message stringlengths 2 48.6k | message_type stringclasses 2 values | message_id int64 0 1 | conversation_id int64 318 108k | cluster float64 8 8 | __index_level_0__ int64 636 217k |
|---|---|---|---|---|---|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
The Fair Nut likes kvass very much. On his birthday parents presented him n kegs of kvass. There are v_i liters of kvass in the i-th keg. Each keg has a lever. You can pour your glass by exactly 1 liter pulling this lever. The Fair Nut likes this drink very much, so he wants to pour his glass by s liters of kvass. But he wants to do it, so kvass level in the least keg is as much as possible.
Help him find out how much kvass can be in the least keg or define it's not possible to pour his glass by s liters of kvass.
Input
The first line contains two integers n and s (1 β€ n β€ 10^3, 1 β€ s β€ 10^{12}) β the number of kegs and glass volume.
The second line contains n integers v_1, v_2, β¦, v_n (1 β€ v_i β€ 10^9) β the volume of i-th keg.
Output
If the Fair Nut cannot pour his glass by s liters of kvass, print -1. Otherwise, print a single integer β how much kvass in the least keg can be.
Examples
Input
3 3
4 3 5
Output
3
Input
3 4
5 3 4
Output
2
Input
3 7
1 2 3
Output
-1
Note
In the first example, the answer is 3, the Fair Nut can take 1 liter from the first keg and 2 liters from the third keg. There are 3 liters of kvass in each keg.
In the second example, the answer is 2, the Fair Nut can take 3 liters from the first keg and 1 liter from the second keg.
In the third example, the Fair Nut can't pour his cup by 7 liters, so the answer is -1.
Submitted Solution:
```
import math
m,n=map(int,input().split())
li=sorted(list(map(int,input().split())))
l=li[0]
bli=[]
if sum(li)<n:
print(-1)
exit()
for i in range(len(li)):
bli.append(li[i]-li[0])
if sum(bli)>=n:
print(l)
else:
print(l-math.ceil((n-sum(bli))/m))
``` | instruction | 0 | 5,069 | 8 | 10,138 |
Yes | output | 1 | 5,069 | 8 | 10,139 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
The Fair Nut likes kvass very much. On his birthday parents presented him n kegs of kvass. There are v_i liters of kvass in the i-th keg. Each keg has a lever. You can pour your glass by exactly 1 liter pulling this lever. The Fair Nut likes this drink very much, so he wants to pour his glass by s liters of kvass. But he wants to do it, so kvass level in the least keg is as much as possible.
Help him find out how much kvass can be in the least keg or define it's not possible to pour his glass by s liters of kvass.
Input
The first line contains two integers n and s (1 β€ n β€ 10^3, 1 β€ s β€ 10^{12}) β the number of kegs and glass volume.
The second line contains n integers v_1, v_2, β¦, v_n (1 β€ v_i β€ 10^9) β the volume of i-th keg.
Output
If the Fair Nut cannot pour his glass by s liters of kvass, print -1. Otherwise, print a single integer β how much kvass in the least keg can be.
Examples
Input
3 3
4 3 5
Output
3
Input
3 4
5 3 4
Output
2
Input
3 7
1 2 3
Output
-1
Note
In the first example, the answer is 3, the Fair Nut can take 1 liter from the first keg and 2 liters from the third keg. There are 3 liters of kvass in each keg.
In the second example, the answer is 2, the Fair Nut can take 3 liters from the first keg and 1 liter from the second keg.
In the third example, the Fair Nut can't pour his cup by 7 liters, so the answer is -1.
Submitted Solution:
```
# ***********************************************
#
# achie27
#
# ***********************************************
def retarray():
return list(map(int, input().split()))
import math
n, s = retarray()
a = retarray()
min_keg = min(a)
acc = 0
for x in a:
acc += (x - min_keg)
if acc >= s:
print(min_keg)
else:
x = math.ceil((s-acc)/n)
if x > min_keg:
print(-1)
else:
print(min_keg-x)
``` | instruction | 0 | 5,070 | 8 | 10,140 |
Yes | output | 1 | 5,070 | 8 | 10,141 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
The Fair Nut likes kvass very much. On his birthday parents presented him n kegs of kvass. There are v_i liters of kvass in the i-th keg. Each keg has a lever. You can pour your glass by exactly 1 liter pulling this lever. The Fair Nut likes this drink very much, so he wants to pour his glass by s liters of kvass. But he wants to do it, so kvass level in the least keg is as much as possible.
Help him find out how much kvass can be in the least keg or define it's not possible to pour his glass by s liters of kvass.
Input
The first line contains two integers n and s (1 β€ n β€ 10^3, 1 β€ s β€ 10^{12}) β the number of kegs and glass volume.
The second line contains n integers v_1, v_2, β¦, v_n (1 β€ v_i β€ 10^9) β the volume of i-th keg.
Output
If the Fair Nut cannot pour his glass by s liters of kvass, print -1. Otherwise, print a single integer β how much kvass in the least keg can be.
Examples
Input
3 3
4 3 5
Output
3
Input
3 4
5 3 4
Output
2
Input
3 7
1 2 3
Output
-1
Note
In the first example, the answer is 3, the Fair Nut can take 1 liter from the first keg and 2 liters from the third keg. There are 3 liters of kvass in each keg.
In the second example, the answer is 2, the Fair Nut can take 3 liters from the first keg and 1 liter from the second keg.
In the third example, the Fair Nut can't pour his cup by 7 liters, so the answer is -1.
Submitted Solution:
```
import math
n, s = map(int, input().split())
kegs = [int(k) for k in input().split()]
if s > sum(kegs):
print(-1)
exit()
opsleft = s
minimum = min(kegs)
amtleft = 0
for keg in kegs:
amtleft += keg-minimum
if s <= amtleft:
print(minimum)
else:
res = minimum - math.ceil((s - amtleft) / n)
print(res)
``` | instruction | 0 | 5,071 | 8 | 10,142 |
Yes | output | 1 | 5,071 | 8 | 10,143 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
The Fair Nut likes kvass very much. On his birthday parents presented him n kegs of kvass. There are v_i liters of kvass in the i-th keg. Each keg has a lever. You can pour your glass by exactly 1 liter pulling this lever. The Fair Nut likes this drink very much, so he wants to pour his glass by s liters of kvass. But he wants to do it, so kvass level in the least keg is as much as possible.
Help him find out how much kvass can be in the least keg or define it's not possible to pour his glass by s liters of kvass.
Input
The first line contains two integers n and s (1 β€ n β€ 10^3, 1 β€ s β€ 10^{12}) β the number of kegs and glass volume.
The second line contains n integers v_1, v_2, β¦, v_n (1 β€ v_i β€ 10^9) β the volume of i-th keg.
Output
If the Fair Nut cannot pour his glass by s liters of kvass, print -1. Otherwise, print a single integer β how much kvass in the least keg can be.
Examples
Input
3 3
4 3 5
Output
3
Input
3 4
5 3 4
Output
2
Input
3 7
1 2 3
Output
-1
Note
In the first example, the answer is 3, the Fair Nut can take 1 liter from the first keg and 2 liters from the third keg. There are 3 liters of kvass in each keg.
In the second example, the answer is 2, the Fair Nut can take 3 liters from the first keg and 1 liter from the second keg.
In the third example, the Fair Nut can't pour his cup by 7 liters, so the answer is -1.
Submitted Solution:
```
a,b=map(int,input().split())
z=list(map(int,input().split()))
s=sum(z);r=min(z)
if b>s:print(-1)
else:print(min(r,(s-b)//a))
``` | instruction | 0 | 5,072 | 8 | 10,144 |
Yes | output | 1 | 5,072 | 8 | 10,145 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
The Fair Nut likes kvass very much. On his birthday parents presented him n kegs of kvass. There are v_i liters of kvass in the i-th keg. Each keg has a lever. You can pour your glass by exactly 1 liter pulling this lever. The Fair Nut likes this drink very much, so he wants to pour his glass by s liters of kvass. But he wants to do it, so kvass level in the least keg is as much as possible.
Help him find out how much kvass can be in the least keg or define it's not possible to pour his glass by s liters of kvass.
Input
The first line contains two integers n and s (1 β€ n β€ 10^3, 1 β€ s β€ 10^{12}) β the number of kegs and glass volume.
The second line contains n integers v_1, v_2, β¦, v_n (1 β€ v_i β€ 10^9) β the volume of i-th keg.
Output
If the Fair Nut cannot pour his glass by s liters of kvass, print -1. Otherwise, print a single integer β how much kvass in the least keg can be.
Examples
Input
3 3
4 3 5
Output
3
Input
3 4
5 3 4
Output
2
Input
3 7
1 2 3
Output
-1
Note
In the first example, the answer is 3, the Fair Nut can take 1 liter from the first keg and 2 liters from the third keg. There are 3 liters of kvass in each keg.
In the second example, the answer is 2, the Fair Nut can take 3 liters from the first keg and 1 liter from the second keg.
In the third example, the Fair Nut can't pour his cup by 7 liters, so the answer is -1.
Submitted Solution:
```
n, s = map(int, input().split())
v = list(map(int, input().split()))
least = min(v)
if sum(v) < s:
print(-1)
else:
for i in v:
s -= (i-least)
if s != 0:
least -= (s)/n
print(int(least))
``` | instruction | 0 | 5,073 | 8 | 10,146 |
No | output | 1 | 5,073 | 8 | 10,147 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
The Fair Nut likes kvass very much. On his birthday parents presented him n kegs of kvass. There are v_i liters of kvass in the i-th keg. Each keg has a lever. You can pour your glass by exactly 1 liter pulling this lever. The Fair Nut likes this drink very much, so he wants to pour his glass by s liters of kvass. But he wants to do it, so kvass level in the least keg is as much as possible.
Help him find out how much kvass can be in the least keg or define it's not possible to pour his glass by s liters of kvass.
Input
The first line contains two integers n and s (1 β€ n β€ 10^3, 1 β€ s β€ 10^{12}) β the number of kegs and glass volume.
The second line contains n integers v_1, v_2, β¦, v_n (1 β€ v_i β€ 10^9) β the volume of i-th keg.
Output
If the Fair Nut cannot pour his glass by s liters of kvass, print -1. Otherwise, print a single integer β how much kvass in the least keg can be.
Examples
Input
3 3
4 3 5
Output
3
Input
3 4
5 3 4
Output
2
Input
3 7
1 2 3
Output
-1
Note
In the first example, the answer is 3, the Fair Nut can take 1 liter from the first keg and 2 liters from the third keg. There are 3 liters of kvass in each keg.
In the second example, the answer is 2, the Fair Nut can take 3 liters from the first keg and 1 liter from the second keg.
In the third example, the Fair Nut can't pour his cup by 7 liters, so the answer is -1.
Submitted Solution:
```
n, s = list(map(int, input().split()))
a= list(map(int, input().split()))
if s > sum(a):
print(-1)
elif s == sum(a):
print(0)
else:
a = sorted(a)
total = sum(a)
b = [0 for _ in range(len(a))]
b[0] = a[0]
for i in range(1, len(a)):
b[i] = a[i] + b[i-1]
c = [0 for _ in range(len(a))]
c[0] = total - a[0] * (len(a))
for i in range(1, len(a)):
c[i] = total - b[i] - a[i] * (len(a)-i-1)
if s <= c[0]:
print(a[0])
else:
s -= c[0]
add = s // a[0] + bool(s%a[0])
final = a[0] - add
print(final)
``` | instruction | 0 | 5,074 | 8 | 10,148 |
No | output | 1 | 5,074 | 8 | 10,149 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
The Fair Nut likes kvass very much. On his birthday parents presented him n kegs of kvass. There are v_i liters of kvass in the i-th keg. Each keg has a lever. You can pour your glass by exactly 1 liter pulling this lever. The Fair Nut likes this drink very much, so he wants to pour his glass by s liters of kvass. But he wants to do it, so kvass level in the least keg is as much as possible.
Help him find out how much kvass can be in the least keg or define it's not possible to pour his glass by s liters of kvass.
Input
The first line contains two integers n and s (1 β€ n β€ 10^3, 1 β€ s β€ 10^{12}) β the number of kegs and glass volume.
The second line contains n integers v_1, v_2, β¦, v_n (1 β€ v_i β€ 10^9) β the volume of i-th keg.
Output
If the Fair Nut cannot pour his glass by s liters of kvass, print -1. Otherwise, print a single integer β how much kvass in the least keg can be.
Examples
Input
3 3
4 3 5
Output
3
Input
3 4
5 3 4
Output
2
Input
3 7
1 2 3
Output
-1
Note
In the first example, the answer is 3, the Fair Nut can take 1 liter from the first keg and 2 liters from the third keg. There are 3 liters of kvass in each keg.
In the second example, the answer is 2, the Fair Nut can take 3 liters from the first keg and 1 liter from the second keg.
In the third example, the Fair Nut can't pour his cup by 7 liters, so the answer is -1.
Submitted Solution:
```
''' Thruth can only be found at one place - THE CODE '''
''' Copyright 2018, SATYAM KUMAR'''
from sys import stdin, stdout
import cProfile, math
from collections import Counter
from bisect import bisect_left,bisect,bisect_right
import itertools
from copy import deepcopy
from fractions import Fraction
import sys, threading
sys.setrecursionlimit(10**6) # max depth of recursion
threading.stack_size(2**27) # new thread will get stack of such size
printHeap = str()
memory_constrained = False
P = 10**9+7
import sys
sys.setrecursionlimit(10000000)
class Operation:
def __init__(self, name, function, function_on_equal, neutral_value=0):
self.name = name
self.f = function
self.f_on_equal = function_on_equal
def add_multiple(x, count):
return x * count
def min_multiple(x, count):
return x
def max_multiple(x, count):
return x
sum_operation = Operation("sum", sum, add_multiple, 0)
min_operation = Operation("min", min, min_multiple, 1e9)
max_operation = Operation("max", max, max_multiple, -1e9)
class SegmentTree:
def __init__(self,
array,
operations=[sum_operation, min_operation, max_operation]):
self.array = array
if type(operations) != list:
raise TypeError("operations must be a list")
self.operations = {}
for op in operations:
self.operations[op.name] = op
self.root = SegmentTreeNode(0, len(array) - 1, self)
def query(self, start, end, operation_name):
if self.operations.get(operation_name) == None:
raise Exception("This operation is not available")
return self.root._query(start, end, self.operations[operation_name])
def summary(self):
return self.root.values
def update(self, position, value):
self.root._update(position, value)
def update_range(self, start, end, value):
self.root._update_range(start, end, value)
def __repr__(self):
return self.root.__repr__()
class SegmentTreeNode:
def __init__(self, start, end, segment_tree):
self.range = (start, end)
self.parent_tree = segment_tree
self.range_value = None
self.values = {}
self.left = None
self.right = None
if start == end:
self._sync()
return
self.left = SegmentTreeNode(start, start + (end - start) // 2,
segment_tree)
self.right = SegmentTreeNode(start + (end - start) // 2 + 1, end,
segment_tree)
self._sync()
def _query(self, start, end, operation):
if end < self.range[0] or start > self.range[1]:
return None
if start <= self.range[0] and self.range[1] <= end:
return self.values[operation.name]
self._push()
left_res = self.left._query(start, end,
operation) if self.left else None
right_res = self.right._query(start, end,
operation) if self.right else None
if left_res is None:
return right_res
if right_res is None:
return left_res
return operation.f([left_res, right_res])
def _update(self, position, value):
if position < self.range[0] or position > self.range[1]:
return
if position == self.range[0] and self.range[1] == position:
self.parent_tree.array[position] = value
self._sync()
return
self._push()
self.left._update(position, value)
self.right._update(position, value)
self._sync()
def _update_range(self, start, end, value):
if end < self.range[0] or start > self.range[1]:
return
if start <= self.range[0] and self.range[1] <= end:
self.range_value = value
self._sync()
return
self._push()
self.left._update_range(start, end, value)
self.right._update_range(start, end, value)
self._sync()
def _sync(self):
if self.range[0] == self.range[1]:
for op in self.parent_tree.operations.values():
current_value = self.parent_tree.array[self.range[0]]
if self.range_value is not None:
current_value = self.range_value
self.values[op.name] = op.f([current_value])
else:
for op in self.parent_tree.operations.values():
result = op.f(
[self.left.values[op.name], self.right.values[op.name]])
if self.range_value is not None:
bound_length = self.range[1] - self.range[0] + 1
result = op.f_on_equal(self.range_value, bound_length)
self.values[op.name] = result
def _push(self):
if self.range_value is None:
return
if self.left:
self.left.range_value = self.range_value
self.right.range_value = self.range_value
self.left._sync()
self.right._sync()
self.range_value = None
def __repr__(self):
ans = "({}, {}): {}\n".format(self.range[0], self.range[1],
self.values)
if self.left:
ans += self.left.__repr__()
if self.right:
ans += self.right.__repr__()
return ans
def display(string_to_print):
stdout.write(str(string_to_print) + "\n")
def primeFactors(n): #n**0.5 complex
factors = dict()
for i in range(2,math.ceil(math.sqrt(n))+1):
while n % i== 0:
if i in factors:
factors[i]+=1
else: factors[i]=1
n = n // i
if n>2:
factors[n]=1
return (factors)
def isprime(n):
"""Returns True if n is prime."""
if n < 4:
return True
if n % 2 == 0:
return False
if n % 3 == 0:
return False
i = 5
w = 2
while i * i <= n:
if n % i == 0:
return False
i += w
w = 6 - w
return True
def test_print(*args):
if testingMode:
print(args)
def display_list(list1, sep=" "):
stdout.write(sep.join(map(str, list1)) + "\n")
def get_int():
return int(stdin.readline().strip())
def get_tuple():
return map(int, stdin.readline().split())
def get_list():
return list(map(int, stdin.readline().split()))
memory = dict()
def clear_cache():
global memory
memory = dict()
def cached_fn(fn, *args):
global memory
if args in memory:
return memory[args]
else:
result = fn(*args)
memory[args] = result
return result
# -------------------------------------------------------------- MAIN PROGRAM
TestCases = False
testingMode = False
optimiseForReccursion = False #Can not be used clubbed with TestCases
def comp(li):
return len(li)
def main():
n,k=get_tuple()
li = get_list()
sm = sum(li)
res = 0
print((sm-k)//n) if sm-k>=0 else print(-1)
# --------------------------------------------------------------------- END
if TestCases:
for _ in range(get_int()):
cProfile.run('main()') if testingMode else main()
else: (cProfile.run('main()') if testingMode else main()) if not optimiseForReccursion else threading.Thread(target=main).start()
``` | instruction | 0 | 5,075 | 8 | 10,150 |
No | output | 1 | 5,075 | 8 | 10,151 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
The Fair Nut likes kvass very much. On his birthday parents presented him n kegs of kvass. There are v_i liters of kvass in the i-th keg. Each keg has a lever. You can pour your glass by exactly 1 liter pulling this lever. The Fair Nut likes this drink very much, so he wants to pour his glass by s liters of kvass. But he wants to do it, so kvass level in the least keg is as much as possible.
Help him find out how much kvass can be in the least keg or define it's not possible to pour his glass by s liters of kvass.
Input
The first line contains two integers n and s (1 β€ n β€ 10^3, 1 β€ s β€ 10^{12}) β the number of kegs and glass volume.
The second line contains n integers v_1, v_2, β¦, v_n (1 β€ v_i β€ 10^9) β the volume of i-th keg.
Output
If the Fair Nut cannot pour his glass by s liters of kvass, print -1. Otherwise, print a single integer β how much kvass in the least keg can be.
Examples
Input
3 3
4 3 5
Output
3
Input
3 4
5 3 4
Output
2
Input
3 7
1 2 3
Output
-1
Note
In the first example, the answer is 3, the Fair Nut can take 1 liter from the first keg and 2 liters from the third keg. There are 3 liters of kvass in each keg.
In the second example, the answer is 2, the Fair Nut can take 3 liters from the first keg and 1 liter from the second keg.
In the third example, the Fair Nut can't pour his cup by 7 liters, so the answer is -1.
