text stringlengths 699 45.9k | conversation_id int64 254 108k | embedding list | cluster int64 3 3 |
|---|---|---|---|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Cowboy Beblop is a funny little boy who likes sitting at his computer. He somehow obtained two elastic hoops in the shape of 2D polygons, which are not necessarily convex. Since there's no gravity on his spaceship, the hoops are standing still in the air. Since the hoops are very elastic, Cowboy Beblop can stretch, rotate, translate or shorten their edges as much as he wants.
For both hoops, you are given the number of their vertices, as well as the position of each vertex, defined by the X , Y and Z coordinates. The vertices are given in the order they're connected: the 1st vertex is connected to the 2nd, which is connected to the 3rd, etc., and the last vertex is connected to the first one. Two hoops are connected if it's impossible to pull them to infinity in different directions by manipulating their edges, without having their edges or vertices intersect at any point – just like when two links of a chain are connected. The polygons' edges do not intersect or overlap.
To make things easier, we say that two polygons are well-connected, if the edges of one polygon cross the area of the other polygon in two different directions (from the upper and lower sides of the plane defined by that polygon) a different number of times.
Cowboy Beblop is fascinated with the hoops he has obtained and he would like to know whether they are well-connected or not. Since he’s busy playing with his dog, Zwei, he’d like you to figure it out for him. He promised you some sweets if you help him!
Input
The first line of input contains an integer n (3 ≤ n ≤ 100 000), which denotes the number of edges of the first polygon. The next N lines each contain the integers x, y and z ( - 1 000 000 ≤ x, y, z ≤ 1 000 000) — coordinates of the vertices, in the manner mentioned above. The next line contains an integer m (3 ≤ m ≤ 100 000) , denoting the number of edges of the second polygon, followed by m lines containing the coordinates of the second polygon’s vertices.
It is guaranteed that both polygons are simple (no self-intersections), and in general that the obtained polygonal lines do not intersect each other. Also, you can assume that no 3 consecutive points of a polygon lie on the same line.
Output
Your output should contain only one line, with the words "YES" or "NO", depending on whether the two given polygons are well-connected.
Example
Input
4
0 0 0
2 0 0
2 2 0
0 2 0
4
1 1 -1
1 1 1
1 3 1
1 3 -1
Output
YES
Note
On the picture below, the two polygons are well-connected, as the edges of the vertical polygon cross the area of the horizontal one exactly once in one direction (for example, from above to below), and zero times in the other (in this case, from below to above). Note that the polygons do not have to be parallel to any of the xy-,xz-,yz- planes in general. <image>
Submitted Solution:
```
# import numpy as np
import sys
from math import gcd, sqrt
EPS = 0.0000000001
input = sys.stdin.readline
def inp():
return(int(input()))
def inlt():
return(list(map(int,input().split())))
def insr():
s = input()
return s
def invr():
return(list(map(int,input().split())))
def plane(p1, p2, p3):
x1, y1, z1 = p1
x2, y2, z2 = p2
x3, y3, z3 = p3
a1, b1, c1 = x2 - x1, y2 - y1, z2 - z1
a2, b2, c2 = x3 - x1, y3 - y1, z3 - z1
a, b, c = b1 * c2 - b2 * c1, a2 * c1 - a1 * c2, a1 * b2 - b1 * a2
d = (- a * x1 - b * y1 - c * z1)
return a, b, c, d
def intersection_of_two_planes(p1, p2):
A1, B1, C1, D1 = p1
A2, B2, C2, D2 = p2
if (A1*B2-A2*B1) != 0:
x = ((B1*D2-B2*D1,A1*B2-A2*B1), B1*C2-B2*C1)
y = ((A2*D1-A1*D2,A1*B2-A2*B1), A2*C1-A1*C2)
z = ((0,1),1)
elif (B1*C2-B2*C1) != 0:
x = ((0,1),1)
y = ((C1*D2-C2*D1,B1*C2-B2*C1), C1*A2-C2*A1)
z = ((B2*D1-B1*D2,B1*C2-B2*C1), B2*A1-B1*A2)
elif (A1*C2-A2*C1) != 0:
y = ((0,1),1)
x = ((C1*D2-C2*D1,A1*C2-A2*C1), C1*B2-C2*B1)
z = ((A2*D1-A1*D2,A1*C2-A2*C1), A2*B1-A1*B2)
else:
return None
return x, y, z
def line_parametric(p1, p2):
x1, y1, z1 = p1
x2, y2, z2 = p2
return ((x2,1),x1-x2), ((y2,1),y1-y2), ((z2,1),z1-z2)
def solve_2_by_2(a1,b1,c1p,c1q,a2,b2,c2p,c2q):
if a1*b2-b1*a2:
return (c1p*c2q*b2-b1*c2p*c1q,(a1*b2-b1*a2)*c1q*c2q), (a1*c2p*c1q-a2*c1p*c2q,(a1*b2-b1*a2)*c1q*c2q)
else:
return None, None
def intersection_of_two_lines(l1, l2):
res = []
px, py, pz = l1
qx, qy, qz = l2
try:
t1, t2 = solve_2_by_2(px[1],-qx[1],qx[0][0]*px[0][1]-px[0][0]*qx[0][1],qx[0][1]*px[0][1],py[1],-qy[1],qy[0][0]*py[0][1]-py[0][0]*qy[0][1],qy[0][1]*py[0][1])
p1, q1 = t1
p2, q2 = t2
if qz[0][1]*pz[0][1]*(p1*pz[1]*q2 - p2*qz[1]*q1) == q1*q2*(qz[0][0]*pz[0][1]-pz[0][0]*qz[0][1]):
return ((px[0][0]*q1+p1*px[1]*px[0][1],q1*px[0][1]), (py[0][0]*q1+p1*py[1]*py[0][1],q1*py[0][1]), (pz[0][0]*q1+p1*pz[1]*pz[0][1],q1*pz[0][1])), (p1,q1)
else:
return None, None
except:
pass
try:
t1, t2 = solve_2_by_2(px[1],-qx[1],qx[0][0]*px[0][1]-px[0][0]*qx[0][1],qx[0][1]*px[0][1],pz[1],-qz[1],qz[0][0]*pz[0][1]-pz[0][0]*qz[0][1],qz[0][1]*pz[0][1])
p1, q1 = t1
p2, q2 = t2
if qy[0][1]*py[0][1]*(p1*py[1]*q2 - p2*qy[1]*q1) == q1*q2*(qy[0][0]*py[0][1]-py[0][0]*qy[0][1]):
return ((px[0][0]*q1+p1*px[1]*px[0][1],q1*px[0][1]), (py[0][0]*q1+p1*py[1]*py[0][1],q1*py[0][1]), (pz[0][0]*q1+p1*pz[1]*pz[0][1],q1*pz[0][1])), (p1,q1)
else:
return None, None
except:
pass
try:
t1, t2 = solve_2_by_2(py[1],-qy[1],qy[0][0]*py[0][1]-py[0][0]*qy[0][1],qy[0][1]*py[0][1],pz[1],-qz[1],qz[0][0]*pz[0][1]-pz[0][0]*qz[0][1],qz[0][1]*pz[0][1])
p1, q1 = t1
p2, q2 = t2
if qx[0][1]*px[0][1]*(p1*px[1]*q2 - p2*qx[1]*q1) == q1*q2*(qx[0][0]*px[0][1]-px[0][0]*qx[0][1]):
return ((px[0][0]*q1+p1*px[1]*px[0][1],q1*px[0][1]), (py[0][0]*q1+p1*py[1]*py[0][1],q1*py[0][1]), (pz[0][0]*q1+p1*pz[1]*pz[0][1],q1*pz[0][1])), (p1,q1)
else:
return None, None
except:
pass
return None, None
def crop(points):
res = []
prev = None
cnt = 0
for p in points:
if p == prev:
cnt+=1
else:
if cnt & 1:
res.append(prev)
cnt = 1
prev = p
if cnt & 1:
res.append(prev)
return res
def distance(p1, p2):
x1, y1, z1 = p1
x2, y2, z2 = p2
if x2.__class__.__name__=='tuple':
x2p, x2q = x2
y2p, y2q = y2
z2p, z2q = z2
return sqrt((x1-x2p/x2q)**2+(y1-y2p/y2q)**2+(z1-z2p/z2q)**2)
return sqrt((x1-x2)**2+(y1-y2)**2+(z1-z2)**2)
def distinct(p1, p2):
pass
x_poly, y_poly = [], []
for _ in range(inp()):
x_poly.append(inlt())
for _ in range(inp()):
y_poly.append(inlt())
x_plane = plane(*x_poly[:3])
y_plane = plane(*y_poly[:3])
incidence = intersection_of_two_planes(x_plane,y_plane)
if incidence:
points = []
for i in range(len(x_poly)):
line = line_parametric(x_poly[i],x_poly[(i+1)%len(x_poly)])
intersection, t = intersection_of_two_lines(incidence,line)
print(incidence, line)
if intersection:
p1 = x_poly[i]
p2 = x_poly[(i+1)%len(x_poly)]
# print(p1,p2,intersection)
# print(abs(distance(p1,intersection) + distance(p2,intersection) - distance(p1,p2)))
if abs(distance(p1,intersection) + distance(p2,intersection) - distance(p1,p2))<EPS:
if distance(p1,intersection) > EPS and distance(p2,intersection) > EPS:
points.append((t,0))
# print('-------')
for i in range(len(y_poly)):
line = line_parametric(y_poly[i],y_poly[(i+1)%len(y_poly)])
intersection, t = intersection_of_two_lines(incidence,line)
print(incidence, line)
if intersection:
p1 = y_poly[i]
p2 = y_poly[(i+1)%len(y_poly)]
# print(p1,p2,intersection)
# print(abs(distance(p1,intersection) + distance(p2,intersection) - distance(p1,p2)))
if abs(distance(p1,intersection) + distance(p2,intersection) - distance(p1,p2))<EPS:
if distance(p1,intersection) > EPS and distance(p2,intersection) > EPS:
points.append((t,1))
points = [i[1] for i in sorted(points, key=lambda t: t[0])]
prev = len(points) + 1
while prev > len(points):
prev = len(points)
points = crop(points)
#
#
print('NO' if not len(points) else 'YES')
else:
print('NO')
```
No
| 75,544 | [
0.40478515625,
0.1573486328125,
0.1959228515625,
0.360595703125,
-0.431640625,
-0.146240234375,
-0.1641845703125,
-0.004245758056640625,
0.270263671875,
1.0244140625,
0.87548828125,
-0.210693359375,
-0.0288238525390625,
-0.78173828125,
-0.397216796875,
0.397705078125,
-0.83447265625,... | 3 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Cowboy Beblop is a funny little boy who likes sitting at his computer. He somehow obtained two elastic hoops in the shape of 2D polygons, which are not necessarily convex. Since there's no gravity on his spaceship, the hoops are standing still in the air. Since the hoops are very elastic, Cowboy Beblop can stretch, rotate, translate or shorten their edges as much as he wants.
For both hoops, you are given the number of their vertices, as well as the position of each vertex, defined by the X , Y and Z coordinates. The vertices are given in the order they're connected: the 1st vertex is connected to the 2nd, which is connected to the 3rd, etc., and the last vertex is connected to the first one. Two hoops are connected if it's impossible to pull them to infinity in different directions by manipulating their edges, without having their edges or vertices intersect at any point – just like when two links of a chain are connected. The polygons' edges do not intersect or overlap.
To make things easier, we say that two polygons are well-connected, if the edges of one polygon cross the area of the other polygon in two different directions (from the upper and lower sides of the plane defined by that polygon) a different number of times.
Cowboy Beblop is fascinated with the hoops he has obtained and he would like to know whether they are well-connected or not. Since he’s busy playing with his dog, Zwei, he’d like you to figure it out for him. He promised you some sweets if you help him!
Input
The first line of input contains an integer n (3 ≤ n ≤ 100 000), which denotes the number of edges of the first polygon. The next N lines each contain the integers x, y and z ( - 1 000 000 ≤ x, y, z ≤ 1 000 000) — coordinates of the vertices, in the manner mentioned above. The next line contains an integer m (3 ≤ m ≤ 100 000) , denoting the number of edges of the second polygon, followed by m lines containing the coordinates of the second polygon’s vertices.
It is guaranteed that both polygons are simple (no self-intersections), and in general that the obtained polygonal lines do not intersect each other. Also, you can assume that no 3 consecutive points of a polygon lie on the same line.
Output
Your output should contain only one line, with the words "YES" or "NO", depending on whether the two given polygons are well-connected.
Example
Input
4
0 0 0
2 0 0
2 2 0
0 2 0
4
1 1 -1
1 1 1
1 3 1
1 3 -1
Output
YES
Note
On the picture below, the two polygons are well-connected, as the edges of the vertical polygon cross the area of the horizontal one exactly once in one direction (for example, from above to below), and zero times in the other (in this case, from below to above). Note that the polygons do not have to be parallel to any of the xy-,xz-,yz- planes in general. <image>
Submitted Solution:
```
# import numpy as np
import sys
from math import gcd, sqrt
EPS = 0.00000000001
input = sys.stdin.readline
def inp():
return(int(input()))
def inlt():
return(list(map(int,input().split())))
def insr():
s = input()
return s
def invr():
return(list(map(int,input().split())))
def plane(p1, p2, p3):
x1, y1, z1 = p1
x2, y2, z2 = p2
x3, y3, z3 = p3
a1, b1, c1 = x2 - x1, y2 - y1, z2 - z1
a2, b2, c2 = x3 - x1, y3 - y1, z3 - z1
a, b, c = b1 * c2 - b2 * c1, a2 * c1 - a1 * c2, a1 * b2 - b1 * a2
d = (- a * x1 - b * y1 - c * z1)
return a, b, c, d
def intersection_of_two_planes(p1, p2):
A1, B1, C1, D1 = p1
A2, B2, C2, D2 = p2
if (A1*B2-A2*B1) != 0:
x = ((B1*D2-B2*D1)/(A1*B2-A2*B1), (B1*C2-B2*C1))
y = ((A2*D1-A1*D2)/(A1*B2-A2*B1), (A2*C1-A1*C2))
z = (0,1)
elif (B1*C2-B2*C1) != 0:
x = (0,1)
y = ((C1*D2-C2*D1)/(B1*C2-B2*C1), (C1*A2-C2*A1))
z = ((B2*D1-B1*D2)/(B1*C2-B2*C1), (B2*A1-B1*A2))
elif (A1*C2-A2*C1) != 0:
y = (0,1)
x = ((C1*D2-C2*D1)/(A1*C2-A2*C1), (C1*B2-C2*B1))
z = ((A2*D1-A1*D2)/(A1*C2-A2*C1), (A2*B1-A1*B2))
else:
return None
return x, y, z
def line_parametric(p1, p2):
x1, y1, z1 = p1
x2, y2, z2 = p2
return (x2,x1-x2), (y2,y1-y2), (z2,z1-z2)
def solve_2_by_2(a1,b1,c1,a2,b2,c2):
if a1*b2-b1*a2:
return (c1*b2-b1*c2)/(a1*b2-b1*a2), (a1*c2-c1*a2)/(a1*b2-b1*a2)
else:
return None
def intersection_of_two_lines(l1, l2):
res = []
px, py, pz = l1
qx, qy, qz = l2
try:
t1, t2 = solve_2_by_2(px[1],-qx[1],qx[0]-px[0],py[1],-qy[1],qy[0]-py[0])
if t1*pz[1] - t2*qz[1] == qz[0]-pz[0]:
return (px[0]+t1*px[1], py[0]+t1*py[1], pz[0]+t1*pz[1]), t1
else:
return None, None
except:
pass
try:
t1, t2 = solve_2_by_2(px[1],-qx[1],qx[0]-px[0],pz[1],-qz[1],qz[0]-pz[0])
if t1*py[1] - t2*qy[1] == qy[0]-py[0]:
return (px[0]+t1*px[1], py[0]+t1*py[1], pz[0]+t1*pz[1]), t1
else:
return None, None
except:
pass
try:
t1, t2 = solve_2_by_2(py[1],-qy[1],qy[0]-py[0],pz[1],-qz[1],qz[0]-pz[0])
if t1*px[1] - t2*qx[1] == qx[0]-px[0]:
return (px[0]+t1*px[1], py[0]+t1*py[1], pz[0]+t1*pz[1]), t1
else:
return None, None
except:
pass
return None, None
def crop(points):
res = []
prev = None
cnt = 0
for p in points:
if p == prev:
cnt+=1
else:
if cnt & 1:
res.append(prev)
cnt = 1
prev = p
if cnt & 1:
res.append(prev)
return res
def distance(p1, p2):
x1, y1, z1 = p1
x2, y2, z2 = p2
return sqrt((x1-x2)**2+(y1-y2)**2+(z1-z2)**2)
x_poly, y_poly = [], []
for _ in range(inp()):
x_poly.append(inlt())
for _ in range(inp()):
y_poly.append(inlt())
x_plane = plane(*x_poly[:3])
y_plane = plane(*y_poly[:3])
incidence = intersection_of_two_planes(x_plane,y_plane)
if incidence:
points = []
for i in range(len(x_poly)):
line = line_parametric(x_poly[i],x_poly[(i+1)%len(x_poly)])
intersection, t = intersection_of_two_lines(incidence,line)
if intersection:
p1 = x_poly[i]
p2 = x_poly[(i+1)%len(x_poly)]
# print(p1,p2,intersection)
# print(abs(distance(p1,intersection) + distance(p2,intersection) - distance(p1,p2)))
if abs(distance(p1,intersection) + distance(p2,intersection) - distance(p1,p2))<EPS:
if distance(p1,p2) > distance(p1,intersection):
points.append((t,0))
# print('-------')
for i in range(len(y_poly)):
line = line_parametric(y_poly[i],y_poly[(i+1)%len(y_poly)])
intersection, t = intersection_of_two_lines(incidence,line)
if intersection:
p1 = y_poly[i]
p2 = y_poly[(i+1)%len(y_poly)]
# print(p1,p2,intersection)
# print(abs(distance(p1,intersection) + distance(p2,intersection) - distance(p1,p2)))
if abs(distance(p1,intersection) + distance(p2,intersection) - distance(p1,p2))<EPS:
if distance(p1,p2) > distance(p1,intersection):
points.append((t,1))
points = [i[1] for i in sorted(points, key=lambda t: t[0])]
prev = len(points) + 1
while prev > len(points):
prev = len(points)
points = crop(points)
#
#
print('NO' if not len(points) else 'YES')
else:
print('NO')
```
No
| 75,545 | [
0.40478515625,
0.1573486328125,
0.1959228515625,
0.360595703125,
-0.431640625,
-0.146240234375,
-0.1641845703125,
-0.004245758056640625,
0.270263671875,
1.0244140625,
0.87548828125,
-0.210693359375,
-0.0288238525390625,
-0.78173828125,
-0.397216796875,
0.397705078125,
-0.83447265625,... | 3 |
Provide a correct Python 3 solution for this coding contest problem.
It’s still hot every day, but September has already come. It’s autumn according to the calendar. Looking around, I see two red dragonflies at rest on the wall in front of me. It’s autumn indeed.
When two red dragonflies’ positional information as measured from the end of the wall is given, make a program to calculate the distance between their heads.
Input
The input is given in the following format.
$x_1$ $x_2$
The input line provides dragonflies’ head positions $x_1$ and $x_2$ ($0 \leq x_1, x_2 \leq 100$) as integers.
Output
Output the distance between the two red dragonflies in a line.
Examples
Input
20 30
Output
10
Input
50 25
Output
25
Input
25 25
Output
0
"Correct Solution:
```
import math
a , b = map(int,input().split())
if a<b:
print(b-a)
elif a>b:
print(a-b)
else:
print("0")
```
| 75,810 | [
-0.00569915771484375,
0.034576416015625,
-0.2025146484375,
-0.116455078125,
-0.40869140625,
-0.463623046875,
0.05609130859375,
0.29248046875,
0.17626953125,
1.158203125,
0.219970703125,
0.25341796875,
0.1868896484375,
-0.4951171875,
-0.026397705078125,
-0.301513671875,
-0.54248046875... | 3 |
Provide a correct Python 3 solution for this coding contest problem.
It’s still hot every day, but September has already come. It’s autumn according to the calendar. Looking around, I see two red dragonflies at rest on the wall in front of me. It’s autumn indeed.
When two red dragonflies’ positional information as measured from the end of the wall is given, make a program to calculate the distance between their heads.
Input
The input is given in the following format.
$x_1$ $x_2$
The input line provides dragonflies’ head positions $x_1$ and $x_2$ ($0 \leq x_1, x_2 \leq 100$) as integers.
Output
Output the distance between the two red dragonflies in a line.
Examples
Input
20 30
Output
10
Input
50 25
Output
25
Input
25 25
Output
0
"Correct Solution:
```
x1,x2 = map(int,input().split())
y = x1 - x2
Y = abs(y)
print(Y)
```
| 75,811 | [
-0.030029296875,
0.06671142578125,
-0.12890625,
-0.080078125,
-0.376953125,
-0.479736328125,
-0.007965087890625,
0.1986083984375,
0.19091796875,
1.1572265625,
0.220458984375,
0.2308349609375,
0.2183837890625,
-0.51953125,
-0.029815673828125,
-0.309326171875,
-0.51513671875,
-0.4396... | 3 |
Provide a correct Python 3 solution for this coding contest problem.
It’s still hot every day, but September has already come. It’s autumn according to the calendar. Looking around, I see two red dragonflies at rest on the wall in front of me. It’s autumn indeed.
When two red dragonflies’ positional information as measured from the end of the wall is given, make a program to calculate the distance between their heads.
Input
The input is given in the following format.
$x_1$ $x_2$
The input line provides dragonflies’ head positions $x_1$ and $x_2$ ($0 \leq x_1, x_2 \leq 100$) as integers.
Output
Output the distance between the two red dragonflies in a line.
Examples
Input
20 30
Output
10
Input
50 25
Output
25
Input
25 25
Output
0
"Correct Solution:
```
#!/usr/bin/python3
# coding: utf-8
x,y = map(int,input().split())
s = x-y
if x<y:
s = -s
print(s)
```
| 75,812 | [
-0.043426513671875,
0.018524169921875,
-0.1259765625,
-0.10870361328125,
-0.37353515625,
-0.496826171875,
0.119384765625,
0.2236328125,
0.1212158203125,
1.1259765625,
0.1776123046875,
0.2498779296875,
0.213623046875,
-0.468505859375,
-0.04132080078125,
-0.431640625,
-0.5126953125,
... | 3 |
Provide a correct Python 3 solution for this coding contest problem.
It’s still hot every day, but September has already come. It’s autumn according to the calendar. Looking around, I see two red dragonflies at rest on the wall in front of me. It’s autumn indeed.
When two red dragonflies’ positional information as measured from the end of the wall is given, make a program to calculate the distance between their heads.
Input
The input is given in the following format.
$x_1$ $x_2$
The input line provides dragonflies’ head positions $x_1$ and $x_2$ ($0 \leq x_1, x_2 \leq 100$) as integers.
Output
Output the distance between the two red dragonflies in a line.
Examples
Input
20 30
Output
10
Input
50 25
Output
25
Input
25 25
Output
0
"Correct Solution:
```
x1, x2 = map(int, input().split())
if x1> x2:
print(x1 - x2)
elif x1< x2:
print(x2 - x1)
else:
print("0")
```
| 75,813 | [
-0.03594970703125,
0.0248565673828125,
-0.1573486328125,
-0.0916748046875,
-0.414306640625,
-0.5185546875,
0.01959228515625,
0.299072265625,
0.21923828125,
1.12890625,
0.2322998046875,
0.279296875,
0.2138671875,
-0.52734375,
-0.0472412109375,
-0.329345703125,
-0.54443359375,
-0.439... | 3 |
Provide a correct Python 3 solution for this coding contest problem.
It’s still hot every day, but September has already come. It’s autumn according to the calendar. Looking around, I see two red dragonflies at rest on the wall in front of me. It’s autumn indeed.
When two red dragonflies’ positional information as measured from the end of the wall is given, make a program to calculate the distance between their heads.
Input
The input is given in the following format.
$x_1$ $x_2$
The input line provides dragonflies’ head positions $x_1$ and $x_2$ ($0 \leq x_1, x_2 \leq 100$) as integers.
Output
Output the distance between the two red dragonflies in a line.
Examples
Input
20 30
Output
10
Input
50 25
Output
25
Input
25 25
Output
0
"Correct Solution:
```
x1,x2=map(int,input().split())
if x2>x1:
print(int(x2-x1))
elif x2==x1:
print(int(x2-x1))
else:
print(int(x1-x2))
```
| 75,814 | [
-0.0184783935546875,
0.0188140869140625,
-0.1361083984375,
-0.09539794921875,
-0.411376953125,
-0.52783203125,
0.0145111083984375,
0.275146484375,
0.1907958984375,
1.134765625,
0.272216796875,
0.2998046875,
0.2022705078125,
-0.54638671875,
-0.038116455078125,
-0.32763671875,
-0.55566... | 3 |
Provide a correct Python 3 solution for this coding contest problem.
It’s still hot every day, but September has already come. It’s autumn according to the calendar. Looking around, I see two red dragonflies at rest on the wall in front of me. It’s autumn indeed.
When two red dragonflies’ positional information as measured from the end of the wall is given, make a program to calculate the distance between their heads.
Input
The input is given in the following format.
$x_1$ $x_2$
The input line provides dragonflies’ head positions $x_1$ and $x_2$ ($0 \leq x_1, x_2 \leq 100$) as integers.
Output
Output the distance between the two red dragonflies in a line.
Examples
Input
20 30
Output
10
Input
50 25
Output
25
Input
25 25
Output
0
"Correct Solution:
```
a,b =map(int,input().split())
if a>b:
print(a-b)
elif a==b:
print("0")
else:
print(b-a)
```
| 75,815 | [
-0.017059326171875,
0.0133514404296875,
-0.17822265625,
-0.12213134765625,
-0.42822265625,
-0.50927734375,
0.031463623046875,
0.29443359375,
0.15869140625,
1.1640625,
0.224609375,
0.2412109375,
0.1961669921875,
-0.5205078125,
-0.0478515625,
-0.34375,
-0.57080078125,
-0.393798828125... | 3 |
Provide a correct Python 3 solution for this coding contest problem.
It’s still hot every day, but September has already come. It’s autumn according to the calendar. Looking around, I see two red dragonflies at rest on the wall in front of me. It’s autumn indeed.
When two red dragonflies’ positional information as measured from the end of the wall is given, make a program to calculate the distance between their heads.
Input
The input is given in the following format.
$x_1$ $x_2$
The input line provides dragonflies’ head positions $x_1$ and $x_2$ ($0 \leq x_1, x_2 \leq 100$) as integers.
Output
Output the distance between the two red dragonflies in a line.
Examples
Input
20 30
Output
10
Input
50 25
Output
25
Input
25 25
Output
0
"Correct Solution:
```
a,b=map(int,input().split())
if a<b:
t=a
a=b
b=t
print(a-b)
```
| 75,816 | [
-0.017242431640625,
0.02142333984375,
-0.13671875,
-0.11297607421875,
-0.362060546875,
-0.4833984375,
0.03411865234375,
0.226806640625,
0.13232421875,
1.1806640625,
0.21240234375,
0.2369384765625,
0.195068359375,
-0.5537109375,
-0.0272979736328125,
-0.328125,
-0.53564453125,
-0.434... | 3 |
Provide a correct Python 3 solution for this coding contest problem.
It’s still hot every day, but September has already come. It’s autumn according to the calendar. Looking around, I see two red dragonflies at rest on the wall in front of me. It’s autumn indeed.
When two red dragonflies’ positional information as measured from the end of the wall is given, make a program to calculate the distance between their heads.
Input
The input is given in the following format.
$x_1$ $x_2$
The input line provides dragonflies’ head positions $x_1$ and $x_2$ ($0 \leq x_1, x_2 \leq 100$) as integers.
Output
Output the distance between the two red dragonflies in a line.
Examples
Input
20 30
Output
10
Input
50 25
Output
25
Input
25 25
Output
0
"Correct Solution:
```
a,b=map(int,input().split())
if a>b:
print(a-b)
else:
print(b-a)
```
| 75,817 | [
-0.047393798828125,
0.03509521484375,
-0.178466796875,
-0.120849609375,
-0.379150390625,
-0.5009765625,
0.03143310546875,
0.25146484375,
0.142822265625,
1.185546875,
0.222412109375,
0.24072265625,
0.1983642578125,
-0.55419921875,
-0.030792236328125,
-0.33642578125,
-0.546875,
-0.43... | 3 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
It’s still hot every day, but September has already come. It’s autumn according to the calendar. Looking around, I see two red dragonflies at rest on the wall in front of me. It’s autumn indeed.
When two red dragonflies’ positional information as measured from the end of the wall is given, make a program to calculate the distance between their heads.
Input
The input is given in the following format.
$x_1$ $x_2$
The input line provides dragonflies’ head positions $x_1$ and $x_2$ ($0 \leq x_1, x_2 \leq 100$) as integers.
Output
Output the distance between the two red dragonflies in a line.
Examples
Input
20 30
Output
10
Input
50 25
Output
25
Input
25 25
Output
0
Submitted Solution:
```
a, b= map(int, input().split())
if b > a :
a, b = b, a
print(a - b)
```
Yes
| 75,818 | [
0.0772705078125,
0.025909423828125,
-0.159912109375,
-0.0082855224609375,
-0.419677734375,
-0.4501953125,
-0.10302734375,
0.280517578125,
0.138427734375,
1.138671875,
0.2041015625,
0.2293701171875,
0.1611328125,
-0.458740234375,
-0.08306884765625,
-0.48486328125,
-0.484619140625,
-... | 3 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
It’s still hot every day, but September has already come. It’s autumn according to the calendar. Looking around, I see two red dragonflies at rest on the wall in front of me. It’s autumn indeed.
When two red dragonflies’ positional information as measured from the end of the wall is given, make a program to calculate the distance between their heads.
Input
The input is given in the following format.
$x_1$ $x_2$
The input line provides dragonflies’ head positions $x_1$ and $x_2$ ($0 \leq x_1, x_2 \leq 100$) as integers.
Output
Output the distance between the two red dragonflies in a line.
Examples
Input
20 30
Output
10
Input
50 25
Output
25
Input
25 25
Output
0
Submitted Solution:
```
a,b=map(int,input().split())
if a-b==0:
print(0)
elif a>b:
print(a-b)
else:
print(b-a)
```
Yes
| 75,819 | [
0.07550048828125,
0.0283660888671875,
-0.1944580078125,
-0.03045654296875,
-0.427001953125,
-0.458251953125,
-0.0863037109375,
0.31591796875,
0.1326904296875,
1.158203125,
0.18408203125,
0.24267578125,
0.1607666015625,
-0.440185546875,
-0.0760498046875,
-0.46875,
-0.50732421875,
-0... | 3 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
It’s still hot every day, but September has already come. It’s autumn according to the calendar. Looking around, I see two red dragonflies at rest on the wall in front of me. It’s autumn indeed.
