message stringlengths 2 48.6k | message_type stringclasses 2 values | message_id int64 0 1 | conversation_id int64 318 108k | cluster float64 8 8 | __index_level_0__ int64 636 217k |
|---|---|---|---|---|---|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
B-Mansion and courier
Problem Statement
Taro lives alone in a mansion. Taro, who loves studying, intends to study in his study in the house today. Taro can't concentrate outside the study, so he always studies in the study.
However, on this day, $ N $ of courier service to Taro arrived. $ i $ ($ 1 \ leq i \ leq N $) The arrival time of the third courier is $ a_i $. It is painful to have the delivery person wait at the front door, so Taro decided to be at the front door by the time the courier arrives. Due to the large size of the mansion, it takes $ M $ one way to move between the study and the entrance.
On the other hand, Taro wants to study for as long as possible. Find the maximum amount of time Taro can study in the study from time $ 0 $ to time $ T $.
Taro is in the study at time $ 0 $, and the courier does not arrive earlier than the time $ M $, and the courier does not arrive later than the time $ T $. Also, the time it takes for Taro to receive the courier can be ignored.
Input
Each dataset consists of two lines. The first line consists of three integers $ N, M, T $ separated by blanks. These integers satisfy $ 1 \ leq N \ leq 100 $, $ 1 \ leq M \ leq 10 {,} 000 $, $ 1 \ leq T \ leq 10 {,} 000 $. The second line consists of $ N $ integers $ a_1, a_2, \ dots, a_N $ separated by blanks. Each $ a_i $ fills $ M \ leq a_i \ leq T $ and is also $ a_i <a_ {i + 1} $ ($ 1 \ leq i <N $).
Output
Output an integer representing the maximum amount of time Taro can study on one line.
Sample Input 1
1 1 5
3
Output for the Sample Input 1
3
Sample Input 2
2 1 10
2 7
Output for the Sample Input 2
6
Sample Input 3
2 4 10
6 8
Output for the Sample Input 3
2
Example
Input
1 1 5
3
Output
3
Submitted Solution:
```
N,M,T=map(int,input().split())
l=list(map(int,input().split()))
t=0
for (i,j) in enumerate(l):
if not i:
t+=(l[i]-M)
elif (l[i]-l[i-1])>2*M:
t+=(l[i]-l[i-1])-2*M
if T-l[-1]>M:
t+=(T-l[-1])-M
print(t)
``` | instruction | 0 | 30,013 | 8 | 60,026 |
Yes | output | 1 | 30,013 | 8 | 60,027 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
B-Mansion and courier
Problem Statement
Taro lives alone in a mansion. Taro, who loves studying, intends to study in his study in the house today. Taro can't concentrate outside the study, so he always studies in the study.
However, on this day, $ N $ of courier service to Taro arrived. $ i $ ($ 1 \ leq i \ leq N $) The arrival time of the third courier is $ a_i $. It is painful to have the delivery person wait at the front door, so Taro decided to be at the front door by the time the courier arrives. Due to the large size of the mansion, it takes $ M $ one way to move between the study and the entrance.
On the other hand, Taro wants to study for as long as possible. Find the maximum amount of time Taro can study in the study from time $ 0 $ to time $ T $.
Taro is in the study at time $ 0 $, and the courier does not arrive earlier than the time $ M $, and the courier does not arrive later than the time $ T $. Also, the time it takes for Taro to receive the courier can be ignored.
Input
Each dataset consists of two lines. The first line consists of three integers $ N, M, T $ separated by blanks. These integers satisfy $ 1 \ leq N \ leq 100 $, $ 1 \ leq M \ leq 10 {,} 000 $, $ 1 \ leq T \ leq 10 {,} 000 $. The second line consists of $ N $ integers $ a_1, a_2, \ dots, a_N $ separated by blanks. Each $ a_i $ fills $ M \ leq a_i \ leq T $ and is also $ a_i <a_ {i + 1} $ ($ 1 \ leq i <N $).
Output
Output an integer representing the maximum amount of time Taro can study on one line.
Sample Input 1
1 1 5
3
Output for the Sample Input 1
3
Sample Input 2
2 1 10
2 7
Output for the Sample Input 2
6
Sample Input 3
2 4 10
6 8
Output for the Sample Input 3
2
Example
Input
1 1 5
3
Output
3
Submitted Solution:
```
n,m,t=map(int,input().split())
d=0
a=[int(i) for i in input().split()]
b=a[0]-m
c=a[0]+m
for i in range(1,n):
if a[i]>c+m:
d+=c-b
b=a[i]-m
c=a[i]+m
if c<t:d+=c-b
else:d+=t-b
print(t-d)
``` | instruction | 0 | 30,014 | 8 | 60,028 |
Yes | output | 1 | 30,014 | 8 | 60,029 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
B-Mansion and courier
Problem Statement
Taro lives alone in a mansion. Taro, who loves studying, intends to study in his study in the house today. Taro can't concentrate outside the study, so he always studies in the study.
However, on this day, $ N $ of courier service to Taro arrived. $ i $ ($ 1 \ leq i \ leq N $) The arrival time of the third courier is $ a_i $. It is painful to have the delivery person wait at the front door, so Taro decided to be at the front door by the time the courier arrives. Due to the large size of the mansion, it takes $ M $ one way to move between the study and the entrance.
On the other hand, Taro wants to study for as long as possible. Find the maximum amount of time Taro can study in the study from time $ 0 $ to time $ T $.
Taro is in the study at time $ 0 $, and the courier does not arrive earlier than the time $ M $, and the courier does not arrive later than the time $ T $. Also, the time it takes for Taro to receive the courier can be ignored.
Input
Each dataset consists of two lines. The first line consists of three integers $ N, M, T $ separated by blanks. These integers satisfy $ 1 \ leq N \ leq 100 $, $ 1 \ leq M \ leq 10 {,} 000 $, $ 1 \ leq T \ leq 10 {,} 000 $. The second line consists of $ N $ integers $ a_1, a_2, \ dots, a_N $ separated by blanks. Each $ a_i $ fills $ M \ leq a_i \ leq T $ and is also $ a_i <a_ {i + 1} $ ($ 1 \ leq i <N $).
Output
Output an integer representing the maximum amount of time Taro can study on one line.
Sample Input 1
1 1 5
3
Output for the Sample Input 1
3
Sample Input 2
2 1 10
2 7
Output for the Sample Input 2
6
Sample Input 3
2 4 10
6 8
Output for the Sample Input 3
2
Example
Input
1 1 5
3
Output
3
Submitted Solution:
```
# coding: utf-8
# Your code here!
a = list(map(int,input().split()))
b = list(map(int,input().split()))
d = 0
k = 0
su = 0
for i in range(a[2]):
k += 1
su = max(su,k)
if b[d] = i + 1:
k = 0
print(su)
``` | instruction | 0 | 30,015 | 8 | 60,030 |
No | output | 1 | 30,015 | 8 | 60,031 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
B-Mansion and courier
Problem Statement
Taro lives alone in a mansion. Taro, who loves studying, intends to study in his study in the house today. Taro can't concentrate outside the study, so he always studies in the study.
However, on this day, $ N $ of courier service to Taro arrived. $ i $ ($ 1 \ leq i \ leq N $) The arrival time of the third courier is $ a_i $. It is painful to have the delivery person wait at the front door, so Taro decided to be at the front door by the time the courier arrives. Due to the large size of the mansion, it takes $ M $ one way to move between the study and the entrance.
On the other hand, Taro wants to study for as long as possible. Find the maximum amount of time Taro can study in the study from time $ 0 $ to time $ T $.
Taro is in the study at time $ 0 $, and the courier does not arrive earlier than the time $ M $, and the courier does not arrive later than the time $ T $. Also, the time it takes for Taro to receive the courier can be ignored.
Input
Each dataset consists of two lines. The first line consists of three integers $ N, M, T $ separated by blanks. These integers satisfy $ 1 \ leq N \ leq 100 $, $ 1 \ leq M \ leq 10 {,} 000 $, $ 1 \ leq T \ leq 10 {,} 000 $. The second line consists of $ N $ integers $ a_1, a_2, \ dots, a_N $ separated by blanks. Each $ a_i $ fills $ M \ leq a_i \ leq T $ and is also $ a_i <a_ {i + 1} $ ($ 1 \ leq i <N $).
Output
Output an integer representing the maximum amount of time Taro can study on one line.
Sample Input 1
1 1 5
3
Output for the Sample Input 1
3
Sample Input 2
2 1 10
2 7
Output for the Sample Input 2
6
Sample Input 3
2 4 10
6 8
Output for the Sample Input 3
2
Example
Input
1 1 5
3
Output
3
Submitted Solution:
```
# coding: utf-8
# Your code here!
a = list(map(int,input().split()))
b = list(map(int,input().split()))
k = [0]*a[2]
pp = 0
su = 0
for i in b:
for j in range(a[2]):
if j >= i - a[1] + 1 and j <= i + a[1] :
k[j] += 1
for i in range(a[2]):
if k[i] == 0:
su += 1
print(su)
``` | instruction | 0 | 30,016 | 8 | 60,032 |
No | output | 1 | 30,016 | 8 | 60,033 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
B-Mansion and courier
Problem Statement
Taro lives alone in a mansion. Taro, who loves studying, intends to study in his study in the house today. Taro can't concentrate outside the study, so he always studies in the study.
However, on this day, $ N $ of courier service to Taro arrived. $ i $ ($ 1 \ leq i \ leq N $) The arrival time of the third courier is $ a_i $. It is painful to have the delivery person wait at the front door, so Taro decided to be at the front door by the time the courier arrives. Due to the large size of the mansion, it takes $ M $ one way to move between the study and the entrance.
On the other hand, Taro wants to study for as long as possible. Find the maximum amount of time Taro can study in the study from time $ 0 $ to time $ T $.
Taro is in the study at time $ 0 $, and the courier does not arrive earlier than the time $ M $, and the courier does not arrive later than the time $ T $. Also, the time it takes for Taro to receive the courier can be ignored.
Input
Each dataset consists of two lines. The first line consists of three integers $ N, M, T $ separated by blanks. These integers satisfy $ 1 \ leq N \ leq 100 $, $ 1 \ leq M \ leq 10 {,} 000 $, $ 1 \ leq T \ leq 10 {,} 000 $. The second line consists of $ N $ integers $ a_1, a_2, \ dots, a_N $ separated by blanks. Each $ a_i $ fills $ M \ leq a_i \ leq T $ and is also $ a_i <a_ {i + 1} $ ($ 1 \ leq i <N $).
Output
Output an integer representing the maximum amount of time Taro can study on one line.
Sample Input 1
1 1 5
3
Output for the Sample Input 1
3
Sample Input 2
2 1 10
2 7
Output for the Sample Input 2
6
Sample Input 3
2 4 10
6 8
Output for the Sample Input 3
2
Example
Input
1 1 5
3
Output
3
Submitted Solution:
```
# coding: utf-8
# Your code here!
a = list(map(int,input().split()))
b = list(map(int,input().split()))
k = [0]*a[2]
su = 0
for i in b:
for j in range(a[2]):
if j >= i - a[1] and j <= i + a[1]:
k[j] += 1
for i in range(a[2]):
if k[i] > 0:
su += 1
print(su)
``` | instruction | 0 | 30,017 | 8 | 60,034 |
No | output | 1 | 30,017 | 8 | 60,035 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
B-Mansion and courier
Problem Statement
Taro lives alone in a mansion. Taro, who loves studying, intends to study in his study in the house today. Taro can't concentrate outside the study, so he always studies in the study.
However, on this day, $ N $ of courier service to Taro arrived. $ i $ ($ 1 \ leq i \ leq N $) The arrival time of the third courier is $ a_i $. It is painful to have the delivery person wait at the front door, so Taro decided to be at the front door by the time the courier arrives. Due to the large size of the mansion, it takes $ M $ one way to move between the study and the entrance.
On the other hand, Taro wants to study for as long as possible. Find the maximum amount of time Taro can study in the study from time $ 0 $ to time $ T $.
Taro is in the study at time $ 0 $, and the courier does not arrive earlier than the time $ M $, and the courier does not arrive later than the time $ T $. Also, the time it takes for Taro to receive the courier can be ignored.
Input
Each dataset consists of two lines. The first line consists of three integers $ N, M, T $ separated by blanks. These integers satisfy $ 1 \ leq N \ leq 100 $, $ 1 \ leq M \ leq 10 {,} 000 $, $ 1 \ leq T \ leq 10 {,} 000 $. The second line consists of $ N $ integers $ a_1, a_2, \ dots, a_N $ separated by blanks. Each $ a_i $ fills $ M \ leq a_i \ leq T $ and is also $ a_i <a_ {i + 1} $ ($ 1 \ leq i <N $).
Output
Output an integer representing the maximum amount of time Taro can study on one line.
Sample Input 1
1 1 5
3
Output for the Sample Input 1
3
Sample Input 2
2 1 10
2 7
Output for the Sample Input 2
6
Sample Input 3
2 4 10
6 8
Output for the Sample Input 3
2
Example
Input
1 1 5
3
Output
3
Submitted Solution:
```
# coding: utf-8
# Your code here!
a = list(map(int,input().split()))
b = list(map(int,input().split()))
k = [0]*a[2]
su = 0
for i in b:
for j in range(a[2]):
if j >= i - a[1] and j <= i + a[1]:
k[j] += 1
for i in range(a[2]):
if k[i] == 0:
su += 1
print(su)
``` | instruction | 0 | 30,018 | 8 | 60,036 |
No | output | 1 | 30,018 | 8 | 60,037 |
Provide tags and a correct Python 3 solution for this coding contest problem.
There are n pillars aligned in a row and numbered from 1 to n.
Initially each pillar contains exactly one disk. The i-th pillar contains a disk having radius a_i.
You can move these disks from one pillar to another. You can take a disk from pillar i and place it on top of pillar j if all these conditions are met:
1. there is no other pillar between pillars i and j. Formally, it means that |i - j| = 1;
2. pillar i contains exactly one disk;
3. either pillar j contains no disks, or the topmost disk on pillar j has radius strictly greater than the radius of the disk you move.
When you place a disk on a pillar that already has some disks on it, you put the new disk on top of previously placed disks, so the new disk will be used to check the third condition if you try to place another disk on the same pillar.
You may take any disk and place it on other pillar any number of times, provided that every time you do it, all three aforementioned conditions are met. Now you wonder, is it possible to place all n disks on the same pillar simultaneously?
Input
The first line contains one integer n (3 β€ n β€ 2 β
10^5) β the number of pillars.
The second line contains n integers a_1, a_2, ..., a_i (1 β€ a_i β€ n), where a_i is the radius of the disk initially placed on the i-th pillar. All numbers a_i are distinct.
Output
Print YES if it is possible to place all the disks on the same pillar simultaneously, and NO otherwise. You may print each letter in any case (YES, yes, Yes will all be recognized as positive answer, NO, no and nO will all be recognized as negative answer).
Examples
Input
4
1 3 4 2
Output
YES
Input
3
3 1 2
Output
NO
Note
In the first case it is possible to place all disks on pillar 3 using the following sequence of actions:
1. take the disk with radius 3 from pillar 2 and place it on top of pillar 3;
2. take the disk with radius 1 from pillar 1 and place it on top of pillar 2;
3. take the disk with radius 2 from pillar 4 and place it on top of pillar 3;
4. take the disk with radius 1 from pillar 2 and place it on top of pillar 3. | instruction | 0 | 30,129 | 8 | 60,258 |
Tags: greedy, implementation
Correct Solution:
```
n=int(input())
a=list(map(int,input().split()))
x=a.index(max(a))
l=a[:x]
r=a[x:]
ans='YES'
if len(l)>=2:
for i in range(len(l)-1):
if l[i+1]-l[i]<0:
ans='NO'
break
if len(r)>=2:
for i in range(len(r)-1):
if r[i+1]-r[i]>0:
ans='NO'
break
print(ans)
``` | output | 1 | 30,129 | 8 | 60,259 |
Provide tags and a correct Python 3 solution for this coding contest problem.
There are n pillars aligned in a row and numbered from 1 to n.
Initially each pillar contains exactly one disk. The i-th pillar contains a disk having radius a_i.
You can move these disks from one pillar to another. You can take a disk from pillar i and place it on top of pillar j if all these conditions are met:
1. there is no other pillar between pillars i and j. Formally, it means that |i - j| = 1;
2. pillar i contains exactly one disk;
3. either pillar j contains no disks, or the topmost disk on pillar j has radius strictly greater than the radius of the disk you move.
