message stringlengths 2 48.6k | message_type stringclasses 2 values | message_id int64 0 1 | conversation_id int64 318 108k | cluster float64 8 8 | __index_level_0__ int64 636 217k |
|---|---|---|---|---|---|
Provide a correct Python 3 solution for this coding contest problem.
Takahashi wants to gain muscle, and decides to work out at AtCoder Gym.
The exercise machine at the gym has N buttons, and exactly one of the buttons is lighten up. These buttons are numbered 1 through N. When Button i is lighten up and you press it, the light is turned off, and then Button a_i will be lighten up. It is possible that i=a_i. When Button i is not lighten up, nothing will happen by pressing it.
Initially, Button 1 is lighten up. Takahashi wants to quit pressing buttons when Button 2 is lighten up.
Determine whether this is possible. If the answer is positive, find the minimum number of times he needs to press buttons.
Constraints
* 2 ≤ N ≤ 10^5
* 1 ≤ a_i ≤ N
Input
Input is given from Standard Input in the following format:
N
a_1
a_2
:
a_N
Output
Print -1 if it is impossible to lighten up Button 2. Otherwise, print the minimum number of times we need to press buttons in order to lighten up Button 2.
Examples
Input
3
3
1
2
Output
2
Input
4
3
4
1
2
Output
-1
Input
5
3
3
4
2
4
Output
3 | instruction | 0 | 91,570 | 8 | 183,140 |
"Correct Solution:
```
n=int(input())
a=list(int(input())-1 for _ in[0]*n)
s=0
c=0
while s!=1 and c<n:
c+=1
s=a[s]
print(c if c<n else -1)
``` | output | 1 | 91,570 | 8 | 183,141 |
Provide a correct Python 3 solution for this coding contest problem.
International Center for Picassonian Cubism is a Spanish national museum of cubist artworks, dedicated to Pablo Picasso. The center held a competition for an artwork that will be displayed in front of the facade of the museum building. The artwork is a collection of cubes that are piled up on the ground and is intended to amuse visitors, who will be curious how the shape of the collection of cubes changes when it is seen from the front and the sides.
The artwork is a collection of cubes with edges of one foot long and is built on a flat ground that is divided into a grid of one foot by one foot squares. Due to some technical reasons, cubes of the artwork must be either put on the ground, fitting into a unit square in the grid, or put on another cube in the way that the bottom face of the upper cube exactly meets the top face of the lower cube. No other way of putting cubes is possible.
You are a member of the judging committee responsible for selecting one out of a plenty of artwork proposals submitted to the competition. The decision is made primarily based on artistic quality but the cost for installing the artwork is another important factor. Your task is to investigate the installation cost for each proposal. The cost is proportional to the number of cubes, so you have to figure out the minimum number of cubes needed for installation.
Each design proposal of an artwork consists of the front view and the side view (the view seen from the right-hand side), as shown in Figure 1.
<image>
Figure 1: An example of an artwork proposal
The front view (resp., the side view) indicates the maximum heights of piles of cubes for each column line (resp., row line) of the grid.
There are several ways to install this proposal of artwork, such as the following figures.
<image>
In these figures, the dotted lines on the ground indicate the grid lines. The left figure makes use of 16 cubes, which is not optimal. That is, the artwork can be installed with a fewer number of cubes. Actually, the right one is optimal and only uses 13 cubes. Note that, a single pile of height three in the right figure plays the roles of two such piles in the left one.
Notice that swapping columns of cubes does not change the side view. Similarly, swapping rows does not change the front view. Thus, such swaps do not change the costs of building the artworks.
For example, consider the artwork proposal given in Figure 2.
<image>
Figure 2: Another example of artwork proposal
An optimal installation of this proposal of artwork can be achieved with 13 cubes, as shown in the following figure, which can be obtained by exchanging the rightmost two columns of the optimal installation of the artwork of Figure 1.
<image>
Input
The input is a sequence of datasets. The end of the input is indicated by a line containing two zeros separated by a space. Each dataset is formatted as follows.
w d
h1 h2 ... hw
h'1 h'2 ... h'd
The integers w and d separated by a space are the numbers of columns and rows of the grid, respectively. You may assume 1 ≤ w ≤ 10 and 1 ≤ d ≤ 10. The integers separated by a space in the second and third lines specify the shape of the artwork. The integers hi (1 ≤ hi ≤ 20, 1 ≤ i ≤ w) in the second line give the front view, i.e., the maximum heights of cubes per each column line, ordered from left to right (seen from the front). The integers hi (1 ≤ hi ≤ 20, 1 ≤ i ≤ d) in the third line give the side view, i.e., the maximum heights of cubes per each row line, ordered from left to right (seen from the right-hand side).
Output
For each dataset, output a line containing the minimum number of cubes. The output should not contain any other extra characters.
You can assume that, for each dataset, there is at least one way to install the artwork.
Example
Input
5 5
1 2 3 4 5
1 2 3 4 5
5 5
2 5 4 1 3
4 1 5 3 2
5 5
1 2 3 4 5
3 3 3 4 5
3 3
7 7 7
7 7 7
3 3
4 4 4
4 3 4
4 3
4 2 2 4
4 2 1
4 4
2 8 8 8
2 3 8 3
10 10
9 9 9 9 9 9 9 9 9 9
9 9 9 9 9 9 9 9 9 9
10 9
20 1 20 20 20 20 20 18 20 20
20 20 20 20 7 20 20 20 20
0 0
Output
15
15
21
21
15
13
32
90
186 | instruction | 0 | 91,669 | 8 | 183,338 |
"Correct Solution:
```
from collections import Counter
while True:
W, D = map(int, input().split())
if not (W | D):
break
hw = [int(x) for x in input().split()]
hd = [int(x) for x in input().split()]
cnt_hw = Counter(hw)
cnt_hd = Counter(hd)
print(min(sum(hw) + sum(max(0, v - cnt_hw[k]) * k for k, v in cnt_hd.items()),
sum(hd) + sum(max(0, v - cnt_hd[k]) * k for k, v in cnt_hw.items())))
``` | output | 1 | 91,669 | 8 | 183,339 |
Provide a correct Python 3 solution for this coding contest problem.
International Center for Picassonian Cubism is a Spanish national museum of cubist artworks, dedicated to Pablo Picasso. The center held a competition for an artwork that will be displayed in front of the facade of the museum building. The artwork is a collection of cubes that are piled up on the ground and is intended to amuse visitors, who will be curious how the shape of the collection of cubes changes when it is seen from the front and the sides.
The artwork is a collection of cubes with edges of one foot long and is built on a flat ground that is divided into a grid of one foot by one foot squares. Due to some technical reasons, cubes of the artwork must be either put on the ground, fitting into a unit square in the grid, or put on another cube in the way that the bottom face of the upper cube exactly meets the top face of the lower cube. No other way of putting cubes is possible.
You are a member of the judging committee responsible for selecting one out of a plenty of artwork proposals submitted to the competition. The decision is made primarily based on artistic quality but the cost for installing the artwork is another important factor. Your task is to investigate the installation cost for each proposal. The cost is proportional to the number of cubes, so you have to figure out the minimum number of cubes needed for installation.
Each design proposal of an artwork consists of the front view and the side view (the view seen from the right-hand side), as shown in Figure 1.
<image>
Figure 1: An example of an artwork proposal
The front view (resp., the side view) indicates the maximum heights of piles of cubes for each column line (resp., row line) of the grid.
There are several ways to install this proposal of artwork, such as the following figures.
<image>
In these figures, the dotted lines on the ground indicate the grid lines. The left figure makes use of 16 cubes, which is not optimal. That is, the artwork can be installed with a fewer number of cubes. Actually, the right one is optimal and only uses 13 cubes. Note that, a single pile of height three in the right figure plays the roles of two such piles in the left one.
Notice that swapping columns of cubes does not change the side view. Similarly, swapping rows does not change the front view. Thus, such swaps do not change the costs of building the artworks.
For example, consider the artwork proposal given in Figure 2.
<image>
Figure 2: Another example of artwork proposal
An optimal installation of this proposal of artwork can be achieved with 13 cubes, as shown in the following figure, which can be obtained by exchanging the rightmost two columns of the optimal installation of the artwork of Figure 1.
<image>
Input
The input is a sequence of datasets. The end of the input is indicated by a line containing two zeros separated by a space. Each dataset is formatted as follows.
w d
h1 h2 ... hw
h'1 h'2 ... h'd
The integers w and d separated by a space are the numbers of columns and rows of the grid, respectively. You may assume 1 ≤ w ≤ 10 and 1 ≤ d ≤ 10. The integers separated by a space in the second and third lines specify the shape of the artwork. The integers hi (1 ≤ hi ≤ 20, 1 ≤ i ≤ w) in the second line give the front view, i.e., the maximum heights of cubes per each column line, ordered from left to right (seen from the front). The integers hi (1 ≤ hi ≤ 20, 1 ≤ i ≤ d) in the third line give the side view, i.e., the maximum heights of cubes per each row line, ordered from left to right (seen from the right-hand side).
Output
For each dataset, output a line containing the minimum number of cubes. The output should not contain any other extra characters.
You can assume that, for each dataset, there is at least one way to install the artwork.
Example
Input
5 5
1 2 3 4 5
1 2 3 4 5
5 5
2 5 4 1 3
4 1 5 3 2
5 5
1 2 3 4 5
3 3 3 4 5
3 3
7 7 7
7 7 7
3 3
4 4 4
4 3 4
4 3
4 2 2 4
4 2 1
4 4
2 8 8 8
2 3 8 3
10 10
9 9 9 9 9 9 9 9 9 9
9 9 9 9 9 9 9 9 9 9
10 9
20 1 20 20 20 20 20 18 20 20
20 20 20 20 7 20 20 20 20
0 0
Output
15
15
21
21
15
13
32
90
186 | instruction | 0 | 91,670 | 8 | 183,340 |
"Correct Solution:
```
while 1:
w,d=map(int,input().split())
if w==d==0:break
a=[0]*21
ans=0
b=list(map(int,input().split()))
for i in range(w):
a[b[i]]+=1
ans=sum(b)
b=list(map(int,input().split()))
for i in range(d):
if a[b[i]]:a[b[i]]-=1
else: ans+=b[i]
print(ans)
``` | output | 1 | 91,670 | 8 | 183,341 |
Provide a correct Python 3 solution for this coding contest problem.
International Center for Picassonian Cubism is a Spanish national museum of cubist artworks, dedicated to Pablo Picasso. The center held a competition for an artwork that will be displayed in front of the facade of the museum building. The artwork is a collection of cubes that are piled up on the ground and is intended to amuse visitors, who will be curious how the shape of the collection of cubes changes when it is seen from the front and the sides.
The artwork is a collection of cubes with edges of one foot long and is built on a flat ground that is divided into a grid of one foot by one foot squares. Due to some technical reasons, cubes of the artwork must be either put on the ground, fitting into a unit square in the grid, or put on another cube in the way that the bottom face of the upper cube exactly meets the top face of the lower cube. No other way of putting cubes is possible.
You are a member of the judging committee responsible for selecting one out of a plenty of artwork proposals submitted to the competition. The decision is made primarily based on artistic quality but the cost for installing the artwork is another important factor. Your task is to investigate the installation cost for each proposal. The cost is proportional to the number of cubes, so you have to figure out the minimum number of cubes needed for installation.
Each design proposal of an artwork consists of the front view and the side view (the view seen from the right-hand side), as shown in Figure 1.
<image>
Figure 1: An example of an artwork proposal
The front view (resp., the side view) indicates the maximum heights of piles of cubes for each column line (resp., row line) of the grid.
There are several ways to install this proposal of artwork, such as the following figures.
<image>
In these figures, the dotted lines on the ground indicate the grid lines. The left figure makes use of 16 cubes, which is not optimal. That is, the artwork can be installed with a fewer number of cubes. Actually, the right one is optimal and only uses 13 cubes. Note that, a single pile of height three in the right figure plays the roles of two such piles in the left one.
Notice that swapping columns of cubes does not change the side view. Similarly, swapping rows does not change the front view. Thus, such swaps do not change the costs of building the artworks.
For example, consider the artwork proposal given in Figure 2.
<image>
Figure 2: Another example of artwork proposal
An optimal installation of this proposal of artwork can be achieved with 13 cubes, as shown in the following figure, which can be obtained by exchanging the rightmost two columns of the optimal installation of the artwork of Figure 1.
<image>
Input
The input is a sequence of datasets. The end of the input is indicated by a line containing two zeros separated by a space. Each dataset is formatted as follows.
w d
h1 h2 ... hw
h'1 h'2 ... h'd
The integers w and d separated by a space are the numbers of columns and rows of the grid, respectively. You may assume 1 ≤ w ≤ 10 and 1 ≤ d ≤ 10. The integers separated by a space in the second and third lines specify the shape of the artwork. The integers hi (1 ≤ hi ≤ 20, 1 ≤ i ≤ w) in the second line give the front view, i.e., the maximum heights of cubes per each column line, ordered from left to right (seen from the front). The integers hi (1 ≤ hi ≤ 20, 1 ≤ i ≤ d) in the third line give the side view, i.e., the maximum heights of cubes per each row line, ordered from left to right (seen from the right-hand side).
Output
For each dataset, output a line containing the minimum number of cubes. The output should not contain any other extra characters.
You can assume that, for each dataset, there is at least one way to install the artwork.
Example
Input
5 5
1 2 3 4 5
1 2 3 4 5
5 5
2 5 4 1 3
4 1 5 3 2
5 5
1 2 3 4 5
3 3 3 4 5
3 3
7 7 7
7 7 7
3 3
4 4 4
4 3 4
4 3
4 2 2 4
4 2 1
4 4
2 8 8 8
2 3 8 3
10 10
9 9 9 9 9 9 9 9 9 9
9 9 9 9 9 9 9 9 9 9
10 9
20 1 20 20 20 20 20 18 20 20
20 20 20 20 7 20 20 20 20
0 0
Output
15
15
21
21
15
13
32
90
186 | instruction | 0 | 91,671 | 8 | 183,342 |
"Correct Solution:
```
while 1:
w,d=map(int,input().split())
if w==0:
break
W=list(map(int,input().split()))
D=list(map(int,input().split()))
ans=0
table=[0 for i in range(21)]
for n in W:
ans+=n
table[n]+=1
for n in D:
if table[n]>0:
table[n]-=1
else:
ans+=n
print(ans)
``` | output | 1 | 91,671 | 8 | 183,343 |
Provide a correct Python 3 solution for this coding contest problem.
International Center for Picassonian Cubism is a Spanish national museum of cubist artworks, dedicated to Pablo Picasso. The center held a competition for an artwork that will be displayed in front of the facade of the museum building. The artwork is a collection of cubes that are piled up on the ground and is intended to amuse visitors, who will be curious how the shape of the collection of cubes changes when it is seen from the front and the sides.
The artwork is a collection of cubes with edges of one foot long and is built on a flat ground that is divided into a grid of one foot by one foot squares. Due to some technical reasons, cubes of the artwork must be either put on the ground, fitting into a unit square in the grid, or put on another cube in the way that the bottom face of the upper cube exactly meets the top face of the lower cube. No other way of putting cubes is possible.
You are a member of the judging committee responsible for selecting one out of a plenty of artwork proposals submitted to the competition. The decision is made primarily based on artistic quality but the cost for installing the artwork is another important factor. Your task is to investigate the installation cost for each proposal. The cost is proportional to the number of cubes, so you have to figure out the minimum number of cubes needed for installation.
Each design proposal of an artwork consists of the front view and the side view (the view seen from the right-hand side), as shown in Figure 1.
<image>
Figure 1: An example of an artwork proposal
The front view (resp., the side view) indicates the maximum heights of piles of cubes for each column line (resp., row line) of the grid.
There are several ways to install this proposal of artwork, such as the following figures.
<image>
In these figures, the dotted lines on the ground indicate the grid lines. The left figure makes use of 16 cubes, which is not optimal. That is, the artwork can be installed with a fewer number of cubes. Actually, the right one is optimal and only uses 13 cubes. Note that, a single pile of height three in the right figure plays the roles of two such piles in the left one.
Notice that swapping columns of cubes does not change the side view. Similarly, swapping rows does not change the front view. Thus, such swaps do not change the costs of building the artworks.
For example, consider the artwork proposal given in Figure 2.
<image>
Figure 2: Another example of artwork proposal
An optimal installation of this proposal of artwork can be achieved with 13 cubes, as shown in the following figure, which can be obtained by exchanging the rightmost two columns of the optimal installation of the artwork of Figure 1.
<image>
Input
The input is a sequence of datasets. The end of the input is indicated by a line containing two zeros separated by a space. Each dataset is formatted as follows.
w d
h1 h2 ... hw
h'1 h'2 ... h'd
The integers w and d separated by a space are the numbers of columns and rows of the grid, respectively. You may assume 1 ≤ w ≤ 10 and 1 ≤ d ≤ 10. The integers separated by a space in the second and third lines specify the shape of the artwork. The integers hi (1 ≤ hi ≤ 20, 1 ≤ i ≤ w) in the second line give the front view, i.e., the maximum heights of cubes per each column line, ordered from left to right (seen from the front). The integers hi (1 ≤ hi ≤ 20, 1 ≤ i ≤ d) in the third line give the side view, i.e., the maximum heights of cubes per each row line, ordered from left to right (seen from the right-hand side).
Output
For each dataset, output a line containing the minimum number of cubes. The output should not contain any other extra characters.
You can assume that, for each dataset, there is at least one way to install the artwork.
Example
Input
5 5
1 2 3 4 5
1 2 3 4 5
5 5
2 5 4 1 3
4 1 5 3 2
5 5
1 2 3 4 5
3 3 3 4 5
3 3
7 7 7
7 7 7
3 3
4 4 4
4 3 4
4 3
4 2 2 4
4 2 1
4 4
2 8 8 8
2 3 8 3
10 10
9 9 9 9 9 9 9 9 9 9
9 9 9 9 9 9 9 9 9 9
10 9
20 1 20 20 20 20 20 18 20 20
20 20 20 20 7 20 20 20 20
0 0
Output
15
15
21
21
15
13
32
90
186 | instruction | 0 | 91,672 | 8 | 183,344 |
"Correct Solution:
```
from collections import Counter
while True:
W, D = map(int, input().split())
if not (W | D):
break
hw = [int(x) for x in input().split()]
hd = [int(x) for x in input().split()]
print(sum(hw) + sum(hd) - sum(k * v for k, v in (Counter(hw) & Counter(hd)).items()))
``` | output | 1 | 91,672 | 8 | 183,345 |
Provide a correct Python 3 solution for this coding contest problem.
International Center for Picassonian Cubism is a Spanish national museum of cubist artworks, dedicated to Pablo Picasso. The center held a competition for an artwork that will be displayed in front of the facade of the museum building. The artwork is a collection of cubes that are piled up on the ground and is intended to amuse visitors, who will be curious how the shape of the collection of cubes changes when it is seen from the front and the sides.
