message stringlengths 2 44.5k | message_type stringclasses 2 values | message_id int64 0 1 | conversation_id int64 276 109k | cluster float64 23 23 | __index_level_0__ int64 552 217k |
|---|---|---|---|---|---|
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given n segments on a Cartesian plane. Each segment's endpoints have integer coordinates. Segments can intersect with each other. No two segments lie on the same line.
Count the number of distinct points with integer coordinates, which are covered by at least one segment.
Input
The first line contains a single integer n (1 ≤ n ≤ 1000) — the number of segments.
Each of the next n lines contains four integers Ax_i, Ay_i, Bx_i, By_i (-10^6 ≤ Ax_i, Ay_i, Bx_i, By_i ≤ 10^6) — the coordinates of the endpoints A, B (A ≠ B) of the i-th segment.
It is guaranteed that no two segments lie on the same line.
Output
Print a single integer — the number of distinct points with integer coordinates, which are covered by at least one segment.
Examples
Input
9
0 0 4 4
-1 5 4 0
4 0 4 4
5 2 11 2
6 1 6 7
5 6 11 6
10 1 10 7
7 0 9 8
10 -1 11 -1
Output
42
Input
4
-1 2 1 2
-1 0 1 0
-1 0 0 3
0 3 1 0
Output
7
Note
The image for the first example:
<image>
Several key points are marked blue, the answer contains some non-marked points as well.
The image for the second example:
<image> | instruction | 0 | 1,662 | 23 | 3,324 |
Tags: fft, geometry, number theory
Correct Solution:
```
def gcd(a, b):
return a if b == 0 else gcd(b, a % b)
def cross(x1, y1, x2, y2):
return x1 * y2 - x2 * y1
def online(line, x, y):
a, b, c, d = line
if min(a, c) <= x <= max(a, c) and min(b, d) <= y <= max(b, d):
return True
else:
return False
def CF1036E():
N = int(input())
lines = []
count = 0
# Find the lines covered by each line segment
for _ in range(N):
x1, y1, x2, y2 = map(int, input().split())
count += gcd(abs(x1 - x2), abs(y1 - y2)) + 1
lines.append((x1, y1, x2, y2))
# Deal with the intersecting points
for i in range(N):
d = set()
for j in range(i+1, N):
px, py, qx, qy = lines[i]
rx, ry, sx, sy = lines[j]
vecx = (px - qx, rx - sx)
vecy = (py - qy, ry - sy)
# Cross of two lines
area = cross(vecx[0], vecx[1], vecy[0], vecy[1])
# Parallel line has no intersecting points
if area == 0: continue
# Computation of the exact point
# This has been referenced from : https://stackoverflow.com/questions/563198/how-do-you-detect-where-two-line-segments-intersect
lineA = cross(px, py, qx, qy)
lineB = cross(rx, ry, sx, sy)
x = cross(lineA, lineB, vecx[0], vecx[1]) / area
y = cross(lineA, lineB, vecy[0], vecy[1]) / area
# Verify the points are good.
# If the points are integers and lie of the lines they are valid.
if not (x % 1 == 0 and y % 1 == 0): continue
if not (online(lines[i], x, y) and online(lines[j], x, y)): continue
d.add((x, y))
count -= len(d)
return count
if __name__ == '__main__':
res = CF1036E()
print(res)
``` | output | 1 | 1,662 | 23 | 3,325 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given n segments on a Cartesian plane. Each segment's endpoints have integer coordinates. Segments can intersect with each other. No two segments lie on the same line.
Count the number of distinct points with integer coordinates, which are covered by at least one segment.
Input
The first line contains a single integer n (1 ≤ n ≤ 1000) — the number of segments.
Each of the next n lines contains four integers Ax_i, Ay_i, Bx_i, By_i (-10^6 ≤ Ax_i, Ay_i, Bx_i, By_i ≤ 10^6) — the coordinates of the endpoints A, B (A ≠ B) of the i-th segment.
It is guaranteed that no two segments lie on the same line.
Output
Print a single integer — the number of distinct points with integer coordinates, which are covered by at least one segment.
Examples
Input
9
0 0 4 4
-1 5 4 0
4 0 4 4
5 2 11 2
6 1 6 7
5 6 11 6
10 1 10 7
7 0 9 8
10 -1 11 -1
Output
42
Input
4
-1 2 1 2
-1 0 1 0
-1 0 0 3
0 3 1 0
Output
7
Note
The image for the first example:
<image>
Several key points are marked blue, the answer contains some non-marked points as well.
The image for the second example:
<image> | instruction | 0 | 1,663 | 23 | 3,326 |
Tags: fft, geometry, number theory
Correct Solution:
```
def gcd(a, b):
return a if b == 0 else gcd(b, a % b)
def cross(x1, y1, x2, y2):
return x1 * y2 - x2 * y1
def line_intersection(line1, line2):
px, py, qx, qy = line1
rx, ry, sx, sy = line2
xdiff = (px - qx, rx - sx)
ydiff = (py - qy, ry - sy)
div = cross(px - qx, rx - sx, py - qy, ry - sy)
if div == 0:
raise Exception('lines do not intersect')
pq, rs = cross(px, py, qx, qy), cross(rx, ry, sx, sy)
x = cross(pq, rs, px - qx, rx - sx) / div
y = cross(pq, rs, py - qy, ry - sy) / div
return x, y
def online(line, x, y):
a, b, c, d = line
if min(a, c) <= x <= max(a, c) and min(b, d) <= y <= max(b, d):
return True
else:
return False
def CF1036E():
N = int(input())
lines = []
count = 0
for _ in range(N):
x1, y1, x2, y2 = map(int, input().split())
points = gcd(abs(x1 - x2), abs(y1 - y2)) + 1
count += points
lines.append((x1, y1, x2, y2))
for i in range(N):
d = set()
for j in range(i+1, N):
px = lines[i][0]
py = lines[i][1]
qx = lines[j][0]
qy = lines[j][1]
rx = lines[i][2] - lines[i][0]
ry = lines[i][3] - lines[i][1]
sx = lines[j][2] - lines[j][0]
sy = lines[j][3] - lines[j][1]
rs = cross(rx, ry, sx, sy)
# qpr = cross(qx - px, qy - py, rx, ry)
if rs == 0: continue
# qpr = cross(qx - px, qy - py, rx, ry)
# qps = cross(qx - px, qy - py, sx, sy)
# t = qps / rs
# u = qpr / rs
x, y = line_intersection(lines[i], lines[j])
if not (x % 1 == 0 and y % 1 == 0): continue
if not (online(lines[i], x, y) and online(lines[j], x, y)): continue
d.add((x, y))
count = count - len(d)
return int(count)
if __name__ == '__main__':
res = CF1036E()
print(res)
``` | output | 1 | 1,663 | 23 | 3,327 |
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 a Cartesian plane. Each segment's endpoints have integer coordinates. Segments can intersect with each other. No two segments lie on the same line.
Count the number of distinct points with integer coordinates, which are covered by at least one segment.
Input
The first line contains a single integer n (1 ≤ n ≤ 1000) — the number of segments.
Each of the next n lines contains four integers Ax_i, Ay_i, Bx_i, By_i (-10^6 ≤ Ax_i, Ay_i, Bx_i, By_i ≤ 10^6) — the coordinates of the endpoints A, B (A ≠ B) of the i-th segment.
It is guaranteed that no two segments lie on the same line.
Output
Print a single integer — the number of distinct points with integer coordinates, which are covered by at least one segment.
Examples
Input
9
0 0 4 4
-1 5 4 0
4 0 4 4
5 2 11 2
6 1 6 7
5 6 11 6
10 1 10 7
7 0 9 8
10 -1 11 -1
Output
42
Input
4
-1 2 1 2
-1 0 1 0
-1 0 0 3
0 3 1 0
Output
7
Note
The image for the first example:
<image>
Several key points are marked blue, the answer contains some non-marked points as well.
The image for the second example:
<image>
Submitted Solution:
```
from math import gcd
from bisect import *
class Point:
def __init__(self, x, y):
self.x, self.y = x, y
def __repr__(self):
return 'Point' + repr((self.x, self.y))
def __add__(self, val):
return Point(self.x + val.x, self.y + val.y)
def __sub__(self, val):
return Point(self.x - val.x, self.y - val.y)
def __mul__(self, ratio):
return Point(self.x * ratio, self.y * ratio)
@staticmethod
def det(A, B):
return A.x * B.y - A.y * B.x
@staticmethod
def dot(A, B):
return A.x * B.x + A.y * B.y
def onSegment(self, A, B):
if self.det(B-A, self-A) != 0:
return False
if self.dot(B-A, self-A) < 0:
return False
if self.dot(A-B, self-B) < 0:
return False
return True
def intPoints(x1, y1, x2, y2):
dx, dy = abs(x2 - x1), abs(y2 - y1)
return gcd(dx, dy) + 1
def crosspoint(L1, L2):
A, B = Point(L1[0], L1[1]), Point(L1[2], L1[3])
C, D = Point(L2[0], L2[1]), Point(L2[2], L2[3])
S1, S2 = Point.det(C-A, D-A), Point.det(C-D, B-A)
delta = (B - A) * S1
if S2 == 0 or delta.x % S2 != 0 or delta.y % S2 != 0:
return None
delta.x = delta.x // S2
delta.y = delta.y // S2
P = A + delta
if not P.onSegment(A, B) or not P.onSegment(C, D):
return None
return (P.x, P.y)
'''
while True:
x1, y1, x2, y2 = map(int, input().split())
A, B = Point(x1, y1), Point(x2, y2)
x1, y1, x2, y2 = map(int, input().split())
C, D = Point(x1, y1), Point(x2, y2)
print(crosspoint(A, B, C, D))
'''
n = int(input())
lines = [ tuple(int(z) for z in input().split()) \
for i in range(n) ]
count = dict()
for i in range(n):
for j in range(i):
P = crosspoint(lines[i], lines[j])
if P == None:
continue
if not P in count:
count[P] = 1
else:
count[P] += 1
answer = sum(intPoints(*L) for L in lines)
tri = [ x*(x+1)//2 for x in range(n+1) ]
for z in count:
k = bisect_right(tri, count[z])
answer -= k - 1;
print(answer)
``` | instruction | 0 | 1,664 | 23 | 3,328 |
Yes | output | 1 | 1,664 | 23 | 3,329 |
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 a Cartesian plane. Each segment's endpoints have integer coordinates. Segments can intersect with each other. No two segments lie on the same line.
Count the number of distinct points with integer coordinates, which are covered by at least one segment.
Input
The first line contains a single integer n (1 ≤ n ≤ 1000) — the number of segments.
Each of the next n lines contains four integers Ax_i, Ay_i, Bx_i, By_i (-10^6 ≤ Ax_i, Ay_i, Bx_i, By_i ≤ 10^6) — the coordinates of the endpoints A, B (A ≠ B) of the i-th segment.
It is guaranteed that no two segments lie on the same line.
Output
Print a single integer — the number of distinct points with integer coordinates, which are covered by at least one segment.
Examples
Input
9
0 0 4 4
-1 5 4 0
4 0 4 4
5 2 11 2
6 1 6 7
5 6 11 6
10 1 10 7
7 0 9 8
10 -1 11 -1
Output
42
Input
4
-1 2 1 2
-1 0 1 0
-1 0 0 3
0 3 1 0
Output
7
Note
The image for the first example:
<image>
Several key points are marked blue, the answer contains some non-marked points as well.
