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 |
|---|---|---|---|---|---|
Provide a correct Python 3 solution for this coding contest problem.
Akaki, a patissier, can make N kinds of doughnut using only a certain powder called "Okashi no Moto" (literally "material of pastry", simply called Moto below) as ingredient. These doughnuts are called Doughnut 1, Doughnut 2, ..., Doughnut N. In order to make one Doughnut i (1 ≤ i ≤ N), she needs to consume m_i grams of Moto. She cannot make a non-integer number of doughnuts, such as 0.5 doughnuts.
Now, she has X grams of Moto. She decides to make as many doughnuts as possible for a party tonight. However, since the tastes of the guests differ, she will obey the following condition:
* For each of the N kinds of doughnuts, make at least one doughnut of that kind.
At most how many doughnuts can be made here? She does not necessarily need to consume all of her Moto. Also, under the constraints of this problem, it is always possible to obey the condition.
Constraints
* 2 ≤ N ≤ 100
* 1 ≤ m_i ≤ 1000
* m_1 + m_2 + ... + m_N ≤ X ≤ 10^5
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
N X
m_1
m_2
:
m_N
Output
Print the maximum number of doughnuts that can be made under the condition.
Examples
Input
3 1000
120
100
140
Output
9
Input
4 360
90
90
90
90
Output
4
Input
5 3000
150
130
150
130
110
Output
26 | instruction | 0 | 89,079 | 9 | 178,158 |
"Correct Solution:
```
N,X=map(int,input().split())
m=[int(input()) for _ in range(N)]
#print(m)
print(N+(X-sum(m))//min(m))
``` | output | 1 | 89,079 | 9 | 178,159 |
Provide a correct Python 3 solution for this coding contest problem.
Akaki, a patissier, can make N kinds of doughnut using only a certain powder called "Okashi no Moto" (literally "material of pastry", simply called Moto below) as ingredient. These doughnuts are called Doughnut 1, Doughnut 2, ..., Doughnut N. In order to make one Doughnut i (1 ≤ i ≤ N), she needs to consume m_i grams of Moto. She cannot make a non-integer number of doughnuts, such as 0.5 doughnuts.
Now, she has X grams of Moto. She decides to make as many doughnuts as possible for a party tonight. However, since the tastes of the guests differ, she will obey the following condition:
* For each of the N kinds of doughnuts, make at least one doughnut of that kind.
At most how many doughnuts can be made here? She does not necessarily need to consume all of her Moto. Also, under the constraints of this problem, it is always possible to obey the condition.
Constraints
* 2 ≤ N ≤ 100
* 1 ≤ m_i ≤ 1000
* m_1 + m_2 + ... + m_N ≤ X ≤ 10^5
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
N X
m_1
m_2
:
m_N
Output
Print the maximum number of doughnuts that can be made under the condition.
Examples
Input
3 1000
120
100
140
Output
9
Input
4 360
90
90
90
90
Output
4
Input
5 3000
150
130
150
130
110
Output
26 | instruction | 0 | 89,080 | 9 | 178,160 |
"Correct Solution:
```
N, X = map(int, input().split())
L = [int(input()) for i in range(N)]
print((X-sum(L))//min(L)+N)
``` | output | 1 | 89,080 | 9 | 178,161 |
Provide a correct Python 3 solution for this coding contest problem.
Akaki, a patissier, can make N kinds of doughnut using only a certain powder called "Okashi no Moto" (literally "material of pastry", simply called Moto below) as ingredient. These doughnuts are called Doughnut 1, Doughnut 2, ..., Doughnut N. In order to make one Doughnut i (1 ≤ i ≤ N), she needs to consume m_i grams of Moto. She cannot make a non-integer number of doughnuts, such as 0.5 doughnuts.
Now, she has X grams of Moto. She decides to make as many doughnuts as possible for a party tonight. However, since the tastes of the guests differ, she will obey the following condition:
* For each of the N kinds of doughnuts, make at least one doughnut of that kind.
At most how many doughnuts can be made here? She does not necessarily need to consume all of her Moto. Also, under the constraints of this problem, it is always possible to obey the condition.
Constraints
* 2 ≤ N ≤ 100
* 1 ≤ m_i ≤ 1000
* m_1 + m_2 + ... + m_N ≤ X ≤ 10^5
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
N X
m_1
m_2
:
m_N
Output
Print the maximum number of doughnuts that can be made under the condition.
Examples
Input
3 1000
120
100
140
Output
9
Input
4 360
90
90
90
90
Output
4
Input
5 3000
150
130
150
130
110
Output
26 | instruction | 0 | 89,081 | 9 | 178,162 |
"Correct Solution:
```
import sys
N,X=map(int,input().split())
m=[int(input()) for i in range(N)]
X-=sum(m)
print(N+X//min(m))
``` | output | 1 | 89,081 | 9 | 178,163 |
Provide a correct Python 3 solution for this coding contest problem.
Akaki, a patissier, can make N kinds of doughnut using only a certain powder called "Okashi no Moto" (literally "material of pastry", simply called Moto below) as ingredient. These doughnuts are called Doughnut 1, Doughnut 2, ..., Doughnut N. In order to make one Doughnut i (1 ≤ i ≤ N), she needs to consume m_i grams of Moto. She cannot make a non-integer number of doughnuts, such as 0.5 doughnuts.
Now, she has X grams of Moto. She decides to make as many doughnuts as possible for a party tonight. However, since the tastes of the guests differ, she will obey the following condition:
* For each of the N kinds of doughnuts, make at least one doughnut of that kind.
At most how many doughnuts can be made here? She does not necessarily need to consume all of her Moto. Also, under the constraints of this problem, it is always possible to obey the condition.
Constraints
* 2 ≤ N ≤ 100
* 1 ≤ m_i ≤ 1000
* m_1 + m_2 + ... + m_N ≤ X ≤ 10^5
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
N X
m_1
m_2
:
m_N
Output
Print the maximum number of doughnuts that can be made under the condition.
Examples
Input
3 1000
120
100
140
Output
9
Input
4 360
90
90
90
90
Output
4
Input
5 3000
150
130
150
130
110
Output
26 | instruction | 0 | 89,082 | 9 | 178,164 |
"Correct Solution:
```
n,x=map(int,input().split())
M=sorted([int(input()) for i in range(n)])
print(len(M)+(x-sum(M))//M[0])
``` | output | 1 | 89,082 | 9 | 178,165 |
Provide a correct Python 3 solution for this coding contest problem.
Akaki, a patissier, can make N kinds of doughnut using only a certain powder called "Okashi no Moto" (literally "material of pastry", simply called Moto below) as ingredient. These doughnuts are called Doughnut 1, Doughnut 2, ..., Doughnut N. In order to make one Doughnut i (1 ≤ i ≤ N), she needs to consume m_i grams of Moto. She cannot make a non-integer number of doughnuts, such as 0.5 doughnuts.
Now, she has X grams of Moto. She decides to make as many doughnuts as possible for a party tonight. However, since the tastes of the guests differ, she will obey the following condition:
* For each of the N kinds of doughnuts, make at least one doughnut of that kind.
At most how many doughnuts can be made here? She does not necessarily need to consume all of her Moto. Also, under the constraints of this problem, it is always possible to obey the condition.
Constraints
* 2 ≤ N ≤ 100
* 1 ≤ m_i ≤ 1000
* m_1 + m_2 + ... + m_N ≤ X ≤ 10^5
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
N X
m_1
m_2
:
m_N
Output
Print the maximum number of doughnuts that can be made under the condition.
Examples
Input
3 1000
120
100
140
Output
9
Input
4 360
90
90
90
90
Output
4
Input
5 3000
150
130
150
130
110
Output
26 | instruction | 0 | 89,083 | 9 | 178,166 |
"Correct Solution:
```
n,x=map(int,input().split())
m=[int(input()) for i in range(n)]
print(int((x-sum(m))/min(m))+len(m))
``` | output | 1 | 89,083 | 9 | 178,167 |
Provide a correct Python 3 solution for this coding contest problem.
Akaki, a patissier, can make N kinds of doughnut using only a certain powder called "Okashi no Moto" (literally "material of pastry", simply called Moto below) as ingredient. These doughnuts are called Doughnut 1, Doughnut 2, ..., Doughnut N. In order to make one Doughnut i (1 ≤ i ≤ N), she needs to consume m_i grams of Moto. She cannot make a non-integer number of doughnuts, such as 0.5 doughnuts.
Now, she has X grams of Moto. She decides to make as many doughnuts as possible for a party tonight. However, since the tastes of the guests differ, she will obey the following condition:
* For each of the N kinds of doughnuts, make at least one doughnut of that kind.
At most how many doughnuts can be made here? She does not necessarily need to consume all of her Moto. Also, under the constraints of this problem, it is always possible to obey the condition.
Constraints
* 2 ≤ N ≤ 100
* 1 ≤ m_i ≤ 1000
* m_1 + m_2 + ... + m_N ≤ X ≤ 10^5
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
N X
m_1
m_2
:
m_N
Output
Print the maximum number of doughnuts that can be made under the condition.
Examples
Input
3 1000
120
100
140
Output
9
Input
4 360
90
90
90
90
Output
4
Input
5 3000
150
130
150
130
110
Output
26 | instruction | 0 | 89,084 | 9 | 178,168 |
"Correct Solution:
```
n, x = map(int, input().split())
a = [int(input()) for i in range(n)]
print(n+(x-sum(a))//(min(a)))
``` | output | 1 | 89,084 | 9 | 178,169 |
Provide a correct Python 3 solution for this coding contest problem.
Akaki, a patissier, can make N kinds of doughnut using only a certain powder called "Okashi no Moto" (literally "material of pastry", simply called Moto below) as ingredient. These doughnuts are called Doughnut 1, Doughnut 2, ..., Doughnut N. In order to make one Doughnut i (1 ≤ i ≤ N), she needs to consume m_i grams of Moto. She cannot make a non-integer number of doughnuts, such as 0.5 doughnuts.
Now, she has X grams of Moto. She decides to make as many doughnuts as possible for a party tonight. However, since the tastes of the guests differ, she will obey the following condition:
* For each of the N kinds of doughnuts, make at least one doughnut of that kind.
At most how many doughnuts can be made here? She does not necessarily need to consume all of her Moto. Also, under the constraints of this problem, it is always possible to obey the condition.
Constraints
* 2 ≤ N ≤ 100
* 1 ≤ m_i ≤ 1000
* m_1 + m_2 + ... + m_N ≤ X ≤ 10^5
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
N X
m_1
m_2
:
m_N
Output
Print the maximum number of doughnuts that can be made under the condition.
Examples
Input
3 1000
120
100
140
Output
9
Input
4 360
90
90
90
90
Output
4
Input
5 3000
150
130
150
130
110
Output
26 | instruction | 0 | 89,085 | 9 | 178,170 |
"Correct Solution:
```
n,x = map(int,input().split())
M = [int(input()) for i in range(n)]
print(n+(x-sum(M))//min(M))
``` | output | 1 | 89,085 | 9 | 178,171 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Akaki, a patissier, can make N kinds of doughnut using only a certain powder called "Okashi no Moto" (literally "material of pastry", simply called Moto below) as ingredient. These doughnuts are called Doughnut 1, Doughnut 2, ..., Doughnut N. In order to make one Doughnut i (1 ≤ i ≤ N), she needs to consume m_i grams of Moto. She cannot make a non-integer number of doughnuts, such as 0.5 doughnuts.
Now, she has X grams of Moto. She decides to make as many doughnuts as possible for a party tonight. However, since the tastes of the guests differ, she will obey the following condition:
* For each of the N kinds of doughnuts, make at least one doughnut of that kind.
At most how many doughnuts can be made here? She does not necessarily need to consume all of her Moto. Also, under the constraints of this problem, it is always possible to obey the condition.
Constraints
* 2 ≤ N ≤ 100
* 1 ≤ m_i ≤ 1000
* m_1 + m_2 + ... + m_N ≤ X ≤ 10^5
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
N X
m_1
m_2
:
m_N
Output
Print the maximum number of doughnuts that can be made under the condition.
Examples
Input
3 1000
120
100
140
Output
9
Input
4 360
90
90
90
90
Output
4
Input
5 3000
150
130
150
130
110
Output
26
Submitted Solution:
```
N,X=map(int,input().split())
donut=[int(input()) for i in range(N)]
print((X-sum(donut))//min(donut)+N)
``` | instruction | 0 | 89,086 | 9 | 178,172 |
Yes | output | 1 | 89,086 | 9 | 178,173 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Akaki, a patissier, can make N kinds of doughnut using only a certain powder called "Okashi no Moto" (literally "material of pastry", simply called Moto below) as ingredient. These doughnuts are called Doughnut 1, Doughnut 2, ..., Doughnut N. In order to make one Doughnut i (1 ≤ i ≤ N), she needs to consume m_i grams of Moto. She cannot make a non-integer number of doughnuts, such as 0.5 doughnuts.
Now, she has X grams of Moto. She decides to make as many doughnuts as possible for a party tonight. However, since the tastes of the guests differ, she will obey the following condition:
* For each of the N kinds of doughnuts, make at least one doughnut of that kind.
At most how many doughnuts can be made here? She does not necessarily need to consume all of her Moto. Also, under the constraints of this problem, it is always possible to obey the condition.
Constraints
* 2 ≤ N ≤ 100
* 1 ≤ m_i ≤ 1000
* m_1 + m_2 + ... + m_N ≤ X ≤ 10^5
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
N X
m_1
m_2
:
m_N
Output
Print the maximum number of doughnuts that can be made under the condition.