The artwork is a collection of cubes with edges of one foot long and is built on a flat ground that is divided into a grid of one foot by one foot squares. Due to some technical reasons, cubes of the artwork must be either put on the ground, fitting into a unit square in the grid, or put on another cube in the way that the bottom face of the upper cube exactly meets the top face of the lower cube. No other way of putting cubes is possible.
You are a member of the judging committee responsible for selecting one out of a plenty of artwork proposals submitted to the competition. The decision is made primarily based on artistic quality but the cost for installing the artwork is another important factor. Your task is to investigate the installation cost for each proposal. The cost is proportional to the number of cubes, so you have to figure out the minimum number of cubes needed for installation.
Each design proposal of an artwork consists of the front view and the side view (the view seen from the right-hand side), as shown in Figure 1.
<image>
Figure 1: An example of an artwork proposal
The front view (resp., the side view) indicates the maximum heights of piles of cubes for each column line (resp., row line) of the grid.
There are several ways to install this proposal of artwork, such as the following figures.
<image>
In these figures, the dotted lines on the ground indicate the grid lines. The left figure makes use of 16 cubes, which is not optimal. That is, the artwork can be installed with a fewer number of cubes. Actually, the right one is optimal and only uses 13 cubes. Note that, a single pile of height three in the right figure plays the roles of two such piles in the left one.
Notice that swapping columns of cubes does not change the side view. Similarly, swapping rows does not change the front view. Thus, such swaps do not change the costs of building the artworks.
For example, consider the artwork proposal given in Figure 2.
<image>
Figure 2: Another example of artwork proposal
An optimal installation of this proposal of artwork can be achieved with 13 cubes, as shown in the following figure, which can be obtained by exchanging the rightmost two columns of the optimal installation of the artwork of Figure 1.
<image>
Input
The input is a sequence of datasets. The end of the input is indicated by a line containing two zeros separated by a space. Each dataset is formatted as follows.
w d
h1 h2 ... hw
h'1 h'2 ... h'd
The integers w and d separated by a space are the numbers of columns and rows of the grid, respectively. You may assume 1 ≤ w ≤ 10 and 1 ≤ d ≤ 10. The integers separated by a space in the second and third lines specify the shape of the artwork. The integers hi (1 ≤ hi ≤ 20, 1 ≤ i ≤ w) in the second line give the front view, i.e., the maximum heights of cubes per each column line, ordered from left to right (seen from the front). The integers hi (1 ≤ hi ≤ 20, 1 ≤ i ≤ d) in the third line give the side view, i.e., the maximum heights of cubes per each row line, ordered from left to right (seen from the right-hand side).
Output
For each dataset, output a line containing the minimum number of cubes. The output should not contain any other extra characters.
You can assume that, for each dataset, there is at least one way to install the artwork.
Example
Input
5 5
1 2 3 4 5
1 2 3 4 5
5 5
2 5 4 1 3
4 1 5 3 2
5 5
1 2 3 4 5
3 3 3 4 5
3 3
7 7 7
7 7 7
3 3
4 4 4
4 3 4
4 3
4 2 2 4
4 2 1
4 4
2 8 8 8
2 3 8 3
10 10
9 9 9 9 9 9 9 9 9 9
9 9 9 9 9 9 9 9 9 9
10 9
20 1 20 20 20 20 20 18 20 20
20 20 20 20 7 20 20 20 20
0 0
Output
15
15
21
21
15
13
32
90
186 | instruction | 0 | 91,673 | 8 | 183,346 |
"Correct Solution:
```
while True:
w,d = map(int,input().split())
if w == 0 and d == 0:
break
h = list(map(int,input().split()))
m = list(map(int,input().split()))
hl = [0] * 30
ml = [0] * 30
for i in h:
hl[i] += 1
for i in m:
ml[i] += 1
ans = 0
for i in range(30):
ans += i * max(hl[i],ml[i])
print (ans)
``` | output | 1 | 91,673 | 8 | 183,347 |
Provide a correct Python 3 solution for this coding contest problem.
International Center for Picassonian Cubism is a Spanish national museum of cubist artworks, dedicated to Pablo Picasso. The center held a competition for an artwork that will be displayed in front of the facade of the museum building. The artwork is a collection of cubes that are piled up on the ground and is intended to amuse visitors, who will be curious how the shape of the collection of cubes changes when it is seen from the front and the sides.
The artwork is a collection of cubes with edges of one foot long and is built on a flat ground that is divided into a grid of one foot by one foot squares. Due to some technical reasons, cubes of the artwork must be either put on the ground, fitting into a unit square in the grid, or put on another cube in the way that the bottom face of the upper cube exactly meets the top face of the lower cube. No other way of putting cubes is possible.
You are a member of the judging committee responsible for selecting one out of a plenty of artwork proposals submitted to the competition. The decision is made primarily based on artistic quality but the cost for installing the artwork is another important factor. Your task is to investigate the installation cost for each proposal. The cost is proportional to the number of cubes, so you have to figure out the minimum number of cubes needed for installation.
Each design proposal of an artwork consists of the front view and the side view (the view seen from the right-hand side), as shown in Figure 1.
<image>
Figure 1: An example of an artwork proposal
The front view (resp., the side view) indicates the maximum heights of piles of cubes for each column line (resp., row line) of the grid.
There are several ways to install this proposal of artwork, such as the following figures.
<image>
In these figures, the dotted lines on the ground indicate the grid lines. The left figure makes use of 16 cubes, which is not optimal. That is, the artwork can be installed with a fewer number of cubes. Actually, the right one is optimal and only uses 13 cubes. Note that, a single pile of height three in the right figure plays the roles of two such piles in the left one.
Notice that swapping columns of cubes does not change the side view. Similarly, swapping rows does not change the front view. Thus, such swaps do not change the costs of building the artworks.
For example, consider the artwork proposal given in Figure 2.
<image>
Figure 2: Another example of artwork proposal
An optimal installation of this proposal of artwork can be achieved with 13 cubes, as shown in the following figure, which can be obtained by exchanging the rightmost two columns of the optimal installation of the artwork of Figure 1.
<image>
Input
The input is a sequence of datasets. The end of the input is indicated by a line containing two zeros separated by a space. Each dataset is formatted as follows.
w d
h1 h2 ... hw
h'1 h'2 ... h'd
The integers w and d separated by a space are the numbers of columns and rows of the grid, respectively. You may assume 1 ≤ w ≤ 10 and 1 ≤ d ≤ 10. The integers separated by a space in the second and third lines specify the shape of the artwork. The integers hi (1 ≤ hi ≤ 20, 1 ≤ i ≤ w) in the second line give the front view, i.e., the maximum heights of cubes per each column line, ordered from left to right (seen from the front). The integers hi (1 ≤ hi ≤ 20, 1 ≤ i ≤ d) in the third line give the side view, i.e., the maximum heights of cubes per each row line, ordered from left to right (seen from the right-hand side).
Output
For each dataset, output a line containing the minimum number of cubes. The output should not contain any other extra characters.
You can assume that, for each dataset, there is at least one way to install the artwork.
Example
Input
5 5
1 2 3 4 5
1 2 3 4 5
5 5
2 5 4 1 3
4 1 5 3 2
5 5
1 2 3 4 5
3 3 3 4 5
3 3
7 7 7
7 7 7
3 3
4 4 4
4 3 4
4 3
4 2 2 4
4 2 1
4 4
2 8 8 8
2 3 8 3
10 10
9 9 9 9 9 9 9 9 9 9
9 9 9 9 9 9 9 9 9 9
10 9
20 1 20 20 20 20 20 18 20 20
20 20 20 20 7 20 20 20 20
0 0
Output
15
15
21
21
15
13
32
90
186 | instruction | 0 | 91,674 | 8 | 183,348 |
"Correct Solution:
```
while True:
w,d = [int(i) for i in input().split()]
if w == 0:
break
h = [int(i) for i in input().split()]
h_ = [int(i) for i in input().split()]
h = sorted(h)
h_ = sorted(h_)
hdic = {}
h_dic = {}
for i in h:
if i in hdic.keys():
hdic[i]+= 1
else:
hdic[i] = 1
for i in h_:
if i in h_dic.keys():
h_dic[i]+= 1
else:
h_dic[i] = 1
ans = sum(h)+ sum(h_)
for num in hdic.keys():
if num in h_dic.keys():
ans -= num*min(hdic[num],h_dic[num])
print(ans)
``` | output | 1 | 91,674 | 8 | 183,349 |
Provide a correct Python 3 solution for this coding contest problem.
International Center for Picassonian Cubism is a Spanish national museum of cubist artworks, dedicated to Pablo Picasso. The center held a competition for an artwork that will be displayed in front of the facade of the museum building. The artwork is a collection of cubes that are piled up on the ground and is intended to amuse visitors, who will be curious how the shape of the collection of cubes changes when it is seen from the front and the sides.
The artwork is a collection of cubes with edges of one foot long and is built on a flat ground that is divided into a grid of one foot by one foot squares. Due to some technical reasons, cubes of the artwork must be either put on the ground, fitting into a unit square in the grid, or put on another cube in the way that the bottom face of the upper cube exactly meets the top face of the lower cube. No other way of putting cubes is possible.
You are a member of the judging committee responsible for selecting one out of a plenty of artwork proposals submitted to the competition. The decision is made primarily based on artistic quality but the cost for installing the artwork is another important factor. Your task is to investigate the installation cost for each proposal. The cost is proportional to the number of cubes, so you have to figure out the minimum number of cubes needed for installation.
Each design proposal of an artwork consists of the front view and the side view (the view seen from the right-hand side), as shown in Figure 1.
<image>
Figure 1: An example of an artwork proposal
The front view (resp., the side view) indicates the maximum heights of piles of cubes for each column line (resp., row line) of the grid.
There are several ways to install this proposal of artwork, such as the following figures.
<image>
In these figures, the dotted lines on the ground indicate the grid lines. The left figure makes use of 16 cubes, which is not optimal. That is, the artwork can be installed with a fewer number of cubes. Actually, the right one is optimal and only uses 13 cubes. Note that, a single pile of height three in the right figure plays the roles of two such piles in the left one.
Notice that swapping columns of cubes does not change the side view. Similarly, swapping rows does not change the front view. Thus, such swaps do not change the costs of building the artworks.
For example, consider the artwork proposal given in Figure 2.
<image>
Figure 2: Another example of artwork proposal
An optimal installation of this proposal of artwork can be achieved with 13 cubes, as shown in the following figure, which can be obtained by exchanging the rightmost two columns of the optimal installation of the artwork of Figure 1.
<image>
Input
The input is a sequence of datasets. The end of the input is indicated by a line containing two zeros separated by a space. Each dataset is formatted as follows.
w d
h1 h2 ... hw
h'1 h'2 ... h'd
The integers w and d separated by a space are the numbers of columns and rows of the grid, respectively. You may assume 1 ≤ w ≤ 10 and 1 ≤ d ≤ 10. The integers separated by a space in the second and third lines specify the shape of the artwork. The integers hi (1 ≤ hi ≤ 20, 1 ≤ i ≤ w) in the second line give the front view, i.e., the maximum heights of cubes per each column line, ordered from left to right (seen from the front). The integers hi (1 ≤ hi ≤ 20, 1 ≤ i ≤ d) in the third line give the side view, i.e., the maximum heights of cubes per each row line, ordered from left to right (seen from the right-hand side).
Output
For each dataset, output a line containing the minimum number of cubes. The output should not contain any other extra characters.
You can assume that, for each dataset, there is at least one way to install the artwork.
Example
Input
5 5
1 2 3 4 5
1 2 3 4 5
5 5
2 5 4 1 3
4 1 5 3 2
5 5
1 2 3 4 5
3 3 3 4 5
3 3
7 7 7
7 7 7
3 3
4 4 4
4 3 4
4 3
4 2 2 4
4 2 1
4 4
2 8 8 8
2 3 8 3
10 10
9 9 9 9 9 9 9 9 9 9
9 9 9 9 9 9 9 9 9 9
10 9
20 1 20 20 20 20 20 18 20 20
20 20 20 20 7 20 20 20 20
0 0
Output
15
15
21
21
15
13
32
90
186 | instruction | 0 | 91,675 | 8 | 183,350 |
"Correct Solution:
```
import math,string,itertools,fractions,heapq,collections,re,array,bisect,sys,random,time,copy,functools
sys.setrecursionlimit(10**7)
inf = 10**3
eps = 1.0 / 10**10
mod = 10**9+7
def LI(): return [int(x) for x in sys.stdin.readline().split()]
def LI_(): return [int(x)-1 for x in sys.stdin.readline().split()]
def LF(): return [float(x) for x in sys.stdin.readline().split()]
def LS(): return sys.stdin.readline().split()
def I(): return int(sys.stdin.readline())
def F(): return float(sys.stdin.readline())
def S(): return input()
def pf(s): return print(s, flush=True)
def main():
rr = []
while True:
w,d = LI()
if w == 0:
break
a = sorted(LI())
b = sorted(LI())
ai = 0
bi = 0
r = 0
while ai < w or bi < d:
if ai >= w:
r += sum(b[bi:])
break
if bi >= d:
r += sum(a[ai:])
break
if a[ai] == b[bi]:
r += a[ai]
ai += 1
bi += 1
elif a[ai] < b[bi]:
r += a[ai]
ai += 1
else:
r += b[bi]
bi += 1
rr.append(r)
return '\n'.join(map(str, rr))
print(main())
``` | output | 1 | 91,675 | 8 | 183,351 |
Provide a correct Python 3 solution for this coding contest problem.
International Center for Picassonian Cubism is a Spanish national museum of cubist artworks, dedicated to Pablo Picasso. The center held a competition for an artwork that will be displayed in front of the facade of the museum building. The artwork is a collection of cubes that are piled up on the ground and is intended to amuse visitors, who will be curious how the shape of the collection of cubes changes when it is seen from the front and the sides.
The artwork is a collection of cubes with edges of one foot long and is built on a flat ground that is divided into a grid of one foot by one foot squares. Due to some technical reasons, cubes of the artwork must be either put on the ground, fitting into a unit square in the grid, or put on another cube in the way that the bottom face of the upper cube exactly meets the top face of the lower cube. No other way of putting cubes is possible.
You are a member of the judging committee responsible for selecting one out of a plenty of artwork proposals submitted to the competition. The decision is made primarily based on artistic quality but the cost for installing the artwork is another important factor. Your task is to investigate the installation cost for each proposal. The cost is proportional to the number of cubes, so you have to figure out the minimum number of cubes needed for installation.
Each design proposal of an artwork consists of the front view and the side view (the view seen from the right-hand side), as shown in Figure 1.
<image>
Figure 1: An example of an artwork proposal
The front view (resp., the side view) indicates the maximum heights of piles of cubes for each column line (resp., row line) of the grid.
There are several ways to install this proposal of artwork, such as the following figures.
<image>
In these figures, the dotted lines on the ground indicate the grid lines. The left figure makes use of 16 cubes, which is not optimal. That is, the artwork can be installed with a fewer number of cubes. Actually, the right one is optimal and only uses 13 cubes. Note that, a single pile of height three in the right figure plays the roles of two such piles in the left one.
Notice that swapping columns of cubes does not change the side view. Similarly, swapping rows does not change the front view. Thus, such swaps do not change the costs of building the artworks.
For example, consider the artwork proposal given in Figure 2.
<image>
Figure 2: Another example of artwork proposal
An optimal installation of this proposal of artwork can be achieved with 13 cubes, as shown in the following figure, which can be obtained by exchanging the rightmost two columns of the optimal installation of the artwork of Figure 1.
<image>
Input
The input is a sequence of datasets. The end of the input is indicated by a line containing two zeros separated by a space. Each dataset is formatted as follows.
w d
h1 h2 ... hw
h'1 h'2 ... h'd
The integers w and d separated by a space are the numbers of columns and rows of the grid, respectively. You may assume 1 ≤ w ≤ 10 and 1 ≤ d ≤ 10. The integers separated by a space in the second and third lines specify the shape of the artwork. The integers hi (1 ≤ hi ≤ 20, 1 ≤ i ≤ w) in the second line give the front view, i.e., the maximum heights of cubes per each column line, ordered from left to right (seen from the front). The integers hi (1 ≤ hi ≤ 20, 1 ≤ i ≤ d) in the third line give the side view, i.e., the maximum heights of cubes per each row line, ordered from left to right (seen from the right-hand side).
Output
For each dataset, output a line containing the minimum number of cubes. The output should not contain any other extra characters.
You can assume that, for each dataset, there is at least one way to install the artwork.
Example
Input
5 5
1 2 3 4 5
1 2 3 4 5
5 5
2 5 4 1 3
4 1 5 3 2
5 5
1 2 3 4 5
3 3 3 4 5
3 3
7 7 7
7 7 7
3 3
4 4 4
4 3 4
4 3
4 2 2 4
4 2 1
4 4
2 8 8 8
2 3 8 3
10 10
9 9 9 9 9 9 9 9 9 9
9 9 9 9 9 9 9 9 9 9
10 9
20 1 20 20 20 20 20 18 20 20
20 20 20 20 7 20 20 20 20
0 0
Output
15
15
21
21
15
13
32
90
186 | instruction | 0 | 91,676 | 8 | 183,352 |
"Correct Solution:
```
def solve():
from sys import stdin
from collections import Counter
f_i = stdin
while True:
w, d = map(int, f_i.readline().split())
if w == 0:
break
h_w = Counter(map(int, f_i.readline().split()))
h_d = Counter(map(int, f_i.readline().split()))
for k, v in h_w.items():
if v > h_d.setdefault(k, v):
h_d[k] = v
print(sum(k * v for k, v in h_d.items()))
solve()
``` | output | 1 | 91,676 | 8 | 183,353 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
International Center for Picassonian Cubism is a Spanish national museum of cubist artworks, dedicated to Pablo Picasso. The center held a competition for an artwork that will be displayed in front of the facade of the museum building. The artwork is a collection of cubes that are piled up on the ground and is intended to amuse visitors, who will be curious how the shape of the collection of cubes changes when it is seen from the front and the sides.
The artwork is a collection of cubes with edges of one foot long and is built on a flat ground that is divided into a grid of one foot by one foot squares. Due to some technical reasons, cubes of the artwork must be either put on the ground, fitting into a unit square in the grid, or put on another cube in the way that the bottom face of the upper cube exactly meets the top face of the lower cube. No other way of putting cubes is possible.
You are a member of the judging committee responsible for selecting one out of a plenty of artwork proposals submitted to the competition. The decision is made primarily based on artistic quality but the cost for installing the artwork is another important factor. Your task is to investigate the installation cost for each proposal. The cost is proportional to the number of cubes, so you have to figure out the minimum number of cubes needed for installation.
Each design proposal of an artwork consists of the front view and the side view (the view seen from the right-hand side), as shown in Figure 1.
