message stringlengths 2 44.5k | message_type stringclasses 2
values | message_id int64 0 1 | conversation_id int64 276 109k | cluster float64 23 23 | __index_level_0__ int64 552 217k |
|---|---|---|---|---|---|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
IT City administration has no rest because of the fame of the Pyramids in Egypt. There is a project of construction of pyramid complex near the city in the place called Emerald Walley. The distinction of the complex is that its pyramids will be not only quadrangular as in Egypt but also triangular and pentagonal. Of course the amount of the city budget funds for the construction depends on the pyramids' volume. Your task is to calculate the volume of the pilot project consisting of three pyramids — one triangular, one quadrangular and one pentagonal.
The first pyramid has equilateral triangle as its base, and all 6 edges of the pyramid have equal length. The second pyramid has a square as its base and all 8 edges of the pyramid have equal length. The third pyramid has a regular pentagon as its base and all 10 edges of the pyramid have equal length.
<image>
Input
The only line of the input contains three integers l3, l4, l5 (1 ≤ l3, l4, l5 ≤ 1000) — the edge lengths of triangular, quadrangular and pentagonal pyramids correspondingly.
Output
Output one number — the total volume of the pyramids. Absolute or relative error should not be greater than 10 - 9.
Examples
Input
2 5 3
Output
38.546168065709
Submitted Solution:
```
import os, sys, pdb
import time, calendar, datetime
import math, itertools
import operator as op
from functools import reduce
def ncr(n, r):
r = min(r, n-r)
numer = reduce(op.mul, range(n, n-r, -1), 1)
denom = reduce(op.mul, range(1, r+1), 1)
return numer // denom
l3, l4, l5 = list(map(int, input().split()))
a3 = l3 / (2 * math.tan(math.pi/3))
a4 = l4 / (2 * math.tan(math.pi/4))
a5 = l5 / (2 * math.tan(math.pi/5))
B3 = 3 * l3 * a3 / 2
B4 = 4 * l4 * a4 / 2
B5 = 5 * l5 * a5 / 2
h3 = math.sqrt(3*l3**2/4 - a3**2)
h4 = math.sqrt(3*l4**2/4 - a4**2)
h5 = math.sqrt(3*l5**2/4 - a5**2)
print((B3*h3+B4*h4+B5*h5)/3)
``` | instruction | 0 | 56,764 | 23 | 113,528 |
Yes | output | 1 | 56,764 | 23 | 113,529 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
IT City administration has no rest because of the fame of the Pyramids in Egypt. There is a project of construction of pyramid complex near the city in the place called Emerald Walley. The distinction of the complex is that its pyramids will be not only quadrangular as in Egypt but also triangular and pentagonal. Of course the amount of the city budget funds for the construction depends on the pyramids' volume. Your task is to calculate the volume of the pilot project consisting of three pyramids — one triangular, one quadrangular and one pentagonal.
The first pyramid has equilateral triangle as its base, and all 6 edges of the pyramid have equal length. The second pyramid has a square as its base and all 8 edges of the pyramid have equal length. The third pyramid has a regular pentagon as its base and all 10 edges of the pyramid have equal length.
<image>
Input
The only line of the input contains three integers l3, l4, l5 (1 ≤ l3, l4, l5 ≤ 1000) — the edge lengths of triangular, quadrangular and pentagonal pyramids correspondingly.
Output
Output one number — the total volume of the pyramids. Absolute or relative error should not be greater than 10 - 9.
Examples
Input
2 5 3
Output
38.546168065709
Submitted Solution:
```
sq2 = 2**0.5
def PyraVolume(number, edge_length):
el3 = edge_length**3
if number == 3:
return el3 / (6 * sq2)
elif number == 4:
return el3 / (3 * sq2)
elif number == 5:
return el3 * (5 + 5**0.5) / 24
a, b, c = map(int, input().split())
print(PyraVolume(3, a) + PyraVolume(4, b) + PyraVolume(5, c))
``` | instruction | 0 | 56,765 | 23 | 113,530 |
Yes | output | 1 | 56,765 | 23 | 113,531 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
IT City administration has no rest because of the fame of the Pyramids in Egypt. There is a project of construction of pyramid complex near the city in the place called Emerald Walley. The distinction of the complex is that its pyramids will be not only quadrangular as in Egypt but also triangular and pentagonal. Of course the amount of the city budget funds for the construction depends on the pyramids' volume. Your task is to calculate the volume of the pilot project consisting of three pyramids — one triangular, one quadrangular and one pentagonal.
The first pyramid has equilateral triangle as its base, and all 6 edges of the pyramid have equal length. The second pyramid has a square as its base and all 8 edges of the pyramid have equal length. The third pyramid has a regular pentagon as its base and all 10 edges of the pyramid have equal length.
<image>
Input
The only line of the input contains three integers l3, l4, l5 (1 ≤ l3, l4, l5 ≤ 1000) — the edge lengths of triangular, quadrangular and pentagonal pyramids correspondingly.
Output
Output one number — the total volume of the pyramids. Absolute or relative error should not be greater than 10 - 9.
Examples
Input
2 5 3
Output
38.546168065709
Submitted Solution:
```
__author__ = 'panag'
import sys
import math
l1, l2, l3 = map(int, sys.stdin.readline().split(' '))
#print(l1, l2, l3)
#1
v1 = (math.sqrt(3)/4*l1**2) * 1/3 * (math.sqrt(l1**2 - (math.sqrt(3)/6*l1)**2)) + (l2 ** 2 * 1/3 * (math.sqrt(l2**2 - (l2/math.sqrt(2))**2))) + (((math.sqrt(5)*math.sqrt(5+2*math.sqrt(5)))/4 * l3**2) * 1/3 * (math.sqrt(l3**2 - (math.sqrt(10)*math.sqrt(5+math.sqrt(5)))/10 * l3**2)))
print(v1)
``` | instruction | 0 | 56,766 | 23 | 113,532 |
No | output | 1 | 56,766 | 23 | 113,533 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
IT City administration has no rest because of the fame of the Pyramids in Egypt. There is a project of construction of pyramid complex near the city in the place called Emerald Walley. The distinction of the complex is that its pyramids will be not only quadrangular as in Egypt but also triangular and pentagonal. Of course the amount of the city budget funds for the construction depends on the pyramids' volume. Your task is to calculate the volume of the pilot project consisting of three pyramids — one triangular, one quadrangular and one pentagonal.
The first pyramid has equilateral triangle as its base, and all 6 edges of the pyramid have equal length. The second pyramid has a square as its base and all 8 edges of the pyramid have equal length. The third pyramid has a regular pentagon as its base and all 10 edges of the pyramid have equal length.
<image>
Input
The only line of the input contains three integers l3, l4, l5 (1 ≤ l3, l4, l5 ≤ 1000) — the edge lengths of triangular, quadrangular and pentagonal pyramids correspondingly.
Output
Output one number — the total volume of the pyramids. Absolute or relative error should not be greater than 10 - 9.
Examples
Input
2 5 3
Output
38.546168065709
Submitted Solution:
```
from math import *
l3,l4,l5 = map(int, input().split())
a = (l3*l3*l3*sqrt(2))/12
b = (l4*l4*l4*sqrt(2))/6
c = (l5*l5)/12
d = sqrt(25+10*sqrt(5))
e = (l5/(2*sin(radians(36))))*l5/(2*sin(radians(36)))
print(a+b+c*d*(l5*l5/e))
``` | instruction | 0 | 56,767 | 23 | 113,534 |
No | output | 1 | 56,767 | 23 | 113,535 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
IT City administration has no rest because of the fame of the Pyramids in Egypt. There is a project of construction of pyramid complex near the city in the place called Emerald Walley. The distinction of the complex is that its pyramids will be not only quadrangular as in Egypt but also triangular and pentagonal. Of course the amount of the city budget funds for the construction depends on the pyramids' volume. Your task is to calculate the volume of the pilot project consisting of three pyramids — one triangular, one quadrangular and one pentagonal.
The first pyramid has equilateral triangle as its base, and all 6 edges of the pyramid have equal length. The second pyramid has a square as its base and all 8 edges of the pyramid have equal length. The third pyramid has a regular pentagon as its base and all 10 edges of the pyramid have equal length.
<image>
Input
The only line of the input contains three integers l3, l4, l5 (1 ≤ l3, l4, l5 ≤ 1000) — the edge lengths of triangular, quadrangular and pentagonal pyramids correspondingly.
Output
Output one number — the total volume of the pyramids. Absolute or relative error should not be greater than 10 - 9.
Examples
Input
2 5 3
Output
38.546168065709
Submitted Solution:
```
from decimal import Decimal
l3, l4, l5 = map(Decimal, input().split())
h3 = (Decimal('2') / Decimal('3')).sqrt()
h4c = Decimal('3').sqrt() / Decimal('2')
h4 = (h4c * h4c - Decimal('1') / Decimal('4')).sqrt()
h5 = (Decimal('38.546168065709') - (h3 * 2 * 2 + h4 * 5 * 5) / 3) / 3
print((l3*l3*h3 + l4*l4*h4 + l5*l5*h5) / 3)
``` | instruction | 0 | 56,768 | 23 | 113,536 |
No | output | 1 | 56,768 | 23 | 113,537 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
IT City administration has no rest because of the fame of the Pyramids in Egypt. There is a project of construction of pyramid complex near the city in the place called Emerald Walley. The distinction of the complex is that its pyramids will be not only quadrangular as in Egypt but also triangular and pentagonal. Of course the amount of the city budget funds for the construction depends on the pyramids' volume. Your task is to calculate the volume of the pilot project consisting of three pyramids — one triangular, one quadrangular and one pentagonal.
The first pyramid has equilateral triangle as its base, and all 6 edges of the pyramid have equal length. The second pyramid has a square as its base and all 8 edges of the pyramid have equal length. The third pyramid has a regular pentagon as its base and all 10 edges of the pyramid have equal length.
<image>
Input
The only line of the input contains three integers l3, l4, l5 (1 ≤ l3, l4, l5 ≤ 1000) — the edge lengths of triangular, quadrangular and pentagonal pyramids correspondingly.
Output
Output one number — the total volume of the pyramids. Absolute or relative error should not be greater than 10 - 9.
Examples
Input
2 5 3
Output
38.546168065709
Submitted Solution:
```
l3,l4,l5=map(int,input().split())
V3=2**0.5/12*l3**3
V4=3**0.5/12*l4**3
V5=0.7244875226648567*l5**3
print(V3+V4+V5)
``` | instruction | 0 | 56,769 | 23 | 113,538 |
No | output | 1 | 56,769 | 23 | 113,539 |
Provide tags and a correct Python 2 solution for this coding contest problem.
You are given a convex polygon P with n distinct vertices p1, p2, ..., pn. Vertex pi has coordinates (xi, yi) in the 2D plane. These vertices are listed in clockwise order.
You can choose a real number D and move each vertex of the polygon a distance of at most D from their original positions.
Find the maximum value of D such that no matter how you move the vertices, the polygon does not intersect itself and stays convex.
Input
The first line has one integer n (4 ≤ n ≤ 1 000) — the number of vertices.
The next n lines contain the coordinates of the vertices. Line i contains two integers xi and yi ( - 109 ≤ xi, yi ≤ 109) — the coordinates of the i-th vertex. These points are guaranteed to be given in clockwise order, and will form a strictly convex polygon (in particular, no three consecutive points lie on the same straight line).
Output
Print one real number D, which is the maximum real number such that no matter how you move the vertices, the polygon stays convex.
Your answer will be considered correct if its absolute or relative error does not exceed 10 - 6.
Namely, let's assume that your answer is a and the answer of the jury is b. The checker program will consider your answer correct if <image>.
Examples
Input
4
0 0
0 1
1 1
1 0
Output
0.3535533906
Input
6
5 0
10 0
12 -4
10 -8
5 -8
3 -4
Output
1.0000000000
Note
Here is a picture of the first sample
<image>
Here is an example of making the polygon non-convex.
<image>
This is not an optimal solution, since the maximum distance we moved one point is ≈ 0.4242640687, whereas we can make it non-convex by only moving each point a distance of at most ≈ 0.3535533906. | instruction | 0 | 56,816 | 23 | 113,632 |
Tags: geometry
Correct Solution:
```
from sys import stdin, stdout
from collections import Counter, defaultdict
from itertools import permutations, combinations
raw_input = stdin.readline
from math import sqrt
pr = stdout.write
def in_num():
return int(raw_input())
def in_arr():
return map(int,raw_input().split())
def pr_num(n):
stdout.write(str(n)+'\n')
def pr_arr(arr):
pr(' '.join(map(str,arr))+'\n')
# fast read function for total integer input
def inp():
# this function returns whole input of
# space/line seperated integers
# Use Ctrl+D to flush stdin.
return map(int,stdin.read().split())
range = xrange # not for python 3.0+
def get_par(x1,y1,x2,y2):
if x1==x2:
return 1,0,-x1
elif y1==y2:
return 0,1,-y1
else:
return y2-y1,x1-x2,(y1*(x2-x1))-(x1*(y2-y1))
def get_dis2(x1,y1,x2,y2,x3,y3):
a,b,c=get_par(x2,y2,x3,y3)
#print a,b,c
a1=-b
b1=a
c1=-(a1*x1+b1*y1)
return (int(abs(((a*x1)+(b*y1)+c)))/sqrt(float(a**2+b**2)))
inp=inp()
n=inp[0]
pos=1
l=[(inp[pos],inp[pos+1]) for pos in range(1,2*n+1,2)]
l=[l[-1]]+l
l.append(l[1])
ans=10**18
for i in range(1,n+1):
#print i,ans
#print l[i][0],l[i][1],l[i-1][0],l[i-1][1],l[i+1][0],l[i+1][1]
ans=min(ans,get_dis2(l[i][0],l[i][1],l[i-1][0],l[i-1][1],
l[i+1][0],l[i+1][1]))
pr_num(ans/2.0)
``` | output | 1 | 56,816 | 23 | 113,633 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given a convex polygon P with n distinct vertices p1, p2, ..., pn. Vertex pi has coordinates (xi, yi) in the 2D plane. These vertices are listed in clockwise order.
You can choose a real number D and move each vertex of the polygon a distance of at most D from their original positions.
Find the maximum value of D such that no matter how you move the vertices, the polygon does not intersect itself and stays convex.
Input
The first line has one integer n (4 ≤ n ≤ 1 000) — the number of vertices.
The next n lines contain the coordinates of the vertices. Line i contains two integers xi and yi ( - 109 ≤ xi, yi ≤ 109) — the coordinates of the i-th vertex. These points are guaranteed to be given in clockwise order, and will form a strictly convex polygon (in particular, no three consecutive points lie on the same straight line).
Output
Print one real number D, which is the maximum real number such that no matter how you move the vertices, the polygon stays convex.
Your answer will be considered correct if its absolute or relative error does not exceed 10 - 6.
Namely, let's assume that your answer is a and the answer of the jury is b. The checker program will consider your answer correct if <image>.
Examples
Input
4
0 0
0 1
1 1
1 0
Output
0.3535533906
Input
6
5 0
10 0
12 -4
10 -8
5 -8
3 -4
Output
1.0000000000
Note
Here is a picture of the first sample
<image>
Here is an example of making the polygon non-convex.
<image>
This is not an optimal solution, since the maximum distance we moved one point is ≈ 0.4242640687, whereas we can make it non-convex by only moving each point a distance of at most ≈ 0.3535533906. | instruction | 0 | 56,817 | 23 | 113,634 |
Tags: geometry
Correct Solution:
```
from collections import Counter
n=int(input())
a=[]
for i in range(n):
a.append(list(map(int,input().split())))
#a.append(a[0])
m=10000000000
for i in range(-1,n-1):
b=a[i+1][0]-a[i-1][0]
c=a[i-1][1]-a[i+1][1]
d=-a[i-1][1]*b-a[i-1][0]*c
m=min(m,((a[i-1][0]-a[i][0])**2+(a[i-1][1]-a[i][1])**2)**0.5,\
((a[i+1][0]-a[i][0])**2+(a[i+1][1]-a[i][1])**2)**0.5,\
abs(b*a[i][1]+\
c*a[i][0]+\
d)/(b**2+c**2)**0.5)
print(m/2)
``` | output | 1 | 56,817 | 23 | 113,635 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given a convex polygon P with n distinct vertices p1, p2, ..., pn. Vertex pi has coordinates (xi, yi) in the 2D plane. These vertices are listed in clockwise order.
You can choose a real number D and move each vertex of the polygon a distance of at most D from their original positions.
Find the maximum value of D such that no matter how you move the vertices, the polygon does not intersect itself and stays convex.
Input
The first line has one integer n (4 ≤ n ≤ 1 000) — the number of vertices.
The next n lines contain the coordinates of the vertices. Line i contains two integers xi and yi ( - 109 ≤ xi, yi ≤ 109) — the coordinates of the i-th vertex. These points are guaranteed to be given in clockwise order, and will form a strictly convex polygon (in particular, no three consecutive points lie on the same straight line).
Output
Print one real number D, which is the maximum real number such that no matter how you move the vertices, the polygon stays convex.
Your answer will be considered correct if its absolute or relative error does not exceed 10 - 6.
Namely, let's assume that your answer is a and the answer of the jury is b. The checker program will consider your answer correct if <image>.
Examples
Input
4
0 0
0 1
1 1
1 0
Output
0.3535533906
Input
6
5 0
10 0
12 -4
10 -8
5 -8
3 -4
Output
1.0000000000
Note
Here is a picture of the first sample
<image>
Here is an example of making the polygon non-convex.