The image for the second example:
<image>
Submitted Solution:
```
def gcd(a, b):
return a if b == 0 else gcd(b, a % b)
def cross(x1, y1, x2, y2):
return x1 * y2 - x2 * y1
def online(line, x, y):
a, b, c, d = line
if min(a, c) <= x <= max(a, c) and min(b, d) <= y <= max(b, d):
return True
return False
def CF1036E():
N = int(input())
lines = []
count = 0
# Find the lines covered by each line segment
for _ in range(N):
x1, y1, x2, y2 = map(int, input().split())
count += gcd(abs(x1 - x2), abs(y1 - y2)) + 1
lines.append((x1, y1, x2, y2))
# Deal with the intersecting points
for i in range(N):
d = set()
for j in range(i+1, N):
px, py, qx, qy = lines[i]
rx, ry, sx, sy = lines[j]
vecx = (px - qx, rx - sx)
vecy = (py - qy, ry - sy)
# Cross of two lines
area = cross(vecx[0], vecx[1], vecy[0], vecy[1])
# Parallel line has no intersecting points
if area == 0: continue
# Computation of the exact point
# This has been referenced from : https://stackoverflow.com/questions/563198/how-do-you-detect-where-two-line-segments-intersect
lineA = cross(px, py, qx, qy)
lineB = cross(rx, ry, sx, sy)
x = cross(lineA, lineB, vecx[0], vecx[1]) / area
y = cross(lineA, lineB, vecy[0], vecy[1]) / area
# Verify the points are good.
# If the points are integers and lie of the lines they are valid.
if not (x % 1 == 0 and y % 1 == 0): continue
if not (online(lines[i], x, y) and online(lines[j], x, y)): continue
d.add((x, y))
count -= len(d)
return count
if __name__ == '__main__':
res = CF1036E()
print(res)
``` | instruction | 0 | 1,665 | 23 | 3,330 |
Yes | output | 1 | 1,665 | 23 | 3,331 |
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 a Cartesian plane. Each segment's endpoints have integer coordinates. Segments can intersect with each other. No two segments lie on the same line.
Count the number of distinct points with integer coordinates, which are covered by at least one segment.
Input
The first line contains a single integer n (1 ≤ n ≤ 1000) — the number of segments.
Each of the next n lines contains four integers Ax_i, Ay_i, Bx_i, By_i (-10^6 ≤ Ax_i, Ay_i, Bx_i, By_i ≤ 10^6) — the coordinates of the endpoints A, B (A ≠ B) of the i-th segment.
It is guaranteed that no two segments lie on the same line.
Output
Print a single integer — the number of distinct points with integer coordinates, which are covered by at least one segment.
Examples
Input
9
0 0 4 4
-1 5 4 0
4 0 4 4
5 2 11 2
6 1 6 7
5 6 11 6
10 1 10 7
7 0 9 8
10 -1 11 -1
Output
42
Input
4
-1 2 1 2
-1 0 1 0
-1 0 0 3
0 3 1 0
Output
7
Note
The image for the first example:
<image>
Several key points are marked blue, the answer contains some non-marked points as well.
The image for the second example:
<image>
Submitted Solution:
```
from math import gcd
from bisect import *
class Point:
def __init__(self, x, y):
self.x, self.y = x, y
def __add__(self, val):
return Point(self.x + val.x, self.y + val.y)
def __sub__(self, val):
return Point(self.x - val.x, self.y - val.y)
def __mul__(self, ratio):
return Point(self.x * ratio, self.y * ratio)
def __truediv__(self, ratio):
return Point(self.x / ratio, self.y / ratio)
@staticmethod
def det(A, B):
return A.x * B.y - A.y * B.x
@staticmethod
def dot(A, B):
return A.x * B.x + A.y * B.y
def onSegment(self, A, B):
if self.det(B-A, self-A) != 0:
return False
if self.dot(B-A, self-A) < 0:
return False
if self.dot(A-B, self-B) < 0:
return False
return True
def intPoints(x1, y1, x2, y2):
dx, dy = abs(x2 - x1), abs(y2 - y1)
return gcd(dx, dy) + 1
def crosspoint(L1, L2):
A, B = Point(L1[0], L1[1]), Point(L1[2], L1[3])
C, D = Point(L2[0], L2[1]), Point(L2[2], L2[3])
S1, S2 = Point.det(C-A, D-A), Point.det(C-D, B-A)
delta = (B - A) * S1
if S2 == 0 or delta.x % S2 != 0 or delta.y % S2 != 0:
return None
delta = delta / S2;
P = A + delta
if not P.onSegment(A, B) or not P.onSegment(C, D):
return None
return (P.x, P.y)
n = int(input())
lines = [ tuple(int(z) for z in input().split()) \
for i in range(n) ]
count = dict()
for i in range(n):
for j in range(i):
P = crosspoint(lines[i], lines[j])
if P != None:
count[P] = count.get(P, 0) + 1
answer = sum(intPoints(*L) for L in lines)
tri = [ x*(x+1)//2 for x in range(n+1) ]
for z in count:
k = bisect_right(tri, count[z])
answer -= k - 1;
print(answer)
``` | instruction | 0 | 1,666 | 23 | 3,332 |
Yes | output | 1 | 1,666 | 23 | 3,333 |
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 a Cartesian plane. Each segment's endpoints have integer coordinates. Segments can intersect with each other. No two segments lie on the same line.
Count the number of distinct points with integer coordinates, which are covered by at least one segment.
Input
The first line contains a single integer n (1 ≤ n ≤ 1000) — the number of segments.
Each of the next n lines contains four integers Ax_i, Ay_i, Bx_i, By_i (-10^6 ≤ Ax_i, Ay_i, Bx_i, By_i ≤ 10^6) — the coordinates of the endpoints A, B (A ≠ B) of the i-th segment.
It is guaranteed that no two segments lie on the same line.
Output
Print a single integer — the number of distinct points with integer coordinates, which are covered by at least one segment.
Examples
Input
9
0 0 4 4
-1 5 4 0
4 0 4 4
5 2 11 2
6 1 6 7
5 6 11 6
10 1 10 7
7 0 9 8
10 -1 11 -1
Output
42
Input
4
-1 2 1 2
-1 0 1 0
-1 0 0 3
0 3 1 0
Output
7
Note
The image for the first example:
<image>
Several key points are marked blue, the answer contains some non-marked points as well.
The image for the second example:
<image>
Submitted Solution:
```
# Borrows primarily from : https://stackoverflow.com/questions/563198/how-do-you-detect-where-two-line-segments-intersect
def gcd(a, b):
return a if b == 0 else gcd(b, a % b)
def cross(x1, y1, x2, y2):
return x1 * y2 - x2 * y1
def online(line, x, y):
a, b, c, d = line
if min(a, c) <= x <= max(a, c) and min(b, d) <= y <= max(b, d):
return True
return False
def CF1036E():
N = int(input())
lines = []
count = 0
# Find the lines covered by each line segment
for _ in range(N):
x1, y1, x2, y2 = map(int, input().split())
count += gcd(abs(x1 - x2), abs(y1 - y2)) + 1
lines.append((x1, y1, x2, y2))
# Deal with the intersecting points
for i in range(N):
d = set()
for j in range(i+1, N):
px, py, qx, qy = lines[i]
rx, ry, sx, sy = lines[j]
vecx = (px - qx, rx - sx)
vecy = (py - qy, ry - sy)
# Cross of two lines
area = cross(vecx[0], vecx[1], vecy[0], vecy[1])
# Not interested in overlapping parallel lines
if area == 0: continue
# Computation of the exact point
# This has been referenced from : https://stackoverflow.com/questions/563198/how-do-you-detect-where-two-line-segments-intersect
lineA = cross(px, py, qx, qy)
lineB = cross(rx, ry, sx, sy)
x = cross(lineA, lineB, vecx[0], vecx[1]) / area
y = cross(lineA, lineB, vecy[0], vecy[1]) / area
# Verify the points are good.
# If the points are integers and lie of the lines they are valid.
if not (x % 1 == 0 and y % 1 == 0): continue
if not (online(lines[i], x, y) and online(lines[j], x, y)): continue
d.add((x, y))
count -= len(d)
return count
if __name__ == '__main__':
res = CF1036E()
print(res)
``` | instruction | 0 | 1,667 | 23 | 3,334 |
Yes | output | 1 | 1,667 | 23 | 3,335 |
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 a Cartesian plane. Each segment's endpoints have integer coordinates. Segments can intersect with each other. No two segments lie on the same line.
Count the number of distinct points with integer coordinates, which are covered by at least one segment.
Input
The first line contains a single integer n (1 ≤ n ≤ 1000) — the number of segments.
Each of the next n lines contains four integers Ax_i, Ay_i, Bx_i, By_i (-10^6 ≤ Ax_i, Ay_i, Bx_i, By_i ≤ 10^6) — the coordinates of the endpoints A, B (A ≠ B) of the i-th segment.
It is guaranteed that no two segments lie on the same line.
Output
Print a single integer — the number of distinct points with integer coordinates, which are covered by at least one segment.
Examples
Input
9
0 0 4 4
-1 5 4 0
4 0 4 4
5 2 11 2
6 1 6 7
5 6 11 6
10 1 10 7
7 0 9 8
10 -1 11 -1
Output
42
Input
4
-1 2 1 2
-1 0 1 0
-1 0 0 3
0 3 1 0
Output
7
Note
The image for the first example:
<image>
Several key points are marked blue, the answer contains some non-marked points as well.
The image for the second example:
<image>
Submitted Solution:
```
import math
def gcd(a, b):
return a if b == 0 else gcd(b, a % b)
def cross(x1, y1, x2, y2):
return x1 * y2 - x2 * y1
def dot(x1, y1, x2, y2):
return x1 * x2 + y1 * y2
def intersect01(a, b):
if a < 0 and b >= 0: return True
if a >= 0 and a <= 1: return True
return False
def CF1036E():
N = int(input())
lines = []
count = 0
for _ in range(N):
x1, y1, x2, y2 = map(int, input().split())
points = gcd(abs(x1 - x2), abs(y1 - y2)) + 1
count += points
lines.append((x1, y1, x2, y2))
if lines[0] == (-100, 51, 100, 11): return 5470
duplicates = []
for i in range(N):
for j in range(i+1, N):
px = lines[i][0]
py = lines[i][1]
qx = lines[j][0]
qy = lines[j][1]
rx = lines[i][2] - lines[i][0]
ry = lines[i][3] - lines[i][1]
sx = lines[j][2] - lines[j][0]
sy = lines[j][3] - lines[j][1]
rs = cross(rx, ry, sx, sy)
qpr = cross(qx - px, qy - py, rx, ry)
if rs == 0 and qpr == 0:
# We have a collinear line. Do they overlap?
t0 = dot(qx - px, qy - py, rx, ry) / dot(rx, ry, rx, ry)
t1 = dot(qx - px + sx, qy - py + sy, rx, ry) / dot(rx, ry, rx, ry)
if intersect01(t0, t1) or intersect01(t1, t0):
print("collinear ", segment)
segment = sorted([
(lines[i][0], lines[i][1]),
(lines[j][0], lines[j][1]),
(lines[i][2], lines[i][3]),
(lines[j][2], lines[j][3])
])
x1, y1 = segment[1]
x2, y2 = segment[2]
if segment[1] != segment[2]:
points = gcd(abs(x1 - x2), abs(y1 - y2)) + 1
count -= points
else:
duplicates.append(segment[1])
else:
pass # Collinear and disjoint
elif rs == 0 and qpr != 0:
pass # Parallel and non-intersecting lines
elif rs != 0:
qpr = cross(qx - px, qy - py, rx, ry)
qps = cross(qx - px, qy - py, sx, sy)
t = qps / rs
u = qpr / rs
if 0 <= t <= 1 and 0 <= u <= 1:
x = (int)(px + t * rx)
y = (int)(py + t * ry)
x1 = (qx + u * sx)
y1 = (qy + u * sy)
if x % 1 == 0 and y % 1 == 0 and x == x1 and y == y1:
duplicates.append((x, y))
else:
pass # non-intersecting and non-parallel
# Deal with Intersecting duplicates
counter = {}
for x, y in duplicates:
counter[(x, y)] = counter.get((x, y), 0) + 1
# print(count, counter)
for key, value in counter.items():
if value > 1: value = math.floor((2 * value) ** 0.5)
count = count - value
print(count)
if __name__ == '__main__':
CF1036E()
``` | instruction | 0 | 1,668 | 23 | 3,336 |
No | output | 1 | 1,668 | 23 | 3,337 |
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 a Cartesian plane. Each segment's endpoints have integer coordinates. Segments can intersect with each other. No two segments lie on the same line.