Examples
Input
3 1000
120
100
140
Output
9
Input
4 360
90
90
90
90
Output
4
Input
5 3000
150
130
150
130
110
Output
26
Submitted Solution:
```
n,x=map(int, input().split())
a=[int(input()) for i in range(n)]
x=x-sum(a)
print(x//min(a)+n)
``` | instruction | 0 | 89,087 | 9 | 178,174 |
Yes | output | 1 | 89,087 | 9 | 178,175 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Akaki, a patissier, can make N kinds of doughnut using only a certain powder called "Okashi no Moto" (literally "material of pastry", simply called Moto below) as ingredient. These doughnuts are called Doughnut 1, Doughnut 2, ..., Doughnut N. In order to make one Doughnut i (1 ≤ i ≤ N), she needs to consume m_i grams of Moto. She cannot make a non-integer number of doughnuts, such as 0.5 doughnuts.
Now, she has X grams of Moto. She decides to make as many doughnuts as possible for a party tonight. However, since the tastes of the guests differ, she will obey the following condition:
* For each of the N kinds of doughnuts, make at least one doughnut of that kind.
At most how many doughnuts can be made here? She does not necessarily need to consume all of her Moto. Also, under the constraints of this problem, it is always possible to obey the condition.
Constraints
* 2 ≤ N ≤ 100
* 1 ≤ m_i ≤ 1000
* m_1 + m_2 + ... + m_N ≤ X ≤ 10^5
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
N X
m_1
m_2
:
m_N
Output
Print the maximum number of doughnuts that can be made under the condition.
Examples
Input
3 1000
120
100
140
Output
9
Input
4 360
90
90
90
90
Output
4
Input
5 3000
150
130
150
130
110
Output
26
Submitted Solution:
```
n,x = map(int,input().split())
d = [int(input()) for i in range(n)]
x -= sum(d)
print(n + x//min(d))
``` | instruction | 0 | 89,088 | 9 | 178,176 |
Yes | output | 1 | 89,088 | 9 | 178,177 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Akaki, a patissier, can make N kinds of doughnut using only a certain powder called "Okashi no Moto" (literally "material of pastry", simply called Moto below) as ingredient. These doughnuts are called Doughnut 1, Doughnut 2, ..., Doughnut N. In order to make one Doughnut i (1 ≤ i ≤ N), she needs to consume m_i grams of Moto. She cannot make a non-integer number of doughnuts, such as 0.5 doughnuts.
Now, she has X grams of Moto. She decides to make as many doughnuts as possible for a party tonight. However, since the tastes of the guests differ, she will obey the following condition:
* For each of the N kinds of doughnuts, make at least one doughnut of that kind.
At most how many doughnuts can be made here? She does not necessarily need to consume all of her Moto. Also, under the constraints of this problem, it is always possible to obey the condition.
Constraints
* 2 ≤ N ≤ 100
* 1 ≤ m_i ≤ 1000
* m_1 + m_2 + ... + m_N ≤ X ≤ 10^5
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
N X
m_1
m_2
:
m_N
Output
Print the maximum number of doughnuts that can be made under the condition.
Examples
Input
3 1000
120
100
140
Output
9
Input
4 360
90
90
90
90
Output
4
Input
5 3000
150
130
150
130
110
Output
26
Submitted Solution:
```
N, X = map(int, input().split())
M = [int(input()) for _ in range(N)]
print(len(M) + (X-sum(M))//min(M))
``` | instruction | 0 | 89,089 | 9 | 178,178 |
Yes | output | 1 | 89,089 | 9 | 178,179 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Akaki, a patissier, can make N kinds of doughnut using only a certain powder called "Okashi no Moto" (literally "material of pastry", simply called Moto below) as ingredient. These doughnuts are called Doughnut 1, Doughnut 2, ..., Doughnut N. In order to make one Doughnut i (1 ≤ i ≤ N), she needs to consume m_i grams of Moto. She cannot make a non-integer number of doughnuts, such as 0.5 doughnuts.
Now, she has X grams of Moto. She decides to make as many doughnuts as possible for a party tonight. However, since the tastes of the guests differ, she will obey the following condition:
* For each of the N kinds of doughnuts, make at least one doughnut of that kind.
At most how many doughnuts can be made here? She does not necessarily need to consume all of her Moto. Also, under the constraints of this problem, it is always possible to obey the condition.
Constraints
* 2 ≤ N ≤ 100
* 1 ≤ m_i ≤ 1000
* m_1 + m_2 + ... + m_N ≤ X ≤ 10^5
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
N X
m_1
m_2
:
m_N
Output
Print the maximum number of doughnuts that can be made under the condition.
Examples
Input
3 1000
120
100
140
Output
9
Input
4 360
90
90
90
90
Output
4
Input
5 3000
150
130
150
130
110
Output
26
Submitted Solution:
```
N,X = map(int,input().split())
m = []
for i in range(N):
m.append(int(input()))
X-=sum(m)
cnt=len(m)
while X>min(m):
X-=min(m)
cnt+=1
print(cnt)
``` | instruction | 0 | 89,090 | 9 | 178,180 |
No | output | 1 | 89,090 | 9 | 178,181 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Akaki, a patissier, can make N kinds of doughnut using only a certain powder called "Okashi no Moto" (literally "material of pastry", simply called Moto below) as ingredient. These doughnuts are called Doughnut 1, Doughnut 2, ..., Doughnut N. In order to make one Doughnut i (1 ≤ i ≤ N), she needs to consume m_i grams of Moto. She cannot make a non-integer number of doughnuts, such as 0.5 doughnuts.
Now, she has X grams of Moto. She decides to make as many doughnuts as possible for a party tonight. However, since the tastes of the guests differ, she will obey the following condition:
* For each of the N kinds of doughnuts, make at least one doughnut of that kind.
At most how many doughnuts can be made here? She does not necessarily need to consume all of her Moto. Also, under the constraints of this problem, it is always possible to obey the condition.
Constraints
* 2 ≤ N ≤ 100
* 1 ≤ m_i ≤ 1000
* m_1 + m_2 + ... + m_N ≤ X ≤ 10^5
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
N X
m_1
m_2
:
m_N
Output
Print the maximum number of doughnuts that can be made under the condition.
Examples
Input
3 1000
120
100
140
Output
9
Input
4 360
90
90
90
90
Output
4
Input
5 3000
150
130
150
130
110
Output
26
Submitted Solution:
```
N, X = map(int, input().split())
M = [int(input()) for _ in range(N)]
ans = len(M)
X = X - sum(M)
while X >= min(M):
X = X - min(M)
ans += 1
print(ans, X)
``` | instruction | 0 | 89,091 | 9 | 178,182 |
No | output | 1 | 89,091 | 9 | 178,183 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Akaki, a patissier, can make N kinds of doughnut using only a certain powder called "Okashi no Moto" (literally "material of pastry", simply called Moto below) as ingredient. These doughnuts are called Doughnut 1, Doughnut 2, ..., Doughnut N. In order to make one Doughnut i (1 ≤ i ≤ N), she needs to consume m_i grams of Moto. She cannot make a non-integer number of doughnuts, such as 0.5 doughnuts.
Now, she has X grams of Moto. She decides to make as many doughnuts as possible for a party tonight. However, since the tastes of the guests differ, she will obey the following condition:
* For each of the N kinds of doughnuts, make at least one doughnut of that kind.
At most how many doughnuts can be made here? She does not necessarily need to consume all of her Moto. Also, under the constraints of this problem, it is always possible to obey the condition.
Constraints
* 2 ≤ N ≤ 100
* 1 ≤ m_i ≤ 1000
* m_1 + m_2 + ... + m_N ≤ X ≤ 10^5
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
N X
m_1
m_2
:
m_N
Output
Print the maximum number of doughnuts that can be made under the condition.
Examples
Input
3 1000
120
100
140
Output
9
Input
4 360
90
90
90
90
Output
4
Input
5 3000
150
130
150
130
110
Output
26
Submitted Solution:
```
n,maxg=map(int,input().split())
lista =[int(input()) for i in range(n)]
count=count+n
nokori=maxg-sum(lista)
print(count)
while nokori-min(lista)>=0:
nokori=nokori-min(lista)
count+=1
print(count)
``` | instruction | 0 | 89,092 | 9 | 178,184 |
No | output | 1 | 89,092 | 9 | 178,185 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Akaki, a patissier, can make N kinds of doughnut using only a certain powder called "Okashi no Moto" (literally "material of pastry", simply called Moto below) as ingredient. These doughnuts are called Doughnut 1, Doughnut 2, ..., Doughnut N. In order to make one Doughnut i (1 ≤ i ≤ N), she needs to consume m_i grams of Moto. She cannot make a non-integer number of doughnuts, such as 0.5 doughnuts.
Now, she has X grams of Moto. She decides to make as many doughnuts as possible for a party tonight. However, since the tastes of the guests differ, she will obey the following condition:
* For each of the N kinds of doughnuts, make at least one doughnut of that kind.
At most how many doughnuts can be made here? She does not necessarily need to consume all of her Moto. Also, under the constraints of this problem, it is always possible to obey the condition.
Constraints
* 2 ≤ N ≤ 100
* 1 ≤ m_i ≤ 1000
* m_1 + m_2 + ... + m_N ≤ X ≤ 10^5
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
N X
m_1
m_2
:
m_N
Output
Print the maximum number of doughnuts that can be made under the condition.
Examples
Input
3 1000
120
100
140
Output
9
Input
4 360
90
90
90
90
Output
4
Input
5 3000
150
130
150
130
110
Output
26
Submitted Solution:
```
# -*- coding: utf-8 -*-
def main():
N, X = map(int, input().split())
m = []
for i in range(N):
m.append(int(input()))
ans = 0
sumM = 0
flag = False
for i in m:
sumM += i
ans += 1
m.sort()
while sumM < X:
sumM += m[0]
if sumM < X:
ans += 1
print(ans)
if __name__ == "__main__":
main()
``` | instruction | 0 | 89,093 | 9 | 178,186 |
No | output | 1 | 89,093 | 9 | 178,187 |
Provide tags and a correct Python 3 solution for this coding contest problem.
It's another Start[c]up finals, and that means there is pizza to order for the onsite contestants. There are only 2 types of pizza (obviously not, but let's just pretend for the sake of the problem), and all pizzas contain exactly S slices.
It is known that the i-th contestant will eat si slices of pizza, and gain ai happiness for each slice of type 1 pizza they eat, and bi happiness for each slice of type 2 pizza they eat. We can order any number of type 1 and type 2 pizzas, but we want to buy the minimum possible number of pizzas for all of the contestants to be able to eat their required number of slices. Given that restriction, what is the maximum possible total happiness that can be achieved?
Input
The first line of input will contain integers N and S (1 ≤ N ≤ 105, 1 ≤ S ≤ 105), the number of contestants and the number of slices per pizza, respectively. N lines follow.
The i-th such line contains integers si, ai, and bi (1 ≤ si ≤ 105, 1 ≤ ai ≤ 105, 1 ≤ bi ≤ 105), the number of slices the i-th contestant will eat, the happiness they will gain from each type 1 slice they eat, and the happiness they will gain from each type 2 slice they eat, respectively.
Output
Print the maximum total happiness that can be achieved.
Examples
Input
3 12
3 5 7
4 6 7
5 9 5
Output
84
Input
6 10
7 4 7
5 8 8
12 5 8
6 11 6
3 3 7
5 9 6
Output
314
Note
In the first example, you only need to buy one pizza. If you buy a type 1 pizza, the total happiness will be 3·5 + 4·6 + 5·9 = 84, and if you buy a type 2 pizza, the total happiness will be 3·7 + 4·7 + 5·5 = 74. | instruction | 0 | 89,818 | 9 | 179,636 |
Tags: binary search, sortings, ternary search
Correct Solution:
```
import math
class Person:
def __init__(self, slices, a_happiness, b_happiness):
self.slices = slices
self.a_happiness = a_happiness
self.b_happiness = b_happiness
self.choice = None
def make_greedy_choice(self):
if self.a_happiness > self.b_happiness:
self.choice = (self.slices, 0)
else:
self.choice = (0, self.slices)
def happiness_for(self, a_slices, b_slices):
return a_slices * self.a_happiness + b_slices * self.b_happiness
@property
def happiness(self):
if self.choice is None:
raise RuntimeError("need pizza choice to compute happiness")
return self.happiness_for(self.choice[0], self.choice[1])
def choose(self, a, b):
self.choice = (a, b)
def __repr__(self):
return "Person({}, {}, {})".format(self.slices, self.a_happiness, self.b_happiness)
if __name__ == "__main__":
n, slices = [int(x) for x in input().split()]
people = []
for _ in range(n):
people.append(Person(*[int(x) for x in input().split()]))
required_slices = sum([person.slices for person in people])
quantity = math.ceil(required_slices / slices)
#print("Ordering {} pizzas (min quantity) for {} people: ".format(quantity, len(people)), people)
# first make greedy choice
for person in people:
person.make_greedy_choice()
greedy_happiness = sum([person.happiness for person in people])
greedy_a_slices = sum([person.choice[0] for person in people])
greedy_b_slices = sum([person.choice[1] for person in people])
greedy_quantity = math.ceil(greedy_a_slices / slices) + math.ceil(greedy_b_slices / slices)
#print("{} total happiness for greedy choices ({}, {}) resulting in {} pizzas".format(greedy_happiness, greedy_a_slices, greedy_b_slices, greedy_quantity))
if greedy_quantity <= quantity:
print(greedy_happiness)
else:
# Need to either change slice choices in a way that causes minimum reduction in happiness
## considering A to B
slices_to_change = greedy_a_slices % slices
a_people = filter(lambda person: person.choice[0] > 0, people)
a_reduction = 0
# sort a_people from least opposed to changing to most opposed
a_people = sorted(a_people, key=lambda person: person.a_happiness - person.b_happiness)
#print(a_people)
for person in a_people:
if slices_to_change == 0:
break
can_change = min(person.choice[0], slices_to_change)
a_reduction += (person.a_happiness - person.b_happiness) * can_change
slices_to_change -= can_change
#print("minimum happiness reduction from changing {} A slice choices to B was {}".format(greedy_a_slices % slices, a_reduction))
## considering B to A
slices_to_change = greedy_b_slices % slices
b_people = filter(lambda person: person.choice[1] > 0, people)
b_reduction = 0
# sort b_people from least opposed to changing to most opposed
b_people = sorted(b_people, key=lambda person: person.b_happiness - person.a_happiness)
#print(b_people)
for person in b_people:
if slices_to_change == 0:
break
can_change = min(person.choice[1], slices_to_change)
b_reduction += (person.b_happiness - person.a_happiness) * can_change
slices_to_change -= can_change
#print("minimum happiness reduction from changing {} B slice choices to A was {}".format(greedy_b_slices % slices, b_reduction))
print(greedy_happiness - min(a_reduction, b_reduction))
``` | output | 1 | 89,818 | 9 | 179,637 |
Provide tags and a correct Python 3 solution for this coding contest problem.
