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 |
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Provide tags and a correct Python 3 solution for this coding contest problem.
Nowadays all circuses in Berland have a round arena with diameter 13 meters, but in the past things were different.
In Ancient Berland arenas in circuses were shaped as a regular (equiangular) polygon, the size and the number of angles could vary from one circus to another. In each corner of the arena there was a special pillar, and the rope strung between the pillars marked the arena edges.
Recently the scientists from Berland have discovered the remains of the ancient circus arena. They found only three pillars, the others were destroyed by the time.
You are given the coordinates of these three pillars. Find out what is the smallest area that the arena could have.
Input
The input file consists of three lines, each of them contains a pair of numbers –– coordinates of the pillar. Any coordinate doesn't exceed 1000 by absolute value, and is given with at most six digits after decimal point.
Output
Output the smallest possible area of the ancient arena. This number should be accurate to at least 6 digits after the decimal point. It's guaranteed that the number of angles in the optimal polygon is not larger than 100.
Examples
Input
0.000000 0.000000
1.000000 1.000000
0.000000 1.000000
Output
1.00000000 | instruction | 0 | 93,781 | 23 | 187,562 |
Tags: geometry, math
Correct Solution:
```
from math import *
p =[list(map(float,input().split())) for i in range(3)]
a,b,c=[hypot(x1-x2,y1-y2) for (x1,y1),(x2,y2) in [(p[0],p[1]),(p[0],p[2]),(p[1],p[2])]]
A,B,C=[acos((y*y+z*z-x*x)/2/z/y) for x,y,z in [(a,b,c),(b,c,a),(c,a,b)]]
R=a/2/sin(A)
def g(x,y):return x if y<1e-3 else g(y,fmod(x,y))
u=2*g(A,g(B,C))
print(round(R * R * pi / u * sin(u),7))
``` | output | 1 | 93,781 | 23 | 187,563 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Nowadays all circuses in Berland have a round arena with diameter 13 meters, but in the past things were different.
In Ancient Berland arenas in circuses were shaped as a regular (equiangular) polygon, the size and the number of angles could vary from one circus to another. In each corner of the arena there was a special pillar, and the rope strung between the pillars marked the arena edges.
Recently the scientists from Berland have discovered the remains of the ancient circus arena. They found only three pillars, the others were destroyed by the time.
You are given the coordinates of these three pillars. Find out what is the smallest area that the arena could have.
Input
The input file consists of three lines, each of them contains a pair of numbers –– coordinates of the pillar. Any coordinate doesn't exceed 1000 by absolute value, and is given with at most six digits after decimal point.
Output
Output the smallest possible area of the ancient arena. This number should be accurate to at least 6 digits after the decimal point. It's guaranteed that the number of angles in the optimal polygon is not larger than 100.
Examples
Input
0.000000 0.000000
1.000000 1.000000
0.000000 1.000000
Output
1.00000000 | instruction | 0 | 93,782 | 23 | 187,564 |
Tags: geometry, math
Correct Solution:
```
from math import sqrt, asin, pi, sin, cos
eps = 1e-6
def dist2(ax, ay, bx, by):
return (ax-bx)**2 + (ay-by)**2
def cross_prod(ax, ay, bx, by):
return ax*by - ay*bx
def inner_prod(ax, ay, bx, by):
return ax*bx + ay*by
def find(x, y, s, c):
for i in range(len(s)):
if abs(x-s[i]) < eps and abs(y-c[i]) < eps:
return True
return False
def pool(x):
if abs(x) < eps:
return 1.
if x < 0:
return x/pi+2
return x/pi
def main():
Ax, Ay = map(float, input().split())
Bx, By = map(float, input().split())
Cx, Cy = map(float, input().split())
#print(Ax, Ay, Bx, By, Cx, Cy)
#for _ in range(n):
# print(ans(input()))
D = 2*(Ax*(By-Cy) + Bx*(Cy-Ay) + Cx*(Ay-By))
Ox = ( (Ax**2+Ay**2)*(By-Cy) + (Bx**2+By**2)*(Cy-Ay) + (Cx**2+Cy**2)*(Ay-By) ) / D
Oy = ( (Ax**2+Ay**2)*(Cx-Bx) + (Bx**2+By**2)*(Ax-Cx) + (Cx**2+Cy**2)*(Bx-Ax) ) / D
R2 = dist2(Ox,Oy,Ax,Ay)
#print(R)
#print(Ox, Oy)
#print(dist(Ox,Oy,Ax,Ay))
#print(dist(Ox,Oy,Bx,By))
#print(dist(Ox,Oy,Cx,Cy))
s1 = cross_prod(Ax-Ox,Ay-Oy,Bx-Ox,By-Oy)/R2
c1 = inner_prod(Ax-Ox,Ay-Oy,Bx-Ox,By-Oy)/R2
s2 = cross_prod(Ax-Ox,Ay-Oy,Cx-Ox,Cy-Oy)/R2
c2 = inner_prod(Ax-Ox,Ay-Oy,Cx-Ox,Cy-Oy)/R2
#angle1 = pool(angle1)
#angle2 = pool(angle2)
#print([s1, s2])
#print([c1, c2])
for n in range(3, 101):
x = list(range(n))
for j in range(len(x)):
x[j] = x[j] * (2*pi/n)
s = list(map(sin, x))
c = list(map(cos, x))
#print(s)
#print(c)
#print(find(s1, c1, s, c))
#print(find(s2, c2, s, c))
if find(s1, c1, s, c) and find(s2, c2, s, c):
area = .5*n*R2*sin(2*pi/n)
print("%.8f" % area)
break
if __name__ == "__main__":
main()
``` | output | 1 | 93,782 | 23 | 187,565 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Nowadays all circuses in Berland have a round arena with diameter 13 meters, but in the past things were different.
In Ancient Berland arenas in circuses were shaped as a regular (equiangular) polygon, the size and the number of angles could vary from one circus to another. In each corner of the arena there was a special pillar, and the rope strung between the pillars marked the arena edges.
Recently the scientists from Berland have discovered the remains of the ancient circus arena. They found only three pillars, the others were destroyed by the time.
You are given the coordinates of these three pillars. Find out what is the smallest area that the arena could have.
Input
The input file consists of three lines, each of them contains a pair of numbers –– coordinates of the pillar. Any coordinate doesn't exceed 1000 by absolute value, and is given with at most six digits after decimal point.
Output
Output the smallest possible area of the ancient arena. This number should be accurate to at least 6 digits after the decimal point. It's guaranteed that the number of angles in the optimal polygon is not larger than 100.
Examples
Input
0.000000 0.000000
1.000000 1.000000
0.000000 1.000000
Output
1.00000000 | instruction | 0 | 93,783 | 23 | 187,566 |
Tags: geometry, math
Correct Solution:
```
import math
from fractions import Fraction
line = input()
data = line.split()
x1, y1 = float(data[0]), float(data[1])
line = input()
data = line.split()
x2, y2 = float(data[0]), float(data[1])
line = input()
data = line.split()
x3, y3 = float(data[0]), float(data[1])
dis12 = (x1 - x2) * (x1 - x2) + (y1 - y2) * (y1 - y2)
dis23 = (x2 - x3) * (x2 - x3) + (y2 - y3) * (y2 - y3)
dis13 = (x1 - x3) * (x1 - x3) + (y1 - y3) * (y1 - y3)
'''
print(dis12)
print(dis13)
print(dis23)
'''
cos1 = (dis13 + dis12 - dis23) / (2 * math.sqrt(dis12 * dis13))
cos2 = (dis23 + dis12 - dis13) / (2 * math.sqrt(dis23 * dis12))
cos3 = (dis13 + dis23 - dis12) / (2 * math.sqrt(dis13 * dis23))
'''
print('*',cos1)
print('*',cos2)
print('*',cos3)
'''
angle1 = math.acos(cos1) * 2
angle2 = math.acos(cos2) * 2
angle3 = math.acos(cos3) * 2
'''
print('#',angle1)
print('#',angle2)
print('#',angle3)
'''
number1 = Fraction.from_float(angle1 / angle2).limit_denominator(101)
#print(number1.numerator)
#print(number1.denominator)
l1 = angle1 / number1.numerator
#print(l1)
n1 = 2 * math.pi / l1
n1_int = int(n1 + 0.5)
#print(n1_int)
number2 = Fraction.from_float(angle1 / angle3).limit_denominator(101)
#print(number2.numerator)
#print(number2.denominator)
l2 = angle1 / number2.numerator
#print(l2)
n2 = 2 * math.pi / l2
n2_int = int(n2 + 0.5)
#print(n2_int)
if abs(n1_int - n1) < abs(n2_int - n2):
n = n1_int
else:
n = n2_int
sin1 = math.sqrt(1 - cos1 * cos1)
#print(sin1)
#print(dis23)
r = math.sqrt(dis23) / (2 * sin1)
#print(r)
area = 1 / 2 * n * math.sin(2 * math.pi / n) * r * r
print('%.6f' % area)
``` | output | 1 | 93,783 | 23 | 187,567 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Nowadays all circuses in Berland have a round arena with diameter 13 meters, but in the past things were different.
In Ancient Berland arenas in circuses were shaped as a regular (equiangular) polygon, the size and the number of angles could vary from one circus to another. In each corner of the arena there was a special pillar, and the rope strung between the pillars marked the arena edges.
Recently the scientists from Berland have discovered the remains of the ancient circus arena. They found only three pillars, the others were destroyed by the time.
You are given the coordinates of these three pillars. Find out what is the smallest area that the arena could have.
Input
The input file consists of three lines, each of them contains a pair of numbers –– coordinates of the pillar. Any coordinate doesn't exceed 1000 by absolute value, and is given with at most six digits after decimal point.
Output
Output the smallest possible area of the ancient arena. This number should be accurate to at least 6 digits after the decimal point. It's guaranteed that the number of angles in the optimal polygon is not larger than 100.
Examples
Input
0.000000 0.000000
1.000000 1.000000
0.000000 1.000000
Output
1.00000000 | instruction | 0 | 93,784 | 23 | 187,568 |
Tags: geometry, math
Correct Solution:
```
import math
x1, y1 = map(float, input().split())
x2, y2 = map(float, input().split())
x3, y3 = map(float, input().split())
c1 = (x2 ** 2 - x1 ** 2 + y2 ** 2 - y1 ** 2) / 2
c2 = (x3 ** 2 - x2 ** 2 + y3 ** 2 - y2 ** 2) / 2
# dist x1-x2
r1 = (c1 * (y3 - y2) - c2 * (y2 - y1)) / ((x2 - x1) * (y3 - y2) - (x3 - x2) * (y2 - y1))
# dist x1-x3
r2 = (c1 * (x3 - x2) - c2 * (x2 - x1)) / ((y2 - y1) * (x3 - x2) - (y3 - y2) * (x2 - x1))
# ipotenuza
r = ((r1 - x1) ** 2 + (r2 - y1) ** 2) ** 0.5
foobar = 0
for i in range(3, 101):
foo = False
bar = False
for ii in range(1, i):
x4 = (x1 - r1) * math.cos(2 * math.pi * ii / i) - math.sin(2 * math.pi * ii / i) * (y1 - r2) + r1
y4 = (y1 - r2) * math.cos(2 * math.pi * ii / i) + math.sin(2 * math.pi * ii / i) * (x1 - r1) + r2
# if i pillars yields a 'full circle'/regular polyogn; print ans
if (x4 - x2) ** 2 + (y4 - y2) ** 2 < 0.00001:
foo = True
if (x4 - x3) ** 2 + (y4 - y3) ** 2 < 0.00001:
bar = True
if foo and bar:
foobar = i
break
# formula for area of equiangular polygon
print(foobar * r ** 2 * math.sin(2 * math.pi / foobar) / 2)
``` | output | 1 | 93,784 | 23 | 187,569 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Nowadays all circuses in Berland have a round arena with diameter 13 meters, but in the past things were different.
In Ancient Berland arenas in circuses were shaped as a regular (equiangular) polygon, the size and the number of angles could vary from one circus to another. In each corner of the arena there was a special pillar, and the rope strung between the pillars marked the arena edges.
Recently the scientists from Berland have discovered the remains of the ancient circus arena. They found only three pillars, the others were destroyed by the time.
You are given the coordinates of these three pillars. Find out what is the smallest area that the arena could have.
Input
The input file consists of three lines, each of them contains a pair of numbers –– coordinates of the pillar. Any coordinate doesn't exceed 1000 by absolute value, and is given with at most six digits after decimal point.
Output
Output the smallest possible area of the ancient arena. This number should be accurate to at least 6 digits after the decimal point. It's guaranteed that the number of angles in the optimal polygon is not larger than 100.
Examples
Input
0.000000 0.000000
1.000000 1.000000
0.000000 1.000000
Output
1.00000000 | instruction | 0 | 93,785 | 23 | 187,570 |
Tags: geometry, math
Correct Solution:
```
import math
import sys
eps=1e-4
def getABC(x1,y1,x2,y2):
A=x2-x1
B=y2-y1
C=(x1*x1-x2*x2+y1*y1-y2*y2)/2
return A,B,C
# assume they are not parallel
def getIntersection(a1,b1,c1,a2,b2,c2):
x=(c2*b1-c1*b2)/(a1*b2-a2*b1)
y=(a1*c2-a2*c1)/(a2*b1-a1*b2)
return x,y
def getAngel(x1,y1,x2,y2):
return math.acos((x1*x2+y1*y2)/(math.sqrt(x1*x1+y1*y1)*math.sqrt(x2*x2+y2*y2)))
def check(num):
return abs(num-int(num+eps))<eps
x=[]
y=[]
for line in sys.stdin.readlines():
t=line.strip().split()
x.append(float(t[0]))
y.append(float(t[1]))
A1,B1,C1=getABC(x[0],y[0],x[1],y[1])
A2,B2,C2=getABC(x[2],y[2],x[1],y[1])
x0,y0=getIntersection(A1,B1,C1,A2,B2,C2)
an=[]
an.append(getAngel(x[0]-x0,y[0]-y0,x[1]-x0,y[1]-y0))
an.append(getAngel(x[0]-x0,y[0]-y0,x[2]-x0,y[2]-y0))
an.append(getAngel(x[2]-x0,y[2]-y0,x[1]-x0,y[1]-y0))
total = 2*math.pi
for i in range(3,101):
flag = True
angel = total/i
for j in range(0,3):
if not check(an[j]/angel):
flag = False
break
if flag:
break
n=i
m=total/n
ans = ((x[0]-x0)*(x[0]-x0)+(y[0]-y0)*(y[0]-y0))*math.sin(m)*n/2
print('%.6f' %ans)
``` | output | 1 | 93,785 | 23 | 187,571 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Nowadays all circuses in Berland have a round arena with diameter 13 meters, but in the past things were different.
In Ancient Berland arenas in circuses were shaped as a regular (equiangular) polygon, the size and the number of angles could vary from one circus to another. In each corner of the arena there was a special pillar, and the rope strung between the pillars marked the arena edges.
Recently the scientists from Berland have discovered the remains of the ancient circus arena. They found only three pillars, the others were destroyed by the time.
You are given the coordinates of these three pillars. Find out what is the smallest area that the arena could have.
Input
The input file consists of three lines, each of them contains a pair of numbers –– coordinates of the pillar. Any coordinate doesn't exceed 1000 by absolute value, and is given with at most six digits after decimal point.
Output
Output the smallest possible area of the ancient arena. This number should be accurate to at least 6 digits after the decimal point. It's guaranteed that the number of angles in the optimal polygon is not larger than 100.
