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 |
|---|---|---|---|---|---|
Provide tags and a correct Python 3 solution for this coding contest problem.
The math faculty of Berland State University has suffered the sudden drop in the math skills of enrolling students. This year the highest grade on the entrance math test was 8. Out of 100! Thus, the decision was made to make the test easier.
Future students will be asked just a single question. They are given a sequence of integer numbers a_1, a_2, ..., a_n, each number is from 1 to 3 and a_i ≠ a_{i + 1} for each valid i. The i-th number represents a type of the i-th figure:
1. circle;
2. isosceles triangle with the length of height equal to the length of base;
3. square.
The figures of the given sequence are placed somewhere on a Cartesian plane in such a way that:
* (i + 1)-th figure is inscribed into the i-th one;
* each triangle base is parallel to OX;
* the triangle is oriented in such a way that the vertex opposite to its base is at the top;
* each square sides are parallel to the axes;
* for each i from 2 to n figure i has the maximum possible length of side for triangle and square and maximum radius for circle.
Note that the construction is unique for some fixed position and size of just the first figure.
The task is to calculate the number of distinct points (not necessarily with integer coordinates) where figures touch. The trick is, however, that the number is sometimes infinite. But that won't make the task difficult for you, will it?
So can you pass the math test and enroll into Berland State University?
Input
The first line contains a single integer n (2 ≤ n ≤ 100) — the number of figures.
The second line contains n integer numbers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 3, a_i ≠ a_{i + 1}) — types of the figures.
Output
The first line should contain either the word "Infinite" if the number of distinct points where figures touch is infinite or "Finite" otherwise.
If the number is finite than print it in the second line. It's guaranteed that the number fits into 32-bit integer type.
Examples
Input
3
2 1 3
Output
Finite
7
Input
3
1 2 3
Output
Infinite
Note
Here are the glorious pictures for the examples. Note that the triangle is not equilateral but just isosceles with the length of height equal to the length of base. Thus it fits into a square in a unique way.
The distinct points where figures touch are marked red.
In the second example the triangle and the square touch each other for the whole segment, it contains infinite number of points.
<image> | instruction | 0 | 12,506 | 23 | 25,012 |
Tags: geometry
Correct Solution:
```
n = int(input())
l = list(map(int,input().split()))
w = 0
c = 0
for i in range(n-1):
if l[i] == 2:
if i!=0 and i!=n-1:
if l[i+1] == 3 or l[i-1] == 3:
w = 1
break
else:
if i == 0:
if l[i+1] == 3:
w = 1
break
else:
if l[i-1] == 3:
w = 1
break
c+=3
if l[i] == 1:
if l[i+1] == 2:
if l[i-1] == 3 and i!=0:
c+=2
else:
c+=3
if l[i+1] == 3:
c+=4
if l[i] == 3:
if l[i+1] == 2:
w = 1
break
c+=4
if w == 1:
print("Infinite")
else:
print("Finite")
print(c)
``` | output | 1 | 12,506 | 23 | 25,013 |
Provide tags and a correct Python 3 solution for this coding contest problem.
The math faculty of Berland State University has suffered the sudden drop in the math skills of enrolling students. This year the highest grade on the entrance math test was 8. Out of 100! Thus, the decision was made to make the test easier.
Future students will be asked just a single question. They are given a sequence of integer numbers a_1, a_2, ..., a_n, each number is from 1 to 3 and a_i ≠ a_{i + 1} for each valid i. The i-th number represents a type of the i-th figure:
1. circle;
2. isosceles triangle with the length of height equal to the length of base;
3. square.
The figures of the given sequence are placed somewhere on a Cartesian plane in such a way that:
* (i + 1)-th figure is inscribed into the i-th one;
* each triangle base is parallel to OX;
* the triangle is oriented in such a way that the vertex opposite to its base is at the top;
* each square sides are parallel to the axes;
* for each i from 2 to n figure i has the maximum possible length of side for triangle and square and maximum radius for circle.
Note that the construction is unique for some fixed position and size of just the first figure.
The task is to calculate the number of distinct points (not necessarily with integer coordinates) where figures touch. The trick is, however, that the number is sometimes infinite. But that won't make the task difficult for you, will it?
So can you pass the math test and enroll into Berland State University?
Input
The first line contains a single integer n (2 ≤ n ≤ 100) — the number of figures.
The second line contains n integer numbers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 3, a_i ≠ a_{i + 1}) — types of the figures.
Output
The first line should contain either the word "Infinite" if the number of distinct points where figures touch is infinite or "Finite" otherwise.
If the number is finite than print it in the second line. It's guaranteed that the number fits into 32-bit integer type.
Examples
Input
3
2 1 3
Output
Finite
7
Input
3
1 2 3
Output
Infinite
Note
Here are the glorious pictures for the examples. Note that the triangle is not equilateral but just isosceles with the length of height equal to the length of base. Thus it fits into a square in a unique way.
The distinct points where figures touch are marked red.
In the second example the triangle and the square touch each other for the whole segment, it contains infinite number of points.
<image> | instruction | 0 | 12,507 | 23 | 25,014 |
Tags: geometry
Correct Solution:
```
n = int(input())
a = list(map(int,input().split()))
cnt,f = 0,0
for i in range(1,n):
if (a[i-1]-1)*(a[i]-1) > 0:
f = 1
cnt += a[i-1]+a[i]
if i > 1 and a[i-2] == 3 and a[i-1] == 1 and a[i] == 2:
cnt -= 1
if f == 1:
print("Infinite")
else:
print("Finite")
print(cnt)
``` | output | 1 | 12,507 | 23 | 25,015 |
Provide tags and a correct Python 3 solution for this coding contest problem.
The math faculty of Berland State University has suffered the sudden drop in the math skills of enrolling students. This year the highest grade on the entrance math test was 8. Out of 100! Thus, the decision was made to make the test easier.
Future students will be asked just a single question. They are given a sequence of integer numbers a_1, a_2, ..., a_n, each number is from 1 to 3 and a_i ≠ a_{i + 1} for each valid i. The i-th number represents a type of the i-th figure:
1. circle;
2. isosceles triangle with the length of height equal to the length of base;
3. square.
The figures of the given sequence are placed somewhere on a Cartesian plane in such a way that:
* (i + 1)-th figure is inscribed into the i-th one;
* each triangle base is parallel to OX;
* the triangle is oriented in such a way that the vertex opposite to its base is at the top;
* each square sides are parallel to the axes;
* for each i from 2 to n figure i has the maximum possible length of side for triangle and square and maximum radius for circle.
Note that the construction is unique for some fixed position and size of just the first figure.
The task is to calculate the number of distinct points (not necessarily with integer coordinates) where figures touch. The trick is, however, that the number is sometimes infinite. But that won't make the task difficult for you, will it?
So can you pass the math test and enroll into Berland State University?
Input
The first line contains a single integer n (2 ≤ n ≤ 100) — the number of figures.
The second line contains n integer numbers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 3, a_i ≠ a_{i + 1}) — types of the figures.
Output
The first line should contain either the word "Infinite" if the number of distinct points where figures touch is infinite or "Finite" otherwise.
If the number is finite than print it in the second line. It's guaranteed that the number fits into 32-bit integer type.
Examples
Input
3
2 1 3
Output
Finite
7
Input
3
1 2 3
Output
Infinite
Note
Here are the glorious pictures for the examples. Note that the triangle is not equilateral but just isosceles with the length of height equal to the length of base. Thus it fits into a square in a unique way.
The distinct points where figures touch are marked red.
In the second example the triangle and the square touch each other for the whole segment, it contains infinite number of points.
<image> | instruction | 0 | 12,508 | 23 | 25,016 |
Tags: geometry
Correct Solution:
```
n = int(input())
a = [int(i) for i in input().split()]
total = 0
suma = dict()
suma[1, 2] = 3
suma[1, 3] = 4
suma[2, 1] = 3
suma[2, 3] = 9999999999999999999999999999999999999999999999999
suma[3, 1] = 4
suma[3, 2] = 9999999999999999999999999999999999999999999999999
for i, j in zip(a[:-1], a[1:]):
total += suma[i, j]
for i in range(len(a)-2):
if a[i] == 3 and a[i+1] == 1 and a[i+2] == 2:
total -= 1
if total < 99999999999999999999999999999:
print("Finite")
print(total)
else:
print("Infinite")
``` | output | 1 | 12,508 | 23 | 25,017 |
Provide tags and a correct Python 3 solution for this coding contest problem.
The math faculty of Berland State University has suffered the sudden drop in the math skills of enrolling students. This year the highest grade on the entrance math test was 8. Out of 100! Thus, the decision was made to make the test easier.
Future students will be asked just a single question. They are given a sequence of integer numbers a_1, a_2, ..., a_n, each number is from 1 to 3 and a_i ≠ a_{i + 1} for each valid i. The i-th number represents a type of the i-th figure:
1. circle;
2. isosceles triangle with the length of height equal to the length of base;
3. square.
The figures of the given sequence are placed somewhere on a Cartesian plane in such a way that:
* (i + 1)-th figure is inscribed into the i-th one;
* each triangle base is parallel to OX;
* the triangle is oriented in such a way that the vertex opposite to its base is at the top;
* each square sides are parallel to the axes;
* for each i from 2 to n figure i has the maximum possible length of side for triangle and square and maximum radius for circle.
Note that the construction is unique for some fixed position and size of just the first figure.
The task is to calculate the number of distinct points (not necessarily with integer coordinates) where figures touch. The trick is, however, that the number is sometimes infinite. But that won't make the task difficult for you, will it?
So can you pass the math test and enroll into Berland State University?
Input
The first line contains a single integer n (2 ≤ n ≤ 100) — the number of figures.
The second line contains n integer numbers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 3, a_i ≠ a_{i + 1}) — types of the figures.
Output
The first line should contain either the word "Infinite" if the number of distinct points where figures touch is infinite or "Finite" otherwise.
If the number is finite than print it in the second line. It's guaranteed that the number fits into 32-bit integer type.
Examples
Input
3
2 1 3
Output
Finite
7
Input
3
1 2 3
Output
Infinite
Note
Here are the glorious pictures for the examples. Note that the triangle is not equilateral but just isosceles with the length of height equal to the length of base. Thus it fits into a square in a unique way.
The distinct points where figures touch are marked red.
In the second example the triangle and the square touch each other for the whole segment, it contains infinite number of points.
<image> | instruction | 0 | 12,509 | 23 | 25,018 |
Tags: geometry
Correct Solution:
```
input()
a = ''.join(input().split())
res = 0
for aij in zip(a, a[1:]):
if '1' in aij:
res += sum(map(int, aij))
else:
print('Infinite')
break
else:
print('Finite')
print(res - a.count('312'))
``` | output | 1 | 12,509 | 23 | 25,019 |
Provide tags and a correct Python 3 solution for this coding contest problem.
The math faculty of Berland State University has suffered the sudden drop in the math skills of enrolling students. This year the highest grade on the entrance math test was 8. Out of 100! Thus, the decision was made to make the test easier.
Future students will be asked just a single question. They are given a sequence of integer numbers a_1, a_2, ..., a_n, each number is from 1 to 3 and a_i ≠ a_{i + 1} for each valid i. The i-th number represents a type of the i-th figure:
1. circle;
2. isosceles triangle with the length of height equal to the length of base;
3. square.
The figures of the given sequence are placed somewhere on a Cartesian plane in such a way that:
* (i + 1)-th figure is inscribed into the i-th one;
* each triangle base is parallel to OX;
* the triangle is oriented in such a way that the vertex opposite to its base is at the top;
* each square sides are parallel to the axes;
* for each i from 2 to n figure i has the maximum possible length of side for triangle and square and maximum radius for circle.
Note that the construction is unique for some fixed position and size of just the first figure.
The task is to calculate the number of distinct points (not necessarily with integer coordinates) where figures touch. The trick is, however, that the number is sometimes infinite. But that won't make the task difficult for you, will it?
So can you pass the math test and enroll into Berland State University?
Input
The first line contains a single integer n (2 ≤ n ≤ 100) — the number of figures.
The second line contains n integer numbers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 3, a_i ≠ a_{i + 1}) — types of the figures.
Output
The first line should contain either the word "Infinite" if the number of distinct points where figures touch is infinite or "Finite" otherwise.
If the number is finite than print it in the second line. It's guaranteed that the number fits into 32-bit integer type.
Examples
Input
3
2 1 3
Output
Finite
7
Input
3
1 2 3
Output
Infinite
Note
Here are the glorious pictures for the examples. Note that the triangle is not equilateral but just isosceles with the length of height equal to the length of base. Thus it fits into a square in a unique way.
The distinct points where figures touch are marked red.
In the second example the triangle and the square touch each other for the whole segment, it contains infinite number of points.
<image> | instruction | 0 | 12,510 | 23 | 25,020 |
Tags: geometry
Correct Solution:
```
def solve(n,A):
cnt=0
for i in range(n-1):
a0,a1,a2=A[i],A[i+1],A[i+2]
if a2==1:
if a1==2:
cnt+=3
else:
cnt+=4
elif a2==2:
if a1==1:
if a0==3:
cnt+=2
else:
cnt+=3
else:
print('Infinite')
return
else:
if a1==1:
cnt+=4
else:
print('Infinite')
return
print('Finite')
print(cnt)
return
n=int(input())
A=[0]+list(map(int,input().split()))
solve(n,A)
``` | output | 1 | 12,510 | 23 | 25,021 |
Provide tags and a correct Python 3 solution for this coding contest problem.
The math faculty of Berland State University has suffered the sudden drop in the math skills of enrolling students. This year the highest grade on the entrance math test was 8. Out of 100! Thus, the decision was made to make the test easier.
Future students will be asked just a single question. They are given a sequence of integer numbers a_1, a_2, ..., a_n, each number is from 1 to 3 and a_i ≠ a_{i + 1} for each valid i. The i-th number represents a type of the i-th figure:
1. circle;
2. isosceles triangle with the length of height equal to the length of base;
3. square.
The figures of the given sequence are placed somewhere on a Cartesian plane in such a way that:
* (i + 1)-th figure is inscribed into the i-th one;
* each triangle base is parallel to OX;
* the triangle is oriented in such a way that the vertex opposite to its base is at the top;
* each square sides are parallel to the axes;
* for each i from 2 to n figure i has the maximum possible length of side for triangle and square and maximum radius for circle.
Note that the construction is unique for some fixed position and size of just the first figure.
The task is to calculate the number of distinct points (not necessarily with integer coordinates) where figures touch. The trick is, however, that the number is sometimes infinite. But that won't make the task difficult for you, will it?
So can you pass the math test and enroll into Berland State University?
Input
The first line contains a single integer n (2 ≤ n ≤ 100) — the number of figures.
The second line contains n integer numbers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 3, a_i ≠ a_{i + 1}) — types of the figures.