Submitted Solution:
```
if __name__ == "__main__":
(n,s) = [ int(x) for x in input().split()]
#print(n,m)
val = [int(x) for x in input().split()]
sum = 0
for x in val:
sum+=x
ans = int((sum-s)/n)
if sum-s <0:
ans = -1
print(ans)
``` | instruction | 0 | 5,076 | 8 | 10,152 |
No | output | 1 | 5,076 | 8 | 10,153 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Mike is the president of country What-The-Fatherland. There are n bears living in this country besides Mike. All of them are standing in a line and they are numbered from 1 to n from left to right. i-th bear is exactly ai feet high.
<image>
A group of bears is a non-empty contiguous segment of the line. The size of a group is the number of bears in that group. The strength of a group is the minimum height of the bear in that group.
Mike is a curious to know for each x such that 1 β€ x β€ n the maximum strength among all groups of size x.
Input
The first line of input contains integer n (1 β€ n β€ 2 Γ 105), the number of bears.
The second line contains n integers separated by space, a1, a2, ..., an (1 β€ ai β€ 109), heights of bears.
Output
Print n integers in one line. For each x from 1 to n, print the maximum strength among all groups of size x.
Examples
Input
10
1 2 3 4 5 4 3 2 1 6
Output
6 4 4 3 3 2 2 1 1 1 | instruction | 0 | 5,430 | 8 | 10,860 |
Tags: binary search, data structures, dp, dsu
Correct Solution:
```
n=int(input())
l=[int(i) for i in input().split()]
stack=[]
ans=[0]*n
left=[-1]*n
right=[n]*n
for i in range(n):
while(stack!=[] and l[i]<=l[stack[-1]]):
stack.pop()
if(stack!=[]):
left[i]=stack[-1]
stack.append(i)
stack=[]
for i in range(n-1,-1,-1):
while(stack!=[] and l[i]<=l[stack[-1]]):
stack.pop()
if(stack!=[]):
right[i]=stack[-1]
stack.append(i)
for i in range(n):
k=right[i]-left[i]-1
ans[k-1]=max(ans[k-1],l[i])
for i in range(n-2,-1,-1):
ans[i]=max(ans[i],ans[i+1])
print(*ans)
``` | output | 1 | 5,430 | 8 | 10,861 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Mike is the president of country What-The-Fatherland. There are n bears living in this country besides Mike. All of them are standing in a line and they are numbered from 1 to n from left to right. i-th bear is exactly ai feet high.
<image>
A group of bears is a non-empty contiguous segment of the line. The size of a group is the number of bears in that group. The strength of a group is the minimum height of the bear in that group.
Mike is a curious to know for each x such that 1 β€ x β€ n the maximum strength among all groups of size x.
Input
The first line of input contains integer n (1 β€ n β€ 2 Γ 105), the number of bears.
The second line contains n integers separated by space, a1, a2, ..., an (1 β€ ai β€ 109), heights of bears.
Output
Print n integers in one line. For each x from 1 to n, print the maximum strength among all groups of size x.
Examples
Input
10
1 2 3 4 5 4 3 2 1 6
Output
6 4 4 3 3 2 2 1 1 1
Submitted Solution:
```
from sys import stdin, stdout
def input(): return stdin.readline().strip()
def print(x, end='\n'): stdout.write(str(x) + end)
n, lst = int(input()), list(map(int, input().split()))
nse, pse, stk, ans = [n for i in range(n)], [-1 for i in range(n)], [], [0 for i in range(n)]
for i in range(n):
while stk and lst[stk[-1]] > lst[i]: nse[stk.pop()] = i
stk.append(i)
stk.clear()
for i in range(n-1, -1, -1):
while stk and lst[stk[-1]] > lst[i]: pse[stk.pop()] = i
stk.append(i)
for i in range(n):
now = nse[i] - pse[i] - 1
ans[now-1] = max(lst[i], ans[now-1])
for i in range(n-2, -1, -1): ans[i] = ans[i+1] if ans[i] == 0 else ans[i]
print(' '.join(map(str, ans)))
``` | instruction | 0 | 5,432 | 8 | 10,864 |
No | output | 1 | 5,432 | 8 | 10,865 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Mike is the president of country What-The-Fatherland. There are n bears living in this country besides Mike. All of them are standing in a line and they are numbered from 1 to n from left to right. i-th bear is exactly ai feet high.
<image>
A group of bears is a non-empty contiguous segment of the line. The size of a group is the number of bears in that group. The strength of a group is the minimum height of the bear in that group.
Mike is a curious to know for each x such that 1 β€ x β€ n the maximum strength among all groups of size x.
Input
The first line of input contains integer n (1 β€ n β€ 2 Γ 105), the number of bears.
The second line contains n integers separated by space, a1, a2, ..., an (1 β€ ai β€ 109), heights of bears.
Output
Print n integers in one line. For each x from 1 to n, print the maximum strength among all groups of size x.
Examples
Input
10
1 2 3 4 5 4 3 2 1 6
Output
6 4 4 3 3 2 2 1 1 1
Submitted Solution:
```
from sys import stdin, stdout
def input(): return stdin.readline().strip()
def print(x, end='\n'): stdout.write(str(x) + end)
n, lst = int(input()), list(map(int, input().split()))
nse, pse, stk, ans = [n for i in range(n)], [-1 for i in range(n)], [], [0 for i in range(n)]
for i in range(n):
while stk and lst[stk[-1]] > lst[i]: nse[stk.pop()] = i
stk.append(i)
stk.clear()
for i in range(n-1, -1, -1):
while stk and lst[stk[-1]] > lst[i]: pse[stk.pop()] = i
stk.append(i)
for i in range(n):
now = nse[i] - pse[i] - 1
ans[now-1] = lst[i]
for i in range(n-2, -1, -1): ans[i] = ans[i+1] if ans[i] == 0 else ans[i]
print(' '.join(map(str, ans)))
``` | instruction | 0 | 5,433 | 8 | 10,866 |
No | output | 1 | 5,433 | 8 | 10,867 |
Provide tags and a correct Python 3 solution for this coding contest problem.
In the year of 30XX participants of some world programming championship live in a single large hotel. The hotel has n floors. Each floor has m sections with a single corridor connecting all of them. The sections are enumerated from 1 to m along the corridor, and all sections with equal numbers on different floors are located exactly one above the other. Thus, the hotel can be represented as a rectangle of height n and width m. We can denote sections with pairs of integers (i, j), where i is the floor, and j is the section number on the floor.
The guests can walk along the corridor on each floor, use stairs and elevators. Each stairs or elevator occupies all sections (1, x), (2, x), β¦, (n, x) for some x between 1 and m. All sections not occupied with stairs or elevators contain guest rooms. It takes one time unit to move between neighboring sections on the same floor or to move one floor up or down using stairs. It takes one time unit to move up to v floors in any direction using an elevator. You can assume you don't have to wait for an elevator, and the time needed to enter or exit an elevator is negligible.
You are to process q queries. Each query is a question "what is the minimum time needed to go from a room in section (x_1, y_1) to a room in section (x_2, y_2)?"
Input
The first line contains five integers n, m, c_l, c_e, v (2 β€ n, m β€ 10^8, 0 β€ c_l, c_e β€ 10^5, 1 β€ c_l + c_e β€ m - 1, 1 β€ v β€ n - 1) β the number of floors and section on each floor, the number of stairs, the number of elevators and the maximum speed of an elevator, respectively.
The second line contains c_l integers l_1, β¦, l_{c_l} in increasing order (1 β€ l_i β€ m), denoting the positions of the stairs. If c_l = 0, the second line is empty.
The third line contains c_e integers e_1, β¦, e_{c_e} in increasing order, denoting the elevators positions in the same format. It is guaranteed that all integers l_i and e_i are distinct.
The fourth line contains a single integer q (1 β€ q β€ 10^5) β the number of queries.
The next q lines describe queries. Each of these lines contains four integers x_1, y_1, x_2, y_2 (1 β€ x_1, x_2 β€ n, 1 β€ y_1, y_2 β€ m) β the coordinates of starting and finishing sections for the query. It is guaranteed that the starting and finishing sections are distinct. It is also guaranteed that these sections contain guest rooms, i. e. y_1 and y_2 are not among l_i and e_i.
Output
Print q integers, one per line β the answers for the queries.
Example
Input
5 6 1 1 3
2
5
3
1 1 5 6
1 3 5 4
3 3 5 3
Output
7
5
4
Note
In the first query the optimal way is to go to the elevator in the 5-th section in four time units, use it to go to the fifth floor in two time units and go to the destination in one more time unit.
In the second query it is still optimal to use the elevator, but in the third query it is better to use the stairs in the section 2. | instruction | 0 | 5,571 | 8 | 11,142 |
Tags: binary search
Correct Solution:
```
import bisect as bs
import sys
inp = sys.stdin.readlines()
n, m, ladders, elevators, v = [int(x) for x in inp[0].strip().split()]
ladders = [int(x) for x in inp[1].strip().split()]
elevators = [int(x) for x in inp[2].strip().split()]
q = int(inp[3].strip())
qs = []
for i in range(q):
qs.append([int(x) for x in inp[4+i].strip().split()])
output = []
for query in qs:
x1, y1, x2, y2 = query
if x1 == x2:
output.append(abs(y1-y2))
continue
cur_ld = []
if ladders and (y1 > ladders[0]):
cur_ld.append(ladders[bs.bisect_left(ladders, y1)-1])
if ladders and (y1 < ladders[-1]):
cur_ld.append(ladders[bs.bisect_right(ladders, y1)])
cur_elev = []
if elevators and (y1 > elevators[0]):
cur_elev.append(elevators[bs.bisect_left(elevators, y1)-1])
if elevators and (y1 < elevators[-1]):
cur_elev.append(elevators[bs.bisect_right(elevators, y1)])
ans = []
for lad in cur_ld:
ans.append(abs(y1 - lad) + abs(y2 - lad) + abs(x1 - x2))
for elev in cur_elev:
height = abs(x1-x2)
elspeed = height // v
if height % v != 0: elspeed+=1
ans.append(abs(y1 - elev) + abs(y2 - elev) + elspeed)
ans = min(ans)
output.append(ans)
output = '\n'.join(map(str, output))
print(output)
``` | output | 1 | 5,571 | 8 | 11,143 |
Provide tags and a correct Python 3 solution for this coding contest problem.
In the year of 30XX participants of some world programming championship live in a single large hotel. The hotel has n floors. Each floor has m sections with a single corridor connecting all of them. The sections are enumerated from 1 to m along the corridor, and all sections with equal numbers on different floors are located exactly one above the other. Thus, the hotel can be represented as a rectangle of height n and width m. We can denote sections with pairs of integers (i, j), where i is the floor, and j is the section number on the floor.
The guests can walk along the corridor on each floor, use stairs and elevators. Each stairs or elevator occupies all sections (1, x), (2, x), β¦, (n, x) for some x between 1 and m. All sections not occupied with stairs or elevators contain guest rooms. It takes one time unit to move between neighboring sections on the same floor or to move one floor up or down using stairs. It takes one time unit to move up to v floors in any direction using an elevator. You can assume you don't have to wait for an elevator, and the time needed to enter or exit an elevator is negligible.
You are to process q queries. Each query is a question "what is the minimum time needed to go from a room in section (x_1, y_1) to a room in section (x_2, y_2)?"
Input
The first line contains five integers n, m, c_l, c_e, v (2 β€ n, m β€ 10^8, 0 β€ c_l, c_e β€ 10^5, 1 β€ c_l + c_e β€ m - 1, 1 β€ v β€ n - 1) β the number of floors and section on each floor, the number of stairs, the number of elevators and the maximum speed of an elevator, respectively.
The second line contains c_l integers l_1, β¦, l_{c_l} in increasing order (1 β€ l_i β€ m), denoting the positions of the stairs. If c_l = 0, the second line is empty.
The third line contains c_e integers e_1, β¦, e_{c_e} in increasing order, denoting the elevators positions in the same format. It is guaranteed that all integers l_i and e_i are distinct.
The fourth line contains a single integer q (1 β€ q β€ 10^5) β the number of queries.
The next q lines describe queries. Each of these lines contains four integers x_1, y_1, x_2, y_2 (1 β€ x_1, x_2 β€ n, 1 β€ y_1, y_2 β€ m) β the coordinates of starting and finishing sections for the query. It is guaranteed that the starting and finishing sections are distinct. It is also guaranteed that these sections contain guest rooms, i. e. y_1 and y_2 are not among l_i and e_i.
Output
Print q integers, one per line β the answers for the queries.
Example
Input
5 6 1 1 3
2
5
3
1 1 5 6
1 3 5 4
3 3 5 3
Output
7
5
4
Note
In the first query the optimal way is to go to the elevator in the 5-th section in four time units, use it to go to the fifth floor in two time units and go to the destination in one more time unit.
In the second query it is still optimal to use the elevator, but in the third query it is better to use the stairs in the section 2. | instruction | 0 | 5,572 | 8 | 11,144 |
Tags: binary search
Correct Solution:
```
# python3
import sys
from bisect import bisect
def readline(): return tuple(map(int, input().split()))
def readlines():
return (tuple(map(int, line.split())) for line in sys.stdin.readlines())
def bisect_bounds(arr, val):
idx = bisect(arr, val)
if idx: yield idx - 1
if idx < len(arr): yield idx
class Minimizator:
def __init__(self, value=float('inf')):
self.value = value
def eat(self, value):
self.value = min(self.value, value)
def main():
n, m, cl, ce, v = readline()
l = readline()
e = readline()
assert len(l) == cl
assert len(e) == ce
q, = readline()
answers = list()
for (x1, y1, x2, y2) in readlines():
if x1 == x2:
answers.append(abs(y1 - y2))
else:
ans = Minimizator(n + 2*m)
for idx in bisect_bounds(l, y1):
ladder = l[idx]
ans.eat(abs(x1 - x2) + abs(ladder - y1) + abs(ladder - y2))
for idx in bisect_bounds(e, y1):
elevator = e[idx]
ans.eat((abs(x1 - x2) - 1) // v + 1
+ abs(elevator - y1) + abs(elevator - y2))
answers.append(ans.value)
print("\n".join(map(str, answers)))
main()
``` | output | 1 | 5,572 | 8 | 11,145 |
Provide tags and a correct Python 3 solution for this coding contest problem.
In the year of 30XX participants of some world programming championship live in a single large hotel. The hotel has n floors. Each floor has m sections with a single corridor connecting all of them. The sections are enumerated from 1 to m along the corridor, and all sections with equal numbers on different floors are located exactly one above the other. Thus, the hotel can be represented as a rectangle of height n and width m. We can denote sections with pairs of integers (i, j), where i is the floor, and j is the section number on the floor.
The guests can walk along the corridor on each floor, use stairs and elevators. Each stairs or elevator occupies all sections (1, x), (2, x), β¦, (n, x) for some x between 1 and m. All sections not occupied with stairs or elevators contain guest rooms. It takes one time unit to move between neighboring sections on the same floor or to move one floor up or down using stairs. It takes one time unit to move up to v floors in any direction using an elevator. You can assume you don't have to wait for an elevator, and the time needed to enter or exit an elevator is negligible.
You are to process q queries. Each query is a question "what is the minimum time needed to go from a room in section (x_1, y_1) to a room in section (x_2, y_2)?"
Input
The first line contains five integers n, m, c_l, c_e, v (2 β€ n, m β€ 10^8, 0 β€ c_l, c_e β€ 10^5, 1 β€ c_l + c_e β€ m - 1, 1 β€ v β€ n - 1) β the number of floors and section on each floor, the number of stairs, the number of elevators and the maximum speed of an elevator, respectively.
The second line contains c_l integers l_1, β¦, l_{c_l} in increasing order (1 β€ l_i β€ m), denoting the positions of the stairs. If c_l = 0, the second line is empty.
The third line contains c_e integers e_1, β¦, e_{c_e} in increasing order, denoting the elevators positions in the same format. It is guaranteed that all integers l_i and e_i are distinct.
The fourth line contains a single integer q (1 β€ q β€ 10^5) β the number of queries.
The next q lines describe queries. Each of these lines contains four integers x_1, y_1, x_2, y_2 (1 β€ x_1, x_2 β€ n, 1 β€ y_1, y_2 β€ m) β the coordinates of starting and finishing sections for the query. It is guaranteed that the starting and finishing sections are distinct. It is also guaranteed that these sections contain guest rooms, i. e. y_1 and y_2 are not among l_i and e_i.
Output
Print q integers, one per line β the answers for the queries.
Example
Input
5 6 1 1 3
2
5
3
1 1 5 6
1 3 5 4
3 3 5 3
Output
7
5
4
Note
In the first query the optimal way is to go to the elevator in the 5-th section in four time units, use it to go to the fifth floor in two time units and go to the destination in one more time unit.
In the second query it is still optimal to use the elevator, but in the third query it is better to use the stairs in the section 2. | instruction | 0 | 5,573 | 8 | 11,146 |
Tags: binary search
Correct Solution:
```
from bisect import bisect_left
input = __import__('sys').stdin.readline
MIS = lambda: map(int,input().split())
def walk(x1, y1, x2, y2, L, v):
if y1 == y2: return abs(x2-x1)
dy = abs(y1-y2)
vertical = dy // v
if dy%v: vertical+= 1
i = bisect_left(L, x1)
xs1 = L[i-1] if 0<=i-1<len(L) else float('inf')
xs2 = L[i] if 0<=i<len(L) else float('inf')
xs3 = L[i+1] if 0<=i+1<len(L) else float('inf')
d = min(abs(x1-xs1)+abs(xs1-x2), abs(x1-xs2)+abs(xs2-x2), abs(x1-xs3)+abs(xs3-x2))
return d + vertical
n, m, cs, ce, v = MIS()
sta = list(MIS())
ele = list(MIS())
for TEST in range(int(input())):
y1, x1, y2, x2 = MIS()
if x1 > x2:
x1, x2 = x2, x1
y1, y2 = y2, y1
print(min(walk(x1, y1, x2, y2, sta, 1), walk(x1, y1, x2, y2, ele, v)))
``` | output | 1 | 5,573 | 8 | 11,147 |
Provide tags and a correct Python 3 solution for this coding contest problem.
In the year of 30XX participants of some world programming championship live in a single large hotel. The hotel has n floors. Each floor has m sections with a single corridor connecting all of them. The sections are enumerated from 1 to m along the corridor, and all sections with equal numbers on different floors are located exactly one above the other. Thus, the hotel can be represented as a rectangle of height n and width m. We can denote sections with pairs of integers (i, j), where i is the floor, and j is the section number on the floor.
The guests can walk along the corridor on each floor, use stairs and elevators. Each stairs or elevator occupies all sections (1, x), (2, x), β¦, (n, x) for some x between 1 and m. All sections not occupied with stairs or elevators contain guest rooms. It takes one time unit to move between neighboring sections on the same floor or to move one floor up or down using stairs. It takes one time unit to move up to v floors in any direction using an elevator. You can assume you don't have to wait for an elevator, and the time needed to enter or exit an elevator is negligible.
You are to process q queries. Each query is a question "what is the minimum time needed to go from a room in section (x_1, y_1) to a room in section (x_2, y_2)?"
Input
The first line contains five integers n, m, c_l, c_e, v (2 β€ n, m β€ 10^8, 0 β€ c_l, c_e β€ 10^5, 1 β€ c_l + c_e β€ m - 1, 1 β€ v β€ n - 1) β the number of floors and section on each floor, the number of stairs, the number of elevators and the maximum speed of an elevator, respectively.
The second line contains c_l integers l_1, β¦, l_{c_l} in increasing order (1 β€ l_i β€ m), denoting the positions of the stairs. If c_l = 0, the second line is empty.
The third line contains c_e integers e_1, β¦, e_{c_e} in increasing order, denoting the elevators positions in the same format. It is guaranteed that all integers l_i and e_i are distinct.
The fourth line contains a single integer q (1 β€ q β€ 10^5) β the number of queries.
The next q lines describe queries. Each of these lines contains four integers x_1, y_1, x_2, y_2 (1 β€ x_1, x_2 β€ n, 1 β€ y_1, y_2 β€ m) β the coordinates of starting and finishing sections for the query. It is guaranteed that the starting and finishing sections are distinct. It is also guaranteed that these sections contain guest rooms, i. e. y_1 and y_2 are not among l_i and e_i.