When you place a disk on a pillar that already has some disks on it, you put the new disk on top of previously placed disks, so the new disk will be used to check the third condition if you try to place another disk on the same pillar.
You may take any disk and place it on other pillar any number of times, provided that every time you do it, all three aforementioned conditions are met. Now you wonder, is it possible to place all n disks on the same pillar simultaneously?
Input
The first line contains one integer n (3 β€ n β€ 2 β
10^5) β the number of pillars.
The second line contains n integers a_1, a_2, ..., a_i (1 β€ a_i β€ n), where a_i is the radius of the disk initially placed on the i-th pillar. All numbers a_i are distinct.
Output
Print YES if it is possible to place all the disks on the same pillar simultaneously, and NO otherwise. You may print each letter in any case (YES, yes, Yes will all be recognized as positive answer, NO, no and nO will all be recognized as negative answer).
Examples
Input
4
1 3 4 2
Output
YES
Input
3
3 1 2
Output
NO
Note
In the first case it is possible to place all disks on pillar 3 using the following sequence of actions:
1. take the disk with radius 3 from pillar 2 and place it on top of pillar 3;
2. take the disk with radius 1 from pillar 1 and place it on top of pillar 2;
3. take the disk with radius 2 from pillar 4 and place it on top of pillar 3;
4. take the disk with radius 1 from pillar 2 and place it on top of pillar 3. | instruction | 0 | 30,130 | 8 | 60,260 |
Tags: greedy, implementation
Correct Solution:
```
def pro(arr):
ind=0
maxi=0
for i in range(len(arr)):
if(arr[i]>maxi):
maxi=arr[i]
ind=i
for i in range(ind-1):
if(arr[i]>arr[i+1]):
print('NO')
return
for i in range(ind+1,len(arr)-1):
if(arr[i]<arr[i+1]):
print('NO')
return
print('YES')
n=int(input())
arr=list(map(int,input().split()))
pro(arr)
``` | output | 1 | 30,130 | 8 | 60,261 |
Provide tags and a correct Python 3 solution for this coding contest problem.
There are n pillars aligned in a row and numbered from 1 to n.
Initially each pillar contains exactly one disk. The i-th pillar contains a disk having radius a_i.
You can move these disks from one pillar to another. You can take a disk from pillar i and place it on top of pillar j if all these conditions are met:
1. there is no other pillar between pillars i and j. Formally, it means that |i - j| = 1;
2. pillar i contains exactly one disk;
3. either pillar j contains no disks, or the topmost disk on pillar j has radius strictly greater than the radius of the disk you move.
When you place a disk on a pillar that already has some disks on it, you put the new disk on top of previously placed disks, so the new disk will be used to check the third condition if you try to place another disk on the same pillar.
You may take any disk and place it on other pillar any number of times, provided that every time you do it, all three aforementioned conditions are met. Now you wonder, is it possible to place all n disks on the same pillar simultaneously?
Input
The first line contains one integer n (3 β€ n β€ 2 β
10^5) β the number of pillars.
The second line contains n integers a_1, a_2, ..., a_i (1 β€ a_i β€ n), where a_i is the radius of the disk initially placed on the i-th pillar. All numbers a_i are distinct.
Output
Print YES if it is possible to place all the disks on the same pillar simultaneously, and NO otherwise. You may print each letter in any case (YES, yes, Yes will all be recognized as positive answer, NO, no and nO will all be recognized as negative answer).
Examples
Input
4
1 3 4 2
Output
YES
Input
3
3 1 2
Output
NO
Note
In the first case it is possible to place all disks on pillar 3 using the following sequence of actions:
1. take the disk with radius 3 from pillar 2 and place it on top of pillar 3;
2. take the disk with radius 1 from pillar 1 and place it on top of pillar 2;
3. take the disk with radius 2 from pillar 4 and place it on top of pillar 3;
4. take the disk with radius 1 from pillar 2 and place it on top of pillar 3. | instruction | 0 | 30,131 | 8 | 60,262 |
Tags: greedy, implementation
Correct Solution:
```
import sys
input=sys.stdin.readline
n=int(input())
a=list(map(int, input().split()))
index=a.index(n)
flag=1
for i in range(1, index):
if a[i]<a[i-1]:
flag=0
for i in reversed(range(index, n - 1)):
if a[i]<a[i+1]:
flag=0
if flag:
print("YES")
else:
print("NO")
``` | output | 1 | 30,131 | 8 | 60,263 |
Provide tags and a correct Python 3 solution for this coding contest problem.
There are n pillars aligned in a row and numbered from 1 to n.
Initially each pillar contains exactly one disk. The i-th pillar contains a disk having radius a_i.
You can move these disks from one pillar to another. You can take a disk from pillar i and place it on top of pillar j if all these conditions are met:
1. there is no other pillar between pillars i and j. Formally, it means that |i - j| = 1;
2. pillar i contains exactly one disk;
3. either pillar j contains no disks, or the topmost disk on pillar j has radius strictly greater than the radius of the disk you move.
When you place a disk on a pillar that already has some disks on it, you put the new disk on top of previously placed disks, so the new disk will be used to check the third condition if you try to place another disk on the same pillar.
You may take any disk and place it on other pillar any number of times, provided that every time you do it, all three aforementioned conditions are met. Now you wonder, is it possible to place all n disks on the same pillar simultaneously?
Input
The first line contains one integer n (3 β€ n β€ 2 β
10^5) β the number of pillars.
The second line contains n integers a_1, a_2, ..., a_i (1 β€ a_i β€ n), where a_i is the radius of the disk initially placed on the i-th pillar. All numbers a_i are distinct.
Output
Print YES if it is possible to place all the disks on the same pillar simultaneously, and NO otherwise. You may print each letter in any case (YES, yes, Yes will all be recognized as positive answer, NO, no and nO will all be recognized as negative answer).
Examples
Input
4
1 3 4 2
Output
YES
Input
3
3 1 2
Output
NO
Note
In the first case it is possible to place all disks on pillar 3 using the following sequence of actions:
1. take the disk with radius 3 from pillar 2 and place it on top of pillar 3;
2. take the disk with radius 1 from pillar 1 and place it on top of pillar 2;
3. take the disk with radius 2 from pillar 4 and place it on top of pillar 3;
4. take the disk with radius 1 from pillar 2 and place it on top of pillar 3. | instruction | 0 | 30,132 | 8 | 60,264 |
Tags: greedy, implementation
Correct Solution:
```
pillars = int(input())
values = [int(x) for x in input().split()]
coorMax = values.index(max(values))
left = values[:coorMax]
right = values[coorMax+1:]
leftSize = len(left)
rightSize = len(right)
first = True
if leftSize > 1:
first = False
for i in range(1,leftSize):
if left[i] < left[i-1]:
break
if i == leftSize -1:
first = True
second = True
if rightSize > 1:
second = False
for i in range(0,rightSize-1):
if right[i] < right[i+1]:
break
if i == rightSize -2:
second = True
if first and second:
print('YES')
else:
print('NO')
``` | output | 1 | 30,132 | 8 | 60,265 |
Provide tags and a correct Python 3 solution for this coding contest problem.
There are n pillars aligned in a row and numbered from 1 to n.
Initially each pillar contains exactly one disk. The i-th pillar contains a disk having radius a_i.
You can move these disks from one pillar to another. You can take a disk from pillar i and place it on top of pillar j if all these conditions are met:
1. there is no other pillar between pillars i and j. Formally, it means that |i - j| = 1;
2. pillar i contains exactly one disk;
3. either pillar j contains no disks, or the topmost disk on pillar j has radius strictly greater than the radius of the disk you move.
When you place a disk on a pillar that already has some disks on it, you put the new disk on top of previously placed disks, so the new disk will be used to check the third condition if you try to place another disk on the same pillar.
You may take any disk and place it on other pillar any number of times, provided that every time you do it, all three aforementioned conditions are met. Now you wonder, is it possible to place all n disks on the same pillar simultaneously?
Input
The first line contains one integer n (3 β€ n β€ 2 β
10^5) β the number of pillars.
The second line contains n integers a_1, a_2, ..., a_i (1 β€ a_i β€ n), where a_i is the radius of the disk initially placed on the i-th pillar. All numbers a_i are distinct.
Output
Print YES if it is possible to place all the disks on the same pillar simultaneously, and NO otherwise. You may print each letter in any case (YES, yes, Yes will all be recognized as positive answer, NO, no and nO will all be recognized as negative answer).
Examples
Input
4
1 3 4 2
Output
YES
Input
3
3 1 2
Output
NO
Note
In the first case it is possible to place all disks on pillar 3 using the following sequence of actions:
1. take the disk with radius 3 from pillar 2 and place it on top of pillar 3;
2. take the disk with radius 1 from pillar 1 and place it on top of pillar 2;
3. take the disk with radius 2 from pillar 4 and place it on top of pillar 3;
4. take the disk with radius 1 from pillar 2 and place it on top of pillar 3. | instruction | 0 | 30,133 | 8 | 60,266 |
Tags: greedy, implementation
Correct Solution:
```
'''input
3
2 1 3
'''
n = int(input())
arr = list(map(int,input().split()))
m = max(arr)
bol = True
num = 0
for i in range(n):
if arr[i]==m:
num = i
for i in range(num,n-1):
if arr[i]<arr[i+1]:
bol = False
break
for i in range(1,num):
if (arr[i-1]>arr[i]):
bol = False
break
if (bol):
print("YES")
else:
print("NO")
``` | output | 1 | 30,133 | 8 | 60,267 |
Provide tags and a correct Python 3 solution for this coding contest problem.
There are n pillars aligned in a row and numbered from 1 to n.
Initially each pillar contains exactly one disk. The i-th pillar contains a disk having radius a_i.
You can move these disks from one pillar to another. You can take a disk from pillar i and place it on top of pillar j if all these conditions are met:
1. there is no other pillar between pillars i and j. Formally, it means that |i - j| = 1;
2. pillar i contains exactly one disk;
3. either pillar j contains no disks, or the topmost disk on pillar j has radius strictly greater than the radius of the disk you move.
When you place a disk on a pillar that already has some disks on it, you put the new disk on top of previously placed disks, so the new disk will be used to check the third condition if you try to place another disk on the same pillar.
You may take any disk and place it on other pillar any number of times, provided that every time you do it, all three aforementioned conditions are met. Now you wonder, is it possible to place all n disks on the same pillar simultaneously?
Input
The first line contains one integer n (3 β€ n β€ 2 β
10^5) β the number of pillars.
The second line contains n integers a_1, a_2, ..., a_i (1 β€ a_i β€ n), where a_i is the radius of the disk initially placed on the i-th pillar. All numbers a_i are distinct.
Output
Print YES if it is possible to place all the disks on the same pillar simultaneously, and NO otherwise. You may print each letter in any case (YES, yes, Yes will all be recognized as positive answer, NO, no and nO will all be recognized as negative answer).
Examples
Input
4
1 3 4 2
Output
YES
Input
3
3 1 2
Output
NO
Note
In the first case it is possible to place all disks on pillar 3 using the following sequence of actions:
1. take the disk with radius 3 from pillar 2 and place it on top of pillar 3;
2. take the disk with radius 1 from pillar 1 and place it on top of pillar 2;
3. take the disk with radius 2 from pillar 4 and place it on top of pillar 3;
4. take the disk with radius 1 from pillar 2 and place it on top of pillar 3. | instruction | 0 | 30,134 | 8 | 60,268 |
Tags: greedy, implementation
Correct Solution:
```
n = int(input())
a = [int(x) for x in input().split()]
ind = max(range(n), key=lambda x:a[x])
fail = False
mn = a[ind]
for i in range(ind, n):
if a[i] > mn:
fail = True
break
mn = min(mn, a[i])
if not fail:
mn = a[ind]
for i in range(ind, -1, -1):
if a[i] > mn:
fail = True
break
mn = min(mn, a[i])
if fail:
print("NO")
else:
print("YES")
``` | output | 1 | 30,134 | 8 | 60,269 |
Provide tags and a correct Python 3 solution for this coding contest problem.
There are n pillars aligned in a row and numbered from 1 to n.
Initially each pillar contains exactly one disk. The i-th pillar contains a disk having radius a_i.
You can move these disks from one pillar to another. You can take a disk from pillar i and place it on top of pillar j if all these conditions are met:
1. there is no other pillar between pillars i and j. Formally, it means that |i - j| = 1;
2. pillar i contains exactly one disk;
3. either pillar j contains no disks, or the topmost disk on pillar j has radius strictly greater than the radius of the disk you move.
When you place a disk on a pillar that already has some disks on it, you put the new disk on top of previously placed disks, so the new disk will be used to check the third condition if you try to place another disk on the same pillar.
You may take any disk and place it on other pillar any number of times, provided that every time you do it, all three aforementioned conditions are met. Now you wonder, is it possible to place all n disks on the same pillar simultaneously?
Input
The first line contains one integer n (3 β€ n β€ 2 β
10^5) β the number of pillars.
The second line contains n integers a_1, a_2, ..., a_i (1 β€ a_i β€ n), where a_i is the radius of the disk initially placed on the i-th pillar. All numbers a_i are distinct.
Output
Print YES if it is possible to place all the disks on the same pillar simultaneously, and NO otherwise. You may print each letter in any case (YES, yes, Yes will all be recognized as positive answer, NO, no and nO will all be recognized as negative answer).
Examples
Input
4
1 3 4 2
Output
YES
Input
3
3 1 2
Output
NO
Note
In the first case it is possible to place all disks on pillar 3 using the following sequence of actions:
1. take the disk with radius 3 from pillar 2 and place it on top of pillar 3;
2. take the disk with radius 1 from pillar 1 and place it on top of pillar 2;
3. take the disk with radius 2 from pillar 4 and place it on top of pillar 3;
4. take the disk with radius 1 from pillar 2 and place it on top of pillar 3. | instruction | 0 | 30,135 | 8 | 60,270 |
Tags: greedy, implementation
Correct Solution:
```
n = int(input())
a= list(map(int, input().split(" ")))
idx = 0
for i in range(n):
if a[i] > a[idx]:
idx = i
for i in range(idx, n-1):
if a[i] < a[i+1]:
print("NO")
exit()
for i in range(idx-1):
if a[i] > a[i+1]:
print("NO")
exit()
print("YES")
``` | output | 1 | 30,135 | 8 | 60,271 |
Provide tags and a correct Python 3 solution for this coding contest problem.
There are n pillars aligned in a row and numbered from 1 to n.
Initially each pillar contains exactly one disk. The i-th pillar contains a disk having radius a_i.
You can move these disks from one pillar to another. You can take a disk from pillar i and place it on top of pillar j if all these conditions are met:
1. there is no other pillar between pillars i and j. Formally, it means that |i - j| = 1;
2. pillar i contains exactly one disk;
3. either pillar j contains no disks, or the topmost disk on pillar j has radius strictly greater than the radius of the disk you move.
When you place a disk on a pillar that already has some disks on it, you put the new disk on top of previously placed disks, so the new disk will be used to check the third condition if you try to place another disk on the same pillar.
You may take any disk and place it on other pillar any number of times, provided that every time you do it, all three aforementioned conditions are met. Now you wonder, is it possible to place all n disks on the same pillar simultaneously?
Input
The first line contains one integer n (3 β€ n β€ 2 β
10^5) β the number of pillars.
The second line contains n integers a_1, a_2, ..., a_i (1 β€ a_i β€ n), where a_i is the radius of the disk initially placed on the i-th pillar. All numbers a_i are distinct.
Output
Print YES if it is possible to place all the disks on the same pillar simultaneously, and NO otherwise. You may print each letter in any case (YES, yes, Yes will all be recognized as positive answer, NO, no and nO will all be recognized as negative answer).
Examples
Input
4
1 3 4 2
Output
YES
Input
3
3 1 2
Output
NO
Note
In the first case it is possible to place all disks on pillar 3 using the following sequence of actions:
1. take the disk with radius 3 from pillar 2 and place it on top of pillar 3;
2. take the disk with radius 1 from pillar 1 and place it on top of pillar 2;
3. take the disk with radius 2 from pillar 4 and place it on top of pillar 3;
4. take the disk with radius 1 from pillar 2 and place it on top of pillar 3. | instruction | 0 | 30,136 | 8 | 60,272 |
Tags: greedy, implementation
Correct Solution:
```
n = int(input())
lis = [int(x) for x in input().split()]
maxIndex = lis.index(n)
res = True
for i in range(1, maxIndex):
if(lis[i] < lis[i - 1]):
res = False
break
for i in range(maxIndex, n - 1):
if(lis[i] < lis[i + 1]):
res = False
break
print("YES" if res else "NO")
``` | output | 1 | 30,136 | 8 | 60,273 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There are n pillars aligned in a row and numbered from 1 to n.