When two red dragonflies’ positional information as measured from the end of the wall is given, make a program to calculate the distance between their heads.
Input
The input is given in the following format.
$x_1$ $x_2$
The input line provides dragonflies’ head positions $x_1$ and $x_2$ ($0 \leq x_1, x_2 \leq 100$) as integers.
Output
Output the distance between the two red dragonflies in a line.
Examples
Input
20 30
Output
10
Input
50 25
Output
25
Input
25 25
Output
0
Submitted Solution:
```
a,b=map(int,input().split())
if a<b:
x=b-a
else:
x=a-b
print(x)
```
Yes
| 75,820 | [
0.080078125,
0.023223876953125,
-0.1427001953125,
-0.00569915771484375,
-0.39794921875,
-0.456787109375,
-0.10791015625,
0.270263671875,
0.1378173828125,
1.1572265625,
0.2374267578125,
0.237060546875,
0.1533203125,
-0.45751953125,
-0.07476806640625,
-0.457763671875,
-0.49267578125,
... | 3 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
It’s still hot every day, but September has already come. It’s autumn according to the calendar. Looking around, I see two red dragonflies at rest on the wall in front of me. It’s autumn indeed.
When two red dragonflies’ positional information as measured from the end of the wall is given, make a program to calculate the distance between their heads.
Input
The input is given in the following format.
$x_1$ $x_2$
The input line provides dragonflies’ head positions $x_1$ and $x_2$ ($0 \leq x_1, x_2 \leq 100$) as integers.
Output
Output the distance between the two red dragonflies in a line.
Examples
Input
20 30
Output
10
Input
50 25
Output
25
Input
25 25
Output
0
Submitted Solution:
```
a,b = input().split()
c=int(int(a)-int(b))
if c<0 : c=c*(-1)
print(c)
```
Yes
| 75,821 | [
0.0867919921875,
0.0048370361328125,
-0.098388671875,
-0.0122833251953125,
-0.426025390625,
-0.481689453125,
-0.0628662109375,
0.295654296875,
0.1209716796875,
1.1201171875,
0.20166015625,
0.2344970703125,
0.1044921875,
-0.46923828125,
-0.1005859375,
-0.474609375,
-0.501953125,
-0.... | 3 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
It’s still hot every day, but September has already come. It’s autumn according to the calendar. Looking around, I see two red dragonflies at rest on the wall in front of me. It’s autumn indeed.
When two red dragonflies’ positional information as measured from the end of the wall is given, make a program to calculate the distance between their heads.
Input
The input is given in the following format.
$x_1$ $x_2$
The input line provides dragonflies’ head positions $x_1$ and $x_2$ ($0 \leq x_1, x_2 \leq 100$) as integers.
Output
Output the distance between the two red dragonflies in a line.
Examples
Input
20 30
Output
10
Input
50 25
Output
25
Input
25 25
Output
0
Submitted Solution:
```
a,b=map(int,split())
print(abs(a-b))
```
No
| 75,822 | [
0.08612060546875,
0.001949310302734375,
-0.1678466796875,
-0.017120361328125,
-0.380126953125,
-0.461669921875,
-0.13037109375,
0.259765625,
0.15283203125,
1.1220703125,
0.2188720703125,
0.17626953125,
0.17333984375,
-0.427734375,
-0.056793212890625,
-0.46240234375,
-0.4765625,
-0.... | 3 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
In recent years John has very successfully settled at his new job at the office. But John doesn't like to idly sit around while his code is compiling, so he immediately found himself an interesting distraction. The point of his distraction was to maintain a water level in the water cooler used by other zebras.
<image>
Originally the cooler contained exactly k liters of water. John decided that the amount of water must always be at least l liters of water but no more than r liters. John will stay at the office for exactly t days. He knows that each day exactly x liters of water will be used by his colleagues. At the beginning of each day he can add exactly y liters of water to the cooler, but at any point in time the amount of water in the cooler must be in the range [l, r].
Now John wants to find out whether he will be able to maintain the water level at the necessary level for t days. Help him answer this question!
Input
The first line of the input contains six integers k, l, r, t, x and y (1 ≤ l ≤ k ≤ r ≤ 10^{18}; 1 ≤ t ≤ 10^{18}; 1 ≤ x ≤ 10^6; 1 ≤ y ≤ 10^{18}) — initial water level, the required range, the number of days, daily water usage and the exact amount of water that can be added, respectively.
Output
Print "Yes" if John can maintain the water level for t days and "No" otherwise.
Examples
Input
8 1 10 2 6 4
Output
No
Input
8 1 10 2 6 5
Output
Yes
Input
9 1 10 9 2 9
Output
No
Input
20 15 25 3 5 7
Output
Yes
Note
In the first example, John can't increase the amount of water at the beginning of the first day, since it would exceed the limit r. That is why after the first day the cooler will contain 2 liters of water. The next day John adds 4 liters to the cooler but loses 6 liters, leaving John with 0 liters, which is outside the range [1, 10].
In the second example, after the first day John is left with 2 liters of water. At the beginning of the next day he adds 5 liters, then 6 liters get used, leaving John with 1 liter of water which is in range [1, 10].
In the third example, after the first day John is left with 7 liters, after the second day — 5 liters, after the fourth — 1 liter. At the beginning of the fifth day John will add 9 liters and lose 2 liters. Meaning, after the fifth day he will have 8 liters left. Then each day the water level will decrease by 2 liters and after the eighth day John will have 2 liters and after the ninth day — 0 liters. 0 is outside range [1, 10], so the answer is "No".
In the fourth example, after the first day John is left with 15 liters of water. At the beginning of the second day he adds 7 liters and loses 5, so after the second day he is left with 17 liters. At the beginning of the third day he adds 7 more liters of water and loses 5, so after the third day he is left with 19 liters. 19 is in range [15, 25] so the answer is "Yes".
Submitted Solution:
```
mystr = input()
counter = 0
result = mystr.split(" ")
k = int(result[0])
l = int(result[1])
r = int(result[2])
t = int(result[3])
x = int(result[4])
y = int(result[5])
max_step_up = (l-r-1)//y
max_step_down = (l-r-1)//x
#print("max_step_up", max_step_up)
#print("max_step_down", max_step_down)
water_level = k
difference = y - x
days_last = (k - l)//x
max_level_can_add = r - y
if k+y <=r:
end_level = k + t*difference
else:
end_level = k - x + (t-1)*difference
if y <= x:
if (end_level >= l) or ():
print("YES")
else:
print("NO")
else:
if ((k + y) > r) and ((k - x) < l):
print("NO")
elif (l + y > r) and (x*t > (k-l)):
print("NO")
elif k == 999984:
if t<999984:
print("YES")
else:
print("NO")
elif k == 753375750984333178:
print("YES")
elif k == 306065266455142157:
print("YES")
elif k == 647896210045108564:
print("YES") #71
elif k == 376674986861816384:
print("YES") #76
elif k == 422834873810910204:
print("YES")
elif (k == 1) and (l ==1) and (r == 1000000000000000000) and (t == 1000000000000000000):
print("YES") #88
elif k == 999999999999999900:
print("YES")
elif k == 512151145295769976:
print("YES")
else:
counter = 0
overload_flag = False
while (water_level <= r) and (water_level>=l) and (counter < t) and overload_flag != True:
counter += 1
step = min([t-counter+1,(water_level-l)//x])
if step > 0:
water_level = water_level - x*step
counter += step-1
#print("step = ", step)
#print("liter step = ", (step*x))
#print("counter = ", counter)
else:
water_level = water_level + y
if water_level > r: overload_flag = True
if l + y > r: overload_flag = True
water_level = water_level - x
if (counter == t) and (water_level <= r) and overload_flag != True:
print("YES")
else:
print("NO")
```
Yes
| 76,096 | [
0.457763671875,
0.2509765625,
0.0120849609375,
0.10491943359375,
-0.62939453125,
-0.3173828125,
0.2626953125,
0.44140625,
0.145263671875,
1.25,
0.3330078125,
-0.302001953125,
-0.1368408203125,
-0.6103515625,
-0.07318115234375,
-0.0904541015625,
-0.734375,
-0.80029296875,
-0.59667... | 3 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
In recent years John has very successfully settled at his new job at the office. But John doesn't like to idly sit around while his code is compiling, so he immediately found himself an interesting distraction. The point of his distraction was to maintain a water level in the water cooler used by other zebras.
<image>
Originally the cooler contained exactly k liters of water. John decided that the amount of water must always be at least l liters of water but no more than r liters. John will stay at the office for exactly t days. He knows that each day exactly x liters of water will be used by his colleagues. At the beginning of each day he can add exactly y liters of water to the cooler, but at any point in time the amount of water in the cooler must be in the range [l, r].
Now John wants to find out whether he will be able to maintain the water level at the necessary level for t days. Help him answer this question!
Input
The first line of the input contains six integers k, l, r, t, x and y (1 ≤ l ≤ k ≤ r ≤ 10^{18}; 1 ≤ t ≤ 10^{18}; 1 ≤ x ≤ 10^6; 1 ≤ y ≤ 10^{18}) — initial water level, the required range, the number of days, daily water usage and the exact amount of water that can be added, respectively.
Output
Print "Yes" if John can maintain the water level for t days and "No" otherwise.
Examples
Input
8 1 10 2 6 4
Output
No
Input
8 1 10 2 6 5
Output
Yes
Input
9 1 10 9 2 9
Output
No
Input
20 15 25 3 5 7
Output
Yes
Note
In the first example, John can't increase the amount of water at the beginning of the first day, since it would exceed the limit r. That is why after the first day the cooler will contain 2 liters of water. The next day John adds 4 liters to the cooler but loses 6 liters, leaving John with 0 liters, which is outside the range [1, 10].
In the second example, after the first day John is left with 2 liters of water. At the beginning of the next day he adds 5 liters, then 6 liters get used, leaving John with 1 liter of water which is in range [1, 10].
In the third example, after the first day John is left with 7 liters, after the second day — 5 liters, after the fourth — 1 liter. At the beginning of the fifth day John will add 9 liters and lose 2 liters. Meaning, after the fifth day he will have 8 liters left. Then each day the water level will decrease by 2 liters and after the eighth day John will have 2 liters and after the ninth day — 0 liters. 0 is outside range [1, 10], so the answer is "No".
In the fourth example, after the first day John is left with 15 liters of water. At the beginning of the second day he adds 7 liters and loses 5, so after the second day he is left with 17 liters. At the beginning of the third day he adds 7 more liters of water and loses 5, so after the third day he is left with 19 liters. 19 is in range [15, 25] so the answer is "Yes".
Submitted Solution:
```
k, l, r, t, x, y = list(map(int, input().split()))
import sys
if y < x:
if k + y <= r:
k += y
k -= x
nb_days = int((k - l) / (x - y)) + 1
print("Yes" if nb_days >= t else "No")
elif x < y:
memo = {}
nb_days = 0
if (k - l) % x == k - l:
k += y
if k > r:
print("No")
sys.exit()
t2 = t
while True:
nb_days += (k - l) // x
k = (k - l) % x + l
if k in memo:
print("Yes")
break
memo[k] = True
if k + y > r:
if nb_days < t:
print("No")
break
else:
print("Yes")
break
else:
k += y
else:
if k + y <= r:
print("Yes")
elif k - y >= l:
print("Yes")
else:
print("No")
```
Yes
| 76,097 | [
0.457763671875,
0.2509765625,
0.0120849609375,
0.10491943359375,
-0.62939453125,
-0.3173828125,
0.2626953125,
0.44140625,
0.145263671875,
1.25,
0.3330078125,
-0.302001953125,
-0.1368408203125,
-0.6103515625,
-0.07318115234375,
-0.0904541015625,
-0.734375,
-0.80029296875,
-0.59667... | 3 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
In recent years John has very successfully settled at his new job at the office. But John doesn't like to idly sit around while his code is compiling, so he immediately found himself an interesting distraction. The point of his distraction was to maintain a water level in the water cooler used by other zebras.
<image>
Originally the cooler contained exactly k liters of water. John decided that the amount of water must always be at least l liters of water but no more than r liters. John will stay at the office for exactly t days. He knows that each day exactly x liters of water will be used by his colleagues. At the beginning of each day he can add exactly y liters of water to the cooler, but at any point in time the amount of water in the cooler must be in the range [l, r].
Now John wants to find out whether he will be able to maintain the water level at the necessary level for t days. Help him answer this question!
Input
The first line of the input contains six integers k, l, r, t, x and y (1 ≤ l ≤ k ≤ r ≤ 10^{18}; 1 ≤ t ≤ 10^{18}; 1 ≤ x ≤ 10^6; 1 ≤ y ≤ 10^{18}) — initial water level, the required range, the number of days, daily water usage and the exact amount of water that can be added, respectively.
Output
Print "Yes" if John can maintain the water level for t days and "No" otherwise.
Examples
Input
8 1 10 2 6 4
Output
No
Input
8 1 10 2 6 5
Output
Yes
Input
9 1 10 9 2 9
Output
No
Input
20 15 25 3 5 7
Output
Yes
Note
In the first example, John can't increase the amount of water at the beginning of the first day, since it would exceed the limit r. That is why after the first day the cooler will contain 2 liters of water. The next day John adds 4 liters to the cooler but loses 6 liters, leaving John with 0 liters, which is outside the range [1, 10].
In the second example, after the first day John is left with 2 liters of water. At the beginning of the next day he adds 5 liters, then 6 liters get used, leaving John with 1 liter of water which is in range [1, 10].
In the third example, after the first day John is left with 7 liters, after the second day — 5 liters, after the fourth — 1 liter. At the beginning of the fifth day John will add 9 liters and lose 2 liters. Meaning, after the fifth day he will have 8 liters left. Then each day the water level will decrease by 2 liters and after the eighth day John will have 2 liters and after the ninth day — 0 liters. 0 is outside range [1, 10], so the answer is "No".
In the fourth example, after the first day John is left with 15 liters of water. At the beginning of the second day he adds 7 liters and loses 5, so after the second day he is left with 17 liters. At the beginning of the third day he adds 7 more liters of water and loses 5, so after the third day he is left with 19 liters. 19 is in range [15, 25] so the answer is "Yes".
Submitted Solution:
```
def solve():
k, l, r, t, x, y = map(int, input().split());k -= l;r -= l
if y < x:
if k + y > r:k -= x;t -= 1
k -= (x - y) * t;return k >= 0
if y + x - 1 <= r:return True
if y > r:k -= x * t;return k >= 0
t -= k // x;k %= x;seen = {k}
while t > 0:
k += y
if k > r:return False
t -= k // x;k %= x
if k in seen:return True
seen.add(k)
return True
print("Yes" if solve() else "No")
```
Yes
| 76,098 | [
0.457763671875,
0.2509765625,
0.0120849609375,
0.10491943359375,
-0.62939453125,
-0.3173828125,
0.2626953125,
0.44140625,
0.145263671875,
1.25,
0.3330078125,
-0.302001953125,
-0.1368408203125,
-0.6103515625,
-0.07318115234375,
-0.0904541015625,
-0.734375,
-0.80029296875,
-0.59667... | 3 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
In recent years John has very successfully settled at his new job at the office. But John doesn't like to idly sit around while his code is compiling, so he immediately found himself an interesting distraction. The point of his distraction was to maintain a water level in the water cooler used by other zebras.
<image>
Originally the cooler contained exactly k liters of water. John decided that the amount of water must always be at least l liters of water but no more than r liters. John will stay at the office for exactly t days. He knows that each day exactly x liters of water will be used by his colleagues. At the beginning of each day he can add exactly y liters of water to the cooler, but at any point in time the amount of water in the cooler must be in the range [l, r].
Now John wants to find out whether he will be able to maintain the water level at the necessary level for t days. Help him answer this question!
Input
The first line of the input contains six integers k, l, r, t, x and y (1 ≤ l ≤ k ≤ r ≤ 10^{18}; 1 ≤ t ≤ 10^{18}; 1 ≤ x ≤ 10^6; 1 ≤ y ≤ 10^{18}) — initial water level, the required range, the number of days, daily water usage and the exact amount of water that can be added, respectively.
Output
Print "Yes" if John can maintain the water level for t days and "No" otherwise.
Examples
Input
8 1 10 2 6 4
Output
No
Input
8 1 10 2 6 5
Output
Yes
Input
9 1 10 9 2 9
Output
No
Input
20 15 25 3 5 7
Output
Yes
Note
In the first example, John can't increase the amount of water at the beginning of the first day, since it would exceed the limit r. That is why after the first day the cooler will contain 2 liters of water. The next day John adds 4 liters to the cooler but loses 6 liters, leaving John with 0 liters, which is outside the range [1, 10].
In the second example, after the first day John is left with 2 liters of water. At the beginning of the next day he adds 5 liters, then 6 liters get used, leaving John with 1 liter of water which is in range [1, 10].
In the third example, after the first day John is left with 7 liters, after the second day — 5 liters, after the fourth — 1 liter. At the beginning of the fifth day John will add 9 liters and lose 2 liters. Meaning, after the fifth day he will have 8 liters left. Then each day the water level will decrease by 2 liters and after the eighth day John will have 2 liters and after the ninth day — 0 liters. 0 is outside range [1, 10], so the answer is "No".
In the fourth example, after the first day John is left with 15 liters of water. At the beginning of the second day he adds 7 liters and loses 5, so after the second day he is left with 17 liters. At the beginning of the third day he adds 7 more liters of water and loses 5, so after the third day he is left with 19 liters. 19 is in range [15, 25] so the answer is "Yes".
Submitted Solution:
```
vis = [0] * (1000010)
k, l, r, t, x, y = [int(i) for i in input().split()]
if x > y:
if k < l or k > r:
print("No")
exit(0)
if k + y > r:
k -= x
t -= 1
if k < l:
print("No")
exit(0)
if (k - l) // (x - y) >= t: print("Yes")
else: print("NO")
else:
k -= l
r -= l
now = k
if k < 0 or k > r:
print("No")
exit(0)
while 1:
t1 = now // x
now -= t1 * x
t -= t1
if t <= 0 or vis[now]:
print("Yes")
exit(0)
vis[now] = 1
now += y
if now > r:
print("No")
exit(0)
print("No")
```
Yes
| 76,099 | [
0.457763671875,
0.2509765625,
0.0120849609375,
0.10491943359375,
-0.62939453125,
-0.3173828125,
0.2626953125,
0.44140625,
0.145263671875,
1.25,
0.3330078125,
-0.302001953125,
-0.1368408203125,
-0.6103515625,
-0.07318115234375,
-0.0904541015625,
-0.734375,
-0.80029296875,
-0.59667... | 3 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
In recent years John has very successfully settled at his new job at the office. But John doesn't like to idly sit around while his code is compiling, so he immediately found himself an interesting distraction. The point of his distraction was to maintain a water level in the water cooler used by other zebras.
<image>
Originally the cooler contained exactly k liters of water. John decided that the amount of water must always be at least l liters of water but no more than r liters. John will stay at the office for exactly t days. He knows that each day exactly x liters of water will be used by his colleagues. At the beginning of each day he can add exactly y liters of water to the cooler, but at any point in time the amount of water in the cooler must be in the range [l, r].
Now John wants to find out whether he will be able to maintain the water level at the necessary level for t days. Help him answer this question!
Input
The first line of the input contains six integers k, l, r, t, x and y (1 ≤ l ≤ k ≤ r ≤ 10^{18}; 1 ≤ t ≤ 10^{18}; 1 ≤ x ≤ 10^6; 1 ≤ y ≤ 10^{18}) — initial water level, the required range, the number of days, daily water usage and the exact amount of water that can be added, respectively.
Output
Print "Yes" if John can maintain the water level for t days and "No" otherwise.
Examples
Input
8 1 10 2 6 4
Output
No
Input
8 1 10 2 6 5
Output
Yes
Input
9 1 10 9 2 9
Output
No
Input
20 15 25 3 5 7
Output
Yes
Note
In the first example, John can't increase the amount of water at the beginning of the first day, since it would exceed the limit r. That is why after the first day the cooler will contain 2 liters of water. The next day John adds 4 liters to the cooler but loses 6 liters, leaving John with 0 liters, which is outside the range [1, 10].
In the second example, after the first day John is left with 2 liters of water. At the beginning of the next day he adds 5 liters, then 6 liters get used, leaving John with 1 liter of water which is in range [1, 10].
In the third example, after the first day John is left with 7 liters, after the second day — 5 liters, after the fourth — 1 liter. At the beginning of the fifth day John will add 9 liters and lose 2 liters. Meaning, after the fifth day he will have 8 liters left. Then each day the water level will decrease by 2 liters and after the eighth day John will have 2 liters and after the ninth day — 0 liters. 0 is outside range [1, 10], so the answer is "No".
In the fourth example, after the first day John is left with 15 liters of water. At the beginning of the second day he adds 7 liters and loses 5, so after the second day he is left with 17 liters. At the beginning of the third day he adds 7 more liters of water and loses 5, so after the third day he is left with 19 liters. 19 is in range [15, 25] so the answer is "Yes".
Submitted Solution:
```
import math
k,l,r,t,x,y = map(int,input().split(' '))
k-=l
r-=l
if y>x:
t1 = k//x
k%=x
while t1 <= t:
if r-k<y:
print('no')
break
else:
k = (k+y)%x
t1 += (k+y)//x
else: print('yes')
elif y==x:
if k//x>=t or r-k%x>=y: print('yes')
else: print('no')
else:
t1 = math.ceil((y-r+k)/x)
t1 += (k-t1*x)//(x-y)
if t1>=t: print('yes')
else: print('no')
```
No
| 76,100 | [
0.457763671875,
0.2509765625,
0.0120849609375,
0.10491943359375,
-0.62939453125,
-0.3173828125,
0.2626953125,
0.44140625,
0.145263671875,
1.25,
0.3330078125,
-0.302001953125,
-0.1368408203125,
-0.6103515625,
-0.07318115234375,
-0.0904541015625,
-0.734375,
-0.80029296875,
-0.59667... | 3 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
In recent years John has very successfully settled at his new job at the office. But John doesn't like to idly sit around while his code is compiling, so he immediately found himself an interesting distraction. The point of his distraction was to maintain a water level in the water cooler used by other zebras.
<image>
Originally the cooler contained exactly k liters of water. John decided that the amount of water must always be at least l liters of water but no more than r liters. John will stay at the office for exactly t days. He knows that each day exactly x liters of water will be used by his colleagues. At the beginning of each day he can add exactly y liters of water to the cooler, but at any point in time the amount of water in the cooler must be in the range [l, r].
Now John wants to find out whether he will be able to maintain the water level at the necessary level for t days. Help him answer this question!
Input
The first line of the input contains six integers k, l, r, t, x and y (1 ≤ l ≤ k ≤ r ≤ 10^{18}; 1 ≤ t ≤ 10^{18}; 1 ≤ x ≤ 10^6; 1 ≤ y ≤ 10^{18}) — initial water level, the required range, the number of days, daily water usage and the exact amount of water that can be added, respectively.
Output
Print "Yes" if John can maintain the water level for t days and "No" otherwise.
Examples
Input
8 1 10 2 6 4
Output
No
Input
8 1 10 2 6 5
Output
Yes
Input
9 1 10 9 2 9
Output
No
Input
20 15 25 3 5 7
Output
Yes
Note
In the first example, John can't increase the amount of water at the beginning of the first day, since it would exceed the limit r. That is why after the first day the cooler will contain 2 liters of water. The next day John adds 4 liters to the cooler but loses 6 liters, leaving John with 0 liters, which is outside the range [1, 10].
In the second example, after the first day John is left with 2 liters of water. At the beginning of the next day he adds 5 liters, then 6 liters get used, leaving John with 1 liter of water which is in range [1, 10].
In the third example, after the first day John is left with 7 liters, after the second day — 5 liters, after the fourth — 1 liter. At the beginning of the fifth day John will add 9 liters and lose 2 liters. Meaning, after the fifth day he will have 8 liters left. Then each day the water level will decrease by 2 liters and after the eighth day John will have 2 liters and after the ninth day — 0 liters. 0 is outside range [1, 10], so the answer is "No".
In the fourth example, after the first day John is left with 15 liters of water. At the beginning of the second day he adds 7 liters and loses 5, so after the second day he is left with 17 liters. At the beginning of the third day he adds 7 more liters of water and loses 5, so after the third day he is left with 19 liters. 19 is in range [15, 25] so the answer is "Yes".
Submitted Solution:
```
# -*- coding: utf-8 -*-
import sys
from collections import deque, defaultdict, namedtuple
import heapq
from math import sqrt, factorial, gcd, ceil, atan, pi
from itertools import permutations
# def input(): return sys.stdin.readline().strip()
# def input(): return sys.stdin.buffer.readline()[:-1] # warning bytes
# def input(): return sys.stdin.buffer.readline().strip() # warning bytes
def input(): return sys.stdin.buffer.readline().decode('utf-8').strip()
import string
import operator
import random
# string.ascii_lowercase
from bisect import bisect_left, bisect_right
from functools import lru_cache, reduce
MOD = int(1e9)+7
INF = float('inf')
# sys.setrecursionlimit(MOD)
def solve():
k, l, r, t, x, y = [int(x) for x in input().split()]
ok = 1
if k + y > r and k - x < l:
ok = 0
if x > y:
if k + y > r:
t -= 1
k -= x
d = y - x
if k + (d * t) < l:
ok = 0
else:
d = k - l
if l + x - 1 + y > r:
m = d % x
if l + m + y > r:
ok = 0
cnt = d // x
if cnt < t:
t -= cnt
cnt = (m + y) // x
m = (m + y) % x
if cnt < t and l + m + y > r:
ok = 0
if ok:
print("Yes")
else:
print("No")
T = 1
# T = int(input())
for case in range(1,T+1):
ans = solve()
"""
"""
```
No
| 76,101 | [
0.457763671875,
0.2509765625,
0.0120849609375,
0.10491943359375,
-0.62939453125,
-0.3173828125,
0.2626953125,
0.44140625,
0.145263671875,
1.25,
0.3330078125,
-0.302001953125,
-0.1368408203125,
-0.6103515625,
-0.07318115234375,
-0.0904541015625,
-0.734375,
-0.80029296875,
-0.59667... | 3 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
In recent years John has very successfully settled at his new job at the office. But John doesn't like to idly sit around while his code is compiling, so he immediately found himself an interesting distraction. The point of his distraction was to maintain a water level in the water cooler used by other zebras.
<image>
Originally the cooler contained exactly k liters of water. John decided that the amount of water must always be at least l liters of water but no more than r liters. John will stay at the office for exactly t days. He knows that each day exactly x liters of water will be used by his colleagues. At the beginning of each day he can add exactly y liters of water to the cooler, but at any point in time the amount of water in the cooler must be in the range [l, r].
Now John wants to find out whether he will be able to maintain the water level at the necessary level for t days. Help him answer this question!
Input
The first line of the input contains six integers k, l, r, t, x and y (1 ≤ l ≤ k ≤ r ≤ 10^{18}; 1 ≤ t ≤ 10^{18}; 1 ≤ x ≤ 10^6; 1 ≤ y ≤ 10^{18}) — initial water level, the required range, the number of days, daily water usage and the exact amount of water that can be added, respectively.
Output
Print "Yes" if John can maintain the water level for t days and "No" otherwise.
Examples
Input
8 1 10 2 6 4
Output
No
Input
8 1 10 2 6 5
Output
Yes
Input
9 1 10 9 2 9
Output
No
Input
20 15 25 3 5 7
Output
Yes
Note
In the first example, John can't increase the amount of water at the beginning of the first day, since it would exceed the limit r. That is why after the first day the cooler will contain 2 liters of water. The next day John adds 4 liters to the cooler but loses 6 liters, leaving John with 0 liters, which is outside the range [1, 10].
In the second example, after the first day John is left with 2 liters of water. At the beginning of the next day he adds 5 liters, then 6 liters get used, leaving John with 1 liter of water which is in range [1, 10].
In the third example, after the first day John is left with 7 liters, after the second day — 5 liters, after the fourth — 1 liter. At the beginning of the fifth day John will add 9 liters and lose 2 liters. Meaning, after the fifth day he will have 8 liters left. Then each day the water level will decrease by 2 liters and after the eighth day John will have 2 liters and after the ninth day — 0 liters. 0 is outside range [1, 10], so the answer is "No".
In the fourth example, after the first day John is left with 15 liters of water. At the beginning of the second day he adds 7 liters and loses 5, so after the second day he is left with 17 liters. At the beginning of the third day he adds 7 more liters of water and loses 5, so after the third day he is left with 19 liters. 19 is in range [15, 25] so the answer is "Yes".
Submitted Solution:
```
k, l, r, t, x, y = list(map(int,input().split()))
border = min(1000000, t)
t -= border
for i in range(border):
if k + y <= r:
k += y
k -= x
if k < l or k > r:
print('No')
exit(0)
if x > y:
k -= (x - y) * t
if k < l:
print('No')
else:
print('Yes')
exit(0)
if t == 0:
print('Yes')
exit(0)
itr = 0
while t > 0 and itr < 1000000:
rem = r - k
itr += 1
dlt = max(0, y - rem)
cnt = (dlt + x - 1) // x
if cnt >= t:
if k - x * t < l:
print('No')
else:
print('Yes')
exit(0)
t -= cnt
k -= x * cnt
cur_value = k
k += y
print('Yes')
```
No
| 76,102 | [
0.457763671875,
0.2509765625,
0.0120849609375,
0.10491943359375,
-0.62939453125,
-0.3173828125,
0.2626953125,
0.44140625,
0.145263671875,
1.25,
0.3330078125,
-0.302001953125,
-0.1368408203125,
-0.6103515625,
-0.07318115234375,
-0.0904541015625,
-0.734375,
-0.80029296875,
-0.59667... | 3 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
In recent years John has very successfully settled at his new job at the office. But John doesn't like to idly sit around while his code is compiling, so he immediately found himself an interesting distraction. The point of his distraction was to maintain a water level in the water cooler used by other zebras.
<image>
Originally the cooler contained exactly k liters of water. John decided that the amount of water must always be at least l liters of water but no more than r liters. John will stay at the office for exactly t days. He knows that each day exactly x liters of water will be used by his colleagues. At the beginning of each day he can add exactly y liters of water to the cooler, but at any point in time the amount of water in the cooler must be in the range [l, r].
Now John wants to find out whether he will be able to maintain the water level at the necessary level for t days. Help him answer this question!
Input
The first line of the input contains six integers k, l, r, t, x and y (1 ≤ l ≤ k ≤ r ≤ 10^{18}; 1 ≤ t ≤ 10^{18}; 1 ≤ x ≤ 10^6; 1 ≤ y ≤ 10^{18}) — initial water level, the required range, the number of days, daily water usage and the exact amount of water that can be added, respectively.
Output
Print "Yes" if John can maintain the water level for t days and "No" otherwise.
Examples
Input
8 1 10 2 6 4
Output
No
Input
8 1 10 2 6 5
Output
Yes
Input
9 1 10 9 2 9
Output
No
Input
20 15 25 3 5 7
Output
Yes
Note
In the first example, John can't increase the amount of water at the beginning of the first day, since it would exceed the limit r. That is why after the first day the cooler will contain 2 liters of water. The next day John adds 4 liters to the cooler but loses 6 liters, leaving John with 0 liters, which is outside the range [1, 10].
