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.
Dima's got a staircase that consists of n stairs. The first stair is at height a1, the second one is at a2, the last one is at an (1 β€ a1 β€ a2 β€ ... β€ an).
Dima decided to play with the staircase, so he is throwing rectangular boxes at the staircase from above. The i-th box has width wi and height hi. Dima throws each box vertically down on the first wi stairs of the staircase, that is, the box covers stairs with numbers 1, 2, ..., wi. Each thrown box flies vertically down until at least one of the two following events happen:
* the bottom of the box touches the top of a stair;
* the bottom of the box touches the top of a box, thrown earlier.
We only consider touching of the horizontal sides of stairs and boxes, at that touching with the corners isn't taken into consideration. Specifically, that implies that a box with width wi cannot touch the stair number wi + 1.
You are given the description of the staircase and the sequence in which Dima threw the boxes at it. For each box, determine how high the bottom of the box after landing will be. Consider a box to fall after the previous one lands.
Input
The first line contains integer n (1 β€ n β€ 105) β the number of stairs in the staircase. The second line contains a non-decreasing sequence, consisting of n integers, a1, a2, ..., an (1 β€ ai β€ 109; ai β€ ai + 1).
The next line contains integer m (1 β€ m β€ 105) β the number of boxes. Each of the following m lines contains a pair of integers wi, hi (1 β€ wi β€ n; 1 β€ hi β€ 109) β the size of the i-th thrown box.
The numbers in the lines are separated by spaces.
Output
Print m integers β for each box the height, where the bottom of the box will be after landing. Print the answers for the boxes in the order, in which the boxes are given in the input.
Please, do not use the %lld specifier to read or write 64-bit integers in C++. It is preferred to use the cin, cout streams or the %I64d specifier.
Examples
Input
5
1 2 3 6 6
4
1 1
3 1
1 1
4 3
Output
1
3
4
6
Input
3
1 2 3
2
1 1
3 1
Output
1
3
Input
1
1
5
1 2
1 10
1 10
1 10
1 10
Output
1
3
13
23
33
Note
The first sample are shown on the picture.
<image>
Submitted Solution:
```
import sys
input = sys.stdin.readline
n, staircase = int(input()), [int(i) for i in input().split()]
boxes, best = [[int(i) for i in input().split()] for _ in range(int(input()))], 0
for box in boxes:
best = max(staircase[box[0] - 1], best)
print(best)
best += box[1] # only track leftmost stair
``` | instruction | 0 | 104,760 | 8 | 209,520 |
Yes | output | 1 | 104,760 | 8 | 209,521 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Dima's got a staircase that consists of n stairs. The first stair is at height a1, the second one is at a2, the last one is at an (1 β€ a1 β€ a2 β€ ... β€ an).
Dima decided to play with the staircase, so he is throwing rectangular boxes at the staircase from above. The i-th box has width wi and height hi. Dima throws each box vertically down on the first wi stairs of the staircase, that is, the box covers stairs with numbers 1, 2, ..., wi. Each thrown box flies vertically down until at least one of the two following events happen:
* the bottom of the box touches the top of a stair;
* the bottom of the box touches the top of a box, thrown earlier.
We only consider touching of the horizontal sides of stairs and boxes, at that touching with the corners isn't taken into consideration. Specifically, that implies that a box with width wi cannot touch the stair number wi + 1.
You are given the description of the staircase and the sequence in which Dima threw the boxes at it. For each box, determine how high the bottom of the box after landing will be. Consider a box to fall after the previous one lands.
Input
The first line contains integer n (1 β€ n β€ 105) β the number of stairs in the staircase. The second line contains a non-decreasing sequence, consisting of n integers, a1, a2, ..., an (1 β€ ai β€ 109; ai β€ ai + 1).
The next line contains integer m (1 β€ m β€ 105) β the number of boxes. Each of the following m lines contains a pair of integers wi, hi (1 β€ wi β€ n; 1 β€ hi β€ 109) β the size of the i-th thrown box.
The numbers in the lines are separated by spaces.
Output
Print m integers β for each box the height, where the bottom of the box will be after landing. Print the answers for the boxes in the order, in which the boxes are given in the input.
Please, do not use the %lld specifier to read or write 64-bit integers in C++. It is preferred to use the cin, cout streams or the %I64d specifier.
Examples
Input
5
1 2 3 6 6
4
1 1
3 1
1 1
4 3
Output
1
3
4
6
Input
3
1 2 3
2
1 1
3 1
Output
1
3
Input
1
1
5
1 2
1 10
1 10
1 10
1 10
Output
1
3
13
23
33
Note
The first sample are shown on the picture.
<image>
Submitted Solution:
```
n = int(input())
a = list(map(int,input().split()))
m = int(input())
l = [list(map(int,input().split())) for _ in range(m)]
ans = [a[l[0][0]-1]]
lasth = l[0][1]
for _,[w,h] in enumerate(l[1:]):
ans.append(max(ans[-1] + lasth,a[w-1]))
lasth = h
print(*ans,sep='\n')
``` | instruction | 0 | 104,761 | 8 | 209,522 |
Yes | output | 1 | 104,761 | 8 | 209,523 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Dima's got a staircase that consists of n stairs. The first stair is at height a1, the second one is at a2, the last one is at an (1 β€ a1 β€ a2 β€ ... β€ an).
Dima decided to play with the staircase, so he is throwing rectangular boxes at the staircase from above. The i-th box has width wi and height hi. Dima throws each box vertically down on the first wi stairs of the staircase, that is, the box covers stairs with numbers 1, 2, ..., wi. Each thrown box flies vertically down until at least one of the two following events happen:
* the bottom of the box touches the top of a stair;
* the bottom of the box touches the top of a box, thrown earlier.
We only consider touching of the horizontal sides of stairs and boxes, at that touching with the corners isn't taken into consideration. Specifically, that implies that a box with width wi cannot touch the stair number wi + 1.
You are given the description of the staircase and the sequence in which Dima threw the boxes at it. For each box, determine how high the bottom of the box after landing will be. Consider a box to fall after the previous one lands.
Input
The first line contains integer n (1 β€ n β€ 105) β the number of stairs in the staircase. The second line contains a non-decreasing sequence, consisting of n integers, a1, a2, ..., an (1 β€ ai β€ 109; ai β€ ai + 1).
The next line contains integer m (1 β€ m β€ 105) β the number of boxes. Each of the following m lines contains a pair of integers wi, hi (1 β€ wi β€ n; 1 β€ hi β€ 109) β the size of the i-th thrown box.
The numbers in the lines are separated by spaces.
Output
Print m integers β for each box the height, where the bottom of the box will be after landing. Print the answers for the boxes in the order, in which the boxes are given in the input.
Please, do not use the %lld specifier to read or write 64-bit integers in C++. It is preferred to use the cin, cout streams or the %I64d specifier.
Examples
Input
5
1 2 3 6 6
4
1 1
3 1
1 1
4 3
Output
1
3
4
6
Input
3
1 2 3
2
1 1
3 1
Output
1
3
Input
1
1
5
1 2
1 10
1 10
1 10
1 10
Output
1
3
13
23
33
Note
The first sample are shown on the picture.
<image>
Submitted Solution:
```
__author__ = "runekri3"
stairs_amount = int(input())
stair_heights = list(map(int, input().split()))
boxes_amount = int(input())
boxes = []
for _ in range(boxes_amount):
boxes.append(list(map(int, input().split())))
for width, height in boxes:
box_bottom = max(stair_heights[0], stair_heights[width - 1])
print(box_bottom)
stair_heights[0] = box_bottom + height
``` | instruction | 0 | 104,762 | 8 | 209,524 |
Yes | output | 1 | 104,762 | 8 | 209,525 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Dima's got a staircase that consists of n stairs. The first stair is at height a1, the second one is at a2, the last one is at an (1 β€ a1 β€ a2 β€ ... β€ an).
Dima decided to play with the staircase, so he is throwing rectangular boxes at the staircase from above. The i-th box has width wi and height hi. Dima throws each box vertically down on the first wi stairs of the staircase, that is, the box covers stairs with numbers 1, 2, ..., wi. Each thrown box flies vertically down until at least one of the two following events happen:
* the bottom of the box touches the top of a stair;
* the bottom of the box touches the top of a box, thrown earlier.
We only consider touching of the horizontal sides of stairs and boxes, at that touching with the corners isn't taken into consideration. Specifically, that implies that a box with width wi cannot touch the stair number wi + 1.
You are given the description of the staircase and the sequence in which Dima threw the boxes at it. For each box, determine how high the bottom of the box after landing will be. Consider a box to fall after the previous one lands.
Input
The first line contains integer n (1 β€ n β€ 105) β the number of stairs in the staircase. The second line contains a non-decreasing sequence, consisting of n integers, a1, a2, ..., an (1 β€ ai β€ 109; ai β€ ai + 1).
The next line contains integer m (1 β€ m β€ 105) β the number of boxes. Each of the following m lines contains a pair of integers wi, hi (1 β€ wi β€ n; 1 β€ hi β€ 109) β the size of the i-th thrown box.
The numbers in the lines are separated by spaces.
Output
Print m integers β for each box the height, where the bottom of the box will be after landing. Print the answers for the boxes in the order, in which the boxes are given in the input.
Please, do not use the %lld specifier to read or write 64-bit integers in C++. It is preferred to use the cin, cout streams or the %I64d specifier.
Examples
Input
5
1 2 3 6 6
4
1 1
3 1
1 1
4 3
Output
1
3
4
6
Input
3
1 2 3
2
1 1
3 1
Output
1
3
Input
1
1
5
1 2
1 10
1 10
1 10
1 10
Output
1
3
13
23
33
Note
The first sample are shown on the picture.
<image>
Submitted Solution:
```
from sys import stdin, stdout
number = int(stdin.readline())
stair_case = list(map(int, stdin.readline().split()))
boxes = int(stdin.readline())
current = []
levels = {}
mx_width = 0
cnt = 0
for index, x in enumerate(stair_case):
levels[index + 1] = x
for x in range(boxes):
w, h = map(int, input().split())
if w > mx_width and cnt < stair_case[-1]:
cnt = levels[w]
stdout.write(str(cnt) + '\n')
cnt += h
mx_width = w
else:
stdout.write(str(cnt) + '\n')
cnt += h
``` | instruction | 0 | 104,763 | 8 | 209,526 |
No | output | 1 | 104,763 | 8 | 209,527 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Dima's got a staircase that consists of n stairs. The first stair is at height a1, the second one is at a2, the last one is at an (1 β€ a1 β€ a2 β€ ... β€ an).
Dima decided to play with the staircase, so he is throwing rectangular boxes at the staircase from above. The i-th box has width wi and height hi. Dima throws each box vertically down on the first wi stairs of the staircase, that is, the box covers stairs with numbers 1, 2, ..., wi. Each thrown box flies vertically down until at least one of the two following events happen:
* the bottom of the box touches the top of a stair;
* the bottom of the box touches the top of a box, thrown earlier.
We only consider touching of the horizontal sides of stairs and boxes, at that touching with the corners isn't taken into consideration. Specifically, that implies that a box with width wi cannot touch the stair number wi + 1.
You are given the description of the staircase and the sequence in which Dima threw the boxes at it. For each box, determine how high the bottom of the box after landing will be. Consider a box to fall after the previous one lands.
Input
The first line contains integer n (1 β€ n β€ 105) β the number of stairs in the staircase. The second line contains a non-decreasing sequence, consisting of n integers, a1, a2, ..., an (1 β€ ai β€ 109; ai β€ ai + 1).
The next line contains integer m (1 β€ m β€ 105) β the number of boxes. Each of the following m lines contains a pair of integers wi, hi (1 β€ wi β€ n; 1 β€ hi β€ 109) β the size of the i-th thrown box.
The numbers in the lines are separated by spaces.
Output
Print m integers β for each box the height, where the bottom of the box will be after landing. Print the answers for the boxes in the order, in which the boxes are given in the input.
Please, do not use the %lld specifier to read or write 64-bit integers in C++. It is preferred to use the cin, cout streams or the %I64d specifier.
Examples
Input
5
1 2 3 6 6
4
1 1
3 1
1 1
4 3
Output
1
3
4
6
Input
3
1 2 3
2
1 1
3 1
Output
1
3
Input
1
1
5
1 2
1 10
1 10
1 10
1 10
Output
1
3
13
23
33
Note
The first sample are shown on the picture.
<image>
Submitted Solution:
```
n = int(input())
a = list(map(int, input().split()))
m = int(input())
boxes = []
for i in range(m):
w, h = list(map(int, input().split()))
boxes.append([w,h])
for i in range(m):
w, h = boxes[i]
t = a[w-1]
print(t)
for i in range(w):
a[i] = t+h
``` | instruction | 0 | 104,764 | 8 | 209,528 |
No | output | 1 | 104,764 | 8 | 209,529 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Dima's got a staircase that consists of n stairs. The first stair is at height a1, the second one is at a2, the last one is at an (1 β€ a1 β€ a2 β€ ... β€ an).
Dima decided to play with the staircase, so he is throwing rectangular boxes at the staircase from above. The i-th box has width wi and height hi. Dima throws each box vertically down on the first wi stairs of the staircase, that is, the box covers stairs with numbers 1, 2, ..., wi. Each thrown box flies vertically down until at least one of the two following events happen:
* the bottom of the box touches the top of a stair;
* the bottom of the box touches the top of a box, thrown earlier.
We only consider touching of the horizontal sides of stairs and boxes, at that touching with the corners isn't taken into consideration. Specifically, that implies that a box with width wi cannot touch the stair number wi + 1.
You are given the description of the staircase and the sequence in which Dima threw the boxes at it. For each box, determine how high the bottom of the box after landing will be. Consider a box to fall after the previous one lands.
Input
The first line contains integer n (1 β€ n β€ 105) β the number of stairs in the staircase. The second line contains a non-decreasing sequence, consisting of n integers, a1, a2, ..., an (1 β€ ai β€ 109; ai β€ ai + 1).
The next line contains integer m (1 β€ m β€ 105) β the number of boxes. Each of the following m lines contains a pair of integers wi, hi (1 β€ wi β€ n; 1 β€ hi β€ 109) β the size of the i-th thrown box.
The numbers in the lines are separated by spaces.
Output
Print m integers β for each box the height, where the bottom of the box will be after landing. Print the answers for the boxes in the order, in which the boxes are given in the input.
Please, do not use the %lld specifier to read or write 64-bit integers in C++. It is preferred to use the cin, cout streams or the %I64d specifier.
Examples
Input
5
1 2 3 6 6
4
1 1
3 1
1 1
4 3
Output
1
3
4
6
Input
3
1 2 3
2
1 1
3 1
Output
1
3
Input
1
1
5
1 2
1 10
1 10
1 10
1 10
Output
1
3
13
23
33
Note
The first sample are shown on the picture.
<image>
Submitted Solution:
```
#pyrival orz
import os
import sys
from io import BytesIO, IOBase
input = sys.stdin.readline
############ ---- Input Functions ---- ############
def inp():
return(int(input()))
def inlt():
return(list(map(int,input().split())))
def insr():
s = input()
return(list(s[:len(s) - 1]))
def invr():
return(map(int,input().split()))
def gcd(a, b):
if b == 0:
return a
else:
return gcd(b, a%b)
def seive(n):
primes = [True]*(n+1)
for i in range(2, n):
if not primes[i]:
continue
j = 2*i
while j <= n:
primes[j] = False
j += i
return primes
def factors(n):
factors = []
x = 2
while x*x <= n:
while n % x == 0:
factors.append(x)
n //= x
if n > 1:
factors.append(x)
return factors
# Functions: list of factors, seive of primes, gcd of two numbers
def main():
try:
n = inp()
a = inlt()
m = inp()
mn = a[0]
for i in range(m):
w, h = invr()
print(max(mn,a[w-1]))
mn = a[w-1] + h
a[w-1] = max(mn, a[w-1])
except Exception as e:
print(e)
# region fastio
BUFSIZE = 8192
class FastIO(IOBase):
newlines = 0
def __init__(self, file):
self._fd = file.fileno()
self.buffer = BytesIO()
self.writable = "x" in file.mode or "r" not in file.mode
self.write = self.buffer.write if self.writable else None
def read(self):
while True:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
if not b:
break
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines = 0
return self.buffer.read()
def readline(self):
while self.newlines == 0:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
self.newlines = b.count(b"\n") + (not b)
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines -= 1
return self.buffer.readline()
def flush(self):
if self.writable:
os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
self.write = lambda s: self.buffer.write(s.encode("ascii"))
self.read = lambda: self.buffer.read().decode("ascii")
self.readline = lambda: self.buffer.readline().decode("ascii")
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")
# endregion
if __name__ == "__main__":
main()
``` | instruction | 0 | 104,765 | 8 | 209,530 |
No | output | 1 | 104,765 | 8 | 209,531 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Dima's got a staircase that consists of n stairs. The first stair is at height a1, the second one is at a2, the last one is at an (1 β€ a1 β€ a2 β€ ... β€ an).