Output
The first line should contain either the word "Infinite" if the number of distinct points where figures touch is infinite or "Finite" otherwise.
If the number is finite than print it in the second line. It's guaranteed that the number fits into 32-bit integer type.
Examples
Input
3
2 1 3
Output
Finite
7
Input
3
1 2 3
Output
Infinite
Note
Here are the glorious pictures for the examples. Note that the triangle is not equilateral but just isosceles with the length of height equal to the length of base. Thus it fits into a square in a unique way.
The distinct points where figures touch are marked red.
In the second example the triangle and the square touch each other for the whole segment, it contains infinite number of points.
<image> | instruction | 0 | 12,511 | 23 | 25,022 |
Tags: geometry
Correct Solution:
```
n = int(input())
a = list(map(int,input().split()))
flag = 0
count = 0
for i in range(1,n):
if a[i] == 1:
if a[i-1] == 1:
flag = 1
break
else:
if a[i-1] == 2:
count += 3
else:
count += 4
elif a[i] == 2:
if a[i-1] == 3 or a[i-1] == 2:
flag = 1
break
else:
count += 3
else:
if a[i-1] == 2 or a[i-1] == 3:
flag = 1
break
else:
count += 4
for i in range(n-2):
if(a[i] == 3 and a[i+1] == 1 and a[i+2] == 2):
count -= 1
if flag == 1:
print("Infinite")
else:
print("Finite")
print(count)
``` | output | 1 | 12,511 | 23 | 25,023 |
Provide tags and a correct Python 3 solution for this coding contest problem.
The math faculty of Berland State University has suffered the sudden drop in the math skills of enrolling students. This year the highest grade on the entrance math test was 8. Out of 100! Thus, the decision was made to make the test easier.
Future students will be asked just a single question. They are given a sequence of integer numbers a_1, a_2, ..., a_n, each number is from 1 to 3 and a_i ≠ a_{i + 1} for each valid i. The i-th number represents a type of the i-th figure:
1. circle;
2. isosceles triangle with the length of height equal to the length of base;
3. square.
The figures of the given sequence are placed somewhere on a Cartesian plane in such a way that:
* (i + 1)-th figure is inscribed into the i-th one;
* each triangle base is parallel to OX;
* the triangle is oriented in such a way that the vertex opposite to its base is at the top;
* each square sides are parallel to the axes;
* for each i from 2 to n figure i has the maximum possible length of side for triangle and square and maximum radius for circle.
Note that the construction is unique for some fixed position and size of just the first figure.
The task is to calculate the number of distinct points (not necessarily with integer coordinates) where figures touch. The trick is, however, that the number is sometimes infinite. But that won't make the task difficult for you, will it?
So can you pass the math test and enroll into Berland State University?
Input
The first line contains a single integer n (2 ≤ n ≤ 100) — the number of figures.
The second line contains n integer numbers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 3, a_i ≠ a_{i + 1}) — types of the figures.
Output
The first line should contain either the word "Infinite" if the number of distinct points where figures touch is infinite or "Finite" otherwise.
If the number is finite than print it in the second line. It's guaranteed that the number fits into 32-bit integer type.
Examples
Input
3
2 1 3
Output
Finite
7
Input
3
1 2 3
Output
Infinite
Note
Here are the glorious pictures for the examples. Note that the triangle is not equilateral but just isosceles with the length of height equal to the length of base. Thus it fits into a square in a unique way.
The distinct points where figures touch are marked red.
In the second example the triangle and the square touch each other for the whole segment, it contains infinite number of points.
<image> | instruction | 0 | 12,512 | 23 | 25,024 |
Tags: geometry
Correct Solution:
```
n = int(input())
vals = input().replace(" ", "")
if "23" in vals or "32" in vals:
print("Infinite")
exit(0)
else:
result = 0
for i in range(len(vals)-1):
temp = vals[i] + vals[i+1]
if temp == "12" or temp == "21":
result += 3
if temp == "31" or temp == "13":
result += 4
result -= vals.count("312")
print("Finite")
print(result)
``` | output | 1 | 12,512 | 23 | 25,025 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
The math faculty of Berland State University has suffered the sudden drop in the math skills of enrolling students. This year the highest grade on the entrance math test was 8. Out of 100! Thus, the decision was made to make the test easier.
Future students will be asked just a single question. They are given a sequence of integer numbers a_1, a_2, ..., a_n, each number is from 1 to 3 and a_i ≠ a_{i + 1} for each valid i. The i-th number represents a type of the i-th figure:
1. circle;
2. isosceles triangle with the length of height equal to the length of base;
3. square.
The figures of the given sequence are placed somewhere on a Cartesian plane in such a way that:
* (i + 1)-th figure is inscribed into the i-th one;
* each triangle base is parallel to OX;
* the triangle is oriented in such a way that the vertex opposite to its base is at the top;
* each square sides are parallel to the axes;
* for each i from 2 to n figure i has the maximum possible length of side for triangle and square and maximum radius for circle.
Note that the construction is unique for some fixed position and size of just the first figure.
The task is to calculate the number of distinct points (not necessarily with integer coordinates) where figures touch. The trick is, however, that the number is sometimes infinite. But that won't make the task difficult for you, will it?
So can you pass the math test and enroll into Berland State University?
Input
The first line contains a single integer n (2 ≤ n ≤ 100) — the number of figures.
The second line contains n integer numbers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 3, a_i ≠ a_{i + 1}) — types of the figures.
Output
The first line should contain either the word "Infinite" if the number of distinct points where figures touch is infinite or "Finite" otherwise.
If the number is finite than print it in the second line. It's guaranteed that the number fits into 32-bit integer type.
Examples
Input
3
2 1 3
Output
Finite
7
Input
3
1 2 3
Output
Infinite
Note
Here are the glorious pictures for the examples. Note that the triangle is not equilateral but just isosceles with the length of height equal to the length of base. Thus it fits into a square in a unique way.
The distinct points where figures touch are marked red.
In the second example the triangle and the square touch each other for the whole segment, it contains infinite number of points.
<image>
Submitted Solution:
```
import sys
n = int(input())
nums = [int(x) for x in input().split()]
fig = []
lst = 1
ans = 0
for i in range(0, len(nums)):
if(lst != 1 and nums[i] != 1):
print("Infinite")
sys.exit()
if(nums[i] != 1):
fig.append(nums[i])
ans += nums[i]+1
if(i != 0 and i != n-1):
ans += nums[i]+1;
lst = nums[i]
for i in range(0, len(fig)-1):
if(fig[i] == 3 and fig[i+1] == 2):
ans -= 1
print("Finite\n",ans,sep='')
``` | instruction | 0 | 12,513 | 23 | 25,026 |
Yes | output | 1 | 12,513 | 23 | 25,027 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
The math faculty of Berland State University has suffered the sudden drop in the math skills of enrolling students. This year the highest grade on the entrance math test was 8. Out of 100! Thus, the decision was made to make the test easier.
Future students will be asked just a single question. They are given a sequence of integer numbers a_1, a_2, ..., a_n, each number is from 1 to 3 and a_i ≠ a_{i + 1} for each valid i. The i-th number represents a type of the i-th figure:
1. circle;
2. isosceles triangle with the length of height equal to the length of base;
3. square.
The figures of the given sequence are placed somewhere on a Cartesian plane in such a way that:
* (i + 1)-th figure is inscribed into the i-th one;
* each triangle base is parallel to OX;
* the triangle is oriented in such a way that the vertex opposite to its base is at the top;
* each square sides are parallel to the axes;
* for each i from 2 to n figure i has the maximum possible length of side for triangle and square and maximum radius for circle.
Note that the construction is unique for some fixed position and size of just the first figure.
The task is to calculate the number of distinct points (not necessarily with integer coordinates) where figures touch. The trick is, however, that the number is sometimes infinite. But that won't make the task difficult for you, will it?
So can you pass the math test and enroll into Berland State University?
Input
The first line contains a single integer n (2 ≤ n ≤ 100) — the number of figures.
The second line contains n integer numbers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 3, a_i ≠ a_{i + 1}) — types of the figures.
Output
The first line should contain either the word "Infinite" if the number of distinct points where figures touch is infinite or "Finite" otherwise.
If the number is finite than print it in the second line. It's guaranteed that the number fits into 32-bit integer type.
Examples
Input
3
2 1 3
Output
Finite
7
Input
3
1 2 3
Output
Infinite
Note
Here are the glorious pictures for the examples. Note that the triangle is not equilateral but just isosceles with the length of height equal to the length of base. Thus it fits into a square in a unique way.
The distinct points where figures touch are marked red.
In the second example the triangle and the square touch each other for the whole segment, it contains infinite number of points.
<image>
Submitted Solution:
```
import io, sys, atexit, os
import math as ma
from sys import exit
from decimal import Decimal as dec
from itertools import permutations
from random import randint as rand
def li ():
return list (map (int, input ().split ()))
def num ():
return map (int, input ().split ())
def nu ():
return int (input ())
def find_gcd ( x, y ):
while (y):
x, y = y, x % y
return x
def lcm ( x, y ):
gg = find_gcd (x, y)
return (x * y // gg)
mm = 1000000007
yp = 0
def solve ():
t = 1
for _ in range (t):
n = nu ()
a = li ()
cc = 0
fl = True
for i in range (1, n):
if ((a [ i ] == 2 and a [ i - 1 ] == 3) or (a [ i ] == 3 and a [ i - 1 ] == 2)):
fl = False
break
if ((a [ i ] == 1 and a [ i - 1 ] == 2) or ((a [ i ] == 2 and a [ i - 1 ] == 1))):
cc += 3
else:
if((a [ i ] == 1 and a [ i - 1 ] == 3)):
if((i+1)<n):
if(a[i+1]==2):
cc+=3
else:
cc+=4
else:
cc+=4
else:
cc+=4
if (fl):
print ("Finite")
print (cc)
else:
print ("Infinite")
if __name__ == "__main__":
solve ()
``` | instruction | 0 | 12,514 | 23 | 25,028 |
Yes | output | 1 | 12,514 | 23 | 25,029 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
The math faculty of Berland State University has suffered the sudden drop in the math skills of enrolling students. This year the highest grade on the entrance math test was 8. Out of 100! Thus, the decision was made to make the test easier.
Future students will be asked just a single question. They are given a sequence of integer numbers a_1, a_2, ..., a_n, each number is from 1 to 3 and a_i ≠ a_{i + 1} for each valid i. The i-th number represents a type of the i-th figure:
1. circle;
2. isosceles triangle with the length of height equal to the length of base;
3. square.
The figures of the given sequence are placed somewhere on a Cartesian plane in such a way that:
* (i + 1)-th figure is inscribed into the i-th one;
* each triangle base is parallel to OX;
* the triangle is oriented in such a way that the vertex opposite to its base is at the top;
* each square sides are parallel to the axes;
* for each i from 2 to n figure i has the maximum possible length of side for triangle and square and maximum radius for circle.
Note that the construction is unique for some fixed position and size of just the first figure.
The task is to calculate the number of distinct points (not necessarily with integer coordinates) where figures touch. The trick is, however, that the number is sometimes infinite. But that won't make the task difficult for you, will it?
So can you pass the math test and enroll into Berland State University?
Input
The first line contains a single integer n (2 ≤ n ≤ 100) — the number of figures.
The second line contains n integer numbers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 3, a_i ≠ a_{i + 1}) — types of the figures.
Output
The first line should contain either the word "Infinite" if the number of distinct points where figures touch is infinite or "Finite" otherwise.
If the number is finite than print it in the second line. It's guaranteed that the number fits into 32-bit integer type.
Examples
Input
3
2 1 3
Output
Finite
7
Input
3
1 2 3
Output
Infinite
Note
Here are the glorious pictures for the examples. Note that the triangle is not equilateral but just isosceles with the length of height equal to the length of base. Thus it fits into a square in a unique way.
The distinct points where figures touch are marked red.
In the second example the triangle and the square touch each other for the whole segment, it contains infinite number of points.
<image>
Submitted Solution:
```
def main():
n = int(input())
arr = list(map(int,input().split()))
for i in range(n-1):
if arr[i] == 2 and arr[i+1] == 3:
print('Infinite')
return
if arr[i] == 3 and arr[i+1] == 2:
print('Infinite')
return
if arr[i] == arr[i+1]:
print('Infinite')
return
ans = 0
for i in range(n-1):
if arr[i] == 1 and arr[i+1] == 2:
ans += 3
if i > 0 and arr[i-1] == 3:
ans -= 1
elif arr[i] == 1 and arr[i+1] == 3:
ans += 4
elif arr[i] == 2 and arr[i+1] == 1:
ans += 3
elif arr[i] == 3 and arr[i+1] == 1:
ans += 4
print('Finite')
print(ans)
main()
``` | instruction | 0 | 12,515 | 23 | 25,030 |
Yes | output | 1 | 12,515 | 23 | 25,031 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
The math faculty of Berland State University has suffered the sudden drop in the math skills of enrolling students. This year the highest grade on the entrance math test was 8. Out of 100! Thus, the decision was made to make the test easier.
Future students will be asked just a single question. They are given a sequence of integer numbers a_1, a_2, ..., a_n, each number is from 1 to 3 and a_i ≠ a_{i + 1} for each valid i. The i-th number represents a type of the i-th figure:
1. circle;
2. isosceles triangle with the length of height equal to the length of base;
3. square.
The figures of the given sequence are placed somewhere on a Cartesian plane in such a way that:
* (i + 1)-th figure is inscribed into the i-th one;
* each triangle base is parallel to OX;
* the triangle is oriented in such a way that the vertex opposite to its base is at the top;
* each square sides are parallel to the axes;
* for each i from 2 to n figure i has the maximum possible length of side for triangle and square and maximum radius for circle.
Note that the construction is unique for some fixed position and size of just the first figure.
The task is to calculate the number of distinct points (not necessarily with integer coordinates) where figures touch. The trick is, however, that the number is sometimes infinite. But that won't make the task difficult for you, will it?
So can you pass the math test and enroll into Berland State University?
Input
The first line contains a single integer n (2 ≤ n ≤ 100) — the number of figures.
The second line contains n integer numbers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 3, a_i ≠ a_{i + 1}) — types of the figures.
Output
The first line should contain either the word "Infinite" if the number of distinct points where figures touch is infinite or "Finite" otherwise.
If the number is finite than print it in the second line. It's guaranteed that the number fits into 32-bit integer type.
Examples
Input
3
2 1 3
Output
Finite
7
Input
3
1 2 3
Output
Infinite
Note
Here are the glorious pictures for the examples. Note that the triangle is not equilateral but just isosceles with the length of height equal to the length of base. Thus it fits into a square in a unique way.
The distinct points where figures touch are marked red.
In the second example the triangle and the square touch each other for the whole segment, it contains infinite number of points.