<image>
This is not an optimal solution, since the maximum distance we moved one point is ≈ 0.4242640687, whereas we can make it non-convex by only moving each point a distance of at most ≈ 0.3535533906. | instruction | 0 | 56,818 | 23 | 113,636 |
Tags: geometry
Correct Solution:
```
def calc(distance_1,distance_2,distance_3):
return ((2*distance_1*distance_3 + 2*distance_2*distance_3 - distance_3*distance_3 - (distance_1-distance_2)**2)/(16*distance_3))**0.5
def distance(point_1,point_2):
point_1,y1 = point_1
point_2,y2 = point_2
return (point_1-point_2)**2 + (y1-y2)**2
n = int(input().strip())
points = []
for _ in range(n):
points.append(tuple([int(x) for x in input().strip().split()]))
min_ = float("inf")
for i in range(n):
distance_1 = distance(points[i],points[(i+1)%n])
distance_2 = distance(points[(i+2)%n],points[(i+1)%n])
distance_3 = distance(points[(i+2)%n],points[i])
min_= min(min_,calc(distance_1,distance_2,distance_3))
print("%.9f"%(min_))
``` | output | 1 | 56,818 | 23 | 113,637 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given a convex polygon P with n distinct vertices p1, p2, ..., pn. Vertex pi has coordinates (xi, yi) in the 2D plane. These vertices are listed in clockwise order.
You can choose a real number D and move each vertex of the polygon a distance of at most D from their original positions.
Find the maximum value of D such that no matter how you move the vertices, the polygon does not intersect itself and stays convex.
Input
The first line has one integer n (4 ≤ n ≤ 1 000) — the number of vertices.
The next n lines contain the coordinates of the vertices. Line i contains two integers xi and yi ( - 109 ≤ xi, yi ≤ 109) — the coordinates of the i-th vertex. These points are guaranteed to be given in clockwise order, and will form a strictly convex polygon (in particular, no three consecutive points lie on the same straight line).
Output
Print one real number D, which is the maximum real number such that no matter how you move the vertices, the polygon stays convex.
Your answer will be considered correct if its absolute or relative error does not exceed 10 - 6.
Namely, let's assume that your answer is a and the answer of the jury is b. The checker program will consider your answer correct if <image>.
Examples
Input
4
0 0
0 1
1 1
1 0
Output
0.3535533906
Input
6
5 0
10 0
12 -4
10 -8
5 -8
3 -4
Output
1.0000000000
Note
Here is a picture of the first sample
<image>
Here is an example of making the polygon non-convex.
<image>
This is not an optimal solution, since the maximum distance we moved one point is ≈ 0.4242640687, whereas we can make it non-convex by only moving each point a distance of at most ≈ 0.3535533906. | instruction | 0 | 56,819 | 23 | 113,638 |
Tags: geometry
Correct Solution:
```
from math import inf
def vect(x, y):
return abs(sum([x[i]*(y[(i+1)%3]-y[(i+2)%3]) for i in range(3)]))
def l(x, y):
return ((x[0]-x[2])**2 + (y[0]-y[2])**2)**0.5
def h(x, y):
return vect(x, y) / l(x, y)
n = int(input())
x = []
y = []
for i in range(n):
a, b = [int(x) for x in input().split()]
x.append(a)
y.append(b)
x += x[:2]
y += y[:2]
dmin = inf
for i in range(n):
d = h(x[i:i+3], y[i:i+3])/2
if dmin > d:
dmin = d
print(dmin)
``` | output | 1 | 56,819 | 23 | 113,639 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given a convex polygon P with n distinct vertices p1, p2, ..., pn. Vertex pi has coordinates (xi, yi) in the 2D plane. These vertices are listed in clockwise order.
You can choose a real number D and move each vertex of the polygon a distance of at most D from their original positions.
Find the maximum value of D such that no matter how you move the vertices, the polygon does not intersect itself and stays convex.
Input
The first line has one integer n (4 ≤ n ≤ 1 000) — the number of vertices.
The next n lines contain the coordinates of the vertices. Line i contains two integers xi and yi ( - 109 ≤ xi, yi ≤ 109) — the coordinates of the i-th vertex. These points are guaranteed to be given in clockwise order, and will form a strictly convex polygon (in particular, no three consecutive points lie on the same straight line).
Output
Print one real number D, which is the maximum real number such that no matter how you move the vertices, the polygon stays convex.
Your answer will be considered correct if its absolute or relative error does not exceed 10 - 6.
Namely, let's assume that your answer is a and the answer of the jury is b. The checker program will consider your answer correct if <image>.
Examples
Input
4
0 0
0 1
1 1
1 0
Output
0.3535533906
Input
6
5 0
10 0
12 -4
10 -8
5 -8
3 -4
Output
1.0000000000
Note
Here is a picture of the first sample
<image>
Here is an example of making the polygon non-convex.
<image>
This is not an optimal solution, since the maximum distance we moved one point is ≈ 0.4242640687, whereas we can make it non-convex by only moving each point a distance of at most ≈ 0.3535533906. | instruction | 0 | 56,820 | 23 | 113,640 |
Tags: geometry
Correct Solution:
```
import sys
import math
data = sys.stdin.read().split()
data_ptr = 0
def data_next():
global data_ptr, data
data_ptr += 1
return data[data_ptr - 1]
N = int(data_next())
arr = list(zip(map(int, data[1::2]), map(int, data[2::2])))
def cross(x1, y1, x2, y2):
return x1 * y2 - x2 * y1
def dist(x, y):
return math.sqrt(x * x + y * y)
ans = 4000000000.0
for i in range(N):
x1, y1 = arr[i - 1]
x2, y2 = arr[i]
x1 -= arr[i - 2][0]
x2 -= arr[i - 2][0]
y1 -= arr[i - 2][1]
y2 -= arr[i - 2][1]
ans = min(ans, abs(cross(x1, y1, x2, y2) / dist(x2, y2) / 2))
print(ans)
``` | output | 1 | 56,820 | 23 | 113,641 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given a convex polygon P with n distinct vertices p1, p2, ..., pn. Vertex pi has coordinates (xi, yi) in the 2D plane. These vertices are listed in clockwise order.
You can choose a real number D and move each vertex of the polygon a distance of at most D from their original positions.
Find the maximum value of D such that no matter how you move the vertices, the polygon does not intersect itself and stays convex.
Input
The first line has one integer n (4 ≤ n ≤ 1 000) — the number of vertices.
The next n lines contain the coordinates of the vertices. Line i contains two integers xi and yi ( - 109 ≤ xi, yi ≤ 109) — the coordinates of the i-th vertex. These points are guaranteed to be given in clockwise order, and will form a strictly convex polygon (in particular, no three consecutive points lie on the same straight line).
Output
Print one real number D, which is the maximum real number such that no matter how you move the vertices, the polygon stays convex.
Your answer will be considered correct if its absolute or relative error does not exceed 10 - 6.
Namely, let's assume that your answer is a and the answer of the jury is b. The checker program will consider your answer correct if <image>.
Examples
Input
4
0 0
0 1
1 1
1 0
Output
0.3535533906
Input
6
5 0
10 0
12 -4
10 -8
5 -8
3 -4
Output
1.0000000000
Note
Here is a picture of the first sample
<image>
Here is an example of making the polygon non-convex.
<image>
This is not an optimal solution, since the maximum distance we moved one point is ≈ 0.4242640687, whereas we can make it non-convex by only moving each point a distance of at most ≈ 0.3535533906. | instruction | 0 | 56,821 | 23 | 113,642 |
Tags: geometry
Correct Solution:
```
n = int(input())
P = []
def h(p1, p2, m):
return abs(((p2[0] - p1[0])*(m[1] - p1[1])-(p2[1] - p1[1])*(m[0] - p1[0]))\
/((p2[0] - p1[0]) ** 2 + (p2[1] - p1[1])**2) ** 0.5)
for i in range(n):
P.append(list(map(int, input().split())))
answer = float("inf")
for i in range(n):
cw = P[(i + 1) % n]
ucw = P[i - 1]
answer = min(answer, h(cw, ucw, P[i])/2)
print(answer)
``` | output | 1 | 56,821 | 23 | 113,643 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given a convex polygon P with n distinct vertices p1, p2, ..., pn. Vertex pi has coordinates (xi, yi) in the 2D plane. These vertices are listed in clockwise order.
You can choose a real number D and move each vertex of the polygon a distance of at most D from their original positions.
Find the maximum value of D such that no matter how you move the vertices, the polygon does not intersect itself and stays convex.
Input
The first line has one integer n (4 ≤ n ≤ 1 000) — the number of vertices.
The next n lines contain the coordinates of the vertices. Line i contains two integers xi and yi ( - 109 ≤ xi, yi ≤ 109) — the coordinates of the i-th vertex. These points are guaranteed to be given in clockwise order, and will form a strictly convex polygon (in particular, no three consecutive points lie on the same straight line).
Output
Print one real number D, which is the maximum real number such that no matter how you move the vertices, the polygon stays convex.
Your answer will be considered correct if its absolute or relative error does not exceed 10 - 6.
Namely, let's assume that your answer is a and the answer of the jury is b. The checker program will consider your answer correct if <image>.
Examples
Input
4
0 0
0 1
1 1
1 0
Output
0.3535533906
Input
6
5 0
10 0
12 -4
10 -8
5 -8
3 -4
Output
1.0000000000
Note
Here is a picture of the first sample
<image>
Here is an example of making the polygon non-convex.
<image>
This is not an optimal solution, since the maximum distance we moved one point is ≈ 0.4242640687, whereas we can make it non-convex by only moving each point a distance of at most ≈ 0.3535533906. | instruction | 0 | 56,822 | 23 | 113,644 |
Tags: geometry
Correct Solution:
```
import math
def dist(a, b, c):
return abs((c[1] - a[1])*b[0] - (c[0] - a[0])*b[1] + c[0]*a[1] - c[1]*a[0])/math.hypot(c[0]-a[0], c[1]- a[1]);
N = int(input())
pts = [tuple(map(int, input().split())) for _ in range(N)]
MIN = 10000000000.0
for i in range(N-2):
MIN = min(MIN, dist(pts[i], pts[i+1], pts[i+2]))
MIN = min(MIN, dist(pts[N-2], pts[N-1], pts[0]))
MIN = min(MIN, dist(pts[N-1], pts[0], pts[1]))
print(MIN/2)
# Made By Mostafa_Khaled
``` | output | 1 | 56,822 | 23 | 113,645 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given a convex polygon P with n distinct vertices p1, p2, ..., pn. Vertex pi has coordinates (xi, yi) in the 2D plane. These vertices are listed in clockwise order.
You can choose a real number D and move each vertex of the polygon a distance of at most D from their original positions.
Find the maximum value of D such that no matter how you move the vertices, the polygon does not intersect itself and stays convex.
Input
The first line has one integer n (4 ≤ n ≤ 1 000) — the number of vertices.
The next n lines contain the coordinates of the vertices. Line i contains two integers xi and yi ( - 109 ≤ xi, yi ≤ 109) — the coordinates of the i-th vertex. These points are guaranteed to be given in clockwise order, and will form a strictly convex polygon (in particular, no three consecutive points lie on the same straight line).
Output
Print one real number D, which is the maximum real number such that no matter how you move the vertices, the polygon stays convex.
Your answer will be considered correct if its absolute or relative error does not exceed 10 - 6.
Namely, let's assume that your answer is a and the answer of the jury is b. The checker program will consider your answer correct if <image>.
Examples
Input
4
0 0
0 1
1 1
1 0
Output
0.3535533906
Input
6
5 0
10 0
12 -4
10 -8
5 -8
3 -4
Output
1.0000000000
Note
Here is a picture of the first sample
<image>
Here is an example of making the polygon non-convex.
<image>
This is not an optimal solution, since the maximum distance we moved one point is ≈ 0.4242640687, whereas we can make it non-convex by only moving each point a distance of at most ≈ 0.3535533906. | instruction | 0 | 56,823 | 23 | 113,646 |
Tags: geometry
Correct Solution:
```
import math
class Vector:
@staticmethod
def from_vectors(v1, v2):
return Vector(v2.x - v1.x, v2.y - v1.y)
def __init__(self, x, y):
self.x = x
self.y = y
def len(self):
return math.sqrt(self.x ** 2 + self.y ** 2)
def cross_product(v1, v2):
return v1.x * v2.y - v1.y * v2.x
n = int(input())
points = [Vector(*map(int, input().split())) for i in range(n)]
ans = float('inf')
for cur in range(n):
prev = (cur - 1) % n
nxt = (cur + 1) % n
left = Vector.from_vectors(points[cur], points[prev])
right = Vector.from_vectors(points[cur], points[nxt])
base_len = Vector.from_vectors(points[prev], points[nxt]).len()
area = abs(cross_product(left, right)) / 2
ans = min(ans, area / base_len)
print(ans)
``` | output | 1 | 56,823 | 23 | 113,647 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given a convex polygon P with n distinct vertices p1, p2, ..., pn. Vertex pi has coordinates (xi, yi) in the 2D plane. These vertices are listed in clockwise order.
You can choose a real number D and move each vertex of the polygon a distance of at most D from their original positions.
Find the maximum value of D such that no matter how you move the vertices, the polygon does not intersect itself and stays convex.
Input
The first line has one integer n (4 ≤ n ≤ 1 000) — the number of vertices.
The next n lines contain the coordinates of the vertices. Line i contains two integers xi and yi ( - 109 ≤ xi, yi ≤ 109) — the coordinates of the i-th vertex. These points are guaranteed to be given in clockwise order, and will form a strictly convex polygon (in particular, no three consecutive points lie on the same straight line).
Output
Print one real number D, which is the maximum real number such that no matter how you move the vertices, the polygon stays convex.
Your answer will be considered correct if its absolute or relative error does not exceed 10 - 6.
Namely, let's assume that your answer is a and the answer of the jury is b. The checker program will consider your answer correct if <image>.
Examples
Input
4
0 0
0 1
1 1
1 0
Output
0.3535533906
Input
6
5 0
10 0
12 -4
10 -8
5 -8
3 -4
Output
1.0000000000
Note
Here is a picture of the first sample
<image>
Here is an example of making the polygon non-convex.
<image>
This is not an optimal solution, since the maximum distance we moved one point is ≈ 0.4242640687, whereas we can make it non-convex by only moving each point a distance of at most ≈ 0.3535533906. | instruction | 0 | 56,824 | 23 | 113,648 |
Tags: geometry
Correct Solution:
```
class Vector:
def __init__(self, x, y):
self.x = x
self.y = y
def __add__(self, other):
return Vector(self.x + other.x, self.y + other.y)
def __sub__(self, other):
return Vector(self.x - other.x, self.y - other.y)
def to(self, other):
return other - self
def __repr__(self):
return "(%s %s)" % (self.x, self.y)
def dot(self, other):
return self.x * other.y - self.y * other.x
def lensq(self):
return self.x ** 2 + self.y ** 2
Vec = Vector
def getH(p, a, b):
s2 = p.to(a).dot(p.to(b))
# a * h / 2 = s
# h = s * 2 / a
return s2 / (a.to(b).lensq() ** 0.5)
pts = [Vec(*map(int, input().split())) for i in range(int(input()))]
n = len(pts)
pts.append(pts[0])
pts.append(pts[1])
ans = 12351513153155135135
for i in range(n):
ans = min(ans, getH(pts[i + 1], pts[i], pts[i + 2])/2)
print(ans)
``` | output | 1 | 56,824 | 23 | 113,649 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a convex polygon P with n distinct vertices p1, p2, ..., pn. Vertex pi has coordinates (xi, yi) in the 2D plane. These vertices are listed in clockwise order.
You can choose a real number D and move each vertex of the polygon a distance of at most D from their original positions.
Find the maximum value of D such that no matter how you move the vertices, the polygon does not intersect itself and stays convex.
Input
The first line has one integer n (4 ≤ n ≤ 1 000) — the number of vertices.
The next n lines contain the coordinates of the vertices. Line i contains two integers xi and yi ( - 109 ≤ xi, yi ≤ 109) — the coordinates of the i-th vertex. These points are guaranteed to be given in clockwise order, and will form a strictly convex polygon (in particular, no three consecutive points lie on the same straight line).
Output
Print one real number D, which is the maximum real number such that no matter how you move the vertices, the polygon stays convex.
Your answer will be considered correct if its absolute or relative error does not exceed 10 - 6.
Namely, let's assume that your answer is a and the answer of the jury is b. The checker program will consider your answer correct if <image>.
Examples
Input
4
0 0
0 1
1 1
1 0
Output
0.3535533906
Input
6
5 0
10 0
12 -4
10 -8
5 -8
3 -4
Output
1.0000000000
Note
Here is a picture of the first sample
<image>
Here is an example of making the polygon non-convex.
<image>
This is not an optimal solution, since the maximum distance we moved one point is ≈ 0.4242640687, whereas we can make it non-convex by only moving each point a distance of at most ≈ 0.3535533906.