Dima decided to play with the staircase, so he is throwing rectangular boxes at the staircase from above. The i-th box has width wi and height hi. Dima throws each box vertically down on the first wi stairs of the staircase, that is, the box covers stairs with numbers 1, 2, ..., wi. Each thrown box flies vertically down until at least one of the two following events happen:
* the bottom of the box touches the top of a stair;
* the bottom of the box touches the top of a box, thrown earlier.
We only consider touching of the horizontal sides of stairs and boxes, at that touching with the corners isn't taken into consideration. Specifically, that implies that a box with width wi cannot touch the stair number wi + 1.
You are given the description of the staircase and the sequence in which Dima threw the boxes at it. For each box, determine how high the bottom of the box after landing will be. Consider a box to fall after the previous one lands.
Input
The first line contains integer n (1 β€ n β€ 105) β the number of stairs in the staircase. The second line contains a non-decreasing sequence, consisting of n integers, a1, a2, ..., an (1 β€ ai β€ 109; ai β€ ai + 1).
The next line contains integer m (1 β€ m β€ 105) β the number of boxes. Each of the following m lines contains a pair of integers wi, hi (1 β€ wi β€ n; 1 β€ hi β€ 109) β the size of the i-th thrown box.
The numbers in the lines are separated by spaces.
Output
Print m integers β for each box the height, where the bottom of the box will be after landing. Print the answers for the boxes in the order, in which the boxes are given in the input.
Please, do not use the %lld specifier to read or write 64-bit integers in C++. It is preferred to use the cin, cout streams or the %I64d specifier.
Examples
Input
5
1 2 3 6 6
4
1 1
3 1
1 1
4 3
Output
1
3
4
6
Input
3
1 2 3
2
1 1
3 1
Output
1
3
Input
1
1
5
1 2
1 10
1 10
1 10
1 10
Output
1
3
13
23
33
Note
The first sample are shown on the picture.
<image>
Submitted Solution:
```
#pyrival orz
import os
import sys
from io import BytesIO, IOBase
input = sys.stdin.readline
############ ---- Input Functions ---- ############
def inp():
return(int(input()))
def inlt():
return(list(map(int,input().split())))
def insr():
s = input()
return(list(s[:len(s) - 1]))
def invr():
return(map(int,input().split()))
def gcd(a, b):
if b == 0:
return a
else:
return gcd(b, a%b)
def seive(n):
primes = [True]*(n+1)
for i in range(2, n):
if not primes[i]:
continue
j = 2*i
while j <= n:
primes[j] = False
j += i
return primes
def factors(n):
factors = []
x = 2
while x*x <= n:
while n % x == 0:
factors.append(x)
n //= x
if n > 1:
factors.append(x)
return factors
# Functions: list of factors, seive of primes, gcd of two numbers
def main():
try:
n = inp()
a = inlt()
m = inp()
mn = a[0]
for i in range(m):
w, h = invr()
print(max(mn,a[w-1]))
mn = a[w-1] + h
a[w-1] = mn
except Exception as e:
print(e)
# region fastio
BUFSIZE = 8192
class FastIO(IOBase):
newlines = 0
def __init__(self, file):
self._fd = file.fileno()
self.buffer = BytesIO()
self.writable = "x" in file.mode or "r" not in file.mode
self.write = self.buffer.write if self.writable else None
def read(self):
while True:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
if not b:
break
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines = 0
return self.buffer.read()
def readline(self):
while self.newlines == 0:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
self.newlines = b.count(b"\n") + (not b)
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines -= 1
return self.buffer.readline()
def flush(self):
if self.writable:
os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
self.write = lambda s: self.buffer.write(s.encode("ascii"))
self.read = lambda: self.buffer.read().decode("ascii")
self.readline = lambda: self.buffer.readline().decode("ascii")
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")
# endregion
if __name__ == "__main__":
main()
``` | instruction | 0 | 104,766 | 8 | 209,532 |
No | output | 1 | 104,766 | 8 | 209,533 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Kalila and Dimna are two jackals living in a huge jungle. One day they decided to join a logging factory in order to make money.
The manager of logging factory wants them to go to the jungle and cut n trees with heights a1, a2, ..., an. They bought a chain saw from a shop. Each time they use the chain saw on the tree number i, they can decrease the height of this tree by one unit. Each time that Kalila and Dimna use the chain saw, they need to recharge it. Cost of charging depends on the id of the trees which have been cut completely (a tree is cut completely if its height equal to 0). If the maximum id of a tree which has been cut completely is i (the tree that have height ai in the beginning), then the cost of charging the chain saw would be bi. If no tree is cut completely, Kalila and Dimna cannot charge the chain saw. The chainsaw is charged in the beginning. We know that for each i < j, ai < aj and bi > bj and also bn = 0 and a1 = 1. Kalila and Dimna want to cut all the trees completely, with minimum cost.
They want you to help them! Will you?
Input
The first line of input contains an integer n (1 β€ n β€ 105). The second line of input contains n integers a1, a2, ..., an (1 β€ ai β€ 109). The third line of input contains n integers b1, b2, ..., bn (0 β€ bi β€ 109).
It's guaranteed that a1 = 1, bn = 0, a1 < a2 < ... < an and b1 > b2 > ... > bn.
Output
The only line of output must contain the minimum cost of cutting all the trees completely.
Please, do not write the %lld specifier to read or write 64-bit integers in Π‘++. It is preferred to use the cin, cout streams or the %I64d specifier.
Examples
Input
5
1 2 3 4 5
5 4 3 2 0
Output
25
Input
6
1 2 3 10 20 30
6 5 4 3 2 0
Output
138 | instruction | 0 | 104,767 | 8 | 209,534 |
Tags: dp, geometry
Correct Solution:
```
N = 100100
def main():
n = int(input())
d=[0]*N
a = list(map(int, input().split()))
b = list(map(int, input().split()))
m,c=[0]*N,[0]*N
m[0] = b[0]
c[0] = d[0]
z = 1
p = 0
for i in range(1,n):
p = min(p,z)
while(p+1<z and m[p+1]*a[i]+c[p+1]<=m[p]*a[i]+c[p]):
p=p+1
d[i] = m[p]*a[i]+c[p]
while(z>=2 and (c[z-2]-c[z-1])*(b[i]-m[z-1])>=(c[z-1]-d[i])*(m[z-1]-m[z-2]) ):
z=z-1
m[z],c[z]=b[i],d[i]
z=z+1
#for j in range(n):print(d[j])
print(d[n-1])
main()
# Made By Mostafa_Khaled
``` | output | 1 | 104,767 | 8 | 209,535 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Kalila and Dimna are two jackals living in a huge jungle. One day they decided to join a logging factory in order to make money.
The manager of logging factory wants them to go to the jungle and cut n trees with heights a1, a2, ..., an. They bought a chain saw from a shop. Each time they use the chain saw on the tree number i, they can decrease the height of this tree by one unit. Each time that Kalila and Dimna use the chain saw, they need to recharge it. Cost of charging depends on the id of the trees which have been cut completely (a tree is cut completely if its height equal to 0). If the maximum id of a tree which has been cut completely is i (the tree that have height ai in the beginning), then the cost of charging the chain saw would be bi. If no tree is cut completely, Kalila and Dimna cannot charge the chain saw. The chainsaw is charged in the beginning. We know that for each i < j, ai < aj and bi > bj and also bn = 0 and a1 = 1. Kalila and Dimna want to cut all the trees completely, with minimum cost.
They want you to help them! Will you?
Input
The first line of input contains an integer n (1 β€ n β€ 105). The second line of input contains n integers a1, a2, ..., an (1 β€ ai β€ 109). The third line of input contains n integers b1, b2, ..., bn (0 β€ bi β€ 109).
It's guaranteed that a1 = 1, bn = 0, a1 < a2 < ... < an and b1 > b2 > ... > bn.
Output
The only line of output must contain the minimum cost of cutting all the trees completely.
Please, do not write the %lld specifier to read or write 64-bit integers in Π‘++. It is preferred to use the cin, cout streams or the %I64d specifier.
Examples
Input
5
1 2 3 4 5
5 4 3 2 0
Output
25
Input
6
1 2 3 10 20 30
6 5 4 3 2 0
Output
138 | instruction | 0 | 104,768 | 8 | 209,536 |
Tags: dp, geometry
Correct Solution:
```
read = lambda: map(int, input().split())
n = int(input())
a = list(read())
b = list(read())
dp = [0] * n
ch = [0]
def get(i, x):
return b[i] * x + dp[i]
def f1():
if len(ch) < 2:
return 0
return get(ch[0], a[i]) >= get(ch[1], a[i])
def f2():
if len(ch) < 2:
return 0
i1 = ch[-1]
x = (dp[i1] - dp[i]) / (b[i] - b[i1])
return get(ch[-2], x) <= get(i, x)
for i in range(1, n):
while f1():
ch.pop(0)
dp[i] = get(ch[0], a[i])
while f2():
ch.pop()
ch.append(i)
print(dp[n - 1])
``` | output | 1 | 104,768 | 8 | 209,537 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Kalila and Dimna are two jackals living in a huge jungle. One day they decided to join a logging factory in order to make money.
The manager of logging factory wants them to go to the jungle and cut n trees with heights a1, a2, ..., an. They bought a chain saw from a shop. Each time they use the chain saw on the tree number i, they can decrease the height of this tree by one unit. Each time that Kalila and Dimna use the chain saw, they need to recharge it. Cost of charging depends on the id of the trees which have been cut completely (a tree is cut completely if its height equal to 0). If the maximum id of a tree which has been cut completely is i (the tree that have height ai in the beginning), then the cost of charging the chain saw would be bi. If no tree is cut completely, Kalila and Dimna cannot charge the chain saw. The chainsaw is charged in the beginning. We know that for each i < j, ai < aj and bi > bj and also bn = 0 and a1 = 1. Kalila and Dimna want to cut all the trees completely, with minimum cost.
They want you to help them! Will you?
Input
The first line of input contains an integer n (1 β€ n β€ 105). The second line of input contains n integers a1, a2, ..., an (1 β€ ai β€ 109). The third line of input contains n integers b1, b2, ..., bn (0 β€ bi β€ 109).
It's guaranteed that a1 = 1, bn = 0, a1 < a2 < ... < an and b1 > b2 > ... > bn.
Output
The only line of output must contain the minimum cost of cutting all the trees completely.
Please, do not write the %lld specifier to read or write 64-bit integers in Π‘++. It is preferred to use the cin, cout streams or the %I64d specifier.
Examples
Input
5
1 2 3 4 5
5 4 3 2 0
Output
25
Input
6
1 2 3 10 20 30
6 5 4 3 2 0
Output
138 | instruction | 0 | 104,769 | 8 | 209,538 |
Tags: dp, geometry
Correct Solution:
```
n = int(input())
a = list(map(int, input().split(' ')))
b = list(map(int, input().split(' ')))
c = [0]*n
st = [0]*n
first=0
second=0
for i in range(1,n):
#Delete from front until the value of lines is increasing w.r.to value a[i]
while(second-first>0 and a[i] * b[st[first]] + c[st[first]] >= a[i] * b[st[first+1]] + c[st[first+1]]):
first = first + 1;
c[i] = a[i] * b[st[first]] + c[st[first]];
#Delete from back until the intersection of lines last two lines in stack is to the left of the new line and the last line
while(second-first>0 and (b[st[second]] - b[i])*(c[st[second]] - c[st[second-1]]) > (c[i] - c[st[second]])*(b[st[second-1]] - b[st[second]])):
second = second-1;
second = second+1
st[second] = i;
print(c[n-1]);
``` | output | 1 | 104,769 | 8 | 209,539 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Kalila and Dimna are two jackals living in a huge jungle. One day they decided to join a logging factory in order to make money.
The manager of logging factory wants them to go to the jungle and cut n trees with heights a1, a2, ..., an. They bought a chain saw from a shop. Each time they use the chain saw on the tree number i, they can decrease the height of this tree by one unit. Each time that Kalila and Dimna use the chain saw, they need to recharge it. Cost of charging depends on the id of the trees which have been cut completely (a tree is cut completely if its height equal to 0). If the maximum id of a tree which has been cut completely is i (the tree that have height ai in the beginning), then the cost of charging the chain saw would be bi. If no tree is cut completely, Kalila and Dimna cannot charge the chain saw. The chainsaw is charged in the beginning. We know that for each i < j, ai < aj and bi > bj and also bn = 0 and a1 = 1. Kalila and Dimna want to cut all the trees completely, with minimum cost.
They want you to help them! Will you?
Input
The first line of input contains an integer n (1 β€ n β€ 105). The second line of input contains n integers a1, a2, ..., an (1 β€ ai β€ 109). The third line of input contains n integers b1, b2, ..., bn (0 β€ bi β€ 109).
It's guaranteed that a1 = 1, bn = 0, a1 < a2 < ... < an and b1 > b2 > ... > bn.
Output
The only line of output must contain the minimum cost of cutting all the trees completely.
Please, do not write the %lld specifier to read or write 64-bit integers in Π‘++. It is preferred to use the cin, cout streams or the %I64d specifier.
Examples
Input
5
1 2 3 4 5
5 4 3 2 0
Output
25
Input
6
1 2 3 10 20 30
6 5 4 3 2 0
Output
138 | instruction | 0 | 104,770 | 8 | 209,540 |
Tags: dp, geometry
Correct Solution:
```
n=int(input())
A=list(map(int,input().strip().split()))
B=list(map(int,input().strip().split()))
def tree_cutting(n,A,B):
C=[0 for _ in range(n)]
hullx=[0 for _ in range(n)]
hully=[0 for _ in range(n)]
sz=0
p=0
C[0]=0
hullx[0]=B[0]
hully[0]=C[0]
for i in range(1,n):
p=min(sz,p)
while sz>0 and p<sz and (hully[p+1]-hully[p])/(hullx[p]-hullx[p+1])<=A[i]:
p+=1
C[i]=hully[p]+hullx[p]*A[i]
sz+=1
hullx[sz]=B[i]
hully[sz]=C[i]
while sz>1 and (hully[sz-2]-hully[sz-1])/(hullx[sz-1]-hullx[sz-2])>=(hully[sz-2]-hully[sz])/(hullx[sz]-hullx[sz-2]):
hullx[sz-1]=hullx[sz]
hully[sz-1]=hully[sz]
sz-=1
return C[n-1]
print(tree_cutting(n,A,B))
``` | output | 1 | 104,770 | 8 | 209,541 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Kalila and Dimna are two jackals living in a huge jungle. One day they decided to join a logging factory in order to make money.
The manager of logging factory wants them to go to the jungle and cut n trees with heights a1, a2, ..., an. They bought a chain saw from a shop. Each time they use the chain saw on the tree number i, they can decrease the height of this tree by one unit. Each time that Kalila and Dimna use the chain saw, they need to recharge it. Cost of charging depends on the id of the trees which have been cut completely (a tree is cut completely if its height equal to 0). If the maximum id of a tree which has been cut completely is i (the tree that have height ai in the beginning), then the cost of charging the chain saw would be bi. If no tree is cut completely, Kalila and Dimna cannot charge the chain saw. The chainsaw is charged in the beginning. We know that for each i < j, ai < aj and bi > bj and also bn = 0 and a1 = 1. Kalila and Dimna want to cut all the trees completely, with minimum cost.
They want you to help them! Will you?
Input
The first line of input contains an integer n (1 β€ n β€ 105). The second line of input contains n integers a1, a2, ..., an (1 β€ ai β€ 109). The third line of input contains n integers b1, b2, ..., bn (0 β€ bi β€ 109).
It's guaranteed that a1 = 1, bn = 0, a1 < a2 < ... < an and b1 > b2 > ... > bn.