Output
Print q integers, one per line β the answers for the queries.
Example
Input
5 6 1 1 3
2
5
3
1 1 5 6
1 3 5 4
3 3 5 3
Output
7
5
4
Note
In the first query the optimal way is to go to the elevator in the 5-th section in four time units, use it to go to the fifth floor in two time units and go to the destination in one more time unit.
In the second query it is still optimal to use the elevator, but in the third query it is better to use the stairs in the section 2. | instruction | 0 | 5,574 | 8 | 11,148 |
Tags: binary search
Correct Solution:
```
def takeClosest(myList, myNumber):
"""
Assumes myList is sorted. Returns closest value to myNumber.
If two numbers are equally close, return the smallest number.
"""
if len(myList) == 0:
return 9e10
pos = bisect_left(myList, myNumber)
if pos == 0:
return myList[0]
if pos == len(myList):
return myList[-1]
before = myList[pos - 1]
after = myList[pos]
if after - myNumber < myNumber - before:
return after
else:
return before
from bisect import bisect_left
from math import ceil
n, m, n_stairs, n_elevators, v = map(int, input().split(" "))
if n_stairs > 0:
stairs = list(map(int, input().split(" ")))
else:
stairs = []
input()
if n_elevators > 0:
elevators = list(map(int, input().split(" ")))
else:
elevators = []
input()
queries = int(input())
res = []
for i in range(queries):
x1, y1, x2, y2 = map(int, input().split(" "))
next_elevator = takeClosest(elevators, (y1 + y2) / 2)
next_stairs = takeClosest(stairs, (y1 + y2) / 2)
time_elevator = abs(x1 - x2) / v
time_stairs = abs(x1 - x2)
mi = min(y1, y2)
ma = max(y1, y2)
if next_elevator < mi:
time_elevator += (mi - next_elevator) * 2
elif next_elevator > ma:
time_elevator += (next_elevator - ma) * 2
if next_stairs < mi:
time_stairs += (mi - next_stairs) * 2
elif next_stairs > ma:
time_stairs += (next_stairs - ma) * 2
dis = abs(y1 - y2)
if time_elevator < time_stairs:
dis += time_elevator
else:
dis += time_stairs
if x1 == x2:
res.append(abs(y1 - y2))
else:
res.append(ceil(dis))
print(*res, sep="\n")
``` | output | 1 | 5,574 | 8 | 11,149 |
Provide tags and a correct Python 3 solution for this coding contest problem.
In the year of 30XX participants of some world programming championship live in a single large hotel. The hotel has n floors. Each floor has m sections with a single corridor connecting all of them. The sections are enumerated from 1 to m along the corridor, and all sections with equal numbers on different floors are located exactly one above the other. Thus, the hotel can be represented as a rectangle of height n and width m. We can denote sections with pairs of integers (i, j), where i is the floor, and j is the section number on the floor.
The guests can walk along the corridor on each floor, use stairs and elevators. Each stairs or elevator occupies all sections (1, x), (2, x), β¦, (n, x) for some x between 1 and m. All sections not occupied with stairs or elevators contain guest rooms. It takes one time unit to move between neighboring sections on the same floor or to move one floor up or down using stairs. It takes one time unit to move up to v floors in any direction using an elevator. You can assume you don't have to wait for an elevator, and the time needed to enter or exit an elevator is negligible.
You are to process q queries. Each query is a question "what is the minimum time needed to go from a room in section (x_1, y_1) to a room in section (x_2, y_2)?"
Input
The first line contains five integers n, m, c_l, c_e, v (2 β€ n, m β€ 10^8, 0 β€ c_l, c_e β€ 10^5, 1 β€ c_l + c_e β€ m - 1, 1 β€ v β€ n - 1) β the number of floors and section on each floor, the number of stairs, the number of elevators and the maximum speed of an elevator, respectively.
The second line contains c_l integers l_1, β¦, l_{c_l} in increasing order (1 β€ l_i β€ m), denoting the positions of the stairs. If c_l = 0, the second line is empty.
The third line contains c_e integers e_1, β¦, e_{c_e} in increasing order, denoting the elevators positions in the same format. It is guaranteed that all integers l_i and e_i are distinct.
The fourth line contains a single integer q (1 β€ q β€ 10^5) β the number of queries.
The next q lines describe queries. Each of these lines contains four integers x_1, y_1, x_2, y_2 (1 β€ x_1, x_2 β€ n, 1 β€ y_1, y_2 β€ m) β the coordinates of starting and finishing sections for the query. It is guaranteed that the starting and finishing sections are distinct. It is also guaranteed that these sections contain guest rooms, i. e. y_1 and y_2 are not among l_i and e_i.
Output
Print q integers, one per line β the answers for the queries.
Example
Input
5 6 1 1 3
2
5
3
1 1 5 6
1 3 5 4
3 3 5 3
Output
7
5
4
Note
In the first query the optimal way is to go to the elevator in the 5-th section in four time units, use it to go to the fifth floor in two time units and go to the destination in one more time unit.
In the second query it is still optimal to use the elevator, but in the third query it is better to use the stairs in the section 2. | instruction | 0 | 5,575 | 8 | 11,150 |
Tags: binary search
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=0, func=lambda a, b: 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(10)]
pp = [0] * 10
def SieveOfEratosthenes(n=10):
p = 2
c = 0
while (p * p <= n):
if (prime[p] == True):
c += 1
for i in range(p, n + 1, p):
pp[i] += 1
prime[i] = False
p += 1
# ---------------------------------Binary Search------------------------------------------
def binarySearch(arr, n, key):
left = 0
right = n - 1
res = -1
while (left <= right):
mid = (right + left) // 2
if (arr[mid] >= key):
res = arr[mid]
right = mid - 1
else:
left = mid + 1
return res
def binarySearch1(arr, n, key):
left = 0
right = n - 1
res = -1
while (left <= right):
mid = (right + left) // 2
if (arr[mid]>=key):
right = mid - 1
else:
res = arr[mid]
left = mid + 1
return res
# ---------------------------------running code------------------------------------------
n, m, cl, ce, v = map(int, input().split())
l = list(map(int, input().split()))
e = list(map(int, input().split()))
l.sort()
e.sort()
q = int(input())
for i in range(q):
res = 10**10
x1, y1, x2, y2 = map(int, input().split())
if x1==x2:
print(abs(y1-y2))
continue
lx1 = binarySearch(l, cl, y1)
lx2 = binarySearch1(l, cl, y1)
ex1 = binarySearch(e, ce, y1)
ex2 = binarySearch1(e, ce, y1)
if lx1!=-1:
res = min(res,abs(lx1 - y1) + abs(lx1 - y2) + abs(x1 - x2))
if lx2!=-1:
res = min(res, abs(lx2 - y1) + abs(lx2 - y2) + abs(x1 - x2))
if ex1!=-1:
res = min(res, abs(ex1 - y1) + abs(ex1 - y2) + math.ceil(abs(x1 - x2) / v))
if ex2!=-1:
res = min(res, abs(ex2 - y1) + abs(ex2 - y2) + math.ceil(abs(x1 - x2) / v))
print(res)
``` | output | 1 | 5,575 | 8 | 11,151 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
In the year of 30XX participants of some world programming championship live in a single large hotel. The hotel has n floors. Each floor has m sections with a single corridor connecting all of them. The sections are enumerated from 1 to m along the corridor, and all sections with equal numbers on different floors are located exactly one above the other. Thus, the hotel can be represented as a rectangle of height n and width m. We can denote sections with pairs of integers (i, j), where i is the floor, and j is the section number on the floor.
The guests can walk along the corridor on each floor, use stairs and elevators. Each stairs or elevator occupies all sections (1, x), (2, x), β¦, (n, x) for some x between 1 and m. All sections not occupied with stairs or elevators contain guest rooms. It takes one time unit to move between neighboring sections on the same floor or to move one floor up or down using stairs. It takes one time unit to move up to v floors in any direction using an elevator. You can assume you don't have to wait for an elevator, and the time needed to enter or exit an elevator is negligible.
You are to process q queries. Each query is a question "what is the minimum time needed to go from a room in section (x_1, y_1) to a room in section (x_2, y_2)?"
Input
The first line contains five integers n, m, c_l, c_e, v (2 β€ n, m β€ 10^8, 0 β€ c_l, c_e β€ 10^5, 1 β€ c_l + c_e β€ m - 1, 1 β€ v β€ n - 1) β the number of floors and section on each floor, the number of stairs, the number of elevators and the maximum speed of an elevator, respectively.
The second line contains c_l integers l_1, β¦, l_{c_l} in increasing order (1 β€ l_i β€ m), denoting the positions of the stairs. If c_l = 0, the second line is empty.
The third line contains c_e integers e_1, β¦, e_{c_e} in increasing order, denoting the elevators positions in the same format. It is guaranteed that all integers l_i and e_i are distinct.
The fourth line contains a single integer q (1 β€ q β€ 10^5) β the number of queries.
The next q lines describe queries. Each of these lines contains four integers x_1, y_1, x_2, y_2 (1 β€ x_1, x_2 β€ n, 1 β€ y_1, y_2 β€ m) β the coordinates of starting and finishing sections for the query. It is guaranteed that the starting and finishing sections are distinct. It is also guaranteed that these sections contain guest rooms, i. e. y_1 and y_2 are not among l_i and e_i.
Output
Print q integers, one per line β the answers for the queries.
Example
Input
5 6 1 1 3
2
5
3
1 1 5 6
1 3 5 4
3 3 5 3
Output
7
5
4
Note
In the first query the optimal way is to go to the elevator in the 5-th section in four time units, use it to go to the fifth floor in two time units and go to the destination in one more time unit.
In the second query it is still optimal to use the elevator, but in the third query it is better to use the stairs in the section 2.
Submitted 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=0, func=lambda a, b: 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(10)]
pp=[0]*10
def SieveOfEratosthenes(n=10):
p = 2
c=0
while (p * p <= n):
if (prime[p] == True):
c+=1
for i in range(p, n+1, p):
pp[i]+=1
prime[i] = False
p += 1
#---------------------------------Binary Search------------------------------------------
def binarySearch(arr, n, key):
left = 0
right = n-1
mid = 0
res=0
while (left <= right):
mid = (right + left)//2
if (arr[mid] >= key):
res=arr[mid]
right = mid-1
else:
left = mid + 1
return res
def binarySearch1(arr, n, key):
left = 0
right = n-1
mid = 0
res=0
while (left <= right):
mid = (right + left)//2
if (arr[mid] > key):
right = mid-1
else:
res=arr[mid]
left = mid + 1
return res
#---------------------------------running code------------------------------------------
n,m,cl,ce,v=map(int,input().split())
l=list(map(int,input().split()))
e=list(map(int,input().split()))
q=int(input())
for i in range (q):
res=10**9
x1,y1,x2,y2=map(int,input().split())
lx1=binarySearch(l, cl, x1)
lx2=binarySearch1(l, cl, x1)
ex1=binarySearch(e, ce, x1)
ex2=binarySearch1(e, ce, x1)
res=abs(lx1-y1)+abs(lx1-y2)+abs(x1-x2)
res=min(res,abs(lx2-y1)+abs(lx2-y2)+abs(x1-x2))
res=min(res,abs(ex1-y1)+abs(ex1-y2)+math.ceil(abs(x1-x2)/v))
res=min(res,abs(ex2-y1)+abs(ex2-y2)+math.ceil(abs(x1-x2)/v))
print(res)
``` | instruction | 0 | 5,576 | 8 | 11,152 |
No | output | 1 | 5,576 | 8 | 11,153 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
In the year of 30XX participants of some world programming championship live in a single large hotel. The hotel has n floors. Each floor has m sections with a single corridor connecting all of them. The sections are enumerated from 1 to m along the corridor, and all sections with equal numbers on different floors are located exactly one above the other. Thus, the hotel can be represented as a rectangle of height n and width m. We can denote sections with pairs of integers (i, j), where i is the floor, and j is the section number on the floor.
The guests can walk along the corridor on each floor, use stairs and elevators. Each stairs or elevator occupies all sections (1, x), (2, x), β¦, (n, x) for some x between 1 and m. All sections not occupied with stairs or elevators contain guest rooms. It takes one time unit to move between neighboring sections on the same floor or to move one floor up or down using stairs. It takes one time unit to move up to v floors in any direction using an elevator. You can assume you don't have to wait for an elevator, and the time needed to enter or exit an elevator is negligible.
You are to process q queries. Each query is a question "what is the minimum time needed to go from a room in section (x_1, y_1) to a room in section (x_2, y_2)?"
Input
The first line contains five integers n, m, c_l, c_e, v (2 β€ n, m β€ 10^8, 0 β€ c_l, c_e β€ 10^5, 1 β€ c_l + c_e β€ m - 1, 1 β€ v β€ n - 1) β the number of floors and section on each floor, the number of stairs, the number of elevators and the maximum speed of an elevator, respectively.
The second line contains c_l integers l_1, β¦, l_{c_l} in increasing order (1 β€ l_i β€ m), denoting the positions of the stairs. If c_l = 0, the second line is empty.
The third line contains c_e integers e_1, β¦, e_{c_e} in increasing order, denoting the elevators positions in the same format. It is guaranteed that all integers l_i and e_i are distinct.
The fourth line contains a single integer q (1 β€ q β€ 10^5) β the number of queries.
The next q lines describe queries. Each of these lines contains four integers x_1, y_1, x_2, y_2 (1 β€ x_1, x_2 β€ n, 1 β€ y_1, y_2 β€ m) β the coordinates of starting and finishing sections for the query. It is guaranteed that the starting and finishing sections are distinct. It is also guaranteed that these sections contain guest rooms, i. e. y_1 and y_2 are not among l_i and e_i.
Output
Print q integers, one per line β the answers for the queries.
Example
Input
5 6 1 1 3
2
5
3
1 1 5 6
1 3 5 4
3 3 5 3
Output
7
5
4
Note
In the first query the optimal way is to go to the elevator in the 5-th section in four time units, use it to go to the fifth floor in two time units and go to the destination in one more time unit.
In the second query it is still optimal to use the elevator, but in the third query it is better to use the stairs in the section 2.
Submitted Solution:
```
def takeClosest(myList, myNumber):
"""
Assumes myList is sorted. Returns closest value to myNumber.
If two numbers are equally close, return the smallest number.
"""
if len(myList) == 0:
return 9e10
pos = bisect_left(myList, myNumber)
if pos == 0:
return myList[0]
if pos == len(myList):
return myList[-1]
before = myList[pos - 1]
after = myList[pos]
if after - myNumber < myNumber - before:
return after
else:
return before
from bisect import bisect_left
from math import ceil
n, m, n_stairs, n_elevators, v = map(int, input().split(" "))
if n_stairs > 0:
stairs = list(map(int, input().split(" ")))
else:
stairs = []
input()
if n_elevators > 0:
elevators = list(map(int, input().split(" ")))
else:
elevators = []
input()
queries = int(input())
res = []
for i in range(queries):
x1, y1, x2, y2 = map(int, input().split(" "))
next_elevator = takeClosest(elevators, (y1 + y2) // 2)
next_stairs = takeClosest(stairs, (y1 + y2) // 2)
time_elevator = abs(x1 - x2) / v
time_stairs = abs(x1 - x2)
mi = min(y1, y2)
ma = max(y1, y2)
if next_elevator < mi:
time_elevator += (mi - next_elevator) * 2
elif next_elevator > ma:
time_elevator += (next_elevator - ma) * 2
if next_stairs < mi:
time_stairs += (mi - next_stairs) * 2
elif next_stairs > ma:
time_stairs += (next_stairs - ma) * 2
dis = abs(y1 - y2)
if time_elevator < time_stairs:
dis += time_elevator
else:
dis += time_stairs
res.append(ceil(dis))
print(*res, sep="\n")
``` | instruction | 0 | 5,577 | 8 | 11,154 |
No | output | 1 | 5,577 | 8 | 11,155 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
In the year of 30XX participants of some world programming championship live in a single large hotel. The hotel has n floors. Each floor has m sections with a single corridor connecting all of them. The sections are enumerated from 1 to m along the corridor, and all sections with equal numbers on different floors are located exactly one above the other. Thus, the hotel can be represented as a rectangle of height n and width m. We can denote sections with pairs of integers (i, j), where i is the floor, and j is the section number on the floor.
The guests can walk along the corridor on each floor, use stairs and elevators. Each stairs or elevator occupies all sections (1, x), (2, x), β¦, (n, x) for some x between 1 and m. All sections not occupied with stairs or elevators contain guest rooms. It takes one time unit to move between neighboring sections on the same floor or to move one floor up or down using stairs. It takes one time unit to move up to v floors in any direction using an elevator. You can assume you don't have to wait for an elevator, and the time needed to enter or exit an elevator is negligible.
You are to process q queries. Each query is a question "what is the minimum time needed to go from a room in section (x_1, y_1) to a room in section (x_2, y_2)?"
Input
The first line contains five integers n, m, c_l, c_e, v (2 β€ n, m β€ 10^8, 0 β€ c_l, c_e β€ 10^5, 1 β€ c_l + c_e β€ m - 1, 1 β€ v β€ n - 1) β the number of floors and section on each floor, the number of stairs, the number of elevators and the maximum speed of an elevator, respectively.
The second line contains c_l integers l_1, β¦, l_{c_l} in increasing order (1 β€ l_i β€ m), denoting the positions of the stairs. If c_l = 0, the second line is empty.
The third line contains c_e integers e_1, β¦, e_{c_e} in increasing order, denoting the elevators positions in the same format. It is guaranteed that all integers l_i and e_i are distinct.
The fourth line contains a single integer q (1 β€ q β€ 10^5) β the number of queries.
The next q lines describe queries. Each of these lines contains four integers x_1, y_1, x_2, y_2 (1 β€ x_1, x_2 β€ n, 1 β€ y_1, y_2 β€ m) β the coordinates of starting and finishing sections for the query. It is guaranteed that the starting and finishing sections are distinct. It is also guaranteed that these sections contain guest rooms, i. e. y_1 and y_2 are not among l_i and e_i.
Output
Print q integers, one per line β the answers for the queries.
Example
Input
5 6 1 1 3
2
5
3
1 1 5 6
1 3 5 4
3 3 5 3
Output
7
5
4
Note
In the first query the optimal way is to go to the elevator in the 5-th section in four time units, use it to go to the fifth floor in two time units and go to the destination in one more time unit.
In the second query it is still optimal to use the elevator, but in the third query it is better to use the stairs in the section 2.
Submitted Solution:
```
def takeClosest(myList, myNumber):
"""
Assumes myList is sorted. Returns closest value to myNumber.
If two numbers are equally close, return the smallest number.
"""
if len(myList) == 0:
return 9e10
pos = bisect_left(myList, myNumber)
if pos == 0:
return myList[0]
if pos == len(myList):
return myList[-1]
before = myList[pos - 1]
after = myList[pos]
if after - myNumber < myNumber - before:
return after
else:
return before
from bisect import bisect_left
from math import ceil
n, m, n_stairs, n_elevators, v = map(int, input().split(" "))
if n_stairs > 0:
stairs = list(map(int, input().split(" ")))
else:
stairs = []
input()
if n_elevators > 0:
elevators = list(map(int, input().split(" ")))
else:
elevators = []
input()
queries = int(input())
res = []
for i in range(queries):
x1, y1, x2, y2 = map(int, input().split(" "))
next_elevator = takeClosest(elevators, (y1 + y2) / 2)
next_stairs = takeClosest(stairs, (y1 + y2) / 2)
time_elevator = abs(x1 - x2) / v
time_stairs = abs(x1 - x2)
mi = min(y1, y2)
ma = max(y1, y2)
if next_elevator < mi:
time_elevator += (mi - next_elevator) * 2
elif next_elevator > ma:
time_elevator += (next_elevator - ma) * 2
if next_stairs < mi:
time_stairs += (mi - next_stairs) * 2
elif next_stairs > ma:
time_stairs += (next_stairs - ma) * 2
dis = abs(y1 - y2)
if time_elevator < time_stairs:
dis += time_elevator
else:
dis += time_stairs
res.append(ceil(dis))
print(*res, sep="\n")
``` | instruction | 0 | 5,578 | 8 | 11,156 |
No | output | 1 | 5,578 | 8 | 11,157 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
In the year of 30XX participants of some world programming championship live in a single large hotel. The hotel has n floors. Each floor has m sections with a single corridor connecting all of them. The sections are enumerated from 1 to m along the corridor, and all sections with equal numbers on different floors are located exactly one above the other. Thus, the hotel can be represented as a rectangle of height n and width m. We can denote sections with pairs of integers (i, j), where i is the floor, and j is the section number on the floor.
