message stringlengths 2 44.5k | message_type stringclasses 2 values | message_id int64 0 1 | conversation_id int64 276 109k | cluster float64 23 23 | __index_level_0__ int64 552 217k |
|---|---|---|---|---|---|
Provide tags and a correct Python 3 solution for this coding contest problem.
A revolution took place on the Buka Island. New government replaced the old one. The new government includes n parties and each of them is entitled to some part of the island according to their contribution to the revolution. However, they can't divide the island.
The island can be conventionally represented as two rectangles a × b and c × d unit squares in size correspondingly. The rectangles are located close to each other. At that, one of the sides with the length of a and one of the sides with the length of c lie on one line. You can see this in more details on the picture.
<image>
The i-th party is entitled to a part of the island equal to xi unit squares. Every such part should fully cover several squares of the island (it is not allowed to cover the squares partially) and be a connected figure. A "connected figure" presupposes that from any square of this party one can move to any other square of the same party moving through edge-adjacent squares also belonging to that party.
Your task is to divide the island between parties.
Input
The first line contains 5 space-separated integers — a, b, c, d and n (1 ≤ a, b, c, d ≤ 50, b ≠ d, 1 ≤ n ≤ 26). The second line contains n space-separated numbers. The i-th of them is equal to number xi (1 ≤ xi ≤ a × b + c × d). It is guaranteed that <image>.
Output
If dividing the island between parties in the required manner is impossible, print "NO" (without the quotes). Otherwise, print "YES" (also without the quotes) and, starting from the next line, print max(b, d) lines each containing a + c characters. To mark what square should belong to what party, use lowercase Latin letters. For the party that is first in order in the input data, use "a", for the second one use "b" and so on. Use "." for the squares that belong to the sea. The first symbol of the second line of the output data should correspond to the square that belongs to the rectangle a × b. The last symbol of the second line should correspond to the square that belongs to the rectangle c × d.
If there are several solutions output any.
Examples
Input
3 4 2 2 3
5 8 3
Output
YES
aaabb
aabbb
cbb..
ccb..
Input
3 2 1 4 4
1 2 3 4
Output
YES
abbd
cccd
...d
...d | instruction | 0 | 81,260 | 23 | 162,520 |
Tags: constructive algorithms
Correct Solution:
```
#Code by Sounak, IIESTS
#------------------------------warmup----------------------------
import os
import sys
import math
from io import BytesIO, IOBase
from fractions import Fraction
import collections
from itertools import permutations
from collections import defaultdict
from collections import deque
import threading
#sys.setrecursionlimit(300000)
#threading.stack_size(10**8)
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")
#-------------------------------------------------------------------------
#mod = 9223372036854775807
class SegmentTree:
def __init__(self, data, default=-10**6, func=lambda a, b: max(a,b)):
"""initialize the segment tree with data"""
self._default = default
self._func = func
self._len = len(data)
self._size = _size = 1 << (self._len - 1).bit_length()
self.data = [default] * (2 * _size)
self.data[_size:_size + self._len] = data
for i in reversed(range(_size)):
self.data[i] = func(self.data[i + i], self.data[i + i + 1])
def __delitem__(self, idx):
self[idx] = self._default
def __getitem__(self, idx):
return self.data[idx + self._size]
def __setitem__(self, idx, value):
idx += self._size
self.data[idx] = value
idx >>= 1
while idx:
self.data[idx] = self._func(self.data[2 * idx], self.data[2 * idx + 1])
idx >>= 1
def __len__(self):
return self._len
def query(self, start, stop):
if start == stop:
return self.__getitem__(start)
stop += 1
start += self._size
stop += self._size
res = self._default
while start < stop:
if start & 1:
res = self._func(res, self.data[start])
start += 1
if stop & 1:
stop -= 1
res = self._func(res, self.data[stop])
start >>= 1
stop >>= 1
return res
def __repr__(self):
return "SegmentTree({0})".format(self.data)
class SegmentTree1:
def __init__(self, data, default=10**6, func=lambda a, b: min(a,b)):
"""initialize the segment tree with data"""
self._default = default
self._func = func
self._len = len(data)
self._size = _size = 1 << (self._len - 1).bit_length()
self.data = [default] * (2 * _size)
self.data[_size:_size + self._len] = data
for i in reversed(range(_size)):
self.data[i] = func(self.data[i + i], self.data[i + i + 1])
def __delitem__(self, idx):
self[idx] = self._default
def __getitem__(self, idx):
return self.data[idx + self._size]
def __setitem__(self, idx, value):
idx += self._size
self.data[idx] = value
idx >>= 1
while idx:
self.data[idx] = self._func(self.data[2 * idx], self.data[2 * idx + 1])
idx >>= 1
def __len__(self):
return self._len
def query(self, start, stop):
if start == stop:
return self.__getitem__(start)
stop += 1
start += self._size
stop += self._size
res = self._default
while start < stop:
if start & 1:
res = self._func(res, self.data[start])
start += 1
if stop & 1:
stop -= 1
res = self._func(res, self.data[stop])
start >>= 1
stop >>= 1
return res
def __repr__(self):
return "SegmentTree({0})".format(self.data)
MOD=10**9+7
class Factorial:
def __init__(self, MOD):
self.MOD = MOD
self.factorials = [1, 1]
self.invModulos = [0, 1]
self.invFactorial_ = [1, 1]
def calc(self, n):
if n <= -1:
print("Invalid argument to calculate n!")
print("n must be non-negative value. But the argument was " + str(n))
exit()
if n < len(self.factorials):
return self.factorials[n]
nextArr = [0] * (n + 1 - len(self.factorials))
initialI = len(self.factorials)
prev = self.factorials[-1]
m = self.MOD
for i in range(initialI, n + 1):
prev = nextArr[i - initialI] = prev * i % m
self.factorials += nextArr
return self.factorials[n]
def inv(self, n):
if n <= -1:
print("Invalid argument to calculate n^(-1)")
print("n must be non-negative value. But the argument was " + str(n))
exit()
p = self.MOD
pi = n % p
if pi < len(self.invModulos):
return self.invModulos[pi]
nextArr = [0] * (n + 1 - len(self.invModulos))
initialI = len(self.invModulos)
for i in range(initialI, min(p, n + 1)):
next = -self.invModulos[p % i] * (p // i) % p
self.invModulos.append(next)
return self.invModulos[pi]
def invFactorial(self, n):
if n <= -1:
print("Invalid argument to calculate (n^(-1))!")
print("n must be non-negative value. But the argument was " + str(n))
exit()
if n < len(self.invFactorial_):
return self.invFactorial_[n]
self.inv(n) # To make sure already calculated n^-1
nextArr = [0] * (n + 1 - len(self.invFactorial_))
initialI = len(self.invFactorial_)
prev = self.invFactorial_[-1]
p = self.MOD
for i in range(initialI, n + 1):
prev = nextArr[i - initialI] = (prev * self.invModulos[i % p]) % p
self.invFactorial_ += nextArr
return self.invFactorial_[n]
class Combination:
def __init__(self, MOD):
self.MOD = MOD
self.factorial = Factorial(MOD)
def ncr(self, n, k):
if k < 0 or n < k:
return 0
k = min(k, n - k)
f = self.factorial
return f.calc(n) * f.invFactorial(max(n - k, k)) * f.invFactorial(min(k, n - k)) % self.MOD
mod=10**9+7
omod=998244353
#-------------------------------------------------------------------------
prime = [True for i in range(200001)]
pp=[0]*200001
def SieveOfEratosthenes(n=200000):
# Create a boolean array "prime[0..n]" and initialize
# all entries it as true. A value in prime[i] will
# finally be false if i is Not a prime, else true.
p = 2
while (p * p <= n):
# If prime[p] is not changed, then it is a prime
if (prime[p] == True):
# Update all multiples of p
for i in range(p * p, n+1, p):
prime[i] = False
p += 1
#---------------------------------running code------------------------------------------
a, b, c, d, n = map(int,input().split())
*arr, = map(int, input().split())
print('YES')
cur_i = i = 0
grid = [['.' for i in range(a + c)] for j in range(max(b, d))]
if (min(b, d) ^ (d < b)) & 1:
cur_j, cur_d = 0, 1
else:
cur_j, cur_d = a + c - 1, -1
while i < n:
grid[cur_i][cur_j] = chr(97 + i)
cur_j += cur_d
if cur_i >= b and cur_j < a or cur_i >= d and cur_j >= a or cur_j < 0 or cur_j >= a + c:
cur_i += 1
cur_j -= cur_d
cur_d = -cur_d
arr[i] -= 1
if not arr[i]:
i += 1
for i in grid:
print(''.join(i))
``` | output | 1 | 81,260 | 23 | 162,521 |
Provide tags and a correct Python 3 solution for this coding contest problem.
A revolution took place on the Buka Island. New government replaced the old one. The new government includes n parties and each of them is entitled to some part of the island according to their contribution to the revolution. However, they can't divide the island.
The island can be conventionally represented as two rectangles a × b and c × d unit squares in size correspondingly. The rectangles are located close to each other. At that, one of the sides with the length of a and one of the sides with the length of c lie on one line. You can see this in more details on the picture.
<image>
The i-th party is entitled to a part of the island equal to xi unit squares. Every such part should fully cover several squares of the island (it is not allowed to cover the squares partially) and be a connected figure. A "connected figure" presupposes that from any square of this party one can move to any other square of the same party moving through edge-adjacent squares also belonging to that party.
Your task is to divide the island between parties.
Input
The first line contains 5 space-separated integers — a, b, c, d and n (1 ≤ a, b, c, d ≤ 50, b ≠ d, 1 ≤ n ≤ 26). The second line contains n space-separated numbers. The i-th of them is equal to number xi (1 ≤ xi ≤ a × b + c × d). It is guaranteed that <image>.
Output
If dividing the island between parties in the required manner is impossible, print "NO" (without the quotes). Otherwise, print "YES" (also without the quotes) and, starting from the next line, print max(b, d) lines each containing a + c characters. To mark what square should belong to what party, use lowercase Latin letters. For the party that is first in order in the input data, use "a", for the second one use "b" and so on. Use "." for the squares that belong to the sea. The first symbol of the second line of the output data should correspond to the square that belongs to the rectangle a × b. The last symbol of the second line should correspond to the square that belongs to the rectangle c × d.
If there are several solutions output any.
Examples
Input
3 4 2 2 3
5 8 3
Output
YES
aaabb
aabbb
cbb..
ccb..
Input
3 2 1 4 4
1 2 3 4
Output
YES
abbd
cccd
...d
...d | instruction | 0 | 81,261 | 23 | 162,522 |
Tags: constructive algorithms
Correct Solution:
```
import os,io
from sys import stdout
import collections
# import random
import math
# from operator import itemgetter
input = io.BytesIO(os.read(0,os.fstat(0).st_size)).readline
from collections import Counter
# from decimal import Decimal
# import heapq
# from functools import lru_cache
# import sys
# sys.setrecursionlimit(10**6)
# from functools import lru_cache
# @lru_cache(maxsize=None)
def primes(n):
sieve = [True] * n
for i in range(3,int(n**0.5)+1,2):
if sieve[i]:
sieve[i*i::2*i]=[False]*((n-i*i-1)//(2*i)+1)
return [2] + [i for i in range(3,n,2) if sieve[i]]
def binomial_coefficient(n, k):
if 0 <= k <= n:
ntok = 1
ktok = 1
for t in range(1, min(k, n - k) + 1):
ntok *= n
ktok *= t
n -= 1
return ntok // ktok
else:
return 0
def powerOfK(k, max):
if k == 1:
return [1]
if k == -1:
return [-1, 1]
result = []
n = 1
while n <= max:
result.append(n)
n *= k
return result
def prefixSum(arr):
for i in range(1, len(arr)):
arr[i] = arr[i] + arr[i-1]
return arr
def divisors(n):
i = 1
result = []
while i*i <= n:
if n%i == 0:
if n/i == i:
result.append(i)
else:
result.append(i)
result.append(n/i)
i+=1
return result
def kadane(a,size):
max_so_far = 0
max_ending_here = 0
for i in range(0, size):
max_ending_here = max_ending_here + a[i]
if (max_so_far < max_ending_here):
max_so_far = max_ending_here
if max_ending_here < 0:
max_ending_here = 0
return max_so_far
# @lru_cache(maxsize=None)
def digitsSum(n):
if n == 0:
return 0
r = 0
while n > 0:
r += n % 10
n //= 10
return r
def print_grid(grid):
for line in grid:
print("".join(line))
# INPUTS --------------------------
# s = input().decode('utf-8').strip()
# n = int(input())
# l = list(map(int, input().split()))
# t = int(input())
# for _ in range(t):
a, b, c, d, n = list(map(int, input().split()))
letters = list(map(int, input().split()))
x = a + c
y = max(b, d)
grid = []
for i in range(y):
grid.append(['_']*x)
if d > b:
for i in range(b, d):
for j in range(0, a):
grid[i][j] = '.'
elif d < b:
for i in range(d, b):
for j in range(a, a+c):
grid[i][j] = '.'
if b < d:
space = d - b
if space % 2 == 0:
r1 = range(a+c-1, -1, -1)
r2 = range(0, a+c)
else:
r1 = range(0, a+c)
r2 = range(a+c-1, -1, -1)
else:
space = b - d
if space % 2 == 0:
r1 = range(0, a+c)
r2 = range(a+c-1, -1, -1)
else:
r1 = range(a+c-1, -1, -1)
r2 = range(0, a+c)
li = 0
lr = letters[li]
for i in range(y-1, -1, -1):
if ((y-1)-i) % 2 == 0:
r = r1
else:
r = r2
for j in r:
if grid[i][j] == '.':
continue
grid[i][j] = chr(97 + li)
lr -= 1
if lr == 0:
li += 1
if li < len(letters):
lr += letters[li]
else:
print("YES")
print_grid(grid)
exit()
``` | output | 1 | 81,261 | 23 | 162,523 |
Provide tags and a correct Python 3 solution for this coding contest problem.
A revolution took place on the Buka Island. New government replaced the old one. The new government includes n parties and each of them is entitled to some part of the island according to their contribution to the revolution. However, they can't divide the island.
The island can be conventionally represented as two rectangles a × b and c × d unit squares in size correspondingly. The rectangles are located close to each other. At that, one of the sides with the length of a and one of the sides with the length of c lie on one line. You can see this in more details on the picture.
<image>
The i-th party is entitled to a part of the island equal to xi unit squares. Every such part should fully cover several squares of the island (it is not allowed to cover the squares partially) and be a connected figure. A "connected figure" presupposes that from any square of this party one can move to any other square of the same party moving through edge-adjacent squares also belonging to that party.
Your task is to divide the island between parties.
Input
The first line contains 5 space-separated integers — a, b, c, d and n (1 ≤ a, b, c, d ≤ 50, b ≠ d, 1 ≤ n ≤ 26). The second line contains n space-separated numbers. The i-th of them is equal to number xi (1 ≤ xi ≤ a × b + c × d). It is guaranteed that <image>.
Output
If dividing the island between parties in the required manner is impossible, print "NO" (without the quotes). Otherwise, print "YES" (also without the quotes) and, starting from the next line, print max(b, d) lines each containing a + c characters. To mark what square should belong to what party, use lowercase Latin letters. For the party that is first in order in the input data, use "a", for the second one use "b" and so on. Use "." for the squares that belong to the sea. The first symbol of the second line of the output data should correspond to the square that belongs to the rectangle a × b. The last symbol of the second line should correspond to the square that belongs to the rectangle c × d.
If there are several solutions output any.
Examples
Input
3 4 2 2 3
5 8 3
Output
YES
aaabb
aabbb
cbb..
ccb..
