message stringlengths 2 22.7k | message_type stringclasses 2 values | message_id int64 0 1 | conversation_id int64 145 109k | cluster float64 9 9 | __index_level_0__ int64 290 217k |
|---|---|---|---|---|---|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
In one little known, but very beautiful country called Waterland, lives a lovely shark Valerie. Like all the sharks, she has several rows of teeth, and feeds on crucians. One of Valerie's distinguishing features is that while eating one crucian she uses only one row of her teeth, the rest of the teeth are "relaxing".
For a long time our heroine had been searching the sea for crucians, but a great misfortune happened. Her teeth started to ache, and she had to see the local dentist, lobster Ashot. As a professional, Ashot quickly relieved Valerie from her toothache. Moreover, he managed to determine the cause of Valerie's developing caries (for what he was later nicknamed Cap).
It turned that Valerie eats too many crucians. To help Valerie avoid further reoccurrence of toothache, Ashot found for each Valerie's tooth its residual viability. Residual viability of a tooth is a value equal to the amount of crucians that Valerie can eat with this tooth. Every time Valerie eats a crucian, viability of all the teeth used for it will decrease by one. When the viability of at least one tooth becomes negative, the shark will have to see the dentist again.
Unhappy, Valerie came back home, where a portion of crucians was waiting for her. For sure, the shark couldn't say no to her favourite meal, but she had no desire to go back to the dentist. That's why she decided to eat the maximum amount of crucians from the portion but so that the viability of no tooth becomes negative.
As Valerie is not good at mathematics, she asked you to help her to find out the total amount of crucians that she can consume for dinner.
We should remind you that while eating one crucian Valerie uses exactly one row of teeth and the viability of each tooth from this row decreases by one.
Input
The first line contains three integers n, m, k (1 ≤ m ≤ n ≤ 1000, 0 ≤ k ≤ 106) — total amount of Valerie's teeth, amount of tooth rows and amount of crucians in Valerie's portion for dinner. Then follow n lines, each containing two integers: r (1 ≤ r ≤ m) — index of the row, where belongs the corresponding tooth, and c (0 ≤ c ≤ 106) — its residual viability.
It's guaranteed that each tooth row has positive amount of teeth.
Output
In the first line output the maximum amount of crucians that Valerie can consume for dinner.
Examples
Input
4 3 18
2 3
1 2
3 6
2 3
Output
11
Input
2 2 13
1 13
2 12
Output
13
Submitted Solution:
```
import sys
n,c,b = map(int,input().split())
m = [10000000]*c
mi= 0
for i in range(n):
s , k=map(int,sys.stdin.readline().split())
if m[s-1] > k:
m[s-1] = k
for i in range(c):
if m[i] != 10000000:
mi+=m[i]
if mi > b:
print(b)
else:
print(mi)
``` | instruction | 0 | 71,077 | 9 | 142,154 |
Yes | output | 1 | 71,077 | 9 | 142,155 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
In one little known, but very beautiful country called Waterland, lives a lovely shark Valerie. Like all the sharks, she has several rows of teeth, and feeds on crucians. One of Valerie's distinguishing features is that while eating one crucian she uses only one row of her teeth, the rest of the teeth are "relaxing".
For a long time our heroine had been searching the sea for crucians, but a great misfortune happened. Her teeth started to ache, and she had to see the local dentist, lobster Ashot. As a professional, Ashot quickly relieved Valerie from her toothache. Moreover, he managed to determine the cause of Valerie's developing caries (for what he was later nicknamed Cap).
It turned that Valerie eats too many crucians. To help Valerie avoid further reoccurrence of toothache, Ashot found for each Valerie's tooth its residual viability. Residual viability of a tooth is a value equal to the amount of crucians that Valerie can eat with this tooth. Every time Valerie eats a crucian, viability of all the teeth used for it will decrease by one. When the viability of at least one tooth becomes negative, the shark will have to see the dentist again.
Unhappy, Valerie came back home, where a portion of crucians was waiting for her. For sure, the shark couldn't say no to her favourite meal, but she had no desire to go back to the dentist. That's why she decided to eat the maximum amount of crucians from the portion but so that the viability of no tooth becomes negative.
As Valerie is not good at mathematics, she asked you to help her to find out the total amount of crucians that she can consume for dinner.
We should remind you that while eating one crucian Valerie uses exactly one row of teeth and the viability of each tooth from this row decreases by one.
Input
The first line contains three integers n, m, k (1 ≤ m ≤ n ≤ 1000, 0 ≤ k ≤ 106) — total amount of Valerie's teeth, amount of tooth rows and amount of crucians in Valerie's portion for dinner. Then follow n lines, each containing two integers: r (1 ≤ r ≤ m) — index of the row, where belongs the corresponding tooth, and c (0 ≤ c ≤ 106) — its residual viability.
It's guaranteed that each tooth row has positive amount of teeth.
Output
In the first line output the maximum amount of crucians that Valerie can consume for dinner.
Examples
Input
4 3 18
2 3
1 2
3 6
2 3
Output
11
Input
2 2 13
1 13
2 12
Output
13
Submitted Solution:
```
import sys
import math
n, m, k = [int(x) for x in (sys.stdin.readline()).split()]
t = [1000000] * (m)
for i in range(n):
r, c = [int(x) for x in (sys.stdin.readline()).split()]
if(c < t[r - 1]):
t[r - 1] = c
vsum = sum(t)
if(vsum <= k):
print(vsum)
else:
print(k)
``` | instruction | 0 | 71,078 | 9 | 142,156 |
Yes | output | 1 | 71,078 | 9 | 142,157 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
In one little known, but very beautiful country called Waterland, lives a lovely shark Valerie. Like all the sharks, she has several rows of teeth, and feeds on crucians. One of Valerie's distinguishing features is that while eating one crucian she uses only one row of her teeth, the rest of the teeth are "relaxing".
For a long time our heroine had been searching the sea for crucians, but a great misfortune happened. Her teeth started to ache, and she had to see the local dentist, lobster Ashot. As a professional, Ashot quickly relieved Valerie from her toothache. Moreover, he managed to determine the cause of Valerie's developing caries (for what he was later nicknamed Cap).
It turned that Valerie eats too many crucians. To help Valerie avoid further reoccurrence of toothache, Ashot found for each Valerie's tooth its residual viability. Residual viability of a tooth is a value equal to the amount of crucians that Valerie can eat with this tooth. Every time Valerie eats a crucian, viability of all the teeth used for it will decrease by one. When the viability of at least one tooth becomes negative, the shark will have to see the dentist again.
Unhappy, Valerie came back home, where a portion of crucians was waiting for her. For sure, the shark couldn't say no to her favourite meal, but she had no desire to go back to the dentist. That's why she decided to eat the maximum amount of crucians from the portion but so that the viability of no tooth becomes negative.
As Valerie is not good at mathematics, she asked you to help her to find out the total amount of crucians that she can consume for dinner.
We should remind you that while eating one crucian Valerie uses exactly one row of teeth and the viability of each tooth from this row decreases by one.
Input
The first line contains three integers n, m, k (1 ≤ m ≤ n ≤ 1000, 0 ≤ k ≤ 106) — total amount of Valerie's teeth, amount of tooth rows and amount of crucians in Valerie's portion for dinner. Then follow n lines, each containing two integers: r (1 ≤ r ≤ m) — index of the row, where belongs the corresponding tooth, and c (0 ≤ c ≤ 106) — its residual viability.
It's guaranteed that each tooth row has positive amount of teeth.
Output
In the first line output the maximum amount of crucians that Valerie can consume for dinner.
Examples
Input
4 3 18
2 3
1 2
3 6
2 3
Output
11
Input
2 2 13
1 13
2 12
Output
13
Submitted Solution:
```
s=input()
a=""
s=s[:len(s):2]
for i in sorted(s):
a+=i
print("+".join(a))
``` | instruction | 0 | 71,079 | 9 | 142,158 |
No | output | 1 | 71,079 | 9 | 142,159 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
In one little known, but very beautiful country called Waterland, lives a lovely shark Valerie. Like all the sharks, she has several rows of teeth, and feeds on crucians. One of Valerie's distinguishing features is that while eating one crucian she uses only one row of her teeth, the rest of the teeth are "relaxing".
For a long time our heroine had been searching the sea for crucians, but a great misfortune happened. Her teeth started to ache, and she had to see the local dentist, lobster Ashot. As a professional, Ashot quickly relieved Valerie from her toothache. Moreover, he managed to determine the cause of Valerie's developing caries (for what he was later nicknamed Cap).
It turned that Valerie eats too many crucians. To help Valerie avoid further reoccurrence of toothache, Ashot found for each Valerie's tooth its residual viability. Residual viability of a tooth is a value equal to the amount of crucians that Valerie can eat with this tooth. Every time Valerie eats a crucian, viability of all the teeth used for it will decrease by one. When the viability of at least one tooth becomes negative, the shark will have to see the dentist again.
Unhappy, Valerie came back home, where a portion of crucians was waiting for her. For sure, the shark couldn't say no to her favourite meal, but she had no desire to go back to the dentist. That's why she decided to eat the maximum amount of crucians from the portion but so that the viability of no tooth becomes negative.
As Valerie is not good at mathematics, she asked you to help her to find out the total amount of crucians that she can consume for dinner.
We should remind you that while eating one crucian Valerie uses exactly one row of teeth and the viability of each tooth from this row decreases by one.
Input
The first line contains three integers n, m, k (1 ≤ m ≤ n ≤ 1000, 0 ≤ k ≤ 106) — total amount of Valerie's teeth, amount of tooth rows and amount of crucians in Valerie's portion for dinner. Then follow n lines, each containing two integers: r (1 ≤ r ≤ m) — index of the row, where belongs the corresponding tooth, and c (0 ≤ c ≤ 106) — its residual viability.
It's guaranteed that each tooth row has positive amount of teeth.
Output
In the first line output the maximum amount of crucians that Valerie can consume for dinner.
Examples
Input
4 3 18
2 3
1 2
3 6
2 3
Output
11
Input
2 2 13
1 13
2 12
Output
13
Submitted Solution:
```
print("TRANGLE")
``` | instruction | 0 | 71,080 | 9 | 142,160 |
No | output | 1 | 71,080 | 9 | 142,161 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
In one little known, but very beautiful country called Waterland, lives a lovely shark Valerie. Like all the sharks, she has several rows of teeth, and feeds on crucians. One of Valerie's distinguishing features is that while eating one crucian she uses only one row of her teeth, the rest of the teeth are "relaxing".
For a long time our heroine had been searching the sea for crucians, but a great misfortune happened. Her teeth started to ache, and she had to see the local dentist, lobster Ashot. As a professional, Ashot quickly relieved Valerie from her toothache. Moreover, he managed to determine the cause of Valerie's developing caries (for what he was later nicknamed Cap).
It turned that Valerie eats too many crucians. To help Valerie avoid further reoccurrence of toothache, Ashot found for each Valerie's tooth its residual viability. Residual viability of a tooth is a value equal to the amount of crucians that Valerie can eat with this tooth. Every time Valerie eats a crucian, viability of all the teeth used for it will decrease by one. When the viability of at least one tooth becomes negative, the shark will have to see the dentist again.
Unhappy, Valerie came back home, where a portion of crucians was waiting for her. For sure, the shark couldn't say no to her favourite meal, but she had no desire to go back to the dentist. That's why she decided to eat the maximum amount of crucians from the portion but so that the viability of no tooth becomes negative.
As Valerie is not good at mathematics, she asked you to help her to find out the total amount of crucians that she can consume for dinner.
We should remind you that while eating one crucian Valerie uses exactly one row of teeth and the viability of each tooth from this row decreases by one.
Input
The first line contains three integers n, m, k (1 ≤ m ≤ n ≤ 1000, 0 ≤ k ≤ 106) — total amount of Valerie's teeth, amount of tooth rows and amount of crucians in Valerie's portion for dinner. Then follow n lines, each containing two integers: r (1 ≤ r ≤ m) — index of the row, where belongs the corresponding tooth, and c (0 ≤ c ≤ 106) — its residual viability.
It's guaranteed that each tooth row has positive amount of teeth.
Output
In the first line output the maximum amount of crucians that Valerie can consume for dinner.
Examples
Input
4 3 18
2 3
1 2
3 6
2 3
Output
11
Input
2 2 13
1 13
2 12
Output
13
Submitted Solution:
```
n, m, k = list(map(int, input().split(' ')))
my_dict = {}
while n > 0:
r, c = list(map(int, input().split(' ')))
key = r
value = c
if key in my_dict and my_dict[key] > value:
my_dict[key] = value
my_dict[key] = value
n -= 1
ans = 0
for key, value in my_dict.items():
ans += my_dict[key]
if ans > k:
print(k)
else:
print(ans)
``` | instruction | 0 | 71,081 | 9 | 142,162 |
No | output | 1 | 71,081 | 9 | 142,163 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
In one little known, but very beautiful country called Waterland, lives a lovely shark Valerie. Like all the sharks, she has several rows of teeth, and feeds on crucians. One of Valerie's distinguishing features is that while eating one crucian she uses only one row of her teeth, the rest of the teeth are "relaxing".
For a long time our heroine had been searching the sea for crucians, but a great misfortune happened. Her teeth started to ache, and she had to see the local dentist, lobster Ashot. As a professional, Ashot quickly relieved Valerie from her toothache. Moreover, he managed to determine the cause of Valerie's developing caries (for what he was later nicknamed Cap).
It turned that Valerie eats too many crucians. To help Valerie avoid further reoccurrence of toothache, Ashot found for each Valerie's tooth its residual viability. Residual viability of a tooth is a value equal to the amount of crucians that Valerie can eat with this tooth. Every time Valerie eats a crucian, viability of all the teeth used for it will decrease by one. When the viability of at least one tooth becomes negative, the shark will have to see the dentist again.
Unhappy, Valerie came back home, where a portion of crucians was waiting for her. For sure, the shark couldn't say no to her favourite meal, but she had no desire to go back to the dentist. That's why she decided to eat the maximum amount of crucians from the portion but so that the viability of no tooth becomes negative.
As Valerie is not good at mathematics, she asked you to help her to find out the total amount of crucians that she can consume for dinner.
We should remind you that while eating one crucian Valerie uses exactly one row of teeth and the viability of each tooth from this row decreases by one.