<image>
Submitted Solution:
```
n= int(input())
list= [int (x) for x in input().split()]
a= list[0]
cnt = 0
prev = a
preprev = -1;
inf = False;
for i in range(1,n):
if(list[i] == 1):
if(prev == 3) :cnt += 4;
else: cnt += 3;
elif(list[i]== 2):
if(prev == 3): inf = True;
else: cnt += 3;
else:
if(prev == 2): inf = True;
else: cnt += 4;
if(preprev == 3 and prev == 1 and list[i] == 2) :cnt-=1
if(preprev == 1 and prev == 3 and list[i]== 2) :cnt -= 2;
preprev = prev;
prev = list[i]
if(inf): print("Infinite")
else:
print("Finite")
print(cnt)
``` | instruction | 0 | 12,516 | 23 | 25,032 |
Yes | output | 1 | 12,516 | 23 | 25,033 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
The math faculty of Berland State University has suffered the sudden drop in the math skills of enrolling students. This year the highest grade on the entrance math test was 8. Out of 100! Thus, the decision was made to make the test easier.
Future students will be asked just a single question. They are given a sequence of integer numbers a_1, a_2, ..., a_n, each number is from 1 to 3 and a_i ≠ a_{i + 1} for each valid i. The i-th number represents a type of the i-th figure:
1. circle;
2. isosceles triangle with the length of height equal to the length of base;
3. square.
The figures of the given sequence are placed somewhere on a Cartesian plane in such a way that:
* (i + 1)-th figure is inscribed into the i-th one;
* each triangle base is parallel to OX;
* the triangle is oriented in such a way that the vertex opposite to its base is at the top;
* each square sides are parallel to the axes;
* for each i from 2 to n figure i has the maximum possible length of side for triangle and square and maximum radius for circle.
Note that the construction is unique for some fixed position and size of just the first figure.
The task is to calculate the number of distinct points (not necessarily with integer coordinates) where figures touch. The trick is, however, that the number is sometimes infinite. But that won't make the task difficult for you, will it?
So can you pass the math test and enroll into Berland State University?
Input
The first line contains a single integer n (2 ≤ n ≤ 100) — the number of figures.
The second line contains n integer numbers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 3, a_i ≠ a_{i + 1}) — types of the figures.
Output
The first line should contain either the word "Infinite" if the number of distinct points where figures touch is infinite or "Finite" otherwise.
If the number is finite than print it in the second line. It's guaranteed that the number fits into 32-bit integer type.
Examples
Input
3
2 1 3
Output
Finite
7
Input
3
1 2 3
Output
Infinite
Note
Here are the glorious pictures for the examples. Note that the triangle is not equilateral but just isosceles with the length of height equal to the length of base. Thus it fits into a square in a unique way.
The distinct points where figures touch are marked red.
In the second example the triangle and the square touch each other for the whole segment, it contains infinite number of points.
<image>
Submitted Solution:
```
def solve(a):
res = 0
for i in range(1, len(a)):
u, v = min(a[i-1], a[i]), max(a[i-1], a[i])
if u == 2 and v == 3:
print("Infinite")
return
if v == 3:
res += 4
else:
res += 3
print("Finite")
print(res)
input()
a = list(map(int, input().split()))
solve(a)
``` | instruction | 0 | 12,517 | 23 | 25,034 |
No | output | 1 | 12,517 | 23 | 25,035 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
The math faculty of Berland State University has suffered the sudden drop in the math skills of enrolling students. This year the highest grade on the entrance math test was 8. Out of 100! Thus, the decision was made to make the test easier.
Future students will be asked just a single question. They are given a sequence of integer numbers a_1, a_2, ..., a_n, each number is from 1 to 3 and a_i ≠ a_{i + 1} for each valid i. The i-th number represents a type of the i-th figure:
1. circle;
2. isosceles triangle with the length of height equal to the length of base;
3. square.
The figures of the given sequence are placed somewhere on a Cartesian plane in such a way that:
* (i + 1)-th figure is inscribed into the i-th one;
* each triangle base is parallel to OX;
* the triangle is oriented in such a way that the vertex opposite to its base is at the top;
* each square sides are parallel to the axes;
* for each i from 2 to n figure i has the maximum possible length of side for triangle and square and maximum radius for circle.
Note that the construction is unique for some fixed position and size of just the first figure.
The task is to calculate the number of distinct points (not necessarily with integer coordinates) where figures touch. The trick is, however, that the number is sometimes infinite. But that won't make the task difficult for you, will it?
So can you pass the math test and enroll into Berland State University?
Input
The first line contains a single integer n (2 ≤ n ≤ 100) — the number of figures.
The second line contains n integer numbers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 3, a_i ≠ a_{i + 1}) — types of the figures.
Output
The first line should contain either the word "Infinite" if the number of distinct points where figures touch is infinite or "Finite" otherwise.
If the number is finite than print it in the second line. It's guaranteed that the number fits into 32-bit integer type.
Examples
Input
3
2 1 3
Output
Finite
7
Input
3
1 2 3
Output
Infinite
Note
Here are the glorious pictures for the examples. Note that the triangle is not equilateral but just isosceles with the length of height equal to the length of base. Thus it fits into a square in a unique way.
The distinct points where figures touch are marked red.
In the second example the triangle and the square touch each other for the whole segment, it contains infinite number of points.
<image>
Submitted Solution:
```
# -*- coding: utf-8 -*-
"""
@Project : CodeForces
@File : 1.py
@Time : 2019/5/1 21:13
@Author : Koushiro
"""
if __name__ == "__main__":
n = int(input())
nums = list(map(int, input().split()))
last=nums[0]
ret=0
Infinite=False
for i in range(1,len(nums)):
if last==1:
if nums[i]==1:
Infinite=True
break
elif nums[i]==2:
ret+=3
else:
ret+=4
last=nums[i]
elif last==2:
if nums[i]==1:
ret+=3
elif nums[i]==2:
ret+=3
else:
Infinite=True
break
last=nums[i]
else:
if nums[i]==1:
ret+=4
elif nums[i]==2:
Infinite=True
break
else:
Infinite=True
break
last=nums[i]
if Infinite:
print('Infinite')
else:
print('Finite')
print(ret)
``` | instruction | 0 | 12,518 | 23 | 25,036 |
No | output | 1 | 12,518 | 23 | 25,037 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
The math faculty of Berland State University has suffered the sudden drop in the math skills of enrolling students. This year the highest grade on the entrance math test was 8. Out of 100! Thus, the decision was made to make the test easier.
Future students will be asked just a single question. They are given a sequence of integer numbers a_1, a_2, ..., a_n, each number is from 1 to 3 and a_i ≠ a_{i + 1} for each valid i. The i-th number represents a type of the i-th figure:
1. circle;
2. isosceles triangle with the length of height equal to the length of base;
3. square.
The figures of the given sequence are placed somewhere on a Cartesian plane in such a way that:
* (i + 1)-th figure is inscribed into the i-th one;
* each triangle base is parallel to OX;
* the triangle is oriented in such a way that the vertex opposite to its base is at the top;
* each square sides are parallel to the axes;
* for each i from 2 to n figure i has the maximum possible length of side for triangle and square and maximum radius for circle.
Note that the construction is unique for some fixed position and size of just the first figure.
The task is to calculate the number of distinct points (not necessarily with integer coordinates) where figures touch. The trick is, however, that the number is sometimes infinite. But that won't make the task difficult for you, will it?
So can you pass the math test and enroll into Berland State University?
Input
The first line contains a single integer n (2 ≤ n ≤ 100) — the number of figures.
The second line contains n integer numbers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 3, a_i ≠ a_{i + 1}) — types of the figures.
Output
The first line should contain either the word "Infinite" if the number of distinct points where figures touch is infinite or "Finite" otherwise.
If the number is finite than print it in the second line. It's guaranteed that the number fits into 32-bit integer type.
Examples
Input
3
2 1 3
Output
Finite
7
Input
3
1 2 3
Output
Infinite
Note
Here are the glorious pictures for the examples. Note that the triangle is not equilateral but just isosceles with the length of height equal to the length of base. Thus it fits into a square in a unique way.
The distinct points where figures touch are marked red.
In the second example the triangle and the square touch each other for the whole segment, it contains infinite number of points.
<image>
Submitted Solution:
```
n = int(input())
A = list(map(int, input().split()))
ans = 0
for i in range(n-1):
if A[i] == 1 and A[i+1] == 2:
ans += 3
elif A[i] == 1 and A[i+1] == 3:
ans += 4
elif A[i] == 2 and A[i+1] == 1:
ans += 3
elif A[i] == 2 and A[i+1] == 3:
print('Infinite')
exit()
elif A[i] == 3 and A[i+1] == 1:
ans += 4
else:
print('Infinite')
exit()
print('Finite')
print(ans)
``` | instruction | 0 | 12,519 | 23 | 25,038 |
No | output | 1 | 12,519 | 23 | 25,039 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
The math faculty of Berland State University has suffered the sudden drop in the math skills of enrolling students. This year the highest grade on the entrance math test was 8. Out of 100! Thus, the decision was made to make the test easier.
Future students will be asked just a single question. They are given a sequence of integer numbers a_1, a_2, ..., a_n, each number is from 1 to 3 and a_i ≠ a_{i + 1} for each valid i. The i-th number represents a type of the i-th figure:
1. circle;
2. isosceles triangle with the length of height equal to the length of base;
3. square.
The figures of the given sequence are placed somewhere on a Cartesian plane in such a way that:
* (i + 1)-th figure is inscribed into the i-th one;
* each triangle base is parallel to OX;
* the triangle is oriented in such a way that the vertex opposite to its base is at the top;
* each square sides are parallel to the axes;
* for each i from 2 to n figure i has the maximum possible length of side for triangle and square and maximum radius for circle.
Note that the construction is unique for some fixed position and size of just the first figure.
The task is to calculate the number of distinct points (not necessarily with integer coordinates) where figures touch. The trick is, however, that the number is sometimes infinite. But that won't make the task difficult for you, will it?
So can you pass the math test and enroll into Berland State University?
Input
The first line contains a single integer n (2 ≤ n ≤ 100) — the number of figures.
The second line contains n integer numbers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 3, a_i ≠ a_{i + 1}) — types of the figures.
Output
The first line should contain either the word "Infinite" if the number of distinct points where figures touch is infinite or "Finite" otherwise.
If the number is finite than print it in the second line. It's guaranteed that the number fits into 32-bit integer type.
Examples
Input
3
2 1 3
Output
Finite
7
Input
3
1 2 3
Output
Infinite
Note
Here are the glorious pictures for the examples. Note that the triangle is not equilateral but just isosceles with the length of height equal to the length of base. Thus it fits into a square in a unique way.
The distinct points where figures touch are marked red.
In the second example the triangle and the square touch each other for the whole segment, it contains infinite number of points.
<image>
Submitted Solution:
```
# import sys
# sys.stdin = open("#input.txt", "r")
def solve(n,ls):
ans = 0
for i in range(n-1):
s = ls[i] + ls[i+1]
if s in [3,4]:
ans += s
else:
return False,0
return True,ans
n = int(input())
ls = list(map(int, input().split()))
fin, an = solve(n,ls)
if fin:
print("Finite")
print(an)
else:
print("Infinite")
``` | instruction | 0 | 12,520 | 23 | 25,040 |
No | output | 1 | 12,520 | 23 | 25,041 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Guy-Manuel and Thomas are going to build a polygon spaceship.
You're given a strictly convex (i. e. no three points are collinear) polygon P which is defined by coordinates of its vertices. Define P(x,y) as a polygon obtained by translating P by vector \overrightarrow {(x,y)}. The picture below depicts an example of the translation:
<image>
Define T as a set of points which is the union of all P(x,y) such that the origin (0,0) lies in P(x,y) (both strictly inside and on the boundary). There is also an equivalent definition: a point (x,y) lies in T only if there are two points A,B in P such that \overrightarrow {AB} = \overrightarrow {(x,y)}. One can prove T is a polygon too. For example, if P is a regular triangle then T is a regular hexagon. At the picture below P is drawn in black and some P(x,y) which contain the origin are drawn in colored:
<image>
The spaceship has the best aerodynamic performance if P and T are similar. Your task is to check whether the polygons P and T are [similar](https://tinyurl.com/vp5m7vl).
Input
The first line of input will contain a single integer n (3 ≤ n ≤ 10^5) — the number of points.
The i-th of the next n lines contains two integers x_i, y_i (|x_i|, |y_i| ≤ 10^9), denoting the coordinates of the i-th vertex.
It is guaranteed that these points are listed in counterclockwise order and these points form a strictly convex polygon.
Output
Output "YES" in a separate line, if P and T are similar. Otherwise, output "NO" in a separate line. You can print each letter in any case (upper or lower).
Examples
Input
4
1 0
4 1
3 4
0 3
Output
YES
Input
3
100 86
50 0
150 0
Output
nO
Input
8
0 0
1 0
2 1
3 3
4 6
3 6
2 5
1 3
Output
YES
Note
The following image shows the first sample: both P and T are squares. The second sample was shown in the statements.
<image> | instruction | 0 | 12,586 | 23 | 25,172 |
Tags: geometry
Correct Solution:
```
import math
n = int(input())
kords = list()
for _ in range(n):
kords.append(list(map(int, input().split())))
if n % 2 == 0:
first_part = kords[:len(kords) // 2 + 1]
second_part = kords[len(kords) // 2:] + [kords[0]]
for i in range(n // 2):
if abs( first_part[i][0] - second_part[i + 1][0] ) == abs( first_part[i + 1][0] - second_part[i][0] ) and abs( first_part[i + 1][1] - second_part[i][1] ) == abs( first_part[i][1] - second_part[i + 1][1] ):
continue
else:
print('NO')
break
else:
print('YES')
else:
print('NO')
``` | output | 1 | 12,586 | 23 | 25,173 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Guy-Manuel and Thomas are going to build a polygon spaceship.
You're given a strictly convex (i. e. no three points are collinear) polygon P which is defined by coordinates of its vertices. Define P(x,y) as a polygon obtained by translating P by vector \overrightarrow {(x,y)}. The picture below depicts an example of the translation:
<image>
Define T as a set of points which is the union of all P(x,y) such that the origin (0,0) lies in P(x,y) (both strictly inside and on the boundary). There is also an equivalent definition: a point (x,y) lies in T only if there are two points A,B in P such that \overrightarrow {AB} = \overrightarrow {(x,y)}. One can prove T is a polygon too. For example, if P is a regular triangle then T is a regular hexagon. At the picture below P is drawn in black and some P(x,y) which contain the origin are drawn in colored:
<image>
The spaceship has the best aerodynamic performance if P and T are similar. Your task is to check whether the polygons P and T are [similar](https://tinyurl.com/vp5m7vl).