Output
The only line of output must contain the minimum cost of cutting all the trees completely.
Please, do not write the %lld specifier to read or write 64-bit integers in Π‘++. It is preferred to use the cin, cout streams or the %I64d specifier.
Examples
Input
5
1 2 3 4 5
5 4 3 2 0
Output
25
Input
6
1 2 3 10 20 30
6 5 4 3 2 0
Output
138 | instruction | 0 | 104,771 | 8 | 209,542 |
Tags: dp, geometry
Correct Solution:
```
n = int(input())
a = [0 for i in range(0, n + 1)]
b = [0 for i in range(0, n + 1)]
a = list(map(int, input().split()))
b = list(map(int, input().split()))
dp = [0 for i in range(0, n + 1)]
c = [[0 for i in range(0, 3)] for j in range(0, n + 1)]
stack = []
stack.append(0)
stack.append(1)
dp[0] = 0
dp[1] = a[1] * b[0]
def intersection(x, y):
return int((dp[y] - dp[x]) / (b[x] - b[y]))
last = 0
c[last] = [0, intersection(0, 1), 0]
last+=1
c[last] = [intersection(0, 1), 1000000001, 1]
last1=0
for i in range(2, n):
while (last1 >= 0):
if (c[last1][0] <= a[i] and c[last1][1] >= a[i]):
dp[i] = dp[c[last1][2]] + b[c[last1][2]] * a[i]
#print(i,dp[i])
break
elif c[last1][0] > a[i]:
last1 -= 1
else:
last1 += 1
while stack:
top = stack[-1]
if len(stack) >= 2:
second_top = stack[-2]
if intersection(second_top, i) < intersection(top,second_top):
stack.pop()
last -= 1
else:
break
else:
break
stack.append(i)
last += 1
c[last] = [intersection(top, i),1000000001,i]
c[last-1]= [c[last-1][0],intersection(top,i),c[last-1][2]]
print(dp[n-1])
``` | output | 1 | 104,771 | 8 | 209,543 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Kalila and Dimna are two jackals living in a huge jungle. One day they decided to join a logging factory in order to make money.
The manager of logging factory wants them to go to the jungle and cut n trees with heights a1, a2, ..., an. They bought a chain saw from a shop. Each time they use the chain saw on the tree number i, they can decrease the height of this tree by one unit. Each time that Kalila and Dimna use the chain saw, they need to recharge it. Cost of charging depends on the id of the trees which have been cut completely (a tree is cut completely if its height equal to 0). If the maximum id of a tree which has been cut completely is i (the tree that have height ai in the beginning), then the cost of charging the chain saw would be bi. If no tree is cut completely, Kalila and Dimna cannot charge the chain saw. The chainsaw is charged in the beginning. We know that for each i < j, ai < aj and bi > bj and also bn = 0 and a1 = 1. Kalila and Dimna want to cut all the trees completely, with minimum cost.
They want you to help them! Will you?
Input
The first line of input contains an integer n (1 β€ n β€ 105). The second line of input contains n integers a1, a2, ..., an (1 β€ ai β€ 109). The third line of input contains n integers b1, b2, ..., bn (0 β€ bi β€ 109).
It's guaranteed that a1 = 1, bn = 0, a1 < a2 < ... < an and b1 > b2 > ... > bn.
Output
The only line of output must contain the minimum cost of cutting all the trees completely.
Please, do not write the %lld specifier to read or write 64-bit integers in Π‘++. It is preferred to use the cin, cout streams or the %I64d specifier.
Examples
Input
5
1 2 3 4 5
5 4 3 2 0
Output
25
Input
6
1 2 3 10 20 30
6 5 4 3 2 0
Output
138 | instruction | 0 | 104,772 | 8 | 209,544 |
Tags: dp, geometry
Correct Solution:
```
n = int(input())
a = [0] + [int(x) for x in input().split()]
b = [0] + [int(x) for x in input().split()]
q = [0 for i in range(n + 1)]
f = [0 for _ in range(n + 1)]
l, r, q[1] = 1, 1, 1
for i in range(2, n + 1):
while l < r and f[q[l + 1]] - f[q[l]] < a[i] * (b[q[l]] - b[q[l + 1]]):
l += 1
f[i] = f[q[l]] + b[q[l]] * a[i]
while l < r and (f[q[r]] - f[q[r - 1]]) * (b[q[r]] - b[i]) >= (f[i] - f[q[r]]) * (b[q[r - 1]] - b[q[r]]):
r -= 1
r += 1
q[r] = i
print(f[n])
``` | output | 1 | 104,772 | 8 | 209,545 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Kalila and Dimna are two jackals living in a huge jungle. One day they decided to join a logging factory in order to make money.
The manager of logging factory wants them to go to the jungle and cut n trees with heights a1, a2, ..., an. They bought a chain saw from a shop. Each time they use the chain saw on the tree number i, they can decrease the height of this tree by one unit. Each time that Kalila and Dimna use the chain saw, they need to recharge it. Cost of charging depends on the id of the trees which have been cut completely (a tree is cut completely if its height equal to 0). If the maximum id of a tree which has been cut completely is i (the tree that have height ai in the beginning), then the cost of charging the chain saw would be bi. If no tree is cut completely, Kalila and Dimna cannot charge the chain saw. The chainsaw is charged in the beginning. We know that for each i < j, ai < aj and bi > bj and also bn = 0 and a1 = 1. Kalila and Dimna want to cut all the trees completely, with minimum cost.
They want you to help them! Will you?
Input
The first line of input contains an integer n (1 β€ n β€ 105). The second line of input contains n integers a1, a2, ..., an (1 β€ ai β€ 109). The third line of input contains n integers b1, b2, ..., bn (0 β€ bi β€ 109).
It's guaranteed that a1 = 1, bn = 0, a1 < a2 < ... < an and b1 > b2 > ... > bn.
Output
The only line of output must contain the minimum cost of cutting all the trees completely.
Please, do not write the %lld specifier to read or write 64-bit integers in Π‘++. It is preferred to use the cin, cout streams or the %I64d specifier.
Examples
Input
5
1 2 3 4 5
5 4 3 2 0
Output
25
Input
6
1 2 3 10 20 30
6 5 4 3 2 0
Output
138 | instruction | 0 | 104,773 | 8 | 209,546 |
Tags: dp, geometry
Correct Solution:
```
def cross(i, j, k, b, cost):
# k, i, j
return (cost[i] - cost[j]) * (b[i] - b[k]) - (b[i] - b[j]) * (cost[i] - cost[k])
def dot(i, j, a, b, cost):
return cost[j] + a[i] * b[j]
def CF319C():
N = int(input())
a = tuple(map(int, input().split()))
b = tuple(map(int, input().split()))
cost = [0] * N
hull = [0] * N # Make hull considering total minimum cost and charge of individual trees
left = 0
right = 1 # Index of items in hull
for i in range(1, N):
# Find the last cut tree which incurs minimum cost for ith tree
while left + 1 < right and dot(i, hull[left], a, b, cost) >= dot(i, hull[left + 1], a, b, cost):
# Only increase left when it decreases cutting cost
left += 1
cost[i] = dot(i, hull[left], a, b, cost)
while right >= 2 and cross(hull[right], hull[right - 1], i, b, cost) >= 0:
right -= 1
right += 1
hull[right] = i
return cost[-1]
if __name__ == '__main__':
res = CF319C()
print(res)
``` | output | 1 | 104,773 | 8 | 209,547 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Kalila and Dimna are two jackals living in a huge jungle. One day they decided to join a logging factory in order to make money.
The manager of logging factory wants them to go to the jungle and cut n trees with heights a1, a2, ..., an. They bought a chain saw from a shop. Each time they use the chain saw on the tree number i, they can decrease the height of this tree by one unit. Each time that Kalila and Dimna use the chain saw, they need to recharge it. Cost of charging depends on the id of the trees which have been cut completely (a tree is cut completely if its height equal to 0). If the maximum id of a tree which has been cut completely is i (the tree that have height ai in the beginning), then the cost of charging the chain saw would be bi. If no tree is cut completely, Kalila and Dimna cannot charge the chain saw. The chainsaw is charged in the beginning. We know that for each i < j, ai < aj and bi > bj and also bn = 0 and a1 = 1. Kalila and Dimna want to cut all the trees completely, with minimum cost.
They want you to help them! Will you?
Input
The first line of input contains an integer n (1 β€ n β€ 105). The second line of input contains n integers a1, a2, ..., an (1 β€ ai β€ 109). The third line of input contains n integers b1, b2, ..., bn (0 β€ bi β€ 109).
It's guaranteed that a1 = 1, bn = 0, a1 < a2 < ... < an and b1 > b2 > ... > bn.
Output
The only line of output must contain the minimum cost of cutting all the trees completely.
Please, do not write the %lld specifier to read or write 64-bit integers in Π‘++. It is preferred to use the cin, cout streams or the %I64d specifier.
Examples
Input
5
1 2 3 4 5
5 4 3 2 0
Output
25
Input
6
1 2 3 10 20 30
6 5 4 3 2 0
Output
138 | instruction | 0 | 104,774 | 8 | 209,548 |
Tags: dp, geometry
Correct Solution:
```
N = 100100
def main():
n = int(input())
d=[0]*N
a = list(map(int, input().split()))
b = list(map(int, input().split()))
m,c=[0]*N,[0]*N
m[0] = b[0]
c[0] = d[0]
z = 1
p = 0
for i in range(1,n):
p = min(p,z)
while(p+1<z and m[p+1]*a[i]+c[p+1]<=m[p]*a[i]+c[p]):
p=p+1
d[i] = m[p]*a[i]+c[p]
while(z>=2 and (c[z-2]-c[z-1])*(b[i]-m[z-1])>=(c[z-1]-d[i])*(m[z-1]-m[z-2]) ):
z=z-1
m[z],c[z]=b[i],d[i]
z=z+1
#for j in range(n):print(d[j])
print(d[n-1])
main()
``` | output | 1 | 104,774 | 8 | 209,549 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Kalila and Dimna are two jackals living in a huge jungle. One day they decided to join a logging factory in order to make money.
The manager of logging factory wants them to go to the jungle and cut n trees with heights a1, a2, ..., an. They bought a chain saw from a shop. Each time they use the chain saw on the tree number i, they can decrease the height of this tree by one unit. Each time that Kalila and Dimna use the chain saw, they need to recharge it. Cost of charging depends on the id of the trees which have been cut completely (a tree is cut completely if its height equal to 0). If the maximum id of a tree which has been cut completely is i (the tree that have height ai in the beginning), then the cost of charging the chain saw would be bi. If no tree is cut completely, Kalila and Dimna cannot charge the chain saw. The chainsaw is charged in the beginning. We know that for each i < j, ai < aj and bi > bj and also bn = 0 and a1 = 1. Kalila and Dimna want to cut all the trees completely, with minimum cost.
They want you to help them! Will you?
Input
The first line of input contains an integer n (1 β€ n β€ 105). The second line of input contains n integers a1, a2, ..., an (1 β€ ai β€ 109). The third line of input contains n integers b1, b2, ..., bn (0 β€ bi β€ 109).
It's guaranteed that a1 = 1, bn = 0, a1 < a2 < ... < an and b1 > b2 > ... > bn.
Output
The only line of output must contain the minimum cost of cutting all the trees completely.
Please, do not write the %lld specifier to read or write 64-bit integers in Π‘++. It is preferred to use the cin, cout streams or the %I64d specifier.
Examples
Input
5
1 2 3 4 5
5 4 3 2 0
Output
25
Input
6
1 2 3 10 20 30
6 5 4 3 2 0
Output
138
Submitted Solution:
```
n = int(input())
a = list(map(int, input().split()))
b = list(map(int, input().split()))
c = [0] * n
stk = [0]
for i in range(1, n):
while len(stk) > 1 and c[stk[1]] - c[stk[0]] <= a[i] * (b[stk[0]] -
b[stk[1]]):
del stk[0]
c[i] = c[stk[0]] + a[i] * b[stk[0]]
while len(stk) > 1 and ((c[stk[-1]] - c[stk[-2]]) * (b[stk[-1]] - b[i]) >
(b[stk[-2]] - b[stk[-1]]) * (c[i] - c[stk[-1]])):
del stk[-1]
stk.append(i)
print(c[n - 1])
``` | instruction | 0 | 104,775 | 8 | 209,550 |
Yes | output | 1 | 104,775 | 8 | 209,551 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Kalila and Dimna are two jackals living in a huge jungle. One day they decided to join a logging factory in order to make money.
The manager of logging factory wants them to go to the jungle and cut n trees with heights a1, a2, ..., an. They bought a chain saw from a shop. Each time they use the chain saw on the tree number i, they can decrease the height of this tree by one unit. Each time that Kalila and Dimna use the chain saw, they need to recharge it. Cost of charging depends on the id of the trees which have been cut completely (a tree is cut completely if its height equal to 0). If the maximum id of a tree which has been cut completely is i (the tree that have height ai in the beginning), then the cost of charging the chain saw would be bi. If no tree is cut completely, Kalila and Dimna cannot charge the chain saw. The chainsaw is charged in the beginning. We know that for each i < j, ai < aj and bi > bj and also bn = 0 and a1 = 1. Kalila and Dimna want to cut all the trees completely, with minimum cost.
They want you to help them! Will you?
Input
The first line of input contains an integer n (1 β€ n β€ 105). The second line of input contains n integers a1, a2, ..., an (1 β€ ai β€ 109). The third line of input contains n integers b1, b2, ..., bn (0 β€ bi β€ 109).
It's guaranteed that a1 = 1, bn = 0, a1 < a2 < ... < an and b1 > b2 > ... > bn.
Output
The only line of output must contain the minimum cost of cutting all the trees completely.
Please, do not write the %lld specifier to read or write 64-bit integers in Π‘++. It is preferred to use the cin, cout streams or the %I64d specifier.
Examples
Input
5
1 2 3 4 5
5 4 3 2 0
Output
25
Input
6
1 2 3 10 20 30
6 5 4 3 2 0
Output
138
Submitted Solution:
```
def dot(a, b, c):
return a + c*b
def main():
n = int(input())
a = list(map(int, input().split(' ')))
b = list(map(int, input().split(' ')))
dp = [ 0 for _ in range(n) ]
st = [0 for _ in range(n) ]
i, j = 0, 0
for k in range(1,n):
while(j - i > 0 and dot(dp[st[i]],b[st[i]],a[k]) >= dot(dp[st[i+1]],b[st[i+1]],a[k])):
i+=1
dp[k] = a[k]*b[st[i]] + dp[st[i]]
while(j - i > 0 and (b[st[j]] - b[k])*(dp[st[j]] - dp[st[j-1]]) > (dp[k] - dp[st[j]])*(b[st[j-1]] - b[st[j]])):
j-=1
j+=1
st[j] = k
print(dp[-1])
if __name__ == "__main__":
main()
``` | instruction | 0 | 104,776 | 8 | 209,552 |
Yes | output | 1 | 104,776 | 8 | 209,553 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Kalila and Dimna are two jackals living in a huge jungle. One day they decided to join a logging factory in order to make money.
The manager of logging factory wants them to go to the jungle and cut n trees with heights a1, a2, ..., an. They bought a chain saw from a shop. Each time they use the chain saw on the tree number i, they can decrease the height of this tree by one unit. Each time that Kalila and Dimna use the chain saw, they need to recharge it. Cost of charging depends on the id of the trees which have been cut completely (a tree is cut completely if its height equal to 0). If the maximum id of a tree which has been cut completely is i (the tree that have height ai in the beginning), then the cost of charging the chain saw would be bi. If no tree is cut completely, Kalila and Dimna cannot charge the chain saw. The chainsaw is charged in the beginning. We know that for each i < j, ai < aj and bi > bj and also bn = 0 and a1 = 1. Kalila and Dimna want to cut all the trees completely, with minimum cost.
They want you to help them! Will you?
Input
The first line of input contains an integer n (1 β€ n β€ 105). The second line of input contains n integers a1, a2, ..., an (1 β€ ai β€ 109). The third line of input contains n integers b1, b2, ..., bn (0 β€ bi β€ 109).
It's guaranteed that a1 = 1, bn = 0, a1 < a2 < ... < an and b1 > b2 > ... > bn.
Output
The only line of output must contain the minimum cost of cutting all the trees completely.