Initially each pillar contains exactly one disk. The i-th pillar contains a disk having radius a_i.
You can move these disks from one pillar to another. You can take a disk from pillar i and place it on top of pillar j if all these conditions are met:
1. there is no other pillar between pillars i and j. Formally, it means that |i - j| = 1;
2. pillar i contains exactly one disk;
3. either pillar j contains no disks, or the topmost disk on pillar j has radius strictly greater than the radius of the disk you move.
When you place a disk on a pillar that already has some disks on it, you put the new disk on top of previously placed disks, so the new disk will be used to check the third condition if you try to place another disk on the same pillar.
You may take any disk and place it on other pillar any number of times, provided that every time you do it, all three aforementioned conditions are met. Now you wonder, is it possible to place all n disks on the same pillar simultaneously?
Input
The first line contains one integer n (3 β€ n β€ 2 β
10^5) β the number of pillars.
The second line contains n integers a_1, a_2, ..., a_i (1 β€ a_i β€ n), where a_i is the radius of the disk initially placed on the i-th pillar. All numbers a_i are distinct.
Output
Print YES if it is possible to place all the disks on the same pillar simultaneously, and NO otherwise. You may print each letter in any case (YES, yes, Yes will all be recognized as positive answer, NO, no and nO will all be recognized as negative answer).
Examples
Input
4
1 3 4 2
Output
YES
Input
3
3 1 2
Output
NO
Note
In the first case it is possible to place all disks on pillar 3 using the following sequence of actions:
1. take the disk with radius 3 from pillar 2 and place it on top of pillar 3;
2. take the disk with radius 1 from pillar 1 and place it on top of pillar 2;
3. take the disk with radius 2 from pillar 4 and place it on top of pillar 3;
4. take the disk with radius 1 from pillar 2 and place it on top of pillar 3.
Submitted Solution:
```
nb_pillars = int(input())
wheels = [int(s) for s in input().split()]
maxi = wheels.index(max(wheels))
could_be = True
for i in range(maxi):
could_be &= wheels[i] < wheels[i + 1]
for i in range(maxi, nb_pillars - 1):
could_be &= wheels[i] > wheels[i + 1]
print(["NO", "YES"][could_be])
``` | instruction | 0 | 30,137 | 8 | 60,274 |
Yes | output | 1 | 30,137 | 8 | 60,275 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There are n pillars aligned in a row and numbered from 1 to n.
Initially each pillar contains exactly one disk. The i-th pillar contains a disk having radius a_i.
You can move these disks from one pillar to another. You can take a disk from pillar i and place it on top of pillar j if all these conditions are met:
1. there is no other pillar between pillars i and j. Formally, it means that |i - j| = 1;
2. pillar i contains exactly one disk;
3. either pillar j contains no disks, or the topmost disk on pillar j has radius strictly greater than the radius of the disk you move.
When you place a disk on a pillar that already has some disks on it, you put the new disk on top of previously placed disks, so the new disk will be used to check the third condition if you try to place another disk on the same pillar.
You may take any disk and place it on other pillar any number of times, provided that every time you do it, all three aforementioned conditions are met. Now you wonder, is it possible to place all n disks on the same pillar simultaneously?
Input
The first line contains one integer n (3 β€ n β€ 2 β
10^5) β the number of pillars.
The second line contains n integers a_1, a_2, ..., a_i (1 β€ a_i β€ n), where a_i is the radius of the disk initially placed on the i-th pillar. All numbers a_i are distinct.
Output
Print YES if it is possible to place all the disks on the same pillar simultaneously, and NO otherwise. You may print each letter in any case (YES, yes, Yes will all be recognized as positive answer, NO, no and nO will all be recognized as negative answer).
Examples
Input
4
1 3 4 2
Output
YES
Input
3
3 1 2
Output
NO
Note
In the first case it is possible to place all disks on pillar 3 using the following sequence of actions:
1. take the disk with radius 3 from pillar 2 and place it on top of pillar 3;
2. take the disk with radius 1 from pillar 1 and place it on top of pillar 2;
3. take the disk with radius 2 from pillar 4 and place it on top of pillar 3;
4. take the disk with radius 1 from pillar 2 and place it on top of pillar 3.
Submitted Solution:
```
n=int(input())
a=*map(int,input().split()),
i=a.index(n)
print('YNEOS'[a!=(*sorted(a[:i]),n,*sorted(a[i+1:])[::-1])::2])
``` | instruction | 0 | 30,138 | 8 | 60,276 |
Yes | output | 1 | 30,138 | 8 | 60,277 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There are n pillars aligned in a row and numbered from 1 to n.
Initially each pillar contains exactly one disk. The i-th pillar contains a disk having radius a_i.
You can move these disks from one pillar to another. You can take a disk from pillar i and place it on top of pillar j if all these conditions are met:
1. there is no other pillar between pillars i and j. Formally, it means that |i - j| = 1;
2. pillar i contains exactly one disk;
3. either pillar j contains no disks, or the topmost disk on pillar j has radius strictly greater than the radius of the disk you move.
When you place a disk on a pillar that already has some disks on it, you put the new disk on top of previously placed disks, so the new disk will be used to check the third condition if you try to place another disk on the same pillar.
You may take any disk and place it on other pillar any number of times, provided that every time you do it, all three aforementioned conditions are met. Now you wonder, is it possible to place all n disks on the same pillar simultaneously?
Input
The first line contains one integer n (3 β€ n β€ 2 β
10^5) β the number of pillars.
The second line contains n integers a_1, a_2, ..., a_i (1 β€ a_i β€ n), where a_i is the radius of the disk initially placed on the i-th pillar. All numbers a_i are distinct.
Output
Print YES if it is possible to place all the disks on the same pillar simultaneously, and NO otherwise. You may print each letter in any case (YES, yes, Yes will all be recognized as positive answer, NO, no and nO will all be recognized as negative answer).
Examples
Input
4
1 3 4 2
Output
YES
Input
3
3 1 2
Output
NO
Note
In the first case it is possible to place all disks on pillar 3 using the following sequence of actions:
1. take the disk with radius 3 from pillar 2 and place it on top of pillar 3;
2. take the disk with radius 1 from pillar 1 and place it on top of pillar 2;
3. take the disk with radius 2 from pillar 4 and place it on top of pillar 3;
4. take the disk with radius 1 from pillar 2 and place it on top of pillar 3.
Submitted Solution:
```
import sys
strInp = lambda : input().strip().split()
intInp = lambda : list(map(int,strInp()))
n = int(input())
arr = intInp()
inc = True
dec = True
failed = False
for i in range(n - 1):
if arr[i] > arr[i+1]:
inc = False
if inc == False:
if arr[i] < arr[i + 1]:
failed = True
if failed :
print('No')
else:
print('Yes')
``` | instruction | 0 | 30,139 | 8 | 60,278 |
Yes | output | 1 | 30,139 | 8 | 60,279 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There are n pillars aligned in a row and numbered from 1 to n.
Initially each pillar contains exactly one disk. The i-th pillar contains a disk having radius a_i.
You can move these disks from one pillar to another. You can take a disk from pillar i and place it on top of pillar j if all these conditions are met:
1. there is no other pillar between pillars i and j. Formally, it means that |i - j| = 1;
2. pillar i contains exactly one disk;
3. either pillar j contains no disks, or the topmost disk on pillar j has radius strictly greater than the radius of the disk you move.
When you place a disk on a pillar that already has some disks on it, you put the new disk on top of previously placed disks, so the new disk will be used to check the third condition if you try to place another disk on the same pillar.
You may take any disk and place it on other pillar any number of times, provided that every time you do it, all three aforementioned conditions are met. Now you wonder, is it possible to place all n disks on the same pillar simultaneously?
Input
The first line contains one integer n (3 β€ n β€ 2 β
10^5) β the number of pillars.
The second line contains n integers a_1, a_2, ..., a_i (1 β€ a_i β€ n), where a_i is the radius of the disk initially placed on the i-th pillar. All numbers a_i are distinct.
Output
Print YES if it is possible to place all the disks on the same pillar simultaneously, and NO otherwise. You may print each letter in any case (YES, yes, Yes will all be recognized as positive answer, NO, no and nO will all be recognized as negative answer).
Examples
Input
4
1 3 4 2
Output
YES
Input
3
3 1 2
Output
NO
Note
In the first case it is possible to place all disks on pillar 3 using the following sequence of actions:
1. take the disk with radius 3 from pillar 2 and place it on top of pillar 3;
2. take the disk with radius 1 from pillar 1 and place it on top of pillar 2;
3. take the disk with radius 2 from pillar 4 and place it on top of pillar 3;
4. take the disk with radius 1 from pillar 2 and place it on top of pillar 3.
Submitted Solution:
```
n=int(input())
s=input().split()
s=list(s)
for i in range(n):
s[i]=int(s[i])
c=0
for i in range(1,n-1):
if(s[i-1]>s[i] and s[i]<s[i+1]):
print("no")
break
else:
c+=1
if(c==n-2):
print("yes")
``` | instruction | 0 | 30,140 | 8 | 60,280 |
Yes | output | 1 | 30,140 | 8 | 60,281 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There are n pillars aligned in a row and numbered from 1 to n.
Initially each pillar contains exactly one disk. The i-th pillar contains a disk having radius a_i.
You can move these disks from one pillar to another. You can take a disk from pillar i and place it on top of pillar j if all these conditions are met:
1. there is no other pillar between pillars i and j. Formally, it means that |i - j| = 1;
2. pillar i contains exactly one disk;
3. either pillar j contains no disks, or the topmost disk on pillar j has radius strictly greater than the radius of the disk you move.
When you place a disk on a pillar that already has some disks on it, you put the new disk on top of previously placed disks, so the new disk will be used to check the third condition if you try to place another disk on the same pillar.
You may take any disk and place it on other pillar any number of times, provided that every time you do it, all three aforementioned conditions are met. Now you wonder, is it possible to place all n disks on the same pillar simultaneously?
Input
The first line contains one integer n (3 β€ n β€ 2 β
10^5) β the number of pillars.
The second line contains n integers a_1, a_2, ..., a_i (1 β€ a_i β€ n), where a_i is the radius of the disk initially placed on the i-th pillar. All numbers a_i are distinct.
Output
Print YES if it is possible to place all the disks on the same pillar simultaneously, and NO otherwise. You may print each letter in any case (YES, yes, Yes will all be recognized as positive answer, NO, no and nO will all be recognized as negative answer).
Examples
Input
4
1 3 4 2
Output
YES
Input
3
3 1 2
Output
NO
Note
In the first case it is possible to place all disks on pillar 3 using the following sequence of actions:
1. take the disk with radius 3 from pillar 2 and place it on top of pillar 3;
2. take the disk with radius 1 from pillar 1 and place it on top of pillar 2;
3. take the disk with radius 2 from pillar 4 and place it on top of pillar 3;
4. take the disk with radius 1 from pillar 2 and place it on top of pillar 3.
Submitted Solution:
```
import sys
input = sys.stdin.readline
n=int(input())
A=list(map(int,input().split()))
compression_dict={a: ind for ind, a in enumerate(sorted(set(A)))}
A=[compression_dict[a] for a in A]
from collections import deque
QUE=deque()
for a in A:
QUE.append(a)
while len(QUE)>=2 and abs(QUE[-1]-QUE[-2])==1:
x=QUE.pop()
y=QUE.pop()
QUE.append(min(x,y))
if len(QUE)==1:
print("YES")
else:
print("NO")
``` | instruction | 0 | 30,141 | 8 | 60,282 |
No | output | 1 | 30,141 | 8 | 60,283 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There are n pillars aligned in a row and numbered from 1 to n.
Initially each pillar contains exactly one disk. The i-th pillar contains a disk having radius a_i.
You can move these disks from one pillar to another. You can take a disk from pillar i and place it on top of pillar j if all these conditions are met:
1. there is no other pillar between pillars i and j. Formally, it means that |i - j| = 1;
2. pillar i contains exactly one disk;
3. either pillar j contains no disks, or the topmost disk on pillar j has radius strictly greater than the radius of the disk you move.
When you place a disk on a pillar that already has some disks on it, you put the new disk on top of previously placed disks, so the new disk will be used to check the third condition if you try to place another disk on the same pillar.
You may take any disk and place it on other pillar any number of times, provided that every time you do it, all three aforementioned conditions are met. Now you wonder, is it possible to place all n disks on the same pillar simultaneously?
Input
The first line contains one integer n (3 β€ n β€ 2 β
10^5) β the number of pillars.
The second line contains n integers a_1, a_2, ..., a_i (1 β€ a_i β€ n), where a_i is the radius of the disk initially placed on the i-th pillar. All numbers a_i are distinct.
Output
Print YES if it is possible to place all the disks on the same pillar simultaneously, and NO otherwise. You may print each letter in any case (YES, yes, Yes will all be recognized as positive answer, NO, no and nO will all be recognized as negative answer).
Examples
Input
4
1 3 4 2
Output
YES
Input
3
3 1 2
Output
NO
Note
In the first case it is possible to place all disks on pillar 3 using the following sequence of actions:
1. take the disk with radius 3 from pillar 2 and place it on top of pillar 3;
2. take the disk with radius 1 from pillar 1 and place it on top of pillar 2;
3. take the disk with radius 2 from pillar 4 and place it on top of pillar 3;
4. take the disk with radius 1 from pillar 2 and place it on top of pillar 3.
Submitted Solution:
```
n = int(input())
arr = list(map(int,input().split(' ')))
k = arr.index(max(arr))
k+=1
ok = True
for i in range(1,k-1):
ok &= arr[i-1]<arr[i]
for i in range(k+1,n-1):
ok &= arr[i+1]<arr[i]
if ok :
print('YES')
else:
print('NO')
``` | instruction | 0 | 30,142 | 8 | 60,284 |
No | output | 1 | 30,142 | 8 | 60,285 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There are n pillars aligned in a row and numbered from 1 to n.
Initially each pillar contains exactly one disk. The i-th pillar contains a disk having radius a_i.
You can move these disks from one pillar to another. You can take a disk from pillar i and place it on top of pillar j if all these conditions are met:
1. there is no other pillar between pillars i and j. Formally, it means that |i - j| = 1;
2. pillar i contains exactly one disk;
3. either pillar j contains no disks, or the topmost disk on pillar j has radius strictly greater than the radius of the disk you move.
When you place a disk on a pillar that already has some disks on it, you put the new disk on top of previously placed disks, so the new disk will be used to check the third condition if you try to place another disk on the same pillar.
You may take any disk and place it on other pillar any number of times, provided that every time you do it, all three aforementioned conditions are met. Now you wonder, is it possible to place all n disks on the same pillar simultaneously?
Input
The first line contains one integer n (3 β€ n β€ 2 β
10^5) β the number of pillars.
The second line contains n integers a_1, a_2, ..., a_i (1 β€ a_i β€ n), where a_i is the radius of the disk initially placed on the i-th pillar. All numbers a_i are distinct.
Output
Print YES if it is possible to place all the disks on the same pillar simultaneously, and NO otherwise. You may print each letter in any case (YES, yes, Yes will all be recognized as positive answer, NO, no and nO will all be recognized as negative answer).
Examples
Input
4
1 3 4 2
Output
YES
Input
3
3 1 2
Output
NO
Note
In the first case it is possible to place all disks on pillar 3 using the following sequence of actions:
1. take the disk with radius 3 from pillar 2 and place it on top of pillar 3;
2. take the disk with radius 1 from pillar 1 and place it on top of pillar 2;
3. take the disk with radius 2 from pillar 4 and place it on top of pillar 3;
4. take the disk with radius 1 from pillar 2 and place it on top of pillar 3.