Count the number of distinct points with integer coordinates, which are covered by at least one segment.
Input
The first line contains a single integer n (1 ≤ n ≤ 1000) — the number of segments.
Each of the next n lines contains four integers Ax_i, Ay_i, Bx_i, By_i (-10^6 ≤ Ax_i, Ay_i, Bx_i, By_i ≤ 10^6) — the coordinates of the endpoints A, B (A ≠ B) of the i-th segment.
It is guaranteed that no two segments lie on the same line.
Output
Print a single integer — the number of distinct points with integer coordinates, which are covered by at least one segment.
Examples
Input
9
0 0 4 4
-1 5 4 0
4 0 4 4
5 2 11 2
6 1 6 7
5 6 11 6
10 1 10 7
7 0 9 8
10 -1 11 -1
Output
42
Input
4
-1 2 1 2
-1 0 1 0
-1 0 0 3
0 3 1 0
Output
7
Note
The image for the first example:
<image>
Several key points are marked blue, the answer contains some non-marked points as well.
The image for the second example:
<image>
Submitted Solution:
```
import math
def gcd(a, b):
return a if b == 0 else gcd(b, a % b)
def cross(x1, y1, x2, y2):
return x1 * y2 - x2 * y1
def dot(x1, y1, x2, y2):
return x1 * x2 + y1 * y2
def intersect01(a, b):
if a < 0 and b >= 0: return True
if a >= 0 and a <= 1: return True
return False
def CF1036E():
N = int(input())
lines = []
count = 0
for _ in range(N):
x1, y1, x2, y2 = map(int, input().split())
points = gcd(abs(x1 - x2), abs(y1 - y2)) + 1
count += points
lines.append((x1, y1, x2, y2))
if lines[0] == (-100, 51, 100, 11):
print(lines)
return 5470
duplicates = []
for i in range(N):
for j in range(i+1, N):
px = lines[i][0]
py = lines[i][1]
qx = lines[j][0]
qy = lines[j][1]
rx = lines[i][2] - lines[i][0]
ry = lines[i][3] - lines[i][1]
sx = lines[j][2] - lines[j][0]
sy = lines[j][3] - lines[j][1]
rs = cross(rx, ry, sx, sy)
qpr = cross(qx - px, qy - py, rx, ry)
if rs == 0 and qpr == 0:
# We have a collinear line. Do they overlap?
t0 = dot(qx - px, qy - py, rx, ry) / dot(rx, ry, rx, ry)
t1 = dot(qx - px + sx, qy - py + sy, rx, ry) / dot(rx, ry, rx, ry)
if intersect01(t0, t1) or intersect01(t1, t0):
print("collinear ", segment)
segment = sorted([
(lines[i][0], lines[i][1]),
(lines[j][0], lines[j][1]),
(lines[i][2], lines[i][3]),
(lines[j][2], lines[j][3])
])
x1, y1 = segment[1]
x2, y2 = segment[2]
if segment[1] != segment[2]:
points = gcd(abs(x1 - x2), abs(y1 - y2)) + 1
count -= points
else:
duplicates.append(segment[1])
else:
pass # Collinear and disjoint
elif rs == 0 and qpr != 0:
pass # Parallel and non-intersecting lines
elif rs != 0:
qpr = cross(qx - px, qy - py, rx, ry)
qps = cross(qx - px, qy - py, sx, sy)
t = qps / rs
u = qpr / rs
if 0 <= t <= 1 and 0 <= u <= 1:
x = (int)(px + t * rx)
y = (int)(py + t * ry)
x1 = (qx + u * sx)
y1 = (qy + u * sy)
if x % 1 == 0 and y % 1 == 0 and x == x1 and y == y1:
duplicates.append((x, y))
else:
pass # non-intersecting and non-parallel
# Deal with Intersecting duplicates
counter = {}
for x, y in duplicates:
counter[(x, y)] = counter.get((x, y), 0) + 1
# print(count, counter)
for key, value in counter.items():
if value > 1: value = math.floor((2 * value) ** 0.5)
count = count - value
return count
if __name__ == '__main__':
res = CF1036E()
print(res)
``` | instruction | 0 | 1,669 | 23 | 3,338 |
No | output | 1 | 1,669 | 23 | 3,339 |
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 a Cartesian plane. Each segment's endpoints have integer coordinates. Segments can intersect with each other. No two segments lie on the same line.
Count the number of distinct points with integer coordinates, which are covered by at least one segment.
Input
The first line contains a single integer n (1 ≤ n ≤ 1000) — the number of segments.
Each of the next n lines contains four integers Ax_i, Ay_i, Bx_i, By_i (-10^6 ≤ Ax_i, Ay_i, Bx_i, By_i ≤ 10^6) — the coordinates of the endpoints A, B (A ≠ B) of the i-th segment.
It is guaranteed that no two segments lie on the same line.
Output
Print a single integer — the number of distinct points with integer coordinates, which are covered by at least one segment.
Examples
Input
9
0 0 4 4
-1 5 4 0
4 0 4 4
5 2 11 2
6 1 6 7
5 6 11 6
10 1 10 7
7 0 9 8
10 -1 11 -1
Output
42
Input
4
-1 2 1 2
-1 0 1 0
-1 0 0 3
0 3 1 0
Output
7
Note
The image for the first example:
<image>
Several key points are marked blue, the answer contains some non-marked points as well.
The image for the second example:
<image>
Submitted Solution:
```
import math
def gcd(a, b):
return a if b == 0 else gcd(b, a % b)
def cross(x1, y1, x2, y2):
return x1 * y2 - x2 * y1
def dot(x1, y1, x2, y2):
return x1 * x2 + y1 * y2
def intersect01(a, b):
if a < 0 and b >= 0: return True
if a >= 0 and a <= 1: return True
return False
def findn(k):
n1 = (1 - (8 * k + 1) ** 0.5)/2
n2 = (1 + (8 * k + 1) ** 0.5)/2
# print(k, n1, n2)
if n1 > 0 and n1 * (n1 - 1) == 2 * k: return n1
else: return n2
return n
def CF1036E():
N = int(input())
lines = []
count = 0
for _ in range(N):
x1, y1, x2, y2 = map(int, input().split())
points = gcd(abs(x1 - x2), abs(y1 - y2)) + 1
count += points
lines.append((x1, y1, x2, y2))
duplicates = []
for i in range(N):
for j in range(i+1, N):
px = lines[i][0]
py = lines[i][1]
qx = lines[j][0]
qy = lines[j][1]
rx = lines[i][2] - lines[i][0]
ry = lines[i][3] - lines[i][1]
sx = lines[j][2] - lines[j][0]
sy = lines[j][3] - lines[j][1]
rs = cross(rx, ry, sx, sy)
qpr = cross(qx - px, qy - py, rx, ry)
if rs == 0 and qpr == 0:
# We have a collinear line. Do they have a common point?
pass
elif rs == 0 and qpr != 0:
pass # Parallel and non-intersecting lines
elif rs != 0:
qpr = cross(qx - px, qy - py, rx, ry)
qps = cross(qx - px, qy - py, sx, sy)
t = qps / rs
u = qpr / rs
if 0 <= t <= 1 and 0 <= u <= 1:
x = (int)(px + t * rx)
y = (int)(py + t * ry)
x1 = (qx + u * sx)
y1 = (qy + u * sy)
if x % 1 == 0 and y % 1 == 0 and x == x1 and y == y1:
duplicates.append((x, y))
else:
pass # non-intersecting and non-parallel
# Deal with Intersecting duplicates
counter = {}
for x, y in duplicates:
counter[(x, y)] = counter.get((x, y), 0) + 1
for key, value in counter.items():
if value > 1:
n = abs(findn(value))
# print(value, n)
value = n - 1
count = count - (value)
return int(count)
if __name__ == '__main__':
res = CF1036E()
print(res)
``` | instruction | 0 | 1,670 | 23 | 3,340 |
No | output | 1 | 1,670 | 23 | 3,341 |
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 a Cartesian plane. Each segment's endpoints have integer coordinates. Segments can intersect with each other. No two segments lie on the same line.
Count the number of distinct points with integer coordinates, which are covered by at least one segment.
Input
The first line contains a single integer n (1 ≤ n ≤ 1000) — the number of segments.
Each of the next n lines contains four integers Ax_i, Ay_i, Bx_i, By_i (-10^6 ≤ Ax_i, Ay_i, Bx_i, By_i ≤ 10^6) — the coordinates of the endpoints A, B (A ≠ B) of the i-th segment.
It is guaranteed that no two segments lie on the same line.
Output
Print a single integer — the number of distinct points with integer coordinates, which are covered by at least one segment.
Examples
Input
9
0 0 4 4
-1 5 4 0
4 0 4 4
5 2 11 2
6 1 6 7
5 6 11 6
10 1 10 7
7 0 9 8
10 -1 11 -1
Output
42
Input
4
-1 2 1 2
-1 0 1 0
-1 0 0 3
0 3 1 0
Output
7
Note
The image for the first example:
<image>
Several key points are marked blue, the answer contains some non-marked points as well.
The image for the second example:
<image>
Submitted Solution:
```
points = []
already_been = []
for i in range(int(input())):
points.append(list(map(int, input().split())))
for point in points:
p1 = [point[0], point[1]]
p2 = [point[2], point[3]]
if p1 not in already_been:
already_been.append(p1)
if p2 not in already_been:
already_been.append(p2)
if p1[0] == p2[0]:
for y in range(p1[1], p2[1] + 1):
if [p1[0], y] not in already_been:
already_been.append([p1[0], y])
elif p1[1] == p2[1]:
for x in range(p1[0], p2[0] + 1):
if [x, p1[1]] not in already_been:
already_been.append([x, p1[1]])
else:
k = (p1[1] - p2[1]) / (p1[0] - p2[0])
b = p2[1] - k * p2[0]
for y in range(min(p1[1], p2[1]), max(p2[1], p1[1]) + 1):
x = (y - b) / k
if x == int(x) and [x, y] not in already_been:
already_been.append([x, y])
print(len(already_been))
``` | instruction | 0 | 1,671 | 23 | 3,342 |
No | output | 1 | 1,671 | 23 | 3,343 |
Evaluate the correctness of the submitted Python 2 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given n segments on a Cartesian plane. Each segment's endpoints have integer coordinates. Segments can intersect with each other. No two segments lie on the same line.
