message stringlengths 2 44.5k | message_type stringclasses 2 values | message_id int64 0 1 | conversation_id int64 276 109k | cluster float64 23 23 | __index_level_0__ int64 552 217k |
|---|---|---|---|---|---|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given N points in the xy-plane. You have a circle of radius one and move it on the xy-plane, so as to enclose as many of the points as possible. Find how many points can be simultaneously enclosed at the maximum. A point is considered enclosed by a circle when it is inside or on the circle.
<image>
Fig 1. Circle and Points
Input
The input consists of a series of data sets, followed by a single line only containing a single character '0', which indicates the end of the input. Each data set begins with a line containing an integer N, which indicates the number of points in the data set. It is followed by N lines describing the coordinates of the points. Each of the N lines has two decimal fractions X and Y, describing the x- and y-coordinates of a point, respectively. They are given with five digits after the decimal point.
You may assume 1 <= N <= 300, 0.0 <= X <= 10.0, and 0.0 <= Y <= 10.0. No two points are closer than 0.0001. No two points in a data set are approximately at a distance of 2.0. More precisely, for any two points in a data set, the distance d between the two never satisfies 1.9999 <= d <= 2.0001. Finally, no three points in a data set are simultaneously very close to a single circle of radius one. More precisely, let P1, P2, and P3 be any three points in a data set, and d1, d2, and d3 the distances from an arbitrarily selected point in the xy-plane to each of them respectively. Then it never simultaneously holds that 0.9999 <= di <= 1.0001 (i = 1, 2, 3).
Output
For each data set, print a single line containing the maximum number of points in the data set that can be simultaneously enclosed by a circle of radius one. No other characters including leading and trailing spaces should be printed.
Example
Input
3
6.47634 7.69628
5.16828 4.79915
6.69533 6.20378
6
7.15296 4.08328
6.50827 2.69466
5.91219 3.86661
5.29853 4.16097
6.10838 3.46039
6.34060 2.41599
8
7.90650 4.01746
4.10998 4.18354
4.67289 4.01887
6.33885 4.28388
4.98106 3.82728
5.12379 5.16473
7.84664 4.67693
4.02776 3.87990
20
6.65128 5.47490
6.42743 6.26189
6.35864 4.61611
6.59020 4.54228
4.43967 5.70059
4.38226 5.70536
5.50755 6.18163
7.41971 6.13668
6.71936 3.04496
5.61832 4.23857
5.99424 4.29328
5.60961 4.32998
6.82242 5.79683
5.44693 3.82724
6.70906 3.65736
7.89087 5.68000
6.23300 4.59530
5.92401 4.92329
6.24168 3.81389
6.22671 3.62210
0
Output
2
5
5
11
Submitted Solution:
```
def cent(p1,p2):
if abs(p1-p2) >= 2:
return None,None
else:
q = (p1+p2)/2
r = (p1-p2)/2
n = abs(r)
t = r * 1j * ((1-n**2)**0.5/n)
return(q + t, q - t)
while(True):
ans = 1
n = int(input())
if n == 0:
quit()
ps = []
for _ in range(n):
x,y = map(float,input().split())
ps.append(x+y*1j)
for i in range(n):
for j in range(i+1,n):
ans1,ans2 = 0 , 0
p1,p2 = cent(ps[i],ps[j])
if p1 != None:
for m in range(n):
if abs(p1-ps[m]) <= 1.001:
ans1 += 1
if abs(p2-ps[m]) <= 1.001:
ans2 += 1
ans = max(ans,ans1,ans2)
print(ans)
``` | instruction | 0 | 14,080 | 23 | 28,160 |
No | output | 1 | 14,080 | 23 | 28,161 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given N points in the xy-plane. You have a circle of radius one and move it on the xy-plane, so as to enclose as many of the points as possible. Find how many points can be simultaneously enclosed at the maximum. A point is considered enclosed by a circle when it is inside or on the circle.
<image>
Fig 1. Circle and Points
Input
The input consists of a series of data sets, followed by a single line only containing a single character '0', which indicates the end of the input. Each data set begins with a line containing an integer N, which indicates the number of points in the data set. It is followed by N lines describing the coordinates of the points. Each of the N lines has two decimal fractions X and Y, describing the x- and y-coordinates of a point, respectively. They are given with five digits after the decimal point.
You may assume 1 <= N <= 300, 0.0 <= X <= 10.0, and 0.0 <= Y <= 10.0. No two points are closer than 0.0001. No two points in a data set are approximately at a distance of 2.0. More precisely, for any two points in a data set, the distance d between the two never satisfies 1.9999 <= d <= 2.0001. Finally, no three points in a data set are simultaneously very close to a single circle of radius one. More precisely, let P1, P2, and P3 be any three points in a data set, and d1, d2, and d3 the distances from an arbitrarily selected point in the xy-plane to each of them respectively. Then it never simultaneously holds that 0.9999 <= di <= 1.0001 (i = 1, 2, 3).
Output
For each data set, print a single line containing the maximum number of points in the data set that can be simultaneously enclosed by a circle of radius one. No other characters including leading and trailing spaces should be printed.
Example
Input
3
6.47634 7.69628
5.16828 4.79915
6.69533 6.20378
6
7.15296 4.08328
6.50827 2.69466
5.91219 3.86661
5.29853 4.16097
6.10838 3.46039
6.34060 2.41599
8
7.90650 4.01746
4.10998 4.18354
4.67289 4.01887
6.33885 4.28388
4.98106 3.82728
5.12379 5.16473
7.84664 4.67693
4.02776 3.87990
20
6.65128 5.47490
6.42743 6.26189
6.35864 4.61611
6.59020 4.54228
4.43967 5.70059
4.38226 5.70536
5.50755 6.18163
7.41971 6.13668
6.71936 3.04496
5.61832 4.23857
5.99424 4.29328
5.60961 4.32998
6.82242 5.79683
5.44693 3.82724
6.70906 3.65736
7.89087 5.68000
6.23300 4.59530
5.92401 4.92329
6.24168 3.81389
6.22671 3.62210
0
Output
2
5
5
11
Submitted Solution:
```
#!/usr/bin/env python
import math
def func():
lst = []
while True:
S = input()
lst.append(S)
if S == '0':
break
lst = list(reversed(lst))
while True:
N = int(lst.pop())
if N == 0:
break
R = range(N)
pos = []
for loop in R:
S = list(map(float, lst.pop().split()))
pos.append(S)
ans = 0
for i in R:
for j in R:
v = [pos[j][0] - pos[i][0], pos[j][1] - pos[i][1]]
norm = math.sqrt(v[0] * v[0] + v[1] * v[1])
if norm > 2 or norm == 0:
continue
vn = [v[0] / norm, v[1] / norm]
norm /= 2
norm2 = math.sqrt(1 - norm * norm)
v90 = [-vn[1], vn[0]]
cent = [pos[i][0] + v[0] / 2 + v90[0] * norm2,
pos[i][1] + v[1] / 2 + v90[1] * norm2]
S = 2
for k in R:
if i == k or j == k:
continue
x = cent[0] - pos[k][0]
y = cent[1] - pos[k][1]
if x * x + y * y <= 1:
S += 1
ans = max(S, ans)
print(max(1, ans))
if __name__ == '__main__':
func()
``` | instruction | 0 | 14,081 | 23 | 28,162 |
No | output | 1 | 14,081 | 23 | 28,163 |
Provide a correct Python 3 solution for this coding contest problem.
For given three points p1, p2, p, find the reflection point x of p onto p1p2.
<image>
Constraints
* 1 ≤ q ≤ 1000
* -10000 ≤ xi, yi ≤ 10000
* p1 and p2 are not identical.
Input
xp1 yp1 xp2 yp2
q
xp0 yp0
xp1 yp1
...
xpq−1 ypq−1
In the first line, integer coordinates of p1 and p2 are given. Then, q queries are given for integer coordinates of p.
Output
For each query, print the coordinate of the reflection point x. The output values should be in a decimal fraction with an error less than 0.00000001.
Examples
Input
0 0 2 0
3
-1 1
0 1
1 1
Output
-1.0000000000 -1.0000000000
0.0000000000 -1.0000000000
1.0000000000 -1.0000000000
Input
0 0 3 4
3
2 5
1 4
0 3
Output
4.2400000000 3.3200000000
3.5600000000 2.0800000000
2.8800000000 0.8400000000 | instruction | 0 | 14,114 | 23 | 28,228 |
"Correct Solution:
```
x1,y1,x2,y2=map(int,input().split())
p1,p2=complex(x1,y1),complex(x2,y2)
a=p2-p1
dis=(x1-x2)**2+(y1-y2)**2
for i in range(int(input())):
p=complex(*map(int,input().split()))
b=p-p1
co=(a.real*b.real+a.imag*b.imag)/dis
ans=p+2*(co*a-b)
print(ans.real,ans.imag)
``` | output | 1 | 14,114 | 23 | 28,229 |
Provide a correct Python 3 solution for this coding contest problem.
For given three points p1, p2, p, find the reflection point x of p onto p1p2.
<image>
Constraints
* 1 ≤ q ≤ 1000
* -10000 ≤ xi, yi ≤ 10000
* p1 and p2 are not identical.
Input
xp1 yp1 xp2 yp2
q
xp0 yp0
xp1 yp1
...
xpq−1 ypq−1
In the first line, integer coordinates of p1 and p2 are given. Then, q queries are given for integer coordinates of p.
Output
For each query, print the coordinate of the reflection point x. The output values should be in a decimal fraction with an error less than 0.00000001.
Examples
Input
0 0 2 0
3
-1 1
0 1
1 1
Output
-1.0000000000 -1.0000000000
0.0000000000 -1.0000000000
1.0000000000 -1.0000000000
Input
0 0 3 4
3
2 5
1 4
0 3
Output
4.2400000000 3.3200000000
3.5600000000 2.0800000000
2.8800000000 0.8400000000 | instruction | 0 | 14,115 | 23 | 28,230 |
"Correct Solution:
```
import sys
input = sys.stdin.readline
class Vector():
def __init__(self, x, y):
self.x = x
self.y = y
def __add__(self, vec):
return Vector(self.x+vec.x, self.y+vec.y)
def __sub__(self, vec):
return Vector(self.x-vec.x, self.y-vec.y)
def __mul__(self, sc):
return Vector(self.x*sc, self.y*sc)
def __truediv__(self, sc):
return Vector(self.x/sc, self.y/sc)
def __iadd__(self, vec):
self.x += vec.x
self.y += vec.y
return self
def __isub__(self, vec):
self.x -= vec.x
self.y -= vec.y
return self
def __imul__(self, sc):
self.x *= sc
self.y *= sc
return self
def __itruediv__(self, sc):
self.x /= sc
self.y /= sc
return self
def __str__(self):
return '{:.9f} {:.9f}'.format(self.x, self.y)
def __eq__(self, vec):
return self.x == vec.x and self.y == vec.y
def dot(self, vec):
return self.x * vec.x + self.y * vec.y
def abs(self):
return (self.x*self.x + self.y*self.y)**0.5
def ortho(self):
return Vector(-self.y, self.x)
x1, y1, x2, y2 = map(int, input().split())
v1 = Vector(x1, y1)
v2 = Vector(x2, y2)
n = v2-v1
n /= n.abs()
m = n.ortho()
for _ in [0]*int(input()):
x, y = map(int, input().split())
v = Vector(x, y) - v1
u = v - m * v.dot(m) * 2
print(v1 + u)
``` | output | 1 | 14,115 | 23 | 28,231 |
Provide a correct Python 3 solution for this coding contest problem.
