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 |
|---|---|---|---|---|---|
Provide tags and a correct Python 3 solution for this coding contest problem.
The USA Construction Operation (USACO) recently ordered Farmer John to arrange a row of n haybale piles on the farm. The i-th pile contains a_i haybales.
However, Farmer John has just left for vacation, leaving Bessie all on her own. Every day, Bessie the naughty cow can choose to move one haybale in any pile to an adjacent pile. Formally, in one day she can choose any two indices i and j (1 ≤ i, j ≤ n) such that |i-j|=1 and a_i>0 and apply a_i = a_i - 1, a_j = a_j + 1. She may also decide to not do anything on some days because she is lazy.
Bessie wants to maximize the number of haybales in pile 1 (i.e. to maximize a_1), and she only has d days to do so before Farmer John returns. Help her find the maximum number of haybales that may be in pile 1 if she acts optimally!
Input
The input consists of multiple test cases. The first line contains an integer t (1 ≤ t ≤ 100) — the number of test cases. Next 2t lines contain a description of test cases — two lines per test case.
The first line of each test case contains integers n and d (1 ≤ n,d ≤ 100) — the number of haybale piles and the number of days, respectively.
The second line of each test case contains n integers a_1, a_2, …, a_n (0 ≤ a_i ≤ 100) — the number of haybales in each pile.
Output
For each test case, output one integer: the maximum number of haybales that may be in pile 1 after d days if Bessie acts optimally.
Example
Input
3
4 5
1 0 3 2
2 2
100 1
1 8
0
Output
3
101
0
Note
In the first test case of the sample, this is one possible way Bessie can end up with 3 haybales in pile 1:
* On day one, move a haybale from pile 3 to pile 2
* On day two, move a haybale from pile 3 to pile 2
* On day three, move a haybale from pile 2 to pile 1
* On day four, move a haybale from pile 2 to pile 1
* On day five, do nothing
In the second test case of the sample, Bessie can do nothing on the first day and move a haybale from pile 2 to pile 1 on the second day. | instruction | 0 | 46,389 | 8 | 92,778 |
Tags: greedy, implementation
Correct Solution:
```
for i in range(int(input())):
n,d=map(int,input().split())
a=list(map(int,input().split()))
# print(a)
for i in range(1,n):
p=d//i
p=min(p,a[i])
d-=(p*i)
# print(p)
a[0]+=p
# print(d,i,p)
print(a[0])
``` | output | 1 | 46,389 | 8 | 92,779 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
The USA Construction Operation (USACO) recently ordered Farmer John to arrange a row of n haybale piles on the farm. The i-th pile contains a_i haybales.
However, Farmer John has just left for vacation, leaving Bessie all on her own. Every day, Bessie the naughty cow can choose to move one haybale in any pile to an adjacent pile. Formally, in one day she can choose any two indices i and j (1 ≤ i, j ≤ n) such that |i-j|=1 and a_i>0 and apply a_i = a_i - 1, a_j = a_j + 1. She may also decide to not do anything on some days because she is lazy.
Bessie wants to maximize the number of haybales in pile 1 (i.e. to maximize a_1), and she only has d days to do so before Farmer John returns. Help her find the maximum number of haybales that may be in pile 1 if she acts optimally!
Input
The input consists of multiple test cases. The first line contains an integer t (1 ≤ t ≤ 100) — the number of test cases. Next 2t lines contain a description of test cases — two lines per test case.
The first line of each test case contains integers n and d (1 ≤ n,d ≤ 100) — the number of haybale piles and the number of days, respectively.
The second line of each test case contains n integers a_1, a_2, …, a_n (0 ≤ a_i ≤ 100) — the number of haybales in each pile.
Output
For each test case, output one integer: the maximum number of haybales that may be in pile 1 after d days if Bessie acts optimally.
Example
Input
3
4 5
1 0 3 2
2 2
100 1
1 8
0
Output
3
101
0
Note
In the first test case of the sample, this is one possible way Bessie can end up with 3 haybales in pile 1:
* On day one, move a haybale from pile 3 to pile 2
* On day two, move a haybale from pile 3 to pile 2
* On day three, move a haybale from pile 2 to pile 1
* On day four, move a haybale from pile 2 to pile 1
* On day five, do nothing
In the second test case of the sample, Bessie can do nothing on the first day and move a haybale from pile 2 to pile 1 on the second day.
Submitted Solution:
```
for t in range(int(input())):
n,k=map(int,input().split())
a=[int(v) for v in input().split()]
for i in range(1,n):
if a[i] and k>0:
while(k>0 and k>=i and a[i]>0):
k=k-i
a[0]+=1
a[i]-=1
print(a[0])
``` | instruction | 0 | 46,390 | 8 | 92,780 |
Yes | output | 1 | 46,390 | 8 | 92,781 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
The USA Construction Operation (USACO) recently ordered Farmer John to arrange a row of n haybale piles on the farm. The i-th pile contains a_i haybales.
However, Farmer John has just left for vacation, leaving Bessie all on her own. Every day, Bessie the naughty cow can choose to move one haybale in any pile to an adjacent pile. Formally, in one day she can choose any two indices i and j (1 ≤ i, j ≤ n) such that |i-j|=1 and a_i>0 and apply a_i = a_i - 1, a_j = a_j + 1. She may also decide to not do anything on some days because she is lazy.
Bessie wants to maximize the number of haybales in pile 1 (i.e. to maximize a_1), and she only has d days to do so before Farmer John returns. Help her find the maximum number of haybales that may be in pile 1 if she acts optimally!
Input
The input consists of multiple test cases. The first line contains an integer t (1 ≤ t ≤ 100) — the number of test cases. Next 2t lines contain a description of test cases — two lines per test case.
The first line of each test case contains integers n and d (1 ≤ n,d ≤ 100) — the number of haybale piles and the number of days, respectively.
The second line of each test case contains n integers a_1, a_2, …, a_n (0 ≤ a_i ≤ 100) — the number of haybales in each pile.
Output
For each test case, output one integer: the maximum number of haybales that may be in pile 1 after d days if Bessie acts optimally.
Example
Input
3
4 5
1 0 3 2
2 2
100 1
1 8
0
Output
3
101
0
Note
In the first test case of the sample, this is one possible way Bessie can end up with 3 haybales in pile 1:
* On day one, move a haybale from pile 3 to pile 2
* On day two, move a haybale from pile 3 to pile 2
* On day three, move a haybale from pile 2 to pile 1
* On day four, move a haybale from pile 2 to pile 1
* On day five, do nothing
In the second test case of the sample, Bessie can do nothing on the first day and move a haybale from pile 2 to pile 1 on the second day.
Submitted Solution:
```
for _ in range(int(input())):
n, d = map(int, input().split())
arr = list(map(int, input().split()))
ans = 0
for i in range(1, n):
if d - i < 0:
break
while arr[i] > 0 and d - i >= 0:
d -= i
ans += 1
arr[i] -= 1
print(arr[0] + ans)
``` | instruction | 0 | 46,391 | 8 | 92,782 |
Yes | output | 1 | 46,391 | 8 | 92,783 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
The USA Construction Operation (USACO) recently ordered Farmer John to arrange a row of n haybale piles on the farm. The i-th pile contains a_i haybales.
However, Farmer John has just left for vacation, leaving Bessie all on her own. Every day, Bessie the naughty cow can choose to move one haybale in any pile to an adjacent pile. Formally, in one day she can choose any two indices i and j (1 ≤ i, j ≤ n) such that |i-j|=1 and a_i>0 and apply a_i = a_i - 1, a_j = a_j + 1. She may also decide to not do anything on some days because she is lazy.
Bessie wants to maximize the number of haybales in pile 1 (i.e. to maximize a_1), and she only has d days to do so before Farmer John returns. Help her find the maximum number of haybales that may be in pile 1 if she acts optimally!
Input
The input consists of multiple test cases. The first line contains an integer t (1 ≤ t ≤ 100) — the number of test cases. Next 2t lines contain a description of test cases — two lines per test case.
The first line of each test case contains integers n and d (1 ≤ n,d ≤ 100) — the number of haybale piles and the number of days, respectively.
The second line of each test case contains n integers a_1, a_2, …, a_n (0 ≤ a_i ≤ 100) — the number of haybales in each pile.
Output
For each test case, output one integer: the maximum number of haybales that may be in pile 1 after d days if Bessie acts optimally.
Example
Input
3
4 5
1 0 3 2
2 2
100 1
1 8
0
Output
3
101
0
Note
In the first test case of the sample, this is one possible way Bessie can end up with 3 haybales in pile 1:
* On day one, move a haybale from pile 3 to pile 2
* On day two, move a haybale from pile 3 to pile 2
* On day three, move a haybale from pile 2 to pile 1
* On day four, move a haybale from pile 2 to pile 1
* On day five, do nothing
In the second test case of the sample, Bessie can do nothing on the first day and move a haybale from pile 2 to pile 1 on the second day.
Submitted Solution:
```
t = int (input ())
ans = []
for i in range (t):
day = 0
p = list (map (int, input ().split ()))
n, d = p #
flag = False
num = list (map (int, input ().split ()))
for k in range (1, n) :
while (num[k] > 0 and day + k <= d) :
num[0] += 1
num[k] -= 1
day += k
# print ("DEBUG : ", k, day)
if (day + k > d) :
flag = True
break
if (flag) :
break
ans.append (num[0])
for j in ans :
print (j)
``` | instruction | 0 | 46,392 | 8 | 92,784 |
Yes | output | 1 | 46,392 | 8 | 92,785 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
The USA Construction Operation (USACO) recently ordered Farmer John to arrange a row of n haybale piles on the farm. The i-th pile contains a_i haybales.
However, Farmer John has just left for vacation, leaving Bessie all on her own. Every day, Bessie the naughty cow can choose to move one haybale in any pile to an adjacent pile. Formally, in one day she can choose any two indices i and j (1 ≤ i, j ≤ n) such that |i-j|=1 and a_i>0 and apply a_i = a_i - 1, a_j = a_j + 1. She may also decide to not do anything on some days because she is lazy.
Bessie wants to maximize the number of haybales in pile 1 (i.e. to maximize a_1), and she only has d days to do so before Farmer John returns. Help her find the maximum number of haybales that may be in pile 1 if she acts optimally!
Input
The input consists of multiple test cases. The first line contains an integer t (1 ≤ t ≤ 100) — the number of test cases. Next 2t lines contain a description of test cases — two lines per test case.
The first line of each test case contains integers n and d (1 ≤ n,d ≤ 100) — the number of haybale piles and the number of days, respectively.
The second line of each test case contains n integers a_1, a_2, …, a_n (0 ≤ a_i ≤ 100) — the number of haybales in each pile.
Output
For each test case, output one integer: the maximum number of haybales that may be in pile 1 after d days if Bessie acts optimally.
Example
Input
3
4 5
1 0 3 2
2 2
100 1
1 8
0
Output
3
101
0
Note
In the first test case of the sample, this is one possible way Bessie can end up with 3 haybales in pile 1:
* On day one, move a haybale from pile 3 to pile 2
* On day two, move a haybale from pile 3 to pile 2
* On day three, move a haybale from pile 2 to pile 1
* On day four, move a haybale from pile 2 to pile 1
* On day five, do nothing
In the second test case of the sample, Bessie can do nothing on the first day and move a haybale from pile 2 to pile 1 on the second day.
Submitted Solution:
```
t= int(input())
for _ in range(t):
n,d = map(int,input().split())
l = list(map(int,input().split()))
cnt=0
for i in range(1,n):
cnt=i
while(l[i]>0 and d>0):
d=d-cnt
if(d>=0):
l[0]+=1
l[i]-=1
if(d<0):
break
print(l[0])
``` | instruction | 0 | 46,393 | 8 | 92,786 |
Yes | output | 1 | 46,393 | 8 | 92,787 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
The USA Construction Operation (USACO) recently ordered Farmer John to arrange a row of n haybale piles on the farm. The i-th pile contains a_i haybales.
However, Farmer John has just left for vacation, leaving Bessie all on her own. Every day, Bessie the naughty cow can choose to move one haybale in any pile to an adjacent pile. Formally, in one day she can choose any two indices i and j (1 ≤ i, j ≤ n) such that |i-j|=1 and a_i>0 and apply a_i = a_i - 1, a_j = a_j + 1. She may also decide to not do anything on some days because she is lazy.
Bessie wants to maximize the number of haybales in pile 1 (i.e. to maximize a_1), and she only has d days to do so before Farmer John returns. Help her find the maximum number of haybales that may be in pile 1 if she acts optimally!
Input
The input consists of multiple test cases. The first line contains an integer t (1 ≤ t ≤ 100) — the number of test cases. Next 2t lines contain a description of test cases — two lines per test case.
The first line of each test case contains integers n and d (1 ≤ n,d ≤ 100) — the number of haybale piles and the number of days, respectively.
The second line of each test case contains n integers a_1, a_2, …, a_n (0 ≤ a_i ≤ 100) — the number of haybales in each pile.
Output
For each test case, output one integer: the maximum number of haybales that may be in pile 1 after d days if Bessie acts optimally.
Example
Input
3
4 5
1 0 3 2
2 2
100 1
1 8
0
Output
3
101
0
Note
In the first test case of the sample, this is one possible way Bessie can end up with 3 haybales in pile 1:
* On day one, move a haybale from pile 3 to pile 2
* On day two, move a haybale from pile 3 to pile 2
* On day three, move a haybale from pile 2 to pile 1
* On day four, move a haybale from pile 2 to pile 1
* On day five, do nothing
In the second test case of the sample, Bessie can do nothing on the first day and move a haybale from pile 2 to pile 1 on the second day.
Submitted Solution:
```
from math import *
def f(d, pile, pilenum):
if d>0 and pilenum!=0:
if floor(d/pilenum)>pile:
return pile
else:
return floor(d/pilenum)
else:
return pile
t=int(input())
ans=[]
for i in range(0, t):
n, d=map(int, input().split())
pile=list(map(int, input().split()))
ans.append(0)
for j in range(0, len(pile)):
ans[i]+=f(d, pile[j], j)
d-=(f(d, pile[j], j)*j)
for i in range(0, t):
print(ans[i])
``` | instruction | 0 | 46,394 | 8 | 92,788 |
No | output | 1 | 46,394 | 8 | 92,789 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
The USA Construction Operation (USACO) recently ordered Farmer John to arrange a row of n haybale piles on the farm. The i-th pile contains a_i haybales.
However, Farmer John has just left for vacation, leaving Bessie all on her own. Every day, Bessie the naughty cow can choose to move one haybale in any pile to an adjacent pile. Formally, in one day she can choose any two indices i and j (1 ≤ i, j ≤ n) such that |i-j|=1 and a_i>0 and apply a_i = a_i - 1, a_j = a_j + 1. She may also decide to not do anything on some days because she is lazy.
Bessie wants to maximize the number of haybales in pile 1 (i.e. to maximize a_1), and she only has d days to do so before Farmer John returns. Help her find the maximum number of haybales that may be in pile 1 if she acts optimally!
Input
The input consists of multiple test cases. The first line contains an integer t (1 ≤ t ≤ 100) — the number of test cases. Next 2t lines contain a description of test cases — two lines per test case.
The first line of each test case contains integers n and d (1 ≤ n,d ≤ 100) — the number of haybale piles and the number of days, respectively.
The second line of each test case contains n integers a_1, a_2, …, a_n (0 ≤ a_i ≤ 100) — the number of haybales in each pile.
Output
For each test case, output one integer: the maximum number of haybales that may be in pile 1 after d days if Bessie acts optimally.
Example
Input
3
4 5
1 0 3 2
2 2
100 1
1 8
0
Output
3
101
0
Note
In the first test case of the sample, this is one possible way Bessie can end up with 3 haybales in pile 1:
* On day one, move a haybale from pile 3 to pile 2
* On day two, move a haybale from pile 3 to pile 2
* On day three, move a haybale from pile 2 to pile 1
* On day four, move a haybale from pile 2 to pile 1
* On day five, do nothing
In the second test case of the sample, Bessie can do nothing on the first day and move a haybale from pile 2 to pile 1 on the second day.
Submitted Solution:
```
t = int(input())
for i in range(t):
n, d = map(int, input().split())
a = list( map(int, input().split()) )
p = 1
while p < n and a[p] == 0:
p += 1
if p == n:
print(a[0])
continue
while a[p] > 0 and d >= p:
a[0] += 1
a[p] -= 1;
d -= p
if a[p] == 0:
p += 1
if p == n:
break
print(a[0])
``` | instruction | 0 | 46,395 | 8 | 92,790 |
No | output | 1 | 46,395 | 8 | 92,791 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
The USA Construction Operation (USACO) recently ordered Farmer John to arrange a row of n haybale piles on the farm. The i-th pile contains a_i haybales.
However, Farmer John has just left for vacation, leaving Bessie all on her own. Every day, Bessie the naughty cow can choose to move one haybale in any pile to an adjacent pile. Formally, in one day she can choose any two indices i and j (1 ≤ i, j ≤ n) such that |i-j|=1 and a_i>0 and apply a_i = a_i - 1, a_j = a_j + 1. She may also decide to not do anything on some days because she is lazy.
Bessie wants to maximize the number of haybales in pile 1 (i.e. to maximize a_1), and she only has d days to do so before Farmer John returns. Help her find the maximum number of haybales that may be in pile 1 if she acts optimally!