Input
The first line contains three integers n, m, k (1 ≤ m ≤ n ≤ 1000, 0 ≤ k ≤ 106) — total amount of Valerie's teeth, amount of tooth rows and amount of crucians in Valerie's portion for dinner. Then follow n lines, each containing two integers: r (1 ≤ r ≤ m) — index of the row, where belongs the corresponding tooth, and c (0 ≤ c ≤ 106) — its residual viability.
It's guaranteed that each tooth row has positive amount of teeth.
Output
In the first line output the maximum amount of crucians that Valerie can consume for dinner.
Examples
Input
4 3 18
2 3
1 2
3 6
2 3
Output
11
Input
2 2 13
1 13
2 12
Output
13
Submitted Solution:
```
n,m,k=input().split()
n=int(n)
m=int(m)
k=int(k)
l=[]
s=0
for i in range(n):
l.insert(i,input())
l=list(set(l))
for i in l:
a,b=i.split()
b=int(b)
s=s+b
if(s>k):
print(k)
else:
print(s)
``` | instruction | 0 | 71,082 | 9 | 142,164 |
No | output | 1 | 71,082 | 9 | 142,165 |
Provide a correct Python 3 solution for this coding contest problem.
F: Tea Party
Yun decided to hold a tea party at the company.
The shop sells $ N $ sets of bread, each containing $ A_1, A_2, A_3, \ dots, A_N $.
Yun decided to make a sandwich by combining two breads into a pair.
Yun-san is very careful, so I want to make sure that I don't have any leftover bread.
Calculate how many sandwiches you can make at most.
input
The first line is given the integer $ N $.
On the second line, $ N $ integers $ A_1, A_2, A_3, \ dots, A_N $ are given, separated by blanks.
output
Output the maximum number of sandwiches you can make. However, insert a line break at the end.
Constraint
* $ N $ is an integer greater than or equal to $ 1 $ and less than or equal to $ 100 $
* $ A_1, A_2, A_3, \ dots, A_N $ are integers between $ 1 $ and $ 100 $
Input example 1
Five
2 3 5 6 7
Output example 1
Ten
Buy the first, third, fourth, and fifth sets to get $ 20 $ in bread.
If you buy all the sets, you will end up with $ 23 $ of bread and you will have a surplus.
Input example 2
Four
3 5 6 8
Output example 2
11
Example
Input
5
2 3 5 6 7
Output
10 | instruction | 0 | 71,578 | 9 | 143,156 |
"Correct Solution:
```
N = int(input())
A = list(map(int, input().split()))
oc = 0
mi = 1e9
for a in A:
if a % 2:
oc += 1
mi = min(mi, a)
print((sum(A) - [0, mi][oc % 2]) // 2)
``` | output | 1 | 71,578 | 9 | 143,157 |
Provide a correct Python 3 solution for this coding contest problem.
F: Tea Party
Yun decided to hold a tea party at the company.
The shop sells $ N $ sets of bread, each containing $ A_1, A_2, A_3, \ dots, A_N $.
Yun decided to make a sandwich by combining two breads into a pair.
Yun-san is very careful, so I want to make sure that I don't have any leftover bread.
Calculate how many sandwiches you can make at most.
input
The first line is given the integer $ N $.
On the second line, $ N $ integers $ A_1, A_2, A_3, \ dots, A_N $ are given, separated by blanks.
output
Output the maximum number of sandwiches you can make. However, insert a line break at the end.
Constraint
* $ N $ is an integer greater than or equal to $ 1 $ and less than or equal to $ 100 $
* $ A_1, A_2, A_3, \ dots, A_N $ are integers between $ 1 $ and $ 100 $
Input example 1
Five
2 3 5 6 7
Output example 1
Ten
Buy the first, third, fourth, and fifth sets to get $ 20 $ in bread.
If you buy all the sets, you will end up with $ 23 $ of bread and you will have a surplus.
Input example 2
Four
3 5 6 8
Output example 2
11
Example
Input
5
2 3 5 6 7
Output
10 | instruction | 0 | 71,579 | 9 | 143,158 |
"Correct Solution:
```
N = int(input())
A = [int(x) for x in input().split()]
A.sort()
ans = sum(A)
for a in A:
if ans % 2 == 0 or a % 2 == 0:
continue
ans -= a
ans = ans//2
print(ans)
``` | output | 1 | 71,579 | 9 | 143,159 |
Provide a correct Python 3 solution for this coding contest problem.
F: Tea Party
Yun decided to hold a tea party at the company.
The shop sells $ N $ sets of bread, each containing $ A_1, A_2, A_3, \ dots, A_N $.
Yun decided to make a sandwich by combining two breads into a pair.
Yun-san is very careful, so I want to make sure that I don't have any leftover bread.
Calculate how many sandwiches you can make at most.
input
The first line is given the integer $ N $.
On the second line, $ N $ integers $ A_1, A_2, A_3, \ dots, A_N $ are given, separated by blanks.
output
Output the maximum number of sandwiches you can make. However, insert a line break at the end.
Constraint
* $ N $ is an integer greater than or equal to $ 1 $ and less than or equal to $ 100 $
* $ A_1, A_2, A_3, \ dots, A_N $ are integers between $ 1 $ and $ 100 $
Input example 1
Five
2 3 5 6 7
Output example 1
Ten
Buy the first, third, fourth, and fifth sets to get $ 20 $ in bread.
If you buy all the sets, you will end up with $ 23 $ of bread and you will have a surplus.
Input example 2
Four
3 5 6 8
Output example 2
11
Example
Input
5
2 3 5 6 7
Output
10 | instruction | 0 | 71,580 | 9 | 143,160 |
"Correct Solution:
```
n = int(input())
li = list(map(int,input().split()))
li.sort(reverse = True)
ki = 0
kifl = False
ans = 0
for a in li:
if a%2 == 0:
ans += a
else:
if(kifl == True):
ans += ki+a
kifl = False
else:
ki = a
kifl = True
print(ans//2)
``` | output | 1 | 71,580 | 9 | 143,161 |
Provide a correct Python 3 solution for this coding contest problem.
F: Tea Party
Yun decided to hold a tea party at the company.
The shop sells $ N $ sets of bread, each containing $ A_1, A_2, A_3, \ dots, A_N $.
Yun decided to make a sandwich by combining two breads into a pair.
Yun-san is very careful, so I want to make sure that I don't have any leftover bread.
Calculate how many sandwiches you can make at most.
input
The first line is given the integer $ N $.
On the second line, $ N $ integers $ A_1, A_2, A_3, \ dots, A_N $ are given, separated by blanks.
output
Output the maximum number of sandwiches you can make. However, insert a line break at the end.
Constraint
* $ N $ is an integer greater than or equal to $ 1 $ and less than or equal to $ 100 $
* $ A_1, A_2, A_3, \ dots, A_N $ are integers between $ 1 $ and $ 100 $
Input example 1
Five
2 3 5 6 7
Output example 1
Ten
Buy the first, third, fourth, and fifth sets to get $ 20 $ in bread.
If you buy all the sets, you will end up with $ 23 $ of bread and you will have a surplus.
Input example 2
Four
3 5 6 8
Output example 2
11
Example
Input
5
2 3 5 6 7
Output
10 | instruction | 0 | 71,581 | 9 | 143,162 |
"Correct Solution:
```
n = int(input())
a = list(map(int, input().split()))
r = sum(a)
if r % 2 == 0:
print(r // 2)
else:
a.sort()
for j in a:
if (r - j) % 2 == 0:
print((r-j) // 2)
break
``` | output | 1 | 71,581 | 9 | 143,163 |
Provide a correct Python 3 solution for this coding contest problem.
F: Tea Party
Yun decided to hold a tea party at the company.
The shop sells $ N $ sets of bread, each containing $ A_1, A_2, A_3, \ dots, A_N $.
Yun decided to make a sandwich by combining two breads into a pair.
Yun-san is very careful, so I want to make sure that I don't have any leftover bread.
Calculate how many sandwiches you can make at most.
input
The first line is given the integer $ N $.
On the second line, $ N $ integers $ A_1, A_2, A_3, \ dots, A_N $ are given, separated by blanks.
output
Output the maximum number of sandwiches you can make. However, insert a line break at the end.
Constraint
* $ N $ is an integer greater than or equal to $ 1 $ and less than or equal to $ 100 $
* $ A_1, A_2, A_3, \ dots, A_N $ are integers between $ 1 $ and $ 100 $
Input example 1
Five
2 3 5 6 7
Output example 1
Ten
Buy the first, third, fourth, and fifth sets to get $ 20 $ in bread.
If you buy all the sets, you will end up with $ 23 $ of bread and you will have a surplus.
Input example 2
Four
3 5 6 8
Output example 2
11
Example
Input
5
2 3 5 6 7
Output
10 | instruction | 0 | 71,582 | 9 | 143,164 |
"Correct Solution:
```
n = int(input())
a = list(map(int,input().split()))
if sum(a) % 2 == 0:
print(sum(a) // 2)
else:
a.sort()
mi = 10000
for i in a:
if i % 2 == 1:
mi = i
break
print((sum(a) - mi) // 2)
``` | output | 1 | 71,582 | 9 | 143,165 |
Provide a correct Python 3 solution for this coding contest problem.
F: Tea Party
Yun decided to hold a tea party at the company.
The shop sells $ N $ sets of bread, each containing $ A_1, A_2, A_3, \ dots, A_N $.
Yun decided to make a sandwich by combining two breads into a pair.
Yun-san is very careful, so I want to make sure that I don't have any leftover bread.
Calculate how many sandwiches you can make at most.
input
The first line is given the integer $ N $.
On the second line, $ N $ integers $ A_1, A_2, A_3, \ dots, A_N $ are given, separated by blanks.
output
Output the maximum number of sandwiches you can make. However, insert a line break at the end.
Constraint
* $ N $ is an integer greater than or equal to $ 1 $ and less than or equal to $ 100 $
* $ A_1, A_2, A_3, \ dots, A_N $ are integers between $ 1 $ and $ 100 $
Input example 1
Five
2 3 5 6 7
Output example 1
Ten
Buy the first, third, fourth, and fifth sets to get $ 20 $ in bread.
If you buy all the sets, you will end up with $ 23 $ of bread and you will have a surplus.
Input example 2
Four
3 5 6 8
Output example 2
11
Example
Input
5
2 3 5 6 7
Output
10 | instruction | 0 | 71,583 | 9 | 143,166 |
"Correct Solution:
```
N=int(input())
A=list(map(int,input().split()))
A.sort()
if sum(A)%2==0:
print(sum(A)//2)
else:
for i in range(N):
if A[i]%2==1:
break
print((sum(A)-A[i])//2)
``` | output | 1 | 71,583 | 9 | 143,167 |
Provide a correct Python 3 solution for this coding contest problem.
F: Tea Party
Yun decided to hold a tea party at the company.
The shop sells $ N $ sets of bread, each containing $ A_1, A_2, A_3, \ dots, A_N $.
Yun decided to make a sandwich by combining two breads into a pair.
Yun-san is very careful, so I want to make sure that I don't have any leftover bread.
Calculate how many sandwiches you can make at most.
input
The first line is given the integer $ N $.
On the second line, $ N $ integers $ A_1, A_2, A_3, \ dots, A_N $ are given, separated by blanks.
output
Output the maximum number of sandwiches you can make. However, insert a line break at the end.
Constraint
* $ N $ is an integer greater than or equal to $ 1 $ and less than or equal to $ 100 $
* $ A_1, A_2, A_3, \ dots, A_N $ are integers between $ 1 $ and $ 100 $
Input example 1
Five
2 3 5 6 7
Output example 1
Ten
Buy the first, third, fourth, and fifth sets to get $ 20 $ in bread.
If you buy all the sets, you will end up with $ 23 $ of bread and you will have a surplus.
Input example 2
Four
3 5 6 8
Output example 2
11
Example
Input
5
2 3 5 6 7
Output
10 | instruction | 0 | 71,584 | 9 | 143,168 |
"Correct Solution:
```
_ = input()
A = [int(x) for x in input().split()]
S = sum(A)
if S % 2 == 0:
print(S // 2)
else:
print((S - min(x for x in A if x % 2)) // 2)
``` | output | 1 | 71,584 | 9 | 143,169 |
Provide a correct Python 3 solution for this coding contest problem.
F: Tea Party
Yun decided to hold a tea party at the company.
The shop sells $ N $ sets of bread, each containing $ A_1, A_2, A_3, \ dots, A_N $.
Yun decided to make a sandwich by combining two breads into a pair.
Yun-san is very careful, so I want to make sure that I don't have any leftover bread.
Calculate how many sandwiches you can make at most.
input
The first line is given the integer $ N $.
On the second line, $ N $ integers $ A_1, A_2, A_3, \ dots, A_N $ are given, separated by blanks.
output
Output the maximum number of sandwiches you can make. However, insert a line break at the end.
Constraint
* $ N $ is an integer greater than or equal to $ 1 $ and less than or equal to $ 100 $
* $ A_1, A_2, A_3, \ dots, A_N $ are integers between $ 1 $ and $ 100 $
Input example 1
Five
2 3 5 6 7
Output example 1
Ten
Buy the first, third, fourth, and fifth sets to get $ 20 $ in bread.
If you buy all the sets, you will end up with $ 23 $ of bread and you will have a surplus.
Input example 2
Four
3 5 6 8
Output example 2
11
Example
Input
5
2 3 5 6 7
Output
10 | instruction | 0 | 71,585 | 9 | 143,170 |
"Correct Solution:
```
n = int(input())
li = list(map(int,input().split()))
li.sort(reverse = True)
ans = acc = 0
for x in li:
if x%2 == 0:
ans += x//2
elif acc != 0:
ans += (acc+x)//2
acc = 0
else:
acc = x
print(ans)
``` | output | 1 | 71,585 | 9 | 143,171 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
F: Tea Party
Yun decided to hold a tea party at the company.
The shop sells $ N $ sets of bread, each containing $ A_1, A_2, A_3, \ dots, A_N $.
Yun decided to make a sandwich by combining two breads into a pair.
Yun-san is very careful, so I want to make sure that I don't have any leftover bread.
Calculate how many sandwiches you can make at most.
input
The first line is given the integer $ N $.
On the second line, $ N $ integers $ A_1, A_2, A_3, \ dots, A_N $ are given, separated by blanks.
output
Output the maximum number of sandwiches you can make. However, insert a line break at the end.
Constraint
* $ N $ is an integer greater than or equal to $ 1 $ and less than or equal to $ 100 $
* $ A_1, A_2, A_3, \ dots, A_N $ are integers between $ 1 $ and $ 100 $
Input example 1
Five
2 3 5 6 7
Output example 1
Ten
Buy the first, third, fourth, and fifth sets to get $ 20 $ in bread.
