000, 1 ≤ k ≤ min(6, n)) — the number of friends, the maximum number of displayed online friends and the number of queries, respectively.
The second line contains n integers t1, t2, ..., tn (1 ≤ ti ≤ 109) where ti describes how good is Limak's relation with the i-th friend.
The i-th of the following q lines contains two integers typei and idi (1 ≤ typei ≤ 2, 1 ≤ idi ≤ n) — the i-th query. If typei = 1 then a friend idi becomes online. If typei = 2 then you should check whether a friend idi is displayed.
It's guaranteed that no two queries of the first type will have the same idi becuase one friend can't become online twice. Also, it's guaranteed that at least one query will be of the second type (typei = 2) so the output won't be empty.
Output
For each query of the second type print one line with the answer — "YES" (without quotes) if the given friend is displayed and "NO" (without quotes) otherwise.
Examples
Input
4 2 8
300 950 500 200
1 3
2 4
2 3
1 1
1 2
2 1
2 2
2 3
Output
NO
YES
NO
YES
YES
Input
6 3 9
50 20 51 17 99 24
1 3
1 4
1 5
1 2
2 4
2 2
1 1
2 4
2 3
Output
NO
YES
NO
YES
Note
In the first sample, Limak has 4 friends who all sleep initially. At first, the system displays nobody because nobody is online. There are the following 8 queries:
1. "1 3" — Friend 3 becomes online.
2. "2 4" — We should check if friend 4 is displayed. He isn't even online and thus we print "NO".
3. "2 3" — We should check if friend 3 is displayed. Right now he is the only friend online and the system displays him. We should print "YES".
4. "1 1" — Friend 1 becomes online. The system now displays both friend 1 and friend 3.
5. "1 2" — Friend 2 becomes online. There are 3 friends online now but we were given k = 2 so only two friends can be displayed. Limak has worse relation with friend 1 than with other two online friends (t1 < t2, t3) so friend 1 won't be displayed
6. "2 1" — Print "NO".
7. "2 2" — Print "YES".
8. "2 3" — Print "YES".
Submitted Solution:
```
n, k, q = map(int, input().split())
data = list(map(int, input().split()))
data1 = set()
for i in range(q):
typ, fr = map(int, input().split())
fr -= 1
if typ == 2:
if data[fr] in data1:
print("YES")
else:
print("NO")
else:
if len(list(data1)) < k:
data1.add(data[fr])
elif data[fr] >= min(list(data1)):
data1.add(data[fr])
data1.remove(min(list(data1)))
``` | instruction | 0 | 58,479 | 14 | 116,958 |
Yes | output | 1 | 58,479 | 14 | 116,959 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Limak is a little polar bear. He loves connecting with other bears via social networks. He has n friends and his relation with the i-th of them is described by a unique integer ti. The bigger this value is, the better the friendship is. No two friends have the same value ti.
Spring is starting and the Winter sleep is over for bears. Limak has just woken up and logged in. All his friends still sleep and thus none of them is online. Some (maybe all) of them will appear online in the next hours, one at a time.
The system displays friends who are online. On the screen there is space to display at most k friends. If there are more than k friends online then the system displays only k best of them — those with biggest ti.
Your task is to handle queries of two types:
* "1 id" — Friend id becomes online. It's guaranteed that he wasn't online before.
* "2 id" — Check whether friend id is displayed by the system. Print "YES" or "NO" in a separate line.
Are you able to help Limak and answer all queries of the second type?
Input
The first line contains three integers n, k and q (1 ≤ n, q ≤ 150 000, 1 ≤ k ≤ min(6, n)) — the number of friends, the maximum number of displayed online friends and the number of queries, respectively.
The second line contains n integers t1, t2, ..., tn (1 ≤ ti ≤ 109) where ti describes how good is Limak's relation with the i-th friend.
The i-th of the following q lines contains two integers typei and idi (1 ≤ typei ≤ 2, 1 ≤ idi ≤ n) — the i-th query. If typei = 1 then a friend idi becomes online. If typei = 2 then you should check whether a friend idi is displayed.
It's guaranteed that no two queries of the first type will have the same idi becuase one friend can't become online twice. Also, it's guaranteed that at least one query will be of the second type (typei = 2) so the output won't be empty.
Output
For each query of the second type print one line with the answer — "YES" (without quotes) if the given friend is displayed and "NO" (without quotes) otherwise.
Examples
Input
4 2 8
300 950 500 200
1 3
2 4
2 3
1 1
1 2
2 1
2 2
2 3
Output
NO
YES
NO
YES
YES
Input
6 3 9
50 20 51 17 99 24
1 3
1 4
1 5
1 2
2 4
2 2
1 1
2 4
2 3
Output
NO
YES
NO
YES
Note
In the first sample, Limak has 4 friends who all sleep initially. At first, the system displays nobody because nobody is online. There are the following 8 queries:
1. "1 3" — Friend 3 becomes online.
2. "2 4" — We should check if friend 4 is displayed. He isn't even online and thus we print "NO".
3. "2 3" — We should check if friend 3 is displayed. Right now he is the only friend online and the system displays him. We should print "YES".
4. "1 1" — Friend 1 becomes online. The system now displays both friend 1 and friend 3.
5. "1 2" — Friend 2 becomes online. There are 3 friends online now but we were given k = 2 so only two friends can be displayed. Limak has worse relation with friend 1 than with other two online friends (t1 < t2, t3) so friend 1 won't be displayed
6. "2 1" — Print "NO".
7. "2 2" — Print "YES".
8. "2 3" — Print "YES".
Submitted Solution:
```
n, k, q = [int(x) for x in input().split()]
T = [int(x) for x in input().split()]
S = set()
for i in range(q):
t, _id = [int(x) for x in input().split()]
if t == 1:
m = min(S, default = 0)
if len(S) == k:
if m < T[_id-1]:
S.remove(m)
S.add(T[_id-1])
else:
S.add(T[_id-1])
else:
print('YES' if T[_id-1] in S else 'NO')
``` | instruction | 0 | 58,480 | 14 | 116,960 |
Yes | output | 1 | 58,480 | 14 | 116,961 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Limak is a little polar bear. He loves connecting with other bears via social networks. He has n friends and his relation with the i-th of them is described by a unique integer ti. The bigger this value is, the better the friendship is. No two friends have the same value ti.
Spring is starting and the Winter sleep is over for bears. Limak has just woken up and logged in. All his friends still sleep and thus none of them is online. Some (maybe all) of them will appear online in the next hours, one at a time.