In the second example, after the first day John is left with 2 liters of water. At the beginning of the next day he adds 5 liters, then 6 liters get used, leaving John with 1 liter of water which is in range [1, 10].
In the third example, after the first day John is left with 7 liters, after the second day — 5 liters, after the fourth — 1 liter. At the beginning of the fifth day John will add 9 liters and lose 2 liters. Meaning, after the fifth day he will have 8 liters left. Then each day the water level will decrease by 2 liters and after the eighth day John will have 2 liters and after the ninth day — 0 liters. 0 is outside range [1, 10], so the answer is "No".
In the fourth example, after the first day John is left with 15 liters of water. At the beginning of the second day he adds 7 liters and loses 5, so after the second day he is left with 17 liters. At the beginning of the third day he adds 7 more liters of water and loses 5, so after the third day he is left with 19 liters. 19 is in range [15, 25] so the answer is "Yes".
Submitted Solution:
```
s = input().split(' ')
k = int(s[0])
l = int(s[1])
r = int(s[2])
t = int(s[3])
x = int(s[4])
y = int(s[5])
def exgcd(a, b):
if b == 0:
return (a, 1, 0)
d, x, y = exgcd(b, a % b)
return (d, y, t - a // b * y)
if x >= y:
if k + y > r:
print("No" if k + y * (t - 1) - x * t < l else "Yes")
else:
print("No" if k + y * t - x * t < l else "Yes")
else:
if r - l - x >= y:
print("Yes")
else:
L = ((l - k + x) % y + y) % y
if L != 0 and L + r - l - x < y:
print("No")
exit(0)
for g in range(1, x + 1):
if (g if g else x) + r - l - x >= y:
continue
#a * x + b * y == g - l + k
d, a, b = exgcd(x, y)
if (g - l + k) % d:
continue
if a <= 0:
a = (a % (y // d) + (y // d)) % (y // d)
if a == 0:
a += y // d
if g == 0:
a += 1
if a <= t:
print("No")
exit(0)
print("Yes")
```
No
| 76,103 | [
0.457763671875,
0.2509765625,
0.0120849609375,
0.10491943359375,
-0.62939453125,
-0.3173828125,
0.2626953125,
0.44140625,
0.145263671875,
1.25,
0.3330078125,
-0.302001953125,
-0.1368408203125,
-0.6103515625,
-0.07318115234375,
-0.0904541015625,
-0.734375,
-0.80029296875,
-0.59667... | 3 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
An army of n droids is lined up in one row. Each droid is described by m integers a1, a2, ..., am, where ai is the number of details of the i-th type in this droid's mechanism. R2-D2 wants to destroy the sequence of consecutive droids of maximum length. He has m weapons, the i-th weapon can affect all the droids in the army by destroying one detail of the i-th type (if the droid doesn't have details of this type, nothing happens to it).
A droid is considered to be destroyed when all of its details are destroyed. R2-D2 can make at most k shots. How many shots from the weapon of what type should R2-D2 make to destroy the sequence of consecutive droids of maximum length?
Input
The first line contains three integers n, m, k (1 ≤ n ≤ 105, 1 ≤ m ≤ 5, 0 ≤ k ≤ 109) — the number of droids, the number of detail types and the number of available shots, respectively.
Next n lines follow describing the droids. Each line contains m integers a1, a2, ..., am (0 ≤ ai ≤ 108), where ai is the number of details of the i-th type for the respective robot.
Output
Print m space-separated integers, where the i-th number is the number of shots from the weapon of the i-th type that the robot should make to destroy the subsequence of consecutive droids of the maximum length.
If there are multiple optimal solutions, print any of them.
It is not necessary to make exactly k shots, the number of shots can be less.
Examples
Input
5 2 4
4 0
1 2
2 1
0 2
1 3
Output
2 2
Input
3 2 4
1 2
1 3
2 2
Output
1 3
Note
In the first test the second, third and fourth droids will be destroyed.
In the second test the first and second droids will be destroyed.
Submitted Solution:
```
f = lambda: map(int, input().split())
n, m, k = f()
p = [0] * m
d, x = -1, 0
q = [[] for i in range(m)]
for y in range(n):
t = list(f())
s = 0
for a, b in zip(q, t):
while a and a[-1][0] < b: a.pop()
a.append((b, y))
s += a[0][0]
if s > k:
while s > k:
s = 0
for a in q:
if a and a[0][1] == x: a.pop(0)
if a: s += a[0][0]
x += 1
elif d < y - x:
d = y - x
p = [a[0][0] for a in q]
for i in p: print(i)
```
Yes
| 76,259 | [
0.353759765625,
0.3310546875,
0.1290283203125,
0.5390625,
-0.7265625,
-0.873046875,
-0.50048828125,
0.1832275390625,
0.264892578125,
1.0185546875,
0.46484375,
0.190673828125,
-0.0782470703125,
-0.42724609375,
-0.331787109375,
0.429931640625,
-0.314208984375,
-0.53173828125,
-0.10... | 3 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
An army of n droids is lined up in one row. Each droid is described by m integers a1, a2, ..., am, where ai is the number of details of the i-th type in this droid's mechanism. R2-D2 wants to destroy the sequence of consecutive droids of maximum length. He has m weapons, the i-th weapon can affect all the droids in the army by destroying one detail of the i-th type (if the droid doesn't have details of this type, nothing happens to it).
A droid is considered to be destroyed when all of its details are destroyed. R2-D2 can make at most k shots. How many shots from the weapon of what type should R2-D2 make to destroy the sequence of consecutive droids of maximum length?
Input
The first line contains three integers n, m, k (1 ≤ n ≤ 105, 1 ≤ m ≤ 5, 0 ≤ k ≤ 109) — the number of droids, the number of detail types and the number of available shots, respectively.
Next n lines follow describing the droids. Each line contains m integers a1, a2, ..., am (0 ≤ ai ≤ 108), where ai is the number of details of the i-th type for the respective robot.
Output
Print m space-separated integers, where the i-th number is the number of shots from the weapon of the i-th type that the robot should make to destroy the subsequence of consecutive droids of the maximum length.
If there are multiple optimal solutions, print any of them.
It is not necessary to make exactly k shots, the number of shots can be less.
Examples
Input
5 2 4
4 0
1 2
2 1
0 2
1 3
Output
2 2
Input
3 2 4
1 2
1 3
2 2
Output
1 3
Note
In the first test the second, third and fourth droids will be destroyed.
In the second test the first and second droids will be destroyed.
Submitted Solution:
```
from heapq import heappush, heappop
from sys import setrecursionlimit
from sys import stdin
from collections import defaultdict
setrecursionlimit(1000000007)
_data = iter(stdin.read().split('\n'))
def input():
return next(_data)
n, m, k = [int(x) for x in input().split()]
a = tuple(tuple(-int(x) for x in input().split()) for i in range(n))
heaps = tuple([0] for _ in range(m))
removed = tuple(defaultdict(int) for _ in range(m))
rv = -1
rt = (0,) * m
t = [0] * m
p = 0
for i in range(n):
ai = a[i]
for j, v, heap in zip(range(m), ai, heaps):
heappush(heap, v)
t[j] = heap[0]
while -sum(t) > k:
ap = a[p]
for j, v, heap, remd in zip(range(m), ap, heaps, removed):
remd[v] += 1
while heap[0] in remd:
top = heappop(heap)
if remd[top] == 1:
del remd[top]
else:
remd[top] -= 1
t[j] = heap[0]
p += 1
if rv < (i + 1) - p:
rv = (i + 1) - p
rt = tuple(t)
print(*map(lambda x: -x, rt))
```
Yes
| 76,260 | [
0.367919921875,
0.30224609375,
0.213623046875,
0.53125,
-0.73291015625,
-0.84814453125,
-0.493896484375,
0.05242919921875,
0.310546875,
1.0078125,
0.402587890625,
0.10211181640625,
-0.05535888671875,
-0.38134765625,
-0.332763671875,
0.480224609375,
-0.370849609375,
-0.56982421875,
... | 3 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
An army of n droids is lined up in one row. Each droid is described by m integers a1, a2, ..., am, where ai is the number of details of the i-th type in this droid's mechanism. R2-D2 wants to destroy the sequence of consecutive droids of maximum length. He has m weapons, the i-th weapon can affect all the droids in the army by destroying one detail of the i-th type (if the droid doesn't have details of this type, nothing happens to it).
A droid is considered to be destroyed when all of its details are destroyed. R2-D2 can make at most k shots. How many shots from the weapon of what type should R2-D2 make to destroy the sequence of consecutive droids of maximum length?
Input
The first line contains three integers n, m, k (1 ≤ n ≤ 105, 1 ≤ m ≤ 5, 0 ≤ k ≤ 109) — the number of droids, the number of detail types and the number of available shots, respectively.
Next n lines follow describing the droids. Each line contains m integers a1, a2, ..., am (0 ≤ ai ≤ 108), where ai is the number of details of the i-th type for the respective robot.
Output
Print m space-separated integers, where the i-th number is the number of shots from the weapon of the i-th type that the robot should make to destroy the subsequence of consecutive droids of the maximum length.
If there are multiple optimal solutions, print any of them.
It is not necessary to make exactly k shots, the number of shots can be less.
Examples
Input
5 2 4
4 0
1 2
2 1
0 2
1 3
Output
2 2
Input
3 2 4
1 2
1 3
2 2
Output
1 3
Note
In the first test the second, third and fourth droids will be destroyed.
In the second test the first and second droids will be destroyed.
Submitted Solution:
```
from heapq import heappush, heappop
from sys import setrecursionlimit
from sys import stdin
setrecursionlimit(1000000007)
_data = iter(stdin.read().split('\n'))
def input():
return next(_data)
n, m, k = [int(x) for x in input().split()]
a = tuple(tuple(-int(x) for x in input().split()) for i in range(n))
heaps1 = tuple([0] for _ in range(m))
heaps2 = tuple([1] for _ in range(m))
rv = -1
rt = (0,) * m
t = [0] * m
p = 0
for i in range(n):
ai = a[i]
for j, v, heap1 in zip(range(m), ai, heaps1):
heappush(heap1, v)
t[j] = heap1[0]
while -sum(t) > k:
ap = a[p]
for j, v, heap1, heap2 in zip(range(m), ap, heaps1, heaps2):
heappush(heap2, v)
while heap1[0] == heap2[0]:
heappop(heap1)
heappop(heap2)
t[j] = heap1[0]
p += 1
if rv < (i + 1) - p:
rv = (i + 1) - p
rt = tuple(t)
print(*map(lambda x: -x, rt))
```
Yes
| 76,261 | [
0.368408203125,
0.30810546875,
0.1971435546875,
0.552734375,
-0.7265625,
-0.8525390625,
-0.47607421875,
0.06268310546875,
0.3115234375,
1.0126953125,
0.408935546875,
0.134521484375,
-0.054412841796875,
-0.393310546875,
-0.306396484375,
0.4853515625,
-0.358154296875,
-0.580078125,
... | 3 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
An army of n droids is lined up in one row. Each droid is described by m integers a1, a2, ..., am, where ai is the number of details of the i-th type in this droid's mechanism. R2-D2 wants to destroy the sequence of consecutive droids of maximum length. He has m weapons, the i-th weapon can affect all the droids in the army by destroying one detail of the i-th type (if the droid doesn't have details of this type, nothing happens to it).
A droid is considered to be destroyed when all of its details are destroyed. R2-D2 can make at most k shots. How many shots from the weapon of what type should R2-D2 make to destroy the sequence of consecutive droids of maximum length?
Input
The first line contains three integers n, m, k (1 ≤ n ≤ 105, 1 ≤ m ≤ 5, 0 ≤ k ≤ 109) — the number of droids, the number of detail types and the number of available shots, respectively.
Next n lines follow describing the droids. Each line contains m integers a1, a2, ..., am (0 ≤ ai ≤ 108), where ai is the number of details of the i-th type for the respective robot.
Output
Print m space-separated integers, where the i-th number is the number of shots from the weapon of the i-th type that the robot should make to destroy the subsequence of consecutive droids of the maximum length.
If there are multiple optimal solutions, print any of them.
It is not necessary to make exactly k shots, the number of shots can be less.
Examples
Input
5 2 4
4 0
1 2
2 1
0 2
1 3
Output
2 2
Input
3 2 4
1 2
1 3
2 2
Output
1 3
Note
In the first test the second, third and fourth droids will be destroyed.
In the second test the first and second droids will be destroyed.
Submitted Solution:
```
from bisect import bisect_left as bl
from bisect import bisect_right as br
import heapq
import math
from collections import *
from functools import reduce,cmp_to_key
import sys
input = sys.stdin.readline
# M = mod = 998244353
# def factors(n):return sorted(set(reduce(list.__add__, ([i, n//i] for i in range(1, int(n**0.5) + 1) if n % i == 0))))
# def inv_mod(n):return pow(n, mod - 2, mod)
def li():return [int(i) for i in input().rstrip('\n').split()]
def st():return input().rstrip('\n')
def val():return int(input().rstrip('\n'))
def li2():return [i for i in input().rstrip('\n').split(' ')]
def li3():return [int(i) for i in input().rstrip('\n')]
def givediff(a,b):
return sum(max(i,j) for i,j in zip(b,a))
n, m, k = li()
l = []
for i in range(n):l.append(li())
l1 = [deque() for i in range(m)]
for i in range(m):l1[i].append([0,l[0][i]])
i, j = 0, 1
ans = 0
perm = [0]*m if sum(l[0]) > k else l[0][:]
curr = l[0][:]
while j != n:
for itr in range(m):
while len(l1[itr]) and l1[itr][-1][-1] <= l[j][itr]:
l1[itr].pop()
l1[itr].append([j,l[j][itr]])
while i < j and givediff(curr,l[j]) > k:
i += 1
for itr in range(m):
while l1[itr][0][0] < i:l1[itr].popleft()
curr[itr] = l1[itr][0][-1]
for itr in range(m):curr[itr] = l1[itr][0][-1]
if ans < j - i + 1 and givediff(l[j],curr) <= k:
ans = j - i + 1
perm = [max(a,b) for a,b in zip(l[j],curr)]
j += 1
# print(l1,'\n\n\n\n',l[j-1],curr,j,i,ans)
# print(ans)
perm[0] += k - sum(perm)
print(*perm)
```
Yes
| 76,262 | [
0.3408203125,
0.308349609375,
0.1424560546875,
0.529296875,
-0.7802734375,
-0.87353515625,
-0.49365234375,
0.0703125,
0.2724609375,
1.095703125,
0.4453125,
0.15576171875,
-0.03863525390625,
-0.4736328125,
-0.3837890625,
0.484375,
-0.360595703125,
-0.64697265625,
-0.231689453125,
... | 3 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
An army of n droids is lined up in one row. Each droid is described by m integers a1, a2, ..., am, where ai is the number of details of the i-th type in this droid's mechanism. R2-D2 wants to destroy the sequence of consecutive droids of maximum length. He has m weapons, the i-th weapon can affect all the droids in the army by destroying one detail of the i-th type (if the droid doesn't have details of this type, nothing happens to it).
A droid is considered to be destroyed when all of its details are destroyed. R2-D2 can make at most k shots. How many shots from the weapon of what type should R2-D2 make to destroy the sequence of consecutive droids of maximum length?
Input
The first line contains three integers n, m, k (1 ≤ n ≤ 105, 1 ≤ m ≤ 5, 0 ≤ k ≤ 109) — the number of droids, the number of detail types and the number of available shots, respectively.
Next n lines follow describing the droids. Each line contains m integers a1, a2, ..., am (0 ≤ ai ≤ 108), where ai is the number of details of the i-th type for the respective robot.
Output
Print m space-separated integers, where the i-th number is the number of shots from the weapon of the i-th type that the robot should make to destroy the subsequence of consecutive droids of the maximum length.
If there are multiple optimal solutions, print any of them.
It is not necessary to make exactly k shots, the number of shots can be less.
Examples
Input
5 2 4
4 0
1 2
2 1
0 2
1 3
Output
2 2
Input
3 2 4
1 2
1 3
2 2
Output
1 3
Note
In the first test the second, third and fourth droids will be destroyed.
In the second test the first and second droids will be destroyed.
Submitted Solution:
```
n, m, k = map(int, input().split())
dt = [ list(map(int, input().split())) for i in range(n) ]
ba = -float('inf')
M, N = 0, 1
maximums = dt[0].copy()
BESTmaximums = [0 for i in range(m)]
added = 0
ismaxatthefirst = [True for i in range(m)]
while M < N and max(N, M) <= n:
if added >= 0:
for i in range(m):
if maximums[i] <= dt[added][i]:
maximums[i] = dt[added][i]
ismaxatthefirst[i] = True
else:
#maximums = dt[M]
for j in range(m):
if maximums[j] != dt[M-1][j] or ismaxatthefirst[j] == False: continue
else: maximums[j] = dt[M][j]
for i in range(M, N):
if maximums[j] < dt[i][j]:
maximums[j] = dt[i][j]
ismaxatthefirst[j] = False
if sum(maximums) <= k:
if N-M > ba:
ba = N-M
BESTmaximums = maximums.copy()
added = N
N += 1
else:
added = -1
M += 1
if ba < 0:
BESTmaximums[0] = k
print(" ".join(map(str, BESTmaximums)))
```
No
| 76,263 | [
0.3955078125,
0.2861328125,
0.1348876953125,
0.47412109375,
-0.744140625,
-0.8828125,
-0.49609375,
0.16259765625,
0.2734375,
1.025390625,
0.447021484375,
0.191162109375,
-0.0184783935546875,
-0.43896484375,
-0.3017578125,
0.425537109375,
-0.32958984375,
-0.54052734375,
-0.1408691... | 3 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
An army of n droids is lined up in one row. Each droid is described by m integers a1, a2, ..., am, where ai is the number of details of the i-th type in this droid's mechanism. R2-D2 wants to destroy the sequence of consecutive droids of maximum length. He has m weapons, the i-th weapon can affect all the droids in the army by destroying one detail of the i-th type (if the droid doesn't have details of this type, nothing happens to it).
A droid is considered to be destroyed when all of its details are destroyed. R2-D2 can make at most k shots. How many shots from the weapon of what type should R2-D2 make to destroy the sequence of consecutive droids of maximum length?
Input
The first line contains three integers n, m, k (1 ≤ n ≤ 105, 1 ≤ m ≤ 5, 0 ≤ k ≤ 109) — the number of droids, the number of detail types and the number of available shots, respectively.
Next n lines follow describing the droids. Each line contains m integers a1, a2, ..., am (0 ≤ ai ≤ 108), where ai is the number of details of the i-th type for the respective robot.
Output
Print m space-separated integers, where the i-th number is the number of shots from the weapon of the i-th type that the robot should make to destroy the subsequence of consecutive droids of the maximum length.
If there are multiple optimal solutions, print any of them.
It is not necessary to make exactly k shots, the number of shots can be less.
Examples
Input
5 2 4
4 0
1 2
2 1
0 2
1 3
Output
2 2
Input
3 2 4
1 2
1 3
2 2
Output
1 3
Note
In the first test the second, third and fourth droids will be destroyed.
In the second test the first and second droids will be destroyed.
Submitted Solution:
```
n, m, k = map(int, input().split())
dt = [ list(map(int, input().split())) for i in range(n) ]
ba = -float('inf')
M, N = 0, 1
maximums = dt[0].copy()
BESTmaximums = [0 for i in range(m)]
added = 0
ismaxatthefirst = [True for i in range(m)]
while M < N and max(N, M) <= n:
if added >= 0:
for i in range(m):
if maximums[i] <= dt[added][i]:
maximums[i] = dt[added][i]
if N != 1: ismaxatthefirst[i] = False
else:
#maximums = dt[M]
for j in range(m):
if maximums[j] != dt[M-1][j] or ismaxatthefirst[j] == False: continue
else: maximums[j] = dt[M][j]
for i in range(M, N):
if maximums[j] < dt[i][j]:
maximums[j] = dt[i][j]
ismaxatthefirst[j] = True
if sum(maximums) <= k:
if N-M > ba:
ba = N-M
BESTmaximums = maximums.copy()
added = N
N += 1
else:
added = -1
M += 1
if ba < 0:
BESTmaximums[0] = k
print(" ".join(map(str, BESTmaximums)))
```
No
| 76,264 | [
0.3955078125,
0.2861328125,
0.1348876953125,
0.47412109375,
-0.744140625,
-0.8828125,
-0.49609375,
0.16259765625,
0.2734375,
1.025390625,
0.447021484375,
0.191162109375,
-0.0184783935546875,
-0.43896484375,
-0.3017578125,
0.425537109375,
-0.32958984375,
-0.54052734375,
-0.1408691... | 3 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
An army of n droids is lined up in one row. Each droid is described by m integers a1, a2, ..., am, where ai is the number of details of the i-th type in this droid's mechanism. R2-D2 wants to destroy the sequence of consecutive droids of maximum length. He has m weapons, the i-th weapon can affect all the droids in the army by destroying one detail of the i-th type (if the droid doesn't have details of this type, nothing happens to it).
A droid is considered to be destroyed when all of its details are destroyed. R2-D2 can make at most k shots. How many shots from the weapon of what type should R2-D2 make to destroy the sequence of consecutive droids of maximum length?
Input
The first line contains three integers n, m, k (1 ≤ n ≤ 105, 1 ≤ m ≤ 5, 0 ≤ k ≤ 109) — the number of droids, the number of detail types and the number of available shots, respectively.
Next n lines follow describing the droids. Each line contains m integers a1, a2, ..., am (0 ≤ ai ≤ 108), where ai is the number of details of the i-th type for the respective robot.
Output
Print m space-separated integers, where the i-th number is the number of shots from the weapon of the i-th type that the robot should make to destroy the subsequence of consecutive droids of the maximum length.
If there are multiple optimal solutions, print any of them.
It is not necessary to make exactly k shots, the number of shots can be less.
Examples
Input
5 2 4
4 0
1 2
2 1
0 2
1 3
Output
2 2
Input
3 2 4
1 2
1 3
2 2
Output
1 3
Note
In the first test the second, third and fourth droids will be destroyed.
In the second test the first and second droids will be destroyed.
Submitted Solution:
```
from sys import stdin
_data = iter(stdin.read().split('\n'))
def input():
return next(_data)
class SegTree:
def __init__(self, n, zero, f):
while n & (n - 1):
n -= n & -n
self.n = 2 * n
self.zero = zero
self.f = f
self.t = [self.zero for _ in range(2 * self.n - 1)]
def __getitem__(self, k):
return self.t[k + self.n - 1]
def __setitem__(self, k, v):
k += self.n - 1
self.t[k] = v
while k > 0:
k = (k - 1) >> 1
self.t[k] = self.f(self.t[2 * k + 1], self.t[2 * k + 2])
def _query(self, a, b, k, l, r):
if b <= l or r <= a:
return self.zero
elif a <= l and r <= b:
return self.t[k]
else:
return self.f(self._query(a, b, 2 * k + 1, l, (l + r) >> 1),
self._query(a, b, 2 * k + 2, (l + r) >> 1, r))
def query(self, a, b):
return self._query(a, b, 0, 0, self.n);
n, m, k = [int(x) for x in input().split()]
zero = tuple(0 for i in range(m))
st = SegTree(n, zero, lambda x, y: map(max, zip(x, y)))
rv = -1
rt = zero
t = [0] * m
p = 0
for i in range(n):
a = tuple(int(x) for x in input().split())
st[i] = a
for j in range(m):
t[j] += a[j]
while sum(t) > k:
p += 1
t = list(st.query(p, i + 1))
if rv < (i + 1) - p:
rv = (i + 1) - p
rt = t
print(*rt)
```
No
| 76,265 | [
0.398681640625,
0.309814453125,
0.161376953125,
0.494140625,
-0.75,
-0.84521484375,
-0.55224609375,
0.177001953125,
0.2646484375,
0.9873046875,
0.30322265625,
0.127197265625,
-0.01313018798828125,
-0.404052734375,
-0.294189453125,
0.4638671875,
-0.326171875,
-0.513671875,
-0.1638... | 3 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
An army of n droids is lined up in one row. Each droid is described by m integers a1, a2, ..., am, where ai is the number of details of the i-th type in this droid's mechanism. R2-D2 wants to destroy the sequence of consecutive droids of maximum length. He has m weapons, the i-th weapon can affect all the droids in the army by destroying one detail of the i-th type (if the droid doesn't have details of this type, nothing happens to it).
A droid is considered to be destroyed when all of its details are destroyed. R2-D2 can make at most k shots. How many shots from the weapon of what type should R2-D2 make to destroy the sequence of consecutive droids of maximum length?
Input
The first line contains three integers n, m, k (1 ≤ n ≤ 105, 1 ≤ m ≤ 5, 0 ≤ k ≤ 109) — the number of droids, the number of detail types and the number of available shots, respectively.
Next n lines follow describing the droids. Each line contains m integers a1, a2, ..., am (0 ≤ ai ≤ 108), where ai is the number of details of the i-th type for the respective robot.
Output
Print m space-separated integers, where the i-th number is the number of shots from the weapon of the i-th type that the robot should make to destroy the subsequence of consecutive droids of the maximum length.
If there are multiple optimal solutions, print any of them.
It is not necessary to make exactly k shots, the number of shots can be less.
Examples
Input
5 2 4
4 0
1 2
2 1
0 2
1 3
Output
2 2
Input
3 2 4
1 2
1 3
2 2
Output
1 3
Note
In the first test the second, third and fourth droids will be destroyed.
In the second test the first and second droids will be destroyed.
Submitted Solution:
```
from sys import stdin
_data = iter(stdin.read().split('\n'))
def input():
return next(_data)
class SegTree:
def __init__(self, n, zero, f):
while n & (n - 1):
n -= n & -n
self.n = 2 * n
self.zero = zero
self.f = f
self.t = [self.zero for _ in range(2 * self.n - 1)]
def __getitem__(self, k):
return self.t[k + self.n - 1]
def __setitem__(self, k, v):
k += self.n - 1
self.t[k] = v
while k > 0:
k = (k - 1) >> 1
self.t[k] = self.f(self.t[2 * k + 1], self.t[2 * k + 2])
def _query(self, a, b, k, l, r):
if b <= l or r <= a:
return self.zero
elif a <= l and r <= b:
return self.t[k]
else:
return self.f(self._query(a, b, 2 * k + 1, l, (l + r) >> 1),
self._query(a, b, 2 * k + 2, (l + r) >> 1, r))
def query(self, a, b):
return self._query(a, b, 0, 0, self.n);
n, m, k = [int(x) for x in input().split()]
zero = tuple(0 for i in range(m))
st = SegTree(n, zero, lambda x, y: map(max, zip(x, y)))
rv = -1
rt = zero
t = [0] * m
p = 0
for i in range(n):
a = tuple(int(x) for x in input().split())
st[i] = a
for j in range(m):
t[j] += a[j]
while sum(t) > k:
p += 1
t = list(st.query(p, i + 1))
if rv < (i + 1) - p:
rv = (i + 1) - p
rt = tuple(t)
print(*rt)
```
No
| 76,266 | [
0.398681640625,
0.309814453125,
0.161376953125,
0.494140625,
-0.75,
-0.84521484375,
-0.55224609375,
0.177001953125,
0.2646484375,
0.9873046875,
0.30322265625,
0.127197265625,
-0.01313018798828125,
-0.404052734375,
-0.294189453125,
0.4638671875,
-0.326171875,
-0.513671875,
-0.1638... | 3 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
One day Masha came home and noticed n mice in the corridor of her flat. Of course, she shouted loudly, so scared mice started to run to the holes in the corridor.
The corridor can be represeted as a numeric axis with n mice and m holes on it. ith mouse is at the coordinate xi, and jth hole — at coordinate pj. jth hole has enough room for cj mice, so not more than cj mice can enter this hole.
What is the minimum sum of distances that mice have to go through so that they all can hide in the holes? If ith mouse goes to the hole j, then its distance is |xi - pj|.
Print the minimum sum of distances.
Input
The first line contains two integer numbers n, m (1 ≤ n, m ≤ 5000) — the number of mice and the number of holes, respectively.
The second line contains n integers x1, x2, ..., xn ( - 109 ≤ xi ≤ 109), where xi is the coordinate of ith mouse.
Next m lines contain pairs of integer numbers pj, cj ( - 109 ≤ pj ≤ 109, 1 ≤ cj ≤ 5000), where pj is the coordinate of jth hole, and cj is the maximum number of mice that can hide in the hole j.
Output
Print one integer number — the minimum sum of distances. If there is no solution, print -1 instead.
Examples
Input
4 5
6 2 8 9
3 6
2 1
3 6
4 7
4 7
Output
11
Input
7 2
10 20 30 40 50 45 35
-1000000000 10
1000000000 1
Output
7000000130
Submitted Solution:
```
from sys import stdin , stdout
mice_num , hole_num = [int(t) for t in stdin.readline().rstrip().split()]
hole = []
mice = [int(t) for t in stdin.readline().rstrip().split()]
for i in range(hole_num):
hole_idx , capacity = [int(t) for t in stdin.readline().rstrip().split()]
hole += [hole_idx] * capacity
sorted_mice = sorted(mice)
sorted_hole = sorted(hole)
def min_distance(sorted_mice , sorted_hole):
min_distance = 0
temp_distance = 0
len_mice = len(sorted_mice)
len_hole = len(sorted_hole)
if len_mice > len_hole:
return -1
else:
for i in range(len_mice):
temp_distance += abs(sorted_mice[i] - sorted_hole[i])
if len_mice == len_hole:
return temp_distance
else:
min_distance = temp_distance
for j in range(1,len_hole - len_mice+1):
temp_distance = 0
for k in range(len_mice):
temp_distance += abs(sorted_mice[k] - sorted_hole[k+j])
if temp_distance < min_distance:
min_distance = temp_distance
return min_distance
print(min_distance(sorted_mice , sorted_hole))
```
No
| 76,380 | [
0.4931640625,
0.056976318359375,
-0.12060546875,
0.0986328125,
-0.69677734375,
-0.41552734375,
-0.1243896484375,
0.5615234375,
-0.07000732421875,
0.74951171875,
0.6845703125,
0.2286376953125,
-0.15869140625,
-0.6435546875,
-0.5693359375,
0.25537109375,
-0.423583984375,
-0.647460937... | 3 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
One day Masha came home and noticed n mice in the corridor of her flat. Of course, she shouted loudly, so scared mice started to run to the holes in the corridor.
The corridor can be represeted as a numeric axis with n mice and m holes on it. ith mouse is at the coordinate xi, and jth hole — at coordinate pj. jth hole has enough room for cj mice, so not more than cj mice can enter this hole.
What is the minimum sum of distances that mice have to go through so that they all can hide in the holes? If ith mouse goes to the hole j, then its distance is |xi - pj|.
Print the minimum sum of distances.
Input
The first line contains two integer numbers n, m (1 ≤ n, m ≤ 5000) — the number of mice and the number of holes, respectively.
