message stringlengths 2 44.5k | message_type stringclasses 2 values | message_id int64 0 1 | conversation_id int64 276 109k | cluster float64 23 23 | __index_level_0__ int64 552 217k |
|---|---|---|---|---|---|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There is a factory that inputs the data of the side and diagonal lengths to the machine and cuts out the plastic plate. At this factory, we cut out only parallelogram molds, although they vary in size. You have been ordered by your boss to count the number of rectangles and rhombuses produced among the parallelograms that are cut out.
Create a program that reads "Data to be input to the machine" and outputs the number of rectangles and diamonds manufactured.
<image>
Input
The input is given in the following format:
a1, b1, c1
a2, b2, c2
::
The data to be entered into the machine is given in multiple lines. On line i, the integers ai, bi, which represent the lengths of two adjacent sides of the i-th parallelogram, and the integer ci, which represents the length of the diagonal, are given, separated by commas (1 ≤ ai, bi, ci ≤ 1000). , ai + bi> ci). The number of data does not exceed 100.
Output
The first line outputs the number of rectangles manufactured, and the second line outputs the number of diamonds manufactured.
Example
Input
3,4,5
5,5,8
4,4,4
5,4,3
Output
1
2
Submitted Solution:
```
hisi=0
tyou=0
while True:
try:
a,b,c=map(int,input().split(","))
if a**2+b**2==c**2:
tyou +=1
elif a==b:
hisi +=1
except:
print(tyou)
print(hisi)
``` | instruction | 0 | 73,209 | 23 | 146,418 |
No | output | 1 | 73,209 | 23 | 146,419 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There is a factory that inputs the data of the side and diagonal lengths to the machine and cuts out the plastic plate. At this factory, we cut out only parallelogram molds, although they vary in size. You have been ordered by your boss to count the number of rectangles and rhombuses produced among the parallelograms that are cut out.
Create a program that reads "Data to be input to the machine" and outputs the number of rectangles and diamonds manufactured.
<image>
Input
The input is given in the following format:
a1, b1, c1
a2, b2, c2
::
The data to be entered into the machine is given in multiple lines. On line i, the integers ai, bi, which represent the lengths of two adjacent sides of the i-th parallelogram, and the integer ci, which represents the length of the diagonal, are given, separated by commas (1 ≤ ai, bi, ci ≤ 1000). , ai + bi> ci). The number of data does not exceed 100.
Output
The first line outputs the number of rectangles manufactured, and the second line outputs the number of diamonds manufactured.
Example
Input
3,4,5
5,5,8
4,4,4
5,4,3
Output
1
2
Submitted Solution:
```
import sys
j = [int(a)**2 + int(b)**2 - int(c)**2 for line in sys.stdin for a,b,c in line.split(",")]
print(len([i for i in j if not i]))
print(len([i for i in j if i]))
``` | instruction | 0 | 73,210 | 23 | 146,420 |
No | output | 1 | 73,210 | 23 | 146,421 |
Provide a correct Python 3 solution for this coding contest problem.
The educational program (AHK Education) of the Aiz Broadcasting Corporation broadcasts a program called "Play with Tsukuro" for children. Today is the time to make a box with drawing paper, but I would like to see if the rectangular drawing paper I prepared can make a rectangular parallelepiped. However, do not cut or fold the drawing paper.
Given six rectangles, write a program to determine if you can make a rectangular parallelepiped using them.
Input
The input is given in the following format.
h1 w1
h2 w2
h3 w3
h4 w4
h5 w5
h6 w6
The input consists of 6 lines, each line given the integer hi (1 ≤ hi ≤ 1000) for the vertical length of each rectangle and the integer wi (1 ≤ wi ≤ 1000) for the horizontal length.
Output
If a rectangular parallelepiped can be created, "yes" is output, and if it cannot be created, "no" is output. However, since a cube is a kind of rectangular parallelepiped, "yes" is output even if it is a cube.
Examples
Input
2 2
2 3
2 3
2 3
2 2
3 2
Output
yes
Input
2 2
2 3
2 3
2 3
2 2
2 2
Output
no | instruction | 0 | 73,219 | 23 | 146,438 |
"Correct Solution:
```
lst = [set(map(int, input().split())) for _ in range(6)]
rec = []
while lst:
x = lst[0]
count = lst.count(x)
if count % 2 == 1:
print("no")
break
rec.append((count, x))
for _ in range(count):
lst.pop(lst.index(x))
else:
if len(rec) == 1:
if len(rec[0][1]) == 1:
print("yes")
else:
print("no")
elif len(rec) == 2:
rec.sort()
if rec[0][1] & rec[1][1] == rec[0][1]:
print("yes")
else:
print("no")
elif len(rec) == 3:
if len(rec[0][1]) == len(rec[1][1]) == len(rec[2][1]) == 2 and (rec[2][1] & (rec[0][1] | rec[1][1]) == rec[2][1]):
print("yes")
else:
print("no")
``` | output | 1 | 73,219 | 23 | 146,439 |
Provide a correct Python 3 solution for this coding contest problem.
The educational program (AHK Education) of the Aiz Broadcasting Corporation broadcasts a program called "Play with Tsukuro" for children. Today is the time to make a box with drawing paper, but I would like to see if the rectangular drawing paper I prepared can make a rectangular parallelepiped. However, do not cut or fold the drawing paper.
Given six rectangles, write a program to determine if you can make a rectangular parallelepiped using them.
Input
The input is given in the following format.
h1 w1
h2 w2
h3 w3
h4 w4
h5 w5
h6 w6
The input consists of 6 lines, each line given the integer hi (1 ≤ hi ≤ 1000) for the vertical length of each rectangle and the integer wi (1 ≤ wi ≤ 1000) for the horizontal length.
Output
If a rectangular parallelepiped can be created, "yes" is output, and if it cannot be created, "no" is output. However, since a cube is a kind of rectangular parallelepiped, "yes" is output even if it is a cube.
Examples
Input
2 2
2 3
2 3
2 3
2 2
3 2
Output
yes
Input
2 2
2 3
2 3
2 3
2 2
2 2
Output
no | instruction | 0 | 73,220 | 23 | 146,440 |
"Correct Solution:
```
a = []
for _ in range(6):a.append(sorted(list(map(int, input().split()))))
a.sort(key = lambda x:x[1])
a.sort(key = lambda x:x[0])
print("yes" if a[0] == a[1] and a[2] == a[3] and a[4] == a[5] and a[0][0] == a[2][0] and a[0][1] == a[4][0] and a[2][1] == a[4][1] else "no")
``` | output | 1 | 73,220 | 23 | 146,441 |
Provide a correct Python 3 solution for this coding contest problem.
The educational program (AHK Education) of the Aiz Broadcasting Corporation broadcasts a program called "Play with Tsukuro" for children. Today is the time to make a box with drawing paper, but I would like to see if the rectangular drawing paper I prepared can make a rectangular parallelepiped. However, do not cut or fold the drawing paper.
Given six rectangles, write a program to determine if you can make a rectangular parallelepiped using them.
Input
The input is given in the following format.
h1 w1
h2 w2
h3 w3
h4 w4
h5 w5
h6 w6
The input consists of 6 lines, each line given the integer hi (1 ≤ hi ≤ 1000) for the vertical length of each rectangle and the integer wi (1 ≤ wi ≤ 1000) for the horizontal length.
Output
If a rectangular parallelepiped can be created, "yes" is output, and if it cannot be created, "no" is output. However, since a cube is a kind of rectangular parallelepiped, "yes" is output even if it is a cube.
Examples
Input
2 2
2 3
2 3
2 3
2 2
3 2
Output
yes
Input
2 2
2 3
2 3
2 3
2 2
2 2
Output
no | instruction | 0 | 73,221 | 23 | 146,442 |
"Correct Solution:
```
c=[[0,0]]*6
for i in range(6):
a,b=sorted(map(int,input().split()))
c[i]=[a,b]
c.sort()
for i in range(0,6,2):
if c[i]!=c[i+1]:print('no');break
else:print(['no','yes'][c[0][0]==c[2][0] and c[0][1]==c[4][0] and c[2][1]==c[4][1]])
``` | output | 1 | 73,221 | 23 | 146,443 |
Provide a correct Python 3 solution for this coding contest problem.
The educational program (AHK Education) of the Aiz Broadcasting Corporation broadcasts a program called "Play with Tsukuro" for children. Today is the time to make a box with drawing paper, but I would like to see if the rectangular drawing paper I prepared can make a rectangular parallelepiped. However, do not cut or fold the drawing paper.
Given six rectangles, write a program to determine if you can make a rectangular parallelepiped using them.
Input
The input is given in the following format.
h1 w1
h2 w2
h3 w3
h4 w4
h5 w5
h6 w6
The input consists of 6 lines, each line given the integer hi (1 ≤ hi ≤ 1000) for the vertical length of each rectangle and the integer wi (1 ≤ wi ≤ 1000) for the horizontal length.
Output
If a rectangular parallelepiped can be created, "yes" is output, and if it cannot be created, "no" is output. However, since a cube is a kind of rectangular parallelepiped, "yes" is output even if it is a cube.
Examples
Input
2 2
2 3
2 3
2 3
2 2
3 2
Output
yes
Input
2 2
2 3
2 3
2 3
2 2
2 2
Output
no | instruction | 0 | 73,222 | 23 | 146,444 |
"Correct Solution:
```
A = sorted(sorted(map(int, input().split())) for i in range(6))
if all(a==b for a, b in zip(A[::2], A[1::2])):
p, q, r = A[::2]
if (p[0] in q and p[1] in r and q[q.index(p[0])^1] == r[r.index(p[1])^1]) or (p[0] in r and p[1] in q and r[r.index(p[0])^1] == q[q.index(p[1])^1]):
print("yes")
else:
print("no")
else:
print("no")
``` | output | 1 | 73,222 | 23 | 146,445 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
The educational program (AHK Education) of the Aiz Broadcasting Corporation broadcasts a program called "Play with Tsukuro" for children. Today is the time to make a box with drawing paper, but I would like to see if the rectangular drawing paper I prepared can make a rectangular parallelepiped. However, do not cut or fold the drawing paper.
Given six rectangles, write a program to determine if you can make a rectangular parallelepiped using them.
Input
The input is given in the following format.
h1 w1
h2 w2
h3 w3
h4 w4
h5 w5
h6 w6
The input consists of 6 lines, each line given the integer hi (1 ≤ hi ≤ 1000) for the vertical length of each rectangle and the integer wi (1 ≤ wi ≤ 1000) for the horizontal length.
Output
If a rectangular parallelepiped can be created, "yes" is output, and if it cannot be created, "no" is output. However, since a cube is a kind of rectangular parallelepiped, "yes" is output even if it is a cube.
Examples
Input
2 2
2 3
2 3
2 3
2 2
3 2
Output
yes
Input
2 2
2 3
2 3
2 3
2 2
2 2
Output
no
Submitted Solution:
```
a = []
for _ in range(6):a.append(sorted(list(map(int, input().split()))))
a.sort(key = lambda x:x[1])
a.sort(key = lambda x:x[0])
print("yes" if a[0] == a[1] and a[2] == a[3] and a[4] == a[5] and a[0][0] == a[2][0] and a[0][1] == a[3][0] and a[2][1] == a[3][1] else "no")
``` | instruction | 0 | 73,223 | 23 | 146,446 |
No | output | 1 | 73,223 | 23 | 146,447 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
The educational program (AHK Education) of the Aiz Broadcasting Corporation broadcasts a program called "Play with Tsukuro" for children. Today is the time to make a box with drawing paper, but I would like to see if the rectangular drawing paper I prepared can make a rectangular parallelepiped. However, do not cut or fold the drawing paper.
Given six rectangles, write a program to determine if you can make a rectangular parallelepiped using them.
Input
The input is given in the following format.
h1 w1
h2 w2
h3 w3
h4 w4
h5 w5
h6 w6
The input consists of 6 lines, each line given the integer hi (1 ≤ hi ≤ 1000) for the vertical length of each rectangle and the integer wi (1 ≤ wi ≤ 1000) for the horizontal length.
Output
If a rectangular parallelepiped can be created, "yes" is output, and if it cannot be created, "no" is output. However, since a cube is a kind of rectangular parallelepiped, "yes" is output even if it is a cube.
Examples
Input
2 2
2 3
2 3
2 3
2 2
3 2
Output
yes
Input
2 2
2 3
2 3
2 3
2 2
2 2
Output
no
Submitted Solution:
```
A = sorted(sorted(map(int, input().split())) for i in range(6))
if all(a==b for a, b in zip(A[::2], A[1::2])):
p, q, r = A[::2]
if (p[0] in q and p[1] in r and q[q.index(p[0])^1] == r[r.index(p[1])^1]) or (p[0] in r and p[1] in q and r[r.index(p[0])^1] == q[q.index(p[1])^1]):
print("yes")
else:
print("no")
``` | instruction | 0 | 73,224 | 23 | 146,448 |
No | output | 1 | 73,224 | 23 | 146,449 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
The educational program (AHK Education) of the Aiz Broadcasting Corporation broadcasts a program called "Play with Tsukuro" for children. Today is the time to make a box with drawing paper, but I would like to see if the rectangular drawing paper I prepared can make a rectangular parallelepiped. However, do not cut or fold the drawing paper.
Given six rectangles, write a program to determine if you can make a rectangular parallelepiped using them.
Input
The input is given in the following format.
h1 w1
h2 w2
h3 w3
h4 w4
h5 w5
h6 w6
The input consists of 6 lines, each line given the integer hi (1 ≤ hi ≤ 1000) for the vertical length of each rectangle and the integer wi (1 ≤ wi ≤ 1000) for the horizontal length.
Output
If a rectangular parallelepiped can be created, "yes" is output, and if it cannot be created, "no" is output. However, since a cube is a kind of rectangular parallelepiped, "yes" is output even if it is a cube.