Submitted Solution:
```
n=int(input())
a=list(map(int,input().split()))
j=a.index(max(a))
j+=1
i,k=j-1,j+1
a.insert(0,99999)
a.append(99999)
def valid(i,k):
return i>0 and k<n+1
while i>0 or k<n+1:
if valid(i,k) and a[i]>a[k] and a[i]<a[j]:
a[j]=a[i]
a[i]=0
i-=1
elif valid(i,k) and a[i]<a[k] and a[k]<a[j]:
a[j]=a[k]
a[k]=0
k+=1
elif i==0:
a[j]=a[k]
a[k]=0
k+=1
elif k==n+1:
a[j]=a[i]
a[i]=0
i-=1
else:
print("NO")
break
#print(a,i,j,k)
else:
print("YES")
``` | instruction | 0 | 30,143 | 8 | 60,286 |
No | output | 1 | 30,143 | 8 | 60,287 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There are n pillars aligned in a row and numbered from 1 to n.
Initially each pillar contains exactly one disk. The i-th pillar contains a disk having radius a_i.
You can move these disks from one pillar to another. You can take a disk from pillar i and place it on top of pillar j if all these conditions are met:
1. there is no other pillar between pillars i and j. Formally, it means that |i - j| = 1;
2. pillar i contains exactly one disk;
3. either pillar j contains no disks, or the topmost disk on pillar j has radius strictly greater than the radius of the disk you move.
When you place a disk on a pillar that already has some disks on it, you put the new disk on top of previously placed disks, so the new disk will be used to check the third condition if you try to place another disk on the same pillar.
You may take any disk and place it on other pillar any number of times, provided that every time you do it, all three aforementioned conditions are met. Now you wonder, is it possible to place all n disks on the same pillar simultaneously?
Input
The first line contains one integer n (3 β€ n β€ 2 β
10^5) β the number of pillars.
The second line contains n integers a_1, a_2, ..., a_i (1 β€ a_i β€ n), where a_i is the radius of the disk initially placed on the i-th pillar. All numbers a_i are distinct.
Output
Print YES if it is possible to place all the disks on the same pillar simultaneously, and NO otherwise. You may print each letter in any case (YES, yes, Yes will all be recognized as positive answer, NO, no and nO will all be recognized as negative answer).
Examples
Input
4
1 3 4 2
Output
YES
Input
3
3 1 2
Output
NO
Note
In the first case it is possible to place all disks on pillar 3 using the following sequence of actions:
1. take the disk with radius 3 from pillar 2 and place it on top of pillar 3;
2. take the disk with radius 1 from pillar 1 and place it on top of pillar 2;
3. take the disk with radius 2 from pillar 4 and place it on top of pillar 3;
4. take the disk with radius 1 from pillar 2 and place it on top of pillar 3.
Submitted Solution:
```
z=int(input())
l=list(map(int,input().split()))
l1=[0]+l+[0]
c=1
x=l.index(max(l))+1
for i in range(x,z):
if l1[i+1]>l1[i]:
print('NO')
c=0
break
for i in range(x,0,-1):
if l1[i-1]>l1[i]:
print('NO')
c=0
break
if c==1:
print('YEs')
``` | instruction | 0 | 30,144 | 8 | 60,288 |
No | output | 1 | 30,144 | 8 | 60,289 |
Provide tags and a correct Python 3 solution for this coding contest problem.
In an attempt to escape the Mischievous Mess Makers' antics, Farmer John has abandoned his farm and is traveling to the other side of Bovinia. During the journey, he and his k cows have decided to stay at the luxurious Grand Moo-dapest Hotel. The hotel consists of n rooms located in a row, some of which are occupied.
Farmer John wants to book a set of k + 1 currently unoccupied rooms for him and his cows. He wants his cows to stay as safe as possible, so he wishes to minimize the maximum distance from his room to the room of his cow. The distance between rooms i and j is defined as |j - i|. Help Farmer John protect his cows by calculating this minimum possible distance.
Input
The first line of the input contains two integers n and k (1 β€ k < n β€ 100 000) β the number of rooms in the hotel and the number of cows travelling with Farmer John.
The second line contains a string of length n describing the rooms. The i-th character of the string will be '0' if the i-th room is free, and '1' if the i-th room is occupied. It is guaranteed that at least k + 1 characters of this string are '0', so there exists at least one possible choice of k + 1 rooms for Farmer John and his cows to stay in.
Output
Print the minimum possible distance between Farmer John's room and his farthest cow.
Examples
Input
7 2
0100100
Output
2
Input
5 1
01010
Output
2
Input
3 2
000
Output
1
Note
In the first sample, Farmer John can book room 3 for himself, and rooms 1 and 4 for his cows. The distance to the farthest cow is 2. Note that it is impossible to make this distance 1, as there is no block of three consecutive unoccupied rooms.
In the second sample, Farmer John can book room 1 for himself and room 3 for his single cow. The distance between him and his cow is 2.
In the third sample, Farmer John books all three available rooms, taking the middle room for himself so that both cows are next to him. His distance from the farthest cow is 1. | instruction | 0 | 30,513 | 8 | 61,026 |
Tags: binary search, two pointers
Correct Solution:
```
n, k = map(int, input().split())
rooms = input()
count = [0] * (n + 1)
for i in range(n):
count[i] = count[i - 1]
if rooms[i] == '0':
count[i] += 1
ans = float('inf')
for i in range(n):
l = 0
r = n
while l + 1 < r:
m = (r + l) // 2
if rooms[i] == '0' and (count[min(i + m, n - 1)] - count[max(-1, i - m - 1)]) > k:
r = m
else:
l = m
ans = min(r, ans)
print(ans)
``` | output | 1 | 30,513 | 8 | 61,027 |
Provide tags and a correct Python 3 solution for this coding contest problem.
In an attempt to escape the Mischievous Mess Makers' antics, Farmer John has abandoned his farm and is traveling to the other side of Bovinia. During the journey, he and his k cows have decided to stay at the luxurious Grand Moo-dapest Hotel. The hotel consists of n rooms located in a row, some of which are occupied.
Farmer John wants to book a set of k + 1 currently unoccupied rooms for him and his cows. He wants his cows to stay as safe as possible, so he wishes to minimize the maximum distance from his room to the room of his cow. The distance between rooms i and j is defined as |j - i|. Help Farmer John protect his cows by calculating this minimum possible distance.
Input
The first line of the input contains two integers n and k (1 β€ k < n β€ 100 000) β the number of rooms in the hotel and the number of cows travelling with Farmer John.
The second line contains a string of length n describing the rooms. The i-th character of the string will be '0' if the i-th room is free, and '1' if the i-th room is occupied. It is guaranteed that at least k + 1 characters of this string are '0', so there exists at least one possible choice of k + 1 rooms for Farmer John and his cows to stay in.
Output
Print the minimum possible distance between Farmer John's room and his farthest cow.
Examples
Input
7 2
0100100
Output
2
Input
5 1
01010
Output
2
Input
3 2
000
Output
1
Note
In the first sample, Farmer John can book room 3 for himself, and rooms 1 and 4 for his cows. The distance to the farthest cow is 2. Note that it is impossible to make this distance 1, as there is no block of three consecutive unoccupied rooms.
In the second sample, Farmer John can book room 1 for himself and room 3 for his single cow. The distance between him and his cow is 2.
In the third sample, Farmer John books all three available rooms, taking the middle room for himself so that both cows are next to him. His distance from the farthest cow is 1. | instruction | 0 | 30,514 | 8 | 61,028 |
Tags: binary search, two pointers
Correct Solution:
```
[n, k] = [int(x) for x in input().split()]
room_free = [bool(1-int(x)) for x in input()]
# maintain queue closest before
# maintain queue of closest after
# maintain two pointers (l, r)
# if l == 0 check
closest_before = []
closest_after = []
# note, he needs k+1 rooms
before = -1
for i in range(n):
if room_free[i]:
before = i
closest_before.append(before)
after = 11**8
for i in range(n-1, -1, -1):
if room_free[i]:
after = i
closest_after.append(after)
closest_after.reverse()
l = -1
r = -1
taken = 0
ans = 10**8
while True:
if taken != k+1:
r += 1
if r >= n:
break
if room_free[r]:
taken += 1
else:
l += 1
if room_free[l]:
mid = int((l+r)/2)
# print(l, r, mid, closest_after[mid])
ans = min(ans, closest_after[mid+1] - l, r - closest_before[mid])
taken -= 1
print(ans)
``` | output | 1 | 30,514 | 8 | 61,029 |
Provide tags and a correct Python 3 solution for this coding contest problem.
In an attempt to escape the Mischievous Mess Makers' antics, Farmer John has abandoned his farm and is traveling to the other side of Bovinia. During the journey, he and his k cows have decided to stay at the luxurious Grand Moo-dapest Hotel. The hotel consists of n rooms located in a row, some of which are occupied.
Farmer John wants to book a set of k + 1 currently unoccupied rooms for him and his cows. He wants his cows to stay as safe as possible, so he wishes to minimize the maximum distance from his room to the room of his cow. The distance between rooms i and j is defined as |j - i|. Help Farmer John protect his cows by calculating this minimum possible distance.
Input
The first line of the input contains two integers n and k (1 β€ k < n β€ 100 000) β the number of rooms in the hotel and the number of cows travelling with Farmer John.
The second line contains a string of length n describing the rooms. The i-th character of the string will be '0' if the i-th room is free, and '1' if the i-th room is occupied. It is guaranteed that at least k + 1 characters of this string are '0', so there exists at least one possible choice of k + 1 rooms for Farmer John and his cows to stay in.
Output
Print the minimum possible distance between Farmer John's room and his farthest cow.
Examples
Input
7 2
0100100
Output
2
Input
5 1
01010
Output
2
Input
3 2
000
Output
1
Note
In the first sample, Farmer John can book room 3 for himself, and rooms 1 and 4 for his cows. The distance to the farthest cow is 2. Note that it is impossible to make this distance 1, as there is no block of three consecutive unoccupied rooms.
In the second sample, Farmer John can book room 1 for himself and room 3 for his single cow. The distance between him and his cow is 2.
In the third sample, Farmer John books all three available rooms, taking the middle room for himself so that both cows are next to him. His distance from the farthest cow is 1. | instruction | 0 | 30,515 | 8 | 61,030 |
Tags: binary search, two pointers
Correct Solution:
```
import sys
r=sys.stdin.readline
n,k=map(int,r().split())
rooms=list(r().strip())
def nextRoom(i):
i+=1
while i<n and rooms[i]=='1':
i+=1
return i
l=nextRoom(-1)
r,m=l,l
ans=sys.maxsize
for i in range(k):
r=nextRoom(r)
while r<n:
while max(m-l,r-m)>max(nextRoom(m)-l,r-nextRoom(m)):
m=nextRoom(m)
ans=min(ans,max(m-l,r-m))
l,r=nextRoom(l),nextRoom(r)
print(ans)
``` | output | 1 | 30,515 | 8 | 61,031 |
Provide tags and a correct Python 3 solution for this coding contest problem.
In an attempt to escape the Mischievous Mess Makers' antics, Farmer John has abandoned his farm and is traveling to the other side of Bovinia. During the journey, he and his k cows have decided to stay at the luxurious Grand Moo-dapest Hotel. The hotel consists of n rooms located in a row, some of which are occupied.
Farmer John wants to book a set of k + 1 currently unoccupied rooms for him and his cows. He wants his cows to stay as safe as possible, so he wishes to minimize the maximum distance from his room to the room of his cow. The distance between rooms i and j is defined as |j - i|. Help Farmer John protect his cows by calculating this minimum possible distance.
Input
The first line of the input contains two integers n and k (1 β€ k < n β€ 100 000) β the number of rooms in the hotel and the number of cows travelling with Farmer John.
The second line contains a string of length n describing the rooms. The i-th character of the string will be '0' if the i-th room is free, and '1' if the i-th room is occupied. It is guaranteed that at least k + 1 characters of this string are '0', so there exists at least one possible choice of k + 1 rooms for Farmer John and his cows to stay in.
Output
Print the minimum possible distance between Farmer John's room and his farthest cow.
Examples
Input
7 2
0100100
Output
2
Input
5 1
01010
Output
2
Input
3 2
000
Output
1
Note
In the first sample, Farmer John can book room 3 for himself, and rooms 1 and 4 for his cows. The distance to the farthest cow is 2. Note that it is impossible to make this distance 1, as there is no block of three consecutive unoccupied rooms.
In the second sample, Farmer John can book room 1 for himself and room 3 for his single cow. The distance between him and his cow is 2.
In the third sample, Farmer John books all three available rooms, taking the middle room for himself so that both cows are next to him. His distance from the farthest cow is 1. | instruction | 0 | 30,516 | 8 | 61,032 |
Tags: binary search, two pointers
Correct Solution:
```
n, k = map(int, input().split())
t = [i for i, v in enumerate(input()) if v == '0']
s, m = n, 0
f = lambda m: max(r - t[m], t[m] - l)
for l, r in zip(t, t[k:]):
while f(m) > f(m + 1): m += 1
s = min(s, f(m))
print(s)
# Made By Mostafa_Khaled
``` | output | 1 | 30,516 | 8 | 61,033 |
Provide tags and a correct Python 3 solution for this coding contest problem.
In an attempt to escape the Mischievous Mess Makers' antics, Farmer John has abandoned his farm and is traveling to the other side of Bovinia. During the journey, he and his k cows have decided to stay at the luxurious Grand Moo-dapest Hotel. The hotel consists of n rooms located in a row, some of which are occupied.
Farmer John wants to book a set of k + 1 currently unoccupied rooms for him and his cows. He wants his cows to stay as safe as possible, so he wishes to minimize the maximum distance from his room to the room of his cow. The distance between rooms i and j is defined as |j - i|. Help Farmer John protect his cows by calculating this minimum possible distance.
Input
The first line of the input contains two integers n and k (1 β€ k < n β€ 100 000) β the number of rooms in the hotel and the number of cows travelling with Farmer John.
The second line contains a string of length n describing the rooms. The i-th character of the string will be '0' if the i-th room is free, and '1' if the i-th room is occupied. It is guaranteed that at least k + 1 characters of this string are '0', so there exists at least one possible choice of k + 1 rooms for Farmer John and his cows to stay in.
Output
Print the minimum possible distance between Farmer John's room and his farthest cow.
Examples
Input
7 2
0100100
Output
2
Input
5 1
01010
Output
2
Input
3 2
000
Output
1
Note
In the first sample, Farmer John can book room 3 for himself, and rooms 1 and 4 for his cows. The distance to the farthest cow is 2. Note that it is impossible to make this distance 1, as there is no block of three consecutive unoccupied rooms.
In the second sample, Farmer John can book room 1 for himself and room 3 for his single cow. The distance between him and his cow is 2.
In the third sample, Farmer John books all three available rooms, taking the middle room for himself so that both cows are next to him. His distance from the farthest cow is 1. | instruction | 0 | 30,517 | 8 | 61,034 |
Tags: binary search, two pointers
Correct Solution:
```
import sys
input = sys.stdin.readline
def judge(i, x):
return acc[min(n, i+x+1)]-acc[max(0, i-x)]>=k+1
def binary_search(i):
l, r = 0, n
while l<=r:
mid = (l+r)//2
if judge(i, mid):
r = mid-1
else:
l = mid+1
return l
n, k = map(int, input().split())
S = input()[:-1]
acc = [0]
for Si in S:
acc.append(acc[-1]+(1 if Si=='0' else 0))
ans = n
for i in range(n):
if S[i]=='0':
ans = min(ans, binary_search(i))
print(ans)
``` | output | 1 | 30,517 | 8 | 61,035 |
Provide tags and a correct Python 3 solution for this coding contest problem.
In an attempt to escape the Mischievous Mess Makers' antics, Farmer John has abandoned his farm and is traveling to the other side of Bovinia. During the journey, he and his k cows have decided to stay at the luxurious Grand Moo-dapest Hotel. The hotel consists of n rooms located in a row, some of which are occupied.
Farmer John wants to book a set of k + 1 currently unoccupied rooms for him and his cows. He wants his cows to stay as safe as possible, so he wishes to minimize the maximum distance from his room to the room of his cow. The distance between rooms i and j is defined as |j - i|. Help Farmer John protect his cows by calculating this minimum possible distance.
Input
The first line of the input contains two integers n and k (1 β€ k < n β€ 100 000) β the number of rooms in the hotel and the number of cows travelling with Farmer John.
The second line contains a string of length n describing the rooms. The i-th character of the string will be '0' if the i-th room is free, and '1' if the i-th room is occupied. It is guaranteed that at least k + 1 characters of this string are '0', so there exists at least one possible choice of k + 1 rooms for Farmer John and his cows to stay in.