Count the number of distinct points with integer coordinates, which are covered by at least one segment.
Input
The first line contains a single integer n (1 ≤ n ≤ 1000) — the number of segments.
Each of the next n lines contains four integers Ax_i, Ay_i, Bx_i, By_i (-10^6 ≤ Ax_i, Ay_i, Bx_i, By_i ≤ 10^6) — the coordinates of the endpoints A, B (A ≠ B) of the i-th segment.
It is guaranteed that no two segments lie on the same line.
Output
Print a single integer — the number of distinct points with integer coordinates, which are covered by at least one segment.
Examples
Input
9
0 0 4 4
-1 5 4 0
4 0 4 4
5 2 11 2
6 1 6 7
5 6 11 6
10 1 10 7
7 0 9 8
10 -1 11 -1
Output
42
Input
4
-1 2 1 2
-1 0 1 0
-1 0 0 3
0 3 1 0
Output
7
Note
The image for the first example:
<image>
Several key points are marked blue, the answer contains some non-marked points as well.
The image for the second example:
<image>
Submitted Solution:
```
""" Python 3 compatibility tools. """
from __future__ import division, print_function
import sys
import os
if 'XXYYZZ' in os.environ:
from typing import List, Set, Dict, Tuple, Text, Optional, Callable, Any, Union
from collections import deque
import collections
from types import GeneratorType
import itertools
import operator
import functools
import random
import copy
import heapq
import math
from atexit import register
from io import BytesIO, IOBase
import __pypy__ # type: ignore
EPS = 10**-12
#########
# INPUT #
#########
class Input(object):
def __init__(self):
if 'CPH' not in os.environ:
sys.stdin = BytesIO(os.read(0, os.fstat(0).st_size))
sys.stdout = BytesIO()
register(lambda: os.write(1, sys.stdout.getvalue()))
def rawInput(self):
# type: () -> str
return sys.stdin.readline().rstrip('\r\n')
def readSplit(self, mapper):
return map(mapper, self.rawInput().split())
def readInt(self):
return int(self.rawInput())
##########
# OUTPUT #
##########
class Output(object):
def __init__(self):
self.out = __pypy__.builders.StringBuilder()
def write(self, text):
# type: (str) -> None
self.out.append(str(text))
def writeLine(self, text):
# type: (str) -> None
self.write(str(text) + '\n')
def finalize(self):
if sys.version_info[0] < 3:
os.write(1, self.out.build())
else:
os.write(1, self.out.build().encode())
###########
# LIBRARY #
###########
def bootstrap(f, stack=[]):
# Deep Recursion helper.
# From: https://github.com/cheran-senthil/PyRival/blob/c1972da95d102d95b9fea7c5c8e0474d61a54378/docs/bootstrap.rst
# Usage:
# @bootstrap
# def recur(n):
# if n == 0:
# yield 1
# yield (yield recur(n-1)) * n
def wrappedfunc(*args, **kwargs):
if stack:
return f(*args, **kwargs)
else:
to = f(*args, **kwargs)
while True:
if type(to) is GeneratorType:
stack.append(to)
to = next(to)
else:
stack.pop()
if not stack:
break
to = stack[-1].send(to)
return to
return wrappedfunc
int_add = __pypy__.intop.int_add
int_sub = __pypy__.intop.int_sub
int_mul = __pypy__.intop.int_mul
def make_mod_mul(mod):
fmod_inv = 1.0 / mod
def mod_mul(a, b, c=0):
res = int_sub(int_add(int_mul(a, b), c), int_mul(
mod, int(fmod_inv * a * b + fmod_inv * c)))
if res >= mod:
return res - mod
elif res < 0:
return res + mod
else:
return res
return mod_mul
class Point(object):
"""
>>> x = Point(-2.0, 3.0)
>>> x
Point(-2.0, 3.0)
>>> x.norm()
13.0
>>> x.length()
3.605551275463989
>>> x = Point(-2, 3)
>>> x
Point(-2, 3)
>>> x.norm()
13
>>> x.length()
3.605551275463989
>>> x = Point(0.0, 0.0)
>>> x.norm()
0.0
>>> x.length()
0.0
"""
def __init__(self, x, y):
# type: (float, float) -> None
self.x = x
self.y = y
def norm(self):
return self.x * self.x + self.y * self.y
def length(self):
return math.sqrt(self.norm())
def __repr__(self):
return "Point({0}, {1})".format(self.x, self.y)
class Segment(object):
"""
Line segment from point p1 to point p2
>>> x = Segment(Point(-2.0, 3.0), Point(2.0, -2.0))
>>> x
Segment(Point(-2.0, 3.0), Point(2.0, -2.0))
>>> x.to_line()
Line(-5.0, -4.0, 2.0)
>>> x.to_point()
Point(4.0, -5.0)
>>> x = Segment(Point(-2, 3), Point(2, -2))
>>> x
Segment(Point(-2, 3), Point(2, -2))
>>> x.to_line()
Line(-5, -4, 2)
>>> x.to_point()
Point(4, -5)
"""
def __init__(self, p1, p2):
# type: (Point, Point) -> None
self.p1 = p1
self.p2 = p2
def to_line(self):
a = self.p2.y - self.p1.y
b = self.p1.x - self.p2.x
return Line(
a,
b,
-a * self.p1.x - b * self.p1.y
)
def to_point(self):
return Point(self.p2.x - self.p1.x, self.p2.y - self.p1.y)
def __repr__(self):
return "Segment({0}, {1})".format(self.p1, self.p2)
class Line(object):
"""
Represents the line: ax + by + c = 0
"""
def __init__(self, a, b, c):
# type: (float, float, float) -> None
self.a = a
self.b = b
self.c = c
def __repr__(self):
return "Line({0}, {1}, {2})".format(self.a, self.b, self.c)
def cross(p1, p2):
# type: (Point, Point) -> float
"""
>>> cross(Point(13.0, 8.0), Point(-1.0, 2.0))
34.0
>>> cross(Point(13, 8), Point(-1, 2))
34
"""
return p1.x * p2.y - p1.y * p2.x
def dot(p1, p2):
# type: (Point, Point) -> float
"""
>>> dot(Point(2.0, 3.0), Point(-4.0, 5.0))
7.0
>>> dot(Point(2, 3), Point(-4, 5))
7
"""
return p1.x * p2.x + p1.y * p2.y
def gcd(x, y):
# (int, int) -> int
# Returns the greatest common divisor of x and y
'''
>>> gcd(50, 70)
10
>>> gcd(70, 50)
10
>>> gcd(1, 99)
1
>>> gcd(2, 99)
1
>>> gcd(99, 1)
1
'''
while y:
x, y = y, x % y
return x
def line_is_parallel(line1, line2):
# type: (Line, Line) -> bool
"""
>>> line_is_parallel(Line(1, 1, -1), Line(1, -1, 0))
False
>>> line_is_parallel(Line(1, 1, -1), Line(1, 1, -1))
True
>>> line_is_parallel(Line(1, 1, 1), Line(2, 2, 2))
True
"""
return abs(line1.a * line2.b - line1.b * line2.a) < EPS
def line_intersection(line1, line2):
# type: (Line, Line) -> Optional[Point]
"""
>>> line_intersection(Line(1, 1, -1), Line(1, -1, 0))
Point(0.5, 0.5)
>>> line_intersection(Line(1, 1, -1), Line(1, 1, -1))
>>> line_intersection(Line(1, 1, 1), Line(2, 2, 2))
"""
zn = line1.a * line2.b - line1.b * line2.a
if abs(zn) < EPS:
return None
return Point(
(line1.c * line2.b - line1.b * line2.c) * -1.0 / zn,
(line1.a * line2.c - line1.c * line2.a) * -1.0 / zn,
)
def segment_intersection(segment1, segment2):
# type: (Segment, Segment) -> Optional[Point]
"""
>>> segment_intersection(Segment(Point(0, 0), Point(1, 1)), Segment(Point(0, 1), Point(1, 0)),)
Point(0.5, 0.5)
>>> segment_intersection( Segment(Point(0, 0), Point(1, 1)), Segment(Point(1, 1), Point(2, 0)),)
Point(1.0, 1.0)
>>> segment_intersection( Segment(Point(0, 0), Point(1, 1)), Segment(Point(2, 0), Point(2, 1)),)
>>> segment_intersection( Segment(Point(0, 0), Point(1, 1)), Segment(Point(0, 1), Point(1, 2)),)
>>> segment_intersection( Segment(Point(0, 0), Point(1, 1)), Segment(Point(0, 0), Point(1, 1)),)
>>> segment_intersection( Segment(Point(0, 0), Point(4, 4)), Segment(Point(5, 2), Point(11, 2)),)
"""
intersection = line_intersection(
segment1.to_line(),
segment2.to_line(),
)
if not intersection:
return None
xmin = min(segment1.p1.x, segment1.p2.x)
xmax = max(segment1.p1.x, segment1.p2.x)
ymin = min(segment1.p1.y, segment1.p2.y)
ymax = max(segment1.p1.y, segment1.p2.y)
if (
xmin - EPS < intersection.x < xmax + EPS and
ymin - EPS < intersection.y < ymax + EPS
):
xmin2 = min(segment2.p1.x, segment2.p2.x)
xmax2 = max(segment2.p1.x, segment2.p2.x)
ymin2 = min(segment2.p1.y, segment2.p2.y)
ymax2 = max(segment2.p1.y, segment2.p2.y)
if (
xmin2 - EPS < intersection.x < xmax2 + EPS and
ymin2 - EPS < intersection.y < ymax2 + EPS
):
return intersection
return None
#########
# LOGIC #
#########
def main(inp, out):
# type: (Input, Output) -> None
n = inp.readInt()
segments = [] # type: List[Segment]
for _ in range(n):
a, b, c, d = inp.readSplit(int)
segments.append(Segment(Point(a, b), Point(c, d)))
ans = 0
for i in range(n):
segment = segments[i]
vec = segment.to_point()
ans += gcd(abs(vec.x), abs(vec.y)) + 1
for j in range(i+1, n):
ot = segments[j]
intersection = segment_intersection(segment, ot)
if intersection:
if (
abs(round(intersection.x) - intersection.x) < EPS and
abs(round(intersection.y) - intersection.y) < EPS):
ans -= 1
out.writeLine(ans)
###############
# BOILERPLATE #
###############
output_obj = Output()
main(Input(), output_obj)
output_obj.finalize()
``` | instruction | 0 | 1,672 | 23 | 3,344 |
No | output | 1 | 1,672 | 23 | 3,345 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Arkady reached the n-th level in Township game, so Masha decided to bake a pie for him! Of course, the pie has a shape of convex n-gon, i.e. a polygon with n vertices.
Arkady decided to cut the pie in two equal in area parts by cutting it by a straight line, so that he can eat one of them and give the other to Masha. There is a difficulty because Arkady has already put a knife at some point of the pie, so he now has to cut the pie by a straight line passing trough this point.
Help Arkady: find a line that passes through the point Arkady has put a knife into and cuts the pie into two parts of equal area, or determine that it's impossible. Your program has to quickly answer many queries with the same pie, but different points in which Arkady puts a knife.
Input
The first line contains two integers n and q (3 ≤ n ≤ 104, 1 ≤ q ≤ 105) — the number of vertices in the pie and the number of queries.
n line follow describing the polygon vertices in clockwise order. The i-th of these line contains two integers xi and yi ( - 106 ≤ xi, yi ≤ 106) — the coordinates of the i-th vertex. It is guaranteed that the polygon is strictly convex, in particular, no three vertices line on the same line.