The guests can walk along the corridor on each floor, use stairs and elevators. Each stairs or elevator occupies all sections (1, x), (2, x), β¦, (n, x) for some x between 1 and m. All sections not occupied with stairs or elevators contain guest rooms. It takes one time unit to move between neighboring sections on the same floor or to move one floor up or down using stairs. It takes one time unit to move up to v floors in any direction using an elevator. You can assume you don't have to wait for an elevator, and the time needed to enter or exit an elevator is negligible.
You are to process q queries. Each query is a question "what is the minimum time needed to go from a room in section (x_1, y_1) to a room in section (x_2, y_2)?"
Input
The first line contains five integers n, m, c_l, c_e, v (2 β€ n, m β€ 10^8, 0 β€ c_l, c_e β€ 10^5, 1 β€ c_l + c_e β€ m - 1, 1 β€ v β€ n - 1) β the number of floors and section on each floor, the number of stairs, the number of elevators and the maximum speed of an elevator, respectively.
The second line contains c_l integers l_1, β¦, l_{c_l} in increasing order (1 β€ l_i β€ m), denoting the positions of the stairs. If c_l = 0, the second line is empty.
The third line contains c_e integers e_1, β¦, e_{c_e} in increasing order, denoting the elevators positions in the same format. It is guaranteed that all integers l_i and e_i are distinct.
The fourth line contains a single integer q (1 β€ q β€ 10^5) β the number of queries.
The next q lines describe queries. Each of these lines contains four integers x_1, y_1, x_2, y_2 (1 β€ x_1, x_2 β€ n, 1 β€ y_1, y_2 β€ m) β the coordinates of starting and finishing sections for the query. It is guaranteed that the starting and finishing sections are distinct. It is also guaranteed that these sections contain guest rooms, i. e. y_1 and y_2 are not among l_i and e_i.
Output
Print q integers, one per line β the answers for the queries.
Example
Input
5 6 1 1 3
2
5
3
1 1 5 6
1 3 5 4
3 3 5 3
Output
7
5
4
Note
In the first query the optimal way is to go to the elevator in the 5-th section in four time units, use it to go to the fifth floor in two time units and go to the destination in one more time unit.
In the second query it is still optimal to use the elevator, but in the third query it is better to use the stairs in the section 2.
Submitted 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=0, func=lambda a, b: 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(10)]
pp=[0]*10
def SieveOfEratosthenes(n=10):
p = 2
c=0
while (p * p <= n):
if (prime[p] == True):
c+=1
for i in range(p, n+1, p):
pp[i]+=1
prime[i] = False
p += 1
#---------------------------------Binary Search------------------------------------------
def binarySearch(arr, n, key):
left = 0
right = n-1
mid = 0
res=n-1
while (left <= right):
mid = (right + left)//2
if (arr[mid] >= key):
res=arr[mid]
right = mid-1
else:
left = mid + 1
return res
def binarySearch1(arr, n, key):
left = 0
right = n-1
mid = 0
res=0
while (left <= right):
mid = (right + left)//2
if (arr[mid] > key):
right = mid-1
else:
res=arr[mid]
left = mid + 1
return res
#---------------------------------running code------------------------------------------
n,m,cl,ce,v=map(int,input().split())
l=list(map(int,input().split()))
e=list(map(int,input().split()))
q=int(input())
for i in range (q):
res=10**9
x1,y1,x2,y2=map(int,input().split())
lx1=binarySearch(l, cl, x1)
lx2=binarySearch1(l, cl, x1)
ex1=binarySearch(e, ce, x1)
ex2=binarySearch1(e, ce, x1)
res=abs(lx1-y1)+abs(lx1-y2)+abs(x1-x2)
res=min(res,abs(lx2-y1)+abs(lx2-y2)+abs(x1-x2))
res=min(res,abs(ex1-y1)+abs(ex1-y2)+math.ceil(abs(x1-x2)/v))
res=min(res,abs(ex2-y1)+abs(ex2-y2)+math.ceil(abs(x1-x2)/v))
print(res)
``` | instruction | 0 | 5,579 | 8 | 11,158 |
No | output | 1 | 5,579 | 8 | 11,159 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Recently in school Alina has learned what are the persistent data structures: they are data structures that always preserves the previous version of itself and access to it when it is modified.
After reaching home Alina decided to invent her own persistent data structure. Inventing didn't take long: there is a bookcase right behind her bed. Alina thinks that the bookcase is a good choice for a persistent data structure. Initially the bookcase is empty, thus there is no book at any position at any shelf.
The bookcase consists of n shelves, and each shelf has exactly m positions for books at it. Alina enumerates shelves by integers from 1 to n and positions at shelves β from 1 to m. Initially the bookcase is empty, thus there is no book at any position at any shelf in it.
Alina wrote down q operations, which will be consecutively applied to the bookcase. Each of the operations has one of four types:
* 1 i j β Place a book at position j at shelf i if there is no book at it.
* 2 i j β Remove the book from position j at shelf i if there is a book at it.
* 3 i β Invert book placing at shelf i. This means that from every position at shelf i which has a book at it, the book should be removed, and at every position at shelf i which has not book at it, a book should be placed.
* 4 k β Return the books in the bookcase in a state they were after applying k-th operation. In particular, k = 0 means that the bookcase should be in initial state, thus every book in the bookcase should be removed from its position.
After applying each of operation Alina is interested in the number of books in the bookcase. Alina got 'A' in the school and had no problem finding this values. Will you do so?
Input
The first line of the input contains three integers n, m and q (1 β€ n, m β€ 103, 1 β€ q β€ 105) β the bookcase dimensions and the number of operations respectively.
The next q lines describes operations in chronological order β i-th of them describes i-th operation in one of the four formats described in the statement.
It is guaranteed that shelf indices and position indices are correct, and in each of fourth-type operation the number k corresponds to some operation before it or equals to 0.
Output
For each operation, print the number of books in the bookcase after applying it in a separate line. The answers should be printed in chronological order.
Examples
Input
2 3 3
1 1 1
3 2
4 0
Output
1
4
0
Input
4 2 6
3 2
2 2 2
3 3
3 2
2 2 2
3 2
Output
2
1
3
3
2
4
Input
2 2 2
3 2
2 2 1
Output
2
1
Note
<image>
This image illustrates the second sample case.
Submitted Solution:
```
n, m, q = map(int, input().split())
ans = [0] * q
op = [(-1, -1)]
cur = [[0 for _ in range(m)] for _ in range(n)]
hist, histcnt = [], []
import math, copy
sq = int(math.sqrt(q) + 0.5)
hist.append(copy.deepcopy(cur))
histcnt.append(0)
def apply(mat, cnt, u, v):
if u == 1 and mat[v[0] - 1][v[1] - 1] == 0:
mat[v[0] - 1][v[1] - 1] = 1
cnt += 1
elif u == 2 and mat[v[0] - 1][v[1] - 1] == 1:
mat[v[0] - 1][v[1] - 1] = 0
cnt -= 1
else:
for j in range(m):
if mat[v[0] - 1][j] == 0:
mat[v[0] - 1][j] = 1
cnt += 1
else:
mat[v[0] - 1][j] = 0
cnt -= 1
return cnt
def run(mat, cnt, f, t):
if f > t:
return cnt
for i in range(f, t + 1):
#print(mat, cnt, f, t, op[i])
if op[i][0] < 4:
cnt = apply(mat, cnt, op[i][0], op[i][1])
else:
k = (op[i][1][0] // sq) * sq
mat, cnt = copy.deepcopy(hist[k]), histcnt[k]
cnt = run(mat, cnt, k + 1, op[i][1][0])
return cnt
for i in range(q):
u, *v = map(int, input().split())
op.append((u, v))
if u < 4:
ans[i] = apply(cur, ans[i - 1], u, v)
else:
k = (v[0] // sq) * sq
ans[i] = run(copy.deepcopy(hist[k]), histcnt[k], k + 1, v[0])
if i % sq == 0:
hist.append(copy.deepcopy(cur))
histcnt.append(ans[i])
print("\n".join(map(str, ans)))
``` | instruction | 0 | 6,253 | 8 | 12,506 |
No | output | 1 | 6,253 | 8 | 12,507 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Recently in school Alina has learned what are the persistent data structures: they are data structures that always preserves the previous version of itself and access to it when it is modified.
After reaching home Alina decided to invent her own persistent data structure. Inventing didn't take long: there is a bookcase right behind her bed. Alina thinks that the bookcase is a good choice for a persistent data structure. Initially the bookcase is empty, thus there is no book at any position at any shelf.
The bookcase consists of n shelves, and each shelf has exactly m positions for books at it. Alina enumerates shelves by integers from 1 to n and positions at shelves β from 1 to m. Initially the bookcase is empty, thus there is no book at any position at any shelf in it.
Alina wrote down q operations, which will be consecutively applied to the bookcase. Each of the operations has one of four types:
* 1 i j β Place a book at position j at shelf i if there is no book at it.
* 2 i j β Remove the book from position j at shelf i if there is a book at it.
* 3 i β Invert book placing at shelf i. This means that from every position at shelf i which has a book at it, the book should be removed, and at every position at shelf i which has not book at it, a book should be placed.
* 4 k β Return the books in the bookcase in a state they were after applying k-th operation. In particular, k = 0 means that the bookcase should be in initial state, thus every book in the bookcase should be removed from its position.
After applying each of operation Alina is interested in the number of books in the bookcase. Alina got 'A' in the school and had no problem finding this values. Will you do so?
Input
The first line of the input contains three integers n, m and q (1 β€ n, m β€ 103, 1 β€ q β€ 105) β the bookcase dimensions and the number of operations respectively.
The next q lines describes operations in chronological order β i-th of them describes i-th operation in one of the four formats described in the statement.
It is guaranteed that shelf indices and position indices are correct, and in each of fourth-type operation the number k corresponds to some operation before it or equals to 0.
Output
For each operation, print the number of books in the bookcase after applying it in a separate line. The answers should be printed in chronological order.
Examples
Input
2 3 3
1 1 1
3 2
4 0
Output
1
4
0
Input
4 2 6
3 2
2 2 2
3 3
3 2
2 2 2
3 2
Output
2
1
3
3
2
4
Input
2 2 2
3 2
2 2 1
Output
2
1
Note
<image>
This image illustrates the second sample case.
Submitted Solution:
```
A = list(map(int,input().split()))
C = [0 for k in range(A[0])]
G = [0]
for i in range(A[2]):
B = list(map(int,input().split()))
if B[0] == 1:
C[B[1]-1] += 1
G.append(G[-1]+1)
elif B[0] == 2:
C[B[1]-1] -= 1
G.append(G[-1]-1)
elif B[0] == 3:
x = C[B[1]-1]
C[B[1]-1] = A[1]-C[B[1]-1]
G.append(G[-1]-x+(C[B[1]-1]))
else:
G.append(G[B[1]])
print(G[-1])
``` | instruction | 0 | 6,254 | 8 | 12,508 |
No | output | 1 | 6,254 | 8 | 12,509 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Recently in school Alina has learned what are the persistent data structures: they are data structures that always preserves the previous version of itself and access to it when it is modified.
After reaching home Alina decided to invent her own persistent data structure. Inventing didn't take long: there is a bookcase right behind her bed. Alina thinks that the bookcase is a good choice for a persistent data structure. Initially the bookcase is empty, thus there is no book at any position at any shelf.
The bookcase consists of n shelves, and each shelf has exactly m positions for books at it. Alina enumerates shelves by integers from 1 to n and positions at shelves β from 1 to m. Initially the bookcase is empty, thus there is no book at any position at any shelf in it.
Alina wrote down q operations, which will be consecutively applied to the bookcase. Each of the operations has one of four types:
* 1 i j β Place a book at position j at shelf i if there is no book at it.
* 2 i j β Remove the book from position j at shelf i if there is a book at it.
* 3 i β Invert book placing at shelf i. This means that from every position at shelf i which has a book at it, the book should be removed, and at every position at shelf i which has not book at it, a book should be placed.
* 4 k β Return the books in the bookcase in a state they were after applying k-th operation. In particular, k = 0 means that the bookcase should be in initial state, thus every book in the bookcase should be removed from its position.
After applying each of operation Alina is interested in the number of books in the bookcase. Alina got 'A' in the school and had no problem finding this values. Will you do so?
Input
The first line of the input contains three integers n, m and q (1 β€ n, m β€ 103, 1 β€ q β€ 105) β the bookcase dimensions and the number of operations respectively.
The next q lines describes operations in chronological order β i-th of them describes i-th operation in one of the four formats described in the statement.
It is guaranteed that shelf indices and position indices are correct, and in each of fourth-type operation the number k corresponds to some operation before it or equals to 0.
Output
For each operation, print the number of books in the bookcase after applying it in a separate line. The answers should be printed in chronological order.
Examples
Input
2 3 3
1 1 1
3 2
4 0
Output
1
4
0
Input
4 2 6
3 2
2 2 2
3 3
3 2
2 2 2
3 2
Output
2
1
3
3
2
4
Input
2 2 2
3 2
2 2 1
Output
2
1
Note
<image>
This image illustrates the second sample case.
Submitted Solution:
```
def bsch(a, x):
n = len(a)
l, r = -1, n-1
while (r-l >= 2):
mid = (r+l) // 2
if a[mid] <= x:
l = mid
else:
r = mid
return r
if __name__ == '__main__':
n, m, q = map(int, input().split())
mask1 = []
for i in range(m):
mask1.append(1 << i)
invert = sum(mask1)
mask2 = []
for i in range(m):
mask2.append(invert-mask1[i])
state = [[0] for _ in range(n)]
sumstate = [[0] for _ in range(n)]
curr = [0 for _ in range(n)]
currsum = [0 for _ in range(n)]
currtot = 0
change = [[0] for _ in range(n)]
for st in range(1,q+1):
s = input().split()
if s[0] == '1':
i, j = map(lambda x:int(x)-1, s[1:])
if curr[i] & mask1[j] == 0:
state[i].append(curr[i])
sumstate[i].append(currsum[i])
change[i].append(st)
curr[i] |= mask1[j]
currsum[i] += 1
currtot += 1
elif s[0] == '2':
i, j = map(lambda x:int(x)-1, s[1:])
if curr[i] & mask1[j] != 0:
state[i].append(curr[i])
sumstate[i].append(currsum[i])
change[i].append(st)
curr[i] &= mask2[j]
currsum[i] -= 1
currtot -= 1
elif s[0] == '3':
i = int(s[1])-1
state[i].append(curr[i])
sumstate[i].append(currsum[i])
change[i].append(st)
curr[i] ^= invert
currtot += m - 2 * currsum[i]
currsum[i] = m - currsum[i]
else:
k = int(s[1])
for i in range(n):
if change[i][-1] <= k:
continue
t = bsch(change[i], k)
if state[i][t] != curr[i]:
state[i].append(state[i][t])
sumstate[i].append(currsum[i])
change[i].append(st)
curr[i] = state[i][t]
currtot += sumstate[i][t] - currsum[i]
currsum[i] = sumstate[i][t]
print(currtot)
``` | instruction | 0 | 6,255 | 8 | 12,510 |
No | output | 1 | 6,255 | 8 | 12,511 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Recently in school Alina has learned what are the persistent data structures: they are data structures that always preserves the previous version of itself and access to it when it is modified.
After reaching home Alina decided to invent her own persistent data structure. Inventing didn't take long: there is a bookcase right behind her bed. Alina thinks that the bookcase is a good choice for a persistent data structure. Initially the bookcase is empty, thus there is no book at any position at any shelf.
The bookcase consists of n shelves, and each shelf has exactly m positions for books at it. Alina enumerates shelves by integers from 1 to n and positions at shelves β from 1 to m. Initially the bookcase is empty, thus there is no book at any position at any shelf in it.
Alina wrote down q operations, which will be consecutively applied to the bookcase. Each of the operations has one of four types:
* 1 i j β Place a book at position j at shelf i if there is no book at it.
* 2 i j β Remove the book from position j at shelf i if there is a book at it.
* 3 i β Invert book placing at shelf i. This means that from every position at shelf i which has a book at it, the book should be removed, and at every position at shelf i which has not book at it, a book should be placed.
* 4 k β Return the books in the bookcase in a state they were after applying k-th operation. In particular, k = 0 means that the bookcase should be in initial state, thus every book in the bookcase should be removed from its position.
After applying each of operation Alina is interested in the number of books in the bookcase. Alina got 'A' in the school and had no problem finding this values. Will you do so?
Input
The first line of the input contains three integers n, m and q (1 β€ n, m β€ 103, 1 β€ q β€ 105) β the bookcase dimensions and the number of operations respectively.
The next q lines describes operations in chronological order β i-th of them describes i-th operation in one of the four formats described in the statement.
It is guaranteed that shelf indices and position indices are correct, and in each of fourth-type operation the number k corresponds to some operation before it or equals to 0.
Output
For each operation, print the number of books in the bookcase after applying it in a separate line. The answers should be printed in chronological order.
Examples
Input
2 3 3
1 1 1
3 2
4 0
Output
1
4
0
Input
4 2 6
3 2
2 2 2
3 3
3 2
2 2 2
3 2
Output
2
1
3
3
2
4
Input
2 2 2
3 2
2 2 1
Output
2
1
Note
<image>
This image illustrates the second sample case.
Submitted Solution:
```
n, m, q = map(int, input().split())
ans = [0] * q
op = []
cur = [[0 for _ in range(m)] for _ in range(n)]
hist, histcnt = [], []
import math, copy
sq = int(math.sqrt(q) + 0.5)
hist.append(copy.deepcopy(cur))
histcnt.append(0)
def apply(mat, cnt, u, v):
if u == 1 and mat[v[0] - 1][v[1] - 1] == 0:
mat[v[0] - 1][v[1] - 1] = 1
cnt += 1
elif u == 2 and mat[v[0] - 1][v[1] - 1] == 1:
mat[v[0] - 1][v[1] - 1] = 0
cnt -= 1
else:
for j in range(m):
if mat[v[0] - 1][j] == 0:
mat[v[0] - 1][j] = 1
cnt += 1
else:
mat[v[0] - 1][j] = 0
cnt -= 1
return cnt
def run(mat, cnt, f, t):
if f > t:
return cnt
for i in range(f, t + 1):
if op[i][0] < 4:
cnt = apply(mat, cnt, op[i][0], op[i][1])
else:
k = (op[i][1][0] // sq) * sq
mat, cnt = hist[k], histcnt[k]
cnt = run(mat, cnt, k + 1, op[i][1][0])
return cnt
for i in range(q):
u, *v = map(int, input().split())
op.append((u, v))
if u < 4:
ans[i] = apply(cur, ans[i - 1], u, v)
else:
k = (v[0] // sq) * sq
ans[i] = run(hist[k], histcnt[k], k + 1, v[0])
if i % sq == 0:
hist.append(copy.deepcopy(cur))
histcnt.append(ans[i])
print("\n".join(map(str, ans)))
``` | instruction | 0 | 6,256 | 8 | 12,512 |
No | output | 1 | 6,256 | 8 | 12,513 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Santa has prepared boxes with presents for n kids, one box for each kid. There are m kinds of presents: balloons, sweets, chocolate bars, toy cars... A child would be disappointed to receive two presents of the same kind, so all kinds of presents in one box are distinct.
Having packed all the presents, Santa realized that different boxes can contain different number of presents. It would be unfair to the children, so he decided to move some presents between boxes, and make their sizes similar. After all movements, the difference between the maximal and the minimal number of presents in a box must be as small as possible. All presents in each box should still be distinct. Santa wants to finish the job as fast as possible, so he wants to minimize the number of movements required to complete the task.
Given the sets of presents in each box, find the shortest sequence of movements of presents between boxes that minimizes the difference of sizes of the smallest and the largest box, and keeps all presents in each box distinct.