<image>
Figure 1: An example of an artwork proposal
The front view (resp., the side view) indicates the maximum heights of piles of cubes for each column line (resp., row line) of the grid.
There are several ways to install this proposal of artwork, such as the following figures.
<image>
In these figures, the dotted lines on the ground indicate the grid lines. The left figure makes use of 16 cubes, which is not optimal. That is, the artwork can be installed with a fewer number of cubes. Actually, the right one is optimal and only uses 13 cubes. Note that, a single pile of height three in the right figure plays the roles of two such piles in the left one.
Notice that swapping columns of cubes does not change the side view. Similarly, swapping rows does not change the front view. Thus, such swaps do not change the costs of building the artworks.
For example, consider the artwork proposal given in Figure 2.
<image>
Figure 2: Another example of artwork proposal
An optimal installation of this proposal of artwork can be achieved with 13 cubes, as shown in the following figure, which can be obtained by exchanging the rightmost two columns of the optimal installation of the artwork of Figure 1.
<image>
Input
The input is a sequence of datasets. The end of the input is indicated by a line containing two zeros separated by a space. Each dataset is formatted as follows.
w d
h1 h2 ... hw
h'1 h'2 ... h'd
The integers w and d separated by a space are the numbers of columns and rows of the grid, respectively. You may assume 1 ≤ w ≤ 10 and 1 ≤ d ≤ 10. The integers separated by a space in the second and third lines specify the shape of the artwork. The integers hi (1 ≤ hi ≤ 20, 1 ≤ i ≤ w) in the second line give the front view, i.e., the maximum heights of cubes per each column line, ordered from left to right (seen from the front). The integers hi (1 ≤ hi ≤ 20, 1 ≤ i ≤ d) in the third line give the side view, i.e., the maximum heights of cubes per each row line, ordered from left to right (seen from the right-hand side).
Output
For each dataset, output a line containing the minimum number of cubes. The output should not contain any other extra characters.
You can assume that, for each dataset, there is at least one way to install the artwork.
Example
Input
5 5
1 2 3 4 5
1 2 3 4 5
5 5
2 5 4 1 3
4 1 5 3 2
5 5
1 2 3 4 5
3 3 3 4 5
3 3
7 7 7
7 7 7
3 3
4 4 4
4 3 4
4 3
4 2 2 4
4 2 1
4 4
2 8 8 8
2 3 8 3
10 10
9 9 9 9 9 9 9 9 9 9
9 9 9 9 9 9 9 9 9 9
10 9
20 1 20 20 20 20 20 18 20 20
20 20 20 20 7 20 20 20 20
0 0
Output
15
15
21
21
15
13
32
90
186
Submitted Solution:
```
from collections import defaultdict
while True:
a, b = [int(i) for i in input().split()]
if a == 0 and b == 0:
exit()
H = [int(i) for i in input().split()]
W = [int(i) for i in input().split()]
h_cnt = defaultdict(int)
w_cnt = defaultdict(int)
for h in H:
h_cnt[h] += 1
for w in W:
w_cnt[w] += 1
ans = 0
for k, v in h_cnt.items():
ans += k * max(v, w_cnt[k])
w_cnt[k] = 0
for k, v in w_cnt.items():
ans += k * v
print(ans)
``` | instruction | 0 | 91,677 | 8 | 183,354 |
Yes | output | 1 | 91,677 | 8 | 183,355 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
International Center for Picassonian Cubism is a Spanish national museum of cubist artworks, dedicated to Pablo Picasso. The center held a competition for an artwork that will be displayed in front of the facade of the museum building. The artwork is a collection of cubes that are piled up on the ground and is intended to amuse visitors, who will be curious how the shape of the collection of cubes changes when it is seen from the front and the sides.
The artwork is a collection of cubes with edges of one foot long and is built on a flat ground that is divided into a grid of one foot by one foot squares. Due to some technical reasons, cubes of the artwork must be either put on the ground, fitting into a unit square in the grid, or put on another cube in the way that the bottom face of the upper cube exactly meets the top face of the lower cube. No other way of putting cubes is possible.
You are a member of the judging committee responsible for selecting one out of a plenty of artwork proposals submitted to the competition. The decision is made primarily based on artistic quality but the cost for installing the artwork is another important factor. Your task is to investigate the installation cost for each proposal. The cost is proportional to the number of cubes, so you have to figure out the minimum number of cubes needed for installation.
Each design proposal of an artwork consists of the front view and the side view (the view seen from the right-hand side), as shown in Figure 1.
<image>
Figure 1: An example of an artwork proposal
The front view (resp., the side view) indicates the maximum heights of piles of cubes for each column line (resp., row line) of the grid.
There are several ways to install this proposal of artwork, such as the following figures.
<image>
In these figures, the dotted lines on the ground indicate the grid lines. The left figure makes use of 16 cubes, which is not optimal. That is, the artwork can be installed with a fewer number of cubes. Actually, the right one is optimal and only uses 13 cubes. Note that, a single pile of height three in the right figure plays the roles of two such piles in the left one.
Notice that swapping columns of cubes does not change the side view. Similarly, swapping rows does not change the front view. Thus, such swaps do not change the costs of building the artworks.
For example, consider the artwork proposal given in Figure 2.
<image>
Figure 2: Another example of artwork proposal
An optimal installation of this proposal of artwork can be achieved with 13 cubes, as shown in the following figure, which can be obtained by exchanging the rightmost two columns of the optimal installation of the artwork of Figure 1.
<image>
Input
The input is a sequence of datasets. The end of the input is indicated by a line containing two zeros separated by a space. Each dataset is formatted as follows.
w d
h1 h2 ... hw
h'1 h'2 ... h'd
The integers w and d separated by a space are the numbers of columns and rows of the grid, respectively. You may assume 1 ≤ w ≤ 10 and 1 ≤ d ≤ 10. The integers separated by a space in the second and third lines specify the shape of the artwork. The integers hi (1 ≤ hi ≤ 20, 1 ≤ i ≤ w) in the second line give the front view, i.e., the maximum heights of cubes per each column line, ordered from left to right (seen from the front). The integers hi (1 ≤ hi ≤ 20, 1 ≤ i ≤ d) in the third line give the side view, i.e., the maximum heights of cubes per each row line, ordered from left to right (seen from the right-hand side).
Output
For each dataset, output a line containing the minimum number of cubes. The output should not contain any other extra characters.
You can assume that, for each dataset, there is at least one way to install the artwork.
Example
Input
5 5
1 2 3 4 5
1 2 3 4 5
5 5
2 5 4 1 3
4 1 5 3 2
5 5
1 2 3 4 5
3 3 3 4 5
3 3
7 7 7
7 7 7
3 3
4 4 4
4 3 4
4 3
4 2 2 4
4 2 1
4 4
2 8 8 8
2 3 8 3
10 10
9 9 9 9 9 9 9 9 9 9
9 9 9 9 9 9 9 9 9 9
10 9
20 1 20 20 20 20 20 18 20 20
20 20 20 20 7 20 20 20 20
0 0
Output
15
15
21
21
15
13
32
90
186
Submitted Solution:
```
from collections import defaultdict
# python template for atcoder1
import sys
sys.setrecursionlimit(10**9)
input = sys.stdin.readline
def solve():
H, W = map(int, input().split())
if H == 0 and W == 0:
exit()
A = list(map(int, input().split()))
B = list(map(int, input().split()))
d_a = defaultdict(int)
for a in A:
d_a[a] += 1
d_b = defaultdict(int)
for b in B:
d_b[b] += 1
ans = 0
for v in set(A + B):
ans += v*max(d_a[v], d_b[v])
print(ans)
while True:
solve()
``` | instruction | 0 | 91,678 | 8 | 183,356 |
Yes | output | 1 | 91,678 | 8 | 183,357 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
International Center for Picassonian Cubism is a Spanish national museum of cubist artworks, dedicated to Pablo Picasso. The center held a competition for an artwork that will be displayed in front of the facade of the museum building. The artwork is a collection of cubes that are piled up on the ground and is intended to amuse visitors, who will be curious how the shape of the collection of cubes changes when it is seen from the front and the sides.
The artwork is a collection of cubes with edges of one foot long and is built on a flat ground that is divided into a grid of one foot by one foot squares. Due to some technical reasons, cubes of the artwork must be either put on the ground, fitting into a unit square in the grid, or put on another cube in the way that the bottom face of the upper cube exactly meets the top face of the lower cube. No other way of putting cubes is possible.
You are a member of the judging committee responsible for selecting one out of a plenty of artwork proposals submitted to the competition. The decision is made primarily based on artistic quality but the cost for installing the artwork is another important factor. Your task is to investigate the installation cost for each proposal. The cost is proportional to the number of cubes, so you have to figure out the minimum number of cubes needed for installation.
Each design proposal of an artwork consists of the front view and the side view (the view seen from the right-hand side), as shown in Figure 1.
<image>
Figure 1: An example of an artwork proposal
The front view (resp., the side view) indicates the maximum heights of piles of cubes for each column line (resp., row line) of the grid.
There are several ways to install this proposal of artwork, such as the following figures.
<image>
In these figures, the dotted lines on the ground indicate the grid lines. The left figure makes use of 16 cubes, which is not optimal. That is, the artwork can be installed with a fewer number of cubes. Actually, the right one is optimal and only uses 13 cubes. Note that, a single pile of height three in the right figure plays the roles of two such piles in the left one.
Notice that swapping columns of cubes does not change the side view. Similarly, swapping rows does not change the front view. Thus, such swaps do not change the costs of building the artworks.
For example, consider the artwork proposal given in Figure 2.
<image>
Figure 2: Another example of artwork proposal
An optimal installation of this proposal of artwork can be achieved with 13 cubes, as shown in the following figure, which can be obtained by exchanging the rightmost two columns of the optimal installation of the artwork of Figure 1.
<image>
Input
The input is a sequence of datasets. The end of the input is indicated by a line containing two zeros separated by a space. Each dataset is formatted as follows.
w d
h1 h2 ... hw
h'1 h'2 ... h'd
The integers w and d separated by a space are the numbers of columns and rows of the grid, respectively. You may assume 1 ≤ w ≤ 10 and 1 ≤ d ≤ 10. The integers separated by a space in the second and third lines specify the shape of the artwork. The integers hi (1 ≤ hi ≤ 20, 1 ≤ i ≤ w) in the second line give the front view, i.e., the maximum heights of cubes per each column line, ordered from left to right (seen from the front). The integers hi (1 ≤ hi ≤ 20, 1 ≤ i ≤ d) in the third line give the side view, i.e., the maximum heights of cubes per each row line, ordered from left to right (seen from the right-hand side).
Output
For each dataset, output a line containing the minimum number of cubes. The output should not contain any other extra characters.
You can assume that, for each dataset, there is at least one way to install the artwork.
Example
Input
5 5
1 2 3 4 5
1 2 3 4 5
5 5
2 5 4 1 3
4 1 5 3 2
5 5
1 2 3 4 5
3 3 3 4 5
3 3
7 7 7
7 7 7
3 3
4 4 4
4 3 4
4 3
4 2 2 4
4 2 1
4 4
2 8 8 8
2 3 8 3
10 10
9 9 9 9 9 9 9 9 9 9
9 9 9 9 9 9 9 9 9 9
10 9
20 1 20 20 20 20 20 18 20 20
20 20 20 20 7 20 20 20 20
0 0
Output
15
15
21
21
15
13
32
90
186
Submitted Solution:
```
#include <bits/stdc++.h>
using namespace std;
int main() {
while (true) {
int W, D;
cin >> W >> D;
if (W == 0 or D == 0) break;
static int HW[10], HD[10];
for (int w = 0; w < (W); ++w) cin >> HW[w];
for (int d = 0; d < (D); ++d) cin >> HD[d];
sort(HW, HW + W);
sort(HD, HD + D);
int result = 0;
{
int w = 0, d = 0;
while (w < W and d < D) {
if (HW[w] < HD[d]) {
result += HW[w];
++w;
} else if (HW[w] > HD[d]) {
result += HD[d];
++d;
} else {
result += HW[w];
++w;
++d;
}
}
while (w < W) {
result += HW[w];
++w;
}
while (d < D) {
result += HD[d];
++d;
}
}
cout << result << endl;
}
return 0;
}
``` | instruction | 0 | 91,679 | 8 | 183,358 |
No | output | 1 | 91,679 | 8 | 183,359 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
International Center for Picassonian Cubism is a Spanish national museum of cubist artworks, dedicated to Pablo Picasso. The center held a competition for an artwork that will be displayed in front of the facade of the museum building. The artwork is a collection of cubes that are piled up on the ground and is intended to amuse visitors, who will be curious how the shape of the collection of cubes changes when it is seen from the front and the sides.
The artwork is a collection of cubes with edges of one foot long and is built on a flat ground that is divided into a grid of one foot by one foot squares. Due to some technical reasons, cubes of the artwork must be either put on the ground, fitting into a unit square in the grid, or put on another cube in the way that the bottom face of the upper cube exactly meets the top face of the lower cube. No other way of putting cubes is possible.
You are a member of the judging committee responsible for selecting one out of a plenty of artwork proposals submitted to the competition. The decision is made primarily based on artistic quality but the cost for installing the artwork is another important factor. Your task is to investigate the installation cost for each proposal. The cost is proportional to the number of cubes, so you have to figure out the minimum number of cubes needed for installation.
Each design proposal of an artwork consists of the front view and the side view (the view seen from the right-hand side), as shown in Figure 1.
<image>
Figure 1: An example of an artwork proposal
The front view (resp., the side view) indicates the maximum heights of piles of cubes for each column line (resp., row line) of the grid.
There are several ways to install this proposal of artwork, such as the following figures.
<image>
In these figures, the dotted lines on the ground indicate the grid lines. The left figure makes use of 16 cubes, which is not optimal. That is, the artwork can be installed with a fewer number of cubes. Actually, the right one is optimal and only uses 13 cubes. Note that, a single pile of height three in the right figure plays the roles of two such piles in the left one.
Notice that swapping columns of cubes does not change the side view. Similarly, swapping rows does not change the front view. Thus, such swaps do not change the costs of building the artworks.
For example, consider the artwork proposal given in Figure 2.
<image>
Figure 2: Another example of artwork proposal
An optimal installation of this proposal of artwork can be achieved with 13 cubes, as shown in the following figure, which can be obtained by exchanging the rightmost two columns of the optimal installation of the artwork of Figure 1.
<image>
Input
The input is a sequence of datasets. The end of the input is indicated by a line containing two zeros separated by a space. Each dataset is formatted as follows.
w d
h1 h2 ... hw
h'1 h'2 ... h'd
The integers w and d separated by a space are the numbers of columns and rows of the grid, respectively. You may assume 1 ≤ w ≤ 10 and 1 ≤ d ≤ 10. The integers separated by a space in the second and third lines specify the shape of the artwork. The integers hi (1 ≤ hi ≤ 20, 1 ≤ i ≤ w) in the second line give the front view, i.e., the maximum heights of cubes per each column line, ordered from left to right (seen from the front). The integers hi (1 ≤ hi ≤ 20, 1 ≤ i ≤ d) in the third line give the side view, i.e., the maximum heights of cubes per each row line, ordered from left to right (seen from the right-hand side).
Output
For each dataset, output a line containing the minimum number of cubes. The output should not contain any other extra characters.
You can assume that, for each dataset, there is at least one way to install the artwork.
Example
Input
5 5
1 2 3 4 5
1 2 3 4 5
5 5
2 5 4 1 3
4 1 5 3 2
5 5
1 2 3 4 5
3 3 3 4 5
3 3
7 7 7
7 7 7
3 3
4 4 4
4 3 4
4 3
4 2 2 4
4 2 1
4 4
2 8 8 8
2 3 8 3
10 10
9 9 9 9 9 9 9 9 9 9
9 9 9 9 9 9 9 9 9 9
10 9
20 1 20 20 20 20 20 18 20 20
20 20 20 20 7 20 20 20 20
0 0
Output
15
15
21
21
15
13
32
90
186
Submitted Solution:
```
while True:
w,d = [int(i) for i in input().split()]
if w == 0:
break
h = [int(i) for i in input().split()]
h_ = [int(i) for i in input().split()]
h = sorted(h)
h_ = sorted(h_)
hdic = {}
h_dic = {}
for i in h:
if i in hdic.keys():
hdic[i]+= 1
else:
hdic[i] = 1
for i in h_:
if i in h_dic.keys():
h_dic[i]+= 1
else:
h_dic[i] = 1
ans = sum(h)+ sum(h_)
print(ans)
for num in hdic.keys():
if num in h_dic.keys():
ans -= num*min(hdic[i],h_dic[i])
print(hdic,h_dic)
print(ans)
``` | instruction | 0 | 91,680 | 8 | 183,360 |
No | output | 1 | 91,680 | 8 | 183,361 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
International Center for Picassonian Cubism is a Spanish national museum of cubist artworks, dedicated to Pablo Picasso. The center held a competition for an artwork that will be displayed in front of the facade of the museum building. The artwork is a collection of cubes that are piled up on the ground and is intended to amuse visitors, who will be curious how the shape of the collection of cubes changes when it is seen from the front and the sides.
The artwork is a collection of cubes with edges of one foot long and is built on a flat ground that is divided into a grid of one foot by one foot squares. Due to some technical reasons, cubes of the artwork must be either put on the ground, fitting into a unit square in the grid, or put on another cube in the way that the bottom face of the upper cube exactly meets the top face of the lower cube. No other way of putting cubes is possible.
You are a member of the judging committee responsible for selecting one out of a plenty of artwork proposals submitted to the competition. The decision is made primarily based on artistic quality but the cost for installing the artwork is another important factor. Your task is to investigate the installation cost for each proposal. The cost is proportional to the number of cubes, so you have to figure out the minimum number of cubes needed for installation.
Each design proposal of an artwork consists of the front view and the side view (the view seen from the right-hand side), as shown in Figure 1.
<image>
Figure 1: An example of an artwork proposal
The front view (resp., the side view) indicates the maximum heights of piles of cubes for each column line (resp., row line) of the grid.
There are several ways to install this proposal of artwork, such as the following figures.
<image>
In these figures, the dotted lines on the ground indicate the grid lines. The left figure makes use of 16 cubes, which is not optimal. That is, the artwork can be installed with a fewer number of cubes. Actually, the right one is optimal and only uses 13 cubes. Note that, a single pile of height three in the right figure plays the roles of two such piles in the left one.
Notice that swapping columns of cubes does not change the side view. Similarly, swapping rows does not change the front view. Thus, such swaps do not change the costs of building the artworks.
For example, consider the artwork proposal given in Figure 2.
<image>
Figure 2: Another example of artwork proposal
An optimal installation of this proposal of artwork can be achieved with 13 cubes, as shown in the following figure, which can be obtained by exchanging the rightmost two columns of the optimal installation of the artwork of Figure 1.