For given three points p1, p2, p, find the reflection point x of p onto p1p2.
<image>
Constraints
* 1 ≤ q ≤ 1000
* -10000 ≤ xi, yi ≤ 10000
* p1 and p2 are not identical.
Input
xp1 yp1 xp2 yp2
q
xp0 yp0
xp1 yp1
...
xpq−1 ypq−1
In the first line, integer coordinates of p1 and p2 are given. Then, q queries are given for integer coordinates of p.
Output
For each query, print the coordinate of the reflection point x. The output values should be in a decimal fraction with an error less than 0.00000001.
Examples
Input
0 0 2 0
3
-1 1
0 1
1 1
Output
-1.0000000000 -1.0000000000
0.0000000000 -1.0000000000
1.0000000000 -1.0000000000
Input
0 0 3 4
3
2 5
1 4
0 3
Output
4.2400000000 3.3200000000
3.5600000000 2.0800000000
2.8800000000 0.8400000000 | instruction | 0 | 14,116 | 23 | 28,232 |
"Correct Solution:
```
xp1,yp1,xp2,yp2=map(int,input().split())
q=int(input())
if xp1==xp2:
for i in range(q):
x0,y0=map(int,input().split())
print('{:.10f}'.format(2*xp1-x0),'{:.10f}'.format(y0))
else:
a=float((yp2-yp1)/(xp2-xp1))
for i in range(q):
x0,y0=map(int,input().split())
x=a*((y0-yp1)-a*(x0-xp1))/(1+a**2)+x0
y=(a*(a*y0+x0-xp1)+yp1)/(1+a**2)
print('{:.10f}'.format(2*x-x0),'{:.10f}'.format(2*y-y0))
``` | output | 1 | 14,116 | 23 | 28,233 |
Provide a correct Python 3 solution for this coding contest problem.
For given three points p1, p2, p, find the reflection point x of p onto p1p2.
<image>
Constraints
* 1 ≤ q ≤ 1000
* -10000 ≤ xi, yi ≤ 10000
* p1 and p2 are not identical.
Input
xp1 yp1 xp2 yp2
q
xp0 yp0
xp1 yp1
...
xpq−1 ypq−1
In the first line, integer coordinates of p1 and p2 are given. Then, q queries are given for integer coordinates of p.
Output
For each query, print the coordinate of the reflection point x. The output values should be in a decimal fraction with an error less than 0.00000001.
Examples
Input
0 0 2 0
3
-1 1
0 1
1 1
Output
-1.0000000000 -1.0000000000
0.0000000000 -1.0000000000
1.0000000000 -1.0000000000
Input
0 0 3 4
3
2 5
1 4
0 3
Output
4.2400000000 3.3200000000
3.5600000000 2.0800000000
2.8800000000 0.8400000000 | instruction | 0 | 14,117 | 23 | 28,234 |
"Correct Solution:
```
import sys
input = sys.stdin.readline
def main():
x1, y1, x2, y2 = map(int, input().split())
q = int(input())
if x2 - x1 == 0:# x = x1 の直線
for i in range(q):
xp, yp = map(int, input().split())
print(xp + 2 * (x1 - xp), yp)
else:
m = (y2 - y1) / (x2 - x1)
for i in range(q):
xp, yp = map(int, input().split())
yq = (2 * m * (xp - x1) + yp * (m * m - 1) + 2 * y1) / (1 + m * m)
xq = xp - m * (yq - yp)
print(xq, yq)
if __name__ == "__main__":
main()
``` | output | 1 | 14,117 | 23 | 28,235 |
Provide a correct Python 3 solution for this coding contest problem.
For given three points p1, p2, p, find the reflection point x of p onto p1p2.
<image>
Constraints
* 1 ≤ q ≤ 1000
* -10000 ≤ xi, yi ≤ 10000
* p1 and p2 are not identical.
Input
xp1 yp1 xp2 yp2
q
xp0 yp0
xp1 yp1
...
xpq−1 ypq−1
In the first line, integer coordinates of p1 and p2 are given. Then, q queries are given for integer coordinates of p.
Output
For each query, print the coordinate of the reflection point x. The output values should be in a decimal fraction with an error less than 0.00000001.
Examples
Input
0 0 2 0
3
-1 1
0 1
1 1
Output
-1.0000000000 -1.0000000000
0.0000000000 -1.0000000000
1.0000000000 -1.0000000000
Input
0 0 3 4
3
2 5
1 4
0 3
Output
4.2400000000 3.3200000000
3.5600000000 2.0800000000
2.8800000000 0.8400000000 | instruction | 0 | 14,118 | 23 | 28,236 |
"Correct Solution:
```
import math
class Point():
def __init__(self, x=None, y=None):
self.x = x
self.y = y
class Line():
def __init__(self, x1, y1, x2, y2):
self.p1 = Point(x1, y1)
self.p2 = Point(x2, y2)
def projection(self, p):
a = (self.p2.y - self.p1.y)/(self.p2.x - self.p1.x) if self.p2.x != self.p1.x else float('inf')
if a == 0:
return (p.x, 2*self.p1.y - p.y)
elif a == float('inf'):
return (2*self.p1.x - p.x, p.y)
else:
b = self.p2.y - a*self.p2.x
d = abs(-1*a*p.x + p.y - b)/math.sqrt(1 + a*a)
ans_y = [p.y + math.sqrt(4*d*d/(a*a+1)), p.y - math.sqrt(4*d*d/(a*a+1))]
ans_x = [p.x - a*(_y - p.y) for _y in ans_y]
rst_x, rst_y = None, None
tmp = abs(-1*a*ans_x[0] + ans_y[0] - b)/math.sqrt(1 + a*a) - d
if tmp < abs(-1*a*ans_x[1] + ans_y[1] - b)/math.sqrt(1 + a*a) - d:
rst_x, rst_y = ans_x[0], ans_y[0]
else:
rst_x, rst_y = ans_x[1], ans_y[1]
return round(rst_x, 9), round(rst_y, 9)
x1, y1, x2, y2 = list(map(int, input().split(' ')))
line = Line(x1, y1, x2, y2)
q = int(input())
for i in range(q):
x, y = list(map(int, input().split(' ')))
p = Point(x, y)
rst_x, rst_y = line.projection(p)
print(rst_x, rst_y)
``` | output | 1 | 14,118 | 23 | 28,237 |
Provide a correct Python 3 solution for this coding contest problem.
For given three points p1, p2, p, find the reflection point x of p onto p1p2.
<image>
Constraints
* 1 ≤ q ≤ 1000
* -10000 ≤ xi, yi ≤ 10000
* p1 and p2 are not identical.
Input
xp1 yp1 xp2 yp2
q
xp0 yp0
xp1 yp1
...
xpq−1 ypq−1
In the first line, integer coordinates of p1 and p2 are given. Then, q queries are given for integer coordinates of p.
Output
For each query, print the coordinate of the reflection point x. The output values should be in a decimal fraction with an error less than 0.00000001.
Examples
Input
0 0 2 0
3
-1 1
0 1
1 1
Output
-1.0000000000 -1.0000000000
0.0000000000 -1.0000000000
1.0000000000 -1.0000000000
Input
0 0 3 4
3
2 5
1 4
0 3
Output
4.2400000000 3.3200000000
3.5600000000 2.0800000000
2.8800000000 0.8400000000 | instruction | 0 | 14,119 | 23 | 28,238 |
"Correct Solution:
```
# -*- coding: utf-8 -*-
import collections
import math
class Vector2(collections.namedtuple("Vector2", ["x", "y"])):
def __add__(self, other):
return Vector2(self.x + other.x, self.y + other.y)
def __sub__(self, other):
return Vector2(self.x - other.x, self.y - other.y)
def __mul__(self, scalar): # cross product
return Vector2(self.x * scalar, self.y * scalar)
def __neg__(self):
return Vector2(-self.x, -self.y)
def __pos__(self):
return Vector2(+self.x, +self.y)
def __abs__(self): # norm
return math.sqrt(float(self.x * self.x + self.y * self.y))
def dot(self, other): # dot product
return self.x * other.x + self.y * other.y
def cross(self, other): # cross product
return self.x * other.y - self.y * other.x
if __name__ == '__main__':
a, b, c, d = map(int, input().split())
p1 = Vector2(a, b)
p2 = Vector2(c, d)
base = p2 - p1
q = int(input())
ans = []
for _ in range(q):
e, f = map(int, input().split())
p = Vector2(e, f)
hypo = Vector2(e - a, f - b)
proj = p1 + base * (hypo.dot(base) / float(base.x**2 + base.y**2))
x = p + (proj - p) * 2
ans.append(x)
for x in ans:
print(x.x, x.y)
``` | output | 1 | 14,119 | 23 | 28,239 |
Provide a correct Python 3 solution for this coding contest problem.
For given three points p1, p2, p, find the reflection point x of p onto p1p2.
<image>
Constraints
* 1 ≤ q ≤ 1000
* -10000 ≤ xi, yi ≤ 10000
* p1 and p2 are not identical.
Input
xp1 yp1 xp2 yp2
q
xp0 yp0
xp1 yp1
...
xpq−1 ypq−1
In the first line, integer coordinates of p1 and p2 are given. Then, q queries are given for integer coordinates of p.
Output
For each query, print the coordinate of the reflection point x. The output values should be in a decimal fraction with an error less than 0.00000001.
Examples
Input
0 0 2 0
3
-1 1
0 1
1 1
Output
-1.0000000000 -1.0000000000
0.0000000000 -1.0000000000
1.0000000000 -1.0000000000
Input
0 0 3 4
3
2 5
1 4
0 3
Output
4.2400000000 3.3200000000
3.5600000000 2.0800000000
2.8800000000 0.8400000000 | instruction | 0 | 14,120 | 23 | 28,240 |
"Correct Solution:
```
from sys import stdin
class Vector:
def __init__(self, x=None, y=None):
self.x = x
self.y = y
def __add__(self, other):
return Vector(self.x + other.x, self.y + other.y)
def __sub__(self, other):
return Vector(self.x - other.x, self.y - other.y)
def __mul__(self, k):
return Vector(self.x * k, self.y * k)
def __gt__(self, other):
return self.x > other.x and self.y > other.y
def __eq__(self, other):
return self.x == other.x and self.y == other.y
def dot(self, other):
return self.x * other.x + self.y * other.y
# usually cross operation return Vector but it returns scalor
def cross(self, other):
return self.x * other.y - self.y * other.x
def norm(self):
return self.x * self.x + self.y * self.y
def abs(self):
return math.sqrt(self.norm())
class Point(Vector):
def __init__(self, *args, **kargs):
return super().__init__(*args, **kargs)
class Segment:
def __init__(self, p1=Point(0, 0), p2=Point(1, 1)):
self.p1 = p1
self.p2 = p2
def project(s, p):
base = s.p2 - s.p1
hypo = p - s.p1
r = hypo.dot(base) / base.norm()
return s.p1 + base * r
def reflect(s, p):
return p + (project(s, p) - p) * 2
def read_and_print_results(s, n):
for _ in range(n):
line = stdin.readline().strip().split()
p = Vector(int(line[0]), int(line[1]))
x = reflect(s, p)
print('{0:0.10f} {1:0.10f}'.format(x.x, x.y))
x1, y1, x2, y2 = input().split()
p1 = Point(int(x1), int(y1))
p2 = Point(int(x2), int(y2))
s = Segment(p1, p2)
n = int(input())
read_and_print_results(s, n)
``` | output | 1 | 14,120 | 23 | 28,241 |
Provide a correct Python 3 solution for this coding contest problem.