Input
The input consists of multiple test cases. The first line contains an integer t (1 ≤ t ≤ 100) — the number of test cases. Next 2t lines contain a description of test cases — two lines per test case.
The first line of each test case contains integers n and d (1 ≤ n,d ≤ 100) — the number of haybale piles and the number of days, respectively.
The second line of each test case contains n integers a_1, a_2, …, a_n (0 ≤ a_i ≤ 100) — the number of haybales in each pile.
Output
For each test case, output one integer: the maximum number of haybales that may be in pile 1 after d days if Bessie acts optimally.
Example
Input
3
4 5
1 0 3 2
2 2
100 1
1 8
0
Output
3
101
0
Note
In the first test case of the sample, this is one possible way Bessie can end up with 3 haybales in pile 1:
* On day one, move a haybale from pile 3 to pile 2
* On day two, move a haybale from pile 3 to pile 2
* On day three, move a haybale from pile 2 to pile 1
* On day four, move a haybale from pile 2 to pile 1
* On day five, do nothing
In the second test case of the sample, Bessie can do nothing on the first day and move a haybale from pile 2 to pile 1 on the second day.
Submitted Solution:
```
try:
for _ in range(int(input())):
a,b=map(int,input().split())
l=list(map(int,input().split()))
k=l[0]
l.pop(0)
for i in range(a-1):
if((b-(l[i]*2))<=0):
k=k+(b//2)
break
else:
k=k+l[i]
b=b-2*l[i]
print(k)
except:
pass
``` | instruction | 0 | 46,396 | 8 | 92,792 |
No | output | 1 | 46,396 | 8 | 92,793 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
The USA Construction Operation (USACO) recently ordered Farmer John to arrange a row of n haybale piles on the farm. The i-th pile contains a_i haybales.
However, Farmer John has just left for vacation, leaving Bessie all on her own. Every day, Bessie the naughty cow can choose to move one haybale in any pile to an adjacent pile. Formally, in one day she can choose any two indices i and j (1 ≤ i, j ≤ n) such that |i-j|=1 and a_i>0 and apply a_i = a_i - 1, a_j = a_j + 1. She may also decide to not do anything on some days because she is lazy.
Bessie wants to maximize the number of haybales in pile 1 (i.e. to maximize a_1), and she only has d days to do so before Farmer John returns. Help her find the maximum number of haybales that may be in pile 1 if she acts optimally!
Input
The input consists of multiple test cases. The first line contains an integer t (1 ≤ t ≤ 100) — the number of test cases. Next 2t lines contain a description of test cases — two lines per test case.
The first line of each test case contains integers n and d (1 ≤ n,d ≤ 100) — the number of haybale piles and the number of days, respectively.
The second line of each test case contains n integers a_1, a_2, …, a_n (0 ≤ a_i ≤ 100) — the number of haybales in each pile.
Output
For each test case, output one integer: the maximum number of haybales that may be in pile 1 after d days if Bessie acts optimally.
Example
Input
3
4 5
1 0 3 2
2 2
100 1
1 8
0
Output
3
101
0
Note
In the first test case of the sample, this is one possible way Bessie can end up with 3 haybales in pile 1:
* On day one, move a haybale from pile 3 to pile 2
* On day two, move a haybale from pile 3 to pile 2
* On day three, move a haybale from pile 2 to pile 1
* On day four, move a haybale from pile 2 to pile 1
* On day five, do nothing
In the second test case of the sample, Bessie can do nothing on the first day and move a haybale from pile 2 to pile 1 on the second day.
Submitted Solution:
```
testcases = int(input())
for testcase in range(testcases):
getInput = str(input())
getInput = getInput.split()
piles = int(getInput[0])
days = int(getInput[1])
arr = []
getInput = str(input())
getInput = getInput.split()
count = 0
for i in range(piles):
arr.append( int(getInput[i]) )
times = 1
for i in range(piles):
if days <= 0:
break
if i == 0 :
continue
if arr[i] == 0 :
times += 1
continue
number = days// times
if number <= arr[i]:
count += min( number, arr[i])
break
days -= arr[i]
times += 1
count += arr[i]
print(count + arr[0])
``` | instruction | 0 | 46,397 | 8 | 92,794 |
No | output | 1 | 46,397 | 8 | 92,795 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Edo has got a collection of n refrigerator magnets!
He decided to buy a refrigerator and hang the magnets on the door. The shop can make the refrigerator with any size of the door that meets the following restrictions: the refrigerator door must be rectangle, and both the length and the width of the door must be positive integers.
Edo figured out how he wants to place the magnets on the refrigerator. He introduced a system of coordinates on the plane, where each magnet is represented as a rectangle with sides parallel to the coordinate axes.
Now he wants to remove no more than k magnets (he may choose to keep all of them) and attach all remaining magnets to the refrigerator door, and the area of the door should be as small as possible. A magnet is considered to be attached to the refrigerator door if its center lies on the door or on its boundary. The relative positions of all the remaining magnets must correspond to the plan.
Let us explain the last two sentences. Let's suppose we want to hang two magnets on the refrigerator. If the magnet in the plan has coordinates of the lower left corner (x1, y1) and the upper right corner (x2, y2), then its center is located at (<image>, <image>) (may not be integers). By saying the relative position should correspond to the plan we mean that the only available operation is translation, i.e. the vector connecting the centers of two magnets in the original plan, must be equal to the vector connecting the centers of these two magnets on the refrigerator.
The sides of the refrigerator door must also be parallel to coordinate axes.
Input
The first line contains two integers n and k (1 ≤ n ≤ 100 000, 0 ≤ k ≤ min(10, n - 1)) — the number of magnets that Edo has and the maximum number of magnets Edo may not place on the refrigerator.
Next n lines describe the initial plan of placing magnets. Each line contains four integers x1, y1, x2, y2 (1 ≤ x1 < x2 ≤ 109, 1 ≤ y1 < y2 ≤ 109) — the coordinates of the lower left and upper right corners of the current magnet. The magnets can partially overlap or even fully coincide.
Output
Print a single integer — the minimum area of the door of refrigerator, which can be used to place at least n - k magnets, preserving the relative positions.
Examples
Input
3 1
1 1 2 2
2 2 3 3
3 3 4 4
Output
1
Input
4 1
1 1 2 2
1 9 2 10
9 9 10 10
9 1 10 2
Output
64
Input
3 0
1 1 2 2
1 1 1000000000 1000000000
1 3 8 12
Output
249999999000000001
Note
In the first test sample it is optimal to remove either the first or the third magnet. If we remove the first magnet, the centers of two others will lie at points (2.5, 2.5) and (3.5, 3.5). Thus, it is enough to buy a fridge with door width 1 and door height 1, the area of the door also equals one, correspondingly.
In the second test sample it doesn't matter which magnet to remove, the answer will not change — we need a fridge with door width 8 and door height 8.
In the third sample you cannot remove anything as k = 0. | instruction | 0 | 47,478 | 8 | 94,956 |
Tags: brute force, greedy, implementation, two pointers
Correct Solution:
```
from sys import*
#
def check(u, d, l, r):
used = [pointsx[i][1] for i in range(l)]
used += [pointsx[-1 - i][1] for i in range(r)]
used += [pointsy[i][1] for i in range(u)]
used += [pointsy[-1 - i][1] for i in range(d)]
if len(set(used)) > k:
return DOHERA
dx = pointsx[-1 - r][0] - pointsx[l][0]
dy = pointsy[-1 - d][0] - pointsy[u][0]
dx += dx & 1
dy += dy & 1
dx = max(2, dx)
dy = max(2, dy)
return dx * dy
#
(n, k) = map(int, input().split())
pointsx = []
pointsy = []
DOHERA = 10 ** 228
for i in range(n):
a = list(map(int, input().split()))
pointsx += [(a[0] + a[2], i)]
pointsy += [(a[1] + a[3], i)]
(pointsx, pointsy) = (sorted(pointsx), sorted(pointsy))
ans = DOHERA
for u in range(0, k + 1):
for d in range(0, k + 1):
for l in range(0, k + 1):
for r in range(0, k + 1):
ans = min(ans, check(u, d, l, r))
print(ans // 4)
``` | output | 1 | 47,478 | 8 | 94,957 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Edo has got a collection of n refrigerator magnets!
He decided to buy a refrigerator and hang the magnets on the door. The shop can make the refrigerator with any size of the door that meets the following restrictions: the refrigerator door must be rectangle, and both the length and the width of the door must be positive integers.
Edo figured out how he wants to place the magnets on the refrigerator. He introduced a system of coordinates on the plane, where each magnet is represented as a rectangle with sides parallel to the coordinate axes.
Now he wants to remove no more than k magnets (he may choose to keep all of them) and attach all remaining magnets to the refrigerator door, and the area of the door should be as small as possible. A magnet is considered to be attached to the refrigerator door if its center lies on the door or on its boundary. The relative positions of all the remaining magnets must correspond to the plan.
Let us explain the last two sentences. Let's suppose we want to hang two magnets on the refrigerator. If the magnet in the plan has coordinates of the lower left corner (x1, y1) and the upper right corner (x2, y2), then its center is located at (<image>, <image>) (may not be integers). By saying the relative position should correspond to the plan we mean that the only available operation is translation, i.e. the vector connecting the centers of two magnets in the original plan, must be equal to the vector connecting the centers of these two magnets on the refrigerator.
The sides of the refrigerator door must also be parallel to coordinate axes.
Input
The first line contains two integers n and k (1 ≤ n ≤ 100 000, 0 ≤ k ≤ min(10, n - 1)) — the number of magnets that Edo has and the maximum number of magnets Edo may not place on the refrigerator.
Next n lines describe the initial plan of placing magnets. Each line contains four integers x1, y1, x2, y2 (1 ≤ x1 < x2 ≤ 109, 1 ≤ y1 < y2 ≤ 109) — the coordinates of the lower left and upper right corners of the current magnet. The magnets can partially overlap or even fully coincide.
Output
Print a single integer — the minimum area of the door of refrigerator, which can be used to place at least n - k magnets, preserving the relative positions.
Examples
Input
3 1
1 1 2 2
2 2 3 3
3 3 4 4
Output
1
Input
4 1
1 1 2 2
1 9 2 10
9 9 10 10
9 1 10 2
Output
64
Input
3 0
1 1 2 2
1 1 1000000000 1000000000
1 3 8 12
Output
249999999000000001
Note
In the first test sample it is optimal to remove either the first or the third magnet. If we remove the first magnet, the centers of two others will lie at points (2.5, 2.5) and (3.5, 3.5). Thus, it is enough to buy a fridge with door width 1 and door height 1, the area of the door also equals one, correspondingly.
In the second test sample it doesn't matter which magnet to remove, the answer will not change — we need a fridge with door width 8 and door height 8.
In the third sample you cannot remove anything as k = 0. | instruction | 0 | 47,479 | 8 | 94,958 |
Tags: brute force, greedy, implementation, two pointers
Correct Solution:
```
from sys import*
#
def check(u, d, l, r):
used = [pointsx[i][1] for i in range(l)]
used += [pointsx[-1 - i][1] for i in range(r)]
used += [pointsy[i][1] for i in range(u)]
used += [pointsy[-1 - i][1] for i in range(d)]
if len(set(used)) > k:
return DOHERA
dx = pointsx[-1 - r][0] - pointsx[l][0]
dy = pointsy[-1 - d][0] - pointsy[u][0]
dx += dx & 1
dy += dy & 1
dx = max(2, dx)
dy = max(2, dy)
return dx * dy
#
(n, k) = map(int, input().split())
pointsx = []
pointsy = []
DOHERA = 10 ** 228
for i in range(n):
a = list(map(int, input().split()))
pointsx += [(a[0] + a[2], i)]
pointsy += [(a[1] + a[3], i)]
(pointsx, pointsy) = (sorted(pointsx), sorted(pointsy))
ans = DOHERA
for u in range(0, k + 1):
for d in range(0, k + 1):
for l in range(0, k + 1):
for r in range(0, k + 1):
if l + r <= k and u + d <= k:
ans = min(ans, check(u, d, l, r))
print(ans // 4)
``` | output | 1 | 47,479 | 8 | 94,959 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Edo has got a collection of n refrigerator magnets!
He decided to buy a refrigerator and hang the magnets on the door. The shop can make the refrigerator with any size of the door that meets the following restrictions: the refrigerator door must be rectangle, and both the length and the width of the door must be positive integers.
Edo figured out how he wants to place the magnets on the refrigerator. He introduced a system of coordinates on the plane, where each magnet is represented as a rectangle with sides parallel to the coordinate axes.
Now he wants to remove no more than k magnets (he may choose to keep all of them) and attach all remaining magnets to the refrigerator door, and the area of the door should be as small as possible. A magnet is considered to be attached to the refrigerator door if its center lies on the door or on its boundary. The relative positions of all the remaining magnets must correspond to the plan.
Let us explain the last two sentences. Let's suppose we want to hang two magnets on the refrigerator. If the magnet in the plan has coordinates of the lower left corner (x1, y1) and the upper right corner (x2, y2), then its center is located at (<image>, <image>) (may not be integers). By saying the relative position should correspond to the plan we mean that the only available operation is translation, i.e. the vector connecting the centers of two magnets in the original plan, must be equal to the vector connecting the centers of these two magnets on the refrigerator.
The sides of the refrigerator door must also be parallel to coordinate axes.
Input
The first line contains two integers n and k (1 ≤ n ≤ 100 000, 0 ≤ k ≤ min(10, n - 1)) — the number of magnets that Edo has and the maximum number of magnets Edo may not place on the refrigerator.
Next n lines describe the initial plan of placing magnets. Each line contains four integers x1, y1, x2, y2 (1 ≤ x1 < x2 ≤ 109, 1 ≤ y1 < y2 ≤ 109) — the coordinates of the lower left and upper right corners of the current magnet. The magnets can partially overlap or even fully coincide.
Output
Print a single integer — the minimum area of the door of refrigerator, which can be used to place at least n - k magnets, preserving the relative positions.
Examples
Input
3 1
1 1 2 2
2 2 3 3
3 3 4 4
Output
1
Input
4 1
1 1 2 2
1 9 2 10
9 9 10 10
9 1 10 2
Output
64
Input
3 0
1 1 2 2
1 1 1000000000 1000000000
1 3 8 12
Output
249999999000000001
Note
In the first test sample it is optimal to remove either the first or the third magnet. If we remove the first magnet, the centers of two others will lie at points (2.5, 2.5) and (3.5, 3.5). Thus, it is enough to buy a fridge with door width 1 and door height 1, the area of the door also equals one, correspondingly.
In the second test sample it doesn't matter which magnet to remove, the answer will not change — we need a fridge with door width 8 and door height 8.
In the third sample you cannot remove anything as k = 0. | instruction | 0 | 47,480 | 8 | 94,960 |
Tags: brute force, greedy, implementation, two pointers
Correct Solution:
```
from sys import*
#
def check(u, d, l, r):
used = [pointsx[i][1] for i in range(l)]
used += [pointsx[-1 - i][1] for i in range(r)]
used += [pointsy[i][1] for i in range(u)]
used += [pointsy[-1 - i][1] for i in range(d)]
if len(set(used)) > k:
return DOHERA
dx = pointsx[-1 - r][0] - pointsx[l][0]
dy = pointsy[-1 - d][0] - pointsy[u][0]
dx += dx & 1
dy += dy & 1
dx = max(2, dx)
dy = max(2, dy)
return dx * dy
#
(n, k) = map(int, input().split())
pointsx = []
pointsy = []
DOHERA = 10 ** 228
for i in range(n):
a = list(map(int, input().split()))
pointsx += [(a[0] + a[2], i)]
pointsy += [(a[1] + a[3], i)]
(pointsx, pointsy) = (sorted(pointsx), sorted(pointsy))
ans = DOHERA
for u in range(0, k + 1):
for d in range(0, k + 1):
for l in range(0, k + 1):
for r in range(0, k + 1):
if l + r <= k and u + d <= k:
ans = min(ans, check(u, d, l, r))
print(ans // 4)
# Made By Mostafa_Khaled
``` | output | 1 | 47,480 | 8 | 94,961 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Edo has got a collection of n refrigerator magnets!
He decided to buy a refrigerator and hang the magnets on the door. The shop can make the refrigerator with any size of the door that meets the following restrictions: the refrigerator door must be rectangle, and both the length and the width of the door must be positive integers.
Edo figured out how he wants to place the magnets on the refrigerator. He introduced a system of coordinates on the plane, where each magnet is represented as a rectangle with sides parallel to the coordinate axes.
Now he wants to remove no more than k magnets (he may choose to keep all of them) and attach all remaining magnets to the refrigerator door, and the area of the door should be as small as possible. A magnet is considered to be attached to the refrigerator door if its center lies on the door or on its boundary. The relative positions of all the remaining magnets must correspond to the plan.
Let us explain the last two sentences. Let's suppose we want to hang two magnets on the refrigerator. If the magnet in the plan has coordinates of the lower left corner (x1, y1) and the upper right corner (x2, y2), then its center is located at (<image>, <image>) (may not be integers). By saying the relative position should correspond to the plan we mean that the only available operation is translation, i.e. the vector connecting the centers of two magnets in the original plan, must be equal to the vector connecting the centers of these two magnets on the refrigerator.
The sides of the refrigerator door must also be parallel to coordinate axes.
Input
The first line contains two integers n and k (1 ≤ n ≤ 100 000, 0 ≤ k ≤ min(10, n - 1)) — the number of magnets that Edo has and the maximum number of magnets Edo may not place on the refrigerator.