Input
The first line of input will contain a single integer n (3 ≤ n ≤ 10^5) — the number of points.
The i-th of the next n lines contains two integers x_i, y_i (|x_i|, |y_i| ≤ 10^9), denoting the coordinates of the i-th vertex.
It is guaranteed that these points are listed in counterclockwise order and these points form a strictly convex polygon.
Output
Output "YES" in a separate line, if P and T are similar. Otherwise, output "NO" in a separate line. You can print each letter in any case (upper or lower).
Examples
Input
4
1 0
4 1
3 4
0 3
Output
YES
Input
3
100 86
50 0
150 0
Output
nO
Input
8
0 0
1 0
2 1
3 3
4 6
3 6
2 5
1 3
Output
YES
Note
The following image shows the first sample: both P and T are squares. The second sample was shown in the statements.
<image> | instruction | 0 | 12,587 | 23 | 25,174 |
Tags: geometry
Correct Solution:
```
import sys
n = int(sys.stdin.readline().strip())
X = []
Y = []
for i in range (0, n):
x, y = list(map(int, sys.stdin.readline().strip().split()))
X.append(x)
Y.append(y)
X.append(X[0])
Y.append(Y[0])
if n % 2 == 1:
print("NO")
else:
v = True
m = n // 2
for i in range (0, m):
if X[i+1]-X[i] != X[i+m]-X[i+m+1]:
v = False
if Y[i+1]-Y[i] != Y[i+m]-Y[i+m+1]:
v = False
if v == True:
print("YES")
else:
print("NO")
``` | output | 1 | 12,587 | 23 | 25,175 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Guy-Manuel and Thomas are going to build a polygon spaceship.
You're given a strictly convex (i. e. no three points are collinear) polygon P which is defined by coordinates of its vertices. Define P(x,y) as a polygon obtained by translating P by vector \overrightarrow {(x,y)}. The picture below depicts an example of the translation:
<image>
Define T as a set of points which is the union of all P(x,y) such that the origin (0,0) lies in P(x,y) (both strictly inside and on the boundary). There is also an equivalent definition: a point (x,y) lies in T only if there are two points A,B in P such that \overrightarrow {AB} = \overrightarrow {(x,y)}. One can prove T is a polygon too. For example, if P is a regular triangle then T is a regular hexagon. At the picture below P is drawn in black and some P(x,y) which contain the origin are drawn in colored:
<image>
The spaceship has the best aerodynamic performance if P and T are similar. Your task is to check whether the polygons P and T are [similar](https://tinyurl.com/vp5m7vl).
Input
The first line of input will contain a single integer n (3 ≤ n ≤ 10^5) — the number of points.
The i-th of the next n lines contains two integers x_i, y_i (|x_i|, |y_i| ≤ 10^9), denoting the coordinates of the i-th vertex.
It is guaranteed that these points are listed in counterclockwise order and these points form a strictly convex polygon.
Output
Output "YES" in a separate line, if P and T are similar. Otherwise, output "NO" in a separate line. You can print each letter in any case (upper or lower).
Examples
Input
4
1 0
4 1
3 4
0 3
Output
YES
Input
3
100 86
50 0
150 0
Output
nO
Input
8
0 0
1 0
2 1
3 3
4 6
3 6
2 5
1 3
Output
YES
Note
The following image shows the first sample: both P and T are squares. The second sample was shown in the statements.
<image> | instruction | 0 | 12,588 | 23 | 25,176 |
Tags: geometry
Correct Solution:
```
from sys import stdin
from collections import deque
mod = 10**9 + 7
import sys
# sys.setrecursionlimit(10**6)
from queue import PriorityQueue
# def rl():
# return [int(w) for w in stdin.readline().split()]
from bisect import bisect_right
from bisect import bisect_left
from collections import defaultdict
from math import sqrt,factorial,gcd,log2,inf,ceil
# map(int,input().split())
# # l = list(map(int,input().split()))
# from itertools import permutations
import heapq
# input = lambda: sys.stdin.readline().rstrip()
input = lambda : sys.stdin.readline().rstrip()
from sys import stdin, stdout
from heapq import heapify, heappush, heappop
from itertools import permutations
n = int(input())
ba = []
if n%2!=0:
print('NO')
exit()
for i in range(n):
a,b = map(int,input().split())
ba.append([a,b])
slope = []
seti = set()
for i in range(n):
a,b = ba[i]
c,d = ba[(i+n//2)%n]
x,y = (a+c)/2,(b+d)/2
seti.add((x,y))
# print(seti)
if len(seti) == 1:
print('YES')
else:
print('NO')
``` | output | 1 | 12,588 | 23 | 25,177 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Guy-Manuel and Thomas are going to build a polygon spaceship.
You're given a strictly convex (i. e. no three points are collinear) polygon P which is defined by coordinates of its vertices. Define P(x,y) as a polygon obtained by translating P by vector \overrightarrow {(x,y)}. The picture below depicts an example of the translation:
<image>
Define T as a set of points which is the union of all P(x,y) such that the origin (0,0) lies in P(x,y) (both strictly inside and on the boundary). There is also an equivalent definition: a point (x,y) lies in T only if there are two points A,B in P such that \overrightarrow {AB} = \overrightarrow {(x,y)}. One can prove T is a polygon too. For example, if P is a regular triangle then T is a regular hexagon. At the picture below P is drawn in black and some P(x,y) which contain the origin are drawn in colored:
<image>
The spaceship has the best aerodynamic performance if P and T are similar. Your task is to check whether the polygons P and T are [similar](https://tinyurl.com/vp5m7vl).
Input
The first line of input will contain a single integer n (3 ≤ n ≤ 10^5) — the number of points.
The i-th of the next n lines contains two integers x_i, y_i (|x_i|, |y_i| ≤ 10^9), denoting the coordinates of the i-th vertex.
It is guaranteed that these points are listed in counterclockwise order and these points form a strictly convex polygon.
Output
Output "YES" in a separate line, if P and T are similar. Otherwise, output "NO" in a separate line. You can print each letter in any case (upper or lower).
Examples
Input
4
1 0
4 1
3 4
0 3
Output
YES
Input
3
100 86
50 0
150 0
Output
nO
Input
8
0 0
1 0
2 1
3 3
4 6
3 6
2 5
1 3
Output
YES
Note
The following image shows the first sample: both P and T are squares. The second sample was shown in the statements.
<image> | instruction | 0 | 12,589 | 23 | 25,178 |
Tags: geometry
Correct Solution:
```
# https://codeforces.com/contest/1300/problem/D
n = int(input())
p = [list(map(int, input().split())) for _ in range(n)]
def is_parallel(x1, y1, x2, y2):
if x1 == -x2 and y1 == -y2:
return True
return False
flg = True
if n % 2 == 1:
flg=False
else:
for i in range(n//2):
p1, p2 = p[i], p[i+1]
p3, p4 = p[i+n//2], p[(i+n//2+1)%n]
flg = is_parallel(p2[0]-p1[0], p2[1]-p1[1], p4[0]-p3[0], p4[1]-p3[1])
if flg==False:
break
if flg==False:
print('no')
else:
print('yes')
``` | output | 1 | 12,589 | 23 | 25,179 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Guy-Manuel and Thomas are going to build a polygon spaceship.
You're given a strictly convex (i. e. no three points are collinear) polygon P which is defined by coordinates of its vertices. Define P(x,y) as a polygon obtained by translating P by vector \overrightarrow {(x,y)}. The picture below depicts an example of the translation:
<image>
Define T as a set of points which is the union of all P(x,y) such that the origin (0,0) lies in P(x,y) (both strictly inside and on the boundary). There is also an equivalent definition: a point (x,y) lies in T only if there are two points A,B in P such that \overrightarrow {AB} = \overrightarrow {(x,y)}. One can prove T is a polygon too. For example, if P is a regular triangle then T is a regular hexagon. At the picture below P is drawn in black and some P(x,y) which contain the origin are drawn in colored:
<image>
The spaceship has the best aerodynamic performance if P and T are similar. Your task is to check whether the polygons P and T are [similar](https://tinyurl.com/vp5m7vl).
Input
The first line of input will contain a single integer n (3 ≤ n ≤ 10^5) — the number of points.
The i-th of the next n lines contains two integers x_i, y_i (|x_i|, |y_i| ≤ 10^9), denoting the coordinates of the i-th vertex.
It is guaranteed that these points are listed in counterclockwise order and these points form a strictly convex polygon.
Output
Output "YES" in a separate line, if P and T are similar. Otherwise, output "NO" in a separate line. You can print each letter in any case (upper or lower).
Examples
Input
4
1 0
4 1
3 4
0 3
Output
YES
Input
3
100 86
50 0
150 0
Output
nO
Input
8
0 0
1 0
2 1
3 3
4 6
3 6
2 5
1 3
Output
YES
Note
The following image shows the first sample: both P and T are squares. The second sample was shown in the statements.
<image> | instruction | 0 | 12,590 | 23 | 25,180 |
Tags: geometry
Correct Solution:
```
def main():
N = int(input())
if N % 2:
print("NO")
return
pts = [complex(*map(int, input().strip().split())) for _ in range(N)]
for i in range(N // 2):
if pts[i] + pts[i + N // 2] != pts[0] + pts[N // 2]:
print("NO")
return
print("YES")
main()
``` | output | 1 | 12,590 | 23 | 25,181 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Guy-Manuel and Thomas are going to build a polygon spaceship.
You're given a strictly convex (i. e. no three points are collinear) polygon P which is defined by coordinates of its vertices. Define P(x,y) as a polygon obtained by translating P by vector \overrightarrow {(x,y)}. The picture below depicts an example of the translation:
<image>
Define T as a set of points which is the union of all P(x,y) such that the origin (0,0) lies in P(x,y) (both strictly inside and on the boundary). There is also an equivalent definition: a point (x,y) lies in T only if there are two points A,B in P such that \overrightarrow {AB} = \overrightarrow {(x,y)}. One can prove T is a polygon too. For example, if P is a regular triangle then T is a regular hexagon. At the picture below P is drawn in black and some P(x,y) which contain the origin are drawn in colored:
<image>
The spaceship has the best aerodynamic performance if P and T are similar. Your task is to check whether the polygons P and T are [similar](https://tinyurl.com/vp5m7vl).
Input
The first line of input will contain a single integer n (3 ≤ n ≤ 10^5) — the number of points.
The i-th of the next n lines contains two integers x_i, y_i (|x_i|, |y_i| ≤ 10^9), denoting the coordinates of the i-th vertex.
It is guaranteed that these points are listed in counterclockwise order and these points form a strictly convex polygon.
Output
Output "YES" in a separate line, if P and T are similar. Otherwise, output "NO" in a separate line. You can print each letter in any case (upper or lower).
Examples
Input
4
1 0
4 1
3 4
0 3
Output
YES
Input
3
100 86
50 0
150 0
Output
nO
Input
8
0 0
1 0
2 1
3 3
4 6
3 6
2 5
1 3
Output
YES
Note
The following image shows the first sample: both P and T are squares. The second sample was shown in the statements.
<image> | instruction | 0 | 12,591 | 23 | 25,182 |
Tags: geometry
Correct Solution:
```
def main():
from sys import stdin, stdout
n = int(stdin.readline())
inp = tuple((tuple(map(float, stdin.readline().split())) for _ in range(n)))
avg_xt2 = sum((inp_i[0] for inp_i in inp)) / n * 2
avg_yt2 = sum((inp_i[1] for inp_i in inp)) / n * 2
sym = tuple(((avg_xt2 - x, avg_yt2 - y) for (x, y) in inp))
nd2 = n // 2
stdout.write("YES" if inp == sym[nd2:] + sym[:nd2] else "NO")
main()
``` | output | 1 | 12,591 | 23 | 25,183 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Guy-Manuel and Thomas are going to build a polygon spaceship.
You're given a strictly convex (i. e. no three points are collinear) polygon P which is defined by coordinates of its vertices. Define P(x,y) as a polygon obtained by translating P by vector \overrightarrow {(x,y)}. The picture below depicts an example of the translation:
<image>
Define T as a set of points which is the union of all P(x,y) such that the origin (0,0) lies in P(x,y) (both strictly inside and on the boundary). There is also an equivalent definition: a point (x,y) lies in T only if there are two points A,B in P such that \overrightarrow {AB} = \overrightarrow {(x,y)}. One can prove T is a polygon too. For example, if P is a regular triangle then T is a regular hexagon. At the picture below P is drawn in black and some P(x,y) which contain the origin are drawn in colored:
<image>
The spaceship has the best aerodynamic performance if P and T are similar. Your task is to check whether the polygons P and T are [similar](https://tinyurl.com/vp5m7vl).
Input
The first line of input will contain a single integer n (3 ≤ n ≤ 10^5) — the number of points.
The i-th of the next n lines contains two integers x_i, y_i (|x_i|, |y_i| ≤ 10^9), denoting the coordinates of the i-th vertex.
It is guaranteed that these points are listed in counterclockwise order and these points form a strictly convex polygon.
Output
Output "YES" in a separate line, if P and T are similar. Otherwise, output "NO" in a separate line. You can print each letter in any case (upper or lower).
Examples
Input
4
1 0
4 1
3 4
0 3
Output
YES
Input
3
100 86
50 0
150 0
Output
nO
Input
8
0 0
1 0
2 1
3 3
4 6
3 6
2 5
1 3
Output
YES
Note
The following image shows the first sample: both P and T are squares. The second sample was shown in the statements.
<image> | instruction | 0 | 12,592 | 23 | 25,184 |
Tags: geometry
Correct Solution:
```
from __future__ import division
N = int(input())
points = []
for p in range(N):
x, y = [int(x) for x in input().split()]
points.append((x, y))
# T always has central symmetry
# To determine: if P has a point of central symmetry
# function to determine the bottom and top most points of our polygon
def bot_top(points):
y_val = []
for p in points:
y_val.append(p[1])
minn = y_val[0]
maxx = y_val[1]
for y in y_val:
if y < minn:
minn = y
if y > maxx:
maxx = y
return minn, maxx
def left_right(points):
y_val = []
for p in points:
y_val.append(p[0])
minn = y_val[0]
maxx = y_val[0]
for y in y_val:
if y < minn:
minn = y
if y > maxx:
maxx = y
return minn, maxx
# point of central symmetry
def cs():
x_1, x_2 = left_right(points)
y_1, y_2 = bot_top(points)
return (x_2 + x_1)/2, (y_1 + y_2)/2
def midpoint(p1, p2):
x_1, y_1 = p1
x_2, y_2 = p2
return (x_2 + x_1)/2, (y_2 + y_1)/2
CS = cs()
zz = True
if N % 2 == 0:
for pos in range(int(N/2)):
p1 = points[pos]
p2 = points[pos + int(N/2)]
mp = midpoint(p1, p2)
if mp != CS:
zz = False
break
if zz:
print ('YES')
else:
print ('NO')
else:
print ('NO')
``` | output | 1 | 12,592 | 23 | 25,185 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Guy-Manuel and Thomas are going to build a polygon spaceship.