Examples
Input
2 2
2 3
2 3
2 3
2 2
3 2
Output
yes
Input
2 2
2 3
2 3
2 3
2 2
2 2
Output
no
Submitted Solution:
```
lst = [set(map(int, input().split())) for _ in range(6)]
rec = []
while lst:
x = lst[0]
count = lst.count(x)
if count % 2 == 1:
print("no")
break
rec.append((count, x))
for _ in range(count):
lst.pop(lst.index(x))
else:
if len(rec) == 1:
if len(rec[0][1]) == 1:
print("yes")
else:
print("no")
elif len(rec) == 2:
rec.sort()
if rec[0][1] & rec[1][1] == rec[0][1]:
print("yes")
else:
print("no")
elif len(rec) == 3:
if len(rec[0][1]) == len(rec[1][1]) == len(rec[2][1]) == 2 and (rec[2][1] in rec[0][1] | rec[1][1]):
print("yes")
else:
print("no")
``` | instruction | 0 | 73,225 | 23 | 146,450 |
No | output | 1 | 73,225 | 23 | 146,451 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
The educational program (AHK Education) of the Aiz Broadcasting Corporation broadcasts a program called "Play with Tsukuro" for children. Today is the time to make a box with drawing paper, but I would like to see if the rectangular drawing paper I prepared can make a rectangular parallelepiped. However, do not cut or fold the drawing paper.
Given six rectangles, write a program to determine if you can make a rectangular parallelepiped using them.
Input
The input is given in the following format.
h1 w1
h2 w2
h3 w3
h4 w4
h5 w5
h6 w6
The input consists of 6 lines, each line given the integer hi (1 ≤ hi ≤ 1000) for the vertical length of each rectangle and the integer wi (1 ≤ wi ≤ 1000) for the horizontal length.
Output
If a rectangular parallelepiped can be created, "yes" is output, and if it cannot be created, "no" is output. However, since a cube is a kind of rectangular parallelepiped, "yes" is output even if it is a cube.
Examples
Input
2 2
2 3
2 3
2 3
2 2
3 2
Output
yes
Input
2 2
2 3
2 3
2 3
2 2
2 2
Output
no
Submitted Solution:
```
A = sorted(sorted(map(int, input().split())) for i in range(6))
if all(a==b for a, b in zip(A[::2], A[1::2])):
print("yes")
else:
print("no")
``` | instruction | 0 | 73,226 | 23 | 146,452 |
No | output | 1 | 73,226 | 23 | 146,453 |
Provide a correct Python 3 solution for this coding contest problem.
Problem
KND is a student programmer at the University of Aizu. His chest is known to be very sexy.
<image>
For simplicity, the part of the skin that can be seen from the chest is represented by the isosceles triangle ABC in the figure. However, due to the slack in the clothes, the two sides AC and BC (where these lengths are l), which have the same length, actually have an additional length x minutes. In order to increase the area of the open part, let's make two new triangular ADCs and BECs by pulling the slack part. Points D and E exist outside the triangle ABC. These two new triangles are caused by slack, and the sum of the lengths of side BE and side EC and the sum of the lengths of side AD and side DC must be l + x. You determine the points D and E so that the sum M of the areas of these three triangles is maximized. As KND's neighbor, you decide to write a program to calculate the maximum area of skin (M) to look out of your clothes, using a, l, x as inputs to find out how sexy his chest is. did.
Constraints
The input satisfies the following conditions.
* All inputs are integers.
* 1 ≤ a ≤ 1000
* 1 ≤ l ≤ 1000
* 1 ≤ x ≤ 1000
Input
The input consists of multiple test cases. One test case is given in the following format. The end of input is indicated by EOF.
a l x
here,
* a: Length of side AB of triangle ABC
* l: Length of two sides AC and BC of triangle ABC
* x: Slack on two sides AC, BC
Is.
Output
Output the maximum area for each test case on one line. This value should not differ more than 10-5 from the value of the judge output.
Example
Input
2 2 1
2 3 1
3 2 3
2 3 5
Output
3.9681187851
6.7970540913
6.5668891783
13.9527248554 | instruction | 0 | 73,239 | 23 | 146,478 |
"Correct Solution:
```
import sys
for i in sys.stdin:
a,l,x=map(int,i.split())
print((a*(4*l*l-a*a)**.5+2*l*((l+x)**2-l*l)**.5)/4)
``` | output | 1 | 73,239 | 23 | 146,479 |
Provide a correct Python 3 solution for this coding contest problem.
Problem
KND is a student programmer at the University of Aizu. His chest is known to be very sexy.
<image>
For simplicity, the part of the skin that can be seen from the chest is represented by the isosceles triangle ABC in the figure. However, due to the slack in the clothes, the two sides AC and BC (where these lengths are l), which have the same length, actually have an additional length x minutes. In order to increase the area of the open part, let's make two new triangular ADCs and BECs by pulling the slack part. Points D and E exist outside the triangle ABC. These two new triangles are caused by slack, and the sum of the lengths of side BE and side EC and the sum of the lengths of side AD and side DC must be l + x. You determine the points D and E so that the sum M of the areas of these three triangles is maximized. As KND's neighbor, you decide to write a program to calculate the maximum area of skin (M) to look out of your clothes, using a, l, x as inputs to find out how sexy his chest is. did.
Constraints
The input satisfies the following conditions.
* All inputs are integers.
* 1 ≤ a ≤ 1000
* 1 ≤ l ≤ 1000
* 1 ≤ x ≤ 1000
Input
The input consists of multiple test cases. One test case is given in the following format. The end of input is indicated by EOF.
a l x
here,
* a: Length of side AB of triangle ABC
* l: Length of two sides AC and BC of triangle ABC
* x: Slack on two sides AC, BC
Is.
Output
Output the maximum area for each test case on one line. This value should not differ more than 10-5 from the value of the judge output.
Example
Input
2 2 1
2 3 1
3 2 3
2 3 5
Output
3.9681187851
6.7970540913
6.5668891783
13.9527248554 | instruction | 0 | 73,240 | 23 | 146,480 |
"Correct Solution:
```
import sys
for s in sys.stdin:a,l,x=map(float,s.split());print(a/2*(l*l-a*a/4)**.5+l/2*(2*l*x+x*x)**.5)
``` | output | 1 | 73,240 | 23 | 146,481 |
Provide a correct Python 3 solution for this coding contest problem.
Problem
KND is a student programmer at the University of Aizu. His chest is known to be very sexy.
<image>
For simplicity, the part of the skin that can be seen from the chest is represented by the isosceles triangle ABC in the figure. However, due to the slack in the clothes, the two sides AC and BC (where these lengths are l), which have the same length, actually have an additional length x minutes. In order to increase the area of the open part, let's make two new triangular ADCs and BECs by pulling the slack part. Points D and E exist outside the triangle ABC. These two new triangles are caused by slack, and the sum of the lengths of side BE and side EC and the sum of the lengths of side AD and side DC must be l + x. You determine the points D and E so that the sum M of the areas of these three triangles is maximized. As KND's neighbor, you decide to write a program to calculate the maximum area of skin (M) to look out of your clothes, using a, l, x as inputs to find out how sexy his chest is. did.
Constraints
The input satisfies the following conditions.
* All inputs are integers.
* 1 ≤ a ≤ 1000
* 1 ≤ l ≤ 1000
* 1 ≤ x ≤ 1000
Input
The input consists of multiple test cases. One test case is given in the following format. The end of input is indicated by EOF.
a l x
here,
* a: Length of side AB of triangle ABC
* l: Length of two sides AC and BC of triangle ABC
* x: Slack on two sides AC, BC
Is.
Output
Output the maximum area for each test case on one line. This value should not differ more than 10-5 from the value of the judge output.
Example
Input
2 2 1
2 3 1
3 2 3
2 3 5
Output
3.9681187851
6.7970540913
6.5668891783
13.9527248554 | instruction | 0 | 73,241 | 23 | 146,482 |
"Correct Solution:
```
import sys
for s in sys.stdin:
a,l,x=map(float,s.split())
print(a/2*(l*l-a*a/4)**.5+l/2*(2*l*x+x*x)**.5)
``` | output | 1 | 73,241 | 23 | 146,483 |
Provide a correct Python 3 solution for this coding contest problem.
Problem
KND is a student programmer at the University of Aizu. His chest is known to be very sexy.
<image>
For simplicity, the part of the skin that can be seen from the chest is represented by the isosceles triangle ABC in the figure. However, due to the slack in the clothes, the two sides AC and BC (where these lengths are l), which have the same length, actually have an additional length x minutes. In order to increase the area of the open part, let's make two new triangular ADCs and BECs by pulling the slack part. Points D and E exist outside the triangle ABC. These two new triangles are caused by slack, and the sum of the lengths of side BE and side EC and the sum of the lengths of side AD and side DC must be l + x. You determine the points D and E so that the sum M of the areas of these three triangles is maximized. As KND's neighbor, you decide to write a program to calculate the maximum area of skin (M) to look out of your clothes, using a, l, x as inputs to find out how sexy his chest is. did.
Constraints
The input satisfies the following conditions.
* All inputs are integers.
* 1 ≤ a ≤ 1000
* 1 ≤ l ≤ 1000
* 1 ≤ x ≤ 1000
Input
The input consists of multiple test cases. One test case is given in the following format. The end of input is indicated by EOF.
a l x
here,
* a: Length of side AB of triangle ABC
* l: Length of two sides AC and BC of triangle ABC
* x: Slack on two sides AC, BC
Is.
Output
Output the maximum area for each test case on one line. This value should not differ more than 10-5 from the value of the judge output.
Example
Input
2 2 1
2 3 1
3 2 3
2 3 5
Output
3.9681187851
6.7970540913
6.5668891783
13.9527248554 | instruction | 0 | 73,242 | 23 | 146,484 |
"Correct Solution:
```
from math import sqrt
try:
while 1:
a, l, x = map(int, input().split())
print("%.10f" % (l*sqrt(x*(2*l+x))/2 + a*sqrt(4*l**2 - a**2)/4))
except EOFError:
...
``` | output | 1 | 73,242 | 23 | 146,485 |
Provide a correct Python 3 solution for this coding contest problem.
Problem
KND is a student programmer at the University of Aizu. His chest is known to be very sexy.
<image>
For simplicity, the part of the skin that can be seen from the chest is represented by the isosceles triangle ABC in the figure. However, due to the slack in the clothes, the two sides AC and BC (where these lengths are l), which have the same length, actually have an additional length x minutes. In order to increase the area of the open part, let's make two new triangular ADCs and BECs by pulling the slack part. Points D and E exist outside the triangle ABC. These two new triangles are caused by slack, and the sum of the lengths of side BE and side EC and the sum of the lengths of side AD and side DC must be l + x. You determine the points D and E so that the sum M of the areas of these three triangles is maximized. As KND's neighbor, you decide to write a program to calculate the maximum area of skin (M) to look out of your clothes, using a, l, x as inputs to find out how sexy his chest is. did.
Constraints
The input satisfies the following conditions.
* All inputs are integers.
* 1 ≤ a ≤ 1000
* 1 ≤ l ≤ 1000
* 1 ≤ x ≤ 1000
Input
The input consists of multiple test cases. One test case is given in the following format. The end of input is indicated by EOF.
a l x
here,
* a: Length of side AB of triangle ABC
* l: Length of two sides AC and BC of triangle ABC
* x: Slack on two sides AC, BC
Is.
Output
Output the maximum area for each test case on one line. This value should not differ more than 10-5 from the value of the judge output.
Example
Input
2 2 1
2 3 1
3 2 3
2 3 5
Output
3.9681187851
6.7970540913
6.5668891783
13.9527248554 | instruction | 0 | 73,243 | 23 | 146,486 |
"Correct Solution:
```
import math
while True :
try :
a, l, x = map(int, input().split())
except EOFError :
break
HC = math.sqrt(l**2 - a**2 / 4)
I = math.sqrt(((l+x)/2)**2 - l**2 / 4)
S1 = a * HC / 2
S2 = l * I / 2
print('{:.8f}'.format(S1 + S2*2))
``` | output | 1 | 73,243 | 23 | 146,487 |
Provide a correct Python 3 solution for this coding contest problem.
Problem
KND is a student programmer at the University of Aizu. His chest is known to be very sexy.
<image>
For simplicity, the part of the skin that can be seen from the chest is represented by the isosceles triangle ABC in the figure. However, due to the slack in the clothes, the two sides AC and BC (where these lengths are l), which have the same length, actually have an additional length x minutes. In order to increase the area of the open part, let's make two new triangular ADCs and BECs by pulling the slack part. Points D and E exist outside the triangle ABC. These two new triangles are caused by slack, and the sum of the lengths of side BE and side EC and the sum of the lengths of side AD and side DC must be l + x. You determine the points D and E so that the sum M of the areas of these three triangles is maximized. As KND's neighbor, you decide to write a program to calculate the maximum area of skin (M) to look out of your clothes, using a, l, x as inputs to find out how sexy his chest is. did.
Constraints
The input satisfies the following conditions.
* All inputs are integers.
* 1 ≤ a ≤ 1000
* 1 ≤ l ≤ 1000
* 1 ≤ x ≤ 1000
Input
The input consists of multiple test cases. One test case is given in the following format. The end of input is indicated by EOF.
a l x
here,
* a: Length of side AB of triangle ABC
* l: Length of two sides AC and BC of triangle ABC
* x: Slack on two sides AC, BC
Is.
Output
Output the maximum area for each test case on one line. This value should not differ more than 10-5 from the value of the judge output.
Example
Input
2 2 1
2 3 1
3 2 3
2 3 5
Output
3.9681187851
6.7970540913
6.5668891783
13.9527248554 | instruction | 0 | 73,244 | 23 | 146,488 |
"Correct Solution:
```
import math
while True:
try:
a,l,x=map(int, input().split())
temp=(l+x)/2
except EOFError:
break
def heron(i,j,k):
d = (i+j+k)/2
return math.sqrt(d*(d-i)*(d-j)*(d-k))
print((str(heron(a,l,l)+heron(l,temp,temp)*2)))
``` | output | 1 | 73,244 | 23 | 146,489 |
Provide a correct Python 3 solution for this coding contest problem.