The system displays friends who are online. On the screen there is space to display at most k friends. If there are more than k friends online then the system displays only k best of them — those with biggest ti.
Your task is to handle queries of two types:
* "1 id" — Friend id becomes online. It's guaranteed that he wasn't online before.
* "2 id" — Check whether friend id is displayed by the system. Print "YES" or "NO" in a separate line.
Are you able to help Limak and answer all queries of the second type?
Input
The first line contains three integers n, k and q (1 ≤ n, q ≤ 150 000, 1 ≤ k ≤ min(6, n)) — the number of friends, the maximum number of displayed online friends and the number of queries, respectively.
The second line contains n integers t1, t2, ..., tn (1 ≤ ti ≤ 109) where ti describes how good is Limak's relation with the i-th friend.
The i-th of the following q lines contains two integers typei and idi (1 ≤ typei ≤ 2, 1 ≤ idi ≤ n) — the i-th query. If typei = 1 then a friend idi becomes online. If typei = 2 then you should check whether a friend idi is displayed.
It's guaranteed that no two queries of the first type will have the same idi becuase one friend can't become online twice. Also, it's guaranteed that at least one query will be of the second type (typei = 2) so the output won't be empty.
Output
For each query of the second type print one line with the answer — "YES" (without quotes) if the given friend is displayed and "NO" (without quotes) otherwise.
Examples
Input
4 2 8
300 950 500 200
1 3
2 4
2 3
1 1
1 2
2 1
2 2
2 3
Output
NO
YES
NO
YES
YES
Input
6 3 9
50 20 51 17 99 24
1 3
1 4
1 5
1 2
2 4
2 2
1 1
2 4
2 3
Output
NO
YES
NO
YES
Note
In the first sample, Limak has 4 friends who all sleep initially. At first, the system displays nobody because nobody is online. There are the following 8 queries:
1. "1 3" — Friend 3 becomes online.
2. "2 4" — We should check if friend 4 is displayed. He isn't even online and thus we print "NO".
3. "2 3" — We should check if friend 3 is displayed. Right now he is the only friend online and the system displays him. We should print "YES".
4. "1 1" — Friend 1 becomes online. The system now displays both friend 1 and friend 3.
5. "1 2" — Friend 2 becomes online. There are 3 friends online now but we were given k = 2 so only two friends can be displayed. Limak has worse relation with friend 1 than with other two online friends (t1 < t2, t3) so friend 1 won't be displayed
6. "2 1" — Print "NO".
7. "2 2" — Print "YES".
8. "2 3" — Print "YES".
Submitted Solution:
```
n, k, q = map(int, input().split())
rel = [int(x) for x in input().split()]
display = []
num = 0
for i in range(q):
type, id = map(int, input().split())
if type == 1:
if num < k:
display.append(id)
num += 1
else:
tmp = []
for num in display:
tmp.append(rel[num-1])
m = min(tmp)
ix = tmp.index(m)
if rel[id-1] > m:
display.append(id)
display.pop(ix)
elif type == 2:
if id in display: print('YES')
else: print('NO')
``` | instruction | 0 | 58,481 | 14 | 116,962 |
No | output | 1 | 58,481 | 14 | 116,963 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Limak is a little polar bear. He loves connecting with other bears via social networks. He has n friends and his relation with the i-th of them is described by a unique integer ti. The bigger this value is, the better the friendship is. No two friends have the same value ti.
Spring is starting and the Winter sleep is over for bears. Limak has just woken up and logged in. All his friends still sleep and thus none of them is online. Some (maybe all) of them will appear online in the next hours, one at a time.
The system displays friends who are online. On the screen there is space to display at most k friends. If there are more than k friends online then the system displays only k best of them — those with biggest ti.
Your task is to handle queries of two types:
* "1 id" — Friend id becomes online. It's guaranteed that he wasn't online before.
* "2 id" — Check whether friend id is displayed by the system. Print "YES" or "NO" in a separate line.
Are you able to help Limak and answer all queries of the second type?
Input
The first line contains three integers n, k and q (1 ≤ n, q ≤ 150 000, 1 ≤ k ≤ min(6, n)) — the number of friends, the maximum number of displayed online friends and the number of queries, respectively.
The second line contains n integers t1, t2, ..., tn (1 ≤ ti ≤ 109) where ti describes how good is Limak's relation with the i-th friend.
The i-th of the following q lines contains two integers typei and idi (1 ≤ typei ≤ 2, 1 ≤ idi ≤ n) — the i-th query. If typei = 1 then a friend idi becomes online. If typei = 2 then you should check whether a friend idi is displayed.
It's guaranteed that no two queries of the first type will have the same idi becuase one friend can't become online twice. Also, it's guaranteed that at least one query will be of the second type (typei = 2) so the output won't be empty.
Output
For each query of the second type print one line with the answer — "YES" (without quotes) if the given friend is displayed and "NO" (without quotes) otherwise.
Examples
Input
4 2 8
300 950 500 200
1 3
2 4
2 3
1 1
1 2
2 1
2 2
2 3
Output
NO
YES
NO
YES
YES
Input
6 3 9
50 20 51 17 99 24
1 3
1 4
1 5
1 2
2 4
2 2
1 1
2 4
2 3
Output
NO
YES
NO
YES
Note
In the first sample, Limak has 4 friends who all sleep initially. At first, the system displays nobody because nobody is online. There are the following 8 queries:
1. "1 3" — Friend 3 becomes online.
2. "2 4" — We should check if friend 4 is displayed. He isn't even online and thus we print "NO".
3. "2 3" — We should check if friend 3 is displayed. Right now he is the only friend online and the system displays him. We should print "YES".
4. "1 1" — Friend 1 becomes online. The system now displays both friend 1 and friend 3.
5. "1 2" — Friend 2 becomes online. There are 3 friends online now but we were given k = 2 so only two friends can be displayed. Limak has worse relation with friend 1 than with other two online friends (t1 < t2, t3) so friend 1 won't be displayed
6. "2 1" — Print "NO".
7. "2 2" — Print "YES".
8. "2 3" — Print "YES".