Examples
Input
0.000000 0.000000
1.000000 1.000000
0.000000 1.000000
Output
1.00000000 | instruction | 0 | 93,786 | 23 | 187,572 |
Tags: geometry, math
Correct Solution:
```
import math
EPS = 1e-6
class Point:
def __init__(self, x, y):
self.x = x
self.y = y
def dist(self, p:"Point"):
return math.sqrt((self.x - p.x) ** 2 + (self.y - p.y) ** 2)
def dist_square(self, p:"Point"):
return (self.x - p.x) ** 2 + (self.y - p.y) ** 2
def getAreafromPolygon(points):
A = 0.0
for p1, p2 in zip(points[:-1], points[1:]):
A += 0.5 * (p1.x*p2.y - p2.x*p1.y)
A += 0.5 * (points[-1].x * points[0].y - points[0].x * points[-1].y)
return abs(A)
def getRadiusCircleFromThreePoints(p1:'Point', p2:'Point', p3:'Point'):
triangle_area = getAreafromPolygon([p1, p2, p3])
if triangle_area < EPS:
raise Exception('three points are in the same line')
r = p1.dist(p2) * p1.dist(p3) * p2.dist(p3) / 4.0 / triangle_area
return r
def get_angle_from_cosine_law(l1:float, l2: float, length_opposite_angle:float):
l3 = length_opposite_angle
if l1 < EPS or l2 < EPS:
raise Exception('side length must be positive real number')
if abs(l1-l2) < EPS and l3 <EPS:
return 0
if abs(l1+l2-l3) < EPS:
return math.pi
if l1+l2-l3 < -EPS:
raise Exception('invalid input; length of three sides cannot form triangle')
return math.acos( (l1**2 + l2**2 - l3**2)/ (2*l1*l2) )
points = []
for i in range(3):
x, y = map(float, input().split(' '))
points.append(Point(x,y))
radius = getRadiusCircleFromThreePoints(points[0], points[1], points[2])
l01 = points[0].dist(points[1])
l02 = points[0].dist(points[2])
angle_p0Op1 = get_angle_from_cosine_law(radius, radius, l01)
angle_p0Op2 = get_angle_from_cosine_law(radius, radius, l02)
i = 3
while i:
angle_per_side = 2*math.pi / i
mod_angle_p0Op1 = math.fmod(angle_p0Op1 ,angle_per_side)
mod_angle_p0Op2 = math.fmod(angle_p0Op2 ,angle_per_side)
if (
(mod_angle_p0Op1 < EPS or mod_angle_p0Op1 > angle_per_side-EPS)
and (mod_angle_p0Op2 < EPS or mod_angle_p0Op2 > angle_per_side-EPS)
):
print(i * 0.5 * radius * radius * math.sin(angle_per_side))
break
i += 1
``` | output | 1 | 93,786 | 23 | 187,573 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Nowadays all circuses in Berland have a round arena with diameter 13 meters, but in the past things were different.
In Ancient Berland arenas in circuses were shaped as a regular (equiangular) polygon, the size and the number of angles could vary from one circus to another. In each corner of the arena there was a special pillar, and the rope strung between the pillars marked the arena edges.
Recently the scientists from Berland have discovered the remains of the ancient circus arena. They found only three pillars, the others were destroyed by the time.
You are given the coordinates of these three pillars. Find out what is the smallest area that the arena could have.
Input
The input file consists of three lines, each of them contains a pair of numbers –– coordinates of the pillar. Any coordinate doesn't exceed 1000 by absolute value, and is given with at most six digits after decimal point.
Output
Output the smallest possible area of the ancient arena. This number should be accurate to at least 6 digits after the decimal point. It's guaranteed that the number of angles in the optimal polygon is not larger than 100.
Examples
Input
0.000000 0.000000
1.000000 1.000000
0.000000 1.000000
Output
1.00000000
Submitted Solution:
```
s1 = input().split()
s2 = input().split()
s3 = input().split()
ax = float(s1[0])
ay = float(s1[1])
bx = float(s2[0])
by = float(s2[1])
cx = float(s3[0])
cy = float(s3[1])
det = (bx-ax)*(cy-ay) - (by-ay)*(cx-ax)
det_x = (by-ay)*(ax*ax+ay*ay-cx*cx-cy*cy)-(cy-ay)*(ax*ax+ay*ay-bx*bx-by*by)
det_y = (cx-ax)*(ax*ax+ay*ay-bx*bx-by*by)-(bx-ax)*(ax*ax+ay*ay-cx*cx-cy*cy)
ox = det_x/det/2.0
oy = det_y/det/2.0
def shortest(x, y, pts):
d2_min = None
for pt in pts:
dx = pt[0]-x
dy = pt[1]-y
d2 = dx*dx + dy* dy
if d2_min is None or d2 < d2_min:
d2_min = d2
return d2_min
import math
d_min = None
num_min = None
oax = ax - ox
oay = ay - oy
r = math.sqrt(oax*oax+oay*oay)
if(abs(oax) < 1.e-8):
alpha = math.pi/2 if oay > 0 else -math.pi/2
else:
alpha = math.atan(oay/oax) if oax > 0 else math.pi + math.atan(oay/oax)
for num in range(3, 101):
pts = []
for i in range(1, num):
rad = i * 2 * math.pi / num
pts.append((ox+r*math.cos(rad+alpha), oy+r*math.sin(rad+alpha)))
s2 = shortest(bx, by, pts)
s3 = shortest(cx, cy, pts)
s = s2 + s3
if d_min is None or s < d_min - 1.e-8:
d_min = s
num_min = num
beta = 2*math.pi / num_min
area = num_min * r * r * math.sin(beta) /2.0
print(f"{area:.8f}")
``` | instruction | 0 | 93,787 | 23 | 187,574 |
Yes | output | 1 | 93,787 | 23 | 187,575 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Nowadays all circuses in Berland have a round arena with diameter 13 meters, but in the past things were different.
In Ancient Berland arenas in circuses were shaped as a regular (equiangular) polygon, the size and the number of angles could vary from one circus to another. In each corner of the arena there was a special pillar, and the rope strung between the pillars marked the arena edges.
Recently the scientists from Berland have discovered the remains of the ancient circus arena. They found only three pillars, the others were destroyed by the time.
You are given the coordinates of these three pillars. Find out what is the smallest area that the arena could have.
Input
The input file consists of three lines, each of them contains a pair of numbers –– coordinates of the pillar. Any coordinate doesn't exceed 1000 by absolute value, and is given with at most six digits after decimal point.
Output
Output the smallest possible area of the ancient arena. This number should be accurate to at least 6 digits after the decimal point. It's guaranteed that the number of angles in the optimal polygon is not larger than 100.
Examples
Input
0.000000 0.000000
1.000000 1.000000
0.000000 1.000000
Output
1.00000000
Submitted Solution:
```
#!/usr/bin/env python
# -*- coding: utf-8 -*-
import math
DEBUG = False
def area(d):
if len(d) < 3:
return False
a = 0.0
p = 0.5 * (d[0] + d[1] + d[2])
a = math.sqrt(p*(p-d[0])*(p-d[1])*(p-d[2]))
return a
def circumcircle(x,y):
distance = euclidean(x,y)
triangle_area = area(distance)
return distance[0] * distance[1] * distance[2] / (4 * triangle_area)
def euclidean(x,y):
d = []
d.append(math.sqrt((x[0]-x[1])**2 + (y[0]-y[1])**2))
d.append(math.sqrt((x[0]-x[2])**2 + (y[0]-y[2])**2))
d.append(math.sqrt((x[1]-x[2])**2 + (y[1]-y[2])**2))
return d
def normal_sin_value(sin_value):
if sin_value > 1:
return 1.0
elif sin_value < -1:
return -1.0
else:
return sin_value
def centerAngle(d,radius):
angle = []
#print(type(d[0]), type(radius))
#print(d[0]/(2*radius))
#print(2 * math.asin( d[0] / (2*radius) ) )
angle.append(2*math.asin(normal_sin_value(d[0]/(2*radius))))
angle.append(2*math.asin(normal_sin_value(d[1]/(2*radius))))
angle.append(2*math.pi - angle[0] - angle[1])
if DEBUG:
print(sum([a for a in angle]))
return angle
def gcd(a,b):
if a < 0.01:
return b
else:
return gcd(math.fmod(b,a),a)
def main():
# get the input data
x = []
y = []
for i in range(3):
temp_input = input().split()
x.append(float(temp_input[0]))
y.append(float(temp_input[1]))
# 1. calculate the length of the edge
# 2. calculate the area of the triangle
# 3. calculate the radius of the circumcircle
# 4. calculate the area of the Berland Circus
edge_length = euclidean(x,y)
triangle_area = area(edge_length)
circumcircle_radius = circumcircle(x,y)
#print("circumcircle_radius: {0[0]}:{1[0]},{0[1]}:{1[1]}, {0[2]}:{1[2]} \n {2}".format(x,y,circumcircle_radius))
# 5. calculat the cetral angle and their gcd
angle = centerAngle(edge_length, circumcircle_radius)
gcd_angle = gcd(gcd(angle[0], angle[1]), angle[2])
result = 2 * math.pi / gcd_angle * circumcircle_radius * math.sin(gcd_angle) * circumcircle_radius * 0.5
if DEBUG:
print("circumcircle_radius",circumcircle_radius)
print("totoal_angle",angle)
print("gcd_angle",gcd_angle)
print("len",edge_length)
print("{:.8f}".format(result))
if __name__ == "__main__":
main()
``` | instruction | 0 | 93,788 | 23 | 187,576 |
Yes | output | 1 | 93,788 | 23 | 187,577 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Nowadays all circuses in Berland have a round arena with diameter 13 meters, but in the past things were different.
In Ancient Berland arenas in circuses were shaped as a regular (equiangular) polygon, the size and the number of angles could vary from one circus to another. In each corner of the arena there was a special pillar, and the rope strung between the pillars marked the arena edges.
Recently the scientists from Berland have discovered the remains of the ancient circus arena. They found only three pillars, the others were destroyed by the time.
You are given the coordinates of these three pillars. Find out what is the smallest area that the arena could have.
Input
The input file consists of three lines, each of them contains a pair of numbers –– coordinates of the pillar. Any coordinate doesn't exceed 1000 by absolute value, and is given with at most six digits after decimal point.
Output
Output the smallest possible area of the ancient arena. This number should be accurate to at least 6 digits after the decimal point. It's guaranteed that the number of angles in the optimal polygon is not larger than 100.
Examples
Input
0.000000 0.000000
1.000000 1.000000
0.000000 1.000000
Output
1.00000000
Submitted Solution:
```
import math
def distance(p1, p2):
return math.sqrt((p1[0] - p2[0]) ** 2 + (p1[1] - p2[1]) ** 2)
def gcd(a, b):
if a < b:
return gcd(b, a)
if abs(b) < 0.001:
return a
else:
return gcd(b, a - math.floor(a / b) * b)
points = []
for i in range(3):
(x, y) = [float(x) for x in input().split()]
point = (x, y)
points.append(point)
a = distance(points[0], points[1])
b = distance(points[1], points[2])
c = distance(points[2], points[0])
s = (a + b + c) / 2
R = (a * b * c) / (4 * math.sqrt(s * (a + b - s) * (a + c - s) * (b + c - s)))
A = math.acos((b * b + c * c - a * a) / (2 * b * c))
B = math.acos((a * a + c * c - b * b) / (2 * a * c))
C = math.acos((b * b + a * a - c * c) / (2 * b * a))
n = math.pi / gcd(A, gcd(B, C))
area = (n * R * R * math.sin((2 * math.pi) / n)) / 2
print(area)
``` | instruction | 0 | 93,789 | 23 | 187,578 |
Yes | output | 1 | 93,789 | 23 | 187,579 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Nowadays all circuses in Berland have a round arena with diameter 13 meters, but in the past things were different.
In Ancient Berland arenas in circuses were shaped as a regular (equiangular) polygon, the size and the number of angles could vary from one circus to another. In each corner of the arena there was a special pillar, and the rope strung between the pillars marked the arena edges.
Recently the scientists from Berland have discovered the remains of the ancient circus arena. They found only three pillars, the others were destroyed by the time.
You are given the coordinates of these three pillars. Find out what is the smallest area that the arena could have.
Input
The input file consists of three lines, each of them contains a pair of numbers –– coordinates of the pillar. Any coordinate doesn't exceed 1000 by absolute value, and is given with at most six digits after decimal point.
Output
Output the smallest possible area of the ancient arena. This number should be accurate to at least 6 digits after the decimal point. It's guaranteed that the number of angles in the optimal polygon is not larger than 100.
Examples
Input
0.000000 0.000000
1.000000 1.000000
0.000000 1.000000
Output
1.00000000
Submitted Solution:
```
import math
def gcd(x,y):
if abs(y) < 1e-4 :
return x
else :
return gcd(y, math.fmod(x, y))
Ax,Ay = list(map(float,input().split()))
Bx,By = list(map(float,input().split()))
Cx,Cy = list(map(float,input().split()))
a = math.sqrt((Bx - Cx) * (Bx - Cx) + (By - Cy) * (By - Cy))
b = math.sqrt((Ax - Cx) * (Ax - Cx) + (Ay - Cy) * (Ay - Cy))
c = math.sqrt((Ax - Bx) * (Ax - Bx) + (Ay - By) * (Ay - By))
s = (a + b + c) / 2
area_triangle = math.sqrt(s * (s - a) * (s - b) * (s - c))
r = (a * b * c) / (4 * area_triangle)
Angle_A = math.acos((b * b + c * c - a * a) / (2 * b * c))
Angle_B = math.acos((a * a + c * c - b * b) / (2 * a * c))
Angle_C = math.acos((b * b + a * a - c * c) / (2 * b * a))
n = math.pi / gcd(gcd(Angle_A, Angle_B), Angle_C)
print ((n * r * r * math.sin((2 * math.pi) / n)) / 2)
``` | instruction | 0 | 93,790 | 23 | 187,580 |
Yes | output | 1 | 93,790 | 23 | 187,581 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Nowadays all circuses in Berland have a round arena with diameter 13 meters, but in the past things were different.
In Ancient Berland arenas in circuses were shaped as a regular (equiangular) polygon, the size and the number of angles could vary from one circus to another. In each corner of the arena there was a special pillar, and the rope strung between the pillars marked the arena edges.
Recently the scientists from Berland have discovered the remains of the ancient circus arena. They found only three pillars, the others were destroyed by the time.
You are given the coordinates of these three pillars. Find out what is the smallest area that the arena could have.
Input
The input file consists of three lines, each of them contains a pair of numbers –– coordinates of the pillar. Any coordinate doesn't exceed 1000 by absolute value, and is given with at most six digits after decimal point.
Output
Output the smallest possible area of the ancient arena. This number should be accurate to at least 6 digits after the decimal point. It's guaranteed that the number of angles in the optimal polygon is not larger than 100.
Examples
Input
0.000000 0.000000
1.000000 1.000000
0.000000 1.000000
Output
1.00000000
Submitted Solution:
```
import math
input_var = []
for n in range(3):
input_var.append(input().split())
P1, P2, P3 = input_var
P1 = [float(num) for num in P1]
P2 = [float(num) for num in P2]
P3 = [float(num) for num in P3]
def triangle_side_lengths(P1, P2, P3):
return [((P1[0] - P2[0])**2 + (P1[1] - P2[1])**2)**(1 / 2), ((P1[0] - P3[0])**2 + (P1[1] - P3[1])**2)**(1 / 2), ((P2[0] - P3[0])**2 + (P2[1] - P3[1])**2)**(1 / 2)]
def circum_radius(P1, P2, P3):
a, b, c = triangle_side_lengths(P1, P2, P3)
return (a * b * c) / ((a + b + c) * (b + c - a) * (c + a - b) * (a + b - c))**(1 / 2)
def side_length(P1, P2, P3):
for i in range(3, 101):
if math.fabs(circum_radius(P1, P2, P3)*2*math.sin(math.pi/i)- round(circum_radius(P1, P2, P3)*2*math.sin(math.pi/i),4))< 1E-6 and i==3:
return (1/2)*i*(circum_radius(P1, P2, P3)**2)*math.sin((2*math.pi)/i)
elif math.fabs(circum_radius(P1, P2, P3)*2*math.sin(math.pi/i)- round(circum_radius(P1, P2, P3)*2*math.sin(math.pi/i),4))< 1E-6 and 3>i:
return (1/2)*i*(circum_radius(P1, P2, P3)**2)*math.sin((2*math.pi)/i)/2
print(side_length(P1, P2, P3))
``` | instruction | 0 | 93,791 | 23 | 187,582 |
No | output | 1 | 93,791 | 23 | 187,583 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Nowadays all circuses in Berland have a round arena with diameter 13 meters, but in the past things were different.
In Ancient Berland arenas in circuses were shaped as a regular (equiangular) polygon, the size and the number of angles could vary from one circus to another. In each corner of the arena there was a special pillar, and the rope strung between the pillars marked the arena edges.