Problem
KND is a student programmer at the University of Aizu. His chest is known to be very sexy.
<image>
For simplicity, the part of the skin that can be seen from the chest is represented by the isosceles triangle ABC in the figure. However, due to the slack in the clothes, the two sides AC and BC (where these lengths are l), which have the same length, actually have an additional length x minutes. In order to increase the area of the open part, let's make two new triangular ADCs and BECs by pulling the slack part. Points D and E exist outside the triangle ABC. These two new triangles are caused by slack, and the sum of the lengths of side BE and side EC and the sum of the lengths of side AD and side DC must be l + x. You determine the points D and E so that the sum M of the areas of these three triangles is maximized. As KND's neighbor, you decide to write a program to calculate the maximum area of skin (M) to look out of your clothes, using a, l, x as inputs to find out how sexy his chest is. did.
Constraints
The input satisfies the following conditions.
* All inputs are integers.
* 1 ≤ a ≤ 1000
* 1 ≤ l ≤ 1000
* 1 ≤ x ≤ 1000
Input
The input consists of multiple test cases. One test case is given in the following format. The end of input is indicated by EOF.
a l x
here,
* a: Length of side AB of triangle ABC
* l: Length of two sides AC and BC of triangle ABC
* x: Slack on two sides AC, BC
Is.
Output
Output the maximum area for each test case on one line. This value should not differ more than 10-5 from the value of the judge output.
Example
Input
2 2 1
2 3 1
3 2 3
2 3 5
Output
3.9681187851
6.7970540913
6.5668891783
13.9527248554 | instruction | 0 | 73,245 | 23 | 146,490 |
"Correct Solution:
```
while True:
try:
a, l, x = map(int, input().split())
except EOFError:
break
s1 = (l + l + x) / 2
v1 = (s1 * (l / 2) * (l / 2) * (x / 2)) ** (1 / 2)
s2 = (l + l + a) / 2
v2 = (s2 * (s2 - l) * (s2 - l) * (s2 - a)) ** (1 / 2)
print(v1 * 2 + v2)
``` | output | 1 | 73,245 | 23 | 146,491 |
Provide a correct Python 3 solution for this coding contest problem.
Problem
KND is a student programmer at the University of Aizu. His chest is known to be very sexy.
<image>
For simplicity, the part of the skin that can be seen from the chest is represented by the isosceles triangle ABC in the figure. However, due to the slack in the clothes, the two sides AC and BC (where these lengths are l), which have the same length, actually have an additional length x minutes. In order to increase the area of the open part, let's make two new triangular ADCs and BECs by pulling the slack part. Points D and E exist outside the triangle ABC. These two new triangles are caused by slack, and the sum of the lengths of side BE and side EC and the sum of the lengths of side AD and side DC must be l + x. You determine the points D and E so that the sum M of the areas of these three triangles is maximized. As KND's neighbor, you decide to write a program to calculate the maximum area of skin (M) to look out of your clothes, using a, l, x as inputs to find out how sexy his chest is. did.
Constraints
The input satisfies the following conditions.
* All inputs are integers.
* 1 ≤ a ≤ 1000
* 1 ≤ l ≤ 1000
* 1 ≤ x ≤ 1000
Input
The input consists of multiple test cases. One test case is given in the following format. The end of input is indicated by EOF.
a l x
here,
* a: Length of side AB of triangle ABC
* l: Length of two sides AC and BC of triangle ABC
* x: Slack on two sides AC, BC
Is.
Output
Output the maximum area for each test case on one line. This value should not differ more than 10-5 from the value of the judge output.
Example
Input
2 2 1
2 3 1
3 2 3
2 3 5
Output
3.9681187851
6.7970540913
6.5668891783
13.9527248554 | instruction | 0 | 73,246 | 23 | 146,492 |
"Correct Solution:
```
import sys
for line in sys.stdin:
a,l,x=map(float,line.split())
center=1/2*a*(l**2-(a/2)**2)**0.5
side=1/2*l*(((l+x)/2)**2-(l/2)**2)**0.5
print("{0:.10f}".format(center+side*2))
``` | output | 1 | 73,246 | 23 | 146,493 |
Provide a correct Python 3 solution for this coding contest problem.
Draw a chessboard which has a height of H cm and a width of W cm. For example, the following figure shows a chessboard which has a height of 6 cm and a width of 10 cm.
.#.#.#.#.
.#.#.#.#.#
.#.#.#.#.
.#.#.#.#.#
.#.#.#.#.
.#.#.#.#.#
Note that the top left corner should be drawn by '#'.
Constraints
* 1 ≤ H ≤ 300
* 1 ≤ W ≤ 300
Input
The input consists of multiple datasets. Each dataset consists of two integers H and W separated by a single space.
The input ends with two 0 (when both H and W are zero).
Output
For each dataset, print the chessboard made of '#' and '.'.
Print a blank line after each dataset.
Example
Input
3 4
5 6
3 3
2 2
1 1
0 0
Output
#.#.
.#.#
#.#.
#.#.#.
.#.#.#
#.#.#.
.#.#.#
#.#.#.
#.#
.#.
#.#
#.
.#
# | instruction | 0 | 73,287 | 23 | 146,574 |
"Correct Solution:
```
while True:
h,w = map(int, input().split())
if (h == 0 and w == 0):
break
for i in range(h):
if (i % 2 == 0):
print("#."*(w//2) + "#"*(w%2))
elif (i % 2 == 1):
print(".#"*(w//2) + "."*(w%2))
print()
``` | output | 1 | 73,287 | 23 | 146,575 |
Provide a correct Python 3 solution for this coding contest problem.
Draw a chessboard which has a height of H cm and a width of W cm. For example, the following figure shows a chessboard which has a height of 6 cm and a width of 10 cm.
.#.#.#.#.
.#.#.#.#.#
.#.#.#.#.
.#.#.#.#.#
.#.#.#.#.
.#.#.#.#.#
Note that the top left corner should be drawn by '#'.
Constraints
* 1 ≤ H ≤ 300
* 1 ≤ W ≤ 300
Input
The input consists of multiple datasets. Each dataset consists of two integers H and W separated by a single space.
The input ends with two 0 (when both H and W are zero).
Output
For each dataset, print the chessboard made of '#' and '.'.
Print a blank line after each dataset.
Example
Input
3 4
5 6
3 3
2 2
1 1
0 0
Output
#.#.
.#.#
#.#.
#.#.#.
.#.#.#
#.#.#.
.#.#.#
#.#.#.
#.#
.#.
#.#
#.
.#
# | instruction | 0 | 73,288 | 23 | 146,576 |
"Correct Solution:
```
while 1:
h,w=map(int,input().split())
if h==0:break
for i in range(h):print(('#.'*w)[i%2:][:w])
print()
``` | output | 1 | 73,288 | 23 | 146,577 |
Provide a correct Python 3 solution for this coding contest problem.
Draw a chessboard which has a height of H cm and a width of W cm. For example, the following figure shows a chessboard which has a height of 6 cm and a width of 10 cm.
.#.#.#.#.
.#.#.#.#.#
.#.#.#.#.
.#.#.#.#.#
.#.#.#.#.
.#.#.#.#.#
Note that the top left corner should be drawn by '#'.
Constraints
* 1 ≤ H ≤ 300
* 1 ≤ W ≤ 300
Input
The input consists of multiple datasets. Each dataset consists of two integers H and W separated by a single space.
The input ends with two 0 (when both H and W are zero).
Output
For each dataset, print the chessboard made of '#' and '.'.
Print a blank line after each dataset.
Example
Input
3 4
5 6
3 3
2 2
1 1
0 0
Output
#.#.
.#.#
#.#.
#.#.#.
.#.#.#
#.#.#.
.#.#.#
#.#.#.
#.#
.#.
#.#
#.
.#
# | instruction | 0 | 73,289 | 23 | 146,578 |
"Correct Solution:
```
while True:
h,w = map(int,input().split())
if h+w == 0:
break;
for i in range(h):
for j in range(w):
print("." if (i+j)&1 else "#",end='')
print()
print()
``` | output | 1 | 73,289 | 23 | 146,579 |
Provide a correct Python 3 solution for this coding contest problem.
Draw a chessboard which has a height of H cm and a width of W cm. For example, the following figure shows a chessboard which has a height of 6 cm and a width of 10 cm.
.#.#.#.#.
.#.#.#.#.#
.#.#.#.#.
.#.#.#.#.#
.#.#.#.#.
.#.#.#.#.#
Note that the top left corner should be drawn by '#'.
Constraints
* 1 ≤ H ≤ 300
* 1 ≤ W ≤ 300
Input
The input consists of multiple datasets. Each dataset consists of two integers H and W separated by a single space.
The input ends with two 0 (when both H and W are zero).
Output
For each dataset, print the chessboard made of '#' and '.'.
Print a blank line after each dataset.
Example
Input
3 4
5 6
3 3
2 2
1 1
0 0
Output
#.#.
.#.#
#.#.
#.#.#.
.#.#.#
#.#.#.
.#.#.#
#.#.#.
#.#
.#.
#.#
#.
.#
# | instruction | 0 | 73,290 | 23 | 146,580 |
"Correct Solution:
```
kk=0
while kk==0:
h,w=map(int,input().split())
if h==0 and w==0:
break
else:
for i in range(h):
if i%2==0:
s="#"
else:
s="."
for k in range(w-1):
if s[-1]=="#":
s=s+"."
else:
s=s+"#"
print(s)
print("")
``` | output | 1 | 73,290 | 23 | 146,581 |
Provide a correct Python 3 solution for this coding contest problem.
Draw a chessboard which has a height of H cm and a width of W cm. For example, the following figure shows a chessboard which has a height of 6 cm and a width of 10 cm.
.#.#.#.#.
.#.#.#.#.#
.#.#.#.#.
.#.#.#.#.#
.#.#.#.#.
.#.#.#.#.#
Note that the top left corner should be drawn by '#'.
Constraints
* 1 ≤ H ≤ 300
* 1 ≤ W ≤ 300
Input
The input consists of multiple datasets. Each dataset consists of two integers H and W separated by a single space.
The input ends with two 0 (when both H and W are zero).
Output
For each dataset, print the chessboard made of '#' and '.'.
Print a blank line after each dataset.
Example
Input
3 4
5 6
3 3
2 2
1 1
0 0
Output
#.#.
.#.#
#.#.
#.#.#.
.#.#.#
#.#.#.
.#.#.#
#.#.#.
#.#
.#.
#.#
#.
.#
# | instruction | 0 | 73,291 | 23 | 146,582 |
"Correct Solution:
```
import sys
while True:
y,x = map(int,input().split())
if x==y==0:
break
for i in range(y):
for j in range(x):
if (i+j)%2==0:
sys.stdout.write("#")
else:
sys.stdout.write(".")
print()
print()
``` | output | 1 | 73,291 | 23 | 146,583 |
Provide a correct Python 3 solution for this coding contest problem.
Draw a chessboard which has a height of H cm and a width of W cm. For example, the following figure shows a chessboard which has a height of 6 cm and a width of 10 cm.
.#.#.#.#.
.#.#.#.#.#
.#.#.#.#.
.#.#.#.#.#
.#.#.#.#.
.#.#.#.#.#
Note that the top left corner should be drawn by '#'.
Constraints
* 1 ≤ H ≤ 300
* 1 ≤ W ≤ 300
Input
The input consists of multiple datasets. Each dataset consists of two integers H and W separated by a single space.
The input ends with two 0 (when both H and W are zero).
Output
For each dataset, print the chessboard made of '#' and '.'.
Print a blank line after each dataset.
Example
Input
3 4
5 6
3 3
2 2
1 1
0 0
Output
#.#.
.#.#
#.#.
#.#.#.
.#.#.#
#.#.#.
.#.#.#
#.#.#.
#.#
.#.
#.#
#.
.#
# | instruction | 0 | 73,292 | 23 | 146,584 |
"Correct Solution:
```
while True:
h,w=map(int,input().split())
if h==0:break
for i in range(h):
if i%2==0:print("".join(['#' if i%2==0 else '.' for i in range(w)]))
else:print("".join(['.' if i%2==0 else '#' for i in range(w)]))
print()
``` | output | 1 | 73,292 | 23 | 146,585 |
Provide a correct Python 3 solution for this coding contest problem.
Draw a chessboard which has a height of H cm and a width of W cm. For example, the following figure shows a chessboard which has a height of 6 cm and a width of 10 cm.
.#.#.#.#.
.#.#.#.#.#
.#.#.#.#.
.#.#.#.#.#
.#.#.#.#.
.#.#.#.#.#
Note that the top left corner should be drawn by '#'.
Constraints
* 1 ≤ H ≤ 300
* 1 ≤ W ≤ 300
Input
The input consists of multiple datasets. Each dataset consists of two integers H and W separated by a single space.
The input ends with two 0 (when both H and W are zero).
Output
For each dataset, print the chessboard made of '#' and '.'.
Print a blank line after each dataset.
Example
Input
3 4
5 6
3 3
2 2
1 1
0 0
Output
#.#.
.#.#
#.#.
#.#.#.
.#.#.#
#.#.#.
.#.#.#
#.#.#.
#.#
.#.
#.#
#.
.#
# | instruction | 0 | 73,293 | 23 | 146,586 |
"Correct Solution:
```
while True:
h, w = map(int, input().split())
if h == 0 and w == 0:
break
for i in range(1, h+1):
if i % 2 == 0:
print(('.#'*w)[:w])
else:
print(('#.'*w)[:w])
print()
``` | output | 1 | 73,293 | 23 | 146,587 |
Provide a correct Python 3 solution for this coding contest problem.
Draw a chessboard which has a height of H cm and a width of W cm. For example, the following figure shows a chessboard which has a height of 6 cm and a width of 10 cm.
.#.#.#.#.
.#.#.#.#.#
.#.#.#.#.
.#.#.#.#.#
.#.#.#.#.
.#.#.#.#.#
Note that the top left corner should be drawn by '#'.