Submitted Solution:
```
import math
class Pt:
def __init__(self, x, y):
self.x, self.y = x, y
def __sub__(self, other):
return Pt(self.x - other.x, self.y - other.y)
def __mul__(self, other):
return self.x * other.x + self.y * other.y
def __xor__(self, other):
return self.x * other.y - self.y * other.x
def len(self):
return math.sqrt(self.x ** 2 + self.y ** 2)
n = int(input())
data = []
for i in range(n):
data.append(Pt(*list(map(int, input().split()))))
min_ = 10 ** 10
for i in range(1, n - 1):
A = data[i - 1]
B = data[i + 1]
C = data[i]
r = abs((C - A) ^ (B - A)) / (B - A).len()
min_ = min(min_, r)
A = data[0]
B = data[n - 2]
C = data[n - 1]
r = abs((C - A) ^ (B - A)) / (B - A).len()
min_ = min(min_, r)
A = data[1]
B = data[n - 1]
C = data[0]
r = abs((C - A) ^ (B - A)) / (B - A).len()
min_ = min(min_, r)
print(min_ / 2)
``` | instruction | 0 | 56,825 | 23 | 113,650 |
Yes | output | 1 | 56,825 | 23 | 113,651 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a convex polygon P with n distinct vertices p1, p2, ..., pn. Vertex pi has coordinates (xi, yi) in the 2D plane. These vertices are listed in clockwise order.
You can choose a real number D and move each vertex of the polygon a distance of at most D from their original positions.
Find the maximum value of D such that no matter how you move the vertices, the polygon does not intersect itself and stays convex.
Input
The first line has one integer n (4 ≤ n ≤ 1 000) — the number of vertices.
The next n lines contain the coordinates of the vertices. Line i contains two integers xi and yi ( - 109 ≤ xi, yi ≤ 109) — the coordinates of the i-th vertex. These points are guaranteed to be given in clockwise order, and will form a strictly convex polygon (in particular, no three consecutive points lie on the same straight line).
Output
Print one real number D, which is the maximum real number such that no matter how you move the vertices, the polygon stays convex.
Your answer will be considered correct if its absolute or relative error does not exceed 10 - 6.
Namely, let's assume that your answer is a and the answer of the jury is b. The checker program will consider your answer correct if <image>.
Examples
Input
4
0 0
0 1
1 1
1 0
Output
0.3535533906
Input
6
5 0
10 0
12 -4
10 -8
5 -8
3 -4
Output
1.0000000000
Note
Here is a picture of the first sample
<image>
Here is an example of making the polygon non-convex.
<image>
This is not an optimal solution, since the maximum distance we moved one point is ≈ 0.4242640687, whereas we can make it non-convex by only moving each point a distance of at most ≈ 0.3535533906.
Submitted Solution:
```
# from decimal import *
# getcontext().prec=16
# from math import sqrt
# from scipy.special import binom
# from collections import defaultdict
# from math import sin,pi,sqrt
from math import sqrt,hypot
def dist(a, b, c):
return abs((c[1] - a[1])*b[0] - (c[0] - a[0])*b[1] + c[0]*a[1] - c[1]*a[0])/hypot(c[0]-a[0], c[1]- a[1])/2
n=int(input())
liste=[ list(map(int,input().split(" "))) for _ in range(n)]
d=-1
for i in range(n):
if i==n-1:
a,b,c=liste[n-2],liste[n-1],liste[0]
else:
a,b,c=liste[i-1],liste[i],liste[i+1]
if d!=-1:
d=min(d,dist(a,b,c))
else:
d=dist(a,b,c)
print(d)
``` | instruction | 0 | 56,826 | 23 | 113,652 |
Yes | output | 1 | 56,826 | 23 | 113,653 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a convex polygon P with n distinct vertices p1, p2, ..., pn. Vertex pi has coordinates (xi, yi) in the 2D plane. These vertices are listed in clockwise order.
You can choose a real number D and move each vertex of the polygon a distance of at most D from their original positions.
Find the maximum value of D such that no matter how you move the vertices, the polygon does not intersect itself and stays convex.
Input
The first line has one integer n (4 ≤ n ≤ 1 000) — the number of vertices.
The next n lines contain the coordinates of the vertices. Line i contains two integers xi and yi ( - 109 ≤ xi, yi ≤ 109) — the coordinates of the i-th vertex. These points are guaranteed to be given in clockwise order, and will form a strictly convex polygon (in particular, no three consecutive points lie on the same straight line).
Output
Print one real number D, which is the maximum real number such that no matter how you move the vertices, the polygon stays convex.
Your answer will be considered correct if its absolute or relative error does not exceed 10 - 6.
Namely, let's assume that your answer is a and the answer of the jury is b. The checker program will consider your answer correct if <image>.
Examples
Input
4
0 0
0 1
1 1
1 0
Output
0.3535533906
Input
6
5 0
10 0
12 -4
10 -8
5 -8
3 -4
Output
1.0000000000
Note
Here is a picture of the first sample
<image>
Here is an example of making the polygon non-convex.
<image>
This is not an optimal solution, since the maximum distance we moved one point is ≈ 0.4242640687, whereas we can make it non-convex by only moving each point a distance of at most ≈ 0.3535533906.
Submitted Solution:
```
import math
def dist(a, b, c):
return abs((c[1] - a[1])*b[0] - (c[0] - a[0])*b[1] + c[0]*a[1] - c[1]*a[0])/math.hypot(c[0]-a[0], c[1]- a[1]);
N = int(input())
pts = [tuple(map(int, input().split())) for _ in range(N)]
MIN = 10000000000.0
for i in range(N-2):
MIN = min(MIN, dist(pts[i], pts[i+1], pts[i+2]))
MIN = min(MIN, dist(pts[N-2], pts[N-1], pts[0]))
MIN = min(MIN, dist(pts[N-1], pts[0], pts[1]))
print(MIN/2)
``` | instruction | 0 | 56,827 | 23 | 113,654 |
Yes | output | 1 | 56,827 | 23 | 113,655 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a convex polygon P with n distinct vertices p1, p2, ..., pn. Vertex pi has coordinates (xi, yi) in the 2D plane. These vertices are listed in clockwise order.
You can choose a real number D and move each vertex of the polygon a distance of at most D from their original positions.
Find the maximum value of D such that no matter how you move the vertices, the polygon does not intersect itself and stays convex.
Input
The first line has one integer n (4 ≤ n ≤ 1 000) — the number of vertices.
The next n lines contain the coordinates of the vertices. Line i contains two integers xi and yi ( - 109 ≤ xi, yi ≤ 109) — the coordinates of the i-th vertex. These points are guaranteed to be given in clockwise order, and will form a strictly convex polygon (in particular, no three consecutive points lie on the same straight line).
Output
Print one real number D, which is the maximum real number such that no matter how you move the vertices, the polygon stays convex.
Your answer will be considered correct if its absolute or relative error does not exceed 10 - 6.
Namely, let's assume that your answer is a and the answer of the jury is b. The checker program will consider your answer correct if <image>.
Examples
Input
4
0 0
0 1
1 1
1 0
Output
0.3535533906
Input
6
5 0
10 0
12 -4
10 -8
5 -8
3 -4
Output
1.0000000000
Note
Here is a picture of the first sample
<image>
Here is an example of making the polygon non-convex.
<image>
This is not an optimal solution, since the maximum distance we moved one point is ≈ 0.4242640687, whereas we can make it non-convex by only moving each point a distance of at most ≈ 0.3535533906.
Submitted Solution:
```
#!/usr/bin/env python3
from decimal import Decimal
def dist(a, b):
x1, y1 = a
x2, y2 = b
return Decimal((x1-x2)**2+(y1-y2)**2).sqrt()
def minh(a, b, c):
m = dist(a, b)
n = dist(b, c)
k = dist(a, c)
p = Decimal(m + n + k)/2
sqp = (p*(p-m)*(p-n)*(p-k)).sqrt()
hm = (Decimal(2)/m)*sqp
hn = (Decimal(2)/n)*sqp
hk = (Decimal(2)/k)*sqp
return min([hm, hn, hk])
def solve():
n = int(input())
coords = []
for i in range(n):
coords.append(tuple(map(int, input().split())))
coords += coords
res = min(
minh(coords[i], coords[i+1], coords[i+2])
for i in range(n))
print(res/2)
if __name__ == '__main__':
solve()
``` | instruction | 0 | 56,828 | 23 | 113,656 |
Yes | output | 1 | 56,828 | 23 | 113,657 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a convex polygon P with n distinct vertices p1, p2, ..., pn. Vertex pi has coordinates (xi, yi) in the 2D plane. These vertices are listed in clockwise order.
You can choose a real number D and move each vertex of the polygon a distance of at most D from their original positions.
Find the maximum value of D such that no matter how you move the vertices, the polygon does not intersect itself and stays convex.
Input
The first line has one integer n (4 ≤ n ≤ 1 000) — the number of vertices.
The next n lines contain the coordinates of the vertices. Line i contains two integers xi and yi ( - 109 ≤ xi, yi ≤ 109) — the coordinates of the i-th vertex. These points are guaranteed to be given in clockwise order, and will form a strictly convex polygon (in particular, no three consecutive points lie on the same straight line).
Output
Print one real number D, which is the maximum real number such that no matter how you move the vertices, the polygon stays convex.
Your answer will be considered correct if its absolute or relative error does not exceed 10 - 6.
Namely, let's assume that your answer is a and the answer of the jury is b. The checker program will consider your answer correct if <image>.
Examples
Input
4
0 0
0 1
1 1
1 0
Output
0.3535533906
Input
6
5 0
10 0
12 -4
10 -8
5 -8
3 -4
Output
1.0000000000
Note
Here is a picture of the first sample
<image>
Here is an example of making the polygon non-convex.
<image>
This is not an optimal solution, since the maximum distance we moved one point is ≈ 0.4242640687, whereas we can make it non-convex by only moving each point a distance of at most ≈ 0.3535533906.
Submitted Solution:
```
import math
def length(point1, point2, point3):
a1 = int(point1[0])
a2 = int(point1[1])
b1 = int(point2[0])
b2 = int(point2[1])
c1 = int(point3[0])
c2 = int(point3[1])
AB = math.sqrt((a1 - b1)**2 + (a2 - b2)**2)
BC = math.sqrt((b1 - c1)**2 + (b2 - c2)**2)
AC = math.sqrt((a1 - c1)**2 + (a2 - c2)**2)
cosAngle = (AB**2 + AC**2 - BC**2) / (2*AB*AC)
AD = cosAngle*AB
length = math.sqrt(AB**2 - AD**2)
return length
lyst = []
minLyst = []
num = int(input())
for i in range(0,num):
x = input()
lyst.append(x.split())
for i in range(0,num):
minLyst.append(length(lyst[i%num], lyst[(i+1)%num], lyst[(i+2)%num])/2)
print(round(min(minLyst),10))
``` | instruction | 0 | 56,829 | 23 | 113,658 |
No | output | 1 | 56,829 | 23 | 113,659 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a convex polygon P with n distinct vertices p1, p2, ..., pn. Vertex pi has coordinates (xi, yi) in the 2D plane. These vertices are listed in clockwise order.
You can choose a real number D and move each vertex of the polygon a distance of at most D from their original positions.
Find the maximum value of D such that no matter how you move the vertices, the polygon does not intersect itself and stays convex.
Input
The first line has one integer n (4 ≤ n ≤ 1 000) — the number of vertices.
The next n lines contain the coordinates of the vertices. Line i contains two integers xi and yi ( - 109 ≤ xi, yi ≤ 109) — the coordinates of the i-th vertex. These points are guaranteed to be given in clockwise order, and will form a strictly convex polygon (in particular, no three consecutive points lie on the same straight line).
Output
Print one real number D, which is the maximum real number such that no matter how you move the vertices, the polygon stays convex.
Your answer will be considered correct if its absolute or relative error does not exceed 10 - 6.
Namely, let's assume that your answer is a and the answer of the jury is b. The checker program will consider your answer correct if <image>.
Examples
Input
4
0 0
0 1
1 1
1 0
Output
0.3535533906
Input
6
5 0
10 0
12 -4
10 -8
5 -8
3 -4
Output
1.0000000000
Note
Here is a picture of the first sample
<image>
Here is an example of making the polygon non-convex.
<image>
This is not an optimal solution, since the maximum distance we moved one point is ≈ 0.4242640687, whereas we can make it non-convex by only moving each point a distance of at most ≈ 0.3535533906.
Submitted Solution:
```
# from decimal import *
# getcontext().prec=16
# from math import sqrt
# from scipy.special import binom
# from collections import defaultdict
# from math import sin,pi,sqrt
from math import sqrt
def dist(a,b,c):
if c[0]!=a[0]:
slope=(c[1]-a[1])/(c[0]-a[0])
if slope==0:
distance=abs(b[1]-a[1])/2
else:
ori=(a[1]-slope*a[0])
distance=abs(b[0]-slope*b[1]-ori)/sqrt(1+slope**2)/2
else:
distance=abs(b[0]-a[0])/2
return distance
n=int(input())
liste=[ list(map(int,input().split(" "))) for _ in range(n)]
d=-1
for i in range(n):
if i==n-1:
a,b,c=liste[n-2],liste[n-1],liste[0]
else:
a,b,c=liste[i-1],liste[i],liste[i+1]
if d!=-1:
d=min(d,dist(a,b,c))
else:
d=dist(a,b,c)
print(d)
``` | instruction | 0 | 56,830 | 23 | 113,660 |
No | output | 1 | 56,830 | 23 | 113,661 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a convex polygon P with n distinct vertices p1, p2, ..., pn. Vertex pi has coordinates (xi, yi) in the 2D plane. These vertices are listed in clockwise order.
You can choose a real number D and move each vertex of the polygon a distance of at most D from their original positions.
Find the maximum value of D such that no matter how you move the vertices, the polygon does not intersect itself and stays convex.
Input
The first line has one integer n (4 ≤ n ≤ 1 000) — the number of vertices.
The next n lines contain the coordinates of the vertices. Line i contains two integers xi and yi ( - 109 ≤ xi, yi ≤ 109) — the coordinates of the i-th vertex. These points are guaranteed to be given in clockwise order, and will form a strictly convex polygon (in particular, no three consecutive points lie on the same straight line).
Output
Print one real number D, which is the maximum real number such that no matter how you move the vertices, the polygon stays convex.
Your answer will be considered correct if its absolute or relative error does not exceed 10 - 6.
Namely, let's assume that your answer is a and the answer of the jury is b. The checker program will consider your answer correct if <image>.
Examples
Input
4
0 0
0 1
1 1
1 0
Output
0.3535533906
Input
6
5 0
10 0
12 -4
10 -8
5 -8
3 -4
Output
1.0000000000
Note
Here is a picture of the first sample
<image>
Here is an example of making the polygon non-convex.
<image>
This is not an optimal solution, since the maximum distance we moved one point is ≈ 0.4242640687, whereas we can make it non-convex by only moving each point a distance of at most ≈ 0.3535533906.
Submitted Solution:
```
n=int(input())
covex=[]
for i in range (n):
tmp=input()
node=tmp.split(" ")
covex.append([int(node[0]),int(node[1])])
d=-1
for i in range(n):
a=((covex[i][0]-covex[i-1][0])**2+(covex[i][1]-covex[i-1][1])**2)**0.5
b=((covex[i][0]-covex[(i+1)%n][0])**2+(covex[i][1]-covex[(i+1)%n][1])**2)**0.5
c=((covex[i-1][0]-covex[(i+1)%n][0])**2+(covex[i-1][1]-covex[(i+1)%n][1])**2)**0.5
cos=((covex[i-1][0]-covex[i][0])*(covex[(i+1)%n][0]-covex[i][0])+(covex[i-1][1]-covex[i][1])*(covex[(i+1)%n][1]-covex[i][1]))/(a*b)
sin=(1-cos**2)**0.5
dd=a*b*sin/(2*c)
if d==-1:
d=dd
elif d>dd:
d=dd
print(round(d,10))
``` | instruction | 0 | 56,831 | 23 | 113,662 |
No | output | 1 | 56,831 | 23 | 113,663 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a convex polygon P with n distinct vertices p1, p2, ..., pn. Vertex pi has coordinates (xi, yi) in the 2D plane. These vertices are listed in clockwise order.
You can choose a real number D and move each vertex of the polygon a distance of at most D from their original positions.
Find the maximum value of D such that no matter how you move the vertices, the polygon does not intersect itself and stays convex.
Input
The first line has one integer n (4 ≤ n ≤ 1 000) — the number of vertices.
The next n lines contain the coordinates of the vertices. Line i contains two integers xi and yi ( - 109 ≤ xi, yi ≤ 109) — the coordinates of the i-th vertex. These points are guaranteed to be given in clockwise order, and will form a strictly convex polygon (in particular, no three consecutive points lie on the same straight line).
Output
Print one real number D, which is the maximum real number such that no matter how you move the vertices, the polygon stays convex.
Your answer will be considered correct if its absolute or relative error does not exceed 10 - 6.
Namely, let's assume that your answer is a and the answer of the jury is b. The checker program will consider your answer correct if <image>.
Examples
Input
4
0 0
0 1
1 1
1 0
Output
0.3535533906
Input
6
5 0
10 0
12 -4
10 -8
5 -8
3 -4
Output
1.0000000000
Note
Here is a picture of the first sample
<image>
Here is an example of making the polygon non-convex.
<image>
This is not an optimal solution, since the maximum distance we moved one point is ≈ 0.4242640687, whereas we can make it non-convex by only moving each point a distance of at most ≈ 0.3535533906.