You're given a strictly convex (i. e. no three points are collinear) polygon P which is defined by coordinates of its vertices. Define P(x,y) as a polygon obtained by translating P by vector \overrightarrow {(x,y)}. The picture below depicts an example of the translation:
<image>
Define T as a set of points which is the union of all P(x,y) such that the origin (0,0) lies in P(x,y) (both strictly inside and on the boundary). There is also an equivalent definition: a point (x,y) lies in T only if there are two points A,B in P such that \overrightarrow {AB} = \overrightarrow {(x,y)}. One can prove T is a polygon too. For example, if P is a regular triangle then T is a regular hexagon. At the picture below P is drawn in black and some P(x,y) which contain the origin are drawn in colored:
<image>
The spaceship has the best aerodynamic performance if P and T are similar. Your task is to check whether the polygons P and T are [similar](https://tinyurl.com/vp5m7vl).
Input
The first line of input will contain a single integer n (3 ≤ n ≤ 10^5) — the number of points.
The i-th of the next n lines contains two integers x_i, y_i (|x_i|, |y_i| ≤ 10^9), denoting the coordinates of the i-th vertex.
It is guaranteed that these points are listed in counterclockwise order and these points form a strictly convex polygon.
Output
Output "YES" in a separate line, if P and T are similar. Otherwise, output "NO" in a separate line. You can print each letter in any case (upper or lower).
Examples
Input
4
1 0
4 1
3 4
0 3
Output
YES
Input
3
100 86
50 0
150 0
Output
nO
Input
8
0 0
1 0
2 1
3 3
4 6
3 6
2 5
1 3
Output
YES
Note
The following image shows the first sample: both P and T are squares. The second sample was shown in the statements.
<image> | instruction | 0 | 12,593 | 23 | 25,186 |
Tags: geometry
Correct Solution:
```
def Input():
tem = input().split()
ans = []
for it in tem:
ans.append(int(it))
return ans
n = Input()[0]
a = []
ma = {}
for i in range(n):
x, y = Input()
a.append((x,y))
a.append(a[0])
flag = True
for i in range(1,n+1):
if (a[i][0]-a[i-1][0])==0:
v = None
else:
v = (a[i][1]-a[i-1][1])/(a[i][0]-a[i-1][0])
dis = ((a[i][1]-a[i-1][1])**2+(a[i][0]-a[i-1][0])**2)**0.5
if v not in ma:
ma[v]=dis
else:
if abs(dis-ma[v])<0.00000001:
del ma[v]
else:
flag = False
break
if len(ma)>0:
flag = False
if flag:
print('YES')
else:
print('NO')
``` | output | 1 | 12,593 | 23 | 25,187 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Guy-Manuel and Thomas are going to build a polygon spaceship.
You're given a strictly convex (i. e. no three points are collinear) polygon P which is defined by coordinates of its vertices. Define P(x,y) as a polygon obtained by translating P by vector \overrightarrow {(x,y)}. The picture below depicts an example of the translation:
<image>
Define T as a set of points which is the union of all P(x,y) such that the origin (0,0) lies in P(x,y) (both strictly inside and on the boundary). There is also an equivalent definition: a point (x,y) lies in T only if there are two points A,B in P such that \overrightarrow {AB} = \overrightarrow {(x,y)}. One can prove T is a polygon too. For example, if P is a regular triangle then T is a regular hexagon. At the picture below P is drawn in black and some P(x,y) which contain the origin are drawn in colored:
<image>
The spaceship has the best aerodynamic performance if P and T are similar. Your task is to check whether the polygons P and T are [similar](https://tinyurl.com/vp5m7vl).
Input
The first line of input will contain a single integer n (3 ≤ n ≤ 10^5) — the number of points.
The i-th of the next n lines contains two integers x_i, y_i (|x_i|, |y_i| ≤ 10^9), denoting the coordinates of the i-th vertex.
It is guaranteed that these points are listed in counterclockwise order and these points form a strictly convex polygon.
Output
Output "YES" in a separate line, if P and T are similar. Otherwise, output "NO" in a separate line. You can print each letter in any case (upper or lower).
Examples
Input
4
1 0
4 1
3 4
0 3
Output
YES
Input
3
100 86
50 0
150 0
Output
nO
Input
8
0 0
1 0
2 1
3 3
4 6
3 6
2 5
1 3
Output
YES
Note
The following image shows the first sample: both P and T are squares. The second sample was shown in the statements.
<image>
Submitted Solution:
```
def mid(a,b):
return [(a[0]+b[0])/2,(a[1]+b[1])/2]
n=int(input())
l=[]
for i in range(n):
l.append(list(map(int,input().split())))
if n%2==0:
k=mid(l[0],l[n//2])
flag=True
for i in range(1,n//2):
if k!=mid(l[i],l[n//2+i]):
flag=False
break
if flag==True:
print("YES")
else:
print("NO")
else:
print("NO")
``` | instruction | 0 | 12,594 | 23 | 25,188 |
Yes | output | 1 | 12,594 | 23 | 25,189 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Guy-Manuel and Thomas are going to build a polygon spaceship.
You're given a strictly convex (i. e. no three points are collinear) polygon P which is defined by coordinates of its vertices. Define P(x,y) as a polygon obtained by translating P by vector \overrightarrow {(x,y)}. The picture below depicts an example of the translation:
<image>
Define T as a set of points which is the union of all P(x,y) such that the origin (0,0) lies in P(x,y) (both strictly inside and on the boundary). There is also an equivalent definition: a point (x,y) lies in T only if there are two points A,B in P such that \overrightarrow {AB} = \overrightarrow {(x,y)}. One can prove T is a polygon too. For example, if P is a regular triangle then T is a regular hexagon. At the picture below P is drawn in black and some P(x,y) which contain the origin are drawn in colored:
<image>
The spaceship has the best aerodynamic performance if P and T are similar. Your task is to check whether the polygons P and T are [similar](https://tinyurl.com/vp5m7vl).
Input
The first line of input will contain a single integer n (3 ≤ n ≤ 10^5) — the number of points.
The i-th of the next n lines contains two integers x_i, y_i (|x_i|, |y_i| ≤ 10^9), denoting the coordinates of the i-th vertex.
It is guaranteed that these points are listed in counterclockwise order and these points form a strictly convex polygon.
Output
Output "YES" in a separate line, if P and T are similar. Otherwise, output "NO" in a separate line. You can print each letter in any case (upper or lower).
Examples
Input
4
1 0
4 1
3 4
0 3
Output
YES
Input
3
100 86
50 0
150 0
Output
nO
Input
8
0 0
1 0
2 1
3 3
4 6
3 6
2 5
1 3
Output
YES
Note
The following image shows the first sample: both P and T are squares. The second sample was shown in the statements.
<image>
Submitted Solution:
```
from sys import stdin
from os import getenv
if not getenv('PYCHARM'):
def input():
return next(stdin)[:-1]
def main():
n = int(input())
pp = []
for _ in range(n):
pp.append(list(map(int, input().split())))
if n%2 != 0:
print("NO")
return
for i in range(n//2):
x1 = pp[i+1][0] - pp[i][0]
y1 = pp[i+1][1] - pp[i][1]
x2 = pp[(i+1+n//2) % n][0] - pp[i+n//2][0]
y2 = pp[(i+1+n//2) % n][1] - pp[i+n//2][1]
if x1 != -x2 or y1 != -y2:
print("NO")
return
print("YES")
if __name__ == "__main__":
main()
``` | instruction | 0 | 12,595 | 23 | 25,190 |
Yes | output | 1 | 12,595 | 23 | 25,191 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Guy-Manuel and Thomas are going to build a polygon spaceship.
You're given a strictly convex (i. e. no three points are collinear) polygon P which is defined by coordinates of its vertices. Define P(x,y) as a polygon obtained by translating P by vector \overrightarrow {(x,y)}. The picture below depicts an example of the translation:
<image>
Define T as a set of points which is the union of all P(x,y) such that the origin (0,0) lies in P(x,y) (both strictly inside and on the boundary). There is also an equivalent definition: a point (x,y) lies in T only if there are two points A,B in P such that \overrightarrow {AB} = \overrightarrow {(x,y)}. One can prove T is a polygon too. For example, if P is a regular triangle then T is a regular hexagon. At the picture below P is drawn in black and some P(x,y) which contain the origin are drawn in colored:
<image>
The spaceship has the best aerodynamic performance if P and T are similar. Your task is to check whether the polygons P and T are [similar](https://tinyurl.com/vp5m7vl).
Input
The first line of input will contain a single integer n (3 ≤ n ≤ 10^5) — the number of points.
The i-th of the next n lines contains two integers x_i, y_i (|x_i|, |y_i| ≤ 10^9), denoting the coordinates of the i-th vertex.
It is guaranteed that these points are listed in counterclockwise order and these points form a strictly convex polygon.
Output
Output "YES" in a separate line, if P and T are similar. Otherwise, output "NO" in a separate line. You can print each letter in any case (upper or lower).
Examples
Input
4
1 0
4 1
3 4
0 3
Output
YES
Input
3
100 86
50 0
150 0
Output
nO
Input
8
0 0
1 0
2 1
3 3
4 6
3 6
2 5
1 3
Output
YES
Note
The following image shows the first sample: both P and T are squares. The second sample was shown in the statements.
<image>
Submitted Solution:
```
import sys
input=sys.stdin.buffer.readline #FOR READING PURE INTEGER INPUTS (space separation ok)
def isParallel(p1a,p1b,p2a,p2b):
dx1=p1a[0]-p1b[0]
dy1=p1a[1]-p1b[1]
dx2=p2a[0]-p2b[0]
dy2=p2a[1]-p2b[1]
return dx1==dx2 and dy1==dy2
n=int(input())
xy=[] #[[x,y]]
for _ in range(n):
a,b=[int(z) for z in input().split()]
xy.append([a,b])
#YES if each line is parallel to another line
xy.sort(key=lambda z:(z[0],z[1])) #sort by x then y.
#xy[0] will match with xy[n-1],xy[1] will match with xy[n-2] etc.
if n%2==1:
print('NO')
else:
ans='YES'
for i in range(n//2):
p1a=xy[i]
p1b=xy[i+1]
j=n-1-i
p2a=xy[j-1]
p2b=xy[j]
if not isParallel(p1a,p1b,p2a,p2b):
ans='NO'
break
print(ans)
``` | instruction | 0 | 12,596 | 23 | 25,192 |
Yes | output | 1 | 12,596 | 23 | 25,193 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Guy-Manuel and Thomas are going to build a polygon spaceship.
You're given a strictly convex (i. e. no three points are collinear) polygon P which is defined by coordinates of its vertices. Define P(x,y) as a polygon obtained by translating P by vector \overrightarrow {(x,y)}. The picture below depicts an example of the translation:
<image>
Define T as a set of points which is the union of all P(x,y) such that the origin (0,0) lies in P(x,y) (both strictly inside and on the boundary). There is also an equivalent definition: a point (x,y) lies in T only if there are two points A,B in P such that \overrightarrow {AB} = \overrightarrow {(x,y)}. One can prove T is a polygon too. For example, if P is a regular triangle then T is a regular hexagon. At the picture below P is drawn in black and some P(x,y) which contain the origin are drawn in colored:
<image>
The spaceship has the best aerodynamic performance if P and T are similar. Your task is to check whether the polygons P and T are [similar](https://tinyurl.com/vp5m7vl).
Input
The first line of input will contain a single integer n (3 ≤ n ≤ 10^5) — the number of points.
The i-th of the next n lines contains two integers x_i, y_i (|x_i|, |y_i| ≤ 10^9), denoting the coordinates of the i-th vertex.
It is guaranteed that these points are listed in counterclockwise order and these points form a strictly convex polygon.
Output
Output "YES" in a separate line, if P and T are similar. Otherwise, output "NO" in a separate line. You can print each letter in any case (upper or lower).
Examples
Input
4
1 0
4 1
3 4
0 3
Output
YES
Input
3
100 86
50 0
150 0
Output
nO
Input
8
0 0
1 0
2 1
3 3
4 6
3 6
2 5
1 3
Output
YES
Note
The following image shows the first sample: both P and T are squares. The second sample was shown in the statements.
<image>
Submitted Solution:
```
import os,io
import sys
input=io.BytesIO(os.read(0,os.fstat(0).st_size)).readline
n=int(input())
shape=[]
for _ in range(n):
x,y=map(int,input().split())
shape.append([x,y])
if n%2==1:
print('NO')
sys.exit()
for i in range(n):
if shape[i][0]-shape[i-1][0]!=shape[(n//2+i-1)%n][0]-shape[(n//2+i)%n][0]:
print('NO')
sys.exit()
if shape[i][1]-shape[i-1][1]!=shape[(n//2+i-1)%n][1]-shape[(n//2+i)%n][1]:
print('NO')
sys.exit()
print('YES')
``` | instruction | 0 | 12,597 | 23 | 25,194 |
Yes | output | 1 | 12,597 | 23 | 25,195 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Guy-Manuel and Thomas are going to build a polygon spaceship.
You're given a strictly convex (i. e. no three points are collinear) polygon P which is defined by coordinates of its vertices. Define P(x,y) as a polygon obtained by translating P by vector \overrightarrow {(x,y)}. The picture below depicts an example of the translation:
<image>
Define T as a set of points which is the union of all P(x,y) such that the origin (0,0) lies in P(x,y) (both strictly inside and on the boundary). There is also an equivalent definition: a point (x,y) lies in T only if there are two points A,B in P such that \overrightarrow {AB} = \overrightarrow {(x,y)}. One can prove T is a polygon too. For example, if P is a regular triangle then T is a regular hexagon. At the picture below P is drawn in black and some P(x,y) which contain the origin are drawn in colored:
<image>
The spaceship has the best aerodynamic performance if P and T are similar. Your task is to check whether the polygons P and T are [similar](https://tinyurl.com/vp5m7vl).
Input
The first line of input will contain a single integer n (3 ≤ n ≤ 10^5) — the number of points.
The i-th of the next n lines contains two integers x_i, y_i (|x_i|, |y_i| ≤ 10^9), denoting the coordinates of the i-th vertex.
It is guaranteed that these points are listed in counterclockwise order and these points form a strictly convex polygon.
Output
Output "YES" in a separate line, if P and T are similar. Otherwise, output "NO" in a separate line. You can print each letter in any case (upper or lower).
Examples
Input
4
1 0
4 1
3 4
0 3
Output
YES
Input
3
100 86
50 0
150 0
Output
nO
Input
8
0 0
1 0
2 1
3 3
4 6
3 6
2 5
1 3
Output
YES
Note
The following image shows the first sample: both P and T are squares. The second sample was shown in the statements.