It's another Start[c]up finals, and that means there is pizza to order for the onsite contestants. There are only 2 types of pizza (obviously not, but let's just pretend for the sake of the problem), and all pizzas contain exactly S slices.
It is known that the i-th contestant will eat si slices of pizza, and gain ai happiness for each slice of type 1 pizza they eat, and bi happiness for each slice of type 2 pizza they eat. We can order any number of type 1 and type 2 pizzas, but we want to buy the minimum possible number of pizzas for all of the contestants to be able to eat their required number of slices. Given that restriction, what is the maximum possible total happiness that can be achieved?
Input
The first line of input will contain integers N and S (1 ≤ N ≤ 105, 1 ≤ S ≤ 105), the number of contestants and the number of slices per pizza, respectively. N lines follow.
The i-th such line contains integers si, ai, and bi (1 ≤ si ≤ 105, 1 ≤ ai ≤ 105, 1 ≤ bi ≤ 105), the number of slices the i-th contestant will eat, the happiness they will gain from each type 1 slice they eat, and the happiness they will gain from each type 2 slice they eat, respectively.
Output
Print the maximum total happiness that can be achieved.
Examples
Input
3 12
3 5 7
4 6 7
5 9 5
Output
84
Input
6 10
7 4 7
5 8 8
12 5 8
6 11 6
3 3 7
5 9 6
Output
314
Note
In the first example, you only need to buy one pizza. If you buy a type 1 pizza, the total happiness will be 3·5 + 4·6 + 5·9 = 84, and if you buy a type 2 pizza, the total happiness will be 3·7 + 4·7 + 5·5 = 74. | instruction | 0 | 89,819 | 9 | 179,638 |
Tags: binary search, sortings, ternary search
Correct Solution:
```
def solve(arr):
arr.sort(key=lambda q: q[1] - q[0])
m = sum(si for a, b, si in arr)
k = s * (m // s)
n = m - k
x, y, z = 0, 0, 0
for a, b, si in arr:
if k >= si:
k -= si
z += si * a
elif k > 0:
z += k * a
x = (si - k) * a
y = (si - k) * b
k = 0
else:
x += si * a
y += si * b
return x, y, z, n
n, s = map(int, input().split())
arr1, arr2 = [], []
for i in range(n):
si, ai, bi = map(int, input().split())
if ai > bi: arr1.append((ai, bi, si))
else: arr2.append((bi, ai, si))
x1, y1, z1, n1 = solve(arr1)
x2, y2, z2, n2 = solve(arr2)
d = x1 + x2 if n1 + n2 > s else max(x1 + y2, x2 + y1)
print(z1 + z2 + d)
``` | output | 1 | 89,819 | 9 | 179,639 |
Provide tags and a correct Python 3 solution for this coding contest problem.
It's another Start[c]up finals, and that means there is pizza to order for the onsite contestants. There are only 2 types of pizza (obviously not, but let's just pretend for the sake of the problem), and all pizzas contain exactly S slices.
It is known that the i-th contestant will eat si slices of pizza, and gain ai happiness for each slice of type 1 pizza they eat, and bi happiness for each slice of type 2 pizza they eat. We can order any number of type 1 and type 2 pizzas, but we want to buy the minimum possible number of pizzas for all of the contestants to be able to eat their required number of slices. Given that restriction, what is the maximum possible total happiness that can be achieved?
Input
The first line of input will contain integers N and S (1 ≤ N ≤ 105, 1 ≤ S ≤ 105), the number of contestants and the number of slices per pizza, respectively. N lines follow.
The i-th such line contains integers si, ai, and bi (1 ≤ si ≤ 105, 1 ≤ ai ≤ 105, 1 ≤ bi ≤ 105), the number of slices the i-th contestant will eat, the happiness they will gain from each type 1 slice they eat, and the happiness they will gain from each type 2 slice they eat, respectively.
Output
Print the maximum total happiness that can be achieved.
Examples
Input
3 12
3 5 7
4 6 7
5 9 5
Output
84
Input
6 10
7 4 7
5 8 8
12 5 8
6 11 6
3 3 7
5 9 6
Output
314
Note
In the first example, you only need to buy one pizza. If you buy a type 1 pizza, the total happiness will be 3·5 + 4·6 + 5·9 = 84, and if you buy a type 2 pizza, the total happiness will be 3·7 + 4·7 + 5·5 = 74. | instruction | 0 | 89,820 | 9 | 179,640 |
Tags: binary search, sortings, ternary search
Correct Solution:
```
import sys
n, s = tuple(map(int, input().split()))
x = []
for i in range(n):
si, a, b = tuple(map(int, input().split()))
x.append([si, a, b])
x = sorted(x, key=lambda t: abs(t[1] - t[2]))
labels = []
sum1 = 0
sum2 = 0
res = 0
for i in range(len(x)):
res += x[i][0] * max(x[i][1], x[i][2])
if x[i][1] > x[i][2]:
sum1 += x[i][0]
labels.append(1)
elif x[i][1] < x[i][2]:
sum2 += x[i][0]
labels.append(2)
else:
if sum1 > sum2:
sum2 += x[i][0]
labels.append(2)
else:
sum1 = x[i][0]
labels.append(1)
if ((sum1 - 1) // s + 1 + (sum2 - 1) // s + 1) == ((sum1 + sum2 - 1) // s + 1):
print(res)
sys.exit(0)
s1 = sum1
s2 = sum2
res1 = res
for i in range(len(labels)):
c = False
if labels[i] == 1:
j = 0
while x[i][0] - j != 0:
if s1 % s == 0 or s2 % s == 0:
c = True
break
j += 1
s1 -= 1
s2 += 1
res1 -= abs(x[i][1] - x[i][2])
if c:
break
s1 = sum1
s2 = sum2
for i in range(len(labels)):
c = False
if labels[i] == 2:
j = 0
while x[i][0] - j != 0:
if s1 % s == 0 or s2 % s == 0:
c = True
break
j += 1
s1 += 1
s2 -= 1
res -= abs(x[i][1] - x[i][2])
if c:
break
print(max(res1, res))
``` | output | 1 | 89,820 | 9 | 179,641 |
Provide tags and a correct Python 3 solution for this coding contest problem.
It's another Start[c]up finals, and that means there is pizza to order for the onsite contestants. There are only 2 types of pizza (obviously not, but let's just pretend for the sake of the problem), and all pizzas contain exactly S slices.
It is known that the i-th contestant will eat si slices of pizza, and gain ai happiness for each slice of type 1 pizza they eat, and bi happiness for each slice of type 2 pizza they eat. We can order any number of type 1 and type 2 pizzas, but we want to buy the minimum possible number of pizzas for all of the contestants to be able to eat their required number of slices. Given that restriction, what is the maximum possible total happiness that can be achieved?
Input
The first line of input will contain integers N and S (1 ≤ N ≤ 105, 1 ≤ S ≤ 105), the number of contestants and the number of slices per pizza, respectively. N lines follow.
The i-th such line contains integers si, ai, and bi (1 ≤ si ≤ 105, 1 ≤ ai ≤ 105, 1 ≤ bi ≤ 105), the number of slices the i-th contestant will eat, the happiness they will gain from each type 1 slice they eat, and the happiness they will gain from each type 2 slice they eat, respectively.
Output
Print the maximum total happiness that can be achieved.
Examples
Input
3 12
3 5 7
4 6 7
5 9 5
Output
84
Input
6 10
7 4 7
5 8 8
12 5 8
6 11 6
3 3 7
5 9 6
Output
314
Note
In the first example, you only need to buy one pizza. If you buy a type 1 pizza, the total happiness will be 3·5 + 4·6 + 5·9 = 84, and if you buy a type 2 pizza, the total happiness will be 3·7 + 4·7 + 5·5 = 74. | instruction | 0 | 89,821 | 9 | 179,642 |
Tags: binary search, sortings, ternary search
Correct Solution:
```
def solve(ls):
ls.sort(key=lambda q: q[1] - q[0])
m = sum(si for a, b, si in ls)
k = s * (m // s)
n = m - k
x = y = z = 0
for a, b, si in ls:
if k >= si:
k -= si
z += si * a
elif k:
z += k * a
x = (si - k) * a
y = (si - k) * b
k = 0
else:
x += si * a
y += si * b
return x, y, z, n
n,s = map(int,input().split())
first=[]
second=[]
for i in range(n):
si, ai, bi = map(int,input().split())
if ai>bi:
first.append((ai,bi,si))
else:
second.append((bi,ai,si))
x1,y1,z1,n1 = solve(first)
x2,y2,z2,n2 = solve(second)
d = x1+x2 if n1+n2>s else max(x1+y2,x2+y1)
print(z1+z2+d)
``` | output | 1 | 89,821 | 9 | 179,643 |
Provide tags and a correct Python 3 solution for this coding contest problem.
It's another Start[c]up finals, and that means there is pizza to order for the onsite contestants. There are only 2 types of pizza (obviously not, but let's just pretend for the sake of the problem), and all pizzas contain exactly S slices.
It is known that the i-th contestant will eat si slices of pizza, and gain ai happiness for each slice of type 1 pizza they eat, and bi happiness for each slice of type 2 pizza they eat. We can order any number of type 1 and type 2 pizzas, but we want to buy the minimum possible number of pizzas for all of the contestants to be able to eat their required number of slices. Given that restriction, what is the maximum possible total happiness that can be achieved?
Input
The first line of input will contain integers N and S (1 ≤ N ≤ 105, 1 ≤ S ≤ 105), the number of contestants and the number of slices per pizza, respectively. N lines follow.
The i-th such line contains integers si, ai, and bi (1 ≤ si ≤ 105, 1 ≤ ai ≤ 105, 1 ≤ bi ≤ 105), the number of slices the i-th contestant will eat, the happiness they will gain from each type 1 slice they eat, and the happiness they will gain from each type 2 slice they eat, respectively.
Output
Print the maximum total happiness that can be achieved.
Examples
Input
3 12
3 5 7
4 6 7
5 9 5
Output
84
Input
6 10
7 4 7
5 8 8
12 5 8
6 11 6
3 3 7
5 9 6
Output
314
Note
In the first example, you only need to buy one pizza. If you buy a type 1 pizza, the total happiness will be 3·5 + 4·6 + 5·9 = 84, and if you buy a type 2 pizza, the total happiness will be 3·7 + 4·7 + 5·5 = 74. | instruction | 0 | 89,822 | 9 | 179,644 |
Tags: binary search, sortings, ternary search
Correct Solution:
```
def get_losts(persons, count):
persons.sort(key = lambda p : p.lost)
losts = 0
i = 0
while count > 0:
df = min(count, persons[i].s)
losts += df * persons[i].lost
count -= df
i += 1
return losts
class Person:
def __init__(self, _s, _a, _b, _lost):
self.s = _s
self.a = _a
self.b = _b
self.lost = _lost
n, m = map(int, input().split())
s_count = 0
a_pizza = list()
a_count = 0
a_points = 0
b_pizza = list()
b_count = 0
b_points = 0
neutral_points = 0
neutral_count = 0
for i in range(n):
s, a, b = map(int, input().split())
s_count += s
if a == b:
neutral_points += s*a
s_count -= s
neutral_count += s
elif a > b:
a_pizza.append(Person(s, a, b, a - b))
a_count += s
a_points += s*a
else:
b_pizza.append(Person(s, a, b, b - a))
b_count += s
b_points += s*b
a_lost = a_count % m
b_lost = b_count % m
if a_lost + b_lost + neutral_count > m or a_lost == 0 or b_lost == 0:
print(neutral_points + a_points + b_points)
else:
a_lost = get_losts(a_pizza, a_lost)
b_lost = get_losts(b_pizza, b_lost)
print(neutral_points + a_points + b_points - min(a_lost, b_lost))
``` | output | 1 | 89,822 | 9 | 179,645 |
Provide tags and a correct Python 3 solution for this coding contest problem.
It's another Start[c]up finals, and that means there is pizza to order for the onsite contestants. There are only 2 types of pizza (obviously not, but let's just pretend for the sake of the problem), and all pizzas contain exactly S slices.
It is known that the i-th contestant will eat si slices of pizza, and gain ai happiness for each slice of type 1 pizza they eat, and bi happiness for each slice of type 2 pizza they eat. We can order any number of type 1 and type 2 pizzas, but we want to buy the minimum possible number of pizzas for all of the contestants to be able to eat their required number of slices. Given that restriction, what is the maximum possible total happiness that can be achieved?