Submitted Solution:
```
def bin(x):
if (x in z) or (len(b) == 0) or (len(b) == 1 and b[0] != x):
print('NO')
return
if b[min(len(b) - 1, k - 1)] <= x:
print('YES')
return
else:
print('NO')
return
b = []
n, k, q = map(int, input().split())
z = dict()
a = [int(x) for x in input().split()]
for i in range(q):
x, y = map(int, input().split())
if x == 1:
if len(b) != 0 and b[-1] > a[y - 1] and len(b) >= k:
z[a[y - 1]] = 1
else:
b.append(a[y - 1])
for j in range(len(b) - 1, 0, -1):
if b[j] > b[j - 1]:
b[j], b[j - 1] = b[j - 1], b[j]
else:
bin(a[y - 1])
``` | instruction | 0 | 58,482 | 14 | 116,964 |
No | output | 1 | 58,482 | 14 | 116,965 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Limak is a little polar bear. He loves connecting with other bears via social networks. He has n friends and his relation with the i-th of them is described by a unique integer ti. The bigger this value is, the better the friendship is. No two friends have the same value ti.
Spring is starting and the Winter sleep is over for bears. Limak has just woken up and logged in. All his friends still sleep and thus none of them is online. Some (maybe all) of them will appear online in the next hours, one at a time.
The system displays friends who are online. On the screen there is space to display at most k friends. If there are more than k friends online then the system displays only k best of them — those with biggest ti.
Your task is to handle queries of two types:
* "1 id" — Friend id becomes online. It's guaranteed that he wasn't online before.
* "2 id" — Check whether friend id is displayed by the system. Print "YES" or "NO" in a separate line.
Are you able to help Limak and answer all queries of the second type?
Input
The first line contains three integers n, k and q (1 ≤ n, q ≤ 150 000, 1 ≤ k ≤ min(6, n)) — the number of friends, the maximum number of displayed online friends and the number of queries, respectively.
The second line contains n integers t1, t2, ..., tn (1 ≤ ti ≤ 109) where ti describes how good is Limak's relation with the i-th friend.
The i-th of the following q lines contains two integers typei and idi (1 ≤ typei ≤ 2, 1 ≤ idi ≤ n) — the i-th query. If typei = 1 then a friend idi becomes online. If typei = 2 then you should check whether a friend idi is displayed.
It's guaranteed that no two queries of the first type will have the same idi becuase one friend can't become online twice. Also, it's guaranteed that at least one query will be of the second type (typei = 2) so the output won't be empty.
Output
For each query of the second type print one line with the answer — "YES" (without quotes) if the given friend is displayed and "NO" (without quotes) otherwise.
Examples
Input
4 2 8
300 950 500 200
1 3
2 4
2 3
1 1
1 2
2 1
2 2
2 3
Output
NO
YES
NO
YES
YES
Input
6 3 9
50 20 51 17 99 24
1 3
1 4
1 5
1 2
2 4
2 2
1 1
2 4
2 3
Output
NO
YES
NO
YES
Note
In the first sample, Limak has 4 friends who all sleep initially. At first, the system displays nobody because nobody is online. There are the following 8 queries:
1. "1 3" — Friend 3 becomes online.
2. "2 4" — We should check if friend 4 is displayed. He isn't even online and thus we print "NO".
3. "2 3" — We should check if friend 3 is displayed. Right now he is the only friend online and the system displays him. We should print "YES".
4. "1 1" — Friend 1 becomes online. The system now displays both friend 1 and friend 3.
5. "1 2" — Friend 2 becomes online. There are 3 friends online now but we were given k = 2 so only two friends can be displayed. Limak has worse relation with friend 1 than with other two online friends (t1 < t2, t3) so friend 1 won't be displayed
6. "2 1" — Print "NO".
7. "2 2" — Print "YES".
8. "2 3" — Print "YES".
Submitted Solution:
```
from sys import stdin as Si
from operator import itemgetter as ig
if __name__=='__main__':
n,k,q = map(int,Si.readline().split())
T = tuple(map(int,Si.readline().split()))
stack = {}
for i in range(q):
a,b = map(int,Si.readline().split())
#print(a,b,stack)
if a==1:
if len(stack)<k: stack[b]=T[b-1]
else:
while len(stack)>=k:
x,y = min(stack.items(),key=ig(1))
del stack[x]
stack[b] = T[b-1]
elif a==2:
if b in stack: print('YES')
else: print('NO')
'''
A. Bear and Reverse Radewoosh
time limit per test
2 seconds
memory limit per test
256 megabytes
input
standard input
output
standard output
Limak and Radewoosh are going to compete against each other in the upcoming algorithmic contest. They are equally skilled but they won't solve problems in the same order.
There will be n problems. The i-th problem has initial score pi and it takes exactly ti minutes to solve it. Problems are sorted by difficulty — it's guaranteed that pi < pi + 1 and ti < ti + 1.
A constant c is given too, representing the speed of loosing points. Then, submitting the i-th problem at time x (x minutes after the start of the contest) gives max(0, pi - c·x) points.
Limak is going to solve problems in order 1, 2, ..., n (sorted increasingly by pi). Radewoosh is going to solve them in order n, n - 1, ..., 1 (sorted decreasingly by pi). Your task is to predict the outcome — print the name of the winner (person who gets more points at the end) or a word "Tie" in case of a tie.
You may assume that the duration of the competition is greater or equal than the sum of all ti. That means both Limak and Radewoosh will accept all n problems.
Input
The first line contains two integers n and c (1 ≤ n ≤ 50, 1 ≤ c ≤ 1000) — the number of problems and the constant representing the speed of loosing points.
The second line contains n integers p1, p2, ..., pn (1 ≤ pi ≤ 1000, pi < pi + 1) — initial scores.
The third line contains n integers t1, t2, ..., tn (1 ≤ ti ≤ 1000, ti < ti + 1) where ti denotes the number of minutes one needs to solve the i-th problem.
Output
Print "Limak" (without quotes) if Limak will get more points in total. Print "Radewoosh" (without quotes) if Radewoosh will get more points in total. Print "Tie" (without quotes) if Limak and Radewoosh will get the same total number of points.
Examples
Input
3 2
50 85 250
10 15 25
Output
Limak
Input
3 6
50 85 250
10 15 25
Output
Radewoosh
Input
8 1
10 20 30 40 50 60 70 80
8 10 58 63 71 72 75 76
Output
Tie
Note
In the first sample, there are 3 problems. Limak solves them as follows:
Limak spends 10 minutes on the 1-st problem and he gets 50 - c·10 = 50 - 2·10 = 30 points.
Limak spends 15 minutes on the 2-nd problem so he submits it 10 + 15 = 25 minutes after the start of the contest. For the 2-nd problem he gets 85 - 2·25 = 35 points.