Please, do not write the %lld specifier to read or write 64-bit integers in Π‘++. It is preferred to use the cin, cout streams or the %I64d specifier.
Examples
Input
5
1 2 3 4 5
5 4 3 2 0
Output
25
Input
6
1 2 3 10 20 30
6 5 4 3 2 0
Output
138
Submitted Solution:
```
#d_n= [C_m+A_n*B_m] for m in range()
#A=[ ->> ]
#B=[ <<- ]
def intersection(p,q):
return (q[1]-p[1])/(p[0]-q[0])
def tree_cutting(n,A,B):
I=[[0,0] for _ in range(n)]
C=[0 for _ in range(n)]
C[0]=0
I[0][0]=-float("inf")
I[0][1]=float("inf")
C[1]=C[0]+A[1]*B[0]
hull=[[B[0],C[0]],[B[1],C[1]]]
I[0][1]=intersection(hull[-1],hull[-2])
I[1][0]=I[0][1]
I[1][1]=float("inf")
curr=1
k=0
for i in range(2,n):
k=min(k,curr)-1
while True:
k+=1
if I[k][0]<=A[i] and A[i]<=I[k][1]:
j=k
break
C[i]=hull[k][1]+A[i]*hull[k][0]
p=[B[i],C[i]]
while intersection(p,hull[-2])<=intersection(hull[-1],hull[-2]):
hull.pop()
curr-=1
if len(hull)<2: break
if B[i]!=hull[-1][0]:
hull.append(p)
I[curr][1]=intersection(hull[-1],hull[-2])
curr+=1
I[curr][0]=intersection(hull[-1],hull[-2])
I[curr][1]=+float("inf")
else:
I[curr][1]=+float("inf")
return C[n-1]
if __name__=="__main__":
n=int(input())
A=list(map(int,input().strip().split()))
B=list(map(int,input().strip().split()))
print(tree_cutting(n,A,B))
``` | instruction | 0 | 104,777 | 8 | 209,554 |
Yes | output | 1 | 104,777 | 8 | 209,555 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Kalila and Dimna are two jackals living in a huge jungle. One day they decided to join a logging factory in order to make money.
The manager of logging factory wants them to go to the jungle and cut n trees with heights a1, a2, ..., an. They bought a chain saw from a shop. Each time they use the chain saw on the tree number i, they can decrease the height of this tree by one unit. Each time that Kalila and Dimna use the chain saw, they need to recharge it. Cost of charging depends on the id of the trees which have been cut completely (a tree is cut completely if its height equal to 0). If the maximum id of a tree which has been cut completely is i (the tree that have height ai in the beginning), then the cost of charging the chain saw would be bi. If no tree is cut completely, Kalila and Dimna cannot charge the chain saw. The chainsaw is charged in the beginning. We know that for each i < j, ai < aj and bi > bj and also bn = 0 and a1 = 1. Kalila and Dimna want to cut all the trees completely, with minimum cost.
They want you to help them! Will you?
Input
The first line of input contains an integer n (1 β€ n β€ 105). The second line of input contains n integers a1, a2, ..., an (1 β€ ai β€ 109). The third line of input contains n integers b1, b2, ..., bn (0 β€ bi β€ 109).
It's guaranteed that a1 = 1, bn = 0, a1 < a2 < ... < an and b1 > b2 > ... > bn.
Output
The only line of output must contain the minimum cost of cutting all the trees completely.
Please, do not write the %lld specifier to read or write 64-bit integers in Π‘++. It is preferred to use the cin, cout streams or the %I64d specifier.
Examples
Input
5
1 2 3 4 5
5 4 3 2 0
Output
25
Input
6
1 2 3 10 20 30
6 5 4 3 2 0
Output
138
Submitted Solution:
```
def cross(i, j, k, b, cost):
# k, i, j
return (cost[i] - cost[j]) * (b[i] - b[k]) - (b[i] - b[j]) * (cost[i] - cost[k])
def dot(i, j, a, b, cost):
return cost[j] + a[i] * b[j]
def CF319C():
N = int(input())
a = tuple(map(int, input().split()))
b = tuple(map(int, input().split()))
cost = [0] * N
hull = [0] * N # Make hull considering total minimum cost and charge of individual trees
left = 0
right = 1 # Index of items in hull
for i in range(1, N):
# Find the last cut tree which incurs minimum cost for ith tree
while left + 1 < right and dot(i, hull[left], a, b, cost) >= dot(i, hull[left + 1], a, b, cost):
# Only increase left when it decreases cutting cost
left += 1
cost[i] = dot(i, hull[left], a, b, cost)
while right >= 2 and cross(hull[right], hull[right - 1], i, b, cost) >= 0:
right -= 1
if left >= right: left = right - 1
right += 1
hull[right] = i
return cost[-1]
if __name__ == '__main__':
res = CF319C()
print(res)
``` | instruction | 0 | 104,778 | 8 | 209,556 |
Yes | output | 1 | 104,778 | 8 | 209,557 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Kalila and Dimna are two jackals living in a huge jungle. One day they decided to join a logging factory in order to make money.
The manager of logging factory wants them to go to the jungle and cut n trees with heights a1, a2, ..., an. They bought a chain saw from a shop. Each time they use the chain saw on the tree number i, they can decrease the height of this tree by one unit. Each time that Kalila and Dimna use the chain saw, they need to recharge it. Cost of charging depends on the id of the trees which have been cut completely (a tree is cut completely if its height equal to 0). If the maximum id of a tree which has been cut completely is i (the tree that have height ai in the beginning), then the cost of charging the chain saw would be bi. If no tree is cut completely, Kalila and Dimna cannot charge the chain saw. The chainsaw is charged in the beginning. We know that for each i < j, ai < aj and bi > bj and also bn = 0 and a1 = 1. Kalila and Dimna want to cut all the trees completely, with minimum cost.
They want you to help them! Will you?
Input
The first line of input contains an integer n (1 β€ n β€ 105). The second line of input contains n integers a1, a2, ..., an (1 β€ ai β€ 109). The third line of input contains n integers b1, b2, ..., bn (0 β€ bi β€ 109).
It's guaranteed that a1 = 1, bn = 0, a1 < a2 < ... < an and b1 > b2 > ... > bn.
Output
The only line of output must contain the minimum cost of cutting all the trees completely.
Please, do not write the %lld specifier to read or write 64-bit integers in Π‘++. It is preferred to use the cin, cout streams or the %I64d specifier.
Examples
Input
5
1 2 3 4 5
5 4 3 2 0
Output
25
Input
6
1 2 3 10 20 30
6 5 4 3 2 0
Output
138
Submitted Solution:
```
def cross(i, j, k, b, cost):
# k, i, j
return (cost[i] - cost[j]) * (b[i] - b[k]) - (b[i] - b[j]) * (cost[i] - cost[k])
def dot(i, j, a, b, cost):
return cost[j] + a[i] * b[j]
def CF319C():
N = int(input())
a = tuple(map(int, input().split()))
b = tuple(map(int, input().split()))
cost = [0] * N
hull = [0] * N # Make hull considering total minimum cost and charge of individual trees
left = 0
right = 0 # Index of items in hull
for i in range(1, N):
# Find the last cut tree which incurs minimum cost for ith tree
while left + 1 < right and dot(i, hull[left], a, b, cost) >= dot(i, hull[left + 1], a, b, cost):
# Only increase left when it decreases cutting cost
left += 1
cost[i] = dot(i, hull[left], a, b, cost)
while right >= 2 and cross(hull[right], hull[right - 1], i, b, cost) >= 0:
right -= 1
if left >= right: left = right - 1
right += 1
hull[right] = i
return cost[-1]
if __name__ == '__main__':
res = CF319C()
print(res)
``` | instruction | 0 | 104,779 | 8 | 209,558 |
No | output | 1 | 104,779 | 8 | 209,559 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Kalila and Dimna are two jackals living in a huge jungle. One day they decided to join a logging factory in order to make money.
The manager of logging factory wants them to go to the jungle and cut n trees with heights a1, a2, ..., an. They bought a chain saw from a shop. Each time they use the chain saw on the tree number i, they can decrease the height of this tree by one unit. Each time that Kalila and Dimna use the chain saw, they need to recharge it. Cost of charging depends on the id of the trees which have been cut completely (a tree is cut completely if its height equal to 0). If the maximum id of a tree which has been cut completely is i (the tree that have height ai in the beginning), then the cost of charging the chain saw would be bi. If no tree is cut completely, Kalila and Dimna cannot charge the chain saw. The chainsaw is charged in the beginning. We know that for each i < j, ai < aj and bi > bj and also bn = 0 and a1 = 1. Kalila and Dimna want to cut all the trees completely, with minimum cost.
They want you to help them! Will you?
Input
The first line of input contains an integer n (1 β€ n β€ 105). The second line of input contains n integers a1, a2, ..., an (1 β€ ai β€ 109). The third line of input contains n integers b1, b2, ..., bn (0 β€ bi β€ 109).
It's guaranteed that a1 = 1, bn = 0, a1 < a2 < ... < an and b1 > b2 > ... > bn.
Output
The only line of output must contain the minimum cost of cutting all the trees completely.
Please, do not write the %lld specifier to read or write 64-bit integers in Π‘++. It is preferred to use the cin, cout streams or the %I64d specifier.
Examples
Input
5
1 2 3 4 5
5 4 3 2 0
Output
25
Input
6
1 2 3 10 20 30
6 5 4 3 2 0
Output
138
Submitted Solution:
```
def cross(i, j, k, b, cost):
return (cost[i] - cost[j]) * (b[i] - b[k]) - (b[i] - b[j]) * (cost[i] - cost[k])
def dot(i, j, a, b, cost):
return cost[j] + a[i] * b[j]
def CF319C():
N = int(input())
a = tuple(map(int, input().split()))
b = tuple(map(int, input().split()))
cost = [0] * N
hull = [0] * N # Make hull considering total minimum cost and charge of individual trees
left = 0
right = 0 # Index of items in hull
for i in range(1, N):
# Find the last cut tree which incurs minimum cost for ith tree
while left + 1 <= right and dot(i, hull[left], a, b, cost) >= dot(i, hull[left + 1], a, b, cost):
# Only increase left when it decreases cutting cost
left += 1
cost[i] = dot(i, hull[left], a, b, cost)
# print("Before ", hull[:right+1])
while right >= 2 and cross(hull[right], hull[right - 1], i, b, cost) >= 0:
# print("removed ", i, hull[right])
right -= 1
right += 1
hull[right] = i
return cost[-1]
if __name__ == '__main__':
res = CF319C()
print(res)
``` | instruction | 0 | 104,780 | 8 | 209,560 |
No | output | 1 | 104,780 | 8 | 209,561 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Kalila and Dimna are two jackals living in a huge jungle. One day they decided to join a logging factory in order to make money.
The manager of logging factory wants them to go to the jungle and cut n trees with heights a1, a2, ..., an. They bought a chain saw from a shop. Each time they use the chain saw on the tree number i, they can decrease the height of this tree by one unit. Each time that Kalila and Dimna use the chain saw, they need to recharge it. Cost of charging depends on the id of the trees which have been cut completely (a tree is cut completely if its height equal to 0). If the maximum id of a tree which has been cut completely is i (the tree that have height ai in the beginning), then the cost of charging the chain saw would be bi. If no tree is cut completely, Kalila and Dimna cannot charge the chain saw. The chainsaw is charged in the beginning. We know that for each i < j, ai < aj and bi > bj and also bn = 0 and a1 = 1. Kalila and Dimna want to cut all the trees completely, with minimum cost.
They want you to help them! Will you?
Input
The first line of input contains an integer n (1 β€ n β€ 105). The second line of input contains n integers a1, a2, ..., an (1 β€ ai β€ 109). The third line of input contains n integers b1, b2, ..., bn (0 β€ bi β€ 109).
It's guaranteed that a1 = 1, bn = 0, a1 < a2 < ... < an and b1 > b2 > ... > bn.
Output
The only line of output must contain the minimum cost of cutting all the trees completely.
Please, do not write the %lld specifier to read or write 64-bit integers in Π‘++. It is preferred to use the cin, cout streams or the %I64d specifier.
Examples
Input
5
1 2 3 4 5
5 4 3 2 0
Output
25
Input
6
1 2 3 10 20 30
6 5 4 3 2 0
Output
138
Submitted Solution:
```
MD = 0x7fffffff
aa, bb, dp = [], [], []
def cross2(x1,y1,x2,y2):
h = x1*(y1/MD) - x2*(y1/MD)
l = x1*(y2%MD) - x2*(y1%MD)
h += l/MD
l %= MD
if h != 0:
if h < 0:
return -1
else:
return 1
else:
if l == 0:
return 0
else:
if l < 0:
return -1
else:
return 1
def cross(i,j,k):
return cross2(bb[j] - bb[i], dp[j] - dp[i], bb[k] - bb[i], dp[k] - dp[i])
def dot(i, a):
return dp[i] + a * bb[i]
def main():
n = int(input())
a = list(map(int, input().split(' ')))
b = list(map(int, input().split(' ')))
qq = []
cnt, h = 1, 0
for i in range(len(a)):
aa.append(a[i])
for i in range(len(b)):
bb.append(b[i])
for _ in range(n):
dp.append(0)
qq.append(0)
del a,b
for i in range(1,n):
a = aa[i]
while h+1 < cnt and dot(qq[h+1], a) <= dot(qq[h], a):
h+=1
dp[i] = dot(qq[h],a)
while cnt >= 2 and cross(qq[cnt-2], qq[cnt-1], i) >= 0:
cnt-=1
if h >= cnt:
h = cnt-1
qq[cnt] = i
cnt+=1
# print(dp[-1])
print(aa)
print(bb)
print(dp)
if __name__ == "__main__":
main()
``` | instruction | 0 | 104,781 | 8 | 209,562 |
No | output | 1 | 104,781 | 8 | 209,563 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Kalila and Dimna are two jackals living in a huge jungle. One day they decided to join a logging factory in order to make money.
The manager of logging factory wants them to go to the jungle and cut n trees with heights a1, a2, ..., an. They bought a chain saw from a shop. Each time they use the chain saw on the tree number i, they can decrease the height of this tree by one unit. Each time that Kalila and Dimna use the chain saw, they need to recharge it. Cost of charging depends on the id of the trees which have been cut completely (a tree is cut completely if its height equal to 0). If the maximum id of a tree which has been cut completely is i (the tree that have height ai in the beginning), then the cost of charging the chain saw would be bi. If no tree is cut completely, Kalila and Dimna cannot charge the chain saw. The chainsaw is charged in the beginning. We know that for each i < j, ai < aj and bi > bj and also bn = 0 and a1 = 1. Kalila and Dimna want to cut all the trees completely, with minimum cost.
They want you to help them! Will you?
Input
The first line of input contains an integer n (1 β€ n β€ 105). The second line of input contains n integers a1, a2, ..., an (1 β€ ai β€ 109). The third line of input contains n integers b1, b2, ..., bn (0 β€ bi β€ 109).
It's guaranteed that a1 = 1, bn = 0, a1 < a2 < ... < an and b1 > b2 > ... > bn.
Output
The only line of output must contain the minimum cost of cutting all the trees completely.
Please, do not write the %lld specifier to read or write 64-bit integers in Π‘++. It is preferred to use the cin, cout streams or the %I64d specifier.
Examples
Input
5
1 2 3 4 5
5 4 3 2 0
Output
25
Input
6
1 2 3 10 20 30
6 5 4 3 2 0
Output
138
Submitted Solution:
```
#d_n= [C_m+A_n*B_m] for m in range()
#A=[ ->> ]
#B=[ <<- ]
def intersection(p,q):
return (q[1]-p[1])/(p[0]-q[0])
def tree_cutting(n,A,B):
I=[[0,0] for _ in range(n)]
C=[0]*(n+1)
C[0]=0
I[0][0]=-float("inf")
I[0][1]=float("inf")
C[1]=C[0]+A[0]*B[0]
hull=[[B[0],C[0]],[B[1],C[1]]]
I[0][1]=intersection(hull[-1],hull[-2])
I[1][0]=I[0][1]
I[1][1]=float("inf")
curr=1
k=0
for i in range(2,n):
k=min(k,curr)-1
while True:
k+=1
if I[k][0]<=A[i] and A[i]<=I[k][1]:
j=k
break
C[i]=hull[k][1]+A[i]*hull[k][0]
p=[B[i],C[i]]
while intersection(p,hull[-2])<=intersection(hull[-1],hull[-2]):
hull.pop()
curr-=1
if len(hull)<2: break
if B[i]!=hull[-1][0]:
hull.append(p)
I[curr][1]=intersection(hull[-1],hull[-2])
curr+=1
I[curr][0]=intersection(hull[-1],hull[-2])
I[curr][1]=+float("inf")
else:
I[curr][1]=+float("inf")
return C[n-1]
if __name__=="__main__":
n=int(input())
A=list(map(int,input().strip().split()))
B=list(map(int,input().strip().split()))
print(tree_cutting(n,A,B))
``` | instruction | 0 | 104,782 | 8 | 209,564 |
No | output | 1 | 104,782 | 8 | 209,565 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Once Vasya played bricks. All the bricks in the set had regular cubical shape. Vasya vas a talented architect, however the tower he built kept falling apart.