If you buy all the sets, you will end up with $ 23 $ of bread and you will have a surplus.
Input example 2
Four
3 5 6 8
Output example 2
11
Example
Input
5
2 3 5 6 7
Output
10
Submitted Solution:
```
n=int(input())
a=list(map(int,input().split()))
if sum(a)%2==0:
print(sum(a)//2)
else:
a.sort()
for i in a:
if i%2==1:
print((sum(a)-i)//2)
break
``` | instruction | 0 | 71,586 | 9 | 143,172 |
Yes | output | 1 | 71,586 | 9 | 143,173 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
F: Tea Party
Yun decided to hold a tea party at the company.
The shop sells $ N $ sets of bread, each containing $ A_1, A_2, A_3, \ dots, A_N $.
Yun decided to make a sandwich by combining two breads into a pair.
Yun-san is very careful, so I want to make sure that I don't have any leftover bread.
Calculate how many sandwiches you can make at most.
input
The first line is given the integer $ N $.
On the second line, $ N $ integers $ A_1, A_2, A_3, \ dots, A_N $ are given, separated by blanks.
output
Output the maximum number of sandwiches you can make. However, insert a line break at the end.
Constraint
* $ N $ is an integer greater than or equal to $ 1 $ and less than or equal to $ 100 $
* $ A_1, A_2, A_3, \ dots, A_N $ are integers between $ 1 $ and $ 100 $
Input example 1
Five
2 3 5 6 7
Output example 1
Ten
Buy the first, third, fourth, and fifth sets to get $ 20 $ in bread.
If you buy all the sets, you will end up with $ 23 $ of bread and you will have a surplus.
Input example 2
Four
3 5 6 8
Output example 2
11
Example
Input
5
2 3 5 6 7
Output
10
Submitted Solution:
```
n = int(input())
a = list(map(int,input().split()))
o = []
ans = 0
for i in a:
if i%2==1:
o.append(i)
else:
ans += i//2
o.sort()
o.reverse()
for i in range(0,len(o)-1,2):
b = o[i]+o[i+1]
ans += b//2
print(ans)
``` | instruction | 0 | 71,587 | 9 | 143,174 |
Yes | output | 1 | 71,587 | 9 | 143,175 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
F: Tea Party
Yun decided to hold a tea party at the company.
The shop sells $ N $ sets of bread, each containing $ A_1, A_2, A_3, \ dots, A_N $.
Yun decided to make a sandwich by combining two breads into a pair.
Yun-san is very careful, so I want to make sure that I don't have any leftover bread.
Calculate how many sandwiches you can make at most.
input
The first line is given the integer $ N $.
On the second line, $ N $ integers $ A_1, A_2, A_3, \ dots, A_N $ are given, separated by blanks.
output
Output the maximum number of sandwiches you can make. However, insert a line break at the end.
Constraint
* $ N $ is an integer greater than or equal to $ 1 $ and less than or equal to $ 100 $
* $ A_1, A_2, A_3, \ dots, A_N $ are integers between $ 1 $ and $ 100 $
Input example 1
Five
2 3 5 6 7
Output example 1
Ten
Buy the first, third, fourth, and fifth sets to get $ 20 $ in bread.
If you buy all the sets, you will end up with $ 23 $ of bread and you will have a surplus.
Input example 2
Four
3 5 6 8
Output example 2
11
Example
Input
5
2 3 5 6 7
Output
10
Submitted Solution:
```
n = input()
x = [int(i) for i in input().split()]
if sum(x) % 2 == 1:
y = [i for i in x if i%2 == 1]
print(int((sum(x) - min(y)) / 2))
else:
print(int(sum(x)/2))
``` | instruction | 0 | 71,588 | 9 | 143,176 |
Yes | output | 1 | 71,588 | 9 | 143,177 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
F: Tea Party
Yun decided to hold a tea party at the company.
The shop sells $ N $ sets of bread, each containing $ A_1, A_2, A_3, \ dots, A_N $.
Yun decided to make a sandwich by combining two breads into a pair.
Yun-san is very careful, so I want to make sure that I don't have any leftover bread.
Calculate how many sandwiches you can make at most.
input
The first line is given the integer $ N $.
On the second line, $ N $ integers $ A_1, A_2, A_3, \ dots, A_N $ are given, separated by blanks.
output
Output the maximum number of sandwiches you can make. However, insert a line break at the end.
Constraint
* $ N $ is an integer greater than or equal to $ 1 $ and less than or equal to $ 100 $
* $ A_1, A_2, A_3, \ dots, A_N $ are integers between $ 1 $ and $ 100 $
Input example 1
Five
2 3 5 6 7
Output example 1
Ten
Buy the first, third, fourth, and fifth sets to get $ 20 $ in bread.
If you buy all the sets, you will end up with $ 23 $ of bread and you will have a surplus.
Input example 2
Four
3 5 6 8
Output example 2
11
Example
Input
5
2 3 5 6 7
Output
10
Submitted Solution:
```
n=int(input())
a=[int(x) for x in input().split()]
if sum(a)%2==0:
print(sum(a)//2)
else:
s=0
a.sort()
for i in a:
if i%2==1:
s=i
break
print((sum(a)-s)//2)
``` | instruction | 0 | 71,589 | 9 | 143,178 |
Yes | output | 1 | 71,589 | 9 | 143,179 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Automatic Bakery of Cyberland (ABC) recently bought an n × m rectangle table. To serve the diners, ABC placed seats around the table. The size of each seat is equal to a unit square, so there are 2(n + m) seats in total.
ABC placed conveyor belts on each unit square on the table. There are three types of conveyor belts: "^", "<" and ">". A "^" belt can bring things upwards. "<" can bring leftwards and ">" can bring rightwards.
Let's number the rows with 1 to n from top to bottom, the columns with 1 to m from left to right. We consider the seats above and below the top of the table are rows 0 and n + 1 respectively. Also we define seats to the left of the table and to the right of the table to be column 0 and m + 1. Due to the conveyor belts direction restriction there are currently no way for a diner sitting in the row n + 1 to be served.
Given the initial table, there will be q events in order. There are two types of events:
* "A x y" means, a piece of bread will appear at row x and column y (we will denote such position as (x, y)). The bread will follow the conveyor belt, until arriving at a seat of a diner. It is possible that the bread gets stuck in an infinite loop. Your task is to simulate the process, and output the final position of the bread, or determine that there will be an infinite loop.
* "C x y c" means that the type of the conveyor belt at (x, y) is changed to c.
Queries are performed separately meaning that even if the bread got stuck in an infinite loop, it won't affect further queries.
Input
The first line of input contains three integers n, m and q (1 ≤ n ≤ 105, 1 ≤ m ≤ 10, 1 ≤ q ≤ 105), separated by a space.
Next n lines, each line contains m characters, describing the table. The characters can only be one of "<^>".
Next q lines, each line describes an event. The format is "C x y c" or "A x y" (Consecutive elements are separated by a space). It's guaranteed that 1 ≤ x ≤ n, 1 ≤ y ≤ m. c is a character from the set "<^>".
There are at most 10000 queries of "C" type.
Output
For each event of type "A", output two integers tx, ty in a line, separated by a space, denoting the destination of (x, y) is (tx, ty).
If there is an infinite loop, you should output tx = ty = - 1.
Examples
Input
2 2 3
>>
^^
A 2 1
C 1 2 <
A 2 1
Output
1 3
-1 -1
Input
4 5 7
><<^<
^<^^>
>>>^>
>^>>^
A 3 1
A 2 2
C 1 4 <
A 3 1
C 1 2 ^
A 3 1
A 2 2
Output
0 4
-1 -1
-1 -1
0 2
0 2
Note
For the first sample:
If the bread goes from (2, 1), it will go out of the table at (1, 3).
After changing the conveyor belt of (1, 2) to "<", when the bread goes from (2, 1) again, it will get stuck at "><", so output is ( - 1, - 1).
Submitted Solution:
```
from math import ceil, sqrt
n, m, qq = map(int, input().split())
nCell = ceil(sqrt(n))
# n x m.
# idx = m * i + j
# because no 'down', infinite loop means ><
# horizontal stripes works because I don't have to wory aboyt >< situations going up/down
# m << n s n, m and q (1 ≤ n ≤ 10^5, 1 ≤ m ≤ 10, 1 ≤ q ≤ 10^5)
# sqrt decomp by n. sqrt(n) >> sqrt(m), means a bigger speedup this way
class table_sector:
up = (-1, 0)
left = (0, -1)
right = (0, 1)
walking = (up, left, right)
def __init__(self, xStart, width, height):
self.xStart = xStart
self.width = width
self.height = height
self.grid = [[None] * width] * self.height
def walk(self, x, y, table):
"""
Walk the sector
param x: global table row param
param y: global table col param
"""
local_x = x % self.height
if self.grid[local_x][y]:
return self.grid[local_x][y]
startX, posX, startY, posY = local_x, local_x, y, y
idx = 0
while True:
delta = table_sector.walking[table[posX + self.xStart][posY]]
posX = posX + delta[0]
posY = posY + delta[1]
if posX == startX and posY == startY:
self.grid[local_x][y] = (-2, -2)
return -2, -2
if posX == -1 or posY == -1 or posY == self.width:
self.grid[local_x][y] = (posX + self.xStart, posY)
return self.grid[local_x][y]
idx += 1
if idx == 2:
startX, startY = posX, posY
idx = 0
table = []
def translate_op(k):
return 0 if k == '^' else 1 if k == '<' else 2
for i in range(n):
row = list(map(lambda k: translate_op(k), input()))
table.append(row)
sectors = {}
def update_query(query):
global sectors, table, nCell, n, m
x = int(query[1]) - 1
y = int(query[2]) - 1
idxSector = x // nCell
nameSector = str(idxSector)
# pop a sector, the sector busted now
sectors.pop(nameSector, None)
table[x][y] = translate_op(query[3])
def do_query(x, y):
global sectors, table, nCell, n, m
while True:
idxSector = x // nCell
nameSector = str(idxSector)
if not nameSector in sectors:
sectors[nameSector] = table_sector(idxSector * nCell, m, nCell)
pos = sectors[nameSector].walk(x, y, table)
if pos == (-2, -2):
print("-1 -1")
return
if (idxSector == 0 and pos[0] == -1) or pos[1] == -1 or pos[1] == n:
print(pos[0] + 1, pos[1] + 1)
return
x = pos[0]
y = pos[1]
def flush_queries(queries):
queries.sort(key=lambda x:int(x[1]))
for query in queries:
x = int(query[1]) - 1
y = int(query[2]) - 1
do_query(x, y)
queries = []
for q in range(qq):
query = input().split()
if query[0] == 'C':
flush_queries(queries)
queries = []
update_query(query)
continue
queries.append(query)
flush_queries(queries)
queries = []
``` | instruction | 0 | 72,039 | 9 | 144,078 |
No | output | 1 | 72,039 | 9 | 144,079 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A teacher decides to give toffees to his students. He asks n students to stand in a queue. Since the teacher is very partial, he follows the following rule to distribute toffees.
He looks at the first two students and gives more toffees to the student having higher marks than the other one. If they have the same marks they get the same number of toffees. The same procedure is followed for each pair of adjacent students starting from the first one to the last one.
It is given that each student receives at least one toffee. You have to find the number of toffees given to each student by the teacher such that the total number of toffees is minimum.
Input
The first line of input contains the number of students n (2 ≤ n ≤ 1000). The second line gives (n - 1) characters consisting of "L", "R" and "=". For each pair of adjacent students "L" means that the left student has higher marks, "R" means that the right student has higher marks and "=" means that both have equal marks.
Output
Output consists of n integers separated by a space representing the number of toffees each student receives in the queue starting from the first one to the last one.
Examples
Input
5
LRLR
Output
2 1 2 1 2
Input
5
=RRR
Output
1 1 2 3 4
Submitted Solution:
```
n = int(input())
s = input()
ans = [0 for i in range(n)]
for i in range(n):
min_val = 1
if i > 0:
if s[i - 1] == '=':
ans[i] = ans[i - 1]
continue
if s[i - 1] == 'R':
min_val = ans[i - 1] + 1
cnt_low = 0
for j in range(i, n - 1):
if s[j] == '=':
continue
if s[j] == 'L':
cnt_low += 1
if s[j] == 'R':
break
min_val = max(min_val, cnt_low + 1)
ans[i] = min_val
print(*ans)
``` | instruction | 0 | 72,129 | 9 | 144,258 |
Yes | output | 1 | 72,129 | 9 | 144,259 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A teacher decides to give toffees to his students. He asks n students to stand in a queue. Since the teacher is very partial, he follows the following rule to distribute toffees.
He looks at the first two students and gives more toffees to the student having higher marks than the other one. If they have the same marks they get the same number of toffees. The same procedure is followed for each pair of adjacent students starting from the first one to the last one.
It is given that each student receives at least one toffee. You have to find the number of toffees given to each student by the teacher such that the total number of toffees is minimum.
Input
The first line of input contains the number of students n (2 ≤ n ≤ 1000). The second line gives (n - 1) characters consisting of "L", "R" and "=". For each pair of adjacent students "L" means that the left student has higher marks, "R" means that the right student has higher marks and "=" means that both have equal marks.
Output
Output consists of n integers separated by a space representing the number of toffees each student receives in the queue starting from the first one to the last one.
Examples
Input
5
LRLR
Output
2 1 2 1 2
Input
5
=RRR
Output
1 1 2 3 4
Submitted Solution:
```
from collections import deque
n = int(input())
s = input()
ans = [1]*n
for i in range(1,n):
if s[i-1]=='R':
ans[i]=ans[i-1]+1
elif s[i-1]=='=':
ans[i]=ans[i-1]
for i in range(n-2, -1, -1):
if s[i]=='L':
ans[i]=max(ans[i+1]+1, ans[i])
elif s[i]=='=':
ans[i]=max(ans[i], ans[i+1])
print(*ans)
``` | instruction | 0 | 72,130 | 9 | 144,260 |
Yes | output | 1 | 72,130 | 9 | 144,261 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A teacher decides to give toffees to his students. He asks n students to stand in a queue. Since the teacher is very partial, he follows the following rule to distribute toffees.
He looks at the first two students and gives more toffees to the student having higher marks than the other one. If they have the same marks they get the same number of toffees. The same procedure is followed for each pair of adjacent students starting from the first one to the last one.