<image>
Submitted Solution:
```
import sys
input = sys.stdin.readline
import math
import copy
import collections
from collections import deque
n = int(input())
points = []
for i in range(n):
points.append(list(map(int,input().split())))
vec = []
for i in range(n):
a = points[i]
b = points[(i+1)%n]
vec.append([b[0]-a[0],b[1]-a[1]])
dirs = set()
for ele in vec:
x,y = ele
d = math.atan(y/x)
if d<0:
d+=math.pi
if d in dirs:
dirs.remove(d)
else:
dirs.add(d)
if dirs:
print("NO")
else:
print("YES")
``` | instruction | 0 | 12,598 | 23 | 25,196 |
No | output | 1 | 12,598 | 23 | 25,197 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Guy-Manuel and Thomas are going to build a polygon spaceship.
You're given a strictly convex (i. e. no three points are collinear) polygon P which is defined by coordinates of its vertices. Define P(x,y) as a polygon obtained by translating P by vector \overrightarrow {(x,y)}. The picture below depicts an example of the translation:
<image>
Define T as a set of points which is the union of all P(x,y) such that the origin (0,0) lies in P(x,y) (both strictly inside and on the boundary). There is also an equivalent definition: a point (x,y) lies in T only if there are two points A,B in P such that \overrightarrow {AB} = \overrightarrow {(x,y)}. One can prove T is a polygon too. For example, if P is a regular triangle then T is a regular hexagon. At the picture below P is drawn in black and some P(x,y) which contain the origin are drawn in colored:
<image>
The spaceship has the best aerodynamic performance if P and T are similar. Your task is to check whether the polygons P and T are [similar](https://tinyurl.com/vp5m7vl).
Input
The first line of input will contain a single integer n (3 ≤ n ≤ 10^5) — the number of points.
The i-th of the next n lines contains two integers x_i, y_i (|x_i|, |y_i| ≤ 10^9), denoting the coordinates of the i-th vertex.
It is guaranteed that these points are listed in counterclockwise order and these points form a strictly convex polygon.
Output
Output "YES" in a separate line, if P and T are similar. Otherwise, output "NO" in a separate line. You can print each letter in any case (upper or lower).
Examples
Input
4
1 0
4 1
3 4
0 3
Output
YES
Input
3
100 86
50 0
150 0
Output
nO
Input
8
0 0
1 0
2 1
3 3
4 6
3 6
2 5
1 3
Output
YES
Note
The following image shows the first sample: both P and T are squares. The second sample was shown in the statements.
<image>
Submitted Solution:
```
# from math import factorial as fac
from collections import defaultdict
# from copy import deepcopy
import sys, math
f = None
try:
f = open('q1.input', 'r')
except IOError:
f = sys.stdin
if 'xrange' in dir(__builtins__):
range = xrange
# print(f.readline())
sys.setrecursionlimit(10**2)
def print_case_iterable(case_num, iterable):
print("Case #{}: {}".format(case_num," ".join(map(str,iterable))))
def print_case_number(case_num, iterable):
print("Case #{}: {}".format(case_num,iterable))
def print_iterable(A):
print (' '.join(A))
def read_int():
return int(f.readline().strip())
def read_int_array():
return [int(x) for x in f.readline().strip().split(" ")]
def rns():
a = [x for x in f.readline().split(" ")]
return int(a[0]), a[1].strip()
def read_string():
return list(f.readline().strip())
def ri():
return int(f.readline().strip())
def ria():
return [int(x) for x in f.readline().strip().split(" ")]
def rns():
a = [x for x in f.readline().split(" ")]
return int(a[0]), a[1].strip()
def rs():
return list(f.readline().strip())
def bi(x):
return bin(x)[2:]
from collections import deque
import math
NUMBER = 10**9 + 7
# NUMBER = 998244353
def factorial(n) :
M = NUMBER
f = 1
for i in range(1, n + 1):
f = (f * i) % M # Now f never can
# exceed 10^9+7
return f
def mult(a,b):
return (a * b) % NUMBER
def minus(a , b):
return (a - b) % NUMBER
def plus(a , b):
return (a + b) % NUMBER
def egcd(a, b):
if a == 0:
return (b, 0, 1)
else:
g, y, x = egcd(b % a, a)
return (g, x - (b // a) * y, y)
def modinv(a):
m = NUMBER
g, x, y = egcd(a, m)
if g != 1:
raise Exception('modular inverse does not exist')
else:
return x % m
def choose(n,k):
if n < k:
assert false
return mult(factorial(n), modinv(mult(factorial(k),factorial(n-k)))) % NUMBER
from collections import deque, defaultdict
import heapq
from types import GeneratorType
def bootstrap(f, stack=[]):
def wrappedfunc(*args, **kwargs):
if stack:
return f(*args, **kwargs)
else:
to = f(*args, **kwargs)
while True:
if type(to) is GeneratorType:
stack.append(to)
to = next(to)
else:
stack.pop()
if not stack:
break
to = stack[-1].send(to)
return to
return wrappedfunc
def dfs(g, timeIn, timeOut,depths,parents):
# assign In-time to node u
cnt = 0
# node, neig_i, parent, depth
stack = [[1,0,0,0]]
while stack:
v,neig_i,parent,depth = stack[-1]
parents[v] = parent
depths[v] = depth
# print (v)
if neig_i == 0:
timeIn[v] = cnt
cnt+=1
while neig_i < len(g[v]):
u = g[v][neig_i]
if u == parent:
neig_i+=1
continue
stack[-1][1] = neig_i + 1
stack.append([u,0,v,depth+1])
break
if neig_i == len(g[v]):
stack.pop()
timeOut[v] = cnt
cnt += 1
# def isAncestor(u: int, v: int, timeIn: list, timeOut: list) -> str:
# return timeIn[u] <= timeIn[v] and timeOut[v] <= timeOut[u]
cnt = 0
@bootstrap
def dfs(v,adj,timeIn, timeOut,depths,parents,parent=0,depth=0):
global cnt
parents[v] = parent
depths[v] = depth
timeIn[v] = cnt
cnt+=1
for u in adj[v]:
if u == parent:
continue
yield dfs(u,adj,timeIn,timeOut,depths,parents,v,depth+1)
timeOut[v] = cnt
cnt+=1
yield
def gcd(a,b):
if a == 0:
return b
return gcd(b % a, a)
# Function to return LCM of two numbers
def lcm(a,b):
return (a*b) / gcd(a,b)
def get_num_2_5(n):
twos = 0
fives = 0
while n>0 and n%2 == 0:
n//=2
twos+=1
while n>0 and n%5 == 0:
n//=5
fives+=1
return (twos,fives)
def shift(a,i,num):
for _ in range(num):
a[i],a[i+1],a[i+2] = a[i+2],a[i],a[i+1]
def equal(x,y):
return abs(x-y) <= 1e-9
# def leq(x,y):
# return x-y <= 1e-9
def solution(a,n):
if n > 3:
return 'yes'
return 'no'
def main():
T = 1
# T = ri()
for i in range(T):
n = ri()
# s = rs()
data = []
for j in range(n):
data.append(ria())
# n,m = ria()
# a = ria()
# b = ria()
x = solution(data,n)
# continue
if 'xrange' not in dir(__builtins__):
print(x)
else:
print >>output,str(x)# "Case #"+str(i+1)+':',
if 'xrange' in dir(__builtins__):
print(output.getvalue())
output.close()
if 'xrange' in dir(__builtins__):
import cStringIO
output = cStringIO.StringIO()
#example usage:
# for l in res:
# print >>output, str(len(l)) + ' ' + ' '.join(l)
if __name__ == '__main__':
main()
``` | instruction | 0 | 12,599 | 23 | 25,198 |
No | output | 1 | 12,599 | 23 | 25,199 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Guy-Manuel and Thomas are going to build a polygon spaceship.
You're given a strictly convex (i. e. no three points are collinear) polygon P which is defined by coordinates of its vertices. Define P(x,y) as a polygon obtained by translating P by vector \overrightarrow {(x,y)}. The picture below depicts an example of the translation:
<image>
Define T as a set of points which is the union of all P(x,y) such that the origin (0,0) lies in P(x,y) (both strictly inside and on the boundary). There is also an equivalent definition: a point (x,y) lies in T only if there are two points A,B in P such that \overrightarrow {AB} = \overrightarrow {(x,y)}. One can prove T is a polygon too. For example, if P is a regular triangle then T is a regular hexagon. At the picture below P is drawn in black and some P(x,y) which contain the origin are drawn in colored:
<image>
The spaceship has the best aerodynamic performance if P and T are similar. Your task is to check whether the polygons P and T are [similar](https://tinyurl.com/vp5m7vl).
Input
The first line of input will contain a single integer n (3 ≤ n ≤ 10^5) — the number of points.
The i-th of the next n lines contains two integers x_i, y_i (|x_i|, |y_i| ≤ 10^9), denoting the coordinates of the i-th vertex.
It is guaranteed that these points are listed in counterclockwise order and these points form a strictly convex polygon.
Output
Output "YES" in a separate line, if P and T are similar. Otherwise, output "NO" in a separate line. You can print each letter in any case (upper or lower).
Examples
Input
4
1 0
4 1
3 4
0 3
Output
YES
Input
3
100 86
50 0
150 0
Output
nO
Input
8
0 0
1 0
2 1
3 3
4 6
3 6
2 5
1 3
Output
YES
Note
The following image shows the first sample: both P and T are squares. The second sample was shown in the statements.
<image>
Submitted Solution:
```
from collections import defaultdict
from math import gcd
n = int(input())
points = []
for i in range(n):
x, y = (int(i) for i in input().split(' '))
points.append((x, y))
points.append(points[0])
if n % 2 == 1:
print("NO")
else:
d = defaultdict(lambda: 0)
for i in range(len(points) - 1):
x1, y1 = points[i]
x2, y2 = points[i + 1]
lcm = gcd((y2 - y1), (x2 - x1))
k1 = (y2 - y1) // lcm
k2 = (x2 - x1) // lcm
d[(k1, k2)] += 1
flag = True
for i in d:
if d[i] % 2 != 0:
flag = False
break
if flag:
print("YES")
else:
print("NO")
``` | instruction | 0 | 12,600 | 23 | 25,200 |
No | output | 1 | 12,600 | 23 | 25,201 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Guy-Manuel and Thomas are going to build a polygon spaceship.
You're given a strictly convex (i. e. no three points are collinear) polygon P which is defined by coordinates of its vertices. Define P(x,y) as a polygon obtained by translating P by vector \overrightarrow {(x,y)}. The picture below depicts an example of the translation:
<image>
Define T as a set of points which is the union of all P(x,y) such that the origin (0,0) lies in P(x,y) (both strictly inside and on the boundary). There is also an equivalent definition: a point (x,y) lies in T only if there are two points A,B in P such that \overrightarrow {AB} = \overrightarrow {(x,y)}. One can prove T is a polygon too. For example, if P is a regular triangle then T is a regular hexagon. At the picture below P is drawn in black and some P(x,y) which contain the origin are drawn in colored:
<image>
The spaceship has the best aerodynamic performance if P and T are similar. Your task is to check whether the polygons P and T are [similar](https://tinyurl.com/vp5m7vl).
Input
The first line of input will contain a single integer n (3 ≤ n ≤ 10^5) — the number of points.
The i-th of the next n lines contains two integers x_i, y_i (|x_i|, |y_i| ≤ 10^9), denoting the coordinates of the i-th vertex.
It is guaranteed that these points are listed in counterclockwise order and these points form a strictly convex polygon.
Output
Output "YES" in a separate line, if P and T are similar. Otherwise, output "NO" in a separate line. You can print each letter in any case (upper or lower).
Examples
Input
4
1 0
4 1
3 4
0 3
Output
YES
Input
3
100 86
50 0
150 0
Output
nO
Input
8
0 0
1 0
2 1
3 3
4 6
3 6
2 5
1 3
Output
YES
Note
The following image shows the first sample: both P and T are squares. The second sample was shown in the statements.
<image>
Submitted Solution:
```
q = int(input())
vertexP = []
z = True
for x in range(q):
x, y = input().split()
x, y = float(x), float(y)
vertexP.append([x, y])
minX = min([each[0] for each in vertexP])
maxX = max([each[0] for each in vertexP])
minY = min([each[1] for each in vertexP])
maxY = max([each[1] for each in vertexP])
cOmX = minX+(maxX-minX)/2
cOmY = minY+(maxY-minY)/2
print(cOmX)
print(cOmY)
for each in vertexP:
each[0] -= cOmX
each[1] -= cOmY
#print(vertexP)
for each in vertexP:
#print([each[0]*-1, each[1]*-1])
if not [each[0]*-1, each[1]*-1] in vertexP:
z = False
break
if z:
print("YES")
else:
print("nO")
``` | instruction | 0 | 12,601 | 23 | 25,202 |
No | output | 1 | 12,601 | 23 | 25,203 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given two set of points. The first set is determined by the equation A1x + B1y + C1 = 0, and the second one is determined by the equation A2x + B2y + C2 = 0.
Write the program which finds the number of points in the intersection of two given sets.
Input
The first line of the input contains three integer numbers A1, B1, C1 separated by space. The second line contains three integer numbers A2, B2, C2 separated by space. All the numbers are between -100 and 100, inclusive.
Output
Print the number of points in the intersection or -1 if there are infinite number of points.
Examples
Input
1 1 0
2 2 0
Output
-1
Input
1 1 0
2 -2 0
Output
1 | instruction | 0 | 12,732 | 23 | 25,464 |
Tags: implementation, math
Correct Solution:
```
R=lambda:map(int,input().split())
a1,b1,c1=R()
a2,b2,c2=R()
Z=lambda x,y,z:not x and not y and z
if a1*b2-a2*b1!=0: print("1")
elif Z(a1,b1,c1) or Z(a2,b2,c2) or c1*b2-b1*c2 or c1*a2-c2*a1: print("0")
else: print("-1")
``` | output | 1 | 12,732 | 23 | 25,465 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given two set of points. The first set is determined by the equation A1x + B1y + C1 = 0, and the second one is determined by the equation A2x + B2y + C2 = 0.
Write the program which finds the number of points in the intersection of two given sets.
Input
The first line of the input contains three integer numbers A1, B1, C1 separated by space. The second line contains three integer numbers A2, B2, C2 separated by space. All the numbers are between -100 and 100, inclusive.
Output
Print the number of points in the intersection or -1 if there are infinite number of points.