The second line contains n integers x1, x2, ..., xn ( - 109 ≤ xi ≤ 109), where xi is the coordinate of ith mouse.
Next m lines contain pairs of integer numbers pj, cj ( - 109 ≤ pj ≤ 109, 1 ≤ cj ≤ 5000), where pj is the coordinate of jth hole, and cj is the maximum number of mice that can hide in the hole j.
Output
Print one integer number — the minimum sum of distances. If there is no solution, print -1 instead.
Examples
Input
4 5
6 2 8 9
3 6
2 1
3 6
4 7
4 7
Output
11
Input
7 2
10 20 30 40 50 45 35
-1000000000 10
1000000000 1
Output
7000000130
Submitted Solution:
```
def get_distance(distance_tuple):
return distance_tuple[2]
"""
mice is a list of mice positions(integers)
holes is a dict where dict[hole index] = (hole position, hole size(quantity of mice that fit in))
"""
def modified_min_spanning_tree(mice, holes):
result_mice = set()
result_holes = {} #dict[holePosition] = quantity of mice currently in it
for hole in holes.keys():
result_holes[hole] = 0
result = 0
distances = [] #list of (mouse position, hole index, distance mouse hole)
for i in holes.keys():
for m in range(len(mice)):
distances.append((m, i, abs(holes[i][0] - mice[m])))
distances.sort(key=get_distance)
for i in distances:
if len(result_mice) == len(mice):
return result
if i[0] in result_mice:
continue
if result_holes[i[1]] < holes[i[1]][1]:
result += i[2]
result_holes[i[1]] += 1
result_mice.add(i[0])
if len(result_mice) < len(mice):
return -1
return result
if __name__ == "__main__":
quantities = input().split(' ');
mice_quantity = int(quantities[0])
holes_quantity = int(quantities[1])
mice = [int(x) for x in input().split(' ')]
holes = {}
for i in range(holes_quantity):
hole = [int (x) for x in input().split(' ')]
holes[i] = (hole[0], hole[1])
print(modified_min_spanning_tree(mice, holes))
```
No
| 76,381 | [
0.486328125,
0.02801513671875,
-0.25146484375,
0.1171875,
-0.6689453125,
-0.422607421875,
-0.2265625,
0.56787109375,
-0.0111846923828125,
0.6865234375,
0.669921875,
0.2008056640625,
-0.1517333984375,
-0.72119140625,
-0.6083984375,
0.330078125,
-0.485595703125,
-0.61669921875,
-0.... | 3 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
One day Masha came home and noticed n mice in the corridor of her flat. Of course, she shouted loudly, so scared mice started to run to the holes in the corridor.
The corridor can be represeted as a numeric axis with n mice and m holes on it. ith mouse is at the coordinate xi, and jth hole — at coordinate pj. jth hole has enough room for cj mice, so not more than cj mice can enter this hole.
What is the minimum sum of distances that mice have to go through so that they all can hide in the holes? If ith mouse goes to the hole j, then its distance is |xi - pj|.
Print the minimum sum of distances.
Input
The first line contains two integer numbers n, m (1 ≤ n, m ≤ 5000) — the number of mice and the number of holes, respectively.
The second line contains n integers x1, x2, ..., xn ( - 109 ≤ xi ≤ 109), where xi is the coordinate of ith mouse.
Next m lines contain pairs of integer numbers pj, cj ( - 109 ≤ pj ≤ 109, 1 ≤ cj ≤ 5000), where pj is the coordinate of jth hole, and cj is the maximum number of mice that can hide in the hole j.
Output
Print one integer number — the minimum sum of distances. If there is no solution, print -1 instead.
Examples
Input
4 5
6 2 8 9
3 6
2 1
3 6
4 7
4 7
Output
11
Input
7 2
10 20 30 40 50 45 35
-1000000000 10
1000000000 1
Output
7000000130
Submitted Solution:
```
def get_distance(distance_tuple):
return distance_tuple[2]
"""
mice is a list of mice positions(integers)
holes is a dict where dict[hole index] = (hole position, hole size(quantity of mice that fit in))
"""
def modified_min_spanning_tree(mice, holes):
result_mice = set()
result_holes = {} #dict[holePosition] = quantity of mice currently in it
for hole in holes.keys():
result_holes[hole] = 0
result = 0
distances = [] #list of (mouse position, hole index, distance mouse hole)
for i in holes.keys():
for m in mice:
distances.append((m, i, abs(holes[i][0] - m)))
distances.sort(key=get_distance)
for i in distances:
if len(result_mice) == len(mice):
return result
if i[0] in result_mice:
continue
if result_holes[i[1]] < holes[i[1]][1]:
result += i[2]
result_holes[i[1]] += 1
result_mice.add(i[0])
if len(result_mice) < len(mice):
return -1
return result
if __name__ == "__main__":
quantities = input().split(' ');
mice_quantity = int(quantities[0])
holes_quantity = int(quantities[1])
mice = [int(x) for x in input().split(' ')]
holes = {}
for i in range(holes_quantity):
hole = [int (x) for x in input().split(' ')]
holes[i] = (hole[0], hole[1])
print(modified_min_spanning_tree(mice, holes))
```
No
| 76,382 | [
0.486328125,
0.02801513671875,
-0.25146484375,
0.1171875,
-0.6689453125,
-0.422607421875,
-0.2265625,
0.56787109375,
-0.0111846923828125,
0.6865234375,
0.669921875,
0.2008056640625,
-0.1517333984375,
-0.72119140625,
-0.6083984375,
0.330078125,
-0.485595703125,
-0.61669921875,
-0.... | 3 |
Provide a correct Python 3 solution for this coding contest problem.
The surveyor starship Hakodate-maru is famous for her two fuel containers with unbounded capacities. They hold the same type of atomic fuel balls.
There, however, is an inconvenience. The shapes of the fuel containers #1 and #2 are always cubic and regular tetrahedral respectively. Both of the fuel containers should be either empty or filled according to their shapes. Otherwise, the fuel balls become extremely unstable and may explode in the fuel containers. Thus, the number of fuel balls for the container #1 should be a cubic number (n3 for some n = 0, 1, 2, 3,... ) and that for the container #2 should be a tetrahedral number ( n(n + 1)(n + 2)/6 for some n = 0, 1, 2, 3,... ).
Hakodate-maru is now at the star base Goryokaku preparing for the next mission to create a precise and detailed chart of stars and interstellar matters. Both of the fuel containers are now empty. Commander Parus of Goryokaku will soon send a message to Captain Future of Hakodate-maru on how many fuel balls Goryokaku can supply. Captain Future should quickly answer to Commander Parus on how many fuel balls she requests before her ship leaves Goryokaku. Of course, Captain Future and her omcers want as many fuel balls as possible.
For example, consider the case Commander Parus offers 151200 fuel balls. If only the fuel container #1 were available (i.e. ifthe fuel container #2 were unavailable), at most 148877 fuel balls could be put into the fuel container since 148877 = 53 × 53 × 53 < 151200 < 54 × 54 × 54 . If only the fuel container #2 were available, at most 147440 fuel balls could be put into the fuel container since 147440 = 95 × 96 × 97/6 < 151200 < 96 × 97 × 98/6 . Using both of the fuel containers #1 and #2, 151200 fuel balls can be put into the fuel containers since 151200 = 39 × 39 × 39 + 81 × 82 × 83/6 . In this case, Captain Future's answer should be "151200".
Commander Parus's offer cannot be greater than 151200 because of the capacity of the fuel storages of Goryokaku. Captain Future and her omcers know that well.
You are a fuel engineer assigned to Hakodate-maru. Your duty today is to help Captain Future with calculating the number of fuel balls she should request.
Input
The input is a sequence of at most 1024 positive integers. Each line contains a single integer. The sequence is followed by a zero, which indicates the end of data and should not be treated as input. You may assume that none of the input integers is greater than 151200.
Output
The output is composed of lines, each containing a single integer. Each output integer should be the greatest integer that is the sum of a nonnegative cubic number and a nonnegative tetrahedral number and that is not greater than the corresponding input number. No other characters should appear in the output.
Example
Input
100
64
50
20
151200
0
Output
99
64
47
20
151200
"Correct Solution:
```
"""
Problem A: Starship Hakodate-maru
https://onlinejudge.u-aizu.ac.jp/problems/1224
cannot be greater than 151200
Sample Input
100
64
50
20
151200
0
Output for the Sample Input
99
64
47
20
151200
maximum: 151200 = 39 × 39 × 39 + 81 × 82 × 83/6
n = 54 n^3 = 157464
n = 96 tetrahedral = 152096
"""
#cube and tetrahedral
cube = []
tetrahedral = []
def solve(n):
min = 0
diff_min = pow(10, 6)
for i in cube:
for j in tetrahedral:
sum_value = i+j
now_diff = n - sum_value
if now_diff < diff_min and now_diff >= 0:
diff_min = now_diff
min = sum_value
#print(n,i,j,sum_value,diff_min)
return min
if __name__ == '__main__':
for i in range(0, 55):
cube.append(i**3)
for i in range(0, 97):
tetrahedral.append(i*(i+1)*(i+2)//6)
ans = []
while(True):
n = int(input())
if n == 0:
break
ans.append(solve(n))
print(*ans, sep='\n')
```
| 76,636 | [
0.413330078125,
0.1265869140625,
0.301025390625,
0.230712890625,
-0.66455078125,
-0.256591796875,
-0.058502197265625,
0.120361328125,
0.333740234375,
0.67822265625,
0.74609375,
0.264892578125,
0.259033203125,
-0.66552734375,
-0.5634765625,
0.35595703125,
-0.06414794921875,
-0.59082... | 3 |
Provide a correct Python 3 solution for this coding contest problem.
The surveyor starship Hakodate-maru is famous for her two fuel containers with unbounded capacities. They hold the same type of atomic fuel balls.
There, however, is an inconvenience. The shapes of the fuel containers #1 and #2 are always cubic and regular tetrahedral respectively. Both of the fuel containers should be either empty or filled according to their shapes. Otherwise, the fuel balls become extremely unstable and may explode in the fuel containers. Thus, the number of fuel balls for the container #1 should be a cubic number (n3 for some n = 0, 1, 2, 3,... ) and that for the container #2 should be a tetrahedral number ( n(n + 1)(n + 2)/6 for some n = 0, 1, 2, 3,... ).
Hakodate-maru is now at the star base Goryokaku preparing for the next mission to create a precise and detailed chart of stars and interstellar matters. Both of the fuel containers are now empty. Commander Parus of Goryokaku will soon send a message to Captain Future of Hakodate-maru on how many fuel balls Goryokaku can supply. Captain Future should quickly answer to Commander Parus on how many fuel balls she requests before her ship leaves Goryokaku. Of course, Captain Future and her omcers want as many fuel balls as possible.
For example, consider the case Commander Parus offers 151200 fuel balls. If only the fuel container #1 were available (i.e. ifthe fuel container #2 were unavailable), at most 148877 fuel balls could be put into the fuel container since 148877 = 53 × 53 × 53 < 151200 < 54 × 54 × 54 . If only the fuel container #2 were available, at most 147440 fuel balls could be put into the fuel container since 147440 = 95 × 96 × 97/6 < 151200 < 96 × 97 × 98/6 . Using both of the fuel containers #1 and #2, 151200 fuel balls can be put into the fuel containers since 151200 = 39 × 39 × 39 + 81 × 82 × 83/6 . In this case, Captain Future's answer should be "151200".
Commander Parus's offer cannot be greater than 151200 because of the capacity of the fuel storages of Goryokaku. Captain Future and her omcers know that well.
You are a fuel engineer assigned to Hakodate-maru. Your duty today is to help Captain Future with calculating the number of fuel balls she should request.
Input
The input is a sequence of at most 1024 positive integers. Each line contains a single integer. The sequence is followed by a zero, which indicates the end of data and should not be treated as input. You may assume that none of the input integers is greater than 151200.
Output
The output is composed of lines, each containing a single integer. Each output integer should be the greatest integer that is the sum of a nonnegative cubic number and a nonnegative tetrahedral number and that is not greater than the corresponding input number. No other characters should appear in the output.
Example
Input
100
64
50
20
151200
0
Output
99
64
47
20
151200
"Correct Solution:
```
#!/usr/bin/env python
# -*- coding: utf-8 -*-
"""
Problems 1224
2001年 アジア地区予選 函館大会 問題A Starship Hakodate-maru
"""
while True:
balls = int(input())
ans = 0
if balls == 0: # ボールの数が0個なら終了する
break
for cs in range(54):
for ts in range(96):
if pow(cs,3) <= balls:
ans = max(ans, pow(cs,3))
if ts*(ts+1)*(ts+2)//6 <= balls:
ans = max(ans, ts*(ts+1)*(ts+2)//6)
if pow(cs,3) + ts*(ts+1)*(ts+2)//6 <= balls:
ans = max(ans, pow(cs,3) + ts*(ts+1)*(ts+2)//6)
print(ans)
```
| 76,637 | [
0.413330078125,
0.1265869140625,
0.301025390625,
0.230712890625,
-0.66455078125,
-0.256591796875,
-0.058502197265625,
0.120361328125,
0.333740234375,
0.67822265625,
0.74609375,
0.264892578125,
0.259033203125,
-0.66552734375,
-0.5634765625,
0.35595703125,
-0.06414794921875,
-0.59082... | 3 |
Provide a correct Python 3 solution for this coding contest problem.
The surveyor starship Hakodate-maru is famous for her two fuel containers with unbounded capacities. They hold the same type of atomic fuel balls.
There, however, is an inconvenience. The shapes of the fuel containers #1 and #2 are always cubic and regular tetrahedral respectively. Both of the fuel containers should be either empty or filled according to their shapes. Otherwise, the fuel balls become extremely unstable and may explode in the fuel containers. Thus, the number of fuel balls for the container #1 should be a cubic number (n3 for some n = 0, 1, 2, 3,... ) and that for the container #2 should be a tetrahedral number ( n(n + 1)(n + 2)/6 for some n = 0, 1, 2, 3,... ).
Hakodate-maru is now at the star base Goryokaku preparing for the next mission to create a precise and detailed chart of stars and interstellar matters. Both of the fuel containers are now empty. Commander Parus of Goryokaku will soon send a message to Captain Future of Hakodate-maru on how many fuel balls Goryokaku can supply. Captain Future should quickly answer to Commander Parus on how many fuel balls she requests before her ship leaves Goryokaku. Of course, Captain Future and her omcers want as many fuel balls as possible.
For example, consider the case Commander Parus offers 151200 fuel balls. If only the fuel container #1 were available (i.e. ifthe fuel container #2 were unavailable), at most 148877 fuel balls could be put into the fuel container since 148877 = 53 × 53 × 53 < 151200 < 54 × 54 × 54 . If only the fuel container #2 were available, at most 147440 fuel balls could be put into the fuel container since 147440 = 95 × 96 × 97/6 < 151200 < 96 × 97 × 98/6 . Using both of the fuel containers #1 and #2, 151200 fuel balls can be put into the fuel containers since 151200 = 39 × 39 × 39 + 81 × 82 × 83/6 . In this case, Captain Future's answer should be "151200".
Commander Parus's offer cannot be greater than 151200 because of the capacity of the fuel storages of Goryokaku. Captain Future and her omcers know that well.
You are a fuel engineer assigned to Hakodate-maru. Your duty today is to help Captain Future with calculating the number of fuel balls she should request.
Input
The input is a sequence of at most 1024 positive integers. Each line contains a single integer. The sequence is followed by a zero, which indicates the end of data and should not be treated as input. You may assume that none of the input integers is greater than 151200.
Output
The output is composed of lines, each containing a single integer. Each output integer should be the greatest integer that is the sum of a nonnegative cubic number and a nonnegative tetrahedral number and that is not greater than the corresponding input number. No other characters should appear in the output.
Example
Input
100
64
50
20
151200
0
Output
99
64
47
20
151200
"Correct Solution:
```
ans = [] # 答え
while True:
N = int(input())
if not N:
break
now_cube = int(N ** (1 / 3 + 0.000001))
now_pyramid = 0
tmp_ans = now_cube ** 3
# 立方体の一辺を小さくしていく、立方体の辺ごとに四角錐の一辺の長さを求め、容量を求める
for i in range(now_cube, -1, -1):
while True:
# もし次の値が最大容量を超えるならば
if (now_pyramid + 1) * (now_pyramid + 2) * (now_pyramid + 3) // 6 + i ** 3 > N:
# 超えない値の時にこれまでの最大値と比較して大きい方を答えとする
tmp_ans = max(tmp_ans, now_pyramid * (now_pyramid + 1) * (now_pyramid + 2) // 6 + i ** 3)
break
# 四角錐の一辺を大きくしていく
now_pyramid += 1
ans.append(tmp_ans)
# 出力
[print(i) for i in ans]
```
| 76,638 | [
0.413330078125,
0.1265869140625,
0.301025390625,
0.230712890625,
-0.66455078125,
-0.256591796875,
-0.058502197265625,
0.120361328125,
0.333740234375,
0.67822265625,
0.74609375,
0.264892578125,
0.259033203125,
-0.66552734375,
-0.5634765625,
0.35595703125,
-0.06414794921875,
-0.59082... | 3 |
Provide a correct Python 3 solution for this coding contest problem.
The surveyor starship Hakodate-maru is famous for her two fuel containers with unbounded capacities. They hold the same type of atomic fuel balls.
There, however, is an inconvenience. The shapes of the fuel containers #1 and #2 are always cubic and regular tetrahedral respectively. Both of the fuel containers should be either empty or filled according to their shapes. Otherwise, the fuel balls become extremely unstable and may explode in the fuel containers. Thus, the number of fuel balls for the container #1 should be a cubic number (n3 for some n = 0, 1, 2, 3,... ) and that for the container #2 should be a tetrahedral number ( n(n + 1)(n + 2)/6 for some n = 0, 1, 2, 3,... ).
Hakodate-maru is now at the star base Goryokaku preparing for the next mission to create a precise and detailed chart of stars and interstellar matters. Both of the fuel containers are now empty. Commander Parus of Goryokaku will soon send a message to Captain Future of Hakodate-maru on how many fuel balls Goryokaku can supply. Captain Future should quickly answer to Commander Parus on how many fuel balls she requests before her ship leaves Goryokaku. Of course, Captain Future and her omcers want as many fuel balls as possible.
For example, consider the case Commander Parus offers 151200 fuel balls. If only the fuel container #1 were available (i.e. ifthe fuel container #2 were unavailable), at most 148877 fuel balls could be put into the fuel container since 148877 = 53 × 53 × 53 < 151200 < 54 × 54 × 54 . If only the fuel container #2 were available, at most 147440 fuel balls could be put into the fuel container since 147440 = 95 × 96 × 97/6 < 151200 < 96 × 97 × 98/6 . Using both of the fuel containers #1 and #2, 151200 fuel balls can be put into the fuel containers since 151200 = 39 × 39 × 39 + 81 × 82 × 83/6 . In this case, Captain Future's answer should be "151200".
Commander Parus's offer cannot be greater than 151200 because of the capacity of the fuel storages of Goryokaku. Captain Future and her omcers know that well.
You are a fuel engineer assigned to Hakodate-maru. Your duty today is to help Captain Future with calculating the number of fuel balls she should request.
Input
The input is a sequence of at most 1024 positive integers. Each line contains a single integer. The sequence is followed by a zero, which indicates the end of data and should not be treated as input. You may assume that none of the input integers is greater than 151200.
Output
The output is composed of lines, each containing a single integer. Each output integer should be the greatest integer that is the sum of a nonnegative cubic number and a nonnegative tetrahedral number and that is not greater than the corresponding input number. No other characters should appear in the output.
Example
Input
100
64
50
20
151200
0
Output
99
64
47
20
151200
"Correct Solution:
```
answer = []
while True:
n = int(input())
if n == 0:
break
ans = 0
for i in range(55):
for j in range(96):
temp = i * i * i + (j * (j + 1) * (j + 2) // 6)
if temp <= n:
if ans < temp:
ans = temp
else:
break
answer.append(ans)
for i in answer:
print(i)
```
| 76,639 | [
0.413330078125,
0.1265869140625,
0.301025390625,
0.230712890625,
-0.66455078125,
-0.256591796875,
-0.058502197265625,
0.120361328125,
0.333740234375,
0.67822265625,
0.74609375,
0.264892578125,
0.259033203125,
-0.66552734375,
-0.5634765625,
0.35595703125,
-0.06414794921875,
-0.59082... | 3 |
Provide a correct Python 3 solution for this coding contest problem.
The surveyor starship Hakodate-maru is famous for her two fuel containers with unbounded capacities. They hold the same type of atomic fuel balls.
There, however, is an inconvenience. The shapes of the fuel containers #1 and #2 are always cubic and regular tetrahedral respectively. Both of the fuel containers should be either empty or filled according to their shapes. Otherwise, the fuel balls become extremely unstable and may explode in the fuel containers. Thus, the number of fuel balls for the container #1 should be a cubic number (n3 for some n = 0, 1, 2, 3,... ) and that for the container #2 should be a tetrahedral number ( n(n + 1)(n + 2)/6 for some n = 0, 1, 2, 3,... ).
Hakodate-maru is now at the star base Goryokaku preparing for the next mission to create a precise and detailed chart of stars and interstellar matters. Both of the fuel containers are now empty. Commander Parus of Goryokaku will soon send a message to Captain Future of Hakodate-maru on how many fuel balls Goryokaku can supply. Captain Future should quickly answer to Commander Parus on how many fuel balls she requests before her ship leaves Goryokaku. Of course, Captain Future and her omcers want as many fuel balls as possible.
For example, consider the case Commander Parus offers 151200 fuel balls. If only the fuel container #1 were available (i.e. ifthe fuel container #2 were unavailable), at most 148877 fuel balls could be put into the fuel container since 148877 = 53 × 53 × 53 < 151200 < 54 × 54 × 54 . If only the fuel container #2 were available, at most 147440 fuel balls could be put into the fuel container since 147440 = 95 × 96 × 97/6 < 151200 < 96 × 97 × 98/6 . Using both of the fuel containers #1 and #2, 151200 fuel balls can be put into the fuel containers since 151200 = 39 × 39 × 39 + 81 × 82 × 83/6 . In this case, Captain Future's answer should be "151200".
Commander Parus's offer cannot be greater than 151200 because of the capacity of the fuel storages of Goryokaku. Captain Future and her omcers know that well.
You are a fuel engineer assigned to Hakodate-maru. Your duty today is to help Captain Future with calculating the number of fuel balls she should request.
Input
The input is a sequence of at most 1024 positive integers. Each line contains a single integer. The sequence is followed by a zero, which indicates the end of data and should not be treated as input. You may assume that none of the input integers is greater than 151200.
Output
The output is composed of lines, each containing a single integer. Each output integer should be the greatest integer that is the sum of a nonnegative cubic number and a nonnegative tetrahedral number and that is not greater than the corresponding input number. No other characters should appear in the output.
Example
Input
100
64
50
20
151200
0
Output
99
64
47
20
151200
"Correct Solution:
```
if __name__ == '__main__':
while True:
num = int(input())
if num == 0:
break
ans = 999999999
for c in range(54):
cubic = c * c * c
if cubic > num:
break
for t in range(96):
tetra = (t * (t+1) * (t+2)) // 6
if cubic + tetra > num:
break
if abs(cubic + tetra - num) < abs(num - ans):
ans = cubic + tetra
print(ans)
```
| 76,640 | [
0.413330078125,
0.1265869140625,
0.301025390625,
0.230712890625,
-0.66455078125,
-0.256591796875,
-0.058502197265625,
0.120361328125,
0.333740234375,
0.67822265625,
0.74609375,
0.264892578125,
0.259033203125,
-0.66552734375,
-0.5634765625,
0.35595703125,
-0.06414794921875,
-0.59082... | 3 |
Provide tags and a correct Python 3 solution for this coding contest problem.
A frog is currently at the point 0 on a coordinate axis Ox. It jumps by the following algorithm: the first jump is a units to the right, the second jump is b units to the left, the third jump is a units to the right, the fourth jump is b units to the left, and so on.
Formally:
* if the frog has jumped an even number of times (before the current jump), it jumps from its current position x to position x+a;
* otherwise it jumps from its current position x to position x-b.
Your task is to calculate the position of the frog after k jumps.
But... One more thing. You are watching t different frogs so you have to answer t independent queries.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 1000) — the number of queries.
Each of the next t lines contain queries (one query per line).
The query is described as three space-separated integers a, b, k (1 ≤ a, b, k ≤ 10^9) — the lengths of two types of jumps and the number of jumps, respectively.
Output
Print t integers. The i-th integer should be the answer for the i-th query.
Example
Input
6
5 2 3
100 1 4
1 10 5
1000000000 1 6
1 1 1000000000
1 1 999999999
Output
8
198
-17
2999999997
0
1
Note
In the first query frog jumps 5 to the right, 2 to the left and 5 to the right so the answer is 5 - 2 + 5 = 8.
In the second query frog jumps 100 to the right, 1 to the left, 100 to the right and 1 to the left so the answer is 100 - 1 + 100 - 1 = 198.
In the third query the answer is 1 - 10 + 1 - 10 + 1 = -17.
In the fourth query the answer is 10^9 - 1 + 10^9 - 1 + 10^9 - 1 = 2999999997.
In the fifth query all frog's jumps are neutralized by each other so the answer is 0.
The sixth query is the same as the fifth but without the last jump so the answer is 1.
Tags: math
Correct Solution:
```
t = int(input())
for i in range(t):
list1 = input().split()
a = int(list1[0])
b = int(list1[1])
k = int(list1[2])
x = 0
if k % 2 == 0:
print((a * (k //2)) - (b * (k //2)))
else:
x += a
k -= 1
print(x + (a * (k //2)) - (b * (k //2)))
```
| 76,720 | [
0.360595703125,
-0.050079345703125,
-0.4248046875,
0.377685546875,
-0.161376953125,
-0.52001953125,
0.185791015625,
0.43310546875,
0.456787109375,
1.080078125,
0.40869140625,
-0.118408203125,
-0.0176544189453125,
-0.50537109375,
-0.231689453125,
0.034149169921875,
-0.390869140625,
... | 3 |
Provide tags and a correct Python 3 solution for this coding contest problem.
A frog is currently at the point 0 on a coordinate axis Ox. It jumps by the following algorithm: the first jump is a units to the right, the second jump is b units to the left, the third jump is a units to the right, the fourth jump is b units to the left, and so on.
Formally:
* if the frog has jumped an even number of times (before the current jump), it jumps from its current position x to position x+a;
* otherwise it jumps from its current position x to position x-b.
Your task is to calculate the position of the frog after k jumps.
But... One more thing. You are watching t different frogs so you have to answer t independent queries.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 1000) — the number of queries.
Each of the next t lines contain queries (one query per line).
The query is described as three space-separated integers a, b, k (1 ≤ a, b, k ≤ 10^9) — the lengths of two types of jumps and the number of jumps, respectively.
Output
Print t integers. The i-th integer should be the answer for the i-th query.
Example
Input
6
5 2 3
100 1 4
1 10 5
1000000000 1 6
1 1 1000000000
1 1 999999999
Output
8
198
-17
2999999997
0
1
Note
In the first query frog jumps 5 to the right, 2 to the left and 5 to the right so the answer is 5 - 2 + 5 = 8.
In the second query frog jumps 100 to the right, 1 to the left, 100 to the right and 1 to the left so the answer is 100 - 1 + 100 - 1 = 198.
In the third query the answer is 1 - 10 + 1 - 10 + 1 = -17.
In the fourth query the answer is 10^9 - 1 + 10^9 - 1 + 10^9 - 1 = 2999999997.
In the fifth query all frog's jumps are neutralized by each other so the answer is 0.
The sixth query is the same as the fifth but without the last jump so the answer is 1.
Tags: math
Correct Solution:
```
t = int(input())
for _ in range(t):
a, b, k = map(int, input().split())
r = a * ((k + 1) // 2)
l = b * (k // 2)
print(r - l)
# 5 - 2 + 5 = 2 * 5 - 1 * 2
```
| 76,721 | [
0.351806640625,
-0.0615234375,
-0.41796875,
0.380126953125,
-0.1568603515625,
-0.521484375,
0.2005615234375,
0.43017578125,
0.447509765625,
1.083984375,
0.4091796875,
-0.132568359375,
-0.0144805908203125,
-0.50244140625,
-0.24658203125,
0.039947509765625,
-0.405029296875,
-0.610351... | 3 |
Provide tags and a correct Python 3 solution for this coding contest problem.
A frog is currently at the point 0 on a coordinate axis Ox. It jumps by the following algorithm: the first jump is a units to the right, the second jump is b units to the left, the third jump is a units to the right, the fourth jump is b units to the left, and so on.
Formally:
* if the frog has jumped an even number of times (before the current jump), it jumps from its current position x to position x+a;
* otherwise it jumps from its current position x to position x-b.
Your task is to calculate the position of the frog after k jumps.
But... One more thing. You are watching t different frogs so you have to answer t independent queries.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 1000) — the number of queries.
Each of the next t lines contain queries (one query per line).
The query is described as three space-separated integers a, b, k (1 ≤ a, b, k ≤ 10^9) — the lengths of two types of jumps and the number of jumps, respectively.
Output
Print t integers. The i-th integer should be the answer for the i-th query.
Example
Input
6
5 2 3
100 1 4
1 10 5
1000000000 1 6
1 1 1000000000
1 1 999999999
Output
8
198
-17
2999999997
0
1
Note
In the first query frog jumps 5 to the right, 2 to the left and 5 to the right so the answer is 5 - 2 + 5 = 8.
In the second query frog jumps 100 to the right, 1 to the left, 100 to the right and 1 to the left so the answer is 100 - 1 + 100 - 1 = 198.
In the third query the answer is 1 - 10 + 1 - 10 + 1 = -17.
In the fourth query the answer is 10^9 - 1 + 10^9 - 1 + 10^9 - 1 = 2999999997.
In the fifth query all frog's jumps are neutralized by each other so the answer is 0.
The sixth query is the same as the fifth but without the last jump so the answer is 1.
Tags: math
Correct Solution:
```
T = int(input())
for t in range(T):
a, b, c = map(int, input().split())
ans = a * ((c + 1) // 2) - b * (c // 2)
print(ans)
```
| 76,722 | [
0.358642578125,
-0.050262451171875,
-0.423095703125,
0.37939453125,
-0.158935546875,
-0.521484375,
0.1868896484375,
0.43896484375,
0.45166015625,
1.083984375,
0.4033203125,
-0.11907958984375,
-0.01277923583984375,
-0.50537109375,
-0.234619140625,
0.03369140625,
-0.387939453125,
-0.... | 3 |
Provide tags and a correct Python 3 solution for this coding contest problem.
A frog is currently at the point 0 on a coordinate axis Ox. It jumps by the following algorithm: the first jump is a units to the right, the second jump is b units to the left, the third jump is a units to the right, the fourth jump is b units to the left, and so on.
Formally:
* if the frog has jumped an even number of times (before the current jump), it jumps from its current position x to position x+a;
* otherwise it jumps from its current position x to position x-b.
Your task is to calculate the position of the frog after k jumps.