He spends 25 minutes on the 3-rd problem so he submits it 10 + 15 + 25 = 50 minutes after the start. For this problem he gets 250 - 2·50 = 150 points.
So, Limak got 30 + 35 + 150 = 215 points.
Radewoosh solves problem in the reversed order:
Radewoosh solves 3-rd problem after 25 minutes so he gets 250 - 2·25 = 200 points.
He spends 15 minutes on the 2-nd problem so he submits it 25 + 15 = 40 minutes after the start. He gets 85 - 2·40 = 5 points for this problem.
He spends 10 minutes on the 1-st problem so he submits it 25 + 15 + 10 = 50 minutes after the start. He gets max(0, 50 - 2·50) = max(0, - 50) = 0 points.
Radewoosh got 200 + 5 + 0 = 205 points in total. Limak has 215 points so Limak wins.
In the second sample, Limak will get 0 points for each problem and Radewoosh will first solve the hardest problem and he will get 250 - 6·25 = 100 points for that. Radewoosh will get 0 points for other two problems but he is the winner anyway.
In the third sample, Limak will get 2 points for the 1-st problem and 2 points for the 2-nd problem. Radewoosh will get 4 points for the 8-th problem. They won't get points for other problems and thus there is a tie because 2 + 2 = 4.
'''
``` | instruction | 0 | 58,483 | 14 | 116,966 |
No | output | 1 | 58,483 | 14 | 116,967 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Limak is a little polar bear. He loves connecting with other bears via social networks. He has n friends and his relation with the i-th of them is described by a unique integer ti. The bigger this value is, the better the friendship is. No two friends have the same value ti.
Spring is starting and the Winter sleep is over for bears. Limak has just woken up and logged in. All his friends still sleep and thus none of them is online. Some (maybe all) of them will appear online in the next hours, one at a time.
The system displays friends who are online. On the screen there is space to display at most k friends. If there are more than k friends online then the system displays only k best of them — those with biggest ti.
Your task is to handle queries of two types:
* "1 id" — Friend id becomes online. It's guaranteed that he wasn't online before.
* "2 id" — Check whether friend id is displayed by the system. Print "YES" or "NO" in a separate line.
Are you able to help Limak and answer all queries of the second type?
Input
The first line contains three integers n, k and q (1 ≤ n, q ≤ 150 000, 1 ≤ k ≤ min(6, n)) — the number of friends, the maximum number of displayed online friends and the number of queries, respectively.
The second line contains n integers t1, t2, ..., tn (1 ≤ ti ≤ 109) where ti describes how good is Limak's relation with the i-th friend.
The i-th of the following q lines contains two integers typei and idi (1 ≤ typei ≤ 2, 1 ≤ idi ≤ n) — the i-th query. If typei = 1 then a friend idi becomes online. If typei = 2 then you should check whether a friend idi is displayed.
It's guaranteed that no two queries of the first type will have the same idi becuase one friend can't become online twice. Also, it's guaranteed that at least one query will be of the second type (typei = 2) so the output won't be empty.
Output
For each query of the second type print one line with the answer — "YES" (without quotes) if the given friend is displayed and "NO" (without quotes) otherwise.
Examples
Input
4 2 8
300 950 500 200
1 3
2 4
2 3
1 1
1 2
2 1
2 2
2 3
Output
NO
YES
NO
YES
YES
Input
6 3 9
50 20 51 17 99 24
1 3
1 4
1 5
1 2
2 4
2 2
1 1
2 4
2 3
Output
NO
YES
NO
YES
Note
In the first sample, Limak has 4 friends who all sleep initially. At first, the system displays nobody because nobody is online. There are the following 8 queries:
1. "1 3" — Friend 3 becomes online.
2. "2 4" — We should check if friend 4 is displayed. He isn't even online and thus we print "NO".
3. "2 3" — We should check if friend 3 is displayed. Right now he is the only friend online and the system displays him. We should print "YES".
4. "1 1" — Friend 1 becomes online. The system now displays both friend 1 and friend 3.
5. "1 2" — Friend 2 becomes online. There are 3 friends online now but we were given k = 2 so only two friends can be displayed. Limak has worse relation with friend 1 than with other two online friends (t1 < t2, t3) so friend 1 won't be displayed
6. "2 1" — Print "NO".
7. "2 2" — Print "YES".
8. "2 3" — Print "YES".
Submitted Solution:
```
import sys
def siftup(ind):
global tree
while (ind != 1) and (frendly[tree[ind]] < frendly[tree[ind // 2]]):
tree[ind], tree[ind // 2] = tree[ind // 2], tree[ind]
ind //= 2
def siftdown(ind):
global tree
while (ind * 2 + 1 <= len(tree) - 1) and (frendly[tree[ind]] > frendly[tree[ind * 2]] or frendly[tree[ind]] > frendly[tree[ind * 2 + 1]]):
if frendly[tree[ind * 2]] > frendly[tree[ind * 2 + 1]]:
tree[ind], tree[ind * 2 + 1] = tree[ind * 2 + 1], tree[ind]
ind = ind * 2 + 1
else:
tree[ind], tree[ind * 2] = tree[ind * 2], tree[ind]
ind *= 2
if (ind * 2 <= len(tree) - 1) and (frendly[tree[ind]] > frendly[tree[ind * 2]]):
tree[ind], tree[ind * 2] = tree[ind * 2], tree[ind]
ind *= 2
n, k, q = map(int, sys.stdin.readline().split())
frendly = [0] + list(map(int, sys.stdin.readline().split()))
bearsout = set([i for i in range(1, n + 1)])
tree = [0]
for i in range(q):
lab, b = map(int, sys.stdin.readline().split())
if lab == 1:
if len(tree) > k:
if frendly[b] > frendly[tree[1]]:
bearsout.discard(b)
bearsout.add(tree[1])
tree[1] = b
siftdown(1)
else:
tree.append(b)
bearsout.discard(b)
siftup(len(tree) - 1)
else:
if b in bearsout:
sys.stdout.write('NO')
else:
sys.stdout.write('YES')
``` | instruction | 0 | 58,484 | 14 | 116,968 |
No | output | 1 | 58,484 | 14 | 116,969 |
Provide tags and a correct Python 3 solution for this coding contest problem.
There are n servers in a laboratory, each of them can perform tasks. Each server has a unique id — integer from 1 to n.
It is known that during the day q tasks will come, the i-th of them is characterized with three integers: ti — the moment in seconds in which the task will come, ki — the number of servers needed to perform it, and di — the time needed to perform this task in seconds. All ti are distinct.