Let us consider the building process. Vasya takes a brick and puts it on top of the already built tower so that the sides of the brick are parallel to the sides of the bricks he has already used. Let's introduce a Cartesian coordinate system on the horizontal plane, where Vasya puts the first brick. Then the projection of brick number i on the plane is a square with sides parallel to the axes of coordinates with opposite corners in points (xi, 1, yi, 1) and (xi, 2, yi, 2). The bricks are cast from homogeneous plastic and the weight of a brick a Γ a Γ a is a3 grams.
It is guaranteed that Vasya puts any brick except the first one on the previous one, that is the area of intersection of the upper side of the previous brick and the lower side of the next brick is always positive.
We (Vasya included) live in a normal world where the laws of physical statics work. And that is why, perhaps, if we put yet another brick, the tower will collapse under its own weight. Vasya puts the cubes consecutively one on top of the other until at least one cube loses the balance and falls down. If it happens, Vasya gets upset and stops the construction. Print the number of bricks in the maximal stable tower, that is the maximal number m satisfying the condition that all the towers consisting of bricks 1, 2, ..., k for every integer k from 1 to m remain stable.
Input
The first input file contains an integer n (1 β€ n β€ 100) which is the number of bricks. Each of the next n lines contains four numbers xi, 1, yi, 1, xi, 2, yi, 2 (xi, 1 β xi, 2, |xi, 1 - xi, 2| = |yi, 1 - yi, 2|) which are the coordinates of the opposite angles of the base of the brick number i. The coordinates are integers and their absolute value does not exceed 50.
The cubes are given in the order Vasya puts them. It is guaranteed that the area of intersection of the upper side of the brick number i - 1 and the lower side of the brick number i is strictly strictly greater than zero for all i β₯ 2.
Output
Print the number of bricks in the maximal stable tower.
Examples
Input
2
0 0 3 3
1 0 4 3
Output
2
Input
2
0 0 3 3
2 0 5 3
Output
1
Input
3
0 0 3 3
1 0 4 3
2 0 5 3
Output
3 | instruction | 0 | 104,796 | 8 | 209,592 |
Tags: implementation
Correct Solution:
```
n = int(input())
l = []
m = []
for i in range(n):
X1, Y1, X2, Y2 = map(int, input().split())
m1 = (abs(X2 - X1)) ** 3
l.append([X1, Y1, X2, Y2])
m.append(m1)
X1, Y1, X2, Y2 = l[0][0], l[0][1], l[0][2], l[0][3]
if X1 > X2: X2, X1 = X1, X2
if Y1 > Y2: Y1, Y2 = Y2, Y1
ans = n
for i in range(1, n):
for j in range(i, 0, -1):
x1, y1, x2, y2 = l[j - 1][0], l[j - 1][1], l[j - 1][2], l[j - 1][3]
if x1 > x2: x2, x1 = x1, x2
if y1 > y2: y1, y2 = y2, y1
numx, numy, den = 0, 0 , 0
for k in range(i, j - 1, -1):
a = (l[k][0] + l[k][2]) / 2
b = (l[k][1] + l[k][3]) / 2
m = (abs(l[k][2] - l[k][0])) ** 3
numx += m * a
numy += m * b
den += m
xcm = numx / den
ycm = numy / den
if x1 <= xcm <= x2 and y1 <= ycm <= y2: pass
else:
ans = i
break
if ans != n: break
print(ans)
``` | output | 1 | 104,796 | 8 | 209,593 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Once Vasya played bricks. All the bricks in the set had regular cubical shape. Vasya vas a talented architect, however the tower he built kept falling apart.
Let us consider the building process. Vasya takes a brick and puts it on top of the already built tower so that the sides of the brick are parallel to the sides of the bricks he has already used. Let's introduce a Cartesian coordinate system on the horizontal plane, where Vasya puts the first brick. Then the projection of brick number i on the plane is a square with sides parallel to the axes of coordinates with opposite corners in points (xi, 1, yi, 1) and (xi, 2, yi, 2). The bricks are cast from homogeneous plastic and the weight of a brick a Γ a Γ a is a3 grams.
It is guaranteed that Vasya puts any brick except the first one on the previous one, that is the area of intersection of the upper side of the previous brick and the lower side of the next brick is always positive.
We (Vasya included) live in a normal world where the laws of physical statics work. And that is why, perhaps, if we put yet another brick, the tower will collapse under its own weight. Vasya puts the cubes consecutively one on top of the other until at least one cube loses the balance and falls down. If it happens, Vasya gets upset and stops the construction. Print the number of bricks in the maximal stable tower, that is the maximal number m satisfying the condition that all the towers consisting of bricks 1, 2, ..., k for every integer k from 1 to m remain stable.
Input
The first input file contains an integer n (1 β€ n β€ 100) which is the number of bricks. Each of the next n lines contains four numbers xi, 1, yi, 1, xi, 2, yi, 2 (xi, 1 β xi, 2, |xi, 1 - xi, 2| = |yi, 1 - yi, 2|) which are the coordinates of the opposite angles of the base of the brick number i. The coordinates are integers and their absolute value does not exceed 50.
The cubes are given in the order Vasya puts them. It is guaranteed that the area of intersection of the upper side of the brick number i - 1 and the lower side of the brick number i is strictly strictly greater than zero for all i β₯ 2.
Output
Print the number of bricks in the maximal stable tower.
Examples
Input
2
0 0 3 3
1 0 4 3
Output
2
Input
2
0 0 3 3
2 0 5 3
Output
1
Input
3
0 0 3 3
1 0 4 3
2 0 5 3
Output
3 | instruction | 0 | 104,797 | 8 | 209,594 |
Tags: implementation
Correct Solution:
```
a=[list(map(int,input().split())) for i in range(int(input()))]
n=len(a)
z=n
for i in range(n):
if i!=0:
c=[(a[i][0]+a[i][2])/2,(a[i][1]+a[i][3])/2]
m=abs(a[i][0]-a[i][2])**3
for j in range(i-1,-1,-1):
x1,y1,x2,y2=min(a[j][0],a[j][2]),min(a[j][1],a[j][3]),max(a[j][0],a[j][2]),max(a[j][1],a[j][3])
if not(x1<=c[0]<=x2 and y1<=c[1]<=y2):
z=i
break
x=c[0]+((a[j][0]+a[j][2])/2-c[0])*((abs(a[j][0]-a[j][2])**3)/(m+abs(a[j][0]-a[j][2])**3))
y=c[1]+((a[j][1]+a[j][3])/2-c[1])*((abs(a[j][0]-a[j][2])**3)/(m+abs(a[j][0]-a[j][2])**3))
c=[x,y]
m+=abs(a[j][0]-a[j][2])**3
if z!=n:
break
print(z)
``` | output | 1 | 104,797 | 8 | 209,595 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Once Vasya played bricks. All the bricks in the set had regular cubical shape. Vasya vas a talented architect, however the tower he built kept falling apart.
Let us consider the building process. Vasya takes a brick and puts it on top of the already built tower so that the sides of the brick are parallel to the sides of the bricks he has already used. Let's introduce a Cartesian coordinate system on the horizontal plane, where Vasya puts the first brick. Then the projection of brick number i on the plane is a square with sides parallel to the axes of coordinates with opposite corners in points (xi, 1, yi, 1) and (xi, 2, yi, 2). The bricks are cast from homogeneous plastic and the weight of a brick a Γ a Γ a is a3 grams.
It is guaranteed that Vasya puts any brick except the first one on the previous one, that is the area of intersection of the upper side of the previous brick and the lower side of the next brick is always positive.
We (Vasya included) live in a normal world where the laws of physical statics work. And that is why, perhaps, if we put yet another brick, the tower will collapse under its own weight. Vasya puts the cubes consecutively one on top of the other until at least one cube loses the balance and falls down. If it happens, Vasya gets upset and stops the construction. Print the number of bricks in the maximal stable tower, that is the maximal number m satisfying the condition that all the towers consisting of bricks 1, 2, ..., k for every integer k from 1 to m remain stable.
Input
The first input file contains an integer n (1 β€ n β€ 100) which is the number of bricks. Each of the next n lines contains four numbers xi, 1, yi, 1, xi, 2, yi, 2 (xi, 1 β xi, 2, |xi, 1 - xi, 2| = |yi, 1 - yi, 2|) which are the coordinates of the opposite angles of the base of the brick number i. The coordinates are integers and their absolute value does not exceed 50.
The cubes are given in the order Vasya puts them. It is guaranteed that the area of intersection of the upper side of the brick number i - 1 and the lower side of the brick number i is strictly strictly greater than zero for all i β₯ 2.
Output
Print the number of bricks in the maximal stable tower.
Examples
Input
2
0 0 3 3
1 0 4 3
Output
2
Input
2
0 0 3 3
2 0 5 3
Output
1
Input
3
0 0 3 3
1 0 4 3
2 0 5 3
Output
3
Submitted Solution:
```
n = int(input())
a = list(map(int,input().split()))
k = 1
ob = r = 0
for i in range(n-1):
b = list(map(int,input().split()))
if b[1]==a[1]:
if a[0]<b[0] and a[2]<b[2]:
if a[2]-b[0]>=b[2]-a[2]:
k+=1
else: break
else:
if b[2]-a[0]>=a[0]-b[0]:
k+=1
else:
break
else:
if a[1]<b[1] and a[3]<b[3]:
if a[3]-b[1]>=b[3]-a[3]:
k+=1
else: break
else:
if b[3]-a[1]>=a[1]-b[1]:
k+=1
else:
break
a = b
print(k)
``` | instruction | 0 | 104,798 | 8 | 209,596 |
No | output | 1 | 104,798 | 8 | 209,597 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Once Vasya played bricks. All the bricks in the set had regular cubical shape. Vasya vas a talented architect, however the tower he built kept falling apart.
Let us consider the building process. Vasya takes a brick and puts it on top of the already built tower so that the sides of the brick are parallel to the sides of the bricks he has already used. Let's introduce a Cartesian coordinate system on the horizontal plane, where Vasya puts the first brick. Then the projection of brick number i on the plane is a square with sides parallel to the axes of coordinates with opposite corners in points (xi, 1, yi, 1) and (xi, 2, yi, 2). The bricks are cast from homogeneous plastic and the weight of a brick a Γ a Γ a is a3 grams.
It is guaranteed that Vasya puts any brick except the first one on the previous one, that is the area of intersection of the upper side of the previous brick and the lower side of the next brick is always positive.
We (Vasya included) live in a normal world where the laws of physical statics work. And that is why, perhaps, if we put yet another brick, the tower will collapse under its own weight. Vasya puts the cubes consecutively one on top of the other until at least one cube loses the balance and falls down. If it happens, Vasya gets upset and stops the construction. Print the number of bricks in the maximal stable tower, that is the maximal number m satisfying the condition that all the towers consisting of bricks 1, 2, ..., k for every integer k from 1 to m remain stable.
Input
The first input file contains an integer n (1 β€ n β€ 100) which is the number of bricks. Each of the next n lines contains four numbers xi, 1, yi, 1, xi, 2, yi, 2 (xi, 1 β xi, 2, |xi, 1 - xi, 2| = |yi, 1 - yi, 2|) which are the coordinates of the opposite angles of the base of the brick number i. The coordinates are integers and their absolute value does not exceed 50.
The cubes are given in the order Vasya puts them. It is guaranteed that the area of intersection of the upper side of the brick number i - 1 and the lower side of the brick number i is strictly strictly greater than zero for all i β₯ 2.
Output
Print the number of bricks in the maximal stable tower.
Examples
Input
2
0 0 3 3
1 0 4 3
Output
2
Input
2
0 0 3 3
2 0 5 3
Output
1
Input
3
0 0 3 3
1 0 4 3
2 0 5 3
Output
3
Submitted Solution:
```
a=[list(map(int,input().split())) for i in range(int(input()))]
n=len(a)
z=n
for i in range(n):
if i!=0:
c=[(a[i][0]+a[i][2])/2,(a[i][1]+a[i][3])/2]
m=abs(a[i][0]-a[i][2])**3
for j in range(i-1,-1,-1):
x1,y1,x2,y2=min(a[j][0],a[j][2]),min(a[j][1],a[j][3]),max(a[j][0],a[j][2]),max(a[j][1],a[j][3])
if not(x1<=c[0]<=x2 and y1<=c[1]<=y2):
z=i
break
x=c[0]+((a[j][0]+a[j][2])/2-c[0])*(m/(m+abs(a[j][0]-a[j][2])**3))
y=c[1]+((a[j][1]+a[j][3])/2-c[1])*(m/(m+abs(a[j][0]-a[j][2])**3))
c=[x,y]
m+=abs(a[j][0]-a[j][2])**3
if z!=n:
break
print(z)
``` | instruction | 0 | 104,799 | 8 | 209,598 |
No | output | 1 | 104,799 | 8 | 209,599 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Once Vasya played bricks. All the bricks in the set had regular cubical shape. Vasya vas a talented architect, however the tower he built kept falling apart.
Let us consider the building process. Vasya takes a brick and puts it on top of the already built tower so that the sides of the brick are parallel to the sides of the bricks he has already used. Let's introduce a Cartesian coordinate system on the horizontal plane, where Vasya puts the first brick. Then the projection of brick number i on the plane is a square with sides parallel to the axes of coordinates with opposite corners in points (xi, 1, yi, 1) and (xi, 2, yi, 2). The bricks are cast from homogeneous plastic and the weight of a brick a Γ a Γ a is a3 grams.
It is guaranteed that Vasya puts any brick except the first one on the previous one, that is the area of intersection of the upper side of the previous brick and the lower side of the next brick is always positive.
We (Vasya included) live in a normal world where the laws of physical statics work. And that is why, perhaps, if we put yet another brick, the tower will collapse under its own weight. Vasya puts the cubes consecutively one on top of the other until at least one cube loses the balance and falls down. If it happens, Vasya gets upset and stops the construction. Print the number of bricks in the maximal stable tower, that is the maximal number m satisfying the condition that all the towers consisting of bricks 1, 2, ..., k for every integer k from 1 to m remain stable.
Input
The first input file contains an integer n (1 β€ n β€ 100) which is the number of bricks. Each of the next n lines contains four numbers xi, 1, yi, 1, xi, 2, yi, 2 (xi, 1 β xi, 2, |xi, 1 - xi, 2| = |yi, 1 - yi, 2|) which are the coordinates of the opposite angles of the base of the brick number i. The coordinates are integers and their absolute value does not exceed 50.
The cubes are given in the order Vasya puts them. It is guaranteed that the area of intersection of the upper side of the brick number i - 1 and the lower side of the brick number i is strictly strictly greater than zero for all i β₯ 2.
Output
Print the number of bricks in the maximal stable tower.
Examples
Input
2
0 0 3 3
1 0 4 3
Output
2
Input
2
0 0 3 3
2 0 5 3
Output
1
Input
3
0 0 3 3
1 0 4 3
2 0 5 3
Output
3
Submitted Solution:
```
a=[list(map(int,input().split())) for i in range(int(input()))]
n=len(a)
z=n
for i in range(n):
if i!=0:
c=[(a[i][0]+a[i][2])/2,(a[i][1]+a[i][3])/2]
m=abs(a[i][0]-a[i][2])**3
for j in range(i-1,-1,-1):
x1,y1,x2,y2=min(a[j][0],a[j][2]),min(a[j][1],a[j][2]),max(a[j][0],a[j][2]),max(a[j][1],a[j][2])
if not(x1<=c[0]<=x2 and y1<=c[1]<=y2):
z=i
break
x=c[0]+((a[j][0]+a[j][2])/2-c[0])*(m/(m+abs(a[j][0]-a[j][2])**3))
y=c[1]+((a[j][1]+a[j][1])/2-c[1])*(m/(m+abs(a[j][0]-a[j][2])**3))
c=[x,y]
m+=abs(a[j][0]-a[j][2])**3
if z!=n:
break
print(z)
``` | instruction | 0 | 104,800 | 8 | 209,600 |
No | output | 1 | 104,800 | 8 | 209,601 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Once Vasya played bricks. All the bricks in the set had regular cubical shape. Vasya vas a talented architect, however the tower he built kept falling apart.