It is given that each student receives at least one toffee. You have to find the number of toffees given to each student by the teacher such that the total number of toffees is minimum.
Input
The first line of input contains the number of students n (2 ≤ n ≤ 1000). The second line gives (n - 1) characters consisting of "L", "R" and "=". For each pair of adjacent students "L" means that the left student has higher marks, "R" means that the right student has higher marks and "=" means that both have equal marks.
Output
Output consists of n integers separated by a space representing the number of toffees each student receives in the queue starting from the first one to the last one.
Examples
Input
5
LRLR
Output
2 1 2 1 2
Input
5
=RRR
Output
1 1 2 3 4
Submitted Solution:
```
import copy
n = int(input())
difference = input()
answer = [0 for i in range(n)]
def is_local_min(i, difference):
for j in range(i - 1, -1, -1):
if difference[j] == 'R':
return False
if difference[j] == 'L':
break
for j in difference[i:]:
if j == 'L':
return False
if j == 'R':
break
return True
loc_min_lst = []
for i in range(n):
if is_local_min(i, difference):
loc_min_lst.append(i)
value = 0
for i in loc_min_lst:
value = 1
for j in range(i, -1, -1):
if j == 0:
answer[0] = value
value = 1
break
if answer[j] >= value:
break
else:
answer[j] = value
if difference[j - 1] == 'R':
break
if difference[j - 1] == 'L':
value += 1
value = 1
for j in range(i, n):
if j == n - 1:
answer[n - 1] = value
value = 1
break
if answer[j] > value:
break
else:
answer[j] = value
if difference[j] == 'L':
break
if difference[j] == 'R':
value += 1
print(' '.join([str(x) for x in answer]))
``` | instruction | 0 | 72,132 | 9 | 144,264 |
Yes | output | 1 | 72,132 | 9 | 144,265 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A teacher decides to give toffees to his students. He asks n students to stand in a queue. Since the teacher is very partial, he follows the following rule to distribute toffees.
He looks at the first two students and gives more toffees to the student having higher marks than the other one. If they have the same marks they get the same number of toffees. The same procedure is followed for each pair of adjacent students starting from the first one to the last one.
It is given that each student receives at least one toffee. You have to find the number of toffees given to each student by the teacher such that the total number of toffees is minimum.
Input
The first line of input contains the number of students n (2 ≤ n ≤ 1000). The second line gives (n - 1) characters consisting of "L", "R" and "=". For each pair of adjacent students "L" means that the left student has higher marks, "R" means that the right student has higher marks and "=" means that both have equal marks.
Output
Output consists of n integers separated by a space representing the number of toffees each student receives in the queue starting from the first one to the last one.
Examples
Input
5
LRLR
Output
2 1 2 1 2
Input
5
=RRR
Output
1 1 2 3 4
Submitted Solution:
```
from collections import deque
n = int(input())
s = input()
q = deque()
candidate = True
vis,ans = [0]*n, [0]*n
for i in range(n-1):
if s[i]=='R':
if candidate:
q.append((i, 1))
vis[i]=1
candidate = False
elif s[i]=='L':
candidate = True
if candidate:
q.append((n-1, 1))
vis[n-1]=1
while q:
i,t = q.popleft()
ans[i]=t
if i>0 and vis[i-1]==0:
if s[i-1]=='L':
q.append((i-1,t+1))
else:
q.appendleft((i-1,t))
vis[i-1]=1
if i+1<n and vis[i+1]==0:
if s[i]=='R':
q.append((i+1,t+1))
else:
q.appendleft((i+1,t))
vis[i+1]=1
print(*ans)
``` | instruction | 0 | 72,133 | 9 | 144,266 |
No | output | 1 | 72,133 | 9 | 144,267 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A teacher decides to give toffees to his students. He asks n students to stand in a queue. Since the teacher is very partial, he follows the following rule to distribute toffees.
He looks at the first two students and gives more toffees to the student having higher marks than the other one. If they have the same marks they get the same number of toffees. The same procedure is followed for each pair of adjacent students starting from the first one to the last one.
It is given that each student receives at least one toffee. You have to find the number of toffees given to each student by the teacher such that the total number of toffees is minimum.
Input
The first line of input contains the number of students n (2 ≤ n ≤ 1000). The second line gives (n - 1) characters consisting of "L", "R" and "=". For each pair of adjacent students "L" means that the left student has higher marks, "R" means that the right student has higher marks and "=" means that both have equal marks.
Output
Output consists of n integers separated by a space representing the number of toffees each student receives in the queue starting from the first one to the last one.
Examples
Input
5
LRLR
Output
2 1 2 1 2
Input
5
=RRR
Output
1 1 2 3 4
Submitted Solution:
```
n = int(input())
a = list(input())
ans = []
num = 0
flag = False
if a[0] == 'L':
flag = True
for i in range(n-1):
if a[i] == 'L':
flag = True
if flag:
if a[i] == 'L' or a[i] == '=':
num += 1
else:
flag = False
if num != 0:
ans.append([i-num, num])
num = 0
if flag:
if num != 0:
ans.append([n-1-num, num])
answer = [1 for i in range(n)]
for i in ans:
y = 0
if i[0]+i[1] < n-1:
y = 1
x = i[1] + y
for j in range(i[0], i[0]+i[1]):
answer[j] = x
x -= 1
j = 1
for i in a:
if i == 'R':
answer[j] = answer[j-1] + 1
if i == '=':
answer[j] = answer[j-1]
#if i == 'L':
# answer[j] = answer[j-1] - 1
j += 1
print(*answer)
'''
5
LRLR
1 2 1 2 1 1
'''
``` | instruction | 0 | 72,134 | 9 | 144,268 |
No | output | 1 | 72,134 | 9 | 144,269 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A teacher decides to give toffees to his students. He asks n students to stand in a queue. Since the teacher is very partial, he follows the following rule to distribute toffees.
He looks at the first two students and gives more toffees to the student having higher marks than the other one. If they have the same marks they get the same number of toffees. The same procedure is followed for each pair of adjacent students starting from the first one to the last one.
It is given that each student receives at least one toffee. You have to find the number of toffees given to each student by the teacher such that the total number of toffees is minimum.
Input
The first line of input contains the number of students n (2 ≤ n ≤ 1000). The second line gives (n - 1) characters consisting of "L", "R" and "=". For each pair of adjacent students "L" means that the left student has higher marks, "R" means that the right student has higher marks and "=" means that both have equal marks.
Output
Output consists of n integers separated by a space representing the number of toffees each student receives in the queue starting from the first one to the last one.
Examples
Input
5
LRLR
Output
2 1 2 1 2
Input
5
=RRR
Output
1 1 2 3 4
Submitted Solution:
```
n = int(input())
s = input()
now = 1
if s[0] == 'L':
now += 1
print(now, end=' ')
for c in s:
if c == 'R':
now += 1
elif c == 'L':
now = 1
print(now, end=' ')
``` | instruction | 0 | 72,135 | 9 | 144,270 |
No | output | 1 | 72,135 | 9 | 144,271 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A teacher decides to give toffees to his students. He asks n students to stand in a queue. Since the teacher is very partial, he follows the following rule to distribute toffees.
He looks at the first two students and gives more toffees to the student having higher marks than the other one. If they have the same marks they get the same number of toffees. The same procedure is followed for each pair of adjacent students starting from the first one to the last one.
It is given that each student receives at least one toffee. You have to find the number of toffees given to each student by the teacher such that the total number of toffees is minimum.
Input
The first line of input contains the number of students n (2 ≤ n ≤ 1000). The second line gives (n - 1) characters consisting of "L", "R" and "=". For each pair of adjacent students "L" means that the left student has higher marks, "R" means that the right student has higher marks and "=" means that both have equal marks.
Output
Output consists of n integers separated by a space representing the number of toffees each student receives in the queue starting from the first one to the last one.
Examples
Input
5
LRLR
Output
2 1 2 1 2
Input
5
=RRR
Output
1 1 2 3 4
Submitted Solution:
```
size = int(input())
adj = [0, 0]
first = True
res = ''
students = input()
for i in students:
if first:
if i=='=':
adj = [1, 1]
else:
if i=='L':
adj = [2, 1]
else:
adj = [1, 2]
first=False
res += str(adj[0]) + ' ' + str(adj[1]) + ' '
else:
if i=='=':
adj = [adj[-1], adj[-1]]
elif i== 'L':
x = max(1, adj[-1]-1)
adj = [adj[-1], x]
else:
adj = [adj[-1], adj[-1] + 1]
res+=str(adj[-1]) + ' '
print(res.strip())
``` | instruction | 0 | 72,136 | 9 | 144,272 |
No | output | 1 | 72,136 | 9 | 144,273 |
Provide a correct Python 3 solution for this coding contest problem.
problem
Chairman K is a regular customer of the JOI pizza shop in the center of JOI city. For some reason, he decided to start a life-saving life this month. So he wanted to order the pizza with the highest calories per dollar among the pizzas he could order at the JOI pizza store. Let's call such a pizza the "best pizza". The "best pizza" is not limited to one type.
At JOI Pizza, you can freely choose from N types of toppings and order the ones placed on the basic dough. You cannot put more than one topping of the same type. You can also order a pizza that doesn't have any toppings on the dough. The price of the dough is $ A and the price of the toppings is $ B. The price of pizza is the sum of the price of the dough and the price of the toppings. That is, the price of a pizza with k types of toppings (0 ≤ k ≤ N) is A + k x B dollars. The total calorie of the pizza is the sum of the calories of the dough and the calories of the toppings placed.
Create a program to find the number of calories per dollar for the "best pizza" given the price of the dough and the price of the toppings, and the calorie value of the dough and each topping.
input
The input consists of N + 3 lines.
On the first line, one integer N (1 ≤ N ≤ 100) representing the number of topping types is written. On the second line, two integers A and B (1 ≤ A ≤ 1000, 1 ≤ B ≤ 1000) are written with a blank as a delimiter. A is the price of the dough and B is the price of the toppings. On the third line, one integer C (1 ≤ C ≤ 10000) representing the number of calories in the dough is written.
On the 3 + i line (1 ≤ i ≤ N), one integer Di (1 ≤ Di ≤ 10000) representing the number of calories in the i-th topping is written.
output
Print the number of calories per dollar for the "best pizza" in one line. However, round down the numbers after the decimal point and output as an integer value.
Input / output example
Input example 1
3
12 2
200
50
300
100
Output example 1
37
In I / O Example 1, with the second and third toppings, 200 + 300 + 100 = 600 calories gives a pizza of $ 12 + 2 x 2 = $ 16.
This pizza has 600/16 = 37.5 calories per dollar. Since this is the "best pizza", we output 37, rounded down to the nearest whole number of 37.5.
Input example 2
Four
20 3
900
300
100
400
1300
Output example 2
100
The question text and the data used for the automatic referee are the question text and the test data for scoring, which are created and published by the Japan Committee for Information Olympics.
Example
Input
3
12 2
200
50
300
100
Output
37 | instruction | 0 | 73,227 | 9 | 146,454 |
"Correct Solution:
```
#C
N = int(input())
A,B = map(int,input().split())
C = int(input())
T = [int(input()) for i in range(N)]
T.sort(reverse=True)
cal = C
cost = A
for t in T:
if cal/cost < (cal+t)/(cost+B):
cal+=t
cost+=B
else:
break
print(cal//cost)
``` | output | 1 | 73,227 | 9 | 146,455 |
Provide a correct Python 3 solution for this coding contest problem.
problem
Chairman K is a regular customer of the JOI pizza shop in the center of JOI city. For some reason, he decided to start a life-saving life this month. So he wanted to order the pizza with the highest calories per dollar among the pizzas he could order at the JOI pizza store. Let's call such a pizza the "best pizza". The "best pizza" is not limited to one type.
At JOI Pizza, you can freely choose from N types of toppings and order the ones placed on the basic dough. You cannot put more than one topping of the same type. You can also order a pizza that doesn't have any toppings on the dough. The price of the dough is $ A and the price of the toppings is $ B. The price of pizza is the sum of the price of the dough and the price of the toppings. That is, the price of a pizza with k types of toppings (0 ≤ k ≤ N) is A + k x B dollars. The total calorie of the pizza is the sum of the calories of the dough and the calories of the toppings placed.
Create a program to find the number of calories per dollar for the "best pizza" given the price of the dough and the price of the toppings, and the calorie value of the dough and each topping.
input
The input consists of N + 3 lines.
On the first line, one integer N (1 ≤ N ≤ 100) representing the number of topping types is written. On the second line, two integers A and B (1 ≤ A ≤ 1000, 1 ≤ B ≤ 1000) are written with a blank as a delimiter. A is the price of the dough and B is the price of the toppings. On the third line, one integer C (1 ≤ C ≤ 10000) representing the number of calories in the dough is written.
On the 3 + i line (1 ≤ i ≤ N), one integer Di (1 ≤ Di ≤ 10000) representing the number of calories in the i-th topping is written.
output
Print the number of calories per dollar for the "best pizza" in one line. However, round down the numbers after the decimal point and output as an integer value.
Input / output example
Input example 1
3
12 2
200
50
300
100
Output example 1
37
In I / O Example 1, with the second and third toppings, 200 + 300 + 100 = 600 calories gives a pizza of $ 12 + 2 x 2 = $ 16.
This pizza has 600/16 = 37.5 calories per dollar. Since this is the "best pizza", we output 37, rounded down to the nearest whole number of 37.5.
Input example 2
Four
20 3
900
300
100
400
1300
Output example 2
100
The question text and the data used for the automatic referee are the question text and the test data for scoring, which are created and published by the Japan Committee for Information Olympics.
Example
Input
3
12 2
200
50
300
100
Output
37 | instruction | 0 | 73,228 | 9 | 146,456 |
"Correct Solution:
```
n=int(input())
a,b=map(int,input().split())
c=int(input())
d=sorted([int(input()) for _ in range(n)])[::-1]
e=c//a
for x in d:
a+=b;c+=x
e=max(e,c//a)
print(e)
``` | output | 1 | 73,228 | 9 | 146,457 |
Provide a correct Python 3 solution for this coding contest problem.
problem
Chairman K is a regular customer of the JOI pizza shop in the center of JOI city. For some reason, he decided to start a life-saving life this month. So he wanted to order the pizza with the highest calories per dollar among the pizzas he could order at the JOI pizza store. Let's call such a pizza the "best pizza". The "best pizza" is not limited to one type.