To perform the i-th task you need ki servers which are unoccupied in the second ti. After the servers begin to perform the task, each of them will be busy over the next di seconds. Thus, they will be busy in seconds ti, ti + 1, ..., ti + di - 1. For performing the task, ki servers with the smallest ids will be chosen from all the unoccupied servers. If in the second ti there are not enough unoccupied servers, the task is ignored.
Write the program that determines which tasks will be performed and which will be ignored.
Input
The first line contains two positive integers n and q (1 ≤ n ≤ 100, 1 ≤ q ≤ 105) — the number of servers and the number of tasks.
Next q lines contains three integers each, the i-th line contains integers ti, ki and di (1 ≤ ti ≤ 106, 1 ≤ ki ≤ n, 1 ≤ di ≤ 1000) — the moment in seconds in which the i-th task will come, the number of servers needed to perform it, and the time needed to perform this task in seconds. The tasks are given in a chronological order and they will come in distinct seconds.
Output
Print q lines. If the i-th task will be performed by the servers, print in the i-th line the sum of servers' ids on which this task will be performed. Otherwise, print -1.
Examples
Input
4 3
1 3 2
2 2 1
3 4 3
Output
6
-1
10
Input
3 2
3 2 3
5 1 2
Output
3
3
Input
8 6
1 3 20
4 2 1
6 5 5
10 1 1
15 3 6
21 8 8
Output
6
9
30
-1
15
36
Note
In the first example in the second 1 the first task will come, it will be performed on the servers with ids 1, 2 and 3 (the sum of the ids equals 6) during two seconds. In the second 2 the second task will come, it will be ignored, because only the server 4 will be unoccupied at that second. In the second 3 the third task will come. By this time, servers with the ids 1, 2 and 3 will be unoccupied again, so the third task will be done on all the servers with the ids 1, 2, 3 and 4 (the sum of the ids is 10).
In the second example in the second 3 the first task will come, it will be performed on the servers with ids 1 and 2 (the sum of the ids is 3) during three seconds. In the second 5 the second task will come, it will be performed on the server 3, because the first two servers will be busy performing the first task. | instruction | 0 | 58,533 | 14 | 117,066 |
Tags: implementation
Correct Solution:
```
import sys
try:
sys.stdin = open('input.txt', 'r')
sys.stdout = open('output.txt', 'w')
except:
pass
input = sys.stdin.readline
n, q = map(int, input().split())
server = [-1]*n
for _ in range(q):
t, k, d = map(int, input().split())
cnt = 0
for i in range(n):
if t >= server[i]:
server[i] = -1
if server[i] == -1:
cnt += 1
if k <= cnt:
ans = 0
for i in range(n):
if server[i] == -1:
server[i] = t + d
ans += i+1
k -= 1
if not k:
break
print(ans)
else:
print(-1)
``` | output | 1 | 58,533 | 14 | 117,067 |
Provide tags and a correct Python 3 solution for this coding contest problem.
There are n servers in a laboratory, each of them can perform tasks. Each server has a unique id — integer from 1 to n.
It is known that during the day q tasks will come, the i-th of them is characterized with three integers: ti — the moment in seconds in which the task will come, ki — the number of servers needed to perform it, and di — the time needed to perform this task in seconds. All ti are distinct.
To perform the i-th task you need ki servers which are unoccupied in the second ti. After the servers begin to perform the task, each of them will be busy over the next di seconds. Thus, they will be busy in seconds ti, ti + 1, ..., ti + di - 1. For performing the task, ki servers with the smallest ids will be chosen from all the unoccupied servers. If in the second ti there are not enough unoccupied servers, the task is ignored.
Write the program that determines which tasks will be performed and which will be ignored.
Input
The first line contains two positive integers n and q (1 ≤ n ≤ 100, 1 ≤ q ≤ 105) — the number of servers and the number of tasks.
Next q lines contains three integers each, the i-th line contains integers ti, ki and di (1 ≤ ti ≤ 106, 1 ≤ ki ≤ n, 1 ≤ di ≤ 1000) — the moment in seconds in which the i-th task will come, the number of servers needed to perform it, and the time needed to perform this task in seconds. The tasks are given in a chronological order and they will come in distinct seconds.
Output
Print q lines. If the i-th task will be performed by the servers, print in the i-th line the sum of servers' ids on which this task will be performed. Otherwise, print -1.
Examples
Input
4 3
1 3 2
2 2 1
3 4 3
Output
6
-1
10
Input
3 2
3 2 3
5 1 2
Output
3
3
Input
8 6
1 3 20
4 2 1
6 5 5
10 1 1
15 3 6
21 8 8
Output
6
9
30
-1
15
36
Note
In the first example in the second 1 the first task will come, it will be performed on the servers with ids 1, 2 and 3 (the sum of the ids equals 6) during two seconds. In the second 2 the second task will come, it will be ignored, because only the server 4 will be unoccupied at that second. In the second 3 the third task will come. By this time, servers with the ids 1, 2 and 3 will be unoccupied again, so the third task will be done on all the servers with the ids 1, 2, 3 and 4 (the sum of the ids is 10).
In the second example in the second 3 the first task will come, it will be performed on the servers with ids 1 and 2 (the sum of the ids is 3) during three seconds. In the second 5 the second task will come, it will be performed on the server 3, because the first two servers will be busy performing the first task. | instruction | 0 | 58,534 | 14 | 117,068 |
Tags: implementation
Correct Solution:
```
import heapq as pq
from sys import stdin, stdout
#Heap method written by Engineermind.cho
#Testing if heap + fast IO can make python pass this question
n, q = map(int, stdin.readline().split())
server = [i for i in range(1, n+1)]
running = []
is_chg = False
for x in stdin.readlines():
t, k, d = map(int, x.split())
while running:
if t >= running[0][0]:
server += pq.heappop(running)[1]
is_chg = True
else:
break
if is_chg:
server = sorted(server)
is_chg = False
if len(server) < k:
print(-1)
continue
else:
print(sum(server[:k]))
if d != 1:
pq.heappush(running, (t+d, server[:k].copy()))
server = server[k:]
``` | output | 1 | 58,534 | 14 | 117,069 |
Provide tags and a correct Python 3 solution for this coding contest problem.
There are n servers in a laboratory, each of them can perform tasks. Each server has a unique id — integer from 1 to n.