Input
3 2 1 4 4
1 2 3 4
Output
YES
abbd
cccd
...d
...d | instruction | 0 | 81,262 | 23 | 162,524 |
Tags: constructive algorithms
Correct Solution:
```
#------------------------template--------------------------#
import os
import sys
from math import *
from collections import *
from fractions import *
from bisect import *
from heapq import*
from io import BytesIO, IOBase
def vsInput():
sys.stdin = open('input.txt', 'r')
sys.stdout = open('output.txt', 'w')
BUFSIZE = 8192
class FastIO(IOBase):
newlines = 0
def __init__(self, file):
self._fd = file.fileno()
self.buffer = BytesIO()
self.writable = "x" in file.mode or "r" not in file.mode
self.write = self.buffer.write if self.writable else None
def read(self):
while True:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
if not b:
break
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines = 0
return self.buffer.read()
def readline(self):
while self.newlines == 0:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
self.newlines = b.count(b"\n") + (not b)
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines -= 1
return self.buffer.readline()
def flush(self):
if self.writable:
os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
self.write = lambda s: self.buffer.write(s.encode("ascii"))
self.read = lambda: self.buffer.read().decode("ascii")
self.readline = lambda: self.buffer.readline().decode("ascii")
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")
ALPHA='abcdefghijklmnopqrstuvwxyz'
M=1000000007
EPS=1e-6
def value():return tuple(map(int,input().split()))
def array():return [int(i) for i in input().split()]
def Int():return int(input())
def Str():return input()
def arrayS():return [i for i in input().split()]
#-------------------------code---------------------------#
# vsInput()
a,b,c,d,n=value()
area=array()
ans=[['.' for i in range(a+c)] for j in range(max(b,d))]
rev=False
if(d<b):
a,c=c,a
b,d=d,b
rev=True
# print(a,b)
ind=0
for j in range(a):
for i in range(b):
if(area[ind]==0):
ind+=1
if(a%2==0):
if(j%2==0): ans[i][j]=ALPHA[ind]
else: ans[b-i-1][j]=ALPHA[ind]
else:
if(j%2): ans[i][j]=ALPHA[ind]
else: ans[b-i-1][j]=ALPHA[ind]
area[ind]-=1
_j=-1
for j in range(a,a+c):
_j+=1
for i in range(d):
if(area[ind]==0):
ind+=1
if(_j%2==0): ans[i][j]=ALPHA[ind]
else: ans[d-i-1][j]=ALPHA[ind]
area[ind]-=1
print("YES")
for i in ans:
if(rev):
print(*i[::-1],sep="")
else:
print(*i,sep="")
``` | output | 1 | 81,262 | 23 | 162,525 |
Provide tags and a correct Python 3 solution for this coding contest problem.
A revolution took place on the Buka Island. New government replaced the old one. The new government includes n parties and each of them is entitled to some part of the island according to their contribution to the revolution. However, they can't divide the island.
The island can be conventionally represented as two rectangles a × b and c × d unit squares in size correspondingly. The rectangles are located close to each other. At that, one of the sides with the length of a and one of the sides with the length of c lie on one line. You can see this in more details on the picture.
<image>
The i-th party is entitled to a part of the island equal to xi unit squares. Every such part should fully cover several squares of the island (it is not allowed to cover the squares partially) and be a connected figure. A "connected figure" presupposes that from any square of this party one can move to any other square of the same party moving through edge-adjacent squares also belonging to that party.
Your task is to divide the island between parties.
Input
The first line contains 5 space-separated integers — a, b, c, d and n (1 ≤ a, b, c, d ≤ 50, b ≠ d, 1 ≤ n ≤ 26). The second line contains n space-separated numbers. The i-th of them is equal to number xi (1 ≤ xi ≤ a × b + c × d). It is guaranteed that <image>.
Output
If dividing the island between parties in the required manner is impossible, print "NO" (without the quotes). Otherwise, print "YES" (also without the quotes) and, starting from the next line, print max(b, d) lines each containing a + c characters. To mark what square should belong to what party, use lowercase Latin letters. For the party that is first in order in the input data, use "a", for the second one use "b" and so on. Use "." for the squares that belong to the sea. The first symbol of the second line of the output data should correspond to the square that belongs to the rectangle a × b. The last symbol of the second line should correspond to the square that belongs to the rectangle c × d.
If there are several solutions output any.
Examples
Input
3 4 2 2 3
5 8 3
Output
YES
aaabb
aabbb
cbb..
ccb..
Input
3 2 1 4 4
1 2 3 4
Output
YES
abbd
cccd
...d
...d | instruction | 0 | 81,263 | 23 | 162,526 |
Tags: constructive algorithms
Correct Solution:
```
import sys
input=sys.stdin.readline
l=input().split()
a=int(l[0])
b=int(l[1])
c=int(l[2])
d=int(l[3])
n=int(l[4])
l=[[0 for i in range(a+c)] for i in range(max(b,d))]
if(b>=d):
for i in range(min(b,d),max(b,d)):
for j in range(a,a+c):
l[i][j]='.'
else:
for i in range(min(b,d),max(b,d)):
for j in range(a):
l[i][j]='.'
currx=0
curry=0
curr=0
p=input().split()
li=[int(i) for i in p]
print("YES")
if(b>=d):
if(d%2==0):
curr=0
for i in range(b):
if(i%2==0):
for j in range(a+c):
if(l[i][j]=='.'):
continue
l[i][j]=curr
li[curr]-=1
if(li[curr]==0):
curr+=1
else:
for j in range(a+c-1,-1,-1):
if(l[i][j]=='.'):
continue
l[i][j]=curr
li[curr]-=1
if(li[curr]==0):
curr+=1
else:
curr=0
for i in range(b):
if(i%2==1):
for j in range(a+c):
if(l[i][j]=='.'):
continue
l[i][j]=curr
li[curr]-=1
if(li[curr]==0):
curr+=1
else:
for j in range(a+c-1,-1,-1):
if(l[i][j]=='.'):
continue
l[i][j]=curr
li[curr]-=1
if(li[curr]==0):
curr+=1
else:
if(b%2==1):
curr=0
for i in range(d):
if(i%2==0):
for j in range(a+c):
if(l[i][j]=='.'):
continue
l[i][j]=curr
li[curr]-=1
if(li[curr]==0):
curr+=1
else:
for j in range(a+c-1,-1,-1):
if(l[i][j]=='.'):
continue
l[i][j]=curr
li[curr]-=1
if(li[curr]==0):
curr+=1
else:
curr=0
for i in range(d):
if(i%2==1):
for j in range(a+c):
if(l[i][j]=='.'):
continue
l[i][j]=curr
li[curr]-=1
if(li[curr]==0):
curr+=1
else:
for j in range(a+c-1,-1,-1):
if(l[i][j]=='.'):
continue
l[i][j]=curr
li[curr]-=1
if(li[curr]==0):
curr+=1
for i in l:
for j in i:
if(j=='.'):
print('.',end="")
else:
print(chr(97+j),end="")
print()
``` | output | 1 | 81,263 | 23 | 162,527 |
Provide tags and a correct Python 3 solution for this coding contest problem.
A revolution took place on the Buka Island. New government replaced the old one. The new government includes n parties and each of them is entitled to some part of the island according to their contribution to the revolution. However, they can't divide the island.
The island can be conventionally represented as two rectangles a × b and c × d unit squares in size correspondingly. The rectangles are located close to each other. At that, one of the sides with the length of a and one of the sides with the length of c lie on one line. You can see this in more details on the picture.
<image>
The i-th party is entitled to a part of the island equal to xi unit squares. Every such part should fully cover several squares of the island (it is not allowed to cover the squares partially) and be a connected figure. A "connected figure" presupposes that from any square of this party one can move to any other square of the same party moving through edge-adjacent squares also belonging to that party.
Your task is to divide the island between parties.
Input
The first line contains 5 space-separated integers — a, b, c, d and n (1 ≤ a, b, c, d ≤ 50, b ≠ d, 1 ≤ n ≤ 26). The second line contains n space-separated numbers. The i-th of them is equal to number xi (1 ≤ xi ≤ a × b + c × d). It is guaranteed that <image>.
Output
If dividing the island between parties in the required manner is impossible, print "NO" (without the quotes). Otherwise, print "YES" (also without the quotes) and, starting from the next line, print max(b, d) lines each containing a + c characters. To mark what square should belong to what party, use lowercase Latin letters. For the party that is first in order in the input data, use "a", for the second one use "b" and so on. Use "." for the squares that belong to the sea. The first symbol of the second line of the output data should correspond to the square that belongs to the rectangle a × b. The last symbol of the second line should correspond to the square that belongs to the rectangle c × d.
If there are several solutions output any.
Examples
Input
3 4 2 2 3
5 8 3
Output
YES
aaabb
aabbb
cbb..
ccb..
Input
3 2 1 4 4
1 2 3 4
Output
YES
abbd
cccd
...d
...d | instruction | 0 | 81,264 | 23 | 162,528 |
Tags: constructive algorithms
Correct Solution:
```
# https://codeforces.com/problemset/problem/63/D
def solve(x0, y0, dx, X):
for i, x in enumerate(X):
while x > 0:
x-=1
m[y0][x0] = i
x0 += dx
if x == 0 and i == len(X)-1:
break
if x0 == -1:
y0 += 1
x0 = 0
dx *= -1
if m[y0][x0] == -1:
return False
elif x0 == a+c:
y0 += 1
x0 = a+c-1
dx *=-1
if m[y0][x0] == -1:
return False
elif m[y0][x0] == -1:
y0 +=1
x0 -=dx
dx *=-1
return True
a, b, c, d, n = map(int, input().split())
X = list(map(int, input().split()))
m = [[0] * (a+c) for _ in range(max(b, d))]
if b < d:
for i in range(b, d):
for j in range(a):
m[i][j] = -1
else:
for i in range(d, b):
for j in range(a, a+c):
m[i][j] = -1
if solve(a+c-1, 0, -1, X) == False:
solve(0, 0, 1, X)
print('YES')
for x in m:
print(''.join([chr(c+97) if c>=0 else '.' for c in x]))
#3 2 1 4 4
#1 2 3 4
``` | output | 1 | 81,264 | 23 | 162,529 |
Provide tags and a correct Python 3 solution for this coding contest problem.
A revolution took place on the Buka Island. New government replaced the old one. The new government includes n parties and each of them is entitled to some part of the island according to their contribution to the revolution. However, they can't divide the island.
The island can be conventionally represented as two rectangles a × b and c × d unit squares in size correspondingly. The rectangles are located close to each other. At that, one of the sides with the length of a and one of the sides with the length of c lie on one line. You can see this in more details on the picture.
<image>
The i-th party is entitled to a part of the island equal to xi unit squares. Every such part should fully cover several squares of the island (it is not allowed to cover the squares partially) and be a connected figure. A "connected figure" presupposes that from any square of this party one can move to any other square of the same party moving through edge-adjacent squares also belonging to that party.
Your task is to divide the island between parties.
Input
The first line contains 5 space-separated integers — a, b, c, d and n (1 ≤ a, b, c, d ≤ 50, b ≠ d, 1 ≤ n ≤ 26). The second line contains n space-separated numbers. The i-th of them is equal to number xi (1 ≤ xi ≤ a × b + c × d). It is guaranteed that <image>.
Output
If dividing the island between parties in the required manner is impossible, print "NO" (without the quotes). Otherwise, print "YES" (also without the quotes) and, starting from the next line, print max(b, d) lines each containing a + c characters. To mark what square should belong to what party, use lowercase Latin letters. For the party that is first in order in the input data, use "a", for the second one use "b" and so on. Use "." for the squares that belong to the sea. The first symbol of the second line of the output data should correspond to the square that belongs to the rectangle a × b. The last symbol of the second line should correspond to the square that belongs to the rectangle c × d.
If there are several solutions output any.
Examples
Input
3 4 2 2 3
5 8 3
Output
YES
aaabb
aabbb
cbb..
ccb..
Input
3 2 1 4 4
1 2 3 4
Output
YES
abbd
cccd
...d
...d | instruction | 0 | 81,265 | 23 | 162,530 |
Tags: constructive algorithms
Correct Solution:
```
import sys
from functools import lru_cache, cmp_to_key
from heapq import merge, heapify, heappop, heappush
from math import *
from collections import defaultdict as dd, deque, Counter as C
from itertools import combinations as comb, permutations as perm
from bisect import bisect_left as bl, bisect_right as br, bisect
from time import perf_counter
from fractions import Fraction
import copy
import time
starttime = time.time()
mod = int(pow(10, 9) + 7)
mod2 = 998244353
def data(): return sys.stdin.readline().strip()
def out(*var, end="\n"): sys.stdout.write(' '.join(map(str, var))+end)
def L(): return list(sp())
def sl(): return list(ssp())
def sp(): return map(int, data().split())
def ssp(): return map(str, data().split())
def l1d(n, val=0): return [val for i in range(n)]
def l2d(n, m, val=0): return [l1d(n, val) for j in range(m)]
try:
# sys.setrecursionlimit(int(pow(10,4)))
sys.stdin = open("input.txt", "r")
# sys.stdout = open("../output.txt", "w")
except:
pass
def pmat(A):
for ele in A:
print(*ele,end="\n")
# def seive():
# prime=[1 for i in range(10**6+1)]
# prime[0]=0
# prime[1]=0
# for i in range(10**6+1):
# if(prime[i]):
# for j in range(2*i,10**6+1,i):
# prime[j]=0
# return prime
a, b, c, d, n = L()
arr= L()
print('YES')
cur_i = i = 0
grid = [['.' for i in range(a + c)] for j in range(max(b, d))]
if (min(b, d) ^ (d < b)) & 1:
cur_j, cur_d = 0, 1
else:
cur_j, cur_d = a + c - 1, -1
while i < n:
# print(grid)
grid[cur_i][cur_j] = chr(97 + i)
cur_j += cur_d
if cur_i >= b and cur_j < a or cur_i >= d and cur_j >= a or cur_j < 0 or cur_j >= a + c:
cur_i += 1
cur_j -= cur_d
cur_d = -cur_d
arr[i] -= 1
if not arr[i]:
i += 1
for i in grid:
print(''.join(i))
endtime = time.time()
# print(f"Runtime of the program is {endtime - starttime}")
``` | output | 1 | 81,265 | 23 | 162,531 |
Provide tags and a correct Python 3 solution for this coding contest problem.
A revolution took place on the Buka Island. New government replaced the old one. The new government includes n parties and each of them is entitled to some part of the island according to their contribution to the revolution. However, they can't divide the island.
The island can be conventionally represented as two rectangles a × b and c × d unit squares in size correspondingly. The rectangles are located close to each other. At that, one of the sides with the length of a and one of the sides with the length of c lie on one line. You can see this in more details on the picture.
<image>
The i-th party is entitled to a part of the island equal to xi unit squares. Every such part should fully cover several squares of the island (it is not allowed to cover the squares partially) and be a connected figure. A "connected figure" presupposes that from any square of this party one can move to any other square of the same party moving through edge-adjacent squares also belonging to that party.
Your task is to divide the island between parties.
Input
The first line contains 5 space-separated integers — a, b, c, d and n (1 ≤ a, b, c, d ≤ 50, b ≠ d, 1 ≤ n ≤ 26). The second line contains n space-separated numbers. The i-th of them is equal to number xi (1 ≤ xi ≤ a × b + c × d). It is guaranteed that <image>.
Output
If dividing the island between parties in the required manner is impossible, print "NO" (without the quotes). Otherwise, print "YES" (also without the quotes) and, starting from the next line, print max(b, d) lines each containing a + c characters. To mark what square should belong to what party, use lowercase Latin letters. For the party that is first in order in the input data, use "a", for the second one use "b" and so on. Use "." for the squares that belong to the sea. The first symbol of the second line of the output data should correspond to the square that belongs to the rectangle a × b. The last symbol of the second line should correspond to the square that belongs to the rectangle c × d.
If there are several solutions output any.
Examples
Input
3 4 2 2 3
5 8 3
Output
YES
aaabb
aabbb
cbb..
ccb..
Input
3 2 1 4 4
1 2 3 4
Output
YES
abbd
cccd
...d
...d | instruction | 0 | 81,266 | 23 | 162,532 |
Tags: constructive algorithms
Correct Solution:
```
a, b, c, d,n = [int(x) for x in input().split()]
p = [int(x) for x in input().split()]
grid = [['o' for _ in range(a+c)] for _ in range(max(b,d))]
e, f = 0,0
dr = 1
if b > d:
for i in range(d,b):
for j in range(a,a+c):
grid[i][j] = '.'
if d%2 == 1:
e, f = 0, (a+c-1)
dr = -1
elif d > b:
for i in range(b,d):
for j in range(0,a):
grid[i][j] = '.'
if b%2 == 0:
e, f = 0, (a+c-1)
dr = -1
k = 0
for _ in range((a+c)*max(d,b)):
if grid[e][f] != '.':
grid[e][f] = chr(ord('a')+k)
p[k] -= 1
if p[k] == 0: k+=1
f += dr
if f == -1:
e, f = e+1, 0
dr = 1
if f == a+c:
e, f = e+1, a+c-1
dr = -1
print("YES")
for i in range(max(b,d)):
print("".join(grid[i]))
``` | output | 1 | 81,266 | 23 | 162,533 |
Provide tags and a correct Python 3 solution for this coding contest problem.
A revolution took place on the Buka Island. New government replaced the old one. The new government includes n parties and each of them is entitled to some part of the island according to their contribution to the revolution. However, they can't divide the island.
The island can be conventionally represented as two rectangles a × b and c × d unit squares in size correspondingly. The rectangles are located close to each other. At that, one of the sides with the length of a and one of the sides with the length of c lie on one line. You can see this in more details on the picture.
<image>
The i-th party is entitled to a part of the island equal to xi unit squares. Every such part should fully cover several squares of the island (it is not allowed to cover the squares partially) and be a connected figure. A "connected figure" presupposes that from any square of this party one can move to any other square of the same party moving through edge-adjacent squares also belonging to that party.
Your task is to divide the island between parties.