Let us consider the building process. Vasya takes a brick and puts it on top of the already built tower so that the sides of the brick are parallel to the sides of the bricks he has already used. Let's introduce a Cartesian coordinate system on the horizontal plane, where Vasya puts the first brick. Then the projection of brick number i on the plane is a square with sides parallel to the axes of coordinates with opposite corners in points (xi, 1, yi, 1) and (xi, 2, yi, 2). The bricks are cast from homogeneous plastic and the weight of a brick a Γ a Γ a is a3 grams.
It is guaranteed that Vasya puts any brick except the first one on the previous one, that is the area of intersection of the upper side of the previous brick and the lower side of the next brick is always positive.
We (Vasya included) live in a normal world where the laws of physical statics work. And that is why, perhaps, if we put yet another brick, the tower will collapse under its own weight. Vasya puts the cubes consecutively one on top of the other until at least one cube loses the balance and falls down. If it happens, Vasya gets upset and stops the construction. Print the number of bricks in the maximal stable tower, that is the maximal number m satisfying the condition that all the towers consisting of bricks 1, 2, ..., k for every integer k from 1 to m remain stable.
Input
The first input file contains an integer n (1 β€ n β€ 100) which is the number of bricks. Each of the next n lines contains four numbers xi, 1, yi, 1, xi, 2, yi, 2 (xi, 1 β xi, 2, |xi, 1 - xi, 2| = |yi, 1 - yi, 2|) which are the coordinates of the opposite angles of the base of the brick number i. The coordinates are integers and their absolute value does not exceed 50.
The cubes are given in the order Vasya puts them. It is guaranteed that the area of intersection of the upper side of the brick number i - 1 and the lower side of the brick number i is strictly strictly greater than zero for all i β₯ 2.
Output
Print the number of bricks in the maximal stable tower.
Examples
Input
2
0 0 3 3
1 0 4 3
Output
2
Input
2
0 0 3 3
2 0 5 3
Output
1
Input
3
0 0 3 3
1 0 4 3
2 0 5 3
Output
3
Submitted Solution:
```
a=[list(map(int,input().split())) for i in range(int(input()))]
n=len(a)
z=n
for i in range(n):
if i!=0:
c=[(a[i][0]+a[i][2])/2,(a[i][1]+a[i][3])/2]
m=abs(a[i][0]-a[i][2])**3
for j in range(i-1,-1,-1):
x1,y1,x2,y2=min(a[j][0],a[j][2]),min(a[j][1],a[j][3]),max(a[j][0],a[j][2]),max(a[j][1],a[j][3])
if not(x1<=c[0]<=x2 and y1<=c[1]<=y2):
z=i
break
x=c[0]+((a[j][0]+a[j][2])/2-c[0])*(m/(m+abs(a[j][0]-a[j][2])**3))
y=c[1]+((a[j][1]+a[j][1])/2-c[1])*(m/(m+abs(a[j][0]-a[j][2])**3))
c=[x,y]
m+=abs(a[j][0]-a[j][2])**3
if z!=n:
break
print(z)
``` | instruction | 0 | 104,801 | 8 | 209,602 |
No | output | 1 | 104,801 | 8 | 209,603 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Kevin Sun wants to move his precious collection of n cowbells from Naperthrill to Exeter, where there is actually grass instead of corn. Before moving, he must pack his cowbells into k boxes of a fixed size. In order to keep his collection safe during transportation, he won't place more than two cowbells into a single box. Since Kevin wishes to minimize expenses, he is curious about the smallest size box he can use to pack his entire collection.
Kevin is a meticulous cowbell collector and knows that the size of his i-th (1 β€ i β€ n) cowbell is an integer si. In fact, he keeps his cowbells sorted by size, so si - 1 β€ si for any i > 1. Also an expert packer, Kevin can fit one or two cowbells into a box of size s if and only if the sum of their sizes does not exceed s. Given this information, help Kevin determine the smallest s for which it is possible to put all of his cowbells into k boxes of size s.
Input
The first line of the input contains two space-separated integers n and k (1 β€ n β€ 2Β·k β€ 100 000), denoting the number of cowbells and the number of boxes, respectively.
The next line contains n space-separated integers s1, s2, ..., sn (1 β€ s1 β€ s2 β€ ... β€ sn β€ 1 000 000), the sizes of Kevin's cowbells. It is guaranteed that the sizes si are given in non-decreasing order.
Output
Print a single integer, the smallest s for which it is possible for Kevin to put all of his cowbells into k boxes of size s.
Examples
Input
2 1
2 5
Output
7
Input
4 3
2 3 5 9
Output
9
Input
3 2
3 5 7
Output
8
Note
In the first sample, Kevin must pack his two cowbells into the same box.
In the second sample, Kevin can pack together the following sets of cowbells: {2, 3}, {5} and {9}.
In the third sample, the optimal solution is {3, 5} and {7}. | instruction | 0 | 104,874 | 8 | 209,748 |
Tags: binary search, greedy
Correct Solution:
```
n, k = map(int, input().split())
arr = list(map(int, input().split()))
arr.reverse()
idx = 0
ans = 0
while idx < n and n - idx - 1 <= (k - 1) * 2:
ans = max(ans, arr[idx])
idx += 1
k -= 1
left = idx
right = n - 1
while left < right:
ans = max(arr[left] + arr[right], ans)
left += 1
right -= 1
print(ans)
``` | output | 1 | 104,874 | 8 | 209,749 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Kevin Sun wants to move his precious collection of n cowbells from Naperthrill to Exeter, where there is actually grass instead of corn. Before moving, he must pack his cowbells into k boxes of a fixed size. In order to keep his collection safe during transportation, he won't place more than two cowbells into a single box. Since Kevin wishes to minimize expenses, he is curious about the smallest size box he can use to pack his entire collection.
Kevin is a meticulous cowbell collector and knows that the size of his i-th (1 β€ i β€ n) cowbell is an integer si. In fact, he keeps his cowbells sorted by size, so si - 1 β€ si for any i > 1. Also an expert packer, Kevin can fit one or two cowbells into a box of size s if and only if the sum of their sizes does not exceed s. Given this information, help Kevin determine the smallest s for which it is possible to put all of his cowbells into k boxes of size s.
Input
The first line of the input contains two space-separated integers n and k (1 β€ n β€ 2Β·k β€ 100 000), denoting the number of cowbells and the number of boxes, respectively.
The next line contains n space-separated integers s1, s2, ..., sn (1 β€ s1 β€ s2 β€ ... β€ sn β€ 1 000 000), the sizes of Kevin's cowbells. It is guaranteed that the sizes si are given in non-decreasing order.
Output
Print a single integer, the smallest s for which it is possible for Kevin to put all of his cowbells into k boxes of size s.
Examples
Input
2 1
2 5
Output
7
Input
4 3
2 3 5 9
Output
9
Input
3 2
3 5 7
Output
8
Note
In the first sample, Kevin must pack his two cowbells into the same box.
In the second sample, Kevin can pack together the following sets of cowbells: {2, 3}, {5} and {9}.
In the third sample, the optimal solution is {3, 5} and {7}. | instruction | 0 | 104,875 | 8 | 209,750 |
Tags: binary search, greedy
Correct Solution:
```
# https://codeforces.com/problemset/problem/604/B
import math
def needed_box(size):
current_count_box = 0
i, j = 0, number_of_cowbells-1
while i < j:
if list_cowbells[i] + list_cowbells[j] <= size:
i+=1
j-=1
elif list_cowbells[j] <= size:
j-=1
else:
return int(1e9)
current_count_box += 1
return current_count_box if i > j else current_count_box+1
number_of_cowbells, number_of_box = map(int, input().strip().split(' '))
list_cowbells = list(map(int, input().strip().split(' ')))
if number_of_cowbells == 1:
print(list_cowbells[0])
else:
left, right = 0, list_cowbells[-1] + list_cowbells[-2]
while left < right:
middle = math.ceil((right+left+1)/2)
number_of_needed_box = needed_box(middle)
if number_of_needed_box > number_of_box:
left = middle
elif number_of_needed_box <= number_of_box:
right = middle-1
print(right+1)
``` | output | 1 | 104,875 | 8 | 209,751 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Kevin Sun wants to move his precious collection of n cowbells from Naperthrill to Exeter, where there is actually grass instead of corn. Before moving, he must pack his cowbells into k boxes of a fixed size. In order to keep his collection safe during transportation, he won't place more than two cowbells into a single box. Since Kevin wishes to minimize expenses, he is curious about the smallest size box he can use to pack his entire collection.
Kevin is a meticulous cowbell collector and knows that the size of his i-th (1 β€ i β€ n) cowbell is an integer si. In fact, he keeps his cowbells sorted by size, so si - 1 β€ si for any i > 1. Also an expert packer, Kevin can fit one or two cowbells into a box of size s if and only if the sum of their sizes does not exceed s. Given this information, help Kevin determine the smallest s for which it is possible to put all of his cowbells into k boxes of size s.
Input
The first line of the input contains two space-separated integers n and k (1 β€ n β€ 2Β·k β€ 100 000), denoting the number of cowbells and the number of boxes, respectively.
The next line contains n space-separated integers s1, s2, ..., sn (1 β€ s1 β€ s2 β€ ... β€ sn β€ 1 000 000), the sizes of Kevin's cowbells. It is guaranteed that the sizes si are given in non-decreasing order.
Output
Print a single integer, the smallest s for which it is possible for Kevin to put all of his cowbells into k boxes of size s.
Examples
Input
2 1
2 5
Output
7
Input
4 3
2 3 5 9
Output
9
Input
3 2
3 5 7
Output
8
Note
In the first sample, Kevin must pack his two cowbells into the same box.
In the second sample, Kevin can pack together the following sets of cowbells: {2, 3}, {5} and {9}.
In the third sample, the optimal solution is {3, 5} and {7}. | instruction | 0 | 104,876 | 8 | 209,752 |
Tags: binary search, greedy
Correct Solution:
```
a = list(map(int,input().split()))
n,k = a[0], a[1]
l = k
bells = list(map(int,input().split()))
mymax = bells[-1]
while l > n/2 and l > 1 and n > 1:
del bells[-1]
l-=1
n-=1
while n > 1:
mymax = max(mymax,bells[0] + bells[-1])
del bells[0]
del bells[-1]
n -= 2
if len(bells)> 0:
mymax = max(mymax,bells[0])
print(mymax)
``` | output | 1 | 104,876 | 8 | 209,753 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Kevin Sun wants to move his precious collection of n cowbells from Naperthrill to Exeter, where there is actually grass instead of corn. Before moving, he must pack his cowbells into k boxes of a fixed size. In order to keep his collection safe during transportation, he won't place more than two cowbells into a single box. Since Kevin wishes to minimize expenses, he is curious about the smallest size box he can use to pack his entire collection.
Kevin is a meticulous cowbell collector and knows that the size of his i-th (1 β€ i β€ n) cowbell is an integer si. In fact, he keeps his cowbells sorted by size, so si - 1 β€ si for any i > 1. Also an expert packer, Kevin can fit one or two cowbells into a box of size s if and only if the sum of their sizes does not exceed s. Given this information, help Kevin determine the smallest s for which it is possible to put all of his cowbells into k boxes of size s.
Input
The first line of the input contains two space-separated integers n and k (1 β€ n β€ 2Β·k β€ 100 000), denoting the number of cowbells and the number of boxes, respectively.
The next line contains n space-separated integers s1, s2, ..., sn (1 β€ s1 β€ s2 β€ ... β€ sn β€ 1 000 000), the sizes of Kevin's cowbells. It is guaranteed that the sizes si are given in non-decreasing order.
Output
Print a single integer, the smallest s for which it is possible for Kevin to put all of his cowbells into k boxes of size s.
Examples
Input
2 1
2 5
Output
7
Input
4 3
2 3 5 9
Output
9
Input
3 2
3 5 7
Output
8
Note
In the first sample, Kevin must pack his two cowbells into the same box.
In the second sample, Kevin can pack together the following sets of cowbells: {2, 3}, {5} and {9}.
In the third sample, the optimal solution is {3, 5} and {7}. | instruction | 0 | 104,877 | 8 | 209,754 |
Tags: binary search, greedy
Correct Solution:
```
n, k = map(int, input().split())
s = list(map(int, input().split()))
total = 0
s.sort()
s.reverse()
if k >= n:
print(s[0])
exit()
x = n - k
y = k - x
for it in range(y):
if s[it] > total:
total = s[it]
ij = -1
for it in range(y, k):
total = max(total, s[it] + s[ij])
ij -= 1
print(total)
``` | output | 1 | 104,877 | 8 | 209,755 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Kevin Sun wants to move his precious collection of n cowbells from Naperthrill to Exeter, where there is actually grass instead of corn. Before moving, he must pack his cowbells into k boxes of a fixed size. In order to keep his collection safe during transportation, he won't place more than two cowbells into a single box. Since Kevin wishes to minimize expenses, he is curious about the smallest size box he can use to pack his entire collection.
Kevin is a meticulous cowbell collector and knows that the size of his i-th (1 β€ i β€ n) cowbell is an integer si. In fact, he keeps his cowbells sorted by size, so si - 1 β€ si for any i > 1. Also an expert packer, Kevin can fit one or two cowbells into a box of size s if and only if the sum of their sizes does not exceed s. Given this information, help Kevin determine the smallest s for which it is possible to put all of his cowbells into k boxes of size s.
Input
The first line of the input contains two space-separated integers n and k (1 β€ n β€ 2Β·k β€ 100 000), denoting the number of cowbells and the number of boxes, respectively.
The next line contains n space-separated integers s1, s2, ..., sn (1 β€ s1 β€ s2 β€ ... β€ sn β€ 1 000 000), the sizes of Kevin's cowbells. It is guaranteed that the sizes si are given in non-decreasing order.
Output
Print a single integer, the smallest s for which it is possible for Kevin to put all of his cowbells into k boxes of size s.
Examples
Input
2 1
2 5
Output
7
Input
4 3
2 3 5 9
Output
9
Input
3 2
3 5 7
Output
8
Note
In the first sample, Kevin must pack his two cowbells into the same box.
In the second sample, Kevin can pack together the following sets of cowbells: {2, 3}, {5} and {9}.
In the third sample, the optimal solution is {3, 5} and {7}. | instruction | 0 | 104,878 | 8 | 209,756 |
Tags: binary search, greedy
Correct Solution:
```
n, k = list(map(int, input().split()))
ss = list(map(int, input().split()))
if k >= n:
print(ss[n-1])
else:
top = 2 * (n - k ) - 1
bottom = 0
m = ss[n - 1]
while top > bottom:
v = ss[top] + ss[bottom]
if v > m:
m = v
bottom += 1
top -= 1
print(m)
``` | output | 1 | 104,878 | 8 | 209,757 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Kevin Sun wants to move his precious collection of n cowbells from Naperthrill to Exeter, where there is actually grass instead of corn. Before moving, he must pack his cowbells into k boxes of a fixed size. In order to keep his collection safe during transportation, he won't place more than two cowbells into a single box. Since Kevin wishes to minimize expenses, he is curious about the smallest size box he can use to pack his entire collection.
Kevin is a meticulous cowbell collector and knows that the size of his i-th (1 β€ i β€ n) cowbell is an integer si. In fact, he keeps his cowbells sorted by size, so si - 1 β€ si for any i > 1. Also an expert packer, Kevin can fit one or two cowbells into a box of size s if and only if the sum of their sizes does not exceed s. Given this information, help Kevin determine the smallest s for which it is possible to put all of his cowbells into k boxes of size s.
Input
The first line of the input contains two space-separated integers n and k (1 β€ n β€ 2Β·k β€ 100 000), denoting the number of cowbells and the number of boxes, respectively.