Next n lines describe the initial plan of placing magnets. Each line contains four integers x1, y1, x2, y2 (1 ≤ x1 < x2 ≤ 109, 1 ≤ y1 < y2 ≤ 109) — the coordinates of the lower left and upper right corners of the current magnet. The magnets can partially overlap or even fully coincide.
Output
Print a single integer — the minimum area of the door of refrigerator, which can be used to place at least n - k magnets, preserving the relative positions.
Examples
Input
3 1
1 1 2 2
2 2 3 3
3 3 4 4
Output
1
Input
4 1
1 1 2 2
1 9 2 10
9 9 10 10
9 1 10 2
Output
64
Input
3 0
1 1 2 2
1 1 1000000000 1000000000
1 3 8 12
Output
249999999000000001
Note
In the first test sample it is optimal to remove either the first or the third magnet. If we remove the first magnet, the centers of two others will lie at points (2.5, 2.5) and (3.5, 3.5). Thus, it is enough to buy a fridge with door width 1 and door height 1, the area of the door also equals one, correspondingly.
In the second test sample it doesn't matter which magnet to remove, the answer will not change — we need a fridge with door width 8 and door height 8.
In the third sample you cannot remove anything as k = 0. | instruction | 0 | 47,481 | 8 | 94,962 |
Tags: brute force, greedy, implementation, two pointers
Correct Solution:
```
from math import ceil
def add(c, k, s, i0, di):
i = i0
while len(s) < k:
s.add(c[i])
i += di
def fst(c, s, i0, di):
i = i0
while c[i] in s:
i += di
return c[i]
def doors(ms, k):
cs = [((m[2]+m[0])/2, (m[3]+m[1])/2, i) for i, m in enumerate(ms)]
cx = sorted(cs, key=lambda t: t[0])
cy = sorted(cs, key=lambda t: t[1])
for kl in range(k+1):
s1 = set(cx[:kl])
for kr in range(k+1-kl):
s2 = s1 | set(cx[-kr:] if kr else [])
add(cy, kl+kr, s2, -kr-1, -1)
for ku in range(k+1-kl-kr):
s3 = s2 | set(cy[:ku])
add(cy, kl+kr+ku, s3, ku, 1)
kd = k-kl-kr-ku
s4 = s3 | set(cy[-kd:] if kd else [])
add(cy, kl+kr+ku+kd, s4, -kd-1, -1)
minx = fst(cx, s4, kl, 1)[0]
maxx = fst(cx, s4, -kr-1, -1)[0]
miny = fst(cy, s4, ku, 1)[1]
maxy = fst(cy, s4, -kd-1, -1)[1]
a = max(ceil(maxx-minx), 1) * max(ceil(maxy-miny), 1)
yield a
def min_door(ms, k):
return min(doors(ms, k))
def quad(i):
a, b, c, d = i
return a, b, c, d
if __name__ == '__main__':
n, k = map(int, input().split())
ms = (quad(map(int, input().split())) for _ in range(n))
print(min_door(ms, k))
``` | output | 1 | 47,481 | 8 | 94,963 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Edo has got a collection of n refrigerator magnets!
He decided to buy a refrigerator and hang the magnets on the door. The shop can make the refrigerator with any size of the door that meets the following restrictions: the refrigerator door must be rectangle, and both the length and the width of the door must be positive integers.
Edo figured out how he wants to place the magnets on the refrigerator. He introduced a system of coordinates on the plane, where each magnet is represented as a rectangle with sides parallel to the coordinate axes.
Now he wants to remove no more than k magnets (he may choose to keep all of them) and attach all remaining magnets to the refrigerator door, and the area of the door should be as small as possible. A magnet is considered to be attached to the refrigerator door if its center lies on the door or on its boundary. The relative positions of all the remaining magnets must correspond to the plan.
Let us explain the last two sentences. Let's suppose we want to hang two magnets on the refrigerator. If the magnet in the plan has coordinates of the lower left corner (x1, y1) and the upper right corner (x2, y2), then its center is located at (<image>, <image>) (may not be integers). By saying the relative position should correspond to the plan we mean that the only available operation is translation, i.e. the vector connecting the centers of two magnets in the original plan, must be equal to the vector connecting the centers of these two magnets on the refrigerator.
The sides of the refrigerator door must also be parallel to coordinate axes.
Input
The first line contains two integers n and k (1 ≤ n ≤ 100 000, 0 ≤ k ≤ min(10, n - 1)) — the number of magnets that Edo has and the maximum number of magnets Edo may not place on the refrigerator.
Next n lines describe the initial plan of placing magnets. Each line contains four integers x1, y1, x2, y2 (1 ≤ x1 < x2 ≤ 109, 1 ≤ y1 < y2 ≤ 109) — the coordinates of the lower left and upper right corners of the current magnet. The magnets can partially overlap or even fully coincide.
Output
Print a single integer — the minimum area of the door of refrigerator, which can be used to place at least n - k magnets, preserving the relative positions.
Examples
Input
3 1
1 1 2 2
2 2 3 3
3 3 4 4
Output
1
Input
4 1
1 1 2 2
1 9 2 10
9 9 10 10
9 1 10 2
Output
64
Input
3 0
1 1 2 2
1 1 1000000000 1000000000
1 3 8 12
Output
249999999000000001
Note
In the first test sample it is optimal to remove either the first or the third magnet. If we remove the first magnet, the centers of two others will lie at points (2.5, 2.5) and (3.5, 3.5). Thus, it is enough to buy a fridge with door width 1 and door height 1, the area of the door also equals one, correspondingly.
In the second test sample it doesn't matter which magnet to remove, the answer will not change — we need a fridge with door width 8 and door height 8.
In the third sample you cannot remove anything as k = 0.
Submitted Solution:
```
def read_data():
n, k = map(int, input().split())
xs = []
ys = []
for i in range(n):
x1, y1, x2, y2 = map(int, input().split())
x = (x1 + x2)/2
y = (y1 + y2)/2
xs.append((x, i))
ys.append((y, i))
return n, k, xs, ys
def solve(n, k, xs, ys):
xs.sort()
ys.sort()
min_area = float('inf')
for L, R, B, T in generate_candidates(n, k, xs, ys):
min_area = min(min_area, int(R - L + 0.5) * int(T - B + 0.5))
return min_area
def generate_candidates(n, k, xs, ys):
for i_x0 in range(k + 1):
for i_x1 in range(n-1, n-k-2, -1):
if i_x0 > i_x1:
break
for i_y0 in range(k + 1):
for i_y1 in range(n-1, n-k-2, -1):
if i_y0 > i_y1:
break
LRBT = find_LRBT(i_x0, i_x1, i_y0, i_y1, n, k, xs, ys)
if LRBT:
yield LRBT
def find_LRBT(i_x0, i_x1, i_y0, i_y1, n, k, xs, ys):
removed = set()
for x0, j_x0 in xs[:i_x0]:
removed.add(j_x0)
for x1, j_x1 in xs[i_x1+1:]:
removed.add(j_x1)
for y0, j_y0 in ys[:i_y0]:
removed.add(j_y0)
for y1, j_y1 in ys[i_y1+1:]:
removed.add(j_y1)
if len(removed) > k:
return False
for x0, j_x0 in xs:
if j_x0 not in removed:
L = x0
break
for x1, j_x1 in xs[::-1]:
if j_x1 not in removed:
R = x1
break
for y0, j_y0 in ys:
if j_y0 not in removed:
B = y0
break
for y1, j_y1 in ys[::-1]:
if j_y1 not in removed:
T = y1
break
return L, R, B, T
n, k, xs, ys = read_data()
print(solve(n, k, xs, ys))
``` | instruction | 0 | 47,482 | 8 | 94,964 |
No | output | 1 | 47,482 | 8 | 94,965 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Edo has got a collection of n refrigerator magnets!
He decided to buy a refrigerator and hang the magnets on the door. The shop can make the refrigerator with any size of the door that meets the following restrictions: the refrigerator door must be rectangle, and both the length and the width of the door must be positive integers.
Edo figured out how he wants to place the magnets on the refrigerator. He introduced a system of coordinates on the plane, where each magnet is represented as a rectangle with sides parallel to the coordinate axes.
Now he wants to remove no more than k magnets (he may choose to keep all of them) and attach all remaining magnets to the refrigerator door, and the area of the door should be as small as possible. A magnet is considered to be attached to the refrigerator door if its center lies on the door or on its boundary. The relative positions of all the remaining magnets must correspond to the plan.
Let us explain the last two sentences. Let's suppose we want to hang two magnets on the refrigerator. If the magnet in the plan has coordinates of the lower left corner (x1, y1) and the upper right corner (x2, y2), then its center is located at (<image>, <image>) (may not be integers). By saying the relative position should correspond to the plan we mean that the only available operation is translation, i.e. the vector connecting the centers of two magnets in the original plan, must be equal to the vector connecting the centers of these two magnets on the refrigerator.
The sides of the refrigerator door must also be parallel to coordinate axes.
Input
The first line contains two integers n and k (1 ≤ n ≤ 100 000, 0 ≤ k ≤ min(10, n - 1)) — the number of magnets that Edo has and the maximum number of magnets Edo may not place on the refrigerator.
Next n lines describe the initial plan of placing magnets. Each line contains four integers x1, y1, x2, y2 (1 ≤ x1 < x2 ≤ 109, 1 ≤ y1 < y2 ≤ 109) — the coordinates of the lower left and upper right corners of the current magnet. The magnets can partially overlap or even fully coincide.
Output
Print a single integer — the minimum area of the door of refrigerator, which can be used to place at least n - k magnets, preserving the relative positions.
Examples
Input
3 1
1 1 2 2
2 2 3 3
3 3 4 4
Output
1
Input
4 1
1 1 2 2
1 9 2 10
9 9 10 10
9 1 10 2
Output
64
Input
3 0
1 1 2 2
1 1 1000000000 1000000000
1 3 8 12
Output
249999999000000001
Note
In the first test sample it is optimal to remove either the first or the third magnet. If we remove the first magnet, the centers of two others will lie at points (2.5, 2.5) and (3.5, 3.5). Thus, it is enough to buy a fridge with door width 1 and door height 1, the area of the door also equals one, correspondingly.
In the second test sample it doesn't matter which magnet to remove, the answer will not change — we need a fridge with door width 8 and door height 8.
In the third sample you cannot remove anything as k = 0.
Submitted Solution:
```
from sys import*
#
def check(u, d, l, r):
used = [pointsx[i][1] for i in range(l)]
used += [pointsx[-1 - i][1] for i in range(r)]
used += [pointsy[i][1] for i in range(u)]
used += [pointsy[-1 - i][1] for i in range(d)]
if len(set(used)) > k:
return 3 * 10 ** 18
dx = pointsx[-1 - r][0] - pointsx[l][0]
dy = pointsy[-1 - d][0] - pointsy[u][0]
dx += dx & 1
dy += dy & 1
dx = max(2, dx)
dy = max(2, dy)
return dx * dy
#
(n, k) = map(int, input().split())
pointsx = []
pointsy = []
for i in range(n):
a = list(map(int, input().split()))
pointsx += [(a[0] + a[2], i)]
pointsy += [(a[1] + a[3], i)]
(pointsx, pointsy) = (sorted(pointsx), sorted(pointsy))
ans = 3 * 10 ** 18
for u in range(0, k + 1):
for d in range(0, k + 1):
for l in range(0, k + 1):
for r in range(0, k + 1):
if l + r <= k and u + d <= k:
ans = min(ans, check(u, d, l, r))
print(ans // 4)
``` | instruction | 0 | 47,483 | 8 | 94,966 |
No | output | 1 | 47,483 | 8 | 94,967 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Edo has got a collection of n refrigerator magnets!
He decided to buy a refrigerator and hang the magnets on the door. The shop can make the refrigerator with any size of the door that meets the following restrictions: the refrigerator door must be rectangle, and both the length and the width of the door must be positive integers.
Edo figured out how he wants to place the magnets on the refrigerator. He introduced a system of coordinates on the plane, where each magnet is represented as a rectangle with sides parallel to the coordinate axes.
Now he wants to remove no more than k magnets (he may choose to keep all of them) and attach all remaining magnets to the refrigerator door, and the area of the door should be as small as possible. A magnet is considered to be attached to the refrigerator door if its center lies on the door or on its boundary. The relative positions of all the remaining magnets must correspond to the plan.
Let us explain the last two sentences. Let's suppose we want to hang two magnets on the refrigerator. If the magnet in the plan has coordinates of the lower left corner (x1, y1) and the upper right corner (x2, y2), then its center is located at (<image>, <image>) (may not be integers). By saying the relative position should correspond to the plan we mean that the only available operation is translation, i.e. the vector connecting the centers of two magnets in the original plan, must be equal to the vector connecting the centers of these two magnets on the refrigerator.
The sides of the refrigerator door must also be parallel to coordinate axes.
Input
The first line contains two integers n and k (1 ≤ n ≤ 100 000, 0 ≤ k ≤ min(10, n - 1)) — the number of magnets that Edo has and the maximum number of magnets Edo may not place on the refrigerator.
Next n lines describe the initial plan of placing magnets. Each line contains four integers x1, y1, x2, y2 (1 ≤ x1 < x2 ≤ 109, 1 ≤ y1 < y2 ≤ 109) — the coordinates of the lower left and upper right corners of the current magnet. The magnets can partially overlap or even fully coincide.
Output
Print a single integer — the minimum area of the door of refrigerator, which can be used to place at least n - k magnets, preserving the relative positions.
Examples
Input
3 1
1 1 2 2
2 2 3 3
3 3 4 4
Output
1
Input
4 1
1 1 2 2
1 9 2 10
9 9 10 10
9 1 10 2
Output
64
Input
3 0
1 1 2 2
1 1 1000000000 1000000000
1 3 8 12
Output
249999999000000001
Note
In the first test sample it is optimal to remove either the first or the third magnet. If we remove the first magnet, the centers of two others will lie at points (2.5, 2.5) and (3.5, 3.5). Thus, it is enough to buy a fridge with door width 1 and door height 1, the area of the door also equals one, correspondingly.
In the second test sample it doesn't matter which magnet to remove, the answer will not change — we need a fridge with door width 8 and door height 8.
In the third sample you cannot remove anything as k = 0.
Submitted Solution:
```
n,k = map(int, input().split())
MAGS=[]
for i in range(n):
x,y,x2,y2 = map(int, input().split())
MAGS.append([(x+x2)/2, (y+y2)/2, True])
MAGS_H = sorted(MAGS[:], key=lambda x:x[0])
MAGS_V = sorted(MAGS[:], key=lambda x:x[1])
def set_false(x):
if x:
x[2]=False
def set_true(x):
if x:
x[2]=True
def get(ind):
if ind == 0:
for i in MAGS_H:
if i[2]:
return i
elif ind == 1:
for i in MAGS_H[::-1]:
if i[2]:
return i
elif ind == 2:
for i in MAGS_V:
if i[2]:
return i
elif ind == 3:
for i in MAGS_V[::-1]:
if i[2]:
return i
return None
def rec(l):
min_val = 10**20
if l > 0:
for i in range(4):
x = get(i)
set_false(x)
min_val = min(min_val, rec(l-1))
set_true(x)
if l == 0:
# print(*(get(i) for i in range(4)))
return max(1,int(get(1)[0] - get(0)[0]))*max(1,int(get(3)[0] - get(2)[0]))
return min_val
print(int(rec(k)))
``` | instruction | 0 | 47,484 | 8 | 94,968 |
No | output | 1 | 47,484 | 8 | 94,969 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Edo has got a collection of n refrigerator magnets!
He decided to buy a refrigerator and hang the magnets on the door. The shop can make the refrigerator with any size of the door that meets the following restrictions: the refrigerator door must be rectangle, and both the length and the width of the door must be positive integers.
Edo figured out how he wants to place the magnets on the refrigerator. He introduced a system of coordinates on the plane, where each magnet is represented as a rectangle with sides parallel to the coordinate axes.
Now he wants to remove no more than k magnets (he may choose to keep all of them) and attach all remaining magnets to the refrigerator door, and the area of the door should be as small as possible. A magnet is considered to be attached to the refrigerator door if its center lies on the door or on its boundary. The relative positions of all the remaining magnets must correspond to the plan.
Let us explain the last two sentences. Let's suppose we want to hang two magnets on the refrigerator. If the magnet in the plan has coordinates of the lower left corner (x1, y1) and the upper right corner (x2, y2), then its center is located at (<image>, <image>) (may not be integers). By saying the relative position should correspond to the plan we mean that the only available operation is translation, i.e. the vector connecting the centers of two magnets in the original plan, must be equal to the vector connecting the centers of these two magnets on the refrigerator.
The sides of the refrigerator door must also be parallel to coordinate axes.
Input
The first line contains two integers n and k (1 ≤ n ≤ 100 000, 0 ≤ k ≤ min(10, n - 1)) — the number of magnets that Edo has and the maximum number of magnets Edo may not place on the refrigerator.
Next n lines describe the initial plan of placing magnets. Each line contains four integers x1, y1, x2, y2 (1 ≤ x1 < x2 ≤ 109, 1 ≤ y1 < y2 ≤ 109) — the coordinates of the lower left and upper right corners of the current magnet. The magnets can partially overlap or even fully coincide.