Input
The first line contains 5 space-separated integers — a, b, c, d and n (1 ≤ a, b, c, d ≤ 50, b ≠ d, 1 ≤ n ≤ 26). The second line contains n space-separated numbers. The i-th of them is equal to number xi (1 ≤ xi ≤ a × b + c × d). It is guaranteed that <image>.
Output
If dividing the island between parties in the required manner is impossible, print "NO" (without the quotes). Otherwise, print "YES" (also without the quotes) and, starting from the next line, print max(b, d) lines each containing a + c characters. To mark what square should belong to what party, use lowercase Latin letters. For the party that is first in order in the input data, use "a", for the second one use "b" and so on. Use "." for the squares that belong to the sea. The first symbol of the second line of the output data should correspond to the square that belongs to the rectangle a × b. The last symbol of the second line should correspond to the square that belongs to the rectangle c × d.
If there are several solutions output any.
Examples
Input
3 4 2 2 3
5 8 3
Output
YES
aaabb
aabbb
cbb..
ccb..
Input
3 2 1 4 4
1 2 3 4
Output
YES
abbd
cccd
...d
...d | instruction | 0 | 81,267 | 23 | 162,534 |
Tags: constructive algorithms
Correct Solution:
```
def read_int():
return int(input().strip())
def read_ints():
return list(map(int, input().strip().split(' ')))
def solve():
a, b, c, d, n = read_ints()
parties = read_ints()
def parties_iterator():
alphas = 'abcdefghijklmnopqrstuvwxyz'
for party, count in zip(alphas, parties):
for _ in range(count):
yield party
ps = parties_iterator()
get_party = lambda: next(ps)
island = [
['.' for _ in range(a+c)] for _ in range(max(b, d))
]
for i in range(b-1, -1, -1):
if i%2 == 0:
for j in range(a):
island[i][j] = get_party()
else:
for j in range(a):
island[i][a-1-j] = get_party()
for i in range(d):
if i%2 == 0:
for j in range(c):
island[i][a+j] = get_party()
else:
for j in range(c):
island[i][a+c-1-j] = get_party()
print('YES')
for row in island:
print(''.join(row))
if __name__ == '__main__':
solve()
``` | output | 1 | 81,267 | 23 | 162,535 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A revolution took place on the Buka Island. New government replaced the old one. The new government includes n parties and each of them is entitled to some part of the island according to their contribution to the revolution. However, they can't divide the island.
The island can be conventionally represented as two rectangles a × b and c × d unit squares in size correspondingly. The rectangles are located close to each other. At that, one of the sides with the length of a and one of the sides with the length of c lie on one line. You can see this in more details on the picture.
<image>
The i-th party is entitled to a part of the island equal to xi unit squares. Every such part should fully cover several squares of the island (it is not allowed to cover the squares partially) and be a connected figure. A "connected figure" presupposes that from any square of this party one can move to any other square of the same party moving through edge-adjacent squares also belonging to that party.
Your task is to divide the island between parties.
Input
The first line contains 5 space-separated integers — a, b, c, d and n (1 ≤ a, b, c, d ≤ 50, b ≠ d, 1 ≤ n ≤ 26). The second line contains n space-separated numbers. The i-th of them is equal to number xi (1 ≤ xi ≤ a × b + c × d). It is guaranteed that <image>.
Output
If dividing the island between parties in the required manner is impossible, print "NO" (without the quotes). Otherwise, print "YES" (also without the quotes) and, starting from the next line, print max(b, d) lines each containing a + c characters. To mark what square should belong to what party, use lowercase Latin letters. For the party that is first in order in the input data, use "a", for the second one use "b" and so on. Use "." for the squares that belong to the sea. The first symbol of the second line of the output data should correspond to the square that belongs to the rectangle a × b. The last symbol of the second line should correspond to the square that belongs to the rectangle c × d.
If there are several solutions output any.
Examples
Input
3 4 2 2 3
5 8 3
Output
YES
aaabb
aabbb
cbb..
ccb..
Input
3 2 1 4 4
1 2 3 4
Output
YES
abbd
cccd
...d
...d
Submitted Solution:
```
import sys
from array import array # noqa: F401
def input():
return sys.stdin.buffer.readline().decode('utf-8')
a, b, c, d, n = map(int, input().split())
party = list(map(int, input().split()))
island = [['.'] * (a + c) for _ in range(max(b, d))]
i, j, ub, lb = (b - 1 if a % 2 else 0), 0, 0, b
delta = (-1 if a % 2 else 1)
pi = 0
while pi < n:
island[i][j] = chr(97 + pi)
party[pi] -= 1
if party[pi] == 0:
pi += 1
if not (ub <= i + delta < lb):
j += 1
if j == a:
lb = d
delta *= -1
else:
i += delta
print('YES')
for row in island:
print(*row, sep='')
``` | instruction | 0 | 81,268 | 23 | 162,536 |
Yes | output | 1 | 81,268 | 23 | 162,537 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A revolution took place on the Buka Island. New government replaced the old one. The new government includes n parties and each of them is entitled to some part of the island according to their contribution to the revolution. However, they can't divide the island.
The island can be conventionally represented as two rectangles a × b and c × d unit squares in size correspondingly. The rectangles are located close to each other. At that, one of the sides with the length of a and one of the sides with the length of c lie on one line. You can see this in more details on the picture.
<image>
The i-th party is entitled to a part of the island equal to xi unit squares. Every such part should fully cover several squares of the island (it is not allowed to cover the squares partially) and be a connected figure. A "connected figure" presupposes that from any square of this party one can move to any other square of the same party moving through edge-adjacent squares also belonging to that party.
Your task is to divide the island between parties.
Input
The first line contains 5 space-separated integers — a, b, c, d and n (1 ≤ a, b, c, d ≤ 50, b ≠ d, 1 ≤ n ≤ 26). The second line contains n space-separated numbers. The i-th of them is equal to number xi (1 ≤ xi ≤ a × b + c × d). It is guaranteed that <image>.
Output
If dividing the island between parties in the required manner is impossible, print "NO" (without the quotes). Otherwise, print "YES" (also without the quotes) and, starting from the next line, print max(b, d) lines each containing a + c characters. To mark what square should belong to what party, use lowercase Latin letters. For the party that is first in order in the input data, use "a", for the second one use "b" and so on. Use "." for the squares that belong to the sea. The first symbol of the second line of the output data should correspond to the square that belongs to the rectangle a × b. The last symbol of the second line should correspond to the square that belongs to the rectangle c × d.
If there are several solutions output any.
Examples
Input
3 4 2 2 3
5 8 3
Output
YES
aaabb
aabbb
cbb..
ccb..
Input
3 2 1 4 4
1 2 3 4
Output
YES
abbd
cccd
...d
...d
Submitted Solution:
```
import math
import sys
input=sys.stdin.readline
from collections import Counter, defaultdict, deque
def f(a, b, c, d, n, x):
grid = [["." for co in range(a+c)]for r in range(max(b, d))]
alphnum = 97
pos = 0
if (a+1)%2:#c odd start from top
for co in range(a):
if co % 2==0:
r = 0
while r < b:
grid[r][co] = chr(alphnum)
x[pos] -= 1
if x[pos] == 0:
pos += 1
alphnum += 1
r += 1
else:
r = b-1
while r > -1:
grid[r][co] = chr(alphnum)
x[pos] -= 1
if x[pos] == 0:
pos += 1
alphnum += 1
r -= 1
# for r in range(max(b, d)):
# print(grid[r])
for co in range(a, a+c):
if co%2==0:
r = 0
while r < d:
#print(r, co)
grid[r][co] = chr(alphnum)
x[pos] -= 1
if x[pos] == 0:
pos += 1
alphnum += 1
r += 1
#print(r, c)
else:
r = d-1
while r > -1:
#print(r, co)
grid[r][co] = chr(alphnum)
x[pos] -= 1
if x[pos] == 0:
pos += 1
alphnum += 1
r -= 1
else:#c even start from down
for co in range(a):
if co % 2==1:
r = 0
while r < b:
grid[r][co] = chr(alphnum)
x[pos] -= 1
if x[pos] == 0:
pos += 1
alphnum += 1
r += 1
else:
r = b-1
while r > -1:
grid[r][co] = chr(alphnum)
x[pos] -= 1
if x[pos] == 0:
pos += 1
alphnum += 1
r -= 1
# for r in range(max(b, d)):
# print(grid[r])
for co in range(a, a+c):
if co%2==1:
r = 0
while r < d:
#print(r, co)
grid[r][co] = chr(alphnum)
x[pos] -= 1
if x[pos] == 0:
pos += 1
alphnum += 1
r += 1
#print(r, c)
else:
r = d-1
while r > -1:
#print(r, co)
grid[r][co] = chr(alphnum)
x[pos] -= 1
if x[pos] == 0:
pos += 1
alphnum += 1
r -= 1
# for r in range(max(b, d)):
# print(grid[r])
return grid
t = 1
result = []
for i in range(t):
#n = int(input())
a, b, c, d, n = list(map(int, input().split()))
x = list(map(int, input().split()))
result.append(f(a, b, c, d, n, x))
for i in range(t):
print("YES")
for j in range(len(result[i])):
print("".join(result[i][j]))
``` | instruction | 0 | 81,269 | 23 | 162,538 |
Yes | output | 1 | 81,269 | 23 | 162,539 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A revolution took place on the Buka Island. New government replaced the old one. The new government includes n parties and each of them is entitled to some part of the island according to their contribution to the revolution. However, they can't divide the island.
The island can be conventionally represented as two rectangles a × b and c × d unit squares in size correspondingly. The rectangles are located close to each other. At that, one of the sides with the length of a and one of the sides with the length of c lie on one line. You can see this in more details on the picture.
<image>
The i-th party is entitled to a part of the island equal to xi unit squares. Every such part should fully cover several squares of the island (it is not allowed to cover the squares partially) and be a connected figure. A "connected figure" presupposes that from any square of this party one can move to any other square of the same party moving through edge-adjacent squares also belonging to that party.
Your task is to divide the island between parties.
Input
The first line contains 5 space-separated integers — a, b, c, d and n (1 ≤ a, b, c, d ≤ 50, b ≠ d, 1 ≤ n ≤ 26). The second line contains n space-separated numbers. The i-th of them is equal to number xi (1 ≤ xi ≤ a × b + c × d). It is guaranteed that <image>.
Output
If dividing the island between parties in the required manner is impossible, print "NO" (without the quotes). Otherwise, print "YES" (also without the quotes) and, starting from the next line, print max(b, d) lines each containing a + c characters. To mark what square should belong to what party, use lowercase Latin letters. For the party that is first in order in the input data, use "a", for the second one use "b" and so on. Use "." for the squares that belong to the sea. The first symbol of the second line of the output data should correspond to the square that belongs to the rectangle a × b. The last symbol of the second line should correspond to the square that belongs to the rectangle c × d.
If there are several solutions output any.
Examples
Input
3 4 2 2 3
5 8 3
Output
YES
aaabb
aabbb
cbb..
ccb..
Input
3 2 1 4 4
1 2 3 4
Output
YES
abbd
cccd
...d
...d
Submitted Solution:
```
a, b, c, d, n = map(int,input().split())
*arr, = map(int, input().split())
print('YES')
cur_i = i = 0
grid = [['.' for i in range(a + c)] for j in range(max(b, d))]
if (min(b, d) ^ (d < b)) & 1:
cur_j, cur_d = 0, 1
else:
cur_j, cur_d = a + c - 1, -1
while i < n:
grid[cur_i][cur_j] = chr(97 + i)
cur_j += cur_d
if cur_i >= b and cur_j < a or cur_i >= d and cur_j >= a or cur_j < 0 or cur_j >= a + c:
cur_i += 1
cur_j -= cur_d
cur_d = -cur_d
arr[i] -= 1
if not arr[i]:
i += 1
for i in grid:
print(''.join(i))
``` | instruction | 0 | 81,270 | 23 | 162,540 |
Yes | output | 1 | 81,270 | 23 | 162,541 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A revolution took place on the Buka Island. New government replaced the old one. The new government includes n parties and each of them is entitled to some part of the island according to their contribution to the revolution. However, they can't divide the island.
The island can be conventionally represented as two rectangles a × b and c × d unit squares in size correspondingly. The rectangles are located close to each other. At that, one of the sides with the length of a and one of the sides with the length of c lie on one line. You can see this in more details on the picture.
<image>
The i-th party is entitled to a part of the island equal to xi unit squares. Every such part should fully cover several squares of the island (it is not allowed to cover the squares partially) and be a connected figure. A "connected figure" presupposes that from any square of this party one can move to any other square of the same party moving through edge-adjacent squares also belonging to that party.
Your task is to divide the island between parties.
Input
The first line contains 5 space-separated integers — a, b, c, d and n (1 ≤ a, b, c, d ≤ 50, b ≠ d, 1 ≤ n ≤ 26). The second line contains n space-separated numbers. The i-th of them is equal to number xi (1 ≤ xi ≤ a × b + c × d). It is guaranteed that <image>.
Output
If dividing the island between parties in the required manner is impossible, print "NO" (without the quotes). Otherwise, print "YES" (also without the quotes) and, starting from the next line, print max(b, d) lines each containing a + c characters. To mark what square should belong to what party, use lowercase Latin letters. For the party that is first in order in the input data, use "a", for the second one use "b" and so on. Use "." for the squares that belong to the sea. The first symbol of the second line of the output data should correspond to the square that belongs to the rectangle a × b. The last symbol of the second line should correspond to the square that belongs to the rectangle c × d.
If there are several solutions output any.
Examples
Input
3 4 2 2 3
5 8 3
Output
YES
aaabb
aabbb
cbb..
ccb..
Input
3 2 1 4 4
1 2 3 4
Output
YES
abbd
cccd
...d
...d
Submitted Solution:
```
import os
import sys
from io import BytesIO, IOBase
from collections import Counter
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)
def gcd(a, b):
if a == 0:
return b
return gcd(b % a, a)
def lcm(a, b):
return (a * b) / gcd(a, b)
pow2=[1]
d={}
def main():
a,b,c,d,n=map(int, input().split())
x=list(map(int, input().split()))
ans=[['.' for i in range(a+c)] for j in range(max(b,d))]
t=0
# for i in ans:
# print(*i)
now=0
ind=0
if a%2:
t=1
for j in range(a+c):
if j<a:
if t==0:
for i in range(b):
ans[i][j]=chr(ord('a')+now)
x[ind]-=1
if x[ind]==0:
ind+=1
now+=1
else:
for i in range(b-1,-1,-1):
ans[i][j] = chr(ord('a') + now)
x[ind]-=1
if x[ind]==0:
ind+=1
now+=1
else:
if t==0:
for i in range(d):
ans[i][j]=chr(ord('a')+now)
x[ind]-=1
if x[ind]==0:
ind+=1
now+=1
else:
for i in range(d-1,-1,-1):
ans[i][j] = chr(ord('a') + now)
x[ind]-=1
if x[ind]==0:
ind+=1
now+=1
t^=1
print('YES')
for i in ans:
for j in i:
print(j,end='')
print('')
return
if __name__=="__main__":
main()
``` | instruction | 0 | 81,271 | 23 | 162,542 |
Yes | output | 1 | 81,271 | 23 | 162,543 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A revolution took place on the Buka Island. New government replaced the old one. The new government includes n parties and each of them is entitled to some part of the island according to their contribution to the revolution. However, they can't divide the island.
The island can be conventionally represented as two rectangles a × b and c × d unit squares in size correspondingly. The rectangles are located close to each other. At that, one of the sides with the length of a and one of the sides with the length of c lie on one line. You can see this in more details on the picture.
<image>
The i-th party is entitled to a part of the island equal to xi unit squares. Every such part should fully cover several squares of the island (it is not allowed to cover the squares partially) and be a connected figure. A "connected figure" presupposes that from any square of this party one can move to any other square of the same party moving through edge-adjacent squares also belonging to that party.
Your task is to divide the island between parties.
Input
The first line contains 5 space-separated integers — a, b, c, d and n (1 ≤ a, b, c, d ≤ 50, b ≠ d, 1 ≤ n ≤ 26). The second line contains n space-separated numbers. The i-th of them is equal to number xi (1 ≤ xi ≤ a × b + c × d). It is guaranteed that <image>.
Output
If dividing the island between parties in the required manner is impossible, print "NO" (without the quotes). Otherwise, print "YES" (also without the quotes) and, starting from the next line, print max(b, d) lines each containing a + c characters. To mark what square should belong to what party, use lowercase Latin letters. For the party that is first in order in the input data, use "a", for the second one use "b" and so on. Use "." for the squares that belong to the sea. The first symbol of the second line of the output data should correspond to the square that belongs to the rectangle a × b. The last symbol of the second line should correspond to the square that belongs to the rectangle c × d.
If there are several solutions output any.
Examples
Input
3 4 2 2 3
5 8 3
Output
YES
aaabb
aabbb
cbb..
ccb..