<image>
Input
The input is a sequence of datasets. The end of the input is indicated by a line containing two zeros separated by a space. Each dataset is formatted as follows.
w d
h1 h2 ... hw
h'1 h'2 ... h'd
The integers w and d separated by a space are the numbers of columns and rows of the grid, respectively. You may assume 1 ≤ w ≤ 10 and 1 ≤ d ≤ 10. The integers separated by a space in the second and third lines specify the shape of the artwork. The integers hi (1 ≤ hi ≤ 20, 1 ≤ i ≤ w) in the second line give the front view, i.e., the maximum heights of cubes per each column line, ordered from left to right (seen from the front). The integers hi (1 ≤ hi ≤ 20, 1 ≤ i ≤ d) in the third line give the side view, i.e., the maximum heights of cubes per each row line, ordered from left to right (seen from the right-hand side).
Output
For each dataset, output a line containing the minimum number of cubes. The output should not contain any other extra characters.
You can assume that, for each dataset, there is at least one way to install the artwork.
Example
Input
5 5
1 2 3 4 5
1 2 3 4 5
5 5
2 5 4 1 3
4 1 5 3 2
5 5
1 2 3 4 5
3 3 3 4 5
3 3
7 7 7
7 7 7
3 3
4 4 4
4 3 4
4 3
4 2 2 4
4 2 1
4 4
2 8 8 8
2 3 8 3
10 10
9 9 9 9 9 9 9 9 9 9
9 9 9 9 9 9 9 9 9 9
10 9
20 1 20 20 20 20 20 18 20 20
20 20 20 20 7 20 20 20 20
0 0
Output
15
15
21
21
15
13
32
90
186
Submitted Solution:
```
while True:
w,d = [int(i) for i in input().split()]
if w == 0:
break
h = [int(i) for i in input().split()]
h_ = [int(i) for i in input().split()]
h = sorted(h)
h_ = sorted(h_)
hdic = {}
h_dic = {}
for i in h:
if i in hdic.keys():
hdic[i]+= 1
else:
hdic[i] = 1
for i in h_:
if i in h_dic.keys():
h_dic[i]+= 1
else:
h_dic[i] = 1
ans = sum(h)+ sum(h_)
for num in hdic.keys():
if num in h_dic.keys():
ans -= num*min(hdic[i],h_dic[i])
print(ans)
``` | instruction | 0 | 91,681 | 8 | 183,362 |
No | output | 1 | 91,681 | 8 | 183,363 |
Provide tags and a correct Python 3 solution for this coding contest problem.
There are n rectangles in a row. You can either turn each rectangle by 90 degrees or leave it as it is. If you turn a rectangle, its width will be height, and its height will be width. Notice that you can turn any number of rectangles, you also can turn all or none of them. You can not change the order of the rectangles.
Find out if there is a way to make the rectangles go in order of non-ascending height. In other words, after all the turns, a height of every rectangle has to be not greater than the height of the previous rectangle (if it is such).
Input
The first line contains a single integer n (1 ≤ n ≤ 10^5) — the number of rectangles.
Each of the next n lines contains two integers w_i and h_i (1 ≤ w_i, h_i ≤ 10^9) — the width and the height of the i-th rectangle.
Output
Print "YES" (without quotes) if there is a way to make the rectangles go in order of non-ascending height, otherwise print "NO".
You can print each letter in any case (upper or lower).
Examples
Input
3
3 4
4 6
3 5
Output
YES
Input
2
3 4
5 5
Output
NO
Note
In the first test, you can rotate the second and the third rectangles so that the heights will be [4, 4, 3].
In the second test, there is no way the second rectangle will be not higher than the first one. | instruction | 0 | 91,722 | 8 | 183,444 |
Tags: greedy, sortings
Correct Solution:
```
n = int(input())
a, b = input().split(' ')
a, b = int(a), int(b)
m = max(a, b)
bol = True
for i in range(n-1):
x, y = input().split(' ')
x, y = int(x), int(y)
if max(x, y) <= m:
m = max(x, y)
elif min(x, y) <= m:
m = min(x, y)
else:
bol = False
if bol:
print("YES")
else:
print("NO")
``` | output | 1 | 91,722 | 8 | 183,445 |
Provide tags and a correct Python 3 solution for this coding contest problem.
There are n rectangles in a row. You can either turn each rectangle by 90 degrees or leave it as it is. If you turn a rectangle, its width will be height, and its height will be width. Notice that you can turn any number of rectangles, you also can turn all or none of them. You can not change the order of the rectangles.
Find out if there is a way to make the rectangles go in order of non-ascending height. In other words, after all the turns, a height of every rectangle has to be not greater than the height of the previous rectangle (if it is such).
Input
The first line contains a single integer n (1 ≤ n ≤ 10^5) — the number of rectangles.
Each of the next n lines contains two integers w_i and h_i (1 ≤ w_i, h_i ≤ 10^9) — the width and the height of the i-th rectangle.
Output
Print "YES" (without quotes) if there is a way to make the rectangles go in order of non-ascending height, otherwise print "NO".
You can print each letter in any case (upper or lower).
Examples
Input
3
3 4
4 6
3 5
Output
YES
Input
2
3 4
5 5
Output
NO
Note
In the first test, you can rotate the second and the third rectangles so that the heights will be [4, 4, 3].
In the second test, there is no way the second rectangle will be not higher than the first one. | instruction | 0 | 91,724 | 8 | 183,448 |
Tags: greedy, sortings
Correct Solution:
```
def func():
prev=0
for i in range(int(input())):
a=list(map(int,input().split()))
if i==0:
prev=max(a)
else:
if max(a)<=prev:
prev=max(a)
elif min(a)<=prev:
prev=min(a)
else:
print("NO")
return
print("YES")
func()
``` | output | 1 | 91,724 | 8 | 183,449 |
Provide tags and a correct Python 3 solution for this coding contest problem.
There are n rectangles in a row. You can either turn each rectangle by 90 degrees or leave it as it is. If you turn a rectangle, its width will be height, and its height will be width. Notice that you can turn any number of rectangles, you also can turn all or none of them. You can not change the order of the rectangles.
Find out if there is a way to make the rectangles go in order of non-ascending height. In other words, after all the turns, a height of every rectangle has to be not greater than the height of the previous rectangle (if it is such).
Input
The first line contains a single integer n (1 ≤ n ≤ 10^5) — the number of rectangles.
Each of the next n lines contains two integers w_i and h_i (1 ≤ w_i, h_i ≤ 10^9) — the width and the height of the i-th rectangle.
Output
Print "YES" (without quotes) if there is a way to make the rectangles go in order of non-ascending height, otherwise print "NO".
You can print each letter in any case (upper or lower).
Examples
Input
3
3 4
4 6
3 5
Output
YES
Input
2
3 4
5 5
Output
NO
Note
In the first test, you can rotate the second and the third rectangles so that the heights will be [4, 4, 3].
In the second test, there is no way the second rectangle will be not higher than the first one. | instruction | 0 | 91,725 | 8 | 183,450 |
Tags: greedy, sortings
Correct Solution:
```
n = int(input())
w = []
h = []
for _ in range(n):
tempW, tempH = map(int, input().split())
w.append(tempW)
h.append(tempH)
# print(w)
# print(h)
h[0] = max(h[0], w[0])
flag = 0
for i in range(1, len(w)):
if h[i] <= h[i - 1]:
if w[i] <= h[i - 1]:
h[i] = max(h[i], w[i])
else:
if w[i] <= h[i - 1]:
h[i] = w[i]
else:
flag = 1
break
if flag == 0:
print("YES")
else:
print("NO")
``` | output | 1 | 91,725 | 8 | 183,451 |
Provide tags and a correct Python 3 solution for this coding contest problem.
There are n rectangles in a row. You can either turn each rectangle by 90 degrees or leave it as it is. If you turn a rectangle, its width will be height, and its height will be width. Notice that you can turn any number of rectangles, you also can turn all or none of them. You can not change the order of the rectangles.
Find out if there is a way to make the rectangles go in order of non-ascending height. In other words, after all the turns, a height of every rectangle has to be not greater than the height of the previous rectangle (if it is such).
Input
The first line contains a single integer n (1 ≤ n ≤ 10^5) — the number of rectangles.
Each of the next n lines contains two integers w_i and h_i (1 ≤ w_i, h_i ≤ 10^9) — the width and the height of the i-th rectangle.
Output
Print "YES" (without quotes) if there is a way to make the rectangles go in order of non-ascending height, otherwise print "NO".
You can print each letter in any case (upper or lower).
Examples
Input
3
3 4
4 6
3 5
Output
YES
Input
2
3 4
5 5
Output
NO
Note
In the first test, you can rotate the second and the third rectangles so that the heights will be [4, 4, 3].
In the second test, there is no way the second rectangle will be not higher than the first one. | instruction | 0 | 91,726 | 8 | 183,452 |
Tags: greedy, sortings
Correct Solution:
```
n = int(input())
l = []
for i in range(n):
w, h = map(int, input().split())
l.append((min(w, h), max(w, h)))
ok = True
i = 1
p = l[0][1]
while i < n and ok:
if p < l[i][0]:
ok = False
elif p >= l[i][1]:
p = l[i][1]
else:
p = l[i][0]
i += 1
print("YES" if ok else "NO")
``` | output | 1 | 91,726 | 8 | 183,453 |
Provide tags and a correct Python 3 solution for this coding contest problem.
There are n rectangles in a row. You can either turn each rectangle by 90 degrees or leave it as it is. If you turn a rectangle, its width will be height, and its height will be width. Notice that you can turn any number of rectangles, you also can turn all or none of them. You can not change the order of the rectangles.
Find out if there is a way to make the rectangles go in order of non-ascending height. In other words, after all the turns, a height of every rectangle has to be not greater than the height of the previous rectangle (if it is such).
Input
The first line contains a single integer n (1 ≤ n ≤ 10^5) — the number of rectangles.
Each of the next n lines contains two integers w_i and h_i (1 ≤ w_i, h_i ≤ 10^9) — the width and the height of the i-th rectangle.
Output
Print "YES" (without quotes) if there is a way to make the rectangles go in order of non-ascending height, otherwise print "NO".
You can print each letter in any case (upper or lower).
Examples
Input
3
3 4
4 6
3 5
Output
YES
Input
2
3 4
5 5
Output
NO
Note
In the first test, you can rotate the second and the third rectangles so that the heights will be [4, 4, 3].
In the second test, there is no way the second rectangle will be not higher than the first one. | instruction | 0 | 91,727 | 8 | 183,454 |
Tags: greedy, sortings
Correct Solution:
```
#497_B
cases = int(input())
found = True
rnum = 0
wds = ["NO", "YES"]
for i in range(0, cases):
ln = [int(j) for j in input().split(" ")]
num = max(ln)
if i > 0:
if num > rnum:
num = min(ln)
if num > rnum:
found = False
break
rnum = num
print(wds[found])
``` | output | 1 | 91,727 | 8 | 183,455 |
Provide tags and a correct Python 3 solution for this coding contest problem.
There are n rectangles in a row. You can either turn each rectangle by 90 degrees or leave it as it is. If you turn a rectangle, its width will be height, and its height will be width. Notice that you can turn any number of rectangles, you also can turn all or none of them. You can not change the order of the rectangles.
Find out if there is a way to make the rectangles go in order of non-ascending height. In other words, after all the turns, a height of every rectangle has to be not greater than the height of the previous rectangle (if it is such).
Input
The first line contains a single integer n (1 ≤ n ≤ 10^5) — the number of rectangles.
Each of the next n lines contains two integers w_i and h_i (1 ≤ w_i, h_i ≤ 10^9) — the width and the height of the i-th rectangle.
Output
Print "YES" (without quotes) if there is a way to make the rectangles go in order of non-ascending height, otherwise print "NO".
You can print each letter in any case (upper or lower).
Examples
Input
3
3 4
4 6
3 5
Output
YES
Input
2
3 4
5 5
Output
NO
Note
In the first test, you can rotate the second and the third rectangles so that the heights will be [4, 4, 3].
In the second test, there is no way the second rectangle will be not higher than the first one. | instruction | 0 | 91,728 | 8 | 183,456 |
Tags: greedy, sortings
Correct Solution:
```
from math import inf
n = int(input())
last = inf
flag = True
for _ in range(n):
w, h = map(int, input().split(" "))
if h < w:
w, h = h, w
if h <= last:
last = h
elif w <= last:
last = w
else:
flag = False
break
if flag:
print("YES")
else:
print("NO")
``` | output | 1 | 91,728 | 8 | 183,457 |
Provide tags and a correct Python 3 solution for this coding contest problem.
There are n rectangles in a row. You can either turn each rectangle by 90 degrees or leave it as it is. If you turn a rectangle, its width will be height, and its height will be width. Notice that you can turn any number of rectangles, you also can turn all or none of them. You can not change the order of the rectangles.
Find out if there is a way to make the rectangles go in order of non-ascending height. In other words, after all the turns, a height of every rectangle has to be not greater than the height of the previous rectangle (if it is such).
Input
The first line contains a single integer n (1 ≤ n ≤ 10^5) — the number of rectangles.
Each of the next n lines contains two integers w_i and h_i (1 ≤ w_i, h_i ≤ 10^9) — the width and the height of the i-th rectangle.
Output
Print "YES" (without quotes) if there is a way to make the rectangles go in order of non-ascending height, otherwise print "NO".
You can print each letter in any case (upper or lower).
Examples
Input
3
3 4
4 6
3 5
Output
YES
Input
2
3 4
5 5
Output
NO
Note
In the first test, you can rotate the second and the third rectangles so that the heights will be [4, 4, 3].
In the second test, there is no way the second rectangle will be not higher than the first one. | instruction | 0 | 91,729 | 8 | 183,458 |
Tags: greedy, sortings
Correct Solution:
```
n = int(input())
h = [0]*n
w = [0]*n
for i in range(n):
h[i], w[i] = map(int, input().split())
if n == 1:
print("YES")
exit(0)
last = h[0]
flag = True
for i in range(1, n):
if h[i] <= last and w[i] <= last:
last = max(h[i], w[i])
elif h[i] <= last:
last = h[i]
elif w[i] <= last:
last = w[i]
else:
flag = False
if not flag:
flag = True
last = w[0]
for i in range(1, n):
if h[i] <= last and w[i] <= last:
last = max(h[i], w[i])
elif h[i] <= last:
last = h[i]
elif w[i] <= last:
last = w[i]
else:
flag = False
if flag:
print("YES")
else:
print("NO")
``` | output | 1 | 91,729 | 8 | 183,459 |
Provide tags and a correct Python 3 solution for this coding contest problem.
There is a building consisting of 10~000 apartments numbered from 1 to 10~000, inclusive.
Call an apartment boring, if its number consists of the same digit. Examples of boring apartments are 11, 2, 777, 9999 and so on.
Our character is a troublemaker, and he calls the intercoms of all boring apartments, till someone answers the call, in the following order:
* First he calls all apartments consisting of digit 1, in increasing order (1, 11, 111, 1111).
* Next he calls all apartments consisting of digit 2, in increasing order (2, 22, 222, 2222)
* And so on.
The resident of the boring apartment x answers the call, and our character stops calling anyone further.
Our character wants to know how many digits he pressed in total and your task is to help him to count the total number of keypresses.
For example, if the resident of boring apartment 22 answered, then our character called apartments with numbers 1, 11, 111, 1111, 2, 22 and the total number of digits he pressed is 1 + 2 + 3 + 4 + 1 + 2 = 13.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 36) — the number of test cases.
The only line of the test case contains one integer x (1 ≤ x ≤ 9999) — the apartment number of the resident who answered the call. It is guaranteed that x consists of the same digit.
Output
For each test case, print the answer: how many digits our character pressed in total.
Example
Input
4
22
9999
1
777
Output
13
90
1
66 | instruction | 0 | 91,923 | 8 | 183,846 |
Tags: implementation, math
Correct Solution:
```
def sum_dig(n):
return int(n*(n+1)/2)
n=int(input())
casos=list()
for i in range(n):
aux=input()
casos.append(aux)
for caso in casos:
for m in range(1,10):
if str(m) in caso:
print((m-1)*10+sum_dig(caso.count(str(m))))
else:
continue
``` | output | 1 | 91,923 | 8 | 183,847 |
Provide tags and a correct Python 3 solution for this coding contest problem.
There is a building consisting of 10~000 apartments numbered from 1 to 10~000, inclusive.
Call an apartment boring, if its number consists of the same digit. Examples of boring apartments are 11, 2, 777, 9999 and so on.
Our character is a troublemaker, and he calls the intercoms of all boring apartments, till someone answers the call, in the following order:
* First he calls all apartments consisting of digit 1, in increasing order (1, 11, 111, 1111).
* Next he calls all apartments consisting of digit 2, in increasing order (2, 22, 222, 2222)
* And so on.
The resident of the boring apartment x answers the call, and our character stops calling anyone further.
Our character wants to know how many digits he pressed in total and your task is to help him to count the total number of keypresses.
For example, if the resident of boring apartment 22 answered, then our character called apartments with numbers 1, 11, 111, 1111, 2, 22 and the total number of digits he pressed is 1 + 2 + 3 + 4 + 1 + 2 = 13.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 36) — the number of test cases.
The only line of the test case contains one integer x (1 ≤ x ≤ 9999) — the apartment number of the resident who answered the call. It is guaranteed that x consists of the same digit.
Output
For each test case, print the answer: how many digits our character pressed in total.
Example
Input
4
22
9999
1
777
Output
13
90
1
66 | instruction | 0 | 91,924 | 8 | 183,848 |
Tags: implementation, math
Correct Solution:
```
for _ in range(int(input())):
n=input()
## print(n)
x=int(n[0])
##print(x)
ans=0
ans+=10*(x-1)
## ans*=10
y=len(n)
ans+=((y)*(y+1))//2
print(ans)
``` | output | 1 | 91,924 | 8 | 183,849 |
Provide tags and a correct Python 3 solution for this coding contest problem.
There is a building consisting of 10~000 apartments numbered from 1 to 10~000, inclusive.
Call an apartment boring, if its number consists of the same digit. Examples of boring apartments are 11, 2, 777, 9999 and so on.
Our character is a troublemaker, and he calls the intercoms of all boring apartments, till someone answers the call, in the following order:
* First he calls all apartments consisting of digit 1, in increasing order (1, 11, 111, 1111).
* Next he calls all apartments consisting of digit 2, in increasing order (2, 22, 222, 2222)
* And so on.
The resident of the boring apartment x answers the call, and our character stops calling anyone further.
Our character wants to know how many digits he pressed in total and your task is to help him to count the total number of keypresses.
For example, if the resident of boring apartment 22 answered, then our character called apartments with numbers 1, 11, 111, 1111, 2, 22 and the total number of digits he pressed is 1 + 2 + 3 + 4 + 1 + 2 = 13.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 36) — the number of test cases.
The only line of the test case contains one integer x (1 ≤ x ≤ 9999) — the apartment number of the resident who answered the call. It is guaranteed that x consists of the same digit.