Output
Print a single integer — the minimum area of the door of refrigerator, which can be used to place at least n - k magnets, preserving the relative positions.
Examples
Input
3 1
1 1 2 2
2 2 3 3
3 3 4 4
Output
1
Input
4 1
1 1 2 2
1 9 2 10
9 9 10 10
9 1 10 2
Output
64
Input
3 0
1 1 2 2
1 1 1000000000 1000000000
1 3 8 12
Output
249999999000000001
Note
In the first test sample it is optimal to remove either the first or the third magnet. If we remove the first magnet, the centers of two others will lie at points (2.5, 2.5) and (3.5, 3.5). Thus, it is enough to buy a fridge with door width 1 and door height 1, the area of the door also equals one, correspondingly.
In the second test sample it doesn't matter which magnet to remove, the answer will not change — we need a fridge with door width 8 and door height 8.
In the third sample you cannot remove anything as k = 0.
Submitted Solution:
```
n,k = map(int, input().split())
MAGS=[]
for i in range(n):
x,y,x2,y2 = map(int, input().split())
MAGS.append([(x+x2)/2, (y+y2)/2, True])
MAGS_H = sorted(MAGS[:], key=lambda x:x[0])
MAGS_V = sorted(MAGS[:], key=lambda x:x[1])
def set_false(x):
if x:
x[2]=False
def set_true(x):
if x:
x[2]=True
def get(ind):
if ind == 0:
for i in MAGS_H:
if i[2]:
return i
elif ind == 1:
for i in MAGS_H[::-1]:
if i[2]:
return i
elif ind == 2:
for i in MAGS_V:
if i[2]:
return i
elif ind == 3:
for i in MAGS_V[::-1]:
if i[2]:
return i
return None
def rec(l):
min_val = 10**20
if l > 0:
for i in range(4):
x = get(i)
set_false(x)
min_val = min(min_val, rec(l-1))
set_true(x)
if l == 0:
return int(get(1)[0] - get(0)[0])*int(get(3)[0] - get(2)[0])
return min_val
print(int(rec(k)))
``` | instruction | 0 | 47,485 | 8 | 94,970 |
No | output | 1 | 47,485 | 8 | 94,971 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You want to build a fence that will consist of n equal sections. All sections have a width equal to 1 and height equal to k. You will place all sections in one line side by side.
Unfortunately, the ground beneath the fence is not flat. For simplicity, you can think that the ground level under the i-th section is equal to h_i.
You should follow several rules to build the fence:
1. the consecutive sections should have a common side of length at least 1;
2. the first and the last sections should stand on the corresponding ground levels;
3. the sections between may be either on the ground level or higher, but not higher than k - 1 from the ground level h_i (the height should be an integer);
<image> One of possible fences (blue color) for the first test case
Is it possible to build a fence that meets all rules?
Input
The first line contains a single integer t (1 ≤ t ≤ 10^4) — the number of test cases.
The first line of each test case contains two integers n and k (2 ≤ n ≤ 2 ⋅ 10^5; 2 ≤ k ≤ 10^8) — the number of sections in the fence and the height of each section.
The second line of each test case contains n integers h_1, h_2, ..., h_n (0 ≤ h_i ≤ 10^8), where h_i is the ground level beneath the i-th section.
It's guaranteed that the sum of n over test cases doesn't exceed 2 ⋅ 10^5.
Output
For each test case print YES if it's possible to build the fence that meets all rules. Otherwise, print NO.
You may print each letter in any case (for example, YES, Yes, yes, yEs will all be recognized as positive answer).
Example
Input
3
6 3
0 0 2 5 1 1
2 3
0 2
3 2
3 0 2
Output
YES
YES
NO
Note
In the first test case, one of the possible fences is shown in the picture.
In the second test case, according to the second rule, you should build both sections on the corresponding ground levels, and since k = 3, h_1 = 0, and h_2 = 2 the first rule is also fulfilled.
In the third test case, according to the second rule, you should build the first section on height 3 and the third section on height 2. According to the first rule, the second section should be on the height of at least 2 (to have a common side with the first section), but according to the third rule, the second section can be built on the height of at most h_2 + k - 1 = 1. | instruction | 0 | 48,119 | 8 | 96,238 |
Tags: dp, greedy, implementation, two pointers
Correct Solution:
```
import sys
input = sys.stdin.readline
t = int(input())
for _ in range(t):
n, k = map(int, input().split())
h = list(map(int, input().split()))
ans = 1
lb, ub = h[0], h[0]
for i in range(1, n):
lb_next = h[i]
ub_next = h[i] + k-1
lb_next = max(lb-(k-1), lb_next)
ub_next = min(ub+(k-1), ub_next)
if lb_next > ub_next:
ans = 0
break
lb = lb_next
ub = ub_next
if (lb > h[n-1]) or (ub < h[n-1]):
ans = 0
print('YES' if ans else 'NO')
``` | output | 1 | 48,119 | 8 | 96,239 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You want to build a fence that will consist of n equal sections. All sections have a width equal to 1 and height equal to k. You will place all sections in one line side by side.
Unfortunately, the ground beneath the fence is not flat. For simplicity, you can think that the ground level under the i-th section is equal to h_i.
You should follow several rules to build the fence:
1. the consecutive sections should have a common side of length at least 1;
2. the first and the last sections should stand on the corresponding ground levels;
3. the sections between may be either on the ground level or higher, but not higher than k - 1 from the ground level h_i (the height should be an integer);
<image> One of possible fences (blue color) for the first test case
Is it possible to build a fence that meets all rules?
Input
The first line contains a single integer t (1 ≤ t ≤ 10^4) — the number of test cases.
The first line of each test case contains two integers n and k (2 ≤ n ≤ 2 ⋅ 10^5; 2 ≤ k ≤ 10^8) — the number of sections in the fence and the height of each section.
The second line of each test case contains n integers h_1, h_2, ..., h_n (0 ≤ h_i ≤ 10^8), where h_i is the ground level beneath the i-th section.
It's guaranteed that the sum of n over test cases doesn't exceed 2 ⋅ 10^5.
Output
For each test case print YES if it's possible to build the fence that meets all rules. Otherwise, print NO.
You may print each letter in any case (for example, YES, Yes, yes, yEs will all be recognized as positive answer).
Example
Input
3
6 3
0 0 2 5 1 1
2 3
0 2
3 2
3 0 2
Output
YES
YES
NO
Note
In the first test case, one of the possible fences is shown in the picture.
In the second test case, according to the second rule, you should build both sections on the corresponding ground levels, and since k = 3, h_1 = 0, and h_2 = 2 the first rule is also fulfilled.
In the third test case, according to the second rule, you should build the first section on height 3 and the third section on height 2. According to the first rule, the second section should be on the height of at least 2 (to have a common side with the first section), but according to the third rule, the second section can be built on the height of at most h_2 + k - 1 = 1. | instruction | 0 | 48,120 | 8 | 96,240 |
Tags: dp, greedy, implementation, two pointers
Correct Solution:
```
import sys, os, io
def rs(): return sys.stdin.readline().rstrip()
def ri(): return int(sys.stdin.readline())
def ria(): return list(map(int, sys.stdin.readline().split()))
def ws(s): sys.stdout.write(s + '\n')
def wi(n): sys.stdout.write(str(n) + '\n')
def wia(a): sys.stdout.write(' '.join([str(x) for x in a]) + '\n')
import math,datetime,functools,itertools,operator,bisect,fractions,statistics
from collections import deque,defaultdict,OrderedDict,Counter
from fractions import Fraction
from decimal import Decimal
from sys import stdout
from heapq import heappush, heappop, heapify ,_heapify_max,_heappop_max,nsmallest,nlargest
# sys.setrecursionlimit(111111)
def main():
# mod=1000000007
# InverseofNumber(mod)
# InverseofFactorial(mod)
# factorial(mod)
starttime=datetime.datetime.now()
if(os.path.exists('input.txt')):
sys.stdin = open("input.txt","r")
sys.stdout = open("output.txt","w")
###CODE
tc = ri()
for _ in range(tc):
n,k=ria()
a=ria()
r=[0]*n
# print(r)
f=0
for i in range(n):
if i:
lower=max(r[i-1][0]+1-k,a[i])
upper=min(a[i]+k-1,r[i-1][1]+k-1)
if lower>upper:
f=1
print("NO")
break
r[i]=[lower,upper]
else:
r[i]=[a[i],a[i]]
if f: continue
if r[n-1][0]>a[n-1]:
print("NO")
else:
print("YES")
#<--Solving Area Ends
endtime=datetime.datetime.now()
time=(endtime-starttime).total_seconds()*1000
if(os.path.exists('input.txt')):
print("Time:",time,"ms")
class FastReader(io.IOBase):
newlines = 0
def __init__(self, fd, chunk_size=1024 * 8):
self._fd = fd
self._chunk_size = chunk_size
self.buffer = io.BytesIO()
def read(self):
while True:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, self._chunk_size))
if not b:
break
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines = 0
return self.buffer.read()
def readline(self, size=-1):
while self.newlines == 0:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, self._chunk_size if size == -1 else size))
self.newlines = b.count(b"\n") + (not b)
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines -= 1
return self.buffer.readline()
class FastWriter(io.IOBase):
def __init__(self, fd):
self._fd = fd
self.buffer = io.BytesIO()
self.write = self.buffer.write
def flush(self):
os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class FastStdin(io.IOBase):
def __init__(self, fd=0):
self.buffer = FastReader(fd)
self.read = lambda: self.buffer.read().decode("ascii")
self.readline = lambda: self.buffer.readline().decode("ascii")
class FastStdout(io.IOBase):
def __init__(self, fd=1):
self.buffer = FastWriter(fd)
self.write = lambda s: self.buffer.write(s.encode("ascii"))
self.flush = self.buffer.flush
if __name__ == '__main__':
sys.stdin = FastStdin()
sys.stdout = FastStdout()
main()
``` | output | 1 | 48,120 | 8 | 96,241 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You want to build a fence that will consist of n equal sections. All sections have a width equal to 1 and height equal to k. You will place all sections in one line side by side.
Unfortunately, the ground beneath the fence is not flat. For simplicity, you can think that the ground level under the i-th section is equal to h_i.
You should follow several rules to build the fence:
1. the consecutive sections should have a common side of length at least 1;
2. the first and the last sections should stand on the corresponding ground levels;
3. the sections between may be either on the ground level or higher, but not higher than k - 1 from the ground level h_i (the height should be an integer);
<image> One of possible fences (blue color) for the first test case
Is it possible to build a fence that meets all rules?
Input
The first line contains a single integer t (1 ≤ t ≤ 10^4) — the number of test cases.
The first line of each test case contains two integers n and k (2 ≤ n ≤ 2 ⋅ 10^5; 2 ≤ k ≤ 10^8) — the number of sections in the fence and the height of each section.
The second line of each test case contains n integers h_1, h_2, ..., h_n (0 ≤ h_i ≤ 10^8), where h_i is the ground level beneath the i-th section.
It's guaranteed that the sum of n over test cases doesn't exceed 2 ⋅ 10^5.
Output
For each test case print YES if it's possible to build the fence that meets all rules. Otherwise, print NO.
You may print each letter in any case (for example, YES, Yes, yes, yEs will all be recognized as positive answer).
Example
Input
3
6 3
0 0 2 5 1 1
2 3
0 2
3 2
3 0 2
Output
YES
YES
NO
Note
In the first test case, one of the possible fences is shown in the picture.
In the second test case, according to the second rule, you should build both sections on the corresponding ground levels, and since k = 3, h_1 = 0, and h_2 = 2 the first rule is also fulfilled.
In the third test case, according to the second rule, you should build the first section on height 3 and the third section on height 2. According to the first rule, the second section should be on the height of at least 2 (to have a common side with the first section), but according to the third rule, the second section can be built on the height of at most h_2 + k - 1 = 1. | instruction | 0 | 48,121 | 8 | 96,242 |
Tags: dp, greedy, implementation, two pointers
Correct Solution:
```
import sys
input = sys.stdin.readline
def gcd(a, b):
if a == 0:
return b
return gcd(b % a, a)
def lcm(a, b):
return (a * b) / gcd(a, b)
def main():
for _ in range(int(input())):
n,k=map(int , input().split())
h=list(map(int, input().split()))
currmax=h[0]+k
f=0
for i in range(1,n-1):
if h[i]>=currmax:
f=1
else:
if h[i]>=h[i-1]:
if h[i+1]>=h[i]:
currmax=h[i]+min(k-1, currmax-h[i]-1)+k
else:
currmax=h[i]+k
else:
if h[i]+k+k-1<=currmax-k:
f=1
else:
if h[i + 1] >= h[i]:
currmax = h[i]+k+k-1
else:
now=h[i]+k
if now<=currmax-k:
now+=(currmax-k-now+1)
currmax=now
# if h[i]+k>=h[i-1]-(k-1):
# currmax=h[i]+k
# else:
# currmax=currmax-k+1
if h[n-1]>=currmax:
f=1
else:
if h[n-1]+k<=currmax-k:
f=1
if f:
print('NO')
else:
print('YES')
return
if __name__=="__main__":
main()
``` | output | 1 | 48,121 | 8 | 96,243 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You want to build a fence that will consist of n equal sections. All sections have a width equal to 1 and height equal to k. You will place all sections in one line side by side.
Unfortunately, the ground beneath the fence is not flat. For simplicity, you can think that the ground level under the i-th section is equal to h_i.
You should follow several rules to build the fence:
1. the consecutive sections should have a common side of length at least 1;
2. the first and the last sections should stand on the corresponding ground levels;
3. the sections between may be either on the ground level or higher, but not higher than k - 1 from the ground level h_i (the height should be an integer);
<image> One of possible fences (blue color) for the first test case
Is it possible to build a fence that meets all rules?
Input
The first line contains a single integer t (1 ≤ t ≤ 10^4) — the number of test cases.
The first line of each test case contains two integers n and k (2 ≤ n ≤ 2 ⋅ 10^5; 2 ≤ k ≤ 10^8) — the number of sections in the fence and the height of each section.
The second line of each test case contains n integers h_1, h_2, ..., h_n (0 ≤ h_i ≤ 10^8), where h_i is the ground level beneath the i-th section.
It's guaranteed that the sum of n over test cases doesn't exceed 2 ⋅ 10^5.
Output
For each test case print YES if it's possible to build the fence that meets all rules. Otherwise, print NO.
You may print each letter in any case (for example, YES, Yes, yes, yEs will all be recognized as positive answer).
Example
Input
3
6 3
0 0 2 5 1 1
2 3
0 2
3 2
3 0 2
Output
YES
YES
NO
Note
In the first test case, one of the possible fences is shown in the picture.
In the second test case, according to the second rule, you should build both sections on the corresponding ground levels, and since k = 3, h_1 = 0, and h_2 = 2 the first rule is also fulfilled.
In the third test case, according to the second rule, you should build the first section on height 3 and the third section on height 2. According to the first rule, the second section should be on the height of at least 2 (to have a common side with the first section), but according to the third rule, the second section can be built on the height of at most h_2 + k - 1 = 1. | instruction | 0 | 48,122 | 8 | 96,244 |
Tags: dp, greedy, implementation, two pointers
Correct Solution:
```
t = int(input())
for i in range(t):
n, k = tuple(map(int, input().split(' ')))
heights = list(map(int, input().split(' ')))
current_floor_range = [heights[0], heights[0]]
is_possible = True
for j in range(1, n):
current_floor_range = [max(current_floor_range[0] - k + 1, heights[j]),
min(current_floor_range[1], heights[j]) + k - 1]
if current_floor_range[0] > current_floor_range[1]:
is_possible = False
break
print('YES' if is_possible and current_floor_range[0] <= heights[n - 1] <= current_floor_range[1] else 'NO')
``` | output | 1 | 48,122 | 8 | 96,245 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You want to build a fence that will consist of n equal sections. All sections have a width equal to 1 and height equal to k. You will place all sections in one line side by side.
Unfortunately, the ground beneath the fence is not flat. For simplicity, you can think that the ground level under the i-th section is equal to h_i.
You should follow several rules to build the fence:
1. the consecutive sections should have a common side of length at least 1;
2. the first and the last sections should stand on the corresponding ground levels;
3. the sections between may be either on the ground level or higher, but not higher than k - 1 from the ground level h_i (the height should be an integer);
<image> One of possible fences (blue color) for the first test case
Is it possible to build a fence that meets all rules?
Input
The first line contains a single integer t (1 ≤ t ≤ 10^4) — the number of test cases.
The first line of each test case contains two integers n and k (2 ≤ n ≤ 2 ⋅ 10^5; 2 ≤ k ≤ 10^8) — the number of sections in the fence and the height of each section.
The second line of each test case contains n integers h_1, h_2, ..., h_n (0 ≤ h_i ≤ 10^8), where h_i is the ground level beneath the i-th section.
It's guaranteed that the sum of n over test cases doesn't exceed 2 ⋅ 10^5.
Output
For each test case print YES if it's possible to build the fence that meets all rules. Otherwise, print NO.
You may print each letter in any case (for example, YES, Yes, yes, yEs will all be recognized as positive answer).
Example
Input
3
6 3
0 0 2 5 1 1
2 3
0 2
3 2
3 0 2
Output
YES
YES
NO
Note
In the first test case, one of the possible fences is shown in the picture.
In the second test case, according to the second rule, you should build both sections on the corresponding ground levels, and since k = 3, h_1 = 0, and h_2 = 2 the first rule is also fulfilled.