At JOI Pizza, you can freely choose from N types of toppings and order the ones placed on the basic dough. You cannot put more than one topping of the same type. You can also order a pizza that doesn't have any toppings on the dough. The price of the dough is $ A and the price of the toppings is $ B. The price of pizza is the sum of the price of the dough and the price of the toppings. That is, the price of a pizza with k types of toppings (0 ≤ k ≤ N) is A + k x B dollars. The total calorie of the pizza is the sum of the calories of the dough and the calories of the toppings placed.
Create a program to find the number of calories per dollar for the "best pizza" given the price of the dough and the price of the toppings, and the calorie value of the dough and each topping.
input
The input consists of N + 3 lines.
On the first line, one integer N (1 ≤ N ≤ 100) representing the number of topping types is written. On the second line, two integers A and B (1 ≤ A ≤ 1000, 1 ≤ B ≤ 1000) are written with a blank as a delimiter. A is the price of the dough and B is the price of the toppings. On the third line, one integer C (1 ≤ C ≤ 10000) representing the number of calories in the dough is written.
On the 3 + i line (1 ≤ i ≤ N), one integer Di (1 ≤ Di ≤ 10000) representing the number of calories in the i-th topping is written.
output
Print the number of calories per dollar for the "best pizza" in one line. However, round down the numbers after the decimal point and output as an integer value.
Input / output example
Input example 1
3
12 2
200
50
300
100
Output example 1
37
In I / O Example 1, with the second and third toppings, 200 + 300 + 100 = 600 calories gives a pizza of $ 12 + 2 x 2 = $ 16.
This pizza has 600/16 = 37.5 calories per dollar. Since this is the "best pizza", we output 37, rounded down to the nearest whole number of 37.5.
Input example 2
Four
20 3
900
300
100
400
1300
Output example 2
100
The question text and the data used for the automatic referee are the question text and the test data for scoring, which are created and published by the Japan Committee for Information Olympics.
Example
Input
3
12 2
200
50
300
100
Output
37 | instruction | 0 | 73,229 | 9 | 146,458 |
"Correct Solution:
```
N=int(input())
A,B=map(int,input().split())
C=int(input())
D=sorted(int(input())for _ in[0]*N)[::-1]
print(max((C+sum(D[:i]))//(A+i*B)for i in range(N)))
``` | output | 1 | 73,229 | 9 | 146,459 |
Provide a correct Python 3 solution for this coding contest problem.
problem
Chairman K is a regular customer of the JOI pizza shop in the center of JOI city. For some reason, he decided to start a life-saving life this month. So he wanted to order the pizza with the highest calories per dollar among the pizzas he could order at the JOI pizza store. Let's call such a pizza the "best pizza". The "best pizza" is not limited to one type.
At JOI Pizza, you can freely choose from N types of toppings and order the ones placed on the basic dough. You cannot put more than one topping of the same type. You can also order a pizza that doesn't have any toppings on the dough. The price of the dough is $ A and the price of the toppings is $ B. The price of pizza is the sum of the price of the dough and the price of the toppings. That is, the price of a pizza with k types of toppings (0 ≤ k ≤ N) is A + k x B dollars. The total calorie of the pizza is the sum of the calories of the dough and the calories of the toppings placed.
Create a program to find the number of calories per dollar for the "best pizza" given the price of the dough and the price of the toppings, and the calorie value of the dough and each topping.
input
The input consists of N + 3 lines.
On the first line, one integer N (1 ≤ N ≤ 100) representing the number of topping types is written. On the second line, two integers A and B (1 ≤ A ≤ 1000, 1 ≤ B ≤ 1000) are written with a blank as a delimiter. A is the price of the dough and B is the price of the toppings. On the third line, one integer C (1 ≤ C ≤ 10000) representing the number of calories in the dough is written.
On the 3 + i line (1 ≤ i ≤ N), one integer Di (1 ≤ Di ≤ 10000) representing the number of calories in the i-th topping is written.
output
Print the number of calories per dollar for the "best pizza" in one line. However, round down the numbers after the decimal point and output as an integer value.
Input / output example
Input example 1
3
12 2
200
50
300
100
Output example 1
37
In I / O Example 1, with the second and third toppings, 200 + 300 + 100 = 600 calories gives a pizza of $ 12 + 2 x 2 = $ 16.
This pizza has 600/16 = 37.5 calories per dollar. Since this is the "best pizza", we output 37, rounded down to the nearest whole number of 37.5.
Input example 2
Four
20 3
900
300
100
400
1300
Output example 2
100
The question text and the data used for the automatic referee are the question text and the test data for scoring, which are created and published by the Japan Committee for Information Olympics.
Example
Input
3
12 2
200
50
300
100
Output
37 | instruction | 0 | 73,230 | 9 | 146,460 |
"Correct Solution:
```
n = int(input())
p_cost, t_cost = map(int, input().split())
p_cal = int(input())
t_cal = []
for i in range(n) :
t_cal.append(int(input()))
t_cal.sort(reverse = True)
total_cost = p_cost
total_cal = p_cal
max_total_cal_par_doll = total_cal // total_cost
for i in range(n) :
total_cost += t_cost
total_cal += t_cal[i]
total_cal_par_doll = total_cal // total_cost
if max_total_cal_par_doll < total_cal // total_cost :
max_total_cal_par_doll = total_cal // total_cost
print(max_total_cal_par_doll)
``` | output | 1 | 73,230 | 9 | 146,461 |
Provide a correct Python 3 solution for this coding contest problem.
problem
Chairman K is a regular customer of the JOI pizza shop in the center of JOI city. For some reason, he decided to start a life-saving life this month. So he wanted to order the pizza with the highest calories per dollar among the pizzas he could order at the JOI pizza store. Let's call such a pizza the "best pizza". The "best pizza" is not limited to one type.
At JOI Pizza, you can freely choose from N types of toppings and order the ones placed on the basic dough. You cannot put more than one topping of the same type. You can also order a pizza that doesn't have any toppings on the dough. The price of the dough is $ A and the price of the toppings is $ B. The price of pizza is the sum of the price of the dough and the price of the toppings. That is, the price of a pizza with k types of toppings (0 ≤ k ≤ N) is A + k x B dollars. The total calorie of the pizza is the sum of the calories of the dough and the calories of the toppings placed.
Create a program to find the number of calories per dollar for the "best pizza" given the price of the dough and the price of the toppings, and the calorie value of the dough and each topping.
input
The input consists of N + 3 lines.
On the first line, one integer N (1 ≤ N ≤ 100) representing the number of topping types is written. On the second line, two integers A and B (1 ≤ A ≤ 1000, 1 ≤ B ≤ 1000) are written with a blank as a delimiter. A is the price of the dough and B is the price of the toppings. On the third line, one integer C (1 ≤ C ≤ 10000) representing the number of calories in the dough is written.
On the 3 + i line (1 ≤ i ≤ N), one integer Di (1 ≤ Di ≤ 10000) representing the number of calories in the i-th topping is written.
output
Print the number of calories per dollar for the "best pizza" in one line. However, round down the numbers after the decimal point and output as an integer value.
Input / output example
Input example 1
3
12 2
200
50
300
100
Output example 1
37
In I / O Example 1, with the second and third toppings, 200 + 300 + 100 = 600 calories gives a pizza of $ 12 + 2 x 2 = $ 16.
This pizza has 600/16 = 37.5 calories per dollar. Since this is the "best pizza", we output 37, rounded down to the nearest whole number of 37.5.
Input example 2
Four
20 3
900
300
100
400
1300
Output example 2
100
The question text and the data used for the automatic referee are the question text and the test data for scoring, which are created and published by the Japan Committee for Information Olympics.
Example
Input
3
12 2
200
50
300
100
Output
37 | instruction | 0 | 73,231 | 9 | 146,462 |
"Correct Solution:
```
n = int(input())
a, b = map(int, input().split())
c = int(input())
topping = []
for _ in range(n):
d = int(input())
topping.append(d)
topping.sort(reverse=True)
ans = c // a
total = c
for i, t in enumerate(topping):
total += t
cal = total // (a + (i+1) * b)
if ans <= cal:
ans = cal
else:
break
print(ans)
``` | output | 1 | 73,231 | 9 | 146,463 |
Provide a correct Python 3 solution for this coding contest problem.
problem
Chairman K is a regular customer of the JOI pizza shop in the center of JOI city. For some reason, he decided to start a life-saving life this month. So he wanted to order the pizza with the highest calories per dollar among the pizzas he could order at the JOI pizza store. Let's call such a pizza the "best pizza". The "best pizza" is not limited to one type.
At JOI Pizza, you can freely choose from N types of toppings and order the ones placed on the basic dough. You cannot put more than one topping of the same type. You can also order a pizza that doesn't have any toppings on the dough. The price of the dough is $ A and the price of the toppings is $ B. The price of pizza is the sum of the price of the dough and the price of the toppings. That is, the price of a pizza with k types of toppings (0 ≤ k ≤ N) is A + k x B dollars. The total calorie of the pizza is the sum of the calories of the dough and the calories of the toppings placed.
Create a program to find the number of calories per dollar for the "best pizza" given the price of the dough and the price of the toppings, and the calorie value of the dough and each topping.
input
The input consists of N + 3 lines.
On the first line, one integer N (1 ≤ N ≤ 100) representing the number of topping types is written. On the second line, two integers A and B (1 ≤ A ≤ 1000, 1 ≤ B ≤ 1000) are written with a blank as a delimiter. A is the price of the dough and B is the price of the toppings. On the third line, one integer C (1 ≤ C ≤ 10000) representing the number of calories in the dough is written.
On the 3 + i line (1 ≤ i ≤ N), one integer Di (1 ≤ Di ≤ 10000) representing the number of calories in the i-th topping is written.
output
Print the number of calories per dollar for the "best pizza" in one line. However, round down the numbers after the decimal point and output as an integer value.
Input / output example
Input example 1
3
12 2
200
50
300
100
Output example 1
37
In I / O Example 1, with the second and third toppings, 200 + 300 + 100 = 600 calories gives a pizza of $ 12 + 2 x 2 = $ 16.
This pizza has 600/16 = 37.5 calories per dollar. Since this is the "best pizza", we output 37, rounded down to the nearest whole number of 37.5.
Input example 2
Four
20 3
900
300
100
400
1300
Output example 2
100
The question text and the data used for the automatic referee are the question text and the test data for scoring, which are created and published by the Japan Committee for Information Olympics.
Example
Input
3
12 2
200
50
300
100
Output
37 | instruction | 0 | 73,232 | 9 | 146,464 |
"Correct Solution:
```
n = int(input())
c, b = map(int, input().split())
k = int(input())
A = [int(input()) for _ in range(n)]
A.sort(reverse=True)
r = k/c
for i in A:
k += i
c += b
if k/c<=r:
break
else:
r = k/c
print(int(r))
``` | output | 1 | 73,232 | 9 | 146,465 |
Provide a correct Python 3 solution for this coding contest problem.
problem
Chairman K is a regular customer of the JOI pizza shop in the center of JOI city. For some reason, he decided to start a life-saving life this month. So he wanted to order the pizza with the highest calories per dollar among the pizzas he could order at the JOI pizza store. Let's call such a pizza the "best pizza". The "best pizza" is not limited to one type.
At JOI Pizza, you can freely choose from N types of toppings and order the ones placed on the basic dough. You cannot put more than one topping of the same type. You can also order a pizza that doesn't have any toppings on the dough. The price of the dough is $ A and the price of the toppings is $ B. The price of pizza is the sum of the price of the dough and the price of the toppings. That is, the price of a pizza with k types of toppings (0 ≤ k ≤ N) is A + k x B dollars. The total calorie of the pizza is the sum of the calories of the dough and the calories of the toppings placed.
Create a program to find the number of calories per dollar for the "best pizza" given the price of the dough and the price of the toppings, and the calorie value of the dough and each topping.
input
The input consists of N + 3 lines.
On the first line, one integer N (1 ≤ N ≤ 100) representing the number of topping types is written. On the second line, two integers A and B (1 ≤ A ≤ 1000, 1 ≤ B ≤ 1000) are written with a blank as a delimiter. A is the price of the dough and B is the price of the toppings. On the third line, one integer C (1 ≤ C ≤ 10000) representing the number of calories in the dough is written.
On the 3 + i line (1 ≤ i ≤ N), one integer Di (1 ≤ Di ≤ 10000) representing the number of calories in the i-th topping is written.
output
Print the number of calories per dollar for the "best pizza" in one line. However, round down the numbers after the decimal point and output as an integer value.
Input / output example
Input example 1
3
12 2
200
50
300
100
Output example 1
37
In I / O Example 1, with the second and third toppings, 200 + 300 + 100 = 600 calories gives a pizza of $ 12 + 2 x 2 = $ 16.
This pizza has 600/16 = 37.5 calories per dollar. Since this is the "best pizza", we output 37, rounded down to the nearest whole number of 37.5.
Input example 2
Four
20 3
900
300
100
400
1300
Output example 2
100
The question text and the data used for the automatic referee are the question text and the test data for scoring, which are created and published by the Japan Committee for Information Olympics.
Example
Input
3
12 2
200
50
300
100
Output
37 | instruction | 0 | 73,233 | 9 | 146,466 |
"Correct Solution:
```
n = int(input())
a, b = map(int,input().split())
c = int(input())
ds = [int(input()) for i in range(n)]
ds.sort(reverse=True)
ans = c // a
for i in range(n):
c += ds[i]
ans = max(c // (a + b * (i + 1)), ans)
print(ans)
``` | output | 1 | 73,233 | 9 | 146,467 |
Provide a correct Python 3 solution for this coding contest problem.
problem
Chairman K is a regular customer of the JOI pizza shop in the center of JOI city. For some reason, he decided to start a life-saving life this month. So he wanted to order the pizza with the highest calories per dollar among the pizzas he could order at the JOI pizza store. Let's call such a pizza the "best pizza". The "best pizza" is not limited to one type.
At JOI Pizza, you can freely choose from N types of toppings and order the ones placed on the basic dough. You cannot put more than one topping of the same type. You can also order a pizza that doesn't have any toppings on the dough. The price of the dough is $ A and the price of the toppings is $ B. The price of pizza is the sum of the price of the dough and the price of the toppings. That is, the price of a pizza with k types of toppings (0 ≤ k ≤ N) is A + k x B dollars. The total calorie of the pizza is the sum of the calories of the dough and the calories of the toppings placed.
Create a program to find the number of calories per dollar for the "best pizza" given the price of the dough and the price of the toppings, and the calorie value of the dough and each topping.
input
The input consists of N + 3 lines.