It is known that during the day q tasks will come, the i-th of them is characterized with three integers: ti — the moment in seconds in which the task will come, ki — the number of servers needed to perform it, and di — the time needed to perform this task in seconds. All ti are distinct.
To perform the i-th task you need ki servers which are unoccupied in the second ti. After the servers begin to perform the task, each of them will be busy over the next di seconds. Thus, they will be busy in seconds ti, ti + 1, ..., ti + di - 1. For performing the task, ki servers with the smallest ids will be chosen from all the unoccupied servers. If in the second ti there are not enough unoccupied servers, the task is ignored.
Write the program that determines which tasks will be performed and which will be ignored.
Input
The first line contains two positive integers n and q (1 ≤ n ≤ 100, 1 ≤ q ≤ 105) — the number of servers and the number of tasks.
Next q lines contains three integers each, the i-th line contains integers ti, ki and di (1 ≤ ti ≤ 106, 1 ≤ ki ≤ n, 1 ≤ di ≤ 1000) — the moment in seconds in which the i-th task will come, the number of servers needed to perform it, and the time needed to perform this task in seconds. The tasks are given in a chronological order and they will come in distinct seconds.
Output
Print q lines. If the i-th task will be performed by the servers, print in the i-th line the sum of servers' ids on which this task will be performed. Otherwise, print -1.
Examples
Input
4 3
1 3 2
2 2 1
3 4 3
Output
6
-1
10
Input
3 2
3 2 3
5 1 2
Output
3
3
Input
8 6
1 3 20
4 2 1
6 5 5
10 1 1
15 3 6
21 8 8
Output
6
9
30
-1
15
36
Note
In the first example in the second 1 the first task will come, it will be performed on the servers with ids 1, 2 and 3 (the sum of the ids equals 6) during two seconds. In the second 2 the second task will come, it will be ignored, because only the server 4 will be unoccupied at that second. In the second 3 the third task will come. By this time, servers with the ids 1, 2 and 3 will be unoccupied again, so the third task will be done on all the servers with the ids 1, 2, 3 and 4 (the sum of the ids is 10).
In the second example in the second 3 the first task will come, it will be performed on the servers with ids 1 and 2 (the sum of the ids is 3) during three seconds. In the second 5 the second task will come, it will be performed on the server 3, because the first two servers will be busy performing the first task. | instruction | 0 | 58,535 | 14 | 117,070 |
Tags: implementation
Correct Solution:
```
'''
8 6
1 3 20
4 2 1
6 5 5
10 1 1
15 3 6
21 8 8
'''
p=input().rstrip().split(' ')
n=int(p[0])
q=int(p[1])
L=[0]*n;
Q=[]
W=[]
S=[]
D=[]
for i in range(0,q):
o=input().rstrip().split(' ')
t=int(o[0])
k=int(o[1])
d=int(o[2])
j=0;
while(len(S)>0 and j<len(S)):
if S[j] <= t:
R=W[j];
del(S[j])
del(W[j]);
for M in range(0,len(R)):
L[R[M]-1]=0;
else:
j+=1;
if L.count(0) >= k:
E=[]
ans=0;
for j in range(0,len(L)):
if L[j]==0 and k!=0:
E.append(j+1)
ans+=(j+1)
k-=1;
L[j]=1;
W.append(E)
S.append(t+d)
print(ans)
else:
print(-1)
# print(S,W,L)
``` | output | 1 | 58,535 | 14 | 117,071 |
Provide tags and a correct Python 3 solution for this coding contest problem.
There are n servers in a laboratory, each of them can perform tasks. Each server has a unique id — integer from 1 to n.
It is known that during the day q tasks will come, the i-th of them is characterized with three integers: ti — the moment in seconds in which the task will come, ki — the number of servers needed to perform it, and di — the time needed to perform this task in seconds. All ti are distinct.
To perform the i-th task you need ki servers which are unoccupied in the second ti. After the servers begin to perform the task, each of them will be busy over the next di seconds. Thus, they will be busy in seconds ti, ti + 1, ..., ti + di - 1. For performing the task, ki servers with the smallest ids will be chosen from all the unoccupied servers. If in the second ti there are not enough unoccupied servers, the task is ignored.
Write the program that determines which tasks will be performed and which will be ignored.
Input
The first line contains two positive integers n and q (1 ≤ n ≤ 100, 1 ≤ q ≤ 105) — the number of servers and the number of tasks.
Next q lines contains three integers each, the i-th line contains integers ti, ki and di (1 ≤ ti ≤ 106, 1 ≤ ki ≤ n, 1 ≤ di ≤ 1000) — the moment in seconds in which the i-th task will come, the number of servers needed to perform it, and the time needed to perform this task in seconds. The tasks are given in a chronological order and they will come in distinct seconds.
Output
Print q lines. If the i-th task will be performed by the servers, print in the i-th line the sum of servers' ids on which this task will be performed. Otherwise, print -1.
Examples
Input
4 3
1 3 2
2 2 1
3 4 3
Output
6
-1
10
Input
3 2
3 2 3
5 1 2
Output
3
3
Input
8 6
1 3 20
4 2 1
6 5 5
10 1 1
15 3 6
21 8 8
Output
6
9
30
-1
15
36
Note
In the first example in the second 1 the first task will come, it will be performed on the servers with ids 1, 2 and 3 (the sum of the ids equals 6) during two seconds. In the second 2 the second task will come, it will be ignored, because only the server 4 will be unoccupied at that second. In the second 3 the third task will come. By this time, servers with the ids 1, 2 and 3 will be unoccupied again, so the third task will be done on all the servers with the ids 1, 2, 3 and 4 (the sum of the ids is 10).
In the second example in the second 3 the first task will come, it will be performed on the servers with ids 1 and 2 (the sum of the ids is 3) during three seconds. In the second 5 the second task will come, it will be performed on the server 3, because the first two servers will be busy performing the first task. | instruction | 0 | 58,536 | 14 | 117,072 |
Tags: implementation
Correct Solution:
```
n , q = map(int,input().split())
tasks=[]
servers = [0]*n
for _ in range(q):
tasks.append(tuple(map(int,input().split())))
for t,k,d in tasks:
a = []
for i in range(n):
if t>=servers[i]:
a.append(i)
if len(a)>=k:
total = 0
for s in range(k):
total+=a[s]+1
servers[a[s]]=t+d
print(total)
else:
print(-1)
``` | output | 1 | 58,536 | 14 | 117,073 |
Provide tags and a correct Python 3 solution for this coding contest problem.