For given three points p1, p2, p, find the reflection point x of p onto p1p2.
<image>
Constraints
* 1 ≤ q ≤ 1000
* -10000 ≤ xi, yi ≤ 10000
* p1 and p2 are not identical.
Input
xp1 yp1 xp2 yp2
q
xp0 yp0
xp1 yp1
...
xpq−1 ypq−1
In the first line, integer coordinates of p1 and p2 are given. Then, q queries are given for integer coordinates of p.
Output
For each query, print the coordinate of the reflection point x. The output values should be in a decimal fraction with an error less than 0.00000001.
Examples
Input
0 0 2 0
3
-1 1
0 1
1 1
Output
-1.0000000000 -1.0000000000
0.0000000000 -1.0000000000
1.0000000000 -1.0000000000
Input
0 0 3 4
3
2 5
1 4
0 3
Output
4.2400000000 3.3200000000
3.5600000000 2.0800000000
2.8800000000 0.8400000000 | instruction | 0 | 14,121 | 23 | 28,242 |
"Correct Solution:
```
#!/usr/bin/python3
import array
from fractions import Fraction
import math
import os
import sys
class Vec(object):
def __init__(self, x, y):
self.x = x
self.y = y
super().__init__()
def __add__(self, other):
return Vec(self.x + other.x, self.y + other.y)
def __sub__(self, other):
return Vec(self.x - other.x, self.y - other.y)
def __mul__(self, scalar):
return Vec(self.x * scalar, self.y * scalar)
def __rmul__(self, scalar):
return Vec(self.x * scalar, self.y * scalar)
def __truediv__(self, scalar):
return Vec(self.x / scalar, self.y / scalar)
def __iadd__(self, other):
self.x += other.x
self.y += other.y
return self
def __isub__(self, other):
self.x -= other.x
self.y -= other.y
return self
def __imul__(self, scalar):
self.x *= scalar
self.y *= scalar
return self
def __idiv__(self, scalar):
self.x /= scalar
self.y /= scalar
return self
def __neg__(self):
return Vec(-self.x, -self.y)
def dot(self, other):
return self.x * other.x + self.y * other.y
def cross(self, other):
return self.x * other.y - self.y * other.x
def abs2(self):
return self.x * self.x + self.y * self.y
def __abs__(self):
return math.sqrt(float(self.abs2()))
def __str__(self):
return '({}, {})'.format(self.x, self.y)
def main():
x1, y1, x2, y2 = read_ints()
Q = read_int()
for _ in range(Q):
x, y = read_ints()
print(*solve(Vec(x1, y1), Vec(x2, y2), Vec(x, y)))
def solve(u, v, a):
v -= u
a -= u
b = Fraction(2 * v.dot(a), v.abs2()) * v - a
b += u
return float(b.x), float(b.y)
###############################################################################
# AUXILIARY FUNCTIONS
DEBUG = 'DEBUG' in os.environ
def inp():
return sys.stdin.readline().rstrip()
def read_int():
return int(inp())
def read_ints():
return [int(e) for e in inp().split()]
def dprint(*value, sep=' ', end='\n'):
if DEBUG:
print(*value, sep=sep, end=end)
if __name__ == '__main__':
main()
``` | output | 1 | 14,121 | 23 | 28,243 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
For given three points p1, p2, p, find the reflection point x of p onto p1p2.
<image>
Constraints
* 1 ≤ q ≤ 1000
* -10000 ≤ xi, yi ≤ 10000
* p1 and p2 are not identical.
Input
xp1 yp1 xp2 yp2
q
xp0 yp0
xp1 yp1
...
xpq−1 ypq−1
In the first line, integer coordinates of p1 and p2 are given. Then, q queries are given for integer coordinates of p.
Output
For each query, print the coordinate of the reflection point x. The output values should be in a decimal fraction with an error less than 0.00000001.
Examples
Input
0 0 2 0
3
-1 1
0 1
1 1
Output
-1.0000000000 -1.0000000000
0.0000000000 -1.0000000000
1.0000000000 -1.0000000000
Input
0 0 3 4
3
2 5
1 4
0 3
Output
4.2400000000 3.3200000000
3.5600000000 2.0800000000
2.8800000000 0.8400000000
Submitted Solution:
```
import math
from typing import Union
class Point(object):
__slots__ = ['x', 'y']
def __init__(self, x, y):
self.x = x
self.y = y
def __add__(self, other):
return Point(self.x + other.x, self.y + other.y)
def __sub__(self, other):
return Point(self.x - other.x, self.y - other.y)
def __mul__(self, other: Union[int, float]):
return Point(self.x * other, self.y * other)
def __repr__(self):
return f"({self.x},{self.y})"
class Segment(object):
__slots__ = ['pt1', 'pt2', 'vector']
def __init__(self, x: Point, y: Point):
self.pt1 = x
self.pt2 = y
self.vector = self.pt2 - self.pt1
def dot(self, other):
return self.vector.x * other.vector.x + self.vector.y * other.vector.y
def cross(self, other):
return self.vector.x * other.vector.y - self.vector.y * other.vector.x
def norm(self):
return pow(self.vector.x, 2) + pow(self.vector.y, 2)
def abs(self):
return math.sqrt(self.norm())
def projection(self, pt: Point)-> Point:
t = self.dot(Segment(self.pt1, pt)) / pow(self.abs(), 2)
return Point(self.pt1.x + t * self.vector.x, self.pt1.y + t * self.vector.y)
def reflection(self, pt: Point) -> Point:
return self.projection(pt) * 2 - pt
def __repr__(self):
return f"{self.pt1},{self.pt2},{self.vector}"
def main():
p0_x, p0_y, p1_x, p1_y = map(int, input().split())
seg = Segment(Point(p0_x, p0_y), Point(p1_x, p1_y))
num_query = int(input())
for i in range(num_query):
pt_x, pt_y = map(int, input().split())
reflection = seg.reflection(Point(pt_x, pt_y))
print("{:.10f} {:.10f}".format(reflection.x, reflection.y))
return
main()
``` | instruction | 0 | 14,122 | 23 | 28,244 |
Yes | output | 1 | 14,122 | 23 | 28,245 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
For given three points p1, p2, p, find the reflection point x of p onto p1p2.
<image>
Constraints
* 1 ≤ q ≤ 1000
* -10000 ≤ xi, yi ≤ 10000
* p1 and p2 are not identical.
Input
xp1 yp1 xp2 yp2
q
xp0 yp0
xp1 yp1
...
xpq−1 ypq−1
In the first line, integer coordinates of p1 and p2 are given. Then, q queries are given for integer coordinates of p.
Output
For each query, print the coordinate of the reflection point x. The output values should be in a decimal fraction with an error less than 0.00000001.
Examples
Input
0 0 2 0
3
-1 1
0 1
1 1
Output
-1.0000000000 -1.0000000000
0.0000000000 -1.0000000000
1.0000000000 -1.0000000000
Input
0 0 3 4
3
2 5
1 4
0 3
Output
4.2400000000 3.3200000000
3.5600000000 2.0800000000
2.8800000000 0.8400000000
Submitted Solution:
```
from math import sin, cos, atan2
def sgn(x, eps=1e-10):
if x < -eps: return -1
if -eps <= x <= eps: return 0
if eps < x: return 1
class Vector():
def __init__(self, x=0.0, y=0.0):
self.x = x
self.y = y
def arg(self):
return atan2(self.y, self.x)
def norm(self):
return (self.x**2 + self.y**2)**0.5
def rotate(self, t):
nx = self.x * cos(t) - self.y * sin(t)
ny = self.x * sin(t) + self.y * cos(t)
return Vector(nx, ny)
def counter(self):
nx = -self.x
ny = -self.y
return Vector(nx, ny)
def times(self, k):
nx = self.x * k
ny = self.y * k
return Vector(nx, ny)
def unit(self):
norm = self.norm()
nx = self.x / norm
ny = self.y / norm
return Vector(nx, ny)
def normal(self):
norm = self.norm()
nx = -self.y / norm
ny = self.x / norm
return Vector(nx, ny)
def add(self, other): #Vector, Vector -> Vector
nx = self.x + other.x
ny = self.y + other.y
return Vector(nx, ny)
def sub(self, other):
nx = self.x - other.x
ny = self.y - other.y
return Vector(nx, ny)
def dot(self, other): #Vector, Vector -> int
return self.x * other.x + self.y * other.y
def cross(self, other): #Vector, Vector -> int
return self.x * other.y - self.y * other.x
def __str__(self):
return '{:.9f}'.format(self.x) + ' ' + '{:.9f}'.format(self.y)
class Line():
def __init__(self, bgn=Vector(), end=Vector()):
self.bgn = bgn
self.end = end
def build(self, a, b, c): #ax + by == 1
assert sgn(a) != 0 or sgn(b) != 0
if sgn(b) == 0:
self.bgn = Vector(-c / a, 0.0)
self.end = Vector(-c / a, 1.0)
else:
self.v = Vector(0, -c / b)
self.u = Vector(1.0, -(a + b) / b)
def vec(self):
return self.end.sub(self.bgn)
def projection(self, point):
v = self.vec()
u = point.sub(self.bgn)
k = v.dot(u) / v.norm()
h = v.unit().times(k)
return self.bgn.add(h)
def refrection(self, point):
proj = self.projection(point)
return proj.sub(point).times(2).add(point)
xp1, yp1, xp2, yp2 = map(int, input().split())
q = int(input())
p1 = Vector(xp1, yp1)
p2 = Vector(xp2, yp2)
l = Line(p1, p2)
for _ in range(q):
x, y = map(int, input().split())
p = Vector(x, y)
ref = l.refrection(p)
print(ref)
``` | instruction | 0 | 14,123 | 23 | 28,246 |
Yes | output | 1 | 14,123 | 23 | 28,247 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
For given three points p1, p2, p, find the reflection point x of p onto p1p2.
<image>
Constraints
* 1 ≤ q ≤ 1000
* -10000 ≤ xi, yi ≤ 10000
* p1 and p2 are not identical.
Input
xp1 yp1 xp2 yp2
q
xp0 yp0
xp1 yp1
...
xpq−1 ypq−1
In the first line, integer coordinates of p1 and p2 are given. Then, q queries are given for integer coordinates of p.
Output
For each query, print the coordinate of the reflection point x. The output values should be in a decimal fraction with an error less than 0.00000001.
Examples
Input
0 0 2 0
3
-1 1
0 1
1 1
Output
-1.0000000000 -1.0000000000
0.0000000000 -1.0000000000
1.0000000000 -1.0000000000
Input
0 0 3 4
3
2 5
1 4
0 3
Output
4.2400000000 3.3200000000
3.5600000000 2.0800000000
2.8800000000 0.8400000000
Submitted Solution:
```
x1,y1,x2,y2 = map(float,input().split())
det = x1 - x2
if det != 0:
a = (y1 - y2) / det
b = (-(y1 * x2) + y2 * x1) / det
q = int(input())
x,y = [],[]
for i in range(q):
x_t,y_t = map(float,input().split())
x.append(x_t)
y.append(y_t)
if det == 0:
for i in range(q):
print(-x[i],y[i])
elif y2 == y1:
for i in range(q):
print(x[i],-y[i])
else:
for i in range(q):
k = y[i] + 1/a * x[i]
s = (k - b)/(a + 1/a)
print(s+(s-x[i]),-1/a * (s+(s-x[i])) + k)
``` | instruction | 0 | 14,124 | 23 | 28,248 |
Yes | output | 1 | 14,124 | 23 | 28,249 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
For given three points p1, p2, p, find the reflection point x of p onto p1p2.