The next line contains n space-separated integers s1, s2, ..., sn (1 β€ s1 β€ s2 β€ ... β€ sn β€ 1 000 000), the sizes of Kevin's cowbells. It is guaranteed that the sizes si are given in non-decreasing order.
Output
Print a single integer, the smallest s for which it is possible for Kevin to put all of his cowbells into k boxes of size s.
Examples
Input
2 1
2 5
Output
7
Input
4 3
2 3 5 9
Output
9
Input
3 2
3 5 7
Output
8
Note
In the first sample, Kevin must pack his two cowbells into the same box.
In the second sample, Kevin can pack together the following sets of cowbells: {2, 3}, {5} and {9}.
In the third sample, the optimal solution is {3, 5} and {7}. | instruction | 0 | 104,879 | 8 | 209,758 |
Tags: binary search, greedy
Correct Solution:
```
n, k = map(int, input().split())
A = list(map(int, input().split()))
Max = A[-1]
while 2 * k > len(A) and A != []:
del A[-1]
k -= 1
while A != []:
Max = max(Max, A[0] + A[-1])
del A[0], A[-1]
print(Max)
``` | output | 1 | 104,879 | 8 | 209,759 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Kevin Sun wants to move his precious collection of n cowbells from Naperthrill to Exeter, where there is actually grass instead of corn. Before moving, he must pack his cowbells into k boxes of a fixed size. In order to keep his collection safe during transportation, he won't place more than two cowbells into a single box. Since Kevin wishes to minimize expenses, he is curious about the smallest size box he can use to pack his entire collection.
Kevin is a meticulous cowbell collector and knows that the size of his i-th (1 β€ i β€ n) cowbell is an integer si. In fact, he keeps his cowbells sorted by size, so si - 1 β€ si for any i > 1. Also an expert packer, Kevin can fit one or two cowbells into a box of size s if and only if the sum of their sizes does not exceed s. Given this information, help Kevin determine the smallest s for which it is possible to put all of his cowbells into k boxes of size s.
Input
The first line of the input contains two space-separated integers n and k (1 β€ n β€ 2Β·k β€ 100 000), denoting the number of cowbells and the number of boxes, respectively.
The next line contains n space-separated integers s1, s2, ..., sn (1 β€ s1 β€ s2 β€ ... β€ sn β€ 1 000 000), the sizes of Kevin's cowbells. It is guaranteed that the sizes si are given in non-decreasing order.
Output
Print a single integer, the smallest s for which it is possible for Kevin to put all of his cowbells into k boxes of size s.
Examples
Input
2 1
2 5
Output
7
Input
4 3
2 3 5 9
Output
9
Input
3 2
3 5 7
Output
8
Note
In the first sample, Kevin must pack his two cowbells into the same box.
In the second sample, Kevin can pack together the following sets of cowbells: {2, 3}, {5} and {9}.
In the third sample, the optimal solution is {3, 5} and {7}. | instruction | 0 | 104,880 | 8 | 209,760 |
Tags: binary search, greedy
Correct Solution:
```
n,k = map(int,input().split())
a = list(map(int,input().split()))
x = 2*k-n
b = a[:n-x]
anss = a[n-x:]
for i in range(len(b)//2):
anss.append(b[i]+b[len(b)-i-1])
print(max(anss))
``` | output | 1 | 104,880 | 8 | 209,761 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Kevin Sun wants to move his precious collection of n cowbells from Naperthrill to Exeter, where there is actually grass instead of corn. Before moving, he must pack his cowbells into k boxes of a fixed size. In order to keep his collection safe during transportation, he won't place more than two cowbells into a single box. Since Kevin wishes to minimize expenses, he is curious about the smallest size box he can use to pack his entire collection.
Kevin is a meticulous cowbell collector and knows that the size of his i-th (1 β€ i β€ n) cowbell is an integer si. In fact, he keeps his cowbells sorted by size, so si - 1 β€ si for any i > 1. Also an expert packer, Kevin can fit one or two cowbells into a box of size s if and only if the sum of their sizes does not exceed s. Given this information, help Kevin determine the smallest s for which it is possible to put all of his cowbells into k boxes of size s.
Input
The first line of the input contains two space-separated integers n and k (1 β€ n β€ 2Β·k β€ 100 000), denoting the number of cowbells and the number of boxes, respectively.
The next line contains n space-separated integers s1, s2, ..., sn (1 β€ s1 β€ s2 β€ ... β€ sn β€ 1 000 000), the sizes of Kevin's cowbells. It is guaranteed that the sizes si are given in non-decreasing order.
Output
Print a single integer, the smallest s for which it is possible for Kevin to put all of his cowbells into k boxes of size s.
Examples
Input
2 1
2 5
Output
7
Input
4 3
2 3 5 9
Output
9
Input
3 2
3 5 7
Output
8
Note
In the first sample, Kevin must pack his two cowbells into the same box.
In the second sample, Kevin can pack together the following sets of cowbells: {2, 3}, {5} and {9}.
In the third sample, the optimal solution is {3, 5} and {7}. | instruction | 0 | 104,881 | 8 | 209,762 |
Tags: binary search, greedy
Correct Solution:
```
n,k = [int(i) for i in input().split()]
sizes = [int(i) for i in input().split()]
if k>=n:
print(max(sizes))
else:
x1 = sizes[-1]
sizes = sizes[:2*(n-k)]
x2 = max([a+b for a, b in zip(sizes,sizes[::-1])])
print(max(x1,x2))
``` | output | 1 | 104,881 | 8 | 209,763 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Kevin Sun wants to move his precious collection of n cowbells from Naperthrill to Exeter, where there is actually grass instead of corn. Before moving, he must pack his cowbells into k boxes of a fixed size. In order to keep his collection safe during transportation, he won't place more than two cowbells into a single box. Since Kevin wishes to minimize expenses, he is curious about the smallest size box he can use to pack his entire collection.
Kevin is a meticulous cowbell collector and knows that the size of his i-th (1 β€ i β€ n) cowbell is an integer si. In fact, he keeps his cowbells sorted by size, so si - 1 β€ si for any i > 1. Also an expert packer, Kevin can fit one or two cowbells into a box of size s if and only if the sum of their sizes does not exceed s. Given this information, help Kevin determine the smallest s for which it is possible to put all of his cowbells into k boxes of size s.
Input
The first line of the input contains two space-separated integers n and k (1 β€ n β€ 2Β·k β€ 100 000), denoting the number of cowbells and the number of boxes, respectively.
The next line contains n space-separated integers s1, s2, ..., sn (1 β€ s1 β€ s2 β€ ... β€ sn β€ 1 000 000), the sizes of Kevin's cowbells. It is guaranteed that the sizes si are given in non-decreasing order.
Output
Print a single integer, the smallest s for which it is possible for Kevin to put all of his cowbells into k boxes of size s.
Examples
Input
2 1
2 5
Output
7
Input
4 3
2 3 5 9
Output
9
Input
3 2
3 5 7
Output
8
Note
In the first sample, Kevin must pack his two cowbells into the same box.
In the second sample, Kevin can pack together the following sets of cowbells: {2, 3}, {5} and {9}.
In the third sample, the optimal solution is {3, 5} and {7}.
Submitted Solution:
```
n,k=map(int,input().split())
l1=list(map(int,input().split()))
l1.sort()
l2=[]
if(k>=n):
print(max(l1))
else:
for i in range(n-1,2*(n-k)-1,-1):
l2.append(l1[i])
for i in range(n-k):
l2.append(l1[i]+l1[2*(n-k)-1-i])
print(max(l2))
``` | instruction | 0 | 104,882 | 8 | 209,764 |
Yes | output | 1 | 104,882 | 8 | 209,765 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Kevin Sun wants to move his precious collection of n cowbells from Naperthrill to Exeter, where there is actually grass instead of corn. Before moving, he must pack his cowbells into k boxes of a fixed size. In order to keep his collection safe during transportation, he won't place more than two cowbells into a single box. Since Kevin wishes to minimize expenses, he is curious about the smallest size box he can use to pack his entire collection.
Kevin is a meticulous cowbell collector and knows that the size of his i-th (1 β€ i β€ n) cowbell is an integer si. In fact, he keeps his cowbells sorted by size, so si - 1 β€ si for any i > 1. Also an expert packer, Kevin can fit one or two cowbells into a box of size s if and only if the sum of their sizes does not exceed s. Given this information, help Kevin determine the smallest s for which it is possible to put all of his cowbells into k boxes of size s.
Input
The first line of the input contains two space-separated integers n and k (1 β€ n β€ 2Β·k β€ 100 000), denoting the number of cowbells and the number of boxes, respectively.
The next line contains n space-separated integers s1, s2, ..., sn (1 β€ s1 β€ s2 β€ ... β€ sn β€ 1 000 000), the sizes of Kevin's cowbells. It is guaranteed that the sizes si are given in non-decreasing order.
Output
Print a single integer, the smallest s for which it is possible for Kevin to put all of his cowbells into k boxes of size s.
Examples
Input
2 1
2 5
Output
7
Input
4 3
2 3 5 9
Output
9
Input
3 2
3 5 7
Output
8
Note
In the first sample, Kevin must pack his two cowbells into the same box.
In the second sample, Kevin can pack together the following sets of cowbells: {2, 3}, {5} and {9}.
In the third sample, the optimal solution is {3, 5} and {7}.
Submitted Solution:
```
n,k = list(map(int, input().split()))
s = list(map(int, input().split()))
a = max(n-k, 0)
if a==0:
print(s[-1])
else:
m = s[-1]
for i in range(a):
m = max(m, s[i] + s[2*a-1-i])
print(m)
``` | instruction | 0 | 104,883 | 8 | 209,766 |
Yes | output | 1 | 104,883 | 8 | 209,767 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Kevin Sun wants to move his precious collection of n cowbells from Naperthrill to Exeter, where there is actually grass instead of corn. Before moving, he must pack his cowbells into k boxes of a fixed size. In order to keep his collection safe during transportation, he won't place more than two cowbells into a single box. Since Kevin wishes to minimize expenses, he is curious about the smallest size box he can use to pack his entire collection.
Kevin is a meticulous cowbell collector and knows that the size of his i-th (1 β€ i β€ n) cowbell is an integer si. In fact, he keeps his cowbells sorted by size, so si - 1 β€ si for any i > 1. Also an expert packer, Kevin can fit one or two cowbells into a box of size s if and only if the sum of their sizes does not exceed s. Given this information, help Kevin determine the smallest s for which it is possible to put all of his cowbells into k boxes of size s.
Input
The first line of the input contains two space-separated integers n and k (1 β€ n β€ 2Β·k β€ 100 000), denoting the number of cowbells and the number of boxes, respectively.
The next line contains n space-separated integers s1, s2, ..., sn (1 β€ s1 β€ s2 β€ ... β€ sn β€ 1 000 000), the sizes of Kevin's cowbells. It is guaranteed that the sizes si are given in non-decreasing order.
Output
Print a single integer, the smallest s for which it is possible for Kevin to put all of his cowbells into k boxes of size s.
Examples
Input
2 1
2 5
Output
7
Input
4 3
2 3 5 9
Output
9
Input
3 2
3 5 7
Output
8
Note
In the first sample, Kevin must pack his two cowbells into the same box.
In the second sample, Kevin can pack together the following sets of cowbells: {2, 3}, {5} and {9}.
In the third sample, the optimal solution is {3, 5} and {7}.
Submitted Solution:
```
def main():
n, k = (map(int, input().split()))
s = list(map(int, input().split()))
minSize = s[n - 1]
if n <= k:
print(minSize)
exit()
i = 0
total = n - 1
while 2 * k > n :
n, k = n - 1, k - 1
for i in range(k):
minSize = max(minSize, s[i] + s[n - i - 1])
print(minSize)
if __name__ == '__main__':
main()
``` | instruction | 0 | 104,884 | 8 | 209,768 |
Yes | output | 1 | 104,884 | 8 | 209,769 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Kevin Sun wants to move his precious collection of n cowbells from Naperthrill to Exeter, where there is actually grass instead of corn. Before moving, he must pack his cowbells into k boxes of a fixed size. In order to keep his collection safe during transportation, he won't place more than two cowbells into a single box. Since Kevin wishes to minimize expenses, he is curious about the smallest size box he can use to pack his entire collection.
Kevin is a meticulous cowbell collector and knows that the size of his i-th (1 β€ i β€ n) cowbell is an integer si. In fact, he keeps his cowbells sorted by size, so si - 1 β€ si for any i > 1. Also an expert packer, Kevin can fit one or two cowbells into a box of size s if and only if the sum of their sizes does not exceed s. Given this information, help Kevin determine the smallest s for which it is possible to put all of his cowbells into k boxes of size s.
Input
The first line of the input contains two space-separated integers n and k (1 β€ n β€ 2Β·k β€ 100 000), denoting the number of cowbells and the number of boxes, respectively.
The next line contains n space-separated integers s1, s2, ..., sn (1 β€ s1 β€ s2 β€ ... β€ sn β€ 1 000 000), the sizes of Kevin's cowbells. It is guaranteed that the sizes si are given in non-decreasing order.
Output
Print a single integer, the smallest s for which it is possible for Kevin to put all of his cowbells into k boxes of size s.
Examples
Input
2 1
2 5
Output
7
Input
4 3
2 3 5 9
Output
9
Input
3 2
3 5 7
Output
8
Note
In the first sample, Kevin must pack his two cowbells into the same box.
In the second sample, Kevin can pack together the following sets of cowbells: {2, 3}, {5} and {9}.
In the third sample, the optimal solution is {3, 5} and {7}.
Submitted Solution:
```
n, k = map(int, input().split())
a = list(map(int, input().split()))
b = [0]*k
for i in range(n-1, -1, -1):
b[(n-1-i) % k] += a[i]
if n-1-i == k-1:
b = b[::-1]
print(max(b))
``` | instruction | 0 | 104,885 | 8 | 209,770 |
Yes | output | 1 | 104,885 | 8 | 209,771 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Kevin Sun wants to move his precious collection of n cowbells from Naperthrill to Exeter, where there is actually grass instead of corn. Before moving, he must pack his cowbells into k boxes of a fixed size. In order to keep his collection safe during transportation, he won't place more than two cowbells into a single box. Since Kevin wishes to minimize expenses, he is curious about the smallest size box he can use to pack his entire collection.
Kevin is a meticulous cowbell collector and knows that the size of his i-th (1 β€ i β€ n) cowbell is an integer si. In fact, he keeps his cowbells sorted by size, so si - 1 β€ si for any i > 1. Also an expert packer, Kevin can fit one or two cowbells into a box of size s if and only if the sum of their sizes does not exceed s. Given this information, help Kevin determine the smallest s for which it is possible to put all of his cowbells into k boxes of size s.
Input
The first line of the input contains two space-separated integers n and k (1 β€ n β€ 2Β·k β€ 100 000), denoting the number of cowbells and the number of boxes, respectively.
The next line contains n space-separated integers s1, s2, ..., sn (1 β€ s1 β€ s2 β€ ... β€ sn β€ 1 000 000), the sizes of Kevin's cowbells. It is guaranteed that the sizes si are given in non-decreasing order.
Output
Print a single integer, the smallest s for which it is possible for Kevin to put all of his cowbells into k boxes of size s.
Examples
Input
2 1
2 5
Output
7
Input
4 3
2 3 5 9
Output
9
Input
3 2
3 5 7
Output
8
Note
In the first sample, Kevin must pack his two cowbells into the same box.
In the second sample, Kevin can pack together the following sets of cowbells: {2, 3}, {5} and {9}.
In the third sample, the optimal solution is {3, 5} and {7}.
Submitted Solution:
```
n,k = list(map(int,input().split()))
arr = list(map(int,input().split()))
arr.sort()
if n==2*k:
print(arr[-1]+arr[-2])
elif n<=k:
print(arr[-1])
else:
m = -1
x = (n-k)//2
for i in range(1,k):
m = max(m,arr[i]+arr[i-1])
m = max(m,arr[-1])
print(m)
``` | instruction | 0 | 104,886 | 8 | 209,772 |
No | output | 1 | 104,886 | 8 | 209,773 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Kevin Sun wants to move his precious collection of n cowbells from Naperthrill to Exeter, where there is actually grass instead of corn. Before moving, he must pack his cowbells into k boxes of a fixed size. In order to keep his collection safe during transportation, he won't place more than two cowbells into a single box. Since Kevin wishes to minimize expenses, he is curious about the smallest size box he can use to pack his entire collection.