Input
The first line of input will contain integers N and S (1 ≤ N ≤ 105, 1 ≤ S ≤ 105), the number of contestants and the number of slices per pizza, respectively. N lines follow.
The i-th such line contains integers si, ai, and bi (1 ≤ si ≤ 105, 1 ≤ ai ≤ 105, 1 ≤ bi ≤ 105), the number of slices the i-th contestant will eat, the happiness they will gain from each type 1 slice they eat, and the happiness they will gain from each type 2 slice they eat, respectively.
Output
Print the maximum total happiness that can be achieved.
Examples
Input
3 12
3 5 7
4 6 7
5 9 5
Output
84
Input
6 10
7 4 7
5 8 8
12 5 8
6 11 6
3 3 7
5 9 6
Output
314
Note
In the first example, you only need to buy one pizza. If you buy a type 1 pizza, the total happiness will be 3·5 + 4·6 + 5·9 = 84, and if you buy a type 2 pizza, the total happiness will be 3·7 + 4·7 + 5·5 = 74. | instruction | 0 | 89,823 | 9 | 179,646 |
Tags: binary search, sortings, ternary search
Correct Solution:
```
def solve(ls):
ls.sort(key=lambda q: q[1] - q[0])
m = sum(si for a, b, si in ls)
k = s * (m // s)
n = m - k
x = y = z = 0
for a, b, si in ls:
if k >= si:
k -= si
z += si * a
elif k:
z += k * a
x = (si - k) * a
y = (si - k) * b
k = 0
else:
x += si * a
y += si * b
return x, y, z, n
n,s = map(int,input().split())
first=[]
second=[]
for i in range(n):
si, ai, bi = map(int,input().split())
if ai>bi:
first.append((ai,bi,si))
else:
second.append((bi,ai,si))
x1,y1,z1,n1 = solve(first)
x2,y2,z2,n2 = solve(second)
d = x1+x2 if n1+n2>s else max(x1+y2,x2+y1)
print(z1+z2+d)
# Made By Mostafa_Khaled
``` | output | 1 | 89,823 | 9 | 179,647 |
Provide tags and a correct Python 3 solution for this coding contest problem.
It's another Start[c]up finals, and that means there is pizza to order for the onsite contestants. There are only 2 types of pizza (obviously not, but let's just pretend for the sake of the problem), and all pizzas contain exactly S slices.
It is known that the i-th contestant will eat si slices of pizza, and gain ai happiness for each slice of type 1 pizza they eat, and bi happiness for each slice of type 2 pizza they eat. We can order any number of type 1 and type 2 pizzas, but we want to buy the minimum possible number of pizzas for all of the contestants to be able to eat their required number of slices. Given that restriction, what is the maximum possible total happiness that can be achieved?
Input
The first line of input will contain integers N and S (1 ≤ N ≤ 105, 1 ≤ S ≤ 105), the number of contestants and the number of slices per pizza, respectively. N lines follow.
The i-th such line contains integers si, ai, and bi (1 ≤ si ≤ 105, 1 ≤ ai ≤ 105, 1 ≤ bi ≤ 105), the number of slices the i-th contestant will eat, the happiness they will gain from each type 1 slice they eat, and the happiness they will gain from each type 2 slice they eat, respectively.
Output
Print the maximum total happiness that can be achieved.
Examples
Input
3 12
3 5 7
4 6 7
5 9 5
Output
84
Input
6 10
7 4 7
5 8 8
12 5 8
6 11 6
3 3 7
5 9 6
Output
314
Note
In the first example, you only need to buy one pizza. If you buy a type 1 pizza, the total happiness will be 3·5 + 4·6 + 5·9 = 84, and if you buy a type 2 pizza, the total happiness will be 3·7 + 4·7 + 5·5 = 74. | instruction | 0 | 89,824 | 9 | 179,648 |
Tags: binary search, sortings, ternary search
Correct Solution:
```
n, S = map(int, input().split())
arr = []
suma = 0
happy = 0
sumb = 0
dif = []
for i in range(n):
c, a, b = map(int, input().split())
if a >= b:
suma += c
happy += a * c
else:
sumb += c
happy += b * c
dif.append((a - b, c))
dif.sort()
num = (suma + sumb - 1) // S + 1
if (suma - 1) // S + 1 + (sumb - 1) // S + 1 <= num:
print(happy)
else:
moda = suma % S
modb = sumb % S
#a->b
for i in range(n):
if dif[i][0] >= 0:
ind = i
break
ans1 = happy
ans2 = happy
first = min(S - modb, moda)
if first <= moda:
now = ind
ans1 = 0
while first > 0:
if dif[now][1] > first:
ans1 += dif[now][0] * first
first = 0
else:
ans1 += dif[now][0] * dif[now][1]
first -= dif[now][1]
now += 1
#b->a
second = min(S - moda, modb)
if second <= modb:
now = ind - 1
ans2 = 0
while second > 0:
if dif[now][1] > second:
ans2 -= dif[now][0] * second
second = 0
else:
ans2 -= dif[now][0] * dif[now][1]
second -= dif[now][1]
now -= 1
print(happy - min(ans1, ans2))
``` | output | 1 | 89,824 | 9 | 179,649 |
Provide tags and a correct Python 3 solution for this coding contest problem.
It's another Start[c]up finals, and that means there is pizza to order for the onsite contestants. There are only 2 types of pizza (obviously not, but let's just pretend for the sake of the problem), and all pizzas contain exactly S slices.
It is known that the i-th contestant will eat si slices of pizza, and gain ai happiness for each slice of type 1 pizza they eat, and bi happiness for each slice of type 2 pizza they eat. We can order any number of type 1 and type 2 pizzas, but we want to buy the minimum possible number of pizzas for all of the contestants to be able to eat their required number of slices. Given that restriction, what is the maximum possible total happiness that can be achieved?
Input
The first line of input will contain integers N and S (1 ≤ N ≤ 105, 1 ≤ S ≤ 105), the number of contestants and the number of slices per pizza, respectively. N lines follow.
The i-th such line contains integers si, ai, and bi (1 ≤ si ≤ 105, 1 ≤ ai ≤ 105, 1 ≤ bi ≤ 105), the number of slices the i-th contestant will eat, the happiness they will gain from each type 1 slice they eat, and the happiness they will gain from each type 2 slice they eat, respectively.
Output
Print the maximum total happiness that can be achieved.
Examples
Input
3 12
3 5 7
4 6 7
5 9 5
Output
84
Input
6 10
7 4 7
5 8 8
12 5 8
6 11 6
3 3 7
5 9 6
Output
314
Note
In the first example, you only need to buy one pizza. If you buy a type 1 pizza, the total happiness will be 3·5 + 4·6 + 5·9 = 84, and if you buy a type 2 pizza, the total happiness will be 3·7 + 4·7 + 5·5 = 74. | instruction | 0 | 89,825 | 9 | 179,650 |
Tags: binary search, sortings, ternary search
Correct Solution:
```
f = lambda: map(int, input().split())
n, s = f()
u, v = [], []
for i in range(n):
d, a, b = f()
if a > b:
u.append([a, b, d])
else:
v.append([b, a, d])
def g(t):
t.sort(key=lambda q: q[1] - q[0])
m = sum(d for a, b, d in t)
k = s * (m // s)
n = m - k
x = y = z = 0
for a, b, d in t:
if k >= d:
k -= d
z += d * a
elif k:
z += k * a
x = (d - k) * a
y = (d - k) * b
k = 0
else:
x += d * a
y += d * b
return x, y, z, n
a, b = g(u), g(v)
d = a[0] + b[0] if a[3] + b[3] > s else max(a[0] + b[1], a[1] + b[0])
print(a[2] + b[2] + d)
``` | output | 1 | 89,825 | 9 | 179,651 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
It's another Start[c]up finals, and that means there is pizza to order for the onsite contestants. There are only 2 types of pizza (obviously not, but let's just pretend for the sake of the problem), and all pizzas contain exactly S slices.
It is known that the i-th contestant will eat si slices of pizza, and gain ai happiness for each slice of type 1 pizza they eat, and bi happiness for each slice of type 2 pizza they eat. We can order any number of type 1 and type 2 pizzas, but we want to buy the minimum possible number of pizzas for all of the contestants to be able to eat their required number of slices. Given that restriction, what is the maximum possible total happiness that can be achieved?
Input
The first line of input will contain integers N and S (1 ≤ N ≤ 105, 1 ≤ S ≤ 105), the number of contestants and the number of slices per pizza, respectively. N lines follow.
The i-th such line contains integers si, ai, and bi (1 ≤ si ≤ 105, 1 ≤ ai ≤ 105, 1 ≤ bi ≤ 105), the number of slices the i-th contestant will eat, the happiness they will gain from each type 1 slice they eat, and the happiness they will gain from each type 2 slice they eat, respectively.
Output
Print the maximum total happiness that can be achieved.
Examples
Input
3 12
3 5 7
4 6 7
5 9 5
Output
84
Input
6 10
7 4 7
5 8 8
12 5 8
6 11 6
3 3 7
5 9 6
Output
314
Note
In the first example, you only need to buy one pizza. If you buy a type 1 pizza, the total happiness will be 3·5 + 4·6 + 5·9 = 84, and if you buy a type 2 pizza, the total happiness will be 3·7 + 4·7 + 5·5 = 74.
Submitted Solution:
```
from math import ceil
N, S = input().split()
N, S = int(N), int(S)
C = 0
pC = 0
nC = 0
cArr = []
for i in range(N):
s, a, b = input().split()
s, a, b = int(s), int(a), int(b)
C += s * b
cArr.append((a - b, s))
if a > b:
pC += s
else:
nC += s
cArr.sort(key=lambda k: -k[0])
tP = int(ceil((nC + pC) / S))
nP = int(pC / S)
hAns = C
sItr = nP * S
itr = 0
while sItr > 0 and itr < N:
si = min(cArr[itr][1], sItr)
hAns += si * cArr[itr][0]
sItr -= si
itr += 1
hAns2 = C
nP = int(pC / S) + 1
sItr = nP * S
e = S*(tP - nP) - nC
itr = 0
while itr < N and cArr[itr][0] > 0:
si = min(cArr[itr][1], sItr)
hAns2 += si * cArr[itr][0]
sItr -= si
itr += 1
if e < 0:
sItr = -e
while sItr > 0 and itr < N:
si = min(cArr[itr][1], sItr)
hAns2 += si * cArr[itr][0]
sItr -= si
itr += 1
print(max(hAns, hAns2))
``` | instruction | 0 | 89,826 | 9 | 179,652 |
Yes | output | 1 | 89,826 | 9 | 179,653 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
It's another Start[c]up finals, and that means there is pizza to order for the onsite contestants. There are only 2 types of pizza (obviously not, but let's just pretend for the sake of the problem), and all pizzas contain exactly S slices.
It is known that the i-th contestant will eat si slices of pizza, and gain ai happiness for each slice of type 1 pizza they eat, and bi happiness for each slice of type 2 pizza they eat. We can order any number of type 1 and type 2 pizzas, but we want to buy the minimum possible number of pizzas for all of the contestants to be able to eat their required number of slices. Given that restriction, what is the maximum possible total happiness that can be achieved?
Input
The first line of input will contain integers N and S (1 ≤ N ≤ 105, 1 ≤ S ≤ 105), the number of contestants and the number of slices per pizza, respectively. N lines follow.
The i-th such line contains integers si, ai, and bi (1 ≤ si ≤ 105, 1 ≤ ai ≤ 105, 1 ≤ bi ≤ 105), the number of slices the i-th contestant will eat, the happiness they will gain from each type 1 slice they eat, and the happiness they will gain from each type 2 slice they eat, respectively.
Output
Print the maximum total happiness that can be achieved.
Examples
Input
3 12
3 5 7
4 6 7
5 9 5
Output
84
Input
6 10
7 4 7
5 8 8
12 5 8
6 11 6
3 3 7
5 9 6
Output
314
Note
In the first example, you only need to buy one pizza. If you buy a type 1 pizza, the total happiness will be 3·5 + 4·6 + 5·9 = 84, and if you buy a type 2 pizza, the total happiness will be 3·7 + 4·7 + 5·5 = 74.
Submitted Solution:
```
n, s = map(int, input().split())
data = [[] for i in range(n)]
sec = 0
fir = 0
su = 0
for i in range(n):
a = list(map(int, input().split()))
data[i].append(abs(a[1] - a[2]))
if a[1] >= a[2]:
fir += a[0]
su += a[0] * a[1]
data[i].append(0)
else:
sec += a[0]
su += a[0] * a[2]
data[i].append(1)
data[i] += a
data = sorted(data)
#print(data)
fis = fir % s
sis = sec % s
if (fis + sis) > s or (fis == 0 or sis == 0):
print(su)
else:
cou = fis
k = 0
su1 = su
while cou > 0 and k < n:
if data[k][1] == 1:
k += 1
continue;
su1 -= min(cou, data[k][2]) * data[k][0]
cou -= min(cou, data[k][2])
k += 1
cou = sis
k = 0
su2 = su
while cou > 0 and k < n:
if data[k][1] == 0:
k += 1
continue;
su2 -= min(cou, data[k][2]) * data[k][0]
cou -= min(cou, data[k][2])
k += 1
print(max(su1, su2))
``` | instruction | 0 | 89,827 | 9 | 179,654 |
Yes | output | 1 | 89,827 | 9 | 179,655 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
It's another Start[c]up finals, and that means there is pizza to order for the onsite contestants. There are only 2 types of pizza (obviously not, but let's just pretend for the sake of the problem), and all pizzas contain exactly S slices.