<image>
Constraints
* 1 ≤ q ≤ 1000
* -10000 ≤ xi, yi ≤ 10000
* p1 and p2 are not identical.
Input
xp1 yp1 xp2 yp2
q
xp0 yp0
xp1 yp1
...
xpq−1 ypq−1
In the first line, integer coordinates of p1 and p2 are given. Then, q queries are given for integer coordinates of p.
Output
For each query, print the coordinate of the reflection point x. The output values should be in a decimal fraction with an error less than 0.00000001.
Examples
Input
0 0 2 0
3
-1 1
0 1
1 1
Output
-1.0000000000 -1.0000000000
0.0000000000 -1.0000000000
1.0000000000 -1.0000000000
Input
0 0 3 4
3
2 5
1 4
0 3
Output
4.2400000000 3.3200000000
3.5600000000 2.0800000000
2.8800000000 0.8400000000
Submitted Solution:
```
def project(x,y):
base=[x2-x1,y2-y1]
hypo=[x-x1,y-y1]
r=(base[0]*hypo[0]+base[1]*hypo[1])/(base[0]**2+base[1]**2)
return x1+base[0]*r,y1+base[1]*r
def reflect(x,y):
px,py=project(x,y)
return x+(px-x)*2,y+(py-y)*2
x1,y1,x2,y2=map(int,input().split())
q=int(input())
for i in range(q):
x,y=map(int,input().split())
print(*reflect(x,y))
``` | instruction | 0 | 14,125 | 23 | 28,250 |
Yes | output | 1 | 14,125 | 23 | 28,251 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
For given three points p1, p2, p, find the reflection point x of p onto p1p2.
<image>
Constraints
* 1 ≤ q ≤ 1000
* -10000 ≤ xi, yi ≤ 10000
* p1 and p2 are not identical.
Input
xp1 yp1 xp2 yp2
q
xp0 yp0
xp1 yp1
...
xpq−1 ypq−1
In the first line, integer coordinates of p1 and p2 are given. Then, q queries are given for integer coordinates of p.
Output
For each query, print the coordinate of the reflection point x. The output values should be in a decimal fraction with an error less than 0.00000001.
Examples
Input
0 0 2 0
3
-1 1
0 1
1 1
Output
-1.0000000000 -1.0000000000
0.0000000000 -1.0000000000
1.0000000000 -1.0000000000
Input
0 0 3 4
3
2 5
1 4
0 3
Output
4.2400000000 3.3200000000
3.5600000000 2.0800000000
2.8800000000 0.8400000000
Submitted Solution:
```
# coding=utf-8
from math import sqrt, fsum
def inner_product(vect1, vect2):
return math.fsum([(v1_el*v2_el) for v1_el, v2_el in zip(vect1, vect2)])
def vector_abs(vect):
return sqrt(sum([element**2 for element in vect]))
def direction_unit_vector(p_from, p_to):
d_vector = [(xt - xf) for xt, xf in zip(p_to, p_from)]
d_u_vector = [element/vector_abs(d_vector) for element in d_vector]
return d_u_vector
def projection(origin, line_from, line_to):
direction_unit = direction_unit_vector(line_from, line_to)
origin_d_vector = [(org-lf) for org, lf in zip(origin, line_from)]
inject_dist = inner_product(direction_unit, origin_d_vector)
on_line_vect = [inject_dist*element for element in direction_unit]
return [olv_el + lf_el for olv_el, lf_el in zip(on_line_vect, line_from)]
def reflection(origin, line_from, line_to):
mid_point = projection(origin, line_from, line_to)
direction = [2*(mp_el - or_el) for mp_el, or_el in zip(mid_point, origin)]
reflected = [(dr_el + or_el) for dr_el, or_el in zip(direction, origin)]
return reflected
if __name__ == '__main__':
xy_list = list(map(int, input().split()))
p1_list = xy_list[:2]
p2_list = xy_list[2:]
Q = int(input())
for i in range(Q):
p_list = list(map(int, input().split()))
x_list = reflection(p_list, p1_list, p2_list)
print(' '.join(map(str, x_list)))
``` | instruction | 0 | 14,126 | 23 | 28,252 |
No | output | 1 | 14,126 | 23 | 28,253 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
For given three points p1, p2, p, find the reflection point x of p onto p1p2.
<image>
Constraints
* 1 ≤ q ≤ 1000
* -10000 ≤ xi, yi ≤ 10000
* p1 and p2 are not identical.
Input
xp1 yp1 xp2 yp2
q
xp0 yp0
xp1 yp1
...
xpq−1 ypq−1
In the first line, integer coordinates of p1 and p2 are given. Then, q queries are given for integer coordinates of p.
Output
For each query, print the coordinate of the reflection point x. The output values should be in a decimal fraction with an error less than 0.00000001.
Examples
Input
0 0 2 0
3
-1 1
0 1
1 1
Output
-1.0000000000 -1.0000000000
0.0000000000 -1.0000000000
1.0000000000 -1.0000000000
Input
0 0 3 4
3
2 5
1 4
0 3
Output
4.2400000000 3.3200000000
3.5600000000 2.0800000000
2.8800000000 0.8400000000
Submitted Solution:
```
def dot(a, b):
return a.real * b.real + a.imag * b.imag
def projection(s,p):
base=abs(b-a)
r=dot(p-a,base)/(base**2)
return a+base*r
def reflect(p0, p1, p2):
a = p1 - p0
b = p2 - p0
v = projection(a, b)
u = v - b
p3 = p2 + 2 * u
return p3
x0,y0,x1,y1 = map(float,input().split())
a = complex(x0,y0)
b = complex(x1,y1)
n=int(input())
for i in range(n):
x,y=map(int,input().split())
c=complex(x,y)
p=reflect(a,b,c)
print("{:.10} {:.10}".format(p.real, p.imag))
``` | instruction | 0 | 14,127 | 23 | 28,254 |
No | output | 1 | 14,127 | 23 | 28,255 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
For given three points p1, p2, p, find the reflection point x of p onto p1p2.
<image>
Constraints
* 1 ≤ q ≤ 1000
* -10000 ≤ xi, yi ≤ 10000
* p1 and p2 are not identical.
Input
xp1 yp1 xp2 yp2
q
xp0 yp0
xp1 yp1
...
xpq−1 ypq−1
In the first line, integer coordinates of p1 and p2 are given. Then, q queries are given for integer coordinates of p.
Output
For each query, print the coordinate of the reflection point x. The output values should be in a decimal fraction with an error less than 0.00000001.
Examples
Input
0 0 2 0
3
-1 1
0 1
1 1
Output
-1.0000000000 -1.0000000000
0.0000000000 -1.0000000000
1.0000000000 -1.0000000000
Input
0 0 3 4
3
2 5
1 4
0 3
Output
4.2400000000 3.3200000000
3.5600000000 2.0800000000
2.8800000000 0.8400000000
Submitted Solution:
```
a, b,c,d = map(float,input().split())
a = complex(a,b)
b = complex(c,d)
for _ in [0] *int(input()):
c = complex(*map(float, input().split()))
q = b-a
c -= a
p = a+q*(c/q).conjugate()
print("{:.10} {:.10}".format(p.real, p.imag))
``` | instruction | 0 | 14,128 | 23 | 28,256 |
No | output | 1 | 14,128 | 23 | 28,257 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
For given three points p1, p2, p, find the reflection point x of p onto p1p2.
<image>
Constraints
* 1 ≤ q ≤ 1000
* -10000 ≤ xi, yi ≤ 10000
* p1 and p2 are not identical.
Input
xp1 yp1 xp2 yp2
q
xp0 yp0
xp1 yp1
...
xpq−1 ypq−1
In the first line, integer coordinates of p1 and p2 are given. Then, q queries are given for integer coordinates of p.
Output
For each query, print the coordinate of the reflection point x. The output values should be in a decimal fraction with an error less than 0.00000001.
Examples
Input
0 0 2 0
3
-1 1
0 1
1 1
Output
-1.0000000000 -1.0000000000
0.0000000000 -1.0000000000
1.0000000000 -1.0000000000
Input
0 0 3 4
3
2 5
1 4
0 3
Output
4.2400000000 3.3200000000
3.5600000000 2.0800000000
2.8800000000 0.8400000000
Submitted Solution:
```
# coding=utf-8
from math import sqrt
def inner_product(vect1, vect2):
return sum([(v1_el*v2_el) for v1_el, v2_el in zip(vect1, vect2)])
def vector_abs(vect):
return sqrt(sum([element**2 for element in vect]))
def direction_unit_vector(p_from, p_to):
d_vector = [(xt - xf) for xt, xf in zip(p_to, p_from)]
d_u_vector = [element/vector_abs(d_vector) for element in d_vector]
return d_u_vector
def projection(origin, line_from, line_to):
direction_unit = direction_unit_vector(line_from, line_to)
origin_d_vector = [(org-lf) for org, lf in zip(origin, line_from)]
inject_dist = inner_product(direction_unit, origin_d_vector)
on_line_vect = [inject_dist*element for element in direction_unit]
return [olv_el + lf_el for olv_el, lf_el in zip(on_line_vect, line_from)]
def reflection(origin, line_from, line_to):
mid_point = projection(origin, line_from, line_to)
direction = [2*(mp_el - or_el) for mp_el, or_el in zip(mid_point, origin)]
reflected = [(dr_el + or_el) for dr_el, or_el in zip(direction, origin)]
return reflected
if __name__ == '__main__':
xy_list = list(map(int, input().split()))
p1_list = xy_list[:2]
p2_list = xy_list[2:]
Q = int(input())
for i in range(Q):
p_list = list(map(int, input().split()))
x_list = reflection(p_list, p1_list, p2_list)
print(' '.join(map(str, x_list)))
``` | instruction | 0 | 14,129 | 23 | 28,258 |
No | output | 1 | 14,129 | 23 | 28,259 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A boy Bob likes to draw. Not long ago he bought a rectangular graph (checked) sheet with n rows and m columns. Bob shaded some of the squares on the sheet. Having seen his masterpiece, he decided to share it with his elder brother, who lives in Flatland. Now Bob has to send his picture by post, but because of the world economic crisis and high oil prices, he wants to send his creation, but to spend as little money as possible. For each sent square of paper (no matter whether it is shaded or not) Bob has to pay 3.14 burles. Please, help Bob cut out of his masterpiece a rectangle of the minimum cost, that will contain all the shaded squares. The rectangle's sides should be parallel to the sheet's sides.
Input
The first line of the input data contains numbers n and m (1 ≤ n, m ≤ 50), n — amount of lines, and m — amount of columns on Bob's sheet. The following n lines contain m characters each. Character «.» stands for a non-shaded square on the sheet, and «*» — for a shaded square. It is guaranteed that Bob has shaded at least one square.
Output
Output the required rectangle of the minimum cost. Study the output data in the sample tests to understand the output format better.
Examples
Input
6 7
.......
..***..
..*....
..***..
..*....
..***..
Output
***
*..
***
*..