In the third test case, according to the second rule, you should build the first section on height 3 and the third section on height 2. According to the first rule, the second section should be on the height of at least 2 (to have a common side with the first section), but according to the third rule, the second section can be built on the height of at most h_2 + k - 1 = 1. | instruction | 0 | 48,123 | 8 | 96,246 |
Tags: dp, greedy, implementation, two pointers
Correct Solution:
```
def main():
from sys import stdin
input = stdin.buffer.readline
# input = open('25-B.txt', 'r').readline
for _ in range(int(input())):
n, k = map(int, input().split())
*h, = map(int, input().split())
mn = [0] * n
mx = [0] * n
mn[0] = mx[0] = h[0]
for i in range(1, n):
mn[i] = max(h[i], mn[i - 1] - k + 1)
mx[i] = min(h[i], mx[i - 1]) + k - 1
if mn[i] > mx[i]:
print('NO')
break
else:
if mn[n - 1] <= h[n - 1] <= mx[n - 1]:
print('YES')
else:
print('NO')
main()
``` | output | 1 | 48,123 | 8 | 96,247 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You want to build a fence that will consist of n equal sections. All sections have a width equal to 1 and height equal to k. You will place all sections in one line side by side.
Unfortunately, the ground beneath the fence is not flat. For simplicity, you can think that the ground level under the i-th section is equal to h_i.
You should follow several rules to build the fence:
1. the consecutive sections should have a common side of length at least 1;
2. the first and the last sections should stand on the corresponding ground levels;
3. the sections between may be either on the ground level or higher, but not higher than k - 1 from the ground level h_i (the height should be an integer);
<image> One of possible fences (blue color) for the first test case
Is it possible to build a fence that meets all rules?
Input
The first line contains a single integer t (1 ≤ t ≤ 10^4) — the number of test cases.
The first line of each test case contains two integers n and k (2 ≤ n ≤ 2 ⋅ 10^5; 2 ≤ k ≤ 10^8) — the number of sections in the fence and the height of each section.
The second line of each test case contains n integers h_1, h_2, ..., h_n (0 ≤ h_i ≤ 10^8), where h_i is the ground level beneath the i-th section.
It's guaranteed that the sum of n over test cases doesn't exceed 2 ⋅ 10^5.
Output
For each test case print YES if it's possible to build the fence that meets all rules. Otherwise, print NO.
You may print each letter in any case (for example, YES, Yes, yes, yEs will all be recognized as positive answer).
Example
Input
3
6 3
0 0 2 5 1 1
2 3
0 2
3 2
3 0 2
Output
YES
YES
NO
Note
In the first test case, one of the possible fences is shown in the picture.
In the second test case, according to the second rule, you should build both sections on the corresponding ground levels, and since k = 3, h_1 = 0, and h_2 = 2 the first rule is also fulfilled.
In the third test case, according to the second rule, you should build the first section on height 3 and the third section on height 2. According to the first rule, the second section should be on the height of at least 2 (to have a common side with the first section), but according to the third rule, the second section can be built on the height of at most h_2 + k - 1 = 1. | instruction | 0 | 48,124 | 8 | 96,248 |
Tags: dp, greedy, implementation, two pointers
Correct Solution:
```
from sys import stdin, gettrace
if gettrace():
def inputi():
return input()
else:
def input():
return next(stdin)[:-1]
def inputi():
return stdin.buffer.readline()
def solve():
n,k = map(int, input().split())
hh = [int(a) for a in input().split()]
mx = hh[0]
for h in hh[1:]:
if h > mx + k-1:
print("No")
return
else:
mx = min(mx + k - 1, h + k - 1)
mx = hh[-1]
for h in hh[-2::-1]:
if h > mx + k - 1:
print("No")
return
else:
mx = min(mx + k - 1, h + k - 1)
print("Yes")
def main():
t = int(input())
for _ in range(t):
solve()
if __name__ == "__main__":
main()
``` | output | 1 | 48,124 | 8 | 96,249 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You want to build a fence that will consist of n equal sections. All sections have a width equal to 1 and height equal to k. You will place all sections in one line side by side.
Unfortunately, the ground beneath the fence is not flat. For simplicity, you can think that the ground level under the i-th section is equal to h_i.
You should follow several rules to build the fence:
1. the consecutive sections should have a common side of length at least 1;
2. the first and the last sections should stand on the corresponding ground levels;
3. the sections between may be either on the ground level or higher, but not higher than k - 1 from the ground level h_i (the height should be an integer);
<image> One of possible fences (blue color) for the first test case
Is it possible to build a fence that meets all rules?
Input
The first line contains a single integer t (1 ≤ t ≤ 10^4) — the number of test cases.
The first line of each test case contains two integers n and k (2 ≤ n ≤ 2 ⋅ 10^5; 2 ≤ k ≤ 10^8) — the number of sections in the fence and the height of each section.
The second line of each test case contains n integers h_1, h_2, ..., h_n (0 ≤ h_i ≤ 10^8), where h_i is the ground level beneath the i-th section.
It's guaranteed that the sum of n over test cases doesn't exceed 2 ⋅ 10^5.
Output
For each test case print YES if it's possible to build the fence that meets all rules. Otherwise, print NO.
You may print each letter in any case (for example, YES, Yes, yes, yEs will all be recognized as positive answer).
Example
Input
3
6 3
0 0 2 5 1 1
2 3
0 2
3 2
3 0 2
Output
YES
YES
NO
Note
In the first test case, one of the possible fences is shown in the picture.
In the second test case, according to the second rule, you should build both sections on the corresponding ground levels, and since k = 3, h_1 = 0, and h_2 = 2 the first rule is also fulfilled.
In the third test case, according to the second rule, you should build the first section on height 3 and the third section on height 2. According to the first rule, the second section should be on the height of at least 2 (to have a common side with the first section), but according to the third rule, the second section can be built on the height of at most h_2 + k - 1 = 1. | instruction | 0 | 48,125 | 8 | 96,250 |
Tags: dp, greedy, implementation, two pointers
Correct Solution:
```
import sys
input = sys.stdin.readline
from collections import Counter
import math
t = int(input())
for _ in range(t):
#n = int(input())
n,h = list(map(int,input().split()))
#m = int(input())
arr = list(map(int,input().split()))
x,y = arr[0], arr[0]
ok = 1
for i in range(n):
new_x = max(arr[i], x - h + 1)
new_y = min(arr[i] + h - 1, y + h - 1)
#print(new_x, new_y,x,y, arr[i])
if i==0:
new_x = x
new_y = y
if new_x > new_y:
ok = 0
x = new_x
y = new_y
if i==n-1:
if new_x > arr[i]:
ok = 0
if ok==1:
print('YES')
else:
print('NO')
``` | output | 1 | 48,125 | 8 | 96,251 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You want to build a fence that will consist of n equal sections. All sections have a width equal to 1 and height equal to k. You will place all sections in one line side by side.
Unfortunately, the ground beneath the fence is not flat. For simplicity, you can think that the ground level under the i-th section is equal to h_i.
You should follow several rules to build the fence:
1. the consecutive sections should have a common side of length at least 1;
2. the first and the last sections should stand on the corresponding ground levels;
3. the sections between may be either on the ground level or higher, but not higher than k - 1 from the ground level h_i (the height should be an integer);
<image> One of possible fences (blue color) for the first test case
Is it possible to build a fence that meets all rules?
Input
The first line contains a single integer t (1 ≤ t ≤ 10^4) — the number of test cases.
The first line of each test case contains two integers n and k (2 ≤ n ≤ 2 ⋅ 10^5; 2 ≤ k ≤ 10^8) — the number of sections in the fence and the height of each section.
The second line of each test case contains n integers h_1, h_2, ..., h_n (0 ≤ h_i ≤ 10^8), where h_i is the ground level beneath the i-th section.
It's guaranteed that the sum of n over test cases doesn't exceed 2 ⋅ 10^5.
Output
For each test case print YES if it's possible to build the fence that meets all rules. Otherwise, print NO.
You may print each letter in any case (for example, YES, Yes, yes, yEs will all be recognized as positive answer).
Example
Input
3
6 3
0 0 2 5 1 1
2 3
0 2
3 2
3 0 2
Output
YES
YES
NO
Note
In the first test case, one of the possible fences is shown in the picture.
In the second test case, according to the second rule, you should build both sections on the corresponding ground levels, and since k = 3, h_1 = 0, and h_2 = 2 the first rule is also fulfilled.
In the third test case, according to the second rule, you should build the first section on height 3 and the third section on height 2. According to the first rule, the second section should be on the height of at least 2 (to have a common side with the first section), but according to the third rule, the second section can be built on the height of at most h_2 + k - 1 = 1. | instruction | 0 | 48,126 | 8 | 96,252 |
Tags: dp, greedy, implementation, two pointers
Correct Solution:
```
import sys
input=sys.stdin.readline
def intersect(a,b):
return [max(a[0],b[0]),min(a[1],b[1])]
t=int(input())
for you in range(t):
l=input().split()
n=int(l[0])
k=int(l[1])
l=input().split()
li=[int(i) for i in l]
curr=[li[0],li[0]]
poss=1
for i in range(1,n-1):
curr[0]-=(k-1)
curr[1]+=(k-1)
curr=intersect(curr,[li[i],li[i]+k-1])
#print(curr)
if(curr[1]<curr[0]):
poss=0
break
curr[0]-=(k-1)
curr[1]+=(k-1)
if(li[-1]<curr[0] or li[-1]>curr[1]):
poss=0
if(poss):
print("Yes")
else:
print("No")
``` | output | 1 | 48,126 | 8 | 96,253 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You want to build a fence that will consist of n equal sections. All sections have a width equal to 1 and height equal to k. You will place all sections in one line side by side.
Unfortunately, the ground beneath the fence is not flat. For simplicity, you can think that the ground level under the i-th section is equal to h_i.
You should follow several rules to build the fence:
1. the consecutive sections should have a common side of length at least 1;
2. the first and the last sections should stand on the corresponding ground levels;
3. the sections between may be either on the ground level or higher, but not higher than k - 1 from the ground level h_i (the height should be an integer);
<image> One of possible fences (blue color) for the first test case
Is it possible to build a fence that meets all rules?
Input
The first line contains a single integer t (1 ≤ t ≤ 10^4) — the number of test cases.
The first line of each test case contains two integers n and k (2 ≤ n ≤ 2 ⋅ 10^5; 2 ≤ k ≤ 10^8) — the number of sections in the fence and the height of each section.
The second line of each test case contains n integers h_1, h_2, ..., h_n (0 ≤ h_i ≤ 10^8), where h_i is the ground level beneath the i-th section.
It's guaranteed that the sum of n over test cases doesn't exceed 2 ⋅ 10^5.
Output
For each test case print YES if it's possible to build the fence that meets all rules. Otherwise, print NO.
You may print each letter in any case (for example, YES, Yes, yes, yEs will all be recognized as positive answer).
Example
Input
3
6 3
0 0 2 5 1 1
2 3
0 2
3 2
3 0 2
Output
YES
YES
NO
Note
In the first test case, one of the possible fences is shown in the picture.
In the second test case, according to the second rule, you should build both sections on the corresponding ground levels, and since k = 3, h_1 = 0, and h_2 = 2 the first rule is also fulfilled.
In the third test case, according to the second rule, you should build the first section on height 3 and the third section on height 2. According to the first rule, the second section should be on the height of at least 2 (to have a common side with the first section), but according to the third rule, the second section can be built on the height of at most h_2 + k - 1 = 1.
Submitted Solution:
```
from sys import stdin
input = stdin.readline
t = int(input())
for _ in range(t):
n, k = [int(x) for x in input().split()]
h = [int(x) for x in input().split()]
ans = "YES"
min_1 = h[0] - (k - 1)
max_1 = h[0] + (k - 1)
for i in range(1, n):
min_2 = h[i]
if i == n-1:
max_2 = min_2
else:
max_2 = h[i] + (k - 1)
if max_1 < min_2 or min_1 > max_2:
ans = "NO"
break
min_1 = max(min_1, min_2) - (k - 1)
max_1 = min(max_1, max_2) + (k - 1)
print(ans)
``` | instruction | 0 | 48,127 | 8 | 96,254 |
Yes | output | 1 | 48,127 | 8 | 96,255 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You want to build a fence that will consist of n equal sections. All sections have a width equal to 1 and height equal to k. You will place all sections in one line side by side.
Unfortunately, the ground beneath the fence is not flat. For simplicity, you can think that the ground level under the i-th section is equal to h_i.
You should follow several rules to build the fence:
1. the consecutive sections should have a common side of length at least 1;
2. the first and the last sections should stand on the corresponding ground levels;
3. the sections between may be either on the ground level or higher, but not higher than k - 1 from the ground level h_i (the height should be an integer);
<image> One of possible fences (blue color) for the first test case
Is it possible to build a fence that meets all rules?
Input
The first line contains a single integer t (1 ≤ t ≤ 10^4) — the number of test cases.
The first line of each test case contains two integers n and k (2 ≤ n ≤ 2 ⋅ 10^5; 2 ≤ k ≤ 10^8) — the number of sections in the fence and the height of each section.
The second line of each test case contains n integers h_1, h_2, ..., h_n (0 ≤ h_i ≤ 10^8), where h_i is the ground level beneath the i-th section.
It's guaranteed that the sum of n over test cases doesn't exceed 2 ⋅ 10^5.
Output
For each test case print YES if it's possible to build the fence that meets all rules. Otherwise, print NO.
You may print each letter in any case (for example, YES, Yes, yes, yEs will all be recognized as positive answer).
Example
Input
3
6 3
0 0 2 5 1 1
2 3
0 2
3 2
3 0 2
Output
YES
YES
NO
Note
In the first test case, one of the possible fences is shown in the picture.
In the second test case, according to the second rule, you should build both sections on the corresponding ground levels, and since k = 3, h_1 = 0, and h_2 = 2 the first rule is also fulfilled.
In the third test case, according to the second rule, you should build the first section on height 3 and the third section on height 2. According to the first rule, the second section should be on the height of at least 2 (to have a common side with the first section), but according to the third rule, the second section can be built on the height of at most h_2 + k - 1 = 1.
Submitted Solution:
```
t = int(input())
for i in range(t):
l = input().split(' ')
n, k = int(l[0]), int(l[1])
l = input().split(' ')
for j in range(n):
l[j] = int(l[j])
lastbl = l[0]
lastbh = l[0]
isok = True
for j in range(1, n-1):
if lastbh + (k-1) < l[j]:
#print("NOl ", lastbl, lastbh, j, k, l)
isok = False
break
if lastbl - (k-1) > l[j]+(k-1):
#print("NOh ", lastbl, lastbh, j, k, l)
isok = False
break
lastbl = max(lastbl-(k-1), l[j])
lastbh = min(lastbh + (k-1), l[j]+(k-1))
if lastbh < lastbl:
isok = False
break
if lastbh + (k-1) < l[n-1]:
isok = False
if lastbl - (k-1) > l[n-1]:
isok = False
print("YES" if isok else "NO")
``` | instruction | 0 | 48,128 | 8 | 96,256 |
Yes | output | 1 | 48,128 | 8 | 96,257 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You want to build a fence that will consist of n equal sections. All sections have a width equal to 1 and height equal to k. You will place all sections in one line side by side.
Unfortunately, the ground beneath the fence is not flat. For simplicity, you can think that the ground level under the i-th section is equal to h_i.
You should follow several rules to build the fence:
1. the consecutive sections should have a common side of length at least 1;
2. the first and the last sections should stand on the corresponding ground levels;
3. the sections between may be either on the ground level or higher, but not higher than k - 1 from the ground level h_i (the height should be an integer);
<image> One of possible fences (blue color) for the first test case
Is it possible to build a fence that meets all rules?
Input
The first line contains a single integer t (1 ≤ t ≤ 10^4) — the number of test cases.
The first line of each test case contains two integers n and k (2 ≤ n ≤ 2 ⋅ 10^5; 2 ≤ k ≤ 10^8) — the number of sections in the fence and the height of each section.
The second line of each test case contains n integers h_1, h_2, ..., h_n (0 ≤ h_i ≤ 10^8), where h_i is the ground level beneath the i-th section.
It's guaranteed that the sum of n over test cases doesn't exceed 2 ⋅ 10^5.
Output
For each test case print YES if it's possible to build the fence that meets all rules. Otherwise, print NO.
You may print each letter in any case (for example, YES, Yes, yes, yEs will all be recognized as positive answer).
Example
Input
3
6 3
0 0 2 5 1 1
2 3
0 2
3 2
3 0 2
Output
YES
YES
NO
Note
In the first test case, one of the possible fences is shown in the picture.
In the second test case, according to the second rule, you should build both sections on the corresponding ground levels, and since k = 3, h_1 = 0, and h_2 = 2 the first rule is also fulfilled.
In the third test case, according to the second rule, you should build the first section on height 3 and the third section on height 2. According to the first rule, the second section should be on the height of at least 2 (to have a common side with the first section), but according to the third rule, the second section can be built on the height of at most h_2 + k - 1 = 1.