But... One more thing. You are watching t different frogs so you have to answer t independent queries.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 1000) — the number of queries.
Each of the next t lines contain queries (one query per line).
The query is described as three space-separated integers a, b, k (1 ≤ a, b, k ≤ 10^9) — the lengths of two types of jumps and the number of jumps, respectively.
Output
Print t integers. The i-th integer should be the answer for the i-th query.
Example
Input
6
5 2 3
100 1 4
1 10 5
1000000000 1 6
1 1 1000000000
1 1 999999999
Output
8
198
-17
2999999997
0
1
Note
In the first query frog jumps 5 to the right, 2 to the left and 5 to the right so the answer is 5 - 2 + 5 = 8.
In the second query frog jumps 100 to the right, 1 to the left, 100 to the right and 1 to the left so the answer is 100 - 1 + 100 - 1 = 198.
In the third query the answer is 1 - 10 + 1 - 10 + 1 = -17.
In the fourth query the answer is 10^9 - 1 + 10^9 - 1 + 10^9 - 1 = 2999999997.
In the fifth query all frog's jumps are neutralized by each other so the answer is 0.
The sixth query is the same as the fifth but without the last jump so the answer is 1.
Tags: math
Correct Solution:
```
x=int(input())
y=0
for i in range(x):
r,l,n=map(int,input().split())
if n%2==0:
y=(n//2)*r-(n//2)*l
else:
y=(n//2)*r-(n//2)*l+r
print(y)
```
| 76,723 | [
0.361083984375,
-0.049407958984375,
-0.40673828125,
0.3798828125,
-0.16015625,
-0.529296875,
0.175048828125,
0.431640625,
0.468505859375,
1.0947265625,
0.4345703125,
-0.11865234375,
-0.013824462890625,
-0.50048828125,
-0.2109375,
0.05303955078125,
-0.415283203125,
-0.63671875,
-0... | 3 |
Provide tags and a correct Python 3 solution for this coding contest problem.
A frog is currently at the point 0 on a coordinate axis Ox. It jumps by the following algorithm: the first jump is a units to the right, the second jump is b units to the left, the third jump is a units to the right, the fourth jump is b units to the left, and so on.
Formally:
* if the frog has jumped an even number of times (before the current jump), it jumps from its current position x to position x+a;
* otherwise it jumps from its current position x to position x-b.
Your task is to calculate the position of the frog after k jumps.
But... One more thing. You are watching t different frogs so you have to answer t independent queries.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 1000) — the number of queries.
Each of the next t lines contain queries (one query per line).
The query is described as three space-separated integers a, b, k (1 ≤ a, b, k ≤ 10^9) — the lengths of two types of jumps and the number of jumps, respectively.
Output
Print t integers. The i-th integer should be the answer for the i-th query.
Example
Input
6
5 2 3
100 1 4
1 10 5
1000000000 1 6
1 1 1000000000
1 1 999999999
Output
8
198
-17
2999999997
0
1
Note
In the first query frog jumps 5 to the right, 2 to the left and 5 to the right so the answer is 5 - 2 + 5 = 8.
In the second query frog jumps 100 to the right, 1 to the left, 100 to the right and 1 to the left so the answer is 100 - 1 + 100 - 1 = 198.
In the third query the answer is 1 - 10 + 1 - 10 + 1 = -17.
In the fourth query the answer is 10^9 - 1 + 10^9 - 1 + 10^9 - 1 = 2999999997.
In the fifth query all frog's jumps are neutralized by each other so the answer is 0.
The sixth query is the same as the fifth but without the last jump so the answer is 1.
Tags: math
Correct Solution:
```
def jump(a,b,k):
if (k % 2 == 0):
return (a - b) * (k // 2)
else:
return (a - b) * (k // 2) + a
t = int(input())
for i in range(t):
a, b, k = map(int, input().split())
result = jump(a, b, k)
print(result)
```
| 76,724 | [
0.34716796875,
-0.052154541015625,
-0.432373046875,
0.375244140625,
-0.153564453125,
-0.53662109375,
0.1953125,
0.458251953125,
0.475341796875,
1.0830078125,
0.4296875,
-0.1461181640625,
0.02838134765625,
-0.455322265625,
-0.210693359375,
0.09033203125,
-0.461669921875,
-0.64794921... | 3 |
Provide tags and a correct Python 3 solution for this coding contest problem.
A frog is currently at the point 0 on a coordinate axis Ox. It jumps by the following algorithm: the first jump is a units to the right, the second jump is b units to the left, the third jump is a units to the right, the fourth jump is b units to the left, and so on.
Formally:
* if the frog has jumped an even number of times (before the current jump), it jumps from its current position x to position x+a;
* otherwise it jumps from its current position x to position x-b.
Your task is to calculate the position of the frog after k jumps.
But... One more thing. You are watching t different frogs so you have to answer t independent queries.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 1000) — the number of queries.
Each of the next t lines contain queries (one query per line).
The query is described as three space-separated integers a, b, k (1 ≤ a, b, k ≤ 10^9) — the lengths of two types of jumps and the number of jumps, respectively.
Output
Print t integers. The i-th integer should be the answer for the i-th query.
Example
Input
6
5 2 3
100 1 4
1 10 5
1000000000 1 6
1 1 1000000000
1 1 999999999
Output
8
198
-17
2999999997
0
1
Note
In the first query frog jumps 5 to the right, 2 to the left and 5 to the right so the answer is 5 - 2 + 5 = 8.
In the second query frog jumps 100 to the right, 1 to the left, 100 to the right and 1 to the left so the answer is 100 - 1 + 100 - 1 = 198.
In the third query the answer is 1 - 10 + 1 - 10 + 1 = -17.
In the fourth query the answer is 10^9 - 1 + 10^9 - 1 + 10^9 - 1 = 2999999997.
In the fifth query all frog's jumps are neutralized by each other so the answer is 0.
The sixth query is the same as the fifth but without the last jump so the answer is 1.
Tags: math
Correct Solution:
```
# -*- coding: utf-8 -*-
import sys
import math
import os
import itertools
import string
import heapq
import _collections
from collections import Counter
from collections import defaultdict
from functools import lru_cache
import bisect
import re
import queue
from decimal import *
class Scanner():
@staticmethod
def int():
return int(sys.stdin.readline().rstrip())
@staticmethod
def string():
return sys.stdin.readline().rstrip()
@staticmethod
def map_int():
return [int(x) for x in Scanner.string().split()]
@staticmethod
def string_list(n):
return [input() for i in range(n)]
@staticmethod
def int_list_list(n):
return [Scanner.map_int() for i in range(n)]
@staticmethod
def int_cols_list(n):
return [int(input()) for i in range(n)]
class Math():
@staticmethod
def gcd(a, b):
if b == 0:
return a
return Math.gcd(b, a % b)
@staticmethod
def lcm(a, b):
return (a * b) // Math.gcd(a, b)
@staticmethod
def divisor(n):
res = []
i = 1
for i in range(1, int(n ** 0.5) + 1):
if n % i == 0:
res.append(i)
if i != n // i:
res.append(n // i)
return res
@staticmethod
def round_up(a, b):
return -(-a // b)
@staticmethod
def is_prime(n):
if n < 2:
return False
if n == 2:
return True
if n % 2 == 0:
return False
d = int(n ** 0.5) + 1
for i in range(3, d + 1, 2):
if n % i == 0:
return False
return True
class PriorityQueue:
def __init__(self, l=[]):
self.q = l
heapq.heapify(self.q)
return
def push(self, n):
heapq.heappush(self.q, n)
return
def pop(self):
return heapq.heappop(self.q)
def pop_count(x):
x = x - ((x >> 1) & 0x5555555555555555)
x = (x & 0x3333333333333333) + ((x >> 2) & 0x3333333333333333)
x = (x + (x >> 4)) & 0x0f0f0f0f0f0f0f0f
x = x + (x >> 8)
x = x + (x >> 16)
x = x + (x >> 32)
return x & 0x0000007f
MOD = int(1e09) + 7
INF = int(1e15)
def main():
# sys.stdin = open("sample.txt")
T = Scanner.int()
for _ in range(T):
a, b, k = Scanner.map_int()
print(a * Math.round_up(k, 2) - b * (k // 2))
if __name__ == "__main__":
main()
```
| 76,725 | [
0.3720703125,
-0.0222320556640625,
-0.4111328125,
0.379638671875,
-0.1748046875,
-0.5126953125,
0.2088623046875,
0.436279296875,
0.444091796875,
1.05078125,
0.41357421875,
-0.1343994140625,
0.0083465576171875,
-0.418212890625,
-0.28271484375,
0.029754638671875,
-0.382568359375,
-0.... | 3 |
Provide tags and a correct Python 3 solution for this coding contest problem.
A frog is currently at the point 0 on a coordinate axis Ox. It jumps by the following algorithm: the first jump is a units to the right, the second jump is b units to the left, the third jump is a units to the right, the fourth jump is b units to the left, and so on.
Formally:
* if the frog has jumped an even number of times (before the current jump), it jumps from its current position x to position x+a;
* otherwise it jumps from its current position x to position x-b.
Your task is to calculate the position of the frog after k jumps.
But... One more thing. You are watching t different frogs so you have to answer t independent queries.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 1000) — the number of queries.
Each of the next t lines contain queries (one query per line).
The query is described as three space-separated integers a, b, k (1 ≤ a, b, k ≤ 10^9) — the lengths of two types of jumps and the number of jumps, respectively.
Output
Print t integers. The i-th integer should be the answer for the i-th query.
Example
Input
6
5 2 3
100 1 4
1 10 5
1000000000 1 6
1 1 1000000000
1 1 999999999
Output
8
198
-17
2999999997
0
1
Note
In the first query frog jumps 5 to the right, 2 to the left and 5 to the right so the answer is 5 - 2 + 5 = 8.
In the second query frog jumps 100 to the right, 1 to the left, 100 to the right and 1 to the left so the answer is 100 - 1 + 100 - 1 = 198.
In the third query the answer is 1 - 10 + 1 - 10 + 1 = -17.
In the fourth query the answer is 10^9 - 1 + 10^9 - 1 + 10^9 - 1 = 2999999997.
In the fifth query all frog's jumps are neutralized by each other so the answer is 0.
The sixth query is the same as the fifth but without the last jump so the answer is 1.
Tags: math
Correct Solution:
```
t=int(input())
while t!=0:
a,b,k=map(int,input().split(" "))
if k%2==0:
a=a-b
k=int(k/2)
print(a*k)
else:
lol=a
a=a-b
k=int((k-1)/2)
print(a*k+lol)
t=t-1
```
| 76,726 | [
0.359130859375,
-0.04095458984375,
-0.424072265625,
0.374267578125,
-0.15673828125,
-0.52587890625,
0.1839599609375,
0.438232421875,
0.44873046875,
1.078125,
0.412353515625,
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-0.00872802734375,
-0.497314453125,
-0.231201171875,
0.040313720703125,
-0.3935546875,
-0.61... | 3 |
Provide tags and a correct Python 3 solution for this coding contest problem.
A frog is currently at the point 0 on a coordinate axis Ox. It jumps by the following algorithm: the first jump is a units to the right, the second jump is b units to the left, the third jump is a units to the right, the fourth jump is b units to the left, and so on.
Formally:
* if the frog has jumped an even number of times (before the current jump), it jumps from its current position x to position x+a;
* otherwise it jumps from its current position x to position x-b.
Your task is to calculate the position of the frog after k jumps.
But... One more thing. You are watching t different frogs so you have to answer t independent queries.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 1000) — the number of queries.
Each of the next t lines contain queries (one query per line).
The query is described as three space-separated integers a, b, k (1 ≤ a, b, k ≤ 10^9) — the lengths of two types of jumps and the number of jumps, respectively.
Output
Print t integers. The i-th integer should be the answer for the i-th query.
Example
Input
6
5 2 3
100 1 4
1 10 5
1000000000 1 6
1 1 1000000000
1 1 999999999
Output
8
198
-17
2999999997
0
1
Note
In the first query frog jumps 5 to the right, 2 to the left and 5 to the right so the answer is 5 - 2 + 5 = 8.
In the second query frog jumps 100 to the right, 1 to the left, 100 to the right and 1 to the left so the answer is 100 - 1 + 100 - 1 = 198.
In the third query the answer is 1 - 10 + 1 - 10 + 1 = -17.
In the fourth query the answer is 10^9 - 1 + 10^9 - 1 + 10^9 - 1 = 2999999997.
In the fifth query all frog's jumps are neutralized by each other so the answer is 0.
The sixth query is the same as the fifth but without the last jump so the answer is 1.
Tags: math
Correct Solution:
```
import sys
num_queries = int(input())
while num_queries > 0:
a, b, k = map(int, input().split(" "))
right = k // 2
if k % 2 == 1: right += 1
left = k // 2
print(a * right - b * left)
num_queries -= 1
```
| 76,727 | [
0.330810546875,
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-0.394287109375,
0.37744140625,
-0.1832275390625,
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0.15625,
0.45166015625,
0.456787109375,
1.087890625,
0.413330078125,
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-0.0015077590942382812,
-0.490234375,
-0.2734375,
0.0938720703125,
-0.379150390625,
-0.6459... | 3 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A frog is currently at the point 0 on a coordinate axis Ox. It jumps by the following algorithm: the first jump is a units to the right, the second jump is b units to the left, the third jump is a units to the right, the fourth jump is b units to the left, and so on.
Formally:
* if the frog has jumped an even number of times (before the current jump), it jumps from its current position x to position x+a;
* otherwise it jumps from its current position x to position x-b.
Your task is to calculate the position of the frog after k jumps.
But... One more thing. You are watching t different frogs so you have to answer t independent queries.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 1000) — the number of queries.
Each of the next t lines contain queries (one query per line).
The query is described as three space-separated integers a, b, k (1 ≤ a, b, k ≤ 10^9) — the lengths of two types of jumps and the number of jumps, respectively.
Output
Print t integers. The i-th integer should be the answer for the i-th query.
Example
Input
6
5 2 3
100 1 4
1 10 5
1000000000 1 6
1 1 1000000000
1 1 999999999
Output
8
198
-17
2999999997
0
1
Note
In the first query frog jumps 5 to the right, 2 to the left and 5 to the right so the answer is 5 - 2 + 5 = 8.
In the second query frog jumps 100 to the right, 1 to the left, 100 to the right and 1 to the left so the answer is 100 - 1 + 100 - 1 = 198.
In the third query the answer is 1 - 10 + 1 - 10 + 1 = -17.
In the fourth query the answer is 10^9 - 1 + 10^9 - 1 + 10^9 - 1 = 2999999997.
In the fifth query all frog's jumps are neutralized by each other so the answer is 0.
The sixth query is the same as the fifth but without the last jump so the answer is 1.
Submitted Solution:
```
import math
for _ in range(int(input())):
a,b,k = map(int,input().split(" "))
p = int(math.ceil(k/2))
print((a*p)-(b*(k-p)))
```
Yes
| 76,728 | [
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0.265625,
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0.5439453125,
0.38134765625,
1.02734375,
0.469482421875,
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-0.0621337890625,
-0.498291015625,
-0.2685546875,
-0.07110595703125,
-0.34765625,
-0.5703125,
... | 3 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A frog is currently at the point 0 on a coordinate axis Ox. It jumps by the following algorithm: the first jump is a units to the right, the second jump is b units to the left, the third jump is a units to the right, the fourth jump is b units to the left, and so on.
Formally:
* if the frog has jumped an even number of times (before the current jump), it jumps from its current position x to position x+a;
* otherwise it jumps from its current position x to position x-b.
Your task is to calculate the position of the frog after k jumps.
But... One more thing. You are watching t different frogs so you have to answer t independent queries.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 1000) — the number of queries.
Each of the next t lines contain queries (one query per line).
The query is described as three space-separated integers a, b, k (1 ≤ a, b, k ≤ 10^9) — the lengths of two types of jumps and the number of jumps, respectively.
Output
Print t integers. The i-th integer should be the answer for the i-th query.
Example
Input
6
5 2 3
100 1 4
1 10 5
1000000000 1 6
1 1 1000000000
1 1 999999999
Output
8
198
-17
2999999997
0
1
Note
In the first query frog jumps 5 to the right, 2 to the left and 5 to the right so the answer is 5 - 2 + 5 = 8.
In the second query frog jumps 100 to the right, 1 to the left, 100 to the right and 1 to the left so the answer is 100 - 1 + 100 - 1 = 198.
In the third query the answer is 1 - 10 + 1 - 10 + 1 = -17.
In the fourth query the answer is 10^9 - 1 + 10^9 - 1 + 10^9 - 1 = 2999999997.
In the fifth query all frog's jumps are neutralized by each other so the answer is 0.
The sixth query is the same as the fifth but without the last jump so the answer is 1.
Submitted Solution:
```
for i in range(int(input())):
l = [int(a) for a in input().split()]
if l[2] % 2 == 0:
print((l[0]-l[1])*(l[2]//2))
else:
print((l[0]-l[1])*(l[2]//2)+l[0])
```
Yes
| 76,729 | [
0.33349609375,
0.031890869140625,
-0.43994140625,
0.2978515625,
-0.1995849609375,
-0.321533203125,
0.08624267578125,
0.548828125,
0.369140625,
1.044921875,
0.45703125,
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-0.5107421875,
-0.2548828125,
-0.064697265625,
-0.330322265625,
-0.5869140625,
... | 3 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A frog is currently at the point 0 on a coordinate axis Ox. It jumps by the following algorithm: the first jump is a units to the right, the second jump is b units to the left, the third jump is a units to the right, the fourth jump is b units to the left, and so on.
Formally:
* if the frog has jumped an even number of times (before the current jump), it jumps from its current position x to position x+a;
* otherwise it jumps from its current position x to position x-b.
Your task is to calculate the position of the frog after k jumps.
But... One more thing. You are watching t different frogs so you have to answer t independent queries.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 1000) — the number of queries.
Each of the next t lines contain queries (one query per line).
The query is described as three space-separated integers a, b, k (1 ≤ a, b, k ≤ 10^9) — the lengths of two types of jumps and the number of jumps, respectively.
Output
Print t integers. The i-th integer should be the answer for the i-th query.
Example
Input
6
5 2 3
100 1 4
1 10 5
1000000000 1 6
1 1 1000000000
1 1 999999999
Output
8
198
-17
2999999997
0
1
Note
In the first query frog jumps 5 to the right, 2 to the left and 5 to the right so the answer is 5 - 2 + 5 = 8.
In the second query frog jumps 100 to the right, 1 to the left, 100 to the right and 1 to the left so the answer is 100 - 1 + 100 - 1 = 198.
In the third query the answer is 1 - 10 + 1 - 10 + 1 = -17.
In the fourth query the answer is 10^9 - 1 + 10^9 - 1 + 10^9 - 1 = 2999999997.
In the fifth query all frog's jumps are neutralized by each other so the answer is 0.
The sixth query is the same as the fifth but without the last jump so the answer is 1.
Submitted Solution:
```
t = int(input())
for _ in range(t):
a,b,k = map(int,input().split())
k = k - 1
if k%2 == 0:
print(a + (a-b)*(k//2))
else:
print(a + (a - b) * (k//2) - b)
```
Yes
| 76,730 | [
0.33984375,
0.0311431884765625,
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0.292236328125,
-0.1845703125,
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0.08709716796875,
0.5498046875,
0.3759765625,
1.0224609375,
0.4375,
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-0.50341796875,
-0.26708984375,
-0.08111572265625,
-0.3212890625,
-0.59521484375,
... | 3 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A frog is currently at the point 0 on a coordinate axis Ox. It jumps by the following algorithm: the first jump is a units to the right, the second jump is b units to the left, the third jump is a units to the right, the fourth jump is b units to the left, and so on.
Formally:
* if the frog has jumped an even number of times (before the current jump), it jumps from its current position x to position x+a;
* otherwise it jumps from its current position x to position x-b.
Your task is to calculate the position of the frog after k jumps.
But... One more thing. You are watching t different frogs so you have to answer t independent queries.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 1000) — the number of queries.
Each of the next t lines contain queries (one query per line).
The query is described as three space-separated integers a, b, k (1 ≤ a, b, k ≤ 10^9) — the lengths of two types of jumps and the number of jumps, respectively.
Output
Print t integers. The i-th integer should be the answer for the i-th query.
Example
Input
6
5 2 3
100 1 4
1 10 5
1000000000 1 6
1 1 1000000000
1 1 999999999
Output
8
198
-17
2999999997
0
1
Note
In the first query frog jumps 5 to the right, 2 to the left and 5 to the right so the answer is 5 - 2 + 5 = 8.
In the second query frog jumps 100 to the right, 1 to the left, 100 to the right and 1 to the left so the answer is 100 - 1 + 100 - 1 = 198.
In the third query the answer is 1 - 10 + 1 - 10 + 1 = -17.
In the fourth query the answer is 10^9 - 1 + 10^9 - 1 + 10^9 - 1 = 2999999997.
In the fifth query all frog's jumps are neutralized by each other so the answer is 0.
The sixth query is the same as the fifth but without the last jump so the answer is 1.
Submitted Solution:
```
# cook your dish here
t=int(input())
for i in range(t):
a,b,k=map(int,input().split(' '))
if k%2==0:
x=int(k/2)
print(a*x-b*x)
else:
k=k-1
x=int(k/2)
print(a*(x+1)-b*x)
```
Yes
| 76,731 | [
0.314208984375,
0.06561279296875,
-0.5087890625,
0.27587890625,
-0.263671875,
-0.2275390625,
0.09326171875,
0.60205078125,
0.456787109375,
1.046875,
0.44091796875,
-0.1663818359375,
-0.035980224609375,
-0.509765625,
-0.323486328125,
-0.007251739501953125,
-0.311279296875,
-0.643066... | 3 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A frog is currently at the point 0 on a coordinate axis Ox. It jumps by the following algorithm: the first jump is a units to the right, the second jump is b units to the left, the third jump is a units to the right, the fourth jump is b units to the left, and so on.
Formally:
* if the frog has jumped an even number of times (before the current jump), it jumps from its current position x to position x+a;
* otherwise it jumps from its current position x to position x-b.
Your task is to calculate the position of the frog after k jumps.
But... One more thing. You are watching t different frogs so you have to answer t independent queries.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 1000) — the number of queries.
Each of the next t lines contain queries (one query per line).
The query is described as three space-separated integers a, b, k (1 ≤ a, b, k ≤ 10^9) — the lengths of two types of jumps and the number of jumps, respectively.
Output
Print t integers. The i-th integer should be the answer for the i-th query.
Example
Input
6
5 2 3
100 1 4
1 10 5
1000000000 1 6
1 1 1000000000
1 1 999999999
Output
8
198
-17
2999999997
0
1
Note
In the first query frog jumps 5 to the right, 2 to the left and 5 to the right so the answer is 5 - 2 + 5 = 8.
In the second query frog jumps 100 to the right, 1 to the left, 100 to the right and 1 to the left so the answer is 100 - 1 + 100 - 1 = 198.
In the third query the answer is 1 - 10 + 1 - 10 + 1 = -17.
In the fourth query the answer is 10^9 - 1 + 10^9 - 1 + 10^9 - 1 = 2999999997.
In the fifth query all frog's jumps are neutralized by each other so the answer is 0.
The sixth query is the same as the fifth but without the last jump so the answer is 1.
Submitted Solution:
```
t = int(input())
res = list()
for i in range(t):
a, b, k = list(map(int, input().strip().split()))
if k % 2 == 1:
pos = (k - k // 2) * a - (k // 2) * b
else:
pos = k * (a - b) / 2
res.append(int(pos))
for i in res:
print(i)
```
No
| 76,732 | [
0.33984375,
0.0311431884765625,
-0.456298828125,
0.292236328125,
-0.1845703125,
-0.32080078125,
0.08709716796875,
0.5498046875,
0.3759765625,
1.0224609375,
0.4375,
-0.176025390625,
-0.077880859375,
-0.50341796875,
-0.26708984375,
-0.08111572265625,
-0.3212890625,
-0.59521484375,
... | 3 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A frog is currently at the point 0 on a coordinate axis Ox. It jumps by the following algorithm: the first jump is a units to the right, the second jump is b units to the left, the third jump is a units to the right, the fourth jump is b units to the left, and so on.
Formally:
* if the frog has jumped an even number of times (before the current jump), it jumps from its current position x to position x+a;
* otherwise it jumps from its current position x to position x-b.
Your task is to calculate the position of the frog after k jumps.
But... One more thing. You are watching t different frogs so you have to answer t independent queries.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 1000) — the number of queries.
Each of the next t lines contain queries (one query per line).
The query is described as three space-separated integers a, b, k (1 ≤ a, b, k ≤ 10^9) — the lengths of two types of jumps and the number of jumps, respectively.
Output
Print t integers. The i-th integer should be the answer for the i-th query.
Example
Input
6
5 2 3
100 1 4
1 10 5
1000000000 1 6
1 1 1000000000
1 1 999999999
Output
8
198
-17
2999999997
0
1
Note
In the first query frog jumps 5 to the right, 2 to the left and 5 to the right so the answer is 5 - 2 + 5 = 8.
In the second query frog jumps 100 to the right, 1 to the left, 100 to the right and 1 to the left so the answer is 100 - 1 + 100 - 1 = 198.
In the third query the answer is 1 - 10 + 1 - 10 + 1 = -17.
In the fourth query the answer is 10^9 - 1 + 10^9 - 1 + 10^9 - 1 = 2999999997.
In the fifth query all frog's jumps are neutralized by each other so the answer is 0.
The sixth query is the same as the fifth but without the last jump so the answer is 1.
Submitted Solution:
```
n = int(input())
for i in range(n):
a, b, k = map(int, input().split())
print((a - b) * k)
```
No
| 76,733 | [
0.33984375,
0.03265380859375,
-0.455810546875,
0.30712890625,
-0.18603515625,
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0.08294677734375,
0.552734375,
0.379638671875,
1.0390625,
0.459228515625,
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-0.07080078125,
-0.50537109375,
-0.261474609375,
-0.06427001953125,
-0.336181640625,
-0.607910156... | 3 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A frog is currently at the point 0 on a coordinate axis Ox. It jumps by the following algorithm: the first jump is a units to the right, the second jump is b units to the left, the third jump is a units to the right, the fourth jump is b units to the left, and so on.
Formally:
* if the frog has jumped an even number of times (before the current jump), it jumps from its current position x to position x+a;
* otherwise it jumps from its current position x to position x-b.
Your task is to calculate the position of the frog after k jumps.
But... One more thing. You are watching t different frogs so you have to answer t independent queries.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 1000) — the number of queries.
Each of the next t lines contain queries (one query per line).
The query is described as three space-separated integers a, b, k (1 ≤ a, b, k ≤ 10^9) — the lengths of two types of jumps and the number of jumps, respectively.
Output
Print t integers. The i-th integer should be the answer for the i-th query.
Example
Input
6
5 2 3
100 1 4
1 10 5
1000000000 1 6
1 1 1000000000
1 1 999999999
Output
8
198
-17
2999999997
0
1
Note
In the first query frog jumps 5 to the right, 2 to the left and 5 to the right so the answer is 5 - 2 + 5 = 8.
In the second query frog jumps 100 to the right, 1 to the left, 100 to the right and 1 to the left so the answer is 100 - 1 + 100 - 1 = 198.
In the third query the answer is 1 - 10 + 1 - 10 + 1 = -17.
In the fourth query the answer is 10^9 - 1 + 10^9 - 1 + 10^9 - 1 = 2999999997.
In the fifth query all frog's jumps are neutralized by each other so the answer is 0.
The sixth query is the same as the fifth but without the last jump so the answer is 1.
Submitted Solution:
```
t = int(input())
out = [0]*t
for it in range(t):
a, b, k = map(int, input().split())
if k%2==0:
out[it] = out[it] + a*(k/2) - b*(k/2)
else:
k = k-1
out[it] = out[it] + a*(k/2) - b*(k/2) + a
for o in range(t):
print(int(out[o]))
```
No
| 76,734 | [
0.33984375,
0.0311431884765625,
-0.456298828125,
0.292236328125,
-0.1845703125,
-0.32080078125,
0.08709716796875,
0.5498046875,
0.3759765625,
1.0224609375,
0.4375,
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-0.50341796875,
-0.26708984375,
-0.08111572265625,
-0.3212890625,
-0.59521484375,
... | 3 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A frog is currently at the point 0 on a coordinate axis Ox. It jumps by the following algorithm: the first jump is a units to the right, the second jump is b units to the left, the third jump is a units to the right, the fourth jump is b units to the left, and so on.
Formally:
* if the frog has jumped an even number of times (before the current jump), it jumps from its current position x to position x+a;
* otherwise it jumps from its current position x to position x-b.
Your task is to calculate the position of the frog after k jumps.
But... One more thing. You are watching t different frogs so you have to answer t independent queries.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 1000) — the number of queries.
Each of the next t lines contain queries (one query per line).
The query is described as three space-separated integers a, b, k (1 ≤ a, b, k ≤ 10^9) — the lengths of two types of jumps and the number of jumps, respectively.
Output
Print t integers. The i-th integer should be the answer for the i-th query.
Example
Input
6
5 2 3
100 1 4
1 10 5
1000000000 1 6
1 1 1000000000
1 1 999999999
Output
8
198
-17
2999999997
0
1
Note
In the first query frog jumps 5 to the right, 2 to the left and 5 to the right so the answer is 5 - 2 + 5 = 8.
In the second query frog jumps 100 to the right, 1 to the left, 100 to the right and 1 to the left so the answer is 100 - 1 + 100 - 1 = 198.
In the third query the answer is 1 - 10 + 1 - 10 + 1 = -17.
In the fourth query the answer is 10^9 - 1 + 10^9 - 1 + 10^9 - 1 = 2999999997.
In the fifth query all frog's jumps are neutralized by each other so the answer is 0.
The sixth query is the same as the fifth but without the last jump so the answer is 1.
Submitted Solution:
```
n=int(input())
j=0
for i in range(n):
a,b,k=map(int,input().split())
m=k%2
if m == 0:
m = k/2
j=m
else:
j=k//2+1
m=k-j
l=(a*j+b*-m)
print(l)
```
No
| 76,735 | [
0.33984375,
0.03265380859375,
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0.30712890625,
-0.18603515625,
-0.331298828125,
0.08294677734375,
0.552734375,
0.379638671875,
1.0390625,
0.459228515625,
-0.18212890625,
-0.07080078125,
-0.50537109375,
-0.261474609375,
-0.06427001953125,
-0.336181640625,
-0.607910156... | 3 |
Provide tags and a correct Python 3 solution for this coding contest problem.
A robot is standing at the origin of the infinite two-dimensional plane. Each second the robot moves exactly 1 meter in one of the four cardinal directions: north, south, west, and east. For the first step the robot can choose any of the four directions, but then at the end of every second it has to turn 90 degrees left or right with respect to the direction it just moved in. For example, if the robot has just moved north or south, the next step it takes has to be either west or east, and vice versa.
The robot makes exactly n steps from its starting position according to the rules above. How many different points can the robot arrive to at the end? The final orientation of the robot can be ignored.