Examples
Input
1 1 0
2 2 0
Output
-1
Input
1 1 0
2 -2 0
Output
1 | instruction | 0 | 12,733 | 23 | 25,466 |
Tags: implementation, math
Correct Solution:
```
a1, b1, c1 = [int(i) for i in input().split()]
a2, b2, c2 = [int(i) for i in input().split()]
if(b1 != 0 and b2 != 0):
#paralelas:
if(a1/b1 == a2/b2):
if(c1/b1 == c2/b2):
print(-1) #paralelas iguais
else:
print(0) #paralelas diferentes
else:
print(1)
elif(b1 == 0 and b2 == 0):
if(a1 != 0):
if(a2 != 0):
if(c1/a1 == c2/a2):
print(-1)
else:
print(0)
else:
if(c2 == 0):
print(-1)
else:
print(0)
if(a1 == 0 and a2 != 0):
if(c1 == 0):
print(-1)
else:
print(0)
if(a1 == 0 and a2 == 0):
if(c1 == 0 and c2 == 0):
print(-1)
else:
print(0)
elif(b1 == 0 and b2 != 0):
if(a1 != 0 and a2 != 0):
print(1)
elif(a1 == 0 and a2 != 0):
if(c1 == 0):
print(-1)
else:
print(0)
elif(a1 != 0 and a2 == 0):
print(1)
elif(a1 == 0 and a2 == 0):
if(c1 == 0):
print(-1)
else:
print(0)
elif(b1 != 0 and b2 == 0):
if(a1 != 0 and a2 != 0):
print(1)
elif(a1 != 0 and a2 == 0):
if(c2 == 0):
print(-1)
else:
print(0)
elif(a1 == 0 and a2 != 0):
print(1)
elif(a1 == 0 and a2 == 0):
if(c2 == 0):
print(-1)
else:
print(0)
else:
print(1)
``` | output | 1 | 12,733 | 23 | 25,467 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given two set of points. The first set is determined by the equation A1x + B1y + C1 = 0, and the second one is determined by the equation A2x + B2y + C2 = 0.
Write the program which finds the number of points in the intersection of two given sets.
Input
The first line of the input contains three integer numbers A1, B1, C1 separated by space. The second line contains three integer numbers A2, B2, C2 separated by space. All the numbers are between -100 and 100, inclusive.
Output
Print the number of points in the intersection or -1 if there are infinite number of points.
Examples
Input
1 1 0
2 2 0
Output
-1
Input
1 1 0
2 -2 0
Output
1 | instruction | 0 | 12,734 | 23 | 25,468 |
Tags: implementation, math
Correct Solution:
```
R = lambda:map(int, input().split())
a1, b1, c1 = R()
a2, b2, c2 = R()
if a1 == b1 == 0 and c1 != 0 or a2 == b2 == 0 and c2 != 0:
print(0)
elif a1 == b1 == c1 == 0 or a2 == b2 == c2 == 0:
print(-1)
else:
if c1 != 0:
a1 /= c1
b1 /= c1
c1 = 1
if c2 != 0:
a2 /= c2
b2 /= c2
c2 = 1
if a1 * b2 - a2 * b1 == 0:
if c1 == c2 and a1 == a2 and b1 == b2 or c1 == c2 == 0:
print(-1)
else:
print(0)
else:
print(1)
``` | output | 1 | 12,734 | 23 | 25,469 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given two set of points. The first set is determined by the equation A1x + B1y + C1 = 0, and the second one is determined by the equation A2x + B2y + C2 = 0.
Write the program which finds the number of points in the intersection of two given sets.
Input
The first line of the input contains three integer numbers A1, B1, C1 separated by space. The second line contains three integer numbers A2, B2, C2 separated by space. All the numbers are between -100 and 100, inclusive.
Output
Print the number of points in the intersection or -1 if there are infinite number of points.
Examples
Input
1 1 0
2 2 0
Output
-1
Input
1 1 0
2 -2 0
Output
1 | instruction | 0 | 12,735 | 23 | 25,470 |
Tags: implementation, math
Correct Solution:
```
a1,b1,c1=list(map(int,input().split()))
a2,b2,c2=list(map(int,input().split()))
if (a1==0 and b1==0 and c1!=0) or (a2==0 and b2==0 and c2!=0):
c=0
print(c)
elif (a2==1 and b2==-1 and c2==-1 and c1!=-1):
c=-1
print(c)
elif (a2==0 and b2==0 and c2==0):
c=-1
print(c)
elif (a1==0 and b1==0 and c1==0):
c=-1
print(c)
elif (b2==0 and b1==0 and c2!=0 and c1!=0 and a2==7):
c=0
print(c)
elif (b2==0 and b1==0 and c2!=0 and c1!=0 and a2!=2):
c=-1
print(c)
elif (a1==0 and b1==0 and c1 ==0) and (a2==0 and b2==1 and c2==-1):
c=-1
print(c)
elif (a1==1 and b1==0 and c1==0) and (a2==1 and b2==0 and c2==0):
c=-1
print(c)
elif (a1==1 and b1==0 and c1 ==1) and (a2==-1 and b2==0 and c2==-1):
c=-1
print(c)
elif (a1==0 and b1==0 and c1 ==0) and (a2==0 and b2==0 and c2==0):
c=-1
print(c)
elif (a1==1 and b1==0 and c1 ==0) and (a2==1 and b2==0 and c2==1):
c=-1
print(c)
elif (a1==1 and b1==1 and c1 ==1) and (a2==0 and b2==0 and c2==0):
c=-1
print(c)
elif (a1*b2==0 and a2*b1==0) and (b1*c2==0 and b2*c1==0):
c=0
print(c)
elif(a1*b2==a2*b1) and (b1*c2==b2*c1):
c=-1
print(c)
elif(a1*b2==a2*b1) and (b1*c2!=b2*c1):
c=0
print(c)
else:
c=1
print(c)
``` | output | 1 | 12,735 | 23 | 25,471 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given two set of points. The first set is determined by the equation A1x + B1y + C1 = 0, and the second one is determined by the equation A2x + B2y + C2 = 0.
Write the program which finds the number of points in the intersection of two given sets.
Input
The first line of the input contains three integer numbers A1, B1, C1 separated by space. The second line contains three integer numbers A2, B2, C2 separated by space. All the numbers are between -100 and 100, inclusive.
Output
Print the number of points in the intersection or -1 if there are infinite number of points.
Examples
Input
1 1 0
2 2 0
Output
-1
Input
1 1 0
2 -2 0
Output
1 | instruction | 0 | 12,736 | 23 | 25,472 |
Tags: implementation, math
Correct Solution:
```
#codeforces 21b intersection, math
def readGen(transform):
while (True):
n=0
tmp=input().split()
m=len(tmp)
while (n<m):
yield(transform(tmp[n]))
n+=1
readint=readGen(int)
A1,B1,C1=next(readint),next(readint),next(readint)
A2,B2,C2=next(readint),next(readint),next(readint)
def cross(a,b,c,d): return a*d-b*c
def zero(A1,B1): return A1**2 + B1**2 == 0
def lineIntersect(A1,B1,C1,A2,B2,C2):
if (cross(A1,B1,A2,B2)==0):
if (cross(A1,C1,A2,C2)==0 and cross(B1,C1,B2,C2)==0):
# same line
return -1
else:
# parallel
return 0
else:
# cross
return 1
def judge(A1,B1,C1,A2,B2,C2):
if (zero(A1,B1) and C1!=0): return 0
if (zero(A2,B2) and C2!=0): return 0
if (not zero(A1,B1) and not zero(A2,B2)):
# line and line
return lineIntersect(A1,B1,C1,A2,B2,C2)
else:
return -1
print(judge(A1,B1,C1,A2,B2,C2))
``` | output | 1 | 12,736 | 23 | 25,473 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given two set of points. The first set is determined by the equation A1x + B1y + C1 = 0, and the second one is determined by the equation A2x + B2y + C2 = 0.
Write the program which finds the number of points in the intersection of two given sets.
Input
The first line of the input contains three integer numbers A1, B1, C1 separated by space. The second line contains three integer numbers A2, B2, C2 separated by space. All the numbers are between -100 and 100, inclusive.
Output
Print the number of points in the intersection or -1 if there are infinite number of points.
Examples
Input
1 1 0
2 2 0
Output
-1
Input
1 1 0
2 -2 0
Output
1 | instruction | 0 | 12,737 | 23 | 25,474 |
Tags: implementation, math
Correct Solution:
```
a1, b1, c1 = input().split(' ')
a2, b2, c2 = input().split(' ')
a1 = int(a1)
b1 = int(b1)
c1 = int(c1)
a2 = int(a2)
b2 = int(b2)
c2 = int(c2)
if((a1 == 0 and b1 == 0) and (a2 == 0 and b2 == 0) and c1 != c2):
print(0)
exit()
if((a1 == 0 and b1 == 0 and c1 == 0) or (a2 == 0 and b2 == 0 and c2 == 0)):
print(-1)
exit()
if((a1 == 0 and b1 == 0 and c1 != 0) or (a2 == 0 and b2 == 0 and c2 != 0) and (c1 == c2)):
print(0)
exit()
if(a1 * b2 == b1 * a2 and b2 * c1 == c2 * b1 and c1 * a2 == c2 * a1):
print(-1)
exit()
if(((a1 * b2) - (a2 * b1)) == 0):
print(0)
exit()
print(1)
exit()
``` | output | 1 | 12,737 | 23 | 25,475 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given two set of points. The first set is determined by the equation A1x + B1y + C1 = 0, and the second one is determined by the equation A2x + B2y + C2 = 0.
Write the program which finds the number of points in the intersection of two given sets.
Input
The first line of the input contains three integer numbers A1, B1, C1 separated by space. The second line contains three integer numbers A2, B2, C2 separated by space. All the numbers are between -100 and 100, inclusive.
Output
Print the number of points in the intersection or -1 if there are infinite number of points.
Examples
Input
1 1 0
2 2 0
Output
-1
Input
1 1 0
2 -2 0
Output
1 | instruction | 0 | 12,738 | 23 | 25,476 |
Tags: implementation, math
Correct Solution:
```
#! /usr/bin/python3
def solve():
x, y, z = map(int, input().strip().split())
a, b, c = map(int, input().strip().split())
ret = 0
if (x * b == y * a):
if (y == 0 and b == 0):
if c * x == a * z:
ret = -1
else:
ret = 0
elif z * b == y * c:
ret = -1
else:
ret = 0
else:
ret = 1
if (x == y and y == a and a == b and b == 0):
if (z == c and z == 0):
ret = -1
else:
ret = 0
print(ret)
solve()
``` | output | 1 | 12,738 | 23 | 25,477 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given two set of points. The first set is determined by the equation A1x + B1y + C1 = 0, and the second one is determined by the equation A2x + B2y + C2 = 0.
Write the program which finds the number of points in the intersection of two given sets.
Input
The first line of the input contains three integer numbers A1, B1, C1 separated by space. The second line contains three integer numbers A2, B2, C2 separated by space. All the numbers are between -100 and 100, inclusive.
Output
Print the number of points in the intersection or -1 if there are infinite number of points.
Examples
Input
1 1 0
2 2 0
Output
-1
Input
1 1 0
2 -2 0
Output
1 | instruction | 0 | 12,739 | 23 | 25,478 |
Tags: implementation, math
Correct Solution:
```
def fazTudo():
ar,br,cr = map(int,input().split());
aw,bw,cw = map(int,input().split());
res=1;
if(br==0 and bw==0):
if(ar==0 and cr!=0 or aw==0 and cw!=0):
print(0);
return;
if(ar==0 and br==0 and cr==0 or aw==0 and bw==0 and cw==0):
print(-1);
return;
if(ar==0 and cr==0 or aw==0 and cw==0):
print(-1);
return;
if(cr/ar==cw/aw):
print(-1)
return;
if(ar==aw and cr!=cw):
print(0);
return;
if(ar!=aw):
print(0);
return;
if(br==0):
if(ar==0 and cr!=0):
print(0);
return;
if(ar==0 and br==0 and cr==0 or aw==0 and bw==0 and cw==0):
print(-1);
return;
if(ar==0 and cr!=0):
print(0);
return;
if(-aw/bw == ar and -cw/bw==cr):
print(1)
return;
if(-aw/bw == ar and -cw/bw!=cr):
print(1);
return;
print(1);
return;
if(bw==0):
if(aw==0 and cw!=0):
print(0);
return;
if(ar==0 and br==0 and cr==0 or aw==0 and bw==0 and cw==0):
print(-1);
return;
if(aw==0 and cw!=0):
print(0);
return;
if(-ar/br == aw and -cr/br==cw):
print(1)
return;
if(-ar/br == aw and -cr/br!=cw):
print(1);
return;
print(1);
return;
a1 = -(ar/br);
b1 = -(cr/br);
a2 = -(aw/bw);
b2 = -(cw/bw);
if(a1==a2):
if(b1==b2):
res = -1;
else:
res = 0;
print(res);
fazTudo();
``` | output | 1 | 12,739 | 23 | 25,479 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given two set of points. The first set is determined by the equation A1x + B1y + C1 = 0, and the second one is determined by the equation A2x + B2y + C2 = 0.
Write the program which finds the number of points in the intersection of two given sets.
Input
The first line of the input contains three integer numbers A1, B1, C1 separated by space. The second line contains three integer numbers A2, B2, C2 separated by space. All the numbers are between -100 and 100, inclusive.
Output
Print the number of points in the intersection or -1 if there are infinite number of points.
Examples
Input
1 1 0
2 2 0
Output
-1
Input
1 1 0
2 -2 0
Output
1
Submitted Solution:
```
a1,b1,c1=list(map(int,input().split()))
a2,b2,c2=list(map(int,input().split()))
if (a1==0 and b1==0 and c1!=0) or (a2==0 and b2==0 and c2!=0):
c=0
print(c)
elif (a2==1 and b2==-1 and c2==-1 and c1!=-1):
c=-1
print(c)
elif (a2==0 and b2==0 and c2==0):
c=-1
print(c)
elif (a1==0 and b1==0 and c1==0):
c=-1
print(c)
elif (b2==0 and b1==0 and c2!=0 and c1!=0 and a2==7):
c=0
print(c)
elif (b2==0 and b1==0 and c2!=0 and c1!=0 and a2!=2):
c=-1
print(c)
elif (a1==0 and b1==0 and c1 ==0) and (a2==0 and b2==1 and c2==-1):
c=-1
print(c)
elif (a1==1 and b1==0 and c1==0) and (a2==1 and b2==0 and c2==0):
c=-1
print(c)
elif (a1==1 and b1==0 and c1 ==1) and (a2==-1 and b2==0 and c2==-1):
c=-1
print(c)
elif (a1==1 and b1==1 and c1 ==1) and (a2==0 and b2==0 and c2==0):
c=-1
print(c)
elif (a1*b2==0 and a2*b1==0) and (b1*c2==0 and b2*c1==0):
c=0
print(c)
elif(a1*b2==a2*b1) and (b1*c2==b2*c1):
c=-1
print(c)
elif(a1*b2==a2*b1) and (b1*c2!=b2*c1):
c=0
print(c)
else:
c=1
print(c)
``` | instruction | 0 | 12,740 | 23 | 25,480 |
Yes | output | 1 | 12,740 | 23 | 25,481 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given two set of points. The first set is determined by the equation A1x + B1y + C1 = 0, and the second one is determined by the equation A2x + B2y + C2 = 0.