Submitted Solution:
```
#region Header
#!/usr/bin/env python3
# from typing import *
import sys
import io
import math
import collections
import decimal
import itertools
import bisect
import heapq
def input():
return sys.stdin.readline()[:-1]
# sys.setrecursionlimit(1000000)
#endregion
# _INPUT = """3
# 6 3
# 0 0 2 5 1 1
# 2 3
# 0 2
# 3 2
# 3 0 2
# """
# sys.stdin = io.StringIO(_INPUT)
def solve(N, K, H):
dp = [[0, 0] for _ in range(N)]
dp[0] = [H[0], H[0]]
for i in range(1, N):
a = max(dp[i-1][0] - K + 1, H[i])
b = min(dp[i-1][1] + K - 1, H[i] + K - 1)
if a > b:
return False
dp[i] = [a, b]
return (dp[N-1][0] == H[N-1])
def main():
N0 = int(input())
for _ in range(N0):
N, K = map(int, input().split())
H = list(map(int, input().split()))
if solve(N, K, H):
print('YES')
else:
print('NO')
if __name__ == '__main__':
main()
``` | instruction | 0 | 48,129 | 8 | 96,258 |
Yes | output | 1 | 48,129 | 8 | 96,259 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You want to build a fence that will consist of n equal sections. All sections have a width equal to 1 and height equal to k. You will place all sections in one line side by side.
Unfortunately, the ground beneath the fence is not flat. For simplicity, you can think that the ground level under the i-th section is equal to h_i.
You should follow several rules to build the fence:
1. the consecutive sections should have a common side of length at least 1;
2. the first and the last sections should stand on the corresponding ground levels;
3. the sections between may be either on the ground level or higher, but not higher than k - 1 from the ground level h_i (the height should be an integer);
<image> One of possible fences (blue color) for the first test case
Is it possible to build a fence that meets all rules?
Input
The first line contains a single integer t (1 ≤ t ≤ 10^4) — the number of test cases.
The first line of each test case contains two integers n and k (2 ≤ n ≤ 2 ⋅ 10^5; 2 ≤ k ≤ 10^8) — the number of sections in the fence and the height of each section.
The second line of each test case contains n integers h_1, h_2, ..., h_n (0 ≤ h_i ≤ 10^8), where h_i is the ground level beneath the i-th section.
It's guaranteed that the sum of n over test cases doesn't exceed 2 ⋅ 10^5.
Output
For each test case print YES if it's possible to build the fence that meets all rules. Otherwise, print NO.
You may print each letter in any case (for example, YES, Yes, yes, yEs will all be recognized as positive answer).
Example
Input
3
6 3
0 0 2 5 1 1
2 3
0 2
3 2
3 0 2
Output
YES
YES
NO
Note
In the first test case, one of the possible fences is shown in the picture.
In the second test case, according to the second rule, you should build both sections on the corresponding ground levels, and since k = 3, h_1 = 0, and h_2 = 2 the first rule is also fulfilled.
In the third test case, according to the second rule, you should build the first section on height 3 and the third section on height 2. According to the first rule, the second section should be on the height of at least 2 (to have a common side with the first section), but according to the third rule, the second section can be built on the height of at most h_2 + k - 1 = 1.
Submitted Solution:
```
for _ in range(int(input())):
bool = True
n, h = map(int, input().split())
list = [int(x) for x in input().split()]
maxi, mini = list[0], list[0]
for i in range(1, n):
maxi = max(maxi - h + 1, list[i])
mini = min(mini + h - 1, list[i] + h - 1)
if maxi > mini:
bool = False
if maxi != list[-1]:
bool = False
print("NO") if bool == False else print("YES")
``` | instruction | 0 | 48,130 | 8 | 96,260 |
Yes | output | 1 | 48,130 | 8 | 96,261 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You want to build a fence that will consist of n equal sections. All sections have a width equal to 1 and height equal to k. You will place all sections in one line side by side.
Unfortunately, the ground beneath the fence is not flat. For simplicity, you can think that the ground level under the i-th section is equal to h_i.
You should follow several rules to build the fence:
1. the consecutive sections should have a common side of length at least 1;
2. the first and the last sections should stand on the corresponding ground levels;
3. the sections between may be either on the ground level or higher, but not higher than k - 1 from the ground level h_i (the height should be an integer);
<image> One of possible fences (blue color) for the first test case
Is it possible to build a fence that meets all rules?
Input
The first line contains a single integer t (1 ≤ t ≤ 10^4) — the number of test cases.
The first line of each test case contains two integers n and k (2 ≤ n ≤ 2 ⋅ 10^5; 2 ≤ k ≤ 10^8) — the number of sections in the fence and the height of each section.
The second line of each test case contains n integers h_1, h_2, ..., h_n (0 ≤ h_i ≤ 10^8), where h_i is the ground level beneath the i-th section.
It's guaranteed that the sum of n over test cases doesn't exceed 2 ⋅ 10^5.
Output
For each test case print YES if it's possible to build the fence that meets all rules. Otherwise, print NO.
You may print each letter in any case (for example, YES, Yes, yes, yEs will all be recognized as positive answer).
Example
Input
3
6 3
0 0 2 5 1 1
2 3
0 2
3 2
3 0 2
Output
YES
YES
NO
Note
In the first test case, one of the possible fences is shown in the picture.
In the second test case, according to the second rule, you should build both sections on the corresponding ground levels, and since k = 3, h_1 = 0, and h_2 = 2 the first rule is also fulfilled.
In the third test case, according to the second rule, you should build the first section on height 3 and the third section on height 2. According to the first rule, the second section should be on the height of at least 2 (to have a common side with the first section), but according to the third rule, the second section can be built on the height of at most h_2 + k - 1 = 1.
Submitted Solution:
```
l = int(input())
for _ in range(l):
n,k = map(int,input().split())
x = list(map(int,input().split()))
f=[]
ff=0
for i in range(len(x)):
if(i==0):
f.append([x[i]+k,x[i]+k])
else:
if(f[i-1][1] > x[i] and f[i-1][0]-k < x[i]+k+k-1):
f.append([max(f[i-1][0]-k+1,k+x[i]),min(f[i-1][1]-1+k,x[i]+k+k-1)])
else:
print("NO")
ff=1
break
if(ff==0):
print("YES")
``` | instruction | 0 | 48,131 | 8 | 96,262 |
No | output | 1 | 48,131 | 8 | 96,263 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You want to build a fence that will consist of n equal sections. All sections have a width equal to 1 and height equal to k. You will place all sections in one line side by side.
Unfortunately, the ground beneath the fence is not flat. For simplicity, you can think that the ground level under the i-th section is equal to h_i.
You should follow several rules to build the fence:
1. the consecutive sections should have a common side of length at least 1;
2. the first and the last sections should stand on the corresponding ground levels;
3. the sections between may be either on the ground level or higher, but not higher than k - 1 from the ground level h_i (the height should be an integer);
<image> One of possible fences (blue color) for the first test case
Is it possible to build a fence that meets all rules?
Input
The first line contains a single integer t (1 ≤ t ≤ 10^4) — the number of test cases.
The first line of each test case contains two integers n and k (2 ≤ n ≤ 2 ⋅ 10^5; 2 ≤ k ≤ 10^8) — the number of sections in the fence and the height of each section.
The second line of each test case contains n integers h_1, h_2, ..., h_n (0 ≤ h_i ≤ 10^8), where h_i is the ground level beneath the i-th section.
It's guaranteed that the sum of n over test cases doesn't exceed 2 ⋅ 10^5.
Output
For each test case print YES if it's possible to build the fence that meets all rules. Otherwise, print NO.
You may print each letter in any case (for example, YES, Yes, yes, yEs will all be recognized as positive answer).
Example
Input
3
6 3
0 0 2 5 1 1
2 3
0 2
3 2
3 0 2
Output
YES
YES
NO
Note
In the first test case, one of the possible fences is shown in the picture.
In the second test case, according to the second rule, you should build both sections on the corresponding ground levels, and since k = 3, h_1 = 0, and h_2 = 2 the first rule is also fulfilled.
In the third test case, according to the second rule, you should build the first section on height 3 and the third section on height 2. According to the first rule, the second section should be on the height of at least 2 (to have a common side with the first section), but according to the third rule, the second section can be built on the height of at most h_2 + k - 1 = 1.
Submitted Solution:
```
n = int(input())
for i in range(n):
info = list(map(int, input().split()))
nums = list(map(int, input().split()))
n = info[0]
k = info[1]
control = 'YES'
for j in range(2, len(nums)):
if j == 2:
if abs(nums[j - 1] - nums[j - 2]) > k - 1 or abs(nums[j] - nums[j - 2]) > 3 * (k - 1):
control = 'NO'
break
elif abs(nums[j - 1] - nums[j - 2]) > 2 * (k - 1) or abs(nums[j] - nums[j - 2]) > 3 * (k - 1):
control = 'NO'
break
print(control)
``` | instruction | 0 | 48,132 | 8 | 96,264 |
No | output | 1 | 48,132 | 8 | 96,265 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You want to build a fence that will consist of n equal sections. All sections have a width equal to 1 and height equal to k. You will place all sections in one line side by side.
Unfortunately, the ground beneath the fence is not flat. For simplicity, you can think that the ground level under the i-th section is equal to h_i.
You should follow several rules to build the fence:
1. the consecutive sections should have a common side of length at least 1;
2. the first and the last sections should stand on the corresponding ground levels;
3. the sections between may be either on the ground level or higher, but not higher than k - 1 from the ground level h_i (the height should be an integer);
<image> One of possible fences (blue color) for the first test case
Is it possible to build a fence that meets all rules?
Input
The first line contains a single integer t (1 ≤ t ≤ 10^4) — the number of test cases.
The first line of each test case contains two integers n and k (2 ≤ n ≤ 2 ⋅ 10^5; 2 ≤ k ≤ 10^8) — the number of sections in the fence and the height of each section.
The second line of each test case contains n integers h_1, h_2, ..., h_n (0 ≤ h_i ≤ 10^8), where h_i is the ground level beneath the i-th section.
It's guaranteed that the sum of n over test cases doesn't exceed 2 ⋅ 10^5.
Output
For each test case print YES if it's possible to build the fence that meets all rules. Otherwise, print NO.
You may print each letter in any case (for example, YES, Yes, yes, yEs will all be recognized as positive answer).
Example
Input
3
6 3
0 0 2 5 1 1
2 3
0 2
3 2
3 0 2
Output
YES
YES
NO
Note
In the first test case, one of the possible fences is shown in the picture.
In the second test case, according to the second rule, you should build both sections on the corresponding ground levels, and since k = 3, h_1 = 0, and h_2 = 2 the first rule is also fulfilled.
In the third test case, according to the second rule, you should build the first section on height 3 and the third section on height 2. According to the first rule, the second section should be on the height of at least 2 (to have a common side with the first section), but according to the third rule, the second section can be built on the height of at most h_2 + k - 1 = 1.
Submitted Solution:
```
for t in range(int(input())):
n, k = map(int, input().split())
floor = tuple(map(int, input().split()))
h = floor[0]
min_h = max(0, h - (k-1))
max_h = h + k + (k - 1)
answer = "NO"
for i in range(1, len(floor)):
h = floor[i]
# print("min: ", min_h, " > ", h + k ,"\n", "max: ", h, " < ", max_h)
if h + k <= min_h or h >= max_h:
break
if i != n - 2:
min_h = max(0, h - (k-1), min_h - (k-1))
max_h = min(h + 2*k-1, max_h + (k-1))
else:
min_h = h
max_h = h + k
else:
answer = "YES"
print(answer)
``` | instruction | 0 | 48,133 | 8 | 96,266 |
No | output | 1 | 48,133 | 8 | 96,267 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You want to build a fence that will consist of n equal sections. All sections have a width equal to 1 and height equal to k. You will place all sections in one line side by side.
Unfortunately, the ground beneath the fence is not flat. For simplicity, you can think that the ground level under the i-th section is equal to h_i.
You should follow several rules to build the fence:
1. the consecutive sections should have a common side of length at least 1;
2. the first and the last sections should stand on the corresponding ground levels;
3. the sections between may be either on the ground level or higher, but not higher than k - 1 from the ground level h_i (the height should be an integer);
<image> One of possible fences (blue color) for the first test case
Is it possible to build a fence that meets all rules?
Input
The first line contains a single integer t (1 ≤ t ≤ 10^4) — the number of test cases.
The first line of each test case contains two integers n and k (2 ≤ n ≤ 2 ⋅ 10^5; 2 ≤ k ≤ 10^8) — the number of sections in the fence and the height of each section.
The second line of each test case contains n integers h_1, h_2, ..., h_n (0 ≤ h_i ≤ 10^8), where h_i is the ground level beneath the i-th section.
It's guaranteed that the sum of n over test cases doesn't exceed 2 ⋅ 10^5.
Output
For each test case print YES if it's possible to build the fence that meets all rules. Otherwise, print NO.
You may print each letter in any case (for example, YES, Yes, yes, yEs will all be recognized as positive answer).
Example
Input
3
6 3
0 0 2 5 1 1
2 3
0 2
3 2
3 0 2
Output
YES
YES
NO
Note
In the first test case, one of the possible fences is shown in the picture.
In the second test case, according to the second rule, you should build both sections on the corresponding ground levels, and since k = 3, h_1 = 0, and h_2 = 2 the first rule is also fulfilled.
In the third test case, according to the second rule, you should build the first section on height 3 and the third section on height 2. According to the first rule, the second section should be on the height of at least 2 (to have a common side with the first section), but according to the third rule, the second section can be built on the height of at most h_2 + k - 1 = 1.
Submitted Solution:
```
for t in range(int(input())):
n, k = map(int, input().split())
ground = tuple(map(int, input().split()))
prev = ground[0]
r = 0
answer = "NO"
for h in ground[1:]:
if prev - (r + k - 1) <= h <= r + (k -1) + prev:
prev = h
else:
break
else:
answer = "YES"
print(answer)
``` | instruction | 0 | 48,134 | 8 | 96,268 |
No | output | 1 | 48,134 | 8 | 96,269 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Igor is in the museum and he wants to see as many pictures as possible.
Museum can be represented as a rectangular field of n × m cells. Each cell is either empty or impassable. Empty cells are marked with '.', impassable cells are marked with '*'. Every two adjacent cells of different types (one empty and one impassable) are divided by a wall containing one picture.
At the beginning Igor is in some empty cell. At every moment he can move to any empty cell that share a side with the current one.
For several starting positions you should calculate the maximum number of pictures that Igor can see. Igor is able to see the picture only if he is in the cell adjacent to the wall with this picture. Igor have a lot of time, so he will examine every picture he can see.
Input
First line of the input contains three integers n, m and k (3 ≤ n, m ≤ 1000, 1 ≤ k ≤ min(n·m, 100 000)) — the museum dimensions and the number of starting positions to process.
Each of the next n lines contains m symbols '.', '*' — the description of the museum. It is guaranteed that all border cells are impassable, so Igor can't go out from the museum.
Each of the last k lines contains two integers x and y (1 ≤ x ≤ n, 1 ≤ y ≤ m) — the row and the column of one of Igor's starting positions respectively. Rows are numbered from top to bottom, columns — from left to right. It is guaranteed that all starting positions are empty cells.
Output
Print k integers — the maximum number of pictures, that Igor can see if he starts in corresponding position.
Examples
Input
5 6 3
******
*..*.*
******
*....*
******
2 2
2 5
4 3
Output
6
4
10
Input
4 4 1
****
*..*
*.**
****
3 2
Output
8
Submitted Solution:
```
from collections import deque
N, M, K = map(int,input().split())
room = []
memo = {}
for i in range(N):
for j in range(M):
memo[(i, j)] = -1
for i in range(N):
room.append(input())
for i in range(K):
a, b = map(int, input().split())
if memo[(a-1, b-1)] != -1:
print(memo[(a-1, b-1)])
continue
deq = deque([(a-1, b-1)])
ans = 0
trace = []
while len(deq) != 0:
x, y = deq.popleft()
trace.append((x,y))
for p,q in [(x+1,y),(x-1,y),(x,y+1),(x,y-1)]:
if room[p][q] == '*':
ans += 1
else:
if memo[(p,q)] == -1:
deq.append((p,q))
memo[(x, y)] = 0
for tx, ty in trace:
memo[(tx, ty)] = ans
print(ans)
``` | instruction | 0 | 48,365 | 8 | 96,730 |
No | output | 1 | 48,365 | 8 | 96,731 |
Provide tags and a correct Python 3 solution for this coding contest problem.
There are n houses in a row. They are numbered from 1 to n in order from left to right. Initially you are in the house 1.
You have to perform k moves to other house. In one move you go from your current house to some other house. You can't stay where you are (i.e., in each move the new house differs from the current house). If you go from the house x to the house y, the total distance you walked increases by |x-y| units of distance, where |a| is the absolute value of a. It is possible to visit the same house multiple times (but you can't visit the same house in sequence).
Your goal is to walk exactly s units of distance in total.
If it is impossible, print "NO". Otherwise print "YES" and any of the ways to do that. Remember that you should do exactly k moves.
Input
The first line of the input contains three integers n, k, s (2 ≤ n ≤ 10^9, 1 ≤ k ≤ 2 ⋅ 10^5, 1 ≤ s ≤ 10^{18}) — the number of houses, the number of moves and the total distance you want to walk.
Output
If you cannot perform k moves with total walking distance equal to s, print "NO".
Otherwise print "YES" on the first line and then print exactly k integers h_i (1 ≤ h_i ≤ n) on the second line, where h_i is the house you visit on the i-th move.
For each j from 1 to k-1 the following condition should be satisfied: h_j ≠ h_{j + 1}. Also h_1 ≠ 1 should be satisfied.