An empty line follows.
q lines follow describing the query points. The i-th of these lines contain two integers xi and yi ( - 106 ≤ xi, yi ≤ 106) — the coordinates of the point in which Arkady puts the knife in the i-th query. In is guaranteed that in each query the given point is strictly inside the polygon, in particular, is not on its edges.
Output
For each query print single integer — the polar angle of the line that is the answer for the corresponding query, in radians. The angle should be in the segment [0;π], the angles are measured from the direction of OX axis in counter-clockwise order. For example, the polar angle of the OY axis is <image>. If there is no answer in that query, print -1.
If there are several answers, print any of them. Your answer is considered correct if the difference between the areas of the parts divided by the total area of the polygon doesn't exceed 10 - 4 by absolute value. In other words, if a and b are the areas of the parts after the cut, then your answer is correct if and only of <image>.
Examples
Input
3 1
0 0
0 3
3 0
1 1
Output
2.67794504460098710000
Input
5 3
6 5
6 3
5 0
0 0
0 5
5 4
3 3
5 2
Output
0.60228734612690049000
1.27933953226473580000
2.85805511179015910000
Submitted Solution:
```
str1=input()
lst=[]
for i in range(len(str1)):
if i%2 == 0:
lst.append(str1[i])
lst.sort()
print(lst[0],end='')
for i in range(1,len(lst)):
print("+",end='')
print(lst[i],end='')
``` | instruction | 0 | 2,148 | 23 | 4,296 |
No | output | 1 | 2,148 | 23 | 4,297 |
Provide a correct Python 3 solution for this coding contest problem.
There are N balls in a two-dimensional plane. The i-th ball is at coordinates (x_i, y_i).
We will collect all of these balls, by choosing two integers p and q such that p \neq 0 or q \neq 0 and then repeating the following operation:
* Choose a ball remaining in the plane and collect it. Let (a, b) be the coordinates of this ball. If we collected a ball at coordinates (a - p, b - q) in the previous operation, the cost of this operation is 0. Otherwise, including when this is the first time to do this operation, the cost of this operation is 1.
Find the minimum total cost required to collect all the balls when we optimally choose p and q.
Constraints
* 1 \leq N \leq 50
* |x_i|, |y_i| \leq 10^9
* If i \neq j, x_i \neq x_j or y_i \neq y_j.
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
N
x_1 y_1
:
x_N y_N
Output
Print the minimum total cost required to collect all the balls.
Examples
Input
2
1 1
2 2
Output
1
Input
3
1 4
4 6
7 8
Output
1
Input
4
1 1
1 2
2 1
2 2
Output
2 | instruction | 0 | 2,270 | 23 | 4,540 |
"Correct Solution:
```
from collections import Counter
import itertools as it
N=int(input())
l=[]
for i in range(N):
l.append(tuple(map(int,input().split())))
l.sort()
if N==1:
ans=1
else:
d=[(c[1][0]-c[0][0],c[1][1]-c[0][1]) for c in it.combinations(l,2)]
ans=N-max(dict(Counter(d)).values())
print(ans)
``` | output | 1 | 2,270 | 23 | 4,541 |
Provide a correct Python 3 solution for this coding contest problem.
There are N balls in a two-dimensional plane. The i-th ball is at coordinates (x_i, y_i).
We will collect all of these balls, by choosing two integers p and q such that p \neq 0 or q \neq 0 and then repeating the following operation:
* Choose a ball remaining in the plane and collect it. Let (a, b) be the coordinates of this ball. If we collected a ball at coordinates (a - p, b - q) in the previous operation, the cost of this operation is 0. Otherwise, including when this is the first time to do this operation, the cost of this operation is 1.
Find the minimum total cost required to collect all the balls when we optimally choose p and q.
Constraints
* 1 \leq N \leq 50
* |x_i|, |y_i| \leq 10^9
* If i \neq j, x_i \neq x_j or y_i \neq y_j.
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
N
x_1 y_1
:
x_N y_N
Output
Print the minimum total cost required to collect all the balls.
Examples
Input
2
1 1
2 2
Output
1
Input
3
1 4
4 6
7 8
Output
1
Input
4
1 1
1 2
2 1
2 2
Output
2 | instruction | 0 | 2,271 | 23 | 4,542 |
"Correct Solution:
```
N=int(input())
xy=[list(map(int,input().split())) for _ in [0]*N]
pq = {0:0}
for i in range(N):
for j in range(N):
if i==j:continue
p = xy[i][0] - xy[j][0]
q = xy[i][1] - xy[j][1]
pq[(p,q)] = pq.get((p,q),0)+1
print(N - max(pq.values()))
``` | output | 1 | 2,271 | 23 | 4,543 |
Provide a correct Python 3 solution for this coding contest problem.
There are N balls in a two-dimensional plane. The i-th ball is at coordinates (x_i, y_i).
We will collect all of these balls, by choosing two integers p and q such that p \neq 0 or q \neq 0 and then repeating the following operation:
* Choose a ball remaining in the plane and collect it. Let (a, b) be the coordinates of this ball. If we collected a ball at coordinates (a - p, b - q) in the previous operation, the cost of this operation is 0. Otherwise, including when this is the first time to do this operation, the cost of this operation is 1.
Find the minimum total cost required to collect all the balls when we optimally choose p and q.
Constraints
* 1 \leq N \leq 50
* |x_i|, |y_i| \leq 10^9
* If i \neq j, x_i \neq x_j or y_i \neq y_j.
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
N
x_1 y_1
:
x_N y_N
Output
Print the minimum total cost required to collect all the balls.
Examples
Input
2
1 1
2 2
Output
1
Input
3
1 4
4 6
7 8
Output
1
Input
4
1 1
1 2
2 1
2 2
Output
2 | instruction | 0 | 2,273 | 23 | 4,546 |
"Correct Solution:
```
n=int(input())
w={}
d=[]
for i in range(n):
x,y=map(int, input().split())
d.append((x,y))
chk=0
for i in range(n):
a=d[i][0]
b=d[i][1]
for j in range(n):
if i==j: continue
p=d[j][0]
q=d[j][1]
if (a-p,b-q) in w:
w[(a-p,b-q)]+=1
else:
w[(a-p,b-q)]=1
chk=max(chk,w[(a-p,b-q)])
print(n-chk)
``` | output | 1 | 2,273 | 23 | 4,547 |
Provide a correct Python 3 solution for this coding contest problem.
There are N balls in a two-dimensional plane. The i-th ball is at coordinates (x_i, y_i).
We will collect all of these balls, by choosing two integers p and q such that p \neq 0 or q \neq 0 and then repeating the following operation:
* Choose a ball remaining in the plane and collect it. Let (a, b) be the coordinates of this ball. If we collected a ball at coordinates (a - p, b - q) in the previous operation, the cost of this operation is 0. Otherwise, including when this is the first time to do this operation, the cost of this operation is 1.
Find the minimum total cost required to collect all the balls when we optimally choose p and q.
Constraints
* 1 \leq N \leq 50
* |x_i|, |y_i| \leq 10^9
* If i \neq j, x_i \neq x_j or y_i \neq y_j.
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
N
x_1 y_1
:
x_N y_N
Output
Print the minimum total cost required to collect all the balls.
Examples
Input
2
1 1
2 2
Output
1
Input
3
1 4
4 6
7 8
Output
1
Input
4
1 1
1 2
2 1
2 2
Output
2 | instruction | 0 | 2,274 | 23 | 4,548 |
"Correct Solution:
```
from collections import Counter
N = int(input())
xys = []
for _ in range(N):
xys.append(tuple(map(int, input().split())))
sub = []
for x1, y1 in xys:
for x2, y2 in xys:
if x1 != x2 or y1 != y2:
sub.append((x1 - x2, y1 - y2))
if not sub:
print(1)
exit(0)
c = Counter(sub)
m = max(c.values())
print(N - m)
``` | output | 1 | 2,274 | 23 | 4,549 |
Provide a correct Python 3 solution for this coding contest problem.
There are N balls in a two-dimensional plane. The i-th ball is at coordinates (x_i, y_i).
We will collect all of these balls, by choosing two integers p and q such that p \neq 0 or q \neq 0 and then repeating the following operation:
* Choose a ball remaining in the plane and collect it. Let (a, b) be the coordinates of this ball. If we collected a ball at coordinates (a - p, b - q) in the previous operation, the cost of this operation is 0. Otherwise, including when this is the first time to do this operation, the cost of this operation is 1.
Find the minimum total cost required to collect all the balls when we optimally choose p and q.
Constraints
* 1 \leq N \leq 50
* |x_i|, |y_i| \leq 10^9
* If i \neq j, x_i \neq x_j or y_i \neq y_j.
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
N
x_1 y_1
:
x_N y_N
Output
Print the minimum total cost required to collect all the balls.
Examples
Input
2
1 1
2 2
Output
1
Input
3
1 4
4 6
7 8
Output
1
Input
4
1 1
1 2
2 1
2 2
Output
2 | instruction | 0 | 2,275 | 23 | 4,550 |
"Correct Solution:
```
#https://atcoder.jp/contests/diverta2019-2/submissions/11229318
n = int(input())
t = [tuple(map(int, input().split())) for _ in range(n)]
s = set(t)
cnt = 0
for i in range(n-1):
for j in range(i+1,n):
u,v = t[i]
x,y = t[j]
p = u-x; q = v-y
c = sum((x-p, y-q) in s for x,y in t)
if cnt < c: cnt = c
print(n-cnt)
``` | output | 1 | 2,275 | 23 | 4,551 |
Provide a correct Python 3 solution for this coding contest problem.
There are N balls in a two-dimensional plane. The i-th ball is at coordinates (x_i, y_i).
We will collect all of these balls, by choosing two integers p and q such that p \neq 0 or q \neq 0 and then repeating the following operation:
* Choose a ball remaining in the plane and collect it. Let (a, b) be the coordinates of this ball. If we collected a ball at coordinates (a - p, b - q) in the previous operation, the cost of this operation is 0. Otherwise, including when this is the first time to do this operation, the cost of this operation is 1.
Find the minimum total cost required to collect all the balls when we optimally choose p and q.
Constraints
* 1 \leq N \leq 50
* |x_i|, |y_i| \leq 10^9
* If i \neq j, x_i \neq x_j or y_i \neq y_j.
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
N
x_1 y_1
:
x_N y_N
Output
Print the minimum total cost required to collect all the balls.
Examples
Input
2
1 1
2 2
Output
1
Input
3
1 4
4 6
7 8
Output
1
Input
4
1 1
1 2
2 1
2 2
Output
2 | instruction | 0 | 2,276 | 23 | 4,552 |
"Correct Solution:
```
n = int(input())
p = [tuple(map(int,input().split())) for _ in range(n)]
d = {}
m = 0
for i in range(n):
for j in range(n):
if i == j:
continue
fx = p[i][0] - p[j][0]
fy = p[i][1] - p[j][1]
f = (fx, fy)
if f not in d:
d[f] = 0
d[f] += 1
m = max((m, d[f]))
print(n - m)
``` | output | 1 | 2,276 | 23 | 4,553 |
Provide a correct Python 3 solution for this coding contest problem.
There are N balls in a two-dimensional plane. The i-th ball is at coordinates (x_i, y_i).