Constraints
* 1 ≤ H ≤ 300
* 1 ≤ W ≤ 300
Input
The input consists of multiple datasets. Each dataset consists of two integers H and W separated by a single space.
The input ends with two 0 (when both H and W are zero).
Output
For each dataset, print the chessboard made of '#' and '.'.
Print a blank line after each dataset.
Example
Input
3 4
5 6
3 3
2 2
1 1
0 0
Output
#.#.
.#.#
#.#.
#.#.#.
.#.#.#
#.#.#.
.#.#.#
#.#.#.
#.#
.#.
#.#
#.
.#
# | instruction | 0 | 73,294 | 23 | 146,588 |
"Correct Solution:
```
while True:
H,W = map(int, input().split())
if H == 0 and W == 0:
break
for i in range(H):
for j in range(W):
if(i+j)%2 == 0:
print('#',end="")
else:
print('.',end="")
print()
print()
``` | output | 1 | 73,294 | 23 | 146,589 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Draw a chessboard which has a height of H cm and a width of W cm. For example, the following figure shows a chessboard which has a height of 6 cm and a width of 10 cm.
.#.#.#.#.
.#.#.#.#.#
.#.#.#.#.
.#.#.#.#.#
.#.#.#.#.
.#.#.#.#.#
Note that the top left corner should be drawn by '#'.
Constraints
* 1 ≤ H ≤ 300
* 1 ≤ W ≤ 300
Input
The input consists of multiple datasets. Each dataset consists of two integers H and W separated by a single space.
The input ends with two 0 (when both H and W are zero).
Output
For each dataset, print the chessboard made of '#' and '.'.
Print a blank line after each dataset.
Example
Input
3 4
5 6
3 3
2 2
1 1
0 0
Output
#.#.
.#.#
#.#.
#.#.#.
.#.#.#
#.#.#.
.#.#.#
#.#.#.
#.#
.#.
#.#
#.
.#
#
Submitted Solution:
```
# coding: utf-8
board = ['#', '.']
while True:
h, w = map(int, input().split())
if h == 0 and w == 0:
exit()
for row in range(h):
for column in range(w):
print(board[(row + column) % 2], end='')
print()
print()
``` | instruction | 0 | 73,295 | 23 | 146,590 |
Yes | output | 1 | 73,295 | 23 | 146,591 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Draw a chessboard which has a height of H cm and a width of W cm. For example, the following figure shows a chessboard which has a height of 6 cm and a width of 10 cm.
.#.#.#.#.
.#.#.#.#.#
.#.#.#.#.
.#.#.#.#.#
.#.#.#.#.
.#.#.#.#.#
Note that the top left corner should be drawn by '#'.
Constraints
* 1 ≤ H ≤ 300
* 1 ≤ W ≤ 300
Input
The input consists of multiple datasets. Each dataset consists of two integers H and W separated by a single space.
The input ends with two 0 (when both H and W are zero).
Output
For each dataset, print the chessboard made of '#' and '.'.
Print a blank line after each dataset.
Example
Input
3 4
5 6
3 3
2 2
1 1
0 0
Output
#.#.
.#.#
#.#.
#.#.#.
.#.#.#
#.#.#.
.#.#.#
#.#.#.
#.#
.#.
#.#
#.
.#
#
Submitted Solution:
```
while True:
h, w = map(int, input().split())
if h == 0 and w == 0:
break
for i in range(h):
for j in range(w):
print('#.'[(i + j) & 1], end='')
print()
print()
``` | instruction | 0 | 73,296 | 23 | 146,592 |
Yes | output | 1 | 73,296 | 23 | 146,593 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Draw a chessboard which has a height of H cm and a width of W cm. For example, the following figure shows a chessboard which has a height of 6 cm and a width of 10 cm.
.#.#.#.#.
.#.#.#.#.#
.#.#.#.#.
.#.#.#.#.#
.#.#.#.#.
.#.#.#.#.#
Note that the top left corner should be drawn by '#'.
Constraints
* 1 ≤ H ≤ 300
* 1 ≤ W ≤ 300
Input
The input consists of multiple datasets. Each dataset consists of two integers H and W separated by a single space.
The input ends with two 0 (when both H and W are zero).
Output
For each dataset, print the chessboard made of '#' and '.'.
Print a blank line after each dataset.
Example
Input
3 4
5 6
3 3
2 2
1 1
0 0
Output
#.#.
.#.#
#.#.
#.#.#.
.#.#.#
#.#.#.
.#.#.#
#.#.#.
#.#
.#.
#.#
#.
.#
#
Submitted Solution:
```
while True:
H, W = map(int, input().split())
if W == 0 and H == 0:
break
a_array = "#." * max(H, W)
for i in range(H):
print (a_array[i % 2: W + i % 2])
print ()
``` | instruction | 0 | 73,297 | 23 | 146,594 |
Yes | output | 1 | 73,297 | 23 | 146,595 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Draw a chessboard which has a height of H cm and a width of W cm. For example, the following figure shows a chessboard which has a height of 6 cm and a width of 10 cm.
.#.#.#.#.
.#.#.#.#.#
.#.#.#.#.
.#.#.#.#.#
.#.#.#.#.
.#.#.#.#.#
Note that the top left corner should be drawn by '#'.
Constraints
* 1 ≤ H ≤ 300
* 1 ≤ W ≤ 300
Input
The input consists of multiple datasets. Each dataset consists of two integers H and W separated by a single space.
The input ends with two 0 (when both H and W are zero).
Output
For each dataset, print the chessboard made of '#' and '.'.
Print a blank line after each dataset.
Example
Input
3 4
5 6
3 3
2 2
1 1
0 0
Output
#.#.
.#.#
#.#.
#.#.#.
.#.#.#
#.#.#.
.#.#.#
#.#.#.
#.#
.#.
#.#
#.
.#
#
Submitted Solution:
```
while True:
H,W=[int(i) for i in input().split(" ")]
if H==W==0:
break
for h in range(H):
s="#" if h%2==0 else "."
for w in range(W-1):
s+="#" if s[-1]=="." else "."
print(s)
print()
``` | instruction | 0 | 73,298 | 23 | 146,596 |
Yes | output | 1 | 73,298 | 23 | 146,597 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Draw a chessboard which has a height of H cm and a width of W cm. For example, the following figure shows a chessboard which has a height of 6 cm and a width of 10 cm.
.#.#.#.#.
.#.#.#.#.#
.#.#.#.#.
.#.#.#.#.#
.#.#.#.#.
.#.#.#.#.#
Note that the top left corner should be drawn by '#'.
Constraints
* 1 ≤ H ≤ 300
* 1 ≤ W ≤ 300
Input
The input consists of multiple datasets. Each dataset consists of two integers H and W separated by a single space.
The input ends with two 0 (when both H and W are zero).
Output
For each dataset, print the chessboard made of '#' and '.'.
Print a blank line after each dataset.
Example
Input
3 4
5 6
3 3
2 2
1 1
0 0
Output
#.#.
.#.#
#.#.
#.#.#.
.#.#.#
#.#.#.
.#.#.#
#.#.#.
#.#
.#.
#.#
#.
.#
#
Submitted Solution:
```
pattern = {
0:["#", "."],
1:[".", "#"],
}
while True:
(H, W) = [int(i) for i in input().split()]
if H == W == 0:
break
for i in range(H):
p = pattern[i % 2]
print ("join([p[i % 2] for i in range(W)]))
print()
``` | instruction | 0 | 73,299 | 23 | 146,598 |
No | output | 1 | 73,299 | 23 | 146,599 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Draw a chessboard which has a height of H cm and a width of W cm. For example, the following figure shows a chessboard which has a height of 6 cm and a width of 10 cm.
.#.#.#.#.
.#.#.#.#.#
.#.#.#.#.
.#.#.#.#.#
.#.#.#.#.
.#.#.#.#.#
Note that the top left corner should be drawn by '#'.
Constraints
* 1 ≤ H ≤ 300
* 1 ≤ W ≤ 300
Input
The input consists of multiple datasets. Each dataset consists of two integers H and W separated by a single space.
The input ends with two 0 (when both H and W are zero).
Output
For each dataset, print the chessboard made of '#' and '.'.
Print a blank line after each dataset.
Example
Input
3 4
5 6
3 3
2 2
1 1
0 0
Output
#.#.
.#.#
#.#.
#.#.#.
.#.#.#
#.#.#.
.#.#.#
#.#.#.
#.#
.#.
#.#
#.
.#
#
Submitted Solution:
```
while(True):
H, W = map(int, input().split())
if(H == 0 and W == 0):
break
else:
fir = "#"
sec = "."
for i in range(H):
for k in range(W):
if k%2:
print(fir, end = "")
else:
print(sec, end = "")
print("")
tmp = fir
fir = sec
sec = tmp
print("")
``` | instruction | 0 | 73,300 | 23 | 146,600 |
No | output | 1 | 73,300 | 23 | 146,601 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Draw a chessboard which has a height of H cm and a width of W cm. For example, the following figure shows a chessboard which has a height of 6 cm and a width of 10 cm.
.#.#.#.#.
.#.#.#.#.#
.#.#.#.#.
.#.#.#.#.#
.#.#.#.#.
.#.#.#.#.#
Note that the top left corner should be drawn by '#'.
Constraints
* 1 ≤ H ≤ 300
* 1 ≤ W ≤ 300
Input
The input consists of multiple datasets. Each dataset consists of two integers H and W separated by a single space.
The input ends with two 0 (when both H and W are zero).
Output
For each dataset, print the chessboard made of '#' and '.'.
Print a blank line after each dataset.
Example
Input
3 4
5 6
3 3
2 2
1 1
0 0
Output
#.#.
.#.#
#.#.
#.#.#.
.#.#.#
#.#.#.
.#.#.#
#.#.#.
#.#
.#.
#.#
#.
.#
#
Submitted Solution:
```
# Print a Chessboard
end = 0
while end == 0:
rectangle = [int(i) for i in input().rstrip().split()]
height = rectangle[0]
width = rectangle[1]
if [height, width] == [0, 0]:
end += 1
else:
for i in range(height):
for j in range(width):
sum = i + j
if sum % 2 == 0:
print('#', end = "")
elif sum % 2 == 1:
print('.', end = "")
print()
``` | instruction | 0 | 73,301 | 23 | 146,602 |
No | output | 1 | 73,301 | 23 | 146,603 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Draw a chessboard which has a height of H cm and a width of W cm. For example, the following figure shows a chessboard which has a height of 6 cm and a width of 10 cm.
.#.#.#.#.
.#.#.#.#.#
.#.#.#.#.
.#.#.#.#.#
.#.#.#.#.
.#.#.#.#.#
Note that the top left corner should be drawn by '#'.
Constraints
* 1 ≤ H ≤ 300
* 1 ≤ W ≤ 300
Input
The input consists of multiple datasets. Each dataset consists of two integers H and W separated by a single space.
The input ends with two 0 (when both H and W are zero).
Output
For each dataset, print the chessboard made of '#' and '.'.
Print a blank line after each dataset.
Example
Input
3 4
5 6
3 3
2 2
1 1
0 0
Output
#.#.
.#.#
#.#.
#.#.#.
.#.#.#
#.#.#.
.#.#.#
#.#.#.
#.#
.#.
#.#
#.
.#
#
Submitted Solution:
```
while True:
H,W = map(int,input().split())
if H==W==0:
break
num = H * W
for i in range(0,num):
if i % 2 == 0:
for j in range(0,W):
if j % 2 == 0:
print("#")
if j % 2 == 1:
print(".")
if i % 2 == 1:
for j in range(0,W):
if j % 2 == 0:
print(".")
if j % 2 == 1:
print("#")
print()
``` | instruction | 0 | 73,302 | 23 | 146,604 |
No | output | 1 | 73,302 | 23 | 146,605 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Little Petya very much likes rectangles and especially squares. Recently he has received 8 points on the plane as a gift from his mother. The points are pairwise distinct. Petya decided to split them into two sets each containing 4 points so that the points from the first set lay at the vertexes of some square and the points from the second set lay at the vertexes of a rectangle. Each point of initial 8 should belong to exactly one set. It is acceptable for a rectangle from the second set was also a square. If there are several partitions, Petya will be satisfied by any of them. Help him find such partition. Note that the rectangle and the square from the partition should have non-zero areas. The sides of the figures do not have to be parallel to the coordinate axes, though it might be the case.
Input
You are given 8 pairs of integers, a pair per line — the coordinates of the points Petya has. The absolute value of all coordinates does not exceed 104. It is guaranteed that no two points coincide.
Output
Print in the first output line "YES" (without the quotes), if the desired partition exists. In the second line output 4 space-separated numbers — point indexes from the input, which lie at the vertexes of the square. The points are numbered starting from 1. The numbers can be printed in any order. In the third line print the indexes of points lying at the vertexes of a rectangle in the similar format. All printed numbers should be pairwise distinct.
If the required partition does not exist, the first line should contain the word "NO" (without the quotes), after which no output is needed.