Recently the scientists from Berland have discovered the remains of the ancient circus arena. They found only three pillars, the others were destroyed by the time.
You are given the coordinates of these three pillars. Find out what is the smallest area that the arena could have.
Input
The input file consists of three lines, each of them contains a pair of numbers –– coordinates of the pillar. Any coordinate doesn't exceed 1000 by absolute value, and is given with at most six digits after decimal point.
Output
Output the smallest possible area of the ancient arena. This number should be accurate to at least 6 digits after the decimal point. It's guaranteed that the number of angles in the optimal polygon is not larger than 100.
Examples
Input
0.000000 0.000000
1.000000 1.000000
0.000000 1.000000
Output
1.00000000
Submitted Solution:
```
#!/usr/bin/env python3
# -*- coding: utf-8 -*-
import math
def read_int():
return int(input())
def read_array(constructor):
return [constructor(x) for x in input().split()]
def read_int_array():
return read_array(int)
p = read_array(float)
q = read_array(float)
r = read_array(float)
pq = math.hypot(q[0] - p[0], q[1] - p[1])
pr = math.hypot(r[0] - p[0], r[1] - p[1])
qr = math.hypot(r[0] - q[0], r[1] - q[1])
edges = [pq, pr, qr]
edges.sort()
cos_inscribed_angle = (edges[1]**2 + edges[2]**2 - edges[0]**2) / (2 * edges[1] * edges[2])
inscribed_angle = math.acos(cos_inscribed_angle)
central_angle = inscribed_angle * 2
sin_inscribed_angle = math.sin(inscribed_angle)
radius = edges[0] / (2 * sin_inscribed_angle)
# how to determine #angle
num_angle = 0
for n in range(1, 101):
diff = (central_angle * n) % (2 * math.pi)
# print(diff)
if math.isclose(diff, 0, abs_tol=0.01) or \
math.isclose(diff, 2 * math.pi, rel_tol=0.01):
num_angle = n
break
else:
print('ERROR: n={0}, num_angle={1}'.format(n, num_angle))
angle = 2 * math.pi / num_angle
area = radius**2 * math.sin(angle) * num_angle / 2
print('{0:f}'.format(area))
``` | instruction | 0 | 93,792 | 23 | 187,584 |
No | output | 1 | 93,792 | 23 | 187,585 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Nowadays all circuses in Berland have a round arena with diameter 13 meters, but in the past things were different.
In Ancient Berland arenas in circuses were shaped as a regular (equiangular) polygon, the size and the number of angles could vary from one circus to another. In each corner of the arena there was a special pillar, and the rope strung between the pillars marked the arena edges.
Recently the scientists from Berland have discovered the remains of the ancient circus arena. They found only three pillars, the others were destroyed by the time.
You are given the coordinates of these three pillars. Find out what is the smallest area that the arena could have.
Input
The input file consists of three lines, each of them contains a pair of numbers –– coordinates of the pillar. Any coordinate doesn't exceed 1000 by absolute value, and is given with at most six digits after decimal point.
Output
Output the smallest possible area of the ancient arena. This number should be accurate to at least 6 digits after the decimal point. It's guaranteed that the number of angles in the optimal polygon is not larger than 100.
Examples
Input
0.000000 0.000000
1.000000 1.000000
0.000000 1.000000
Output
1.00000000
Submitted Solution:
```
import math
A = input().split()
B = input().split()
C = input().split()
#a1, a2 = 17.288379, 68.223317
#b1, b2 = 48.776683, 71.688379
#c1, c2 = 23.170559, 106.572762
a1, a2 = float(A[0]), float(A[1])
b1, b2 = float(B[0]), float(B[1])
c1, c2 = float(C[0]), float(C[1])
a = ((b1-c1)**2 + (b2-c2)**2) **0.5
b = ((a1-c1)**2 + (a2-c2)**2) **0.5
c = ((b1-a1)**2 + (b2-a2)**2) **0.5
#t = (a+b+c)/2
#area = (t*(t-a)*(t-b)*(t-c))**0.5
#print(area)
#print(a,b,c)
cosa = (b**2+c**2-a**2)/(2*b*c)
sina = (1-cosa**2)**0.5
R = a/(2*sina)
side = min(a,b,c)
s = (R+R+side)/2
#tri_area = 1/2*R*R*sinTheta
tri_area = (s*(s-R)*(s-R)*(s-side))**0.5
sides = [a,b,c]
countSide = []
thetaAll = []
for s in sides:
cosTheta = (R**2+R**2-s**2)/(2*R**2)
theta = math.acos(cosTheta)
thetaAll.append(round(math.degrees(theta)))
count = (2 * math.pi) / theta
countSide.append(count)
#print(countSide)
#print("degrees:", thetaAll)
theta = math.radians(math.gcd(max(thetaAll), min(thetaAll)))
count = 2*math.pi / theta
total_area = 1/2 * math.sin(theta) * R * R * count
#print(count)
print(total_area)
#print(a)
#print(sina)
#print("R:", a/(2*sina))
#print("tri_area:", tri_area)
#print(theta)
#print(count)
#print(total_area)
``` | instruction | 0 | 93,793 | 23 | 187,586 |
No | output | 1 | 93,793 | 23 | 187,587 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Nowadays all circuses in Berland have a round arena with diameter 13 meters, but in the past things were different.
In Ancient Berland arenas in circuses were shaped as a regular (equiangular) polygon, the size and the number of angles could vary from one circus to another. In each corner of the arena there was a special pillar, and the rope strung between the pillars marked the arena edges.
Recently the scientists from Berland have discovered the remains of the ancient circus arena. They found only three pillars, the others were destroyed by the time.
You are given the coordinates of these three pillars. Find out what is the smallest area that the arena could have.
Input
The input file consists of three lines, each of them contains a pair of numbers –– coordinates of the pillar. Any coordinate doesn't exceed 1000 by absolute value, and is given with at most six digits after decimal point.
Output
Output the smallest possible area of the ancient arena. This number should be accurate to at least 6 digits after the decimal point. It's guaranteed that the number of angles in the optimal polygon is not larger than 100.
Examples
Input
0.000000 0.000000
1.000000 1.000000
0.000000 1.000000
Output
1.00000000
Submitted Solution:
```
from math import sqrt,acos,sin
from re import split
def dis(a,b):
return sqrt((a[0]-b[0])**2+(a[1]-b[1])**2)
def ang(a,d,b):
return acos((dis(a,d)**2+dis(b,d)**2-dis(a,b)**2)/(2*dis(a,d)*dis(b,d)))
def S(a,b,d):
s=(1/2)*dis(a,d)*dis(b,d)*sin(ang(a,d,b))
if round(s,6)!=0:
return s
else:
return 10000000000
def l2f(a):
for i in range(len(a)):
a[i]=float(a[i])
a=input()
a=a.split()
l2f(a)
b=input()
b=b.split()
l2f(b)
c=input()
c=c.split()
l2f(c)
A1=2*(b[0]-a[0])
B1=2*(b[1]-a[1])
C1=b[0]**2+b[1]**2-a[0]**2-a[1]**2
A2=2*(c[0]-b[0])
B2=2*(c[1]-b[1])
C2=c[0]**2+c[1]**2-b[0]**2-b[1]**2
d=[]
d.append(((C1*B2)-(C2*B1))/((A1*B2)-(A2*B1)))
d.append(((A1*C2)-(A2*C1))/((A1*B2)-(A2*B1)))
ang_min=min(ang(a,d,b),ang(b,d,c),ang(a,d,c))
n=round(2*acos(-1)/ang_min)
if ang(a,d,b)==ang_min:
s=S(a,b,d)*n
if ang(b,d,c)==ang_min:
s=S(b,c,d)*n
if ang(a,d,c)==ang_min:
s=S(a,c,d)*n
print("%.8f" % s)
``` | instruction | 0 | 93,794 | 23 | 187,588 |
No | output | 1 | 93,794 | 23 | 187,589 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Johnny has a younger sister Anne, who is very clever and smart. As she came home from the kindergarten, she told his brother about the task that her kindergartener asked her to solve. The task was just to construct a triangle out of four sticks of different colours. Naturally, one of the sticks is extra. It is not allowed to break the sticks or use their partial length. Anne has perfectly solved this task, now she is asking Johnny to do the same.
The boy answered that he would cope with it without any difficulty. However, after a while he found out that different tricky things can occur. It can happen that it is impossible to construct a triangle of a positive area, but it is possible to construct a degenerate triangle. It can be so, that it is impossible to construct a degenerate triangle even. As Johnny is very lazy, he does not want to consider such a big amount of cases, he asks you to help him.
Input
The first line of the input contains four space-separated positive integer numbers not exceeding 100 — lengthes of the sticks.
Output
Output TRIANGLE if it is possible to construct a non-degenerate triangle. Output SEGMENT if the first case cannot take place and it is possible to construct a degenerate triangle. Output IMPOSSIBLE if it is impossible to construct any triangle. Remember that you are to use three sticks. It is not allowed to break the sticks or use their partial length.
Examples
Input
4 2 1 3
Output
TRIANGLE
Input
7 2 2 4
Output
SEGMENT
Input
3 5 9 1
Output
IMPOSSIBLE | instruction | 0 | 93,994 | 23 | 187,988 |
Tags: brute force, geometry
Correct Solution:
```
ints,answer = list(reversed(sorted(list(map(int, input().split(' ')))))), "IMPOSSIBLE"
if ints[0] < ints[1] + ints[2]:
#0 did work
answer = "TRIANGLE"
elif ints[0] == ints[1] + ints[2]:
#segment only a triangle is better.
answer = "SEGMENT"
if ints[1] < ints[2] + ints[3]:
#1 did work
answer = "TRIANGLE"
else:
# a triangle or segment is needed.
if ints[1] < ints[2] + ints[3]:
#1 did work
answer = "TRIANGLE"
elif ints[1] == ints[2] + ints[3]:
answer = "SEGMENT"
print(answer)
``` | output | 1 | 93,994 | 23 | 187,989 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Johnny has a younger sister Anne, who is very clever and smart. As she came home from the kindergarten, she told his brother about the task that her kindergartener asked her to solve. The task was just to construct a triangle out of four sticks of different colours. Naturally, one of the sticks is extra. It is not allowed to break the sticks or use their partial length. Anne has perfectly solved this task, now she is asking Johnny to do the same.
The boy answered that he would cope with it without any difficulty. However, after a while he found out that different tricky things can occur. It can happen that it is impossible to construct a triangle of a positive area, but it is possible to construct a degenerate triangle. It can be so, that it is impossible to construct a degenerate triangle even. As Johnny is very lazy, he does not want to consider such a big amount of cases, he asks you to help him.
Input
The first line of the input contains four space-separated positive integer numbers not exceeding 100 — lengthes of the sticks.
Output
Output TRIANGLE if it is possible to construct a non-degenerate triangle. Output SEGMENT if the first case cannot take place and it is possible to construct a degenerate triangle. Output IMPOSSIBLE if it is impossible to construct any triangle. Remember that you are to use three sticks. It is not allowed to break the sticks or use their partial length.
Examples
Input
4 2 1 3
Output
TRIANGLE
Input
7 2 2 4
Output
SEGMENT
Input
3 5 9 1
Output
IMPOSSIBLE | instruction | 0 | 93,995 | 23 | 187,990 |
Tags: brute force, geometry
Correct Solution:
```
def task():
a,b,c,d = input().split()
a,b,c,d = [int(a),int(b),int(c),int(d)]
if (a+b>c and a+c>b and c+b>a) or (d+a>b and d+b>a and a+b>d) or (a+c>d and c+d>a and d+a>c) or (b+c>d and b+d>c and d+c>b):
return("TRIANGLE")
elif a+b==c or a+d==c or d+b==c or a+c==d or a+b==d or c+b==d or c+b==a or c+d==a or d+b==a or a+c==b or a+d==b or c+d==b:
return("SEGMENT")
else:
return("IMPOSSIBLE")
print(task())
``` | output | 1 | 93,995 | 23 | 187,991 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Johnny has a younger sister Anne, who is very clever and smart. As she came home from the kindergarten, she told his brother about the task that her kindergartener asked her to solve. The task was just to construct a triangle out of four sticks of different colours. Naturally, one of the sticks is extra. It is not allowed to break the sticks or use their partial length. Anne has perfectly solved this task, now she is asking Johnny to do the same.
The boy answered that he would cope with it without any difficulty. However, after a while he found out that different tricky things can occur. It can happen that it is impossible to construct a triangle of a positive area, but it is possible to construct a degenerate triangle. It can be so, that it is impossible to construct a degenerate triangle even. As Johnny is very lazy, he does not want to consider such a big amount of cases, he asks you to help him.
Input
The first line of the input contains four space-separated positive integer numbers not exceeding 100 — lengthes of the sticks.
Output
Output TRIANGLE if it is possible to construct a non-degenerate triangle. Output SEGMENT if the first case cannot take place and it is possible to construct a degenerate triangle. Output IMPOSSIBLE if it is impossible to construct any triangle. Remember that you are to use three sticks. It is not allowed to break the sticks or use their partial length.
Examples
Input
4 2 1 3
Output
TRIANGLE
Input
7 2 2 4
Output
SEGMENT
Input
3 5 9 1
Output
IMPOSSIBLE | instruction | 0 | 93,996 | 23 | 187,992 |
Tags: brute force, geometry
Correct Solution:
```
arr = list(map(int, input().strip().split()))
arr.sort()
a, b, c, d = arr
if a+b < c:
if b+c > d:
print('TRIANGLE')
elif b+c == d:
print('SEGMENT')
else:
print('IMPOSSIBLE')
elif a+b == c:
if b+c > d:
print('TRIANGLE')
else:
print('SEGMENT')
else:
print('TRIANGLE')
``` | output | 1 | 93,996 | 23 | 187,993 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Johnny has a younger sister Anne, who is very clever and smart. As she came home from the kindergarten, she told his brother about the task that her kindergartener asked her to solve. The task was just to construct a triangle out of four sticks of different colours. Naturally, one of the sticks is extra. It is not allowed to break the sticks or use their partial length. Anne has perfectly solved this task, now she is asking Johnny to do the same.
The boy answered that he would cope with it without any difficulty. However, after a while he found out that different tricky things can occur. It can happen that it is impossible to construct a triangle of a positive area, but it is possible to construct a degenerate triangle. It can be so, that it is impossible to construct a degenerate triangle even. As Johnny is very lazy, he does not want to consider such a big amount of cases, he asks you to help him.
Input
The first line of the input contains four space-separated positive integer numbers not exceeding 100 — lengthes of the sticks.
Output
Output TRIANGLE if it is possible to construct a non-degenerate triangle. Output SEGMENT if the first case cannot take place and it is possible to construct a degenerate triangle. Output IMPOSSIBLE if it is impossible to construct any triangle. Remember that you are to use three sticks. It is not allowed to break the sticks or use their partial length.
Examples
Input
4 2 1 3
Output
TRIANGLE
Input
7 2 2 4
Output
SEGMENT
Input
3 5 9 1
Output
IMPOSSIBLE | instruction | 0 | 93,997 | 23 | 187,994 |
Tags: brute force, geometry
Correct Solution:
```
a=list(map(int,input().split()))
a.sort()
if a[0]+a[1]>a[2]:
print("TRIANGLE")
elif a[1]+a[2]>a[3]:
print("TRIANGLE")
elif a[0]+a[1]==a[2] or a[1]+a[2]==a[3]:
print("SEGMENT")
else:
print("IMPOSSIBLE")
``` | output | 1 | 93,997 | 23 | 187,995 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Johnny has a younger sister Anne, who is very clever and smart. As she came home from the kindergarten, she told his brother about the task that her kindergartener asked her to solve. The task was just to construct a triangle out of four sticks of different colours. Naturally, one of the sticks is extra. It is not allowed to break the sticks or use their partial length. Anne has perfectly solved this task, now she is asking Johnny to do the same.
The boy answered that he would cope with it without any difficulty. However, after a while he found out that different tricky things can occur. It can happen that it is impossible to construct a triangle of a positive area, but it is possible to construct a degenerate triangle. It can be so, that it is impossible to construct a degenerate triangle even. As Johnny is very lazy, he does not want to consider such a big amount of cases, he asks you to help him.