Input
The first line of input contains two integers n, m (1 β€ n, m β€ 100\ 000), the number of boxes and the number of kinds of the presents. Denote presents with integers from 1 to m.
Each of the following n lines contains the description of one box. It begins with an integer s_i (s_i β₯ 0), the number of presents in the box, s_i distinct integers between 1 and m follow, denoting the kinds of presents in that box.
The total number of presents in all boxes does not exceed 500 000.
Output
Print one integer k at the first line of output, the number of movements in the shortest sequence that makes the sizes of the boxes differ by at most one. Then print k lines that describe movements in the same order in which they should be performed. Each movement is described by three integers from_i, to_i, kind_i. It means that the present of kind kind_i is moved from the box with number from_i to the box with number to_i. Boxes are numbered from one in the order they are given in the input.
At the moment when the movement is performed the present with kind kind_i must be present in the box with number from_i. After performing all moves each box must not contain two presents of the same kind.
If there are several optimal solutions, output any of them.
Example
Input
3 5
5 1 2 3 4 5
2 1 2
2 3 4
Output
2
1 3 5
1 2 3
Submitted Solution:
```
n, m = map(int, input().split())
presents = []
presentNum = 0
for i in range(n):
pred = list(map(int, input().split()))
presents.append((pred[0], i+1, pred[1:]))
presentNum += pred[0]
presents = sorted(presents)
presentAL = int(presentNum / n)
usedPres = []
lastIndex = n - 1
answers = []
for i in range(n):
cNum = len(presents[i][2])
while (cNum < presentAL):
diff = list(set(presents[lastIndex][2]) - set(usedPres) - set(presents[i][2]))
for j in range(presentAL - cNum):
if j >= len(diff):
lastIndex -= 1
usedPres = []
break
answers.append([presents[lastIndex][1], presents[i][1], diff[j]])
usedPres.append(diff[j])
cNum += 1
print(len(answers))
for ans in answers:
print(ans[0], ans[1], ans[2])
``` | instruction | 0 | 6,608 | 8 | 13,216 |
No | output | 1 | 6,608 | 8 | 13,217 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Santa has prepared boxes with presents for n kids, one box for each kid. There are m kinds of presents: balloons, sweets, chocolate bars, toy cars... A child would be disappointed to receive two presents of the same kind, so all kinds of presents in one box are distinct.
Having packed all the presents, Santa realized that different boxes can contain different number of presents. It would be unfair to the children, so he decided to move some presents between boxes, and make their sizes similar. After all movements, the difference between the maximal and the minimal number of presents in a box must be as small as possible. All presents in each box should still be distinct. Santa wants to finish the job as fast as possible, so he wants to minimize the number of movements required to complete the task.
Given the sets of presents in each box, find the shortest sequence of movements of presents between boxes that minimizes the difference of sizes of the smallest and the largest box, and keeps all presents in each box distinct.
Input
The first line of input contains two integers n, m (1 β€ n, m β€ 100\ 000), the number of boxes and the number of kinds of the presents. Denote presents with integers from 1 to m.
Each of the following n lines contains the description of one box. It begins with an integer s_i (s_i β₯ 0), the number of presents in the box, s_i distinct integers between 1 and m follow, denoting the kinds of presents in that box.
The total number of presents in all boxes does not exceed 500 000.
Output
Print one integer k at the first line of output, the number of movements in the shortest sequence that makes the sizes of the boxes differ by at most one. Then print k lines that describe movements in the same order in which they should be performed. Each movement is described by three integers from_i, to_i, kind_i. It means that the present of kind kind_i is moved from the box with number from_i to the box with number to_i. Boxes are numbered from one in the order they are given in the input.
At the moment when the movement is performed the present with kind kind_i must be present in the box with number from_i. After performing all moves each box must not contain two presents of the same kind.
If there are several optimal solutions, output any of them.
Example
Input
3 5
5 1 2 3 4 5
2 1 2
2 3 4
Output
2
1 3 5
1 2 3
Submitted Solution:
```
n, m = map(int, input().split())
presents = []
presentNum = 0
for i in range(n):
pred = list(map(int, input().split()))
presents.append((pred[0], i+1, pred[1:]))
presentNum += pred[0]
presents = sorted(presents)
presentAL = round(presentNum / n)
usedPres = []
lastIndex = n - 1
answers = []
for i in range(n):
cNum = len(presents[i][2])
while (cNum < presentAL):
presentList = set(presents[lastIndex][2]) - set(usedPres)
checked = False
while (len(list(presentList)) <= presentAL):
lastIndex -= 1
usedPres = []
presentList = set(presents[lastIndex][2])
if (lastIndex <= i):
break
diff = list(presentList - set(presents[i][2]))
for j in range(presentAL - cNum):
if j >= len(diff):
lastIndex -= 1
usedPres = []
break
answers.append([presents[lastIndex][1], presents[i][1], diff[j]])
usedPres.append(diff[j])
cNum += 1
print(len(answers))
for ans in answers:
print(ans[0], ans[1], ans[2])
``` | instruction | 0 | 6,609 | 8 | 13,218 |
No | output | 1 | 6,609 | 8 | 13,219 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Santa has prepared boxes with presents for n kids, one box for each kid. There are m kinds of presents: balloons, sweets, chocolate bars, toy cars... A child would be disappointed to receive two presents of the same kind, so all kinds of presents in one box are distinct.
Having packed all the presents, Santa realized that different boxes can contain different number of presents. It would be unfair to the children, so he decided to move some presents between boxes, and make their sizes similar. After all movements, the difference between the maximal and the minimal number of presents in a box must be as small as possible. All presents in each box should still be distinct. Santa wants to finish the job as fast as possible, so he wants to minimize the number of movements required to complete the task.
Given the sets of presents in each box, find the shortest sequence of movements of presents between boxes that minimizes the difference of sizes of the smallest and the largest box, and keeps all presents in each box distinct.
Input
The first line of input contains two integers n, m (1 β€ n, m β€ 100\ 000), the number of boxes and the number of kinds of the presents. Denote presents with integers from 1 to m.
Each of the following n lines contains the description of one box. It begins with an integer s_i (s_i β₯ 0), the number of presents in the box, s_i distinct integers between 1 and m follow, denoting the kinds of presents in that box.
The total number of presents in all boxes does not exceed 500 000.
Output
Print one integer k at the first line of output, the number of movements in the shortest sequence that makes the sizes of the boxes differ by at most one. Then print k lines that describe movements in the same order in which they should be performed. Each movement is described by three integers from_i, to_i, kind_i. It means that the present of kind kind_i is moved from the box with number from_i to the box with number to_i. Boxes are numbered from one in the order they are given in the input.
At the moment when the movement is performed the present with kind kind_i must be present in the box with number from_i. After performing all moves each box must not contain two presents of the same kind.
If there are several optimal solutions, output any of them.
Example
Input
3 5
5 1 2 3 4 5
2 1 2
2 3 4
Output
2
1 3 5
1 2 3
Submitted Solution:
```
n, m = map(int, input().split())
presents = []
presentNum = 0
for i in range(n):
pred = list(map(int, input().split()))
presents.append((pred[0], i+1, pred[1:]))
presentNum += pred[0]
presents = sorted(presents)
presentAL = int(presentNum / n)
usedPres = []
lastIndex = n - 1
for i in range(n):
cNum = len(presents[i][2])
while (cNum < presentAL):
diff = list(set(presents[lastIndex][2]) - set(usedPres) - set(presents[i][2]))
for j in range(presentAL - cNum):
if j >= len(diff):
lastIndex -= 1
usedPres = []
break
print(presents[lastIndex][1], presents[i][1], diff[j])
usedPres.append(diff[j])
cNum += 1
``` | instruction | 0 | 6,610 | 8 | 13,220 |
No | output | 1 | 6,610 | 8 | 13,221 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Santa has prepared boxes with presents for n kids, one box for each kid. There are m kinds of presents: balloons, sweets, chocolate bars, toy cars... A child would be disappointed to receive two presents of the same kind, so all kinds of presents in one box are distinct.
Having packed all the presents, Santa realized that different boxes can contain different number of presents. It would be unfair to the children, so he decided to move some presents between boxes, and make their sizes similar. After all movements, the difference between the maximal and the minimal number of presents in a box must be as small as possible. All presents in each box should still be distinct. Santa wants to finish the job as fast as possible, so he wants to minimize the number of movements required to complete the task.
Given the sets of presents in each box, find the shortest sequence of movements of presents between boxes that minimizes the difference of sizes of the smallest and the largest box, and keeps all presents in each box distinct.
Input
The first line of input contains two integers n, m (1 β€ n, m β€ 100\ 000), the number of boxes and the number of kinds of the presents. Denote presents with integers from 1 to m.
Each of the following n lines contains the description of one box. It begins with an integer s_i (s_i β₯ 0), the number of presents in the box, s_i distinct integers between 1 and m follow, denoting the kinds of presents in that box.
The total number of presents in all boxes does not exceed 500 000.
Output
Print one integer k at the first line of output, the number of movements in the shortest sequence that makes the sizes of the boxes differ by at most one. Then print k lines that describe movements in the same order in which they should be performed. Each movement is described by three integers from_i, to_i, kind_i. It means that the present of kind kind_i is moved from the box with number from_i to the box with number to_i. Boxes are numbered from one in the order they are given in the input.
At the moment when the movement is performed the present with kind kind_i must be present in the box with number from_i. After performing all moves each box must not contain two presents of the same kind.
If there are several optimal solutions, output any of them.
Example
Input
3 5
5 1 2 3 4 5
2 1 2
2 3 4
Output
2
1 3 5
1 2 3
Submitted Solution:
```
n, m = map(int, input().split())
presents = []
presentNum = 0
for i in range(n):
pred = list(map(int, input().split()))
presents.append((pred[0], i+1, pred[1:]))
presentNum += pred[0]
presents = sorted(presents)
presentAL = round(presentNum / n)
usedPres = []
lastIndex = n - 1
answers = []
for i in range(n):
cNum = len(presents[i][2])
while (cNum < presentAL):
presentList = set(presents[lastIndex][2]) - set(usedPres)
checked = False
while (len(list(presentList)) <= presentAL):
lastIndex -= 1
usedPres = []
presentList = set(presents[lastIndex][2])
if (lastIndex < i):
break
diff = list(presentList - set(presents[i][2]))
for j in range(presentAL - cNum):
if j >= len(diff):
lastIndex -= 1
usedPres = []
break
answers.append([presents[lastIndex][1], presents[i][1], diff[j]])
usedPres.append(diff[j])
cNum += 1
print(len(answers))
for ans in answers:
print(ans[0], ans[1], ans[2])
``` | instruction | 0 | 6,611 | 8 | 13,222 |
No | output | 1 | 6,611 | 8 | 13,223 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Vasya plays a computer game with ninjas. At this stage Vasya's ninja should get out of a deep canyon.
The canyon consists of two vertical parallel walls, their height is n meters. Let's imagine that we split these walls into 1 meter-long areas and number them with positive integers from 1 to n from bottom to top. Some areas are safe and the ninja can climb them. Others are spiky and ninja can't be there. Let's call such areas dangerous.
Initially the ninja is on the lower area of the left wall. He can use each second to perform one of the following actions:
* climb one area up;
* climb one area down;
* jump to the opposite wall. That gets the ninja to the area that is exactly k meters higher than the area he jumped from. More formally, if before the jump the ninja is located at area x of one wall, then after the jump he is located at area x + k of the other wall.
If at some point of time the ninja tries to get to an area with a number larger than n, then we can assume that the ninja got out of the canyon.
The canyon gets flooded and each second the water level raises one meter. Initially the water level is at the lower border of the first area. Ninja cannot be on the area covered by water. We can assume that the ninja and the water "move in turns" β first the ninja performs some action, then the water raises for one meter, then the ninja performs one more action and so on.
The level is considered completed if the ninja manages to get out of the canyon.
After several failed attempts Vasya started to doubt whether it is possible to complete the level at all. Help him answer the question.
Input
The first line contains two integers n and k (1 β€ n, k β€ 105) β the height of the canyon and the height of ninja's jump, correspondingly.
The second line contains the description of the left wall β a string with the length of n characters. The i-th character represents the state of the i-th wall area: character "X" represents a dangerous area and character "-" represents a safe area.
The third line describes the right wall in the same format.
It is guaranteed that the first area of the left wall is not dangerous.
Output
Print "YES" (without the quotes) if the ninja can get out from the canyon, otherwise, print "NO" (without the quotes).
Examples
Input
7 3
---X--X
-X--XX-
Output
YES
Input
6 2
--X-X-
X--XX-
Output
NO
Note
In the first sample the ninja should first jump to the right wall, then go one meter down along the right wall, then jump to the left wall. The next jump can get the ninja from the canyon.
In the second sample there's no way the ninja can get out of the canyon. | instruction | 0 | 6,894 | 8 | 13,788 |
Tags: shortest paths
Correct Solution:
```
f = lambda: [q != '-' for q in input()]
n, k = map(int, input().split())
t = [(0, 0, f(), f())]
def g(d, s, a, b):
if d > n - 1:
print('YES')
exit()
if not (a[d] or s > d):
a[d] = 1
t.append((d, s, a, b))
while t:
d, s, a, b = t.pop()
g(d + 1, s + 1, a, b)
g(d - 1, s + 1, a, b)
g(d + k, s + 1, b, a)
print('NO')
``` | output | 1 | 6,894 | 8 | 13,789 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Vasya plays a computer game with ninjas. At this stage Vasya's ninja should get out of a deep canyon.
The canyon consists of two vertical parallel walls, their height is n meters. Let's imagine that we split these walls into 1 meter-long areas and number them with positive integers from 1 to n from bottom to top. Some areas are safe and the ninja can climb them. Others are spiky and ninja can't be there. Let's call such areas dangerous.
Initially the ninja is on the lower area of the left wall. He can use each second to perform one of the following actions:
* climb one area up;
* climb one area down;
* jump to the opposite wall. That gets the ninja to the area that is exactly k meters higher than the area he jumped from. More formally, if before the jump the ninja is located at area x of one wall, then after the jump he is located at area x + k of the other wall.
If at some point of time the ninja tries to get to an area with a number larger than n, then we can assume that the ninja got out of the canyon.
The canyon gets flooded and each second the water level raises one meter. Initially the water level is at the lower border of the first area. Ninja cannot be on the area covered by water. We can assume that the ninja and the water "move in turns" β first the ninja performs some action, then the water raises for one meter, then the ninja performs one more action and so on.
The level is considered completed if the ninja manages to get out of the canyon.
After several failed attempts Vasya started to doubt whether it is possible to complete the level at all. Help him answer the question.
Input
The first line contains two integers n and k (1 β€ n, k β€ 105) β the height of the canyon and the height of ninja's jump, correspondingly.
The second line contains the description of the left wall β a string with the length of n characters. The i-th character represents the state of the i-th wall area: character "X" represents a dangerous area and character "-" represents a safe area.
The third line describes the right wall in the same format.
It is guaranteed that the first area of the left wall is not dangerous.
Output
Print "YES" (without the quotes) if the ninja can get out from the canyon, otherwise, print "NO" (without the quotes).
Examples
Input
7 3
---X--X
-X--XX-
Output
YES
Input
6 2
--X-X-
X--XX-
Output
NO
Note
In the first sample the ninja should first jump to the right wall, then go one meter down along the right wall, then jump to the left wall. The next jump can get the ninja from the canyon.
In the second sample there's no way the ninja can get out of the canyon. | instruction | 0 | 6,895 | 8 | 13,790 |
Tags: shortest paths
Correct Solution:
```
# 198B
__author__ = 'artyom'
n, k = map(int, input().split())
w = [input(), input()]
def neighbours(vertex, time):
vertices = set()
if vertex[1] + 1 >= n or w[vertex[0]][vertex[1] + 1] != 'X':
vertices.add((vertex[0], vertex[1] + 1))
if vertex[1] + k >= n or w[1 - vertex[0]][vertex[1] + k] != 'X':
vertices.add((1 - vertex[0], vertex[1] + k))
if vertex[1] - 1 > time and w[vertex[0]][vertex[1] - 1] != 'X':
vertices.add((vertex[0], vertex[1] - 1))
return vertices
def bfs(*start):
stack, visited = [(start, 0)], set()
while stack:
vertex, time = stack.pop(0)
if vertex[1] >= n:
return 1
if vertex not in visited:
visited.add(vertex)
for neighbour in neighbours(vertex, time):
if neighbour not in visited:
stack.append((neighbour, time + 1))
return 0
print('YES' if bfs(0, 0) else 'NO')
``` | output | 1 | 6,895 | 8 | 13,791 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Vasya plays a computer game with ninjas. At this stage Vasya's ninja should get out of a deep canyon.
The canyon consists of two vertical parallel walls, their height is n meters. Let's imagine that we split these walls into 1 meter-long areas and number them with positive integers from 1 to n from bottom to top. Some areas are safe and the ninja can climb them. Others are spiky and ninja can't be there. Let's call such areas dangerous.
Initially the ninja is on the lower area of the left wall. He can use each second to perform one of the following actions:
* climb one area up;
* climb one area down;
* jump to the opposite wall. That gets the ninja to the area that is exactly k meters higher than the area he jumped from. More formally, if before the jump the ninja is located at area x of one wall, then after the jump he is located at area x + k of the other wall.
If at some point of time the ninja tries to get to an area with a number larger than n, then we can assume that the ninja got out of the canyon.
The canyon gets flooded and each second the water level raises one meter. Initially the water level is at the lower border of the first area. Ninja cannot be on the area covered by water. We can assume that the ninja and the water "move in turns" β first the ninja performs some action, then the water raises for one meter, then the ninja performs one more action and so on.
The level is considered completed if the ninja manages to get out of the canyon.
After several failed attempts Vasya started to doubt whether it is possible to complete the level at all. Help him answer the question.
Input
The first line contains two integers n and k (1 β€ n, k β€ 105) β the height of the canyon and the height of ninja's jump, correspondingly.
The second line contains the description of the left wall β a string with the length of n characters. The i-th character represents the state of the i-th wall area: character "X" represents a dangerous area and character "-" represents a safe area.
The third line describes the right wall in the same format.
It is guaranteed that the first area of the left wall is not dangerous.
Output
Print "YES" (without the quotes) if the ninja can get out from the canyon, otherwise, print "NO" (without the quotes).
Examples
Input
7 3
---X--X
-X--XX-
Output
YES
Input
6 2
--X-X-
X--XX-
Output
NO
Note
In the first sample the ninja should first jump to the right wall, then go one meter down along the right wall, then jump to the left wall. The next jump can get the ninja from the canyon.
In the second sample there's no way the ninja can get out of the canyon. | instruction | 0 | 6,896 | 8 | 13,792 |
Tags: shortest paths
Correct Solution:
```
def bfs(n):
visited=[False]*(2*n)
queue=[]
queue2=[]
queue.append(0)
queue2.append(0)
visited[0]=True
flag=True
water=0
while queue and flag:
s=queue.pop(0)
water= queue2.pop(0)
for i in graph[s]:
if not visited[i]:
a=0
#print(water,i)
if i>=n:
a=i-n
else:
a=i
if a>water:
if i==n-1 or i==2*n-1:
flag=False
else:
queue.append(i)
queue2.append(water+1)
visited[i]=True
return not flag
n,k=map(int,input().split())
l=input()
r=input()
graph=[]
for i in range(n):
graph.append([])
if l[i]=='-':
if i+1<n:
if l[i+1]=='-':
graph[i].append(i+1)
if i-1>=0:
if l[i-1]=='-':
graph[i].append(i-1)
if i+k<n:
if r[i+k]=='-':
graph[i].append(n+i+k)
else:
graph[i].append(2*n-1)
for i in range(n):
graph.append([])
if r[i]=='-':
if i+1<n:
if r[i+1]=='-':
graph[n+i].append(n+i+1)
if i-1>=0:
if r[i-1]=='-':
graph[n+i].append(n+i-1)
if i+k<n:
if l[i+k]=='-':
graph[n+i].append(i+k)
else:
graph[n+i].append(n-1)
#print(graph)
if n==1:
print("YES")
elif bfs(n):
print("YES")
else:
print("NO")
``` | output | 1 | 6,896 | 8 | 13,793 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Vasya plays a computer game with ninjas. At this stage Vasya's ninja should get out of a deep canyon.