On the first line, one integer N (1 ≤ N ≤ 100) representing the number of topping types is written. On the second line, two integers A and B (1 ≤ A ≤ 1000, 1 ≤ B ≤ 1000) are written with a blank as a delimiter. A is the price of the dough and B is the price of the toppings. On the third line, one integer C (1 ≤ C ≤ 10000) representing the number of calories in the dough is written.
On the 3 + i line (1 ≤ i ≤ N), one integer Di (1 ≤ Di ≤ 10000) representing the number of calories in the i-th topping is written.
output
Print the number of calories per dollar for the "best pizza" in one line. However, round down the numbers after the decimal point and output as an integer value.
Input / output example
Input example 1
3
12 2
200
50
300
100
Output example 1
37
In I / O Example 1, with the second and third toppings, 200 + 300 + 100 = 600 calories gives a pizza of $ 12 + 2 x 2 = $ 16.
This pizza has 600/16 = 37.5 calories per dollar. Since this is the "best pizza", we output 37, rounded down to the nearest whole number of 37.5.
Input example 2
Four
20 3
900
300
100
400
1300
Output example 2
100
The question text and the data used for the automatic referee are the question text and the test data for scoring, which are created and published by the Japan Committee for Information Olympics.
Example
Input
3
12 2
200
50
300
100
Output
37 | instruction | 0 | 73,234 | 9 | 146,468 |
"Correct Solution:
```
def main():
t_n = int(input())
price_a, price_b = map(int, input().split())
cal_a = int(input())
t_c = sorted([int(input()) for _ in range(t_n)], reverse=True)
cpd_a = cal_a / price_a
cal_sum = cal_a
price_sum = price_a
b_sum = 0
for i in range(t_n):
b_sum += t_c[i]
new_cpd_a = (cal_a + b_sum) / (price_a + (i + 1) * price_b)
if new_cpd_a > cpd_a:
cal_sum += t_c[i]
price_sum += price_b
cpd_a = new_cpd_a
print(int(cal_sum / price_sum))
if __name__ == '__main__':
main()
``` | output | 1 | 73,234 | 9 | 146,469 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
problem
Chairman K is a regular customer of the JOI pizza shop in the center of JOI city. For some reason, he decided to start a life-saving life this month. So he wanted to order the pizza with the highest calories per dollar among the pizzas he could order at the JOI pizza store. Let's call such a pizza the "best pizza". The "best pizza" is not limited to one type.
At JOI Pizza, you can freely choose from N types of toppings and order the ones placed on the basic dough. You cannot put more than one topping of the same type. You can also order a pizza that doesn't have any toppings on the dough. The price of the dough is $ A and the price of the toppings is $ B. The price of pizza is the sum of the price of the dough and the price of the toppings. That is, the price of a pizza with k types of toppings (0 ≤ k ≤ N) is A + k x B dollars. The total calorie of the pizza is the sum of the calories of the dough and the calories of the toppings placed.
Create a program to find the number of calories per dollar for the "best pizza" given the price of the dough and the price of the toppings, and the calorie value of the dough and each topping.
input
The input consists of N + 3 lines.
On the first line, one integer N (1 ≤ N ≤ 100) representing the number of topping types is written. On the second line, two integers A and B (1 ≤ A ≤ 1000, 1 ≤ B ≤ 1000) are written with a blank as a delimiter. A is the price of the dough and B is the price of the toppings. On the third line, one integer C (1 ≤ C ≤ 10000) representing the number of calories in the dough is written.
On the 3 + i line (1 ≤ i ≤ N), one integer Di (1 ≤ Di ≤ 10000) representing the number of calories in the i-th topping is written.
output
Print the number of calories per dollar for the "best pizza" in one line. However, round down the numbers after the decimal point and output as an integer value.
Input / output example
Input example 1
3
12 2
200
50
300
100
Output example 1
37
In I / O Example 1, with the second and third toppings, 200 + 300 + 100 = 600 calories gives a pizza of $ 12 + 2 x 2 = $ 16.
This pizza has 600/16 = 37.5 calories per dollar. Since this is the "best pizza", we output 37, rounded down to the nearest whole number of 37.5.
Input example 2
Four
20 3
900
300
100
400
1300
Output example 2
100
The question text and the data used for the automatic referee are the question text and the test data for scoring, which are created and published by the Japan Committee for Information Olympics.
Example
Input
3
12 2
200
50
300
100
Output
37
Submitted Solution:
```
# AOJ 0567: Best Pizza
# Python3 2018.6.30 bal4u
n = int(input())
a, b = map(int, input().split())
c = int(input())
ans, p = c/a, a
d = [int(input()) for i in range(n)]
d.sort(reverse=True)
for i in range(n):
t = (c+d[i])/(p+b)
if t > ans:
ans = t
c += d[i]
p += b
else: break;
print(int(ans))
``` | instruction | 0 | 73,235 | 9 | 146,470 |
Yes | output | 1 | 73,235 | 9 | 146,471 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
problem
Chairman K is a regular customer of the JOI pizza shop in the center of JOI city. For some reason, he decided to start a life-saving life this month. So he wanted to order the pizza with the highest calories per dollar among the pizzas he could order at the JOI pizza store. Let's call such a pizza the "best pizza". The "best pizza" is not limited to one type.
At JOI Pizza, you can freely choose from N types of toppings and order the ones placed on the basic dough. You cannot put more than one topping of the same type. You can also order a pizza that doesn't have any toppings on the dough. The price of the dough is $ A and the price of the toppings is $ B. The price of pizza is the sum of the price of the dough and the price of the toppings. That is, the price of a pizza with k types of toppings (0 ≤ k ≤ N) is A + k x B dollars. The total calorie of the pizza is the sum of the calories of the dough and the calories of the toppings placed.
Create a program to find the number of calories per dollar for the "best pizza" given the price of the dough and the price of the toppings, and the calorie value of the dough and each topping.
input
The input consists of N + 3 lines.
On the first line, one integer N (1 ≤ N ≤ 100) representing the number of topping types is written. On the second line, two integers A and B (1 ≤ A ≤ 1000, 1 ≤ B ≤ 1000) are written with a blank as a delimiter. A is the price of the dough and B is the price of the toppings. On the third line, one integer C (1 ≤ C ≤ 10000) representing the number of calories in the dough is written.
On the 3 + i line (1 ≤ i ≤ N), one integer Di (1 ≤ Di ≤ 10000) representing the number of calories in the i-th topping is written.
output
Print the number of calories per dollar for the "best pizza" in one line. However, round down the numbers after the decimal point and output as an integer value.
Input / output example
Input example 1
3
12 2
200
50
300
100
Output example 1
37
In I / O Example 1, with the second and third toppings, 200 + 300 + 100 = 600 calories gives a pizza of $ 12 + 2 x 2 = $ 16.
This pizza has 600/16 = 37.5 calories per dollar. Since this is the "best pizza", we output 37, rounded down to the nearest whole number of 37.5.
Input example 2
Four
20 3
900
300
100
400
1300
Output example 2
100
The question text and the data used for the automatic referee are the question text and the test data for scoring, which are created and published by the Japan Committee for Information Olympics.
Example
Input
3
12 2
200
50
300
100
Output
37
Submitted Solution:
```
# -*- coding: utf-8 -*-
"""
"""
import sys
from sys import stdin
from collections import deque
input = stdin.readline
from collections import namedtuple
item = namedtuple('item', ['cal', 'pri'])
def Cond(x, n, items, pizza):
y = [0] * n
for i in range(n):
y[i] = items[i].cal - x * items[i].pri
y.sort(reverse=True)
total = pizza.cal - x * pizza.pri
for i in range(n):
if y[i] > 0:
total += y[i]
else:
break
return total >= 0
def main(args):
# n = 3
# k = 2
# items = [item(2, 2), item(5, 3), item(2, 1)]
N = int(input())
A, B = map(int, input().split())
C = int(input())
toppings = []
for i in range(N):
toppings.append(item(int(input()), B))
pizza = item(C, A)
lb = 0
ub = 1e5
for i in range(100):
mid = (lb + ub) / 2
if Cond(mid, N, toppings, pizza):
lb = mid
else:
ub = mid
print(int(ub))
if __name__ == '__main__':
main(sys.argv[1:])
``` | instruction | 0 | 73,236 | 9 | 146,472 |
Yes | output | 1 | 73,236 | 9 | 146,473 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
problem
Chairman K is a regular customer of the JOI pizza shop in the center of JOI city. For some reason, he decided to start a life-saving life this month. So he wanted to order the pizza with the highest calories per dollar among the pizzas he could order at the JOI pizza store. Let's call such a pizza the "best pizza". The "best pizza" is not limited to one type.
At JOI Pizza, you can freely choose from N types of toppings and order the ones placed on the basic dough. You cannot put more than one topping of the same type. You can also order a pizza that doesn't have any toppings on the dough. The price of the dough is $ A and the price of the toppings is $ B. The price of pizza is the sum of the price of the dough and the price of the toppings. That is, the price of a pizza with k types of toppings (0 ≤ k ≤ N) is A + k x B dollars. The total calorie of the pizza is the sum of the calories of the dough and the calories of the toppings placed.
Create a program to find the number of calories per dollar for the "best pizza" given the price of the dough and the price of the toppings, and the calorie value of the dough and each topping.
input
The input consists of N + 3 lines.
On the first line, one integer N (1 ≤ N ≤ 100) representing the number of topping types is written. On the second line, two integers A and B (1 ≤ A ≤ 1000, 1 ≤ B ≤ 1000) are written with a blank as a delimiter. A is the price of the dough and B is the price of the toppings. On the third line, one integer C (1 ≤ C ≤ 10000) representing the number of calories in the dough is written.
On the 3 + i line (1 ≤ i ≤ N), one integer Di (1 ≤ Di ≤ 10000) representing the number of calories in the i-th topping is written.
output
Print the number of calories per dollar for the "best pizza" in one line. However, round down the numbers after the decimal point and output as an integer value.
Input / output example
Input example 1
3
12 2
200
50
300
100
Output example 1
37
In I / O Example 1, with the second and third toppings, 200 + 300 + 100 = 600 calories gives a pizza of $ 12 + 2 x 2 = $ 16.
This pizza has 600/16 = 37.5 calories per dollar. Since this is the "best pizza", we output 37, rounded down to the nearest whole number of 37.5.
Input example 2
Four
20 3
900
300
100
400
1300
Output example 2
100
The question text and the data used for the automatic referee are the question text and the test data for scoring, which are created and published by the Japan Committee for Information Olympics.
Example
Input
3
12 2
200
50
300
100
Output
37
Submitted Solution:
```
N, A, B, C, *D = map(int, open(0).read().split())
D.sort(reverse=1)
ans = C // A
s = 0
for i in range(N):
s += D[i]
ans = max(ans, (C + s)//(A + (i+1)*B))
print(ans)
``` | instruction | 0 | 73,237 | 9 | 146,474 |
Yes | output | 1 | 73,237 | 9 | 146,475 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
problem
Chairman K is a regular customer of the JOI pizza shop in the center of JOI city. For some reason, he decided to start a life-saving life this month. So he wanted to order the pizza with the highest calories per dollar among the pizzas he could order at the JOI pizza store. Let's call such a pizza the "best pizza". The "best pizza" is not limited to one type.
At JOI Pizza, you can freely choose from N types of toppings and order the ones placed on the basic dough. You cannot put more than one topping of the same type. You can also order a pizza that doesn't have any toppings on the dough. The price of the dough is $ A and the price of the toppings is $ B. The price of pizza is the sum of the price of the dough and the price of the toppings. That is, the price of a pizza with k types of toppings (0 ≤ k ≤ N) is A + k x B dollars. The total calorie of the pizza is the sum of the calories of the dough and the calories of the toppings placed.
Create a program to find the number of calories per dollar for the "best pizza" given the price of the dough and the price of the toppings, and the calorie value of the dough and each topping.
input
The input consists of N + 3 lines.
On the first line, one integer N (1 ≤ N ≤ 100) representing the number of topping types is written. On the second line, two integers A and B (1 ≤ A ≤ 1000, 1 ≤ B ≤ 1000) are written with a blank as a delimiter. A is the price of the dough and B is the price of the toppings. On the third line, one integer C (1 ≤ C ≤ 10000) representing the number of calories in the dough is written.
On the 3 + i line (1 ≤ i ≤ N), one integer Di (1 ≤ Di ≤ 10000) representing the number of calories in the i-th topping is written.
output
Print the number of calories per dollar for the "best pizza" in one line. However, round down the numbers after the decimal point and output as an integer value.
Input / output example
Input example 1
3
12 2
200
50
300
100
Output example 1
37
In I / O Example 1, with the second and third toppings, 200 + 300 + 100 = 600 calories gives a pizza of $ 12 + 2 x 2 = $ 16.
This pizza has 600/16 = 37.5 calories per dollar. Since this is the "best pizza", we output 37, rounded down to the nearest whole number of 37.5.
Input example 2
Four
20 3
900
300
100
400
1300
Output example 2
100
The question text and the data used for the automatic referee are the question text and the test data for scoring, which are created and published by the Japan Committee for Information Olympics.
Example
Input
3
12 2
200
50
300
100
Output
37
Submitted Solution:
```
N=int(input())
A,B=map(int,input().split())
C=int(input())
D=sorted(int(input())for _ in[0]*N)[::-1]
E=C//A
for x in D:A+=B;C+=x;E=max(E,C//A)
print(E)
``` | instruction | 0 | 73,238 | 9 | 146,476 |
Yes | output | 1 | 73,238 | 9 | 146,477 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Long story short, shashlik is Miroslav's favorite food. Shashlik is prepared on several skewers simultaneously. There are two states for each skewer: initial and turned over.
This time Miroslav laid out n skewers parallel to each other, and enumerated them with consecutive integers from 1 to n in order from left to right. For better cooking, he puts them quite close to each other, so when he turns skewer number i, it leads to turning k closest skewers from each side of the skewer i, that is, skewers number i - k, i - k + 1, ..., i - 1, i + 1, ..., i + k - 1, i + k (if they exist).
For example, let n = 6 and k = 1. When Miroslav turns skewer number 3, then skewers with numbers 2, 3, and 4 will come up turned over. If after that he turns skewer number 1, then skewers number 1, 3, and 4 will be turned over, while skewer number 2 will be in the initial position (because it is turned again).
As we said before, the art of cooking requires perfect timing, so Miroslav wants to turn over all n skewers with the minimal possible number of actions. For example, for the above example n = 6 and k = 1, two turnings are sufficient: he can turn over skewers number 2 and 5.