Input
The first line of the input contains four space-separated positive integer numbers not exceeding 100 — lengthes of the sticks.
Output
Output TRIANGLE if it is possible to construct a non-degenerate triangle. Output SEGMENT if the first case cannot take place and it is possible to construct a degenerate triangle. Output IMPOSSIBLE if it is impossible to construct any triangle. Remember that you are to use three sticks. It is not allowed to break the sticks or use their partial length.
Examples
Input
4 2 1 3
Output
TRIANGLE
Input
7 2 2 4
Output
SEGMENT
Input
3 5 9 1
Output
IMPOSSIBLE | instruction | 0 | 93,998 | 23 | 187,996 |
Tags: brute force, geometry
Correct Solution:
```
x,y,z,w=map(int,input().split())
arr=[]
arr.append(x)
arr.append(y)
arr.append(z)
arr.append(w)
arr.sort()
# print(arr)
if (arr[0]+arr[1]>arr[2]) or (arr[0]+arr[1]>arr[3]) or (arr[1]+arr[2]>arr[3]):
print("TRIANGLE")
elif (arr[0]+arr[1]==arr[2]) or (arr[0]+arr[1]==arr[3]) or (arr[1]+arr[2]==arr[3]):
print("SEGMENT")
else:
print("IMPOSSIBLE")
``` | output | 1 | 93,998 | 23 | 187,997 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Johnny has a younger sister Anne, who is very clever and smart. As she came home from the kindergarten, she told his brother about the task that her kindergartener asked her to solve. The task was just to construct a triangle out of four sticks of different colours. Naturally, one of the sticks is extra. It is not allowed to break the sticks or use their partial length. Anne has perfectly solved this task, now she is asking Johnny to do the same.
The boy answered that he would cope with it without any difficulty. However, after a while he found out that different tricky things can occur. It can happen that it is impossible to construct a triangle of a positive area, but it is possible to construct a degenerate triangle. It can be so, that it is impossible to construct a degenerate triangle even. As Johnny is very lazy, he does not want to consider such a big amount of cases, he asks you to help him.
Input
The first line of the input contains four space-separated positive integer numbers not exceeding 100 — lengthes of the sticks.
Output
Output TRIANGLE if it is possible to construct a non-degenerate triangle. Output SEGMENT if the first case cannot take place and it is possible to construct a degenerate triangle. Output IMPOSSIBLE if it is impossible to construct any triangle. Remember that you are to use three sticks. It is not allowed to break the sticks or use their partial length.
Examples
Input
4 2 1 3
Output
TRIANGLE
Input
7 2 2 4
Output
SEGMENT
Input
3 5 9 1
Output
IMPOSSIBLE | instruction | 0 | 93,999 | 23 | 187,998 |
Tags: brute force, geometry
Correct Solution:
```
l = list(map(int, input().split()))
m = l[0]
l.sort()
if ((l[0]+l[1])>l[2]) or ((l[1]+l[2])>l[3]):
print('TRIANGLE')
elif ((l[0]+l[1]) == l[2]) or ((l[1]+l[2]) == l[3]):
print('SEGMENT')
else:
print('IMPOSSIBLE')
``` | output | 1 | 93,999 | 23 | 187,999 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Johnny has a younger sister Anne, who is very clever and smart. As she came home from the kindergarten, she told his brother about the task that her kindergartener asked her to solve. The task was just to construct a triangle out of four sticks of different colours. Naturally, one of the sticks is extra. It is not allowed to break the sticks or use their partial length. Anne has perfectly solved this task, now she is asking Johnny to do the same.
The boy answered that he would cope with it without any difficulty. However, after a while he found out that different tricky things can occur. It can happen that it is impossible to construct a triangle of a positive area, but it is possible to construct a degenerate triangle. It can be so, that it is impossible to construct a degenerate triangle even. As Johnny is very lazy, he does not want to consider such a big amount of cases, he asks you to help him.
Input
The first line of the input contains four space-separated positive integer numbers not exceeding 100 — lengthes of the sticks.
Output
Output TRIANGLE if it is possible to construct a non-degenerate triangle. Output SEGMENT if the first case cannot take place and it is possible to construct a degenerate triangle. Output IMPOSSIBLE if it is impossible to construct any triangle. Remember that you are to use three sticks. It is not allowed to break the sticks or use their partial length.
Examples
Input
4 2 1 3
Output
TRIANGLE
Input
7 2 2 4
Output
SEGMENT
Input
3 5 9 1
Output
IMPOSSIBLE | instruction | 0 | 94,000 | 23 | 188,000 |
Tags: brute force, geometry
Correct Solution:
```
dic = {"t": False, "s": False}
a, b, c, d = list(map(int, input().split(" ")))
l = [[a, b, c], [a, b, d], [a, c, d], [b, c, d]]
for x, y, z in l:
if(x + y > z and y + z > x and x + z > y):
dic["t"] = True
break
elif(x + y == z or y + z == x or x + z == y):
dic["s"] = True
if(dic["t"]):
print("TRIANGLE")
elif(dic["s"]):
print("SEGMENT")
else:
print("IMPOSSIBLE")
``` | output | 1 | 94,000 | 23 | 188,001 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Johnny has a younger sister Anne, who is very clever and smart. As she came home from the kindergarten, she told his brother about the task that her kindergartener asked her to solve. The task was just to construct a triangle out of four sticks of different colours. Naturally, one of the sticks is extra. It is not allowed to break the sticks or use their partial length. Anne has perfectly solved this task, now she is asking Johnny to do the same.
The boy answered that he would cope with it without any difficulty. However, after a while he found out that different tricky things can occur. It can happen that it is impossible to construct a triangle of a positive area, but it is possible to construct a degenerate triangle. It can be so, that it is impossible to construct a degenerate triangle even. As Johnny is very lazy, he does not want to consider such a big amount of cases, he asks you to help him.
Input
The first line of the input contains four space-separated positive integer numbers not exceeding 100 — lengthes of the sticks.
Output
Output TRIANGLE if it is possible to construct a non-degenerate triangle. Output SEGMENT if the first case cannot take place and it is possible to construct a degenerate triangle. Output IMPOSSIBLE if it is impossible to construct any triangle. Remember that you are to use three sticks. It is not allowed to break the sticks or use their partial length.
Examples
Input
4 2 1 3
Output
TRIANGLE
Input
7 2 2 4
Output
SEGMENT
Input
3 5 9 1
Output
IMPOSSIBLE | instruction | 0 | 94,001 | 23 | 188,002 |
Tags: brute force, geometry
Correct Solution:
```
import sys
from itertools import combinations
sticks = list(map(int,sys.stdin.readline().split()))
for edges in combinations(sticks, 3):
edges = sorted(edges)
# edges.sort()
# print(edges)
if edges[0] + edges[1] > edges[-1] and edges[-1] - edges[0] < edges[1]:
print('TRIANGLE')
exit(0)
for edges in combinations(sticks, 3):
edges = sorted(edges)
# print(edges)
if edges[0] + edges[1] == edges[-1]:
print('SEGMENT')
exit(0)
print('IMPOSSIBLE')
``` | output | 1 | 94,001 | 23 | 188,003 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Johnny has a younger sister Anne, who is very clever and smart. As she came home from the kindergarten, she told his brother about the task that her kindergartener asked her to solve. The task was just to construct a triangle out of four sticks of different colours. Naturally, one of the sticks is extra. It is not allowed to break the sticks or use their partial length. Anne has perfectly solved this task, now she is asking Johnny to do the same.
The boy answered that he would cope with it without any difficulty. However, after a while he found out that different tricky things can occur. It can happen that it is impossible to construct a triangle of a positive area, but it is possible to construct a degenerate triangle. It can be so, that it is impossible to construct a degenerate triangle even. As Johnny is very lazy, he does not want to consider such a big amount of cases, he asks you to help him.
Input
The first line of the input contains four space-separated positive integer numbers not exceeding 100 — lengthes of the sticks.
Output
Output TRIANGLE if it is possible to construct a non-degenerate triangle. Output SEGMENT if the first case cannot take place and it is possible to construct a degenerate triangle. Output IMPOSSIBLE if it is impossible to construct any triangle. Remember that you are to use three sticks. It is not allowed to break the sticks or use their partial length.
Examples
Input
4 2 1 3
Output
TRIANGLE
Input
7 2 2 4
Output
SEGMENT
Input
3 5 9 1
Output
IMPOSSIBLE
Submitted Solution:
```
a,b,c,d=sorted(list(map(int,input().split())))
if a+b>c or b+c>d:
print("TRIANGLE")
elif(a+b==c or b+c==d):
print("SEGMENT")
else:
print("IMPOSSIBLE")
``` | instruction | 0 | 94,002 | 23 | 188,004 |
Yes | output | 1 | 94,002 | 23 | 188,005 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Johnny has a younger sister Anne, who is very clever and smart. As she came home from the kindergarten, she told his brother about the task that her kindergartener asked her to solve. The task was just to construct a triangle out of four sticks of different colours. Naturally, one of the sticks is extra. It is not allowed to break the sticks or use their partial length. Anne has perfectly solved this task, now she is asking Johnny to do the same.
The boy answered that he would cope with it without any difficulty. However, after a while he found out that different tricky things can occur. It can happen that it is impossible to construct a triangle of a positive area, but it is possible to construct a degenerate triangle. It can be so, that it is impossible to construct a degenerate triangle even. As Johnny is very lazy, he does not want to consider such a big amount of cases, he asks you to help him.
Input
The first line of the input contains four space-separated positive integer numbers not exceeding 100 — lengthes of the sticks.
Output
Output TRIANGLE if it is possible to construct a non-degenerate triangle. Output SEGMENT if the first case cannot take place and it is possible to construct a degenerate triangle. Output IMPOSSIBLE if it is impossible to construct any triangle. Remember that you are to use three sticks. It is not allowed to break the sticks or use their partial length.
Examples
Input
4 2 1 3
Output
TRIANGLE
Input
7 2 2 4
Output
SEGMENT
Input
3 5 9 1
Output
IMPOSSIBLE
Submitted Solution:
```
def triangle(x, y, z):
if x + y > z and y + z > x and x + z > y:
return 'T'
if x + y == z or y + z == x or x + z == y:
return 'S'
else:
return 'I'
temp = list(map(int, input().split(' ')))
a, b, c, d = temp[0], temp[1], temp[2], temp[3]
if triangle(a, b, c) == 'T' or \
triangle(b, c, d) == 'T' or \
triangle(c, d, a) == 'T' or \
triangle(d, a, b) == 'T':
print('TRIANGLE')
elif triangle(a, b, c) == 'S' or \
triangle(b, c, d) == 'S' or \
triangle(c, d, a) == 'S' or \
triangle(d, a, b) == 'S':
print('SEGMENT')
else:
print('IMPOSSIBLE')
``` | instruction | 0 | 94,003 | 23 | 188,006 |
Yes | output | 1 | 94,003 | 23 | 188,007 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Johnny has a younger sister Anne, who is very clever and smart. As she came home from the kindergarten, she told his brother about the task that her kindergartener asked her to solve. The task was just to construct a triangle out of four sticks of different colours. Naturally, one of the sticks is extra. It is not allowed to break the sticks or use their partial length. Anne has perfectly solved this task, now she is asking Johnny to do the same.
The boy answered that he would cope with it without any difficulty. However, after a while he found out that different tricky things can occur. It can happen that it is impossible to construct a triangle of a positive area, but it is possible to construct a degenerate triangle. It can be so, that it is impossible to construct a degenerate triangle even. As Johnny is very lazy, he does not want to consider such a big amount of cases, he asks you to help him.
Input
The first line of the input contains four space-separated positive integer numbers not exceeding 100 — lengthes of the sticks.
Output
Output TRIANGLE if it is possible to construct a non-degenerate triangle. Output SEGMENT if the first case cannot take place and it is possible to construct a degenerate triangle. Output IMPOSSIBLE if it is impossible to construct any triangle. Remember that you are to use three sticks. It is not allowed to break the sticks or use their partial length.
Examples
Input
4 2 1 3
Output
TRIANGLE
Input
7 2 2 4
Output
SEGMENT
Input
3 5 9 1
Output
IMPOSSIBLE
Submitted Solution:
```
t = list(map(int,input().split()))
t.sort()
# print(t) # s0, s1, s2, s3
if t[3] < t[2] + t[1] or t[3] < t[0] + t[1] or t[2] < t[1] + t[0]:
print('TRIANGLE')
elif t[3] == t[2] + t[1] or t[3] == t[0] + t[1] or t[2] == t[1] + t[0]:
print('SEGMENT')
else:
print('IMPOSSIBLE')
``` | instruction | 0 | 94,004 | 23 | 188,008 |
Yes | output | 1 | 94,004 | 23 | 188,009 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Johnny has a younger sister Anne, who is very clever and smart. As she came home from the kindergarten, she told his brother about the task that her kindergartener asked her to solve. The task was just to construct a triangle out of four sticks of different colours. Naturally, one of the sticks is extra. It is not allowed to break the sticks or use their partial length. Anne has perfectly solved this task, now she is asking Johnny to do the same.
The boy answered that he would cope with it without any difficulty. However, after a while he found out that different tricky things can occur. It can happen that it is impossible to construct a triangle of a positive area, but it is possible to construct a degenerate triangle. It can be so, that it is impossible to construct a degenerate triangle even. As Johnny is very lazy, he does not want to consider such a big amount of cases, he asks you to help him.
Input
The first line of the input contains four space-separated positive integer numbers not exceeding 100 — lengthes of the sticks.
Output
Output TRIANGLE if it is possible to construct a non-degenerate triangle. Output SEGMENT if the first case cannot take place and it is possible to construct a degenerate triangle. Output IMPOSSIBLE if it is impossible to construct any triangle. Remember that you are to use three sticks. It is not allowed to break the sticks or use their partial length.
Examples
Input
4 2 1 3
Output
TRIANGLE
Input
7 2 2 4
Output
SEGMENT
Input
3 5 9 1
Output
IMPOSSIBLE
Submitted Solution:
```
a = sorted(int(x) for x in input().split())
if a[0] + a[1] > a[2] or a[1] + a[2] > a[3]:
print("TRIANGLE")
elif a[0] + a[1] == a[2] or a[1] + a[2] == a[3]:
print("SEGMENT")
else:
print("IMPOSSIBLE")
``` | instruction | 0 | 94,005 | 23 | 188,010 |
Yes | output | 1 | 94,005 | 23 | 188,011 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Johnny has a younger sister Anne, who is very clever and smart. As she came home from the kindergarten, she told his brother about the task that her kindergartener asked her to solve. The task was just to construct a triangle out of four sticks of different colours. Naturally, one of the sticks is extra. It is not allowed to break the sticks or use their partial length. Anne has perfectly solved this task, now she is asking Johnny to do the same.
The boy answered that he would cope with it without any difficulty. However, after a while he found out that different tricky things can occur. It can happen that it is impossible to construct a triangle of a positive area, but it is possible to construct a degenerate triangle. It can be so, that it is impossible to construct a degenerate triangle even. As Johnny is very lazy, he does not want to consider such a big amount of cases, he asks you to help him.
Input
The first line of the input contains four space-separated positive integer numbers not exceeding 100 — lengthes of the sticks.
Output
Output TRIANGLE if it is possible to construct a non-degenerate triangle. Output SEGMENT if the first case cannot take place and it is possible to construct a degenerate triangle. Output IMPOSSIBLE if it is impossible to construct any triangle. Remember that you are to use three sticks. It is not allowed to break the sticks or use their partial length.
Examples
Input
4 2 1 3
Output
TRIANGLE
Input
7 2 2 4
Output
SEGMENT
Input
3 5 9 1
Output
IMPOSSIBLE
Submitted Solution:
```
s_original = [int(i) for i in input().split()]
for i in range(4):
s = s_original.copy()
s.pop(i)
if(s[0] < s[1] + s[2] and s[1] < s[0] + s[2] and s[2] < s[1] + s[0]):
print('TRIANGLE')
exit(0)
for i in range(4):
s = s_original.copy()
s.pop(i)
if(s[0] == s[1] + s[2] or s[1] == s[0] + s[2] or s[2] == s[1] + s[0]):
print('SEGMENT')
exit(0)
print('IMPLOSSIBLE')
# 1512831717474
``` | instruction | 0 | 94,006 | 23 | 188,012 |
No | output | 1 | 94,006 | 23 | 188,013 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Johnny has a younger sister Anne, who is very clever and smart. As she came home from the kindergarten, she told his brother about the task that her kindergartener asked her to solve. The task was just to construct a triangle out of four sticks of different colours. Naturally, one of the sticks is extra. It is not allowed to break the sticks or use their partial length. Anne has perfectly solved this task, now she is asking Johnny to do the same.