Output
For each test case, print the answer: how many digits our character pressed in total.
Example
Input
4
22
9999
1
777
Output
13
90
1
66 | instruction | 0 | 91,925 | 8 | 183,850 |
Tags: implementation, math
Correct Solution:
```
'''
There is a building consisting of 10 000 apartments numbered from 1 to 10 000, inclusive.
Call an apartment boring, if its number consists of the same digit. Examples of boring apartments are 11,2,777,9999 and so on.
Our character is a troublemaker, and he calls the intercoms of all boring apartments, till someone answers the call, in the following order:
First he calls all apartments consisting of digit 1, in increasing order (1,11,111,1111).
Next he calls all apartments consisting of digit 2, in increasing order (2,22,222,2222)
And so on.
The resident of the boring apartment x answers the call, and our character stops calling anyone further.
Our character wants to know how many digits he pressed in total and your task is to help him to count the total number of keypresses.
For example, if the resident of boring apartment 22 answered, then our character called apartments with numbers 1,11,111,1111,2,22 and the total number of digits he pressed is 1+2+3+4+1+2=13.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1≤t≤36) — the number of test
cases.
The only line of the test case contains one integer x (1≤x≤9999) — the apartment
number of the resident who answered the call. It is guaranteed that x consists of the same digit.
Output
For each test case, print the answer: how many digits our character pressed in
total.
Example
input
4
22
9999
1
777
output
13
90
1
66
'''
def solver(n):
total = 0
x = n%10
nlen, temp = 0, n
while temp:
temp = int(temp/10)
nlen += 1
total = 10*(x-1) + int((nlen*(nlen+1))/2)
return total
if __name__ == "__main__":
t = int(input())
for _ in range(t):
n = int(input())
print(solver(n))
``` | output | 1 | 91,925 | 8 | 183,851 |
Provide tags and a correct Python 3 solution for this coding contest problem.
There is a building consisting of 10~000 apartments numbered from 1 to 10~000, inclusive.
Call an apartment boring, if its number consists of the same digit. Examples of boring apartments are 11, 2, 777, 9999 and so on.
Our character is a troublemaker, and he calls the intercoms of all boring apartments, till someone answers the call, in the following order:
* First he calls all apartments consisting of digit 1, in increasing order (1, 11, 111, 1111).
* Next he calls all apartments consisting of digit 2, in increasing order (2, 22, 222, 2222)
* And so on.
The resident of the boring apartment x answers the call, and our character stops calling anyone further.
Our character wants to know how many digits he pressed in total and your task is to help him to count the total number of keypresses.
For example, if the resident of boring apartment 22 answered, then our character called apartments with numbers 1, 11, 111, 1111, 2, 22 and the total number of digits he pressed is 1 + 2 + 3 + 4 + 1 + 2 = 13.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 36) — the number of test cases.
The only line of the test case contains one integer x (1 ≤ x ≤ 9999) — the apartment number of the resident who answered the call. It is guaranteed that x consists of the same digit.
Output
For each test case, print the answer: how many digits our character pressed in total.
Example
Input
4
22
9999
1
777
Output
13
90
1
66 | instruction | 0 | 91,926 | 8 | 183,852 |
Tags: implementation, math
Correct Solution:
```
n = int(input())
for i in range(n):
x = input()
fl = int(x[0])
ans = (fl - 1) * 10 + sum(range(1, len(x) + 1))
print(ans)
``` | output | 1 | 91,926 | 8 | 183,853 |
Provide tags and a correct Python 3 solution for this coding contest problem.
There is a building consisting of 10~000 apartments numbered from 1 to 10~000, inclusive.
Call an apartment boring, if its number consists of the same digit. Examples of boring apartments are 11, 2, 777, 9999 and so on.
Our character is a troublemaker, and he calls the intercoms of all boring apartments, till someone answers the call, in the following order:
* First he calls all apartments consisting of digit 1, in increasing order (1, 11, 111, 1111).
* Next he calls all apartments consisting of digit 2, in increasing order (2, 22, 222, 2222)
* And so on.
The resident of the boring apartment x answers the call, and our character stops calling anyone further.
Our character wants to know how many digits he pressed in total and your task is to help him to count the total number of keypresses.
For example, if the resident of boring apartment 22 answered, then our character called apartments with numbers 1, 11, 111, 1111, 2, 22 and the total number of digits he pressed is 1 + 2 + 3 + 4 + 1 + 2 = 13.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 36) — the number of test cases.
The only line of the test case contains one integer x (1 ≤ x ≤ 9999) — the apartment number of the resident who answered the call. It is guaranteed that x consists of the same digit.
Output
For each test case, print the answer: how many digits our character pressed in total.
Example
Input
4
22
9999
1
777
Output
13
90
1
66 | instruction | 0 | 91,927 | 8 | 183,854 |
Tags: implementation, math
Correct Solution:
```
t=int(input())
for z in range(t):
n=int(input())
ans=0
ans+=10*(n%10-1)
if n//1000>0:
ans+=10
elif n//100>0:
ans+=6
elif n//10>0:
ans+=3
else:
ans+=1
print(ans)
``` | output | 1 | 91,927 | 8 | 183,855 |
Provide tags and a correct Python 3 solution for this coding contest problem.
There is a building consisting of 10~000 apartments numbered from 1 to 10~000, inclusive.
Call an apartment boring, if its number consists of the same digit. Examples of boring apartments are 11, 2, 777, 9999 and so on.
Our character is a troublemaker, and he calls the intercoms of all boring apartments, till someone answers the call, in the following order:
* First he calls all apartments consisting of digit 1, in increasing order (1, 11, 111, 1111).
* Next he calls all apartments consisting of digit 2, in increasing order (2, 22, 222, 2222)
* And so on.
The resident of the boring apartment x answers the call, and our character stops calling anyone further.
Our character wants to know how many digits he pressed in total and your task is to help him to count the total number of keypresses.
For example, if the resident of boring apartment 22 answered, then our character called apartments with numbers 1, 11, 111, 1111, 2, 22 and the total number of digits he pressed is 1 + 2 + 3 + 4 + 1 + 2 = 13.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 36) — the number of test cases.
The only line of the test case contains one integer x (1 ≤ x ≤ 9999) — the apartment number of the resident who answered the call. It is guaranteed that x consists of the same digit.
Output
For each test case, print the answer: how many digits our character pressed in total.
Example
Input
4
22
9999
1
777
Output
13
90
1
66 | instruction | 0 | 91,928 | 8 | 183,856 |
Tags: implementation, math
Correct Solution:
```
from sys import maxsize, stdout, stdin
mod = int(1e9 + 7)
def I(): return int(stdin.readline())
def lint(): return [int(x) for x in stdin.readline().split()]
def S(): return input().strip()
def grid(r, c): return [lint() for i in range(r)]
for _ in range(I()):
n = input().strip()
d = n[0]
x = len(n)
passed = (int(d)-1)*10
ans = (x*(x+1))//2
print(passed+ans)
``` | output | 1 | 91,928 | 8 | 183,857 |
Provide tags and a correct Python 3 solution for this coding contest problem.
There is a building consisting of 10~000 apartments numbered from 1 to 10~000, inclusive.
Call an apartment boring, if its number consists of the same digit. Examples of boring apartments are 11, 2, 777, 9999 and so on.
Our character is a troublemaker, and he calls the intercoms of all boring apartments, till someone answers the call, in the following order:
* First he calls all apartments consisting of digit 1, in increasing order (1, 11, 111, 1111).
* Next he calls all apartments consisting of digit 2, in increasing order (2, 22, 222, 2222)
* And so on.
The resident of the boring apartment x answers the call, and our character stops calling anyone further.
Our character wants to know how many digits he pressed in total and your task is to help him to count the total number of keypresses.
For example, if the resident of boring apartment 22 answered, then our character called apartments with numbers 1, 11, 111, 1111, 2, 22 and the total number of digits he pressed is 1 + 2 + 3 + 4 + 1 + 2 = 13.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 36) — the number of test cases.
The only line of the test case contains one integer x (1 ≤ x ≤ 9999) — the apartment number of the resident who answered the call. It is guaranteed that x consists of the same digit.
Output
For each test case, print the answer: how many digits our character pressed in total.
Example
Input
4
22
9999
1
777
Output
13
90
1
66 | instruction | 0 | 91,929 | 8 | 183,858 |
Tags: implementation, math
Correct Solution:
```
c=[]
for i in range(1,10):
for j in range(1,4+1):
b=str(i)*j
c.append(b)
for _ in range(int(input())):
s=0
a=input()
for i in c:
s+=len(i)
if a==i:
print(s)
break
``` | output | 1 | 91,929 | 8 | 183,859 |
Provide tags and a correct Python 3 solution for this coding contest problem.
There is a building consisting of 10~000 apartments numbered from 1 to 10~000, inclusive.
Call an apartment boring, if its number consists of the same digit. Examples of boring apartments are 11, 2, 777, 9999 and so on.
Our character is a troublemaker, and he calls the intercoms of all boring apartments, till someone answers the call, in the following order:
* First he calls all apartments consisting of digit 1, in increasing order (1, 11, 111, 1111).
* Next he calls all apartments consisting of digit 2, in increasing order (2, 22, 222, 2222)
* And so on.
The resident of the boring apartment x answers the call, and our character stops calling anyone further.
Our character wants to know how many digits he pressed in total and your task is to help him to count the total number of keypresses.
For example, if the resident of boring apartment 22 answered, then our character called apartments with numbers 1, 11, 111, 1111, 2, 22 and the total number of digits he pressed is 1 + 2 + 3 + 4 + 1 + 2 = 13.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 36) — the number of test cases.
The only line of the test case contains one integer x (1 ≤ x ≤ 9999) — the apartment number of the resident who answered the call. It is guaranteed that x consists of the same digit.
Output
For each test case, print the answer: how many digits our character pressed in total.
Example
Input
4
22
9999
1
777
Output
13
90
1
66 | instruction | 0 | 91,930 | 8 | 183,860 |
Tags: implementation, math
Correct Solution:
```
def solve(x):
summa = (int(x[0]) - 1) * 10
if len(x) == 1:
return summa + 1
elif len(x) == 2:
return summa + 3
elif len(x) == 3:
return summa + 6
else:
return summa + 10
t = int(input())
for i in range(0, t):
print(solve(input()))
``` | output | 1 | 91,930 | 8 | 183,861 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There is a building consisting of 10~000 apartments numbered from 1 to 10~000, inclusive.
Call an apartment boring, if its number consists of the same digit. Examples of boring apartments are 11, 2, 777, 9999 and so on.
Our character is a troublemaker, and he calls the intercoms of all boring apartments, till someone answers the call, in the following order:
* First he calls all apartments consisting of digit 1, in increasing order (1, 11, 111, 1111).
* Next he calls all apartments consisting of digit 2, in increasing order (2, 22, 222, 2222)
* And so on.
The resident of the boring apartment x answers the call, and our character stops calling anyone further.
Our character wants to know how many digits he pressed in total and your task is to help him to count the total number of keypresses.
For example, if the resident of boring apartment 22 answered, then our character called apartments with numbers 1, 11, 111, 1111, 2, 22 and the total number of digits he pressed is 1 + 2 + 3 + 4 + 1 + 2 = 13.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 36) — the number of test cases.
The only line of the test case contains one integer x (1 ≤ x ≤ 9999) — the apartment number of the resident who answered the call. It is guaranteed that x consists of the same digit.
Output
For each test case, print the answer: how many digits our character pressed in total.
Example
Input
4
22
9999
1
777
Output
13
90
1
66
Submitted Solution:
```
for i in range (int(input())):
a=int(input())
if(a%1111)==0:
print(((a//1111)*10))
elif a%111==0:
print(((a//111)*10)-4)
elif a%11==0:
print(((a // 11) * 10) -7)
else :
print((a* 10) -9)
``` | instruction | 0 | 91,931 | 8 | 183,862 |
Yes | output | 1 | 91,931 | 8 | 183,863 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There is a building consisting of 10~000 apartments numbered from 1 to 10~000, inclusive.
Call an apartment boring, if its number consists of the same digit. Examples of boring apartments are 11, 2, 777, 9999 and so on.
Our character is a troublemaker, and he calls the intercoms of all boring apartments, till someone answers the call, in the following order:
* First he calls all apartments consisting of digit 1, in increasing order (1, 11, 111, 1111).
* Next he calls all apartments consisting of digit 2, in increasing order (2, 22, 222, 2222)
* And so on.
The resident of the boring apartment x answers the call, and our character stops calling anyone further.
Our character wants to know how many digits he pressed in total and your task is to help him to count the total number of keypresses.
For example, if the resident of boring apartment 22 answered, then our character called apartments with numbers 1, 11, 111, 1111, 2, 22 and the total number of digits he pressed is 1 + 2 + 3 + 4 + 1 + 2 = 13.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 36) — the number of test cases.
The only line of the test case contains one integer x (1 ≤ x ≤ 9999) — the apartment number of the resident who answered the call. It is guaranteed that x consists of the same digit.
Output
For each test case, print the answer: how many digits our character pressed in total.
Example
Input
4
22
9999
1
777
Output
13
90
1
66
Submitted Solution:
```
# cook your dish here
# cook your dish here
#from functools import reduce
#mod=int(1e9+7)
#import resource
#resource.setrlimit(resource.RLIMIT_STACK, [0x100000000, resource.RLIM_INFINITY])
#import threading
#threading.stack_size(2**26)
"""fact=[1]
#for i in range(1,100001):
# fact.append((fact[-1]*i)%mod)
#ifact=[0]*100001
#ifact[100000]=pow(fact[100000],mod-2,mod)
#for i in range(100000,0,-1):
# ifact[i-1]=(i*ifact[i])%mod"""
#from collections import deque, defaultdict, Counter, OrderedDict
#from math import ceil, sqrt, hypot, factorial, pi, sin, cos, radians, gcd
#from heapq import heappush, heappop, heapify, nlargest, nsmallest
# sys.setrecursionlimit(10**6)
from sys import stdin, stdout
import bisect #c++ upperbound
from bisect import bisect_left as bl #c++ lowerbound bl(array,element)
from bisect import bisect_right as br #c++ upperbound
import itertools
from collections import Counter
from math import sqrt
import collections
import math
import heapq
import re
def modinv(n,p):
return pow(n,p-2,p)
def cin():
return map(int,sin().split())
def ain(): #takes array as input
return list(map(int,sin().split()))
def sin():
return input()
def inin():
return int(input())
def Divisors(n) :
l = []
for i in range(1, int(math.sqrt(n) + 1)) :
if (n % i == 0) :
if (n // i == i) :
l.append(i)
else :
l.append(i)
l.append(n//i)
return l
def most_frequent(list):
return max(set(list), key = list.count)
def GCD(x,y):
while(y):
x, y = y, x % y
return x
def ncr(n,r,p): #To use this, Uncomment 19-25
t=((fact[n])*((ifact[r]*ifact[n-r])%p))%p
return t
def Convert(string):
li = list(string.split(""))
return li
def SieveOfEratosthenes(n):
global prime
prime = [True for i in range(n+1)]
p = 2
while (p * p <= n):
if (prime[p] == True):
for i in range(p * p, n+1, p):
prime[i] = False
p += 1
f=[]
for p in range(2, n):
if prime[p]:
f.append(p)
return f
prime=[]
q=[]
def dfs(n,d,v,c):
global q
v[n]=1
x=d[n]
q.append(n)
j=c
for i in x:
if i not in v:
f=dfs(i,d,v,c+1)
j=max(j,f)
# print(f)
return j
#Implement heapq
#grades = [110, 25, 38, 49, 20, 95, 33, 87, 80, 90]
#print(heapq.nlargest(3, grades)) #top 3 largest
#print(heapq.nsmallest(4, grades))
#Always make a variable of predefined function for ex- fn=len
#n,k=map(int,input().split())
"""*******************************************************"""
def main():
for _ in range(int(input())):
n=inin()
temp=len(str(n))
k=int(str(n)[-1])
ans=10*(k-1)
if temp==1:
ans+=1
elif temp==2:
ans+=3
elif temp==3:
ans+=6
else:
ans+=10
print(ans)
"""*******************************************************"""
######## Python 2 and 3 footer by Pajenegod and c1729
py2 = round(0.5)
if py2:
from future_builtins import ascii, filter, hex, map, oct, zip
range = xrange
import os, sys
from io import IOBase, BytesIO
BUFSIZE = 8192
class FastIO(BytesIO):
newlines = 0
def __init__(self, file):
self._file = file
self._fd = file.fileno()
self.writable = "x" in file.mode or "w" in file.mode
self.write = super(FastIO, self).write if self.writable else None
def _fill(self):
s = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
self.seek((self.tell(), self.seek(0,2), super(FastIO, self).write(s))[0])
return s
def read(self):
while self._fill(): pass
return super(FastIO,self).read()
def readline(self):
while self.newlines == 0:
s = self._fill(); self.newlines = s.count(b"\n") + (not s)
self.newlines -= 1
return super(FastIO, self).readline()
def flush(self):
if self.writable:
os.write(self._fd, self.getvalue())
self.truncate(0), self.seek(0)
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
if py2:
self.write = self.buffer.write
self.read = self.buffer.read
self.readline = self.buffer.readline
else:
self.write = lambda s:self.buffer.write(s.encode('ascii'))
self.read = lambda:self.buffer.read().decode('ascii')
self.readline = lambda:self.buffer.readline().decode('ascii')
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip('\r\n')
if __name__== "__main__":
main()
#threading.Thread(target=main).start()
``` | instruction | 0 | 91,932 | 8 | 183,864 |
Yes | output | 1 | 91,932 | 8 | 183,865 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There is a building consisting of 10~000 apartments numbered from 1 to 10~000, inclusive.
Call an apartment boring, if its number consists of the same digit. Examples of boring apartments are 11, 2, 777, 9999 and so on.
Our character is a troublemaker, and he calls the intercoms of all boring apartments, till someone answers the call, in the following order:
* First he calls all apartments consisting of digit 1, in increasing order (1, 11, 111, 1111).
* Next he calls all apartments consisting of digit 2, in increasing order (2, 22, 222, 2222)
* And so on.
The resident of the boring apartment x answers the call, and our character stops calling anyone further.
Our character wants to know how many digits he pressed in total and your task is to help him to count the total number of keypresses.
For example, if the resident of boring apartment 22 answered, then our character called apartments with numbers 1, 11, 111, 1111, 2, 22 and the total number of digits he pressed is 1 + 2 + 3 + 4 + 1 + 2 = 13.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 36) — the number of test cases.
The only line of the test case contains one integer x (1 ≤ x ≤ 9999) — the apartment number of the resident who answered the call. It is guaranteed that x consists of the same digit.
Output
For each test case, print the answer: how many digits our character pressed in total.