***
Input
3 3
***
*.*
***
Output
***
*.*
***
Submitted Solution:
```
n,m=map(int,input().split())
arr=[]
arr1=[]
for i in range(0,n):
arr.append(input())
for i in range(0,n):
for j in range(0,m):
if arr[i][j]=='*':
arr1.append((i,j))
minx=10000
maxx=-1
miny=10000
maxy=-1
for i in range(0,len(arr1)):
if minx>arr1[i][0]:
minx=arr1[i][0]
if maxx<arr1[i][0]:
maxx=arr1[i][0]
if miny>arr1[i][1]:
miny=arr1[i][1]
if maxy<arr1[i][1]:
maxy=arr1[i][1]
for i in range(minx,maxx+1):
for j in range(miny,maxy+1):
if j!=maxy:
print(arr[i][j],end="")
else:
print(arr[i][j])
# 15 3
# ...
# ...
# ...
# .**
# ...
# ...
# *..
# ...
# .*.
# ..*
# ..*
# *..
# ..*
# ...
# ...
``` | instruction | 0 | 14,409 | 23 | 28,818 |
Yes | output | 1 | 14,409 | 23 | 28,819 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A boy Bob likes to draw. Not long ago he bought a rectangular graph (checked) sheet with n rows and m columns. Bob shaded some of the squares on the sheet. Having seen his masterpiece, he decided to share it with his elder brother, who lives in Flatland. Now Bob has to send his picture by post, but because of the world economic crisis and high oil prices, he wants to send his creation, but to spend as little money as possible. For each sent square of paper (no matter whether it is shaded or not) Bob has to pay 3.14 burles. Please, help Bob cut out of his masterpiece a rectangle of the minimum cost, that will contain all the shaded squares. The rectangle's sides should be parallel to the sheet's sides.
Input
The first line of the input data contains numbers n and m (1 ≤ n, m ≤ 50), n — amount of lines, and m — amount of columns on Bob's sheet. The following n lines contain m characters each. Character «.» stands for a non-shaded square on the sheet, and «*» — for a shaded square. It is guaranteed that Bob has shaded at least one square.
Output
Output the required rectangle of the minimum cost. Study the output data in the sample tests to understand the output format better.
Examples
Input
6 7
.......
..***..
..*....
..***..
..*....
..***..
Output
***
*..
***
*..
***
Input
3 3
***
*.*
***
Output
***
*.*
***
Submitted Solution:
```
x, y = map(int, input().split())
q = [input() for i in range(x)]
t = 0
while q[t] == '.' * y:
t += 1
b = x
while q[b-1] == '.' * y:
b -= 1
l = 0
while sum(i[l] == '*' for i in q) == 0:
l += 1
r = y
while sum(i[r-1] == '*' for i in q) == 0:
r -= 1
print('\n'.join(i[l:r] for i in q[t:b]))
``` | instruction | 0 | 14,410 | 23 | 28,820 |
Yes | output | 1 | 14,410 | 23 | 28,821 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A boy Bob likes to draw. Not long ago he bought a rectangular graph (checked) sheet with n rows and m columns. Bob shaded some of the squares on the sheet. Having seen his masterpiece, he decided to share it with his elder brother, who lives in Flatland. Now Bob has to send his picture by post, but because of the world economic crisis and high oil prices, he wants to send his creation, but to spend as little money as possible. For each sent square of paper (no matter whether it is shaded or not) Bob has to pay 3.14 burles. Please, help Bob cut out of his masterpiece a rectangle of the minimum cost, that will contain all the shaded squares. The rectangle's sides should be parallel to the sheet's sides.
Input
The first line of the input data contains numbers n and m (1 ≤ n, m ≤ 50), n — amount of lines, and m — amount of columns on Bob's sheet. The following n lines contain m characters each. Character «.» stands for a non-shaded square on the sheet, and «*» — for a shaded square. It is guaranteed that Bob has shaded at least one square.
Output
Output the required rectangle of the minimum cost. Study the output data in the sample tests to understand the output format better.
Examples
Input
6 7
.......
..***..
..*....
..***..
..*....
..***..
Output
***
*..
***
*..
***
Input
3 3
***
*.*
***
Output
***
*.*
***
Submitted Solution:
```
n,m = map(int,input().split())
t = []
for i in range(n):
t.append(input())
minx,miny = float("inf"),float("inf")
maxx,maxy = float("-inf"),float("-inf")
for i in range(n):
flag = 0
for j in range(m):
if t[i][j] == "*":
flag = 1
break
if flag == 1:
minx = min(minx,j)
for i in range(n):
flag = 0
for j in range(m-1,-1,-1):
if t[i][j] == "*":
flag = 1
break
if flag == 1:
maxx = max(maxx,j)
for i in range(m):
flag = 0
for j in range(n):
if t[j][i] == "*":
flag = 1
break
if flag == 1:
miny = min(miny,j)
for i in range(m):
flag = 0
for j in range(n-1,-1,-1):
if t[j][i] == "*":
flag = 1
break
if flag == 1:
maxy = max(maxy,j)
for i in range(miny,maxy+1):
for j in range(minx,maxx+1):
print(t[i][j],end="")
print()
``` | instruction | 0 | 14,411 | 23 | 28,822 |
Yes | output | 1 | 14,411 | 23 | 28,823 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A boy Bob likes to draw. Not long ago he bought a rectangular graph (checked) sheet with n rows and m columns. Bob shaded some of the squares on the sheet. Having seen his masterpiece, he decided to share it with his elder brother, who lives in Flatland. Now Bob has to send his picture by post, but because of the world economic crisis and high oil prices, he wants to send his creation, but to spend as little money as possible. For each sent square of paper (no matter whether it is shaded or not) Bob has to pay 3.14 burles. Please, help Bob cut out of his masterpiece a rectangle of the minimum cost, that will contain all the shaded squares. The rectangle's sides should be parallel to the sheet's sides.
Input
The first line of the input data contains numbers n and m (1 ≤ n, m ≤ 50), n — amount of lines, and m — amount of columns on Bob's sheet. The following n lines contain m characters each. Character «.» stands for a non-shaded square on the sheet, and «*» — for a shaded square. It is guaranteed that Bob has shaded at least one square.
Output
Output the required rectangle of the minimum cost. Study the output data in the sample tests to understand the output format better.
Examples
Input
6 7
.......
..***..
..*....
..***..
..*....
..***..
Output
***
*..
***
*..
***
Input
3 3
***
*.*
***
Output
***
*.*
***
Submitted Solution:
```
n, m = map(int, input().split())
grid = [input() for i in range(n)]
minr, maxr, minc, maxc = n, -1, m, -1
for i in range(n):
for j in range(m):
if grid[i][j] == '*':
minr = min(minr, i)
maxr = max(maxr, i)
minc = min(minc, j)
maxc = max(maxc, j)
for i in range(minr, maxr+1):
print(grid[i][minc:maxc+1])
``` | instruction | 0 | 14,412 | 23 | 28,824 |
Yes | output | 1 | 14,412 | 23 | 28,825 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A boy Bob likes to draw. Not long ago he bought a rectangular graph (checked) sheet with n rows and m columns. Bob shaded some of the squares on the sheet. Having seen his masterpiece, he decided to share it with his elder brother, who lives in Flatland. Now Bob has to send his picture by post, but because of the world economic crisis and high oil prices, he wants to send his creation, but to spend as little money as possible. For each sent square of paper (no matter whether it is shaded or not) Bob has to pay 3.14 burles. Please, help Bob cut out of his masterpiece a rectangle of the minimum cost, that will contain all the shaded squares. The rectangle's sides should be parallel to the sheet's sides.
Input
The first line of the input data contains numbers n and m (1 ≤ n, m ≤ 50), n — amount of lines, and m — amount of columns on Bob's sheet. The following n lines contain m characters each. Character «.» stands for a non-shaded square on the sheet, and «*» — for a shaded square. It is guaranteed that Bob has shaded at least one square.
Output
Output the required rectangle of the minimum cost. Study the output data in the sample tests to understand the output format better.
Examples
Input
6 7
.......
..***..
..*....
..***..
..*....
..***..
Output
***
*..
***
*..
***
Input
3 3
***
*.*
***
Output
***
*.*
***
Submitted Solution:
```
a, b = map(int, input().split())
ma = []
mi = []
x = []
for i in range(a):
t = input()
x.append(t)
se = []
c =-1
for i in x:
c = c+1
if "*" in i:
se.append(c)
y = i.find('*')
mi.append(y)
z = i[::-1].find('*')
ma.append(b - 1 - z)
lower = min(mi)
upper = max(ma)
start = min(se)
end = max(se)
for j in range(int(start), int(end)):
print(x[j][lower:upper + 1])
``` | instruction | 0 | 14,413 | 23 | 28,826 |
No | output | 1 | 14,413 | 23 | 28,827 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A boy Bob likes to draw. Not long ago he bought a rectangular graph (checked) sheet with n rows and m columns. Bob shaded some of the squares on the sheet. Having seen his masterpiece, he decided to share it with his elder brother, who lives in Flatland. Now Bob has to send his picture by post, but because of the world economic crisis and high oil prices, he wants to send his creation, but to spend as little money as possible. For each sent square of paper (no matter whether it is shaded or not) Bob has to pay 3.14 burles. Please, help Bob cut out of his masterpiece a rectangle of the minimum cost, that will contain all the shaded squares. The rectangle's sides should be parallel to the sheet's sides.
Input
The first line of the input data contains numbers n and m (1 ≤ n, m ≤ 50), n — amount of lines, and m — amount of columns on Bob's sheet. The following n lines contain m characters each. Character «.» stands for a non-shaded square on the sheet, and «*» — for a shaded square. It is guaranteed that Bob has shaded at least one square.
Output
Output the required rectangle of the minimum cost. Study the output data in the sample tests to understand the output format better.
Examples
Input
6 7
.......
..***..
..*....
..***..
..*....
..***..
Output
***
*..
***
*..
***
Input
3 3
***
*.*
***
Output
***
*.*
***
Submitted Solution:
```
import math,itertools,fractions,heapq,collections,bisect,sys,queue,copy
sys.setrecursionlimit(10**7)
inf=10**20
mod=10**9+7
dd=[(-1,0),(0,1),(1,0),(0,-1)]
ddn=[(-1,0),(-1,1),(0,1),(1,1),(1,0),(1,-1),(0,-1),(-1,-1)]
def LI(): return [int(x) for x in sys.stdin.readline().split()]
# def LF(): return [float(x) for x in sys.stdin.readline().split()]
def I(): return int(sys.stdin.readline())
def F(): return float(sys.stdin.readline())
def LS(): return sys.stdin.readline().split()
def S(): return input()
def main():
h,w=LI()
l=[S() for _ in range(h)]
y1=x1=0
f=False
for i in range(h):
for j in range(w):
if l[i][j]=='*':
y1=i
x1=j
f=True
break
if f:
break
y2=x2=0
f=False
for i in range(h)[::-1]:
for j in range(w)[::-1]:
if l[i][j]=='*':
y2=i
x2=j
f=True
break
if f:
break
# print(x1,y1,x2,y2)
for i in range(y1,y2+1):
ans=''
for j in range(x1,x2+1):
ans+=l[i][j]
print(ans)
main()
# print(main())
``` | instruction | 0 | 14,415 | 23 | 28,830 |
No | output | 1 | 14,415 | 23 | 28,831 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A boy Bob likes to draw. Not long ago he bought a rectangular graph (checked) sheet with n rows and m columns. Bob shaded some of the squares on the sheet. Having seen his masterpiece, he decided to share it with his elder brother, who lives in Flatland. Now Bob has to send his picture by post, but because of the world economic crisis and high oil prices, he wants to send his creation, but to spend as little money as possible. For each sent square of paper (no matter whether it is shaded or not) Bob has to pay 3.14 burles. Please, help Bob cut out of his masterpiece a rectangle of the minimum cost, that will contain all the shaded squares. The rectangle's sides should be parallel to the sheet's sides.