The boy answered that he would cope with it without any difficulty. However, after a while he found out that different tricky things can occur. It can happen that it is impossible to construct a triangle of a positive area, but it is possible to construct a degenerate triangle. It can be so, that it is impossible to construct a degenerate triangle even. As Johnny is very lazy, he does not want to consider such a big amount of cases, he asks you to help him.
Input
The first line of the input contains four space-separated positive integer numbers not exceeding 100 — lengthes of the sticks.
Output
Output TRIANGLE if it is possible to construct a non-degenerate triangle. Output SEGMENT if the first case cannot take place and it is possible to construct a degenerate triangle. Output IMPOSSIBLE if it is impossible to construct any triangle. Remember that you are to use three sticks. It is not allowed to break the sticks or use their partial length.
Examples
Input
4 2 1 3
Output
TRIANGLE
Input
7 2 2 4
Output
SEGMENT
Input
3 5 9 1
Output
IMPOSSIBLE
Submitted Solution:
```
n=list(map(int,input().split()))
o,e=0,0
for i in range(4):
if n[i]%2==0:
e+=1
else:
o+=1
if o==1:
print('SEGMENT')
elif o==2:
print('TRIANGLE')
else:
print('IMPOSSIBLE')
``` | instruction | 0 | 94,007 | 23 | 188,014 |
No | output | 1 | 94,007 | 23 | 188,015 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Johnny has a younger sister Anne, who is very clever and smart. As she came home from the kindergarten, she told his brother about the task that her kindergartener asked her to solve. The task was just to construct a triangle out of four sticks of different colours. Naturally, one of the sticks is extra. It is not allowed to break the sticks or use their partial length. Anne has perfectly solved this task, now she is asking Johnny to do the same.
The boy answered that he would cope with it without any difficulty. However, after a while he found out that different tricky things can occur. It can happen that it is impossible to construct a triangle of a positive area, but it is possible to construct a degenerate triangle. It can be so, that it is impossible to construct a degenerate triangle even. As Johnny is very lazy, he does not want to consider such a big amount of cases, he asks you to help him.
Input
The first line of the input contains four space-separated positive integer numbers not exceeding 100 — lengthes of the sticks.
Output
Output TRIANGLE if it is possible to construct a non-degenerate triangle. Output SEGMENT if the first case cannot take place and it is possible to construct a degenerate triangle. Output IMPOSSIBLE if it is impossible to construct any triangle. Remember that you are to use three sticks. It is not allowed to break the sticks or use their partial length.
Examples
Input
4 2 1 3
Output
TRIANGLE
Input
7 2 2 4
Output
SEGMENT
Input
3 5 9 1
Output
IMPOSSIBLE
Submitted Solution:
```
#!/usr/bin/env python3
def s3(a, b, c):
diff = 2 * max(a, b, c) - (a + b + c)
if (diff < 0):
return 'TRIANGLE'
elif (diff == 0):
return 'SEGMENT'
else:
return 'IMPOSSIBLE'
def s4(a, b, c, d):
s = []
s.append(s3(a, b, c))
s.append(s3(a, b, d))
s.append(s3(a, c, d))
s.append(s3(b, c, d))
print(s)
if 'TRIANGLE' in s:
return 'TRIANGLE'
elif 'SEGMENT' in s:
return 'SEGMENT'
else:
return 'IMPOSSIBLE'
def main():
a, b, c, d = [ int(i) for i in input().split(' ') ]
print(s4(a, b, c, d))
if __name__ == '__main__':
main()
``` | instruction | 0 | 94,008 | 23 | 188,016 |
No | output | 1 | 94,008 | 23 | 188,017 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Johnny has a younger sister Anne, who is very clever and smart. As she came home from the kindergarten, she told his brother about the task that her kindergartener asked her to solve. The task was just to construct a triangle out of four sticks of different colours. Naturally, one of the sticks is extra. It is not allowed to break the sticks or use their partial length. Anne has perfectly solved this task, now she is asking Johnny to do the same.
The boy answered that he would cope with it without any difficulty. However, after a while he found out that different tricky things can occur. It can happen that it is impossible to construct a triangle of a positive area, but it is possible to construct a degenerate triangle. It can be so, that it is impossible to construct a degenerate triangle even. As Johnny is very lazy, he does not want to consider such a big amount of cases, he asks you to help him.
Input
The first line of the input contains four space-separated positive integer numbers not exceeding 100 — lengthes of the sticks.
Output
Output TRIANGLE if it is possible to construct a non-degenerate triangle. Output SEGMENT if the first case cannot take place and it is possible to construct a degenerate triangle. Output IMPOSSIBLE if it is impossible to construct any triangle. Remember that you are to use three sticks. It is not allowed to break the sticks or use their partial length.
Examples
Input
4 2 1 3
Output
TRIANGLE
Input
7 2 2 4
Output
SEGMENT
Input
3 5 9 1
Output
IMPOSSIBLE
Submitted Solution:
```
a, b, c, d = map(int, input().split())
x = [a, b, c, d]
s=0
for i in range(4):
for j in range(4):
for k in range(4):
if (i != j != k):
a = min((x[i], x[j], x[k]))
c = max((x[i], x[j], x[k]))
b = sum((x[i], x[j], x[k]))-a-c
if (a+b<c):
print('TRIANGLE')
exit()
if (a+b==c):
s=1
if s:
print('SEGMENT')
else:
print('IMPOSSIBLE')
``` | instruction | 0 | 94,009 | 23 | 188,018 |
No | output | 1 | 94,009 | 23 | 188,019 |
Provide a correct Python 3 solution for this coding contest problem.
There is a cube which consists of n × n × n small cubes. Small cubes have marks on their surfaces. An example where n = 4 is shown in the following figure.
<image>
Then, as shown in the figure above (right), make a hole that penetrates horizontally or vertically from the marked surface to the opposite surface.
Your job is to create a program that reads the positions marked n and counts the number of small cubes with no holes.
Input
The input consists of several datasets. Each dataset is given in the following format:
n h
c1 a1 b1
c2 a2 b2
..
..
..
ch ah bh
h is an integer indicating the number of marks. The h lines that follow enter the positions of the h marks. The coordinate axes shown in the figure below will be used to specify the position of the mark. (x, y, z) = (1, 1, 1) is the lower left cube, and (x, y, z) = (n, n, n) is the upper right cube.
<image>
ci is a string indicating the plane marked with the i-th. ci is one of "xy", "xz", and "yz", indicating that the i-th mark is on the xy, xz, and yz planes, respectively.
ai and bi indicate the coordinates on the plane indicated by ci. For the xy, xz, and yz planes, ai and bi indicate the plane coordinates (x, y), (x, z), and (y, z), respectively. For example, in the above figure, the values of ci, ai, and bi of marks A, B, and C are "xy 4 4", "xz 1 2", and "yz 2 3", respectively.
When both n and h are 0, it indicates the end of input.
You can assume that n ≤ 500 and h ≤ 200.
Output
For each dataset, print the number of non-perforated cubes on one line.
Example
Input
4 3
xy 4 4
xz 1 2
yz 2 3
4 5
xy 1 1
xy 3 3
xz 3 3
yz 2 1
yz 3 3
0 0
Output
52
46 | instruction | 0 | 94,342 | 23 | 188,684 |
"Correct Solution:
```
# AOJ 1030 Cubes Without Holes
# Python3 2018.7.6 bal4u
import sys
from sys import stdin
input = stdin.readline
# n <= 500, 2^9 = 512
while True:
n, h = map(int, input().split())
if n == 0: break
ans = []
for i in range(h):
c, a, b = input().split()
a, b = int(a)-1, int(b)-1
if c == "xy":
ans += [a+(b<<9)+(z<<18) for z in range(n)]
elif c == "xz":
ans += [a+(y<<9)+(b<<18) for y in range(n)]
else:
ans += [x+(a<<9)+(b<<18) for x in range(n)]
print(n**3-len(set(ans)))
``` | output | 1 | 94,342 | 23 | 188,685 |
Provide a correct Python 3 solution for this coding contest problem.
There is a cube which consists of n × n × n small cubes. Small cubes have marks on their surfaces. An example where n = 4 is shown in the following figure.
<image>
Then, as shown in the figure above (right), make a hole that penetrates horizontally or vertically from the marked surface to the opposite surface.
Your job is to create a program that reads the positions marked n and counts the number of small cubes with no holes.
Input
The input consists of several datasets. Each dataset is given in the following format:
n h
c1 a1 b1
c2 a2 b2
..
..
..
ch ah bh
h is an integer indicating the number of marks. The h lines that follow enter the positions of the h marks. The coordinate axes shown in the figure below will be used to specify the position of the mark. (x, y, z) = (1, 1, 1) is the lower left cube, and (x, y, z) = (n, n, n) is the upper right cube.
<image>
ci is a string indicating the plane marked with the i-th. ci is one of "xy", "xz", and "yz", indicating that the i-th mark is on the xy, xz, and yz planes, respectively.
ai and bi indicate the coordinates on the plane indicated by ci. For the xy, xz, and yz planes, ai and bi indicate the plane coordinates (x, y), (x, z), and (y, z), respectively. For example, in the above figure, the values of ci, ai, and bi of marks A, B, and C are "xy 4 4", "xz 1 2", and "yz 2 3", respectively.
When both n and h are 0, it indicates the end of input.
You can assume that n ≤ 500 and h ≤ 200.
Output
For each dataset, print the number of non-perforated cubes on one line.
Example
Input
4 3
xy 4 4
xz 1 2
yz 2 3
4 5
xy 1 1
xy 3 3
xz 3 3
yz 2 1
yz 3 3
0 0
Output
52
46 | instruction | 0 | 94,343 | 23 | 188,686 |
"Correct Solution:
```
while True:
n, h = map(int, input().split())
if n == 0:
break
hit = set()
for _ in range(h):
c, a, b = input().split()
if c == "xy":
add = {(int(a), int(b), z) for z in range(1, n + 1)}
elif c == "xz":
add = {(int(a), y, int(b)) for y in range(1, n + 1)}
elif c == "yz":
add = {(x, int(a), int(b)) for x in range(1, n + 1)}
hit = hit | add
print(n ** 3 - len(hit))
``` | output | 1 | 94,343 | 23 | 188,687 |
Provide a correct Python 3 solution for this coding contest problem.
There is a cube which consists of n × n × n small cubes. Small cubes have marks on their surfaces. An example where n = 4 is shown in the following figure.
<image>
Then, as shown in the figure above (right), make a hole that penetrates horizontally or vertically from the marked surface to the opposite surface.
Your job is to create a program that reads the positions marked n and counts the number of small cubes with no holes.
Input
The input consists of several datasets. Each dataset is given in the following format:
n h
c1 a1 b1
c2 a2 b2
..
..
..
ch ah bh
h is an integer indicating the number of marks. The h lines that follow enter the positions of the h marks. The coordinate axes shown in the figure below will be used to specify the position of the mark. (x, y, z) = (1, 1, 1) is the lower left cube, and (x, y, z) = (n, n, n) is the upper right cube.
<image>
ci is a string indicating the plane marked with the i-th. ci is one of "xy", "xz", and "yz", indicating that the i-th mark is on the xy, xz, and yz planes, respectively.
ai and bi indicate the coordinates on the plane indicated by ci. For the xy, xz, and yz planes, ai and bi indicate the plane coordinates (x, y), (x, z), and (y, z), respectively. For example, in the above figure, the values of ci, ai, and bi of marks A, B, and C are "xy 4 4", "xz 1 2", and "yz 2 3", respectively.
When both n and h are 0, it indicates the end of input.
You can assume that n ≤ 500 and h ≤ 200.
Output
For each dataset, print the number of non-perforated cubes on one line.
Example
Input
4 3
xy 4 4
xz 1 2
yz 2 3
4 5
xy 1 1
xy 3 3
xz 3 3
yz 2 1
yz 3 3
0 0
Output
52
46 | instruction | 0 | 94,344 | 23 | 188,688 |
"Correct Solution:
```
# AOJ 1030 Cubes Without Holes
# Python3 2018.7.6 bal4u
import sys
from sys import stdin
input = stdin.readline
# n <= 500, 2^9 = 512
while True:
n, h = map(int, input().split())
if n == 0: break
ans = []
for i in range(h):
s = input()
c = s[:2]
a, b = s[3:].split()
a, b = int(a)-1, int(b)-1
if c == "xy":
k = a+(b<<9)
ans += [k+(z<<18) for z in range(n)]
elif c == "xz":
k = a+(b<<18)
ans += [k+(y<<9) for y in range(n)]
else:
k = (a<<9)+(b<<18)
ans += [k+x for x in range(n)]
print(n**3-len(set(ans)))
``` | output | 1 | 94,344 | 23 | 188,689 |
Provide a correct Python 3 solution for this coding contest problem.
There is a cube which consists of n × n × n small cubes. Small cubes have marks on their surfaces. An example where n = 4 is shown in the following figure.
<image>
Then, as shown in the figure above (right), make a hole that penetrates horizontally or vertically from the marked surface to the opposite surface.
Your job is to create a program that reads the positions marked n and counts the number of small cubes with no holes.
Input
The input consists of several datasets. Each dataset is given in the following format:
n h
c1 a1 b1
c2 a2 b2
..
..
..
ch ah bh
h is an integer indicating the number of marks. The h lines that follow enter the positions of the h marks. The coordinate axes shown in the figure below will be used to specify the position of the mark. (x, y, z) = (1, 1, 1) is the lower left cube, and (x, y, z) = (n, n, n) is the upper right cube.
<image>
ci is a string indicating the plane marked with the i-th. ci is one of "xy", "xz", and "yz", indicating that the i-th mark is on the xy, xz, and yz planes, respectively.
ai and bi indicate the coordinates on the plane indicated by ci. For the xy, xz, and yz planes, ai and bi indicate the plane coordinates (x, y), (x, z), and (y, z), respectively. For example, in the above figure, the values of ci, ai, and bi of marks A, B, and C are "xy 4 4", "xz 1 2", and "yz 2 3", respectively.
When both n and h are 0, it indicates the end of input.
You can assume that n ≤ 500 and h ≤ 200.
Output
For each dataset, print the number of non-perforated cubes on one line.
Example
Input
4 3
xy 4 4
xz 1 2
yz 2 3
4 5
xy 1 1
xy 3 3
xz 3 3
yz 2 1
yz 3 3
0 0
Output
52
46 | instruction | 0 | 94,345 | 23 | 188,690 |
"Correct Solution:
```
def solve():
while True:
n, h = map(int, input().split())
if n == 0:
break
s = set()
for _ in [0]*h:
t, *a = input().split()
i, j = map(lambda x: int(x)-1, a)
k = 1
if t == "xy":
j *= n
k = n**2
elif t == "xz":
j *= n**2
k = n
else:
i *= n
j *= (n**2)
s.update(range(i+j, i+j+k*n, k))
print(n**3 - len(s))
if __name__ == "__main__":
solve()
``` | output | 1 | 94,345 | 23 | 188,691 |
Provide a correct Python 3 solution for this coding contest problem.
There is a cube which consists of n × n × n small cubes. Small cubes have marks on their surfaces. An example where n = 4 is shown in the following figure.
<image>
Then, as shown in the figure above (right), make a hole that penetrates horizontally or vertically from the marked surface to the opposite surface.
Your job is to create a program that reads the positions marked n and counts the number of small cubes with no holes.
Input
The input consists of several datasets. Each dataset is given in the following format:
n h
c1 a1 b1
c2 a2 b2
..
..
..
ch ah bh
h is an integer indicating the number of marks. The h lines that follow enter the positions of the h marks. The coordinate axes shown in the figure below will be used to specify the position of the mark. (x, y, z) = (1, 1, 1) is the lower left cube, and (x, y, z) = (n, n, n) is the upper right cube.