There are n servers in a laboratory, each of them can perform tasks. Each server has a unique id — integer from 1 to n.
It is known that during the day q tasks will come, the i-th of them is characterized with three integers: ti — the moment in seconds in which the task will come, ki — the number of servers needed to perform it, and di — the time needed to perform this task in seconds. All ti are distinct.
To perform the i-th task you need ki servers which are unoccupied in the second ti. After the servers begin to perform the task, each of them will be busy over the next di seconds. Thus, they will be busy in seconds ti, ti + 1, ..., ti + di - 1. For performing the task, ki servers with the smallest ids will be chosen from all the unoccupied servers. If in the second ti there are not enough unoccupied servers, the task is ignored.
Write the program that determines which tasks will be performed and which will be ignored.
Input
The first line contains two positive integers n and q (1 ≤ n ≤ 100, 1 ≤ q ≤ 105) — the number of servers and the number of tasks.
Next q lines contains three integers each, the i-th line contains integers ti, ki and di (1 ≤ ti ≤ 106, 1 ≤ ki ≤ n, 1 ≤ di ≤ 1000) — the moment in seconds in which the i-th task will come, the number of servers needed to perform it, and the time needed to perform this task in seconds. The tasks are given in a chronological order and they will come in distinct seconds.
Output
Print q lines. If the i-th task will be performed by the servers, print in the i-th line the sum of servers' ids on which this task will be performed. Otherwise, print -1.
Examples
Input
4 3
1 3 2
2 2 1
3 4 3
Output
6
-1
10
Input
3 2
3 2 3
5 1 2
Output
3
3
Input
8 6
1 3 20
4 2 1
6 5 5
10 1 1
15 3 6
21 8 8
Output
6
9
30
-1
15
36
Note
In the first example in the second 1 the first task will come, it will be performed on the servers with ids 1, 2 and 3 (the sum of the ids equals 6) during two seconds. In the second 2 the second task will come, it will be ignored, because only the server 4 will be unoccupied at that second. In the second 3 the third task will come. By this time, servers with the ids 1, 2 and 3 will be unoccupied again, so the third task will be done on all the servers with the ids 1, 2, 3 and 4 (the sum of the ids is 10).
In the second example in the second 3 the first task will come, it will be performed on the servers with ids 1 and 2 (the sum of the ids is 3) during three seconds. In the second 5 the second task will come, it will be performed on the server 3, because the first two servers will be busy performing the first task. | instruction | 0 | 58,537 | 14 | 117,074 |
Tags: implementation
Correct Solution:
```
import sys
input = sys.stdin.readline
#
server, task = map(int, input().split())
serverstatus = [0]*server
for i in range(task):
moment, serven, time = map(int, input().split())
ava = 0
total = 0
d = serverstatus[:]
for i in range(server):
if ava < serven and serverstatus[i] <= moment:
ava += 1
total += i+1
d[i] = moment + time
elif ava == serven:
break
if ava < serven:
print(-1)
else:
print(total)
serverstatus = d[:]
``` | output | 1 | 58,537 | 14 | 117,075 |
Provide tags and a correct Python 3 solution for this coding contest problem.
There are n servers in a laboratory, each of them can perform tasks. Each server has a unique id — integer from 1 to n.
It is known that during the day q tasks will come, the i-th of them is characterized with three integers: ti — the moment in seconds in which the task will come, ki — the number of servers needed to perform it, and di — the time needed to perform this task in seconds. All ti are distinct.
To perform the i-th task you need ki servers which are unoccupied in the second ti. After the servers begin to perform the task, each of them will be busy over the next di seconds. Thus, they will be busy in seconds ti, ti + 1, ..., ti + di - 1. For performing the task, ki servers with the smallest ids will be chosen from all the unoccupied servers. If in the second ti there are not enough unoccupied servers, the task is ignored.
Write the program that determines which tasks will be performed and which will be ignored.
Input
The first line contains two positive integers n and q (1 ≤ n ≤ 100, 1 ≤ q ≤ 105) — the number of servers and the number of tasks.
Next q lines contains three integers each, the i-th line contains integers ti, ki and di (1 ≤ ti ≤ 106, 1 ≤ ki ≤ n, 1 ≤ di ≤ 1000) — the moment in seconds in which the i-th task will come, the number of servers needed to perform it, and the time needed to perform this task in seconds. The tasks are given in a chronological order and they will come in distinct seconds.
Output
Print q lines. If the i-th task will be performed by the servers, print in the i-th line the sum of servers' ids on which this task will be performed. Otherwise, print -1.
Examples
Input
4 3
1 3 2
2 2 1
3 4 3
Output
6
-1
10
Input
3 2
3 2 3
5 1 2
Output
3
3
Input
8 6
1 3 20
4 2 1
6 5 5
10 1 1
15 3 6
21 8 8
Output
6
9
30
-1
15
36
Note
In the first example in the second 1 the first task will come, it will be performed on the servers with ids 1, 2 and 3 (the sum of the ids equals 6) during two seconds. In the second 2 the second task will come, it will be ignored, because only the server 4 will be unoccupied at that second. In the second 3 the third task will come. By this time, servers with the ids 1, 2 and 3 will be unoccupied again, so the third task will be done on all the servers with the ids 1, 2, 3 and 4 (the sum of the ids is 10).
In the second example in the second 3 the first task will come, it will be performed on the servers with ids 1 and 2 (the sum of the ids is 3) during three seconds. In the second 5 the second task will come, it will be performed on the server 3, because the first two servers will be busy performing the first task. | instruction | 0 | 58,538 | 14 | 117,076 |
Tags: implementation
Correct Solution:
```
import heapq as pq
from sys import stdin, stdout
#Heap method written by Engineermind.cho
#Testing if heap + fast IO can make python pass this question
n, q = map(int, stdin.readline().split())
server = [i for i in range(1, n+1)]
running = []
is_chg = False
for x in stdin.readlines():
t, k, d = map(int, x.split())
while running:
if t >= running[0][0]:
server += pq.heappop(running)[1]
is_chg = True
else:
break
if is_chg:
server = sorted(server)
is_chg = False
if len(server) < k:
stdout.write('-1\n')
continue
else:
stdout.write(str(sum(server[:k]))+'\n')
if d != 1:
pq.heappush(running, (t+d, server[:k].copy()))
server = server[k:]
# Made By Mostafa_Khaled
``` | output | 1 | 58,538 | 14 | 117,077 |
Provide tags and a correct Python 3 solution for this coding contest problem.