Examples
Input
0 0
10 11
10 0
0 11
1 1
2 2
2 1
1 2
Output
YES
5 6 7 8
1 2 3 4
Input
0 0
1 1
2 2
3 3
4 4
5 5
6 6
7 7
Output
NO
Input
0 0
4 4
4 0
0 4
1 2
2 3
3 2
2 1
Output
YES
1 2 3 4
5 6 7 8
Note
Pay attention to the third example: the figures do not necessarily have to be parallel to the coordinate axes. | instruction | 0 | 73,490 | 23 | 146,980 |
Tags: brute force, geometry, math
Correct Solution:
```
from sys import stdin, stdout
from itertools import permutations as p
from itertools import combinations as com
def sqr_or_rec(a, b, c, d):
x = ((a[1] - b[1])**2 + (a[0] - b[0]) **2)
y = ((c[1] - b[1])**2 + (c[0] - b[0]) **2)
w = ((c[1] - d[1])**2 + (c[0] - d[0]) **2)
z = ((a[1] - d[1])**2 + (a[0] - d[0]) **2)
hypo1 = ((a[1] - c[1])**2 + (a[0] - c[0]) **2)
hypo2 = ((b[1] - d[1])**2 + (b[0] - d[0]) **2)
if len(set([x,y,w,z])) == 1 and hypo1 == x + y and hypo2 == hypo1 and hypo2 == w + z:
return 0
elif len(set([x,y,w,z])) == 2 and hypo1 == x + y and hypo2 == hypo1 and hypo2 == w + z:
return 1
else:
return -1
x = []
for i in range(8):
l, m = map(int, input().split())
x.append([l, m])
sqr, rec = [], []
for g in p(list(range(8)), 4):
a, b, c, d = x[g[0]], x[g[1]], x[g[2]], x[g[3]]
if sqr_or_rec(a,b,c,d) == 0:
y = sorted(g)
if y not in sqr:
sqr.append(y)
elif sqr_or_rec(a,b,c,d) == 1:
y = sorted(g)
if y not in rec:
rec.append(y)
if len(sqr) == 0 or (len(sqr) == 1 and len(rec) == 0):
print("NO")
else:
if len(sqr) >= 2:
for q in com(sqr, 2):
if len(set(q[0] + q[1])) == 8:
print("YES")
for k in sqr[0]:
stdout.write(str(k+1)+" ")
print()
for k in sqr[1]:
stdout.write(str(k+1)+" ")
print()
exit()
for q in rec:
if len(set(q + sqr[0])) == 8:
print("YES")
for k in sqr[0]:
stdout.write(str(k+1)+" ")
print()
for k in q:
stdout.write(str(k+1)+" ")
print()
exit()
print("NO")
# Made By Mostafa_Khaled
``` | output | 1 | 73,490 | 23 | 146,981 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Little Petya very much likes rectangles and especially squares. Recently he has received 8 points on the plane as a gift from his mother. The points are pairwise distinct. Petya decided to split them into two sets each containing 4 points so that the points from the first set lay at the vertexes of some square and the points from the second set lay at the vertexes of a rectangle. Each point of initial 8 should belong to exactly one set. It is acceptable for a rectangle from the second set was also a square. If there are several partitions, Petya will be satisfied by any of them. Help him find such partition. Note that the rectangle and the square from the partition should have non-zero areas. The sides of the figures do not have to be parallel to the coordinate axes, though it might be the case.
Input
You are given 8 pairs of integers, a pair per line — the coordinates of the points Petya has. The absolute value of all coordinates does not exceed 104. It is guaranteed that no two points coincide.
Output
Print in the first output line "YES" (without the quotes), if the desired partition exists. In the second line output 4 space-separated numbers — point indexes from the input, which lie at the vertexes of the square. The points are numbered starting from 1. The numbers can be printed in any order. In the third line print the indexes of points lying at the vertexes of a rectangle in the similar format. All printed numbers should be pairwise distinct.
If the required partition does not exist, the first line should contain the word "NO" (without the quotes), after which no output is needed.
Examples
Input
0 0
10 11
10 0
0 11
1 1
2 2
2 1
1 2
Output
YES
5 6 7 8
1 2 3 4
Input
0 0
1 1
2 2
3 3
4 4
5 5
6 6
7 7
Output
NO
Input
0 0
4 4
4 0
0 4
1 2
2 3
3 2
2 1
Output
YES
1 2 3 4
5 6 7 8
Note
Pay attention to the third example: the figures do not necessarily have to be parallel to the coordinate axes. | instruction | 0 | 73,491 | 23 | 146,982 |
Tags: brute force, geometry, math
Correct Solution:
```
from itertools import permutations as p
l = [list(map(int, input().split())) + [_] for _ in range(1, 9)]
def dist(a, b):
return (a[0] - b[0]) ** 2 + (a[1] - b[1]) ** 2
def rect(a, b, c, d):
return dist(a, b) == dist(c, d) and dist(a, c) == dist(b, d) and dist(a, d) == dist(b, c) and dist(a, b) * dist(b, c) != 0
def sq(a, b, c, d):
return rect(a, b, c, d) and dist(a, b) == dist(b, c)
for t in p(l):
if sq(*t[:4]) and rect(*t[4:]):
print("YES")
print(' '.join([str(_[2]) for _ in t[:4]]))
print(' '.join([str(_[2]) for _ in t[4:]]))
exit()
print("NO")
``` | output | 1 | 73,491 | 23 | 146,983 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Little Petya very much likes rectangles and especially squares. Recently he has received 8 points on the plane as a gift from his mother. The points are pairwise distinct. Petya decided to split them into two sets each containing 4 points so that the points from the first set lay at the vertexes of some square and the points from the second set lay at the vertexes of a rectangle. Each point of initial 8 should belong to exactly one set. It is acceptable for a rectangle from the second set was also a square. If there are several partitions, Petya will be satisfied by any of them. Help him find such partition. Note that the rectangle and the square from the partition should have non-zero areas. The sides of the figures do not have to be parallel to the coordinate axes, though it might be the case.
Input
You are given 8 pairs of integers, a pair per line — the coordinates of the points Petya has. The absolute value of all coordinates does not exceed 104. It is guaranteed that no two points coincide.
Output
Print in the first output line "YES" (without the quotes), if the desired partition exists. In the second line output 4 space-separated numbers — point indexes from the input, which lie at the vertexes of the square. The points are numbered starting from 1. The numbers can be printed in any order. In the third line print the indexes of points lying at the vertexes of a rectangle in the similar format. All printed numbers should be pairwise distinct.
If the required partition does not exist, the first line should contain the word "NO" (without the quotes), after which no output is needed.
Examples
Input
0 0
10 11
10 0
0 11
1 1
2 2
2 1
1 2
Output
YES
5 6 7 8
1 2 3 4
Input
0 0
1 1
2 2
3 3
4 4
5 5
6 6
7 7
Output
NO
Input
0 0
4 4
4 0
0 4
1 2
2 3
3 2
2 1
Output
YES
1 2 3 4
5 6 7 8
Note
Pay attention to the third example: the figures do not necessarily have to be parallel to the coordinate axes. | instruction | 0 | 73,492 | 23 | 146,984 |
Tags: brute force, geometry, math
Correct Solution:
```
from itertools import combinations, permutations
from math import sqrt
EPS = 1e-6
def sd(x, y):
return ((x[0] - y[0])**2 + (x[1] - y[1])**2)
def is_square(points):
for perm in permutations(points):
if sd(perm[0], perm[1]) == sd(perm[1], perm[2]) == \
sd(perm[2], perm[3]) == sd(perm[3], perm[0]) and \
sd(perm[0], perm[2]) == 2 * sd(perm[0], perm[1]) and \
sd(perm[1], perm[3]) == 2 * sd(perm[0], perm[1]) and \
sqrt(sd(perm[0], perm[1])) * sqrt(sd(perm[1], perm[2])) > EPS:
return True
return False
def is_rect(points):
for perm in permutations(points):
if sd(perm[0], perm[1]) == sd(perm[2], perm[3]) and \
sd(perm[1], perm[2]) == sd(perm[3], perm[0]) and \
sd(perm[0], perm[2]) == sd(perm[0], perm[1]) + sd(perm[1], perm[2]) and \
sd(perm[1], perm[3]) == sd(perm[1], perm[2]) + sd(perm[2], perm[3]) and \
sqrt(sd(perm[0], perm[1])) * sqrt(sd(perm[1], perm[2])) > EPS:
return True
return False
AMOUNT = 8
points = []
for _ in range(AMOUNT):
points += [list(map(int, input().split()))]
found = False
to_choose = list(range(AMOUNT))
for for_square in combinations(to_choose, r=4):
square, rect = [], []
for i in range(AMOUNT):
if i in for_square:
square += [points[i]]
else:
rect += [points[i]]
if is_square(square) and is_rect(rect):
print("YES")
print(' '.join(map(lambda x: str(x + 1), for_square)))
print(' '.join(map(lambda x: str(x + 1), [y for y in range(AMOUNT) if y not in for_square])))
found = True
break
if not found:
print("NO")
``` | output | 1 | 73,492 | 23 | 146,985 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Little Petya very much likes rectangles and especially squares. Recently he has received 8 points on the plane as a gift from his mother. The points are pairwise distinct. Petya decided to split them into two sets each containing 4 points so that the points from the first set lay at the vertexes of some square and the points from the second set lay at the vertexes of a rectangle. Each point of initial 8 should belong to exactly one set. It is acceptable for a rectangle from the second set was also a square. If there are several partitions, Petya will be satisfied by any of them. Help him find such partition. Note that the rectangle and the square from the partition should have non-zero areas. The sides of the figures do not have to be parallel to the coordinate axes, though it might be the case.
Input
You are given 8 pairs of integers, a pair per line — the coordinates of the points Petya has. The absolute value of all coordinates does not exceed 104. It is guaranteed that no two points coincide.
Output
Print in the first output line "YES" (without the quotes), if the desired partition exists. In the second line output 4 space-separated numbers — point indexes from the input, which lie at the vertexes of the square. The points are numbered starting from 1. The numbers can be printed in any order. In the third line print the indexes of points lying at the vertexes of a rectangle in the similar format. All printed numbers should be pairwise distinct.
If the required partition does not exist, the first line should contain the word "NO" (without the quotes), after which no output is needed.
Examples
Input
0 0
10 11
10 0
0 11
1 1
2 2
2 1
1 2
Output
YES
5 6 7 8
1 2 3 4
Input
0 0
1 1
2 2
3 3
4 4
5 5
6 6
7 7
Output
NO
Input
0 0
4 4
4 0
0 4
1 2
2 3
3 2
2 1
Output
YES
1 2 3 4
5 6 7 8
Note
Pay attention to the third example: the figures do not necessarily have to be parallel to the coordinate axes. | instruction | 0 | 73,493 | 23 | 146,986 |
Tags: brute force, geometry, math
Correct Solution:
```
import sys
import math
MAX = 100000000000000000000000
def is_inverse(a, b):
if b == 0:
return a == MAX or a == -MAX
if a == 0:
return b == MAX or b == -MAX
if a == MAX or a == -MAX:
return b == 0 or b == -0
if b == MAX or b == -MAX:
return a == 0 or a == -0
return a == -(1/b)
def slope(a, b):
a -= 1
b -= 1
y1, y2 = pts[a][1], pts[b][1]
x1, x2 = pts[a][0], pts[b][0]
dx = x1 - x2
dy = y1 - y2
if dx == 0:
return MAX
return dy / dx
def dist(a, b):
a -= 1
b -= 1
y1, y2 = pts[a][1], pts[b][1]
x1, x2 = pts[a][0], pts[b][0]
dx = x1 - x2
dy = y1 - y2
return dx**2 + dy**2
def is_valid_square(vals):
first, second, third, fourth = vals[0], vals[1], vals[2], vals[3]
points = [first, second, third, fourth]
for i in points:
for j in [x for x in points if x != i]:
for k in [z for z in points if z != i and z != j]:
for l in [y for y in points if y != j and y != k and y != i]:
if slope(i, j) == slope(k, l) and \
is_inverse(slope(i, k), slope(i, j)) and \
dist(i, j) == dist(i, k) == dist(k, l) == dist(j, l):
return True
return False
def is_valid_rectangle(vals):
first, second, third, fourth = vals[0], vals[1], vals[2], vals[3]
points = [first, second, third, fourth]
for i in points:
for j in [x for x in points if x != i]:
for k in [z for z in points if z != i and z != j]:
for l in [y for y in points if y != j and y != k and y != i]:
if slope(i, j) == slope(k, l) and \
is_inverse(slope(i, k), slope(i, j)) and \
dist(i, j) == dist(k, l) and dist(i, k) == dist(j, l):
return True
return False
def partition_pts(cur_set):
if len(cur_set) == 4:
square_set, rect_set = cur_set, [x for x in range(1, 9) if x not in cur_set]
if is_valid_square(cur_set) and \
is_valid_rectangle(rect_set):
print("YES")
r1, r2 = "", ""
for x in cur_set:
r1 += str(x) + " "
for x in rect_set:
r2 += str(x) + " "
print(r1)
print(r2)
sys.exit(0)
else:
return
for i in [x for x in range(1, 9) if x not in cur_set]:
partition_pts(cur_set + [i])
pts = []
for i in range(8):
pts.append([int(x) for x in sys.stdin.readline().split()])
partition_pts([])
print("NO")
``` | output | 1 | 73,493 | 23 | 146,987 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Little Petya very much likes rectangles and especially squares. Recently he has received 8 points on the plane as a gift from his mother. The points are pairwise distinct. Petya decided to split them into two sets each containing 4 points so that the points from the first set lay at the vertexes of some square and the points from the second set lay at the vertexes of a rectangle. Each point of initial 8 should belong to exactly one set. It is acceptable for a rectangle from the second set was also a square. If there are several partitions, Petya will be satisfied by any of them. Help him find such partition. Note that the rectangle and the square from the partition should have non-zero areas. The sides of the figures do not have to be parallel to the coordinate axes, though it might be the case.
Input
You are given 8 pairs of integers, a pair per line — the coordinates of the points Petya has. The absolute value of all coordinates does not exceed 104. It is guaranteed that no two points coincide.
Output
Print in the first output line "YES" (without the quotes), if the desired partition exists. In the second line output 4 space-separated numbers — point indexes from the input, which lie at the vertexes of the square. The points are numbered starting from 1. The numbers can be printed in any order. In the third line print the indexes of points lying at the vertexes of a rectangle in the similar format. All printed numbers should be pairwise distinct.
If the required partition does not exist, the first line should contain the word "NO" (without the quotes), after which no output is needed.