<image>
ci is a string indicating the plane marked with the i-th. ci is one of "xy", "xz", and "yz", indicating that the i-th mark is on the xy, xz, and yz planes, respectively.
ai and bi indicate the coordinates on the plane indicated by ci. For the xy, xz, and yz planes, ai and bi indicate the plane coordinates (x, y), (x, z), and (y, z), respectively. For example, in the above figure, the values of ci, ai, and bi of marks A, B, and C are "xy 4 4", "xz 1 2", and "yz 2 3", respectively.
When both n and h are 0, it indicates the end of input.
You can assume that n ≤ 500 and h ≤ 200.
Output
For each dataset, print the number of non-perforated cubes on one line.
Example
Input
4 3
xy 4 4
xz 1 2
yz 2 3
4 5
xy 1 1
xy 3 3
xz 3 3
yz 2 1
yz 3 3
0 0
Output
52
46 | instruction | 0 | 94,346 | 23 | 188,692 |
"Correct Solution:
```
while 1:
N, H = map(int, input().split())
if N == H == 0:
break
s = set()
for i in range(H):
c, a, b = input().split(); a = int(a)-1; b = int(b)-1
if c == "xy":
s.update((a, b, i) for i in range(N))
if c == "xz":
s.update((a, i, b) for i in range(N))
if c == "yz":
s.update((i, a, b) for i in range(N))
print(N**3 - len(s))
``` | output | 1 | 94,346 | 23 | 188,693 |
Provide a correct Python 3 solution for this coding contest problem.
There is a cube which consists of n × n × n small cubes. Small cubes have marks on their surfaces. An example where n = 4 is shown in the following figure.
<image>
Then, as shown in the figure above (right), make a hole that penetrates horizontally or vertically from the marked surface to the opposite surface.
Your job is to create a program that reads the positions marked n and counts the number of small cubes with no holes.
Input
The input consists of several datasets. Each dataset is given in the following format:
n h
c1 a1 b1
c2 a2 b2
..
..
..
ch ah bh
h is an integer indicating the number of marks. The h lines that follow enter the positions of the h marks. The coordinate axes shown in the figure below will be used to specify the position of the mark. (x, y, z) = (1, 1, 1) is the lower left cube, and (x, y, z) = (n, n, n) is the upper right cube.
<image>
ci is a string indicating the plane marked with the i-th. ci is one of "xy", "xz", and "yz", indicating that the i-th mark is on the xy, xz, and yz planes, respectively.
ai and bi indicate the coordinates on the plane indicated by ci. For the xy, xz, and yz planes, ai and bi indicate the plane coordinates (x, y), (x, z), and (y, z), respectively. For example, in the above figure, the values of ci, ai, and bi of marks A, B, and C are "xy 4 4", "xz 1 2", and "yz 2 3", respectively.
When both n and h are 0, it indicates the end of input.
You can assume that n ≤ 500 and h ≤ 200.
Output
For each dataset, print the number of non-perforated cubes on one line.
Example
Input
4 3
xy 4 4
xz 1 2
yz 2 3
4 5
xy 1 1
xy 3 3
xz 3 3
yz 2 1
yz 3 3
0 0
Output
52
46 | instruction | 0 | 94,347 | 23 | 188,694 |
"Correct Solution:
```
# AOJ 1030 Cubes Without Holes
# Python3 2018.7.6 bal4u
import sys
from sys import stdin
input = stdin.readline
while True:
n, h = map(int, input().split())
if n == 0: break
ans = []
for i in range(h):
c, a, b = input().split()
a, b = int(a)-1, int(b)-1
if c == "xy":
ans += [a+b*n+z*n**2 for z in range(n)]
elif c == "xz":
ans += [a+y*n+b*n**2 for y in range(n)]
else:
ans += [x+a*n+b*n**2 for x in range(n)]
print(n**3-len(set(ans)))
``` | output | 1 | 94,347 | 23 | 188,695 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There is a cube which consists of n × n × n small cubes. Small cubes have marks on their surfaces. An example where n = 4 is shown in the following figure.
<image>
Then, as shown in the figure above (right), make a hole that penetrates horizontally or vertically from the marked surface to the opposite surface.
Your job is to create a program that reads the positions marked n and counts the number of small cubes with no holes.
Input
The input consists of several datasets. Each dataset is given in the following format:
n h
c1 a1 b1
c2 a2 b2
..
..
..
ch ah bh
h is an integer indicating the number of marks. The h lines that follow enter the positions of the h marks. The coordinate axes shown in the figure below will be used to specify the position of the mark. (x, y, z) = (1, 1, 1) is the lower left cube, and (x, y, z) = (n, n, n) is the upper right cube.
<image>
ci is a string indicating the plane marked with the i-th. ci is one of "xy", "xz", and "yz", indicating that the i-th mark is on the xy, xz, and yz planes, respectively.
ai and bi indicate the coordinates on the plane indicated by ci. For the xy, xz, and yz planes, ai and bi indicate the plane coordinates (x, y), (x, z), and (y, z), respectively. For example, in the above figure, the values of ci, ai, and bi of marks A, B, and C are "xy 4 4", "xz 1 2", and "yz 2 3", respectively.
When both n and h are 0, it indicates the end of input.
You can assume that n ≤ 500 and h ≤ 200.
Output
For each dataset, print the number of non-perforated cubes on one line.
Example
Input
4 3
xy 4 4
xz 1 2
yz 2 3
4 5
xy 1 1
xy 3 3
xz 3 3
yz 2 1
yz 3 3
0 0
Output
52
46
Submitted Solution:
```
# AOJ 1030 Cubes Without Holes
# Python3 2018.7.6 bal4u
import sys
from sys import stdin
input = stdin.readline
while True:
n, h = map(int, input().split())
if n == 0: break
f = [[[0 for k in range(3)] for c in range(500)] for r in range(500)]
ans = n**3
for i in range(h):
c, a, b = input().split()
a, b = int(a)-1, int(b)-1
ans -= n
if c == "xy":
f[a][b][0] = 1
ans += sum([(f[a][j][1] | f[a][j][2]) for j in range(n)])
elif c == "xz":
f[a][b][1] = 1
ans += sum([(f[a][j][0] | f[i][b][2]) for j in range(n)])
else:
f[a][b][2] = 1
ans += sum([(f[j][a][0] | f[j][b][1]) for j in range(n)])
print(ans)
``` | instruction | 0 | 94,348 | 23 | 188,696 |
No | output | 1 | 94,348 | 23 | 188,697 |
Provide a correct Python 3 solution for this coding contest problem.
Suppose that P1 is an infinite-height prism whose axis is parallel to the z-axis, and P2 is also an infinite-height prism whose axis is parallel to the y-axis. P1 is defined by the polygon C1 which is the cross section of P1 and the xy-plane, and P2 is also defined by the polygon C2 which is the cross section of P2 and the xz-plane.
Figure I.1 shows two cross sections which appear as the first dataset in the sample input, and Figure I.2 shows the relationship between the prisms and their cross sections.
<image>
Figure I.1: Cross sections of Prisms
<image>
Figure I.2: Prisms and their cross sections
<image>
Figure I.3: Intersection of two prisms
Figure I.3 shows the intersection of two prisms in Figure I.2, namely, P1 and P2.
Write a program which calculates the volume of the intersection of two prisms.
Input
The input is a sequence of datasets. The number of datasets is less than 200.
Each dataset is formatted as follows.
m n
x11 y11
x12 y12
.
.
.
x1m y1m
x21 z21
x22 z22
.
.
.
x2n z2n
m and n are integers (3 ≤ m ≤ 100, 3 ≤ n ≤ 100) which represent the numbers of the vertices of the polygons, C1 and C2, respectively.
x1i, y1i, x2j and z2j are integers between -100 and 100, inclusive. (x1i, y1i) and (x2j , z2j) mean the i-th and j-th vertices' positions of C1 and C2 respectively.
The sequences of these vertex positions are given in the counterclockwise order either on the xy-plane or the xz-plane as in Figure I.1.
You may assume that all the polygons are convex, that is, all the interior angles of the polygons are less than 180 degrees. You may also assume that all the polygons are simple, that is, each polygon's boundary does not cross nor touch itself.
The end of the input is indicated by a line containing two zeros.
Output
For each dataset, output the volume of the intersection of the two prisms, P1 and P2, with a decimal representation in a line.
None of the output values may have an error greater than 0.001. The output should not contain any other extra characters.
Example
Input
4 3
7 2
3 3
0 2
3 1
4 2
0 1
8 1
4 4
30 2
30 12
2 12
2 2
15 2
30 8
13 14
2 8
8 5
13 5
21 7
21 9
18 15
11 15
6 10
6 8
8 5
10 12
5 9
15 6
20 10
18 12
3 3
5 5
10 3
10 10
20 8
10 15
10 8
4 4
-98 99
-99 -99
99 -98
99 97
-99 99
-98 -98
99 -99
96 99
0 0
Output
4.708333333333333
1680.0000000000005
491.1500000000007
0.0
7600258.4847715655 | instruction | 0 | 94,365 | 23 | 188,730 |
"Correct Solution:
```
# aoj 1313
import sys
input = sys.stdin.readline
# the width of the polygon sliced at x
def width(X, Y, x):
n = len(X)
lb, ub = float('inf'), -float('inf')
for i in range(n):
x1, y1, x2, y2 = X[i], Y[i], X[(i+1) % n], Y[(i+1) % n]
if (x1-x)*(x2-x) <= 0 and x1 != x2:
y = y1 + (y2-y1) * (x-x1) / (x2-x1)
lb = min(lb, y)
ub = max(ub, y)
return max(0, ub-lb)
while True:
m, n = map(int, input().split())
if m == n == 0:
break
X1 = [0] * m
Y1 = [0] * m
X2 = [0] * n
Z2 = [0] * n
for i in range(m):
X1[i], Y1[i] = map(int, input().split())
for i in range(n):
X2[i], Z2[i] = map(int, input().split())
min1, max1 = min(X1), max(X1)
min2, max2 = min(X2), max(X2)
X = X1 + X2
X.sort()
ans = 0
for i in range(len(X)-1):
a, b = X[i], X[i+1]
c = (a+b) / 2
if min1 <= c <= max1 and min2 <= c <= max2:
fa = width(X1, Y1, a) * width(X2, Z2, a)
fb = width(X1, Y1, b) * width(X2, Z2, b)
fc = width(X1, Y1, c) * width(X2, Z2, c)
ans += (b-a) / 6 * (fa+4*fc+fb)
print(ans)
``` | output | 1 | 94,365 | 23 | 188,731 |
Provide a correct Python 3 solution for this coding contest problem.
Suppose that P1 is an infinite-height prism whose axis is parallel to the z-axis, and P2 is also an infinite-height prism whose axis is parallel to the y-axis. P1 is defined by the polygon C1 which is the cross section of P1 and the xy-plane, and P2 is also defined by the polygon C2 which is the cross section of P2 and the xz-plane.
Figure I.1 shows two cross sections which appear as the first dataset in the sample input, and Figure I.2 shows the relationship between the prisms and their cross sections.
<image>
Figure I.1: Cross sections of Prisms
<image>
Figure I.2: Prisms and their cross sections
<image>
Figure I.3: Intersection of two prisms
Figure I.3 shows the intersection of two prisms in Figure I.2, namely, P1 and P2.
Write a program which calculates the volume of the intersection of two prisms.
Input
The input is a sequence of datasets. The number of datasets is less than 200.
Each dataset is formatted as follows.
m n
x11 y11
x12 y12
.
.
.
x1m y1m
x21 z21
x22 z22
.
.
.
x2n z2n
m and n are integers (3 ≤ m ≤ 100, 3 ≤ n ≤ 100) which represent the numbers of the vertices of the polygons, C1 and C2, respectively.
x1i, y1i, x2j and z2j are integers between -100 and 100, inclusive. (x1i, y1i) and (x2j , z2j) mean the i-th and j-th vertices' positions of C1 and C2 respectively.
The sequences of these vertex positions are given in the counterclockwise order either on the xy-plane or the xz-plane as in Figure I.1.
You may assume that all the polygons are convex, that is, all the interior angles of the polygons are less than 180 degrees. You may also assume that all the polygons are simple, that is, each polygon's boundary does not cross nor touch itself.
The end of the input is indicated by a line containing two zeros.
Output
For each dataset, output the volume of the intersection of the two prisms, P1 and P2, with a decimal representation in a line.
None of the output values may have an error greater than 0.001. The output should not contain any other extra characters.
Example
Input
4 3
7 2
3 3
0 2
3 1
4 2
0 1
8 1
4 4
30 2
30 12
2 12
2 2
15 2
30 8
13 14
2 8
8 5
13 5
21 7
21 9
18 15
11 15
6 10
6 8
8 5
10 12
5 9
15 6
20 10
18 12
3 3
5 5
10 3
10 10
20 8
10 15
10 8
4 4
-98 99
-99 -99
99 -98
99 97
-99 99
-98 -98
99 -99
96 99
0 0
Output
4.708333333333333
1680.0000000000005
491.1500000000007
0.0
7600258.4847715655 | instruction | 0 | 94,366 | 23 | 188,732 |
"Correct Solution:
```
import sys
readline = sys.stdin.readline
write = sys.stdout.write
def solve():
M, N = map(int, readline().split())
if M == N == 0:
return False
P = [list(map(int, readline().split())) for i in range(M)]
Q = [list(map(int, readline().split())) for i in range(N)]
xs = set()
xs.update(x for x, y in P)
xs.update(x for x, z in Q)
*X, = xs
X.sort()
def calc(x):
y_ma = -100; y_mi = 100
for i in range(M):
x0, y0 = P[i-1]; x1, y1 = P[i]
if x0 <= x <= x1 or x1 <= x <= x0:
if x0 == x1:
y_ma = max(y_ma, max(y1, y0))
y_mi = min(y_mi, min(y1, y0))
else:
y = (x - x0)*(y1-y0) / (x1-x0) + y0
y_ma = max(y_ma, y)
y_mi = min(y_mi, y)
if not y_mi <= y_ma:
return 0
z_ma = -100; z_mi = 100
for i in range(N):
x0, z0 = Q[i-1]; x1, z1 = Q[i]
if x0 <= x <= x1 or x1 <= x <= x0:
if x0 == x1:
z_ma = max(z_ma, max(z1, z0))
z_mi = min(z_mi, min(z1, z0))
else:
z = (x - x0)*(z1-z0) / (x1-x0) + z0
z_ma = max(z_ma, z)
z_mi = min(z_mi, z)
if not z_mi <= z_ma:
return 0
return (z_ma - z_mi) * (y_ma - y_mi)
ans = 0
L = len(X)
for i in range(L-1):
x0 = X[i]; x1 = X[i+1]
s0 = calc(x0); s1 = calc((x0 + x1)/2); s2 = calc(x1)
if s1 > 0:
ans += (x1 - x0) * (s0 + s2 + 4*s1) / 6
write("%.16f\n" % ans)
return True
while solve():
...
``` | output | 1 | 94,366 | 23 | 188,733 |
Provide a correct Python 3 solution for this coding contest problem.
Suppose that P1 is an infinite-height prism whose axis is parallel to the z-axis, and P2 is also an infinite-height prism whose axis is parallel to the y-axis. P1 is defined by the polygon C1 which is the cross section of P1 and the xy-plane, and P2 is also defined by the polygon C2 which is the cross section of P2 and the xz-plane.
Figure I.1 shows two cross sections which appear as the first dataset in the sample input, and Figure I.2 shows the relationship between the prisms and their cross sections.
<image>
Figure I.1: Cross sections of Prisms
<image>
Figure I.2: Prisms and their cross sections
<image>
Figure I.3: Intersection of two prisms
Figure I.3 shows the intersection of two prisms in Figure I.2, namely, P1 and P2.
Write a program which calculates the volume of the intersection of two prisms.
Input
The input is a sequence of datasets. The number of datasets is less than 200.
Each dataset is formatted as follows.
m n
x11 y11
x12 y12
.
.
.
x1m y1m
x21 z21
x22 z22
.
.
.
x2n z2n
m and n are integers (3 ≤ m ≤ 100, 3 ≤ n ≤ 100) which represent the numbers of the vertices of the polygons, C1 and C2, respectively.
x1i, y1i, x2j and z2j are integers between -100 and 100, inclusive. (x1i, y1i) and (x2j , z2j) mean the i-th and j-th vertices' positions of C1 and C2 respectively.