There are n servers in a laboratory, each of them can perform tasks. Each server has a unique id — integer from 1 to n.
It is known that during the day q tasks will come, the i-th of them is characterized with three integers: ti — the moment in seconds in which the task will come, ki — the number of servers needed to perform it, and di — the time needed to perform this task in seconds. All ti are distinct.
To perform the i-th task you need ki servers which are unoccupied in the second ti. After the servers begin to perform the task, each of them will be busy over the next di seconds. Thus, they will be busy in seconds ti, ti + 1, ..., ti + di - 1. For performing the task, ki servers with the smallest ids will be chosen from all the unoccupied servers. If in the second ti there are not enough unoccupied servers, the task is ignored.
Write the program that determines which tasks will be performed and which will be ignored.
Input
The first line contains two positive integers n and q (1 ≤ n ≤ 100, 1 ≤ q ≤ 105) — the number of servers and the number of tasks.
Next q lines contains three integers each, the i-th line contains integers ti, ki and di (1 ≤ ti ≤ 106, 1 ≤ ki ≤ n, 1 ≤ di ≤ 1000) — the moment in seconds in which the i-th task will come, the number of servers needed to perform it, and the time needed to perform this task in seconds. The tasks are given in a chronological order and they will come in distinct seconds.
Output
Print q lines. If the i-th task will be performed by the servers, print in the i-th line the sum of servers' ids on which this task will be performed. Otherwise, print -1.
Examples
Input
4 3
1 3 2
2 2 1
3 4 3
Output
6
-1
10
Input
3 2
3 2 3
5 1 2
Output
3
3
Input
8 6
1 3 20
4 2 1
6 5 5
10 1 1
15 3 6
21 8 8
Output
6
9
30
-1
15
36
Note
In the first example in the second 1 the first task will come, it will be performed on the servers with ids 1, 2 and 3 (the sum of the ids equals 6) during two seconds. In the second 2 the second task will come, it will be ignored, because only the server 4 will be unoccupied at that second. In the second 3 the third task will come. By this time, servers with the ids 1, 2 and 3 will be unoccupied again, so the third task will be done on all the servers with the ids 1, 2, 3 and 4 (the sum of the ids is 10).
In the second example in the second 3 the first task will come, it will be performed on the servers with ids 1 and 2 (the sum of the ids is 3) during three seconds. In the second 5 the second task will come, it will be performed on the server 3, because the first two servers will be busy performing the first task. | instruction | 0 | 58,539 | 14 | 117,078 |
Tags: implementation
Correct Solution:
```
num = [int(n) for n in input().split()]
server = [0 for n in range(0,num[0])]
result = list()
num_task = num[1]
temp_ti = 0
for i in range(0,num_task):
temp_task = [int(n) for n in input().split()]
time_passed = temp_task[0]-temp_ti
temp_ti = temp_task[0]
###length
for n in range(0,num[0]):
server[n] -= time_passed
server[n] = 0 if(server[n]<0) else server[n]
if(server.count(0) < temp_task[1]):
result.append(-1)
else:
accumulate_result = 0
for i in range(0,num[0]):
if(not bool(server[i])):
server[i] += temp_task[2]
temp_task[1] -=1
accumulate_result +=(i+1)
if(temp_task[1] ==0):
break
result.append(accumulate_result)
for i in range(0,num_task):
print(result[i])
``` | output | 1 | 58,539 | 14 | 117,079 |
Provide tags and a correct Python 3 solution for this coding contest problem.
There are n servers in a laboratory, each of them can perform tasks. Each server has a unique id — integer from 1 to n.
It is known that during the day q tasks will come, the i-th of them is characterized with three integers: ti — the moment in seconds in which the task will come, ki — the number of servers needed to perform it, and di — the time needed to perform this task in seconds. All ti are distinct.
To perform the i-th task you need ki servers which are unoccupied in the second ti. After the servers begin to perform the task, each of them will be busy over the next di seconds. Thus, they will be busy in seconds ti, ti + 1, ..., ti + di - 1. For performing the task, ki servers with the smallest ids will be chosen from all the unoccupied servers. If in the second ti there are not enough unoccupied servers, the task is ignored.
Write the program that determines which tasks will be performed and which will be ignored.
Input
The first line contains two positive integers n and q (1 ≤ n ≤ 100, 1 ≤ q ≤ 105) — the number of servers and the number of tasks.
Next q lines contains three integers each, the i-th line contains integers ti, ki and di (1 ≤ ti ≤ 106, 1 ≤ ki ≤ n, 1 ≤ di ≤ 1000) — the moment in seconds in which the i-th task will come, the number of servers needed to perform it, and the time needed to perform this task in seconds. The tasks are given in a chronological order and they will come in distinct seconds.
Output
Print q lines. If the i-th task will be performed by the servers, print in the i-th line the sum of servers' ids on which this task will be performed. Otherwise, print -1.
Examples
Input
4 3
1 3 2
2 2 1
3 4 3
Output
6
-1
10
Input
3 2
3 2 3
5 1 2
Output
3
3
Input
8 6
1 3 20
4 2 1
6 5 5
10 1 1
15 3 6
21 8 8
Output
6
9
30
-1
15
36
Note
In the first example in the second 1 the first task will come, it will be performed on the servers with ids 1, 2 and 3 (the sum of the ids equals 6) during two seconds. In the second 2 the second task will come, it will be ignored, because only the server 4 will be unoccupied at that second. In the second 3 the third task will come. By this time, servers with the ids 1, 2 and 3 will be unoccupied again, so the third task will be done on all the servers with the ids 1, 2, 3 and 4 (the sum of the ids is 10).
In the second example in the second 3 the first task will come, it will be performed on the servers with ids 1 and 2 (the sum of the ids is 3) during three seconds. In the second 5 the second task will come, it will be performed on the server 3, because the first two servers will be busy performing the first task. | instruction | 0 | 58,540 | 14 | 117,080 |
Tags: implementation
Correct Solution:
```
n,q=map(int,input().split())
s=[0]*(n+1)
ans = []
for i in range(q):
t,k,d=map(int,input().split())
sm=[]
for j in range(1, n+1):
if s[j] < t:
sm.append(j)
if len(sm)==k:break
if len(sm)!=k:ans.append('-1')
else:
ans.append(str(sum(sm)))
for j in sm: s[j]=d+t-1
print('\n'.join(ans))
``` | output | 1 | 58,540 | 14 | 117,081 |
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