Submitted Solution:
```
d = lambda i, j: (t[i][0] - t[j % n][0]) ** 2 + (t[i][1] - t[j % n][1]) ** 2
n = int(input())
t = [list(map(int, input().split())) for q in range(n)]
h = 1e13
for i in range(n):
a, b, c = d(i - 1, i), d(i, i + 1), d(i - 1, i + 1)
h = min(h, (4 * a * b - (a + b - c) ** 2) / c)
print(h ** 0.5 / 4)
``` | instruction | 0 | 56,832 | 23 | 113,664 |
No | output | 1 | 56,832 | 23 | 113,665 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Everything red frightens Nian the monster. So do red paper and... you, red on Codeforces, potential or real.
Big Banban has got a piece of paper with endless lattice points, where lattice points form squares with the same area. His most favorite closed shape is the circle because of its beauty and simplicity. Once he had obtained this piece of paper, he prepares it for paper-cutting.
<image>
He drew n concentric circles on it and numbered these circles from 1 to n such that the center of each circle is the same lattice point and the radius of the k-th circle is <image> times the length of a lattice edge.
Define the degree of beauty of a lattice point as the summation of the indices of circles such that this lattice point is inside them, or on their bounds. Banban wanted to ask you the total degree of beauty of all the lattice points, but changed his mind.
Defining the total degree of beauty of all the lattice points on a piece of paper with n circles as f(n), you are asked to figure out <image>.
Input
The first line contains one integer m (1 ≤ m ≤ 1012).
Output
In the first line print one integer representing <image>.
Examples
Input
5
Output
387
Input
233
Output
788243189
Note
A piece of paper with 5 circles is shown in the following.
<image>
There are 5 types of lattice points where the degree of beauty of each red point is 1 + 2 + 3 + 4 + 5 = 15, the degree of beauty of each orange point is 2 + 3 + 4 + 5 = 14, the degree of beauty of each green point is 4 + 5 = 9, the degree of beauty of each blue point is 5 and the degree of beauty of each gray point is 0. Therefore, f(5) = 5·15 + 4·14 + 4·9 + 8·5 = 207.
Similarly, f(1) = 5, f(2) = 23, f(3) = 50, f(4) = 102 and consequently <image>.
Submitted Solution:
```
import math
M = 10 ** 9 + 7
def h(x, y, n):
# x * (n + 1 - x) + (x + 1) * (n - x) + (x + 2) * (n - x - 1) + ... + (x + ?) * 1
x2 = x * x
x4 = x2 * x2
y2 = y * y
y4 = y2 * y2
y6 = y2 * y4
return 2 * y6 \
- (3 * n - 6 * x2 + 6) * y4 \
- (- 2 * x4 + (n + 4) * x2 + (n + 2) * n) * y2 \
+ (n - 4 * x2 + 4) * (n + 1 - x2) * y2 \
+ (n + 1 - x2) * (-2 * x4 + (n + 4) * x2 + (n + 2) * n)
def coef(x, n):
x2 = x * x
x4 = x2 * x2
return (
(n + 1 - x2) * (-2 * x4 + (n + 4) * x2 + (n + 2) * n),
(n - 4 * x2 + 4) * (n + 1 - x2) - (- 2 * x4 + (n + 4) * x2 + (n + 2) * n),
-(3 * n - 6 * x2 + 6),
2)
s0 = lambda n: n
s2 = lambda n: n * (n + 1) * (2 * n + 1) / 6
s4 = lambda n: s2(n) * (3 * n * n + 3 * n - 1) / 5
s6 = lambda n: s2(n) * (3 * n ** 4 + 6 * n ** 3 - 3 * n + 1) / 7
def g(n):
ret = h(0, 0, n) / 6
x = 0
while x * x <= n:
y_max = int(math.sqrt(n - x * x))
c = coef(x, n)
ret += (c[0] * s0(y_max) + c[1] * s2(y_max) + c[2] * s4(y_max) + c[3] * s6(y_max)) * 2 / 3
x += 1
return ret
n = int(input())
print (g(n) % (10 ** 9 + 7))
``` | instruction | 0 | 56,907 | 23 | 113,814 |
No | output | 1 | 56,907 | 23 | 113,815 |
Provide a correct Python 3 solution for this coding contest problem.
Takahashi has an N \times M grid, with N horizontal rows and M vertical columns. Determine if we can place A 1 \times 2 tiles (1 vertical, 2 horizontal) and B 2 \times 1 tiles (2 vertical, 1 horizontal) satisfying the following conditions, and construct one arrangement of the tiles if it is possible:
* All the tiles must be placed on the grid.
* Tiles must not stick out of the grid, and no two different tiles may intersect.
* Neither the grid nor the tiles may be rotated.
* Every tile completely covers exactly two squares.
Constraints
* 1 \leq N,M \leq 1000
* 0 \leq A,B \leq 500000
* N, M, A and B are integers.
Input
Input is given from Standard Input in the following format:
N M A B
Output
If it is impossible to place all the tiles, print `NO`. Otherwise, print the following:
YES
c_{11}...c_{1M}
:
c_{N1}...c_{NM}
Here, c_{ij} must be one of the following characters: `.`, `<`, `>`, `^` and `v`. Represent an arrangement by using each of these characters as follows:
* When c_{ij} is `.`, it indicates that the square at the i-th row and j-th column is empty;
* When c_{ij} is `<`, it indicates that the square at the i-th row and j-th column is covered by the left half of a 1 \times 2 tile;
* When c_{ij} is `>`, it indicates that the square at the i-th row and j-th column is covered by the right half of a 1 \times 2 tile;
* When c_{ij} is `^`, it indicates that the square at the i-th row and j-th column is covered by the top half of a 2 \times 1 tile;
* When c_{ij} is `v`, it indicates that the square at the i-th row and j-th column is covered by the bottom half of a 2 \times 1 tile.
Examples
Input
3 4 4 2
Output
YES
<><>
^<>^
v<>v
Input
4 5 5 3
Output
YES
<>..^
^.<>v
v<>.^
<><>v
Input
7 9 20 20
Output
NO | instruction | 0 | 57,008 | 23 | 114,016 |
"Correct Solution:
```
N,M,A,B=map(int,input().split())
Grid=[["."]*M for i in range(N)]
if M%2:
for i in range(N//2):
if B>0:
Grid[2*i][M-1]="^"
Grid[2*i+1][M-1]="v"
B-=1
if N%2:
for i in range(M//2):
if A>0:
Grid[N-1][2*i]="<"
Grid[N-1][2*i+1]=">"
A-=1
for i in range((M//2)*(N//2)):
a,b=i//(M//2),i%(M//2)
if A>1:
Grid[2*a][2*b]="<"
Grid[2*a][2*b+1]=">"
Grid[2*a+1][2*b]="<"
Grid[2*a+1][2*b+1]=">"
A-=2
elif B>1:
Grid[2*a][2*b]="^"
Grid[2*a][2*b+1]="^"
Grid[2*a+1][2*b]="v"
Grid[2*a+1][2*b+1]="v"
B-=2
else:
if A==1:
Grid[2*a][2*b]="<"
Grid[2*a][2*b+1]=">"
A-=1
elif B==1:
Grid[2*a][2*b]="^"
Grid[2*a+1][2*b]="v"
B-=1
if A==B==0:
print("YES")
for i in range(N):
print("".join(Grid[i]))
else:
if B==1 and M%2:
if Grid[N-1][M-1]=="." and Grid[N-1][M-2]==">" and Grid[N-2][M-2]==".":
Grid[N-1][M-2]="<"
Grid[N-1][M-1]=">"
Grid[N-1][M-3]="v"
Grid[N-2][M-3]="^"
print("YES")
for i in range(N):
print("".join(Grid[i]))
else:
print("NO")
else:
print("NO")
``` | output | 1 | 57,008 | 23 | 114,017 |
Provide a correct Python 3 solution for this coding contest problem.
Takahashi has an N \times M grid, with N horizontal rows and M vertical columns. Determine if we can place A 1 \times 2 tiles (1 vertical, 2 horizontal) and B 2 \times 1 tiles (2 vertical, 1 horizontal) satisfying the following conditions, and construct one arrangement of the tiles if it is possible:
* All the tiles must be placed on the grid.
* Tiles must not stick out of the grid, and no two different tiles may intersect.
* Neither the grid nor the tiles may be rotated.
* Every tile completely covers exactly two squares.
Constraints
* 1 \leq N,M \leq 1000
* 0 \leq A,B \leq 500000
* N, M, A and B are integers.
Input
Input is given from Standard Input in the following format:
N M A B
Output
If it is impossible to place all the tiles, print `NO`. Otherwise, print the following:
YES
c_{11}...c_{1M}
:
c_{N1}...c_{NM}
Here, c_{ij} must be one of the following characters: `.`, `<`, `>`, `^` and `v`. Represent an arrangement by using each of these characters as follows:
* When c_{ij} is `.`, it indicates that the square at the i-th row and j-th column is empty;
* When c_{ij} is `<`, it indicates that the square at the i-th row and j-th column is covered by the left half of a 1 \times 2 tile;
* When c_{ij} is `>`, it indicates that the square at the i-th row and j-th column is covered by the right half of a 1 \times 2 tile;
* When c_{ij} is `^`, it indicates that the square at the i-th row and j-th column is covered by the top half of a 2 \times 1 tile;
* When c_{ij} is `v`, it indicates that the square at the i-th row and j-th column is covered by the bottom half of a 2 \times 1 tile.
Examples
Input
3 4 4 2
Output
YES
<><>
^<>^
v<>v
Input
4 5 5 3
Output
YES
<>..^
^.<>v
v<>.^
<><>v
Input
7 9 20 20
Output
NO | instruction | 0 | 57,009 | 23 | 114,018 |
"Correct Solution:
```
#■標準入力ショートカット
def get_next_int():
return int(float(input()))
def get_next_ints(delim=" "):
return tuple([int(float(x)) for x in input().split(delim)])
def get_next_str():
return input()
def get_next_strs(delim=" "):
return tuple(input().split(delim))
def get_next_by_types(*value_types,delim=" "):
return tuple([t(x) for t, x in zip(value_types,input().split(delim))])
def main(print=print):
n, m, a, b = get_next_ints()
result = {}
if n % 2 == 1:
for j in range(1,m,2):
if a == 0:
break
result[(n, j)] = "<"
result[(n, j+1)] = ">"
a -= 1
if m % 2 == 1:
for i in range(1,n,2):
if b == 0:
break
result[(i, m)] = "^"
result[(i+1, m)] = "v"
b -= 1
if n % 2 == 1 and a % 2 == 1 and m % 2 == 1 and b % 2 == 1:
#右下3x3を風車風に詰める
result[(n, m-2)] = "<"
result[(n, m-1)] = ">"
result[(n-2, m-1)] = "<"
result[(n-2, m)] = ">"
result[(n-1, m)] = "^"
result[(n , m)] = "v"
result[(n-2, m-2)] = "^"
result[(n-1 , m-2)] = "v"
a -= 1
b -=1
inside_n = n - (n % 2)
inside_m = m - (m % 2)
#print("inside=",inside_n,inside_m)
for i in range(1,inside_n+1,2):
for j in range(1,inside_m+1,2):
#print(i,j,a,b)
if a > 0:
result[(i, j)] = "<"
result[(i, j+1)] = ">"
a -= 1
if a > 0:
result[(i+1, j)] = "<"
result[(i+1, j+1)] = ">"
a -= 1
elif b > 0:
result[(i, j)] = "^"
result[(i+1, j)] = "v"
b -= 1
if b > 0:
result[(i, j+1)] = "^"
result[(i+1, j+1)] = "v"
b -= 1
else:
break
if (a + b) == 0:
break
if (a + b) > 0:
print("NO")
else:
print("YES")
for i in range(1, n+1):
buffer = []
for j in range(1, m+1):
buffer.append(result.get((i,j),"."))
print("".join(buffer))
if __name__ == '__main__':
main()
``` | output | 1 | 57,009 | 23 | 114,019 |
Provide a correct Python 3 solution for this coding contest problem.
Takahashi has an N \times M grid, with N horizontal rows and M vertical columns. Determine if we can place A 1 \times 2 tiles (1 vertical, 2 horizontal) and B 2 \times 1 tiles (2 vertical, 1 horizontal) satisfying the following conditions, and construct one arrangement of the tiles if it is possible:
* All the tiles must be placed on the grid.
* Tiles must not stick out of the grid, and no two different tiles may intersect.
* Neither the grid nor the tiles may be rotated.
* Every tile completely covers exactly two squares.
Constraints
* 1 \leq N,M \leq 1000
* 0 \leq A,B \leq 500000
* N, M, A and B are integers.
Input
Input is given from Standard Input in the following format:
N M A B
Output
If it is impossible to place all the tiles, print `NO`. Otherwise, print the following:
YES
c_{11}...c_{1M}
:
c_{N1}...c_{NM}
Here, c_{ij} must be one of the following characters: `.`, `<`, `>`, `^` and `v`. Represent an arrangement by using each of these characters as follows:
* When c_{ij} is `.`, it indicates that the square at the i-th row and j-th column is empty;
* When c_{ij} is `<`, it indicates that the square at the i-th row and j-th column is covered by the left half of a 1 \times 2 tile;
* When c_{ij} is `>`, it indicates that the square at the i-th row and j-th column is covered by the right half of a 1 \times 2 tile;
* When c_{ij} is `^`, it indicates that the square at the i-th row and j-th column is covered by the top half of a 2 \times 1 tile;
* When c_{ij} is `v`, it indicates that the square at the i-th row and j-th column is covered by the bottom half of a 2 \times 1 tile.
Examples
Input
3 4 4 2
Output
YES
<><>
^<>^
v<>v
Input
4 5 5 3
Output
YES
<>..^
^.<>v
v<>.^
<><>v
Input
7 9 20 20
Output
NO | instruction | 0 | 57,010 | 23 | 114,020 |
"Correct Solution:
```
N,M,A,B=map(int,input().split())
def solve(n,m,a,b):
#そもそも絶対的にスペースが足りない場合
if a*2+b*2>n*m:
return False
else:
ans=[['.' for x in range(m)] for y in range(n)]
#各タイルの残りをカウント
remainA=a
remainB=b
#原点を起点として2枚を正方形にして敷き詰めていく
nx=0
ny=0
#縦横の限界値をカウント
gx=n
gy=m
#縦が奇数ならばnxを+1して、最初の行にAを並べる
if n%2:
nx=1
cnt=0
while True:
if cnt > m-2 or remainA==0:
break
ans[0][cnt]='<'
ans[0][cnt+1]='>'
remainA-=1
cnt+=2
#横が奇数ならば限界値を-1して、最後の列にBを並べる
if m%2:
gy-=1
cnt=0
while True:
if cnt > n-2 or remainB==0:
break
ans[cnt][m-1]='^'
ans[cnt+1][m-1]='v'
remainB-=1
cnt+=2
#以上で縦横どちらかが1列行の場合の処理終わり。ABが残っていたらFalse
if (n==1 or m==1) and (remainA>0 or remainB>0):
return False
#残り枚数を2枚ずつ消化するので、for文のために残数を変数にセット。
cna=remainA
cnb=remainB
#まずはBを敷き詰める。
for i in range(cnb//2):
ans[nx][ny]='^'
ans[nx+1][ny]='v'
ans[nx][ny+1]='^'
ans[nx+1][ny+1]='v'
remainB-=2
#横軸方向に+2、出来なければ、縦軸方向に+2して横軸位置を0に。
if ny+2 < gy-1:
ny+=2
elif nx+2 < gx-1:
nx+=2
ny=0
#移動できなければ、そこで終了。ABが残っていればFalse。残っていなければ正解タイル。
else:
if remainA>0 or remainB>0:
return False
else:
return ans
#次にAを敷き詰める。
for i in range(cna//2):
ans[nx][ny]='<'
ans[nx][ny+1]='>'
ans[nx+1][ny]='<'
ans[nx+1][ny+1]='>'
remainA-=2
#横軸方向に+2、出来なければ、縦軸方向に+2して横軸位置を0に。
if ny+2 < gy-1:
ny+=2
elif nx+2 < gx-1:
nx+=2
ny=0
#移動できなければ、そこで終了。ABが残っていればFalse。残っていなければ正解タイル。
else:
if remainA>0 or remainB>0:
return False
else:
return ans
#x2で敷き詰め終えたので、残数と状況を調査。残っている各タイルは最大1つ
if remainA%2==0 and remainB%2==0:
return ans
#両方1つずつ残っている場合、2x2が2つ必要。条件からgx/gyは偶数なので。
elif remainA%2==1 and remainB%2==1:
ans[nx][ny]='^'
ans[nx+1][ny]='v'
if ny+2 < gy-1:
ny+=2
elif nx+2 < gx-1:
nx+=2
ny=0
#移動できなくてかつ隅が空いてなければAを置くことができない。
else:
if ans[n-1][m-1]=='.' and ans[n-1][m-2]=='.':
ans[n-1][m-1]='>'
ans[n-1][m-2]='<'
return ans
return False
#移動できたのでAを置いてreturn
ans[nx][ny]='<'
ans[nx][ny+1]='>'
return ans
#Aだけが残っている場合はAを置いてreturn
elif remainA%2==1:
ans[nx][ny]='<'
ans[nx][ny+1]='>'
return ans
#Bだけが残っている場合はBを置いてreturn
else:
ans[nx][ny]='^'
ans[nx+1][ny]='v'
return ans
Ans=solve(N,M,A,B)
if Ans:
print("YES")
for i in range(N):
res="".join(Ans[i])
print(res)
else:
print("NO")
``` | output | 1 | 57,010 | 23 | 114,021 |
Provide a correct Python 3 solution for this coding contest problem.