It is known that the i-th contestant will eat si slices of pizza, and gain ai happiness for each slice of type 1 pizza they eat, and bi happiness for each slice of type 2 pizza they eat. We can order any number of type 1 and type 2 pizzas, but we want to buy the minimum possible number of pizzas for all of the contestants to be able to eat their required number of slices. Given that restriction, what is the maximum possible total happiness that can be achieved?
Input
The first line of input will contain integers N and S (1 ≤ N ≤ 105, 1 ≤ S ≤ 105), the number of contestants and the number of slices per pizza, respectively. N lines follow.
The i-th such line contains integers si, ai, and bi (1 ≤ si ≤ 105, 1 ≤ ai ≤ 105, 1 ≤ bi ≤ 105), the number of slices the i-th contestant will eat, the happiness they will gain from each type 1 slice they eat, and the happiness they will gain from each type 2 slice they eat, respectively.
Output
Print the maximum total happiness that can be achieved.
Examples
Input
3 12
3 5 7
4 6 7
5 9 5
Output
84
Input
6 10
7 4 7
5 8 8
12 5 8
6 11 6
3 3 7
5 9 6
Output
314
Note
In the first example, you only need to buy one pizza. If you buy a type 1 pizza, the total happiness will be 3·5 + 4·6 + 5·9 = 84, and if you buy a type 2 pizza, the total happiness will be 3·7 + 4·7 + 5·5 = 74.
Submitted Solution:
```
n, S = 0, 0
s = []
a = []
b = []
class node:
def __init__(self, x, id):
self.x = x
self.id = id
return
def __lt__(self, p):
return self.x < p.x
c = []
i , f = 0, 0
ans, sum, a1, a2 = 0, 0, 0, 0
s1, s2 = 0, 0
line = input().split()
n, S = int(line[0]), int(line[1])
for i in range(n):
line = input().split()
s.append(int(line[0]))
a.append(int(line[1]))
b.append(int(line[2]))
if a[i] > b[i]:
s1 += s[i]
elif a[i] < b[i]:
s2 += s[i]
sum += s[i]
ans += max(a[i], b[i]) * s[i]
c.append(node(a[i] - b[i], i))
cnt = (sum + S - 1) // S
if (s1 + S - 1) // S + (s2 + S - 1) // S <= cnt:
print(ans)
else:
c.sort()
s1 %= S
s2 %= S
for i in range(n):
if c[i].x <= 0:
f = i
continue
if not s1:
break
t = min(s[c[i].id], s1)
a1 += t * c[i].x
s1 -= t
for i in range(f, -1, -1):
if not c[i].x:
continue
if not s2:
break
t = min(s[c[i].id], s2)
a2 -= t * c[i].x
s2 -= t
print(ans - min(a1, a2))
``` | instruction | 0 | 89,828 | 9 | 179,656 |
Yes | output | 1 | 89,828 | 9 | 179,657 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
It's another Start[c]up finals, and that means there is pizza to order for the onsite contestants. There are only 2 types of pizza (obviously not, but let's just pretend for the sake of the problem), and all pizzas contain exactly S slices.
It is known that the i-th contestant will eat si slices of pizza, and gain ai happiness for each slice of type 1 pizza they eat, and bi happiness for each slice of type 2 pizza they eat. We can order any number of type 1 and type 2 pizzas, but we want to buy the minimum possible number of pizzas for all of the contestants to be able to eat their required number of slices. Given that restriction, what is the maximum possible total happiness that can be achieved?
Input
The first line of input will contain integers N and S (1 ≤ N ≤ 105, 1 ≤ S ≤ 105), the number of contestants and the number of slices per pizza, respectively. N lines follow.
The i-th such line contains integers si, ai, and bi (1 ≤ si ≤ 105, 1 ≤ ai ≤ 105, 1 ≤ bi ≤ 105), the number of slices the i-th contestant will eat, the happiness they will gain from each type 1 slice they eat, and the happiness they will gain from each type 2 slice they eat, respectively.
Output
Print the maximum total happiness that can be achieved.
Examples
Input
3 12
3 5 7
4 6 7
5 9 5
Output
84
Input
6 10
7 4 7
5 8 8
12 5 8
6 11 6
3 3 7
5 9 6
Output
314
Note
In the first example, you only need to buy one pizza. If you buy a type 1 pizza, the total happiness will be 3·5 + 4·6 + 5·9 = 84, and if you buy a type 2 pizza, the total happiness will be 3·7 + 4·7 + 5·5 = 74.
Submitted Solution:
```
def cns(ts,s):
if ts/s==int(ts/s):
return ts
else:
return (int(ts/s)+1)*s
n,spp=[int(i) for i in input().split()]
tsr=0
da=[[] for i in range(100005)]
db=[[] for i in range(100005)]
sl=[]
for i in range(n):
sl.append([int(j) for j in input().split()])
tsr+=sl[i][0]
if sl[i][1]>sl[i][2]:
da[sl[i][1]-sl[i][2]].append(i)
else:
db[sl[i][2]-sl[i][1]].append(i)
tsa=cns(tsr,spp)
a1=0
c1=0
for i in range(100000,-1,-1):
for j in da[i]:
a1+=sl[j][0]*sl[j][1]
c1+=sl[j][0]
c1r=cns(c1,spp)-c1
c2r=tsa-cns(c1,spp)
for i in range(100000,-1,-1):
for j in db[i]:
if sl[j][0]>c2r:
a1+=c2r*sl[j][2]
a1+=(sl[j][0]-c2r)*sl[j][1]
c2r=0
else:
a1+=sl[j][0]*sl[j][2]
c2r-=sl[j][0]
a2=0
c2=0
for i in range(100000,-1,-1):
for j in db[i]:
a2+=sl[j][0]*sl[j][2]
c2+=sl[j][0]
c2r=cns(c2,spp)-c2
c1r=tsa-cns(c2,spp)
for i in range(100000,-1,-1):
for j in da[i]:
if sl[j][0]>c1r:
a2+=c1r*sl[j][1]
a2+=(sl[j][0]-c1r)*sl[j][2]
c1r=0
else:
a2+=sl[j][0]*sl[j][1]
c1r-=sl[j][0]
print(max(a1,a2))
``` | instruction | 0 | 89,829 | 9 | 179,658 |
Yes | output | 1 | 89,829 | 9 | 179,659 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
It's another Start[c]up finals, and that means there is pizza to order for the onsite contestants. There are only 2 types of pizza (obviously not, but let's just pretend for the sake of the problem), and all pizzas contain exactly S slices.
It is known that the i-th contestant will eat si slices of pizza, and gain ai happiness for each slice of type 1 pizza they eat, and bi happiness for each slice of type 2 pizza they eat. We can order any number of type 1 and type 2 pizzas, but we want to buy the minimum possible number of pizzas for all of the contestants to be able to eat their required number of slices. Given that restriction, what is the maximum possible total happiness that can be achieved?
Input
The first line of input will contain integers N and S (1 ≤ N ≤ 105, 1 ≤ S ≤ 105), the number of contestants and the number of slices per pizza, respectively. N lines follow.
The i-th such line contains integers si, ai, and bi (1 ≤ si ≤ 105, 1 ≤ ai ≤ 105, 1 ≤ bi ≤ 105), the number of slices the i-th contestant will eat, the happiness they will gain from each type 1 slice they eat, and the happiness they will gain from each type 2 slice they eat, respectively.
Output
Print the maximum total happiness that can be achieved.
Examples
Input
3 12
3 5 7
4 6 7
5 9 5
Output
84
Input
6 10
7 4 7
5 8 8
12 5 8
6 11 6
3 3 7
5 9 6
Output
314
Note
In the first example, you only need to buy one pizza. If you buy a type 1 pizza, the total happiness will be 3·5 + 4·6 + 5·9 = 84, and if you buy a type 2 pizza, the total happiness will be 3·7 + 4·7 + 5·5 = 74.
Submitted Solution:
```
f = lambda: map(int, input().split())
n, s = f()
u, v = [], []
for i in range(n):
d, a, b = f()
if a > b:
u.append([a, b, d])
else:
v.append([b, a, d])
def g(t):
t.sort(key=lambda q: q[1] - q[0])
k = s * (sum(d for a, b, d in t) // s)
x = y = z = 0
for a, b, d in t:
if k >= d:
k -= d
z += d * a
elif k:
z += k * a
x = (d - k) * a
y = (d - k) * b
k = 0
else:
x += d * a
y += d * b
return x, y, z
a, b = g(u), g(v)
print(a[2] + b[2] + max(a[0] + b[1], a[1] + b[0]))
``` | instruction | 0 | 89,830 | 9 | 179,660 |
No | output | 1 | 89,830 | 9 | 179,661 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
It's another Start[c]up finals, and that means there is pizza to order for the onsite contestants. There are only 2 types of pizza (obviously not, but let's just pretend for the sake of the problem), and all pizzas contain exactly S slices.
It is known that the i-th contestant will eat si slices of pizza, and gain ai happiness for each slice of type 1 pizza they eat, and bi happiness for each slice of type 2 pizza they eat. We can order any number of type 1 and type 2 pizzas, but we want to buy the minimum possible number of pizzas for all of the contestants to be able to eat their required number of slices. Given that restriction, what is the maximum possible total happiness that can be achieved?
Input
The first line of input will contain integers N and S (1 ≤ N ≤ 105, 1 ≤ S ≤ 105), the number of contestants and the number of slices per pizza, respectively. N lines follow.
The i-th such line contains integers si, ai, and bi (1 ≤ si ≤ 105, 1 ≤ ai ≤ 105, 1 ≤ bi ≤ 105), the number of slices the i-th contestant will eat, the happiness they will gain from each type 1 slice they eat, and the happiness they will gain from each type 2 slice they eat, respectively.
Output
Print the maximum total happiness that can be achieved.
Examples
Input
3 12
3 5 7
4 6 7
5 9 5
Output
84
Input
6 10
7 4 7
5 8 8
12 5 8
6 11 6
3 3 7
5 9 6
Output
314
Note
In the first example, you only need to buy one pizza. If you buy a type 1 pizza, the total happiness will be 3·5 + 4·6 + 5·9 = 84, and if you buy a type 2 pizza, the total happiness will be 3·7 + 4·7 + 5·5 = 74.
Submitted Solution:
```
N, S = input().split()
N, S = int(N), int(S)
C = 0
pC = 0
cArr = []
for i in range(N):
s, a, b = input().split()
s, a, b = int(s), int(a), int(b)
C += s * b
cArr.append((a - b, s))
if a > b:
pC += s
cArr.sort(key=lambda k: -k[0])
hAns = 0
for i in range(2):
nP = int(pC / S) + i
hAnsTmp = C
sItr = nP * S
itr = 0
while sItr > 0:
si = min(cArr[itr][1], sItr)
hAnsTmp += si * cArr[itr][0]
sItr -= si
itr += 1
hAns = max(hAns, hAnsTmp)
print(hAns)
``` | instruction | 0 | 89,831 | 9 | 179,662 |
No | output | 1 | 89,831 | 9 | 179,663 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
It's another Start[c]up finals, and that means there is pizza to order for the onsite contestants. There are only 2 types of pizza (obviously not, but let's just pretend for the sake of the problem), and all pizzas contain exactly S slices.
It is known that the i-th contestant will eat si slices of pizza, and gain ai happiness for each slice of type 1 pizza they eat, and bi happiness for each slice of type 2 pizza they eat. We can order any number of type 1 and type 2 pizzas, but we want to buy the minimum possible number of pizzas for all of the contestants to be able to eat their required number of slices. Given that restriction, what is the maximum possible total happiness that can be achieved?
Input
The first line of input will contain integers N and S (1 ≤ N ≤ 105, 1 ≤ S ≤ 105), the number of contestants and the number of slices per pizza, respectively. N lines follow.
The i-th such line contains integers si, ai, and bi (1 ≤ si ≤ 105, 1 ≤ ai ≤ 105, 1 ≤ bi ≤ 105), the number of slices the i-th contestant will eat, the happiness they will gain from each type 1 slice they eat, and the happiness they will gain from each type 2 slice they eat, respectively.
Output
Print the maximum total happiness that can be achieved.
Examples
Input
3 12
3 5 7
4 6 7
5 9 5
Output
84
Input
6 10
7 4 7
5 8 8
12 5 8
6 11 6
3 3 7
5 9 6
Output
314
Note
In the first example, you only need to buy one pizza. If you buy a type 1 pizza, the total happiness will be 3·5 + 4·6 + 5·9 = 84, and if you buy a type 2 pizza, the total happiness will be 3·7 + 4·7 + 5·5 = 74.
Submitted Solution:
```
from math import ceil
N, S = input().split()
N, S = int(N), int(S)
C = 0
pC = 0
sTot = 0
cArr = []
for i in range(N):
s, a, b = input().split()
s, a, b = int(s), int(a), int(b)
C += s * b
sTot += s
cArr.append((a - b, s))
if a > b:
pC += s
cArr.sort(key=lambda k: -k[0])
hAns = 0
for i in range(2):
nP = int(pC / S) + i
hAnsTmp = C
sItr = nP * S
e = S * ceil(sTot / S) - sTot
itr = 0
while sItr > 0 and itr < N:
if not (cArr[itr][0] <= 0 and sItr <= e):
si = min(cArr[itr][1], sItr)
hAnsTmp += si * cArr[itr][0]
sItr -= si
itr += 1
hAns = max(hAns, hAnsTmp)
print(hAns)
``` | instruction | 0 | 89,832 | 9 | 179,664 |
No | output | 1 | 89,832 | 9 | 179,665 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
It's another Start[c]up finals, and that means there is pizza to order for the onsite contestants. There are only 2 types of pizza (obviously not, but let's just pretend for the sake of the problem), and all pizzas contain exactly S slices.