Examples
Input
10 2 15
Output
YES
10 4
Input
10 9 45
Output
YES
10 1 10 1 2 1 2 1 6
Input
10 9 81
Output
YES
10 1 10 1 10 1 10 1 10
Input
10 9 82
Output
NO | instruction | 0 | 48,815 | 8 | 97,630 |
Tags: constructive algorithms, greedy
Correct Solution:
```
from itertools import*
n,k,s=map(int,input().split())
b=k<=s<=k*(n-1)
print(('NO','YES')[b])
if b:
l=s%k*[s//k+1]+(k-s%k)*[s//k];l[0]+=1
for i in range(1,k,2):l[i]=-l[i]
print(*accumulate(l))
``` | output | 1 | 48,815 | 8 | 97,631 |
Provide tags and a correct Python 3 solution for this coding contest problem.
There are n houses in a row. They are numbered from 1 to n in order from left to right. Initially you are in the house 1.
You have to perform k moves to other house. In one move you go from your current house to some other house. You can't stay where you are (i.e., in each move the new house differs from the current house). If you go from the house x to the house y, the total distance you walked increases by |x-y| units of distance, where |a| is the absolute value of a. It is possible to visit the same house multiple times (but you can't visit the same house in sequence).
Your goal is to walk exactly s units of distance in total.
If it is impossible, print "NO". Otherwise print "YES" and any of the ways to do that. Remember that you should do exactly k moves.
Input
The first line of the input contains three integers n, k, s (2 ≤ n ≤ 10^9, 1 ≤ k ≤ 2 ⋅ 10^5, 1 ≤ s ≤ 10^{18}) — the number of houses, the number of moves and the total distance you want to walk.
Output
If you cannot perform k moves with total walking distance equal to s, print "NO".
Otherwise print "YES" on the first line and then print exactly k integers h_i (1 ≤ h_i ≤ n) on the second line, where h_i is the house you visit on the i-th move.
For each j from 1 to k-1 the following condition should be satisfied: h_j ≠ h_{j + 1}. Also h_1 ≠ 1 should be satisfied.
Examples
Input
10 2 15
Output
YES
10 4
Input
10 9 45
Output
YES
10 1 10 1 2 1 2 1 6
Input
10 9 81
Output
YES
10 1 10 1 10 1 10 1 10
Input
10 9 82
Output
NO | instruction | 0 | 48,816 | 8 | 97,632 |
Tags: constructive algorithms, greedy
Correct Solution:
```
n,k,s=map(int,input().split())
if k<=s<=k*(n-1):
print("YES")
l=(s+n-2)//(n-1)
e=k-l
r=[]
for i in range(l):
v=min(s+1,n)
s-=v-1
a=[*range(2,min(e,n-2)+2)]+[v]
e-=n-2
if i%2:a=[n+1-x for x in a]
r.extend(a)
print(*r)
else:
print("NO")
``` | output | 1 | 48,816 | 8 | 97,633 |
Provide tags and a correct Python 3 solution for this coding contest problem.
There are n houses in a row. They are numbered from 1 to n in order from left to right. Initially you are in the house 1.
You have to perform k moves to other house. In one move you go from your current house to some other house. You can't stay where you are (i.e., in each move the new house differs from the current house). If you go from the house x to the house y, the total distance you walked increases by |x-y| units of distance, where |a| is the absolute value of a. It is possible to visit the same house multiple times (but you can't visit the same house in sequence).
Your goal is to walk exactly s units of distance in total.
If it is impossible, print "NO". Otherwise print "YES" and any of the ways to do that. Remember that you should do exactly k moves.
Input
The first line of the input contains three integers n, k, s (2 ≤ n ≤ 10^9, 1 ≤ k ≤ 2 ⋅ 10^5, 1 ≤ s ≤ 10^{18}) — the number of houses, the number of moves and the total distance you want to walk.
Output
If you cannot perform k moves with total walking distance equal to s, print "NO".
Otherwise print "YES" on the first line and then print exactly k integers h_i (1 ≤ h_i ≤ n) on the second line, where h_i is the house you visit on the i-th move.
For each j from 1 to k-1 the following condition should be satisfied: h_j ≠ h_{j + 1}. Also h_1 ≠ 1 should be satisfied.
Examples
Input
10 2 15
Output
YES
10 4
Input
10 9 45
Output
YES
10 1 10 1 2 1 2 1 6
Input
10 9 81
Output
YES
10 1 10 1 10 1 10 1 10
Input
10 9 82
Output
NO | instruction | 0 | 48,817 | 8 | 97,634 |
Tags: constructive algorithms, greedy
Correct Solution:
```
def next(a, b):
global n
if 1 <= a + b <= n:
return a + b
else:
return a - b
n, k, s = [int(i) for i in input().split()]
cur = 1
kk = k
if k <= s <= (n - 1) * k:
print('YES')
for i in range(k):
if n - 1 < s - (kk - 1):
cur = next(cur, n - 1)
print(cur, end = ' ')
s -= n - 1
else:
cur = next(cur, s - (kk - 1))
print(cur, end = ' ')
s -= s - (kk - 1)
kk -= 1
else:
print('NO')
``` | output | 1 | 48,817 | 8 | 97,635 |
Provide tags and a correct Python 3 solution for this coding contest problem.
There are n houses in a row. They are numbered from 1 to n in order from left to right. Initially you are in the house 1.
You have to perform k moves to other house. In one move you go from your current house to some other house. You can't stay where you are (i.e., in each move the new house differs from the current house). If you go from the house x to the house y, the total distance you walked increases by |x-y| units of distance, where |a| is the absolute value of a. It is possible to visit the same house multiple times (but you can't visit the same house in sequence).
Your goal is to walk exactly s units of distance in total.
If it is impossible, print "NO". Otherwise print "YES" and any of the ways to do that. Remember that you should do exactly k moves.
Input
The first line of the input contains three integers n, k, s (2 ≤ n ≤ 10^9, 1 ≤ k ≤ 2 ⋅ 10^5, 1 ≤ s ≤ 10^{18}) — the number of houses, the number of moves and the total distance you want to walk.
Output
If you cannot perform k moves with total walking distance equal to s, print "NO".
Otherwise print "YES" on the first line and then print exactly k integers h_i (1 ≤ h_i ≤ n) on the second line, where h_i is the house you visit on the i-th move.
For each j from 1 to k-1 the following condition should be satisfied: h_j ≠ h_{j + 1}. Also h_1 ≠ 1 should be satisfied.
Examples
Input
10 2 15
Output
YES
10 4
Input
10 9 45
Output
YES
10 1 10 1 2 1 2 1 6
Input
10 9 81
Output
YES
10 1 10 1 10 1 10 1 10
Input
10 9 82
Output
NO | instruction | 0 | 48,818 | 8 | 97,636 |
Tags: constructive algorithms, greedy
Correct Solution:
```
n,m,k=map(int,input().split())
if k<m or k>m*(n-1):
print("NO")
else:
print("YES")
i,j=1,n
while True:
if k-(n-1)<m:
temp=j
j=i
i=temp
break
else:
k-=(n-1)
m-=1
print(j,end=" ")
temp=j
j=i
i=temp
if j==n:
print(j-(k-m+1),end=" ")
j-=k-m+1
k-=m-1
m-=1
i=j+1
while(m>0):
print(i,end=" ")
k-=1
m-=1
temp=j
j=i
i=temp
else:
print(j+(k-m+1),end=" ")
j+=k-m+1
k-=m-1
m-=1
i=j-1
while(m>0):
print(i,end=" ")
k-=1
m-=1
temp=j
j=i
i=temp
``` | output | 1 | 48,818 | 8 | 97,637 |
Provide tags and a correct Python 3 solution for this coding contest problem.
There are n houses in a row. They are numbered from 1 to n in order from left to right. Initially you are in the house 1.
You have to perform k moves to other house. In one move you go from your current house to some other house. You can't stay where you are (i.e., in each move the new house differs from the current house). If you go from the house x to the house y, the total distance you walked increases by |x-y| units of distance, where |a| is the absolute value of a. It is possible to visit the same house multiple times (but you can't visit the same house in sequence).
Your goal is to walk exactly s units of distance in total.
If it is impossible, print "NO". Otherwise print "YES" and any of the ways to do that. Remember that you should do exactly k moves.
Input
The first line of the input contains three integers n, k, s (2 ≤ n ≤ 10^9, 1 ≤ k ≤ 2 ⋅ 10^5, 1 ≤ s ≤ 10^{18}) — the number of houses, the number of moves and the total distance you want to walk.
Output
If you cannot perform k moves with total walking distance equal to s, print "NO".
Otherwise print "YES" on the first line and then print exactly k integers h_i (1 ≤ h_i ≤ n) on the second line, where h_i is the house you visit on the i-th move.
For each j from 1 to k-1 the following condition should be satisfied: h_j ≠ h_{j + 1}. Also h_1 ≠ 1 should be satisfied.
Examples
Input
10 2 15
Output
YES
10 4
Input
10 9 45
Output
YES
10 1 10 1 2 1 2 1 6
Input
10 9 81
Output
YES
10 1 10 1 10 1 10 1 10
Input
10 9 82
Output
NO | instruction | 0 | 48,819 | 8 | 97,638 |
Tags: constructive algorithms, greedy
Correct Solution:
```
n,k,s=map(int,input().split())
if (n-1)*k<s or k>s:
exit(print("NO"))
ans=[]
now=1
while k:
if s==k:
if now==1:
ans.append(now+1)
now+=1
else:
ans.append(now-1)
now-=1
s-=1
else:
t=min(n-1,s-k+1)
if now>t:
ans.append(now-t)
now-=t
else:
ans.append(now+t)
now+=t
s-=t
k-=1
print("YES")
print(*ans)
``` | output | 1 | 48,819 | 8 | 97,639 |
Provide tags and a correct Python 3 solution for this coding contest problem.
There are n houses in a row. They are numbered from 1 to n in order from left to right. Initially you are in the house 1.
You have to perform k moves to other house. In one move you go from your current house to some other house. You can't stay where you are (i.e., in each move the new house differs from the current house). If you go from the house x to the house y, the total distance you walked increases by |x-y| units of distance, where |a| is the absolute value of a. It is possible to visit the same house multiple times (but you can't visit the same house in sequence).
Your goal is to walk exactly s units of distance in total.
If it is impossible, print "NO". Otherwise print "YES" and any of the ways to do that. Remember that you should do exactly k moves.
Input
The first line of the input contains three integers n, k, s (2 ≤ n ≤ 10^9, 1 ≤ k ≤ 2 ⋅ 10^5, 1 ≤ s ≤ 10^{18}) — the number of houses, the number of moves and the total distance you want to walk.
Output
If you cannot perform k moves with total walking distance equal to s, print "NO".
Otherwise print "YES" on the first line and then print exactly k integers h_i (1 ≤ h_i ≤ n) on the second line, where h_i is the house you visit on the i-th move.
For each j from 1 to k-1 the following condition should be satisfied: h_j ≠ h_{j + 1}. Also h_1 ≠ 1 should be satisfied.
Examples
Input
10 2 15
Output
YES
10 4
Input
10 9 45
Output
YES
10 1 10 1 2 1 2 1 6
Input
10 9 81
Output
YES
10 1 10 1 10 1 10 1 10
Input
10 9 82
Output
NO | instruction | 0 | 48,820 | 8 | 97,640 |
Tags: constructive algorithms, greedy
Correct Solution:
```
n, k, s = map(int, input().split())
ans = []
cur = 1
if s < k:
print('NO')
exit(0)
for i in range(k):
prev = cur
if i % 2 == 0:
cur += min(s, n - cur, s - (k - i - 1))
s -= abs(cur - prev)
ans.append(cur)
else:
cur -= min(s, cur - 1, s - (k - i - 1))
s -= abs(cur - prev)
ans.append(cur)
if s == 0:
print('YES')
print(*ans)
else:
print('NO')
``` | output | 1 | 48,820 | 8 | 97,641 |
Provide tags and a correct Python 3 solution for this coding contest problem.
There are n houses in a row. They are numbered from 1 to n in order from left to right. Initially you are in the house 1.
You have to perform k moves to other house. In one move you go from your current house to some other house. You can't stay where you are (i.e., in each move the new house differs from the current house). If you go from the house x to the house y, the total distance you walked increases by |x-y| units of distance, where |a| is the absolute value of a. It is possible to visit the same house multiple times (but you can't visit the same house in sequence).
Your goal is to walk exactly s units of distance in total.
If it is impossible, print "NO". Otherwise print "YES" and any of the ways to do that. Remember that you should do exactly k moves.
Input
The first line of the input contains three integers n, k, s (2 ≤ n ≤ 10^9, 1 ≤ k ≤ 2 ⋅ 10^5, 1 ≤ s ≤ 10^{18}) — the number of houses, the number of moves and the total distance you want to walk.
Output
If you cannot perform k moves with total walking distance equal to s, print "NO".
Otherwise print "YES" on the first line and then print exactly k integers h_i (1 ≤ h_i ≤ n) on the second line, where h_i is the house you visit on the i-th move.
For each j from 1 to k-1 the following condition should be satisfied: h_j ≠ h_{j + 1}. Also h_1 ≠ 1 should be satisfied.
Examples
Input
10 2 15
Output
YES
10 4
Input
10 9 45
Output
YES
10 1 10 1 2 1 2 1 6
Input
10 9 81
Output
YES
10 1 10 1 10 1 10 1 10
Input
10 9 82
Output
NO | instruction | 0 | 48,821 | 8 | 97,642 |
Tags: constructive algorithms, greedy
Correct Solution:
```
n,k,s=map(int,input().split())
l=s%k*[s//k+1]+(k-s%k)*[s//k]
for i in range(1,k,2):l[i]=-l[i]
t=1
for i in range(k):l[i]+=t;t=l[i]
b=k<=s<=k*(n-1)
print(('NO','YES')[b])
if b:print(*l)
``` | output | 1 | 48,821 | 8 | 97,643 |
Provide tags and a correct Python 3 solution for this coding contest problem.
There are n houses in a row. They are numbered from 1 to n in order from left to right. Initially you are in the house 1.
You have to perform k moves to other house. In one move you go from your current house to some other house. You can't stay where you are (i.e., in each move the new house differs from the current house). If you go from the house x to the house y, the total distance you walked increases by |x-y| units of distance, where |a| is the absolute value of a. It is possible to visit the same house multiple times (but you can't visit the same house in sequence).
Your goal is to walk exactly s units of distance in total.
If it is impossible, print "NO". Otherwise print "YES" and any of the ways to do that. Remember that you should do exactly k moves.
Input
The first line of the input contains three integers n, k, s (2 ≤ n ≤ 10^9, 1 ≤ k ≤ 2 ⋅ 10^5, 1 ≤ s ≤ 10^{18}) — the number of houses, the number of moves and the total distance you want to walk.
Output
If you cannot perform k moves with total walking distance equal to s, print "NO".
Otherwise print "YES" on the first line and then print exactly k integers h_i (1 ≤ h_i ≤ n) on the second line, where h_i is the house you visit on the i-th move.
For each j from 1 to k-1 the following condition should be satisfied: h_j ≠ h_{j + 1}. Also h_1 ≠ 1 should be satisfied.
Examples
Input
10 2 15
Output
YES
10 4
Input
10 9 45
Output
YES
10 1 10 1 2 1 2 1 6
Input
10 9 81
Output
YES
10 1 10 1 10 1 10 1 10
Input
10 9 82
Output
NO | instruction | 0 | 48,822 | 8 | 97,644 |
Tags: constructive algorithms, greedy
Correct Solution:
```
# dominator
n,k,s=map(int,input().split())
cur=1
if k>s or k*(n-1)<s:
print("NO")
else:
print("YES")
while k>0:
l=min(n-1,s-(k-1))
if(cur>l):
cur-=l
else:
cur+=l
print(cur,end=" ")
s-=l
k-=1
print()
``` | output | 1 | 48,822 | 8 | 97,645 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There are n houses in a row. They are numbered from 1 to n in order from left to right. Initially you are in the house 1.
You have to perform k moves to other house. In one move you go from your current house to some other house. You can't stay where you are (i.e., in each move the new house differs from the current house). If you go from the house x to the house y, the total distance you walked increases by |x-y| units of distance, where |a| is the absolute value of a. It is possible to visit the same house multiple times (but you can't visit the same house in sequence).
Your goal is to walk exactly s units of distance in total.
If it is impossible, print "NO". Otherwise print "YES" and any of the ways to do that. Remember that you should do exactly k moves.
Input
The first line of the input contains three integers n, k, s (2 ≤ n ≤ 10^9, 1 ≤ k ≤ 2 ⋅ 10^5, 1 ≤ s ≤ 10^{18}) — the number of houses, the number of moves and the total distance you want to walk.
Output
If you cannot perform k moves with total walking distance equal to s, print "NO".
Otherwise print "YES" on the first line and then print exactly k integers h_i (1 ≤ h_i ≤ n) on the second line, where h_i is the house you visit on the i-th move.
For each j from 1 to k-1 the following condition should be satisfied: h_j ≠ h_{j + 1}. Also h_1 ≠ 1 should be satisfied.