Output
Print the minimum possible distance between Farmer John's room and his farthest cow.
Examples
Input
7 2
0100100
Output
2
Input
5 1
01010
Output
2
Input
3 2
000
Output
1
Note
In the first sample, Farmer John can book room 3 for himself, and rooms 1 and 4 for his cows. The distance to the farthest cow is 2. Note that it is impossible to make this distance 1, as there is no block of three consecutive unoccupied rooms.
In the second sample, Farmer John can book room 1 for himself and room 3 for his single cow. The distance between him and his cow is 2.
In the third sample, Farmer John books all three available rooms, taking the middle room for himself so that both cows are next to him. His distance from the farthest cow is 1. | instruction | 0 | 30,518 | 8 | 61,036 |
Tags: binary search, two pointers
Correct Solution:
```
res = 1e9
def next(i):
i += 1
while(i < N and S[i] == '1'):
i += 1
return i
N, K = map(int, input().split())
S = input()
l = next(-1)
m = l
r = l
for j in range(K):
r = next(r)
while(r < N):
while(max(m - l, r - m) > max(next(m) - l, r - next(m))):
m = next(m)
res = min(res, max(m - l, r - m))
l = next(l)
r = next(r)
print(res)
``` | output | 1 | 30,518 | 8 | 61,037 |
Provide tags and a correct Python 3 solution for this coding contest problem.
In an attempt to escape the Mischievous Mess Makers' antics, Farmer John has abandoned his farm and is traveling to the other side of Bovinia. During the journey, he and his k cows have decided to stay at the luxurious Grand Moo-dapest Hotel. The hotel consists of n rooms located in a row, some of which are occupied.
Farmer John wants to book a set of k + 1 currently unoccupied rooms for him and his cows. He wants his cows to stay as safe as possible, so he wishes to minimize the maximum distance from his room to the room of his cow. The distance between rooms i and j is defined as |j - i|. Help Farmer John protect his cows by calculating this minimum possible distance.
Input
The first line of the input contains two integers n and k (1 β€ k < n β€ 100 000) β the number of rooms in the hotel and the number of cows travelling with Farmer John.
The second line contains a string of length n describing the rooms. The i-th character of the string will be '0' if the i-th room is free, and '1' if the i-th room is occupied. It is guaranteed that at least k + 1 characters of this string are '0', so there exists at least one possible choice of k + 1 rooms for Farmer John and his cows to stay in.
Output
Print the minimum possible distance between Farmer John's room and his farthest cow.
Examples
Input
7 2
0100100
Output
2
Input
5 1
01010
Output
2
Input
3 2
000
Output
1
Note
In the first sample, Farmer John can book room 3 for himself, and rooms 1 and 4 for his cows. The distance to the farthest cow is 2. Note that it is impossible to make this distance 1, as there is no block of three consecutive unoccupied rooms.
In the second sample, Farmer John can book room 1 for himself and room 3 for his single cow. The distance between him and his cow is 2.
In the third sample, Farmer John books all three available rooms, taking the middle room for himself so that both cows are next to him. His distance from the farthest cow is 1. | instruction | 0 | 30,519 | 8 | 61,038 |
Tags: binary search, two pointers
Correct Solution:
```
import sys
def can_book(mid, n, k, accum_arr):
for i in range(1, n + 1):
if (accum_arr[i] == accum_arr[i-1]):
continue
max_idx = min(i + mid, n)
min_idx = max(i - mid - 1, 0)
if (accum_arr[max_idx] - accum_arr[min_idx] >= k + 1):
return True
return False
def solve(n, k, arr):
accum_arr = [0]
for i in range(1,n+1):
accum_arr.append(accum_arr[i-1] + (1 - arr[i-1]))
l = 0
h = len(arr)
while l + 1 < h:
mid = (l + h) // 2
if can_book(mid, n, k, accum_arr):
h = mid
else:
l = mid
return h
n, k = input().split()
n = int(n)
k = int(k)
arr = list(map(int, sys.stdin.readline().rstrip()))
print(solve(n,k,arr))
``` | output | 1 | 30,519 | 8 | 61,039 |
Provide tags and a correct Python 3 solution for this coding contest problem.
In an attempt to escape the Mischievous Mess Makers' antics, Farmer John has abandoned his farm and is traveling to the other side of Bovinia. During the journey, he and his k cows have decided to stay at the luxurious Grand Moo-dapest Hotel. The hotel consists of n rooms located in a row, some of which are occupied.
Farmer John wants to book a set of k + 1 currently unoccupied rooms for him and his cows. He wants his cows to stay as safe as possible, so he wishes to minimize the maximum distance from his room to the room of his cow. The distance between rooms i and j is defined as |j - i|. Help Farmer John protect his cows by calculating this minimum possible distance.
Input
The first line of the input contains two integers n and k (1 β€ k < n β€ 100 000) β the number of rooms in the hotel and the number of cows travelling with Farmer John.
The second line contains a string of length n describing the rooms. The i-th character of the string will be '0' if the i-th room is free, and '1' if the i-th room is occupied. It is guaranteed that at least k + 1 characters of this string are '0', so there exists at least one possible choice of k + 1 rooms for Farmer John and his cows to stay in.
Output
Print the minimum possible distance between Farmer John's room and his farthest cow.
Examples
Input
7 2
0100100
Output
2
Input
5 1
01010
Output
2
Input
3 2
000
Output
1
Note
In the first sample, Farmer John can book room 3 for himself, and rooms 1 and 4 for his cows. The distance to the farthest cow is 2. Note that it is impossible to make this distance 1, as there is no block of three consecutive unoccupied rooms.
In the second sample, Farmer John can book room 1 for himself and room 3 for his single cow. The distance between him and his cow is 2.
In the third sample, Farmer John books all three available rooms, taking the middle room for himself so that both cows are next to him. His distance from the farthest cow is 1. | instruction | 0 | 30,520 | 8 | 61,040 |
Tags: binary search, two pointers
Correct Solution:
```
res = 1e9
def next(i):
i+=1
while(i < n and s[i] == '1'):
i+=1
return i
n,k = list(map(int, input().split()))
s = input()
l = next(-1)
m = l
r = l
for i in range(k):
r = next(r)
while(r < n):
while(max(m - l, r - m) > max(next(m) - l, r - next(m))):
m = next(m)
res = min(res, max(m - l, r - m))
l = next(l)
r = next(r)
print(res)
``` | output | 1 | 30,520 | 8 | 61,041 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
In an attempt to escape the Mischievous Mess Makers' antics, Farmer John has abandoned his farm and is traveling to the other side of Bovinia. During the journey, he and his k cows have decided to stay at the luxurious Grand Moo-dapest Hotel. The hotel consists of n rooms located in a row, some of which are occupied.
Farmer John wants to book a set of k + 1 currently unoccupied rooms for him and his cows. He wants his cows to stay as safe as possible, so he wishes to minimize the maximum distance from his room to the room of his cow. The distance between rooms i and j is defined as |j - i|. Help Farmer John protect his cows by calculating this minimum possible distance.
Input
The first line of the input contains two integers n and k (1 β€ k < n β€ 100 000) β the number of rooms in the hotel and the number of cows travelling with Farmer John.
The second line contains a string of length n describing the rooms. The i-th character of the string will be '0' if the i-th room is free, and '1' if the i-th room is occupied. It is guaranteed that at least k + 1 characters of this string are '0', so there exists at least one possible choice of k + 1 rooms for Farmer John and his cows to stay in.
Output
Print the minimum possible distance between Farmer John's room and his farthest cow.
Examples
Input
7 2
0100100
Output
2
Input
5 1
01010
Output
2
Input
3 2
000
Output
1
Note
In the first sample, Farmer John can book room 3 for himself, and rooms 1 and 4 for his cows. The distance to the farthest cow is 2. Note that it is impossible to make this distance 1, as there is no block of three consecutive unoccupied rooms.
In the second sample, Farmer John can book room 1 for himself and room 3 for his single cow. The distance between him and his cow is 2.
In the third sample, Farmer John books all three available rooms, taking the middle room for himself so that both cows are next to him. His distance from the farthest cow is 1.
Submitted Solution:
```
n, k = [int(_) for _ in input().split()]
a = input()
def next_free(x):
x += 1
while x < n and a[x] == '1':
x += 1
return x
l = next_free(-1)
r = -1
k += 1
for _ in range(k):
r = next_free(r)
i = l
dis = max(i-l, r-i)
next_i = next_free(i)
while True:
cur_dis = max(i-l, r-i)
next_dis = max(next_i - l, r - next_i)
if cur_dis <= next_dis:
break
i = next_i
dis = min(dis, next_dis)
next_i = next_free(i)
while True:
r = next_free(r)
if r >= n:
break
l = next_free(l)
if i < l:
i = l
prev_dis = max(i-l, r-i)
dis = min(dis, prev_dis)
m = next_free(i)
while m <= r:
cur_dis = max(m-l, r-m)
if cur_dis >= prev_dis:
break
prev_dis = cur_dis
i = m
dis = min(dis, cur_dis)
m = next_free(m)
print(dis)
# binary search the ANSWER!!
# choose an answer and check if it works
# array [i] keeps track of how many zeros are before it
``` | instruction | 0 | 30,521 | 8 | 61,042 |
Yes | output | 1 | 30,521 | 8 | 61,043 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
In an attempt to escape the Mischievous Mess Makers' antics, Farmer John has abandoned his farm and is traveling to the other side of Bovinia. During the journey, he and his k cows have decided to stay at the luxurious Grand Moo-dapest Hotel. The hotel consists of n rooms located in a row, some of which are occupied.
Farmer John wants to book a set of k + 1 currently unoccupied rooms for him and his cows. He wants his cows to stay as safe as possible, so he wishes to minimize the maximum distance from his room to the room of his cow. The distance between rooms i and j is defined as |j - i|. Help Farmer John protect his cows by calculating this minimum possible distance.
Input
The first line of the input contains two integers n and k (1 β€ k < n β€ 100 000) β the number of rooms in the hotel and the number of cows travelling with Farmer John.
The second line contains a string of length n describing the rooms. The i-th character of the string will be '0' if the i-th room is free, and '1' if the i-th room is occupied. It is guaranteed that at least k + 1 characters of this string are '0', so there exists at least one possible choice of k + 1 rooms for Farmer John and his cows to stay in.
Output
Print the minimum possible distance between Farmer John's room and his farthest cow.
Examples
Input
7 2
0100100
Output
2
Input
5 1
01010
Output
2
Input
3 2
000
Output
1
Note
In the first sample, Farmer John can book room 3 for himself, and rooms 1 and 4 for his cows. The distance to the farthest cow is 2. Note that it is impossible to make this distance 1, as there is no block of three consecutive unoccupied rooms.
In the second sample, Farmer John can book room 1 for himself and room 3 for his single cow. The distance between him and his cow is 2.
In the third sample, Farmer John books all three available rooms, taking the middle room for himself so that both cows are next to him. His distance from the farthest cow is 1.
Submitted Solution:
```
n, k = map(int, input().split())
t = [i for i, v in enumerate(input()) if v == '0']
s, i = n, 0
a, b = t[:2]
for l, r in zip(t, t[k:]):
while max(r - a, a - l) > max(r - b, b - l):
i += 1
a, b = t[i:i + 2]
s = min(s, max(r - a, a - l))
print(s)
``` | instruction | 0 | 30,522 | 8 | 61,044 |
Yes | output | 1 | 30,522 | 8 | 61,045 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
In an attempt to escape the Mischievous Mess Makers' antics, Farmer John has abandoned his farm and is traveling to the other side of Bovinia. During the journey, he and his k cows have decided to stay at the luxurious Grand Moo-dapest Hotel. The hotel consists of n rooms located in a row, some of which are occupied.
Farmer John wants to book a set of k + 1 currently unoccupied rooms for him and his cows. He wants his cows to stay as safe as possible, so he wishes to minimize the maximum distance from his room to the room of his cow. The distance between rooms i and j is defined as |j - i|. Help Farmer John protect his cows by calculating this minimum possible distance.
Input
The first line of the input contains two integers n and k (1 β€ k < n β€ 100 000) β the number of rooms in the hotel and the number of cows travelling with Farmer John.
The second line contains a string of length n describing the rooms. The i-th character of the string will be '0' if the i-th room is free, and '1' if the i-th room is occupied. It is guaranteed that at least k + 1 characters of this string are '0', so there exists at least one possible choice of k + 1 rooms for Farmer John and his cows to stay in.
Output
Print the minimum possible distance between Farmer John's room and his farthest cow.
Examples
Input
7 2
0100100
Output
2
Input
5 1
01010
Output
2
Input
3 2
000
Output
1
Note
In the first sample, Farmer John can book room 3 for himself, and rooms 1 and 4 for his cows. The distance to the farthest cow is 2. Note that it is impossible to make this distance 1, as there is no block of three consecutive unoccupied rooms.
In the second sample, Farmer John can book room 1 for himself and room 3 for his single cow. The distance between him and his cow is 2.
In the third sample, Farmer John books all three available rooms, taking the middle room for himself so that both cows are next to him. His distance from the farthest cow is 1.
Submitted Solution:
```
def main():
from bisect import bisect
n, k = map(int, input().split())
l = [i for i, c in enumerate(input()) if c == "0"]
l.append(n * 3)
for i, j in enumerate(range(k, len(l) - 1)):
lo, hi = l[i], l[j]
m = bisect(l, (lo + hi) // 2)
mid = l[m]
a, b = mid - lo, hi - mid
if a < b:
a = b
mid = l[m - 1]
b, c = mid - lo, hi - mid
if b < c:
b = c
if a > b:
a = b
if n > a:
n = a
print(n)
if __name__ == '__main__':
main()
``` | instruction | 0 | 30,523 | 8 | 61,046 |
Yes | output | 1 | 30,523 | 8 | 61,047 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
In an attempt to escape the Mischievous Mess Makers' antics, Farmer John has abandoned his farm and is traveling to the other side of Bovinia. During the journey, he and his k cows have decided to stay at the luxurious Grand Moo-dapest Hotel. The hotel consists of n rooms located in a row, some of which are occupied.
Farmer John wants to book a set of k + 1 currently unoccupied rooms for him and his cows. He wants his cows to stay as safe as possible, so he wishes to minimize the maximum distance from his room to the room of his cow. The distance between rooms i and j is defined as |j - i|. Help Farmer John protect his cows by calculating this minimum possible distance.
Input
The first line of the input contains two integers n and k (1 β€ k < n β€ 100 000) β the number of rooms in the hotel and the number of cows travelling with Farmer John.
The second line contains a string of length n describing the rooms. The i-th character of the string will be '0' if the i-th room is free, and '1' if the i-th room is occupied. It is guaranteed that at least k + 1 characters of this string are '0', so there exists at least one possible choice of k + 1 rooms for Farmer John and his cows to stay in.
Output
Print the minimum possible distance between Farmer John's room and his farthest cow.
Examples
Input
7 2
0100100
Output
2
Input
5 1
01010
Output
2
Input
3 2
000
Output
1
Note
In the first sample, Farmer John can book room 3 for himself, and rooms 1 and 4 for his cows. The distance to the farthest cow is 2. Note that it is impossible to make this distance 1, as there is no block of three consecutive unoccupied rooms.
In the second sample, Farmer John can book room 1 for himself and room 3 for his single cow. The distance between him and his cow is 2.
In the third sample, Farmer John books all three available rooms, taking the middle room for himself so that both cows are next to him. His distance from the farthest cow is 1.
Submitted Solution:
```
from collections import deque
rooms_number, cows_number = map(int, input().split())
rooms = input()
free_rooms = [i for i in range(rooms_number) if rooms[i] == '0']
def binary_search(left, right, item):
global free_rooms
while right - left > 1:
center = left + (right - left)//2
if free_rooms[center] > item:
right = center
else:
left = center
return left
min_distation_to_farthest_cow = 10**20
best_min_distation_to_farthest_cow = cows_number//2 + cows_number%2
for i in range(len(free_rooms) - cows_number):
left = free_rooms[i]
right = free_rooms[i + cows_number]
center = left + (right - left)//2
j = binary_search(i, i + cows_number, center)
if free_rooms[j] == center:
distation_to_farthest_cow = right - center
else:
distation_to_farthest_cow = min(right - free_rooms[j], free_rooms[j + 1] - left)
if min_distation_to_farthest_cow > distation_to_farthest_cow:
min_distation_to_farthest_cow = distation_to_farthest_cow
if distation_to_farthest_cow == best_min_distation_to_farthest_cow:
break
print(min_distation_to_farthest_cow)
``` | instruction | 0 | 30,524 | 8 | 61,048 |
Yes | output | 1 | 30,524 | 8 | 61,049 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
In an attempt to escape the Mischievous Mess Makers' antics, Farmer John has abandoned his farm and is traveling to the other side of Bovinia. During the journey, he and his k cows have decided to stay at the luxurious Grand Moo-dapest Hotel. The hotel consists of n rooms located in a row, some of which are occupied.