Example
Input
4
22
9999
1
777
Output
13
90
1
66
Submitted Solution:
```
t = int(input())
for _ in range(t):
n = int(input())
count = 0
for i in range(1, int(str(n)[0]) + 1):
s_i = str(i)
while len(s_i) <= 4:
count += len(s_i)
if s_i == str(n):
break
s_i += str(i)
print(count)
``` | instruction | 0 | 91,933 | 8 | 183,866 |
Yes | output | 1 | 91,933 | 8 | 183,867 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There is a building consisting of 10~000 apartments numbered from 1 to 10~000, inclusive.
Call an apartment boring, if its number consists of the same digit. Examples of boring apartments are 11, 2, 777, 9999 and so on.
Our character is a troublemaker, and he calls the intercoms of all boring apartments, till someone answers the call, in the following order:
* First he calls all apartments consisting of digit 1, in increasing order (1, 11, 111, 1111).
* Next he calls all apartments consisting of digit 2, in increasing order (2, 22, 222, 2222)
* And so on.
The resident of the boring apartment x answers the call, and our character stops calling anyone further.
Our character wants to know how many digits he pressed in total and your task is to help him to count the total number of keypresses.
For example, if the resident of boring apartment 22 answered, then our character called apartments with numbers 1, 11, 111, 1111, 2, 22 and the total number of digits he pressed is 1 + 2 + 3 + 4 + 1 + 2 = 13.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 36) — the number of test cases.
The only line of the test case contains one integer x (1 ≤ x ≤ 9999) — the apartment number of the resident who answered the call. It is guaranteed that x consists of the same digit.
Output
For each test case, print the answer: how many digits our character pressed in total.
Example
Input
4
22
9999
1
777
Output
13
90
1
66
Submitted Solution:
```
def solve():
temp = int(input())
return int((temp % 10 - 1)*10 + (len(str(temp)) * (len(str(temp)) + 1)) / 2)
if __name__ == "__main__":
for i in range(int(input())):
print(solve())
``` | instruction | 0 | 91,934 | 8 | 183,868 |
Yes | output | 1 | 91,934 | 8 | 183,869 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There is a building consisting of 10~000 apartments numbered from 1 to 10~000, inclusive.
Call an apartment boring, if its number consists of the same digit. Examples of boring apartments are 11, 2, 777, 9999 and so on.
Our character is a troublemaker, and he calls the intercoms of all boring apartments, till someone answers the call, in the following order:
* First he calls all apartments consisting of digit 1, in increasing order (1, 11, 111, 1111).
* Next he calls all apartments consisting of digit 2, in increasing order (2, 22, 222, 2222)
* And so on.
The resident of the boring apartment x answers the call, and our character stops calling anyone further.
Our character wants to know how many digits he pressed in total and your task is to help him to count the total number of keypresses.
For example, if the resident of boring apartment 22 answered, then our character called apartments with numbers 1, 11, 111, 1111, 2, 22 and the total number of digits he pressed is 1 + 2 + 3 + 4 + 1 + 2 = 13.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 36) — the number of test cases.
The only line of the test case contains one integer x (1 ≤ x ≤ 9999) — the apartment number of the resident who answered the call. It is guaranteed that x consists of the same digit.
Output
For each test case, print the answer: how many digits our character pressed in total.
Example
Input
4
22
9999
1
777
Output
13
90
1
66
Submitted Solution:
```
def getNum(x):
return ((int(x[0]) - 1) * 4) + len(x)
t = input()
for i in t :
num = input()
print(getNum(num))
``` | instruction | 0 | 91,935 | 8 | 183,870 |
No | output | 1 | 91,935 | 8 | 183,871 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There is a building consisting of 10~000 apartments numbered from 1 to 10~000, inclusive.
Call an apartment boring, if its number consists of the same digit. Examples of boring apartments are 11, 2, 777, 9999 and so on.
Our character is a troublemaker, and he calls the intercoms of all boring apartments, till someone answers the call, in the following order:
* First he calls all apartments consisting of digit 1, in increasing order (1, 11, 111, 1111).
* Next he calls all apartments consisting of digit 2, in increasing order (2, 22, 222, 2222)
* And so on.
The resident of the boring apartment x answers the call, and our character stops calling anyone further.
Our character wants to know how many digits he pressed in total and your task is to help him to count the total number of keypresses.
For example, if the resident of boring apartment 22 answered, then our character called apartments with numbers 1, 11, 111, 1111, 2, 22 and the total number of digits he pressed is 1 + 2 + 3 + 4 + 1 + 2 = 13.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 36) — the number of test cases.
The only line of the test case contains one integer x (1 ≤ x ≤ 9999) — the apartment number of the resident who answered the call. It is guaranteed that x consists of the same digit.
Output
For each test case, print the answer: how many digits our character pressed in total.
Example
Input
4
22
9999
1
777
Output
13
90
1
66
Submitted Solution:
```
import sys
if __name__ == '__main__':
t = int(sys.stdin.readline())
sumdict = {
1:1,
2:1+2,
3:1+2+3,
4:1+2+3+4
}
for _ in range(t):
num = int(sys.stdin.readline())
i = 9
while i>0:
if str(num//i)[0] == '1':
break
i -= 1
i-=1
k = len(str(num))
out = (i * 10) + sumdict[k]
sys.__stdout__.write(str(out)+'\n')
``` | instruction | 0 | 91,936 | 8 | 183,872 |
No | output | 1 | 91,936 | 8 | 183,873 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There is a building consisting of 10~000 apartments numbered from 1 to 10~000, inclusive.
Call an apartment boring, if its number consists of the same digit. Examples of boring apartments are 11, 2, 777, 9999 and so on.
Our character is a troublemaker, and he calls the intercoms of all boring apartments, till someone answers the call, in the following order:
* First he calls all apartments consisting of digit 1, in increasing order (1, 11, 111, 1111).
* Next he calls all apartments consisting of digit 2, in increasing order (2, 22, 222, 2222)
* And so on.
The resident of the boring apartment x answers the call, and our character stops calling anyone further.
Our character wants to know how many digits he pressed in total and your task is to help him to count the total number of keypresses.
For example, if the resident of boring apartment 22 answered, then our character called apartments with numbers 1, 11, 111, 1111, 2, 22 and the total number of digits he pressed is 1 + 2 + 3 + 4 + 1 + 2 = 13.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 36) — the number of test cases.
The only line of the test case contains one integer x (1 ≤ x ≤ 9999) — the apartment number of the resident who answered the call. It is guaranteed that x consists of the same digit.
Output
For each test case, print the answer: how many digits our character pressed in total.
Example
Input
4
22
9999
1
777
Output
13
90
1
66
Submitted Solution:
```
import time
I = input
def calc(last):
n = [str(i) * k for i in range(1,10) for k in range(1,5)]
iter = ''
for i in '123456789':
for j in n:
if all([i == l for l in j]):
if int(j) > int(last) and all([last[0] == h for h in j]):
break
iter += j
if all([i == k for k in last]):
break
print(len(iter))
def main():
start = time.time()
n = I()
for i in range(int(n)):
calc(I())
print(time.time()-start)
if __name__ == "__main__":
main()
``` | instruction | 0 | 91,937 | 8 | 183,874 |
No | output | 1 | 91,937 | 8 | 183,875 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There is a building consisting of 10~000 apartments numbered from 1 to 10~000, inclusive.
Call an apartment boring, if its number consists of the same digit. Examples of boring apartments are 11, 2, 777, 9999 and so on.
Our character is a troublemaker, and he calls the intercoms of all boring apartments, till someone answers the call, in the following order:
* First he calls all apartments consisting of digit 1, in increasing order (1, 11, 111, 1111).
* Next he calls all apartments consisting of digit 2, in increasing order (2, 22, 222, 2222)
* And so on.
The resident of the boring apartment x answers the call, and our character stops calling anyone further.
Our character wants to know how many digits he pressed in total and your task is to help him to count the total number of keypresses.
For example, if the resident of boring apartment 22 answered, then our character called apartments with numbers 1, 11, 111, 1111, 2, 22 and the total number of digits he pressed is 1 + 2 + 3 + 4 + 1 + 2 = 13.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 36) — the number of test cases.
The only line of the test case contains one integer x (1 ≤ x ≤ 9999) — the apartment number of the resident who answered the call. It is guaranteed that x consists of the same digit.
Output
For each test case, print the answer: how many digits our character pressed in total.
Example
Input
4
22
9999
1
777
Output
13
90
1
66
Submitted Solution:
```
def main_func(x:str):
return str(int(len(x)*(len(x)+1)/2+(int(x[0])-1)*10))
t = 0
input(t)
r = ''
for i in range(t):
input(r)
print(main_func(r))
``` | instruction | 0 | 91,938 | 8 | 183,876 |
No | output | 1 | 91,938 | 8 | 183,877 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Little Dormi received a histogram with n bars of height a_1, a_2, …, a_n for Christmas. However, the more he played with his new histogram, the more he realized its imperfections, so today he wanted to modify it to his liking.
To modify the histogram, Little Dormi is able to perform the following operation an arbitrary number of times:
* Select an index i (1 ≤ i ≤ n) where a_i>0, and assign a_i := a_i-1.
Little Dormi defines the ugliness score of his histogram (after performing some number of operations) as the sum of the vertical length of its outline and the number of operations he performed on it. And to make the histogram as perfect as possible, he would like to minimize the ugliness score after modifying it with some number of operations.
However, as his histogram is very large, Little Dormi is having trouble minimizing the ugliness score, so as Little Dormi's older brother, help him find the minimal ugliness.
Consider the following example where the histogram has 4 columns of heights 4,8,9,6:
<image>
The blue region represents the histogram, and the red lines represent the vertical portion of the outline. Currently, the vertical length of the outline is 4+4+1+3+6 = 18, so if Little Dormi does not modify the histogram at all, the ugliness would be 18.
However, Little Dormi can apply the operation once on column 2 and twice on column 3, resulting in a histogram with heights 4,7,7,6:
<image>
Now, as the total vertical length of the outline (red lines) is 4+3+1+6=14, the ugliness is 14+3=17 dollars. It can be proven that this is optimal.
Input
Each test contains multiple test cases. The first line contains the number of test cases t (1 ≤ t ≤ 10^4). Description of the test cases follows.
The first line of each test case contains a single integer n (1 ≤ n ≤ 4 ⋅ 10^5).
The second line of each test case contains n integers a_1, a_2, …, a_n (0 ≤ a_i ≤ 10^9).
It is guaranteed that the sum of n over all test cases does not exceed 4 ⋅ 10^5.
Output
For each test case output one integer, the minimal ugliness Little Dormi can achieve with the histogram in that test case.
Example
Input
2
4
4 8 9 6
6
2 1 7 4 0 0
Output
17
12
Note
Example 1 is the example described in the statement.
The initial histogram for example 2 is given below:
<image>
The ugliness is currently 2+1+6+3+4=16.
By applying the operation once on column 1, six times on column 3, and three times on column 4, we can end up with a histogram with heights 1,1,1,1,0,0:
<image>
The vertical length of the outline is now 1+1=2 and Little Dormi made 1+6+3=10 operations, so the final ugliness is 2+10=12, which can be proven to be optimal. | instruction | 0 | 91,975 | 8 | 183,950 |
Tags: greedy, implementation, math
Correct Solution:
```
import os,sys
from io import BytesIO, IOBase
from collections import deque, Counter,defaultdict as dft
from heapq import heappop ,heappush
from math import log,sqrt,factorial,cos,tan,sin,radians,log2,ceil,floor
from bisect import bisect,bisect_left,bisect_right
from decimal import *
import sys,threading
from itertools import permutations, combinations
from copy import deepcopy
input = sys.stdin.readline
ii = lambda: int(input())
si = lambda: input().rstrip()
mp = lambda: map(int, input().split())
ms= lambda: map(str,input().strip().split(" "))
ml = lambda: list(mp())
mf = lambda: map(float, input().split())
alphs = "abcdefghijklmnopqrstuvwxyz"
# stuff you should look for
# int overflow, array bounds
# special cases (n=1?)
# do smth instead of nothing and stay organized
# WRITE STUFF DOWN
# DON'T GET STUCK ON ONE APPROACH
#
def solve():
n=ii()
arr=ml()
res=0
arr=[0]+arr+[0]
for i in range(1,n+1):
mx=max(arr[i-1],arr[i+1])
if arr[i]>mx:
res+=arr[i]-mx
arr[i]=mx
res+=abs(arr[i]-arr[i-1])
#print(arr)
print(res+arr[-2])
BUFSIZE = 8192
class FastIO(IOBase):
newlines = 0
def __init__(self, file):
self._fd = file.fileno()
self.buffer = BytesIO()
self.writable = "x" in file.mode or "r" not in file.mode
self.write = self.buffer.write if self.writable else None
def read(self):
while True:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
if not b:
break
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines = 0
return self.buffer.read()
def readline(self):
while self.newlines == 0:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
self.newlines = b.count(b"\n") + (not b)
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines -= 1
return self.buffer.readline()
def flush(self):
if self.writable:
os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
self.write = lambda s: self.buffer.write(s.encode("ascii"))
self.read = lambda: self.buffer.read().decode("ascii")
self.readline = lambda: self.buffer.readline().decode("ascii")
def print(*args, **kwargs):
"""Prints the values to a stream, or to sys.stdout by default."""
sep, file = kwargs.pop("sep", " "), kwargs.pop("file", sys.stdout)
at_start = True
for x in args:
if not at_start:
file.write(sep)
file.write(str(x))
at_start = False
file.write(kwargs.pop("end", "\n"))
if kwargs.pop("flush", False):
file.flush()
if sys.version_info[0] < 3:
sys.stdin, sys.stdout = FastIO(sys.stdin), FastIO(sys.stdout)
else:
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")
# endregion
if __name__ == "__main__":
tc=1
tc = ii()
for i in range(tc):
solve()
``` | output | 1 | 91,975 | 8 | 183,951 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Little Dormi received a histogram with n bars of height a_1, a_2, …, a_n for Christmas. However, the more he played with his new histogram, the more he realized its imperfections, so today he wanted to modify it to his liking.
To modify the histogram, Little Dormi is able to perform the following operation an arbitrary number of times:
* Select an index i (1 ≤ i ≤ n) where a_i>0, and assign a_i := a_i-1.
Little Dormi defines the ugliness score of his histogram (after performing some number of operations) as the sum of the vertical length of its outline and the number of operations he performed on it. And to make the histogram as perfect as possible, he would like to minimize the ugliness score after modifying it with some number of operations.
However, as his histogram is very large, Little Dormi is having trouble minimizing the ugliness score, so as Little Dormi's older brother, help him find the minimal ugliness.
Consider the following example where the histogram has 4 columns of heights 4,8,9,6:
<image>
The blue region represents the histogram, and the red lines represent the vertical portion of the outline. Currently, the vertical length of the outline is 4+4+1+3+6 = 18, so if Little Dormi does not modify the histogram at all, the ugliness would be 18.
However, Little Dormi can apply the operation once on column 2 and twice on column 3, resulting in a histogram with heights 4,7,7,6:
<image>
Now, as the total vertical length of the outline (red lines) is 4+3+1+6=14, the ugliness is 14+3=17 dollars. It can be proven that this is optimal.
Input
Each test contains multiple test cases. The first line contains the number of test cases t (1 ≤ t ≤ 10^4). Description of the test cases follows.
The first line of each test case contains a single integer n (1 ≤ n ≤ 4 ⋅ 10^5).
The second line of each test case contains n integers a_1, a_2, …, a_n (0 ≤ a_i ≤ 10^9).
It is guaranteed that the sum of n over all test cases does not exceed 4 ⋅ 10^5.
Output
For each test case output one integer, the minimal ugliness Little Dormi can achieve with the histogram in that test case.
Example
Input
2
4
4 8 9 6
6
2 1 7 4 0 0
Output
17
12
Note
Example 1 is the example described in the statement.
The initial histogram for example 2 is given below:
<image>
The ugliness is currently 2+1+6+3+4=16.
By applying the operation once on column 1, six times on column 3, and three times on column 4, we can end up with a histogram with heights 1,1,1,1,0,0:
<image>
The vertical length of the outline is now 1+1=2 and Little Dormi made 1+6+3=10 operations, so the final ugliness is 2+10=12, which can be proven to be optimal. | instruction | 0 | 91,976 | 8 | 183,952 |
Tags: greedy, implementation, math
Correct Solution:
```
def get_sum(a, op):
res = op
for k in range(1,len(a)):
res += abs(a[k] - a[k-1])
return res
for _ in range(int(input())):
n = int(input())
a = list(map(int,input().split()))
a.insert(0, 0)
a.append(0)
mins = 100000000
ops = 0
for k in range(n + 1):
if a[k] > a[k-1] and a[k] > a[k+1]:
raz = a[k] - max(a[k-1], a[k+1])
a[k] = max(a[k-1], a[k+1])
ops += raz
# print(a)
print(get_sum(a, ops))
# n, m = map(int,input().split())
# a = []
# for k in range(n):
# a.append(list(input()))
#
# b = list(map(int,input().split()))
#
#
#
# mn = 10000000000
# for g in range(m):
# res = 0
# f = False
# next = -1
#
#
# for i in range(g, m-1):
# if not f:
# # print(i)
# res += 1
# for j in range(n):
# if a[j][i] == "#":
# break
# f = True
# ff = True
# for k in range(j, -1, -1):
# if a[k][i + 1] == "#":
# next = k
# ff = False
# break
# if ff:
# f = False
# else:
# ff = True
# for k in range(next, -1, -1):
# if a[k][i + 1] == "#":
# next = k
# ff = False
# break
# if ff:
# f = False
#
# f = False
# for i in range(g, 0, -1):
# if not f:
# # print(i)
# res += 1
# for j in range(n):
# if a[j][i] == "#":
# break
# f = True
# ff = True
# for k in range(j, -1, -1):
# if a[k][i - 1] == "#":
# next = k
# ff = False
# break
# if ff:
# f = False
# else:
# ff = True
# for k in range(next, -1, -1):
# if a[k][i - 1] == "#":
# next = k
# ff = False
# break
# if ff:
# f = False
#
#
# res -= 1
# if res < mn:
# print(g)
# mn = min(mn, res)
#
# print(mn)
# import itertools as it
# from random import choice
# while True:
# n = int(input())
# a = []
# for i in range(1, n + 1):
# a.append(i)
# print(*choice(list(it.permutations(a))))
# print(*choice(list(it.permutations(a))))
``` | output | 1 | 91,976 | 8 | 183,953 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Little Dormi received a histogram with n bars of height a_1, a_2, …, a_n for Christmas. However, the more he played with his new histogram, the more he realized its imperfections, so today he wanted to modify it to his liking.
To modify the histogram, Little Dormi is able to perform the following operation an arbitrary number of times:
* Select an index i (1 ≤ i ≤ n) where a_i>0, and assign a_i := a_i-1.