The canyon consists of two vertical parallel walls, their height is n meters. Let's imagine that we split these walls into 1 meter-long areas and number them with positive integers from 1 to n from bottom to top. Some areas are safe and the ninja can climb them. Others are spiky and ninja can't be there. Let's call such areas dangerous.
Initially the ninja is on the lower area of the left wall. He can use each second to perform one of the following actions:
* climb one area up;
* climb one area down;
* jump to the opposite wall. That gets the ninja to the area that is exactly k meters higher than the area he jumped from. More formally, if before the jump the ninja is located at area x of one wall, then after the jump he is located at area x + k of the other wall.
If at some point of time the ninja tries to get to an area with a number larger than n, then we can assume that the ninja got out of the canyon.
The canyon gets flooded and each second the water level raises one meter. Initially the water level is at the lower border of the first area. Ninja cannot be on the area covered by water. We can assume that the ninja and the water "move in turns" β first the ninja performs some action, then the water raises for one meter, then the ninja performs one more action and so on.
The level is considered completed if the ninja manages to get out of the canyon.
After several failed attempts Vasya started to doubt whether it is possible to complete the level at all. Help him answer the question.
Input
The first line contains two integers n and k (1 β€ n, k β€ 105) β the height of the canyon and the height of ninja's jump, correspondingly.
The second line contains the description of the left wall β a string with the length of n characters. The i-th character represents the state of the i-th wall area: character "X" represents a dangerous area and character "-" represents a safe area.
The third line describes the right wall in the same format.
It is guaranteed that the first area of the left wall is not dangerous.
Output
Print "YES" (without the quotes) if the ninja can get out from the canyon, otherwise, print "NO" (without the quotes).
Examples
Input
7 3
---X--X
-X--XX-
Output
YES
Input
6 2
--X-X-
X--XX-
Output
NO
Note
In the first sample the ninja should first jump to the right wall, then go one meter down along the right wall, then jump to the left wall. The next jump can get the ninja from the canyon.
In the second sample there's no way the ninja can get out of the canyon. | instruction | 0 | 6,897 | 8 | 13,794 |
Tags: shortest paths
Correct Solution:
```
n, k = map(int, input().split(' '))
l = input()
r = input()
data = [0, ' '+l, ' '+r]
dist = [[1000000]*(10**5+5) for _ in range(3)]
visited = [[False]*100005 for _ in range(3)]
visited[1][1] = True
dist[1][1] = 0
qx = [1]
qy = [1]
while qx:
x = qx.pop()
y = qy.pop()
if dist[x][y] >= y:
continue
if x == 1:
poss = [[x, y-1], [x, y+1], [x+1, y+k]]
if x == 2:
poss = [[x, y-1], [x, y+1], [x-1, y+k]]
for i in poss:
newx = i[0]
newy = i[1]
if newy > n: print("YES"); quit();
if 1<=newy<=n and not visited[newx][newy] and data[newx][newy] != 'X':
visited[newx][newy] = True
qx = [newx] + qx
qy = [newy] + qy
dist[newx][newy] = dist[x][y] + 1
print("NO")
``` | output | 1 | 6,897 | 8 | 13,795 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Vasya plays a computer game with ninjas. At this stage Vasya's ninja should get out of a deep canyon.
The canyon consists of two vertical parallel walls, their height is n meters. Let's imagine that we split these walls into 1 meter-long areas and number them with positive integers from 1 to n from bottom to top. Some areas are safe and the ninja can climb them. Others are spiky and ninja can't be there. Let's call such areas dangerous.
Initially the ninja is on the lower area of the left wall. He can use each second to perform one of the following actions:
* climb one area up;
* climb one area down;
* jump to the opposite wall. That gets the ninja to the area that is exactly k meters higher than the area he jumped from. More formally, if before the jump the ninja is located at area x of one wall, then after the jump he is located at area x + k of the other wall.
If at some point of time the ninja tries to get to an area with a number larger than n, then we can assume that the ninja got out of the canyon.
The canyon gets flooded and each second the water level raises one meter. Initially the water level is at the lower border of the first area. Ninja cannot be on the area covered by water. We can assume that the ninja and the water "move in turns" β first the ninja performs some action, then the water raises for one meter, then the ninja performs one more action and so on.
The level is considered completed if the ninja manages to get out of the canyon.
After several failed attempts Vasya started to doubt whether it is possible to complete the level at all. Help him answer the question.
Input
The first line contains two integers n and k (1 β€ n, k β€ 105) β the height of the canyon and the height of ninja's jump, correspondingly.
The second line contains the description of the left wall β a string with the length of n characters. The i-th character represents the state of the i-th wall area: character "X" represents a dangerous area and character "-" represents a safe area.
The third line describes the right wall in the same format.
It is guaranteed that the first area of the left wall is not dangerous.
Output
Print "YES" (without the quotes) if the ninja can get out from the canyon, otherwise, print "NO" (without the quotes).
Examples
Input
7 3
---X--X
-X--XX-
Output
YES
Input
6 2
--X-X-
X--XX-
Output
NO
Note
In the first sample the ninja should first jump to the right wall, then go one meter down along the right wall, then jump to the left wall. The next jump can get the ninja from the canyon.
In the second sample there's no way the ninja can get out of the canyon. | instruction | 0 | 6,898 | 8 | 13,796 |
Tags: shortest paths
Correct Solution:
```
n, k = map(int, input().split())
lzid = input()
dzid = input()
zidovi = [lzid, dzid]
q = [[-1, [False,0]]] #[koraci, [zid, visina]]
izasao = 0
bio = [[0 for i in range(n+k+100)], [0 for i in range(n+k+100)]]
while len(q) != 0:
trenutni = q.pop(0)
korak = trenutni[0]
pozicija = trenutni[1]
tren_zid = pozicija[0]
tren_visina = pozicija[1]
if bio[tren_zid][tren_visina] == 0:
bio[tren_zid][tren_visina] = 1
if tren_visina > n-1:
print("YES")
izasao = 1
break
elif tren_visina == n-1 and zidovi[tren_zid][tren_visina] != 'X' and tren_visina > korak:
print("YES")
izasao = 1
break
elif zidovi[tren_zid][tren_visina] != 'X' and tren_visina > korak:
q.append([korak+1, [tren_zid, tren_visina-1]])
q.append([korak+1, [tren_zid, tren_visina+1]])
q.append([korak+1, [not(tren_zid), tren_visina+k]])
## if tren_visina - 1 > korak+1:
## if zidovi[tren_zid][tren_visina-1] != 'X':
## q.append([korak+1, [tren_zid, tren_visina-1]])
## if tren_visina + 1 > korak:
## if tren_visina + k <= n-1:
## if zidovi[tren_zid][tren_visina+1] != 'X':
## q.append([korak+1, [tren_zid, tren_visina+1]])
## else:
## print("YES")
## izasao = 1
## break
## if tren_visina + k > korak:
## if tren_visina + k <= n-1:
## if zidovi[not(tren_zid)][tren_visina+k] != 'X':
## q.append([korak+1, [not(tren_zid), tren_visina+k]])
## else:
## print("YES")
## izasao = 1
if izasao == 0:
print("NO")
``` | output | 1 | 6,898 | 8 | 13,797 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Vasya plays a computer game with ninjas. At this stage Vasya's ninja should get out of a deep canyon.
The canyon consists of two vertical parallel walls, their height is n meters. Let's imagine that we split these walls into 1 meter-long areas and number them with positive integers from 1 to n from bottom to top. Some areas are safe and the ninja can climb them. Others are spiky and ninja can't be there. Let's call such areas dangerous.
Initially the ninja is on the lower area of the left wall. He can use each second to perform one of the following actions:
* climb one area up;
* climb one area down;
* jump to the opposite wall. That gets the ninja to the area that is exactly k meters higher than the area he jumped from. More formally, if before the jump the ninja is located at area x of one wall, then after the jump he is located at area x + k of the other wall.
If at some point of time the ninja tries to get to an area with a number larger than n, then we can assume that the ninja got out of the canyon.
The canyon gets flooded and each second the water level raises one meter. Initially the water level is at the lower border of the first area. Ninja cannot be on the area covered by water. We can assume that the ninja and the water "move in turns" β first the ninja performs some action, then the water raises for one meter, then the ninja performs one more action and so on.
The level is considered completed if the ninja manages to get out of the canyon.
After several failed attempts Vasya started to doubt whether it is possible to complete the level at all. Help him answer the question.
Input
The first line contains two integers n and k (1 β€ n, k β€ 105) β the height of the canyon and the height of ninja's jump, correspondingly.
The second line contains the description of the left wall β a string with the length of n characters. The i-th character represents the state of the i-th wall area: character "X" represents a dangerous area and character "-" represents a safe area.
The third line describes the right wall in the same format.
It is guaranteed that the first area of the left wall is not dangerous.
Output
Print "YES" (without the quotes) if the ninja can get out from the canyon, otherwise, print "NO" (without the quotes).
Examples
Input
7 3
---X--X
-X--XX-
Output
YES
Input
6 2
--X-X-
X--XX-
Output
NO
Note
In the first sample the ninja should first jump to the right wall, then go one meter down along the right wall, then jump to the left wall. The next jump can get the ninja from the canyon.
In the second sample there's no way the ninja can get out of the canyon. | instruction | 0 | 6,899 | 8 | 13,798 |
Tags: shortest paths
Correct Solution:
```
f = lambda: [q == '-' for q in input()]
n, k = map(int, input().split())
l, r = f(), f()
u, v = [0], []
def yes(d):
if d > n - 1:
print('YES')
exit()
for i in range(n):
a, b = [], []
for d in u:
if l[d - 1] and d - 1 > i:
a.append(d - 1)
l[d - 1] = 0
yes(d + 1)
if l[d + 1]:
a.append(d + 1)
l[d + 1] = 0
yes(d + k)
if r[d + k]:
b.append(d + k)
r[d + k] = 0
for d in v:
if r[d - 1] and d - 1 > i:
b.append(d - 1)
r[d - 1] = 0
yes(d + 1)
if r[d + 1]:
b.append(d + 1)
r[d + 1] = 0
yes(d + k)
if l[d + k]:
a.append(d + k)
l[d + k] = 0
u, v = a, b
print('NO')
``` | output | 1 | 6,899 | 8 | 13,799 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Vasya plays a computer game with ninjas. At this stage Vasya's ninja should get out of a deep canyon.
The canyon consists of two vertical parallel walls, their height is n meters. Let's imagine that we split these walls into 1 meter-long areas and number them with positive integers from 1 to n from bottom to top. Some areas are safe and the ninja can climb them. Others are spiky and ninja can't be there. Let's call such areas dangerous.
Initially the ninja is on the lower area of the left wall. He can use each second to perform one of the following actions:
* climb one area up;
* climb one area down;
* jump to the opposite wall. That gets the ninja to the area that is exactly k meters higher than the area he jumped from. More formally, if before the jump the ninja is located at area x of one wall, then after the jump he is located at area x + k of the other wall.
If at some point of time the ninja tries to get to an area with a number larger than n, then we can assume that the ninja got out of the canyon.
The canyon gets flooded and each second the water level raises one meter. Initially the water level is at the lower border of the first area. Ninja cannot be on the area covered by water. We can assume that the ninja and the water "move in turns" β first the ninja performs some action, then the water raises for one meter, then the ninja performs one more action and so on.
The level is considered completed if the ninja manages to get out of the canyon.
After several failed attempts Vasya started to doubt whether it is possible to complete the level at all. Help him answer the question.
Input
The first line contains two integers n and k (1 β€ n, k β€ 105) β the height of the canyon and the height of ninja's jump, correspondingly.
The second line contains the description of the left wall β a string with the length of n characters. The i-th character represents the state of the i-th wall area: character "X" represents a dangerous area and character "-" represents a safe area.
The third line describes the right wall in the same format.
It is guaranteed that the first area of the left wall is not dangerous.
Output
Print "YES" (without the quotes) if the ninja can get out from the canyon, otherwise, print "NO" (without the quotes).
Examples
Input
7 3
---X--X
-X--XX-
Output
YES
Input
6 2
--X-X-
X--XX-
Output
NO
Note
In the first sample the ninja should first jump to the right wall, then go one meter down along the right wall, then jump to the left wall. The next jump can get the ninja from the canyon.
In the second sample there's no way the ninja can get out of the canyon. | instruction | 0 | 6,900 | 8 | 13,800 |
Tags: shortest paths
Correct Solution:
```
from sys import stdin, stdout
from collections import deque
n, k = map(int, stdin.readline().split())
maps = []
maps.append(list(stdin.readline() + '-'))
maps.append(list(stdin.readline() + '-'))
visit = [[0, 0] for i in range(n + 1)]
visit[0][0] = 1
queue = deque()
label = 0
queue.append((0, -1, 0))#ΡΠ²ΠΎΠΉ ΡΡΠΎΠ²Π΅Π½Ρ, ΡΡΠΎΠ²Π΅Π½Ρ Π²ΠΎΠ΄Ρ, Π½ΠΎΠΌΠ΅Ρ ΡΡΠ΅Π½Ρ
while queue:
mine, line, num = queue.popleft()
if line >= mine:
continue
if mine + k >= n:
label = 1
if mine + 1 < n and not visit[mine + 1][num] and maps[num][mine + 1] == '-':
queue.append((mine + 1, line + 1, num))
visit[mine + 1][num] = 1
if mine and mine - line > 1 and not visit[mine - 1][num] and maps[num][mine - 1] == '-':
queue.append((mine - 1, line + 1, num))
visit[mine - 1][num] = 1
if mine + k < n and not visit[mine + k][(num + 1) % 2] and maps[(num + 1) % 2][mine + k] == '-':
queue.append((min(mine + k, n), line + 1, (num + 1) % 2))
visit[min(mine + k, n)][(num + 1) % 2] = 1
if label:
stdout.write('YES')
else:
stdout.write('NO')
``` | output | 1 | 6,900 | 8 | 13,801 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Vasya plays a computer game with ninjas. At this stage Vasya's ninja should get out of a deep canyon.
The canyon consists of two vertical parallel walls, their height is n meters. Let's imagine that we split these walls into 1 meter-long areas and number them with positive integers from 1 to n from bottom to top. Some areas are safe and the ninja can climb them. Others are spiky and ninja can't be there. Let's call such areas dangerous.
Initially the ninja is on the lower area of the left wall. He can use each second to perform one of the following actions:
* climb one area up;
* climb one area down;
* jump to the opposite wall. That gets the ninja to the area that is exactly k meters higher than the area he jumped from. More formally, if before the jump the ninja is located at area x of one wall, then after the jump he is located at area x + k of the other wall.
If at some point of time the ninja tries to get to an area with a number larger than n, then we can assume that the ninja got out of the canyon.
The canyon gets flooded and each second the water level raises one meter. Initially the water level is at the lower border of the first area. Ninja cannot be on the area covered by water. We can assume that the ninja and the water "move in turns" β first the ninja performs some action, then the water raises for one meter, then the ninja performs one more action and so on.
The level is considered completed if the ninja manages to get out of the canyon.
After several failed attempts Vasya started to doubt whether it is possible to complete the level at all. Help him answer the question.
Input
The first line contains two integers n and k (1 β€ n, k β€ 105) β the height of the canyon and the height of ninja's jump, correspondingly.
The second line contains the description of the left wall β a string with the length of n characters. The i-th character represents the state of the i-th wall area: character "X" represents a dangerous area and character "-" represents a safe area.
The third line describes the right wall in the same format.
It is guaranteed that the first area of the left wall is not dangerous.
Output
Print "YES" (without the quotes) if the ninja can get out from the canyon, otherwise, print "NO" (without the quotes).
Examples
Input
7 3
---X--X
-X--XX-
Output
YES
Input
6 2
--X-X-
X--XX-
Output
NO
Note
In the first sample the ninja should first jump to the right wall, then go one meter down along the right wall, then jump to the left wall. The next jump can get the ninja from the canyon.
In the second sample there's no way the ninja can get out of the canyon. | instruction | 0 | 6,901 | 8 | 13,802 |
Tags: shortest paths
Correct Solution:
```
from collections import deque
l, j = [int(i) for i in input().split(' ')]
wallA = list(input())
wallB = list(input())
g = {}
for i in range(l):
# Each 4-tuple represents: (Visited?, Current Height, Current Water Height, Drowned?)
if wallA[i] == '-':
g[(1,i+1)] = (-1, 0, 0, False)
if wallB[i] == '-':
g[(-1,i+1)] = (-1, 0, 0, False)
g[(1, 1)] = ('VISITED', 1, 0, False)
q = deque([(1, 1)])
while q:
c = q.popleft()
up = (c[0], c[1]+1)
down = (c[0], c[1]-1)
jump = (c[0]*-1, c[1] + j)
if g[c][1] <= g[c][2]:
g[c] = (g[c][0], g[c][1], g[c][2], True)
if up in g and g[up][0] == -1:
q.append(up)
g[up] = ('VISITED', g[c][1] + 1, g[c][2] + 1, g[c][3])
if down in g and g[down][0] == -1:
q.append(down)
g[down] = ('VISITED', g[c][1] - 1, g[c][2] + 1, g[c][3])
if jump in g and g[jump][0] == -1:
q.append(jump)
g[jump] = ('VISITED', g[c][1] + j, g[c][2] + 1, g[c][3])
def graphHasEscape(graph):
for node in graph:
result = graph[node]
if result[0] == 'VISITED' and ((result[1] + 1 > l) or (result[1] + j > l)) and not result[3]:
return True
break
return False
if graphHasEscape(g):
print('YES')
else:
print('NO')
``` | output | 1 | 6,901 | 8 | 13,803 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Vasya plays a computer game with ninjas. At this stage Vasya's ninja should get out of a deep canyon.
The canyon consists of two vertical parallel walls, their height is n meters. Let's imagine that we split these walls into 1 meter-long areas and number them with positive integers from 1 to n from bottom to top. Some areas are safe and the ninja can climb them. Others are spiky and ninja can't be there. Let's call such areas dangerous.
Initially the ninja is on the lower area of the left wall. He can use each second to perform one of the following actions:
* climb one area up;
* climb one area down;
* jump to the opposite wall. That gets the ninja to the area that is exactly k meters higher than the area he jumped from. More formally, if before the jump the ninja is located at area x of one wall, then after the jump he is located at area x + k of the other wall.
If at some point of time the ninja tries to get to an area with a number larger than n, then we can assume that the ninja got out of the canyon.
The canyon gets flooded and each second the water level raises one meter. Initially the water level is at the lower border of the first area. Ninja cannot be on the area covered by water. We can assume that the ninja and the water "move in turns" β first the ninja performs some action, then the water raises for one meter, then the ninja performs one more action and so on.
The level is considered completed if the ninja manages to get out of the canyon.
After several failed attempts Vasya started to doubt whether it is possible to complete the level at all. Help him answer the question.
Input
The first line contains two integers n and k (1 β€ n, k β€ 105) β the height of the canyon and the height of ninja's jump, correspondingly.
The second line contains the description of the left wall β a string with the length of n characters. The i-th character represents the state of the i-th wall area: character "X" represents a dangerous area and character "-" represents a safe area.
The third line describes the right wall in the same format.
It is guaranteed that the first area of the left wall is not dangerous.
Output
Print "YES" (without the quotes) if the ninja can get out from the canyon, otherwise, print "NO" (without the quotes).
Examples
Input
7 3
---X--X
-X--XX-
Output
YES
Input
6 2
--X-X-
X--XX-
Output
NO
Note
In the first sample the ninja should first jump to the right wall, then go one meter down along the right wall, then jump to the left wall. The next jump can get the ninja from the canyon.
In the second sample there's no way the ninja can get out of the canyon.