Takahashi has an N \times M grid, with N horizontal rows and M vertical columns. Determine if we can place A 1 \times 2 tiles (1 vertical, 2 horizontal) and B 2 \times 1 tiles (2 vertical, 1 horizontal) satisfying the following conditions, and construct one arrangement of the tiles if it is possible:
* All the tiles must be placed on the grid.
* Tiles must not stick out of the grid, and no two different tiles may intersect.
* Neither the grid nor the tiles may be rotated.
* Every tile completely covers exactly two squares.
Constraints
* 1 \leq N,M \leq 1000
* 0 \leq A,B \leq 500000
* N, M, A and B are integers.
Input
Input is given from Standard Input in the following format:
N M A B
Output
If it is impossible to place all the tiles, print `NO`. Otherwise, print the following:
YES
c_{11}...c_{1M}
:
c_{N1}...c_{NM}
Here, c_{ij} must be one of the following characters: `.`, `<`, `>`, `^` and `v`. Represent an arrangement by using each of these characters as follows:
* When c_{ij} is `.`, it indicates that the square at the i-th row and j-th column is empty;
* When c_{ij} is `<`, it indicates that the square at the i-th row and j-th column is covered by the left half of a 1 \times 2 tile;
* When c_{ij} is `>`, it indicates that the square at the i-th row and j-th column is covered by the right half of a 1 \times 2 tile;
* When c_{ij} is `^`, it indicates that the square at the i-th row and j-th column is covered by the top half of a 2 \times 1 tile;
* When c_{ij} is `v`, it indicates that the square at the i-th row and j-th column is covered by the bottom half of a 2 \times 1 tile.
Examples
Input
3 4 4 2
Output
YES
<><>
^<>^
v<>v
Input
4 5 5 3
Output
YES
<>..^
^.<>v
v<>.^
<><>v
Input
7 9 20 20
Output
NO | instruction | 0 | 57,011 | 23 | 114,022 |
"Correct Solution:
```
import math,string,itertools,fractions,heapq,collections,re,array,bisect,sys,random,time,copy,functools
sys.setrecursionlimit(10**7)
inf = 10**20
eps = 1.0 / 10**15
mod = 10**9+7
def LI(): return [int(x) for x in sys.stdin.readline().split()]
def LI_(): return [int(x)-1 for x in sys.stdin.readline().split()]
def LF(): return [float(x) for x in sys.stdin.readline().split()]
def LS(): return sys.stdin.readline().split()
def I(): return int(sys.stdin.readline())
def F(): return float(sys.stdin.readline())
def S(): return input()
def pf(s): return print(s, flush=True)
def main():
n,m,a,b = LI()
if n * m < (a+b) * 2:
return 'NO'
r = [['.']*m for _ in range(n)]
if n % 2 == 1:
for i in range(min(a,m//2)):
r[-1][i*2] = '<'
r[-1][i*2+1] = '>'
a -= 1
if m % 2 == 1:
for i in range(min(b,n//2)):
r[i*2][-1] = '^'
r[i*2+1][-1] = 'v'
b -= 1
if (a%2) * (b%2) * (n%2) * (m%2) == 1 and n > 2 and m > 2:
r[-1][-1] = 'v'
r[-2][-1] = '^'
r[-3][-1] = '>'
r[-3][-2] = '<'
r[-3][-3] = '^'
r[-2][-3] = 'v'
a -= 1
b -= 1
for i in range(n//2):
for j in range(m//2):
if a > 0:
r[i*2][j*2] = '<'
r[i*2][j*2+1] = '>'
if a > 1:
r[i*2+1][j*2] = '<'
r[i*2+1][j*2+1] = '>'
a -= 2
elif b <= 0:
break
else:
r[i*2][j*2] = '^'
r[i*2+1][j*2] = 'v'
if b > 1:
r[i*2][j*2+1] = '^'
r[i*2+1][j*2+1] = 'v'
b -= 2
if a > 0 or b > 0:
return 'NO'
return 'YES\n' + '\n'.join([''.join(c) for c in r])
print(main())
``` | output | 1 | 57,011 | 23 | 114,023 |
Provide a correct Python 3 solution for this coding contest problem.
Takahashi has an N \times M grid, with N horizontal rows and M vertical columns. Determine if we can place A 1 \times 2 tiles (1 vertical, 2 horizontal) and B 2 \times 1 tiles (2 vertical, 1 horizontal) satisfying the following conditions, and construct one arrangement of the tiles if it is possible:
* All the tiles must be placed on the grid.
* Tiles must not stick out of the grid, and no two different tiles may intersect.
* Neither the grid nor the tiles may be rotated.
* Every tile completely covers exactly two squares.
Constraints
* 1 \leq N,M \leq 1000
* 0 \leq A,B \leq 500000
* N, M, A and B are integers.
Input
Input is given from Standard Input in the following format:
N M A B
Output
If it is impossible to place all the tiles, print `NO`. Otherwise, print the following:
YES
c_{11}...c_{1M}
:
c_{N1}...c_{NM}
Here, c_{ij} must be one of the following characters: `.`, `<`, `>`, `^` and `v`. Represent an arrangement by using each of these characters as follows:
* When c_{ij} is `.`, it indicates that the square at the i-th row and j-th column is empty;
* When c_{ij} is `<`, it indicates that the square at the i-th row and j-th column is covered by the left half of a 1 \times 2 tile;
* When c_{ij} is `>`, it indicates that the square at the i-th row and j-th column is covered by the right half of a 1 \times 2 tile;
* When c_{ij} is `^`, it indicates that the square at the i-th row and j-th column is covered by the top half of a 2 \times 1 tile;
* When c_{ij} is `v`, it indicates that the square at the i-th row and j-th column is covered by the bottom half of a 2 \times 1 tile.
Examples
Input
3 4 4 2
Output
YES
<><>
^<>^
v<>v
Input
4 5 5 3
Output
YES
<>..^
^.<>v
v<>.^
<><>v
Input
7 9 20 20
Output
NO | instruction | 0 | 57,012 | 23 | 114,024 |
"Correct Solution:
```
n,m,a,b = list(map(int,input().split()))
ban = [["."]*m for i in range(n)]
aa = a
bb = b
if n%2 == 1 and m %2 == 1 and n>=3 and m>=3 and a>=2 and b>=2:
aa-=2
bb-=2
ban[-3][-3] = "<"
ban[-3][-2] = ">"
ban[-3][-1] = "^"
ban[-2][-3] = "^"
ban[-2][-1] = "v"
ban[-1][-3] = "v"
ban[-1][-2] = "<"
ban[-1][-1] = ">"
for i in range(0,m-m%2-2,2):
if aa == 0:
break
ban[-1][i] = "<"
ban[-1][i+1] = ">"
aa-=1
for i in range(0,n-n%2-2,2):
if bb == 0:
break
ban[i][-1] = "^"
ban[i+1][-1] = "v"
bb-=1
n-=1
m-=1
yo = aa//2
ta = bb//2
yoa = aa%2
taa = bb%2
if n*m//4 -1 < yo+ta+yoa+taa:
n+=1
m+=1
ban = [["."]*m for i in range(n)]
else:
tu = []
for i in range(0,n,2):
for j in range(0,m,2):
tu.append([i,j])
tu.pop()
for i in range(yo+ta):
x,y = tu[i]
if i<yo:
ban[x][y] = "<"
ban[x+1][y] = "<"
ban[x][y+1] = ">"
ban[x+1][y+1] = ">"
else:
ban[x][y] = "^"
ban[x+1][y] = "v"
ban[x][y+1] = "^"
ban[x+1][y+1] = "v"
if yoa and taa:
x,y = tu[yo+ta]
ban[x][y] = "<"
ban[x][y+1] = ">"
x,y = tu[yo+ta+1]
ban[x][y] = "^"
ban[x+1][y] = "v"
elif yoa:
x,y = tu[yo+ta]
ban[x][y] = "<"
ban[x][y+1] = ">"
elif taa:
x,y = tu[yo+ta]
ban[x][y] = "^"
ban[x+1][y] = "v"
print("YES")
for i in ban:
print("".join(i))
exit()
if n%2 ==1:
for i in range(0,m-m%2,2):
if a == 0:
break
ban[-1][i] = "<"
ban[-1][i+1] = ">"
a-=1
n-=1
if m%2 == 1:
for i in range(0,n,2):
if b == 0:
break
ban[i][-1] = "^"
ban[i+1][-1] = "v"
b-=1
m-=1
if n%2 == 0 and m%2 == 0:
yo = a//2
ta = b//2
yoa = a%2
taa = b%2
if n*m//4 < yo+ta+yoa+taa:
print("NO")
exit()
tu = []
for i in range(0,n,2):
for j in range(0,m,2):
tu.append([i,j])
for i in range(yo+ta):
x,y = tu[i]
if i<yo:
ban[x][y] = "<"
ban[x+1][y] = "<"
ban[x][y+1] = ">"
ban[x+1][y+1] = ">"
else:
ban[x][y] = "^"
ban[x+1][y] = "v"
ban[x][y+1] = "^"
ban[x+1][y+1] = "v"
if yoa and taa:
x,y = tu[yo+ta]
ban[x][y] = "<"
ban[x][y+1] = ">"
x,y = tu[yo+ta+1]
ban[x][y] = "^"
ban[x+1][y] = "v"
elif yoa:
x,y = tu[yo+ta]
ban[x][y] = "<"
ban[x][y+1] = ">"
elif taa:
x,y = tu[yo+ta]
ban[x][y] = "^"
ban[x+1][y] = "v"
print("YES")
for i in ban:
print("".join(i))
``` | output | 1 | 57,012 | 23 | 114,025 |
Provide a correct Python 3 solution for this coding contest problem.
Takahashi has an N \times M grid, with N horizontal rows and M vertical columns. Determine if we can place A 1 \times 2 tiles (1 vertical, 2 horizontal) and B 2 \times 1 tiles (2 vertical, 1 horizontal) satisfying the following conditions, and construct one arrangement of the tiles if it is possible:
* All the tiles must be placed on the grid.
* Tiles must not stick out of the grid, and no two different tiles may intersect.
* Neither the grid nor the tiles may be rotated.
* Every tile completely covers exactly two squares.
Constraints
* 1 \leq N,M \leq 1000
* 0 \leq A,B \leq 500000
* N, M, A and B are integers.
Input
Input is given from Standard Input in the following format:
N M A B
Output
If it is impossible to place all the tiles, print `NO`. Otherwise, print the following:
YES
c_{11}...c_{1M}
:
c_{N1}...c_{NM}
Here, c_{ij} must be one of the following characters: `.`, `<`, `>`, `^` and `v`. Represent an arrangement by using each of these characters as follows:
* When c_{ij} is `.`, it indicates that the square at the i-th row and j-th column is empty;
* When c_{ij} is `<`, it indicates that the square at the i-th row and j-th column is covered by the left half of a 1 \times 2 tile;
* When c_{ij} is `>`, it indicates that the square at the i-th row and j-th column is covered by the right half of a 1 \times 2 tile;
* When c_{ij} is `^`, it indicates that the square at the i-th row and j-th column is covered by the top half of a 2 \times 1 tile;
* When c_{ij} is `v`, it indicates that the square at the i-th row and j-th column is covered by the bottom half of a 2 \times 1 tile.
Examples
Input
3 4 4 2
Output
YES
<><>
^<>^
v<>v
Input
4 5 5 3
Output
YES
<>..^
^.<>v
v<>.^
<><>v
Input
7 9 20 20
Output
NO | instruction | 0 | 57,013 | 23 | 114,026 |
"Correct Solution:
```
import math
def prity(mat):
for i in range(len(mat)):
for j in range(len(mat[0])):
print(mat[i][j], end='')
print()
def tile_a(i, j, mat):
assert mat[i][j] == '.' and mat[i][j+1] == '.'
mat[i][j] = '<'
mat[i][j+1] = '>'
def tile_b(i, j, mat):
assert mat[i][j] == '.' and mat[i+1][j] == '.'
mat[i][j] = '^'
mat[i+1][j] = 'v'
N, M, A, B = [int(_) for _ in input().split()]
a, b = A, B
mat = [['.' for j in range(M)] for i in range(N)]
if N % 2 == 1:
i = N - 1 #tile last row
for w in range(M // 2):
if a > 0:
j = 2 * w + (M % 2)
tile_a(i, j, mat)
a -= 1
if M % 2 == 1:
j = M - 1 # tile last column
for h in range(N // 2):
if b > 0:
i = 2 * h
tile_b(i, j, mat)
b -= 1
height = N // 2
width = M // 2
skip_lst = []
if N % 2 == 1 and M % 2 == 1 and a % 2 == 1 and b % 2 == 1:
tile_a(2 * (height - 1), 0, mat)
tile_b(2 * (height - 1) + 1, 0, mat)
a -= 1
b -= 1
skip_lst.append((2 * (height - 1), 0))
for h in range(height):
for w in range(width):
i = 2 * h
j = 2 * w
if not((i, j) in skip_lst):
if a > 1:
tile_a(i, j, mat)
tile_a(i + 1, j, mat)
a -= 2
elif a == 1:
tile_a(i, j, mat)
a -= 1
elif b > 1:
tile_b(i, j, mat)
tile_b(i, j + 1, mat)
b -= 2
elif b == 1:
tile_b(i, j, mat)
b -= 1
if a == 0 and b == 0:
print("YES")
prity(mat)
else:
print("NO")
``` | output | 1 | 57,013 | 23 | 114,027 |
Provide a correct Python 3 solution for this coding contest problem.
Takahashi has an N \times M grid, with N horizontal rows and M vertical columns. Determine if we can place A 1 \times 2 tiles (1 vertical, 2 horizontal) and B 2 \times 1 tiles (2 vertical, 1 horizontal) satisfying the following conditions, and construct one arrangement of the tiles if it is possible:
* All the tiles must be placed on the grid.
* Tiles must not stick out of the grid, and no two different tiles may intersect.
* Neither the grid nor the tiles may be rotated.
* Every tile completely covers exactly two squares.
Constraints
* 1 \leq N,M \leq 1000
* 0 \leq A,B \leq 500000
* N, M, A and B are integers.
Input
Input is given from Standard Input in the following format:
N M A B
Output
If it is impossible to place all the tiles, print `NO`. Otherwise, print the following:
YES
c_{11}...c_{1M}
:
c_{N1}...c_{NM}
Here, c_{ij} must be one of the following characters: `.`, `<`, `>`, `^` and `v`. Represent an arrangement by using each of these characters as follows:
* When c_{ij} is `.`, it indicates that the square at the i-th row and j-th column is empty;
* When c_{ij} is `<`, it indicates that the square at the i-th row and j-th column is covered by the left half of a 1 \times 2 tile;
* When c_{ij} is `>`, it indicates that the square at the i-th row and j-th column is covered by the right half of a 1 \times 2 tile;
* When c_{ij} is `^`, it indicates that the square at the i-th row and j-th column is covered by the top half of a 2 \times 1 tile;
* When c_{ij} is `v`, it indicates that the square at the i-th row and j-th column is covered by the bottom half of a 2 \times 1 tile.
Examples
Input
3 4 4 2
Output
YES
<><>
^<>^
v<>v
Input
4 5 5 3
Output
YES
<>..^
^.<>v
v<>.^
<><>v
Input
7 9 20 20
Output
NO | instruction | 0 | 57,014 | 23 | 114,028 |
"Correct Solution:
```
def solve(n, m, a, b):
cell = n * m
need = 2 * (a + b)
if cell < need:
return 'NO'
d = '.<>^v'
f = [[0] * m for _ in range(n)]
if n & 1:
for j in range(0, m - 1, 2):
if a:
f[-1][j] = 1
f[-1][j + 1] = 2
a -= 1
else:
break
if m & 1:
for i in range(0, n - 1, 2):
if b:
f[i][-1] = 3
f[i + 1][-1] = 4
b -= 1
else:
break
sqa = a // 2
sqb = b // 2
sq = sqa + sqb
ra, rb = a & 1, b & 1
sqc = (n // 2) * (m // 2)
if sq > sqc:
return 'NO'
if sq == sqc and (ra or rb):
return 'NO'
if sq + 1 == sqc and (ra and rb):
if n & 1 and m & 1:
f[-3][-3] = f[-1][-2] = 1
f[-3][-2] = f[-1][-1] = 2
f[-3][-1] = f[-2][-3] = 3
f[-2][-1] = f[-1][-3] = 4
ra = rb = 0
else:
return 'NO'
i, j = 0, 0
for i in range(0, n - 1, 2):
for j in range(0, m - 1, 2):
if sqa:
f[i][j] = f[i + 1][j] = 1
f[i][j + 1] = f[i + 1][j + 1] = 2
sqa -= 1
elif sqb:
f[i][j] = f[i][j + 1] = 3
f[i + 1][j] = f[i + 1][j + 1] = 4
sqb -= 1
else:
break
# print(sqa, sqb)
# print('\n'.join(''.join(map(d.__getitem__, r)) for r in f))
# print('---')
else:
continue
break
if ra:
f[i][j] = 1
f[i][j + 1] = 2
j += 2
if j >= m - 1:
j = 0
i += 2
if rb:
f[i][j] = 3
f[i + 1][j] = 4
return 'YES\n' + '\n'.join(''.join(map(d.__getitem__, r)) for r in f)
n, m, a, b = map(int, input().split())
print(solve(n, m, a, b))
``` | output | 1 | 57,014 | 23 | 114,029 |
Provide a correct Python 3 solution for this coding contest problem.