Little Dormi defines the ugliness score of his histogram (after performing some number of operations) as the sum of the vertical length of its outline and the number of operations he performed on it. And to make the histogram as perfect as possible, he would like to minimize the ugliness score after modifying it with some number of operations.
However, as his histogram is very large, Little Dormi is having trouble minimizing the ugliness score, so as Little Dormi's older brother, help him find the minimal ugliness.
Consider the following example where the histogram has 4 columns of heights 4,8,9,6:
<image>
The blue region represents the histogram, and the red lines represent the vertical portion of the outline. Currently, the vertical length of the outline is 4+4+1+3+6 = 18, so if Little Dormi does not modify the histogram at all, the ugliness would be 18.
However, Little Dormi can apply the operation once on column 2 and twice on column 3, resulting in a histogram with heights 4,7,7,6:
<image>
Now, as the total vertical length of the outline (red lines) is 4+3+1+6=14, the ugliness is 14+3=17 dollars. It can be proven that this is optimal.
Input
Each test contains multiple test cases. The first line contains the number of test cases t (1 ≤ t ≤ 10^4). Description of the test cases follows.
The first line of each test case contains a single integer n (1 ≤ n ≤ 4 ⋅ 10^5).
The second line of each test case contains n integers a_1, a_2, …, a_n (0 ≤ a_i ≤ 10^9).
It is guaranteed that the sum of n over all test cases does not exceed 4 ⋅ 10^5.
Output
For each test case output one integer, the minimal ugliness Little Dormi can achieve with the histogram in that test case.
Example
Input
2
4
4 8 9 6
6
2 1 7 4 0 0
Output
17
12
Note
Example 1 is the example described in the statement.
The initial histogram for example 2 is given below:
<image>
The ugliness is currently 2+1+6+3+4=16.
By applying the operation once on column 1, six times on column 3, and three times on column 4, we can end up with a histogram with heights 1,1,1,1,0,0:
<image>
The vertical length of the outline is now 1+1=2 and Little Dormi made 1+6+3=10 operations, so the final ugliness is 2+10=12, which can be proven to be optimal. | instruction | 0 | 91,977 | 8 | 183,954 |
Tags: greedy, implementation, math
Correct Solution:
```
def ugly(arr):
n = len(arr)
if(n == 1):
return arr[0]
f = arr[0] - arr[1]
less = 0
if(f > 0):
less += f
l = arr[-1] - arr[-2]
if(l > 0):
less += l
ug = 0
for i in range(n - 1):
ug += abs(arr[i] - arr[i + 1])
ug += arr[0] + arr[-1]
for i in range(1, n - 1):
if(arr[i - 1] < arr[i] and arr[i] > arr[i + 1]):
less += min(arr[i] - arr[i - 1], arr[i] - arr[i + 1])
return ug - less
t = int(input())
for i in range(t):
input()
arr = list(map(int, input().split()))
print(ugly(arr))
``` | output | 1 | 91,977 | 8 | 183,955 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Little Dormi received a histogram with n bars of height a_1, a_2, …, a_n for Christmas. However, the more he played with his new histogram, the more he realized its imperfections, so today he wanted to modify it to his liking.
To modify the histogram, Little Dormi is able to perform the following operation an arbitrary number of times:
* Select an index i (1 ≤ i ≤ n) where a_i>0, and assign a_i := a_i-1.
Little Dormi defines the ugliness score of his histogram (after performing some number of operations) as the sum of the vertical length of its outline and the number of operations he performed on it. And to make the histogram as perfect as possible, he would like to minimize the ugliness score after modifying it with some number of operations.
However, as his histogram is very large, Little Dormi is having trouble minimizing the ugliness score, so as Little Dormi's older brother, help him find the minimal ugliness.
Consider the following example where the histogram has 4 columns of heights 4,8,9,6:
<image>
The blue region represents the histogram, and the red lines represent the vertical portion of the outline. Currently, the vertical length of the outline is 4+4+1+3+6 = 18, so if Little Dormi does not modify the histogram at all, the ugliness would be 18.
However, Little Dormi can apply the operation once on column 2 and twice on column 3, resulting in a histogram with heights 4,7,7,6:
<image>
Now, as the total vertical length of the outline (red lines) is 4+3+1+6=14, the ugliness is 14+3=17 dollars. It can be proven that this is optimal.
Input
Each test contains multiple test cases. The first line contains the number of test cases t (1 ≤ t ≤ 10^4). Description of the test cases follows.
The first line of each test case contains a single integer n (1 ≤ n ≤ 4 ⋅ 10^5).
The second line of each test case contains n integers a_1, a_2, …, a_n (0 ≤ a_i ≤ 10^9).
It is guaranteed that the sum of n over all test cases does not exceed 4 ⋅ 10^5.
Output
For each test case output one integer, the minimal ugliness Little Dormi can achieve with the histogram in that test case.
Example
Input
2
4
4 8 9 6
6
2 1 7 4 0 0
Output
17
12
Note
Example 1 is the example described in the statement.
The initial histogram for example 2 is given below:
<image>
The ugliness is currently 2+1+6+3+4=16.
By applying the operation once on column 1, six times on column 3, and three times on column 4, we can end up with a histogram with heights 1,1,1,1,0,0:
<image>
The vertical length of the outline is now 1+1=2 and Little Dormi made 1+6+3=10 operations, so the final ugliness is 2+10=12, which can be proven to be optimal. | instruction | 0 | 91,978 | 8 | 183,956 |
Tags: greedy, implementation, math
Correct Solution:
```
for _ in range(int(input())):
n = int(input())
a = [int(x) for x in input().split()]
a = [0] + a + [0]
ans = 0
for i in range(1,n+1):
diff = a[i] - max(a[i-1], a[i+1])
if diff > 0:
ans += diff
a[i] = a[i] - diff
for i in range(0,len(a)-1):
ans += abs(a[i]-a[i+1])
print(ans)
``` | output | 1 | 91,978 | 8 | 183,957 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Little Dormi received a histogram with n bars of height a_1, a_2, …, a_n for Christmas. However, the more he played with his new histogram, the more he realized its imperfections, so today he wanted to modify it to his liking.
To modify the histogram, Little Dormi is able to perform the following operation an arbitrary number of times:
* Select an index i (1 ≤ i ≤ n) where a_i>0, and assign a_i := a_i-1.
Little Dormi defines the ugliness score of his histogram (after performing some number of operations) as the sum of the vertical length of its outline and the number of operations he performed on it. And to make the histogram as perfect as possible, he would like to minimize the ugliness score after modifying it with some number of operations.
However, as his histogram is very large, Little Dormi is having trouble minimizing the ugliness score, so as Little Dormi's older brother, help him find the minimal ugliness.
Consider the following example where the histogram has 4 columns of heights 4,8,9,6:
<image>
The blue region represents the histogram, and the red lines represent the vertical portion of the outline. Currently, the vertical length of the outline is 4+4+1+3+6 = 18, so if Little Dormi does not modify the histogram at all, the ugliness would be 18.
However, Little Dormi can apply the operation once on column 2 and twice on column 3, resulting in a histogram with heights 4,7,7,6:
<image>
Now, as the total vertical length of the outline (red lines) is 4+3+1+6=14, the ugliness is 14+3=17 dollars. It can be proven that this is optimal.
Input
Each test contains multiple test cases. The first line contains the number of test cases t (1 ≤ t ≤ 10^4). Description of the test cases follows.
The first line of each test case contains a single integer n (1 ≤ n ≤ 4 ⋅ 10^5).
The second line of each test case contains n integers a_1, a_2, …, a_n (0 ≤ a_i ≤ 10^9).
It is guaranteed that the sum of n over all test cases does not exceed 4 ⋅ 10^5.
Output
For each test case output one integer, the minimal ugliness Little Dormi can achieve with the histogram in that test case.
Example
Input
2
4
4 8 9 6
6
2 1 7 4 0 0
Output
17
12
Note
Example 1 is the example described in the statement.
The initial histogram for example 2 is given below:
<image>
The ugliness is currently 2+1+6+3+4=16.
By applying the operation once on column 1, six times on column 3, and three times on column 4, we can end up with a histogram with heights 1,1,1,1,0,0:
<image>
The vertical length of the outline is now 1+1=2 and Little Dormi made 1+6+3=10 operations, so the final ugliness is 2+10=12, which can be proven to be optimal. | instruction | 0 | 91,979 | 8 | 183,958 |
Tags: greedy, implementation, math
Correct Solution:
```
t = int(input())
for test in range(t):
n = int(input())
arr = list(map(int, input().split()))
if(n == 1):
print(arr[0])
elif(n == 2):
print(arr[0]+arr[1])
else:
ans = 0
for i in range(n):
if(i==0 and arr[i+1] < arr[i]):
ans+=arr[i]-arr[i+1]
arr[i] = arr[i+1]
elif(i == n-1 and arr[i-1]<arr[i]):
ans+=arr[i]-arr[i-1]
arr[i]=arr[i-1]
elif(arr[i] > arr[i-1] and arr[i]>arr[i+1]):
ans+= arr[i]-max(arr[i-1],arr[i+1])
arr[i] = max(arr[i-1],arr[i+1])
ans+=arr[0]+arr[-1]
for i in range(1,n):
ans+= abs(arr[i] - arr[i-1])
print(ans )
``` | output | 1 | 91,979 | 8 | 183,959 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Little Dormi received a histogram with n bars of height a_1, a_2, …, a_n for Christmas. However, the more he played with his new histogram, the more he realized its imperfections, so today he wanted to modify it to his liking.
To modify the histogram, Little Dormi is able to perform the following operation an arbitrary number of times:
* Select an index i (1 ≤ i ≤ n) where a_i>0, and assign a_i := a_i-1.
Little Dormi defines the ugliness score of his histogram (after performing some number of operations) as the sum of the vertical length of its outline and the number of operations he performed on it. And to make the histogram as perfect as possible, he would like to minimize the ugliness score after modifying it with some number of operations.
However, as his histogram is very large, Little Dormi is having trouble minimizing the ugliness score, so as Little Dormi's older brother, help him find the minimal ugliness.
Consider the following example where the histogram has 4 columns of heights 4,8,9,6:
<image>
The blue region represents the histogram, and the red lines represent the vertical portion of the outline. Currently, the vertical length of the outline is 4+4+1+3+6 = 18, so if Little Dormi does not modify the histogram at all, the ugliness would be 18.
However, Little Dormi can apply the operation once on column 2 and twice on column 3, resulting in a histogram with heights 4,7,7,6:
<image>
Now, as the total vertical length of the outline (red lines) is 4+3+1+6=14, the ugliness is 14+3=17 dollars. It can be proven that this is optimal.
Input
Each test contains multiple test cases. The first line contains the number of test cases t (1 ≤ t ≤ 10^4). Description of the test cases follows.
The first line of each test case contains a single integer n (1 ≤ n ≤ 4 ⋅ 10^5).
The second line of each test case contains n integers a_1, a_2, …, a_n (0 ≤ a_i ≤ 10^9).
It is guaranteed that the sum of n over all test cases does not exceed 4 ⋅ 10^5.
Output
For each test case output one integer, the minimal ugliness Little Dormi can achieve with the histogram in that test case.
Example
Input
2
4
4 8 9 6
6
2 1 7 4 0 0
Output
17
12
Note
Example 1 is the example described in the statement.
The initial histogram for example 2 is given below:
<image>
The ugliness is currently 2+1+6+3+4=16.
By applying the operation once on column 1, six times on column 3, and three times on column 4, we can end up with a histogram with heights 1,1,1,1,0,0:
<image>
The vertical length of the outline is now 1+1=2 and Little Dormi made 1+6+3=10 operations, so the final ugliness is 2+10=12, which can be proven to be optimal. | instruction | 0 | 91,980 | 8 | 183,960 |
Tags: greedy, implementation, math
Correct Solution:
```
#du vinner ett p om du kan ta bort 2 outlines för 1 remove operation
#kolla 3 om 3 om mitten är mkt högre, downa till
for _ in range(int(input())):
N = int(input())
n = list(map(int,input().split()))
#om en stapel, rm all
if N == 1:
print(n[0])
continue
#edge cases
rm = 0
if n[0] > n[1]:
rm += n[0] - n[1]
n[0] = n[1]
if n[-1] > n[-2]:
rm += n[-1] - n[-2]
n[-1] = n[-2]
for i in range(N-2):
if n[i+1] > n[i] and n[i+1] > n[i+2]:
rm += n[i+1] - max(n[i],n[i+2])
n[i+1] = max(n[i],n[i+2])
#räkna outline score
#pinpointa hitta extreme points (locals)
tot = 0
for i in range(N-1):
upwards = n[i+1] - n[i] #edgecase med 0 til n[0]
if upwards > 0:
tot += upwards
downwards = n[i]-n[i+1]
if downwards > 0:
tot += downwards
tot += n[0]
tot += n[-1] #edg c
#print(n)
print(tot + rm)
#du kan jämt ut rmma 2 bredvid varandra, för att sedan kunna rmma ? vinst drag?
#nej kan aldrig hända
#wbu 2*max, den måste ju inte alltid sätmma
# 0 3 3 3 0 0 0 3 3 3 t.ex.
``` | output | 1 | 91,980 | 8 | 183,961 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Little Dormi received a histogram with n bars of height a_1, a_2, …, a_n for Christmas. However, the more he played with his new histogram, the more he realized its imperfections, so today he wanted to modify it to his liking.
To modify the histogram, Little Dormi is able to perform the following operation an arbitrary number of times:
* Select an index i (1 ≤ i ≤ n) where a_i>0, and assign a_i := a_i-1.
Little Dormi defines the ugliness score of his histogram (after performing some number of operations) as the sum of the vertical length of its outline and the number of operations he performed on it. And to make the histogram as perfect as possible, he would like to minimize the ugliness score after modifying it with some number of operations.
However, as his histogram is very large, Little Dormi is having trouble minimizing the ugliness score, so as Little Dormi's older brother, help him find the minimal ugliness.
Consider the following example where the histogram has 4 columns of heights 4,8,9,6:
<image>
The blue region represents the histogram, and the red lines represent the vertical portion of the outline. Currently, the vertical length of the outline is 4+4+1+3+6 = 18, so if Little Dormi does not modify the histogram at all, the ugliness would be 18.
However, Little Dormi can apply the operation once on column 2 and twice on column 3, resulting in a histogram with heights 4,7,7,6:
<image>
Now, as the total vertical length of the outline (red lines) is 4+3+1+6=14, the ugliness is 14+3=17 dollars. It can be proven that this is optimal.
Input
Each test contains multiple test cases. The first line contains the number of test cases t (1 ≤ t ≤ 10^4). Description of the test cases follows.
The first line of each test case contains a single integer n (1 ≤ n ≤ 4 ⋅ 10^5).
The second line of each test case contains n integers a_1, a_2, …, a_n (0 ≤ a_i ≤ 10^9).
It is guaranteed that the sum of n over all test cases does not exceed 4 ⋅ 10^5.
Output
For each test case output one integer, the minimal ugliness Little Dormi can achieve with the histogram in that test case.
Example
Input
2
4
4 8 9 6
6
2 1 7 4 0 0
Output
17
12
Note
Example 1 is the example described in the statement.
The initial histogram for example 2 is given below:
<image>
The ugliness is currently 2+1+6+3+4=16.
By applying the operation once on column 1, six times on column 3, and three times on column 4, we can end up with a histogram with heights 1,1,1,1,0,0:
<image>
The vertical length of the outline is now 1+1=2 and Little Dormi made 1+6+3=10 operations, so the final ugliness is 2+10=12, which can be proven to be optimal. | instruction | 0 | 91,981 | 8 | 183,962 |
Tags: greedy, implementation, math
Correct Solution:
```
#from itertools import product
#from itertools import combinations
from collections import Counter
#from collections import defaultdict
#from collections import deque # deque([iterable[, maxlen]]) #appendleft popleft rotate
#from heapq import heapify, heappop, heappush # func(heapifiedlist, item)
# sadness lies below
#from bisect import bisect_left, bisect_right, insort # func(sortedlist, item)
#from sys import setrecursionlimit
from sys import stdin, stderr
input = stdin.readline
def dbp(*args, **kwargs): # calling with dbp(locals()) is perfectly cromulent
print(*args, file=stderr, **kwargs)
def get_int_list():
return [int(x) for x in input().strip().split()]
def do_thing():
n = int(input())
alist = get_int_list()
#dbp(alist)
ugly = 0
last = 0
for idx, a in enumerate(alist):
delta = a-last
alist[idx] = delta
ugly += abs(delta)
last = a
delta = -last
alist.append(delta)
ugly += last
#dbp(alist)
for idx, a in enumerate(alist):
if idx == 0:
last = a
continue
if last > 0 and a < 0:
squash = min(last, -a)
ugly -= squash
last = a
return ugly
def multicase():
maxcc = int(input().strip())
for cc in range(maxcc):
print(do_thing())
if __name__ == "__main__":
multicase()
#print(do_thing())
``` | output | 1 | 91,981 | 8 | 183,963 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Little Dormi received a histogram with n bars of height a_1, a_2, …, a_n for Christmas. However, the more he played with his new histogram, the more he realized its imperfections, so today he wanted to modify it to his liking.
To modify the histogram, Little Dormi is able to perform the following operation an arbitrary number of times:
* Select an index i (1 ≤ i ≤ n) where a_i>0, and assign a_i := a_i-1.
Little Dormi defines the ugliness score of his histogram (after performing some number of operations) as the sum of the vertical length of its outline and the number of operations he performed on it. And to make the histogram as perfect as possible, he would like to minimize the ugliness score after modifying it with some number of operations.
However, as his histogram is very large, Little Dormi is having trouble minimizing the ugliness score, so as Little Dormi's older brother, help him find the minimal ugliness.
Consider the following example where the histogram has 4 columns of heights 4,8,9,6:
<image>
The blue region represents the histogram, and the red lines represent the vertical portion of the outline. Currently, the vertical length of the outline is 4+4+1+3+6 = 18, so if Little Dormi does not modify the histogram at all, the ugliness would be 18.
However, Little Dormi can apply the operation once on column 2 and twice on column 3, resulting in a histogram with heights 4,7,7,6:
<image>
Now, as the total vertical length of the outline (red lines) is 4+3+1+6=14, the ugliness is 14+3=17 dollars. It can be proven that this is optimal.
Input
Each test contains multiple test cases. The first line contains the number of test cases t (1 ≤ t ≤ 10^4). Description of the test cases follows.
The first line of each test case contains a single integer n (1 ≤ n ≤ 4 ⋅ 10^5).
The second line of each test case contains n integers a_1, a_2, …, a_n (0 ≤ a_i ≤ 10^9).
It is guaranteed that the sum of n over all test cases does not exceed 4 ⋅ 10^5.
Output
For each test case output one integer, the minimal ugliness Little Dormi can achieve with the histogram in that test case.
Example
Input
2
4
4 8 9 6
6
2 1 7 4 0 0
Output
17
12
Note
Example 1 is the example described in the statement.