Submitted Solution:
```
from collections import deque
import sys
n, k = map(int, sys.stdin.readline().split())
s1, s2 = sys.stdin.readline(), sys.stdin.readline()
mat = [[] for _ in range(2 * 100000 + 10)]
used = [-1] * (2 * 100000 + 10)
for i in range(n):
if i > 0 and s1[i - 1] != 'X' and s1[i] != 'X':
mat[i].append(i - 1)
used[i - 1] = 0
if i < n - 1 and s1[i] != 'X' and s1[i + 1] != 'X':
mat[i].append(i + 1)
used[i + 1] = 0
used[i] = 0
if i + k >= n:
mat[i].append(2 * 100000 + 9)
used[2 * 100000 + 9] = 0
elif s2[i + k] != 'X':
mat[i].append(i + k + 100001)
used[i + k + 100001] = 0
for i in range(n):
if i > 0 and s2[i - 1] != 'X' and s2[i] != 'X':
mat[i + 100001].append(i - 1 + 100001)
used[i - 1 + 100001] = 0
if i < n - 1 and s2[i] != 'X' and s2[i + 1] != 'X':
mat[i + 100001].append(i + 1 + 100001)
used[i + 1 + 100001] = 0
used[i + 100001] = 0
if i + k >= n:
mat[i + 100001].append(2 * 100000 + 9)
used[2 * 100000 + 9] = 0
elif s1[i + k] != 'X':
mat[i + 100001].append(i + k)
used[i + k] = 0
q = deque([0])
dist = [0] * (2 * 100000 + 10)
while q:
u = q[0]
q.popleft()
used[u] = 1
for v in mat[u]:
if used[v] == 0:
dist[v] = dist[u] + 1
used[v] = 1
if (v <= 100000 and dist[v] <= v) or (v > 100000 and dist[v] <= v - 100001) or v == 2 * 100000 + 9:
q.append(v)
if used[2 * 100000 + 9] == 1:
sys.stdout.write('YES')
else:
sys.stdout.write('NO')
``` | instruction | 0 | 6,902 | 8 | 13,804 |
Yes | output | 1 | 6,902 | 8 | 13,805 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Vasya plays a computer game with ninjas. At this stage Vasya's ninja should get out of a deep canyon.
The canyon consists of two vertical parallel walls, their height is n meters. Let's imagine that we split these walls into 1 meter-long areas and number them with positive integers from 1 to n from bottom to top. Some areas are safe and the ninja can climb them. Others are spiky and ninja can't be there. Let's call such areas dangerous.
Initially the ninja is on the lower area of the left wall. He can use each second to perform one of the following actions:
* climb one area up;
* climb one area down;
* jump to the opposite wall. That gets the ninja to the area that is exactly k meters higher than the area he jumped from. More formally, if before the jump the ninja is located at area x of one wall, then after the jump he is located at area x + k of the other wall.
If at some point of time the ninja tries to get to an area with a number larger than n, then we can assume that the ninja got out of the canyon.
The canyon gets flooded and each second the water level raises one meter. Initially the water level is at the lower border of the first area. Ninja cannot be on the area covered by water. We can assume that the ninja and the water "move in turns" β first the ninja performs some action, then the water raises for one meter, then the ninja performs one more action and so on.
The level is considered completed if the ninja manages to get out of the canyon.
After several failed attempts Vasya started to doubt whether it is possible to complete the level at all. Help him answer the question.
Input
The first line contains two integers n and k (1 β€ n, k β€ 105) β the height of the canyon and the height of ninja's jump, correspondingly.
The second line contains the description of the left wall β a string with the length of n characters. The i-th character represents the state of the i-th wall area: character "X" represents a dangerous area and character "-" represents a safe area.
The third line describes the right wall in the same format.
It is guaranteed that the first area of the left wall is not dangerous.
Output
Print "YES" (without the quotes) if the ninja can get out from the canyon, otherwise, print "NO" (without the quotes).
Examples
Input
7 3
---X--X
-X--XX-
Output
YES
Input
6 2
--X-X-
X--XX-
Output
NO
Note
In the first sample the ninja should first jump to the right wall, then go one meter down along the right wall, then jump to the left wall. The next jump can get the ninja from the canyon.
In the second sample there's no way the ninja can get out of the canyon.
Submitted Solution:
```
left, right = (0, 1)
undiscovered = 0
processed = 1
def up(state):
water = state[0] + 1
point = state[1] + 1
l_or_r = state[2]
return [water, point, l_or_r]
def down(state):
water = state[0] + 1
point = state[1] - 1
l_or_r = state[2]
return [water, point, l_or_r]
def jump(state, k):
water = state[0] + 1
point = state[1] + k
if state[2] == left: l_or_r = right
else: l_or_r = left
return [water, point, l_or_r]
def push_next_states(state, k, states):
states[len(states):] = [up(state), down(state), jump(state, k)]
return states
def solve(n, k, lwall, rwall):
water = -1
point = 0
l_or_r = left
state = [water, point, l_or_r]
table = [[undiscovered]*n, [undiscovered]*n]
states = push_next_states(state, k, [])
while states != []:
state = states.pop()
if state is None: break
water = state[0]
point = state[1]
if n <= point: return True
if water < point:
l_or_r = state[2]
if table[l_or_r][point] != processed:
if l_or_r == left: wall = lwall
else: wall = rwall
if wall[point] == '-':
if n <= point+k: return True
push_next_states(state, k, states)
table[l_or_r][point] = processed
return False
def main():
import sys
n, k = [int(x) for x in sys.stdin.readline().split()]
lwall = sys.stdin.readline().rstrip()
rwall = sys.stdin.readline().rstrip()
result = solve(n, k, lwall, rwall)
if result: print("YES", end="")
else: print("NO", end="")
main()
``` | instruction | 0 | 6,903 | 8 | 13,806 |
Yes | output | 1 | 6,903 | 8 | 13,807 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Vasya plays a computer game with ninjas. At this stage Vasya's ninja should get out of a deep canyon.
The canyon consists of two vertical parallel walls, their height is n meters. Let's imagine that we split these walls into 1 meter-long areas and number them with positive integers from 1 to n from bottom to top. Some areas are safe and the ninja can climb them. Others are spiky and ninja can't be there. Let's call such areas dangerous.
Initially the ninja is on the lower area of the left wall. He can use each second to perform one of the following actions:
* climb one area up;
* climb one area down;
* jump to the opposite wall. That gets the ninja to the area that is exactly k meters higher than the area he jumped from. More formally, if before the jump the ninja is located at area x of one wall, then after the jump he is located at area x + k of the other wall.
If at some point of time the ninja tries to get to an area with a number larger than n, then we can assume that the ninja got out of the canyon.
The canyon gets flooded and each second the water level raises one meter. Initially the water level is at the lower border of the first area. Ninja cannot be on the area covered by water. We can assume that the ninja and the water "move in turns" β first the ninja performs some action, then the water raises for one meter, then the ninja performs one more action and so on.
The level is considered completed if the ninja manages to get out of the canyon.
After several failed attempts Vasya started to doubt whether it is possible to complete the level at all. Help him answer the question.
Input
The first line contains two integers n and k (1 β€ n, k β€ 105) β the height of the canyon and the height of ninja's jump, correspondingly.
The second line contains the description of the left wall β a string with the length of n characters. The i-th character represents the state of the i-th wall area: character "X" represents a dangerous area and character "-" represents a safe area.
The third line describes the right wall in the same format.
It is guaranteed that the first area of the left wall is not dangerous.
Output
Print "YES" (without the quotes) if the ninja can get out from the canyon, otherwise, print "NO" (without the quotes).
Examples
Input
7 3
---X--X
-X--XX-
Output
YES
Input
6 2
--X-X-
X--XX-
Output
NO
Note
In the first sample the ninja should first jump to the right wall, then go one meter down along the right wall, then jump to the left wall. The next jump can get the ninja from the canyon.
In the second sample there's no way the ninja can get out of the canyon.
Submitted Solution:
```
n,k = map(int,input().split())
l = input()
r = input()
data = [0, ' '+l,' '+r]
dist = [[1000000]*100005 for _ in range(3)]
visited = [[False]*100005 for _ in range(3)]
dist[1][1]=0
visited[1][1]=True
qx,qy = [1],[1]
while qy:
x,y = qx.pop(),qy.pop()
if dist[x][y]>=y:
continue
if x==1:
poss = [[1,y+1],[1,y-1],[2,y+k]]
else:
poss = [[2,y+1],[2,y-1],[1,y+k]]
for i,e in enumerate(poss):
newx,newy = e[0],e[1]
if newy>n:
print('YES')
from sys import exit
exit()
if 0<newy<=n and not visited[newx][newy] and data[newx][newy]=='-':
visited[newx][newy]=True
dist[newx][newy]=dist[x][y]+1
qx=[newx]+qx
qy=[newy]+qy
print('NO')
``` | instruction | 0 | 6,904 | 8 | 13,808 |
Yes | output | 1 | 6,904 | 8 | 13,809 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Vasya plays a computer game with ninjas. At this stage Vasya's ninja should get out of a deep canyon.
The canyon consists of two vertical parallel walls, their height is n meters. Let's imagine that we split these walls into 1 meter-long areas and number them with positive integers from 1 to n from bottom to top. Some areas are safe and the ninja can climb them. Others are spiky and ninja can't be there. Let's call such areas dangerous.
Initially the ninja is on the lower area of the left wall. He can use each second to perform one of the following actions:
* climb one area up;
* climb one area down;
* jump to the opposite wall. That gets the ninja to the area that is exactly k meters higher than the area he jumped from. More formally, if before the jump the ninja is located at area x of one wall, then after the jump he is located at area x + k of the other wall.
If at some point of time the ninja tries to get to an area with a number larger than n, then we can assume that the ninja got out of the canyon.
The canyon gets flooded and each second the water level raises one meter. Initially the water level is at the lower border of the first area. Ninja cannot be on the area covered by water. We can assume that the ninja and the water "move in turns" β first the ninja performs some action, then the water raises for one meter, then the ninja performs one more action and so on.
The level is considered completed if the ninja manages to get out of the canyon.
After several failed attempts Vasya started to doubt whether it is possible to complete the level at all. Help him answer the question.
Input
The first line contains two integers n and k (1 β€ n, k β€ 105) β the height of the canyon and the height of ninja's jump, correspondingly.
The second line contains the description of the left wall β a string with the length of n characters. The i-th character represents the state of the i-th wall area: character "X" represents a dangerous area and character "-" represents a safe area.
The third line describes the right wall in the same format.
It is guaranteed that the first area of the left wall is not dangerous.
Output
Print "YES" (without the quotes) if the ninja can get out from the canyon, otherwise, print "NO" (without the quotes).
Examples
Input
7 3
---X--X
-X--XX-
Output
YES
Input
6 2
--X-X-
X--XX-
Output
NO
Note
In the first sample the ninja should first jump to the right wall, then go one meter down along the right wall, then jump to the left wall. The next jump can get the ninja from the canyon.
In the second sample there's no way the ninja can get out of the canyon.
Submitted Solution:
```
from sys import stdin, stdout
from collections import deque
n, k = map(int, stdin.readline().split())
maps = []
maps.append(list(stdin.readline() + '-'))
maps.append(list(stdin.readline() + '-'))
visit = [[0, 0] for i in range(n + 1)]
visit[0][0] = 1
queue = deque()
label = 0
queue.append((0, -1, 0))#Ψ±βΨ°Β²Ψ°ΒΎΨ°ΒΉ Ψ±ΖΨ±β¬Ψ°ΒΎΨ°Β²Ψ°Β΅Ψ°Β½Ψ±Ε, Ψ±ΖΨ±β¬Ψ°ΒΎΨ°Β²Ψ°Β΅Ψ°Β½Ψ±Ε Ψ°Β²Ψ°ΒΎΨ°Β΄Ψ±βΉ, Ψ°Β½Ψ°ΒΎΨ°ΒΌΨ°Β΅Ψ±β¬ Ψ±ΩΎΨ±βΨ°Β΅Ψ°Β½Ψ±βΉ
while queue:
mine, line, num = queue.popleft()
if line >= mine:
continue
if mine + k >= n:
label = 1
if mine + 1 < n and not visit[mine + 1][num] and maps[num][mine + 1] == '-':
queue.append((mine + 1, line + 1, num))
visit[mine + 1][num] = 1
if mine and mine - line > 1 and not visit[mine - 1][num] and maps[num][mine - 1] == '-':
queue.append((mine - 1, line + 1, num))
visit[mine - 1][num] = 1
if mine + k < n and not visit[mine + k][(num + 1) % 2] and maps[(num + 1) % 2][mine + k] == '-':
queue.append((min(mine + k, n), line + 1, (num + 1) % 2))
visit[min(mine + k, n)][(num + 1) % 2] = 1
if label:
stdout.write('YES')
else:
stdout.write('NO')
# Made By Mostafa_Khaled
``` | instruction | 0 | 6,905 | 8 | 13,810 |
Yes | output | 1 | 6,905 | 8 | 13,811 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Vasya plays a computer game with ninjas. At this stage Vasya's ninja should get out of a deep canyon.
The canyon consists of two vertical parallel walls, their height is n meters. Let's imagine that we split these walls into 1 meter-long areas and number them with positive integers from 1 to n from bottom to top. Some areas are safe and the ninja can climb them. Others are spiky and ninja can't be there. Let's call such areas dangerous.
Initially the ninja is on the lower area of the left wall. He can use each second to perform one of the following actions:
* climb one area up;
* climb one area down;
* jump to the opposite wall. That gets the ninja to the area that is exactly k meters higher than the area he jumped from. More formally, if before the jump the ninja is located at area x of one wall, then after the jump he is located at area x + k of the other wall.
If at some point of time the ninja tries to get to an area with a number larger than n, then we can assume that the ninja got out of the canyon.
The canyon gets flooded and each second the water level raises one meter. Initially the water level is at the lower border of the first area. Ninja cannot be on the area covered by water. We can assume that the ninja and the water "move in turns" β first the ninja performs some action, then the water raises for one meter, then the ninja performs one more action and so on.
The level is considered completed if the ninja manages to get out of the canyon.
After several failed attempts Vasya started to doubt whether it is possible to complete the level at all. Help him answer the question.
Input
The first line contains two integers n and k (1 β€ n, k β€ 105) β the height of the canyon and the height of ninja's jump, correspondingly.
The second line contains the description of the left wall β a string with the length of n characters. The i-th character represents the state of the i-th wall area: character "X" represents a dangerous area and character "-" represents a safe area.
The third line describes the right wall in the same format.
It is guaranteed that the first area of the left wall is not dangerous.
Output
Print "YES" (without the quotes) if the ninja can get out from the canyon, otherwise, print "NO" (without the quotes).
Examples
Input
7 3
---X--X
-X--XX-
Output
YES
Input
6 2
--X-X-
X--XX-
Output
NO
Note
In the first sample the ninja should first jump to the right wall, then go one meter down along the right wall, then jump to the left wall. The next jump can get the ninja from the canyon.
In the second sample there's no way the ninja can get out of the canyon.
Submitted Solution:
```
left, right = (0, 1)
undiscovered = 0
processed = 1
def climb_up(state):
s = state[:]
s[1] += 1
return s
def climb_down(state):
s = state[:]
s[1] -= 1
return s
def jump_to_other(state, k):
s = state[:]
s[1] += k
if s[2] == left: s[2] = right
else: s[2] = left
return s
def push_next_states(state, k, states):
s = state[:]
s[0] += 1
states.append(climb_up(s))
states.append(climb_down(s))
states.append(jump_to_other(s, k))
return states
def solve(n, k, lwall, rwall):
water = -1
layer = 0
wall = left
state = [water, layer, wall]
states = push_next_states(state, k, [])
table = [[undiscovered]*n, [undiscovered]*n]
while states != []:
state = states.pop(-1)
if n <= state[1]: return True
if table[state[2]][state[1]] != processed:
water = state[0]
layer = state[1]
if state[2] == left: wall = lwall
else: wall = rwall
if water < layer and wall[layer] == '-':
push_next_states(state, k, states)
table[state[2]][layer] = processed
return False
def main():
import sys
n, k = [int(x) for x in sys.stdin.readline().split()]
lwall = sys.stdin.readline().rstrip()
rwall = sys.stdin.readline().rstrip()
result = solve(n, k, lwall, rwall)
if result: print("YES", end="")
else: print("NO", end="")
main()
``` | instruction | 0 | 6,906 | 8 | 13,812 |
No | output | 1 | 6,906 | 8 | 13,813 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Vasya plays a computer game with ninjas. At this stage Vasya's ninja should get out of a deep canyon.
The canyon consists of two vertical parallel walls, their height is n meters. Let's imagine that we split these walls into 1 meter-long areas and number them with positive integers from 1 to n from bottom to top. Some areas are safe and the ninja can climb them. Others are spiky and ninja can't be there. Let's call such areas dangerous.
Initially the ninja is on the lower area of the left wall. He can use each second to perform one of the following actions:
* climb one area up;
* climb one area down;
* jump to the opposite wall. That gets the ninja to the area that is exactly k meters higher than the area he jumped from. More formally, if before the jump the ninja is located at area x of one wall, then after the jump he is located at area x + k of the other wall.
If at some point of time the ninja tries to get to an area with a number larger than n, then we can assume that the ninja got out of the canyon.
The canyon gets flooded and each second the water level raises one meter. Initially the water level is at the lower border of the first area. Ninja cannot be on the area covered by water. We can assume that the ninja and the water "move in turns" β first the ninja performs some action, then the water raises for one meter, then the ninja performs one more action and so on.
The level is considered completed if the ninja manages to get out of the canyon.
After several failed attempts Vasya started to doubt whether it is possible to complete the level at all. Help him answer the question.
Input
The first line contains two integers n and k (1 β€ n, k β€ 105) β the height of the canyon and the height of ninja's jump, correspondingly.
The second line contains the description of the left wall β a string with the length of n characters. The i-th character represents the state of the i-th wall area: character "X" represents a dangerous area and character "-" represents a safe area.
The third line describes the right wall in the same format.
It is guaranteed that the first area of the left wall is not dangerous.
Output
Print "YES" (without the quotes) if the ninja can get out from the canyon, otherwise, print "NO" (without the quotes).
Examples
Input
7 3
---X--X
-X--XX-
Output
YES
Input
6 2
--X-X-
X--XX-
Output
NO
Note
In the first sample the ninja should first jump to the right wall, then go one meter down along the right wall, then jump to the left wall. The next jump can get the ninja from the canyon.
In the second sample there's no way the ninja can get out of the canyon.
Submitted Solution:
```
n, k = map(int, input().split())
lzid = input()
dzid = input()
zidovi = [lzid, dzid]
q = [[-1, [False,0]]] #[koraci, [zid, visina]]
izasao = 0
while len(q) != 0:
trenutni = q.pop(0)
korak = trenutni[0]
pozicija = trenutni[1]
tren_zid = pozicija[0]
tren_visina = pozicija[1]
print("Korak:", korak)
print("pozicija:", pozicija)
if pozicija[1] == n-1:
print("YES")
izasao = 1
break
if tren_visina - 1 > korak+1:
if zidovi[tren_zid][tren_visina-1] != 'X':
q.append([korak+1, [tren_zid, tren_visina-1]])
if tren_visina + 1 > korak:
if tren_visina + k <= n-1:
if zidovi[tren_zid][tren_visina+1] != 'X':
q.append([korak+1, [tren_zid, tren_visina+1]])
else:
print("YES")
izasao = 1
break
if tren_visina + k > korak:
if tren_visina + k <= n-1:
if zidovi[not(tren_zid)][tren_visina+k] != 'X':
q.append([korak+1, [not(tren_zid), tren_visina+k]])
else:
print("YES")
izasao = 1
if izasao == 0:
print("NO")
``` | instruction | 0 | 6,907 | 8 | 13,814 |
No | output | 1 | 6,907 | 8 | 13,815 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Vasya plays a computer game with ninjas. At this stage Vasya's ninja should get out of a deep canyon.
The canyon consists of two vertical parallel walls, their height is n meters. Let's imagine that we split these walls into 1 meter-long areas and number them with positive integers from 1 to n from bottom to top. Some areas are safe and the ninja can climb them. Others are spiky and ninja can't be there. Let's call such areas dangerous.
Initially the ninja is on the lower area of the left wall. He can use each second to perform one of the following actions:
* climb one area up;
* climb one area down;
* jump to the opposite wall. That gets the ninja to the area that is exactly k meters higher than the area he jumped from. More formally, if before the jump the ninja is located at area x of one wall, then after the jump he is located at area x + k of the other wall.
If at some point of time the ninja tries to get to an area with a number larger than n, then we can assume that the ninja got out of the canyon.