Input
3 2 1 4 4
1 2 3 4
Output
YES
abbd
cccd
...d
...d
Submitted Solution:
```
import sys
from functools import lru_cache, cmp_to_key
from heapq import merge, heapify, heappop, heappush
from math import *
from collections import defaultdict as dd, deque, Counter as C
from itertools import combinations as comb, permutations as perm
from bisect import bisect_left as bl, bisect_right as br, bisect
from time import perf_counter
from fractions import Fraction
import copy
import time
starttime = time.time()
mod = int(pow(10, 9) + 7)
mod2 = 998244353
def data(): return sys.stdin.readline().strip()
def out(*var, end="\n"): sys.stdout.write(' '.join(map(str, var))+end)
def L(): return list(sp())
def sl(): return list(ssp())
def sp(): return map(int, data().split())
def ssp(): return map(str, data().split())
def l1d(n, val=0): return [val for i in range(n)]
def l2d(n, m, val=0): return [l1d(n, val) for j in range(m)]
try:
# sys.setrecursionlimit(int(pow(10,4)))
sys.stdin = open("input.txt", "r")
# sys.stdout = open("../output.txt", "w")
except:
pass
def pmat(A):
for ele in A:
print(*ele,end="\n")
# def seive():
# prime=[1 for i in range(10**6+1)]
# prime[0]=0
# prime[1]=0
# for i in range(10**6+1):
# if(prime[i]):
# for j in range(2*i,10**6+1,i):
# prime[j]=0
# return prime
a,b,c,d,n=L()
A=L()
D=[[1 for i in range(a+c)] for j in range(max(b,d))]
if b>d:
for i in range(d,b):
for j in range(a,a+c):
D[i][j]="."
else:
for i in range(b,d):
for j in range(0,a):
D[i][j]="."
def bfs(x,y,ch):
q=deque()
q.append([x,y])
D[x][y]=chr(ord("a")+ch)
count=1
if A[ch]==1:
return
while(q):
x,y=q.popleft()
row=[-1,0,0,1]
col=[0,-1,1,0]
for i in range(4):
a=x+row[i]
b=y+col[i]
if a>=0 and a<len(D) and b>=0 and b<len(D[0]) and D[a][b]==1:
D[a][b]=chr(ord("a")+ch)
count+=1
if count==A[ch]:
return 1
q.append([a,b])
return 0
for idx in range(n):
for i in range(len(D)):
for j in range(len(D[0])-1,-1,-1):
if D[i][j]==1:
if bfs(i,j,idx)==0:
pmat(D)
print("NO")
exit()
break
else:
continue
break
print("YES")
for ele in D:
for c in ele:
print(c,end="")
print()
endtime = time.time()
# print(f"Runtime of the program is {endtime - starttime}")
``` | instruction | 0 | 81,272 | 23 | 162,544 |
No | output | 1 | 81,272 | 23 | 162,545 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A revolution took place on the Buka Island. New government replaced the old one. The new government includes n parties and each of them is entitled to some part of the island according to their contribution to the revolution. However, they can't divide the island.
The island can be conventionally represented as two rectangles a × b and c × d unit squares in size correspondingly. The rectangles are located close to each other. At that, one of the sides with the length of a and one of the sides with the length of c lie on one line. You can see this in more details on the picture.
<image>
The i-th party is entitled to a part of the island equal to xi unit squares. Every such part should fully cover several squares of the island (it is not allowed to cover the squares partially) and be a connected figure. A "connected figure" presupposes that from any square of this party one can move to any other square of the same party moving through edge-adjacent squares also belonging to that party.
Your task is to divide the island between parties.
Input
The first line contains 5 space-separated integers — a, b, c, d and n (1 ≤ a, b, c, d ≤ 50, b ≠ d, 1 ≤ n ≤ 26). The second line contains n space-separated numbers. The i-th of them is equal to number xi (1 ≤ xi ≤ a × b + c × d). It is guaranteed that <image>.
Output
If dividing the island between parties in the required manner is impossible, print "NO" (without the quotes). Otherwise, print "YES" (also without the quotes) and, starting from the next line, print max(b, d) lines each containing a + c characters. To mark what square should belong to what party, use lowercase Latin letters. For the party that is first in order in the input data, use "a", for the second one use "b" and so on. Use "." for the squares that belong to the sea. The first symbol of the second line of the output data should correspond to the square that belongs to the rectangle a × b. The last symbol of the second line should correspond to the square that belongs to the rectangle c × d.
If there are several solutions output any.
Examples
Input
3 4 2 2 3
5 8 3
Output
YES
aaabb
aabbb
cbb..
ccb..
Input
3 2 1 4 4
1 2 3 4
Output
YES
abbd
cccd
...d
...d
Submitted Solution:
```
a, b, c, d,n = [int(x) for x in input().split()]
p = [int(x) for x in input().split()]
grid = [['o' for _ in range(a+c)] for _ in range(max(b,d))]
e, f = 0,0
dr = 1
if b > d:
for i in range(d,b):
for j in range(a,a+c):
grid[i][j] = '.'
if d%2 == 1:
e, f = 0, (a+c-1)
dr = -1
elif d > b:
for i in range(b,d):
for j in range(0,a):
grid[i][j] = '.'
if d%2 == 0:
e, f = 0, (a+c-1)
dr = -1
k = 0
for _ in range((a+c)*max(d,b)):
if grid[e][f] != '.':
grid[e][f] = chr(ord('a')+k)
p[k] -= 1
if p[k] == 0: k+=1
f += dr
if f == -1:
e, f = e+1, 0
dr = 1
if f == a+c:
e, f = e+1, a+c-1
dr = -1
print("YES")
for i in range(max(b,d)):
print("".join(grid[i]))
``` | instruction | 0 | 81,273 | 23 | 162,546 |
No | output | 1 | 81,273 | 23 | 162,547 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A revolution took place on the Buka Island. New government replaced the old one. The new government includes n parties and each of them is entitled to some part of the island according to their contribution to the revolution. However, they can't divide the island.
The island can be conventionally represented as two rectangles a × b and c × d unit squares in size correspondingly. The rectangles are located close to each other. At that, one of the sides with the length of a and one of the sides with the length of c lie on one line. You can see this in more details on the picture.
<image>
The i-th party is entitled to a part of the island equal to xi unit squares. Every such part should fully cover several squares of the island (it is not allowed to cover the squares partially) and be a connected figure. A "connected figure" presupposes that from any square of this party one can move to any other square of the same party moving through edge-adjacent squares also belonging to that party.
Your task is to divide the island between parties.
Input
The first line contains 5 space-separated integers — a, b, c, d and n (1 ≤ a, b, c, d ≤ 50, b ≠ d, 1 ≤ n ≤ 26). The second line contains n space-separated numbers. The i-th of them is equal to number xi (1 ≤ xi ≤ a × b + c × d). It is guaranteed that <image>.
Output
If dividing the island between parties in the required manner is impossible, print "NO" (without the quotes). Otherwise, print "YES" (also without the quotes) and, starting from the next line, print max(b, d) lines each containing a + c characters. To mark what square should belong to what party, use lowercase Latin letters. For the party that is first in order in the input data, use "a", for the second one use "b" and so on. Use "." for the squares that belong to the sea. The first symbol of the second line of the output data should correspond to the square that belongs to the rectangle a × b. The last symbol of the second line should correspond to the square that belongs to the rectangle c × d.
If there are several solutions output any.
Examples
Input
3 4 2 2 3
5 8 3
Output
YES
aaabb
aabbb
cbb..
ccb..
Input
3 2 1 4 4
1 2 3 4
Output
YES
abbd
cccd
...d
...d
Submitted Solution:
```
import os
import sys
from io import BytesIO, IOBase
from collections import Counter
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)
def gcd(a, b):
if a == 0:
return b
return gcd(b % a, a)
def lcm(a, b):
return (a * b) / gcd(a, b)
pow2=[1]
d={}
def main():
a,b,c,d,n=map(int, input().split())
x=list(map(int, input().split()))
ans=[['.' for i in range(a+c)] for j in range(max(b,d))]
t=0
# for i in ans:
# print(*i)
now=0
ind=0
for j in range(a+c):
if j<a:
if t==0:
for i in range(b):
ans[i][j]=chr(ord('a')+now)
x[ind]-=1
if x[ind]==0:
ind+=1
now+=1
else:
for i in range(b-1,-1,-1):
ans[i][j] = chr(ord('a') + now)
x[ind]-=1
if x[ind]==0:
ind+=1
now+=1
else:
if t==0:
for i in range(d):
ans[i][j]=chr(ord('a')+now)
x[ind]-=1
if x[ind]==0:
ind+=1
now+=1
else:
for i in range(d-1,-1,-1):
ans[i][j] = chr(ord('a') + now)
x[ind]-=1
if x[ind]==0:
ind+=1
now+=1
t^=1
print('YES')
for i in ans:
for j in i:
print(j, end=' ')
print('')
return
if __name__=="__main__":
main()
``` | instruction | 0 | 81,274 | 23 | 162,548 |
No | output | 1 | 81,274 | 23 | 162,549 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A revolution took place on the Buka Island. New government replaced the old one. The new government includes n parties and each of them is entitled to some part of the island according to their contribution to the revolution. However, they can't divide the island.
The island can be conventionally represented as two rectangles a × b and c × d unit squares in size correspondingly. The rectangles are located close to each other. At that, one of the sides with the length of a and one of the sides with the length of c lie on one line. You can see this in more details on the picture.
<image>
The i-th party is entitled to a part of the island equal to xi unit squares. Every such part should fully cover several squares of the island (it is not allowed to cover the squares partially) and be a connected figure. A "connected figure" presupposes that from any square of this party one can move to any other square of the same party moving through edge-adjacent squares also belonging to that party.
Your task is to divide the island between parties.
Input
The first line contains 5 space-separated integers — a, b, c, d and n (1 ≤ a, b, c, d ≤ 50, b ≠ d, 1 ≤ n ≤ 26). The second line contains n space-separated numbers. The i-th of them is equal to number xi (1 ≤ xi ≤ a × b + c × d). It is guaranteed that <image>.
Output
If dividing the island between parties in the required manner is impossible, print "NO" (without the quotes). Otherwise, print "YES" (also without the quotes) and, starting from the next line, print max(b, d) lines each containing a + c characters. To mark what square should belong to what party, use lowercase Latin letters. For the party that is first in order in the input data, use "a", for the second one use "b" and so on. Use "." for the squares that belong to the sea. The first symbol of the second line of the output data should correspond to the square that belongs to the rectangle a × b. The last symbol of the second line should correspond to the square that belongs to the rectangle c × d.
If there are several solutions output any.
Examples
Input
3 4 2 2 3
5 8 3
Output
YES
aaabb
aabbb
cbb..
ccb..
Input
3 2 1 4 4
1 2 3 4
Output
YES
abbd
cccd
...d
...d
Submitted Solution:
```
import math
import sys
input=sys.stdin.readline
from collections import Counter, defaultdict, deque
def f(a, b, c, d, n, x):
grid = [["." for co in range(a+c)]for r in range(max(b, d))]
alphnum = 97
pos = 0
for co in range(a):
if co % 2:
r = 0
while r < b:
grid[r][co] = chr(alphnum)
x[pos] -= 1
if x[pos] == 0:
pos += 1
alphnum += 1
r += 1
else:
r = b-1
while r > -1:
grid[r][co] = chr(alphnum)
x[pos] -= 1
if x[pos] == 0:
pos += 1
alphnum += 1
r -= 1
# for r in range(max(b, d)):
# print(grid[r])
for co in range(a, a+c):
if co%2:
r = 0
while r < d:
#print(r, co)
grid[r][co] = chr(alphnum)
x[pos] -= 1
if x[pos] == 0:
pos += 1
alphnum += 1
r += 1
#print(r, c)
else:
r = d-1
while r > -1:
#print(r, co)
grid[r][co] = chr(alphnum)
x[pos] -= 1
if x[pos] == 0:
pos += 1
alphnum += 1
r -= 1
# for r in range(max(b, d)):
# print(grid[r])
return grid
t = 1
result = []
for i in range(t):
#n = int(input())
a, b, c, d, n = list(map(int, input().split()))
x = list(map(int, input().split()))
result.append(f(a, b, c, d, n, x))
for i in range(t):
print("YES")
for j in range(len(result[i])):
print("".join(result[i][j]))
``` | instruction | 0 | 81,275 | 23 | 162,550 |
No | output | 1 | 81,275 | 23 | 162,551 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Filya just learned new geometry object — rectangle. He is given a field consisting of n × n unit cells. Rows are numbered from bottom to top with integer from 1 to n. Columns are numbered from left to right with integers from 1 to n. Cell, located at the intersection of the row r and column c is denoted as (r, c). Filya has painted two rectangles, such that their sides are parallel to coordinate axes and each cell lies fully inside or fully outside each of them. Moreover, no cell lies in both rectangles.
Later, hedgehog Filya became interested in the location of his rectangles but was unable to find the sheet of paper they were painted on. They were taken by Sonya and now she wants to play a little game with Filya. He tells her a query rectangle and she replies with the number of initial rectangles that lie fully inside the given query rectangle. The query rectangle should match the same conditions as initial rectangles. Rectangle lies fully inside the query if each o its cells lies inside the query.
Filya knows Sonya really well, so is sure that if he asks more than 200 questions she will stop to reply.
Input
The first line of the input contains an integer n (2 ≤ n ≤ 216) — size of the field.
For each query an integer between 0 and 2 is returned — the number of initial rectangles that lie fully inside the query rectangle.
Output
To make a query you have to print "? x1 y1 x2 y2" (without quotes) (1 ≤ x1 ≤ x2 ≤ n, 1 ≤ y1 ≤ y2 ≤ n), where (x1, y1) stands for the position of the bottom left cell of the query and (x2, y2) stands for the up right cell of the query. You are allowed to ask no more than 200 queries. After each query you should perform "flush" operation and read the answer.
In case you suppose you've already determined the location of two rectangles (or run out of queries) you should print "! x11 y11 x12 y12 x21 y21 x22 y22" (without quotes), where first four integers describe the bottom left and up right cells of the first rectangle, and following four describe the corresponding cells of the second rectangle. You can print the rectangles in an arbitrary order. After you have printed the answer, print the end of the line and perform "flush". Your program should terminate immediately after it print the answer.
Interaction
To flush you can use (just after printing an integer and end-of-line):
* fflush(stdout) in C++;
* System.out.flush() in Java;
* stdout.flush() in Python;
* flush(output) in Pascal;
* See the documentation for other languages.
You will get the Wrong Answer verdict if you ask more than 200 queries, or if you print an incorrect coordinates.
You will get the Idleness Limit Exceeded verdict if you don't print anything (but you should) or if you forget about flushing the output (more info below).
Hacking.
The first line should contain an integer n (2 ≤ n ≤ 216).
The second line should contain four integers x1, y1, x2, y2 (1 ≤ x1 ≤ x2 ≤ n, 1 ≤ y1 ≤ y2 ≤ n) — the description of the first rectangle.
The third line contains the description of the second rectangle in the similar way.