We will collect all of these balls, by choosing two integers p and q such that p \neq 0 or q \neq 0 and then repeating the following operation:
* Choose a ball remaining in the plane and collect it. Let (a, b) be the coordinates of this ball. If we collected a ball at coordinates (a - p, b - q) in the previous operation, the cost of this operation is 0. Otherwise, including when this is the first time to do this operation, the cost of this operation is 1.
Find the minimum total cost required to collect all the balls when we optimally choose p and q.
Constraints
* 1 \leq N \leq 50
* |x_i|, |y_i| \leq 10^9
* If i \neq j, x_i \neq x_j or y_i \neq y_j.
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
N
x_1 y_1
:
x_N y_N
Output
Print the minimum total cost required to collect all the balls.
Examples
Input
2
1 1
2 2
Output
1
Input
3
1 4
4 6
7 8
Output
1
Input
4
1 1
1 2
2 1
2 2
Output
2 | instruction | 0 | 2,277 | 23 | 4,554 |
"Correct Solution:
```
import itertools
from collections import Counter
N = int(input())
point = [tuple(map(int, input().split())) for _ in range(N)]
ans = Counter()
if N == 1:
print(1)
exit()
for first, second in list(itertools.permutations(point, 2)):
fx, fy = first
sx, sy = second
ans[(sx-fx, sy-fy)] += 1
print(max(1, N - ans.most_common()[0][1]))
``` | output | 1 | 2,277 | 23 | 4,555 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There are N balls in a two-dimensional plane. The i-th ball is at coordinates (x_i, y_i).
We will collect all of these balls, by choosing two integers p and q such that p \neq 0 or q \neq 0 and then repeating the following operation:
* Choose a ball remaining in the plane and collect it. Let (a, b) be the coordinates of this ball. If we collected a ball at coordinates (a - p, b - q) in the previous operation, the cost of this operation is 0. Otherwise, including when this is the first time to do this operation, the cost of this operation is 1.
Find the minimum total cost required to collect all the balls when we optimally choose p and q.
Constraints
* 1 \leq N \leq 50
* |x_i|, |y_i| \leq 10^9
* If i \neq j, x_i \neq x_j or y_i \neq y_j.
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
N
x_1 y_1
:
x_N y_N
Output
Print the minimum total cost required to collect all the balls.
Examples
Input
2
1 1
2 2
Output
1
Input
3
1 4
4 6
7 8
Output
1
Input
4
1 1
1 2
2 1
2 2
Output
2
Submitted Solution:
```
from collections import Counter
from itertools import combinations
N = int(input())
R = [[int(i) for i in input().split()] for _ in range(N)]
if N == 1:
print(1)
quit()
L = []
for (x1, y1), (x2, y2) in combinations(R, 2):
L.append((x1 - x2, y1 - y2))
L.append((-x1 + x2, -y1 + y2))
C = Counter(L)
print(N - C.most_common(1)[0][1])
``` | instruction | 0 | 2,278 | 23 | 4,556 |
Yes | output | 1 | 2,278 | 23 | 4,557 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There are N balls in a two-dimensional plane. The i-th ball is at coordinates (x_i, y_i).
We will collect all of these balls, by choosing two integers p and q such that p \neq 0 or q \neq 0 and then repeating the following operation:
* Choose a ball remaining in the plane and collect it. Let (a, b) be the coordinates of this ball. If we collected a ball at coordinates (a - p, b - q) in the previous operation, the cost of this operation is 0. Otherwise, including when this is the first time to do this operation, the cost of this operation is 1.
Find the minimum total cost required to collect all the balls when we optimally choose p and q.
Constraints
* 1 \leq N \leq 50
* |x_i|, |y_i| \leq 10^9
* If i \neq j, x_i \neq x_j or y_i \neq y_j.
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
N
x_1 y_1
:
x_N y_N
Output
Print the minimum total cost required to collect all the balls.
Examples
Input
2
1 1
2 2
Output
1
Input
3
1 4
4 6
7 8
Output
1
Input
4
1 1
1 2
2 1
2 2
Output
2
Submitted Solution:
```
from collections import Counter
n = int(input())
XY = [list(map(int,input().split())) for _ in range(n)]
dXY = Counter([])
for i in range(n):
for j in range(n):
if i!=j:
dXY += Counter([str(XY[i][0]-XY[j][0])+'_'+str(XY[i][1]-XY[j][1])])
if n==1:
print(1)
else:
print(n-dXY.most_common()[0][1])
``` | instruction | 0 | 2,279 | 23 | 4,558 |
Yes | output | 1 | 2,279 | 23 | 4,559 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There are N balls in a two-dimensional plane. The i-th ball is at coordinates (x_i, y_i).
We will collect all of these balls, by choosing two integers p and q such that p \neq 0 or q \neq 0 and then repeating the following operation:
* Choose a ball remaining in the plane and collect it. Let (a, b) be the coordinates of this ball. If we collected a ball at coordinates (a - p, b - q) in the previous operation, the cost of this operation is 0. Otherwise, including when this is the first time to do this operation, the cost of this operation is 1.
Find the minimum total cost required to collect all the balls when we optimally choose p and q.
Constraints
* 1 \leq N \leq 50
* |x_i|, |y_i| \leq 10^9
* If i \neq j, x_i \neq x_j or y_i \neq y_j.
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
N
x_1 y_1
:
x_N y_N
Output
Print the minimum total cost required to collect all the balls.
Examples
Input
2
1 1
2 2
Output
1
Input
3
1 4
4 6
7 8
Output
1
Input
4
1 1
1 2
2 1
2 2
Output
2
Submitted Solution:
```
n=int(input())
A=[]
B=[]
for i in range(n):
a,b=map(int,input().split())
A.append([a,b])
for i in range(n):
for j in range(i):
B.append([A[i][0]-A[j][0],A[i][1]-A[j][1]])
for j in range(i+1,n):
B.append([A[i][0]-A[j][0],A[i][1]-A[j][1]])
l=0
for i in B:
l=max(l,B.count(i))
print(n-l)
``` | instruction | 0 | 2,280 | 23 | 4,560 |
Yes | output | 1 | 2,280 | 23 | 4,561 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There are N balls in a two-dimensional plane. The i-th ball is at coordinates (x_i, y_i).
We will collect all of these balls, by choosing two integers p and q such that p \neq 0 or q \neq 0 and then repeating the following operation:
* Choose a ball remaining in the plane and collect it. Let (a, b) be the coordinates of this ball. If we collected a ball at coordinates (a - p, b - q) in the previous operation, the cost of this operation is 0. Otherwise, including when this is the first time to do this operation, the cost of this operation is 1.
Find the minimum total cost required to collect all the balls when we optimally choose p and q.
Constraints
* 1 \leq N \leq 50
* |x_i|, |y_i| \leq 10^9
* If i \neq j, x_i \neq x_j or y_i \neq y_j.
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
N
x_1 y_1
:
x_N y_N
Output
Print the minimum total cost required to collect all the balls.
Examples
Input
2
1 1
2 2
Output
1
Input
3
1 4
4 6
7 8
Output
1
Input
4
1 1
1 2
2 1
2 2
Output
2
Submitted Solution:
```
from itertools import combinations
from collections import Counter
N = int(input())
xy = [list(map(int, input().split())) for _ in range(N)]
count = []
for ((x1,x2),(x3,x4)) in combinations(xy, 2):
count.append((x1-x3,x2-x4))
count.append((x3-x1,x4-x2))
c = Counter(count)
m = 0
for i in c.values():
m = max(m, i)
if N == 1:
print(1)
else:
print(N-m)
``` | instruction | 0 | 2,281 | 23 | 4,562 |
Yes | output | 1 | 2,281 | 23 | 4,563 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There are N balls in a two-dimensional plane. The i-th ball is at coordinates (x_i, y_i).
We will collect all of these balls, by choosing two integers p and q such that p \neq 0 or q \neq 0 and then repeating the following operation:
* Choose a ball remaining in the plane and collect it. Let (a, b) be the coordinates of this ball. If we collected a ball at coordinates (a - p, b - q) in the previous operation, the cost of this operation is 0. Otherwise, including when this is the first time to do this operation, the cost of this operation is 1.
Find the minimum total cost required to collect all the balls when we optimally choose p and q.
Constraints
* 1 \leq N \leq 50
* |x_i|, |y_i| \leq 10^9
* If i \neq j, x_i \neq x_j or y_i \neq y_j.
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
N
x_1 y_1
:
x_N y_N
Output
Print the minimum total cost required to collect all the balls.
Examples
Input
2
1 1
2 2
Output
1
Input
3
1 4
4 6
7 8
Output
1
Input
4
1 1
1 2
2 1
2 2
Output
2
Submitted Solution:
```
N = int(input())
cost = 1
xy_set = []
for i in range(N):
x, y = map(int, input().split())
xy_set.append((x, y))
xy_set_sorted = sorted(xy_set, key=lambda a: a[0])
xy_sub_set = {}
for i in range(1, N):
xs = xy_set_sorted[i][0] - xy_set_sorted[i-1][0]
ys = xy_set_sorted[i][1] - xy_set_sorted[i - 1][1]
if xy_sub_set.get((xs, ys)) is None:
xy_sub_set[(xs, ys)] = 1
else:
xy_sub_set[(xs, ys)] += 1
xy_sub_set_sorted = sorted(xy_sub_set.items(), key=lambda a: a[1], reverse=True)
print(cost + (N-1) - xy_sub_set_sorted[0][1])
``` | instruction | 0 | 2,282 | 23 | 4,564 |
No | output | 1 | 2,282 | 23 | 4,565 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There are N balls in a two-dimensional plane. The i-th ball is at coordinates (x_i, y_i).
We will collect all of these balls, by choosing two integers p and q such that p \neq 0 or q \neq 0 and then repeating the following operation:
* Choose a ball remaining in the plane and collect it. Let (a, b) be the coordinates of this ball. If we collected a ball at coordinates (a - p, b - q) in the previous operation, the cost of this operation is 0. Otherwise, including when this is the first time to do this operation, the cost of this operation is 1.
Find the minimum total cost required to collect all the balls when we optimally choose p and q.
Constraints
* 1 \leq N \leq 50
* |x_i|, |y_i| \leq 10^9
* If i \neq j, x_i \neq x_j or y_i \neq y_j.
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
N
x_1 y_1
:
x_N y_N
Output
Print the minimum total cost required to collect all the balls.
Examples
Input
2
1 1
2 2
Output
1
Input
3
1 4
4 6
7 8
Output
1
Input
4
1 1
1 2
2 1
2 2
Output
2
Submitted Solution:
```
N = int(input())
p = []
dic = {}
sameX = {}
allNum = []
for i in range(N):
x,y = map(int,input().split())
p.append([x,y])
allNum.append(i)
for i in range(N):
for j in range(i):
ji = [j,i]
if p[j][0] - p[i][0] != 0:
a = ( p[j][1] - p[i][1] ) / ( p[j][0] - p[i][0] )
b = p[j][1] - ( a * p[j][1] )
if a == -0.0:
a = 0.0
ab = [a,b]
if ( str (ab) ) in dic:
dic[ str (ab) ].append(ji)
else:
dic[ str(ab) ] = [ji]
elif p[j][0] - p[i][0] == 0:
if p[j][0] in sameX:
sameX[ p[j][0] ].append(ji)
else:
sameX[ p[j][0] ] = [ji]
#print (dic,sameX)
DorS = 0
maxi = 0
keys = ""
ans = 0
while len(allNum) > 0:
DorS = 0
maxi = 0
keys = ""
for i in dic:
eable = {}
for j in dic[i]:
if j[0] in allNum:
eable[j[0]] = 1
if j[1] in allNum:
eable[j[1]] = 1
if len(eable) > maxi:
DorS = 0
maxi = len(dic[i])
keys = i
for i in sameX:
eable = {}
for j in sameX[i]:
if j[0] in allNum:
eable[j[0]] = 1
if j[1] in allNum:
eable[j[1]] = 1
if len(eable) > maxi:
DorS = 1
maxi = len(sameX[i])
keys = i
if DorS == 0:
delPs = dic[keys]
del dic[keys]
else:
delPs = sameX[keys]
del sameX[keys]
#print (delPs)
rmFlag = False
for i in delPs:
if i[0] in allNum :
allNum.remove(i[0])
rmFlag = True
if i[1] in allNum :
allNum.remove(i[1])
rmFlag = True
if rmFlag :
ans += 1
#print (allNum)
print (ans)
``` | instruction | 0 | 2,283 | 23 | 4,566 |
No | output | 1 | 2,283 | 23 | 4,567 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There are N balls in a two-dimensional plane. The i-th ball is at coordinates (x_i, y_i).