The canyon gets flooded and each second the water level raises one meter. Initially the water level is at the lower border of the first area. Ninja cannot be on the area covered by water. We can assume that the ninja and the water "move in turns" β first the ninja performs some action, then the water raises for one meter, then the ninja performs one more action and so on.
The level is considered completed if the ninja manages to get out of the canyon.
After several failed attempts Vasya started to doubt whether it is possible to complete the level at all. Help him answer the question.
Input
The first line contains two integers n and k (1 β€ n, k β€ 105) β the height of the canyon and the height of ninja's jump, correspondingly.
The second line contains the description of the left wall β a string with the length of n characters. The i-th character represents the state of the i-th wall area: character "X" represents a dangerous area and character "-" represents a safe area.
The third line describes the right wall in the same format.
It is guaranteed that the first area of the left wall is not dangerous.
Output
Print "YES" (without the quotes) if the ninja can get out from the canyon, otherwise, print "NO" (without the quotes).
Examples
Input
7 3
---X--X
-X--XX-
Output
YES
Input
6 2
--X-X-
X--XX-
Output
NO
Note
In the first sample the ninja should first jump to the right wall, then go one meter down along the right wall, then jump to the left wall. The next jump can get the ninja from the canyon.
In the second sample there's no way the ninja can get out of the canyon.
Submitted Solution:
```
first_input = []
for i in input().split():
first_input.append(int(i))
walls=[]
left_wall = []
for i in input():
left_wall.append(i)
right_wall = []
for i in input():
right_wall.append(i)
walls.append(left_wall)
walls.append(right_wall)
my_positions = (0,0)
def my_function(position, walls, time):
moves = []
if position[1]+1 < first_input[0] and walls[position[0]][position[1]+1] != 'X':
moves.append(((position[0],position[1]+1), time+1))
elif position[1]+1 > first_input[0]:
return "YES"
if position[1] > 0 and walls[position[0]][position[1]-1] != 'X' and time<position[1]-1:
moves.append(((position[0],position[1]-1), time+1))
if position[1]+first_input[1] < first_input[0] and walls[position[0]^1][position[1]+first_input[1]] != 'X':
moves.append(((position[0]^1,position[1]+first_input[1]), time+1))
elif position[1]+first_input[1] > first_input[0]:
return "YES"
return moves
frontier = []
frontier.append((my_positions,0))
output = "NO"
while len(frontier)>0:
if frontier[0][0][1]==first_input[0]-1:
output="YES"
break
x = my_function(frontier[0][0], walls, frontier[0][1])
del frontier[0]
if x == "YES":
output=x
break
else:
frontier.extend(x)
print(output)
``` | instruction | 0 | 6,908 | 8 | 13,816 |
No | output | 1 | 6,908 | 8 | 13,817 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Vasya plays a computer game with ninjas. At this stage Vasya's ninja should get out of a deep canyon.
The canyon consists of two vertical parallel walls, their height is n meters. Let's imagine that we split these walls into 1 meter-long areas and number them with positive integers from 1 to n from bottom to top. Some areas are safe and the ninja can climb them. Others are spiky and ninja can't be there. Let's call such areas dangerous.
Initially the ninja is on the lower area of the left wall. He can use each second to perform one of the following actions:
* climb one area up;
* climb one area down;
* jump to the opposite wall. That gets the ninja to the area that is exactly k meters higher than the area he jumped from. More formally, if before the jump the ninja is located at area x of one wall, then after the jump he is located at area x + k of the other wall.
If at some point of time the ninja tries to get to an area with a number larger than n, then we can assume that the ninja got out of the canyon.
The canyon gets flooded and each second the water level raises one meter. Initially the water level is at the lower border of the first area. Ninja cannot be on the area covered by water. We can assume that the ninja and the water "move in turns" β first the ninja performs some action, then the water raises for one meter, then the ninja performs one more action and so on.
The level is considered completed if the ninja manages to get out of the canyon.
After several failed attempts Vasya started to doubt whether it is possible to complete the level at all. Help him answer the question.
Input
The first line contains two integers n and k (1 β€ n, k β€ 105) β the height of the canyon and the height of ninja's jump, correspondingly.
The second line contains the description of the left wall β a string with the length of n characters. The i-th character represents the state of the i-th wall area: character "X" represents a dangerous area and character "-" represents a safe area.
The third line describes the right wall in the same format.
It is guaranteed that the first area of the left wall is not dangerous.
Output
Print "YES" (without the quotes) if the ninja can get out from the canyon, otherwise, print "NO" (without the quotes).
Examples
Input
7 3
---X--X
-X--XX-
Output
YES
Input
6 2
--X-X-
X--XX-
Output
NO
Note
In the first sample the ninja should first jump to the right wall, then go one meter down along the right wall, then jump to the left wall. The next jump can get the ninja from the canyon.
In the second sample there's no way the ninja can get out of the canyon.
Submitted Solution:
```
v, st = map(int, input().split())
l = list(input())
p = list(input())
bool_l = [False for i in range(v)]
bool_p = [False for i in range(v)]
def step(poz, pared):
if (poz >= v):
return True
else:
if (poz < 0):
return False
if (pared == 0 and l[poz] == 'X'):
return False
if (pared == 1 and p[poz] == 'X'):
return False
if (pared == 0 and bool_l[poz] == True):
return False
if (pared == 1 and bool_p[poz] == True):
return False
if (pared == 0):
bool_l[poz] = True
else:
bool_p[poz] = True
if (poz >= v):
return True
else:
return (step(poz+1, pared) or step(poz-1,pared) or step(poz+st, 1-pared))
if step(0,0) == True:
print("YES")
else:
print("NO")
``` | instruction | 0 | 6,909 | 8 | 13,818 |
No | output | 1 | 6,909 | 8 | 13,819 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are planning to build housing on a street. There are n spots available on the street on which you can build a house. The spots are labeled from 1 to n from left to right. In each spot, you can build a house with an integer height between 0 and h.
In each spot, if a house has height a, you will gain a^2 dollars from it.
The city has m zoning restrictions. The i-th restriction says that the tallest house from spots l_i to r_i (inclusive) must be at most x_i.
You would like to build houses to maximize your profit. Determine the maximum profit possible.
Input
The first line contains three integers n, h, and m (1 β€ n,h,m β€ 50) β the number of spots, the maximum height, and the number of restrictions.
Each of the next m lines contains three integers l_i, r_i, and x_i (1 β€ l_i β€ r_i β€ n, 0 β€ x_i β€ h) β left and right limits (inclusive) of the i-th restriction and the maximum possible height in that range.
Output
Print a single integer, the maximum profit you can make.
Examples
Input
3 3 3
1 1 1
2 2 3
3 3 2
Output
14
Input
4 10 2
2 3 8
3 4 7
Output
262
Note
In the first example, there are 3 houses, the maximum height of a house is 3, and there are 3 restrictions. The first restriction says the tallest house between 1 and 1 must be at most 1. The second restriction says the tallest house between 2 and 2 must be at most 3. The third restriction says the tallest house between 3 and 3 must be at most 2.
In this case, it is optimal to build houses with heights [1, 3, 2]. This fits within all the restrictions. The total profit in this case is 1^2 + 3^2 + 2^2 = 14.
In the second example, there are 4 houses, the maximum height of a house is 10, and there are 2 restrictions. The first restriction says the tallest house from 2 to 3 must be at most 8. The second restriction says the tallest house from 3 to 4 must be at most 7.
In this case, it's optimal to build houses with heights [10, 8, 7, 7]. We get a profit of 10^2+8^2+7^2+7^2 = 262. Note that there are two restrictions on house 3 and both of them must be satisfied. Also, note that even though there isn't any explicit restrictions on house 1, we must still limit its height to be at most 10 (h=10). | instruction | 0 | 7,630 | 8 | 15,260 |
Tags: implementation
Correct Solution:
```
n, h, m = list(map(int, input().split()))
A = [h] * n
for _ in range(m):
l, r, x = list(map(int, input().split()))
# print(l,r,x)
for i in range(l-1,r):
A[i] = min(A[i], x)
s = 0
for a in A:
s += a * a
# print(A)
print(s)
``` | output | 1 | 7,630 | 8 | 15,261 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are planning to build housing on a street. There are n spots available on the street on which you can build a house. The spots are labeled from 1 to n from left to right. In each spot, you can build a house with an integer height between 0 and h.
In each spot, if a house has height a, you will gain a^2 dollars from it.
The city has m zoning restrictions. The i-th restriction says that the tallest house from spots l_i to r_i (inclusive) must be at most x_i.
You would like to build houses to maximize your profit. Determine the maximum profit possible.
Input
The first line contains three integers n, h, and m (1 β€ n,h,m β€ 50) β the number of spots, the maximum height, and the number of restrictions.
Each of the next m lines contains three integers l_i, r_i, and x_i (1 β€ l_i β€ r_i β€ n, 0 β€ x_i β€ h) β left and right limits (inclusive) of the i-th restriction and the maximum possible height in that range.
Output
Print a single integer, the maximum profit you can make.
Examples
Input
3 3 3
1 1 1
2 2 3
3 3 2
Output
14
Input
4 10 2
2 3 8
3 4 7
Output
262
Note
In the first example, there are 3 houses, the maximum height of a house is 3, and there are 3 restrictions. The first restriction says the tallest house between 1 and 1 must be at most 1. The second restriction says the tallest house between 2 and 2 must be at most 3. The third restriction says the tallest house between 3 and 3 must be at most 2.
In this case, it is optimal to build houses with heights [1, 3, 2]. This fits within all the restrictions. The total profit in this case is 1^2 + 3^2 + 2^2 = 14.
In the second example, there are 4 houses, the maximum height of a house is 10, and there are 2 restrictions. The first restriction says the tallest house from 2 to 3 must be at most 8. The second restriction says the tallest house from 3 to 4 must be at most 7.
In this case, it's optimal to build houses with heights [10, 8, 7, 7]. We get a profit of 10^2+8^2+7^2+7^2 = 262. Note that there are two restrictions on house 3 and both of them must be satisfied. Also, note that even though there isn't any explicit restrictions on house 1, we must still limit its height to be at most 10 (h=10). | instruction | 0 | 7,631 | 8 | 15,262 |
Tags: implementation
Correct Solution:
```
import sys
input = sys.stdin.readline
n,h,m=list(map(int,input().split()))
a = [h]*n
for i in range(m):
b,c,d = list(map(int,input().split()))
a[b-1:c] = [min(x,d) for x in a[b-1:c]]
print(sum([x**2 for x in a]))
``` | output | 1 | 7,631 | 8 | 15,263 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are planning to build housing on a street. There are n spots available on the street on which you can build a house. The spots are labeled from 1 to n from left to right. In each spot, you can build a house with an integer height between 0 and h.
In each spot, if a house has height a, you will gain a^2 dollars from it.
The city has m zoning restrictions. The i-th restriction says that the tallest house from spots l_i to r_i (inclusive) must be at most x_i.
You would like to build houses to maximize your profit. Determine the maximum profit possible.
Input
The first line contains three integers n, h, and m (1 β€ n,h,m β€ 50) β the number of spots, the maximum height, and the number of restrictions.
Each of the next m lines contains three integers l_i, r_i, and x_i (1 β€ l_i β€ r_i β€ n, 0 β€ x_i β€ h) β left and right limits (inclusive) of the i-th restriction and the maximum possible height in that range.
Output
Print a single integer, the maximum profit you can make.
Examples
Input
3 3 3
1 1 1
2 2 3
3 3 2
Output
14
Input
4 10 2
2 3 8
3 4 7
Output
262
Note
In the first example, there are 3 houses, the maximum height of a house is 3, and there are 3 restrictions. The first restriction says the tallest house between 1 and 1 must be at most 1. The second restriction says the tallest house between 2 and 2 must be at most 3. The third restriction says the tallest house between 3 and 3 must be at most 2.
In this case, it is optimal to build houses with heights [1, 3, 2]. This fits within all the restrictions. The total profit in this case is 1^2 + 3^2 + 2^2 = 14.
In the second example, there are 4 houses, the maximum height of a house is 10, and there are 2 restrictions. The first restriction says the tallest house from 2 to 3 must be at most 8. The second restriction says the tallest house from 3 to 4 must be at most 7.
In this case, it's optimal to build houses with heights [10, 8, 7, 7]. We get a profit of 10^2+8^2+7^2+7^2 = 262. Note that there are two restrictions on house 3 and both of them must be satisfied. Also, note that even though there isn't any explicit restrictions on house 1, we must still limit its height to be at most 10 (h=10). | instruction | 0 | 7,632 | 8 | 15,264 |
Tags: implementation
Correct Solution:
```
n,h,m=map(int,input().split())
b=[h]*n
while m:
m-=1
l,r,x=map(int,input().split())
for i in range(l-1,r):
b[i]=min(b[i],x)
print(sum([x**2 for x in b]))
``` | output | 1 | 7,632 | 8 | 15,265 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are planning to build housing on a street. There are n spots available on the street on which you can build a house. The spots are labeled from 1 to n from left to right. In each spot, you can build a house with an integer height between 0 and h.
In each spot, if a house has height a, you will gain a^2 dollars from it.
The city has m zoning restrictions. The i-th restriction says that the tallest house from spots l_i to r_i (inclusive) must be at most x_i.
You would like to build houses to maximize your profit. Determine the maximum profit possible.
Input
The first line contains three integers n, h, and m (1 β€ n,h,m β€ 50) β the number of spots, the maximum height, and the number of restrictions.
Each of the next m lines contains three integers l_i, r_i, and x_i (1 β€ l_i β€ r_i β€ n, 0 β€ x_i β€ h) β left and right limits (inclusive) of the i-th restriction and the maximum possible height in that range.
Output
Print a single integer, the maximum profit you can make.
Examples
Input
3 3 3
1 1 1
2 2 3
3 3 2
Output
14
Input
4 10 2
2 3 8
3 4 7
Output
262
Note
In the first example, there are 3 houses, the maximum height of a house is 3, and there are 3 restrictions. The first restriction says the tallest house between 1 and 1 must be at most 1. The second restriction says the tallest house between 2 and 2 must be at most 3. The third restriction says the tallest house between 3 and 3 must be at most 2.
In this case, it is optimal to build houses with heights [1, 3, 2]. This fits within all the restrictions. The total profit in this case is 1^2 + 3^2 + 2^2 = 14.
In the second example, there are 4 houses, the maximum height of a house is 10, and there are 2 restrictions. The first restriction says the tallest house from 2 to 3 must be at most 8. The second restriction says the tallest house from 3 to 4 must be at most 7.
In this case, it's optimal to build houses with heights [10, 8, 7, 7]. We get a profit of 10^2+8^2+7^2+7^2 = 262. Note that there are two restrictions on house 3 and both of them must be satisfied. Also, note that even though there isn't any explicit restrictions on house 1, we must still limit its height to be at most 10 (h=10). | instruction | 0 | 7,633 | 8 | 15,266 |
Tags: implementation
Correct Solution:
```
n, h, m = map(int,input().split())
ans = [h for i in range(n+1)]
for i in range(m):
l, r, x = map(int,input().split())
for j in range(l,r+1):
ans[j] = min(ans[j],x)
sum = 0
for i in range(1,n+1):
sum += ans[i]*ans[i]
print(sum)
``` | output | 1 | 7,633 | 8 | 15,267 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are planning to build housing on a street. There are n spots available on the street on which you can build a house. The spots are labeled from 1 to n from left to right. In each spot, you can build a house with an integer height between 0 and h.
In each spot, if a house has height a, you will gain a^2 dollars from it.
The city has m zoning restrictions. The i-th restriction says that the tallest house from spots l_i to r_i (inclusive) must be at most x_i.
You would like to build houses to maximize your profit. Determine the maximum profit possible.
Input
The first line contains three integers n, h, and m (1 β€ n,h,m β€ 50) β the number of spots, the maximum height, and the number of restrictions.
Each of the next m lines contains three integers l_i, r_i, and x_i (1 β€ l_i β€ r_i β€ n, 0 β€ x_i β€ h) β left and right limits (inclusive) of the i-th restriction and the maximum possible height in that range.
Output
Print a single integer, the maximum profit you can make.
Examples
Input
3 3 3
1 1 1
2 2 3
3 3 2
Output
14
Input
4 10 2
2 3 8
3 4 7
Output
262
Note
In the first example, there are 3 houses, the maximum height of a house is 3, and there are 3 restrictions. The first restriction says the tallest house between 1 and 1 must be at most 1. The second restriction says the tallest house between 2 and 2 must be at most 3. The third restriction says the tallest house between 3 and 3 must be at most 2.
In this case, it is optimal to build houses with heights [1, 3, 2]. This fits within all the restrictions. The total profit in this case is 1^2 + 3^2 + 2^2 = 14.
In the second example, there are 4 houses, the maximum height of a house is 10, and there are 2 restrictions. The first restriction says the tallest house from 2 to 3 must be at most 8. The second restriction says the tallest house from 3 to 4 must be at most 7.
In this case, it's optimal to build houses with heights [10, 8, 7, 7]. We get a profit of 10^2+8^2+7^2+7^2 = 262. Note that there are two restrictions on house 3 and both of them must be satisfied. Also, note that even though there isn't any explicit restrictions on house 1, we must still limit its height to be at most 10 (h=10). | instruction | 0 | 7,634 | 8 | 15,268 |
Tags: implementation
Correct Solution:
```
n,h,m=map(int,input().split())
a=[h for i in range(n)]
for i in range(m):
l,r,x=map(int,input().split())
for j in range(l-1,r):
a[j]=min(a[j],x)
sum=0
for i in range(len(a)):
sum=sum+a[i]*a[i]
print(sum)
``` | output | 1 | 7,634 | 8 | 15,269 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are planning to build housing on a street. There are n spots available on the street on which you can build a house. The spots are labeled from 1 to n from left to right. In each spot, you can build a house with an integer height between 0 and h.
In each spot, if a house has height a, you will gain a^2 dollars from it.
The city has m zoning restrictions. The i-th restriction says that the tallest house from spots l_i to r_i (inclusive) must be at most x_i.
You would like to build houses to maximize your profit. Determine the maximum profit possible.
Input
The first line contains three integers n, h, and m (1 β€ n,h,m β€ 50) β the number of spots, the maximum height, and the number of restrictions.
Each of the next m lines contains three integers l_i, r_i, and x_i (1 β€ l_i β€ r_i β€ n, 0 β€ x_i β€ h) β left and right limits (inclusive) of the i-th restriction and the maximum possible height in that range.
Output
Print a single integer, the maximum profit you can make.
Examples
Input
3 3 3
1 1 1
2 2 3
3 3 2
Output
14
Input
4 10 2
2 3 8
3 4 7
Output
262
Note
In the first example, there are 3 houses, the maximum height of a house is 3, and there are 3 restrictions. The first restriction says the tallest house between 1 and 1 must be at most 1. The second restriction says the tallest house between 2 and 2 must be at most 3. The third restriction says the tallest house between 3 and 3 must be at most 2.
In this case, it is optimal to build houses with heights [1, 3, 2]. This fits within all the restrictions. The total profit in this case is 1^2 + 3^2 + 2^2 = 14.
In the second example, there are 4 houses, the maximum height of a house is 10, and there are 2 restrictions. The first restriction says the tallest house from 2 to 3 must be at most 8. The second restriction says the tallest house from 3 to 4 must be at most 7.
In this case, it's optimal to build houses with heights [10, 8, 7, 7]. We get a profit of 10^2+8^2+7^2+7^2 = 262. Note that there are two restrictions on house 3 and both of them must be satisfied. Also, note that even though there isn't any explicit restrictions on house 1, we must still limit its height to be at most 10 (h=10). | instruction | 0 | 7,635 | 8 | 15,270 |
Tags: implementation
Correct Solution:
```
n,h,m=map(int,input().split())
import math
ar=[math.inf]*n
for i in range(m):
l,r,x=map(int,input().split())
for k in range(l-1,r):
ar[k]=min(ar[k],x)
for i in range(len(ar)):
if ar[i]==math.inf:
ar[i]=h
s=0
for i in ar:
s=s+i**2
print(s)
``` | output | 1 | 7,635 | 8 | 15,271 |
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