Farmer John wants to book a set of k + 1 currently unoccupied rooms for him and his cows. He wants his cows to stay as safe as possible, so he wishes to minimize the maximum distance from his room to the room of his cow. The distance between rooms i and j is defined as |j - i|. Help Farmer John protect his cows by calculating this minimum possible distance.
Input
The first line of the input contains two integers n and k (1 β€ k < n β€ 100 000) β the number of rooms in the hotel and the number of cows travelling with Farmer John.
The second line contains a string of length n describing the rooms. The i-th character of the string will be '0' if the i-th room is free, and '1' if the i-th room is occupied. It is guaranteed that at least k + 1 characters of this string are '0', so there exists at least one possible choice of k + 1 rooms for Farmer John and his cows to stay in.
Output
Print the minimum possible distance between Farmer John's room and his farthest cow.
Examples
Input
7 2
0100100
Output
2
Input
5 1
01010
Output
2
Input
3 2
000
Output
1
Note
In the first sample, Farmer John can book room 3 for himself, and rooms 1 and 4 for his cows. The distance to the farthest cow is 2. Note that it is impossible to make this distance 1, as there is no block of three consecutive unoccupied rooms.
In the second sample, Farmer John can book room 1 for himself and room 3 for his single cow. The distance between him and his cow is 2.
In the third sample, Farmer John books all three available rooms, taking the middle room for himself so that both cows are next to him. His distance from the farthest cow is 1.
Submitted Solution:
```
import sys
def can_book(mid, n, k, accum_arr):
for i in range(1, n + 1):
if (accum_arr[i] == accum_arr[i-1]):
continue
max_idx = min(i + mid, n)
min_idx = max(i - mid, 0)
if (accum_arr[max_idx] - accum_arr[min_idx] >= k + 1):
return True
return False
def solve(n, k, arr):
accum_arr = [0]
for i in range(1,n+1):
accum_arr.append(accum_arr[i-1] + (1 - arr[i-1]))
l = 0
h = len(arr)
while l + 1 < h:
mid = (l + h) // 2
if can_book(mid, n, k, accum_arr):
h = mid
else:
l = mid
return h
n, k = input().split()
n = int(n)
k = int(k)
arr = list(map(int, sys.stdin.readline().rstrip()))
print(solve(n,k,arr))
``` | instruction | 0 | 30,525 | 8 | 61,050 |
No | output | 1 | 30,525 | 8 | 61,051 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
In an attempt to escape the Mischievous Mess Makers' antics, Farmer John has abandoned his farm and is traveling to the other side of Bovinia. During the journey, he and his k cows have decided to stay at the luxurious Grand Moo-dapest Hotel. The hotel consists of n rooms located in a row, some of which are occupied.
Farmer John wants to book a set of k + 1 currently unoccupied rooms for him and his cows. He wants his cows to stay as safe as possible, so he wishes to minimize the maximum distance from his room to the room of his cow. The distance between rooms i and j is defined as |j - i|. Help Farmer John protect his cows by calculating this minimum possible distance.
Input
The first line of the input contains two integers n and k (1 β€ k < n β€ 100 000) β the number of rooms in the hotel and the number of cows travelling with Farmer John.
The second line contains a string of length n describing the rooms. The i-th character of the string will be '0' if the i-th room is free, and '1' if the i-th room is occupied. It is guaranteed that at least k + 1 characters of this string are '0', so there exists at least one possible choice of k + 1 rooms for Farmer John and his cows to stay in.
Output
Print the minimum possible distance between Farmer John's room and his farthest cow.
Examples
Input
7 2
0100100
Output
2
Input
5 1
01010
Output
2
Input
3 2
000
Output
1
Note
In the first sample, Farmer John can book room 3 for himself, and rooms 1 and 4 for his cows. The distance to the farthest cow is 2. Note that it is impossible to make this distance 1, as there is no block of three consecutive unoccupied rooms.
In the second sample, Farmer John can book room 1 for himself and room 3 for his single cow. The distance between him and his cow is 2.
In the third sample, Farmer John books all three available rooms, taking the middle room for himself so that both cows are next to him. His distance from the farthest cow is 1.
Submitted Solution:
```
# You lost the game.
n,k = map(int, input().split())
s = str(input())
L = []
for i in range(n):
if s[i] == '0':
L = L + [i]
r = 10**7
i = 1
j = 1
while i < len(L)-k+j:
e = max(L[i]-L[i-j],L[i+k-j]-L[i])
if e > r:
if j < k:
j += 1
else:
j = 1
i += 1
else:
r = e
i += 1
print(r)
``` | instruction | 0 | 30,526 | 8 | 61,052 |
No | output | 1 | 30,526 | 8 | 61,053 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
In an attempt to escape the Mischievous Mess Makers' antics, Farmer John has abandoned his farm and is traveling to the other side of Bovinia. During the journey, he and his k cows have decided to stay at the luxurious Grand Moo-dapest Hotel. The hotel consists of n rooms located in a row, some of which are occupied.
Farmer John wants to book a set of k + 1 currently unoccupied rooms for him and his cows. He wants his cows to stay as safe as possible, so he wishes to minimize the maximum distance from his room to the room of his cow. The distance between rooms i and j is defined as |j - i|. Help Farmer John protect his cows by calculating this minimum possible distance.
Input
The first line of the input contains two integers n and k (1 β€ k < n β€ 100 000) β the number of rooms in the hotel and the number of cows travelling with Farmer John.
The second line contains a string of length n describing the rooms. The i-th character of the string will be '0' if the i-th room is free, and '1' if the i-th room is occupied. It is guaranteed that at least k + 1 characters of this string are '0', so there exists at least one possible choice of k + 1 rooms for Farmer John and his cows to stay in.
Output
Print the minimum possible distance between Farmer John's room and his farthest cow.
Examples
Input
7 2
0100100
Output
2
Input
5 1
01010
Output
2
Input
3 2
000
Output
1
Note
In the first sample, Farmer John can book room 3 for himself, and rooms 1 and 4 for his cows. The distance to the farthest cow is 2. Note that it is impossible to make this distance 1, as there is no block of three consecutive unoccupied rooms.
In the second sample, Farmer John can book room 1 for himself and room 3 for his single cow. The distance between him and his cow is 2.
In the third sample, Farmer John books all three available rooms, taking the middle room for himself so that both cows are next to him. His distance from the farthest cow is 1.
Submitted Solution:
```
# cows
inp = input().split()
cows = int(inp[1])
pattern = input()
def maxRooms(pat):
for i in range(len(pattern), 0, -1):
sub = '0' * i
if sub in pattern:
ini = pattern.index(sub)
return i, ini, ini + i
def locateCows(pat, ini, fin, cows, distance):
ini -= 1
fin += 1
if ini > 0:
if pat[ini] == '0':
cows -= 1
if cows == 0:
return distance
if fin < len(pat):
if pat[fin] == '0':
cows -= 1
if cows == 0:
return distance
return locateCows(pat, ini, fin, cows, distance + 1)
together = 0
initial = 0
final = 0
together, initial, final = maxRooms(pattern)
if together == cows + 1:
print(1)
else:
print(locateCows(pattern, initial, final - 1, cows + 1 - together, 1))
``` | instruction | 0 | 30,527 | 8 | 61,054 |
No | output | 1 | 30,527 | 8 | 61,055 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
In an attempt to escape the Mischievous Mess Makers' antics, Farmer John has abandoned his farm and is traveling to the other side of Bovinia. During the journey, he and his k cows have decided to stay at the luxurious Grand Moo-dapest Hotel. The hotel consists of n rooms located in a row, some of which are occupied.
Farmer John wants to book a set of k + 1 currently unoccupied rooms for him and his cows. He wants his cows to stay as safe as possible, so he wishes to minimize the maximum distance from his room to the room of his cow. The distance between rooms i and j is defined as |j - i|. Help Farmer John protect his cows by calculating this minimum possible distance.
Input
The first line of the input contains two integers n and k (1 β€ k < n β€ 100 000) β the number of rooms in the hotel and the number of cows travelling with Farmer John.
The second line contains a string of length n describing the rooms. The i-th character of the string will be '0' if the i-th room is free, and '1' if the i-th room is occupied. It is guaranteed that at least k + 1 characters of this string are '0', so there exists at least one possible choice of k + 1 rooms for Farmer John and his cows to stay in.
Output
Print the minimum possible distance between Farmer John's room and his farthest cow.
Examples
Input
7 2
0100100
Output
2
Input
5 1
01010
Output
2
Input
3 2
000
Output
1
Note
In the first sample, Farmer John can book room 3 for himself, and rooms 1 and 4 for his cows. The distance to the farthest cow is 2. Note that it is impossible to make this distance 1, as there is no block of three consecutive unoccupied rooms.
In the second sample, Farmer John can book room 1 for himself and room 3 for his single cow. The distance between him and his cow is 2.
In the third sample, Farmer John books all three available rooms, taking the middle room for himself so that both cows are next to him. His distance from the farthest cow is 1.
Submitted Solution:
```
#[int(i) for i in input().split()]
n, k = [int(i) for i in input().split()]
k += 1
num = input()
d = [1000000] * n
i = num.find('0')
cnt = 1
end = 0
for j in range(i + 1, n):
if num[j] == '0':
cnt += 1
if cnt == k:
end = j
break
fer = i
last_j = fer
for j in range(i, end + 1):
if num[j] == '0' and max(end - j, j - i) > max(end - last_j, last_j - i):
break
if num[j] == '0' and max(end - fer, fer - i) > max(end - j, j - i):
last_j = j
fer = j
d[i] = max(end - fer, fer - i)
while end != n:
i += 1
end += 1
while i < n and num[i] == '1':
i += 1
while end < n and num[end] == '1':
end += 1
if end < n:
f1 = fer
last_j = f1
for j in range(fer, end + 1):
if num[j] == '0' and max(end - j, j - i) > max(end - last_j, last_j - i):
break
if num[j] == '0' and max(end - f1, f1 - i) > max(end - j, j - i):
last_j = j
f1 = j
d[i] = f1
print(min(d))
``` | instruction | 0 | 30,528 | 8 | 61,056 |
No | output | 1 | 30,528 | 8 | 61,057 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Alice likes snow a lot! Unfortunately, this year's winter is already over, and she can't expect to have any more of it. Bob has thus bought her a gift β a large snow maker. He plans to make some amount of snow every day. On day i he will make a pile of snow of volume Vi and put it in her garden.
Each day, every pile will shrink a little due to melting. More precisely, when the temperature on a given day is Ti, each pile will reduce its volume by Ti. If this would reduce the volume of a pile to or below zero, it disappears forever. All snow piles are independent of each other.
Note that the pile made on day i already loses part of its volume on the same day. In an extreme case, this may mean that there are no piles left at the end of a particular day.
You are given the initial pile sizes and the temperature on each day. Determine the total volume of snow melted on each day.
Input
The first line contains a single integer N (1 β€ N β€ 105) β the number of days.
The second line contains N integers V1, V2, ..., VN (0 β€ Vi β€ 109), where Vi is the initial size of a snow pile made on the day i.
The third line contains N integers T1, T2, ..., TN (0 β€ Ti β€ 109), where Ti is the temperature on the day i.
Output
Output a single line with N integers, where the i-th integer represents the total volume of snow melted on day i.
Examples
Input
3
10 10 5
5 7 2
Output
5 12 4
Input
5
30 25 20 15 10
9 10 12 4 13
Output
9 20 35 11 25
Note
In the first sample, Bob first makes a snow pile of volume 10, which melts to the size of 5 on the same day. On the second day, he makes another pile of size 10. Since it is a bit warmer than the day before, the first pile disappears completely while the second pile shrinks to 3. At the end of the second day, he has only a single pile of size 3. On the third day he makes a smaller pile than usual, but as the temperature dropped too, both piles survive till the end of the day. | instruction | 0 | 30,649 | 8 | 61,298 |
Tags: binary search, data structures
Correct Solution:
```
from heapq import *
n=int(input())
l=list(map(int,input().split()))
l1=list(map(int,input().split()))
L=[]
heapify(L)
l2=l1[:]
qw=[]
for i in range(1,n) :
l2[i]+=l2[i-1]
for i in range(n) :
v=l[i]+l2[i]-l1[i]
heappush(L,v)
mi=heappop(L)
count=0
while mi<=l2[i] :
count+=mi-l2[i]+l1[i]
if not L :
break
mi=heappop(L)
else :
heappush(L,mi)
ost=count+len(L)*l1[i]
qw.append(ost)
print(*qw)
``` | output | 1 | 30,649 | 8 | 61,299 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Alice likes snow a lot! Unfortunately, this year's winter is already over, and she can't expect to have any more of it. Bob has thus bought her a gift β a large snow maker. He plans to make some amount of snow every day. On day i he will make a pile of snow of volume Vi and put it in her garden.
Each day, every pile will shrink a little due to melting. More precisely, when the temperature on a given day is Ti, each pile will reduce its volume by Ti. If this would reduce the volume of a pile to or below zero, it disappears forever. All snow piles are independent of each other.
Note that the pile made on day i already loses part of its volume on the same day. In an extreme case, this may mean that there are no piles left at the end of a particular day.
You are given the initial pile sizes and the temperature on each day. Determine the total volume of snow melted on each day.
Input
The first line contains a single integer N (1 β€ N β€ 105) β the number of days.
The second line contains N integers V1, V2, ..., VN (0 β€ Vi β€ 109), where Vi is the initial size of a snow pile made on the day i.
The third line contains N integers T1, T2, ..., TN (0 β€ Ti β€ 109), where Ti is the temperature on the day i.
Output
Output a single line with N integers, where the i-th integer represents the total volume of snow melted on day i.
Examples
Input
3
10 10 5
5 7 2
Output
5 12 4
Input
5
30 25 20 15 10
9 10 12 4 13
Output
9 20 35 11 25
Note
In the first sample, Bob first makes a snow pile of volume 10, which melts to the size of 5 on the same day. On the second day, he makes another pile of size 10. Since it is a bit warmer than the day before, the first pile disappears completely while the second pile shrinks to 3. At the end of the second day, he has only a single pile of size 3. On the third day he makes a smaller pile than usual, but as the temperature dropped too, both piles survive till the end of the day. | instruction | 0 | 30,650 | 8 | 61,300 |
Tags: binary search, data structures
Correct Solution:
```
from sys import stdin
input = stdin.readline
import heapq
def f(v, t):
ans=0
r=0
q=[]
heapq.heapify(q)
for i in range(len(v)):
ans=0
heapq.heappush(q,v[i]+r)
while q and q[0] <t[i]+r:
ans+=heapq.heappop(q)-r
ans+=len(q)*(t[i])
print(ans,end=" ")
r+=t[i]
n=input()
l=list(map(int,input().strip().split()))
d=list(map(int,input().strip().split()))
f(l,d)
``` | output | 1 | 30,650 | 8 | 61,301 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Alice likes snow a lot! Unfortunately, this year's winter is already over, and she can't expect to have any more of it. Bob has thus bought her a gift β a large snow maker. He plans to make some amount of snow every day. On day i he will make a pile of snow of volume Vi and put it in her garden.
Each day, every pile will shrink a little due to melting. More precisely, when the temperature on a given day is Ti, each pile will reduce its volume by Ti. If this would reduce the volume of a pile to or below zero, it disappears forever. All snow piles are independent of each other.
Note that the pile made on day i already loses part of its volume on the same day. In an extreme case, this may mean that there are no piles left at the end of a particular day.
You are given the initial pile sizes and the temperature on each day. Determine the total volume of snow melted on each day.
Input
The first line contains a single integer N (1 β€ N β€ 105) β the number of days.
The second line contains N integers V1, V2, ..., VN (0 β€ Vi β€ 109), where Vi is the initial size of a snow pile made on the day i.