Input
The only line contains a single integer n (1 ≤ n ≤ 1000) — the number of steps the robot makes.
Output
Print a single integer — the number of different possible locations after exactly n steps.
Examples
Input
1
Output
4
Input
2
Output
4
Input
3
Output
12
Note
In the first sample case, the robot will end up 1 meter north, south, west, or east depending on its initial direction.
In the second sample case, the robot will always end up √{2} meters north-west, north-east, south-west, or south-east.
Tags: dp, math
Correct Solution:
```
n = int(input())
if n%2 == 0:
if n == 2:
print(4)
else:
print((n//2+1)**2)
else:
if n == 1:
print(4)
else:
print(((n//2+1)*(n//2+2))*2)
```
| 76,916 | [
0.15087890625,
0.4658203125,
-0.4970703125,
-0.03326416015625,
-0.5888671875,
-0.51416015625,
-0.00485992431640625,
0.4169921875,
0.12310791015625,
1.166015625,
0.50244140625,
-0.045745849609375,
0.0533447265625,
-0.60888671875,
-0.38232421875,
0.141845703125,
-0.5166015625,
-0.776... | 3 |
Provide tags and a correct Python 3 solution for this coding contest problem.
A robot is standing at the origin of the infinite two-dimensional plane. Each second the robot moves exactly 1 meter in one of the four cardinal directions: north, south, west, and east. For the first step the robot can choose any of the four directions, but then at the end of every second it has to turn 90 degrees left or right with respect to the direction it just moved in. For example, if the robot has just moved north or south, the next step it takes has to be either west or east, and vice versa.
The robot makes exactly n steps from its starting position according to the rules above. How many different points can the robot arrive to at the end? The final orientation of the robot can be ignored.
Input
The only line contains a single integer n (1 ≤ n ≤ 1000) — the number of steps the robot makes.
Output
Print a single integer — the number of different possible locations after exactly n steps.
Examples
Input
1
Output
4
Input
2
Output
4
Input
3
Output
12
Note
In the first sample case, the robot will end up 1 meter north, south, west, or east depending on its initial direction.
In the second sample case, the robot will always end up √{2} meters north-west, north-east, south-west, or south-east.
Tags: dp, math
Correct Solution:
```
from sys import stdin, stdout
import math
from collections import defaultdict
ans = []
if __name__ == '__main__':
n = int(stdin.readline())
if not n%2:
k = n/2
print(int((k+1)**2))
else:
k = math.floor(n/2)
print(2*(k+1)*(k+2))
```
| 76,917 | [
0.161376953125,
0.425537109375,
-0.45361328125,
-0.0258636474609375,
-0.60400390625,
-0.471435546875,
-0.125732421875,
0.389892578125,
0.048095703125,
1.193359375,
0.39599609375,
-0.08056640625,
0.119140625,
-0.56591796875,
-0.390869140625,
0.1533203125,
-0.51513671875,
-0.82275390... | 3 |
Provide tags and a correct Python 3 solution for this coding contest problem.
A robot is standing at the origin of the infinite two-dimensional plane. Each second the robot moves exactly 1 meter in one of the four cardinal directions: north, south, west, and east. For the first step the robot can choose any of the four directions, but then at the end of every second it has to turn 90 degrees left or right with respect to the direction it just moved in. For example, if the robot has just moved north or south, the next step it takes has to be either west or east, and vice versa.
The robot makes exactly n steps from its starting position according to the rules above. How many different points can the robot arrive to at the end? The final orientation of the robot can be ignored.
Input
The only line contains a single integer n (1 ≤ n ≤ 1000) — the number of steps the robot makes.
Output
Print a single integer — the number of different possible locations after exactly n steps.
Examples
Input
1
Output
4
Input
2
Output
4
Input
3
Output
12
Note
In the first sample case, the robot will end up 1 meter north, south, west, or east depending on its initial direction.
In the second sample case, the robot will always end up √{2} meters north-west, north-east, south-west, or south-east.
Tags: dp, math
Correct Solution:
```
import sys
reader = (s.rstrip() for s in sys.stdin)
input = reader.__next__
n= int(input())
ans = [4,4,12,9]
if n<=4:
print(ans[n-1])
else:
for i in range(4,n):
toPlus = ((i)//2+1)*4
if i%2==0:
ans.append(ans[i-2]+toPlus)
else:
ans.append(ans[i-4]+toPlus)
print(ans[-1])
#"{} {} {}".format(maxele,minele,minele)
# 4 4 12 9 24 16 40 25 60
# +8 +12 +12 +16 +16 +20
```
| 76,918 | [
0.1873779296875,
0.439453125,
-0.424560546875,
-0.0114593505859375,
-0.58837890625,
-0.495849609375,
0.027587890625,
0.398193359375,
0.10595703125,
1.20703125,
0.42138671875,
-0.07696533203125,
0.1571044921875,
-0.5947265625,
-0.428955078125,
0.11419677734375,
-0.495361328125,
-0.7... | 3 |
Provide tags and a correct Python 3 solution for this coding contest problem.
A robot is standing at the origin of the infinite two-dimensional plane. Each second the robot moves exactly 1 meter in one of the four cardinal directions: north, south, west, and east. For the first step the robot can choose any of the four directions, but then at the end of every second it has to turn 90 degrees left or right with respect to the direction it just moved in. For example, if the robot has just moved north or south, the next step it takes has to be either west or east, and vice versa.
The robot makes exactly n steps from its starting position according to the rules above. How many different points can the robot arrive to at the end? The final orientation of the robot can be ignored.
Input
The only line contains a single integer n (1 ≤ n ≤ 1000) — the number of steps the robot makes.
Output
Print a single integer — the number of different possible locations after exactly n steps.
Examples
Input
1
Output
4
Input
2
Output
4
Input
3
Output
12
Note
In the first sample case, the robot will end up 1 meter north, south, west, or east depending on its initial direction.
In the second sample case, the robot will always end up √{2} meters north-west, north-east, south-west, or south-east.
Tags: dp, math
Correct Solution:
```
from sys import stdin, stdout
from math import floor, gcd, fabs, factorial, fmod, sqrt, inf, log
from collections import defaultdict as dd, deque
from bisect import bisect_left as bl, bisect_right as br, bisect
mod = pow(10, 9) + 7
mod2 = 998244353
def inp(): return stdin.readline().strip()
def out(var, end="\n"): stdout.write(str(var)+"\n")
def outa(*var, end="\n"): stdout.write(' '.join(map(str, var)) + end)
def lmp(): return list(mp())
def mp(): return map(int, inp().split())
def smp(): return map(str, inp().split())
def l1d(n, val=0): return [val for i in range(n)]
def l2d(n, m, val=0): return [l1d(n, val) for j in range(m)]
def remadd(x, y): return 1 if x%y else 0
def ceil(a,b): return (a+b-1)//b
def chkprime(x):
if x<=1: return False
if x in (2, 3): return True
if x%2 == 0: return False
for i in range(3, int(sqrt(x))+1, 2):
if x%i == 0: return False
return True
n = int(inp())
if n%2:
out((n//2+1)*(n//2+2)*2)
else:
if n%4==0:
out(((n//2)+1)**2)
else: out(4*((n//4)+1)**2)
```
| 76,919 | [
0.09942626953125,
0.450927734375,
-0.418212890625,
0.12298583984375,
-0.689453125,
-0.368896484375,
-0.04571533203125,
0.273193359375,
-0.01506805419921875,
1.251953125,
0.43505859375,
-0.026275634765625,
0.1744384765625,
-0.68359375,
-0.352294921875,
0.1656494140625,
-0.49169921875,... | 3 |
Provide tags and a correct Python 3 solution for this coding contest problem.
A robot is standing at the origin of the infinite two-dimensional plane. Each second the robot moves exactly 1 meter in one of the four cardinal directions: north, south, west, and east. For the first step the robot can choose any of the four directions, but then at the end of every second it has to turn 90 degrees left or right with respect to the direction it just moved in. For example, if the robot has just moved north or south, the next step it takes has to be either west or east, and vice versa.
The robot makes exactly n steps from its starting position according to the rules above. How many different points can the robot arrive to at the end? The final orientation of the robot can be ignored.
Input
The only line contains a single integer n (1 ≤ n ≤ 1000) — the number of steps the robot makes.
Output
Print a single integer — the number of different possible locations after exactly n steps.
Examples
Input
1
Output
4
Input
2
Output
4
Input
3
Output
12
Note
In the first sample case, the robot will end up 1 meter north, south, west, or east depending on its initial direction.
In the second sample case, the robot will always end up √{2} meters north-west, north-east, south-west, or south-east.
Tags: dp, math
Correct Solution:
```
n = int(input())
half = n//2
if n % 2 == 0:
print((half+1)**2)
else:
print(4*(half+1)+half*(half+1)*2)
```
| 76,920 | [
0.1470947265625,
0.478271484375,
-0.50146484375,
-0.03424072265625,
-0.60595703125,
-0.517578125,
0.004169464111328125,
0.42333984375,
0.1334228515625,
1.203125,
0.52392578125,
-0.03118896484375,
0.04351806640625,
-0.62646484375,
-0.368896484375,
0.206787109375,
-0.51416015625,
-0.... | 3 |
Provide tags and a correct Python 3 solution for this coding contest problem.
A robot is standing at the origin of the infinite two-dimensional plane. Each second the robot moves exactly 1 meter in one of the four cardinal directions: north, south, west, and east. For the first step the robot can choose any of the four directions, but then at the end of every second it has to turn 90 degrees left or right with respect to the direction it just moved in. For example, if the robot has just moved north or south, the next step it takes has to be either west or east, and vice versa.
The robot makes exactly n steps from its starting position according to the rules above. How many different points can the robot arrive to at the end? The final orientation of the robot can be ignored.
Input
The only line contains a single integer n (1 ≤ n ≤ 1000) — the number of steps the robot makes.
Output
Print a single integer — the number of different possible locations after exactly n steps.
Examples
Input
1
Output
4
Input
2
Output
4
Input
3
Output
12
Note
In the first sample case, the robot will end up 1 meter north, south, west, or east depending on its initial direction.
In the second sample case, the robot will always end up √{2} meters north-west, north-east, south-west, or south-east.
Tags: dp, math
Correct Solution:
```
n=int(input())
if(n==0):print("1")
elif(n==1):print("4")
elif(n%2==0
):
a=int(n//2)
print((a+1)**2)
else:
a=int(n//2)
print(2*(a+1)*(a+2))
```
| 76,921 | [
0.1798095703125,
0.451171875,
-0.55615234375,
-0.01751708984375,
-0.59228515625,
-0.5029296875,
-0.020172119140625,
0.44677734375,
0.1302490234375,
1.1611328125,
0.4794921875,
-0.0196075439453125,
0.07379150390625,
-0.611328125,
-0.395263671875,
0.127685546875,
-0.5205078125,
-0.81... | 3 |
Provide tags and a correct Python 3 solution for this coding contest problem.
A robot is standing at the origin of the infinite two-dimensional plane. Each second the robot moves exactly 1 meter in one of the four cardinal directions: north, south, west, and east. For the first step the robot can choose any of the four directions, but then at the end of every second it has to turn 90 degrees left or right with respect to the direction it just moved in. For example, if the robot has just moved north or south, the next step it takes has to be either west or east, and vice versa.
The robot makes exactly n steps from its starting position according to the rules above. How many different points can the robot arrive to at the end? The final orientation of the robot can be ignored.
Input
The only line contains a single integer n (1 ≤ n ≤ 1000) — the number of steps the robot makes.
Output
Print a single integer — the number of different possible locations after exactly n steps.
Examples
Input
1
Output
4
Input
2
Output
4
Input
3
Output
12
Note
In the first sample case, the robot will end up 1 meter north, south, west, or east depending on its initial direction.
In the second sample case, the robot will always end up √{2} meters north-west, north-east, south-west, or south-east.
Tags: dp, math
Correct Solution:
```
n=int(input())
if n%2==0:
g=n//2
print((g+1)*(g+1))
else:
g=n//2+1
out=2*(g+1)*g
print(out)
```
| 76,922 | [
0.1893310546875,
0.493896484375,
-0.48486328125,
-0.006008148193359375,
-0.5966796875,
-0.51123046875,
0.00623321533203125,
0.436279296875,
0.143798828125,
1.1796875,
0.53076171875,
-0.029327392578125,
0.031890869140625,
-0.61328125,
-0.35107421875,
0.17138671875,
-0.5234375,
-0.77... | 3 |
Provide tags and a correct Python 3 solution for this coding contest problem.
A robot is standing at the origin of the infinite two-dimensional plane. Each second the robot moves exactly 1 meter in one of the four cardinal directions: north, south, west, and east. For the first step the robot can choose any of the four directions, but then at the end of every second it has to turn 90 degrees left or right with respect to the direction it just moved in. For example, if the robot has just moved north or south, the next step it takes has to be either west or east, and vice versa.
The robot makes exactly n steps from its starting position according to the rules above. How many different points can the robot arrive to at the end? The final orientation of the robot can be ignored.
Input
The only line contains a single integer n (1 ≤ n ≤ 1000) — the number of steps the robot makes.
Output
Print a single integer — the number of different possible locations after exactly n steps.
Examples
Input
1
Output
4
Input
2
Output
4
Input
3
Output
12
Note
In the first sample case, the robot will end up 1 meter north, south, west, or east depending on its initial direction.
In the second sample case, the robot will always end up √{2} meters north-west, north-east, south-west, or south-east.
Tags: dp, math
Correct Solution:
```
n=int(input())
if n==1 or n==2:
print(4)
elif n==0:
print(1)
else:
l=[0 for _ in range(n+1)]
l[0]=1
l[1]=4
l[2]=4
j=2
for i in range(3,n+1):
if i%2==1:
l[i]=l[i-2]+(4*j)
else:
l[i]=l[i-4]+(4*j)
j+=1
print(l[n])
```
| 76,923 | [
0.18994140625,
0.439208984375,
-0.52783203125,
-0.0048675537109375,
-0.56884765625,
-0.49365234375,
0.005100250244140625,
0.4140625,
0.1368408203125,
1.181640625,
0.49462890625,
-0.04852294921875,
0.07086181640625,
-0.61083984375,
-0.387451171875,
0.13427734375,
-0.5107421875,
-0.8... | 3 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A robot is standing at the origin of the infinite two-dimensional plane. Each second the robot moves exactly 1 meter in one of the four cardinal directions: north, south, west, and east. For the first step the robot can choose any of the four directions, but then at the end of every second it has to turn 90 degrees left or right with respect to the direction it just moved in. For example, if the robot has just moved north or south, the next step it takes has to be either west or east, and vice versa.
The robot makes exactly n steps from its starting position according to the rules above. How many different points can the robot arrive to at the end? The final orientation of the robot can be ignored.
Input
The only line contains a single integer n (1 ≤ n ≤ 1000) — the number of steps the robot makes.
Output
Print a single integer — the number of different possible locations after exactly n steps.
Examples
Input
1
Output
4
Input
2
Output
4
Input
3
Output
12
Note
In the first sample case, the robot will end up 1 meter north, south, west, or east depending on its initial direction.
In the second sample case, the robot will always end up √{2} meters north-west, north-east, south-west, or south-east.
Submitted Solution:
```
n= int(input())
if n % 2:
print(((n+1)+2)*(n+1)//2)
else:
print((n+2)**2//4)
```
Yes
| 76,924 | [
0.25,
0.498291015625,
-0.57666015625,
-0.059722900390625,
-0.68896484375,
-0.371826171875,
-0.07366943359375,
0.498779296875,
0.2119140625,
1.1474609375,
0.52685546875,
-0.01435089111328125,
-0.031951904296875,
-0.6171875,
-0.420654296875,
0.091552734375,
-0.42578125,
-0.73828125,
... | 3 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A robot is standing at the origin of the infinite two-dimensional plane. Each second the robot moves exactly 1 meter in one of the four cardinal directions: north, south, west, and east. For the first step the robot can choose any of the four directions, but then at the end of every second it has to turn 90 degrees left or right with respect to the direction it just moved in. For example, if the robot has just moved north or south, the next step it takes has to be either west or east, and vice versa.
The robot makes exactly n steps from its starting position according to the rules above. How many different points can the robot arrive to at the end? The final orientation of the robot can be ignored.
Input
The only line contains a single integer n (1 ≤ n ≤ 1000) — the number of steps the robot makes.
Output
Print a single integer — the number of different possible locations after exactly n steps.
Examples
Input
1
Output
4
Input
2
Output
4
Input
3
Output
12
Note
In the first sample case, the robot will end up 1 meter north, south, west, or east depending on its initial direction.
In the second sample case, the robot will always end up √{2} meters north-west, north-east, south-west, or south-east.
Submitted Solution:
```
from sys import stdin
for _ in range(1):
n=int(input())
if(n%2==0): print((int(n/2)+1)**2)
else: print(2*(int(n/2)+2)*(int(n/2)+1))
```
Yes
| 76,925 | [
0.2183837890625,
0.456298828125,
-0.541015625,
-0.0650634765625,
-0.6904296875,
-0.37060546875,
-0.11077880859375,
0.50634765625,
0.157470703125,
1.1787109375,
0.51171875,
-0.0325927734375,
-0.0264892578125,
-0.6259765625,
-0.4296875,
0.105224609375,
-0.4189453125,
-0.7421875,
-0... | 3 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A robot is standing at the origin of the infinite two-dimensional plane. Each second the robot moves exactly 1 meter in one of the four cardinal directions: north, south, west, and east. For the first step the robot can choose any of the four directions, but then at the end of every second it has to turn 90 degrees left or right with respect to the direction it just moved in. For example, if the robot has just moved north or south, the next step it takes has to be either west or east, and vice versa.
The robot makes exactly n steps from its starting position according to the rules above. How many different points can the robot arrive to at the end? The final orientation of the robot can be ignored.
Input
The only line contains a single integer n (1 ≤ n ≤ 1000) — the number of steps the robot makes.
Output
Print a single integer — the number of different possible locations after exactly n steps.
Examples
Input
1
Output
4
Input
2
Output
4
Input
3
Output
12
Note
In the first sample case, the robot will end up 1 meter north, south, west, or east depending on its initial direction.
In the second sample case, the robot will always end up √{2} meters north-west, north-east, south-west, or south-east.
Submitted Solution:
```
import sys
input=sys.stdin.readline
# for _ in range(int(input())):
n=int(input())
# n,m=map(int,input().split())
# r=(list(input().strip()))
# a=list(map(int,input().split()))
if n&1:
y=n//2
x=y+1
else:
y=n//2
x=n//2
k=(x+1)*(y+1)
if n&1:
k=k*2
print(k)
```
Yes
| 76,926 | [
0.2015380859375,
0.444580078125,
-0.55126953125,
-0.06072998046875,
-0.671875,
-0.337158203125,
-0.08349609375,
0.479248046875,
0.1546630859375,
1.1806640625,
0.481201171875,
0.003681182861328125,
0.026519775390625,
-0.66064453125,
-0.438720703125,
0.0970458984375,
-0.443603515625,
... | 3 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A robot is standing at the origin of the infinite two-dimensional plane. Each second the robot moves exactly 1 meter in one of the four cardinal directions: north, south, west, and east. For the first step the robot can choose any of the four directions, but then at the end of every second it has to turn 90 degrees left or right with respect to the direction it just moved in. For example, if the robot has just moved north or south, the next step it takes has to be either west or east, and vice versa.
The robot makes exactly n steps from its starting position according to the rules above. How many different points can the robot arrive to at the end? The final orientation of the robot can be ignored.
Input
The only line contains a single integer n (1 ≤ n ≤ 1000) — the number of steps the robot makes.
Output
Print a single integer — the number of different possible locations after exactly n steps.
Examples
Input
1
Output
4
Input
2
Output
4
Input
3
Output
12
Note
In the first sample case, the robot will end up 1 meter north, south, west, or east depending on its initial direction.
In the second sample case, the robot will always end up √{2} meters north-west, north-east, south-west, or south-east.
Submitted Solution:
```
n = int(input())
s = n//2+1
if n == 1: print(4)
elif n%2: print(2*s*(s+1))
else: print(s**2)
```
Yes
| 76,927 | [
0.268310546875,
0.486328125,
-0.58935546875,
-0.057281494140625,
-0.7197265625,
-0.3623046875,
-0.07293701171875,
0.5078125,
0.1976318359375,
1.14453125,
0.51904296875,
-0.02484130859375,
-0.031707763671875,
-0.60498046875,
-0.425537109375,
0.09197998046875,
-0.434326171875,
-0.720... | 3 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A robot is standing at the origin of the infinite two-dimensional plane. Each second the robot moves exactly 1 meter in one of the four cardinal directions: north, south, west, and east. For the first step the robot can choose any of the four directions, but then at the end of every second it has to turn 90 degrees left or right with respect to the direction it just moved in. For example, if the robot has just moved north or south, the next step it takes has to be either west or east, and vice versa.
The robot makes exactly n steps from its starting position according to the rules above. How many different points can the robot arrive to at the end? The final orientation of the robot can be ignored.
Input
The only line contains a single integer n (1 ≤ n ≤ 1000) — the number of steps the robot makes.
Output
Print a single integer — the number of different possible locations after exactly n steps.
Examples
Input
1
Output
4
Input
2
Output
4
Input
3
Output
12
Note
In the first sample case, the robot will end up 1 meter north, south, west, or east depending on its initial direction.
In the second sample case, the robot will always end up √{2} meters north-west, north-east, south-west, or south-east.
Submitted Solution:
```
import math
n = input()
n = int(n)
prev = 0
ans = 0
for i in range(1, n+1):
ans = 4 * (2**(i-1)) - prev
prev = ans
print(ans)
```
No
| 76,928 | [
0.25244140625,
0.476806640625,
-0.56884765625,
-0.06640625,
-0.65966796875,
-0.37109375,
-0.056427001953125,
0.5263671875,
0.1697998046875,
1.18359375,
0.56689453125,
-0.0499267578125,
-0.0251312255859375,
-0.62939453125,
-0.4140625,
0.11712646484375,
-0.40185546875,
-0.69970703125... | 3 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A robot is standing at the origin of the infinite two-dimensional plane. Each second the robot moves exactly 1 meter in one of the four cardinal directions: north, south, west, and east. For the first step the robot can choose any of the four directions, but then at the end of every second it has to turn 90 degrees left or right with respect to the direction it just moved in. For example, if the robot has just moved north or south, the next step it takes has to be either west or east, and vice versa.
The robot makes exactly n steps from its starting position according to the rules above. How many different points can the robot arrive to at the end? The final orientation of the robot can be ignored.
Input
The only line contains a single integer n (1 ≤ n ≤ 1000) — the number of steps the robot makes.
Output
Print a single integer — the number of different possible locations after exactly n steps.
Examples
Input
1
Output
4
Input
2
Output
4
Input
3
Output
12
Note
In the first sample case, the robot will end up 1 meter north, south, west, or east depending on its initial direction.
In the second sample case, the robot will always end up √{2} meters north-west, north-east, south-west, or south-east.
Submitted Solution:
```
n = int(input())
if n%2 == 0:
print(n*2)
else:
print(n*4)
```
No
| 76,929 | [
0.247314453125,
0.4931640625,
-0.59228515625,
-0.049346923828125,
-0.6640625,
-0.39599609375,
-0.056884765625,
0.5283203125,
0.2078857421875,
1.1767578125,
0.5556640625,
-0.0222015380859375,
-0.0396728515625,
-0.61328125,
-0.42333984375,
0.113525390625,
-0.43212890625,
-0.729980468... | 3 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A robot is standing at the origin of the infinite two-dimensional plane. Each second the robot moves exactly 1 meter in one of the four cardinal directions: north, south, west, and east. For the first step the robot can choose any of the four directions, but then at the end of every second it has to turn 90 degrees left or right with respect to the direction it just moved in. For example, if the robot has just moved north or south, the next step it takes has to be either west or east, and vice versa.
The robot makes exactly n steps from its starting position according to the rules above. How many different points can the robot arrive to at the end? The final orientation of the robot can be ignored.
Input
The only line contains a single integer n (1 ≤ n ≤ 1000) — the number of steps the robot makes.
Output
Print a single integer — the number of different possible locations after exactly n steps.
Examples
Input
1
Output
4
Input
2
Output
4
Input
3
Output
12
Note
In the first sample case, the robot will end up 1 meter north, south, west, or east depending on its initial direction.
In the second sample case, the robot will always end up √{2} meters north-west, north-east, south-west, or south-east.
Submitted Solution:
```
def solve(N):
if N <= 2:
print(4)
else:
print((N - 1) ** 2 + (N - 1) * 4)
N = int(input())
solve(N)
```
No
| 76,930 | [
0.266845703125,
0.50537109375,
-0.5908203125,
-0.027801513671875,
-0.63818359375,
-0.39111328125,
-0.089111328125,
0.56396484375,
0.2149658203125,
1.1279296875,
0.54638671875,
-0.032745361328125,
-0.01105499267578125,
-0.61328125,
-0.423828125,
0.1151123046875,
-0.404052734375,
-0.... | 3 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A robot is standing at the origin of the infinite two-dimensional plane. Each second the robot moves exactly 1 meter in one of the four cardinal directions: north, south, west, and east. For the first step the robot can choose any of the four directions, but then at the end of every second it has to turn 90 degrees left or right with respect to the direction it just moved in. For example, if the robot has just moved north or south, the next step it takes has to be either west or east, and vice versa.
The robot makes exactly n steps from its starting position according to the rules above. How many different points can the robot arrive to at the end? The final orientation of the robot can be ignored.
Input
The only line contains a single integer n (1 ≤ n ≤ 1000) — the number of steps the robot makes.
Output
Print a single integer — the number of different possible locations after exactly n steps.
Examples
Input
1
Output
4
Input
2
Output
4
Input
3
Output
12
Note
In the first sample case, the robot will end up 1 meter north, south, west, or east depending on its initial direction.
In the second sample case, the robot will always end up √{2} meters north-west, north-east, south-west, or south-east.
Submitted Solution:
```
def read_ints():
return map(int, input().split())
n, = read_ints()
if n == 0:
result = 1
elif n == 1 or n == 2:
result = 4
else:
side = (n+1)//2 * 2 + 1
result = side * (side - 1) // 2 + side // 2 + (n % 2 == 0)
print(result)
```
No
| 76,931 | [
0.20703125,
0.45654296875,
-0.59423828125,
-0.056793212890625,
-0.69873046875,
-0.353759765625,
-0.0860595703125,
0.513671875,
0.1578369140625,
1.1611328125,
0.52099609375,
-0.002567291259765625,
0.0060272216796875,
-0.60693359375,
-0.435546875,
0.05401611328125,
-0.425048828125,
-... | 3 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given n segments on the Ox-axis. You can drive a nail in any integer point on the Ox-axis line nail so, that all segments containing this point, are considered nailed down. If the nail passes through endpoint of some segment, this segment is considered to be nailed too. What is the smallest number of nails needed to nail all the segments down?
Input
The first line of the input contains single integer number n (1 ≤ n ≤ 1000) — amount of segments. Following n lines contain descriptions of the segments. Each description is a pair of integer numbers — endpoints coordinates. All the coordinates don't exceed 10000 by absolute value. Segments can degenarate to points.
Output
The first line should contain one integer number — the smallest number of nails needed to nail all the segments down. The second line should contain coordinates of driven nails separated by space in any order. If the answer is not unique, output any.
Examples
Input
2
0 2
2 5
Output
1
2
Input
5
0 3
4 2
4 8
8 10
7 7
Output
3
7 10 3
Tags: greedy, sortings
Correct Solution:
```
n = int(input())
normal_segments = []
degen_segments = []
nails = {}
def is_nailed(a , b, nails):
for i in range(a, b + 1):
if i in nails:
if nails[i] == True:
return True
return False
for i in range(0, n):
data = input()
data = data.split(' ')
data[0] = int(data[0])
data[1] = int(data[1])
if data[0] - data[1] == 0:
degen_segments.append(data[0])
else:
if data[0] > data[1]:
normal_segments.append((data[1], data[0]))
else:
normal_segments.append((data[0], data[1]))
normal_segments.sort(key=lambda tup: tup[1])
degen_segments.sort()
# need to place a nail in degen segments always.
for i in degen_segments:
nails[i] = True
for i in normal_segments:
temp = is_nailed(i[0], i[1], nails)
if temp == False:
nails[i[1]] = True
print(len(nails.keys()))
for i in nails.keys():
print(i, end=' ')
# Made By Mostafa_Khaled
```
| 76,986 | [
0.34228515625,
0.1497802734375,
-0.1751708984375,
0.1444091796875,
-0.58154296875,
-0.3466796875,
-0.08203125,
0.2900390625,
0.275390625,
1.0341796875,
0.72802734375,
0.14306640625,
0.333984375,
-0.8935546875,
-0.374267578125,
0.51025390625,
-0.36767578125,
-0.81298828125,
-0.242... | 3 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given n segments on the Ox-axis. You can drive a nail in any integer point on the Ox-axis line nail so, that all segments containing this point, are considered nailed down. If the nail passes through endpoint of some segment, this segment is considered to be nailed too. What is the smallest number of nails needed to nail all the segments down?
Input
The first line of the input contains single integer number n (1 ≤ n ≤ 1000) — amount of segments. Following n lines contain descriptions of the segments. Each description is a pair of integer numbers — endpoints coordinates. All the coordinates don't exceed 10000 by absolute value. Segments can degenarate to points.
Output
The first line should contain one integer number — the smallest number of nails needed to nail all the segments down. The second line should contain coordinates of driven nails separated by space in any order. If the answer is not unique, output any.
Examples
Input
2
0 2
2 5
Output
1
2
Input
5
0 3
4 2
4 8
8 10
7 7
Output
3
7 10 3
Tags: greedy, sortings
Correct Solution:
```
import sys
class Seg:
def __init__(self, left, right):
self.left = left
self.right = right
def solve():
a = list()
n = int(input())
for i in range(n):
left, right = sorted(list(map(int, input().split())))
a.append(Seg(left, right))
a.sort(key = lambda x : x.right)
nails = list()
lastnail = -100000
for seg in a:
if lastnail >= seg.left: continue
nails.append(seg.right)
lastnail = seg.right
print(len(nails))
print(' '.join(map(str, nails)))
if sys.hexversion == 50594544 : sys.stdin = open("test.txt")
solve()
```
| 76,987 | [
0.402587890625,
0.130126953125,
-0.1697998046875,
0.294921875,
-0.64404296875,
-0.350830078125,
-0.07464599609375,
0.1956787109375,
0.315673828125,
1.00390625,
0.484619140625,
0.035736083984375,
0.34326171875,
-0.865234375,
-0.332275390625,
0.54541015625,
-0.330322265625,
-0.797363... | 3 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given n segments on the Ox-axis. You can drive a nail in any integer point on the Ox-axis line nail so, that all segments containing this point, are considered nailed down. If the nail passes through endpoint of some segment, this segment is considered to be nailed too. What is the smallest number of nails needed to nail all the segments down?
Input
The first line of the input contains single integer number n (1 ≤ n ≤ 1000) — amount of segments. Following n lines contain descriptions of the segments. Each description is a pair of integer numbers — endpoints coordinates. All the coordinates don't exceed 10000 by absolute value. Segments can degenarate to points.