Example
Input
5
2
1
0
1
1
1
0
1
Output
? 1 1 5 5
? 1 1 3 3
? 1 1 3 1
? 2 2 2 2
? 3 3 5 5
? 3 3 3 5
? 3 3 3 4
? 3 4 3 5
! 2 2 2 2 3 4 3 5
Submitted Solution:
```
import math
def bin_search_left(l, r, pred):
# pred(l) == True
while l < r:
m = math.ceil((l + r) / 2)
if pred(m):
l = m
else:
r = m - 1
return l
def bin_search_right(l, r, pred):
# pred(r) == True
while l < r:
m = (l + r) // 2
if pred(m):
r = m
else:
l = m + 1
return r
def find_rectangle(x1, y1, x2, y2):
#assert num_rects(x1, y1, x2, y2) == 1
x1 = bin_search_left(x1, x2, lambda x: num_rects(x, y1, x2, y2) > 0)
x2 = bin_search_right(x1, x2, lambda x: num_rects(x, y1, x2, y2) > 0)
y1 = bin_search_left(x1, x2, lambda y: num_rects(x1, y, x2, y2) > 0)
x2 = bin_search_right(x1, x2, lambda y: num_rects(x1, y1, x2, y) > 0)
return x1, y1, x2, y2
def solve_hor_separated(n):
h = bin_search_right(1, n, lambda x: num_rects(1, 1, x, n) > 0)
r1 = find_rectangle(1, 1, h, n)
r2 = find_rectangle(h, 1, n, n)
return r1, r2
def solve_ver_separated(n):
v = bin_search_right(1, n, lambda y: num_rects(1, 1, n, y) > 0)
r1 = find_rectangle(1, 1, n, v)
r2 = find_rectangle(1, v, n, n)
return r1, r2
def solve(n):
h = bin_search_right(1, n, lambda x: num_rects(1, 1, x, n) > 0)
r = num_rects(1, 1, h, n)
if r == 2 or h == n or num_rects(h + 1, 1, n, n) != 1:
r1, r2 = solve_ver_separated(n)
else:
r1, r2 = solve_hor_separated(n)
return r1, r2
if True:
def num_rects(x1, y1, x2, y2):
print("? {} {} {} {}".format(x1, y1, x2, y2), flush=True)
return int(input())
n = int(input())
r1, r2 = solve(n)
print("! {} {} {} {} {} {} {} {}".format(*r1, *r2))
else:
tests = [
(2, 1, 2, 1, 2, 2, 1, 2, 1),
(2, 1, 1, 1, 1, 2, 2, 2, 2),
(2, 1, 1, 1, 2, 2, 1, 2, 2),
(2**16, 10, 20, 30, 40, 100, 30, 200, 200),
(2 ** 16, 1, 1, 1, 1, 2 ** 16, 2 ** 16, 2 ** 16, 2 ** 16),
(2 ** 16, 1, 1, 1, 1, 2 ** 16 - 1, 2 ** 16, 2 ** 16, 2 ** 16),
(2 ** 16, 1, 1, 1, 1, 2 ** 16, 2 ** 16 - 1, 2 ** 16, 2 ** 16),
(2 ** 16, 1, 1, 1027, 1, 2 ** 16, 2 ** 16 - 1, 2 ** 16, 2 ** 16),
(2 ** 16, 143, 1, 1027, 1, 2 ** 16, 2 ** 16 - 1, 2 ** 16, 2 ** 16),
]
for n, x11, y11, x12, y12, x21, y21, x22, y22 in tests:
assert 1 <= x11 <= x12
assert 1 <= y11 <= y12
assert 1 <= x21 <= x22
assert 1 <= y21 <= y22
call_count = 0
def num_rects(x1, y1, x2, y2):
global call_count
call_count += 1
res = 0
if x11 <= x1 <= x12 and x11 <= x2 <= x12 and \
y11 <= y1 <= y12 and y11 <= y2 <= y12:
res += 1
if x21 <= x1 <= x22 and x21 <= x2 <= x22 and \
y21 <= y1 <= y22 and y21 <= y2 <= y22:
res += 1
return res
r1, r2 = solve(n)
assert r1, r2 == ((x11, y11, x12, y12), (x21, y21, x22, y22))
assert call_count <= 200
``` | instruction | 0 | 81,307 | 23 | 162,614 |
No | output | 1 | 81,307 | 23 | 162,615 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Filya just learned new geometry object — rectangle. He is given a field consisting of n × n unit cells. Rows are numbered from bottom to top with integer from 1 to n. Columns are numbered from left to right with integers from 1 to n. Cell, located at the intersection of the row r and column c is denoted as (r, c). Filya has painted two rectangles, such that their sides are parallel to coordinate axes and each cell lies fully inside or fully outside each of them. Moreover, no cell lies in both rectangles.
Later, hedgehog Filya became interested in the location of his rectangles but was unable to find the sheet of paper they were painted on. They were taken by Sonya and now she wants to play a little game with Filya. He tells her a query rectangle and she replies with the number of initial rectangles that lie fully inside the given query rectangle. The query rectangle should match the same conditions as initial rectangles. Rectangle lies fully inside the query if each o its cells lies inside the query.
Filya knows Sonya really well, so is sure that if he asks more than 200 questions she will stop to reply.
Input
The first line of the input contains an integer n (2 ≤ n ≤ 216) — size of the field.
For each query an integer between 0 and 2 is returned — the number of initial rectangles that lie fully inside the query rectangle.
Output
To make a query you have to print "? x1 y1 x2 y2" (without quotes) (1 ≤ x1 ≤ x2 ≤ n, 1 ≤ y1 ≤ y2 ≤ n), where (x1, y1) stands for the position of the bottom left cell of the query and (x2, y2) stands for the up right cell of the query. You are allowed to ask no more than 200 queries. After each query you should perform "flush" operation and read the answer.
In case you suppose you've already determined the location of two rectangles (or run out of queries) you should print "! x11 y11 x12 y12 x21 y21 x22 y22" (without quotes), where first four integers describe the bottom left and up right cells of the first rectangle, and following four describe the corresponding cells of the second rectangle. You can print the rectangles in an arbitrary order. After you have printed the answer, print the end of the line and perform "flush". Your program should terminate immediately after it print the answer.
Interaction
To flush you can use (just after printing an integer and end-of-line):
* fflush(stdout) in C++;
* System.out.flush() in Java;
* stdout.flush() in Python;
* flush(output) in Pascal;
* See the documentation for other languages.
You will get the Wrong Answer verdict if you ask more than 200 queries, or if you print an incorrect coordinates.
You will get the Idleness Limit Exceeded verdict if you don't print anything (but you should) or if you forget about flushing the output (more info below).
Hacking.
The first line should contain an integer n (2 ≤ n ≤ 216).
The second line should contain four integers x1, y1, x2, y2 (1 ≤ x1 ≤ x2 ≤ n, 1 ≤ y1 ≤ y2 ≤ n) — the description of the first rectangle.
The third line contains the description of the second rectangle in the similar way.
Example
Input
5
2
1
0
1
1
1
0
1
Output
? 1 1 5 5
? 1 1 3 3
? 1 1 3 1
? 2 2 2 2
? 3 3 5 5
? 3 3 3 5
? 3 3 3 4
? 3 4 3 5
! 2 2 2 2 3 4 3 5
Submitted Solution:
```
def answer(x11, y11, x12, y12, x21, y21, x22, y22):
print("! {0} {1} {2} {3} {4} {5} {6} {7}".format(
x11, y11, x12, y12, x21, y21, x22, y22))
def request(x1, y1, x2, y2):
print("? {0} {1} {2} {3}".format(x1, y1, x2, y2))
return int(input())
def binary_search(add, min_bound, max_bound, x_func, y_func, process_reply):
while min_bound < max_bound:
mid = (min_bound + max_bound - add) // 2
x1, x2 = x_func(mid)
y1, y2 = y_func(mid)
reply = request(x1, y1, x2, y2)
if process_reply(reply):
max_bound = mid + add
else:
min_bound = mid + 1 + add
return min_bound
def find_single_rectangle(x1, y1, x2, y2):
xr = binary_search(0, x1, x2,
lambda mid: (x1, mid),
lambda mid: (y1, y2),
lambda rep: rep == 1)
yr = binary_search(0, y1, y2,
lambda mid: (x1, x2),
lambda mid: (y1, mid),
lambda rep: rep == 1)
xl = binary_search(-1, x1, x2,
lambda mid: (mid, x2),
lambda mid: (y1, y2),
lambda rep: rep == 0)
yl = binary_search(-1, y1, y2,
lambda mid: (x1, x2),
lambda mid: (mid, y2),
lambda rep: rep == 0)
return xl, yl, xr, yr
def find_border(n):
hor = binary_search(0 , 1, n,
lambda mid: (1, n),
lambda mid: (1, mid),
lambda rep: rep >= 1)
if request(1, 1, n, hor) == 1:
return 1, 1, n, hor, 1, hor + 1, n, n
ver = binary_search(0 , 1, n,
lambda mid: (1, mid),
lambda mid: (1, n),
lambda rep: rep >= 1)
if request(1, 1, ver, n) != 1:
raise
return 1, 1, ver, n, 1, ver + 1, n, n
def main():
n = int(input())
x11, y11, x12, y12, x21, y21, x22, y22 = find_border(n)
X11, Y11, X12, Y12 = find_single_rectangle(x11, y11, x12, y12)
X21, Y21, X22, Y22 = find_single_rectangle(x21, y21, x22, y22)
answer(X11, Y11, X12, Y12, X21, Y21, X22, Y22)
if __name__ == "__main__":
main()
``` | instruction | 0 | 81,308 | 23 | 162,616 |
No | output | 1 | 81,308 | 23 | 162,617 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Filya just learned new geometry object — rectangle. He is given a field consisting of n × n unit cells. Rows are numbered from bottom to top with integer from 1 to n. Columns are numbered from left to right with integers from 1 to n. Cell, located at the intersection of the row r and column c is denoted as (r, c). Filya has painted two rectangles, such that their sides are parallel to coordinate axes and each cell lies fully inside or fully outside each of them. Moreover, no cell lies in both rectangles.
Later, hedgehog Filya became interested in the location of his rectangles but was unable to find the sheet of paper they were painted on. They were taken by Sonya and now she wants to play a little game with Filya. He tells her a query rectangle and she replies with the number of initial rectangles that lie fully inside the given query rectangle. The query rectangle should match the same conditions as initial rectangles. Rectangle lies fully inside the query if each o its cells lies inside the query.
Filya knows Sonya really well, so is sure that if he asks more than 200 questions she will stop to reply.
Input
The first line of the input contains an integer n (2 ≤ n ≤ 216) — size of the field.
For each query an integer between 0 and 2 is returned — the number of initial rectangles that lie fully inside the query rectangle.
Output
To make a query you have to print "? x1 y1 x2 y2" (without quotes) (1 ≤ x1 ≤ x2 ≤ n, 1 ≤ y1 ≤ y2 ≤ n), where (x1, y1) stands for the position of the bottom left cell of the query and (x2, y2) stands for the up right cell of the query. You are allowed to ask no more than 200 queries. After each query you should perform "flush" operation and read the answer.
In case you suppose you've already determined the location of two rectangles (or run out of queries) you should print "! x11 y11 x12 y12 x21 y21 x22 y22" (without quotes), where first four integers describe the bottom left and up right cells of the first rectangle, and following four describe the corresponding cells of the second rectangle. You can print the rectangles in an arbitrary order. After you have printed the answer, print the end of the line and perform "flush". Your program should terminate immediately after it print the answer.
Interaction
To flush you can use (just after printing an integer and end-of-line):
* fflush(stdout) in C++;
* System.out.flush() in Java;
* stdout.flush() in Python;
* flush(output) in Pascal;
* See the documentation for other languages.
You will get the Wrong Answer verdict if you ask more than 200 queries, or if you print an incorrect coordinates.
You will get the Idleness Limit Exceeded verdict if you don't print anything (but you should) or if you forget about flushing the output (more info below).
Hacking.
The first line should contain an integer n (2 ≤ n ≤ 216).
The second line should contain four integers x1, y1, x2, y2 (1 ≤ x1 ≤ x2 ≤ n, 1 ≤ y1 ≤ y2 ≤ n) — the description of the first rectangle.
The third line contains the description of the second rectangle in the similar way.
Example
Input
5
2
1
0
1
1
1
0
1
Output
? 1 1 5 5
? 1 1 3 3
? 1 1 3 1
? 2 2 2 2
? 3 3 5 5
? 3 3 3 5
? 3 3 3 4
? 3 4 3 5
! 2 2 2 2 3 4 3 5
Submitted Solution:
```
def answer(x11, y11, x12, y12, x21, y21, x22, y22):
print("! {0} {1} {2} {3} {4} {5} {6} {7}".format(
x11, y11, x12, y12, x21, y21, x22, y22))
def request(x1, y1, x2, y2):
print("? {0} {1} {2} {3}".format(x1, y1, x2, y2))
return int(input())
def binary_search(add, min_bound, max_bound, x_func, y_func, process_reply):
while min_bound < max_bound:
mid = (min_bound + max_bound - add) // 2
x1, x2 = x_func(mid)
y1, y2 = y_func(mid)
reply = request(x1, y1, x2, y2)
if process_reply(reply):
max_bound = mid + add
else:
min_bound = mid + 1 + add
return min_bound
def find_single_rectangle(x1, y1, x2, y2):
xr = binary_search(0, x1, x2,
lambda mid: (x1, mid),
lambda mid: (y1, y2),
lambda rep: rep == 1)
yr = binary_search(0, y1, y2,
lambda mid: (x1, x2),
lambda mid: (y1, mid),
lambda rep: rep == 1)
xl = binary_search(-1, x1, x2,
lambda mid: (mid, x2),
lambda mid: (y1, y2),
lambda rep: rep == 0)
yl = binary_search(-1, y1, y2,
lambda mid: (x1, x2),
lambda mid: (mid, y2),
lambda rep: rep == 0)
return xl, yl, xr, yr
def find_border(n):
hor = binary_search(0 , 1, n,
lambda mid: (1, n),
lambda mid: (1, mid),
lambda rep: rep >= 1)
if request(1, 1, n, hor) == 1:
return 1, 1, n, hor, 1, hor + 1, n, n
ver = binary_search(0 , 1, n,
lambda mid: (1, mid),
lambda mid: (1, n),
lambda rep: rep >= 1)
return 1, 1, ver, n, 1, ver + 1, n, n
def main():
n = int(input())
x11, y11, x12, y12, x21, y21, x22, y22 = find_border(n)
X11, Y11, X12, Y12 = find_single_rectangle(x11, y11, x12, y12)
X21, Y21, X22, Y22 = find_single_rectangle(x21, y21, x22, y22)
answer(X11, Y11, X12, Y12, X21, Y21, X22, Y22)
if __name__ == "__main__":
main()
``` | instruction | 0 | 81,309 | 23 | 162,618 |
No | output | 1 | 81,309 | 23 | 162,619 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Filya just learned new geometry object — rectangle. He is given a field consisting of n × n unit cells. Rows are numbered from bottom to top with integer from 1 to n. Columns are numbered from left to right with integers from 1 to n. Cell, located at the intersection of the row r and column c is denoted as (r, c). Filya has painted two rectangles, such that their sides are parallel to coordinate axes and each cell lies fully inside or fully outside each of them. Moreover, no cell lies in both rectangles.
Later, hedgehog Filya became interested in the location of his rectangles but was unable to find the sheet of paper they were painted on. They were taken by Sonya and now she wants to play a little game with Filya. He tells her a query rectangle and she replies with the number of initial rectangles that lie fully inside the given query rectangle. The query rectangle should match the same conditions as initial rectangles. Rectangle lies fully inside the query if each o its cells lies inside the query.
Filya knows Sonya really well, so is sure that if he asks more than 200 questions she will stop to reply.
Input
The first line of the input contains an integer n (2 ≤ n ≤ 216) — size of the field.
For each query an integer between 0 and 2 is returned — the number of initial rectangles that lie fully inside the query rectangle.
Output
To make a query you have to print "? x1 y1 x2 y2" (without quotes) (1 ≤ x1 ≤ x2 ≤ n, 1 ≤ y1 ≤ y2 ≤ n), where (x1, y1) stands for the position of the bottom left cell of the query and (x2, y2) stands for the up right cell of the query. You are allowed to ask no more than 200 queries. After each query you should perform "flush" operation and read the answer.
In case you suppose you've already determined the location of two rectangles (or run out of queries) you should print "! x11 y11 x12 y12 x21 y21 x22 y22" (without quotes), where first four integers describe the bottom left and up right cells of the first rectangle, and following four describe the corresponding cells of the second rectangle. You can print the rectangles in an arbitrary order. After you have printed the answer, print the end of the line and perform "flush". Your program should terminate immediately after it print the answer.
Interaction
To flush you can use (just after printing an integer and end-of-line):
* fflush(stdout) in C++;
* System.out.flush() in Java;
* stdout.flush() in Python;
* flush(output) in Pascal;
* See the documentation for other languages.
You will get the Wrong Answer verdict if you ask more than 200 queries, or if you print an incorrect coordinates.
You will get the Idleness Limit Exceeded verdict if you don't print anything (but you should) or if you forget about flushing the output (more info below).
Hacking.
The first line should contain an integer n (2 ≤ n ≤ 216).
The second line should contain four integers x1, y1, x2, y2 (1 ≤ x1 ≤ x2 ≤ n, 1 ≤ y1 ≤ y2 ≤ n) — the description of the first rectangle.
The third line contains the description of the second rectangle in the similar way.