The third line contains N integers T1, T2, ..., TN (0 β€ Ti β€ 109), where Ti is the temperature on the day i.
Output
Output a single line with N integers, where the i-th integer represents the total volume of snow melted on day i.
Examples
Input
3
10 10 5
5 7 2
Output
5 12 4
Input
5
30 25 20 15 10
9 10 12 4 13
Output
9 20 35 11 25
Note
In the first sample, Bob first makes a snow pile of volume 10, which melts to the size of 5 on the same day. On the second day, he makes another pile of size 10. Since it is a bit warmer than the day before, the first pile disappears completely while the second pile shrinks to 3. At the end of the second day, he has only a single pile of size 3. On the third day he makes a smaller pile than usual, but as the temperature dropped too, both piles survive till the end of the day. | instruction | 0 | 30,651 | 8 | 61,302 |
Tags: binary search, data structures
Correct Solution:
```
#Code by Sounak, IIESTS
#------------------------------warmup----------------------------
import os
import sys
import math
from io import BytesIO, IOBase
from fractions import Fraction
import collections
from itertools import permutations
BUFSIZE = 8192
class FastIO(IOBase):
newlines = 0
def __init__(self, file):
self._fd = file.fileno()
self.buffer = BytesIO()
self.writable = "x" in file.mode or "r" not in file.mode
self.write = self.buffer.write if self.writable else None
def read(self):
while True:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
if not b:
break
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines = 0
return self.buffer.read()
def readline(self):
while self.newlines == 0:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
self.newlines = b.count(b"\n") + (not b)
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines -= 1
return self.buffer.readline()
def flush(self):
if self.writable:
os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
self.write = lambda s: self.buffer.write(s.encode("ascii"))
self.read = lambda: self.buffer.read().decode("ascii")
self.readline = lambda: self.buffer.readline().decode("ascii")
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")
#-------------------game starts now-----------------------------------------------------
def findMaxTemp(psT, v, start, minus):
l = start - 1
r = len(psT)
while r - l > 1:
m = l + (r - l) // 2
val = psT[m] - minus
if val < v:
l = m
else:
r = m
return l, r
def solve():
N = int(input())
V = list(map(int, input().split()))
T = list(map(int, input().split()))
ans = [0 for _ in range(N)]
partialAns = [0 for _ in range(N)]
cnt = [0 for _ in range(N)]
minus = 0
psT = T[:]
for i in range(1, N):
psT[i] += psT[i - 1]
for i in range(len(V)):
l, r = findMaxTemp(psT, V[i], i, minus)
if r == i:
partialAns[r] += V[i]
elif r == len(psT):
cnt[i] += 1
else:
cnt[i] += 1; cnt[l + 1] -= 1
value = V[i] - (psT[l] - minus)
partialAns[r] += value
minus = psT[i]
for i in range(1, N):
cnt[i] += cnt[i - 1]
for i in range(N):
ans[i] += partialAns[i]
ans[i] += cnt[i] * T[i]
return ans
for x in solve():
print(x, end=' ')
``` | output | 1 | 30,651 | 8 | 61,303 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Alice likes snow a lot! Unfortunately, this year's winter is already over, and she can't expect to have any more of it. Bob has thus bought her a gift β a large snow maker. He plans to make some amount of snow every day. On day i he will make a pile of snow of volume Vi and put it in her garden.
Each day, every pile will shrink a little due to melting. More precisely, when the temperature on a given day is Ti, each pile will reduce its volume by Ti. If this would reduce the volume of a pile to or below zero, it disappears forever. All snow piles are independent of each other.
Note that the pile made on day i already loses part of its volume on the same day. In an extreme case, this may mean that there are no piles left at the end of a particular day.
You are given the initial pile sizes and the temperature on each day. Determine the total volume of snow melted on each day.
Input
The first line contains a single integer N (1 β€ N β€ 105) β the number of days.
The second line contains N integers V1, V2, ..., VN (0 β€ Vi β€ 109), where Vi is the initial size of a snow pile made on the day i.
The third line contains N integers T1, T2, ..., TN (0 β€ Ti β€ 109), where Ti is the temperature on the day i.
Output
Output a single line with N integers, where the i-th integer represents the total volume of snow melted on day i.
Examples
Input
3
10 10 5
5 7 2
Output
5 12 4
Input
5
30 25 20 15 10
9 10 12 4 13
Output
9 20 35 11 25
Note
In the first sample, Bob first makes a snow pile of volume 10, which melts to the size of 5 on the same day. On the second day, he makes another pile of size 10. Since it is a bit warmer than the day before, the first pile disappears completely while the second pile shrinks to 3. At the end of the second day, he has only a single pile of size 3. On the third day he makes a smaller pile than usual, but as the temperature dropped too, both piles survive till the end of the day. | instruction | 0 | 30,652 | 8 | 61,304 |
Tags: binary search, data structures
Correct Solution:
```
import heapq
N = int(input().strip())
V = list(map(int, input().strip().split()))
T = list(map(int, input().strip().split()))
res = []
pq = []
summ = 0
for i in range(N):
cur = 0
heapq.heappush(pq, summ + V[i])
summ += T[i]
while pq:
item = pq[0]
#print(item, type(item))
if summ < item:
cur += (len(pq) * T[i])
break
else:
heapq.heappop(pq)
cur += (item - (summ - T[i]))
res.append(str(cur))
print(" ".join(res).strip())
'''
# PriorityQueue θΆ
ζΆ
from queue import PriorityQueue
N = int(input().strip())
V = list(map(int, input().strip().split()))
T = list(map(int, input().strip().split()))
res = []
pq = PriorityQueue()
summ = 0
for i in range(N):
cur = 0
pq.put(summ + V[i])
summ += T[i]
while not pq.empty():
item = pq.queue[0]
#print(item, type(item))
if summ < item:
cur += (pq.qsize() * T[i])
break
else:
pq.get()
cur += (item - (summ - T[i]))
res.append(str(cur))
print(" ".join(res).strip())
'''
'''
# ζ΄ε解ζ³
for i in range(N):
cur = 0
for j in range(i+1):
cur += min(T[i], V[j])
V[j] -= min(T[i], V[j])
res.append(str(cur))
print(" ".join(res).strip())
'''
# 10 10 5
# 6 7 2
# q: 10, 10 + 6, 5 + 6 + 7
# 6, 4 + 7, 3 + 2
``` | output | 1 | 30,652 | 8 | 61,305 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Alice likes snow a lot! Unfortunately, this year's winter is already over, and she can't expect to have any more of it. Bob has thus bought her a gift β a large snow maker. He plans to make some amount of snow every day. On day i he will make a pile of snow of volume Vi and put it in her garden.
Each day, every pile will shrink a little due to melting. More precisely, when the temperature on a given day is Ti, each pile will reduce its volume by Ti. If this would reduce the volume of a pile to or below zero, it disappears forever. All snow piles are independent of each other.
Note that the pile made on day i already loses part of its volume on the same day. In an extreme case, this may mean that there are no piles left at the end of a particular day.
You are given the initial pile sizes and the temperature on each day. Determine the total volume of snow melted on each day.
Input
The first line contains a single integer N (1 β€ N β€ 105) β the number of days.
The second line contains N integers V1, V2, ..., VN (0 β€ Vi β€ 109), where Vi is the initial size of a snow pile made on the day i.
The third line contains N integers T1, T2, ..., TN (0 β€ Ti β€ 109), where Ti is the temperature on the day i.
Output
Output a single line with N integers, where the i-th integer represents the total volume of snow melted on day i.
Examples
Input
3
10 10 5
5 7 2
Output
5 12 4
Input
5
30 25 20 15 10
9 10 12 4 13
Output
9 20 35 11 25
Note
In the first sample, Bob first makes a snow pile of volume 10, which melts to the size of 5 on the same day. On the second day, he makes another pile of size 10. Since it is a bit warmer than the day before, the first pile disappears completely while the second pile shrinks to 3. At the end of the second day, he has only a single pile of size 3. On the third day he makes a smaller pile than usual, but as the temperature dropped too, both piles survive till the end of the day. | instruction | 0 | 30,653 | 8 | 61,306 |
Tags: binary search, data structures
Correct Solution:
```
import heapq
n = int(input())
v = list(map(int, input().split()))
t = list(map(int, input().split()))
d = []
s = 0
for i in range(n):
ans = 0
heapq.heappush(d, v[i] + s)
while d and d[0] <= s + t[i]:
ans += d[0] - s
heapq.heappop(d)
s += t[i]
ans += len(d) * t[i]
print(ans, end=' ')
``` | output | 1 | 30,653 | 8 | 61,307 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Alice likes snow a lot! Unfortunately, this year's winter is already over, and she can't expect to have any more of it. Bob has thus bought her a gift β a large snow maker. He plans to make some amount of snow every day. On day i he will make a pile of snow of volume Vi and put it in her garden.
Each day, every pile will shrink a little due to melting. More precisely, when the temperature on a given day is Ti, each pile will reduce its volume by Ti. If this would reduce the volume of a pile to or below zero, it disappears forever. All snow piles are independent of each other.
Note that the pile made on day i already loses part of its volume on the same day. In an extreme case, this may mean that there are no piles left at the end of a particular day.
You are given the initial pile sizes and the temperature on each day. Determine the total volume of snow melted on each day.
Input
The first line contains a single integer N (1 β€ N β€ 105) β the number of days.
The second line contains N integers V1, V2, ..., VN (0 β€ Vi β€ 109), where Vi is the initial size of a snow pile made on the day i.
The third line contains N integers T1, T2, ..., TN (0 β€ Ti β€ 109), where Ti is the temperature on the day i.
Output
Output a single line with N integers, where the i-th integer represents the total volume of snow melted on day i.
Examples
Input
3
10 10 5
5 7 2
Output
5 12 4
Input
5
30 25 20 15 10
9 10 12 4 13
Output
9 20 35 11 25
Note
In the first sample, Bob first makes a snow pile of volume 10, which melts to the size of 5 on the same day. On the second day, he makes another pile of size 10. Since it is a bit warmer than the day before, the first pile disappears completely while the second pile shrinks to 3. At the end of the second day, he has only a single pile of size 3. On the third day he makes a smaller pile than usual, but as the temperature dropped too, both piles survive till the end of the day. | instruction | 0 | 30,654 | 8 | 61,308 |
Tags: binary search, data structures
Correct Solution:
```
import sys
import math
import collections
# sys.setrecursionlimit(10**6)
# sys.setrecursionlimit(100000000)
inp =lambda: int(input())
strng =lambda: input().strip()
strl =lambda: list(input().strip()) #
mulf =lambda: map(float,input().strip().split()) #
seq =lambda: list(map(int,input().strip().split())) #
mod = (10**9)+7
input=sys.stdin.readline
# print("Hello, World!", flush=True)
# ord("A") chr(65)
def bs(l,h,v):
if(l<=h):
m = (l+h)//2
if(pret[m] < v):
if(m ==h):
return m
else:
return bs(m+1,h,v)
else:
if(m-1>=l and pret[m-1]>=v):
return bs(l,m-1,v)
return m
return -1
n = inp()
v = seq()
t = seq()
pret = [0]*n
pret[0] = t[0]
for i in range(1,n):
pret[i] = pret[i-1] + t[i]
# print(pret,"ppp")
freq = [0]*n
extra = [0]*n
for i in range(n):
temp = v[i]
if(i>0):
temp += pret[i-1]
val = bs(i,n-1,temp)
# print(val,"indexx")
if(val == i):
if(temp >= pret[i]):
freq[i] += 1
if(i>0):
freq[i-1] -= 1
elif(temp < pret[i]):
diff = pret[i] - temp
extra[i] += t[i] - diff
else:
if(temp >= pret[val]):
freq[val] += 1
if(i>0):
freq[i-1] -= 1
elif(temp < pret[val]):
# print(i)
diff = pret[val] - temp
extra[val] += t[val] - diff
freq[val-1] += 1
# print(val-1)
if(i>0):
freq[i-1] -= 1
# print(freq,"ffff")
# print(extra,"eeeee")
ans = [""]*n
curr = 0
# print(freq)
# print(extra)
for i in range(n-1,-1,-1):
temp = freq[i]
freq[i] += curr
curr += temp
for i in range(n):
temp = freq[i]*t[i] + extra[i]
temp = str(temp)
ans[i] = temp
print(" ".join(ans))
``` | output | 1 | 30,654 | 8 | 61,309 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Alice likes snow a lot! Unfortunately, this year's winter is already over, and she can't expect to have any more of it. Bob has thus bought her a gift β a large snow maker. He plans to make some amount of snow every day. On day i he will make a pile of snow of volume Vi and put it in her garden.
Each day, every pile will shrink a little due to melting. More precisely, when the temperature on a given day is Ti, each pile will reduce its volume by Ti. If this would reduce the volume of a pile to or below zero, it disappears forever. All snow piles are independent of each other.
Note that the pile made on day i already loses part of its volume on the same day. In an extreme case, this may mean that there are no piles left at the end of a particular day.
You are given the initial pile sizes and the temperature on each day. Determine the total volume of snow melted on each day.
Input
The first line contains a single integer N (1 β€ N β€ 105) β the number of days.
The second line contains N integers V1, V2, ..., VN (0 β€ Vi β€ 109), where Vi is the initial size of a snow pile made on the day i.
The third line contains N integers T1, T2, ..., TN (0 β€ Ti β€ 109), where Ti is the temperature on the day i.
Output
Output a single line with N integers, where the i-th integer represents the total volume of snow melted on day i.
Examples
Input
3
10 10 5
5 7 2
Output
5 12 4
Input
5
30 25 20 15 10
9 10 12 4 13
Output
9 20 35 11 25
Note
In the first sample, Bob first makes a snow pile of volume 10, which melts to the size of 5 on the same day. On the second day, he makes another pile of size 10. Since it is a bit warmer than the day before, the first pile disappears completely while the second pile shrinks to 3. At the end of the second day, he has only a single pile of size 3. On the third day he makes a smaller pile than usual, but as the temperature dropped too, both piles survive till the end of the day. | instruction | 0 | 30,655 | 8 | 61,310 |
Tags: binary search, data structures
Correct Solution:
```
import heapq
import sys
heap = []
n = int(input())
V = list(map(int, input().split()))
T = list(map(int, input().split()))
tmp = 0
for i in range(n):
ans = 0
heapq.heappush(heap, tmp+V[i])
while len(heap) and heap[0]<=tmp+T[i]:
ans += heapq.heappop(heap)-tmp
tmp += T[i]
ans += T[i]*len(heap)
print(ans, end=' ')
if i%10000 == 0:
sys.stdout.flush()
``` | output | 1 | 30,655 | 8 | 61,311 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Alice likes snow a lot! Unfortunately, this year's winter is already over, and she can't expect to have any more of it. Bob has thus bought her a gift β a large snow maker. He plans to make some amount of snow every day. On day i he will make a pile of snow of volume Vi and put it in her garden.
Each day, every pile will shrink a little due to melting. More precisely, when the temperature on a given day is Ti, each pile will reduce its volume by Ti. If this would reduce the volume of a pile to or below zero, it disappears forever. All snow piles are independent of each other.
Note that the pile made on day i already loses part of its volume on the same day. In an extreme case, this may mean that there are no piles left at the end of a particular day.
You are given the initial pile sizes and the temperature on each day. Determine the total volume of snow melted on each day.
Input
The first line contains a single integer N (1 β€ N β€ 105) β the number of days.
The second line contains N integers V1, V2, ..., VN (0 β€ Vi β€ 109), where Vi is the initial size of a snow pile made on the day i.
The third line contains N integers T1, T2, ..., TN (0 β€ Ti β€ 109), where Ti is the temperature on the day i.
Output
Output a single line with N integers, where the i-th integer represents the total volume of snow melted on day i.