Input
The first line of the input data contains numbers n and m (1 ≤ n, m ≤ 50), n — amount of lines, and m — amount of columns on Bob's sheet. The following n lines contain m characters each. Character «.» stands for a non-shaded square on the sheet, and «*» — for a shaded square. It is guaranteed that Bob has shaded at least one square.
Output
Output the required rectangle of the minimum cost. Study the output data in the sample tests to understand the output format better.
Examples
Input
6 7
.......
..***..
..*....
..***..
..*....
..***..
Output
***
*..
***
*..
***
Input
3 3
***
*.*
***
Output
***
*.*
***
Submitted Solution:
```
# https://codeforces.com/problemset/problem/14/A
n, m = map(int, input().split())
first = float('inf')
last = float('-inf')
new = list()
for i in range(n):
line = input()
for j in range(m):
if line[j] == "*":
first = min(j, first)
new.append(line)
break
for j in range(m - 1, -1, -1):
if line[j] == "*":
last = max(j + 1, last)
break
for i in new:
print(i[first: last])
``` | instruction | 0 | 14,416 | 23 | 28,832 |
No | output | 1 | 14,416 | 23 | 28,833 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You have a string of decimal digits s. Let's define bij = si·sj. Find in matrix b the number of such rectangles that the sum bij for all cells (i, j) that are the elements of the rectangle equals a in each rectangle.
A rectangle in a matrix is a group of four integers (x, y, z, t) (x ≤ y, z ≤ t). The elements of the rectangle are all cells (i, j) such that x ≤ i ≤ y, z ≤ j ≤ t.
Input
The first line contains integer a (0 ≤ a ≤ 109), the second line contains a string of decimal integers s (1 ≤ |s| ≤ 4000).
Output
Print a single integer — the answer to a problem.
Please, do not write the %lld specifier to read or write 64-bit integers in С++. It is preferred to use the cin, cout streams or the %I64d specifier.
Examples
Input
10
12345
Output
6
Input
16
439873893693495623498263984765
Output
40 | instruction | 0 | 14,534 | 23 | 29,068 |
Tags: combinatorics, data structures, implementation
Correct Solution:
```
a = int(input())
s = input()
d = {}
for i in range(len(s)):
for j in range(i, len(s)):
if j == i: t = int(s[j])
else: t += int(s[j])
d[t] = d.get(t, 0) + 1
if a == 0:
if 0 in d:
cnt_pairs = (len(s) * (len(s) + 1)) // 2
print((d[0] * cnt_pairs) + (d[0] * (cnt_pairs - d[0])))
else:
print(0)
else:
c = 0
for f in d:
if f != 0 and a % f == 0 and (a//f) in d:
c += d[f] * d[a//f]
print(c)
``` | output | 1 | 14,534 | 23 | 29,069 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You have a string of decimal digits s. Let's define bij = si·sj. Find in matrix b the number of such rectangles that the sum bij for all cells (i, j) that are the elements of the rectangle equals a in each rectangle.
A rectangle in a matrix is a group of four integers (x, y, z, t) (x ≤ y, z ≤ t). The elements of the rectangle are all cells (i, j) such that x ≤ i ≤ y, z ≤ j ≤ t.
Input
The first line contains integer a (0 ≤ a ≤ 109), the second line contains a string of decimal integers s (1 ≤ |s| ≤ 4000).
Output
Print a single integer — the answer to a problem.
Please, do not write the %lld specifier to read or write 64-bit integers in С++. It is preferred to use the cin, cout streams or the %I64d specifier.
Examples
Input
10
12345
Output
6
Input
16
439873893693495623498263984765
Output
40 | instruction | 0 | 14,535 | 23 | 29,070 |
Tags: combinatorics, data structures, implementation
Correct Solution:
```
import math
def main(arr,a):
f=[]
for i in range(1,int(math.sqrt(a))+1):
f1=i
f2=a/i
if f2==int(f2):
f.append(f1)
if f2!=f1:
f.append(int(f2))
f.sort()
prefix=[0]
for i in range(len(arr)):
prefix.append(prefix[-1]+arr[i])
diff={}
for i in range(len(prefix)):
for j in range(0,i):
val= prefix[i]-prefix[j]
if val not in diff:
diff[val]=0
diff[val]+=1
if a==0:
for e in diff:
f.append(e)
f.append(0)
f.sort()
ans=0
i=0
j=len(f)-1
while i<=j:
f1=f[i]
f2=f[j]
if f1 in diff and f2 in diff:
if f1==f2:
ans+=diff[f1]*diff[f2]
else:
ans+=2*(diff[f1]*diff[f2])
i+=1
j-=1
return ans
a=int(input())
arr=[int(c) for c in input()]
print(main(arr,a))
``` | output | 1 | 14,535 | 23 | 29,071 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You have a string of decimal digits s. Let's define bij = si·sj. Find in matrix b the number of such rectangles that the sum bij for all cells (i, j) that are the elements of the rectangle equals a in each rectangle.
A rectangle in a matrix is a group of four integers (x, y, z, t) (x ≤ y, z ≤ t). The elements of the rectangle are all cells (i, j) such that x ≤ i ≤ y, z ≤ j ≤ t.
Input
The first line contains integer a (0 ≤ a ≤ 109), the second line contains a string of decimal integers s (1 ≤ |s| ≤ 4000).
Output
Print a single integer — the answer to a problem.
Please, do not write the %lld specifier to read or write 64-bit integers in С++. It is preferred to use the cin, cout streams or the %I64d specifier.
Examples
Input
10
12345
Output
6
Input
16
439873893693495623498263984765
Output
40 | instruction | 0 | 14,536 | 23 | 29,072 |
Tags: combinatorics, data structures, implementation
Correct Solution:
```
import sys,collections as cc
input = sys.stdin.readline
I = lambda : list(map(int,input().split()))
a,=I()
s=list(map(int,[i for i in input().strip()]))
d=cc.Counter([])
n=len(s)
for i in range(1,n+1):
su=sum(s[:i])
for j in range(i,n):
d[su]+=1
su=su-s[j-i]+s[j]
d[su]+=1
an=0
if a==0:
an=(n*(n+1)//2)**2
del d[0]
xx=sum(d.values())
for i in d:
an-=d[i]*(xx)
else:
for i in d.keys():
if i!=0 and a%i==0:
an=an+(d[i]*d[a//i])
print(an)
``` | output | 1 | 14,536 | 23 | 29,073 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You have a string of decimal digits s. Let's define bij = si·sj. Find in matrix b the number of such rectangles that the sum bij for all cells (i, j) that are the elements of the rectangle equals a in each rectangle.
A rectangle in a matrix is a group of four integers (x, y, z, t) (x ≤ y, z ≤ t). The elements of the rectangle are all cells (i, j) such that x ≤ i ≤ y, z ≤ j ≤ t.
Input
The first line contains integer a (0 ≤ a ≤ 109), the second line contains a string of decimal integers s (1 ≤ |s| ≤ 4000).
Output
Print a single integer — the answer to a problem.
Please, do not write the %lld specifier to read or write 64-bit integers in С++. It is preferred to use the cin, cout streams or the %I64d specifier.
Examples
Input
10
12345
Output
6
Input
16
439873893693495623498263984765
Output
40 | instruction | 0 | 14,537 | 23 | 29,074 |
Tags: combinatorics, data structures, implementation
Correct Solution:
```
import sys
from collections import defaultdict as dd
from collections import deque
def eprint(*args):
print(*args, file=sys.stderr)
pl=1
from math import *
import copy
#sys.setrecursionlimit(10**6)
if pl:
input=sys.stdin.readline
def li():
return [int(xxx) for xxx in input().split()]
def fi():
return int(input())
def si():
return list(input().rstrip())
def mi():
return map(int,input().split())
d=[]
from bisect import *
from itertools import permutations
from bisect import *
f=[0 for i in range(100)]
for i in range(1,100):
if i==1:
f[i]=1
else:
f[i]=2*f[i-1]+1
#print(f[:15])
def rec(n,k):
s=[]
while n!=0:
n,r=n//k,n%k
#print(n,r)
if r<0:
r-=k
n+=1
#print(s,n,r)
s.append(r)
return s
n=fi()
s=si()
a=[]
for i in range(len(s)):
a.append(int(s[i]))
d={}
nn=len(s)
for i in range(nn):
c=0
for j in range(i,nn):
c+=a[j]
if c not in d:
d[c]=1
else:
d[c]+=1
ans=0
#print(d)
if n==0 and 0 in d:
for j in d:
if j==0:
ans+=d[0]*d[0]
else:
ans+=d[j]*d[0]*2
for j in range(1,int(sqrt(n))+1):
if n%j==0:
if n//j==j:
if j in d:
ans+=(d[j]*(d[j]))
elif n//j in d and j in d:
ans+=d[j]*d[n//j]*2
print(ans)
``` | output | 1 | 14,537 | 23 | 29,075 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You have a string of decimal digits s. Let's define bij = si·sj. Find in matrix b the number of such rectangles that the sum bij for all cells (i, j) that are the elements of the rectangle equals a in each rectangle.
A rectangle in a matrix is a group of four integers (x, y, z, t) (x ≤ y, z ≤ t). The elements of the rectangle are all cells (i, j) such that x ≤ i ≤ y, z ≤ j ≤ t.
Input
The first line contains integer a (0 ≤ a ≤ 109), the second line contains a string of decimal integers s (1 ≤ |s| ≤ 4000).
Output
Print a single integer — the answer to a problem.
Please, do not write the %lld specifier to read or write 64-bit integers in С++. It is preferred to use the cin, cout streams or the %I64d specifier.
Examples
Input
10
12345
Output
6
Input
16
439873893693495623498263984765
Output
40 | instruction | 0 | 14,538 | 23 | 29,076 |
Tags: combinatorics, data structures, implementation
Correct Solution:
```
def divisors(x):
def f(y, q):
t = -len(r)
while not y % q:
y //= q
for i in range(t, 0):
r.append(r[t] * q)
return y
r, p = [1], 7
x = f(f(f(x, 2), 3), 5)
while x >= p * p:
for s in 4, 2, 4, 2, 4, 6, 2, 6:
if not x % p:
x = f(x, p)
p += s
if x > 1:
f(x, x)
return r
def main():
a, s = int(input()), input()
if not a:
z = sum(x * (x + 1) for x in map(len, s.translate(
str.maketrans('123456789', ' ')).split())) // 2
x = len(s)
print((x * (x + 1) - z) * z)
return
sums, x, cnt = {}, 0, 1
for u in map(int, s):
if u:
sums[x] = cnt
x += u
cnt = 1
else:
cnt += 1
if x * x < a:
print(0)
return
sums[x], u = cnt, a // x
l = [v for v in divisors(a) if v <= x]
z = a // max(l)
d = {x: 0 for x in l if z <= x}
for x in d:
for k, v in sums.items():
u = sums.get(k + x, 0)
if u:
d[x] += v * u
print(sum(u * d[a // x] for x, u in d.items()))
if __name__ == '__main__':
main()
# Made By Mostafa_Khaled
``` | output | 1 | 14,538 | 23 | 29,077 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You have a string of decimal digits s. Let's define bij = si·sj. Find in matrix b the number of such rectangles that the sum bij for all cells (i, j) that are the elements of the rectangle equals a in each rectangle.
A rectangle in a matrix is a group of four integers (x, y, z, t) (x ≤ y, z ≤ t). The elements of the rectangle are all cells (i, j) such that x ≤ i ≤ y, z ≤ j ≤ t.