Write the program which finds the number of points in the intersection of two given sets.
Input
The first line of the input contains three integer numbers A1, B1, C1 separated by space. The second line contains three integer numbers A2, B2, C2 separated by space. All the numbers are between -100 and 100, inclusive.
Output
Print the number of points in the intersection or -1 if there are infinite number of points.
Examples
Input
1 1 0
2 2 0
Output
-1
Input
1 1 0
2 -2 0
Output
1
Submitted Solution:
```
def intersection(lst1, lst2):
if lst1[0] * lst2[1] != lst1[1] * lst2[0]:
return 1
elif lst1[0] == lst1[1] == lst2[0] == lst2[1] == 0 and (
lst1[2] != 0 and lst2[2] == 0 or lst1[2] == 0 and lst2[2] != 0 or lst1[2] != 0 and lst2[2] != 0):
return 0
elif lst1[0] * lst2[1] == lst1[1] * lst2[0] and (
lst1[2] * lst2[1] != lst2[2] * lst1[1] or lst1[0] * lst2[2] != lst1[2] * lst2[0]):
return 0
elif lst1[0] * lst2[1] == lst1[1] * lst2[0] and lst1[2] * lst2[1] == lst2[2] * lst1[1]:
return -1
elif lst2[0] * lst1[1] == - lst1[0] * lst2[1] and lst1[2] == lst1[2] == 0:
return 1
a = [int(i) for i in input().split()]
b = [int(j) for j in input().split()]
print(intersection(a, b))
``` | instruction | 0 | 12,741 | 23 | 25,482 |
Yes | output | 1 | 12,741 | 23 | 25,483 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given two set of points. The first set is determined by the equation A1x + B1y + C1 = 0, and the second one is determined by the equation A2x + B2y + C2 = 0.
Write the program which finds the number of points in the intersection of two given sets.
Input
The first line of the input contains three integer numbers A1, B1, C1 separated by space. The second line contains three integer numbers A2, B2, C2 separated by space. All the numbers are between -100 and 100, inclusive.
Output
Print the number of points in the intersection or -1 if there are infinite number of points.
Examples
Input
1 1 0
2 2 0
Output
-1
Input
1 1 0
2 -2 0
Output
1
Submitted Solution:
```
A1, B1, C1 = tuple(map(int, input().split(" ")))
A2, B2, C2 = tuple(map(int, input().split(" ")))
D = A1*B2 - A2*B1
Dx = C1*B2 - C2*B1
Dy = C2*A1 - C1*A2
if(A1 == 0 and B1 == 0 and C1 != 0):
print(0)
elif(A2 == 0 and B2 == 0 and C2 != 0):
print(0)
elif(D == 0):
if(Dx == 0 and Dy == 0):
print(-1)
else:
print(0)
else:
print(1)
``` | instruction | 0 | 12,742 | 23 | 25,484 |
Yes | output | 1 | 12,742 | 23 | 25,485 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given two set of points. The first set is determined by the equation A1x + B1y + C1 = 0, and the second one is determined by the equation A2x + B2y + C2 = 0.
Write the program which finds the number of points in the intersection of two given sets.
Input
The first line of the input contains three integer numbers A1, B1, C1 separated by space. The second line contains three integer numbers A2, B2, C2 separated by space. All the numbers are between -100 and 100, inclusive.
Output
Print the number of points in the intersection or -1 if there are infinite number of points.
Examples
Input
1 1 0
2 2 0
Output
-1
Input
1 1 0
2 -2 0
Output
1
Submitted Solution:
```
a1, b1, c1 = [int(i) for i in input().split()]
a2, b2, c2 = [int(i) for i in input().split()]
if c1 != c2 and (a1 == 0 and b1 == 0) and (a2 == 0 and b2 == 0) :
print(0)
elif (a1 == 0 and b1 == 0 and c1 == 0) or (a2 == 0 and b2 == 0 and c2 == 0) :
print(-1)
elif (a1 == 0 and b1 == 0 and c1 != 0) or (a2 == 0 and b2 == 0 and c2 != 0) and (c1 == c2):
print(0)
elif a1 * b2 == b1 * a2 and b2 * c1 == c2 * b1 and c1 * a2 == c2 * a1 :
print(-1)
elif ((a1 * b2) - (a2 * b1)) == 0:
print(0)
else:
print(1)
``` | instruction | 0 | 12,743 | 23 | 25,486 |
Yes | output | 1 | 12,743 | 23 | 25,487 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given two set of points. The first set is determined by the equation A1x + B1y + C1 = 0, and the second one is determined by the equation A2x + B2y + C2 = 0.
Write the program which finds the number of points in the intersection of two given sets.
Input
The first line of the input contains three integer numbers A1, B1, C1 separated by space. The second line contains three integer numbers A2, B2, C2 separated by space. All the numbers are between -100 and 100, inclusive.
Output
Print the number of points in the intersection or -1 if there are infinite number of points.
Examples
Input
1 1 0
2 2 0
Output
-1
Input
1 1 0
2 -2 0
Output
1
Submitted Solution:
```
def intersection(lst1, lst2):
if lst1[0] * lst2[1] != lst1[1] * lst2[0]:
return 1
elif lst1[0] * lst2[1] == lst1[1] * lst2[0] and (
lst1[2] * lst2[1] != lst2[2] * lst1[1] or lst1[0] * lst2[2] != lst1[2] * lst2[0]):
return 0
elif lst1[0] * lst2[1] == lst1[1] * lst2[0] and lst1[2] * lst2[1] == lst2[2] * lst1[1]:
return -1
elif lst2[0] * lst1[1] == - lst1[0] * lst2[1] and lst1[2] == lst1[2] == 0:
return 1
elif lst1[0] * lst2[1] == lst1[1] * lst2[0] and (lst2[2] == 0 and lst1[2] != 0 or lst2[2] != 0 and lst1[2] == 0):
return 0
a = [int(i) for i in input().split()]
b = [int(j) for j in input().split()]
print(intersection(a, b))
``` | instruction | 0 | 12,745 | 23 | 25,490 |
No | output | 1 | 12,745 | 23 | 25,491 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given two set of points. The first set is determined by the equation A1x + B1y + C1 = 0, and the second one is determined by the equation A2x + B2y + C2 = 0.
Write the program which finds the number of points in the intersection of two given sets.
Input
The first line of the input contains three integer numbers A1, B1, C1 separated by space. The second line contains three integer numbers A2, B2, C2 separated by space. All the numbers are between -100 and 100, inclusive.
Output
Print the number of points in the intersection or -1 if there are infinite number of points.
Examples
Input
1 1 0
2 2 0
Output
-1
Input
1 1 0
2 -2 0
Output
1
Submitted Solution:
```
R = lambda:map(int, input().split())
a1, b1, c1 = R()
a2, b2, c2 = R()
if a1 * b2 - a2 * b1 == 0:
if c1 == c2 or a1 == b1 == c1 == 0 or a2 == b2 == c2 == 0:
print(-1)
else:
print(0)
else:
print(1)
``` | instruction | 0 | 12,746 | 23 | 25,492 |
No | output | 1 | 12,746 | 23 | 25,493 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given two set of points. The first set is determined by the equation A1x + B1y + C1 = 0, and the second one is determined by the equation A2x + B2y + C2 = 0.
Write the program which finds the number of points in the intersection of two given sets.
Input
The first line of the input contains three integer numbers A1, B1, C1 separated by space. The second line contains three integer numbers A2, B2, C2 separated by space. All the numbers are between -100 and 100, inclusive.
Output
Print the number of points in the intersection or -1 if there are infinite number of points.
Examples
Input
1 1 0
2 2 0
Output
-1
Input
1 1 0
2 -2 0
Output
1
Submitted Solution:
```
a1, b1, c1 = [int(x) for x in input().split()]
a2, b2, c2 = [int(x) for x in input().split()]
det = (a1*b2) - (a2*b1)
if det!=0:
print(1)
elif ((a1*b2)==(b1*a2) and (a1*c2)==(a2*c1) and (b1*c2)==(b2*c1)) and not (a1==0 and b1==0 and a2==0 and b2==0 and c1!=c2):
print(-1)
else:
print(0)
``` | instruction | 0 | 12,747 | 23 | 25,494 |
No | output | 1 | 12,747 | 23 | 25,495 |
Provide tags and a correct Python 3 solution for this coding contest problem.
There is a right triangle with legs of length a and b. Your task is to determine whether it is possible to locate the triangle on the plane in such a way that none of its sides is parallel to the coordinate axes. All the vertices must have integer coordinates. If there exists such a location, you have to output the appropriate coordinates of vertices.
Input
The first line contains two integers a, b (1 ≤ a, b ≤ 1000), separated by a single space.
Output
In the first line print either "YES" or "NO" (without the quotes) depending on whether the required location exists. If it does, print in the next three lines three pairs of integers — the coordinates of the triangle vertices, one pair per line. The coordinates must be integers, not exceeding 109 in their absolute value.
Examples
Input
1 1
Output
NO
Input
5 5
Output
YES
2 1
5 5
-2 4
Input
5 10
Output
YES
-10 4
-2 -2
1 2 | instruction | 0 | 12,809 | 23 | 25,618 |
Tags: brute force, geometry, implementation, math
Correct Solution:
```
a, b = map(int, input().split())
s = [i*i for i in range(1, a)]
t = [i*i for i in range(1, b)]
def r(x):
return int(x**0.5)
for i in s:
if a*a-i in s:
for j in t:
if b*b-j in t:
if i!=j and i*j-(a*a-i)*(b*b-j) == 0:
print('YES')
print(r(i),r(a*a-i))
print(r(j),-r(b*b-j))
print(0,0)
exit()
print('NO')
``` | output | 1 | 12,809 | 23 | 25,619 |
Provide tags and a correct Python 3 solution for this coding contest problem.
There is a right triangle with legs of length a and b. Your task is to determine whether it is possible to locate the triangle on the plane in such a way that none of its sides is parallel to the coordinate axes. All the vertices must have integer coordinates. If there exists such a location, you have to output the appropriate coordinates of vertices.
Input
The first line contains two integers a, b (1 ≤ a, b ≤ 1000), separated by a single space.
Output
In the first line print either "YES" or "NO" (without the quotes) depending on whether the required location exists. If it does, print in the next three lines three pairs of integers — the coordinates of the triangle vertices, one pair per line. The coordinates must be integers, not exceeding 109 in their absolute value.
Examples
Input
1 1
Output
NO
Input
5 5
Output
YES
2 1
5 5
-2 4
Input
5 10
Output
YES
-10 4
-2 -2
1 2 | instruction | 0 | 12,810 | 23 | 25,620 |
Tags: brute force, geometry, implementation, math
Correct Solution:
```
a,b=list(map(int,input().split()))
f=0
x1,y1,x2,y2=0,0,0,0
for x in range(1,a+1):
for y in range(1,a+1):
if (x**2+y**2==a**2) and (x!=0 and y!=0):
m=-x/y
if float(int(b/(1+m**2)**(0.5)))==b/(1+m**2)**(0.5) and float(int((b*m)/(1+m**2)**(0.5)))==(b*m)/(1+m**2)**(0.5):
x1=x
y1=y
x2=int(b/(1+m**2)**(0.5))
y2=int((b*m)/(1+m**2)**(0.5))
if x1==x2 or y1==y2:
x2=-x2
y2=-y2
f=1
break
if f:
break
if f:
print('YES')
print(0,0)
print(x1,y1)
print(x2,y2)
else:
print('NO')
``` | output | 1 | 12,810 | 23 | 25,621 |
Provide tags and a correct Python 3 solution for this coding contest problem.
There is a right triangle with legs of length a and b. Your task is to determine whether it is possible to locate the triangle on the plane in such a way that none of its sides is parallel to the coordinate axes. All the vertices must have integer coordinates. If there exists such a location, you have to output the appropriate coordinates of vertices.
Input
The first line contains two integers a, b (1 ≤ a, b ≤ 1000), separated by a single space.
Output
In the first line print either "YES" or "NO" (without the quotes) depending on whether the required location exists. If it does, print in the next three lines three pairs of integers — the coordinates of the triangle vertices, one pair per line. The coordinates must be integers, not exceeding 109 in their absolute value.
Examples
Input
1 1
Output
NO
Input
5 5
Output
YES
2 1
5 5
-2 4
Input
5 10
Output
YES
-10 4
-2 -2
1 2 | instruction | 0 | 12,811 | 23 | 25,622 |
Tags: brute force, geometry, implementation, math
Correct Solution:
```
a,b=map(int,input().split())
def get(a):
return list([i,j] for i in range(1,a) for j in range(1,a) if i*i+j*j==a*a)
A=get(a)
B=get(b)
for [a,b] in A:
for [c,d] in B:
if a*c==b*d and b!=d:
print("YES\n0 0")
print(-a,b)
print(c,d)
exit(0)
print("NO")
``` | output | 1 | 12,811 | 23 | 25,623 |
Provide tags and a correct Python 3 solution for this coding contest problem.
There is a right triangle with legs of length a and b. Your task is to determine whether it is possible to locate the triangle on the plane in such a way that none of its sides is parallel to the coordinate axes. All the vertices must have integer coordinates. If there exists such a location, you have to output the appropriate coordinates of vertices.
Input
The first line contains two integers a, b (1 ≤ a, b ≤ 1000), separated by a single space.
Output
In the first line print either "YES" or "NO" (without the quotes) depending on whether the required location exists. If it does, print in the next three lines three pairs of integers — the coordinates of the triangle vertices, one pair per line. The coordinates must be integers, not exceeding 109 in their absolute value.
Examples
Input
1 1
Output
NO
Input
5 5
Output
YES
2 1
5 5
-2 4
Input
5 10
Output
YES
-10 4
-2 -2
1 2 | instruction | 0 | 12,812 | 23 | 25,624 |
Tags: brute force, geometry, implementation, math
Correct Solution:
```
sqrt = {i * i: i for i in range(1, 1000)}
a, b = map(int, input().split())
for y in range(1, a):
x2 = a * a - y * y
if x2 in sqrt:
x = sqrt[x2]
if b * y % a == 0 and b * x % a == 0 and b * x // a != y:
print('YES')
print(-x, y)
print(0, 0)
print(b * y // a, b * x // a)
exit()
print('NO')
``` | output | 1 | 12,812 | 23 | 25,625 |
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