Output
The first line should contain one integer number — the smallest number of nails needed to nail all the segments down. The second line should contain coordinates of driven nails separated by space in any order. If the answer is not unique, output any.
Examples
Input
2
0 2
2 5
Output
1
2
Input
5
0 3
4 2
4 8
8 10
7 7
Output
3
7 10 3
Tags: greedy, sortings
Correct Solution:
```
n = int(input())
l = []
for i in range(n):
a, b = map(int, input().split())
a, b = min(a, b), max(a, b)
l.append((a, 0, i))
l.append((b, 1, i))
l.sort()
ans = []
cur = set()
a = [0] * n
for x in l:
if x[1] == 0:
cur.add(x[2])
elif a[x[2]] == 0:
ans.append(x[0])
for e in cur:
a[e] = 1
cur.clear()
print(len(ans))
print(*ans)
```
| 76,988 | [
0.342529296875,
0.1175537109375,
-0.236083984375,
0.116943359375,
-0.64013671875,
-0.4375,
0.01375579833984375,
0.1292724609375,
0.470947265625,
1.025390625,
0.669921875,
0.150390625,
0.1922607421875,
-0.7861328125,
-0.33251953125,
0.50439453125,
-0.3974609375,
-0.93017578125,
-0... | 3 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given n segments on the Ox-axis. You can drive a nail in any integer point on the Ox-axis line nail so, that all segments containing this point, are considered nailed down. If the nail passes through endpoint of some segment, this segment is considered to be nailed too. What is the smallest number of nails needed to nail all the segments down?
Input
The first line of the input contains single integer number n (1 ≤ n ≤ 1000) — amount of segments. Following n lines contain descriptions of the segments. Each description is a pair of integer numbers — endpoints coordinates. All the coordinates don't exceed 10000 by absolute value. Segments can degenarate to points.
Output
The first line should contain one integer number — the smallest number of nails needed to nail all the segments down. The second line should contain coordinates of driven nails separated by space in any order. If the answer is not unique, output any.
Examples
Input
2
0 2
2 5
Output
1
2
Input
5
0 3
4 2
4 8
8 10
7 7
Output
3
7 10 3
Tags: greedy, sortings
Correct Solution:
```
n = int(input())
normal_segments = []
degen_segments = []
nails = {}
def is_nailed(a , b, nails):
for i in range(a, b + 1):
if i in nails:
if nails[i] == True:
return True
return False
for i in range(0, n):
data = input()
data = data.split(' ')
data[0] = int(data[0])
data[1] = int(data[1])
if data[0] - data[1] == 0:
degen_segments.append(data[0])
else:
if data[0] > data[1]:
normal_segments.append((data[1], data[0]))
else:
normal_segments.append((data[0], data[1]))
normal_segments.sort(key=lambda tup: tup[1])
degen_segments.sort()
# need to place a nail in degen segments always.
for i in degen_segments:
nails[i] = True
for i in normal_segments:
temp = is_nailed(i[0], i[1], nails)
if temp == False:
nails[i[1]] = True
print(len(nails.keys()))
for i in nails.keys():
print(i, end=' ')
```
| 76,989 | [
0.34228515625,
0.1497802734375,
-0.1751708984375,
0.1444091796875,
-0.58154296875,
-0.3466796875,
-0.08203125,
0.2900390625,
0.275390625,
1.0341796875,
0.72802734375,
0.14306640625,
0.333984375,
-0.8935546875,
-0.374267578125,
0.51025390625,
-0.36767578125,
-0.81298828125,
-0.242... | 3 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given n segments on the Ox-axis. You can drive a nail in any integer point on the Ox-axis line nail so, that all segments containing this point, are considered nailed down. If the nail passes through endpoint of some segment, this segment is considered to be nailed too. What is the smallest number of nails needed to nail all the segments down?
Input
The first line of the input contains single integer number n (1 ≤ n ≤ 1000) — amount of segments. Following n lines contain descriptions of the segments. Each description is a pair of integer numbers — endpoints coordinates. All the coordinates don't exceed 10000 by absolute value. Segments can degenarate to points.
Output
The first line should contain one integer number — the smallest number of nails needed to nail all the segments down. The second line should contain coordinates of driven nails separated by space in any order. If the answer is not unique, output any.
Examples
Input
2
0 2
2 5
Output
1
2
Input
5
0 3
4 2
4 8
8 10
7 7
Output
3
7 10 3
Tags: greedy, sortings
Correct Solution:
```
import sys
from array import array # noqa: F401
from operator import itemgetter
def input():
return sys.stdin.buffer.readline().decode('utf-8')
n = int(input())
a = [(x, y) if x < y else (y, x) for _ in range(n) for x, y in (map(int, input().split()),)]
a.sort(key=itemgetter(1))
nailed = -10**9
ans = []
for l, r in a:
if l <= nailed <= r:
continue
nailed = r
ans.append(r)
print(len(ans))
print(*ans)
```
| 76,991 | [
0.28125,
0.045684814453125,
-0.32666015625,
0.12066650390625,
-0.67724609375,
-0.35498046875,
0.033905029296875,
0.11749267578125,
0.3828125,
1.0048828125,
0.662109375,
0.09808349609375,
0.205322265625,
-0.751953125,
-0.30126953125,
0.483642578125,
-0.362548828125,
-0.86328125,
-... | 3 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given n segments on the Ox-axis. You can drive a nail in any integer point on the Ox-axis line nail so, that all segments containing this point, are considered nailed down. If the nail passes through endpoint of some segment, this segment is considered to be nailed too. What is the smallest number of nails needed to nail all the segments down?
Input
The first line of the input contains single integer number n (1 ≤ n ≤ 1000) — amount of segments. Following n lines contain descriptions of the segments. Each description is a pair of integer numbers — endpoints coordinates. All the coordinates don't exceed 10000 by absolute value. Segments can degenarate to points.
Output
The first line should contain one integer number — the smallest number of nails needed to nail all the segments down. The second line should contain coordinates of driven nails separated by space in any order. If the answer is not unique, output any.
Examples
Input
2
0 2
2 5
Output
1
2
Input
5
0 3
4 2
4 8
8 10
7 7
Output
3
7 10 3
Submitted Solution:
```
#!/usr/bin/python3
import sys
input = lambda: sys.stdin.readline().strip()
n = int(input())
s = sorted((sorted(int(x) for x in input().split()) for i in range(n)), key=lambda p: p[1])
ans = []
for i, j in s:
if not any(i <= x <= j for x in ans):
ans.append(j)
print(len(ans))
print(*ans)
```
Yes
| 76,992 | [
0.29248046875,
0.12469482421875,
-0.432861328125,
0.026580810546875,
-0.837890625,
-0.215576171875,
-0.22021484375,
0.1995849609375,
0.386962890625,
0.98388671875,
0.67041015625,
0.134033203125,
0.1871337890625,
-0.8203125,
-0.39111328125,
0.505859375,
-0.4521484375,
-0.837890625,
... | 3 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given n segments on the Ox-axis. You can drive a nail in any integer point on the Ox-axis line nail so, that all segments containing this point, are considered nailed down. If the nail passes through endpoint of some segment, this segment is considered to be nailed too. What is the smallest number of nails needed to nail all the segments down?
Input
The first line of the input contains single integer number n (1 ≤ n ≤ 1000) — amount of segments. Following n lines contain descriptions of the segments. Each description is a pair of integer numbers — endpoints coordinates. All the coordinates don't exceed 10000 by absolute value. Segments can degenarate to points.
Output
The first line should contain one integer number — the smallest number of nails needed to nail all the segments down. The second line should contain coordinates of driven nails separated by space in any order. If the answer is not unique, output any.
Examples
Input
2
0 2
2 5
Output
1
2
Input
5
0 3
4 2
4 8
8 10
7 7
Output
3
7 10 3
Submitted Solution:
```
if __name__ == '__main__':
number = int(input())
segments = []
i = 0
# while i < number:
for i in range(0, number):
pos = input().split()
if int(pos[0]) < int(pos[1]):
a = int(pos[0])
b = int(pos[1])
else:
b = int(pos[0])
a = int(pos[1])
segments.append((a, b))
segments.sort()
final = []
ok = -9999999
nr = len(segments)
for i in range(nr):
a = segments[i][0]
b = segments[i][1]
if a <= ok and b >= ok:
continue
verif = 0
for j in segments[i + 1:]:
a = segments[i][1]
b = segments[i][1]
if a >= j[0] and b >= j[1]:
verif = 1
if verif:
continue
else:
ok = segments[i][1]
final.append(ok)
print(len(final))
space = ''
for i in final:
space = space + str(i) + ' '
print(space)
```
Yes
| 76,993 | [
0.36328125,
0.11138916015625,
-0.39453125,
0.062286376953125,
-0.79443359375,
-0.26416015625,
-0.126708984375,
0.2010498046875,
0.40478515625,
0.98046875,
0.69140625,
0.1922607421875,
0.1934814453125,
-0.8564453125,
-0.4052734375,
0.5390625,
-0.456298828125,
-0.830078125,
-0.1697... | 3 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given n segments on the Ox-axis. You can drive a nail in any integer point on the Ox-axis line nail so, that all segments containing this point, are considered nailed down. If the nail passes through endpoint of some segment, this segment is considered to be nailed too. What is the smallest number of nails needed to nail all the segments down?
Input
The first line of the input contains single integer number n (1 ≤ n ≤ 1000) — amount of segments. Following n lines contain descriptions of the segments. Each description is a pair of integer numbers — endpoints coordinates. All the coordinates don't exceed 10000 by absolute value. Segments can degenarate to points.
Output
The first line should contain one integer number — the smallest number of nails needed to nail all the segments down. The second line should contain coordinates of driven nails separated by space in any order. If the answer is not unique, output any.
Examples
Input
2
0 2
2 5
Output
1
2
Input
5
0 3
4 2
4 8
8 10
7 7
Output
3
7 10 3
Submitted Solution:
```
n = int(input())
lis=list()
cou=1
arr=list()
for i in range(n):
a , b = map(int,input().split())
if a>b:
lis.append([b,a])
else:
lis.append([a,b])
lis=sorted(lis,key=lambda l:l[1])
#print(lis)
i=0
j=1
arr.append(lis[i][1])
while(i<n and j<n):
if lis[i][1]<lis[j][0]:
cou+=1
i=j
arr.append(lis[i][1])
# print(lis[i][1])
else:
j+=1
print(cou)
print(*arr)
```
Yes
| 76,994 | [
0.314208984375,
0.1488037109375,
-0.276123046875,
0.042388916015625,
-0.79150390625,
-0.308837890625,
-0.2431640625,
0.26220703125,
0.468994140625,
0.90380859375,
0.578125,
0.2303466796875,
0.2255859375,
-0.791015625,
-0.3984375,
0.414794921875,
-0.44677734375,
-0.634765625,
-0.1... | 3 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given n segments on the Ox-axis. You can drive a nail in any integer point on the Ox-axis line nail so, that all segments containing this point, are considered nailed down. If the nail passes through endpoint of some segment, this segment is considered to be nailed too. What is the smallest number of nails needed to nail all the segments down?
Input
The first line of the input contains single integer number n (1 ≤ n ≤ 1000) — amount of segments. Following n lines contain descriptions of the segments. Each description is a pair of integer numbers — endpoints coordinates. All the coordinates don't exceed 10000 by absolute value. Segments can degenarate to points.
Output
The first line should contain one integer number — the smallest number of nails needed to nail all the segments down. The second line should contain coordinates of driven nails separated by space in any order. If the answer is not unique, output any.
Examples
Input
2
0 2
2 5
Output
1
2
Input
5
0 3
4 2
4 8
8 10
7 7
Output
3
7 10 3
Submitted Solution:
```
n = int(input())
segments = []
for _ in range(n):
segments.append(list(map(int, input().split())))
for i in range(n):
if segments[i][0] > segments[i][1]:
segments[i][0], segments[i][1] = segments[i][1], segments[i][0]
segments.sort(key=lambda x: (x[0], -x[1]))
started = False
i = 0
nails = []
stack = []
while i < n:
stack.append(segments[i])
i += 1
edge = stack[-1][1]
while i < n and segments[i][0] <= edge:
stack.append(segments[i])
i += 1
edge = min(edge, stack[-1][1])
nails.append(edge)
stack = []
print(len(nails))
print(*nails)
```
Yes
| 76,995 | [
0.315673828125,
0.12103271484375,
-0.388671875,
0.0787353515625,
-0.70458984375,
-0.21728515625,
-0.164794921875,
0.27294921875,
0.419189453125,
1.0498046875,
0.6806640625,
0.12939453125,
0.253173828125,
-0.94677734375,
-0.4462890625,
0.564453125,
-0.54248046875,
-0.78857421875,
... | 3 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given n segments on the Ox-axis. You can drive a nail in any integer point on the Ox-axis line nail so, that all segments containing this point, are considered nailed down. If the nail passes through endpoint of some segment, this segment is considered to be nailed too. What is the smallest number of nails needed to nail all the segments down?
Input
The first line of the input contains single integer number n (1 ≤ n ≤ 1000) — amount of segments. Following n lines contain descriptions of the segments. Each description is a pair of integer numbers — endpoints coordinates. All the coordinates don't exceed 10000 by absolute value. Segments can degenarate to points.
Output
The first line should contain one integer number — the smallest number of nails needed to nail all the segments down. The second line should contain coordinates of driven nails separated by space in any order. If the answer is not unique, output any.
Examples
Input
2
0 2
2 5
Output
1
2
Input
5
0 3
4 2
4 8
8 10
7 7
Output
3
7 10 3
Submitted Solution:
```
n = int(input())
l = []
cur = set()
for i in range(n):
a, b = map(int, input().split())
l.append((a, 1, i))
l.append((b, 2, i))
l.sort()
ans = []
for i in l:
if i[1] == 1:
cur.add(i[2])
elif i[2] in cur:
ans.append(i[0])
cur.clear()
print(len(ans))
print(*ans)
```
No
| 76,996 | [
0.30517578125,
0.10467529296875,
-0.420166015625,
0.00856781005859375,
-0.79541015625,
-0.251708984375,
-0.17919921875,
0.19287109375,
0.380615234375,
0.9912109375,
0.71630859375,
0.1358642578125,
0.171875,
-0.82666015625,
-0.398193359375,
0.52001953125,
-0.50244140625,
-0.81884765... | 3 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given n segments on the Ox-axis. You can drive a nail in any integer point on the Ox-axis line nail so, that all segments containing this point, are considered nailed down. If the nail passes through endpoint of some segment, this segment is considered to be nailed too. What is the smallest number of nails needed to nail all the segments down?
Input
The first line of the input contains single integer number n (1 ≤ n ≤ 1000) — amount of segments. Following n lines contain descriptions of the segments. Each description is a pair of integer numbers — endpoints coordinates. All the coordinates don't exceed 10000 by absolute value. Segments can degenarate to points.
Output
The first line should contain one integer number — the smallest number of nails needed to nail all the segments down. The second line should contain coordinates of driven nails separated by space in any order. If the answer is not unique, output any.
Examples
Input
2
0 2
2 5
Output
1
2
Input
5
0 3
4 2
4 8
8 10
7 7
Output
3
7 10 3
Submitted Solution:
```
#!/usr/bin/env python3
# Read data
n = int(input())
segments = [map(int, input().split()) for i in range(n)]
segments = [(min(a,b), max(a,b)) for (a,b) in segments] # Make sure segments run from low to high
# Sort the start and end positions
start_pos = [a for (a,b) in segments]
start_pos.sort()
end_pos = [b for (a,b) in segments]
end_pos.sort()
nail_positions = [] # Positions where the nails will be put
while (n != 0):
# Search for a position covered by the most segments
next_start, next_end = 0,0
best = (0,0)
while (next_start < n):
if (start_pos[next_start] <= end_pos[next_end]):
next_start += 1
best = max(best, (next_start - next_end, start_pos[next_start-1]))
else:
next_end += 1
# Put a nail at the found position
nail_pos = best[1]
nail_positions.append(nail_pos)
# Find the nailed and remaining segments
nailed_segments = [(a,b) for (a,b) in segments if ((a <= nail_pos) and (nail_pos <= b))]
segments = [(a,b) for (a,b) in segments if ((a > nail_pos) or (nail_pos > b))]
n = len(segments)
# Update the start and end positions
for (a,b) in nailed_segments:
start_pos.remove(a)
end_pos.remove(b)
print(len(nail_positions))
print(" ".join(["%d" % pos for pos in nail_positions]))
```
No
| 76,997 | [
0.331787109375,
0.10113525390625,
-0.433837890625,
0.07244873046875,
-0.712890625,
-0.267578125,
-0.1412353515625,
0.27099609375,
0.4697265625,
1.0126953125,
0.6474609375,
0.13134765625,
0.24755859375,
-0.884765625,
-0.417724609375,
0.53857421875,
-0.44580078125,
-0.8134765625,
-... | 3 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given n segments on the Ox-axis. You can drive a nail in any integer point on the Ox-axis line nail so, that all segments containing this point, are considered nailed down. If the nail passes through endpoint of some segment, this segment is considered to be nailed too. What is the smallest number of nails needed to nail all the segments down?
Input
The first line of the input contains single integer number n (1 ≤ n ≤ 1000) — amount of segments. Following n lines contain descriptions of the segments. Each description is a pair of integer numbers — endpoints coordinates. All the coordinates don't exceed 10000 by absolute value. Segments can degenarate to points.
Output
The first line should contain one integer number — the smallest number of nails needed to nail all the segments down. The second line should contain coordinates of driven nails separated by space in any order. If the answer is not unique, output any.
Examples
Input
2
0 2
2 5
Output
1
2
Input
5
0 3
4 2
4 8
8 10
7 7
Output
3
7 10 3
Submitted Solution:
```
n = int(input())
lines = sorted([sorted([int(i) for i in input().split()]) for _ in range(n)])
array = []
bool_array = [0]*n
for i in range(n):
array.append([lines[i][0], i])
array.append([lines[i][1], i])
array.sort()
screws = []
last = -1
last_indx = -1
b_indx = 0
for i in array:
if i[0] == last:
b_indx += 1
last_indx += 1
continue
if i[1] == last_indx + 1:
last_indx += 1
bool_array[i[1]] = 1
elif bool_array[i[1]]:
screws.append(str(i[0]))
last = i[0]
b_indx += 1
for j in range(b_indx, last_indx + 1):
bool_array[j] = 0
print(len(screws))
print(' '.join(screws))
```
No
| 76,998 | [
0.286376953125,
-0.0176544189453125,
-0.38427734375,
0.0750732421875,
-0.828125,
-0.285888671875,
-0.28125,
0.11883544921875,
0.41845703125,
0.9970703125,
0.72900390625,
0.11358642578125,
0.092529296875,
-0.87109375,
-0.320068359375,
0.61669921875,
-0.42431640625,
-0.77978515625,
... | 3 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given n segments on the Ox-axis. You can drive a nail in any integer point on the Ox-axis line nail so, that all segments containing this point, are considered nailed down. If the nail passes through endpoint of some segment, this segment is considered to be nailed too. What is the smallest number of nails needed to nail all the segments down?
Input
The first line of the input contains single integer number n (1 ≤ n ≤ 1000) — amount of segments. Following n lines contain descriptions of the segments. Each description is a pair of integer numbers — endpoints coordinates. All the coordinates don't exceed 10000 by absolute value. Segments can degenarate to points.
Output
The first line should contain one integer number — the smallest number of nails needed to nail all the segments down. The second line should contain coordinates of driven nails separated by space in any order. If the answer is not unique, output any.
Examples
Input
2
0 2
2 5
Output
1
2
Input
5
0 3
4 2
4 8
8 10
7 7
Output
3
7 10 3
Submitted Solution:
```
n = int(input())
lines = sorted([sorted([int(i) for i in input().split()]) for _ in range(n)])
array = []
bool_array = [0]*n
for i in range(n):
array.append([lines[i][0], i])
array.append([lines[i][1], i])
array.sort()
screws = []
last = -1
last_indx = -1
b_indx = 0
for i in array:
if i[0] == last:
b_indx += 1
last_indx += 1
continue
if i[1] == last_indx + 1:
last_indx += 1
bool_array[i[1]] = 1
elif bool_array[i[1]]:
screws.append(str(i[0]))
last = i[0]
for j in range(b_indx, last_indx + 1):
bool_array[j] = 0
b_indx += 1
print(len(screws))
print(' '.join(screws))
```
No
| 76,999 | [
0.293212890625,
-0.0233917236328125,
-0.3798828125,
0.07843017578125,
-0.82421875,
-0.286376953125,
-0.2783203125,
0.1055908203125,
0.417724609375,
0.9970703125,
0.73095703125,
0.10546875,
0.08795166015625,
-0.873046875,
-0.3203125,
0.609375,
-0.428466796875,
-0.78076171875,
-0.2... | 3 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Yaroslav is playing a game called "Time". The game has a timer showing the lifespan he's got left. As soon as the timer shows 0, Yaroslav's character dies and the game ends. Also, the game has n clock stations, station number i is at point (xi, yi) of the plane. As the player visits station number i, he increases the current time on his timer by ai. The stations are for one-time use only, so if the player visits some station another time, the time on his timer won't grow.
A player spends d·dist time units to move between stations, where dist is the distance the player has covered and d is some constant. The distance between stations i and j is determined as |xi - xj| + |yi - yj|.
Initially, the player is at station number 1, and the player has strictly more than zero and strictly less than one units of time. At station number 1 one unit of money can increase the time on the timer by one time unit (you can buy only integer number of time units).
Now Yaroslav is wondering, how much money he needs to get to station n. Help Yaroslav. Consider the time to buy and to increase the timer value negligibly small.
Input
The first line contains integers n and d (3 ≤ n ≤ 100, 103 ≤ d ≤ 105) — the number of stations and the constant from the statement.
The second line contains n - 2 integers: a2, a3, ..., an - 1 (1 ≤ ai ≤ 103). The next n lines contain the coordinates of the stations. The i-th of them contains two integers xi, yi (-100 ≤ xi, yi ≤ 100).
It is guaranteed that no two stations are located at the same point.
Output
In a single line print an integer — the answer to the problem.
Examples
Input
3 1000
1000
0 0
0 1
0 3
Output
2000
Input
3 1000
1000
1 0
1 1
1 2
Output
1000
Tags: binary search, graphs, shortest paths
Correct Solution:
```
from sys import stdin
from math import inf
def readline():
return map(int, stdin.readline().strip().split())
def main():
n, d = readline()
a = [0] + list(readline()) + [0]
x = [0] * n
y = [0] * n
for i in range(n):
x[i], y[i] = readline()
lower_cost = [inf] * n
lower_cost[0] = 0
visited = [False] * n
for i in range(n - 1):
lower_value = inf
position = 0
for j in range(n):
if not visited[j] and lower_value > lower_cost[j]:
lower_value = lower_cost[j]
position = j
visited[position] = True
for k in range(n):
if not visited[k]:
diff = lower_cost[position] + d * (abs(x[k] - x[position]) + abs(y[k] - y[position])) - a[position]
if lower_cost[k] > diff:
lower_cost[k] = diff
return lower_cost[-1]
if __name__ == '__main__':
print(main())
```
| 77,032 | [
0.55322265625,
0.341064453125,
-0.2337646484375,
0.328857421875,
0.01263427734375,
-0.57861328125,
-0.0032806396484375,
0.01226043701171875,
-0.04547119140625,
1.140625,
0.5283203125,
0.06646728515625,
0.2142333984375,
-0.5341796875,
0.03143310546875,
-0.1143798828125,
-0.6181640625,... | 3 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Yaroslav is playing a game called "Time". The game has a timer showing the lifespan he's got left. As soon as the timer shows 0, Yaroslav's character dies and the game ends. Also, the game has n clock stations, station number i is at point (xi, yi) of the plane. As the player visits station number i, he increases the current time on his timer by ai. The stations are for one-time use only, so if the player visits some station another time, the time on his timer won't grow.
A player spends d·dist time units to move between stations, where dist is the distance the player has covered and d is some constant. The distance between stations i and j is determined as |xi - xj| + |yi - yj|.
Initially, the player is at station number 1, and the player has strictly more than zero and strictly less than one units of time. At station number 1 one unit of money can increase the time on the timer by one time unit (you can buy only integer number of time units).
Now Yaroslav is wondering, how much money he needs to get to station n. Help Yaroslav. Consider the time to buy and to increase the timer value negligibly small.
Input
The first line contains integers n and d (3 ≤ n ≤ 100, 103 ≤ d ≤ 105) — the number of stations and the constant from the statement.
The second line contains n - 2 integers: a2, a3, ..., an - 1 (1 ≤ ai ≤ 103). The next n lines contain the coordinates of the stations. The i-th of them contains two integers xi, yi (-100 ≤ xi, yi ≤ 100).
It is guaranteed that no two stations are located at the same point.
Output
In a single line print an integer — the answer to the problem.
Examples
Input
3 1000
1000
0 0
0 1
0 3
Output
2000
Input
3 1000
1000
1 0
1 1
1 2
Output
1000
Tags: binary search, graphs, shortest paths
Correct Solution:
```
from sys import stdin, stdout
from math import inf
def main():
n, d = readline()
a = [0] + list(readline()) + [0]
x = [0] * n
y = [0] * n
# parent = [-1] * n # In case you want to know the path traveled
for i in range(n):
x[i], y[i] = readline()
return dijkstra(n, d, a, x, y) # , parent)
def readline():
return map(int, stdin.readline().strip().split())
def dijkstra(n, d, a, x, y): # , parent):
lower_cost = [inf] * n
lower_cost[0] = 0
visited = [False] * n
for i in range(n - 1):
position = minimum(n, visited, lower_cost)
if position == n - 1:
break
visited[position] = True
for k in range(n):
if not visited[k]:
diff = lower_cost[position] + d * (abs(x[k] - x[position]) + abs(y[k] - y[position])) - a[position]
if lower_cost[k] > diff:
lower_cost[k] = diff
# parent[k] = position
return lower_cost[-1]
def minimum(n, visited, lower_cost):
lower_value = inf
position = 0
for j in range(n):
if not visited[j] and lower_value > lower_cost[j]:
lower_value = lower_cost[j]
position = j
return position
if __name__ == '__main__':
stdout.write("".join(str(main()) + "\n"))
```
| 77,033 | [
0.53173828125,
0.3232421875,
-0.2132568359375,
0.37451171875,
-0.0005841255187988281,
-0.60791015625,
-0.01476287841796875,
0.025299072265625,
-0.0150146484375,
1.1259765625,
0.5244140625,
0.052703857421875,
0.2127685546875,
-0.51806640625,
0.0017938613891601562,
-0.0950927734375,
-0... | 3 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Yaroslav is playing a game called "Time". The game has a timer showing the lifespan he's got left. As soon as the timer shows 0, Yaroslav's character dies and the game ends. Also, the game has n clock stations, station number i is at point (xi, yi) of the plane. As the player visits station number i, he increases the current time on his timer by ai. The stations are for one-time use only, so if the player visits some station another time, the time on his timer won't grow.
A player spends d·dist time units to move between stations, where dist is the distance the player has covered and d is some constant. The distance between stations i and j is determined as |xi - xj| + |yi - yj|.
Initially, the player is at station number 1, and the player has strictly more than zero and strictly less than one units of time. At station number 1 one unit of money can increase the time on the timer by one time unit (you can buy only integer number of time units).
Now Yaroslav is wondering, how much money he needs to get to station n. Help Yaroslav. Consider the time to buy and to increase the timer value negligibly small.
Input
The first line contains integers n and d (3 ≤ n ≤ 100, 103 ≤ d ≤ 105) — the number of stations and the constant from the statement.
The second line contains n - 2 integers: a2, a3, ..., an - 1 (1 ≤ ai ≤ 103). The next n lines contain the coordinates of the stations. The i-th of them contains two integers xi, yi (-100 ≤ xi, yi ≤ 100).
It is guaranteed that no two stations are located at the same point.
Output
In a single line print an integer — the answer to the problem.
Examples
Input
3 1000
1000
0 0
0 1
0 3
Output
2000
Input
3 1000
1000
1 0
1 1
1 2
Output
1000
Tags: binary search, graphs, shortest paths
Correct Solution:
```
from sys import stdin
from math import inf
def main():
n, d = readline()
a = [0] + list(readline()) + [0]
x = [0] * n
y = [0] * n
# parent = [-1] * n # In case you want to know the path traveled
for i in range(n):
x[i], y[i] = readline()
return dijkstra(n, d, a, x, y) # , parent)
def readline():
return map(int, stdin.readline().strip().split())
def dijkstra(n, d, a, x, y): # , parent):
lower_cost = [inf] * n
lower_cost[0] = 0
visited = [False] * n
for i in range(n - 1):
position = minimum(n, visited, lower_cost)
visited[position] = True
for k in range(n):
if not visited[k]:
diff = lower_cost[position] + d * (abs(x[k] - x[position]) + abs(y[k] - y[position])) - a[position]
if lower_cost[k] > diff:
lower_cost[k] = diff
# parent[k] = position
return lower_cost[-1]
def minimum(n, visited, lower_cost):
lower_value = inf
position = 0
for j in range(n):
if visited[j] or lower_value <= lower_cost[j]:
continue
lower_value = lower_cost[j]
position = j
return position
if __name__ == '__main__':
print(main())
```
| 77,037 | [
0.53271484375,
0.323486328125,
-0.20654296875,
0.35498046875,
0.0015869140625,
-0.59326171875,
0.0003743171691894531,
0.022613525390625,
-0.0166473388671875,
1.115234375,
0.54638671875,
0.0631103515625,
0.219970703125,
-0.54541015625,
-0.0024929046630859375,
-0.08404541015625,
-0.616... | 3 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Yaroslav is playing a game called "Time". The game has a timer showing the lifespan he's got left. As soon as the timer shows 0, Yaroslav's character dies and the game ends. Also, the game has n clock stations, station number i is at point (xi, yi) of the plane. As the player visits station number i, he increases the current time on his timer by ai. The stations are for one-time use only, so if the player visits some station another time, the time on his timer won't grow.
A player spends d·dist time units to move between stations, where dist is the distance the player has covered and d is some constant. The distance between stations i and j is determined as |xi - xj| + |yi - yj|.
Initially, the player is at station number 1, and the player has strictly more than zero and strictly less than one units of time. At station number 1 one unit of money can increase the time on the timer by one time unit (you can buy only integer number of time units).
Now Yaroslav is wondering, how much money he needs to get to station n. Help Yaroslav. Consider the time to buy and to increase the timer value negligibly small.
Input
The first line contains integers n and d (3 ≤ n ≤ 100, 103 ≤ d ≤ 105) — the number of stations and the constant from the statement.
The second line contains n - 2 integers: a2, a3, ..., an - 1 (1 ≤ ai ≤ 103). The next n lines contain the coordinates of the stations. The i-th of them contains two integers xi, yi (-100 ≤ xi, yi ≤ 100).
It is guaranteed that no two stations are located at the same point.
Output
In a single line print an integer — the answer to the problem.