It is known that the i-th contestant will eat si slices of pizza, and gain ai happiness for each slice of type 1 pizza they eat, and bi happiness for each slice of type 2 pizza they eat. We can order any number of type 1 and type 2 pizzas, but we want to buy the minimum possible number of pizzas for all of the contestants to be able to eat their required number of slices. Given that restriction, what is the maximum possible total happiness that can be achieved?
Input
The first line of input will contain integers N and S (1 ≤ N ≤ 105, 1 ≤ S ≤ 105), the number of contestants and the number of slices per pizza, respectively. N lines follow.
The i-th such line contains integers si, ai, and bi (1 ≤ si ≤ 105, 1 ≤ ai ≤ 105, 1 ≤ bi ≤ 105), the number of slices the i-th contestant will eat, the happiness they will gain from each type 1 slice they eat, and the happiness they will gain from each type 2 slice they eat, respectively.
Output
Print the maximum total happiness that can be achieved.
Examples
Input
3 12
3 5 7
4 6 7
5 9 5
Output
84
Input
6 10
7 4 7
5 8 8
12 5 8
6 11 6
3 3 7
5 9 6
Output
314
Note
In the first example, you only need to buy one pizza. If you buy a type 1 pizza, the total happiness will be 3·5 + 4·6 + 5·9 = 84, and if you buy a type 2 pizza, the total happiness will be 3·7 + 4·7 + 5·5 = 74.
Submitted Solution:
```
# -*- coding: utf-8 -*-
import math
import collections
import bisect
import heapq
import time
"""
created by shhuan at 2017/10/3 12:35
"""
N, S = map(int, input().split())
M = []
for i in range(N):
M.append([int(x) for x in input().split()])
s = [M[i][0] for i in range(N)]
a = [M[i][1] for i in range(N)]
b = [M[i][2] for i in range(N)]
total = sum(s)
numpizza = int(math.ceil(total/S))
numslice = numpizza * S
pa = 0
pb = 0
pab = 0
sab = sorted([(a[i]-b[i], i) for i in range(N)], reverse=True)
for d, i in sab:
if d < 0:
pb += s[i]
elif d > 0:
pa += s[i]
else:
pab += 1
maxHappiness = 0
sbak = s
for i in range(pa//S, (pa+pab)//S+2):
j = i * S
k = numslice - j
if int(math.ceil(j/S)+math.ceil(k/S)) != numpizza:
continue
h = 0
s = [x for x in sbak]
l = 0
while j >= s[sab[l][1]] and l < N and sab[l][0] >= 0:
h += s[sab[l][1]] * a[sab[l][1]]
j -= s[sab[l][1]]
l += 1
r = N-1
while r >= 0 and k >= s[sab[r][1]] and sab[r][0] <= 0:
h += s[sab[r][1]] * b[sab[r][1]]
k -= s[sab[r][1]]
r -= 1
hm = 0
for jj in range(j+1):
kk = s[sab[l][1]] - jj
if kk <= k:
hm = max(hm, jj*a[sab[l][1]] + kk*b[sab[l][1]])
h += hm
maxHappiness = max(maxHappiness, h)
print(maxHappiness)
``` | instruction | 0 | 89,833 | 9 | 179,666 |
No | output | 1 | 89,833 | 9 | 179,667 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Julia is going to cook a chicken in the kitchen of her dormitory. To save energy, the stove in the kitchen automatically turns off after k minutes after turning on.
During cooking, Julia goes to the kitchen every d minutes and turns on the stove if it is turned off. While the cooker is turned off, it stays warm. The stove switches on and off instantly.
It is known that the chicken needs t minutes to be cooked on the stove, if it is turned on, and 2t minutes, if it is turned off. You need to find out, how much time will Julia have to cook the chicken, if it is considered that the chicken is cooked evenly, with constant speed when the stove is turned on and at a constant speed when it is turned off.
Input
The single line contains three integers k, d and t (1 ≤ k, d, t ≤ 1018).
Output
Print a single number, the total time of cooking in minutes. The relative or absolute error must not exceed 10 - 9.
Namely, let's assume that your answer is x and the answer of the jury is y. The checker program will consider your answer correct if <image>.
Examples
Input
3 2 6
Output
6.5
Input
4 2 20
Output
20.0
Note
In the first example, the chicken will be cooked for 3 minutes on the turned on stove, after this it will be cooked for <image>. Then the chicken will be cooked for one minute on a turned off stove, it will be cooked for <image>. Thus, after four minutes the chicken will be cooked for <image>. Before the fifth minute Julia will turn on the stove and after 2.5 minutes the chicken will be ready <image>.
In the second example, when the stove is turned off, Julia will immediately turn it on, so the stove will always be turned on and the chicken will be cooked in 20 minutes. | instruction | 0 | 89,835 | 9 | 179,670 |
Tags: binary search, implementation, math
Correct Solution:
```
import math
s = input()
s = s.split()
s = list(map(int, s))
k = s[0]
d = s[1]
t = s[2]
i = math.ceil(k / d)
c = i * d
m = (c + k) / 2
r1 = int(t / m)
remain = t - r1 * m
if remain < k:
print(r1 * c + remain)
else:
print(r1 * c + (remain - k) * 2 + k)
``` | output | 1 | 89,835 | 9 | 179,671 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Julia is going to cook a chicken in the kitchen of her dormitory. To save energy, the stove in the kitchen automatically turns off after k minutes after turning on.
During cooking, Julia goes to the kitchen every d minutes and turns on the stove if it is turned off. While the cooker is turned off, it stays warm. The stove switches on and off instantly.
It is known that the chicken needs t minutes to be cooked on the stove, if it is turned on, and 2t minutes, if it is turned off. You need to find out, how much time will Julia have to cook the chicken, if it is considered that the chicken is cooked evenly, with constant speed when the stove is turned on and at a constant speed when it is turned off.
Input
The single line contains three integers k, d and t (1 ≤ k, d, t ≤ 1018).
Output
Print a single number, the total time of cooking in minutes. The relative or absolute error must not exceed 10 - 9.
Namely, let's assume that your answer is x and the answer of the jury is y. The checker program will consider your answer correct if <image>.
Examples
Input
3 2 6
Output
6.5
Input
4 2 20
Output
20.0
Note
In the first example, the chicken will be cooked for 3 minutes on the turned on stove, after this it will be cooked for <image>. Then the chicken will be cooked for one minute on a turned off stove, it will be cooked for <image>. Thus, after four minutes the chicken will be cooked for <image>. Before the fifth minute Julia will turn on the stove and after 2.5 minutes the chicken will be ready <image>.
In the second example, when the stove is turned off, Julia will immediately turn it on, so the stove will always be turned on and the chicken will be cooked in 20 minutes. | instruction | 0 | 89,836 | 9 | 179,672 |
Tags: binary search, implementation, math
Correct Solution:
```
k, d, t = (int(x) for x in input().split())
import math as m
if d >= k:
chunksize = d
chunkspeed = k + (d-k)/2
else:
lcm = k // m.gcd(k, d) * d
lft = 0
rgt = lcm//d
while lft != rgt:
cur = (lft+rgt)//2
if d*cur < k:
lft = cur+1
else:
rgt = cur
chunksize = lft * d
chunkspeed = k + (chunksize-k)/2
chunks = m.floor(t / chunkspeed)
# print(chunksize)
# print(chunkspeed)
ans = chunksize * chunks
rem = t - (chunkspeed * chunks)
if rem <= k:
ans += rem
else:
ans += k
rem -= k
ans += rem*2
print(ans)
``` | output | 1 | 89,836 | 9 | 179,673 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Julia is going to cook a chicken in the kitchen of her dormitory. To save energy, the stove in the kitchen automatically turns off after k minutes after turning on.
During cooking, Julia goes to the kitchen every d minutes and turns on the stove if it is turned off. While the cooker is turned off, it stays warm. The stove switches on and off instantly.
It is known that the chicken needs t minutes to be cooked on the stove, if it is turned on, and 2t minutes, if it is turned off. You need to find out, how much time will Julia have to cook the chicken, if it is considered that the chicken is cooked evenly, with constant speed when the stove is turned on and at a constant speed when it is turned off.
Input
The single line contains three integers k, d and t (1 ≤ k, d, t ≤ 1018).
Output
Print a single number, the total time of cooking in minutes. The relative or absolute error must not exceed 10 - 9.
Namely, let's assume that your answer is x and the answer of the jury is y. The checker program will consider your answer correct if <image>.
Examples
Input
3 2 6
Output
6.5
Input
4 2 20
Output
20.0
Note
In the first example, the chicken will be cooked for 3 minutes on the turned on stove, after this it will be cooked for <image>. Then the chicken will be cooked for one minute on a turned off stove, it will be cooked for <image>. Thus, after four minutes the chicken will be cooked for <image>. Before the fifth minute Julia will turn on the stove and after 2.5 minutes the chicken will be ready <image>.
In the second example, when the stove is turned off, Julia will immediately turn it on, so the stove will always be turned on and the chicken will be cooked in 20 minutes. | instruction | 0 | 89,837 | 9 | 179,674 |
Tags: binary search, implementation, math
Correct Solution:
```
import math
k, d, t = map(int, input().split())
if d >= k:
off = d - k
else:
off = math.ceil(k / d) * d - k
times = t // (off / 2 + k)
t1 = times * (off + k)
t2 = t % (off / 2 + k)
if t2 > k:
t2 = k + (t2 - k) * 2
ans = t1 + t2
print (ans)
``` | output | 1 | 89,837 | 9 | 179,675 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Julia is going to cook a chicken in the kitchen of her dormitory. To save energy, the stove in the kitchen automatically turns off after k minutes after turning on.
During cooking, Julia goes to the kitchen every d minutes and turns on the stove if it is turned off. While the cooker is turned off, it stays warm. The stove switches on and off instantly.
It is known that the chicken needs t minutes to be cooked on the stove, if it is turned on, and 2t minutes, if it is turned off. You need to find out, how much time will Julia have to cook the chicken, if it is considered that the chicken is cooked evenly, with constant speed when the stove is turned on and at a constant speed when it is turned off.
Input
The single line contains three integers k, d and t (1 ≤ k, d, t ≤ 1018).
Output
Print a single number, the total time of cooking in minutes. The relative or absolute error must not exceed 10 - 9.
Namely, let's assume that your answer is x and the answer of the jury is y. The checker program will consider your answer correct if <image>.
Examples
Input
3 2 6
Output
6.5
Input
4 2 20
Output
20.0
Note
In the first example, the chicken will be cooked for 3 minutes on the turned on stove, after this it will be cooked for <image>. Then the chicken will be cooked for one minute on a turned off stove, it will be cooked for <image>. Thus, after four minutes the chicken will be cooked for <image>. Before the fifth minute Julia will turn on the stove and after 2.5 minutes the chicken will be ready <image>.
In the second example, when the stove is turned off, Julia will immediately turn it on, so the stove will always be turned on and the chicken will be cooked in 20 minutes. | instruction | 0 | 89,838 | 9 | 179,676 |
Tags: binary search, implementation, math
Correct Solution:
```
k, d, t = list(map(int, input().split()))
if k % d == 0:
print(t)
exit()
m = (k+d-1) // d * d
if 2*t % (m+k) == 0:
print(m*2*t / (m+k))
exit()
n = 2*t // (m+k)
res = n*m
f = 1 - (m +k) * n/(2*t)
if f <= k/t:
print(res+f*t)
else:
res += k
f -= k / t
print(res + f*2*t)
``` | output | 1 | 89,838 | 9 | 179,677 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Julia is going to cook a chicken in the kitchen of her dormitory. To save energy, the stove in the kitchen automatically turns off after k minutes after turning on.
During cooking, Julia goes to the kitchen every d minutes and turns on the stove if it is turned off. While the cooker is turned off, it stays warm. The stove switches on and off instantly.
It is known that the chicken needs t minutes to be cooked on the stove, if it is turned on, and 2t minutes, if it is turned off. You need to find out, how much time will Julia have to cook the chicken, if it is considered that the chicken is cooked evenly, with constant speed when the stove is turned on and at a constant speed when it is turned off.
Input
The single line contains three integers k, d and t (1 ≤ k, d, t ≤ 1018).
Output
Print a single number, the total time of cooking in minutes. The relative or absolute error must not exceed 10 - 9.
Namely, let's assume that your answer is x and the answer of the jury is y. The checker program will consider your answer correct if <image>.
Examples
Input
3 2 6
Output
6.5
Input
4 2 20
Output
20.0
Note
In the first example, the chicken will be cooked for 3 minutes on the turned on stove, after this it will be cooked for <image>. Then the chicken will be cooked for one minute on a turned off stove, it will be cooked for <image>. Thus, after four minutes the chicken will be cooked for <image>. Before the fifth minute Julia will turn on the stove and after 2.5 minutes the chicken will be ready <image>.
In the second example, when the stove is turned off, Julia will immediately turn it on, so the stove will always be turned on and the chicken will be cooked in 20 minutes. | instruction | 0 | 89,839 | 9 | 179,678 |
Tags: binary search, implementation, math
Correct Solution:
```
k, d, t = [int(x) for x in input().split()]
if d >= k:
off1 = d - k
else:
if k%d == 0:
off1 = 0
else:
off1 = d - k%d
reptime = k + off1/2
times = t // reptime
trest = t - times * reptime
sol = times * (k + off1)
if trest <= k:
sol += trest
else:
sol += k
sol += 2*(trest-k)
print("%.8f"%sol)
``` | output | 1 | 89,839 | 9 | 179,679 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Julia is going to cook a chicken in the kitchen of her dormitory. To save energy, the stove in the kitchen automatically turns off after k minutes after turning on.
During cooking, Julia goes to the kitchen every d minutes and turns on the stove if it is turned off. While the cooker is turned off, it stays warm. The stove switches on and off instantly.