The sequences of these vertex positions are given in the counterclockwise order either on the xy-plane or the xz-plane as in Figure I.1.
You may assume that all the polygons are convex, that is, all the interior angles of the polygons are less than 180 degrees. You may also assume that all the polygons are simple, that is, each polygon's boundary does not cross nor touch itself.
The end of the input is indicated by a line containing two zeros.
Output
For each dataset, output the volume of the intersection of the two prisms, P1 and P2, with a decimal representation in a line.
None of the output values may have an error greater than 0.001. The output should not contain any other extra characters.
Example
Input
4 3
7 2
3 3
0 2
3 1
4 2
0 1
8 1
4 4
30 2
30 12
2 12
2 2
15 2
30 8
13 14
2 8
8 5
13 5
21 7
21 9
18 15
11 15
6 10
6 8
8 5
10 12
5 9
15 6
20 10
18 12
3 3
5 5
10 3
10 10
20 8
10 15
10 8
4 4
-98 99
-99 -99
99 -98
99 97
-99 99
-98 -98
99 -99
96 99
0 0
Output
4.708333333333333
1680.0000000000005
491.1500000000007
0.0
7600258.4847715655 | instruction | 0 | 94,367 | 23 | 188,734 |
"Correct Solution:
```
def width(_X, _Y, _n, _x):
lb = float("inf")
ub = -float("inf")
for j in range(0, _n):
x1 = _X[j]
y1 = _Y[j]
x2 = _X[(j + 1) % _n]
y2 = _Y[(j + 1) % _n]
if (x1 - _x) * (x2 - _x) <= 0 and x1 != x2:
y = y1 + (y2 - y1) * (_x - x1) / (x2 - x1)
lb = min(lb, y)
ub = max(ub, y)
return max(0, ub - lb)
if __name__ == '__main__':
while True:
M, N = list(map(int, input().split()))
if M == 0 and N == 0:
break
X1 = []
Y1 = []
X2 = []
Z2 = []
for i in range(M):
tmp_x1, tmp_y1 = list(map(int, input().split()))
X1.append(tmp_x1)
Y1.append(tmp_y1)
for i in range(N):
tmp_x2, tmp_z2 = list(map(int, input().split()))
X2.append(tmp_x2)
Z2.append(tmp_z2)
min1 = min(X1)
max1 = max(X1)
min2 = min(X2)
max2 = max(X2)
xs = X1 + X2
xs = sorted(xs)
res = 0
for i in range(0, len(xs) - 1):
a = xs[i]
b = xs[i + 1]
c = (a + b) / 2
# print("a" + str(a) + "b" + str(b) + "c" + str(c))
if min1 <= c <= max1 and min2 <= c <= max2:
fa = width(X1, Y1, M, a) * width(X2, Z2, N, a)
fb = width(X1, Y1, M, b) * width(X2, Z2, N, b)
fc = width(X1, Y1, M, c) * width(X2, Z2, N, c)
res += ((b - a) / 6) * (fa + 4 * fc + fb)
# print(res)
print(res)
``` | output | 1 | 94,367 | 23 | 188,735 |
Provide a correct Python 3 solution for this coding contest problem.
Suppose that P1 is an infinite-height prism whose axis is parallel to the z-axis, and P2 is also an infinite-height prism whose axis is parallel to the y-axis. P1 is defined by the polygon C1 which is the cross section of P1 and the xy-plane, and P2 is also defined by the polygon C2 which is the cross section of P2 and the xz-plane.
Figure I.1 shows two cross sections which appear as the first dataset in the sample input, and Figure I.2 shows the relationship between the prisms and their cross sections.
<image>
Figure I.1: Cross sections of Prisms
<image>
Figure I.2: Prisms and their cross sections
<image>
Figure I.3: Intersection of two prisms
Figure I.3 shows the intersection of two prisms in Figure I.2, namely, P1 and P2.
Write a program which calculates the volume of the intersection of two prisms.
Input
The input is a sequence of datasets. The number of datasets is less than 200.
Each dataset is formatted as follows.
m n
x11 y11
x12 y12
.
.
.
x1m y1m
x21 z21
x22 z22
.
.
.
x2n z2n
m and n are integers (3 ≤ m ≤ 100, 3 ≤ n ≤ 100) which represent the numbers of the vertices of the polygons, C1 and C2, respectively.
x1i, y1i, x2j and z2j are integers between -100 and 100, inclusive. (x1i, y1i) and (x2j , z2j) mean the i-th and j-th vertices' positions of C1 and C2 respectively.
The sequences of these vertex positions are given in the counterclockwise order either on the xy-plane or the xz-plane as in Figure I.1.
You may assume that all the polygons are convex, that is, all the interior angles of the polygons are less than 180 degrees. You may also assume that all the polygons are simple, that is, each polygon's boundary does not cross nor touch itself.
The end of the input is indicated by a line containing two zeros.
Output
For each dataset, output the volume of the intersection of the two prisms, P1 and P2, with a decimal representation in a line.
None of the output values may have an error greater than 0.001. The output should not contain any other extra characters.
Example
Input
4 3
7 2
3 3
0 2
3 1
4 2
0 1
8 1
4 4
30 2
30 12
2 12
2 2
15 2
30 8
13 14
2 8
8 5
13 5
21 7
21 9
18 15
11 15
6 10
6 8
8 5
10 12
5 9
15 6
20 10
18 12
3 3
5 5
10 3
10 10
20 8
10 15
10 8
4 4
-98 99
-99 -99
99 -98
99 97
-99 99
-98 -98
99 -99
96 99
0 0
Output
4.708333333333333
1680.0000000000005
491.1500000000007
0.0
7600258.4847715655 | instruction | 0 | 94,368 | 23 | 188,736 |
"Correct Solution:
```
def width(X,Y,x):
n = len(X)
lb,ub = float('inf'),-float('inf')
for i in range(n):
x1,y1,x2,y2 = X[i],Y[i],X[(i+1)%n],Y[(i+1)%n]
if (x1-x)*(x2-x) <= 0 and x1 != x2:
y = y1 + (y2-y1)*(x-x1)/(x2-x1)
lb = min(lb,y)
ub = max(ub,y)
return max(0,ub-lb)
while 1:
M,N = map(int,input().split())
if not (M and N):
break
X1 = [0]*M
Y1 = [0]*M
X2 = [0]*N
Z2 = [0]*N
XS = [0]*N
for i in range(M):
X1[i],Y1[i] = map(int,input().split())
for i in range(N):
X2[i],Z2[i] = map(int,input().split())
XS = X1+X2
XS.sort()
min1,max1 = min(X1),max(X1)
min2,max2 = min(X2),max(X2)
res = 0
for i in range(len(XS)-1):
a = XS[i]
b = XS[i+1]
c = (a+b)/2
if min1 <= c <= max1 and min2 <= c <= max2:
fa = width(X1,Y1,a)*width(X2,Z2,a)
fb = width(X1,Y1,b)*width(X2,Z2,b)
fc = width(X1,Y1,c)*width(X2,Z2,c)
res += (b-a)/6 * (fa+4*fc+fb)
print('%.10f'%res)
``` | output | 1 | 94,368 | 23 | 188,737 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given n intervals in form [l; r] on a number line.
You are also given m queries in form [x; y]. What is the minimal number of intervals you have to take so that every point (not necessarily integer) from x to y is covered by at least one of them?
If you can't choose intervals so that every point from x to y is covered, then print -1 for that query.
Input
The first line contains two integers n and m (1 ≤ n, m ≤ 2 ⋅ 10^5) — the number of intervals and the number of queries, respectively.
Each of the next n lines contains two integer numbers l_i and r_i (0 ≤ l_i < r_i ≤ 5 ⋅ 10^5) — the given intervals.
Each of the next m lines contains two integer numbers x_i and y_i (0 ≤ x_i < y_i ≤ 5 ⋅ 10^5) — the queries.
Output
Print m integer numbers. The i-th number should be the answer to the i-th query: either the minimal number of intervals you have to take so that every point (not necessarily integer) from x_i to y_i is covered by at least one of them or -1 if you can't choose intervals so that every point from x_i to y_i is covered.
Examples
Input
2 3
1 3
2 4
1 3
1 4
3 4
Output
1
2
1
Input
3 4
1 3
1 3
4 5
1 2
1 3
1 4
1 5
Output
1
1
-1
-1
Note
In the first example there are three queries:
1. query [1; 3] can be covered by interval [1; 3];
2. query [1; 4] can be covered by intervals [1; 3] and [2; 4]. There is no way to cover [1; 4] by a single interval;
3. query [3; 4] can be covered by interval [2; 4]. It doesn't matter that the other points are covered besides the given query.
In the second example there are four queries:
1. query [1; 2] can be covered by interval [1; 3]. Note that you can choose any of the two given intervals [1; 3];
2. query [1; 3] can be covered by interval [1; 3];
3. query [1; 4] can't be covered by any set of intervals;
4. query [1; 5] can't be covered by any set of intervals. Note that intervals [1; 3] and [4; 5] together don't cover [1; 5] because even non-integer points should be covered. Here 3.5, for example, isn't covered.
Submitted Solution:
```
import sys
input = sys.stdin.readline
n,m = map(int,input().split())
a = [tuple(map(int,input().split())) for i in range(n)]
q = [tuple(map(int,input().split())) for j in range(m)]
a.sort(key = lambda x:x[1],reverse=True)
rng = 5*10**5+1
g = [0 for i in range(rng)]
idx = 0
for i in range(1,rng)[::-1]:
while idx <= n-1 and i < a[idx][0]:
idx += 1
if idx == n:
break
if a[idx][0] <= i <= a[idx][1]:
g[i] = a[idx][1]
dbl = [g]+[[0 for i in range(rng)] for j in range(20)]
for i in range(1,21):
for j in range(rng):
dbl[i][j] = dbl[i-1][dbl[i-1][j]]
for l,r in q:
if dbl[-1][l] < r:
print(-1)
continue
ans = 0
for i in range(21)[::-1]:
if dbl[i][l] < r:
ans += 2**i
l = dbl[i][l]
print(ans+1)
``` | instruction | 0 | 94,474 | 23 | 188,948 |
No | output | 1 | 94,474 | 23 | 188,949 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given n intervals in form [l; r] on a number line.
You are also given m queries in form [x; y]. What is the minimal number of intervals you have to take so that every point (not necessarily integer) from x to y is covered by at least one of them?
If you can't choose intervals so that every point from x to y is covered, then print -1 for that query.
Input
The first line contains two integers n and m (1 ≤ n, m ≤ 2 ⋅ 10^5) — the number of intervals and the number of queries, respectively.
Each of the next n lines contains two integer numbers l_i and r_i (0 ≤ l_i < r_i ≤ 5 ⋅ 10^5) — the given intervals.
Each of the next m lines contains two integer numbers x_i and y_i (0 ≤ x_i < y_i ≤ 5 ⋅ 10^5) — the queries.
Output
Print m integer numbers. The i-th number should be the answer to the i-th query: either the minimal number of intervals you have to take so that every point (not necessarily integer) from x_i to y_i is covered by at least one of them or -1 if you can't choose intervals so that every point from x_i to y_i is covered.
Examples
Input
2 3
1 3
2 4
1 3
1 4
3 4
Output
1
2
1
Input
3 4
1 3
1 3
4 5
1 2
1 3
1 4
1 5
Output
1
1
-1
-1
Note
In the first example there are three queries:
1. query [1; 3] can be covered by interval [1; 3];
2. query [1; 4] can be covered by intervals [1; 3] and [2; 4]. There is no way to cover [1; 4] by a single interval;
3. query [3; 4] can be covered by interval [2; 4]. It doesn't matter that the other points are covered besides the given query.
In the second example there are four queries:
1. query [1; 2] can be covered by interval [1; 3]. Note that you can choose any of the two given intervals [1; 3];
2. query [1; 3] can be covered by interval [1; 3];
3. query [1; 4] can't be covered by any set of intervals;
4. query [1; 5] can't be covered by any set of intervals. Note that intervals [1; 3] and [4; 5] together don't cover [1; 5] because even non-integer points should be covered. Here 3.5, for example, isn't covered.
Submitted Solution:
```
import io, os
input = io.StringIO(os.read(0, os.fstat(0).st_size).decode()).readline
MX = 5 * 10 ** 5 + 5
LG = 20
n, m = map(int, input().split())
jmp = [[0] * MX for i in range(LG)]
for _ in range(n):
l, r = map(int, input().split())
jmp[0][l] = max(jmp[0][l], r)
for i in range(MX):
jmp[0][i] = max(jmp[0][i - 1], jmp[0][i])
for i in range(MX - 1, 0, -1):
for j in range(1, LG):
jmp[j][i] = jmp[j - 1][jmp[j - 1][i]]
out = []
for _ in range(m):
x, y = map(int, input().split())
ans = n + 1
if x == y:
if jmp[0][x] >= y: ans = 1
elif jmp[-1][x] >= y:
cur = 0
for j in range(LG - 1, -1, -1):
if jmp[j][x] < y:
x = jmp[j][x]
cur += 1 << j
else:
ans = min(ans, cur + (1 << j))
out.append(ans if ans <= n else -1)
print(*out, sep='\n')
``` | instruction | 0 | 94,475 | 23 | 188,950 |
No | output | 1 | 94,475 | 23 | 188,951 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given n intervals in form [l; r] on a number line.
You are also given m queries in form [x; y]. What is the minimal number of intervals you have to take so that every point (not necessarily integer) from x to y is covered by at least one of them?
If you can't choose intervals so that every point from x to y is covered, then print -1 for that query.
Input
The first line contains two integers n and m (1 ≤ n, m ≤ 2 ⋅ 10^5) — the number of intervals and the number of queries, respectively.
Each of the next n lines contains two integer numbers l_i and r_i (0 ≤ l_i < r_i ≤ 5 ⋅ 10^5) — the given intervals.
Each of the next m lines contains two integer numbers x_i and y_i (0 ≤ x_i < y_i ≤ 5 ⋅ 10^5) — the queries.
Output
Print m integer numbers. The i-th number should be the answer to the i-th query: either the minimal number of intervals you have to take so that every point (not necessarily integer) from x_i to y_i is covered by at least one of them or -1 if you can't choose intervals so that every point from x_i to y_i is covered.
Examples
Input
2 3
1 3
2 4
1 3
1 4
3 4
Output
1
2
1
Input
3 4
1 3
1 3
4 5
1 2
1 3
1 4
1 5
Output
1
1
-1
-1
Note
In the first example there are three queries:
1. query [1; 3] can be covered by interval [1; 3];
2. query [1; 4] can be covered by intervals [1; 3] and [2; 4]. There is no way to cover [1; 4] by a single interval;
3. query [3; 4] can be covered by interval [2; 4]. It doesn't matter that the other points are covered besides the given query.
In the second example there are four queries:
1. query [1; 2] can be covered by interval [1; 3]. Note that you can choose any of the two given intervals [1; 3];
2. query [1; 3] can be covered by interval [1; 3];
3. query [1; 4] can't be covered by any set of intervals;
4. query [1; 5] can't be covered by any set of intervals. Note that intervals [1; 3] and [4; 5] together don't cover [1; 5] because even non-integer points should be covered. Here 3.5, for example, isn't covered.
Submitted Solution:
```
import sys
input = sys.stdin.readline
from bisect import *
n,m = map(int,input().split())
a = [list(map(int,input().split())) for i in range(n)]
q = [list(map(int,input().split())) for j in range(m)]
a.sort()
an = [a[0]]
notcovered = []
if a[0][0] > 1:
for i in range(1,a[0][0]):
notcovered.append(i)
for i in range(1,n):
if a[i][1] > an[-1][1]:
if a[i][0] > an[-1][0]:
an.append(a[i])
if an[-1][0] > an[-2][1]:
if an[-1][0] == an[-2][1]+1:
notcovered.append(an[-1][0]-0.5)
else:
notcovered.extend(list(range(an[-2][1]+1,an[-1][0])))
else:
an[-1][1] = a[i][1]
if an[-1][1] < 5*10**5:
notcovered.extend(list(range(an[-1][1]+1,5*10**5+1)))
anl,anr = list(zip(*an))
for i,j in q:
if notcovered:
jdg = bisect_right(notcovered,j)-bisect_left(notcovered,i)
if jdg > 0:
print(-1)
continue
idl = bisect_right(anl,i)
idr = bisect_left(anr,j)
print(idr-idl+2)
``` | instruction | 0 | 94,476 | 23 | 188,952 |
No | output | 1 | 94,476 | 23 | 188,953 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given n intervals in form [l; r] on a number line.