Help Miroslav turn over all n skewers.
Input
The first line contains two integers n and k (1 ≤ n ≤ 1000, 0 ≤ k ≤ 1000) — the number of skewers and the number of skewers from each side that are turned in one step.
Output
The first line should contain integer l — the minimum number of actions needed by Miroslav to turn over all n skewers. After than print l integers from 1 to n denoting the number of the skewer that is to be turned over at the corresponding step.
Examples
Input
7 2
Output
2
1 6
Input
5 1
Output
2
1 4
Note
In the first example the first operation turns over skewers 1, 2 and 3, the second operation turns over skewers 4, 5, 6 and 7.
In the second example it is also correct to turn over skewers 2 and 5, but turning skewers 2 and 4, or 1 and 5 are incorrect solutions because the skewer 3 is in the initial state after these operations. | instruction | 0 | 73,303 | 9 | 146,606 |
Tags: dp, greedy, math
Correct Solution:
```
from functools import reduce
n, k = map(int, input().split())
flip = 2*k+1
l = -(-n//flip)
max = (k+1) + (l-1) * flip
i = k+1
if max > n:
i -= (max-n)
arr = []
# if n%flip > k:
# i = k+1
while i <= n:
arr.append(i)
i += flip
# else:
# i = 1
# while i <= n:
# arr.append(i)
# i += flip
s = reduce(lambda x, y : str(x)+' '+str(y), arr)
print(l)
print(s)
``` | output | 1 | 73,303 | 9 | 146,607 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Long story short, shashlik is Miroslav's favorite food. Shashlik is prepared on several skewers simultaneously. There are two states for each skewer: initial and turned over.
This time Miroslav laid out n skewers parallel to each other, and enumerated them with consecutive integers from 1 to n in order from left to right. For better cooking, he puts them quite close to each other, so when he turns skewer number i, it leads to turning k closest skewers from each side of the skewer i, that is, skewers number i - k, i - k + 1, ..., i - 1, i + 1, ..., i + k - 1, i + k (if they exist).
For example, let n = 6 and k = 1. When Miroslav turns skewer number 3, then skewers with numbers 2, 3, and 4 will come up turned over. If after that he turns skewer number 1, then skewers number 1, 3, and 4 will be turned over, while skewer number 2 will be in the initial position (because it is turned again).
As we said before, the art of cooking requires perfect timing, so Miroslav wants to turn over all n skewers with the minimal possible number of actions. For example, for the above example n = 6 and k = 1, two turnings are sufficient: he can turn over skewers number 2 and 5.
Help Miroslav turn over all n skewers.
Input
The first line contains two integers n and k (1 ≤ n ≤ 1000, 0 ≤ k ≤ 1000) — the number of skewers and the number of skewers from each side that are turned in one step.
Output
The first line should contain integer l — the minimum number of actions needed by Miroslav to turn over all n skewers. After than print l integers from 1 to n denoting the number of the skewer that is to be turned over at the corresponding step.
Examples
Input
7 2
Output
2
1 6
Input
5 1
Output
2
1 4
Note
In the first example the first operation turns over skewers 1, 2 and 3, the second operation turns over skewers 4, 5, 6 and 7.
In the second example it is also correct to turn over skewers 2 and 5, but turning skewers 2 and 4, or 1 and 5 are incorrect solutions because the skewer 3 is in the initial state after these operations. | instruction | 0 | 73,304 | 9 | 146,608 |
Tags: dp, greedy, math
Correct Solution:
```
n, k = map(int, input().split())
t = 2 * k + 1
ans = list(range(min(((n - 1) % t) + 1, k + 1), n + 1, t))
print(len(ans))
print(' '.join(map(str, ans)))
``` | output | 1 | 73,304 | 9 | 146,609 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Long story short, shashlik is Miroslav's favorite food. Shashlik is prepared on several skewers simultaneously. There are two states for each skewer: initial and turned over.
This time Miroslav laid out n skewers parallel to each other, and enumerated them with consecutive integers from 1 to n in order from left to right. For better cooking, he puts them quite close to each other, so when he turns skewer number i, it leads to turning k closest skewers from each side of the skewer i, that is, skewers number i - k, i - k + 1, ..., i - 1, i + 1, ..., i + k - 1, i + k (if they exist).
For example, let n = 6 and k = 1. When Miroslav turns skewer number 3, then skewers with numbers 2, 3, and 4 will come up turned over. If after that he turns skewer number 1, then skewers number 1, 3, and 4 will be turned over, while skewer number 2 will be in the initial position (because it is turned again).
As we said before, the art of cooking requires perfect timing, so Miroslav wants to turn over all n skewers with the minimal possible number of actions. For example, for the above example n = 6 and k = 1, two turnings are sufficient: he can turn over skewers number 2 and 5.
Help Miroslav turn over all n skewers.
Input
The first line contains two integers n and k (1 ≤ n ≤ 1000, 0 ≤ k ≤ 1000) — the number of skewers and the number of skewers from each side that are turned in one step.
Output
The first line should contain integer l — the minimum number of actions needed by Miroslav to turn over all n skewers. After than print l integers from 1 to n denoting the number of the skewer that is to be turned over at the corresponding step.
Examples
Input
7 2
Output
2
1 6
Input
5 1
Output
2
1 4
Note
In the first example the first operation turns over skewers 1, 2 and 3, the second operation turns over skewers 4, 5, 6 and 7.
In the second example it is also correct to turn over skewers 2 and 5, but turning skewers 2 and 4, or 1 and 5 are incorrect solutions because the skewer 3 is in the initial state after these operations. | instruction | 0 | 73,305 | 9 | 146,610 |
Tags: dp, greedy, math
Correct Solution:
```
n, k = map(int, input().split())
ans = n // (k * 2 + 1)
x = n % (k * 2 + 1)
if x == 0:
print(ans)
for i in range(k + 1, n + 1, k * 2 + 1):
print(i, end=" ")
elif x > k:
print(ans + 1)
for i in range(x - k, n + 1, k * 2 + 1):
print(i, end=" ")
else:
print(ans + 1)
for i in range(1, n + 1, k * 2 + 1):
print(i, end=" ")
``` | output | 1 | 73,305 | 9 | 146,611 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Long story short, shashlik is Miroslav's favorite food. Shashlik is prepared on several skewers simultaneously. There are two states for each skewer: initial and turned over.
This time Miroslav laid out n skewers parallel to each other, and enumerated them with consecutive integers from 1 to n in order from left to right. For better cooking, he puts them quite close to each other, so when he turns skewer number i, it leads to turning k closest skewers from each side of the skewer i, that is, skewers number i - k, i - k + 1, ..., i - 1, i + 1, ..., i + k - 1, i + k (if they exist).
For example, let n = 6 and k = 1. When Miroslav turns skewer number 3, then skewers with numbers 2, 3, and 4 will come up turned over. If after that he turns skewer number 1, then skewers number 1, 3, and 4 will be turned over, while skewer number 2 will be in the initial position (because it is turned again).
As we said before, the art of cooking requires perfect timing, so Miroslav wants to turn over all n skewers with the minimal possible number of actions. For example, for the above example n = 6 and k = 1, two turnings are sufficient: he can turn over skewers number 2 and 5.
Help Miroslav turn over all n skewers.
Input
The first line contains two integers n and k (1 ≤ n ≤ 1000, 0 ≤ k ≤ 1000) — the number of skewers and the number of skewers from each side that are turned in one step.
Output
The first line should contain integer l — the minimum number of actions needed by Miroslav to turn over all n skewers. After than print l integers from 1 to n denoting the number of the skewer that is to be turned over at the corresponding step.
Examples
Input
7 2
Output
2
1 6
Input
5 1
Output
2
1 4
Note
In the first example the first operation turns over skewers 1, 2 and 3, the second operation turns over skewers 4, 5, 6 and 7.
In the second example it is also correct to turn over skewers 2 and 5, but turning skewers 2 and 4, or 1 and 5 are incorrect solutions because the skewer 3 is in the initial state after these operations. | instruction | 0 | 73,306 | 9 | 146,612 |
Tags: dp, greedy, math
Correct Solution:
```
import sys
# from collections import deque
# from collections import Counter
# from math import sqrt
# from math import log
from math import ceil
# from bisect import bisect_left, bisect_right
alpha=['a', 'b', 'c', 'd', 'e', 'f', 'g', 'h', 'i', 'j', 'k', 'l', 'm', 'n', 'o', 'p', 'q', 'r', 's', 't', 'u', 'v', 'w', 'x', 'y', 'z']
# mod=10**9+7
# mod=998244353
# def BinarySearch(a,x):
# i=bisect_left(a,x)
# if(i!=len(a) and a[i]==x):
# return i
# else:
# return -1
# def sieve(n):
# prime=[True for i in range(n+1)]
# p=2
# while(p*p<=n):
# if (prime[p]==True):
# for i in range(p*p,n+1,p):
# prime[i]=False
# p+=1
# prime[0]=False
# prime[1]=False
# s=set()
# for i in range(len(prime)):
# if(prime[i]):
# s.add(i)
# return s
# def gcd(a, b):
# if(a==0):
# return b
# return gcd(b%a,a)
fast_reader=sys.stdin.readline
fast_writer=sys.stdout.write
def input():
return fast_reader().strip()
def print(*argv):
fast_writer(' '.join((str(i)) for i in argv))
fast_writer('\n')
#____________________________________________________________________________________________________________________________________
def find(s1,s2):
for i in alpha:
if(i!=s1 and i!=s2):
return i
for _ in range(1):
n,k=map(int, input().split())
ans=ceil(n/(2*k+1))
print(ans)
if(n%(2*k+1)<=k+1 and n%(2*k+1)>0):
ans2=n%(2*k+1)
else:
ans2=k+1
l=[]
while(ans2<=n):
l.append(ans2)
ans2+=2*k+1
print(*l)
``` | output | 1 | 73,306 | 9 | 146,613 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Long story short, shashlik is Miroslav's favorite food. Shashlik is prepared on several skewers simultaneously. There are two states for each skewer: initial and turned over.
This time Miroslav laid out n skewers parallel to each other, and enumerated them with consecutive integers from 1 to n in order from left to right. For better cooking, he puts them quite close to each other, so when he turns skewer number i, it leads to turning k closest skewers from each side of the skewer i, that is, skewers number i - k, i - k + 1, ..., i - 1, i + 1, ..., i + k - 1, i + k (if they exist).
For example, let n = 6 and k = 1. When Miroslav turns skewer number 3, then skewers with numbers 2, 3, and 4 will come up turned over. If after that he turns skewer number 1, then skewers number 1, 3, and 4 will be turned over, while skewer number 2 will be in the initial position (because it is turned again).
As we said before, the art of cooking requires perfect timing, so Miroslav wants to turn over all n skewers with the minimal possible number of actions. For example, for the above example n = 6 and k = 1, two turnings are sufficient: he can turn over skewers number 2 and 5.
Help Miroslav turn over all n skewers.
Input
The first line contains two integers n and k (1 ≤ n ≤ 1000, 0 ≤ k ≤ 1000) — the number of skewers and the number of skewers from each side that are turned in one step.
Output
The first line should contain integer l — the minimum number of actions needed by Miroslav to turn over all n skewers. After than print l integers from 1 to n denoting the number of the skewer that is to be turned over at the corresponding step.
Examples
Input
7 2
Output
2
1 6
Input
5 1
Output
2
1 4
Note
In the first example the first operation turns over skewers 1, 2 and 3, the second operation turns over skewers 4, 5, 6 and 7.
In the second example it is also correct to turn over skewers 2 and 5, but turning skewers 2 and 4, or 1 and 5 are incorrect solutions because the skewer 3 is in the initial state after these operations. | instruction | 0 | 73,307 | 9 | 146,614 |
Tags: dp, greedy, math
Correct Solution:
```
def mi():
return map(int, input().split(' '))
n, k = mi()
t = 2*k + 1
m = n % t
start = 0
ans = []
if m == 0:
print(int(n/t))
start = 2*k
else:
print(int(n/t) + 1)
#ans.append(m-1//k)
start = m
for i in range((start//2)+1,n+1,t):
ans.append(i)
print(*ans)
``` | output | 1 | 73,307 | 9 | 146,615 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Long story short, shashlik is Miroslav's favorite food. Shashlik is prepared on several skewers simultaneously. There are two states for each skewer: initial and turned over.
This time Miroslav laid out n skewers parallel to each other, and enumerated them with consecutive integers from 1 to n in order from left to right. For better cooking, he puts them quite close to each other, so when he turns skewer number i, it leads to turning k closest skewers from each side of the skewer i, that is, skewers number i - k, i - k + 1, ..., i - 1, i + 1, ..., i + k - 1, i + k (if they exist).
For example, let n = 6 and k = 1. When Miroslav turns skewer number 3, then skewers with numbers 2, 3, and 4 will come up turned over. If after that he turns skewer number 1, then skewers number 1, 3, and 4 will be turned over, while skewer number 2 will be in the initial position (because it is turned again).
As we said before, the art of cooking requires perfect timing, so Miroslav wants to turn over all n skewers with the minimal possible number of actions. For example, for the above example n = 6 and k = 1, two turnings are sufficient: he can turn over skewers number 2 and 5.
Help Miroslav turn over all n skewers.
Input
The first line contains two integers n and k (1 ≤ n ≤ 1000, 0 ≤ k ≤ 1000) — the number of skewers and the number of skewers from each side that are turned in one step.
Output
The first line should contain integer l — the minimum number of actions needed by Miroslav to turn over all n skewers. After than print l integers from 1 to n denoting the number of the skewer that is to be turned over at the corresponding step.
Examples
Input
7 2
Output
2
1 6
Input
5 1
Output
2
1 4
Note
In the first example the first operation turns over skewers 1, 2 and 3, the second operation turns over skewers 4, 5, 6 and 7.
In the second example it is also correct to turn over skewers 2 and 5, but turning skewers 2 and 4, or 1 and 5 are incorrect solutions because the skewer 3 is in the initial state after these operations. | instruction | 0 | 73,308 | 9 | 146,616 |
Tags: dp, greedy, math
Correct Solution:
```
n,k=map(int,input().split())
i,p=k+1,2*k+1
t=n//p
st=[]
if n%p!=0:
t+=1
if t*p-k>n:
i-=t*p-k-n
print(t)
while t>0:
st.append(i+(t-1)*p)
t-=1
print(' '.join(list(sorted(map(str,st)))))
``` | output | 1 | 73,308 | 9 | 146,617 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Long story short, shashlik is Miroslav's favorite food. Shashlik is prepared on several skewers simultaneously. There are two states for each skewer: initial and turned over.