Examples
Input
0 0
10 11
10 0
0 11
1 1
2 2
2 1
1 2
Output
YES
5 6 7 8
1 2 3 4
Input
0 0
1 1
2 2
3 3
4 4
5 5
6 6
7 7
Output
NO
Input
0 0
4 4
4 0
0 4
1 2
2 3
3 2
2 1
Output
YES
1 2 3 4
5 6 7 8
Note
Pay attention to the third example: the figures do not necessarily have to be parallel to the coordinate axes. | instruction | 0 | 73,494 | 23 | 146,988 |
Tags: brute force, geometry, math
Correct Solution:
```
# from decimal import *
# getcontext().prec=16
# from math import sqrt
# from scipy.special import binom
# from collections import defaultdict
from math import sin,pi
from copy import deepcopy
def check(a,b,c):
liste=[[a,b,c],[b,c,a],[b,a,c]]
for element in liste:
a,b,c=element
if (b[0]-a[0])*(c[0]-b[0])+(b[1]-a[1])*(c[1]-b[1])==0:
return True
return False
def check_square(liste):
a,b,c=liste
liste=[[a,b,c],[b,c,a],[b,a,c]]
for element in liste:
a,b,c=element
if ( (b[0]-a[0])**2+(b[1]-a[1])**2 )==( (b[0]-c[0])**2+(b[1]-c[1])**2 ):
return True
return False
tempo=[0 for i in range(8)]
perm=[]
for i in range(5):
tempo[i]=1
for j in range(i+1,6):
tempo[j]=1
for k in range(j+1,7):
tempo[k]=1
for l in range(k+1,8):
tempo[l]=1
copy=deepcopy(tempo)
perm.append(copy)
tempo[l]=0
tempo[k]=0
tempo[j]=0
tempo[i]=0
entry=[]
for i in range(8):
x,y=list(map(int,input().split(" ")))
entry.append((x,y))
bool=False
for permutation in perm:
first_set=[]
second_set=[]
for i in range(8):
if permutation[i]==1:
first_set.append(entry[i])
else:
second_set.append(entry[i])
a,b,c,d=first_set
w,x,y,z=second_set
if check(a,b,c) and check(b,c,d) and check(c,d,a):
if check(w,x,y) and check(x,y,z) and check(y,z,w):
if check_square(first_set[:-1]):
bool=True
elif check_square(second_set[:-1]):
first_set,second_set=second_set,first_set
bool=True
if bool:
break
if not bool:
print("NO")
else:
index_square=[]
index_rectangle=[]
for element in first_set:
index_square.append(str(entry.index(element)+1))
for element in second_set:
index_rectangle.append(str(entry.index(element)+1))
print('YES\n'+(" ").join(index_square)+'\n'+(' ').join(index_rectangle))
``` | output | 1 | 73,494 | 23 | 146,989 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Little Petya very much likes rectangles and especially squares. Recently he has received 8 points on the plane as a gift from his mother. The points are pairwise distinct. Petya decided to split them into two sets each containing 4 points so that the points from the first set lay at the vertexes of some square and the points from the second set lay at the vertexes of a rectangle. Each point of initial 8 should belong to exactly one set. It is acceptable for a rectangle from the second set was also a square. If there are several partitions, Petya will be satisfied by any of them. Help him find such partition. Note that the rectangle and the square from the partition should have non-zero areas. The sides of the figures do not have to be parallel to the coordinate axes, though it might be the case.
Input
You are given 8 pairs of integers, a pair per line — the coordinates of the points Petya has. The absolute value of all coordinates does not exceed 104. It is guaranteed that no two points coincide.
Output
Print in the first output line "YES" (without the quotes), if the desired partition exists. In the second line output 4 space-separated numbers — point indexes from the input, which lie at the vertexes of the square. The points are numbered starting from 1. The numbers can be printed in any order. In the third line print the indexes of points lying at the vertexes of a rectangle in the similar format. All printed numbers should be pairwise distinct.
If the required partition does not exist, the first line should contain the word "NO" (without the quotes), after which no output is needed.
Examples
Input
0 0
10 11
10 0
0 11
1 1
2 2
2 1
1 2
Output
YES
5 6 7 8
1 2 3 4
Input
0 0
1 1
2 2
3 3
4 4
5 5
6 6
7 7
Output
NO
Input
0 0
4 4
4 0
0 4
1 2
2 3
3 2
2 1
Output
YES
1 2 3 4
5 6 7 8
Note
Pay attention to the third example: the figures do not necessarily have to be parallel to the coordinate axes. | instruction | 0 | 73,495 | 23 | 146,990 |
Tags: brute force, geometry, math
Correct Solution:
```
import math
from itertools import permutations
def dist_cube(a, b):
return (a[0] - b[0]) ** 2 + (a[1] - b[1]) ** 2
def area(x1, y1, x2, y2, x3, y3):
return abs((x1 * (y2 - y3) + x2 * (y3 - y1) + x3 * (y1 - y2)))
def in_triangle(x1, y1, x2, y2, x3, y3, x, y):
A = area(x1, y1, x2, y2, x3, y3)
A1 = area(x, y, x2, y2, x3, y3)
A2 = area(x1, y1, x, y, x3, y3)
A3 = area(x1, y1, x2, y2, x, y)
#print(abs(A - (A1 + A2 + A3)))
return abs(A - (A1 + A2 + A3)) < 0.000001
class CodeforcesTask135BSolution:
def __init__(self):
self.result = ''
self.points = []
def read_input(self):
for x in range(8):
self.points.append([int(y) for y in input().split(" ")])
def process_task(self):
sol = []
for perm in permutations([x + 1 for x in range(8)]):
sq = []
for p in perm[:4]:
sq.append(self.points[p - 1])
rc = []
for p in perm[4:]:
rc.append(self.points[p - 1])
sq_dists = [dist_cube(sq[0], sq[1]), dist_cube(sq[1], sq[2]), dist_cube(sq[2], sq[3]),
dist_cube(sq[3], sq[1]), dist_cube(sq[0], sq[2]), dist_cube(sq[0], sq[3])]
rc_dists = [dist_cube(rc[0], rc[1]), dist_cube(rc[1], rc[2]), dist_cube(rc[2], rc[3]),
dist_cube(rc[3], rc[1]), dist_cube(rc[0], rc[2]), dist_cube(rc[0], rc[3])]
sq_cnt = [sq_dists.count(x) for x in list(set(sq_dists))]
sq_cnt.sort(reverse=True)
rc_cnt = [rc_dists.count(x) for x in list(set(rc_dists))]
rc_cnt.sort(reverse=True)
if len(set(sq_dists)) == 2 and sq_cnt[0] == 4 and sq_cnt[1] == 2 and ((rc_cnt == [4, 2] and 2 == len(set(rc_dists))) or (len(set(rc_dists)) == 3 and rc_cnt == [2, 2, 2])):
if len(set(rc_dists)) == 3 and rc_cnt == [2, 2, 2]:
do_ = False
for j in range(4):
rcc = rc[:j] + rc[j + 1:]
#print(rcc, rc[j], in_triangle(rcc[0][0], rcc[0][1], rcc[1][0], rcc[1][1], rcc[2][0], rcc[2][1], rc[j][0], rc[j][1]))
if in_triangle(rcc[0][0], rcc[0][1], rcc[1][0], rcc[1][1], rcc[2][0], rcc[2][1], rc[j][0], rc[j][1]):
do_ = True
break
if not do_:
sol = perm
break
else:
sol = perm
break
if sol:
self.result = "YES\n{0}\n{1}".format(" ".join([str(x) for x in sol[:4]]), " ".join([str(x) for x in sol[4:]]))
else:
self.result = "NO"
def get_result(self):
return self.result
if __name__ == "__main__":
Solution = CodeforcesTask135BSolution()
Solution.read_input()
Solution.process_task()
print(Solution.get_result())
``` | output | 1 | 73,495 | 23 | 146,991 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Little Petya very much likes rectangles and especially squares. Recently he has received 8 points on the plane as a gift from his mother. The points are pairwise distinct. Petya decided to split them into two sets each containing 4 points so that the points from the first set lay at the vertexes of some square and the points from the second set lay at the vertexes of a rectangle. Each point of initial 8 should belong to exactly one set. It is acceptable for a rectangle from the second set was also a square. If there are several partitions, Petya will be satisfied by any of them. Help him find such partition. Note that the rectangle and the square from the partition should have non-zero areas. The sides of the figures do not have to be parallel to the coordinate axes, though it might be the case.
Input
You are given 8 pairs of integers, a pair per line — the coordinates of the points Petya has. The absolute value of all coordinates does not exceed 104. It is guaranteed that no two points coincide.
Output
Print in the first output line "YES" (without the quotes), if the desired partition exists. In the second line output 4 space-separated numbers — point indexes from the input, which lie at the vertexes of the square. The points are numbered starting from 1. The numbers can be printed in any order. In the third line print the indexes of points lying at the vertexes of a rectangle in the similar format. All printed numbers should be pairwise distinct.
If the required partition does not exist, the first line should contain the word "NO" (without the quotes), after which no output is needed.
Examples
Input
0 0
10 11
10 0
0 11
1 1
2 2
2 1
1 2
Output
YES
5 6 7 8
1 2 3 4
Input
0 0
1 1
2 2
3 3
4 4
5 5
6 6
7 7
Output
NO
Input
0 0
4 4
4 0
0 4
1 2
2 3
3 2
2 1
Output
YES
1 2 3 4
5 6 7 8
Note
Pay attention to the third example: the figures do not necessarily have to be parallel to the coordinate axes. | instruction | 0 | 73,496 | 23 | 146,992 |
Tags: brute force, geometry, math
Correct Solution:
```
def L(a, b): return (a[0] - b[0]) ** 2 + (a[1] - b[1]) ** 2
def fs(a, b, c, d): return L(a, b) == L(b, c) == L(c, d) == L(d, a) == L(a, c) / 2
def fr(a, b, c, d): return L(a, b) == L(c, d) and L(b, c) == L(d, a) and L(a, c) == L(b, d)
def gs(a, b, c, d): return fs(a, b, c, d) or fs(a, b, d, c) or fs(a, c, b, d)
def gr(a, b, c, d): return fr(a, b, c, d) or fr(a, b, d, c) or fr(a, c, b, d)
def hr(q): return gr(t[q[0]], t[q[1]], t[q[2]], t[q[3]])
t = [(0, 0)] + [list(map(int, input().split())) for i in range(8)]
p = {1, 2, 3, 4, 5, 6, 7, 8}
for i in range(1, 6):
for j in range(i + 1, 7):
for k in range(j + 1, 8):
for l in range(k + 1, 9):
if gs(t[i], t[j], t[k], t[l]):
q = list(p - {i, j, k, l})
if hr(q):
print('YES')
print(i, j, k, l)
print(' '.join(map(str, q)))
exit(0)
print('NO')
``` | output | 1 | 73,496 | 23 | 146,993 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Little Petya very much likes rectangles and especially squares. Recently he has received 8 points on the plane as a gift from his mother. The points are pairwise distinct. Petya decided to split them into two sets each containing 4 points so that the points from the first set lay at the vertexes of some square and the points from the second set lay at the vertexes of a rectangle. Each point of initial 8 should belong to exactly one set. It is acceptable for a rectangle from the second set was also a square. If there are several partitions, Petya will be satisfied by any of them. Help him find such partition. Note that the rectangle and the square from the partition should have non-zero areas. The sides of the figures do not have to be parallel to the coordinate axes, though it might be the case.
Input
You are given 8 pairs of integers, a pair per line — the coordinates of the points Petya has. The absolute value of all coordinates does not exceed 104. It is guaranteed that no two points coincide.
Output
Print in the first output line "YES" (without the quotes), if the desired partition exists. In the second line output 4 space-separated numbers — point indexes from the input, which lie at the vertexes of the square. The points are numbered starting from 1. The numbers can be printed in any order. In the third line print the indexes of points lying at the vertexes of a rectangle in the similar format. All printed numbers should be pairwise distinct.
If the required partition does not exist, the first line should contain the word "NO" (without the quotes), after which no output is needed.
Examples
Input
0 0
10 11
10 0
0 11
1 1
2 2
2 1
1 2
Output
YES
5 6 7 8
1 2 3 4
Input
0 0
1 1
2 2
3 3
4 4
5 5
6 6
7 7
Output
NO
Input
0 0
4 4
4 0
0 4
1 2
2 3
3 2
2 1
Output
YES
1 2 3 4
5 6 7 8
Note
Pay attention to the third example: the figures do not necessarily have to be parallel to the coordinate axes. | instruction | 0 | 73,497 | 23 | 146,994 |
Tags: brute force, geometry, math
Correct Solution:
```
from math import *
def a(x1,y1,x2,y2,x3,y3,x4,y4):
b=sorted([[x1,y1],[x2,y2],[x3,y3],[x4,y4]])
x1,y1=b[0]
x2,y2=b[1]
x3,y3=b[2]
x4,y4=b[3]
a1=sqrt((x1-x2)**2+(y1-y2)**2)
a2=sqrt((x4-x2)**2+(y4-y2)**2)
a3=sqrt((x4-x3)**2+(y4-y3)**2)
a4=sqrt((x1-x3)**2+(y1-y3)**2)
return a1==a2==a3==a4 and a1!=0 and a4!=0 and abs(abs(degrees(asin((y2-y1)/a1)-asin((y3-y1)/a4)))-90)<=10**(-8)
def b(x1,y1,x2,y2,x3,y3,x4,y4):
b=sorted([[x1,y1],[x2,y2],[x3,y3],[x4,y4]])
x1,y1=b[0]
x2,y2=b[1]
x3,y3=b[2]
x4,y4=b[3]
a1=sqrt((x1-x2)**2+(y1-y2)**2)
a2=sqrt((x4-x2)**2+(y4-y2)**2)
a3=sqrt((x4-x3)**2+(y4-y3)**2)
a4=sqrt((x1-x3)**2+(y1-y3)**2)
return a1==a3 and a2==a4 and a1!=0 and a4!=0 and abs(abs(degrees(asin((y2-y1)/a1)-asin((y3-y1)/a4)))-90)<=10**(-8)
c=[list(map(int,input().split())) for i in range(8)]
z=False
for i in range(5):
for j in range(i+1,6):
for k in range(j+1,7):
for l in range(k+1,8):
if a(*c[i]+c[j]+c[k]+c[l]):
d=[]
e=[]
for m in range(8):
if m not in [i,j,k,l]:
d+=c[m]
e.append(m+1)
if b(*d):
print('YES')
print(i+1,j+1,k+1,l+1)
print(*e)
z=True
break
if z:
break
if z:
break
if z:
break
if not(z):
print('NO')
``` | output | 1 | 73,497 | 23 | 146,995 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Little Petya very much likes rectangles and especially squares. Recently he has received 8 points on the plane as a gift from his mother. The points are pairwise distinct. Petya decided to split them into two sets each containing 4 points so that the points from the first set lay at the vertexes of some square and the points from the second set lay at the vertexes of a rectangle. Each point of initial 8 should belong to exactly one set. It is acceptable for a rectangle from the second set was also a square. If there are several partitions, Petya will be satisfied by any of them. Help him find such partition. Note that the rectangle and the square from the partition should have non-zero areas. The sides of the figures do not have to be parallel to the coordinate axes, though it might be the case.