Examples
Input
3 1000
1000
0 0
0 1
0 3
Output
2000
Input
3 1000
1000
1 0
1 1
1 2
Output
1000
Submitted Solution:
```
n, d = map(int, input().split())
a = [0] + list(map(int, input().split())) + [0]
x = []
y = []
for i in range(n):
xx, yy = map(int, input().split())
x += [xx]
y += [yy]
b = [-1] * n
b[0] = 0
c = True
while c:
c = False
for i in range(n):
for j in range(1, n):
if i != j and b[i] != -1:
t = b[i] + (abs(x[i] - x[j]) + abs(y[i] - y[j])) * d - a[j]
if b[j] == -1 or t < b[j]:
b[j] = t
c = True
print(b[-1])
```
Yes
| 77,040 | [
0.59619140625,
0.291259765625,
-0.292724609375,
0.41552734375,
-0.048736572265625,
-0.52392578125,
-0.013427734375,
0.1729736328125,
-0.05157470703125,
1.1318359375,
0.62158203125,
0.058624267578125,
0.2496337890625,
-0.62109375,
-0.029876708984375,
-0.08349609375,
-0.62744140625,
... | 3 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Yaroslav is playing a game called "Time". The game has a timer showing the lifespan he's got left. As soon as the timer shows 0, Yaroslav's character dies and the game ends. Also, the game has n clock stations, station number i is at point (xi, yi) of the plane. As the player visits station number i, he increases the current time on his timer by ai. The stations are for one-time use only, so if the player visits some station another time, the time on his timer won't grow.
A player spends d·dist time units to move between stations, where dist is the distance the player has covered and d is some constant. The distance between stations i and j is determined as |xi - xj| + |yi - yj|.
Initially, the player is at station number 1, and the player has strictly more than zero and strictly less than one units of time. At station number 1 one unit of money can increase the time on the timer by one time unit (you can buy only integer number of time units).
Now Yaroslav is wondering, how much money he needs to get to station n. Help Yaroslav. Consider the time to buy and to increase the timer value negligibly small.
Input
The first line contains integers n and d (3 ≤ n ≤ 100, 103 ≤ d ≤ 105) — the number of stations and the constant from the statement.
The second line contains n - 2 integers: a2, a3, ..., an - 1 (1 ≤ ai ≤ 103). The next n lines contain the coordinates of the stations. The i-th of them contains two integers xi, yi (-100 ≤ xi, yi ≤ 100).
It is guaranteed that no two stations are located at the same point.
Output
In a single line print an integer — the answer to the problem.
Examples
Input
3 1000
1000
0 0
0 1
0 3
Output
2000
Input
3 1000
1000
1 0
1 1
1 2
Output
1000
Submitted Solution:
```
n, d = map(int, input().split())
a = [0] + list(map(int, input().split())) + [0] #aumento de timer según estación
X = []
Y = []
for i in range(n):
x, y = map(int, input().split()) #Coordenadas de la estación i.
X.append(x)
Y.append(y)
mon = [-1] * n #array para el monto necesario para llegar.
mon[0] = 0
Z = 0 #valor que permitirá entrar al loop
while Z == 0:
Z = 1
for i in range(n): #estamos en estación i
for j in range(1, n): #queremos ir a estación j
if i != j and mon[i] != -1: #si no queremos ir a la misma estación y donde estamos pudimos llegar.
costo = mon[i] + (abs(X[i] - X[j]) + abs(Y[i] - Y[j]))*d - a[j] #nuevo costo necesario para ir de i a j.
if mon[j] == -1 or costo < mon[j]: #si el nuevo costo es menor que uno anterior, se guarda este o si antes no había ningun costo guardado.
mon[j] = costo
Z = 0 #volvemos a entrar al loop, esto asegura que todos los mon[] van a dejar de ser '1 y tendran un costo.
print(mon[-1]) #costo de llegar a la última estación.
```
Yes
| 77,041 | [
0.60107421875,
0.305419921875,
-0.255859375,
0.43212890625,
0.04449462890625,
-0.51123046875,
-0.09912109375,
0.135986328125,
-0.021270751953125,
1.14453125,
0.6953125,
0.1319580078125,
0.1669921875,
-0.572265625,
-0.049285888671875,
-0.06317138671875,
-0.51904296875,
-1.005859375,... | 3 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Yaroslav is playing a game called "Time". The game has a timer showing the lifespan he's got left. As soon as the timer shows 0, Yaroslav's character dies and the game ends. Also, the game has n clock stations, station number i is at point (xi, yi) of the plane. As the player visits station number i, he increases the current time on his timer by ai. The stations are for one-time use only, so if the player visits some station another time, the time on his timer won't grow.
A player spends d·dist time units to move between stations, where dist is the distance the player has covered and d is some constant. The distance between stations i and j is determined as |xi - xj| + |yi - yj|.
Initially, the player is at station number 1, and the player has strictly more than zero and strictly less than one units of time. At station number 1 one unit of money can increase the time on the timer by one time unit (you can buy only integer number of time units).
Now Yaroslav is wondering, how much money he needs to get to station n. Help Yaroslav. Consider the time to buy and to increase the timer value negligibly small.
Input
The first line contains integers n and d (3 ≤ n ≤ 100, 103 ≤ d ≤ 105) — the number of stations and the constant from the statement.
The second line contains n - 2 integers: a2, a3, ..., an - 1 (1 ≤ ai ≤ 103). The next n lines contain the coordinates of the stations. The i-th of them contains two integers xi, yi (-100 ≤ xi, yi ≤ 100).
It is guaranteed that no two stations are located at the same point.
Output
In a single line print an integer — the answer to the problem.
Examples
Input
3 1000
1000
0 0
0 1
0 3
Output
2000
Input
3 1000
1000
1 0
1 1
1 2
Output
1000
Submitted Solution:
```
from sys import stdin
from math import inf
def readline():
return map(int, stdin.readline().strip().split())
def dijkstra(): # , parent):
n, d = readline()
a = [0] + list(readline()) + [0]
x = [0] * n
y = [0] * n
# parent = [-1] * n # In case you want to know the path traveled
for i in range(n):
x[i], y[i] = readline()
lower_cost = [inf] * n
lower_cost[0] = 0
visited = [False] * n
for i in range(n - 1):
lower_value = inf
position = 0
for j in range(n):
if visited[j] or lower_value <= lower_cost[j]:
continue
lower_value = lower_cost[j]
position = j
visited[position] = True
for k in range(n):
if not visited[k]:
diff = lower_cost[position] + d * (abs(x[k] - x[position]) + abs(y[k] - y[position])) - a[position]
if lower_cost[k] > diff:
lower_cost[k] = diff
# parent[k] = position
return lower_cost[-1]
if __name__ == '__main__':
print(dijkstra())
```
Yes
| 77,042 | [
0.59765625,
0.307861328125,
-0.272705078125,
0.373046875,
-0.0399169921875,
-0.50439453125,
-0.015838623046875,
0.08746337890625,
-0.0716552734375,
1.189453125,
0.60498046875,
0.00968170166015625,
0.16552734375,
-0.5751953125,
-0.032012939453125,
-0.0841064453125,
-0.65576171875,
-... | 3 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Yaroslav is playing a game called "Time". The game has a timer showing the lifespan he's got left. As soon as the timer shows 0, Yaroslav's character dies and the game ends. Also, the game has n clock stations, station number i is at point (xi, yi) of the plane. As the player visits station number i, he increases the current time on his timer by ai. The stations are for one-time use only, so if the player visits some station another time, the time on his timer won't grow.
A player spends d·dist time units to move between stations, where dist is the distance the player has covered and d is some constant. The distance between stations i and j is determined as |xi - xj| + |yi - yj|.
Initially, the player is at station number 1, and the player has strictly more than zero and strictly less than one units of time. At station number 1 one unit of money can increase the time on the timer by one time unit (you can buy only integer number of time units).
Now Yaroslav is wondering, how much money he needs to get to station n. Help Yaroslav. Consider the time to buy and to increase the timer value negligibly small.
Input
The first line contains integers n and d (3 ≤ n ≤ 100, 103 ≤ d ≤ 105) — the number of stations and the constant from the statement.
The second line contains n - 2 integers: a2, a3, ..., an - 1 (1 ≤ ai ≤ 103). The next n lines contain the coordinates of the stations. The i-th of them contains two integers xi, yi (-100 ≤ xi, yi ≤ 100).
It is guaranteed that no two stations are located at the same point.
Output
In a single line print an integer — the answer to the problem.
Examples
Input
3 1000
1000
0 0
0 1
0 3
Output
2000
Input
3 1000
1000
1 0
1 1
1 2
Output
1000
Submitted Solution:
```
from sys import stdin
from math import inf
def readline():
return map(int, stdin.readline().strip().split())
def dijkstra():
n, d = readline()
a = [0] + list(readline()) + [0]
x = [0] * n
y = [0] * n
for i in range(n):
x[i], y[i] = readline()
lower_cost = [inf] * n
lower_cost[0] = 0
visited = [False] * n
for i in range(n - 1):
lower_value = inf
position = 0
for j in range(n):
if not visited[j] and lower_value > lower_cost[j]:
lower_value = lower_cost[j]
position = j
visited[position] = True
for k in range(n):
if not visited[k]:
diff = lower_cost[position] + d * (abs(x[k] - x[position]) + abs(y[k] - y[position])) - a[position]
if lower_cost[k] > diff:
lower_cost[k] = diff
return lower_cost[-1]
if __name__ == '__main__':
print(dijkstra())
```
Yes
| 77,043 | [
0.60400390625,
0.308837890625,
-0.271484375,
0.366455078125,
-0.0400390625,
-0.50244140625,
-0.021453857421875,
0.081787109375,
-0.06951904296875,
1.1923828125,
0.60595703125,
0.005779266357421875,
0.17431640625,
-0.57666015625,
-0.0390625,
-0.09832763671875,
-0.66015625,
-0.980468... | 3 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Yaroslav is playing a game called "Time". The game has a timer showing the lifespan he's got left. As soon as the timer shows 0, Yaroslav's character dies and the game ends. Also, the game has n clock stations, station number i is at point (xi, yi) of the plane. As the player visits station number i, he increases the current time on his timer by ai. The stations are for one-time use only, so if the player visits some station another time, the time on his timer won't grow.
A player spends d·dist time units to move between stations, where dist is the distance the player has covered and d is some constant. The distance between stations i and j is determined as |xi - xj| + |yi - yj|.
Initially, the player is at station number 1, and the player has strictly more than zero and strictly less than one units of time. At station number 1 one unit of money can increase the time on the timer by one time unit (you can buy only integer number of time units).
Now Yaroslav is wondering, how much money he needs to get to station n. Help Yaroslav. Consider the time to buy and to increase the timer value negligibly small.
Input
The first line contains integers n and d (3 ≤ n ≤ 100, 103 ≤ d ≤ 105) — the number of stations and the constant from the statement.
The second line contains n - 2 integers: a2, a3, ..., an - 1 (1 ≤ ai ≤ 103). The next n lines contain the coordinates of the stations. The i-th of them contains two integers xi, yi (-100 ≤ xi, yi ≤ 100).
It is guaranteed that no two stations are located at the same point.
Output
In a single line print an integer — the answer to the problem.
Examples
Input
3 1000
1000
0 0
0 1
0 3
Output
2000
Input
3 1000
1000
1 0
1 1
1 2
Output
1000
Submitted Solution:
```
from sys import stdin
def readline():
return map(int, stdin.readline().strip().split())
def main():
n, d = readline()
a = [0] + list(readline()) + [0]
x = [0] * n
y = [0] * n
for i in range(n):
x[i], y[i] = readline()
less = [-1] * n
less[0] = 0
for i in range(n):
time = 0
for j in range(1, n):
if i != j:
difference = d * (abs(x[i] - x[j]) + abs(y[i] - y[j])) - (a[i] + time)
if difference < 0:
time = -difference
difference = 0
else:
time = 0
difference += less[i]
if less[j] == -1 or less[j] > difference:
less[j] = difference
return less[-1]
if __name__ == '__main__':
print(main())
```
No
| 77,044 | [
0.59033203125,
0.3076171875,
-0.249755859375,
0.403564453125,
-0.0540771484375,
-0.51953125,
-0.0111846923828125,
0.1270751953125,
-0.08306884765625,
1.1435546875,
0.6083984375,
0.024749755859375,
0.2178955078125,
-0.591796875,
-0.039306640625,
-0.09039306640625,
-0.6279296875,
-0.... | 3 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Yaroslav is playing a game called "Time". The game has a timer showing the lifespan he's got left. As soon as the timer shows 0, Yaroslav's character dies and the game ends. Also, the game has n clock stations, station number i is at point (xi, yi) of the plane. As the player visits station number i, he increases the current time on his timer by ai. The stations are for one-time use only, so if the player visits some station another time, the time on his timer won't grow.
A player spends d·dist time units to move between stations, where dist is the distance the player has covered and d is some constant. The distance between stations i and j is determined as |xi - xj| + |yi - yj|.
Initially, the player is at station number 1, and the player has strictly more than zero and strictly less than one units of time. At station number 1 one unit of money can increase the time on the timer by one time unit (you can buy only integer number of time units).
Now Yaroslav is wondering, how much money he needs to get to station n. Help Yaroslav. Consider the time to buy and to increase the timer value negligibly small.
Input
The first line contains integers n and d (3 ≤ n ≤ 100, 103 ≤ d ≤ 105) — the number of stations and the constant from the statement.
The second line contains n - 2 integers: a2, a3, ..., an - 1 (1 ≤ ai ≤ 103). The next n lines contain the coordinates of the stations. The i-th of them contains two integers xi, yi (-100 ≤ xi, yi ≤ 100).
It is guaranteed that no two stations are located at the same point.
Output
In a single line print an integer — the answer to the problem.
Examples
Input
3 1000
1000
0 0
0 1
0 3
Output
2000
Input
3 1000
1000
1 0
1 1
1 2
Output
1000
Submitted Solution:
```
f = lambda: list(map(int, input().split()))
n, d = f()
a = [0] + f() + [0]
p = [f() for i in range(n)]
s = [1e9] * n
q, s[0] = 1, 0
while q:
q = 0
for i in range(n):
for j in range(i + 1, n):
t = s[i] + (abs(p[i][0] - p[j][0]) + abs(p[i][1] - p[j][1])) * d - a[j]
if t < s[j]: q, s[j] = 1, t
print(s[-1])
```
No
| 77,045 | [
0.60498046875,
0.3173828125,
-0.2822265625,
0.46142578125,
-0.056884765625,
-0.5224609375,
-0.030303955078125,
0.1727294921875,
-0.043609619140625,
1.1064453125,
0.626953125,
0.040679931640625,
0.2259521484375,
-0.6162109375,
-0.0638427734375,
-0.067626953125,
-0.60693359375,
-0.98... | 3 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Yaroslav is playing a game called "Time". The game has a timer showing the lifespan he's got left. As soon as the timer shows 0, Yaroslav's character dies and the game ends. Also, the game has n clock stations, station number i is at point (xi, yi) of the plane. As the player visits station number i, he increases the current time on his timer by ai. The stations are for one-time use only, so if the player visits some station another time, the time on his timer won't grow.
A player spends d·dist time units to move between stations, where dist is the distance the player has covered and d is some constant. The distance between stations i and j is determined as |xi - xj| + |yi - yj|.
Initially, the player is at station number 1, and the player has strictly more than zero and strictly less than one units of time. At station number 1 one unit of money can increase the time on the timer by one time unit (you can buy only integer number of time units).
Now Yaroslav is wondering, how much money he needs to get to station n. Help Yaroslav. Consider the time to buy and to increase the timer value negligibly small.
Input
The first line contains integers n and d (3 ≤ n ≤ 100, 103 ≤ d ≤ 105) — the number of stations and the constant from the statement.
The second line contains n - 2 integers: a2, a3, ..., an - 1 (1 ≤ ai ≤ 103). The next n lines contain the coordinates of the stations. The i-th of them contains two integers xi, yi (-100 ≤ xi, yi ≤ 100).
It is guaranteed that no two stations are located at the same point.
Output
In a single line print an integer — the answer to the problem.
Examples
Input
3 1000
1000
0 0
0 1
0 3
Output
2000
Input
3 1000
1000
1 0
1 1
1 2
Output
1000
Submitted Solution:
```
R = lambda: map(int, input().split())
n, d = R()
a = [0] + list(R()) + [0]
x = []
y = []
INF = float('inf')
def distance(x1, y1, x2, y2):
return (abs(x1 - x2) + abs(y1 - y2)) * d
for i in range(n):
xi, yi = R()
x += [xi]
y += [yi]
b = [INF] * n
b[0] = 0
c = True
while c:
c = False
for i in range(n):
for j in range(i+1, n):
if b[i] != INF:
t = b[i] + distance(x[i], y[i], x[j], y[j]) - a[j]
if t < b[j]:
b[j] = t
# print(b)
c = True
print(b[-1])
```
No
| 77,046 | [
0.609375,
0.32177734375,
-0.3046875,
0.461181640625,
-0.0435791015625,
-0.533203125,
-0.00801849365234375,
0.18017578125,
-0.04443359375,
1.123046875,
0.62744140625,
0.058807373046875,
0.2266845703125,
-0.6103515625,
-0.049591064453125,
-0.07623291015625,
-0.59423828125,
-0.9926757... | 3 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Yaroslav is playing a game called "Time". The game has a timer showing the lifespan he's got left. As soon as the timer shows 0, Yaroslav's character dies and the game ends. Also, the game has n clock stations, station number i is at point (xi, yi) of the plane. As the player visits station number i, he increases the current time on his timer by ai. The stations are for one-time use only, so if the player visits some station another time, the time on his timer won't grow.
A player spends d·dist time units to move between stations, where dist is the distance the player has covered and d is some constant. The distance between stations i and j is determined as |xi - xj| + |yi - yj|.
Initially, the player is at station number 1, and the player has strictly more than zero and strictly less than one units of time. At station number 1 one unit of money can increase the time on the timer by one time unit (you can buy only integer number of time units).
Now Yaroslav is wondering, how much money he needs to get to station n. Help Yaroslav. Consider the time to buy and to increase the timer value negligibly small.
Input
The first line contains integers n and d (3 ≤ n ≤ 100, 103 ≤ d ≤ 105) — the number of stations and the constant from the statement.
The second line contains n - 2 integers: a2, a3, ..., an - 1 (1 ≤ ai ≤ 103). The next n lines contain the coordinates of the stations. The i-th of them contains two integers xi, yi (-100 ≤ xi, yi ≤ 100).
It is guaranteed that no two stations are located at the same point.
Output
In a single line print an integer — the answer to the problem.
Examples
Input
3 1000
1000
0 0
0 1
0 3
Output
2000
Input
3 1000
1000
1 0
1 1
1 2
Output
1000
Submitted Solution:
```
n, d = map(int, input().split())
a = [0] + list(map(int, input().split())) + [0]
x = []
y = []
for i in range(n):
xx, yy = map(int, input().split())
x += [xx]
y += [yy]
b = [10**7] * n
b[0] = 0
for k in range(n - 1):
for i in range(n):
for j in range(1, n):
if i != j and b[i] != 10**7:
b[j] = min(b[j], b[i] + (abs(x[i] - x[j]) + abs(y[i] - y[j])) * d - a[j])
print(b[-1])
```
No
| 77,047 | [
0.59619140625,
0.291259765625,
-0.292724609375,
0.41552734375,
-0.048736572265625,
-0.52392578125,
-0.013427734375,
0.1729736328125,
-0.05157470703125,
1.1318359375,
0.62158203125,
0.058624267578125,
0.2496337890625,
-0.62109375,
-0.029876708984375,
-0.08349609375,
-0.62744140625,
... | 3 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
n soldiers stand in a circle. For each soldier his height ai is known. A reconnaissance unit can be made of such two neighbouring soldiers, whose heights difference is minimal, i.e. |ai - aj| is minimal. So each of them will be less noticeable with the other. Output any pair of soldiers that can form a reconnaissance unit.
Input
The first line contains integer n (2 ≤ n ≤ 100) — amount of soldiers. Then follow the heights of the soldiers in their order in the circle — n space-separated integers a1, a2, ..., an (1 ≤ ai ≤ 1000). The soldier heights are given in clockwise or counterclockwise direction.
Output
Output two integers — indexes of neighbouring soldiers, who should form a reconnaissance unit. If there are many optimum solutions, output any of them. Remember, that the soldiers stand in a circle.
Examples
Input
5
10 12 13 15 10
Output
5 1
Input
4
10 20 30 40
Output
1 2
Submitted Solution:
```
n = int(input())
a = [int(i) for i in input().split()]
b = []
for j in range(n-1):
q = a[j]-a[j+1]
if q > 0:
b.append(q)
else:
b.append(-1*q)
w = a[-1]-a[0]
if w > 0:
b.append(w)
else:
b.append(-1*w)
m = min(b)
e = b.index(m) + 1
if e == n:
print(str(e) + ' ' + str(1))
else:
print(str(e) + ' ' + str(e+1))
```
Yes
| 77,057 | [
0.380615234375,
-0.007526397705078125,
0.0157470703125,
0.08624267578125,
-0.60888671875,
-0.273681640625,
-0.2353515625,
0.32958984375,
0.249755859375,
0.90185546875,
0.66064453125,
0.075927734375,
0.148193359375,
-0.6171875,
-0.2183837890625,
0.330078125,
-0.68798828125,
-0.97363... | 3 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
n soldiers stand in a circle. For each soldier his height ai is known. A reconnaissance unit can be made of such two neighbouring soldiers, whose heights difference is minimal, i.e. |ai - aj| is minimal. So each of them will be less noticeable with the other. Output any pair of soldiers that can form a reconnaissance unit.
Input
The first line contains integer n (2 ≤ n ≤ 100) — amount of soldiers. Then follow the heights of the soldiers in their order in the circle — n space-separated integers a1, a2, ..., an (1 ≤ ai ≤ 1000). The soldier heights are given in clockwise or counterclockwise direction.
Output
Output two integers — indexes of neighbouring soldiers, who should form a reconnaissance unit. If there are many optimum solutions, output any of them. Remember, that the soldiers stand in a circle.
Examples
Input
5
10 12 13 15 10
Output
5 1
Input
4
10 20 30 40
Output
1 2
Submitted Solution:
```
n = int(input())
l = list(map(int,input().strip().split()))
res = 999999999
for i in range(0, n-1) :
if abs(l[i]-l[i+ 1]) < res:
res = abs(l[i] - l[i+1])
inde = [i, i+1]
if abs(l[-1]-l[0]) < res:
print(n, 1)
else:
print(inde[0]+1, inde[1]+1)
```
Yes
| 77,058 | [
0.4345703125,
0.00917816162109375,
-0.015625,
0.07952880859375,
-0.56982421875,
-0.274658203125,
-0.2685546875,
0.339599609375,
0.2841796875,
0.90673828125,
0.6533203125,
0.08453369140625,
0.1650390625,
-0.61328125,
-0.2056884765625,
0.34228515625,
-0.62744140625,
-0.8798828125,
... | 3 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
n soldiers stand in a circle. For each soldier his height ai is known. A reconnaissance unit can be made of such two neighbouring soldiers, whose heights difference is minimal, i.e. |ai - aj| is minimal. So each of them will be less noticeable with the other. Output any pair of soldiers that can form a reconnaissance unit.
Input
The first line contains integer n (2 ≤ n ≤ 100) — amount of soldiers. Then follow the heights of the soldiers in their order in the circle — n space-separated integers a1, a2, ..., an (1 ≤ ai ≤ 1000). The soldier heights are given in clockwise or counterclockwise direction.
Output
Output two integers — indexes of neighbouring soldiers, who should form a reconnaissance unit. If there are many optimum solutions, output any of them. Remember, that the soldiers stand in a circle.
Examples
Input
5
10 12 13 15 10
Output
5 1
Input
4
10 20 30 40
Output
1 2
Submitted Solution:
```
#A. Reconnaissance 2
n = int(input())
arr = list(map(int,input().split()))
x,y = n,1
min_ = abs(arr[0] - arr[n-1])
for i in range(1,n):
if abs(arr[i]-arr[i-1])<min_:
min_ = abs(arr[i]-arr[i-1])
x = i + 1
y = i
print(x,y)
```
Yes
| 77,060 | [
0.455322265625,
0.04791259765625,
-0.006244659423828125,
0.108642578125,
-0.572265625,
-0.26806640625,
-0.34130859375,
0.363525390625,
0.25439453125,
0.91943359375,
0.716796875,
0.090576171875,
0.15625,
-0.6142578125,
-0.1702880859375,
0.369384765625,
-0.619140625,
-0.8916015625,
... | 3 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Dima is living in a dormitory, as well as some cockroaches.
At the moment 0 Dima saw a cockroach running on a table and decided to kill it. Dima needs exactly T seconds for aiming, and after that he will precisely strike the cockroach and finish it.
To survive the cockroach has to run into a shadow, cast by round plates standing on the table, in T seconds. Shadow casted by any of the plates has the shape of a circle. Shadow circles may intersect, nest or overlap arbitrarily.
The cockroach uses the following strategy: first he equiprobably picks a direction to run towards and then runs towards it with the constant speed v. If at some moment t ≤ T it reaches any shadow circle, it immediately stops in the shadow and thus will stay alive. Otherwise the cockroach is killed by the Dima's precise strike. Consider that the Dima's precise strike is instant.
Determine the probability of that the cockroach will stay alive.
Input
In the first line of the input the four integers x0, y0, v, T (|x0|, |y0| ≤ 109, 0 ≤ v, T ≤ 109) are given — the cockroach initial position on the table in the Cartesian system at the moment 0, the cockroach's constant speed and the time in seconds Dima needs for aiming respectively.
In the next line the only number n (1 ≤ n ≤ 100 000) is given — the number of shadow circles casted by plates.
In the next n lines shadow circle description is given: the ith of them consists of three integers xi, yi, ri (|xi|, |yi| ≤ 109, 0 ≤ r ≤ 109) — the ith shadow circle on-table position in the Cartesian system and its radius respectively.
Consider that the table is big enough for the cockroach not to run to the table edges and avoid Dima's precise strike.
Output
Print the only real number p — the probability of that the cockroach will stay alive.
Your answer will be considered correct if its absolute or relative error does not exceed 10 - 4.
Examples
Input
0 0 1 1
3
1 1 1
-1 -1 1
-2 2 1
Output
0.50000000000
Input
0 0 1 0
1
1 0 1
Output
1.00000000000
Note
The picture for the first sample is given below.
<image>
Red color stands for points which being chosen as the cockroach's running direction will cause him being killed, green color for those standing for survival directions. Please note that despite containing a circle centered in ( - 2, 2) a part of zone is colored red because the cockroach is not able to reach it in one second.
Submitted Solution:
```
import math as m
l = [int(x) for x in input().split()]
x0, y0, v, T = tuple(l)
n = int(input())
c = []
theta = []
flag = 0
for i in range(n):
l = [int(x) for x in input().split()]
c.append(l)
for i in range(n):
x, y, r = tuple(c[i])
d = m.sqrt((x-x0)**2 + (y-y0)**2)
s = d - r
if s <= 0:
flag = 1
break
if s >= v*T:
continue
t = m.sqrt(d**2 - r**2)
if t <= v*T:
a = m.asin(r/d)
theta.append(2*a)
else:
a = m.acos((d**2 + (v*T)**2 - r**2)/(2*d*v*T))
theta.append(2*a)
if flag:
print(1.0)
else:
print(sum(theta)/(2*m.pi))
```
No
| 77,183 | [
0.317626953125,
0.08544921875,
-0.050018310546875,
0.244140625,
-0.609375,
-0.35498046875,
-0.1563720703125,
0.12286376953125,
0.029754638671875,
0.9619140625,
0.7607421875,
0.186767578125,
-0.00925445556640625,
-0.52734375,
-0.321533203125,
0.56201171875,
-0.260986328125,
-0.45239... | 3 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Dima is living in a dormitory, as well as some cockroaches.
At the moment 0 Dima saw a cockroach running on a table and decided to kill it. Dima needs exactly T seconds for aiming, and after that he will precisely strike the cockroach and finish it.
To survive the cockroach has to run into a shadow, cast by round plates standing on the table, in T seconds. Shadow casted by any of the plates has the shape of a circle. Shadow circles may intersect, nest or overlap arbitrarily.
The cockroach uses the following strategy: first he equiprobably picks a direction to run towards and then runs towards it with the constant speed v. If at some moment t ≤ T it reaches any shadow circle, it immediately stops in the shadow and thus will stay alive. Otherwise the cockroach is killed by the Dima's precise strike. Consider that the Dima's precise strike is instant.
Determine the probability of that the cockroach will stay alive.
Input
In the first line of the input the four integers x0, y0, v, T (|x0|, |y0| ≤ 109, 0 ≤ v, T ≤ 109) are given — the cockroach initial position on the table in the Cartesian system at the moment 0, the cockroach's constant speed and the time in seconds Dima needs for aiming respectively.
In the next line the only number n (1 ≤ n ≤ 100 000) is given — the number of shadow circles casted by plates.
In the next n lines shadow circle description is given: the ith of them consists of three integers xi, yi, ri (|xi|, |yi| ≤ 109, 0 ≤ r ≤ 109) — the ith shadow circle on-table position in the Cartesian system and its radius respectively.
Consider that the table is big enough for the cockroach not to run to the table edges and avoid Dima's precise strike.
Output
Print the only real number p — the probability of that the cockroach will stay alive.
Your answer will be considered correct if its absolute or relative error does not exceed 10 - 4.
Examples
Input
0 0 1 1
3
1 1 1
-1 -1 1
-2 2 1
Output
0.50000000000
Input
0 0 1 0
1
1 0 1
Output
1.00000000000
Note
The picture for the first sample is given below.
<image>
Red color stands for points which being chosen as the cockroach's running direction will cause him being killed, green color for those standing for survival directions. Please note that despite containing a circle centered in ( - 2, 2) a part of zone is colored red because the cockroach is not able to reach it in one second.
Submitted Solution:
```
from math import acos, pi
def rast_between_dots(xa, ya, xb, yb):
return ((xb-xa)**2 + (yb-ya)**2)**(1/2)
x0, y0, v, T = map(int, input().split())
n = int(input())
possibility = 0
for i in range(n):
xi, yi, ri = map(int, input().split())
r = rast_between_dots(xi, yi, x0, y0)
if rast_between_dots(xi, yi, x0, y0) <= ri:
possibility = 1
break
elif r < T*v+ri:
a = (r**2-ri**2)**(1/2)
angleOfPossibility = acos((2*a**2-4*ri**2)/(2*a**2))
possibility += angleOfPossibility/(4*pi)
print(possibility)
```
No
| 77,184 | [
0.317626953125,
0.08544921875,
-0.050018310546875,
0.244140625,
-0.609375,
-0.35498046875,
-0.1563720703125,
0.12286376953125,
0.029754638671875,
0.9619140625,
0.7607421875,
0.186767578125,
-0.00925445556640625,
-0.52734375,
-0.321533203125,
0.56201171875,
-0.260986328125,
-0.45239... | 3 |
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