This time Miroslav laid out n skewers parallel to each other, and enumerated them with consecutive integers from 1 to n in order from left to right. For better cooking, he puts them quite close to each other, so when he turns skewer number i, it leads to turning k closest skewers from each side of the skewer i, that is, skewers number i - k, i - k + 1, ..., i - 1, i + 1, ..., i + k - 1, i + k (if they exist).
For example, let n = 6 and k = 1. When Miroslav turns skewer number 3, then skewers with numbers 2, 3, and 4 will come up turned over. If after that he turns skewer number 1, then skewers number 1, 3, and 4 will be turned over, while skewer number 2 will be in the initial position (because it is turned again).
As we said before, the art of cooking requires perfect timing, so Miroslav wants to turn over all n skewers with the minimal possible number of actions. For example, for the above example n = 6 and k = 1, two turnings are sufficient: he can turn over skewers number 2 and 5.
Help Miroslav turn over all n skewers.
Input
The first line contains two integers n and k (1 ≤ n ≤ 1000, 0 ≤ k ≤ 1000) — the number of skewers and the number of skewers from each side that are turned in one step.
Output
The first line should contain integer l — the minimum number of actions needed by Miroslav to turn over all n skewers. After than print l integers from 1 to n denoting the number of the skewer that is to be turned over at the corresponding step.
Examples
Input
7 2
Output
2
1 6
Input
5 1
Output
2
1 4
Note
In the first example the first operation turns over skewers 1, 2 and 3, the second operation turns over skewers 4, 5, 6 and 7.
In the second example it is also correct to turn over skewers 2 and 5, but turning skewers 2 and 4, or 1 and 5 are incorrect solutions because the skewer 3 is in the initial state after these operations. | instruction | 0 | 73,309 | 9 | 146,618 |
Tags: dp, greedy, math
Correct Solution:
```
n, k = map(int, input().split())
r = n % (2*k + 1)
out = []
if r >= k + 1:
cnt = r - k
while cnt < n:
out.append(cnt)
cnt += 2*k + 1
else:
cut1 = (r + 2*k + 1) // 2
cut2 = (r + 2*k + 2) // 2
cnt = cut1 - k
if cnt == 0:
cnt += k + 1
while cnt <= n:
out.append(cnt)
cnt += 2*k + 1
print(len(out))
out = map(str, out)
print(' '.join(out))
``` | output | 1 | 73,309 | 9 | 146,619 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Long story short, shashlik is Miroslav's favorite food. Shashlik is prepared on several skewers simultaneously. There are two states for each skewer: initial and turned over.
This time Miroslav laid out n skewers parallel to each other, and enumerated them with consecutive integers from 1 to n in order from left to right. For better cooking, he puts them quite close to each other, so when he turns skewer number i, it leads to turning k closest skewers from each side of the skewer i, that is, skewers number i - k, i - k + 1, ..., i - 1, i + 1, ..., i + k - 1, i + k (if they exist).
For example, let n = 6 and k = 1. When Miroslav turns skewer number 3, then skewers with numbers 2, 3, and 4 will come up turned over. If after that he turns skewer number 1, then skewers number 1, 3, and 4 will be turned over, while skewer number 2 will be in the initial position (because it is turned again).
As we said before, the art of cooking requires perfect timing, so Miroslav wants to turn over all n skewers with the minimal possible number of actions. For example, for the above example n = 6 and k = 1, two turnings are sufficient: he can turn over skewers number 2 and 5.
Help Miroslav turn over all n skewers.
Input
The first line contains two integers n and k (1 ≤ n ≤ 1000, 0 ≤ k ≤ 1000) — the number of skewers and the number of skewers from each side that are turned in one step.
Output
The first line should contain integer l — the minimum number of actions needed by Miroslav to turn over all n skewers. After than print l integers from 1 to n denoting the number of the skewer that is to be turned over at the corresponding step.
Examples
Input
7 2
Output
2
1 6
Input
5 1
Output
2
1 4
Note
In the first example the first operation turns over skewers 1, 2 and 3, the second operation turns over skewers 4, 5, 6 and 7.
In the second example it is also correct to turn over skewers 2 and 5, but turning skewers 2 and 4, or 1 and 5 are incorrect solutions because the skewer 3 is in the initial state after these operations. | instruction | 0 | 73,310 | 9 | 146,620 |
Tags: dp, greedy, math
Correct Solution:
```
n,k=list(map(int,input().split(" ")))
i,D=1,k*2+1
g=[]
while True:
R=k+i
p=(n-R)%D
if p==0 or p>=R:
div=n/D
a = int(div)
b = abs(div)-abs(int(div))
if b:
qw=int(n/D)+1
else:
qw=int(n/D)
for w in range(qw):
g.append(i)
i=i+D
else:
i+=1
if p>=R or p==0:
break;
print(len(g))
for ss in g:
print(ss, end=" ")
print()
``` | output | 1 | 73,310 | 9 | 146,621 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Long story short, shashlik is Miroslav's favorite food. Shashlik is prepared on several skewers simultaneously. There are two states for each skewer: initial and turned over.
This time Miroslav laid out n skewers parallel to each other, and enumerated them with consecutive integers from 1 to n in order from left to right. For better cooking, he puts them quite close to each other, so when he turns skewer number i, it leads to turning k closest skewers from each side of the skewer i, that is, skewers number i - k, i - k + 1, ..., i - 1, i + 1, ..., i + k - 1, i + k (if they exist).
For example, let n = 6 and k = 1. When Miroslav turns skewer number 3, then skewers with numbers 2, 3, and 4 will come up turned over. If after that he turns skewer number 1, then skewers number 1, 3, and 4 will be turned over, while skewer number 2 will be in the initial position (because it is turned again).
As we said before, the art of cooking requires perfect timing, so Miroslav wants to turn over all n skewers with the minimal possible number of actions. For example, for the above example n = 6 and k = 1, two turnings are sufficient: he can turn over skewers number 2 and 5.
Help Miroslav turn over all n skewers.
Input
The first line contains two integers n and k (1 ≤ n ≤ 1000, 0 ≤ k ≤ 1000) — the number of skewers and the number of skewers from each side that are turned in one step.
Output
The first line should contain integer l — the minimum number of actions needed by Miroslav to turn over all n skewers. After than print l integers from 1 to n denoting the number of the skewer that is to be turned over at the corresponding step.
Examples
Input
7 2
Output
2
1 6
Input
5 1
Output
2
1 4
Note
In the first example the first operation turns over skewers 1, 2 and 3, the second operation turns over skewers 4, 5, 6 and 7.
In the second example it is also correct to turn over skewers 2 and 5, but turning skewers 2 and 4, or 1 and 5 are incorrect solutions because the skewer 3 is in the initial state after these operations.
Submitted Solution:
```
## 1040B
import math
(n, k) = (int(_) for _ in input().split(' '))
nr = math.ceil( n / (2*k + 1))
start = 1
while start + (2*k + 1)*(nr - 1) + k < n:
start += 1
toturn = list()
for i in range(nr):
toturn.append(start + (2*k + 1)* i)
print(nr)
print(' '.join(str(_) for _ in toturn))
``` | instruction | 0 | 73,311 | 9 | 146,622 |
Yes | output | 1 | 73,311 | 9 | 146,623 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Long story short, shashlik is Miroslav's favorite food. Shashlik is prepared on several skewers simultaneously. There are two states for each skewer: initial and turned over.
This time Miroslav laid out n skewers parallel to each other, and enumerated them with consecutive integers from 1 to n in order from left to right. For better cooking, he puts them quite close to each other, so when he turns skewer number i, it leads to turning k closest skewers from each side of the skewer i, that is, skewers number i - k, i - k + 1, ..., i - 1, i + 1, ..., i + k - 1, i + k (if they exist).
For example, let n = 6 and k = 1. When Miroslav turns skewer number 3, then skewers with numbers 2, 3, and 4 will come up turned over. If after that he turns skewer number 1, then skewers number 1, 3, and 4 will be turned over, while skewer number 2 will be in the initial position (because it is turned again).
As we said before, the art of cooking requires perfect timing, so Miroslav wants to turn over all n skewers with the minimal possible number of actions. For example, for the above example n = 6 and k = 1, two turnings are sufficient: he can turn over skewers number 2 and 5.
Help Miroslav turn over all n skewers.
Input
The first line contains two integers n and k (1 ≤ n ≤ 1000, 0 ≤ k ≤ 1000) — the number of skewers and the number of skewers from each side that are turned in one step.
Output
The first line should contain integer l — the minimum number of actions needed by Miroslav to turn over all n skewers. After than print l integers from 1 to n denoting the number of the skewer that is to be turned over at the corresponding step.
Examples
Input
7 2
Output
2
1 6
Input
5 1
Output
2
1 4
Note
In the first example the first operation turns over skewers 1, 2 and 3, the second operation turns over skewers 4, 5, 6 and 7.
In the second example it is also correct to turn over skewers 2 and 5, but turning skewers 2 and 4, or 1 and 5 are incorrect solutions because the skewer 3 is in the initial state after these operations.
Submitted Solution:
```
import math
[n, k] = [int(t) for t in input().split(' ')]
res = [0] * math.ceil(n / (2*k+1))
for tstart in range(k+1):
i = 0
pos = tstart
while True:
right = pos + k + 1
res[i] = pos
i += 1
pos += 2 * k + 1
if pos >= n: break
if right >= n:
print(i)
print(*[t + 1 for t in res[:i]])
break
``` | instruction | 0 | 73,312 | 9 | 146,624 |
Yes | output | 1 | 73,312 | 9 | 146,625 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Long story short, shashlik is Miroslav's favorite food. Shashlik is prepared on several skewers simultaneously. There are two states for each skewer: initial and turned over.
This time Miroslav laid out n skewers parallel to each other, and enumerated them with consecutive integers from 1 to n in order from left to right. For better cooking, he puts them quite close to each other, so when he turns skewer number i, it leads to turning k closest skewers from each side of the skewer i, that is, skewers number i - k, i - k + 1, ..., i - 1, i + 1, ..., i + k - 1, i + k (if they exist).
For example, let n = 6 and k = 1. When Miroslav turns skewer number 3, then skewers with numbers 2, 3, and 4 will come up turned over. If after that he turns skewer number 1, then skewers number 1, 3, and 4 will be turned over, while skewer number 2 will be in the initial position (because it is turned again).
As we said before, the art of cooking requires perfect timing, so Miroslav wants to turn over all n skewers with the minimal possible number of actions. For example, for the above example n = 6 and k = 1, two turnings are sufficient: he can turn over skewers number 2 and 5.
Help Miroslav turn over all n skewers.
Input
The first line contains two integers n and k (1 ≤ n ≤ 1000, 0 ≤ k ≤ 1000) — the number of skewers and the number of skewers from each side that are turned in one step.
Output
The first line should contain integer l — the minimum number of actions needed by Miroslav to turn over all n skewers. After than print l integers from 1 to n denoting the number of the skewer that is to be turned over at the corresponding step.
Examples
Input
7 2
Output
2
1 6
Input
5 1
Output
2
1 4
Note
In the first example the first operation turns over skewers 1, 2 and 3, the second operation turns over skewers 4, 5, 6 and 7.
In the second example it is also correct to turn over skewers 2 and 5, but turning skewers 2 and 4, or 1 and 5 are incorrect solutions because the skewer 3 is in the initial state after these operations.
Submitted Solution:
```
n, k = map(int, input().split())
K = 2*k+1
a = [i for i in range(1, n+1, K)]
#print(' '.join(str(x) for x in a))
# print(a[-1])
l = n-a[len(a)-1]
l = int(l/2)
b = [i+l for i in a]
print(len(b))
print(" ".join(str(x) for x in b))
``` | instruction | 0 | 73,313 | 9 | 146,626 |
Yes | output | 1 | 73,313 | 9 | 146,627 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Long story short, shashlik is Miroslav's favorite food. Shashlik is prepared on several skewers simultaneously. There are two states for each skewer: initial and turned over.
This time Miroslav laid out n skewers parallel to each other, and enumerated them with consecutive integers from 1 to n in order from left to right. For better cooking, he puts them quite close to each other, so when he turns skewer number i, it leads to turning k closest skewers from each side of the skewer i, that is, skewers number i - k, i - k + 1, ..., i - 1, i + 1, ..., i + k - 1, i + k (if they exist).
For example, let n = 6 and k = 1. When Miroslav turns skewer number 3, then skewers with numbers 2, 3, and 4 will come up turned over. If after that he turns skewer number 1, then skewers number 1, 3, and 4 will be turned over, while skewer number 2 will be in the initial position (because it is turned again).
As we said before, the art of cooking requires perfect timing, so Miroslav wants to turn over all n skewers with the minimal possible number of actions. For example, for the above example n = 6 and k = 1, two turnings are sufficient: he can turn over skewers number 2 and 5.
Help Miroslav turn over all n skewers.
Input
The first line contains two integers n and k (1 ≤ n ≤ 1000, 0 ≤ k ≤ 1000) — the number of skewers and the number of skewers from each side that are turned in one step.
Output
The first line should contain integer l — the minimum number of actions needed by Miroslav to turn over all n skewers. After than print l integers from 1 to n denoting the number of the skewer that is to be turned over at the corresponding step.
Examples
Input
7 2
Output
2
1 6
Input
5 1
Output
2
1 4
Note
In the first example the first operation turns over skewers 1, 2 and 3, the second operation turns over skewers 4, 5, 6 and 7.
In the second example it is also correct to turn over skewers 2 and 5, but turning skewers 2 and 4, or 1 and 5 are incorrect solutions because the skewer 3 is in the initial state after these operations.
Submitted Solution:
```
n,k=[int(el) for el in input().split()]
if n<=2*k+1:
print(1)
print (int(n/2)+1)
raise SystemExit
if (n)%(2*k+1)==0:
l=int(n/(2*k+1))
else:
l=int(n/(2*k+1))+1
ost=n%(2*k+1)
if ost==0:
start=k+1
else:
start=int((ost-1)/2)+1
print(l)
l=[]
for i in range (start,n+1,2*k+1):
l.append(str(i))
print(' '.join(l))
``` | instruction | 0 | 73,314 | 9 | 146,628 |
Yes | output | 1 | 73,314 | 9 | 146,629 |
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