Input
You are given 8 pairs of integers, a pair per line — the coordinates of the points Petya has. The absolute value of all coordinates does not exceed 104. It is guaranteed that no two points coincide.
Output
Print in the first output line "YES" (without the quotes), if the desired partition exists. In the second line output 4 space-separated numbers — point indexes from the input, which lie at the vertexes of the square. The points are numbered starting from 1. The numbers can be printed in any order. In the third line print the indexes of points lying at the vertexes of a rectangle in the similar format. All printed numbers should be pairwise distinct.
If the required partition does not exist, the first line should contain the word "NO" (without the quotes), after which no output is needed.
Examples
Input
0 0
10 11
10 0
0 11
1 1
2 2
2 1
1 2
Output
YES
5 6 7 8
1 2 3 4
Input
0 0
1 1
2 2
3 3
4 4
5 5
6 6
7 7
Output
NO
Input
0 0
4 4
4 0
0 4
1 2
2 3
3 2
2 1
Output
YES
1 2 3 4
5 6 7 8
Note
Pay attention to the third example: the figures do not necessarily have to be parallel to the coordinate axes.
Submitted Solution:
```
from itertools import permutations as p
l = [list(map(int, input().split())) + [_] for _ in range(1, 9)]
def dist(a, b):
return (a[0] - b[0]) ** 2 + (a[1] - b[1]) ** 2
def rect(a, b, c, d):
return dist(a, b) == dist(c, d) and dist(a, c) == dist(b, d) and dist(a, d) == dist(b, c) and dist(a, b) * dist(b, c) != 0
def sq(a, b, c, d):
# print(rect(a, b, c, d))
return rect(a, b, c, d) and dist(a, b) == dist(b, c)
for t in p(l):
if sq(*t[:4]) and rect(*t[4:]):
print("YES")
print(' '.join([str(_[2]) for _ in t[:4]]))
print(' '.join([str(_[2]) for _ in t[4:]]))
exit()
print("NO")
``` | instruction | 0 | 73,498 | 23 | 146,996 |
Yes | output | 1 | 73,498 | 23 | 146,997 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Little Petya very much likes rectangles and especially squares. Recently he has received 8 points on the plane as a gift from his mother. The points are pairwise distinct. Petya decided to split them into two sets each containing 4 points so that the points from the first set lay at the vertexes of some square and the points from the second set lay at the vertexes of a rectangle. Each point of initial 8 should belong to exactly one set. It is acceptable for a rectangle from the second set was also a square. If there are several partitions, Petya will be satisfied by any of them. Help him find such partition. Note that the rectangle and the square from the partition should have non-zero areas. The sides of the figures do not have to be parallel to the coordinate axes, though it might be the case.
Input
You are given 8 pairs of integers, a pair per line — the coordinates of the points Petya has. The absolute value of all coordinates does not exceed 104. It is guaranteed that no two points coincide.
Output
Print in the first output line "YES" (without the quotes), if the desired partition exists. In the second line output 4 space-separated numbers — point indexes from the input, which lie at the vertexes of the square. The points are numbered starting from 1. The numbers can be printed in any order. In the third line print the indexes of points lying at the vertexes of a rectangle in the similar format. All printed numbers should be pairwise distinct.
If the required partition does not exist, the first line should contain the word "NO" (without the quotes), after which no output is needed.
Examples
Input
0 0
10 11
10 0
0 11
1 1
2 2
2 1
1 2
Output
YES
5 6 7 8
1 2 3 4
Input
0 0
1 1
2 2
3 3
4 4
5 5
6 6
7 7
Output
NO
Input
0 0
4 4
4 0
0 4
1 2
2 3
3 2
2 1
Output
YES
1 2 3 4
5 6 7 8
Note
Pay attention to the third example: the figures do not necessarily have to be parallel to the coordinate axes.
Submitted Solution:
```
from itertools import combinations
def get_center(points):
center = []
for d in range(len(points[0])):
c = sum(p[d] for p in points)
center.append(c / 4)
return tuple(center)
def all_the_same(xs):
if len(xs) <= 1:
return True
first, *ys = xs
return all(y == first for y in ys)
def distance(p1, p2):
return (p1[0] - p2[0])**2 + (p1[1] - p2[1])**2
def all_distance(points):
stack = list(points)
distances = []
while len(stack) > 1:
point = stack.pop()
distances.extend([distance(point, p) for p in stack])
return distances
def is_rect(points):
center = get_center(points)
l = [distance(center, p) for p in points]
return all_the_same(l)
def is_sqr(points):
if not is_rect(points):
return False
distances = all_distance(points)
distances.sort()
return all_the_same(distances[:-2])
def main(points):
idcs = list(range(0, 8))
for group in combinations(idcs, 4):
xs = [points[i] for i in group]
ys = [points[i] for i in idcs if i not in group]
if is_rect(xs) and is_rect(ys):
if is_sqr(xs):
print('YES')
print(' '.join([str(i+1) for i in group]))
print(' '.join([str(i+1) for i in idcs if i not in group]))
break
elif is_sqr(ys):
print('YES')
print(' '.join([str(i+1) for i in idcs if i not in group]))
print(' '.join([str(i+1) for i in group]))
break
else:
print('NO')
all_points = []
for _ in range(8):
all_points.append(list(map(int, input().split())))
main(all_points)
``` | instruction | 0 | 73,499 | 23 | 146,998 |
Yes | output | 1 | 73,499 | 23 | 146,999 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Little Petya very much likes rectangles and especially squares. Recently he has received 8 points on the plane as a gift from his mother. The points are pairwise distinct. Petya decided to split them into two sets each containing 4 points so that the points from the first set lay at the vertexes of some square and the points from the second set lay at the vertexes of a rectangle. Each point of initial 8 should belong to exactly one set. It is acceptable for a rectangle from the second set was also a square. If there are several partitions, Petya will be satisfied by any of them. Help him find such partition. Note that the rectangle and the square from the partition should have non-zero areas. The sides of the figures do not have to be parallel to the coordinate axes, though it might be the case.
Input
You are given 8 pairs of integers, a pair per line — the coordinates of the points Petya has. The absolute value of all coordinates does not exceed 104. It is guaranteed that no two points coincide.
Output
Print in the first output line "YES" (without the quotes), if the desired partition exists. In the second line output 4 space-separated numbers — point indexes from the input, which lie at the vertexes of the square. The points are numbered starting from 1. The numbers can be printed in any order. In the third line print the indexes of points lying at the vertexes of a rectangle in the similar format. All printed numbers should be pairwise distinct.
If the required partition does not exist, the first line should contain the word "NO" (without the quotes), after which no output is needed.
Examples
Input
0 0
10 11
10 0
0 11
1 1
2 2
2 1
1 2
Output
YES
5 6 7 8
1 2 3 4
Input
0 0
1 1
2 2
3 3
4 4
5 5
6 6
7 7
Output
NO
Input
0 0
4 4
4 0
0 4
1 2
2 3
3 2
2 1
Output
YES
1 2 3 4
5 6 7 8
Note
Pay attention to the third example: the figures do not necessarily have to be parallel to the coordinate axes.
Submitted Solution:
```
import itertools
import math
import os
import sys
eps = 1e-8
coord = [[]] + [list(map(int, input().split())) for _ in range(8)]
idx = list(range(1, 9))
def perpendicular(v1, v2):
return sum([x * y for (x, y) in zip(v1, v2)]) < eps
def all_perpendicular(vs):
return all([perpendicular(vs[i], vs[(i+1)%4]) for i in range(4)])
def rect_sides(vs):
ls = list(map(lambda v: math.hypot(*v), vs))
return abs(ls[0] - ls[2]) < eps and abs(ls[1] - ls[3]) < eps
def square_sides(vs):
ls = list(map(lambda v: math.hypot(*v), vs))
l = ls[0]
for lx in ls:
if abs(lx - l) > eps:
return False
return True
def coords_to_vecs(cs):
return [
[cs[(i+1)%4][0] - cs[i][0], cs[(i+1)%4][1] - cs[i][1]]
for i in range(4)]
def is_square(coords):
for p in itertools.permutations(coords):
vs = coords_to_vecs(p)
if all_perpendicular(vs) and square_sides(vs):
return True
return False
def is_rect(coord):
for p in itertools.permutations(coord):
vs = coords_to_vecs(p)
if all_perpendicular(vs) and rect_sides(vs):
return True
return False
for comb in itertools.combinations(idx, 4):
fsi = list(comb)
ssi = list(set(idx) - set(comb))
fs = [coord[i] for i in fsi]
ss = [coord[i] for i in ssi]
if is_square(fs) and is_rect(ss):
print("YES")
print(' '.join(map(str, fsi)))
print(' '.join(map(str, ssi)))
sys.exit(0)
if is_square(ss) and is_rect(fs):
print("YES")
print(' '.join(map(str, ssi)))
print(' '.join(map(str, fsi)))
sys.exit(0)
print("NO")
``` | instruction | 0 | 73,500 | 23 | 147,000 |
Yes | output | 1 | 73,500 | 23 | 147,001 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Little Petya very much likes rectangles and especially squares. Recently he has received 8 points on the plane as a gift from his mother. The points are pairwise distinct. Petya decided to split them into two sets each containing 4 points so that the points from the first set lay at the vertexes of some square and the points from the second set lay at the vertexes of a rectangle. Each point of initial 8 should belong to exactly one set. It is acceptable for a rectangle from the second set was also a square. If there are several partitions, Petya will be satisfied by any of them. Help him find such partition. Note that the rectangle and the square from the partition should have non-zero areas. The sides of the figures do not have to be parallel to the coordinate axes, though it might be the case.
Input
You are given 8 pairs of integers, a pair per line — the coordinates of the points Petya has. The absolute value of all coordinates does not exceed 104. It is guaranteed that no two points coincide.
Output
Print in the first output line "YES" (without the quotes), if the desired partition exists. In the second line output 4 space-separated numbers — point indexes from the input, which lie at the vertexes of the square. The points are numbered starting from 1. The numbers can be printed in any order. In the third line print the indexes of points lying at the vertexes of a rectangle in the similar format. All printed numbers should be pairwise distinct.
If the required partition does not exist, the first line should contain the word "NO" (without the quotes), after which no output is needed.
Examples
Input
0 0
10 11
10 0
0 11
1 1
2 2
2 1
1 2
Output
YES
5 6 7 8
1 2 3 4
Input
0 0
1 1
2 2
3 3
4 4
5 5
6 6
7 7
Output
NO
Input
0 0
4 4
4 0
0 4
1 2
2 3
3 2
2 1
Output
YES
1 2 3 4
5 6 7 8
Note
Pay attention to the third example: the figures do not necessarily have to be parallel to the coordinate axes.
Submitted Solution:
```
from sys import stdin, stdout
from itertools import permutations as p
from itertools import combinations as com
def sqr_or_rec(a, b, c, d):
x = ((a[1] - b[1])**2 + (a[0] - b[0]) **2)
y = ((c[1] - b[1])**2 + (c[0] - b[0]) **2)
w = ((c[1] - d[1])**2 + (c[0] - d[0]) **2)
z = ((a[1] - d[1])**2 + (a[0] - d[0]) **2)
hypo1 = ((a[1] - c[1])**2 + (a[0] - c[0]) **2)
hypo2 = ((b[1] - d[1])**2 + (b[0] - d[0]) **2)
if len(set([x,y,w,z])) == 1 and hypo1 == x + y and hypo2 == hypo1 and hypo2 == w + z:
return 0
elif len(set([x,y,w,z])) == 2 and hypo1 == x + y and hypo2 == hypo1 and hypo2 == w + z:
return 1
else:
return -1
x = []
for i in range(8):
l, m = map(int, input().split())
x.append([l, m])
sqr, rec = [], []
for g in p(list(range(8)), 4):
a, b, c, d = x[g[0]], x[g[1]], x[g[2]], x[g[3]]
if sqr_or_rec(a,b,c,d) == 0:
y = sorted(g)
if y not in sqr:
sqr.append(y)
elif sqr_or_rec(a,b,c,d) == 1:
y = sorted(g)
if y not in rec:
rec.append(y)
if len(sqr) == 0 or (len(sqr) == 1 and len(rec) == 0):
print("NO")
else:
if len(sqr) >= 2:
for q in com(sqr, 2):
if len(set(q[0] + q[1])) == 8:
print("YES")
for k in sqr[0]:
stdout.write(str(k+1)+" ")
print()
for k in sqr[1]:
stdout.write(str(k+1)+" ")
print()
exit()
for q in rec:
if len(set(q + sqr[0])) == 8:
print("YES")
for k in sqr[0]:
stdout.write(str(k+1)+" ")
print()
for k in q:
stdout.write(str(k+1)+" ")
print()
exit()
print("NO")
``` | instruction | 0 | 73,501 | 23 | 147,002 |
Yes | output | 1 | 73,501 | 23 | 147,003 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Little Petya very much likes rectangles and especially squares. Recently he has received 8 points on the plane as a gift from his mother. The points are pairwise distinct. Petya decided to split them into two sets each containing 4 points so that the points from the first set lay at the vertexes of some square and the points from the second set lay at the vertexes of a rectangle. Each point of initial 8 should belong to exactly one set. It is acceptable for a rectangle from the second set was also a square. If there are several partitions, Petya will be satisfied by any of them. Help him find such partition. Note that the rectangle and the square from the partition should have non-zero areas. The sides of the figures do not have to be parallel to the coordinate axes, though it might be the case.