You are also given m queries in form [x; y]. What is the minimal number of intervals you have to take so that every point (not necessarily integer) from x to y is covered by at least one of them?
If you can't choose intervals so that every point from x to y is covered, then print -1 for that query.
Input
The first line contains two integers n and m (1 ≤ n, m ≤ 2 ⋅ 10^5) — the number of intervals and the number of queries, respectively.
Each of the next n lines contains two integer numbers l_i and r_i (0 ≤ l_i < r_i ≤ 5 ⋅ 10^5) — the given intervals.
Each of the next m lines contains two integer numbers x_i and y_i (0 ≤ x_i < y_i ≤ 5 ⋅ 10^5) — the queries.
Output
Print m integer numbers. The i-th number should be the answer to the i-th query: either the minimal number of intervals you have to take so that every point (not necessarily integer) from x_i to y_i is covered by at least one of them or -1 if you can't choose intervals so that every point from x_i to y_i is covered.
Examples
Input
2 3
1 3
2 4
1 3
1 4
3 4
Output
1
2
1
Input
3 4
1 3
1 3
4 5
1 2
1 3
1 4
1 5
Output
1
1
-1
-1
Note
In the first example there are three queries:
1. query [1; 3] can be covered by interval [1; 3];
2. query [1; 4] can be covered by intervals [1; 3] and [2; 4]. There is no way to cover [1; 4] by a single interval;
3. query [3; 4] can be covered by interval [2; 4]. It doesn't matter that the other points are covered besides the given query.
In the second example there are four queries:
1. query [1; 2] can be covered by interval [1; 3]. Note that you can choose any of the two given intervals [1; 3];
2. query [1; 3] can be covered by interval [1; 3];
3. query [1; 4] can't be covered by any set of intervals;
4. query [1; 5] can't be covered by any set of intervals. Note that intervals [1; 3] and [4; 5] together don't cover [1; 5] because even non-integer points should be covered. Here 3.5, for example, isn't covered.
Submitted Solution:
```
n, m = map(int, input().split())
def overlap(i1, i2):
if i1[0] > i2[0]:
i1, i2 = i2, i1
return i2[0] <= i1[1]
def contains(i1, i2):
return i2[0] >= i1[0] and i2[1] <= i1[1]
intervals = {}
for i in range(n):
x, y = map(int, input().split())
if (x, y) in intervals:
continue
new_entries = {(x, y): 1}
for interval in intervals:
if overlap(interval, (x, y)):
old = intervals[interval]
new_entries[(min(interval[0], x), max(interval[1], y))] = old + 1
intervals.update(new_entries)
for i in range(m):
x, y = map(int, input().split())
match = None
for interval in intervals:
if contains(interval, (x, y)):
if match is None:
match = intervals[interval]
else:
match = min(match, intervals[interval])
if match is None:
print('-1')
else:
print(match)
``` | instruction | 0 | 94,477 | 23 | 188,954 |
No | output | 1 | 94,477 | 23 | 188,955 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given n integers. You need to choose a subset and put the chosen numbers in a beautiful rectangle (rectangular matrix). Each chosen number should occupy one of its rectangle cells, each cell must be filled with exactly one chosen number. Some of the n numbers may not be chosen.
A rectangle (rectangular matrix) is called beautiful if in each row and in each column all values are different.
What is the largest (by the total number of cells) beautiful rectangle you can construct? Print the rectangle itself.
Input
The first line contains n (1 ≤ n ≤ 4⋅10^5). The second line contains n integers (1 ≤ a_i ≤ 10^9).
Output
In the first line print x (1 ≤ x ≤ n) — the total number of cells of the required maximum beautiful rectangle. In the second line print p and q (p ⋅ q=x): its sizes. In the next p lines print the required rectangle itself. If there are several answers, print any.
Examples
Input
12
3 1 4 1 5 9 2 6 5 3 5 8
Output
12
3 4
1 2 3 5
3 1 5 4
5 6 8 9
Input
5
1 1 1 1 1
Output
1
1 1
1 | instruction | 0 | 94,494 | 23 | 188,988 |
Tags: brute force, combinatorics, constructive algorithms, data structures, greedy, math
Correct Solution:
```
n = int(input())
l = list(map(int,input().split()))
if n == 1:
print(1)
print(1,1)
print(l[0])
else:
d = {}
for i in l:
d[i] = 0
for i in l:
d[i] += 1
equal = [0] * (n + 1)
for i in d:
equal[d[i]] += 1
atmost = [0] * (n + 1)
atmost[0] = equal[0]
for i in range(1, n+1):
atmost[i] = atmost[i-1] + equal[i]
sumka = 0
best_iloczyn = 0
best_a = 0
best_b = 0
for a in range(1, n):
if a**2 > n:
break
sumka += (len(d) - atmost[a-1])
b_cand = sumka//a
if b_cand < a:
continue
if a * b_cand > best_iloczyn:
best_iloczyn = a * b_cand
best_a = a
best_b = b_cand
print(best_iloczyn)
print(best_a, best_b)
li = []
for i in d:
if d[i] >= best_a:
li += [i]*min(best_a, d[i])
for i in d:
if d[i] < best_a:
li += [i]*min(best_a, d[i])
#print(li)
mat = [[0] * best_b for i in range(best_a)]
for dd in range(1, best_a + 1):
if best_a%dd==0 and best_b%dd==0:
du = dd
i = 0
for st in range(du):
for j in range(best_iloczyn//du):
mat[i%best_a][(st+i)%best_b] = li[i]
i += 1
for i in range(best_a):
print(*mat[i])
``` | output | 1 | 94,494 | 23 | 188,989 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given n integers. You need to choose a subset and put the chosen numbers in a beautiful rectangle (rectangular matrix). Each chosen number should occupy one of its rectangle cells, each cell must be filled with exactly one chosen number. Some of the n numbers may not be chosen.
A rectangle (rectangular matrix) is called beautiful if in each row and in each column all values are different.
What is the largest (by the total number of cells) beautiful rectangle you can construct? Print the rectangle itself.
Input
The first line contains n (1 ≤ n ≤ 4⋅10^5). The second line contains n integers (1 ≤ a_i ≤ 10^9).
Output
In the first line print x (1 ≤ x ≤ n) — the total number of cells of the required maximum beautiful rectangle. In the second line print p and q (p ⋅ q=x): its sizes. In the next p lines print the required rectangle itself. If there are several answers, print any.
Examples
Input
12
3 1 4 1 5 9 2 6 5 3 5 8
Output
12
3 4
1 2 3 5
3 1 5 4
5 6 8 9
Input
5
1 1 1 1 1
Output
1
1 1
1 | instruction | 0 | 94,496 | 23 | 188,992 |
Tags: brute force, combinatorics, constructive algorithms, data structures, greedy, math
Correct Solution:
```
n = int(input())
arr = list(map(int, input().split()))
d = {}
for i in arr:
d[i] = d.get(i, 0) + 1
d2 = {}
for k, v in d.items():
d2.setdefault(v, []).append(k)
s = n
prev = 0
ansp = ansq = anss = 0
for p in range(n, 0, -1):
q = s // p
if p <= q and q * p > anss:
anss = q * p
ansq = q
ansp = p
prev += len(d2.get(p, []))
s -= prev
def get_ans():
cur_i = 0
cur_j = 0
cur = 0
for k, v in d3:
for val in v:
f = min(k, anss - cur, ansp)
cur += f
for i in range(f):
cur_i = (cur_i + 1) % ansp
cur_j = (cur_j + 1) % ansq
if ans[cur_i][cur_j]:
cur_i = (cur_i + 1) % ansp
ans[cur_i][cur_j] = val
print(anss)
print(ansp, ansq)
d3 = sorted(d2.items(), reverse=True)
ans = [[0] * ansq for i in range(ansp)]
get_ans()
for i in range(ansp):
print(*ans[i])
``` | output | 1 | 94,496 | 23 | 188,993 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given n integers. You need to choose a subset and put the chosen numbers in a beautiful rectangle (rectangular matrix). Each chosen number should occupy one of its rectangle cells, each cell must be filled with exactly one chosen number. Some of the n numbers may not be chosen.
A rectangle (rectangular matrix) is called beautiful if in each row and in each column all values are different.
What is the largest (by the total number of cells) beautiful rectangle you can construct? Print the rectangle itself.
Input
The first line contains n (1 ≤ n ≤ 4⋅10^5). The second line contains n integers (1 ≤ a_i ≤ 10^9).
Output
In the first line print x (1 ≤ x ≤ n) — the total number of cells of the required maximum beautiful rectangle. In the second line print p and q (p ⋅ q=x): its sizes. In the next p lines print the required rectangle itself. If there are several answers, print any.
Examples
Input
12
3 1 4 1 5 9 2 6 5 3 5 8
Output
12
3 4
1 2 3 5
3 1 5 4
5 6 8 9
Input
5
1 1 1 1 1
Output
1
1 1
1 | instruction | 0 | 94,497 | 23 | 188,994 |
Tags: brute force, combinatorics, constructive algorithms, data structures, greedy, math
Correct Solution:
```
from sys import stdin, stdout
def getmaxrectangle(n, a):
dic = {}
dicCntVals = {}
for va in a:
if va not in dic:
dic[va] = 0
dic[va] += 1
for val in dic.keys():
cnt = dic[val]
if cnt not in dicCntVals:
dicCntVals[cnt] = []
dicCntVals[cnt].append(val)
geq = [0]*(n+1)
if n in dicCntVals:
geq[n] = len(dicCntVals[n])
for cnt in range(n-1, 0, -1):
geq[cnt] = geq[cnt + 1]
if cnt in dicCntVals:
geq[cnt] += len(dicCntVals[cnt])
#print(geq)
b_pq = 0
b_p = 0
b_q = 0
ttl = 0
for p in range(1, n+1):
ttl += geq[p]
q = int(ttl/p)
if q >= p and q*p > b_pq:
b_pq = q*p
b_p = p
b_q = q
x = 0
y = 0
#print(str(b_pq))
r = [[0 for j in range(b_q)] for i in range(b_p)]
#print(str((b_p)))
#print(str((b_q)))
#print(str(len(r)))
#print(str(len(r[0])))
for i in range(n, 0, -1):
if i not in dicCntVals:
continue
for j in dicCntVals[i]:
for k in range(min(b_p, i)):
if r[x][y] != 0:
x = (x + 1) % b_p;
if r[x][y] == 0:
r[x][y] = j
x = (x + 1) % b_p
y = (y + 1) % b_q
return r
if __name__ == '__main__':
n = int(stdin.readline())
a = list(map(int, stdin.readline().split()))
res = getmaxrectangle(n, a)
stdout.write(str(len(res) * len(res[0])))
stdout.write('\n')
stdout.write(str(len(res)) + ' ' + str(len(res[0])))
stdout.write('\n')
for i in range(len(res)):
for j in range(len(res[i])):
stdout.write(str(res[i][j]))
stdout.write(' ')
stdout.write('\n')
``` | output | 1 | 94,497 | 23 | 188,995 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given n integers. You need to choose a subset and put the chosen numbers in a beautiful rectangle (rectangular matrix). Each chosen number should occupy one of its rectangle cells, each cell must be filled with exactly one chosen number. Some of the n numbers may not be chosen.
A rectangle (rectangular matrix) is called beautiful if in each row and in each column all values are different.
What is the largest (by the total number of cells) beautiful rectangle you can construct? Print the rectangle itself.
Input
The first line contains n (1 ≤ n ≤ 4⋅10^5). The second line contains n integers (1 ≤ a_i ≤ 10^9).
Output
In the first line print x (1 ≤ x ≤ n) — the total number of cells of the required maximum beautiful rectangle. In the second line print p and q (p ⋅ q=x): its sizes. In the next p lines print the required rectangle itself. If there are several answers, print any.
Examples
Input
12
3 1 4 1 5 9 2 6 5 3 5 8
Output
12
3 4
1 2 3 5
3 1 5 4
5 6 8 9
Input
5
1 1 1 1 1
Output
1
1 1
1 | instruction | 0 | 94,498 | 23 | 188,996 |
Tags: brute force, combinatorics, constructive algorithms, data structures, greedy, math
Correct Solution:
```
from itertools import accumulate
import math
from collections import Counter
import sys
input = sys.stdin.readline
n = int(input())
A = list(map(int, input().split()))
D = Counter(A)
MAXV = max(max(D.values()), int(math.sqrt(n)))
VCOUNT = [0] * (MAXV + 1)
for v in D.values():
VCOUNT[v] += 1
SUM = n
ACC = list(accumulate(VCOUNT[::-1]))[::-1]
ANS = 0
for i in range(MAXV, 0, -1):
if SUM // i >= i:
if ANS < i * (SUM // i):
ANS = i * (SUM // i)
ANSX = i, (SUM // i)
SUM -= ACC[i]
print(ANS)
X, Y = ANSX[0], ANSX[1]
print(X, Y)
maxx,a,b=ANS,X,Y
A = []
D = D.most_common()
for key, d in D:
num = min(a, d)
for number in range(num):
A.append(key)
ANS = [[0] * (b) for _ in range(a)]
AIM = A[:maxx]
pos = 0
turn = 0
while True:
posi = 0
posj = turn
for i in range(a):
ANS[(posi + i) % a][(posj + i) % b] = AIM[pos]
pos += 1
if pos == maxx:
break
turn += 1
'''
for i in range(a):
print(*ANS[i])
'''
for ans in ANS:
sys.stdout.write(" ".join(map(str,ans))+"\n")
``` | output | 1 | 94,498 | 23 | 188,997 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given n integers. You need to choose a subset and put the chosen numbers in a beautiful rectangle (rectangular matrix). Each chosen number should occupy one of its rectangle cells, each cell must be filled with exactly one chosen number. Some of the n numbers may not be chosen.
A rectangle (rectangular matrix) is called beautiful if in each row and in each column all values are different.
What is the largest (by the total number of cells) beautiful rectangle you can construct? Print the rectangle itself.
Input
The first line contains n (1 ≤ n ≤ 4⋅10^5). The second line contains n integers (1 ≤ a_i ≤ 10^9).
Output
In the first line print x (1 ≤ x ≤ n) — the total number of cells of the required maximum beautiful rectangle. In the second line print p and q (p ⋅ q=x): its sizes. In the next p lines print the required rectangle itself. If there are several answers, print any.
Examples
Input
12
3 1 4 1 5 9 2 6 5 3 5 8
Output
12
3 4
1 2 3 5
3 1 5 4
5 6 8 9
Input
5
1 1 1 1 1
Output
1
1 1
1 | instruction | 0 | 94,499 | 23 | 188,998 |
Tags: brute force, combinatorics, constructive algorithms, data structures, greedy, math
Correct Solution:
```
from collections import Counter
from sys import stdin
def input():
return next(stdin)[:-1]
def main():
n = int(input())
aa = input().split()
cc = list(Counter(aa).most_common())
if n % cc[0][1] == 0 and cc[0][1] * cc[0][1] <= n:
h = cc[0][1]
w = n//cc[0][1]
best = n
else:
count_count = [0] * (n+1)
for v, c in cc:
count_count[c] += 1
geq = [count_count[n]]
for v in reversed(count_count[:n]):
geq.append(geq[-1] + v)
geq.reverse()
tot = 0
best = 0
for a in range(1,n+1):
tot += geq[a]
b = tot//a
if a <= b and best < a * b:
best = a * b
h = a
w = b
print(best)
print(h,w)
x = 0
y = 0
mat = [[''] * w for _ in range(h) ]
for v, c in cc:
for j in range(min(c, h)):
if mat[x][y] != '':
x = (x+1)%h
if mat[x][y] == '':
mat[x][y] = v
x = (x+1)%h
y = (y+1)%w
for i in range(h):
print(' '.join(mat[i]))
if __name__ == "__main__":
main()
``` | output | 1 | 94,499 | 23 | 188,999 |
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