Takahashi has an N \times M grid, with N horizontal rows and M vertical columns. Determine if we can place A 1 \times 2 tiles (1 vertical, 2 horizontal) and B 2 \times 1 tiles (2 vertical, 1 horizontal) satisfying the following conditions, and construct one arrangement of the tiles if it is possible:
* All the tiles must be placed on the grid.
* Tiles must not stick out of the grid, and no two different tiles may intersect.
* Neither the grid nor the tiles may be rotated.
* Every tile completely covers exactly two squares.
Constraints
* 1 \leq N,M \leq 1000
* 0 \leq A,B \leq 500000
* N, M, A and B are integers.
Input
Input is given from Standard Input in the following format:
N M A B
Output
If it is impossible to place all the tiles, print `NO`. Otherwise, print the following:
YES
c_{11}...c_{1M}
:
c_{N1}...c_{NM}
Here, c_{ij} must be one of the following characters: `.`, `<`, `>`, `^` and `v`. Represent an arrangement by using each of these characters as follows:
* When c_{ij} is `.`, it indicates that the square at the i-th row and j-th column is empty;
* When c_{ij} is `<`, it indicates that the square at the i-th row and j-th column is covered by the left half of a 1 \times 2 tile;
* When c_{ij} is `>`, it indicates that the square at the i-th row and j-th column is covered by the right half of a 1 \times 2 tile;
* When c_{ij} is `^`, it indicates that the square at the i-th row and j-th column is covered by the top half of a 2 \times 1 tile;
* When c_{ij} is `v`, it indicates that the square at the i-th row and j-th column is covered by the bottom half of a 2 \times 1 tile.
Examples
Input
3 4 4 2
Output
YES
<><>
^<>^
v<>v
Input
4 5 5 3
Output
YES
<>..^
^.<>v
v<>.^
<><>v
Input
7 9 20 20
Output
NO | instruction | 0 | 57,015 | 23 | 114,030 |
"Correct Solution:
```
R,C,A,B = map(int,input().split())
if (A+B)*2 > R*C:
print('NO')
exit()
ans = [['.' for j in range(C)] for i in range(R)]
if R%2:
for i in range(C//2):
if A == 0: break
ans[-1][i*2] = '<'
ans[-1][i*2+1] = '>'
A -= 1
if C%2:
for i in range(R//2):
if B == 0: break
ans[i*2][-1] = '^'
ans[i*2+1][-1] = 'v'
B -= 1
for i in range(R//2):
for j in range(C//2):
if A > 1:
for k in range(2):
ans[i*2+k][j*2] = '<'
ans[i*2+k][j*2+1] = '>'
A -= 1
elif B > 1:
for k in range(2):
ans[i*2][j*2+k] = '^'
ans[i*2+1][j*2+k] = 'v'
B -= 1
elif A == 1:
ans[i*2][j*2] = '<'
ans[i*2][j*2+1] = '>'
A -= 1
elif B == 1:
ans[i*2][j*2] = '^'
ans[i*2+1][j*2] = 'v'
B -= 1
if R%2 and C%2 and B and ans[-2][-3] == '.':
if ans[-1][-3] == '<':
ans[-1][-2] = '<'
ans[-1][-1] = '>'
ans[-2][-3] = '^'
ans[-1][-3] = 'v'
B -= 1
if A > 0 or B > 0:
print('NO')
else:
print('YES')
print(*[''.join(row) for row in ans], sep='\n')
``` | output | 1 | 57,015 | 23 | 114,031 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Takahashi has an N \times M grid, with N horizontal rows and M vertical columns. Determine if we can place A 1 \times 2 tiles (1 vertical, 2 horizontal) and B 2 \times 1 tiles (2 vertical, 1 horizontal) satisfying the following conditions, and construct one arrangement of the tiles if it is possible:
* All the tiles must be placed on the grid.
* Tiles must not stick out of the grid, and no two different tiles may intersect.
* Neither the grid nor the tiles may be rotated.
* Every tile completely covers exactly two squares.
Constraints
* 1 \leq N,M \leq 1000
* 0 \leq A,B \leq 500000
* N, M, A and B are integers.
Input
Input is given from Standard Input in the following format:
N M A B
Output
If it is impossible to place all the tiles, print `NO`. Otherwise, print the following:
YES
c_{11}...c_{1M}
:
c_{N1}...c_{NM}
Here, c_{ij} must be one of the following characters: `.`, `<`, `>`, `^` and `v`. Represent an arrangement by using each of these characters as follows:
* When c_{ij} is `.`, it indicates that the square at the i-th row and j-th column is empty;
* When c_{ij} is `<`, it indicates that the square at the i-th row and j-th column is covered by the left half of a 1 \times 2 tile;
* When c_{ij} is `>`, it indicates that the square at the i-th row and j-th column is covered by the right half of a 1 \times 2 tile;
* When c_{ij} is `^`, it indicates that the square at the i-th row and j-th column is covered by the top half of a 2 \times 1 tile;
* When c_{ij} is `v`, it indicates that the square at the i-th row and j-th column is covered by the bottom half of a 2 \times 1 tile.
Examples
Input
3 4 4 2
Output
YES
<><>
^<>^
v<>v
Input
4 5 5 3
Output
YES
<>..^
^.<>v
v<>.^
<><>v
Input
7 9 20 20
Output
NO
Submitted Solution:
```
n,m,a,b=map(int,input().split())
board=[['.']*m for _ in range(n)]
if n*m<2*(a+b):
print('NO')
exit()
if n==1 or m==1:
if n==1 and m==1:
if a==0 and b==0:
print('YES')
print('.')
else:
print('NO')
exit()
elif n==1:
if b>=1:
print('NO')
exit()
else:
if m//2<a:
print('NO')
exit()
else:
for x in range(0,m//2):
board[0][2*x]='<'
board[0][2*x+1]='>'
print('YES')
for i in range(n):
print(''.join(map(str,board[i])))
elif m==1:
if a>=1:
print('NO')
exit()
else:
if n//2<b:
print('NO')
exit()
else:
for y in range(0,n//2):
board[2*y][0]='^'
board[2*y+1][0]='v'
print('YES')
for i in range(n):
print(''.join(map(str,board[i])))
else:
if (n*m)%2!=1:
if n%2==1:
tmp=min(m//2,a)
a-=tmp
for i in range(tmp):
board[n-1][2*i]='<'
board[n-1][2*i+1]='>'
if m%2==1:
tmp=min(n//2,b)
b-=tmp
for i in range(tmp):
board[2*i][m-1]='^'
board[2*i+1][m-1]='v'
for y in range(0,n-(n%2),2):
for x in range(0,m-(m%2),2):
if a>=2:
a-=2
board[y][x]='<'
board[y][x+1]='>'
board[y+1][x]='<'
board[y+1][x+1]='>'
elif a==1:
a-=1
board[y][x]='<'
board[y][x+1]='>'
elif b>=2:
b-=2
board[y][x]='^'
board[y+1][x]='v'
board[y][x+1]='^'
board[y+1][x+1]='v'
elif b==1:
b-=1
board[y][x]='^'
board[y+1][x]='v'
if a!=0 or b!=0:
print('NO')
else:
print('YES')
for i in range(n):
print(''.join(map(str,board[i])))
else:
if n%2==1:
tmp=min(m//2-1,a)
a-=tmp
for i in range(tmp):
board[n-1][2*i]='<'
board[n-1][2*i+1]='>'
if m%2==1:
tmp=min(n//2-1,b)
b-=tmp
for i in range(tmp):
board[2*i][m-1]='^'
board[2*i+1][m-1]='v'
for y in range(0,n-2,2):
for x in range(0,m-2,2):
if y==n-3 and x==m-3:
continue
if a>=b:
if a>=2:
a-=2
board[y][x]='<'
board[y][x+1]='>'
board[y+1][x]='<'
board[y+1][x+1]='>'
elif a==1:
a-=1
board[y][x]='<'
board[y][x+1]='>'
else:
if b>=2:
b-=2
board[y][x]='^'
board[y+1][x]='v'
board[y][x+1]='^'
board[y+1][x+1]='v'
elif b==1:
b-=1
board[y][x]='^'
board[y+1][x]='v'
if a>=4 or b>=4 or a+b>=5:
print('NO')
exit()
x=m-3
y=n-3
if a==2 and b==2:
a-=2
b-=2
board[y][x]='<'
board[y][x+1]='>'
board[y][x+2]='^'
board[y+1][x+2]='v'
board[y+1][x]='^'
board[y+2][x]='v'
board[y+2][x+1]='<'
board[y+2][x+2]='>'
else:
if a!=0:
a-=1
board[y+2][x]='<'
board[y+2][x+1]='>'
if b!=0:
b-=1
board[y][x+2]='^'
board[y+1][x+2]='v'
if a>=2:
a-=2
board[y][x]='<'
board[y][x+1]='>'
board[y+1][x]='<'
board[y+1][x+1]='>'
elif a==1:
a-=1
board[y][x]='<'
board[y][x+1]='>'
elif b>=2:
b-=2
board[y][x]='^'
board[y+1][x]='v'
board[y][x+1]='^'
board[y+1][x+1]='v'
elif b==1:
b-=1
board[y][x]='^'
board[y+1][x]='v'
print('YES')
for i in range(n):
print(''.join(map(str,board[i])))
``` | instruction | 0 | 57,016 | 23 | 114,032 |
Yes | output | 1 | 57,016 | 23 | 114,033 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Takahashi has an N \times M grid, with N horizontal rows and M vertical columns. Determine if we can place A 1 \times 2 tiles (1 vertical, 2 horizontal) and B 2 \times 1 tiles (2 vertical, 1 horizontal) satisfying the following conditions, and construct one arrangement of the tiles if it is possible:
* All the tiles must be placed on the grid.
* Tiles must not stick out of the grid, and no two different tiles may intersect.
* Neither the grid nor the tiles may be rotated.
* Every tile completely covers exactly two squares.
Constraints
* 1 \leq N,M \leq 1000
* 0 \leq A,B \leq 500000
* N, M, A and B are integers.
Input
Input is given from Standard Input in the following format:
N M A B
Output
If it is impossible to place all the tiles, print `NO`. Otherwise, print the following:
YES
c_{11}...c_{1M}
:
c_{N1}...c_{NM}
Here, c_{ij} must be one of the following characters: `.`, `<`, `>`, `^` and `v`. Represent an arrangement by using each of these characters as follows:
* When c_{ij} is `.`, it indicates that the square at the i-th row and j-th column is empty;
* When c_{ij} is `<`, it indicates that the square at the i-th row and j-th column is covered by the left half of a 1 \times 2 tile;
* When c_{ij} is `>`, it indicates that the square at the i-th row and j-th column is covered by the right half of a 1 \times 2 tile;
* When c_{ij} is `^`, it indicates that the square at the i-th row and j-th column is covered by the top half of a 2 \times 1 tile;
* When c_{ij} is `v`, it indicates that the square at the i-th row and j-th column is covered by the bottom half of a 2 \times 1 tile.
Examples
Input
3 4 4 2
Output
YES
<><>
^<>^
v<>v
Input
4 5 5 3
Output
YES
<>..^
^.<>v
v<>.^
<><>v
Input
7 9 20 20
Output
NO
Submitted Solution:
```
n,m,a,b=[int(i) for i in input().split()]
mass=[["." for i in range(m)] for j in range(n)]
i=0
k=0
num=0
count=0
dummy=0
try:
while num<a:
mass[i][2*k]="<"
mass[i][2*k+1]=">"
if n%2==0:
if i<=n-2:
i+=1
else:
i=0
k+=1
else:
if i<=n-4:
i+=1
elif 2*k+3<m:
i=0
k+=1
else:
count=1
break
num+=1
except:
dummy=1
print("NO")
if count==1:
k=0
i+=1
num+=1
try:
while num<a:
mass[i][2*k]="<"
mass[i][2*k+1]=">"
if i<=n-2:
i+=1
else:
i=n-2
k+=1
num+=1
except:
dummy=1
print("NO")
num_b=0
l=n-2
j=m-1
try:
if dummy!=1:
while num_b<b:
if mass[l][j]==".":
mass[l][j]="^"
mass[l+1][j]="v"
else:
dummy=2
break
if mass[l][j-1]==".":
j-=1
else:
l-=2
j=m-1
num_b+=1
except:
dummy=1
print("NO")
if dummy==0:
print("YES")
for i in range(n):
print("".join(mass[i]))
elif dummy==2:
sss=(n-2)*int((m-1)/2)
ssss=(a-sss)%2
sssss=(a-sss)//2
if n*m-1==2*(a+b) and n%2==1 and m%2==1 and sss<a and ssss==1:
mass[n-1][2*sssss]="<"
mass[n-1][2*sssss+1]=">"
mass[1][m-2]="."
mass[1][m-3]="."
mass[0][m-1]=">"
mass[0][m-2]="<"
mass[0][m-3]="."
mass[0][m-3]="^"
mass[1][m-3]="v"
print("YES")
for i in range(n):
print("".join(mass[i]))
else:
print("NO")
# for i in range(n):
# print(mass[i])
``` | instruction | 0 | 57,017 | 23 | 114,034 |
No | output | 1 | 57,017 | 23 | 114,035 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Takahashi has an N \times M grid, with N horizontal rows and M vertical columns. Determine if we can place A 1 \times 2 tiles (1 vertical, 2 horizontal) and B 2 \times 1 tiles (2 vertical, 1 horizontal) satisfying the following conditions, and construct one arrangement of the tiles if it is possible:
* All the tiles must be placed on the grid.
* Tiles must not stick out of the grid, and no two different tiles may intersect.
* Neither the grid nor the tiles may be rotated.
* Every tile completely covers exactly two squares.
Constraints
* 1 \leq N,M \leq 1000
* 0 \leq A,B \leq 500000
* N, M, A and B are integers.
Input
Input is given from Standard Input in the following format:
N M A B
Output
If it is impossible to place all the tiles, print `NO`. Otherwise, print the following:
YES
c_{11}...c_{1M}
:
c_{N1}...c_{NM}
Here, c_{ij} must be one of the following characters: `.`, `<`, `>`, `^` and `v`. Represent an arrangement by using each of these characters as follows:
* When c_{ij} is `.`, it indicates that the square at the i-th row and j-th column is empty;
* When c_{ij} is `<`, it indicates that the square at the i-th row and j-th column is covered by the left half of a 1 \times 2 tile;
* When c_{ij} is `>`, it indicates that the square at the i-th row and j-th column is covered by the right half of a 1 \times 2 tile;
* When c_{ij} is `^`, it indicates that the square at the i-th row and j-th column is covered by the top half of a 2 \times 1 tile;
* When c_{ij} is `v`, it indicates that the square at the i-th row and j-th column is covered by the bottom half of a 2 \times 1 tile.
Examples
Input
3 4 4 2
Output
YES
<><>
^<>^
v<>v
Input
4 5 5 3
Output
YES
<>..^
^.<>v
v<>.^
<><>v
Input
7 9 20 20
Output
NO
Submitted Solution:
```
import sys
input = sys.stdin.readline
import numpy as np
N,M,A,B = map(int,input().split())
def solve(N,M,A,B):
grid = np.full((N,M),'.',dtype='U1')
put_row = np.full((N,M-1),10**9,dtype=np.int32)
if N%2 == 0:
for n in range(0,M//2):
put_row[:,n+n] = np.arange(N*n,N*(n+1))
elif M%2 == 0:
# N-1行目を埋めたあと、2個ずつ
put_row[-1,::2] = np.arange(M//2)
x = M//2
for n in range(M//2):
put_row[:-1,n+n] = np.arange(x,x+N-1)
x += N-1
else:
# N-1行目を右優先で埋めたあと、左上から
put_row[-1,1::2] = np.arange(M//2)
x = M//2
for n in range(M//2):
put_row[:-1,n+n] = np.arange(x,x+N-1)
x += N-1
# 置くべき優先度を定義し終わった
x,y = np.where(put_row < A)
if len(x) != A:
print('NO')
return
grid[x,y] = '<'
grid[x,y+1] = '>'
# 空きます
put_col = (grid == '.')
# 下も空いている（上側としておける）
put_col[:-1] &= put_col[1:]
put_col[-1] = 0
for n in range(1,N):
# ひとつ上から置けるならやめる
put_col[n] &= ~put_col[n-1]
x,y = np.where(put_col)
x = x[:B]; y = y[:B]
if len(x) != B:
print('NO')
return
grid[x,y] = '^'
grid[x+1,y] = 'v'
print('YES')
print('\n'.join(''.join(row) for row in grid))
return
solve(N,M,A,B)
``` | instruction | 0 | 57,018 | 23 | 114,036 |
No | output | 1 | 57,018 | 23 | 114,037 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Takahashi has an N \times M grid, with N horizontal rows and M vertical columns. Determine if we can place A 1 \times 2 tiles (1 vertical, 2 horizontal) and B 2 \times 1 tiles (2 vertical, 1 horizontal) satisfying the following conditions, and construct one arrangement of the tiles if it is possible:
* All the tiles must be placed on the grid.
* Tiles must not stick out of the grid, and no two different tiles may intersect.
* Neither the grid nor the tiles may be rotated.
* Every tile completely covers exactly two squares.
Constraints
* 1 \leq N,M \leq 1000
* 0 \leq A,B \leq 500000
* N, M, A and B are integers.