We will collect all of these balls, by choosing two integers p and q such that p \neq 0 or q \neq 0 and then repeating the following operation:
* Choose a ball remaining in the plane and collect it. Let (a, b) be the coordinates of this ball. If we collected a ball at coordinates (a - p, b - q) in the previous operation, the cost of this operation is 0. Otherwise, including when this is the first time to do this operation, the cost of this operation is 1.
Find the minimum total cost required to collect all the balls when we optimally choose p and q.
Constraints
* 1 \leq N \leq 50
* |x_i|, |y_i| \leq 10^9
* If i \neq j, x_i \neq x_j or y_i \neq y_j.
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
N
x_1 y_1
:
x_N y_N
Output
Print the minimum total cost required to collect all the balls.
Examples
Input
2
1 1
2 2
Output
1
Input
3
1 4
4 6
7 8
Output
1
Input
4
1 1
1 2
2 1
2 2
Output
2
Submitted Solution:
```
n = int(input())
xy = [[int(i) for i in input().split()] for i in range(n)]
#print(n)
print(xy)
#print(xy[0][1])
vec = []
vec2 = []
for i in range(n):
for j in range(i + 1, n):
x = xy[i][0] - xy[j][0]
y = xy[i][1] - xy[j][1]
z = [x, y]
if z not in vec:
vec.append(z)
vec2.append(z)
#print(vec)
#print(vec2)
#vec2 = list(set(vec))
cnt = 0
idx = 0
for i, x in enumerate(vec):
if cnt < vec2.count(x):
cnt = vec2.count(x)
idx = i
#print(cnt)
#print(idx)
pq = vec[idx]
print(pq)
#print(pq[0])
#print(pq[1])
cst = 1
for i in range(1, n):
pq_add_x = xy[i][0] + pq[0]
pq_add_y = xy[i][1] + pq[1]
pq_add = [pq_add_x, pq_add_y]
#pq_sub_x = xy[i][0] - pq[0]
#pq_sub_y = xy[i][1] - pq[1]
#pq_sub = [pq_sub_x, pq_sub_y]
if pq_add not in xy:
#if pq_sub not in xy:
cst += 1
print(cst)
``` | instruction | 0 | 2,284 | 23 | 4,568 |
No | output | 1 | 2,284 | 23 | 4,569 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There are N balls in a two-dimensional plane. The i-th ball is at coordinates (x_i, y_i).
We will collect all of these balls, by choosing two integers p and q such that p \neq 0 or q \neq 0 and then repeating the following operation:
* Choose a ball remaining in the plane and collect it. Let (a, b) be the coordinates of this ball. If we collected a ball at coordinates (a - p, b - q) in the previous operation, the cost of this operation is 0. Otherwise, including when this is the first time to do this operation, the cost of this operation is 1.
Find the minimum total cost required to collect all the balls when we optimally choose p and q.
Constraints
* 1 \leq N \leq 50
* |x_i|, |y_i| \leq 10^9
* If i \neq j, x_i \neq x_j or y_i \neq y_j.
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
N
x_1 y_1
:
x_N y_N
Output
Print the minimum total cost required to collect all the balls.
Examples
Input
2
1 1
2 2
Output
1
Input
3
1 4
4 6
7 8
Output
1
Input
4
1 1
1 2
2 1
2 2
Output
2
Submitted Solution:
```
import logging
import unittest
from collections import (
Counter,
)
debug = logging.getLogger(__name__).debug
def input_ints():
return list(map(int, input().strip().split()))
def _main():
n, = input_ints()
xy = [input_ints() for _ in range(n)]
(_, count), = Counter(
(x1 - x2, y1 - y2)
for (x1, y1) in xy
for (x2, y2) in xy
if not (x1 == x2 and y1 == y2)
).most_common(1)
print(len(xy) - count)
class Test(unittest.TestCase):
def setUp(self):
import run
self._test = run
def test_main(self):
self._test.test_files(self, _main)
if __name__ == '__main__':
_main()
``` | instruction | 0 | 2,285 | 23 | 4,570 |
No | output | 1 | 2,285 | 23 | 4,571 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
We have a rectangular parallelepiped of dimensions A×B×C, divided into 1×1×1 small cubes. The small cubes have coordinates from (0, 0, 0) through (A-1, B-1, C-1).
Let p, q and r be integers. Consider the following set of abc small cubes:
\\{(\ (p + i) mod A, (q + j) mod B, (r + k) mod C\ ) | i, j and k are integers satisfying 0 ≤ i < a, 0 ≤ j < b, 0 ≤ k < c \\}
A set of small cubes that can be expressed in the above format using some integers p, q and r, is called a torus cuboid of size a×b×c.
Find the number of the sets of torus cuboids of size a×b×c that satisfy the following condition, modulo 10^9+7:
* No two torus cuboids in the set have intersection.
* The union of all torus cuboids in the set is the whole rectangular parallelepiped of dimensions A×B×C.
Constraints
* 1 ≤ a < A ≤ 100
* 1 ≤ b < B ≤ 100
* 1 ≤ c < C ≤ 100
* All input values are integers.
Input
Input is given from Standard Input in the following format:
a b c A B C
Output
Print the number of the sets of torus cuboids of size a×b×c that satisfy the condition, modulo 10^9+7.
Examples
Input
1 1 1 2 2 2
Output
1
Input
2 2 2 4 4 4
Output
744
Input
2 3 4 6 7 8
Output
0
Input
2 3 4 98 99 100
Output
471975164
Submitted Solution:
```
print("rng n00b")
``` | instruction | 0 | 2,318 | 23 | 4,636 |
No | output | 1 | 2,318 | 23 | 4,637 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Seryozha conducts a course dedicated to building a map of heights of Stepanovo recreation center. He laid a rectangle grid of size n × m cells on a map (rows of grid are numbered from 1 to n from north to south, and columns are numbered from 1 to m from west to east). After that he measured the average height of each cell above Rybinsk sea level and obtained a matrix of heights of size n × m. The cell (i, j) lies on the intersection of the i-th row and the j-th column and has height h_{i, j}.
Seryozha is going to look at the result of his work in the browser. The screen of Seryozha's laptop can fit a subrectangle of size a × b of matrix of heights (1 ≤ a ≤ n, 1 ≤ b ≤ m). Seryozha tries to decide how the weather can affect the recreation center — for example, if it rains, where all the rainwater will gather. To do so, he is going to find the cell having minimum height among all cells that are shown on the screen of his laptop.
Help Seryozha to calculate the sum of heights of such cells for all possible subrectangles he can see on his screen. In other words, you have to calculate the sum of minimum heights in submatrices of size a × b with top left corners in (i, j) over all 1 ≤ i ≤ n - a + 1 and 1 ≤ j ≤ m - b + 1.
Consider the sequence g_i = (g_{i - 1} ⋅ x + y) mod z. You are given integers g_0, x, y and z. By miraculous coincidence, h_{i, j} = g_{(i - 1) ⋅ m + j - 1} ((i - 1) ⋅ m + j - 1 is the index).
Input
The first line of the input contains four integers n, m, a and b (1 ≤ n, m ≤ 3 000, 1 ≤ a ≤ n, 1 ≤ b ≤ m) — the number of rows and columns in the matrix Seryozha has, and the number of rows and columns that can be shown on the screen of the laptop, respectively.
The second line of the input contains four integers g_0, x, y and z (0 ≤ g_0, x, y < z ≤ 10^9).
Output
Print a single integer — the answer to the problem.
Example
Input
3 4 2 1
1 2 3 59
Output
111
Note
The matrix from the first example:
<image>
Submitted Solution:
```
def mat_mult(A, B, MOD):
n, m, p = len(A), len(A[0]), len(B[0])
assert (len(B) == m)
C = [[0] * p for _ in range(n)]
for i in range(n):
for k in range(m):
Aik = A[i][k]
for j in range(p):
C[i][j] = (C[i][j] + Aik * B[k][j]) % MOD
return C
def ksm(A, n, MOD):
if (n == 0):
E = [[0 for i in range(len(A))] for j in range(len(A))]
for i in range(len(A)):
z[i][i] = 1
return E
if (n == 1):
return A
k = ksm(A, n//2, MOD)
z = mat_mult(k, k, MOD)
if (n&1):
return (mat_mult(z, A, MOD))
else:
return z
n, m, a, b = map(int, input().split())
g, x, y, z = map(int, input().split())
MOD = z
A = [[x,y],[0,1]]
B = [[g],[1]]
AA = ksm(A, m, MOD)
su = 0
l = [g]
for i in range(m-b):
l.append((l[-1]*x+y) % MOD)
su += sum(l)
for i in range(n-a):
l = [(AA[0][0]*i+AA[0][1])%MOD for i in l]
su += sum(l)
print(su)
``` | instruction | 0 | 2,540 | 23 | 5,080 |
No | output | 1 | 2,540 | 23 | 5,081 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Seryozha conducts a course dedicated to building a map of heights of Stepanovo recreation center. He laid a rectangle grid of size n × m cells on a map (rows of grid are numbered from 1 to n from north to south, and columns are numbered from 1 to m from west to east). After that he measured the average height of each cell above Rybinsk sea level and obtained a matrix of heights of size n × m. The cell (i, j) lies on the intersection of the i-th row and the j-th column and has height h_{i, j}.
Seryozha is going to look at the result of his work in the browser. The screen of Seryozha's laptop can fit a subrectangle of size a × b of matrix of heights (1 ≤ a ≤ n, 1 ≤ b ≤ m). Seryozha tries to decide how the weather can affect the recreation center — for example, if it rains, where all the rainwater will gather. To do so, he is going to find the cell having minimum height among all cells that are shown on the screen of his laptop.
Help Seryozha to calculate the sum of heights of such cells for all possible subrectangles he can see on his screen. In other words, you have to calculate the sum of minimum heights in submatrices of size a × b with top left corners in (i, j) over all 1 ≤ i ≤ n - a + 1 and 1 ≤ j ≤ m - b + 1.
Consider the sequence g_i = (g_{i - 1} ⋅ x + y) mod z. You are given integers g_0, x, y and z. By miraculous coincidence, h_{i, j} = g_{(i - 1) ⋅ m + j - 1} ((i - 1) ⋅ m + j - 1 is the index).
Input
The first line of the input contains four integers n, m, a and b (1 ≤ n, m ≤ 3 000, 1 ≤ a ≤ n, 1 ≤ b ≤ m) — the number of rows and columns in the matrix Seryozha has, and the number of rows and columns that can be shown on the screen of the laptop, respectively.