It is known that the chicken needs t minutes to be cooked on the stove, if it is turned on, and 2t minutes, if it is turned off. You need to find out, how much time will Julia have to cook the chicken, if it is considered that the chicken is cooked evenly, with constant speed when the stove is turned on and at a constant speed when it is turned off.
Input
The single line contains three integers k, d and t (1 ≤ k, d, t ≤ 1018).
Output
Print a single number, the total time of cooking in minutes. The relative or absolute error must not exceed 10 - 9.
Namely, let's assume that your answer is x and the answer of the jury is y. The checker program will consider your answer correct if <image>.
Examples
Input
3 2 6
Output
6.5
Input
4 2 20
Output
20.0
Note
In the first example, the chicken will be cooked for 3 minutes on the turned on stove, after this it will be cooked for <image>. Then the chicken will be cooked for one minute on a turned off stove, it will be cooked for <image>. Thus, after four minutes the chicken will be cooked for <image>. Before the fifth minute Julia will turn on the stove and after 2.5 minutes the chicken will be ready <image>.
In the second example, when the stove is turned off, Julia will immediately turn it on, so the stove will always be turned on and the chicken will be cooked in 20 minutes. | instruction | 0 | 89,840 | 9 | 179,680 |
Tags: binary search, implementation, math
Correct Solution:
```
import math
k,d,t = [int(x) for x in input().split(' ')]
v = (d*math.ceil(k/d))
cyc = k + ((v-k)/2)
a = (t//cyc)
c = t-(a*cyc)-k
ans = 0
if c>0:
ans = c
print(t-(a*cyc)+(a*v)+ans)
``` | output | 1 | 89,840 | 9 | 179,681 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Julia is going to cook a chicken in the kitchen of her dormitory. To save energy, the stove in the kitchen automatically turns off after k minutes after turning on.
During cooking, Julia goes to the kitchen every d minutes and turns on the stove if it is turned off. While the cooker is turned off, it stays warm. The stove switches on and off instantly.
It is known that the chicken needs t minutes to be cooked on the stove, if it is turned on, and 2t minutes, if it is turned off. You need to find out, how much time will Julia have to cook the chicken, if it is considered that the chicken is cooked evenly, with constant speed when the stove is turned on and at a constant speed when it is turned off.
Input
The single line contains three integers k, d and t (1 ≤ k, d, t ≤ 1018).
Output
Print a single number, the total time of cooking in minutes. The relative or absolute error must not exceed 10 - 9.
Namely, let's assume that your answer is x and the answer of the jury is y. The checker program will consider your answer correct if <image>.
Examples
Input
3 2 6
Output
6.5
Input
4 2 20
Output
20.0
Note
In the first example, the chicken will be cooked for 3 minutes on the turned on stove, after this it will be cooked for <image>. Then the chicken will be cooked for one minute on a turned off stove, it will be cooked for <image>. Thus, after four minutes the chicken will be cooked for <image>. Before the fifth minute Julia will turn on the stove and after 2.5 minutes the chicken will be ready <image>.
In the second example, when the stove is turned off, Julia will immediately turn it on, so the stove will always be turned on and the chicken will be cooked in 20 minutes. | instruction | 0 | 89,841 | 9 | 179,682 |
Tags: binary search, implementation, math
Correct Solution:
```
k, d, t = map(int, input().split())
if k >= t or k%d == 0:
print(t)
exit()
if d < k:
dd = d*(k//d)
d = dd+d
no_of_cycles = 2*t//(k+d)
cooked = no_of_cycles*(k+d)/(2*t)
remaining = 1-cooked
ans = no_of_cycles*d
# print(no_of_cycles, ans, remaining, cooked)
if remaining <= k/t:
# print("AA")
ans += remaining*t
else:
remaining -= k/t
ans += k
ans += remaining * 2*t
print(ans)
``` | output | 1 | 89,841 | 9 | 179,683 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Julia is going to cook a chicken in the kitchen of her dormitory. To save energy, the stove in the kitchen automatically turns off after k minutes after turning on.
During cooking, Julia goes to the kitchen every d minutes and turns on the stove if it is turned off. While the cooker is turned off, it stays warm. The stove switches on and off instantly.
It is known that the chicken needs t minutes to be cooked on the stove, if it is turned on, and 2t minutes, if it is turned off. You need to find out, how much time will Julia have to cook the chicken, if it is considered that the chicken is cooked evenly, with constant speed when the stove is turned on and at a constant speed when it is turned off.
Input
The single line contains three integers k, d and t (1 ≤ k, d, t ≤ 1018).
Output
Print a single number, the total time of cooking in minutes. The relative or absolute error must not exceed 10 - 9.
Namely, let's assume that your answer is x and the answer of the jury is y. The checker program will consider your answer correct if <image>.
Examples
Input
3 2 6
Output
6.5
Input
4 2 20
Output
20.0
Note
In the first example, the chicken will be cooked for 3 minutes on the turned on stove, after this it will be cooked for <image>. Then the chicken will be cooked for one minute on a turned off stove, it will be cooked for <image>. Thus, after four minutes the chicken will be cooked for <image>. Before the fifth minute Julia will turn on the stove and after 2.5 minutes the chicken will be ready <image>.
In the second example, when the stove is turned off, Julia will immediately turn it on, so the stove will always be turned on and the chicken will be cooked in 20 minutes. | instruction | 0 | 89,842 | 9 | 179,684 |
Tags: binary search, implementation, math
Correct Solution:
```
k, d, t = (int(x) for x in input().split())
if d >= k:
chunksize = d
chunkspeed = k + (d-k)/2
else:
if k % d == 0:
chunksize = k
else:
chunksize = ((k // d) + 1) * d
chunkspeed = k + (chunksize-k)/2
chunks = int(t / chunkspeed)
ans = chunksize * chunks
rem = t - (chunkspeed * chunks)
if rem <= k:
ans += rem
else:
ans += k
rem -= k
ans += rem*2
print(ans)
``` | output | 1 | 89,842 | 9 | 179,685 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Julia is going to cook a chicken in the kitchen of her dormitory. To save energy, the stove in the kitchen automatically turns off after k minutes after turning on.
During cooking, Julia goes to the kitchen every d minutes and turns on the stove if it is turned off. While the cooker is turned off, it stays warm. The stove switches on and off instantly.
It is known that the chicken needs t minutes to be cooked on the stove, if it is turned on, and 2t minutes, if it is turned off. You need to find out, how much time will Julia have to cook the chicken, if it is considered that the chicken is cooked evenly, with constant speed when the stove is turned on and at a constant speed when it is turned off.
Input
The single line contains three integers k, d and t (1 ≤ k, d, t ≤ 1018).
Output
Print a single number, the total time of cooking in minutes. The relative or absolute error must not exceed 10 - 9.
Namely, let's assume that your answer is x and the answer of the jury is y. The checker program will consider your answer correct if <image>.
Examples
Input
3 2 6
Output
6.5
Input
4 2 20
Output
20.0
Note
In the first example, the chicken will be cooked for 3 minutes on the turned on stove, after this it will be cooked for <image>. Then the chicken will be cooked for one minute on a turned off stove, it will be cooked for <image>. Thus, after four minutes the chicken will be cooked for <image>. Before the fifth minute Julia will turn on the stove and after 2.5 minutes the chicken will be ready <image>.
In the second example, when the stove is turned off, Julia will immediately turn it on, so the stove will always be turned on and the chicken will be cooked in 20 minutes.
Submitted Solution:
```
k, d, t = map(int, input().split())
a = 2*k+((k+d-1)//d)*d-k
q, r = divmod(2*t, a)
T = q*(((k+d-1)//d)*d)
if 0 <= r <= 2*k:
T += r/2
else:
T += k+(r-2*k)
print(T)
``` | instruction | 0 | 89,844 | 9 | 179,688 |
Yes | output | 1 | 89,844 | 9 | 179,689 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Julia is going to cook a chicken in the kitchen of her dormitory. To save energy, the stove in the kitchen automatically turns off after k minutes after turning on.
During cooking, Julia goes to the kitchen every d minutes and turns on the stove if it is turned off. While the cooker is turned off, it stays warm. The stove switches on and off instantly.
It is known that the chicken needs t minutes to be cooked on the stove, if it is turned on, and 2t minutes, if it is turned off. You need to find out, how much time will Julia have to cook the chicken, if it is considered that the chicken is cooked evenly, with constant speed when the stove is turned on and at a constant speed when it is turned off.
Input
The single line contains three integers k, d and t (1 ≤ k, d, t ≤ 1018).
Output
Print a single number, the total time of cooking in minutes. The relative or absolute error must not exceed 10 - 9.
Namely, let's assume that your answer is x and the answer of the jury is y. The checker program will consider your answer correct if <image>.
Examples
Input
3 2 6
Output
6.5
Input
4 2 20
Output
20.0
Note
In the first example, the chicken will be cooked for 3 minutes on the turned on stove, after this it will be cooked for <image>. Then the chicken will be cooked for one minute on a turned off stove, it will be cooked for <image>. Thus, after four minutes the chicken will be cooked for <image>. Before the fifth minute Julia will turn on the stove and after 2.5 minutes the chicken will be ready <image>.
In the second example, when the stove is turned off, Julia will immediately turn it on, so the stove will always be turned on and the chicken will be cooked in 20 minutes.
Submitted Solution:
```
from math import *
k,d,t=map(int,input().split())
if d<k:
l=ceil(k/d)*d
w=l-k
else:
w=d-k
div=2*k+w
ans=((2*t)//div)*(w+k)
rem=(2*t)%div
if rem!=0:
if rem<=2*k:
ans+=rem/2
else:
ans+=rem-k
print(ans)
``` | instruction | 0 | 89,845 | 9 | 179,690 |
Yes | output | 1 | 89,845 | 9 | 179,691 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Julia is going to cook a chicken in the kitchen of her dormitory. To save energy, the stove in the kitchen automatically turns off after k minutes after turning on.
During cooking, Julia goes to the kitchen every d minutes and turns on the stove if it is turned off. While the cooker is turned off, it stays warm. The stove switches on and off instantly.
It is known that the chicken needs t minutes to be cooked on the stove, if it is turned on, and 2t minutes, if it is turned off. You need to find out, how much time will Julia have to cook the chicken, if it is considered that the chicken is cooked evenly, with constant speed when the stove is turned on and at a constant speed when it is turned off.
Input
The single line contains three integers k, d and t (1 ≤ k, d, t ≤ 1018).
Output
Print a single number, the total time of cooking in minutes. The relative or absolute error must not exceed 10 - 9.
Namely, let's assume that your answer is x and the answer of the jury is y. The checker program will consider your answer correct if <image>.
Examples
Input
3 2 6
Output
6.5
Input
4 2 20
Output
20.0
Note
In the first example, the chicken will be cooked for 3 minutes on the turned on stove, after this it will be cooked for <image>. Then the chicken will be cooked for one minute on a turned off stove, it will be cooked for <image>. Thus, after four minutes the chicken will be cooked for <image>. Before the fifth minute Julia will turn on the stove and after 2.5 minutes the chicken will be ready <image>.
In the second example, when the stove is turned off, Julia will immediately turn it on, so the stove will always be turned on and the chicken will be cooked in 20 minutes.
Submitted Solution:
```
k,d,t=[int(i)for i in input().split()]
t = 2 * t
if d >= k:
s = 2*k + (d - k)
tt = k + (d - k)
if s >= t:
if 2*k >= t:
print(t//2)
exit(0)
else:
print(k + 2*(t//2-k))
exit(0)
h = t // s
x = t % s
if x <= 2*k:
print(h * tt + x / 2)
else:
print(h * tt + k + (x - 2 *k))
else:
last = 0
if k % d == 0:last = k
else :last = k - k % d + d
s = 2*k + (last - k)
tt = k + (last - k)
if s >= t:
if 2*k >= t:
print(t//2)
exit(0)
else:
print(k + 2*(t//2-k))
exit(0)
h = t // s
x = t % s
if x <= 2*k:
print(h * tt + x / 2)
else:
print(h * tt + k + (x - 2 *k))
``` | instruction | 0 | 89,846 | 9 | 179,692 |
Yes | output | 1 | 89,846 | 9 | 179,693 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Julia is going to cook a chicken in the kitchen of her dormitory. To save energy, the stove in the kitchen automatically turns off after k minutes after turning on.
During cooking, Julia goes to the kitchen every d minutes and turns on the stove if it is turned off. While the cooker is turned off, it stays warm. The stove switches on and off instantly.
It is known that the chicken needs t minutes to be cooked on the stove, if it is turned on, and 2t minutes, if it is turned off. You need to find out, how much time will Julia have to cook the chicken, if it is considered that the chicken is cooked evenly, with constant speed when the stove is turned on and at a constant speed when it is turned off.
Input
The single line contains three integers k, d and t (1 ≤ k, d, t ≤ 1018).
Output
Print a single number, the total time of cooking in minutes. The relative or absolute error must not exceed 10 - 9.
Namely, let's assume that your answer is x and the answer of the jury is y. The checker program will consider your answer correct if <image>.
Examples
Input
3 2 6
Output
6.5
Input
4 2 20
Output
20.0
Note
In the first example, the chicken will be cooked for 3 minutes on the turned on stove, after this it will be cooked for <image>. Then the chicken will be cooked for one minute on a turned off stove, it will be cooked for <image>. Thus, after four minutes the chicken will be cooked for <image>. Before the fifth minute Julia will turn on the stove and after 2.5 minutes the chicken will be ready <image>.
In the second example, when the stove is turned off, Julia will immediately turn it on, so the stove will always be turned on and the chicken will be cooked in 20 minutes.