The initial histogram for example 2 is given below:
<image>
The ugliness is currently 2+1+6+3+4=16.
By applying the operation once on column 1, six times on column 3, and three times on column 4, we can end up with a histogram with heights 1,1,1,1,0,0:
<image>
The vertical length of the outline is now 1+1=2 and Little Dormi made 1+6+3=10 operations, so the final ugliness is 2+10=12, which can be proven to be optimal. | instruction | 0 | 91,982 | 8 | 183,964 |
Tags: greedy, implementation, math
Correct Solution:
```
import sys,math,bisect
from random import randint
inf = float('inf')
mod = 10**9+7
"========================================"
def lcm(a,b):
return int((a/math.gcd(a,b))*b)
def gcd(a,b):
return int(math.gcd(a,b))
def tobinary(n):
return bin(n)[2:]
def binarySearch(a,x):
i = bisect.bisect_left(a,x)
if i!=len(a) and a[i]==x:
return i
else:
return -1
def lowerBound(a, x):
i = bisect.bisect_left(a, x)
if i:
return (i-1)
else:
return -1
def upperBound(a,x):
i = bisect.bisect_right(a,x)
if i!= len(a)+1 and a[i-1]==x:
return (i-1)
else:
return -1
def primesInRange(n):
ans = []
prime = [True for i in range(n+1)]
p = 2
while (p * p <= n):
if (prime[p] == True):
for i in range(p * p, n+1, p):
prime[i] = False
p += 1
for p in range(2, n+1):
if prime[p]:
ans.append(p)
return ans
def primeFactors(n):
factors = []
while n % 2 == 0:
factors.append(2)
n = n // 2
for i in range(3,int(math.sqrt(n))+1,2):
while n % i== 0:
factors.append(i)
n = n // i
if n > 2:
factors.append(n)
return factors
def isPrime(n,k=5):
if (n <2):
return True
for i in range(0,k):
a = randint(1,n-1)
if(pow(a,n-1,n)!=1):
return False
return True
"========================================="
"""
n = int(input())
n,k = map(int,input().split())
arr = list(map(int,input().split()))
"""
from collections import deque,defaultdict,Counter
import heapq,string
for _ in range(int(input())):
n = int(input())
arr = list(map(int,input().split()))
arr.insert(0,0)
arr.append(0)
cost = 0
for i in range(1,n+1):
og=arr[i]
if arr[i]>arr[i-1] and arr[i]>arr[i+1]:
arr[i]=max(arr[i-1],arr[i+1])
cost+=og-(arr[i])
for i in range(1,n+2):
cost+=abs(arr[i]-arr[i-1])
print(cost)
``` | output | 1 | 91,982 | 8 | 183,965 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Little Dormi received a histogram with n bars of height a_1, a_2, …, a_n for Christmas. However, the more he played with his new histogram, the more he realized its imperfections, so today he wanted to modify it to his liking.
To modify the histogram, Little Dormi is able to perform the following operation an arbitrary number of times:
* Select an index i (1 ≤ i ≤ n) where a_i>0, and assign a_i := a_i-1.
Little Dormi defines the ugliness score of his histogram (after performing some number of operations) as the sum of the vertical length of its outline and the number of operations he performed on it. And to make the histogram as perfect as possible, he would like to minimize the ugliness score after modifying it with some number of operations.
However, as his histogram is very large, Little Dormi is having trouble minimizing the ugliness score, so as Little Dormi's older brother, help him find the minimal ugliness.
Consider the following example where the histogram has 4 columns of heights 4,8,9,6:
<image>
The blue region represents the histogram, and the red lines represent the vertical portion of the outline. Currently, the vertical length of the outline is 4+4+1+3+6 = 18, so if Little Dormi does not modify the histogram at all, the ugliness would be 18.
However, Little Dormi can apply the operation once on column 2 and twice on column 3, resulting in a histogram with heights 4,7,7,6:
<image>
Now, as the total vertical length of the outline (red lines) is 4+3+1+6=14, the ugliness is 14+3=17 dollars. It can be proven that this is optimal.
Input
Each test contains multiple test cases. The first line contains the number of test cases t (1 ≤ t ≤ 10^4). Description of the test cases follows.
The first line of each test case contains a single integer n (1 ≤ n ≤ 4 ⋅ 10^5).
The second line of each test case contains n integers a_1, a_2, …, a_n (0 ≤ a_i ≤ 10^9).
It is guaranteed that the sum of n over all test cases does not exceed 4 ⋅ 10^5.
Output
For each test case output one integer, the minimal ugliness Little Dormi can achieve with the histogram in that test case.
Example
Input
2
4
4 8 9 6
6
2 1 7 4 0 0
Output
17
12
Note
Example 1 is the example described in the statement.
The initial histogram for example 2 is given below:
<image>
The ugliness is currently 2+1+6+3+4=16.
By applying the operation once on column 1, six times on column 3, and three times on column 4, we can end up with a histogram with heights 1,1,1,1,0,0:
<image>
The vertical length of the outline is now 1+1=2 and Little Dormi made 1+6+3=10 operations, so the final ugliness is 2+10=12, which can be proven to be optimal.
Submitted Solution:
```
for _ in range(int(input())):
n=int(input())
lst=list(map(int, input().split()))
extra=0
if n==1:
print(lst[0])
continue
for i in range(n):
if i==0:
if lst[i]>lst[i+1]:
#print('A')
extra+=abs(lst[i]-lst[i+1])
lst[i]=lst[i+1]
elif i==n-1:
if lst[i]>lst[i-1]:
#print('B')
extra+=abs(lst[i]-lst[i-1])
lst[i]=lst[i-1]
else:
if lst[i]>lst[i-1] and lst[i]>lst[i+1]:
#print('C')
val1=abs(lst[i]-lst[i+1])
val2=abs(lst[i]-lst[i-1])
extra+=min(val1,val2)
lst[i]=max(lst[i-1],lst[i+1])
#print(extra)
#print(lst)
for i in range(n):
if i==0:
extra+=lst[i]
else:
extra+=abs(lst[i]-lst[i-1])
extra+=lst[-1]
print(extra)
``` | instruction | 0 | 91,983 | 8 | 183,966 |
Yes | output | 1 | 91,983 | 8 | 183,967 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Little Dormi received a histogram with n bars of height a_1, a_2, …, a_n for Christmas. However, the more he played with his new histogram, the more he realized its imperfections, so today he wanted to modify it to his liking.
To modify the histogram, Little Dormi is able to perform the following operation an arbitrary number of times:
* Select an index i (1 ≤ i ≤ n) where a_i>0, and assign a_i := a_i-1.
Little Dormi defines the ugliness score of his histogram (after performing some number of operations) as the sum of the vertical length of its outline and the number of operations he performed on it. And to make the histogram as perfect as possible, he would like to minimize the ugliness score after modifying it with some number of operations.
However, as his histogram is very large, Little Dormi is having trouble minimizing the ugliness score, so as Little Dormi's older brother, help him find the minimal ugliness.
Consider the following example where the histogram has 4 columns of heights 4,8,9,6:
<image>
The blue region represents the histogram, and the red lines represent the vertical portion of the outline. Currently, the vertical length of the outline is 4+4+1+3+6 = 18, so if Little Dormi does not modify the histogram at all, the ugliness would be 18.
However, Little Dormi can apply the operation once on column 2 and twice on column 3, resulting in a histogram with heights 4,7,7,6:
<image>
Now, as the total vertical length of the outline (red lines) is 4+3+1+6=14, the ugliness is 14+3=17 dollars. It can be proven that this is optimal.
Input
Each test contains multiple test cases. The first line contains the number of test cases t (1 ≤ t ≤ 10^4). Description of the test cases follows.
The first line of each test case contains a single integer n (1 ≤ n ≤ 4 ⋅ 10^5).
The second line of each test case contains n integers a_1, a_2, …, a_n (0 ≤ a_i ≤ 10^9).
It is guaranteed that the sum of n over all test cases does not exceed 4 ⋅ 10^5.
Output
For each test case output one integer, the minimal ugliness Little Dormi can achieve with the histogram in that test case.
Example
Input
2
4
4 8 9 6
6
2 1 7 4 0 0
Output
17
12
Note
Example 1 is the example described in the statement.
The initial histogram for example 2 is given below:
<image>
The ugliness is currently 2+1+6+3+4=16.
By applying the operation once on column 1, six times on column 3, and three times on column 4, we can end up with a histogram with heights 1,1,1,1,0,0:
<image>
The vertical length of the outline is now 1+1=2 and Little Dormi made 1+6+3=10 operations, so the final ugliness is 2+10=12, which can be proven to be optimal.
Submitted Solution:
```
for _ in range(int(input())):
l = [0]
n = int(input())
for i in input().split():
l.append(int(i))
l.append(0)
if n == 1:
print(l[1])
continue
ans = 0
for i in range(1, n+1):
if l[i] > l[i-1] and l[i] > l[i+1]:
ans += l[i]-max(l[i-1], l[i+1])
l[i] = max(l[i-1], l[i+1])
for i, j in zip(l, l[1:]):
ans += abs(i-j)
print(ans)
``` | instruction | 0 | 91,984 | 8 | 183,968 |
Yes | output | 1 | 91,984 | 8 | 183,969 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Little Dormi received a histogram with n bars of height a_1, a_2, …, a_n for Christmas. However, the more he played with his new histogram, the more he realized its imperfections, so today he wanted to modify it to his liking.
To modify the histogram, Little Dormi is able to perform the following operation an arbitrary number of times:
* Select an index i (1 ≤ i ≤ n) where a_i>0, and assign a_i := a_i-1.
Little Dormi defines the ugliness score of his histogram (after performing some number of operations) as the sum of the vertical length of its outline and the number of operations he performed on it. And to make the histogram as perfect as possible, he would like to minimize the ugliness score after modifying it with some number of operations.
However, as his histogram is very large, Little Dormi is having trouble minimizing the ugliness score, so as Little Dormi's older brother, help him find the minimal ugliness.
Consider the following example where the histogram has 4 columns of heights 4,8,9,6:
<image>
The blue region represents the histogram, and the red lines represent the vertical portion of the outline. Currently, the vertical length of the outline is 4+4+1+3+6 = 18, so if Little Dormi does not modify the histogram at all, the ugliness would be 18.
However, Little Dormi can apply the operation once on column 2 and twice on column 3, resulting in a histogram with heights 4,7,7,6:
<image>
Now, as the total vertical length of the outline (red lines) is 4+3+1+6=14, the ugliness is 14+3=17 dollars. It can be proven that this is optimal.
Input
Each test contains multiple test cases. The first line contains the number of test cases t (1 ≤ t ≤ 10^4). Description of the test cases follows.
The first line of each test case contains a single integer n (1 ≤ n ≤ 4 ⋅ 10^5).
The second line of each test case contains n integers a_1, a_2, …, a_n (0 ≤ a_i ≤ 10^9).
It is guaranteed that the sum of n over all test cases does not exceed 4 ⋅ 10^5.
Output
For each test case output one integer, the minimal ugliness Little Dormi can achieve with the histogram in that test case.
Example
Input
2
4
4 8 9 6
6
2 1 7 4 0 0
Output
17
12
Note
Example 1 is the example described in the statement.
The initial histogram for example 2 is given below:
<image>
The ugliness is currently 2+1+6+3+4=16.
By applying the operation once on column 1, six times on column 3, and three times on column 4, we can end up with a histogram with heights 1,1,1,1,0,0:
<image>
The vertical length of the outline is now 1+1=2 and Little Dormi made 1+6+3=10 operations, so the final ugliness is 2+10=12, which can be proven to be optimal.
Submitted Solution:
```
# Rishabh Rao (https://github.com/rishabhrao)
import sys
MOD = 1000000007
def inp(): return sys.stdin.readline().strip()
def ii(): return int(inp())
def iis(): return [int(i) for i in inp().split()]
def solve():
n = ii()
a = [0] + iis() + [0]
min_ugliness = 0
for i in range(1, n + 1):
if a[i] > a[i - 1] and a[i] > a[i + 1]:
biggest_neighbour = max(a[i - 1], a[i + 1])
min_ugliness += a[i] - biggest_neighbour
a[i] = biggest_neighbour
if i < n:
min_ugliness += abs(a[i] - a[i - 1])
min_ugliness += abs(a[n] - a[n - 1]) + a[n]
return min_ugliness
t = ii()
for _ in range(t):
print(solve())
``` | instruction | 0 | 91,985 | 8 | 183,970 |
Yes | output | 1 | 91,985 | 8 | 183,971 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Little Dormi received a histogram with n bars of height a_1, a_2, …, a_n for Christmas. However, the more he played with his new histogram, the more he realized its imperfections, so today he wanted to modify it to his liking.
To modify the histogram, Little Dormi is able to perform the following operation an arbitrary number of times:
* Select an index i (1 ≤ i ≤ n) where a_i>0, and assign a_i := a_i-1.
Little Dormi defines the ugliness score of his histogram (after performing some number of operations) as the sum of the vertical length of its outline and the number of operations he performed on it. And to make the histogram as perfect as possible, he would like to minimize the ugliness score after modifying it with some number of operations.
However, as his histogram is very large, Little Dormi is having trouble minimizing the ugliness score, so as Little Dormi's older brother, help him find the minimal ugliness.
Consider the following example where the histogram has 4 columns of heights 4,8,9,6:
<image>
The blue region represents the histogram, and the red lines represent the vertical portion of the outline. Currently, the vertical length of the outline is 4+4+1+3+6 = 18, so if Little Dormi does not modify the histogram at all, the ugliness would be 18.
However, Little Dormi can apply the operation once on column 2 and twice on column 3, resulting in a histogram with heights 4,7,7,6:
<image>
Now, as the total vertical length of the outline (red lines) is 4+3+1+6=14, the ugliness is 14+3=17 dollars. It can be proven that this is optimal.
Input
Each test contains multiple test cases. The first line contains the number of test cases t (1 ≤ t ≤ 10^4). Description of the test cases follows.
The first line of each test case contains a single integer n (1 ≤ n ≤ 4 ⋅ 10^5).
The second line of each test case contains n integers a_1, a_2, …, a_n (0 ≤ a_i ≤ 10^9).
It is guaranteed that the sum of n over all test cases does not exceed 4 ⋅ 10^5.
Output
For each test case output one integer, the minimal ugliness Little Dormi can achieve with the histogram in that test case.
Example
Input
2
4
4 8 9 6
6
2 1 7 4 0 0
Output
17
12
Note
Example 1 is the example described in the statement.
The initial histogram for example 2 is given below:
<image>
The ugliness is currently 2+1+6+3+4=16.
By applying the operation once on column 1, six times on column 3, and three times on column 4, we can end up with a histogram with heights 1,1,1,1,0,0:
<image>
The vertical length of the outline is now 1+1=2 and Little Dormi made 1+6+3=10 operations, so the final ugliness is 2+10=12, which can be proven to be optimal.
Submitted Solution:
```
def solve(arr,n,ans):
if n == 1:
ans.append(str(arr[0]))
return
total = arr[0]+arr[-1]
for i in range(1,n):
total += abs(arr[i]-arr[i-1])
for i in range(n):
if i == 0:
if i+1 < n and arr[i] > arr[i+1]:
total -= (arr[i]-arr[i+1])
arr[i] = arr[i+1]
elif i == n-1:
if arr[i] > arr[i-1]:
total -= (arr[i]-arr[i-1])
else:
if arr[i] > arr[i-1] and arr[i] > arr[i+1]:
total -= (arr[i]-max(arr[i-1],arr[i+1]))
arr[i] = max(arr[i-1],arr[i+1])
ans.append(str(total))
def main():
t = int(input())
ans = []
for i in range(t):
n = int(input())
arr = list(map(int,input().split()))
solve(arr,n,ans)
print('\n'.join(ans))
main()
``` | instruction | 0 | 91,986 | 8 | 183,972 |
Yes | output | 1 | 91,986 | 8 | 183,973 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Little Dormi received a histogram with n bars of height a_1, a_2, …, a_n for Christmas. However, the more he played with his new histogram, the more he realized its imperfections, so today he wanted to modify it to his liking.
To modify the histogram, Little Dormi is able to perform the following operation an arbitrary number of times:
* Select an index i (1 ≤ i ≤ n) where a_i>0, and assign a_i := a_i-1.
Little Dormi defines the ugliness score of his histogram (after performing some number of operations) as the sum of the vertical length of its outline and the number of operations he performed on it. And to make the histogram as perfect as possible, he would like to minimize the ugliness score after modifying it with some number of operations.
However, as his histogram is very large, Little Dormi is having trouble minimizing the ugliness score, so as Little Dormi's older brother, help him find the minimal ugliness.
Consider the following example where the histogram has 4 columns of heights 4,8,9,6:
<image>
The blue region represents the histogram, and the red lines represent the vertical portion of the outline. Currently, the vertical length of the outline is 4+4+1+3+6 = 18, so if Little Dormi does not modify the histogram at all, the ugliness would be 18.
However, Little Dormi can apply the operation once on column 2 and twice on column 3, resulting in a histogram with heights 4,7,7,6:
<image>
Now, as the total vertical length of the outline (red lines) is 4+3+1+6=14, the ugliness is 14+3=17 dollars. It can be proven that this is optimal.
Input
Each test contains multiple test cases. The first line contains the number of test cases t (1 ≤ t ≤ 10^4). Description of the test cases follows.
The first line of each test case contains a single integer n (1 ≤ n ≤ 4 ⋅ 10^5).
The second line of each test case contains n integers a_1, a_2, …, a_n (0 ≤ a_i ≤ 10^9).
It is guaranteed that the sum of n over all test cases does not exceed 4 ⋅ 10^5.
Output
For each test case output one integer, the minimal ugliness Little Dormi can achieve with the histogram in that test case.
Example
Input
2
4
4 8 9 6
6
2 1 7 4 0 0
Output
17
12
Note
Example 1 is the example described in the statement.
The initial histogram for example 2 is given below:
<image>
The ugliness is currently 2+1+6+3+4=16.
By applying the operation once on column 1, six times on column 3, and three times on column 4, we can end up with a histogram with heights 1,1,1,1,0,0:
<image>
The vertical length of the outline is now 1+1=2 and Little Dormi made 1+6+3=10 operations, so the final ugliness is 2+10=12, which can be proven to be optimal.
Submitted Solution:
```
t=int(input())
for _ in range(t):
n=int(input())
ar=[0]+list(map(int,input().split()))+[0]
s=0
for i in range(1,n):
if (ar[i-1]<ar[i]>ar[i+1]):
s+=ar[i]-max(ar[i-1],ar[i+1])
ar[i]=max(ar[i-1],ar[i+1])
for i in range(1,n+2):
s+=abs(ar[i]-ar[i-1])
print(s)
``` | instruction | 0 | 91,987 | 8 | 183,974 |
No | output | 1 | 91,987 | 8 | 183,975 |
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