Input
Input is given from Standard Input in the following format:
N M A B
Output
If it is impossible to place all the tiles, print `NO`. Otherwise, print the following:
YES
c_{11}...c_{1M}
:
c_{N1}...c_{NM}
Here, c_{ij} must be one of the following characters: `.`, `<`, `>`, `^` and `v`. Represent an arrangement by using each of these characters as follows:
* When c_{ij} is `.`, it indicates that the square at the i-th row and j-th column is empty;
* When c_{ij} is `<`, it indicates that the square at the i-th row and j-th column is covered by the left half of a 1 \times 2 tile;
* When c_{ij} is `>`, it indicates that the square at the i-th row and j-th column is covered by the right half of a 1 \times 2 tile;
* When c_{ij} is `^`, it indicates that the square at the i-th row and j-th column is covered by the top half of a 2 \times 1 tile;
* When c_{ij} is `v`, it indicates that the square at the i-th row and j-th column is covered by the bottom half of a 2 \times 1 tile.
Examples
Input
3 4 4 2
Output
YES
<><>
^<>^
v<>v
Input
4 5 5 3
Output
YES
<>..^
^.<>v
v<>.^
<><>v
Input
7 9 20 20
Output
NO
Submitted Solution:
```
n,m,a,b=map(int,input().split())
board=[['.']*m for _ in range(n)]
if n*m<2*(a+b):
print('NO')
exit()
if n==1 or m==1:
if n==1 and m==1:
if a==0 and b==0:
print('YES')
print('.')
else:
print('NO')
exit()
elif n==1:
if b>=1:
print('NO')
exit()
else:
if m//2<a:
print('NO')
exit()
else:
for x in range(0,m//2):
board[0][2*x]='<'
board[0][2*x+1]='>'
for i in range(n):
print(''.join(map(str,board[i])))
elif m==1:
if a>=1:
print('NO')
exit()
else:
if n//2<b:
print('NO')
exit()
else:
for y in range(0,n//2):
board[2*y][0]='^'
board[2*y+1][0]='v'
for i in range(n):
print(''.join(map(str,board[i])))
else:
if (n*m)%2!=1:
if n%2==1:
tmp=min(m//2,a)
a-=tmp
for i in range(tmp):
board[n-1][2*i]='<'
board[n-1][2*i+1]='>'
if m%2==1:
tmp=min(n//2,b)
b-=tmp
for i in range(tmp):
board[2*i][m-1]='^'
board[2*i+1][m-1]='v'
for y in range(0,n-(n%2),2):
for x in range(0,m-(m%2),2):
if a>=2:
a-=2
board[y][x]='<'
board[y][x+1]='>'
board[y+1][x]='<'
board[y+1][x+1]='>'
elif a==1:
a-=1
board[y][x]='<'
board[y][x+1]='>'
elif b>=2:
b-=2
board[y][x]='^'
board[y+1][x]='v'
board[y][x+1]='^'
board[y+1][x+1]='v'
elif b==1:
b-=1
board[y][x]='^'
board[y+1][x]='v'
if a!=0 or b!=0:
print('NO')
else:
print('YES')
for i in range(n):
print(''.join(map(str,board[i])))
else:
if n%2==1:
tmp=min(m//2-1,a)
a-=tmp
for i in range(tmp):
board[n-1][2*i]='<'
board[n-1][2*i+1]='>'
if m%2==1:
tmp=min(n//2-1,b)
b-=tmp
for i in range(tmp):
board[2*i][m-1]='^'
board[2*i+1][m-1]='v'
for y in range(0,n-2,2):
for x in range(0,m-2,2):
if y==n-3 and x==m-3:
continue
if a>=b:
if a>=2:
a-=2
board[y][x]='<'
board[y][x+1]='>'
board[y+1][x]='<'
board[y+1][x+1]='>'
elif a==1:
a-=1
board[y][x]='<'
board[y][x+1]='>'
else:
if b>=2:
b-=2
board[y][x]='^'
board[y+1][x]='v'
board[y][x+1]='^'
board[y+1][x+1]='v'
elif b==1:
b-=1
board[y][x]='^'
board[y+1][x]='v'
if a>=4 or b>=4 or a+b>=5:
print('NO')
exit()
x=m-3
y=n-3
if a==2 and b==2:
a-=2
b-=2
board[y][x]='<'
board[y][x+1]='>'
board[y][x+2]='^'
board[y+1][x+2]='v'
board[y+1][x]='^'
board[y+2][x]='v'
board[y+2][x+1]='<'
board[y+2][x+2]='>'
else:
if a!=0:
a-=1
board[y+2][x]='<'
board[y+2][x+1]='>'
if b!=0:
b-=1
board[y][x+2]='^'
board[y+1][x+2]='v'
if a>=2:
a-=2
board[y][x]='<'
board[y][x+1]='>'
board[y+1][x]='<'
board[y+1][x+1]='>'
elif a==1:
a-=1
board[y][x]='<'
board[y][x+1]='>'
elif b>=2:
b-=2
board[y][x]='^'
board[y+1][x]='v'
board[y][x+1]='^'
board[y+1][x+1]='v'
elif b==1:
b-=1
board[y][x]='^'
board[y+1][x]='v'
print('YES')
for i in range(n):
print(''.join(map(str,board[i])))
``` | instruction | 0 | 57,019 | 23 | 114,038 |
No | output | 1 | 57,019 | 23 | 114,039 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Takahashi has an N \times M grid, with N horizontal rows and M vertical columns. Determine if we can place A 1 \times 2 tiles (1 vertical, 2 horizontal) and B 2 \times 1 tiles (2 vertical, 1 horizontal) satisfying the following conditions, and construct one arrangement of the tiles if it is possible:
* All the tiles must be placed on the grid.
* Tiles must not stick out of the grid, and no two different tiles may intersect.
* Neither the grid nor the tiles may be rotated.
* Every tile completely covers exactly two squares.
Constraints
* 1 \leq N,M \leq 1000
* 0 \leq A,B \leq 500000
* N, M, A and B are integers.
Input
Input is given from Standard Input in the following format:
N M A B
Output
If it is impossible to place all the tiles, print `NO`. Otherwise, print the following:
YES
c_{11}...c_{1M}
:
c_{N1}...c_{NM}
Here, c_{ij} must be one of the following characters: `.`, `<`, `>`, `^` and `v`. Represent an arrangement by using each of these characters as follows:
* When c_{ij} is `.`, it indicates that the square at the i-th row and j-th column is empty;
* When c_{ij} is `<`, it indicates that the square at the i-th row and j-th column is covered by the left half of a 1 \times 2 tile;
* When c_{ij} is `>`, it indicates that the square at the i-th row and j-th column is covered by the right half of a 1 \times 2 tile;
* When c_{ij} is `^`, it indicates that the square at the i-th row and j-th column is covered by the top half of a 2 \times 1 tile;
* When c_{ij} is `v`, it indicates that the square at the i-th row and j-th column is covered by the bottom half of a 2 \times 1 tile.
Examples
Input
3 4 4 2
Output
YES
<><>
^<>^
v<>v
Input
4 5 5 3
Output
YES
<>..^
^.<>v
v<>.^
<><>v
Input
7 9 20 20
Output
NO
Submitted Solution:
```
n,m,a,b=map(int,input().split())
board=[['.']*m for _ in range(n)]
if n*m<2*(a+b):
print('NO')
exit()
if n==1 or m==1:
print(5)
else:
if (n*m)%2!=1:
if n%2==1:
tmp=min(m//2,a)
a-=tmp
for i in range(tmp):
board[n-1][2*i]='<'
board[n-1][2*i+1]='>'
if m%2==1:
tmp=min(n//2,b)
b-=tmp
for i in range(tmp):
board[2*i][m-1]='^'
board[2*i+1][m-1]='v'
for y in range(0,n-(n%2),2):
for x in range(0,m-(m%2),2):
if a>=2:
a-=2
board[y][x]='<'
board[y][x+1]='>'
board[y+1][x]='<'
board[y+1][x+1]='>'
elif a==1:
a-=1
board[y][x]='<'
board[y][x+1]='>'
elif b>=2:
b-=2
board[y][x]='^'
board[y+1][x]='v'
board[y][x+1]='^'
board[y+1][x+1]='v'
elif b==1:
b-=1
board[y][x]='^'
board[y+1][x]='v'
if a!=0 or b!=0:
print('NO')
else:
print('YES')
for i in range(n):
print(''.join(map(str,board[i])))
else:
if n%2==1:
tmp=min(m//2-1,a)
a-=tmp
for i in range(tmp):
board[n-1][2*i]='<'
board[n-1][2*i+1]='>'
if m%2==1:
tmp=min(n//2-1,b)
b-=tmp
for i in range(tmp):
board[2*i][m-1]='^'
board[2*i+1][m-1]='v'
for y in range(0,n-2,2):
for x in range(0,m-2,2):
if y==n-3 and x==m-3:
continue
if a>=b:
if a>=2:
a-=2
board[y][x]='<'
board[y][x+1]='>'
board[y+1][x]='<'
board[y+1][x+1]='>'
elif a==1:
a-=1
board[y][x]='<'
board[y][x+1]='>'
else:
if b>=2:
b-=2
board[y][x]='^'
board[y+1][x]='v'
board[y][x+1]='^'
board[y+1][x+1]='v'
elif b==1:
b-=1
board[y][x]='^'
board[y+1][x]='v'
if a>=4 or b>=4 or a+b>=5:
print('NO')
exit()
x=m-3
y=n-3
if a==2 and b==2:
a-=2
b-=2
board[y][x]='<'
board[y][x+1]='>'
board[y][x+2]='^'
board[y+1][x+2]='v'
board[y+1][x]='^'
board[y+2][x]='v'
board[y+2][x+1]='<'
board[y+2][x+2]='>'
else:
if a!=0:
a-=1
board[y+2][x]='<'
board[y+2][x+1]='>'
if b!=0:
b-=1
board[y][x+2]='^'
board[y+1][x+2]='v'
if a>=2:
a-=2
board[y][x]='<'
board[y][x+1]='>'
board[y+1][x]='<'
board[y+1][x+1]='>'
elif a==1:
a-=1
board[y][x]='<'
board[y][x+1]='>'
elif b>=2:
b-=2
board[y][x]='^'
board[y+1][x]='v'
board[y][x+1]='^'
board[y+1][x+1]='v'
elif b==1:
b-=1
board[y][x]='^'
board[y+1][x]='v'
print('YES')
for i in range(n):
print(''.join(map(str,board[i])))
``` | instruction | 0 | 57,020 | 23 | 114,040 |
No | output | 1 | 57,020 | 23 | 114,041 |
Provide a correct Python 3 solution for this coding contest problem.
Let's play Amidakuji.
In the following example, there are five vertical lines and four horizontal lines. The horizontal lines can intersect (jump across) the vertical lines.
<image>
In the starting points (top of the figure), numbers are assigned to vertical lines in ascending order from left to right. At the first step, 2 and 4 are swaped by the first horizontal line which connects second and fourth vertical lines (we call this operation (2, 4)). Likewise, we perform (3, 5), (1, 2) and (3, 4), then obtain "4 1 2 5 3" in the bottom.
Your task is to write a program which reads the number of vertical lines w and configurations of horizontal lines and prints the final state of the Amidakuji. In the starting pints, numbers 1, 2, 3, ..., w are assigne to the vertical lines from left to right.
Hint
Try it. -->
Input
w
n
a1,b1
a2,b2
.
.
an,bn
w (w ≤ 30) is the number of vertical lines. n (n ≤ 30) is the number of horizontal lines. A pair of two integers ai and bi delimited by a comma represents the i-th horizontal line.
Output
The number which should be under the 1st (leftmost) vertical line
The number which should be under the 2nd vertical line
:
The number which should be under the w-th vertical line
Example
Input
5
4
2,4
3,5
1,2
3,4
Output
4
1
2
5
3 | instruction | 0 | 57,054 | 23 | 114,108 |
"Correct Solution:
```
numList = []
for i in range(int(input())):
numList.append(i + 1)
for i in range(int(input())):
a, b = list(map(int, input().split(',')))
a, b = a - 1, b - 1
tmp = numList[a]
numList[a] = numList[b]
numList[b] = tmp
for num in numList:
print(num)
``` | output | 1 | 57,054 | 23 | 114,109 |
Provide a correct Python 3 solution for this coding contest problem.
Let's play Amidakuji.
In the following example, there are five vertical lines and four horizontal lines. The horizontal lines can intersect (jump across) the vertical lines.
<image>
In the starting points (top of the figure), numbers are assigned to vertical lines in ascending order from left to right. At the first step, 2 and 4 are swaped by the first horizontal line which connects second and fourth vertical lines (we call this operation (2, 4)). Likewise, we perform (3, 5), (1, 2) and (3, 4), then obtain "4 1 2 5 3" in the bottom.
Your task is to write a program which reads the number of vertical lines w and configurations of horizontal lines and prints the final state of the Amidakuji. In the starting pints, numbers 1, 2, 3, ..., w are assigne to the vertical lines from left to right.
Hint
Try it. -->
Input
w
n
a1,b1
a2,b2
.
.
an,bn
w (w ≤ 30) is the number of vertical lines. n (n ≤ 30) is the number of horizontal lines. A pair of two integers ai and bi delimited by a comma represents the i-th horizontal line.
Output
The number which should be under the 1st (leftmost) vertical line
The number which should be under the 2nd vertical line
:
The number which should be under the w-th vertical line
Example
Input
5
4
2,4
3,5
1,2
3,4
Output
4
1
2
5
3 | instruction | 0 | 57,055 | 23 | 114,110 |
"Correct Solution:
```
w = int(input())
n = int(input())
c = list(range(0, w+1))
for i in range(n):
a, b = map(int, input().split(','))
c[a], c[b] = c[b], c[a]
for i in range(1, w + 1):
print(c[i])
``` | output | 1 | 57,055 | 23 | 114,111 |
Provide a correct Python 3 solution for this coding contest problem.
Let's play Amidakuji.
In the following example, there are five vertical lines and four horizontal lines. The horizontal lines can intersect (jump across) the vertical lines.
<image>
In the starting points (top of the figure), numbers are assigned to vertical lines in ascending order from left to right. At the first step, 2 and 4 are swaped by the first horizontal line which connects second and fourth vertical lines (we call this operation (2, 4)). Likewise, we perform (3, 5), (1, 2) and (3, 4), then obtain "4 1 2 5 3" in the bottom.
Your task is to write a program which reads the number of vertical lines w and configurations of horizontal lines and prints the final state of the Amidakuji. In the starting pints, numbers 1, 2, 3, ..., w are assigne to the vertical lines from left to right.
Hint
Try it. -->
Input
w
n
a1,b1
a2,b2
.
.
an,bn
w (w ≤ 30) is the number of vertical lines. n (n ≤ 30) is the number of horizontal lines. A pair of two integers ai and bi delimited by a comma represents the i-th horizontal line.
Output
The number which should be under the 1st (leftmost) vertical line
The number which should be under the 2nd vertical line
:
The number which should be under the w-th vertical line
Example
Input
5
4
2,4
3,5
1,2
3,4
Output
4
1
2
5
3 | instruction | 0 | 57,056 | 23 | 114,112 |
"Correct Solution:
```
w = int(input())
n = int(input())
l = list(range(1, w + 1))
for i in range(n):
a, b = map(int, input().split(','))
a -= 1
b -= 1
l[a], l[b] = l[b], l[a]
for i in range(w):
print(l[i])
``` | output | 1 | 57,056 | 23 | 114,113 |
Provide a correct Python 3 solution for this coding contest problem.
Let's play Amidakuji.
In the following example, there are five vertical lines and four horizontal lines. The horizontal lines can intersect (jump across) the vertical lines.
<image>
In the starting points (top of the figure), numbers are assigned to vertical lines in ascending order from left to right. At the first step, 2 and 4 are swaped by the first horizontal line which connects second and fourth vertical lines (we call this operation (2, 4)). Likewise, we perform (3, 5), (1, 2) and (3, 4), then obtain "4 1 2 5 3" in the bottom.
Your task is to write a program which reads the number of vertical lines w and configurations of horizontal lines and prints the final state of the Amidakuji. In the starting pints, numbers 1, 2, 3, ..., w are assigne to the vertical lines from left to right.
Hint
Try it. -->
Input
w
n
a1,b1
a2,b2
.
.
an,bn
w (w ≤ 30) is the number of vertical lines. n (n ≤ 30) is the number of horizontal lines. A pair of two integers ai and bi delimited by a comma represents the i-th horizontal line.