Submitted Solution:
```
import math
k,d,t=map(int,input().split())
if k>=t:
print(t)
elif k>=d and k%d==0:
print(t)
else:
if k!=0:
p=math.ceil(k/d)
r=(p*d)-k
f=(k/t)+((r)/(2*t))
m=math.ceil(1/f)
n=m-1
req=1-(n*f)
if n*f+(k/t)>1:
s=req*t
ans=n*(p*d)+s
print(ans)
elif n*f+(k/t)==1:
ans=n*(p*d)+k
print(ans)
else:
re=1-(n*f)+(k/t)
s=re*(2*t)
ans=n*(p*d)+k+s
print(ans)
``` | instruction | 0 | 89,847 | 9 | 179,694 |
No | output | 1 | 89,847 | 9 | 179,695 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Julia is going to cook a chicken in the kitchen of her dormitory. To save energy, the stove in the kitchen automatically turns off after k minutes after turning on.
During cooking, Julia goes to the kitchen every d minutes and turns on the stove if it is turned off. While the cooker is turned off, it stays warm. The stove switches on and off instantly.
It is known that the chicken needs t minutes to be cooked on the stove, if it is turned on, and 2t minutes, if it is turned off. You need to find out, how much time will Julia have to cook the chicken, if it is considered that the chicken is cooked evenly, with constant speed when the stove is turned on and at a constant speed when it is turned off.
Input
The single line contains three integers k, d and t (1 ≤ k, d, t ≤ 1018).
Output
Print a single number, the total time of cooking in minutes. The relative or absolute error must not exceed 10 - 9.
Namely, let's assume that your answer is x and the answer of the jury is y. The checker program will consider your answer correct if <image>.
Examples
Input
3 2 6
Output
6.5
Input
4 2 20
Output
20.0
Note
In the first example, the chicken will be cooked for 3 minutes on the turned on stove, after this it will be cooked for <image>. Then the chicken will be cooked for one minute on a turned off stove, it will be cooked for <image>. Thus, after four minutes the chicken will be cooked for <image>. Before the fifth minute Julia will turn on the stove and after 2.5 minutes the chicken will be ready <image>.
In the second example, when the stove is turned off, Julia will immediately turn it on, so the stove will always be turned on and the chicken will be cooked in 20 minutes.
Submitted Solution:
```
k,d,t = map(int,input().split())
if k%d == 0 or t<=k:
print(t)
else:
cyc = k + (d - k%d)
dt = k*2 + (d - k%d)
t *= 2
n = int(t/dt)
if t % dt == 0:
n -= 1
ans = 0
if(t % dt > k*2):
ans = cyc*n + (t%dt)/2
else:
ans = cyc*n + k + (dt - 2*k)
print(ans)
``` | instruction | 0 | 89,848 | 9 | 179,696 |
No | output | 1 | 89,848 | 9 | 179,697 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Julia is going to cook a chicken in the kitchen of her dormitory. To save energy, the stove in the kitchen automatically turns off after k minutes after turning on.
During cooking, Julia goes to the kitchen every d minutes and turns on the stove if it is turned off. While the cooker is turned off, it stays warm. The stove switches on and off instantly.
It is known that the chicken needs t minutes to be cooked on the stove, if it is turned on, and 2t minutes, if it is turned off. You need to find out, how much time will Julia have to cook the chicken, if it is considered that the chicken is cooked evenly, with constant speed when the stove is turned on and at a constant speed when it is turned off.
Input
The single line contains three integers k, d and t (1 ≤ k, d, t ≤ 1018).
Output
Print a single number, the total time of cooking in minutes. The relative or absolute error must not exceed 10 - 9.
Namely, let's assume that your answer is x and the answer of the jury is y. The checker program will consider your answer correct if <image>.
Examples
Input
3 2 6
Output
6.5
Input
4 2 20
Output
20.0
Note
In the first example, the chicken will be cooked for 3 minutes on the turned on stove, after this it will be cooked for <image>. Then the chicken will be cooked for one minute on a turned off stove, it will be cooked for <image>. Thus, after four minutes the chicken will be cooked for <image>. Before the fifth minute Julia will turn on the stove and after 2.5 minutes the chicken will be ready <image>.
In the second example, when the stove is turned off, Julia will immediately turn it on, so the stove will always be turned on and the chicken will be cooked in 20 minutes.
Submitted Solution:
```
k,d,t=input().split()
k,d,t=int(k),int(d),int(t)
time=0
if True:
m=int((k-1)/d)*d+d-k
#print(m)
tc=k+0.5*m
ta=k+m
l=int(t/tc)
time=ta*l
#print(tc)
#print(ta)
#print(time)
#print(l)
if (k>t-l*tc):
time+=t-l*tc
print(t-l*tc)
else:
time+=k
time+=2*(t-l*tc-k)
print(time)
``` | instruction | 0 | 89,849 | 9 | 179,698 |
No | output | 1 | 89,849 | 9 | 179,699 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Julia is going to cook a chicken in the kitchen of her dormitory. To save energy, the stove in the kitchen automatically turns off after k minutes after turning on.
During cooking, Julia goes to the kitchen every d minutes and turns on the stove if it is turned off. While the cooker is turned off, it stays warm. The stove switches on and off instantly.
It is known that the chicken needs t minutes to be cooked on the stove, if it is turned on, and 2t minutes, if it is turned off. You need to find out, how much time will Julia have to cook the chicken, if it is considered that the chicken is cooked evenly, with constant speed when the stove is turned on and at a constant speed when it is turned off.
Input
The single line contains three integers k, d and t (1 ≤ k, d, t ≤ 1018).
Output
Print a single number, the total time of cooking in minutes. The relative or absolute error must not exceed 10 - 9.
Namely, let's assume that your answer is x and the answer of the jury is y. The checker program will consider your answer correct if <image>.
Examples
Input
3 2 6
Output
6.5
Input
4 2 20
Output
20.0
Note
In the first example, the chicken will be cooked for 3 minutes on the turned on stove, after this it will be cooked for <image>. Then the chicken will be cooked for one minute on a turned off stove, it will be cooked for <image>. Thus, after four minutes the chicken will be cooked for <image>. Before the fifth minute Julia will turn on the stove and after 2.5 minutes the chicken will be ready <image>.
In the second example, when the stove is turned off, Julia will immediately turn it on, so the stove will always be turned on and the chicken will be cooked in 20 minutes.
Submitted Solution:
```
k,d,t = map(int,input().split())
if k%d == 0 or t<=k:
print(t)
else:
cyc = k + (d - k%d)
dt = k*2 + (d - k%d)
t *= 2
n = int(t/dt)
ans = 0
if(t % dt <= k*2):
ans = cyc*n + (t%dt)/2
else:
ans = cyc*n + k + (dt - 2*k)
print(ans)
``` | instruction | 0 | 89,850 | 9 | 179,700 |
No | output | 1 | 89,850 | 9 | 179,701 |
Provide a correct Python 3 solution for this coding contest problem.
<image>
My grandmother uses a balance. The balance will balance if you place the same size on both of the two dishes, otherwise it will tilt to the heavier side. The weights of the 10 weights are 1g, 2g, 4g, 8g, 16g, 32g, 64g, 128g, 256g, 512g in order of lightness.
My grandmother says, "Weigh up to about 1 kg in grams." "Then, try to weigh the juice here," and my grandmother put the juice on the left plate and the 8g, 64g, and 128g weights on the right plate to balance. Then he answered, "The total weight is 200g, so the juice is 200g. How is it correct?"
Since the weight of the item to be placed on the left plate is given, create a program that outputs the weight to be placed on the right plate in order of lightness when balancing with the item of the weight given by the balance. However, the weight of the item to be weighed shall be less than or equal to the total weight of all weights (= 1023g).
Hint
The weight of the weight is 2 to the nth power (n = 0, 1, .... 9) g.
Input
Given multiple datasets. For each dataset, the weight of the item to be placed on the left plate is given in one line. Please process until the end of the input. The number of datasets does not exceed 50.
Output
For each data set, separate the weights (ascending order) to be placed on the right plate with one blank and output them on one line.
Example
Input
5
7
127
Output
1 4
1 2 4
1 2 4 8 16 32 64 | instruction | 0 | 90,010 | 9 | 180,020 |
"Correct Solution:
```
if __name__ == '__main__':
while True:
try:
n = int(input())
ans = []
for i in range(9,-1,-1):
if n >= 2**i:
ans.append(2**i)
n -= 2**i
print(*sorted(ans))
except EOFError:
break
``` | output | 1 | 90,010 | 9 | 180,021 |
Provide a correct Python 3 solution for this coding contest problem.
<image>
My grandmother uses a balance. The balance will balance if you place the same size on both of the two dishes, otherwise it will tilt to the heavier side. The weights of the 10 weights are 1g, 2g, 4g, 8g, 16g, 32g, 64g, 128g, 256g, 512g in order of lightness.
My grandmother says, "Weigh up to about 1 kg in grams." "Then, try to weigh the juice here," and my grandmother put the juice on the left plate and the 8g, 64g, and 128g weights on the right plate to balance. Then he answered, "The total weight is 200g, so the juice is 200g. How is it correct?"
Since the weight of the item to be placed on the left plate is given, create a program that outputs the weight to be placed on the right plate in order of lightness when balancing with the item of the weight given by the balance. However, the weight of the item to be weighed shall be less than or equal to the total weight of all weights (= 1023g).
Hint
The weight of the weight is 2 to the nth power (n = 0, 1, .... 9) g.
Input
Given multiple datasets. For each dataset, the weight of the item to be placed on the left plate is given in one line. Please process until the end of the input. The number of datasets does not exceed 50.
Output
For each data set, separate the weights (ascending order) to be placed on the right plate with one blank and output them on one line.
Example
Input
5
7
127
Output
1 4
1 2 4
1 2 4 8 16 32 64 | instruction | 0 | 90,011 | 9 | 180,022 |
"Correct Solution:
```
# AOJ 0031 Weight
# Python3 2018.6.14 bal4u
w = [1, 2, 4, 8, 16, 32, 64, 128, 256, 512];
while True:
try:
a = int(input())
ans = []
i = 0
while a > 0:
if a & 1:
ans.append(w[i])
a >>= 1
i += 1
print(*ans)
except EOFError:
break
``` | output | 1 | 90,011 | 9 | 180,023 |
Provide a correct Python 3 solution for this coding contest problem.
<image>
My grandmother uses a balance. The balance will balance if you place the same size on both of the two dishes, otherwise it will tilt to the heavier side. The weights of the 10 weights are 1g, 2g, 4g, 8g, 16g, 32g, 64g, 128g, 256g, 512g in order of lightness.
My grandmother says, "Weigh up to about 1 kg in grams." "Then, try to weigh the juice here," and my grandmother put the juice on the left plate and the 8g, 64g, and 128g weights on the right plate to balance. Then he answered, "The total weight is 200g, so the juice is 200g. How is it correct?"
Since the weight of the item to be placed on the left plate is given, create a program that outputs the weight to be placed on the right plate in order of lightness when balancing with the item of the weight given by the balance. However, the weight of the item to be weighed shall be less than or equal to the total weight of all weights (= 1023g).
Hint
The weight of the weight is 2 to the nth power (n = 0, 1, .... 9) g.
Input
Given multiple datasets. For each dataset, the weight of the item to be placed on the left plate is given in one line. Please process until the end of the input. The number of datasets does not exceed 50.
Output
For each data set, separate the weights (ascending order) to be placed on the right plate with one blank and output them on one line.
Example
Input
5
7
127
Output
1 4
1 2 4
1 2 4 8 16 32 64 | instruction | 0 | 90,012 | 9 | 180,024 |
"Correct Solution:
```
import sys
for line in sys.stdin:
weight = [1, 2, 4, 8, 16, 32, 64, 128, 256, 512]
ans = []
n = bin(int(line))[2:]
for i, x in enumerate(n[::-1]):
if x == "1":
ans.append(str(weight[i]))
print(" ".join(ans))
``` | output | 1 | 90,012 | 9 | 180,025 |
Provide a correct Python 3 solution for this coding contest problem.
<image>
My grandmother uses a balance. The balance will balance if you place the same size on both of the two dishes, otherwise it will tilt to the heavier side. The weights of the 10 weights are 1g, 2g, 4g, 8g, 16g, 32g, 64g, 128g, 256g, 512g in order of lightness.
My grandmother says, "Weigh up to about 1 kg in grams." "Then, try to weigh the juice here," and my grandmother put the juice on the left plate and the 8g, 64g, and 128g weights on the right plate to balance. Then he answered, "The total weight is 200g, so the juice is 200g. How is it correct?"
Since the weight of the item to be placed on the left plate is given, create a program that outputs the weight to be placed on the right plate in order of lightness when balancing with the item of the weight given by the balance. However, the weight of the item to be weighed shall be less than or equal to the total weight of all weights (= 1023g).
Hint
The weight of the weight is 2 to the nth power (n = 0, 1, .... 9) g.
Input
Given multiple datasets. For each dataset, the weight of the item to be placed on the left plate is given in one line. Please process until the end of the input. The number of datasets does not exceed 50.
Output
For each data set, separate the weights (ascending order) to be placed on the right plate with one blank and output them on one line.
Example
Input
5
7
127
Output
1 4
1 2 4
1 2 4 8 16 32 64 | instruction | 0 | 90,013 | 9 | 180,026 |
"Correct Solution:
```
def solve(x):
B=list(map(int,x))
B.reverse()
for i in range(len(B)):
B[i]=B[i]*(2**i)
#remove 0
while 0 in B:
B.remove(0)
print(" ".join(map(str,B)))
while True:
try:
x=int(input())
solve(format(x,'b'))
except EOFError:
break
``` | output | 1 | 90,013 | 9 | 180,027 |
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