Input
The first line contains integer a (0 ≤ a ≤ 109), the second line contains a string of decimal integers s (1 ≤ |s| ≤ 4000).
Output
Print a single integer — the answer to a problem.
Please, do not write the %lld specifier to read or write 64-bit integers in С++. It is preferred to use the cin, cout streams or the %I64d specifier.
Examples
Input
10
12345
Output
6
Input
16
439873893693495623498263984765
Output
40 | instruction | 0 | 14,539 | 23 | 29,078 |
Tags: combinatorics, data structures, implementation
Correct Solution:
```
def divisors(x):
def f(y, q):
t = -len(r)
while not y % q:
y //= q
for i in range(t, 0):
r.append(r[t] * q)
return y
r, p = [1], 7
x = f(f(f(x, 2), 3), 5)
while x >= p * p:
for s in 4, 2, 4, 2, 4, 6, 2, 6:
if not x % p:
x = f(x, p)
p += s
if x > 1:
f(x, x)
return r
def main():
a, s = int(input()), input()
if not a:
z = sum(x * (x + 1) for x in map(len, s.translate(
str.maketrans('123456789', ' ')).split())) // 2
x = len(s)
print((x * (x + 1) - z) * z)
return
sums, x, cnt = {}, 0, 1
for u in map(int, s):
if u:
sums[x] = cnt
x += u
cnt = 1
else:
cnt += 1
if x * x < a:
print(0)
return
sums[x], u = cnt, a // x
l = [v for v in divisors(a) if v <= x]
z = a // max(l)
d = {x: 0 for x in l if z <= x}
for x in d:
for k, v in sums.items():
u = sums.get(k + x, 0)
if u:
d[x] += v * u
print(sum(u * d[a // x] for x, u in d.items()))
if __name__ == '__main__':
main()
``` | output | 1 | 14,539 | 23 | 29,079 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You have a string of decimal digits s. Let's define bij = si·sj. Find in matrix b the number of such rectangles that the sum bij for all cells (i, j) that are the elements of the rectangle equals a in each rectangle.
A rectangle in a matrix is a group of four integers (x, y, z, t) (x ≤ y, z ≤ t). The elements of the rectangle are all cells (i, j) such that x ≤ i ≤ y, z ≤ j ≤ t.
Input
The first line contains integer a (0 ≤ a ≤ 109), the second line contains a string of decimal integers s (1 ≤ |s| ≤ 4000).
Output
Print a single integer — the answer to a problem.
Please, do not write the %lld specifier to read or write 64-bit integers in С++. It is preferred to use the cin, cout streams or the %I64d specifier.
Examples
Input
10
12345
Output
6
Input
16
439873893693495623498263984765
Output
40
Submitted Solution:
```
import math
def main(arr,a):
f=[]
for i in range(1,int(math.sqrt(a))+1):
f1=i
f2=a/i
if f2==int(f2):
f.append(f1)
if f2!=f1:
f.append(int(f2))
f.sort()
prefix=[0]
for i in range(len(arr)):
prefix.append(prefix[-1]+arr[i])
diff={}
for i in range(len(prefix)):
for j in range(0,i):
val= prefix[i]-prefix[j]
if val not in diff:
diff[val]=0
diff[val]+=1
ans=0
i=0
j=len(f)-1
while i<=j:
f1=f[i]
f2=f[j]
if f1 in diff and f2 in diff:
if f1==f2:
ans+=diff[f1]*diff[f2]
else:
ans+=2*(diff[f1]*diff[f2])
i+=1
j-=1
return ans
a=int(input())
arr=[int(c) for c in input()]
print(main(arr,a))
``` | instruction | 0 | 14,540 | 23 | 29,080 |
No | output | 1 | 14,540 | 23 | 29,081 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You have a string of decimal digits s. Let's define bij = si·sj. Find in matrix b the number of such rectangles that the sum bij for all cells (i, j) that are the elements of the rectangle equals a in each rectangle.
A rectangle in a matrix is a group of four integers (x, y, z, t) (x ≤ y, z ≤ t). The elements of the rectangle are all cells (i, j) such that x ≤ i ≤ y, z ≤ j ≤ t.
Input
The first line contains integer a (0 ≤ a ≤ 109), the second line contains a string of decimal integers s (1 ≤ |s| ≤ 4000).
Output
Print a single integer — the answer to a problem.
Please, do not write the %lld specifier to read or write 64-bit integers in С++. It is preferred to use the cin, cout streams or the %I64d specifier.
Examples
Input
10
12345
Output
6
Input
16
439873893693495623498263984765
Output
40
Submitted Solution:
```
import sys
from collections import defaultdict as dd
from collections import deque
def eprint(*args):
print(*args, file=sys.stderr)
pl=1
from math import *
import copy
#sys.setrecursionlimit(10**6)
if pl:
input=sys.stdin.readline
else:
sys.stdin=open('input.txt', 'r')
sys.stdout=open('all.txt','w')
def li():
return [int(xxx) for xxx in input().split()]
def fi():
return int(input())
def si():
return list(input().rstrip())
def mi():
return map(int,input().split())
d=[]
from bisect import *
from itertools import permutations
from bisect import *
f=[0 for i in range(100)]
for i in range(1,100):
if i==1:
f[i]=1
else:
f[i]=2*f[i-1]+1
#print(f[:15])
def rec(n,k):
s=[]
while n!=0:
n,r=n//k,n%k
#print(n,r)
if r<0:
r-=k
n+=1
#print(s,n,r)
s.append(r)
return s
n=fi()
s=si()
a=[]
for i in range(len(s)):
a.append(int(s[i]))
d={}
nn=len(s)
for i in range(nn):
c=0
for j in range(i,nn):
c+=a[j]
if c not in d:
d[c]=1
else:
d[c]+=1
ans=0
#print(d)
for j in range(1,int(sqrt(n))+1):
if n%j==0:
if n//j==j:
if j in d:
ans+=(d[j]*(d[j]))
elif n//j in d and j in d:
ans+=d[j]*d[n//j]*2
print(ans)
``` | instruction | 0 | 14,541 | 23 | 29,082 |
No | output | 1 | 14,541 | 23 | 29,083 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You have a string of decimal digits s. Let's define bij = si·sj. Find in matrix b the number of such rectangles that the sum bij for all cells (i, j) that are the elements of the rectangle equals a in each rectangle.
A rectangle in a matrix is a group of four integers (x, y, z, t) (x ≤ y, z ≤ t). The elements of the rectangle are all cells (i, j) such that x ≤ i ≤ y, z ≤ j ≤ t.
Input
The first line contains integer a (0 ≤ a ≤ 109), the second line contains a string of decimal integers s (1 ≤ |s| ≤ 4000).
Output
Print a single integer — the answer to a problem.
Please, do not write the %lld specifier to read or write 64-bit integers in С++. It is preferred to use the cin, cout streams or the %I64d specifier.
Examples
Input
10
12345
Output
6
Input
16
439873893693495623498263984765
Output
40
Submitted Solution:
```
import sys,collections as cc
input = sys.stdin.readline
I = lambda : list(map(int,input().split()))
a,=I()
s=list(map(int,[i for i in input().strip()]))
d=cc.Counter([])
n=len(s)
for i in range(1,n+1):
su=sum(s[:i])
for j in range(i,n):
d[su]+=1
su=su-s[j-i]+s[j]
d[su]+=1
an=0
if a==0:
an=d[0]*(sum(d.values()))
else:
for i in d.keys():
if i!=0 and a%i==0:
an=an+(d[i]*d[a//i])
print(an)
``` | instruction | 0 | 14,542 | 23 | 29,084 |
No | output | 1 | 14,542 | 23 | 29,085 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You have a string of decimal digits s. Let's define bij = si·sj. Find in matrix b the number of such rectangles that the sum bij for all cells (i, j) that are the elements of the rectangle equals a in each rectangle.
A rectangle in a matrix is a group of four integers (x, y, z, t) (x ≤ y, z ≤ t). The elements of the rectangle are all cells (i, j) such that x ≤ i ≤ y, z ≤ j ≤ t.
Input
The first line contains integer a (0 ≤ a ≤ 109), the second line contains a string of decimal integers s (1 ≤ |s| ≤ 4000).
Output
Print a single integer — the answer to a problem.
Please, do not write the %lld specifier to read or write 64-bit integers in С++. It is preferred to use the cin, cout streams or the %I64d specifier.
Examples
Input
10
12345
Output
6
Input
16
439873893693495623498263984765
Output
40
Submitted Solution:
```
def genprimes(limit):
lim12 = max(limit, 12)
lim = lim12 // 6
sieve = [False, True, True] * lim
lim = lim * 3 - 1
for i, s in enumerate(sieve):
if s:
p, pp = i * 2 + 3, (i + 3) * i * 2 + 3
le = (lim - pp) // p + 1
if le > 0:
sieve[pp::p] = [False] * le
else:
break
sieve[0] = sieve[3] = True
res = [i for i, s in zip(range(3, lim12, 2), sieve) if s]
for i, p in enumerate((2, 3, 5, 7)):
res[i] = p
while res[-1] >= limit:
del res[-1]
return res
def factor(x):
result = []
for p in genprimes(int(x ** .5) + 1):
cnt = 0
while not x % p:
x //= p
cnt += 1
if cnt:
result.append((p, cnt))
if x == 1:
return result
if x > 1:
result.append((x, 1))
return result
def divisors(x):
l = [1]
for k, v in factor(x):
for x in l[:]:
for j in range(v):
x *= k
l.append(x)
return l
def main():
a, s, l, x = int(input()), input(), [0], 0
if not a:
x, z = len(s), s.count('0')
print((x * (x + 1) - z) * z)
return
d = dict.fromkeys(divisors(a), 0)
for u in map(int, s):
x += u
for y in l:
if x - y in d:
d[x - y] += 1
l.append(x)
print(sum(u * d[a // x] for x, u in d.items()))
if __name__ == '__main__':
main()
``` | instruction | 0 | 14,543 | 23 | 29,086 |
No | output | 1 | 14,543 | 23 | 29,087 |
Provide a correct Python 3 solution for this coding contest problem.
Have you ever heard of the unit "○○ tsubo" that expresses the area of land? Since ancient times, one samurai has said the area for making rice to eat in a day.
There is a land of a [m] x b [m]. Enter a and b and create a program that outputs the tsubo area S [tsubo] of the land. 1 tsubo = 3.305785 [m2], and a and b are integers less than or equal to 100.
input
a b
A and b separated by one space are given on one line.
output
Output the tsubo area S in one line. An error of 0.0001 or less is allowed.
Example
Input
15 25
Output
113.437508 | instruction | 0 | 14,924 | 23 | 29,848 |
"Correct Solution:
```
a,b=(int(x) for x in input().split())
print("{:.6f}".format(a*b/3.305785))
``` | output | 1 | 14,924 | 23 | 29,849 |
Provide a correct Python 3 solution for this coding contest problem.
Have you ever heard of the unit "○○ tsubo" that expresses the area of land? Since ancient times, one samurai has said the area for making rice to eat in a day.
There is a land of a [m] x b [m]. Enter a and b and create a program that outputs the tsubo area S [tsubo] of the land. 1 tsubo = 3.305785 [m2], and a and b are integers less than or equal to 100.
input
a b
A and b separated by one space are given on one line.
output
Output the tsubo area S in one line. An error of 0.0001 or less is allowed.
Example
Input
15 25
Output
113.437508 | instruction | 0 | 14,925 | 23 | 29,850 |
"Correct Solution:
```
a,b=map(int,input().split())
x=3.305785
print((a*b)/x)
``` | output | 1 | 14,925 | 23 | 29,851 |
Provide a correct Python 3 solution for this coding contest problem.