Example
Input
5
2
1
0
1
1
1
0
1
Output
? 1 1 5 5
? 1 1 3 3
? 1 1 3 1
? 2 2 2 2
? 3 3 5 5
? 3 3 3 5
? 3 3 3 4
? 3 4 3 5
! 2 2 2 2 3 4 3 5
Submitted Solution:
```
def answer(x11, y11, x12, y12, x21, y21, x22, y22):
print("! {0} {1} {2} {3} {4} {5} {6} {7}".format(
x11, y11, x12, y12, x21, y21, x22, y22))
def request(x1, y1, x2, y2):
print("? {0} {1} {2} {3}".format(x1, y1, x2, y2))
return int(input())
def binary_search(add, min_bound, max_bound, x_func, y_func, process_reply):
while min_bound < max_bound:
mid = (min_bound + max_bound - add) // 2
x1, x2 = x_func(mid)
y1, y2 = y_func(mid)
reply = request(x1, y1, x2, y2)
if process_reply(reply):
max_bound = mid + add
else:
min_bound = mid + 1 + add
return min_bound
def find_single_rectangle(x1, y1, x2, y2):
xr = binary_search(0, x1, x2,
lambda mid: (x1, mid),
lambda mid: (y1, y2),
lambda rep: rep == 1)
yr = binary_search(0, y1, y2,
lambda mid: (x1, x2),
lambda mid: (y1, mid),
lambda rep: rep == 1)
xl = binary_search(-1, x1, x2,
lambda mid: (mid, x2),
lambda mid: (y1, y2),
lambda rep: rep == 0)
yl = binary_search(-1, y1, y2,
lambda mid: (x1, x2),
lambda mid: (mid, y2),
lambda rep: rep == 0)
return xl, yl, xr, yr
def find_border(n):
hor = binary_search(0, 1, n,
lambda mid: (1, n),
lambda mid: (1, mid),
lambda rep: rep >= 1)
if request(1, 1, n, hor) == 1:
return 1, 1, n, hor, 1, hor + 1, n, n
ver = binary_search(0, 1, n,
lambda mid: (1, mid),
lambda mid: (1, n),
lambda rep: rep >= 1)
return 1, 1, ver, n, ver + 1, 1, n, n
def main():
n = int(input())
x11, y11, x12, y12, x21, y21, x22, y22 = find_border(n)
X11, Y11, X12, Y12 = find_single_rectangle(x11, y11, x12, y12)
X21, Y21, X22, Y22 = find_single_rectangle(x21, y21, x22, y22)
answer(X11, Y11, X12, Y12, X21, Y21, X22, Y22)
if __name__ == "__main__":
main()
``` | instruction | 0 | 81,310 | 23 | 162,620 |
No | output | 1 | 81,310 | 23 | 162,621 |
Provide a correct Python 3 solution for this coding contest problem.
problem
Given the squares of $ R * C $. Each square is either an empty square or a square with a hole. The given square meets the following conditions.
* The cells with holes are connected. (You can move a square with a hole in the cross direction to any square with a hole)
* Empty cells are connected.
You can generate rectangular tiles of any length with a width of $ 1 $. I would like to install multiple tiles to fill all the holes in the square. When installing tiles, the following restrictions must be observed.
* Tiles can only be installed vertically or horizontally in the $ 2 $ direction.
* Do not install more than one tile on one square.
* There should be no tiles on the squares without holes.
Please answer the minimum number of tiles when all the squares with holes are filled with tiles while observing the above restrictions.
output
Output the minimum number of times. Please also output a line break at the end.
Example
Input
5 5
.....
.#.#.
.###.
.#.#.
.....
Output
3 | instruction | 0 | 81,671 | 23 | 163,342 |
"Correct Solution:
```
# -*- coding: utf-8 -*-
import sys
def input(): return sys.stdin.readline().strip()
def list2d(a, b, c): return [[c] * b for i in range(a)]
def list3d(a, b, c, d): return [[[d] * c for j in range(b)] for i in range(a)]
def list4d(a, b, c, d, e): return [[[[e] * d for j in range(c)] for j in range(b)] for i in range(a)]
def ceil(x, y=1): return int(-(-x // y))
def INT(): return int(input())
def MAP(): return map(int, input().split())
def LIST(N=None): return list(MAP()) if N is None else [INT() for i in range(N)]
def Yes(): print('Yes')
def No(): print('No')
def YES(): print('YES')
def NO(): print('NO')
sys.setrecursionlimit(10 ** 9)
INF = 10 ** 18
MOD = 10 ** 9 + 7
class Dinic:
""" 最大流(Dinic) """
INF = 10 ** 18
def __init__(self, n):
self.n = n
self.links = [[] for _ in range(n)]
self.depth = None
self.progress = None
def add_link(self, _from, to, cap):
self.links[_from].append([cap, to, len(self.links[to])])
self.links[to].append([0, _from, len(self.links[_from]) - 1])
def bfs(self, s):
from collections import deque
depth = [-1] * self.n
depth[s] = 0
q = deque([s])
while q:
v = q.popleft()
for cap, to, _ in self.links[v]:
if cap > 0 and depth[to] < 0:
depth[to] = depth[v] + 1
q.append(to)
self.depth = depth
def dfs(self, v, t, flow):
if v == t:
return flow
links_v = self.links[v]
for i in range(self.progress[v], len(links_v)):
self.progress[v] = i
cap, to, rev = link = links_v[i]
if cap == 0 or self.depth[v] >= self.depth[to]:
continue
d = self.dfs(to, t, min(flow, cap))
if d == 0:
continue
link[0] -= d
self.links[to][rev][0] += d
return d
return 0
def max_flow(self, s, t):
INF = Dinic.INF
flow = 0
while True:
self.bfs(s)
if self.depth[t] < 0:
return flow
self.progress = [0] * self.n
current_flow = self.dfs(s, t, INF)
while current_flow > 0:
flow += current_flow
current_flow = self.dfs(s, t, INF)
def build_grid(H, W, intv, _type, space=True, padding=False):
# 入力がスペース区切りかどうか
if space:
_input = lambda: input().split()
else:
_input = lambda: input()
_list = lambda: list(map(_type, _input()))
# 余白の有無
if padding:
offset = 1
else:
offset = 0
grid = list2d(H+offset*2, W+offset*2, intv)
for i in range(offset, H+offset):
row = _list()
for j in range(offset, W+offset):
grid[i][j] = row[j-offset]
return grid
H, W = MAP()
grid = build_grid(H, W, '', str, space=0)
N = H * W
M = 0
# 黒マスの数
total = 0
# 黒マスが隣り合う箇所の数
adjcnt = 0
for i in range(H):
for j in range(W):
if grid[i][j] == '#':
total += 1
if i+1 < H and grid[i+1][j] == '#':
adjcnt += 1
if j+1 < W and grid[i][j+1] == '#':
adjcnt += 1
dinic = Dinic(N+adjcnt+2)
s = N + adjcnt
t = N + adjcnt + 1
k = N
for i in range(H):
for j in range(W):
# 同じ側から(i, j)と(i+1, j)選択する場合に、利得1を与える
if i+1 < H and grid[i][j] == grid[i+1][j] == '#':
dinic.add_link(s, k, 1)
dinic.add_link(k, i*W+j, INF)
dinic.add_link(k, (i+1)*W+j, INF)
k += 1
if j+1 < W and grid[i][j] == grid[i][j+1] == '#':
dinic.add_link(k, t, 1)
dinic.add_link(i*W+j, k, INF)
dinic.add_link(i*W+j+1, k, INF)
k += 1
res = dinic.max_flow(s, t)
ans = total - (adjcnt - res)
print(ans)
``` | output | 1 | 81,671 | 23 | 163,343 |
Provide a correct Python 3 solution for this coding contest problem.
problem
Given the squares of $ R * C $. Each square is either an empty square or a square with a hole. The given square meets the following conditions.
* The cells with holes are connected. (You can move a square with a hole in the cross direction to any square with a hole)
* Empty cells are connected.
You can generate rectangular tiles of any length with a width of $ 1 $. I would like to install multiple tiles to fill all the holes in the square. When installing tiles, the following restrictions must be observed.
* Tiles can only be installed vertically or horizontally in the $ 2 $ direction.
* Do not install more than one tile on one square.
* There should be no tiles on the squares without holes.
Please answer the minimum number of tiles when all the squares with holes are filled with tiles while observing the above restrictions.
output
Output the minimum number of times. Please also output a line break at the end.
Example
Input
5 5
.....
.#.#.
.###.
.#.#.
.....
Output
3 | instruction | 0 | 81,672 | 23 | 163,344 |
"Correct Solution:
```
import sys, re
from collections import deque, defaultdict, Counter
from math import ceil, sqrt, hypot, factorial, pi, sin, cos, radians
from itertools import accumulate, permutations, combinations, product, groupby, combinations_with_replacement
from operator import itemgetter, mul
from copy import deepcopy
from string import ascii_lowercase, ascii_uppercase, digits
from bisect import bisect, bisect_left
from fractions import gcd
from heapq import heappush, heappop
from functools import reduce
def input(): return sys.stdin.readline().strip()
def INT(): return int(input())
def MAP(): return map(int, input().split())
def LIST(): return list(map(int, input().split()))
def ZIP(n): return zip(*(MAP() for _ in range(n)))
sys.setrecursionlimit(10 ** 9)
INF = 10**10
mod = 10 ** 9 + 7
class Dinic:
def __init__(self, v, inf=10**10):
self.v = v
self.inf = inf
self.G = [[] for _ in range(v)]
self.level = [-1]*v # 深さ
self.ite = [0]*v # DFSでの探索が済んでいるか
def add_edge(self, fr, to, cap):
self.G[fr].append([to, cap, len(self.G[to])])
self.G[to].append([fr, 0, len(self.G[fr])-1])
def bfs(self, s): # BFSで深さ決定,sがstart
self.level = [-1]*self.v # 必要
self.level[s] = 0
Q = deque()
Q.append(s)
while Q:
v = Q.popleft()
for i in range(len(self.G[v])):
e = self.G[v][i]
if e[1]>0 and self.level[e[0]]<0: ###capacity>0かつtoの深さ未定
self.level[e[0]] = self.level[v]+1
Q.append(e[0])
def dfs(self, v, t, f): # DFSで増加パス探索,v開始、t終点、総フローf
if v==t:
return f
for i in range(self.ite[v], len(self.G[v])):
self.ite[v] = i
e = self.G[v][i]
if e[1]>0 and self.level[v]<self.level[e[0]]:
d = self.dfs(e[0], t, min(f, e[1]))
if d>0:
e[1] -= d # cap減少
self.G[e[0]][e[2]][1] += d # 逆辺のcap増加
return d
return 0
def max_flow(self, s, t):
flow = 0
while True:
self.bfs(s)
if self.level[t]<0:
return flow
self.ite = [0]*self.v # DFSでの探索が済んでいるか否か
f = self.dfs(s,t,self.inf)
while f>0:
flow += f
f = self.dfs(s,t,self.inf)
R, C = MAP()
S = [input() for _ in range(R)]
D = Dinic(3*R*C+2)
s = 3*R*C
t = s+1
block = 0
adj = 0
for i in range(R):
for j in range(C):
if S[i][j] == ".": continue
block += 1
if i+1 < R and S[i+1][j] == "#": # 縦
adj += 1
newv = R*C+C*i+j
D.add_edge(newv, t, 1)
D.add_edge(C*i+j, newv, INF)
D.add_edge(C*(i+1)+j, newv, INF)
if j+1 < C and S[i][j+1] == "#": # 横
adj += 1
newv = 2*R*C+C*i+j
D.add_edge(s, newv, 1)
D.add_edge(newv, C*i+j, INF)
D.add_edge(newv, C*i+(j+1), INF)
print(block-(adj-D.max_flow(s, t)))
``` | output | 1 | 81,672 | 23 | 163,345 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given n points on a plane. All the points are distinct and no three of them lie on the same line. Find the number of parallelograms with the vertices at the given points.
Input
The first line of the input contains integer n (1 ≤ n ≤ 2000) — the number of points.
Each of the next n lines contains two integers (xi, yi) (0 ≤ xi, yi ≤ 109) — the coordinates of the i-th point.
Output
Print the only integer c — the number of parallelograms with the vertices at the given points.
Example
Input
4
0 1
1 0
1 1
2 0
Output
1 | instruction | 0 | 82,194 | 23 | 164,388 |
Tags: geometry
Correct Solution:
```
import sys
def read_input(input_path=None):
if input_path is None:
f = sys.stdin
else:
f = open(input_path, 'r')
n = int(f.readline())
points = list()
for _ in range(n):
x, y = map(int, f.readline().split())
points.append((x, y))
return n, points
def sol(n, points):
count = dict()
for i in range(n):
for j in range(i+1, n):
mid = (points[i][0] + points[j][0], points[i][1] + points[j][1])
count[mid] = count.get(mid, 0) + 1
res = 0
for p in count:
res += count[p]*(count[p] - 1)//2
return [f"{res}"]
def solve(input_path=None):
return sol(*read_input(input_path))
def main():
for line in sol(*read_input()):
print(f"{line}")
if __name__ == '__main__':
main()
``` | output | 1 | 82,194 | 23 | 164,389 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given n points on a plane. All the points are distinct and no three of them lie on the same line. Find the number of parallelograms with the vertices at the given points.
Input
The first line of the input contains integer n (1 ≤ n ≤ 2000) — the number of points.
Each of the next n lines contains two integers (xi, yi) (0 ≤ xi, yi ≤ 109) — the coordinates of the i-th point.
Output
Print the only integer c — the number of parallelograms with the vertices at the given points.
Example
Input
4
0 1
1 0
1 1
2 0
Output
1 | instruction | 0 | 82,195 | 23 | 164,390 |
Tags: geometry
Correct Solution:
```
import math
n=int(input())
dots=[]
for i in range (n) :
temp=list(map(int,input().split()))
dots.append(temp)
lines={}
for i in range(n) :
for j in range (i+1,n) :
dx=dots[i][0]-dots[j][0]
dy=dots[i][1]-dots[j][1]
if dx<0 :
dx=-dx
dy=-dy
if dx==0 and dy<0 :
dy=-dy
if (dx,dy) in lines :
lines[(dx,dy)]+=1
else :
lines[(dx,dy)]=1
ans=0
for x in lines :
t=lines[x]
ans+=t*(t-1)/2
ans/=2
print(int(ans))
``` | output | 1 | 82,195 | 23 | 164,391 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given n points on a plane. All the points are distinct and no three of them lie on the same line. Find the number of parallelograms with the vertices at the given points.
Input
The first line of the input contains integer n (1 ≤ n ≤ 2000) — the number of points.
Each of the next n lines contains two integers (xi, yi) (0 ≤ xi, yi ≤ 109) — the coordinates of the i-th point.
Output
Print the only integer c — the number of parallelograms with the vertices at the given points.
Example
Input
4
0 1
1 0
1 1
2 0
Output
1 | instruction | 0 | 82,196 | 23 | 164,392 |
Tags: geometry
Correct Solution:
```
n = int(input())
a = []
for _ in range(n):
xy = list(map(int, input().split()))
a.append(xy)
midpoint = {}
for i in range(n):
for j in range(i + 1, n):
x1, y1 = a[i][0], a[i][1]
x2, y2 = a[j][0], a[j][1]
dx = x2 + x1
dy = y2 + y1
midpoint[(dx / 2, dy / 2)] = midpoint.get((dx / 2, dy / 2), 0) + 1
ans = 0
for k, v in midpoint.items():
#print(k, v, sep=' : ')
ans += (v * (v - 1)) // 2
print(ans)
#import matplotlib.pyplot as plt
#import numpy as np
#x, y = zip(*a)
#plt.scatter(x, y)
#plt.grid()
#plt.show()
``` | output | 1 | 82,196 | 23 | 164,393 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given n points on a plane. All the points are distinct and no three of them lie on the same line. Find the number of parallelograms with the vertices at the given points.
Input
The first line of the input contains integer n (1 ≤ n ≤ 2000) — the number of points.
Each of the next n lines contains two integers (xi, yi) (0 ≤ xi, yi ≤ 109) — the coordinates of the i-th point.
Output
Print the only integer c — the number of parallelograms with the vertices at the given points.
Example
Input
4
0 1
1 0
1 1
2 0
Output
1 | instruction | 0 | 82,197 | 23 | 164,394 |
Tags: geometry
Correct Solution:
```
import sys,math,string,bisect
input=sys.stdin.readline
from collections import deque,defaultdict
L=lambda : list(map(int,input().split()))
Ls=lambda : list(input().split())
M=lambda : map(int,input().split())
I=lambda :int(input())
n=I()
d=defaultdict(int)
l=[]
for i in range(n):
p=L()
l.append(p)
for i in range(n-1):
for j in range(i+1,n):
d[(l[i][0]+l[j][0],l[i][1]+l[j][1])]+=1
ans=0
for i in d.keys():
ans+=(d[i]*(d[i]-1))//2
print(ans)
``` | output | 1 | 82,197 | 23 | 164,395 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given n points on a plane. All the points are distinct and no three of them lie on the same line. Find the number of parallelograms with the vertices at the given points.
Input
The first line of the input contains integer n (1 ≤ n ≤ 2000) — the number of points.
Each of the next n lines contains two integers (xi, yi) (0 ≤ xi, yi ≤ 109) — the coordinates of the i-th point.
Output
Print the only integer c — the number of parallelograms with the vertices at the given points.
Example
Input
4
0 1
1 0
1 1
2 0
Output
1 | instruction | 0 | 82,198 | 23 | 164,396 |
Tags: geometry
Correct Solution:
```
from collections import defaultdict
def mp(): return map(int,input().split())
def lt(): return list(map(int,input().split()))
def pt(x): print(x)
def ip(): return input()
def it(): return int(input())
def sl(x): return [t for t in x]
def spl(x): return x.split()
def aj(liste, item): liste.append(item)
def bin(x): return "{0:b}".format(x)
n = it()
P = []
for i in range(n):
a,b = mp()
P.append([a,b])
Mid = defaultdict(int)
result = 0
for i in range(n-1):
for j in range(i+1,n):
x,y = P[i][0]+P[j][0],P[i][1]+P[j][1]
result += Mid[(x,y)]
Mid[(x,y)] += 1
print(result)
``` | output | 1 | 82,198 | 23 | 164,397 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given n points on a plane. All the points are distinct and no three of them lie on the same line. Find the number of parallelograms with the vertices at the given points.