Examples
Input
3
10 10 5
5 7 2
Output
5 12 4
Input
5
30 25 20 15 10
9 10 12 4 13
Output
9 20 35 11 25
Note
In the first sample, Bob first makes a snow pile of volume 10, which melts to the size of 5 on the same day. On the second day, he makes another pile of size 10. Since it is a bit warmer than the day before, the first pile disappears completely while the second pile shrinks to 3. At the end of the second day, he has only a single pile of size 3. On the third day he makes a smaller pile than usual, but as the temperature dropped too, both piles survive till the end of the day. | instruction | 0 | 30,656 | 8 | 61,312 |
Tags: binary search, data structures
Correct Solution:
```
import os
import sys
from io import BytesIO, IOBase
# region fastio
BUFSIZE = 8192
class FastIO(IOBase):
newlines = 0
def __init__(self, file):
self._fd = file.fileno()
self.buffer = BytesIO()
self.writable = "x" in file.mode or "r" not in file.mode
self.write = self.buffer.write if self.writable else None
def read(self):
while True:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
if not b:
break
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines = 0
return self.buffer.read()
def readline(self):
while self.newlines == 0:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
self.newlines = b.count(b"\n") + (not b)
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines -= 1
return self.buffer.readline()
def flush(self):
if self.writable:
os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
self.write = lambda s: self.buffer.write(s.encode("ascii"))
self.read = lambda: self.buffer.read().decode("ascii")
self.readline = lambda: self.buffer.readline().decode("ascii")
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")
# ------------------------------
from math import factorial
from collections import Counter, defaultdict, deque
from heapq import heapify, heappop, heappush
def RL(): return map(int, sys.stdin.readline().rstrip().split())
def RLL(): return list(map(int, sys.stdin.readline().rstrip().split()))
def N(): return int(input())
def comb(n, m): return factorial(n) / (factorial(m) * factorial(n - m)) if n >= m else 0
def perm(n, m): return factorial(n) // (factorial(n - m)) if n >= m else 0
def mdis(x1, y1, x2, y2): return abs(x1 - x2) + abs(y1 - y2)
mod = 998244353
INF = float('inf')
# ------------------------------
from bisect import bisect_right
def main():
n = N()
arrv = RLL()
arrt = RLL()
res = [0]*n
dif = [0]*n
sm = [0]
for i in arrt: sm.append(sm[-1]+i)
for i in range(n):
now = arrv[i]+sm[i]
p = bisect_right(sm, now)
if p-1>=0:
res[i]+=1
if p<=n: res[p-1]-=1
if p<=n: dif[p-1]+=now-sm[p-1]
for i in range(1, n):
res[i]+=res[i-1]
for i in range(n):
res[i]*=arrt[i]
res[i]+=dif[i]
print(*res)
if __name__ == "__main__":
main()
``` | output | 1 | 30,656 | 8 | 61,313 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Alice likes snow a lot! Unfortunately, this year's winter is already over, and she can't expect to have any more of it. Bob has thus bought her a gift β a large snow maker. He plans to make some amount of snow every day. On day i he will make a pile of snow of volume Vi and put it in her garden.
Each day, every pile will shrink a little due to melting. More precisely, when the temperature on a given day is Ti, each pile will reduce its volume by Ti. If this would reduce the volume of a pile to or below zero, it disappears forever. All snow piles are independent of each other.
Note that the pile made on day i already loses part of its volume on the same day. In an extreme case, this may mean that there are no piles left at the end of a particular day.
You are given the initial pile sizes and the temperature on each day. Determine the total volume of snow melted on each day.
Input
The first line contains a single integer N (1 β€ N β€ 105) β the number of days.
The second line contains N integers V1, V2, ..., VN (0 β€ Vi β€ 109), where Vi is the initial size of a snow pile made on the day i.
The third line contains N integers T1, T2, ..., TN (0 β€ Ti β€ 109), where Ti is the temperature on the day i.
Output
Output a single line with N integers, where the i-th integer represents the total volume of snow melted on day i.
Examples
Input
3
10 10 5
5 7 2
Output
5 12 4
Input
5
30 25 20 15 10
9 10 12 4 13
Output
9 20 35 11 25
Note
In the first sample, Bob first makes a snow pile of volume 10, which melts to the size of 5 on the same day. On the second day, he makes another pile of size 10. Since it is a bit warmer than the day before, the first pile disappears completely while the second pile shrinks to 3. At the end of the second day, he has only a single pile of size 3. On the third day he makes a smaller pile than usual, but as the temperature dropped too, both piles survive till the end of the day.
Submitted Solution:
```
#!/usr/bin/env python
#pyrival orz
import os
import sys
from io import BytesIO, IOBase
"""
for _ in range(int(input())):
n,m=map(int,input().split())
n=int(input())
a = [int(x) for x in input().split()]
"""
def main():
n=int(input())
a = [int(x) for x in input().split()]
b = [int(x) for x in input().split()]
a=[0]+a
b=[0]+b
ans=[0]*(n+1)
p=[0]*(n+1)
ans=[0]*(n+1)
ct=[0]*(n+1)
for i in range(1,n+1):
p[i]=p[i-1]+b[i]
# print(p)
from bisect import bisect_right as ub
for i in range(1,n+1):
j=ub(p,p[i-1]+a[i])
# print(i,j)
ct[i]+=1
if j==n+1:
continue
ct[j]-=1
ans[j]+=p[i-1]+a[i]-p[j-1]
# print(*ans,*ct)
for i in range(1,n+1):
ct[i]+=ct[i-1]
for i in range(1,n+1):
ans[i]+=ct[i]*b[i]
print(*ans[1:])
# region fastio
BUFSIZE = 8192
class FastIO(IOBase):
newlines = 0
def __init__(self, file):
self._fd = file.fileno()
self.buffer = BytesIO()
self.writable = "x" in file.mode or "r" not in file.mode
self.write = self.buffer.write if self.writable else None
def read(self):
while True:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
if not b:
break
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines = 0
return self.buffer.read()
def readline(self):
while self.newlines == 0:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
self.newlines = b.count(b"\n") + (not b)
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines -= 1
return self.buffer.readline()
def flush(self):
if self.writable:
os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
self.write = lambda s: self.buffer.write(s.encode("ascii"))
self.read = lambda: self.buffer.read().decode("ascii")
self.readline = lambda: self.buffer.readline().decode("ascii")
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")
# endregion
if __name__ == "__main__":
main()
``` | instruction | 0 | 30,657 | 8 | 61,314 |
Yes | output | 1 | 30,657 | 8 | 61,315 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Alice likes snow a lot! Unfortunately, this year's winter is already over, and she can't expect to have any more of it. Bob has thus bought her a gift β a large snow maker. He plans to make some amount of snow every day. On day i he will make a pile of snow of volume Vi and put it in her garden.
Each day, every pile will shrink a little due to melting. More precisely, when the temperature on a given day is Ti, each pile will reduce its volume by Ti. If this would reduce the volume of a pile to or below zero, it disappears forever. All snow piles are independent of each other.
Note that the pile made on day i already loses part of its volume on the same day. In an extreme case, this may mean that there are no piles left at the end of a particular day.
You are given the initial pile sizes and the temperature on each day. Determine the total volume of snow melted on each day.
Input
The first line contains a single integer N (1 β€ N β€ 105) β the number of days.
The second line contains N integers V1, V2, ..., VN (0 β€ Vi β€ 109), where Vi is the initial size of a snow pile made on the day i.
The third line contains N integers T1, T2, ..., TN (0 β€ Ti β€ 109), where Ti is the temperature on the day i.
Output
Output a single line with N integers, where the i-th integer represents the total volume of snow melted on day i.
Examples
Input
3
10 10 5
5 7 2
Output
5 12 4
Input
5
30 25 20 15 10
9 10 12 4 13
Output
9 20 35 11 25
Note
In the first sample, Bob first makes a snow pile of volume 10, which melts to the size of 5 on the same day. On the second day, he makes another pile of size 10. Since it is a bit warmer than the day before, the first pile disappears completely while the second pile shrinks to 3. At the end of the second day, he has only a single pile of size 3. On the third day he makes a smaller pile than usual, but as the temperature dropped too, both piles survive till the end of the day.
Submitted Solution:
```
# import sys
# from io import StringIO
#
# sys.stdin = StringIO(open(__file__.replace('.py', '.in')).read())
def bin_search_base(arr, val, base_index):
lo = base_index
hi = len(arr) - 1
while lo < hi:
if hi - lo == 1 and arr[lo] - arr[base_index] < val < arr[hi] - arr[base_index]:
return lo
if val >= arr[hi] - arr[base_index]:
return hi
c = (hi + lo) // 2
if val < arr[c] - arr[base_index]:
hi = c
elif val > arr[c] - arr[base_index]:
lo = c
else:
return c
return lo
n = int(input())
volumes = list(map(int, input().split()))
temps = list(map(int, input().split()))
temp_prefix = [0]
for i in range(n):
temp_prefix.append(temps[i] + temp_prefix[i])
gone_count = [0] * (n + 1)
remaining_count = [0] * n
gone_volume = [0] * (n + 1)
for i in range(n):
j = bin_search_base(temp_prefix, volumes[i], i)
gone_count[j] += 1
gone_volume[j] += volumes[i] - (temp_prefix[j] - temp_prefix[i])
for i in range(n):
remaining_count[i] = 1 + remaining_count[i-1] - gone_count[i]
for i in range(n):
gone_volume[i] += remaining_count[i] * temps[i]
print(' '.join(map(str, gone_volume[:n])))
``` | instruction | 0 | 30,658 | 8 | 61,316 |
Yes | output | 1 | 30,658 | 8 | 61,317 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Alice likes snow a lot! Unfortunately, this year's winter is already over, and she can't expect to have any more of it. Bob has thus bought her a gift β a large snow maker. He plans to make some amount of snow every day. On day i he will make a pile of snow of volume Vi and put it in her garden.
Each day, every pile will shrink a little due to melting. More precisely, when the temperature on a given day is Ti, each pile will reduce its volume by Ti. If this would reduce the volume of a pile to or below zero, it disappears forever. All snow piles are independent of each other.
Note that the pile made on day i already loses part of its volume on the same day. In an extreme case, this may mean that there are no piles left at the end of a particular day.
You are given the initial pile sizes and the temperature on each day. Determine the total volume of snow melted on each day.
Input
The first line contains a single integer N (1 β€ N β€ 105) β the number of days.
The second line contains N integers V1, V2, ..., VN (0 β€ Vi β€ 109), where Vi is the initial size of a snow pile made on the day i.
The third line contains N integers T1, T2, ..., TN (0 β€ Ti β€ 109), where Ti is the temperature on the day i.
Output
Output a single line with N integers, where the i-th integer represents the total volume of snow melted on day i.
Examples
Input
3
10 10 5
5 7 2
Output
5 12 4
Input
5
30 25 20 15 10
9 10 12 4 13
Output
9 20 35 11 25
Note
In the first sample, Bob first makes a snow pile of volume 10, which melts to the size of 5 on the same day. On the second day, he makes another pile of size 10. Since it is a bit warmer than the day before, the first pile disappears completely while the second pile shrinks to 3. At the end of the second day, he has only a single pile of size 3. On the third day he makes a smaller pile than usual, but as the temperature dropped too, both piles survive till the end of the day.
Submitted Solution:
```
from collections import Counter
import os
import sys
from io import BytesIO, IOBase
BUFSIZE = 8192
class FastIO(IOBase):
newlines = 0
def __init__(self, file):
self._fd = file.fileno()
self.buffer = BytesIO()
self.writable = "x" in file.mode or "r" not in file.mode
self.write = self.buffer.write if self.writable else None
def read(self):
while True:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
if not b:
break
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines = 0
return self.buffer.read()
def readline(self):
while self.newlines == 0:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
self.newlines = b.count(b"\n") + (not b)
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines -= 1
return self.buffer.readline()
def flush(self):
if self.writable:
os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
self.write = lambda s: self.buffer.write(s.encode("ascii"))
self.read = lambda: self.buffer.read().decode("ascii")
self.readline = lambda: self.buffer.readline().decode("ascii")
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")
##########################################################
def dice():
dp = [0] * (n + 1)
dp[1] = 1
dp[0] = 1
for i in range(2, n + 1):
for j in range(1, 7):
if i - j >= 0:
dp[i] = (dp[i] + dp[i - j]) % (10 ** 9 + 7)
else:
break
#for _ in range(int(input())):
#import math
import sys
# from collections import deque
#from collections import Counter
# ls=list(map(int,input().split()))
# for i in range(m):
# for i in range(int(input())):
#n,k= map(int, input().split())
#arr=list(map(int,input().split()))
#n=sys.stdin.readline()
#n=int(n)
# n = int(inaput())
#sys.setrecursionlimit(10 **8)
def shut():
import os
shutdown = input()
if shutdown == 'no':
exit()
else:
os.system("shutdown /s /t 1")
def cost(l,r,ch,s):
var=0
for i in range(l,r):
if s[i]!=ch:
var+=1
return var
def fxn(l,r,ch,s):
if l+1==r:
if s[l]==ch:
return 0
return 1
mid=(l+r)//2
return min(cost(l,mid,ch,s)+fxn(mid,r,chr(ord(ch)+1),s),fxn(l,mid,chr(ord(ch)+1),s)+cost(mid,r,ch,s))
'''for _ in range(int(input())):
n = int(input())
s=input()
s=s
ans=fxn(0,n,"a",s)
print(ans)''"
import heapq
import sys
from math import gcd
n=int(input())
b,w=map(int,input().split())
heap=[]
heap1=[]
for _ in range(n-1):
ba,w=map(int,input().split())
if ba>b:
heap.append(w-ba+1)
else:
heap1.append((-1*ba,w))
heapq.heapify(heap)
heapq.heapify(heap1)
ans=len(heap)+1
while heap:
need=heap[0]
if need>b:
break
b-=need
heapq.heappop(heap)
while heap1 and -1*heap1[0][0]>b:
t,w=heapq.heappop(heap1)
t=-1*t
w=w
heapq.heappush(heap,w+1-t)
ans=min(ans,len(heap)+1)
print(ans)'''
import heapq
n=int(input())
v=list(map(int,input().split()))
t=list(map(int,input().split()))
heap=[]
sum=0
for i in range(n):
ans=0
heapq.heappush(heap,v[i]+sum)
while len(heap) and heap[0]<sum+t[i]:
ans+=heap[0]-sum
heapq.heappop(heap)
ans+=len(heap)*t[i]
sum+=t[i]
print(ans,end=" ")
#for _ in (int(input())):
#open("output.txt","w").write(s)
#n=int(open("input.txt").read())
#open("output.txt","w").write(s)
#arr=list(map(int,input().split()))
``` | instruction | 0 | 30,659 | 8 | 61,318 |
Yes | output | 1 | 30,659 | 8 | 61,319 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Alice likes snow a lot! Unfortunately, this year's winter is already over, and she can't expect to have any more of it. Bob has thus bought her a gift β a large snow maker. He plans to make some amount of snow every day. On day i he will make a pile of snow of volume Vi and put it in her garden.
Each day, every pile will shrink a little due to melting. More precisely, when the temperature on a given day is Ti, each pile will reduce its volume by Ti. If this would reduce the volume of a pile to or below zero, it disappears forever. All snow piles are independent of each other.
Note that the pile made on day i already loses part of its volume on the same day. In an extreme case, this may mean that there are no piles left at the end of a particular day.
You are given the initial pile sizes and the temperature on each day. Determine the total volume of snow melted on each day.
Input
The first line contains a single integer N (1 β€ N β€ 105) β the number of days.
The second line contains N integers V1, V2, ..., VN (0 β€ Vi β€ 109), where Vi is the initial size of a snow pile made on the day i.
The third line contains N integers T1, T2, ..., TN (0 β€ Ti β€ 109), where Ti is the temperature on the day i.
Output
Output a single line with N integers, where the i-th integer represents the total volume of snow melted on day i.
Examples
Input
3
10 10 5
5 7 2
Output
5 12 4
Input
5
30 25 20 15 10
9 10 12 4 13
Output
9 20 35 11 25
Note
In the first sample, Bob first makes a snow pile of volume 10, which melts to the size of 5 on the same day. On the second day, he makes another pile of size 10. Since it is a bit warmer than the day before, the first pile disappears completely while the second pile shrinks to 3. At the end of the second day, he has only a single pile of size 3. On the third day he makes a smaller pile than usual, but as the temperature dropped too, both piles survive till the end of the day.
Submitted Solution:
```
from sys import stdin, stdout
import heapq
n = int(input())
v = list(map(int, input().split()))
t = list(map(int, input().split()))
heap = []
sum = 0
for i in range(n):
ans = 0
heapq.heappush(heap, sum + v[i])
while len(heap) and heap[0] <= sum + t[i]:
ans += heap[0] - sum
heapq.heappop(heap)
ans += len(heap) * t[i]
sum += t[i]
print(ans, end=' ')
``` | instruction | 0 | 30,660 | 8 | 61,320 |
Yes | output | 1 | 30,660 | 8 | 61,321 |
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