Have you ever heard of the unit "○○ tsubo" that expresses the area of land? Since ancient times, one samurai has said the area for making rice to eat in a day.
There is a land of a [m] x b [m]. Enter a and b and create a program that outputs the tsubo area S [tsubo] of the land. 1 tsubo = 3.305785 [m2], and a and b are integers less than or equal to 100.
input
a b
A and b separated by one space are given on one line.
output
Output the tsubo area S in one line. An error of 0.0001 or less is allowed.
Example
Input
15 25
Output
113.437508 | instruction | 0 | 14,926 | 23 | 29,852 |
"Correct Solution:
```
a, b = map(int,input().split())
print(format(a*b/3.305785,'.5f'))
``` | output | 1 | 14,926 | 23 | 29,853 |
Provide a correct Python 3 solution for this coding contest problem.
Have you ever heard of the unit "○○ tsubo" that expresses the area of land? Since ancient times, one samurai has said the area for making rice to eat in a day.
There is a land of a [m] x b [m]. Enter a and b and create a program that outputs the tsubo area S [tsubo] of the land. 1 tsubo = 3.305785 [m2], and a and b are integers less than or equal to 100.
input
a b
A and b separated by one space are given on one line.
output
Output the tsubo area S in one line. An error of 0.0001 or less is allowed.
Example
Input
15 25
Output
113.437508 | instruction | 0 | 14,927 | 23 | 29,854 |
"Correct Solution:
```
a, b = map(int, input().split())
s=3.305785
print('{:.6f}'.format(a*b/s))
``` | output | 1 | 14,927 | 23 | 29,855 |
Provide a correct Python 3 solution for this coding contest problem.
Have you ever heard of the unit "○○ tsubo" that expresses the area of land? Since ancient times, one samurai has said the area for making rice to eat in a day.
There is a land of a [m] x b [m]. Enter a and b and create a program that outputs the tsubo area S [tsubo] of the land. 1 tsubo = 3.305785 [m2], and a and b are integers less than or equal to 100.
input
a b
A and b separated by one space are given on one line.
output
Output the tsubo area S in one line. An error of 0.0001 or less is allowed.
Example
Input
15 25
Output
113.437508 | instruction | 0 | 14,928 | 23 | 29,856 |
"Correct Solution:
```
def test():
a,b=list(map(int,input().split()))
c=3.305785
s=a*b/c
return s
if __name__ == '__main__':
print(test())
``` | output | 1 | 14,928 | 23 | 29,857 |
Provide a correct Python 3 solution for this coding contest problem.
Have you ever heard of the unit "○○ tsubo" that expresses the area of land? Since ancient times, one samurai has said the area for making rice to eat in a day.
There is a land of a [m] x b [m]. Enter a and b and create a program that outputs the tsubo area S [tsubo] of the land. 1 tsubo = 3.305785 [m2], and a and b are integers less than or equal to 100.
input
a b
A and b separated by one space are given on one line.
output
Output the tsubo area S in one line. An error of 0.0001 or less is allowed.
Example
Input
15 25
Output
113.437508 | instruction | 0 | 14,929 | 23 | 29,858 |
"Correct Solution:
```
a,b=map(int,input().split())
print('{:.6f}'.format((a*b)/3.305785))
``` | output | 1 | 14,929 | 23 | 29,859 |
Provide a correct Python 3 solution for this coding contest problem.
Have you ever heard of the unit "○○ tsubo" that expresses the area of land? Since ancient times, one samurai has said the area for making rice to eat in a day.
There is a land of a [m] x b [m]. Enter a and b and create a program that outputs the tsubo area S [tsubo] of the land. 1 tsubo = 3.305785 [m2], and a and b are integers less than or equal to 100.
input
a b
A and b separated by one space are given on one line.
output
Output the tsubo area S in one line. An error of 0.0001 or less is allowed.
Example
Input
15 25
Output
113.437508 | instruction | 0 | 14,930 | 23 | 29,860 |
"Correct Solution:
```
a,b = map(float,input().split())
S = a*b/3.305785
print(S)
``` | output | 1 | 14,930 | 23 | 29,861 |
Provide a correct Python 3 solution for this coding contest problem.
Have you ever heard of the unit "○○ tsubo" that expresses the area of land? Since ancient times, one samurai has said the area for making rice to eat in a day.
There is a land of a [m] x b [m]. Enter a and b and create a program that outputs the tsubo area S [tsubo] of the land. 1 tsubo = 3.305785 [m2], and a and b are integers less than or equal to 100.
input
a b
A and b separated by one space are given on one line.
output
Output the tsubo area S in one line. An error of 0.0001 or less is allowed.
Example
Input
15 25
Output
113.437508 | instruction | 0 | 14,931 | 23 | 29,862 |
"Correct Solution:
```
a,b=map(int,input().split())
S=round(a*b/3.305785,6)
print(S)
``` | output | 1 | 14,931 | 23 | 29,863 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Have you ever heard of the unit "○○ tsubo" that expresses the area of land? Since ancient times, one samurai has said the area for making rice to eat in a day.
There is a land of a [m] x b [m]. Enter a and b and create a program that outputs the tsubo area S [tsubo] of the land. 1 tsubo = 3.305785 [m2], and a and b are integers less than or equal to 100.
input
a b
A and b separated by one space are given on one line.
output
Output the tsubo area S in one line. An error of 0.0001 or less is allowed.
Example
Input
15 25
Output
113.437508
Submitted Solution:
```
a,b=input().split()
a=int(a)
b=int(b)
c=a*b/3.305785
print(f"{c:.6f}")
``` | instruction | 0 | 14,932 | 23 | 29,864 |
Yes | output | 1 | 14,932 | 23 | 29,865 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Have you ever heard of the unit "○○ tsubo" that expresses the area of land? Since ancient times, one samurai has said the area for making rice to eat in a day.
There is a land of a [m] x b [m]. Enter a and b and create a program that outputs the tsubo area S [tsubo] of the land. 1 tsubo = 3.305785 [m2], and a and b are integers less than or equal to 100.
input
a b
A and b separated by one space are given on one line.
output
Output the tsubo area S in one line. An error of 0.0001 or less is allowed.
Example
Input
15 25
Output
113.437508
Submitted Solution:
```
###
### atcorder test program
###
import sys
### math class
class math:
### pi
pi = 3.14159265358979323846264338
### GCD
def gcd(self, a, b):
if b == 0:
return a
return self.gcd(b, a%b)
### LCM
def lcm(self, a, b):
return (a*b)//self.gcd(a,b)
### Prime number search
def Pnum(self, a):
if a == 1: return False
for i in range(2,int(a**0.5)+1):
if a % i == 0:
return False
return True
### Circle area
def caria(self, r):
return r*r*self.pi
math = math()
### output class
class output:
### list
def list(self, l):
l = list(l)
print(" ", end="")
for i, num in enumerate(l):
print(num, end="")
if i != len(l)-1:
print(" ", end="")
print()
output = output()
### input sample
#i = input()
#A, B, C = [x for x in input().split()]
#inlist = [int(w) for w in input().split()]
#R = float(input())
#A = [int(x) for x in input().split()]
#for line in sys.stdin.readlines():
# x, y = [int(temp) for temp in line.split()]
### output sample
#print("{0} {1} {2:.5f}".format(A//B, A%B, A/B))
#print("{0:.6f} {1:.6f}".format(R*R*math.pi,R*2*math.pi))
#print(" {}".format(i), end="")
#A, B, C = [int(x) for x in input().split()]
a, b = [float(x) for x in input().split()]
print(a*b/3.305785)
``` | instruction | 0 | 14,933 | 23 | 29,866 |
Yes | output | 1 | 14,933 | 23 | 29,867 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Have you ever heard of the unit "○○ tsubo" that expresses the area of land? Since ancient times, one samurai has said the area for making rice to eat in a day.
There is a land of a [m] x b [m]. Enter a and b and create a program that outputs the tsubo area S [tsubo] of the land. 1 tsubo = 3.305785 [m2], and a and b are integers less than or equal to 100.
input
a b
A and b separated by one space are given on one line.
output
Output the tsubo area S in one line. An error of 0.0001 or less is allowed.
Example
Input
15 25
Output
113.437508
Submitted Solution:
```
a, b = list(map(float, input().split()))
print(a*b / 3.305785)
``` | instruction | 0 | 14,934 | 23 | 29,868 |
Yes | output | 1 | 14,934 | 23 | 29,869 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Have you ever heard of the unit "○○ tsubo" that expresses the area of land? Since ancient times, one samurai has said the area for making rice to eat in a day.
There is a land of a [m] x b [m]. Enter a and b and create a program that outputs the tsubo area S [tsubo] of the land. 1 tsubo = 3.305785 [m2], and a and b are integers less than or equal to 100.
input
a b
A and b separated by one space are given on one line.
output
Output the tsubo area S in one line. An error of 0.0001 or less is allowed.
Example
Input
15 25
Output
113.437508
Submitted Solution:
```
a,b=map(float,input().split())
print(f'{(a*b/3.305785):.6f}')
``` | instruction | 0 | 14,935 | 23 | 29,870 |
Yes | output | 1 | 14,935 | 23 | 29,871 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Have you ever heard of the unit "○○ tsubo" that expresses the area of land? Since ancient times, one samurai has said the area for making rice to eat in a day.
There is a land of a [m] x b [m]. Enter a and b and create a program that outputs the tsubo area S [tsubo] of the land. 1 tsubo = 3.305785 [m2], and a and b are integers less than or equal to 100.
input
a b
A and b separated by one space are given on one line.
output
Output the tsubo area S in one line. An error of 0.0001 or less is allowed.
Example
Input
15 25
Output
113.437508
Submitted Solution:
```
def tsubo(x,y):
return (x * y) / 3.305785
S = tsubo(a,b)
print(S)
``` | instruction | 0 | 14,936 | 23 | 29,872 |
No | output | 1 | 14,936 | 23 | 29,873 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Have you ever heard of the unit "○○ tsubo" that expresses the area of land? Since ancient times, one samurai has said the area for making rice to eat in a day.
There is a land of a [m] x b [m]. Enter a and b and create a program that outputs the tsubo area S [tsubo] of the land. 1 tsubo = 3.305785 [m2], and a and b are integers less than or equal to 100.
input
a b
A and b separated by one space are given on one line.
output
Output the tsubo area S in one line. An error of 0.0001 or less is allowed.
Example
Input
15 25
Output
113.437508
Submitted Solution:
```
#coding:utf-8
a, b = map(int, raw_input(). split())
print(a * b / 3.305785)
``` | instruction | 0 | 14,937 | 23 | 29,874 |
No | output | 1 | 14,937 | 23 | 29,875 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Have you ever heard of the unit "○○ tsubo" that expresses the area of land? Since ancient times, one samurai has said the area for making rice to eat in a day.
There is a land of a [m] x b [m]. Enter a and b and create a program that outputs the tsubo area S [tsubo] of the land. 1 tsubo = 3.305785 [m2], and a and b are integers less than or equal to 100.
input
a b
A and b separated by one space are given on one line.
output
Output the tsubo area S in one line. An error of 0.0001 or less is allowed.
Example
Input
15 25
Output
113.437508
Submitted Solution:
```
def tsubo(a,b):
return (a * b) / 3.305785
tsubo(a,b)
``` | instruction | 0 | 14,938 | 23 | 29,876 |
No | output | 1 | 14,938 | 23 | 29,877 |
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