Input
The first line of the input contains integer n (1 ≤ n ≤ 2000) — the number of points.
Each of the next n lines contains two integers (xi, yi) (0 ≤ xi, yi ≤ 109) — the coordinates of the i-th point.
Output
Print the only integer c — the number of parallelograms with the vertices at the given points.
Example
Input
4
0 1
1 0
1 1
2 0
Output
1 | instruction | 0 | 82,199 | 23 | 164,398 |
Tags: geometry
Correct Solution:
```
from collections import Counter
n = int(input())
ls = []
for i in range(n):
a,b = map(int, input().split())
ls.append((a,b))
counter = Counter()
for i in range(n):
for j in range(i +1, n):
medio =( (ls[i][0] + ls[j][ 0])/2, (ls[i][1] + ls[j][ 1])/2)
counter[medio] += 1
sum = 0
for c in counter:
if counter[c] > 1:
sum += counter[c] * (counter[c] -1 ) /2
print(int(sum))
``` | output | 1 | 82,199 | 23 | 164,399 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given n points on a plane. All the points are distinct and no three of them lie on the same line. Find the number of parallelograms with the vertices at the given points.
Input
The first line of the input contains integer n (1 ≤ n ≤ 2000) — the number of points.
Each of the next n lines contains two integers (xi, yi) (0 ≤ xi, yi ≤ 109) — the coordinates of the i-th point.
Output
Print the only integer c — the number of parallelograms with the vertices at the given points.
Example
Input
4
0 1
1 0
1 1
2 0
Output
1 | instruction | 0 | 82,200 | 23 | 164,400 |
Tags: geometry
Correct Solution:
```
from collections import defaultdict
n=int(input())
C=[]
cnt=0
D=defaultdict(int)
for i in range(n) :
C.append([int(i) for i in input().split()])
for i0 in range(n) :
x0=C[i0][0]
y0=C[i0][1]
for Dots in C[i0+1:] :
x=Dots[0]-x0
y=Dots[1]-y0
if x<0 :
x,y=-x,-y
elif x==0 :
if y<0 :
y=-y
cnt+=D[(x,y)]
D[(x,y)]+=1
print(int(cnt/2))
``` | output | 1 | 82,200 | 23 | 164,401 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given n points on a plane. All the points are distinct and no three of them lie on the same line. Find the number of parallelograms with the vertices at the given points.
Input
The first line of the input contains integer n (1 ≤ n ≤ 2000) — the number of points.
Each of the next n lines contains two integers (xi, yi) (0 ≤ xi, yi ≤ 109) — the coordinates of the i-th point.
Output
Print the only integer c — the number of parallelograms with the vertices at the given points.
Example
Input
4
0 1
1 0
1 1
2 0
Output
1 | instruction | 0 | 82,201 | 23 | 164,402 |
Tags: geometry
Correct Solution:
```
# [https://codeforces.com/contest/660/submission/31703169]
p = []
d = {}
s = 0
for i in range(int(input())):
x, y = map(int, input().split())
for a, b in p:
q = (a + x, b + y)
if q not in d:
d[q] = 0
s += d[q]
d[q] += 1
p.append((x, y))
print(s)
``` | output | 1 | 82,201 | 23 | 164,403 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given n points on a plane. All the points are distinct and no three of them lie on the same line. Find the number of parallelograms with the vertices at the given points.
Input
The first line of the input contains integer n (1 ≤ n ≤ 2000) — the number of points.
Each of the next n lines contains two integers (xi, yi) (0 ≤ xi, yi ≤ 109) — the coordinates of the i-th point.
Output
Print the only integer c — the number of parallelograms with the vertices at the given points.
Example
Input
4
0 1
1 0
1 1
2 0
Output
1
Submitted Solution:
```
from collections import defaultdict
import math
def main():
n = int(input())
x = [None] * n
y = [None] * n
dictionary = defaultdict(int)
total = 0
for i in range(n):
x[i], y[i] = map(int, input().split())
for i in range(n):
for j in range(i+1, n):
xdist = x[i] - x[j]
ydist = y[i] - y[j]
if(xdist > 0):
xdist *= -1
ydist *= -1
elif(xdist == 0 and ydist > 0):
xdist *= -1
ydist *= -1
dictionary[(xdist, ydist)] += 1
for i in dictionary.values():
total += math.floor((i * (i - 1)) / 2)
print (math.floor(total / 2))
if __name__ == '__main__':
main()
``` | instruction | 0 | 82,202 | 23 | 164,404 |
Yes | output | 1 | 82,202 | 23 | 164,405 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given n points on a plane. All the points are distinct and no three of them lie on the same line. Find the number of parallelograms with the vertices at the given points.
Input
The first line of the input contains integer n (1 ≤ n ≤ 2000) — the number of points.
Each of the next n lines contains two integers (xi, yi) (0 ≤ xi, yi ≤ 109) — the coordinates of the i-th point.
Output
Print the only integer c — the number of parallelograms with the vertices at the given points.
Example
Input
4
0 1
1 0
1 1
2 0
Output
1
Submitted Solution:
```
#### IMPORTANT LIBRARY ####
############################
### DO NOT USE import random --> 250ms to load the library
############################
### In case of extra libraries: https://github.com/cheran-senthil/PyRival
######################
####### IMPORT #######
######################
from functools import cmp_to_key
from collections import deque, Counter
from heapq import heappush, heappop
from math import log, ceil
######################
#### STANDARD I/O ####
######################
import sys
import os
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")
if sys.version_info[0] < 3:
sys.stdin, sys.stdout = FastIO(sys.stdin), FastIO(sys.stdout)
else:
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
def print(*args, **kwargs):
sep, file = kwargs.pop("sep", " "), kwargs.pop("file", sys.stdout)
at_start = True
for x in args:
if not at_start:
file.write(sep)
file.write(str(x))
at_start = False
file.write(kwargs.pop("end", "\n"))
if kwargs.pop("flush", False):
file.flush()
def inp():
return sys.stdin.readline().rstrip("\r\n") # for fast input
def ii():
return int(inp())
def si():
return str(inp())
def li(lag = 0):
l = list(map(int, inp().split()))
if lag != 0:
for i in range(len(l)):
l[i] += lag
return l
def mi(lag = 0):
matrix = list()
for i in range(n):
matrix.append(li(lag))
return matrix
def lsi(): #string list
return list(map(str, inp().split()))
def print_list(lista, space = " "):
print(space.join(map(str, lista)))
######################
### BISECT METHODS ###
######################
def bisect_left(a, x):
"""i tale che a[i] >= x e a[i-1] < x"""
left = 0
right = len(a)
while left < right:
mid = (left+right)//2
if a[mid] < x:
left = mid+1
else:
right = mid
return left
def bisect_right(a, x):
"""i tale che a[i] > x e a[i-1] <= x"""
left = 0
right = len(a)
while left < right:
mid = (left+right)//2
if a[mid] > x:
right = mid
else:
left = mid+1
return left
def bisect_elements(a, x):
"""elementi pari a x nell'árray sortato"""
return bisect_right(a, x) - bisect_left(a, x)
######################
### MOD OPERATION ####
######################
MOD = 10**9 + 7
maxN = 5
FACT = [0] * maxN
INV_FACT = [0] * maxN
def add(x, y):
return (x+y) % MOD
def multiply(x, y):
return (x*y) % MOD
def power(x, y):
if y == 0:
return 1
elif y % 2:
return multiply(x, power(x, y-1))
else:
a = power(x, y//2)
return multiply(a, a)
def inverse(x):
return power(x, MOD-2)
def divide(x, y):
return multiply(x, inverse(y))
def allFactorials():
FACT[0] = 1
for i in range(1, maxN):
FACT[i] = multiply(i, FACT[i-1])
def inverseFactorials():
n = len(INV_FACT)
INV_FACT[n-1] = inverse(FACT[n-1])
for i in range(n-2, -1, -1):
INV_FACT[i] = multiply(INV_FACT[i+1], i+1)
def coeffBinom(n, k):
if n < k:
return 0
return multiply(FACT[n], multiply(INV_FACT[k], INV_FACT[n-k]))
######################
#### GRAPH ALGOS #####
######################
# ZERO BASED GRAPH
def create_graph(n, m, undirected = 1, unweighted = 1):
graph = [[] for i in range(n)]
if unweighted:
for i in range(m):
[x, y] = li(lag = -1)
graph[x].append(y)
if undirected:
graph[y].append(x)
else:
for i in range(m):
[x, y, w] = li(lag = -1)
w += 1
graph[x].append([y,w])
if undirected:
graph[y].append([x,w])
return graph
def create_tree(n, unweighted = 1):
children = [[] for i in range(n)]
if unweighted:
for i in range(n-1):
[x, y] = li(lag = -1)
children[x].append(y)
children[y].append(x)
else:
for i in range(n-1):
[x, y, w] = li(lag = -1)
w += 1
children[x].append([y, w])
children[y].append([x, w])
return children
def dist(tree, n, A, B = -1):
s = [[A, 0]]
massimo, massimo_nodo = 0, 0
distanza = -1
v = [-1] * n
while s:
el, dis = s.pop()
if dis > massimo:
massimo = dis
massimo_nodo = el
if el == B:
distanza = dis
for child in tree[el]:
if v[child] == -1:
v[child] = 1
s.append([child, dis+1])
return massimo, massimo_nodo, distanza
def diameter(tree):
_, foglia, _ = dist(tree, n, 0)
diam, _, _ = dist(tree, n, foglia)
return diam
def dfs(graph, n, A):
v = [-1] * n
s = [[A, 0]]
v[A] = 0
while s:
el, dis = s.pop()
for child in graph[el]:
if v[child] == -1:
v[child] = dis + 1
s.append([child, dis + 1])
return v #visited: -1 if not visited, otherwise v[B] is the distance in terms of edges
def bfs(graph, n, A):
v = [-1] * n
s = deque()
s.append([A, 0])
v[A] = 0
while s:
el, dis = s.popleft()
for child in graph[el]:
if v[child] == -1:
v[child] = dis + 1
s.append([child, dis + 1])
return v #visited: -1 if not visited, otherwise v[B] is the distance in terms of edges
#FROM A GIVEN ROOT, RECOVER THE STRUCTURE
def parents_children_root_unrooted_tree(tree, n, root = 0):
q = deque()
visited = [0] * n
parent = [-1] * n
children = [[] for i in range(n)]
q.append(root)
while q:
all_done = 1
visited[q[0]] = 1
for child in tree[q[0]]:
if not visited[child]:
all_done = 0
q.appendleft(child)
if all_done:
for child in tree[q[0]]:
if parent[child] == -1:
parent[q[0]] = child
children[child].append(q[0])
q.popleft()
return parent, children
# CALCULATING LONGEST PATH FOR ALL THE NODES
def all_longest_path_passing_from_node(parent, children, n):
q = deque()
visited = [len(children[i]) for i in range(n)]
downwards = [[0,0] for i in range(n)]
upward = [1] * n
longest_path = [1] * n
for i in range(n):
if not visited[i]:
q.append(i)
downwards[i] = [1,0]
while q:
node = q.popleft()
if parent[node] != -1:
visited[parent[node]] -= 1
if not visited[parent[node]]:
q.append(parent[node])
else:
root = node
for child in children[node]:
downwards[node] = sorted([downwards[node][0], downwards[node][1], downwards[child][0] + 1], reverse = True)[0:2]
s = [node]
while s:
node = s.pop()
if parent[node] != -1:
if downwards[parent[node]][0] == downwards[node][0] + 1:
upward[node] = 1 + max(upward[parent[node]], downwards[parent[node]][1])
else:
upward[node] = 1 + max(upward[parent[node]], downwards[parent[node]][0])
longest_path[node] = downwards[node][0] + downwards[node][1] + upward[node] - min([downwards[node][0], downwards[node][1], upward[node]]) - 1
for child in children[node]:
s.append(child)
return longest_path
### TBD SUCCESSOR GRAPH 7.5
### TBD TREE QUERIES 10.2 da 2 a 4
### TBD ADVANCED TREE 10.3
### TBD GRAPHS AND MATRICES 11.3.3 e 11.4.3 e 11.5.3 (ON GAMES)
######################
## END OF LIBRARIES ##
######################
n = ii()
d = dict()
a = list()
for i in range(n):
x,y = li()
a.append([x,y])
for i in range(n-1):
for j in range(i+1,n):
xi,yi = a[i]
xj,yj = a[j]
punto = 2*10**9*(xi+xj)+(yi+yj)
d[punto] = d.get(punto,0)+1
res = 0
for x in d:
res += d[x]*(d[x]-1)//2
print(res)
``` | instruction | 0 | 82,203 | 23 | 164,406 |
Yes | output | 1 | 82,203 | 23 | 164,407 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given n points on a plane. All the points are distinct and no three of them lie on the same line. Find the number of parallelograms with the vertices at the given points.
Input
The first line of the input contains integer n (1 ≤ n ≤ 2000) — the number of points.
Each of the next n lines contains two integers (xi, yi) (0 ≤ xi, yi ≤ 109) — the coordinates of the i-th point.
Output
Print the only integer c — the number of parallelograms with the vertices at the given points.
Example
Input
4
0 1
1 0
1 1
2 0
Output
1
Submitted Solution:
```
def middle(a, b):
return (b[0] + a[0])/2, (b[1] + a[1])/2
def solve(n, points):
middles = dict()
solutions = 0
for i in range(n-1):
for j in range(i+1, n):
s = middle(points[i], points[j])
solutions += middles.setdefault(s, 0)
middles[s] += 1
return solutions
if __name__ == "__main__":
n = int(input())
points = list()
for _ in range(n):
points.append(tuple(map(int, input().split())))
print(solve(n, points))
``` | instruction | 0 | 82,204 | 23 | 164,408 |
Yes | output | 1 | 82,204 | 23 | 164,409 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given n points on a plane. All the points are distinct and no three of them lie on the same line. Find the number of parallelograms with the vertices at the given points.
Input
The first line of the input contains integer n (1 ≤ n ≤ 2000) — the number of points.
Each of the next n lines contains two integers (xi, yi) (0 ≤ xi, yi ≤ 109) — the coordinates of the i-th point.
Output
Print the only integer c — the number of parallelograms with the vertices at the given points.
Example
Input
4
0 1
1 0
1 1
2 0
Output
1
Submitted Solution:
```
n = int(input())
x = []
y = []
for i in range(n):
x.append(0)
y.append(0)
x[i], y[i] = map(int, input().split())
from collections import defaultdict
dict = defaultdict(int)
for i in range(n):
for j in range(i + 1, n):
tx = x[i] - x[j]
ty = y[i] - y[j]
if tx > 0 or tx == 0 and ty > 0:
tx = -tx
ty = -ty
dict[(tx, ty)] += 1
ans = 0
#print(dict)
for i in dict.values():
ans += i * (i - 1) // 2
print(ans // 2)
``` | instruction | 0 | 82,205 | 23 | 164,410 |
Yes | output | 1 | 82,205 | 23 | 164,411 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given n points on a plane. All the points are distinct and no three of them lie on the same line. Find the number of parallelograms with the vertices at the given points.
Input
The first line of the input contains integer n (1 ≤ n ≤ 2000) — the number of points.
Each of the next n lines contains two integers (xi, yi) (0 ≤ xi, yi ≤ 109) — the coordinates of the i-th point.
Output
Print the only integer c — the number of parallelograms with the vertices at the given points.
Example
Input
4
0 1
1 0
1 1
2 0
Output
1
Submitted Solution:
```
from collections import defaultdict
import math
def main():
n = int(input())
x = []
y = []
dictionary = defaultdict(int)
total = 0
for i in range(n):
x.append(i)
y.append(i)
x[i], y[i] = map(int, input().split())
for i in range(n):
for j in range(i+1, n):
xdist = x[i] - x[j]
ydist = y[i] - y[j]
if(xdist > 0):
xdist *= -1
ydist *= -1
elif(xdist == 0 and ydist > 0):
xdist *= -1
ydist *= -1
dictionary[(xdist, ydist)] += 1
for i in dictionary.values():
total = total + (i * (i - 1)) // 2
print (total)
if __name__ == '__main__':
main()
``` | instruction | 0 | 82,206 | 23 | 164,412 |
No | output | 1 | 82,206 | 23 | 164,413 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given n points on a plane. All the points are distinct and no three of them lie on the same line. Find the number of parallelograms with the vertices at the given points.
Input
The first line of the input contains integer n (1 ≤ n ≤ 2000) — the number of points.