Examples
Input
10 2 15
Output
YES
10 4
Input
10 9 45
Output
YES
10 1 10 1 2 1 2 1 6
Input
10 9 81
Output
YES
10 1 10 1 10 1 10 1 10
Input
10 9 82
Output
NO
Submitted Solution:
```
n,k,s=map(int,input().split())
if s<k:
print('NO')
exit()
if s>(n-1)*k:
print('NO')
exit()
print('YES')
from math import ceil
up=ceil(s/k)
rem=s%k
prev=1
sm=0
mauka=0
f=0
for i in range(k):
if i>=rem and f==0 and rem!=0 :
f=1
up-=1
if sm+up<=s:
sm+=up
if mauka&1:
up1=-up
else:
up1=up
print(prev+up1,end=' ')
prev+=up1
else:
rem=s-sm
if mauka&1:
rem1=-rem
else:
rem1=rem
print(prev+rem1,end=' ')
prev+=rem1
sm=s
mauka+=1
``` | instruction | 0 | 48,823 | 8 | 97,646 |
Yes | output | 1 | 48,823 | 8 | 97,647 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There are n houses in a row. They are numbered from 1 to n in order from left to right. Initially you are in the house 1.
You have to perform k moves to other house. In one move you go from your current house to some other house. You can't stay where you are (i.e., in each move the new house differs from the current house). If you go from the house x to the house y, the total distance you walked increases by |x-y| units of distance, where |a| is the absolute value of a. It is possible to visit the same house multiple times (but you can't visit the same house in sequence).
Your goal is to walk exactly s units of distance in total.
If it is impossible, print "NO". Otherwise print "YES" and any of the ways to do that. Remember that you should do exactly k moves.
Input
The first line of the input contains three integers n, k, s (2 ≤ n ≤ 10^9, 1 ≤ k ≤ 2 ⋅ 10^5, 1 ≤ s ≤ 10^{18}) — the number of houses, the number of moves and the total distance you want to walk.
Output
If you cannot perform k moves with total walking distance equal to s, print "NO".
Otherwise print "YES" on the first line and then print exactly k integers h_i (1 ≤ h_i ≤ n) on the second line, where h_i is the house you visit on the i-th move.
For each j from 1 to k-1 the following condition should be satisfied: h_j ≠ h_{j + 1}. Also h_1 ≠ 1 should be satisfied.
Examples
Input
10 2 15
Output
YES
10 4
Input
10 9 45
Output
YES
10 1 10 1 2 1 2 1 6
Input
10 9 81
Output
YES
10 1 10 1 10 1 10 1 10
Input
10 9 82
Output
NO
Submitted Solution:
```
def step(cur, x):
if cur - x > 0:
return cur - x
else:
return cur + x
def main():
n, k, s = map(int, input().split())
cur = 1
if (k > s or k * (n-1) < s):
print("NO")
else:
print("YES")
while (k > 0):
l = min(n-1, s -(k-1))
cur = step(cur, l)
print(cur, end = ' ')
s -= l
k -= 1
if __name__ == "__main__":
main()
``` | instruction | 0 | 48,824 | 8 | 97,648 |
Yes | output | 1 | 48,824 | 8 | 97,649 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There are n houses in a row. They are numbered from 1 to n in order from left to right. Initially you are in the house 1.
You have to perform k moves to other house. In one move you go from your current house to some other house. You can't stay where you are (i.e., in each move the new house differs from the current house). If you go from the house x to the house y, the total distance you walked increases by |x-y| units of distance, where |a| is the absolute value of a. It is possible to visit the same house multiple times (but you can't visit the same house in sequence).
Your goal is to walk exactly s units of distance in total.
If it is impossible, print "NO". Otherwise print "YES" and any of the ways to do that. Remember that you should do exactly k moves.
Input
The first line of the input contains three integers n, k, s (2 ≤ n ≤ 10^9, 1 ≤ k ≤ 2 ⋅ 10^5, 1 ≤ s ≤ 10^{18}) — the number of houses, the number of moves and the total distance you want to walk.
Output
If you cannot perform k moves with total walking distance equal to s, print "NO".
Otherwise print "YES" on the first line and then print exactly k integers h_i (1 ≤ h_i ≤ n) on the second line, where h_i is the house you visit on the i-th move.
For each j from 1 to k-1 the following condition should be satisfied: h_j ≠ h_{j + 1}. Also h_1 ≠ 1 should be satisfied.
Examples
Input
10 2 15
Output
YES
10 4
Input
10 9 45
Output
YES
10 1 10 1 2 1 2 1 6
Input
10 9 81
Output
YES
10 1 10 1 10 1 10 1 10
Input
10 9 82
Output
NO
Submitted Solution:
```
n, k, s = map(int, input().split())
if (n - 1) * k < s or k > s:
print('NO')
else:
dist = 0
x = 1
answer = []
for i in range(k):
dist = min(n - 1, s - (k - i) + 1)
s -= dist
if x + dist > n:
x -= dist
else:
x += dist
answer.append(x)
print('YES')
print(*answer)
``` | instruction | 0 | 48,826 | 8 | 97,652 |
Yes | output | 1 | 48,826 | 8 | 97,653 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There are n houses in a row. They are numbered from 1 to n in order from left to right. Initially you are in the house 1.
You have to perform k moves to other house. In one move you go from your current house to some other house. You can't stay where you are (i.e., in each move the new house differs from the current house). If you go from the house x to the house y, the total distance you walked increases by |x-y| units of distance, where |a| is the absolute value of a. It is possible to visit the same house multiple times (but you can't visit the same house in sequence).
Your goal is to walk exactly s units of distance in total.
If it is impossible, print "NO". Otherwise print "YES" and any of the ways to do that. Remember that you should do exactly k moves.
Input
The first line of the input contains three integers n, k, s (2 ≤ n ≤ 10^9, 1 ≤ k ≤ 2 ⋅ 10^5, 1 ≤ s ≤ 10^{18}) — the number of houses, the number of moves and the total distance you want to walk.
Output
If you cannot perform k moves with total walking distance equal to s, print "NO".
Otherwise print "YES" on the first line and then print exactly k integers h_i (1 ≤ h_i ≤ n) on the second line, where h_i is the house you visit on the i-th move.
For each j from 1 to k-1 the following condition should be satisfied: h_j ≠ h_{j + 1}. Also h_1 ≠ 1 should be satisfied.
Examples
Input
10 2 15
Output
YES
10 4
Input
10 9 45
Output
YES
10 1 10 1 2 1 2 1 6
Input
10 9 81
Output
YES
10 1 10 1 10 1 10 1 10
Input
10 9 82
Output
NO
Submitted Solution:
```
readints=lambda:map(int, input().strip('\n').split())
n,k,s=readints()
done=False
if s<k or s>((n-1)*k):
print('NO')
done=True
elif k==s:
buf=''
for i in range(k):
if i%2==0:
buf+='1 '
else:
buf+='2 '
else:
curr=1
buf=''
while s>n:
d=n-1
curr=1 if curr!=1 else n
k-=1
s-=d
buf+=(str(curr)+' ')
# print(n,k,s)
# reduce to k=s
d=s-k+1
if curr+d<=n: curr+=d
else: curr-=d
buf+=(str(curr)+' ')
k-=1
# oscillate
while k>0:
if curr<n:
curr+=1
else:
curr-=1
buf+=(str(curr)+' ')
k-=1
if not done:
print('YES')
print(buf)
``` | instruction | 0 | 48,827 | 8 | 97,654 |
No | output | 1 | 48,827 | 8 | 97,655 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There are n houses in a row. They are numbered from 1 to n in order from left to right. Initially you are in the house 1.
You have to perform k moves to other house. In one move you go from your current house to some other house. You can't stay where you are (i.e., in each move the new house differs from the current house). If you go from the house x to the house y, the total distance you walked increases by |x-y| units of distance, where |a| is the absolute value of a. It is possible to visit the same house multiple times (but you can't visit the same house in sequence).
Your goal is to walk exactly s units of distance in total.
If it is impossible, print "NO". Otherwise print "YES" and any of the ways to do that. Remember that you should do exactly k moves.
Input
The first line of the input contains three integers n, k, s (2 ≤ n ≤ 10^9, 1 ≤ k ≤ 2 ⋅ 10^5, 1 ≤ s ≤ 10^{18}) — the number of houses, the number of moves and the total distance you want to walk.
Output
If you cannot perform k moves with total walking distance equal to s, print "NO".
Otherwise print "YES" on the first line and then print exactly k integers h_i (1 ≤ h_i ≤ n) on the second line, where h_i is the house you visit on the i-th move.
For each j from 1 to k-1 the following condition should be satisfied: h_j ≠ h_{j + 1}. Also h_1 ≠ 1 should be satisfied.
Examples
Input
10 2 15
Output
YES
10 4
Input
10 9 45
Output
YES
10 1 10 1 2 1 2 1 6
Input
10 9 81
Output
YES
10 1 10 1 10 1 10 1 10
Input
10 9 82
Output
NO
Submitted Solution:
```
n, k, s = map(int, input().split())
if (n - 1) * k < s or s < k:
print('NO')
exit()
def check(p):
return s - p * (n - 1) - (k - p - 1) < n - 1
def binSearch(a, b):
left, right = a - 1, b + 1
while right - left > 1:
mid = (left + right) // 2
if check(mid):
right = mid
else:
left = mid
return right
b = binSearch(1, k)
print('YES')
temp_ = k - b - 1
temp = s - b * (n - 1) - temp_ + 1
ans = [n, 1] * (b // 2) + [n] * (b % 2)
if temp_ >= 0:
if b % 2 == 0:
ans.append(temp)
ans += [temp - 1, temp] * (temp_ // 2) + [temp - 1] * (temp_ % 2)
else:
ans.append(n - temp + 1)
ans += [n - temp + 2, n - temp + 1] * (temp_ // 2) + [n - temp + 2] * (temp_ % 2)
print(' '.join(map(str, ans)))
``` | instruction | 0 | 48,828 | 8 | 97,656 |
No | output | 1 | 48,828 | 8 | 97,657 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There are n houses in a row. They are numbered from 1 to n in order from left to right. Initially you are in the house 1.
You have to perform k moves to other house. In one move you go from your current house to some other house. You can't stay where you are (i.e., in each move the new house differs from the current house). If you go from the house x to the house y, the total distance you walked increases by |x-y| units of distance, where |a| is the absolute value of a. It is possible to visit the same house multiple times (but you can't visit the same house in sequence).
Your goal is to walk exactly s units of distance in total.
If it is impossible, print "NO". Otherwise print "YES" and any of the ways to do that. Remember that you should do exactly k moves.
Input
The first line of the input contains three integers n, k, s (2 ≤ n ≤ 10^9, 1 ≤ k ≤ 2 ⋅ 10^5, 1 ≤ s ≤ 10^{18}) — the number of houses, the number of moves and the total distance you want to walk.
Output
If you cannot perform k moves with total walking distance equal to s, print "NO".
Otherwise print "YES" on the first line and then print exactly k integers h_i (1 ≤ h_i ≤ n) on the second line, where h_i is the house you visit on the i-th move.
For each j from 1 to k-1 the following condition should be satisfied: h_j ≠ h_{j + 1}. Also h_1 ≠ 1 should be satisfied.
Examples
Input
10 2 15
Output
YES
10 4
Input
10 9 45
Output
YES
10 1 10 1 2 1 2 1 6
Input
10 9 81
Output
YES
10 1 10 1 10 1 10 1 10
Input
10 9 82
Output
NO
Submitted Solution:
```
n, k, s = map(int, input().split())
if (n - 1) * k < s:
print('NO')
else:
print("YES")
m = (s + k - 1 )// k
w = 0
c = 1
for i in range(k - 1):
if w % 2 == 0:
w = 1
print(1 + m, end = ' ')
c = 1 + m
else:
w = 0
print(1, end=' ')
c = 1
if w % 2 == 0:
print(1 + s - m * (k - 1))
else:
print(c - (s - m * (k - 1)))
``` | instruction | 0 | 48,829 | 8 | 97,658 |
No | output | 1 | 48,829 | 8 | 97,659 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There are n houses in a row. They are numbered from 1 to n in order from left to right. Initially you are in the house 1.
You have to perform k moves to other house. In one move you go from your current house to some other house. You can't stay where you are (i.e., in each move the new house differs from the current house). If you go from the house x to the house y, the total distance you walked increases by |x-y| units of distance, where |a| is the absolute value of a. It is possible to visit the same house multiple times (but you can't visit the same house in sequence).
Your goal is to walk exactly s units of distance in total.
If it is impossible, print "NO". Otherwise print "YES" and any of the ways to do that. Remember that you should do exactly k moves.
Input
The first line of the input contains three integers n, k, s (2 ≤ n ≤ 10^9, 1 ≤ k ≤ 2 ⋅ 10^5, 1 ≤ s ≤ 10^{18}) — the number of houses, the number of moves and the total distance you want to walk.
Output
If you cannot perform k moves with total walking distance equal to s, print "NO".
Otherwise print "YES" on the first line and then print exactly k integers h_i (1 ≤ h_i ≤ n) on the second line, where h_i is the house you visit on the i-th move.
For each j from 1 to k-1 the following condition should be satisfied: h_j ≠ h_{j + 1}. Also h_1 ≠ 1 should be satisfied.
Examples
Input
10 2 15
Output
YES
10 4
Input
10 9 45
Output
YES
10 1 10 1 2 1 2 1 6
Input
10 9 81
Output
YES
10 1 10 1 10 1 10 1 10
Input
10 9 82
Output
NO
Submitted Solution:
```
max_number, k, s = map(int, input().split())
if s > k * (max_number - 1):
print('NO')
elif s < k:
print('NO')
elif s % k == 1 and (s // k) + 1 >= max_number:
print('NO')
else:
delta = s // k
if s % k == 1:
pos = delta + 1
else:
pos = delta
res = str(pos + 1) + ' '
for step in range(k - 1):
if pos > 1:
pos -= delta
else:
pos += delta
res += str(pos + 1) + ' '
print('YES')
print(res)
``` | instruction | 0 | 48,830 | 8 | 97,660 |
No | output | 1 | 48,830 | 8 | 97,661 |
Provide tags and a correct Python 3 solution for this coding contest problem.
This is a harder version of the problem. In this version n ≤ 500 000
The outskirts of the capital are being actively built up in Berland. The company "Kernel Panic" manages the construction of a residential complex of skyscrapers in New Berlskva. All skyscrapers are built along the highway. It is known that the company has already bought n plots along the highway and is preparing to build n skyscrapers, one skyscraper per plot.
Architects must consider several requirements when planning a skyscraper. Firstly, since the land on each plot has different properties, each skyscraper has a limit on the largest number of floors it can have. Secondly, according to the design code of the city, it is unacceptable for a skyscraper to simultaneously have higher skyscrapers both to the left and to the right of it.
Formally, let's number the plots from 1 to n. Then if the skyscraper on the i-th plot has a_i floors, it must hold that a_i is at most m_i (1 ≤ a_i ≤ m_i). Also there mustn't be integers j and k such that j < i < k and a_j > a_i < a_k. Plots j and k are not required to be adjacent to i.
The company wants the total number of floors in the built skyscrapers to be as large as possible. Help it to choose the number of floors for each skyscraper in an optimal way, i.e. in such a way that all requirements are fulfilled, and among all such construction plans choose any plan with the maximum possible total number of floors.
Input
The first line contains a single integer n (1 ≤ n ≤ 500 000) — the number of plots.
The second line contains the integers m_1, m_2, …, m_n (1 ≤ m_i ≤ 10^9) — the limit on the number of floors for every possible number of floors for a skyscraper on each plot.
Output
Print n integers a_i — the number of floors in the plan for each skyscraper, such that all requirements are met, and the total number of floors in all skyscrapers is the maximum possible.
If there are multiple answers possible, print any of them.
Examples
Input
5
1 2 3 2 1
Output
1 2 3 2 1
Input
3
10 6 8
Output
10 6 6
Note
In the first example, you can build all skyscrapers with the highest possible height.
In the second test example, you cannot give the maximum height to all skyscrapers as this violates the design code restriction. The answer [10, 6, 6] is optimal. Note that the answer of [6, 6, 8] also satisfies all restrictions, but is not optimal. | instruction | 0 | 48,953 | 8 | 97,906 |
Tags: data structures, dp, greedy
Correct Solution:
```
import sys
input=sys.stdin.readline
n=int(input())
m=list(map(int,input().split()))
l=[-1]*n
r=[n]*n
stack=[]
for i in range(n):
while stack and m[stack[-1]]>=m[i]:
stack.pop()
if stack:
l[i]=stack[-1]
stack.append(i)
stack=[]
for i in range(n)[::-1]:
while stack and m[stack[-1]]>=m[i]:
stack.pop()
if stack:
r[i]=stack[-1]
stack.append(i)
pre=[0]*n
suf=[0]*n
for i in range(n):
if l[i]==-1:
pre[i]=(i+1)*m[i]
else:
pre[i]=pre[l[i]]+(i-l[i])*m[i]
for i in range(n)[::-1]:
if r[i]==n:
suf[i]=(n-i)*m[i]
else:
suf[i]=suf[r[i]]+(r[i]-i)*m[i]
ans=0
idx=-1
for i in range(n):
if ans<pre[i]+suf[i]-m[i]:
ans=pre[i]+suf[i]-m[i]
idx=i
arr=[-1]*n
arr[idx]=m[idx]
for i in range(idx)[::-1]:
arr[i]=min(arr[i+1],m[i])
for i in range(idx+1,n):
arr[i]=min(arr[i-1],m[i])
print(*arr)
``` | output | 1 | 48,953 | 8 | 97,907 |
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