Kevin is a meticulous cowbell collector and knows that the size of his i-th (1 β€ i β€ n) cowbell is an integer si. In fact, he keeps his cowbells sorted by size, so si - 1 β€ si for any i > 1. Also an expert packer, Kevin can fit one or two cowbells into a box of size s if and only if the sum of their sizes does not exceed s. Given this information, help Kevin determine the smallest s for which it is possible to put all of his cowbells into k boxes of size s.
Input
The first line of the input contains two space-separated integers n and k (1 β€ n β€ 2Β·k β€ 100 000), denoting the number of cowbells and the number of boxes, respectively.
The next line contains n space-separated integers s1, s2, ..., sn (1 β€ s1 β€ s2 β€ ... β€ sn β€ 1 000 000), the sizes of Kevin's cowbells. It is guaranteed that the sizes si are given in non-decreasing order.
Output
Print a single integer, the smallest s for which it is possible for Kevin to put all of his cowbells into k boxes of size s.
Examples
Input
2 1
2 5
Output
7
Input
4 3
2 3 5 9
Output
9
Input
3 2
3 5 7
Output
8
Note
In the first sample, Kevin must pack his two cowbells into the same box.
In the second sample, Kevin can pack together the following sets of cowbells: {2, 3}, {5} and {9}.
In the third sample, the optimal solution is {3, 5} and {7}.
Submitted Solution:
```
def f(tar,l,k):
s,count=0,0
for i in range(len(l)):
if s+l[i]<=tar:
s+=l[i]
else:
count+=1
s=l[i]
if s<=tar:
count+=1
if count<=k:
return True
else:
return False
n,k=map(int,input().split())
l=list(map(int,input().split()))
low,high,ans=max(l),10**18,10**18
while(low<=high):
mid=low+(high-low)//2
x=f(mid,l,k)
if x:
ans=min(ans,mid)
high=mid-1
else:
low=mid+1
print(ans)
``` | instruction | 0 | 104,887 | 8 | 209,774 |
No | output | 1 | 104,887 | 8 | 209,775 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Kevin Sun wants to move his precious collection of n cowbells from Naperthrill to Exeter, where there is actually grass instead of corn. Before moving, he must pack his cowbells into k boxes of a fixed size. In order to keep his collection safe during transportation, he won't place more than two cowbells into a single box. Since Kevin wishes to minimize expenses, he is curious about the smallest size box he can use to pack his entire collection.
Kevin is a meticulous cowbell collector and knows that the size of his i-th (1 β€ i β€ n) cowbell is an integer si. In fact, he keeps his cowbells sorted by size, so si - 1 β€ si for any i > 1. Also an expert packer, Kevin can fit one or two cowbells into a box of size s if and only if the sum of their sizes does not exceed s. Given this information, help Kevin determine the smallest s for which it is possible to put all of his cowbells into k boxes of size s.
Input
The first line of the input contains two space-separated integers n and k (1 β€ n β€ 2Β·k β€ 100 000), denoting the number of cowbells and the number of boxes, respectively.
The next line contains n space-separated integers s1, s2, ..., sn (1 β€ s1 β€ s2 β€ ... β€ sn β€ 1 000 000), the sizes of Kevin's cowbells. It is guaranteed that the sizes si are given in non-decreasing order.
Output
Print a single integer, the smallest s for which it is possible for Kevin to put all of his cowbells into k boxes of size s.
Examples
Input
2 1
2 5
Output
7
Input
4 3
2 3 5 9
Output
9
Input
3 2
3 5 7
Output
8
Note
In the first sample, Kevin must pack his two cowbells into the same box.
In the second sample, Kevin can pack together the following sets of cowbells: {2, 3}, {5} and {9}.
In the third sample, the optimal solution is {3, 5} and {7}.
Submitted Solution:
```
n, k = [int(elem) for elem in input().split(" ")]
sizes = [int(elem) for elem in input().split(" ")]
if n <=2:
print(sum(sizes))
elif n <= k:
print(sizes[-1])
else:
# if k >= n/2:
# print(max(sizes[n-k-1]+sizes[n-k],
# sizes[0]+sizes[2*(n-k)-1], sizes[-1]))
# else:
# print(max(sizes[n-k-1]+sizes[n-k],
# sizes[0]+sizes[2*(n-k)-1]))
boxes = sizes[n-k:]
for i in range(n-k):
boxes[i] += sizes[(n-k)-i-1]
print(max(boxes))
``` | instruction | 0 | 104,888 | 8 | 209,776 |
No | output | 1 | 104,888 | 8 | 209,777 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Kevin Sun wants to move his precious collection of n cowbells from Naperthrill to Exeter, where there is actually grass instead of corn. Before moving, he must pack his cowbells into k boxes of a fixed size. In order to keep his collection safe during transportation, he won't place more than two cowbells into a single box. Since Kevin wishes to minimize expenses, he is curious about the smallest size box he can use to pack his entire collection.
Kevin is a meticulous cowbell collector and knows that the size of his i-th (1 β€ i β€ n) cowbell is an integer si. In fact, he keeps his cowbells sorted by size, so si - 1 β€ si for any i > 1. Also an expert packer, Kevin can fit one or two cowbells into a box of size s if and only if the sum of their sizes does not exceed s. Given this information, help Kevin determine the smallest s for which it is possible to put all of his cowbells into k boxes of size s.
Input
The first line of the input contains two space-separated integers n and k (1 β€ n β€ 2Β·k β€ 100 000), denoting the number of cowbells and the number of boxes, respectively.
The next line contains n space-separated integers s1, s2, ..., sn (1 β€ s1 β€ s2 β€ ... β€ sn β€ 1 000 000), the sizes of Kevin's cowbells. It is guaranteed that the sizes si are given in non-decreasing order.
Output
Print a single integer, the smallest s for which it is possible for Kevin to put all of his cowbells into k boxes of size s.
Examples
Input
2 1
2 5
Output
7
Input
4 3
2 3 5 9
Output
9
Input
3 2
3 5 7
Output
8
Note
In the first sample, Kevin must pack his two cowbells into the same box.
In the second sample, Kevin can pack together the following sets of cowbells: {2, 3}, {5} and {9}.
In the third sample, the optimal solution is {3, 5} and {7}.
Submitted Solution:
```
no_of_cowbell,no_of_boxe=map(int,input().split())
li1=list(map(int,input().split()))
cows_packed=no_of_boxe*2
if(no_of_cowbell==1):
print(li1[0])
else:
if(no_of_cowbell<cows_packed):
print(max([li1[0]+li1[1],li1[no_of_cowbell-1]]))
else:
print(li1[no_of_cowbell-1]+li1[no_of_cowbell-2])
``` | instruction | 0 | 104,889 | 8 | 209,778 |
No | output | 1 | 104,889 | 8 | 209,779 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Little Petya often travels to his grandmother in the countryside. The grandmother has a large garden, which can be represented as a rectangle 1 Γ n in size, when viewed from above. This rectangle is divided into n equal square sections. The garden is very unusual as each of the square sections possesses its own fixed height and due to the newest irrigation system we can create artificial rain above each section.
Creating artificial rain is an expensive operation. That's why we limit ourselves to creating the artificial rain only above one section. At that, the water from each watered section will flow into its neighbouring sections if their height does not exceed the height of the section. That is, for example, the garden can be represented by a 1 Γ 5 rectangle, where the section heights are equal to 4, 2, 3, 3, 2. Then if we create an artificial rain over any of the sections with the height of 3, the water will flow over all the sections, except the ones with the height of 4. See the illustration of this example at the picture:
<image>
As Petya is keen on programming, he decided to find such a section that if we create artificial rain above it, the number of watered sections will be maximal. Help him.
Input
The first line contains a positive integer n (1 β€ n β€ 1000). The second line contains n positive integers which are the height of the sections. All the numbers are no less than 1 and not more than 1000.
Output
Print a single number, the maximal number of watered sections if we create artificial rain above exactly one section.
Examples
Input
1
2
Output
1
Input
5
1 2 1 2 1
Output
3
Input
8
1 2 1 1 1 3 3 4
Output
6 | instruction | 0 | 105,730 | 8 | 211,460 |
Tags: brute force, implementation
Correct Solution:
```
n = int(input())
sections = list(map(int, input().split()))
scores = []
for i in range(n):
total = 1
curr = sections[i]
for j in range(i+1,n):
if sections[j]<=curr:
total+=1
curr = sections[j]
else:
break
curr = sections[i]
for k in reversed(range(0,i)):
if sections[k]<=curr:
total+=1
curr = sections[k]
else:
break
scores.append(total)
print(max(scores))
``` | output | 1 | 105,730 | 8 | 211,461 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Little Petya often travels to his grandmother in the countryside. The grandmother has a large garden, which can be represented as a rectangle 1 Γ n in size, when viewed from above. This rectangle is divided into n equal square sections. The garden is very unusual as each of the square sections possesses its own fixed height and due to the newest irrigation system we can create artificial rain above each section.
Creating artificial rain is an expensive operation. That's why we limit ourselves to creating the artificial rain only above one section. At that, the water from each watered section will flow into its neighbouring sections if their height does not exceed the height of the section. That is, for example, the garden can be represented by a 1 Γ 5 rectangle, where the section heights are equal to 4, 2, 3, 3, 2. Then if we create an artificial rain over any of the sections with the height of 3, the water will flow over all the sections, except the ones with the height of 4. See the illustration of this example at the picture:
<image>
As Petya is keen on programming, he decided to find such a section that if we create artificial rain above it, the number of watered sections will be maximal. Help him.
Input
The first line contains a positive integer n (1 β€ n β€ 1000). The second line contains n positive integers which are the height of the sections. All the numbers are no less than 1 and not more than 1000.
Output
Print a single number, the maximal number of watered sections if we create artificial rain above exactly one section.
Examples
Input
1
2
Output
1
Input
5
1 2 1 2 1
Output
3
Input
8
1 2 1 1 1 3 3 4
Output
6 | instruction | 0 | 105,731 | 8 | 211,462 |
Tags: brute force, implementation
Correct Solution:
```
n = int(input())
arr = list(map(int, input().split()))
mx = 1
for i in range(n):
count = 1
r = l = i
while l - 1 >= 0:
l -= 1
if arr[l] <= arr[l+1]:
count += 1
else:
break
while r + 1 < n:
r += 1
if arr[r] <= arr[r-1]:
count += 1
else:
break
mx = max(mx, count)
print (mx)
``` | output | 1 | 105,731 | 8 | 211,463 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Little Petya often travels to his grandmother in the countryside. The grandmother has a large garden, which can be represented as a rectangle 1 Γ n in size, when viewed from above. This rectangle is divided into n equal square sections. The garden is very unusual as each of the square sections possesses its own fixed height and due to the newest irrigation system we can create artificial rain above each section.
Creating artificial rain is an expensive operation. That's why we limit ourselves to creating the artificial rain only above one section. At that, the water from each watered section will flow into its neighbouring sections if their height does not exceed the height of the section. That is, for example, the garden can be represented by a 1 Γ 5 rectangle, where the section heights are equal to 4, 2, 3, 3, 2. Then if we create an artificial rain over any of the sections with the height of 3, the water will flow over all the sections, except the ones with the height of 4. See the illustration of this example at the picture:
<image>
As Petya is keen on programming, he decided to find such a section that if we create artificial rain above it, the number of watered sections will be maximal. Help him.
Input
The first line contains a positive integer n (1 β€ n β€ 1000). The second line contains n positive integers which are the height of the sections. All the numbers are no less than 1 and not more than 1000.
Output
Print a single number, the maximal number of watered sections if we create artificial rain above exactly one section.
Examples
Input
1
2
Output
1
Input
5
1 2 1 2 1
Output
3
Input
8
1 2 1 1 1 3 3 4
Output
6 | instruction | 0 | 105,732 | 8 | 211,464 |
Tags: brute force, implementation
Correct Solution:
```
n = int(input())
maxi = 0
arr = [int(x) for x in input().split()]
for i in range(n):
count = 0
for j in range(i,-1,-1):
if j-1>=0:
if arr[j]<arr[j-1]:
break
else:
count += 1
for j in range(i,n):
if j+1<n:
if arr[j]<arr[j+1]:
break
else:
count += 1
if maxi<count:
maxi = count
print(maxi+1)
``` | output | 1 | 105,732 | 8 | 211,465 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Little Petya often travels to his grandmother in the countryside. The grandmother has a large garden, which can be represented as a rectangle 1 Γ n in size, when viewed from above. This rectangle is divided into n equal square sections. The garden is very unusual as each of the square sections possesses its own fixed height and due to the newest irrigation system we can create artificial rain above each section.
Creating artificial rain is an expensive operation. That's why we limit ourselves to creating the artificial rain only above one section. At that, the water from each watered section will flow into its neighbouring sections if their height does not exceed the height of the section. That is, for example, the garden can be represented by a 1 Γ 5 rectangle, where the section heights are equal to 4, 2, 3, 3, 2. Then if we create an artificial rain over any of the sections with the height of 3, the water will flow over all the sections, except the ones with the height of 4. See the illustration of this example at the picture:
<image>
As Petya is keen on programming, he decided to find such a section that if we create artificial rain above it, the number of watered sections will be maximal. Help him.
Input
The first line contains a positive integer n (1 β€ n β€ 1000). The second line contains n positive integers which are the height of the sections. All the numbers are no less than 1 and not more than 1000.
Output
Print a single number, the maximal number of watered sections if we create artificial rain above exactly one section.
Examples
Input
1
2
Output
1
Input
5
1 2 1 2 1
Output
3
Input
8
1 2 1 1 1 3 3 4
Output
6 | instruction | 0 | 105,733 | 8 | 211,466 |
Tags: brute force, implementation
Correct Solution:
```
n = int(input())
l = list(map(int,input().split()))
l2 = list(reversed(l))
ans = 0
for i in range(n):
tmp = l[i]
c = 1
for j in range(i - 1 , - 1, - 1):
if l[j] > tmp:
break
tmp , c = l[j] , c + 1
tmp = l[i]
for j in range(i+1 , n):
if l[j] > tmp :
break
tmp , c = l[j] , c +1
if c > ans :
ans=c
print(ans)
``` | output | 1 | 105,733 | 8 | 211,467 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Little Petya often travels to his grandmother in the countryside. The grandmother has a large garden, which can be represented as a rectangle 1 Γ n in size, when viewed from above. This rectangle is divided into n equal square sections. The garden is very unusual as each of the square sections possesses its own fixed height and due to the newest irrigation system we can create artificial rain above each section.
Creating artificial rain is an expensive operation. That's why we limit ourselves to creating the artificial rain only above one section. At that, the water from each watered section will flow into its neighbouring sections if their height does not exceed the height of the section. That is, for example, the garden can be represented by a 1 Γ 5 rectangle, where the section heights are equal to 4, 2, 3, 3, 2. Then if we create an artificial rain over any of the sections with the height of 3, the water will flow over all the sections, except the ones with the height of 4. See the illustration of this example at the picture:
<image>
As Petya is keen on programming, he decided to find such a section that if we create artificial rain above it, the number of watered sections will be maximal. Help him.
Input
The first line contains a positive integer n (1 β€ n β€ 1000). The second line contains n positive integers which are the height of the sections. All the numbers are no less than 1 and not more than 1000.
Output
Print a single number, the maximal number of watered sections if we create artificial rain above exactly one section.
Examples
Input
1
2
Output
1
Input
5
1 2 1 2 1
Output
3
Input
8
1 2 1 1 1 3 3 4
Output
6 | instruction | 0 | 105,734 | 8 | 211,468 |
Tags: brute force, implementation
Correct Solution:
```
n = int(input())
heights = [int(x) for x in input().split()]
leftCounter = [0] * n
rightCounter = [0] * n
for i in range(1,n):
if heights[i-1] <= heights[i]:
leftCounter[i] = leftCounter[i-1] + 1
if heights[n-i-1] >= heights[n-i]:
rightCounter[n-i-1] = rightCounter[n-i] + 1
maxSections = 0
for i in range(n):
maxSections = max(maxSections,leftCounter[i]+rightCounter[i]+1)
print(maxSections)
``` | output | 1 | 105,734 | 8 | 211,469 |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.