Each of the next n lines contains two integers (xi, yi) (0 ≤ xi, yi ≤ 109) — the coordinates of the i-th point.
Output
Print the only integer c — the number of parallelograms with the vertices at the given points.
Example
Input
4
0 1
1 0
1 1
2 0
Output
1
Submitted Solution:
```
n = int(input())
a = []
for _ in range(n):
xy = list(map(int, input().split()))
a.append(xy)
slopes = {}
for i in range(n):
for j in range(i + 1, n):
slope = 0.0
x1, y1 = a[i][0], a[i][1]
x2, y2 = a[j][0], a[j][1]
if x2 - x1 != 0:
slope = (y2 - y1) / (x2 - x1)
slopes[slope] = slopes.get(slope, 0) + 1
cnt = 0
m, n = 0, 0
#print('-->')
for k, v in slopes.items():
#print('---')
#print(k, v)
#cnt += v // 2
if v == 2:
m += 1
#print(k, v)
#cnt += 1
elif v == 3:
n += 1
print(min(n, m))
#print(cnt)
#print(cnt / 2)
#print(cnt // 2)
#import matplotlib.pyplot as plt
#import numpy as np
#x, y = zip(*a)
#plt.scatter(x, y)
#plt.grid()
#plt.show()
``` | instruction | 0 | 82,207 | 23 | 164,414 |
No | output | 1 | 82,207 | 23 | 164,415 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given n points on a plane. All the points are distinct and no three of them lie on the same line. Find the number of parallelograms with the vertices at the given points.
Input
The first line of the input contains integer n (1 ≤ n ≤ 2000) — the number of points.
Each of the next n lines contains two integers (xi, yi) (0 ≤ xi, yi ≤ 109) — the coordinates of the i-th point.
Output
Print the only integer c — the number of parallelograms with the vertices at the given points.
Example
Input
4
0 1
1 0
1 1
2 0
Output
1
Submitted Solution:
```
def main():
n = int(input())
total = 0
x = [None] * n
y = [None] * n
for i in range(n):
x[i], y[i] = map(int, input().split())
for j in range(n - 2):
for k in range(n - 1):
for m in range(n):
test = x[j] * (y[k] - y[m]) + x[k] * (y[m] - y[j]) + x[m] * (y[j] - y[k])
if (test != 0):
total += 1
print(total)
if __name__ == '__main__':
main()
``` | instruction | 0 | 82,208 | 23 | 164,416 |
No | output | 1 | 82,208 | 23 | 164,417 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given n points on a plane. All the points are distinct and no three of them lie on the same line. Find the number of parallelograms with the vertices at the given points.
Input
The first line of the input contains integer n (1 ≤ n ≤ 2000) — the number of points.
Each of the next n lines contains two integers (xi, yi) (0 ≤ xi, yi ≤ 109) — the coordinates of the i-th point.
Output
Print the only integer c — the number of parallelograms with the vertices at the given points.
Example
Input
4
0 1
1 0
1 1
2 0
Output
1
Submitted Solution:
```
from sys import stdin, stdout
n = int(stdin.readline())
points = []
for i in range(n):
x, y = map(int, stdin.readline().split())
points.append((x, y))
points.sort(key = lambda x: x[1])
pair = {}
for i in range(n):
for j in range(i + 1, n):
if (points[j][0] - points[i][0], points[j][1] - points[i][1]) not in pair:
pair[(points[j][0] - points[i][0], points[j][1] - points[i][1])] = 1
else:
pair[(points[j][0] - points[i][0], points[j][1] - points[i][1])] += 1
ans = 0
for v in pair:
ans += pair[v] * (pair[v] - 1) // 2
stdout.write(str(ans // 2))
``` | instruction | 0 | 82,209 | 23 | 164,418 |
No | output | 1 | 82,209 | 23 | 164,419 |
Provide a correct Python 3 solution for this coding contest problem.
Determine if an N-sided polygon (not necessarily convex) with sides of length L_1, L_2, ..., L_N can be drawn in a two-dimensional plane.
You can use the following theorem:
Theorem: an N-sided polygon satisfying the condition can be drawn if and only if the longest side is strictly shorter than the sum of the lengths of the other N-1 sides.
Constraints
* All values in input are integers.
* 3 \leq N \leq 10
* 1 \leq L_i \leq 100
Input
Input is given from Standard Input in the following format:
N
L_1 L_2 ... L_N
Output
If an N-sided polygon satisfying the condition can be drawn, print `Yes`; otherwise, print `No`.
Examples
Input
4
3 8 5 1
Output
Yes
Input
4
3 8 4 1
Output
No
Input
10
1 8 10 5 8 12 34 100 11 3
Output
No | instruction | 0 | 82,411 | 23 | 164,822 |
"Correct Solution:
```
N=int(input())
l=list(map(int,input().split()))
if 2*max(l)<sum(l):
print("Yes")
else:
print("No")
``` | output | 1 | 82,411 | 23 | 164,823 |
Provide a correct Python 3 solution for this coding contest problem.
Determine if an N-sided polygon (not necessarily convex) with sides of length L_1, L_2, ..., L_N can be drawn in a two-dimensional plane.
You can use the following theorem:
Theorem: an N-sided polygon satisfying the condition can be drawn if and only if the longest side is strictly shorter than the sum of the lengths of the other N-1 sides.
Constraints
* All values in input are integers.
* 3 \leq N \leq 10
* 1 \leq L_i \leq 100
Input
Input is given from Standard Input in the following format:
N
L_1 L_2 ... L_N
Output
If an N-sided polygon satisfying the condition can be drawn, print `Yes`; otherwise, print `No`.
Examples
Input
4
3 8 5 1
Output
Yes
Input
4
3 8 4 1
Output
No
Input
10
1 8 10 5 8 12 34 100 11 3
Output
No | instruction | 0 | 82,412 | 23 | 164,824 |
"Correct Solution:
```
n=int(input())
l=list(map(int,input().split()))
m=max(l)
print('Yes' if sum(l)-m>m else 'No')
``` | output | 1 | 82,412 | 23 | 164,825 |
Provide a correct Python 3 solution for this coding contest problem.
Determine if an N-sided polygon (not necessarily convex) with sides of length L_1, L_2, ..., L_N can be drawn in a two-dimensional plane.
You can use the following theorem:
Theorem: an N-sided polygon satisfying the condition can be drawn if and only if the longest side is strictly shorter than the sum of the lengths of the other N-1 sides.
Constraints
* All values in input are integers.
* 3 \leq N \leq 10
* 1 \leq L_i \leq 100
Input
Input is given from Standard Input in the following format:
N
L_1 L_2 ... L_N
Output
If an N-sided polygon satisfying the condition can be drawn, print `Yes`; otherwise, print `No`.
Examples
Input
4
3 8 5 1
Output
Yes
Input
4
3 8 4 1
Output
No
Input
10
1 8 10 5 8 12 34 100 11 3
Output
No | instruction | 0 | 82,413 | 23 | 164,826 |
"Correct Solution:
```
n=int(input())
l=list(map(int,input().split()))
print(("No","Yes")[((sum(l)-max(l))-max(l))>0])
``` | output | 1 | 82,413 | 23 | 164,827 |
Provide a correct Python 3 solution for this coding contest problem.
Determine if an N-sided polygon (not necessarily convex) with sides of length L_1, L_2, ..., L_N can be drawn in a two-dimensional plane.
You can use the following theorem:
Theorem: an N-sided polygon satisfying the condition can be drawn if and only if the longest side is strictly shorter than the sum of the lengths of the other N-1 sides.
Constraints
* All values in input are integers.
* 3 \leq N \leq 10
* 1 \leq L_i \leq 100
Input
Input is given from Standard Input in the following format:
N
L_1 L_2 ... L_N
Output
If an N-sided polygon satisfying the condition can be drawn, print `Yes`; otherwise, print `No`.
Examples
Input
4
3 8 5 1
Output
Yes
Input
4
3 8 4 1
Output
No
Input
10
1 8 10 5 8 12 34 100 11 3
Output
No | instruction | 0 | 82,414 | 23 | 164,828 |
"Correct Solution:
```
input()
a = list(map(int, input().split()))
print('Yes' if max(a) < sum(a) - max(a) else 'No')
``` | output | 1 | 82,414 | 23 | 164,829 |
Provide a correct Python 3 solution for this coding contest problem.
Determine if an N-sided polygon (not necessarily convex) with sides of length L_1, L_2, ..., L_N can be drawn in a two-dimensional plane.
You can use the following theorem:
Theorem: an N-sided polygon satisfying the condition can be drawn if and only if the longest side is strictly shorter than the sum of the lengths of the other N-1 sides.
Constraints
* All values in input are integers.
* 3 \leq N \leq 10
* 1 \leq L_i \leq 100
Input
Input is given from Standard Input in the following format:
N
L_1 L_2 ... L_N
Output
If an N-sided polygon satisfying the condition can be drawn, print `Yes`; otherwise, print `No`.
Examples
Input
4
3 8 5 1
Output
Yes
Input
4
3 8 4 1
Output
No
Input
10
1 8 10 5 8 12 34 100 11 3
Output
No | instruction | 0 | 82,415 | 23 | 164,830 |
"Correct Solution:
```
input()
L=list(map(int,input().split()))
print("Yes" if sum(L)>max(L)*2 else "No")
``` | output | 1 | 82,415 | 23 | 164,831 |
Provide a correct Python 3 solution for this coding contest problem.
Determine if an N-sided polygon (not necessarily convex) with sides of length L_1, L_2, ..., L_N can be drawn in a two-dimensional plane.
You can use the following theorem:
Theorem: an N-sided polygon satisfying the condition can be drawn if and only if the longest side is strictly shorter than the sum of the lengths of the other N-1 sides.
Constraints
* All values in input are integers.
* 3 \leq N \leq 10
* 1 \leq L_i \leq 100
Input
Input is given from Standard Input in the following format:
N
L_1 L_2 ... L_N
Output
If an N-sided polygon satisfying the condition can be drawn, print `Yes`; otherwise, print `No`.
Examples
Input
4
3 8 5 1
Output
Yes
Input
4
3 8 4 1
Output
No
Input
10
1 8 10 5 8 12 34 100 11 3
Output
No | instruction | 0 | 82,416 | 23 | 164,832 |
"Correct Solution:
```
N = int(input())
L = [int(a) for a in input().split(" ")]
print("Yes" if sum(L)>2*max(L) else "No")
``` | output | 1 | 82,416 | 23 | 164,833 |
Provide a correct Python 3 solution for this coding contest problem.
Determine if an N-sided polygon (not necessarily convex) with sides of length L_1, L_2, ..., L_N can be drawn in a two-dimensional plane.
You can use the following theorem:
Theorem: an N-sided polygon satisfying the condition can be drawn if and only if the longest side is strictly shorter than the sum of the lengths of the other N-1 sides.
Constraints
* All values in input are integers.
* 3 \leq N \leq 10
* 1 \leq L_i \leq 100
Input
Input is given from Standard Input in the following format:
N
L_1 L_2 ... L_N
Output
If an N-sided polygon satisfying the condition can be drawn, print `Yes`; otherwise, print `No`.
Examples
Input
4
3 8 5 1
Output
Yes
Input
4
3 8 4 1
Output
No
Input
10
1 8 10 5 8 12 34 100 11 3
Output
No | instruction | 0 | 82,417 | 23 | 164,834 |
"Correct Solution:
```
n,*a,=map(int,open(0).read().split());print("YNeos"[sum(a)<=max(a)*2::2])
``` | output | 1 | 82,417 | 23 | 164,835 |
Provide a correct Python 3 solution for this coding contest problem.
Determine if an N-sided polygon (not necessarily convex) with sides of length L_1, L_2, ..., L_N can be drawn in a two-dimensional plane.
You can use the following theorem:
Theorem: an N-sided polygon satisfying the condition can be drawn if and only if the longest side is strictly shorter than the sum of the lengths of the other N-1 sides.
Constraints
* All values in input are integers.
* 3 \leq N \leq 10
* 1 \leq L_i \leq 100
Input
Input is given from Standard Input in the following format:
N
L_1 L_2 ... L_N
Output
If an N-sided polygon satisfying the condition can be drawn, print `Yes`; otherwise, print `No`.
Examples
Input
4
3 8 5 1
Output
Yes
Input
4
3 8 4 1
Output
No
Input
10
1 8 10 5 8 12 34 100 11 3
Output
No | instruction | 0 | 82,418 | 23 | 164,836 |
"Correct Solution:
```
n = int(input())
l = sorted(map(int, input().split()))
print("Yes" if l[-1] < sum(l[:-1]) else "No")
``` | output | 1 | 82,418 | 23 | 164,837 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Determine if an N-sided polygon (not necessarily convex) with sides of length L_1, L_2, ..., L_N can be drawn in a two-dimensional plane.
You can use the following theorem:
Theorem: an N-sided polygon satisfying the condition can be drawn if and only if the longest side is strictly shorter than the sum of the lengths of the other N-1 sides.
Constraints
* All values in input are integers.
* 3 \leq N \leq 10
* 1 \leq L_i \leq 100
Input
Input is given from Standard Input in the following format:
N
L_1 L_2 ... L_N
Output
If an N-sided polygon satisfying the condition can be drawn, print `Yes`; otherwise, print `No`.
Examples
Input
4
3 8 5 1
Output
Yes
Input
4
3 8 4 1
Output
No
Input
10
1 8 10 5 8 12 34 100 11 3
Output
No
Submitted Solution:
```
n=int(input())
x=list(map(int,input().split()))
if 2*max(x)<sum(x):
print('Yes')
else:
print('No')
``` | instruction | 0 | 82,419 | 23 | 164,838 |
Yes | output | 1 | 82,419 | 23 | 164,839 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Determine if an N-sided polygon (not necessarily convex) with sides of length L_1, L_2, ..., L_N can be drawn in a two-dimensional plane.
You can use the following theorem:
Theorem: an N-sided polygon satisfying the condition can be drawn if and only if the longest side is strictly shorter than the sum of the lengths of the other N-1 sides.
Constraints
* All values in input are integers.
* 3 \leq N \leq 10
* 1 \leq L_i \leq 100
Input
Input is given from Standard Input in the following format:
N
L_1 L_2 ... L_N
Output
If an N-sided polygon satisfying the condition can be drawn, print `Yes`; otherwise, print `No`.
Examples
Input
4
3 8 5 1
Output
Yes
Input
4
3 8 4 1
Output
No
Input
10
1 8 10 5 8 12 34 100 11 3
Output
No
Submitted Solution:
```
N=int(input())
a=list(map(int,input().split()))
m=max(a)
if 2*m<sum(a): print("Yes")
else: print("No")
``` | instruction | 0 | 82,420 | 23 | 164,840 |
Yes | output | 1 | 82,420 | 23 | 164,841 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Determine if an N-sided polygon (not necessarily convex) with sides of length L_1, L_2, ..., L_N can be drawn in a two-dimensional plane.
You can use the following theorem:
Theorem: an N-sided polygon satisfying the condition can be drawn if and only if the longest side is strictly shorter than the sum of the lengths of the other N-1 sides.
Constraints
* All values in input are integers.
* 3 \leq N \leq 10
* 1 \leq L_i \leq 100
Input
Input is given from Standard Input in the following format:
N
L_1 L_2 ... L_N
Output
If an N-sided polygon satisfying the condition can be drawn, print `Yes`; otherwise, print `No`.
Examples
Input
4
3 8 5 1
Output
Yes
Input
4
3 8 4 1
Output
No
Input
10
1 8 10 5 8 12 34 100 11 3
Output
No
Submitted Solution:
```
N = int(input())
L = list(map(int, input().split()))
print("Yes" if max(L) < sum(L)-max(L) else "No")
``` | instruction | 0 | 82,421 | 23 | 164,842 |
Yes | output | 1 | 82,421 | 23 | 164,843 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Determine if an N-sided polygon (not necessarily convex) with sides of length L_1, L_2, ..., L_N can be drawn in a two-dimensional plane.
You can use the following theorem:
Theorem: an N-sided polygon satisfying the condition can be drawn if and only if the longest side is strictly shorter than the sum of the lengths of the other N-1 sides.
Constraints
* All values in input are integers.
* 3 \leq N \leq 10
* 1 \leq L_i \leq 100
Input
Input is given from Standard Input in the following format:
N
L_1 L_2 ... L_N
Output
If an N-sided polygon satisfying the condition can be drawn, print `Yes`; otherwise, print `No`.
Examples
Input
4
3 8 5 1
Output
Yes
Input
4
3 8 4 1
Output
No
Input
10
1 8 10 5 8 12 34 100 11 3
Output
No
Submitted Solution:
```
_=input()
L=sorted(list(map(int,input().split())))
print('Yes' if L[-1] < sum(L[:-1]) else 'No')
``` | instruction | 0 | 82,422 | 23 | 164,844 |
Yes | output | 1 | 82,422 | 23 | 164,845 |
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