Output
The number which should be under the 1st (leftmost) vertical line
The number which should be under the 2nd vertical line
:
The number which should be under the w-th vertical line
Example
Input
5
4
2,4
3,5
1,2
3,4
Output
4
1
2
5
3 | instruction | 0 | 57,057 | 23 | 114,114 |
"Correct Solution:
```
# coding=utf-8
###
### for atcorder program
###
import sys
import math
import array
# math class
class mymath:
### pi
pi = 3.14159265358979323846264338
### Prime Number
def pnum_eratosthenes(self, n):
ptable = [0 for i in range(n+1)]
plist = []
for i in range(2, n+1):
if ptable[i]==0:
plist.append(i)
for j in range(i+i, n+1, i):
ptable[j] = 1
return plist
### GCD
def gcd(self, a, b):
if b == 0:
return a
return self.gcd(b, a%b)
### LCM
def lcm(self, a, b):
return (a*b)//self.gcd(a,b)
### Mat Multiplication
def mul(self, A, B):
ans = []
for a in A:
c = 0
for j, row in enumerate(a):
c += row*B[j]
ans.append(c)
return ans
mymath = mymath()
### output class
class output:
### list
def list(self, l):
l = list(l)
#print(" ", end="")
for i, num in enumerate(l):
print(num, end="")
if i != len(l)-1:
print(" ", end="")
print()
output = output()
### input sample
#i = input()
#N = int(input())
#A, B, C = [x for x in input().split()]
#N, K = [int(x) for x in input().split()]
#inlist = [int(w) for w in input().split()]
#R = float(input())
#A.append(list(map(int,input().split())))
#for line in sys.stdin.readlines():
# x, y = [int(temp) for temp in line.split()]
### output sample
#print("{0} {1} {2:.5f}".format(A//B, A%B, A/B))
#print("{0:.6f} {1:.6f}".format(R*R*math.pi,R*2*math.pi))
#print(" {}".format(i), end="")
def main():
W = int(input())
N = int(input())
AB = [[int(x)-1 for x in input().split(',')] for i in range(N)]
ans = list(range(1, W+1))
for ab in AB:
ans[ab[0]], ans[ab[1]] = ans[ab[1]], ans[ab[0]]
for i in ans:
print(i)
if __name__ == '__main__':
main()
``` | output | 1 | 57,057 | 23 | 114,115 |
Provide a correct Python 3 solution for this coding contest problem.
Let's play Amidakuji.
In the following example, there are five vertical lines and four horizontal lines. The horizontal lines can intersect (jump across) the vertical lines.
<image>
In the starting points (top of the figure), numbers are assigned to vertical lines in ascending order from left to right. At the first step, 2 and 4 are swaped by the first horizontal line which connects second and fourth vertical lines (we call this operation (2, 4)). Likewise, we perform (3, 5), (1, 2) and (3, 4), then obtain "4 1 2 5 3" in the bottom.
Your task is to write a program which reads the number of vertical lines w and configurations of horizontal lines and prints the final state of the Amidakuji. In the starting pints, numbers 1, 2, 3, ..., w are assigne to the vertical lines from left to right.
Hint
Try it. -->
Input
w
n
a1,b1
a2,b2
.
.
an,bn
w (w ≤ 30) is the number of vertical lines. n (n ≤ 30) is the number of horizontal lines. A pair of two integers ai and bi delimited by a comma represents the i-th horizontal line.
Output
The number which should be under the 1st (leftmost) vertical line
The number which should be under the 2nd vertical line
:
The number which should be under the w-th vertical line
Example
Input
5
4
2,4
3,5
1,2
3,4
Output
4
1
2
5
3 | instruction | 0 | 57,059 | 23 | 114,118 |
"Correct Solution:
```
li = [i for i in range(int(input())+1)]
for i in range(int(input())):
line = list(map(int, input().split(',')))
li[line[0]],li[line[1]] = li[line[1]],li[line[0]]
for i in li[1::]: print(i)
``` | output | 1 | 57,059 | 23 | 114,119 |
Provide a correct Python 3 solution for this coding contest problem.
Let's play Amidakuji.
In the following example, there are five vertical lines and four horizontal lines. The horizontal lines can intersect (jump across) the vertical lines.
<image>
In the starting points (top of the figure), numbers are assigned to vertical lines in ascending order from left to right. At the first step, 2 and 4 are swaped by the first horizontal line which connects second and fourth vertical lines (we call this operation (2, 4)). Likewise, we perform (3, 5), (1, 2) and (3, 4), then obtain "4 1 2 5 3" in the bottom.
Your task is to write a program which reads the number of vertical lines w and configurations of horizontal lines and prints the final state of the Amidakuji. In the starting pints, numbers 1, 2, 3, ..., w are assigne to the vertical lines from left to right.
Hint
Try it. -->
Input
w
n
a1,b1
a2,b2
.
.
an,bn
w (w ≤ 30) is the number of vertical lines. n (n ≤ 30) is the number of horizontal lines. A pair of two integers ai and bi delimited by a comma represents the i-th horizontal line.
Output
The number which should be under the 1st (leftmost) vertical line
The number which should be under the 2nd vertical line
:
The number which should be under the w-th vertical line
Example
Input
5
4
2,4
3,5
1,2
3,4
Output
4
1
2
5
3 | instruction | 0 | 57,060 | 23 | 114,120 |
"Correct Solution:
```
w = int(input())
c = [i for i in range(1, w + 1)]
n = int(input())
for i in range(n):
a, b = map(int, input().split(","))
r = c[a - 1]
c[a - 1] = c[b - 1]
c[b - 1] = r
for i in range(w):
print(c[i])
``` | output | 1 | 57,060 | 23 | 114,121 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Let's play Amidakuji.
In the following example, there are five vertical lines and four horizontal lines. The horizontal lines can intersect (jump across) the vertical lines.
<image>
In the starting points (top of the figure), numbers are assigned to vertical lines in ascending order from left to right. At the first step, 2 and 4 are swaped by the first horizontal line which connects second and fourth vertical lines (we call this operation (2, 4)). Likewise, we perform (3, 5), (1, 2) and (3, 4), then obtain "4 1 2 5 3" in the bottom.
Your task is to write a program which reads the number of vertical lines w and configurations of horizontal lines and prints the final state of the Amidakuji. In the starting pints, numbers 1, 2, 3, ..., w are assigne to the vertical lines from left to right.
Hint
Try it. -->
Input
w
n
a1,b1
a2,b2
.
.
an,bn
w (w ≤ 30) is the number of vertical lines. n (n ≤ 30) is the number of horizontal lines. A pair of two integers ai and bi delimited by a comma represents the i-th horizontal line.
Output
The number which should be under the 1st (leftmost) vertical line
The number which should be under the 2nd vertical line
:
The number which should be under the w-th vertical line
Example
Input
5
4
2,4
3,5
1,2
3,4
Output
4
1
2
5
3
Submitted Solution:
```
w = int(input())
lots = [i for i in range(w)]
n = int(input())
for i in range(n):
a, b = map(int, input().split(","))
lots[a - 1], lots[b - 1] = lots[b - 1], lots[a - 1]
for i in range(w):
print(lots[i] + 1)
``` | instruction | 0 | 57,062 | 23 | 114,124 |
Yes | output | 1 | 57,062 | 23 | 114,125 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Let's play Amidakuji.
In the following example, there are five vertical lines and four horizontal lines. The horizontal lines can intersect (jump across) the vertical lines.
<image>
In the starting points (top of the figure), numbers are assigned to vertical lines in ascending order from left to right. At the first step, 2 and 4 are swaped by the first horizontal line which connects second and fourth vertical lines (we call this operation (2, 4)). Likewise, we perform (3, 5), (1, 2) and (3, 4), then obtain "4 1 2 5 3" in the bottom.
Your task is to write a program which reads the number of vertical lines w and configurations of horizontal lines and prints the final state of the Amidakuji. In the starting pints, numbers 1, 2, 3, ..., w are assigne to the vertical lines from left to right.
Hint
Try it. -->
Input
w
n
a1,b1
a2,b2
.
.
an,bn
w (w ≤ 30) is the number of vertical lines. n (n ≤ 30) is the number of horizontal lines. A pair of two integers ai and bi delimited by a comma represents the i-th horizontal line.
Output
The number which should be under the 1st (leftmost) vertical line
The number which should be under the 2nd vertical line
:
The number which should be under the w-th vertical line
Example
Input
5
4
2,4
3,5
1,2
3,4
Output
4
1
2
5
3
Submitted Solution:
```
# -*- coding:utf-8 -*-
import sys
w = int(input())
n = int(input())
array = []
count = 0
for i in sys.stdin:
array.append(i)
count += 1
if count == n:
break
a, b = [0]*n, [0]*n
for i in range(n):
s = array[i]
a[i], b[i] = s[0], s[2]
a[i], b[i] = int(a[i]), int(b[i])
lines = []
k = 0
for i in range(w):
lines.append(k)
k += 1
for i in range(n):
tmp1 = lines[a[i]-1]
tmp2 = lines[b[i]-1]
lines[a[i]-1] = tmp2
lines[b[i]-1] = tmp1
for i in range(len(lines)):
print(lines[i]+1)
``` | instruction | 0 | 57,065 | 23 | 114,130 |
No | output | 1 | 57,065 | 23 | 114,131 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Let's play Amidakuji.
In the following example, there are five vertical lines and four horizontal lines. The horizontal lines can intersect (jump across) the vertical lines.
<image>
In the starting points (top of the figure), numbers are assigned to vertical lines in ascending order from left to right. At the first step, 2 and 4 are swaped by the first horizontal line which connects second and fourth vertical lines (we call this operation (2, 4)). Likewise, we perform (3, 5), (1, 2) and (3, 4), then obtain "4 1 2 5 3" in the bottom.
Your task is to write a program which reads the number of vertical lines w and configurations of horizontal lines and prints the final state of the Amidakuji. In the starting pints, numbers 1, 2, 3, ..., w are assigne to the vertical lines from left to right.
Hint
Try it. -->
Input
w
n
a1,b1
a2,b2
.
.
an,bn
w (w ≤ 30) is the number of vertical lines. n (n ≤ 30) is the number of horizontal lines. A pair of two integers ai and bi delimited by a comma represents the i-th horizontal line.
Output
The number which should be under the 1st (leftmost) vertical line
The number which should be under the 2nd vertical line
:
The number which should be under the w-th vertical line
Example
Input
5
4
2,4
3,5
1,2
3,4
Output
4
1
2
5
3
Submitted Solution:
```
stick = [i for i in range(1, int(input()) + 1)]
net = int(input())
for _ in range(net):
a, b = list(map(int, input().split(",")))
stick[a - 1], stick[b - 1] = stick[b - 1], stick[a - 1]
print(stick)
print("\n".join(map(str, stick)))
``` | instruction | 0 | 57,066 | 23 | 114,132 |
No | output | 1 | 57,066 | 23 | 114,133 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Let's play Amidakuji.
In the following example, there are five vertical lines and four horizontal lines. The horizontal lines can intersect (jump across) the vertical lines.
<image>
In the starting points (top of the figure), numbers are assigned to vertical lines in ascending order from left to right. At the first step, 2 and 4 are swaped by the first horizontal line which connects second and fourth vertical lines (we call this operation (2, 4)). Likewise, we perform (3, 5), (1, 2) and (3, 4), then obtain "4 1 2 5 3" in the bottom.
Your task is to write a program which reads the number of vertical lines w and configurations of horizontal lines and prints the final state of the Amidakuji. In the starting pints, numbers 1, 2, 3, ..., w are assigne to the vertical lines from left to right.
Hint
Try it. -->
Input
w
n
a1,b1
a2,b2
.
.
an,bn
w (w ≤ 30) is the number of vertical lines. n (n ≤ 30) is the number of horizontal lines. A pair of two integers ai and bi delimited by a comma represents the i-th horizontal line.
Output
The number which should be under the 1st (leftmost) vertical line
The number which should be under the 2nd vertical line
:
The number which should be under the w-th vertical line
Example
Input
5
4
2,4
3,5
1,2
3,4
Output
4
1
2
5
3
Submitted Solution:
```
#!/usr/bin/env python
#-*- coding:utf-8 -*-
import sys
import math
w = int(input())
n = int(input())
s = []
for _ in range(n):
x = input().split(',')
s.append((int(x[0]), int(x[1])))
result = [i for i in range(1, w+1)]
print(len(result))
for x, y in s:
result[x-1], result[y-1] = result[y-1], result[x-1]
for x in result:
print(x)
``` | instruction | 0 | 57,067 | 23 | 114,134 |
No | output | 1 | 57,067 | 23 | 114,135 |
Provide tags and a correct Python 3 solution for this coding contest problem.
At a geometry lesson Gerald was given a task: to get vector B out of vector A. Besides, the teacher permitted him to perform the following operations with vector А:
* Turn the vector by 90 degrees clockwise.
* Add to the vector a certain vector C.
Operations could be performed in any order any number of times.
Can Gerald cope with the task?
Input
The first line contains integers x1 и y1 — the coordinates of the vector A ( - 108 ≤ x1, y1 ≤ 108). The second and the third line contain in the similar manner vectors B and C (their coordinates are integers; their absolute value does not exceed 108).
Output
Print "YES" (without the quotes) if it is possible to get vector B using the given operations. Otherwise print "NO" (without the quotes).
Examples
Input
0 0
1 1
0 1
Output
YES
Input
0 0
1 1
1 1
Output
YES
Input
0 0
1 1
2 2
Output
NO | instruction | 0 | 57,155 | 23 | 114,310 |
Tags: implementation, math
Correct Solution:
```
import math
def ok(xa, ya):
x, y = xb - xa, yb - ya
d = math.gcd(abs(xc), abs(yc))
if xc == 0 and yc == 0:
return x == 0 and y == 0
if xc == 0:
return x % yc == 0 and y % yc == 0
if yc == 0:
return x % xc == 0 and y % xc == 0
if (x % d != 0) or (y % d != 0):
return 0
a, b, c1, c2 = xc // d, yc // d, x // d, -y // d
if a == 0 and b == 0:
return c1 == 0 and c2 == 0
if (c1 * b + c2 * a) % (a * a + b * b) != 0:
return 0
yy = (c1 * b + c2 * a) / (a * a + b * b)
if a == 0:
return (c2 - a * yy) % b == 0
else:
return (c1 - b * yy) % a == 0
xa, ya = map(int,input().split())
xb, yb = map(int,input().split())
xc, yc = map(int,input().split())
if ok(xa, ya) or ok(-ya, xa) or ok(-xa, -ya) or ok(ya, -xa):
print('YES')
else:
print('NO')
``` | output | 1 | 57,155 | 23 | 114,311 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
At a geometry lesson Gerald was given a task: to get vector B out of vector A. Besides, the teacher permitted him to perform the following operations with vector А:
* Turn the vector by 90 degrees clockwise.
* Add to the vector a certain vector C.
Operations could be performed in any order any number of times.
Can Gerald cope with the task?
Input
The first line contains integers x1 и y1 — the coordinates of the vector A ( - 108 ≤ x1, y1 ≤ 108). The second and the third line contain in the similar manner vectors B and C (their coordinates are integers; their absolute value does not exceed 108).
Output
Print "YES" (without the quotes) if it is possible to get vector B using the given operations. Otherwise print "NO" (without the quotes).
Examples
Input
0 0
1 1
0 1
Output
YES
Input
0 0
1 1
1 1
Output
YES
Input
0 0
1 1
2 2
Output
NO
Submitted Solution:
```
import math
def ok(xa, ya):
x, y = xb - xa, yb - ya
d = math.gcd(abs(xc), abs(yc))
if (x % d != 0) or (y % d != 0):
return 0
a, b, c1, c2 = xc // d, yc // d, x // d, -y // d
if a == 0 and b == 0:
return c1 == 0 and c2 == 0
if (c1 * b + c2 * a) % (a * a + b * b) != 0:
return 0
yy = (c1 * b + c2 * a) / (a * a + b * b)
if a != 0:
return (c1 - b * yy) % a == 0
else:
return (c2 - a * yy) % b == 0
xa, ya = map(int,input().split())
xb, yb = map(int,input().split())
xc, yc = map(int,input().split())
if xc == 0 or yc == 0:
if xc == yc:
if xa == xb and ya == yb:
print('YES')
else:
print('NO')
exit(0)
if xc == 0:
if (xb - xa) % yc == 0 and (yb - ya) % yc == 0:
print('YES')
else:
print('NO')
exit(0)
if yc == 0:
if (xb - xa) % xc == 0 and (yb - ya) % xc == 0:
print('YES')
else:
print('NO')
exit(0)
if ok(xa, ya) or ok(-ya, xa) or ok(-xa, -ya) or ok(ya, -xa):
print('YES')
else:
print('NO')
``` | instruction | 0 | 57,156 | 23 | 114,312 |
No | output | 1 | 57,156 | 23 | 114,313 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
At a geometry lesson Gerald was given a task: to get vector B out of vector A. Besides, the teacher permitted him to perform the following operations with vector А:
* Turn the vector by 90 degrees clockwise.
* Add to the vector a certain vector C.
Operations could be performed in any order any number of times.
Can Gerald cope with the task?
Input
The first line contains integers x1 и y1 — the coordinates of the vector A ( - 108 ≤ x1, y1 ≤ 108). The second and the third line contain in the similar manner vectors B and C (their coordinates are integers; their absolute value does not exceed 108).
Output
Print "YES" (without the quotes) if it is possible to get vector B using the given operations. Otherwise print "NO" (without the quotes).
Examples
Input
0 0
1 1
0 1
Output
YES
Input
0 0
1 1
1 1
Output
YES
Input
0 0
1 1
2 2
Output
NO
Submitted Solution:
```
def amob(a,b):
if a<0 and b>0 or a>0 and b<0:return 0
if b==0:return 1
return a%b==0
ax,ay=map(int,input().split())
bx,by=map(int,input().split())
cx,cy=map(int,input().split())
#
for x,y in[(ax,ay),(ay,-ax),(-ax,-ay),(-ay,ax)]:
nx=bx-x
ny=by-y
if(amob(nx,cx) or amob(nx,cy))and (amob(ny,cx)or amob(ny,cy)):exit(print('YES'))
print('NO')
#3,0 -> 0,3 -> -3,0 -> 0,-3
``` | instruction | 0 | 57,157 | 23 | 114,314 |
No | output | 1 | 57,157 | 23 | 114,315 |
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