Input
You are given 8 pairs of integers, a pair per line — the coordinates of the points Petya has. The absolute value of all coordinates does not exceed 104. It is guaranteed that no two points coincide.
Output
Print in the first output line "YES" (without the quotes), if the desired partition exists. In the second line output 4 space-separated numbers — point indexes from the input, which lie at the vertexes of the square. The points are numbered starting from 1. The numbers can be printed in any order. In the third line print the indexes of points lying at the vertexes of a rectangle in the similar format. All printed numbers should be pairwise distinct.
If the required partition does not exist, the first line should contain the word "NO" (without the quotes), after which no output is needed.
Examples
Input
0 0
10 11
10 0
0 11
1 1
2 2
2 1
1 2
Output
YES
5 6 7 8
1 2 3 4
Input
0 0
1 1
2 2
3 3
4 4
5 5
6 6
7 7
Output
NO
Input
0 0
4 4
4 0
0 4
1 2
2 3
3 2
2 1
Output
YES
1 2 3 4
5 6 7 8
Note
Pay attention to the third example: the figures do not necessarily have to be parallel to the coordinate axes.
Submitted Solution:
```
# from decimal import *
# getcontext().prec=16
# from math import sqrt
# from scipy.special import binom
# from collections import defaultdict
from math import sin,pi
from copy import deepcopy
def check(a,b,c):
liste=[[a,b,c],[b,c,a],[b,a,c]]
for element in liste:
a,b,c=element
if (b[0]-a[0])*(c[0]-b[0])+(b[1]-a[1])*(c[1]-b[1])==0:
return True
return False
def check_square(liste):
a,b,c=liste
liste=[[a,b,c],[b,c,a],[b,a,c]]
for element in liste:
a,b,c=element
if ( (b[0]-a[0])**2+(b[1]-a[1])**2 )==( (b[0]-c[0])**2+(b[1]-c[1])**2 ):
return True
return False
tempo=[0 for i in range(8)]
perm=[]
for i in range(5):
tempo[i]=1
for j in range(i+1,6):
tempo[j]=1
for k in range(j+1,7):
tempo[k]=1
for l in range(k+1,8):
tempo[l]=1
copy=deepcopy(tempo)
perm.append(copy)
tempo[l]=0
tempo[k]=0
tempo[j]=0
tempo[i]=0
entry=[]
for i in range(8):
x,y=list(map(int,input().split(" ")))
entry.append((x,y))
bool=False
for permutation in perm:
first_set=[]
second_set=[]
for i in range(8):
if permutation[i]==1:
first_set.append(entry[i])
else:
second_set.append(entry[i])
a,b,c,d=first_set
w,x,y,z=second_set
if check(a,b,c) and check(b,c,d) and check(c,d,a):
if check(w,x,y) and check(x,y,z) and check(y,z,w):
if check_square(first_set[:-1]):
bool=True
elif check_square(second_set[:-1]):
first_set,second_set=second_set,first_set
bool=True
if bool:
break
if not bool:
print("No")
else:
index_square=[]
index_rectangle=[]
for element in first_set:
index_square.append(str(entry.index(element)+1))
for element in second_set:
index_rectangle.append(str(entry.index(element)+1))
print('YES\n'+(" ").join(index_square)+'\n'+(' ').join(index_rectangle))
``` | instruction | 0 | 73,502 | 23 | 147,004 |
No | output | 1 | 73,502 | 23 | 147,005 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Little Petya very much likes rectangles and especially squares. Recently he has received 8 points on the plane as a gift from his mother. The points are pairwise distinct. Petya decided to split them into two sets each containing 4 points so that the points from the first set lay at the vertexes of some square and the points from the second set lay at the vertexes of a rectangle. Each point of initial 8 should belong to exactly one set. It is acceptable for a rectangle from the second set was also a square. If there are several partitions, Petya will be satisfied by any of them. Help him find such partition. Note that the rectangle and the square from the partition should have non-zero areas. The sides of the figures do not have to be parallel to the coordinate axes, though it might be the case.
Input
You are given 8 pairs of integers, a pair per line — the coordinates of the points Petya has. The absolute value of all coordinates does not exceed 104. It is guaranteed that no two points coincide.
Output
Print in the first output line "YES" (without the quotes), if the desired partition exists. In the second line output 4 space-separated numbers — point indexes from the input, which lie at the vertexes of the square. The points are numbered starting from 1. The numbers can be printed in any order. In the third line print the indexes of points lying at the vertexes of a rectangle in the similar format. All printed numbers should be pairwise distinct.
If the required partition does not exist, the first line should contain the word "NO" (without the quotes), after which no output is needed.
Examples
Input
0 0
10 11
10 0
0 11
1 1
2 2
2 1
1 2
Output
YES
5 6 7 8
1 2 3 4
Input
0 0
1 1
2 2
3 3
4 4
5 5
6 6
7 7
Output
NO
Input
0 0
4 4
4 0
0 4
1 2
2 3
3 2
2 1
Output
YES
1 2 3 4
5 6 7 8
Note
Pay attention to the third example: the figures do not necessarily have to be parallel to the coordinate axes.
Submitted Solution:
```
from itertools import combinations
def get_center(points):
center = []
for d in range(len(points[0])):
c = sum(p[d] for p in points)
center.append(c / 4)
return tuple(center)
def all_the_same(xs):
if len(xs) <= 1:
return True
first, *ys = xs
return all(y == first for y in ys)
def distance(p1, p2):
return (p1[0] - p2[0])**2 + (p1[1] - p2[1])**2
def all_distance(points):
stack = list(points)
distances = []
while len(stack) > 1:
point = stack.pop()
distances.extend([distance(point, p) for p in stack])
return distances
def is_rect(points):
center = get_center(points)
l = [distance(center, p) for p in points]
return all_the_same(l)
def is_sqr(points):
if not is_rect(points):
return False
distances = all_distance(points)
distances.sort()
return all_the_same(distances[:-2])
def main(points):
idcs = list(range(0, 8))
for group in combinations(idcs, 4):
xs = [points[i] for i in group]
ys = [points[i] for i in idcs if i not in group]
if is_rect(xs) and is_rect(ys):
if is_sqr(xs):
print('YES')
print(' '.join([str(i+1) for i in idcs if i not in group]))
print(' '.join([str(i+1) for i in group]))
break
elif is_sqr(ys):
print('YES')
print(' '.join([str(i+1) for i in group]))
print(' '.join([str(i+1) for i in idcs if i not in group]))
break
else:
print('NO')
all_points = []
for _ in range(8):
all_points.append(list(map(int, input().split())))
main(all_points)
``` | instruction | 0 | 73,503 | 23 | 147,006 |
No | output | 1 | 73,503 | 23 | 147,007 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Little Petya very much likes rectangles and especially squares. Recently he has received 8 points on the plane as a gift from his mother. The points are pairwise distinct. Petya decided to split them into two sets each containing 4 points so that the points from the first set lay at the vertexes of some square and the points from the second set lay at the vertexes of a rectangle. Each point of initial 8 should belong to exactly one set. It is acceptable for a rectangle from the second set was also a square. If there are several partitions, Petya will be satisfied by any of them. Help him find such partition. Note that the rectangle and the square from the partition should have non-zero areas. The sides of the figures do not have to be parallel to the coordinate axes, though it might be the case.
Input
You are given 8 pairs of integers, a pair per line — the coordinates of the points Petya has. The absolute value of all coordinates does not exceed 104. It is guaranteed that no two points coincide.
Output
Print in the first output line "YES" (without the quotes), if the desired partition exists. In the second line output 4 space-separated numbers — point indexes from the input, which lie at the vertexes of the square. The points are numbered starting from 1. The numbers can be printed in any order. In the third line print the indexes of points lying at the vertexes of a rectangle in the similar format. All printed numbers should be pairwise distinct.
If the required partition does not exist, the first line should contain the word "NO" (without the quotes), after which no output is needed.
Examples
Input
0 0
10 11
10 0
0 11
1 1
2 2
2 1
1 2
Output
YES
5 6 7 8
1 2 3 4
Input
0 0
1 1
2 2
3 3
4 4
5 5
6 6
7 7
Output
NO
Input
0 0
4 4
4 0
0 4
1 2
2 3
3 2
2 1
Output
YES
1 2 3 4
5 6 7 8
Note
Pay attention to the third example: the figures do not necessarily have to be parallel to the coordinate axes.
Submitted Solution:
```
import math
from itertools import permutations
def dist_cube(a, b):
return (a[0] - b[0]) ** 2 + (a[1] - b[1]) ** 2
class CodeforcesTask135BSolution:
def __init__(self):
self.result = ''
self.points = []
def read_input(self):
for x in range(8):
self.points.append([int(y) for y in input().split(" ")])
def process_task(self):
sol = []
for perm in permutations([x + 1 for x in range(8)]):
sq = []
for p in perm[:4]:
sq.append(self.points[p - 1])
rc = []
for p in perm[4:]:
rc.append(self.points[p - 1])
sq_dists = [dist_cube(sq[0], sq[1]), dist_cube(sq[1], sq[2]), dist_cube(sq[2], sq[3]),
dist_cube(sq[3], sq[1]), dist_cube(sq[0], sq[2]), dist_cube(sq[0], sq[3])]
rc_dists = [dist_cube(rc[0], rc[1]), dist_cube(rc[1], rc[2]), dist_cube(rc[2], rc[3]),
dist_cube(rc[3], rc[1]), dist_cube(rc[0], rc[2]), dist_cube(rc[0], rc[3])]
sq_cnt = [sq_dists.count(x) for x in list(set(sq_dists))]
sq_cnt.sort(reverse=True)
rc_cnt = [rc_dists.count(x) for x in list(set(rc_dists))]
rc_cnt.sort(reverse=True)
if len(set(sq_dists)) == 2 and 2 <= len(set(rc_dists)) <= 3 and sq_cnt[0] == 4 and sq_cnt[1] == 2 and (rc_cnt == [4, 2] or rc_cnt == [2, 2, 2]):
sol = perm
break
if sol:
self.result = "YES\n{0}\n{1}".format(" ".join([str(x) for x in sol[:4]]), " ".join([str(x) for x in sol[4:]]))
else:
self.result = "NO"
def get_result(self):
return self.result
if __name__ == "__main__":
Solution = CodeforcesTask135BSolution()
Solution.read_input()
Solution.process_task()
print(Solution.get_result())
``` | instruction | 0 | 73,504 | 23 | 147,008 |
No | output | 1 | 73,504 | 23 | 147,009 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Little Petya very much likes rectangles and especially squares. Recently he has received 8 points on the plane as a gift from his mother. The points are pairwise distinct. Petya decided to split them into two sets each containing 4 points so that the points from the first set lay at the vertexes of some square and the points from the second set lay at the vertexes of a rectangle. Each point of initial 8 should belong to exactly one set. It is acceptable for a rectangle from the second set was also a square. If there are several partitions, Petya will be satisfied by any of them. Help him find such partition. Note that the rectangle and the square from the partition should have non-zero areas. The sides of the figures do not have to be parallel to the coordinate axes, though it might be the case.
Input
You are given 8 pairs of integers, a pair per line — the coordinates of the points Petya has. The absolute value of all coordinates does not exceed 104. It is guaranteed that no two points coincide.
Output
Print in the first output line "YES" (without the quotes), if the desired partition exists. In the second line output 4 space-separated numbers — point indexes from the input, which lie at the vertexes of the square. The points are numbered starting from 1. The numbers can be printed in any order. In the third line print the indexes of points lying at the vertexes of a rectangle in the similar format. All printed numbers should be pairwise distinct.
If the required partition does not exist, the first line should contain the word "NO" (without the quotes), after which no output is needed.
Examples
Input
0 0
10 11
10 0
0 11
1 1
2 2
2 1
1 2
Output
YES
5 6 7 8
1 2 3 4
Input
0 0
1 1
2 2
3 3
4 4
5 5
6 6
7 7
Output
NO
Input
0 0
4 4
4 0
0 4
1 2
2 3
3 2
2 1
Output
YES
1 2 3 4
5 6 7 8
Note
Pay attention to the third example: the figures do not necessarily have to be parallel to the coordinate axes.
Submitted Solution:
```
from itertools import permutations
def is_rectangle(a, b, c, d):
ab = (b[0]-a[0], b[1]-a[1])
ad = (d[0]-a[0], d[1]-a[1])
dp = ab[0]*ad[0] + ab[1]*ad[1]
ne = len(set((a, b, c, d))) == 4
return ne and dp == 0 and (a[0]+ab[0]+ad[0], a[1]+ab[1]+ad[1]) == c
def is_square(a, b, c, d):
abl = (b[0]-a[0])**2 + (b[1]-a[1])**2
adl = (d[0]-a[0])**2 + (d[1]-a[1])**2
return abl == adl and is_rectangle(a, b, c, d)
def is_rectangle_perm(a, b, c, d):
for p in permutations((a, b, c, d)):
if is_rectangle(*p):
return True
return False
def is_square_perm(a, b, c, d):
for p in permutations((a, b, c, d)):
if is_rectangle(*p):
return True
return False
ps = []
for i in range(8):
ps.append(tuple(map(int, input().split())))
possible = False
for i in range(0, 8):
for j in range(i+1, 8):
for k in range(j+1, 8):
for l in range(k+1, 8):
g0 = set(ps)
g1 = set((ps[i], ps[j], ps[k], ps[l]))
g2 = g0 - g1
possible = is_rectangle_perm(*g1) and is_square_perm(*g2)
possible = is_rectangle_perm(*g2) and is_square_perm(*g1)
print("YES" if possible else "NO")
``` | instruction | 0 | 73,505 | 23 | 147,010 |
No | output | 1 | 73,505 | 23 | 147,011 |
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