The second line of the input contains four integers g_0, x, y and z (0 ≤ g_0, x, y < z ≤ 10^9).
Output
Print a single integer — the answer to the problem.
Example
Input
3 4 2 1
1 2 3 59
Output
111
Note
The matrix from the first example:
<image>
Submitted Solution:
```
def mat_mult(A, B, MOD):
n, m, p = len(A), len(A[0]), len(B[0])
assert (len(B) == m)
C = [[0] * p for _ in range(n)]
for i in range(n):
for k in range(m):
Aik = A[i][k]
for j in range(p):
C[i][j] = (C[i][j] + Aik * B[k][j]) % MOD
return C
def ksm(A, n, MOD):
if (n == 0):
E = [[0 for i in range(len(A))] for j in range(len(A))]
for i in range(len(A)):
E[i][i] = 1
return E
if (n == 1):
return A
k = ksm(A, n//2, MOD)
z = mat_mult(k, k, MOD)
if (n&1):
return (mat_mult(z, A, MOD))
else:
return z
n, m, a, b = map(int, input().split())
g, x, y, z = map(int, input().split())
MOD = z
A = [[x,y],[0,1]]
B = [[g],[1]]
AA = ksm(A, m, MOD)
su = 0
l = [[g]]
for i in range(m-1):
l[0].append((l[0][-1]*x+y) % MOD)
for i in range(n-1):
l.append([(AA[0][0]*i+AA[0][1])%MOD for i in l[-1]])
for i in range(n-a+1):
for j in range(m-b+1):
p = []
for k in range(a):
p += l[i][j:j+b]
su += min(p)
print(su)
``` | instruction | 0 | 2,541 | 23 | 5,082 |
No | output | 1 | 2,541 | 23 | 5,083 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Seryozha conducts a course dedicated to building a map of heights of Stepanovo recreation center. He laid a rectangle grid of size n × m cells on a map (rows of grid are numbered from 1 to n from north to south, and columns are numbered from 1 to m from west to east). After that he measured the average height of each cell above Rybinsk sea level and obtained a matrix of heights of size n × m. The cell (i, j) lies on the intersection of the i-th row and the j-th column and has height h_{i, j}.
Seryozha is going to look at the result of his work in the browser. The screen of Seryozha's laptop can fit a subrectangle of size a × b of matrix of heights (1 ≤ a ≤ n, 1 ≤ b ≤ m). Seryozha tries to decide how the weather can affect the recreation center — for example, if it rains, where all the rainwater will gather. To do so, he is going to find the cell having minimum height among all cells that are shown on the screen of his laptop.
Help Seryozha to calculate the sum of heights of such cells for all possible subrectangles he can see on his screen. In other words, you have to calculate the sum of minimum heights in submatrices of size a × b with top left corners in (i, j) over all 1 ≤ i ≤ n - a + 1 and 1 ≤ j ≤ m - b + 1.
Consider the sequence g_i = (g_{i - 1} ⋅ x + y) mod z. You are given integers g_0, x, y and z. By miraculous coincidence, h_{i, j} = g_{(i - 1) ⋅ m + j - 1} ((i - 1) ⋅ m + j - 1 is the index).
Input
The first line of the input contains four integers n, m, a and b (1 ≤ n, m ≤ 3 000, 1 ≤ a ≤ n, 1 ≤ b ≤ m) — the number of rows and columns in the matrix Seryozha has, and the number of rows and columns that can be shown on the screen of the laptop, respectively.
The second line of the input contains four integers g_0, x, y and z (0 ≤ g_0, x, y < z ≤ 10^9).
Output
Print a single integer — the answer to the problem.
Example
Input
3 4 2 1
1 2 3 59
Output
111
Note
The matrix from the first example:
<image>
Submitted Solution:
```
u = list(map(int,input().split()))
v = list(map(int,input().split()))
n=u[0]
m=u[1]
h=[[0]*m for _ in range(n) ]
k=[[0]*m for _ in range(n) ]
g=[]
a=u[2]
b=u[3]
g.append(v[0])
x=int(v[1])
y=int(v[2])
z=int(v[3])
s=[]
for i in range(m*n-1):
g.append(int(g[i]*x+y)%z)
s.append(int(0))
for i in range(n) :
l=int(1)
r=int(0)
for j in range(m) :
h[i][j]=g[(i)*m+j]
while ( r >= l and h[i][s[r]] >= h[i][j] ): r-=1
r+=1
s[r]=j
while ( s[l] <= j-b ): l+=1
k[i][j]=h[i][s[l]]
print(k[i][j],end=' ')
print(" ")
dem=int(0)
for j in range(m):
if ( j < b-1 ): continue
l=1
r=0
for i in range(n):
#print(s[r],i,k[s[r]][j],k[i][j])
while ( r >= l and k[s[r]][j] >= k[i][j] ): r-=1
r+=1
s[r]=i
while ( s[l] <= i-a ): l+=1
if ( i > a-2 ): dem+=k[s[l]][j]
print(dem)
``` | instruction | 0 | 2,542 | 23 | 5,084 |
No | output | 1 | 2,542 | 23 | 5,085 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Seryozha conducts a course dedicated to building a map of heights of Stepanovo recreation center. He laid a rectangle grid of size n × m cells on a map (rows of grid are numbered from 1 to n from north to south, and columns are numbered from 1 to m from west to east). After that he measured the average height of each cell above Rybinsk sea level and obtained a matrix of heights of size n × m. The cell (i, j) lies on the intersection of the i-th row and the j-th column and has height h_{i, j}.
Seryozha is going to look at the result of his work in the browser. The screen of Seryozha's laptop can fit a subrectangle of size a × b of matrix of heights (1 ≤ a ≤ n, 1 ≤ b ≤ m). Seryozha tries to decide how the weather can affect the recreation center — for example, if it rains, where all the rainwater will gather. To do so, he is going to find the cell having minimum height among all cells that are shown on the screen of his laptop.
Help Seryozha to calculate the sum of heights of such cells for all possible subrectangles he can see on his screen. In other words, you have to calculate the sum of minimum heights in submatrices of size a × b with top left corners in (i, j) over all 1 ≤ i ≤ n - a + 1 and 1 ≤ j ≤ m - b + 1.
Consider the sequence g_i = (g_{i - 1} ⋅ x + y) mod z. You are given integers g_0, x, y and z. By miraculous coincidence, h_{i, j} = g_{(i - 1) ⋅ m + j - 1} ((i - 1) ⋅ m + j - 1 is the index).
Input
The first line of the input contains four integers n, m, a and b (1 ≤ n, m ≤ 3 000, 1 ≤ a ≤ n, 1 ≤ b ≤ m) — the number of rows and columns in the matrix Seryozha has, and the number of rows and columns that can be shown on the screen of the laptop, respectively.
The second line of the input contains four integers g_0, x, y and z (0 ≤ g_0, x, y < z ≤ 10^9).
Output
Print a single integer — the answer to the problem.
Example
Input
3 4 2 1
1 2 3 59
Output
111
Note
The matrix from the first example:
<image>
Submitted Solution:
```
from sys import getsizeof
class deque:
def __init__(self):
self.items = []
self.l = 0
self.r = -1
def clear(self):
self.l = 0
self.r = -1
self.items.clear()
def nemp(self):
return self.l <= self.r
def add(self, x):
self.items.append(x)
self.r += 1
def pop(self):
self.r -= 1
def popleft(self):
self.l += 1
def left(self):
return self.items[self.l]
def right(self):
return self.items[self.r]
n, m, a, b = map(int, input().split())
g0, x, y, z = map(int, input().split())
h = [[0]*m for _ in range(n)]
h2 = [[0]*m for _ in range(n)]
for i in range(n):
for j in range(m):
h[i][j], g0 = g0, (g0*x+y)%z
deq = deque()
for i in range(n):
deq.clear()
for j in range(m):
while deq.nemp() and j-deq.left()+1 > b: deq.popleft()
while deq.nemp() and h[i][j] <= h[i][deq.right()]: deq.pop()
deq.add(j)
h2[i][j] = h[i][deq.left()]
ans = 0
for j in range(m):
deq.clear()
for i in range(n):
while deq.nemp() and i-deq.left()+1 > a: deq.popleft()
while deq.nemp() and h2[i][j] <= h2[deq.right()][j]: deq.pop()
deq.add(i)
if i>a-2 and j>b-2: ans += h2[deq.left()][j]
print(ans)
#print('size = ', getsizeof(h)+getsizeof(h2)+getsizeof(deq))
'''
1000 1000 100 100
1 2 3 59
'''
``` | instruction | 0 | 2,543 | 23 | 5,086 |
No | output | 1 | 2,543 | 23 | 5,087 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given n integers. You need to choose a subset and put the chosen numbers in a beautiful rectangle (rectangular matrix). Each chosen number should occupy one of its rectangle cells, each cell must be filled with exactly one chosen number. Some of the n numbers may not be chosen.
A rectangle (rectangular matrix) is called beautiful if in each row and in each column all values are different.
What is the largest (by the total number of cells) beautiful rectangle you can construct? Print the rectangle itself.
Input
The first line contains n (1 ≤ n ≤ 4⋅10^5). The second line contains n integers (1 ≤ a_i ≤ 10^9).
Output
In the first line print x (1 ≤ x ≤ n) — the total number of cells of the required maximum beautiful rectangle. In the second line print p and q (p ⋅ q=x): its sizes. In the next p lines print the required rectangle itself. If there are several answers, print any.
Examples
Input
12
3 1 4 1 5 9 2 6 5 3 5 8
Output
12
3 4
1 2 3 5
3 1 5 4
5 6 8 9
Input
5
1 1 1 1 1
Output
1
1 1
1 | instruction | 0 | 2,592 | 23 | 5,184 |
Tags: brute force, combinatorics, constructive algorithms, data structures, greedy, math
Correct Solution:
```
from collections import Counter
from sys import stdin
def input():
return next(stdin)[:-1]
def main():
n = int(input())
aa = input().split()
cc = list(Counter(aa).most_common())
if n % cc[0][1] == 0 and cc[0][1] * cc[0][1] <= n:
h = cc[0][1]
w = n//cc[0][1]
best = n
else:
count_count = [0] * (n+1)
for v, c in cc:
count_count[c] += 1
geq = [count_count[n]]
for v in reversed(count_count[:n]):
geq.append(geq[-1] + v)
geq.reverse()
tot = 0
best = 0
for a in range(1,n+1):
tot += geq[a]
b = tot//a
if a <= b and best < a * b:
best = a * b
h = a
w = b
print(best)
print(h,w)
x = 0
y = 0
mat = [[''] * w for _ in range(h) ]
for v, c in cc:
for j in range(min(c, h)):
if mat[x][y] != '':
x = (x+1)%h
if mat[x][y] == '':
mat[x][y] = v
x = (x+1)%h
y = (y+1)%w
for i in range(h):
print(' '.join(mat[i]))
if __name__ == "__main__":
main()
``` | output | 1 | 2,592 | 23 | 5,185 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given n integers. You need to choose a subset and put the chosen numbers in a beautiful rectangle (rectangular matrix). Each chosen number should occupy one of its rectangle cells, each cell must be filled with exactly one chosen number. Some of the n numbers may not be chosen.
A rectangle (rectangular matrix) is called beautiful if in each row and in each column all values are different.
What is the largest (by the total number of cells) beautiful rectangle you can construct? Print the rectangle itself.
Input
The first line contains n (1 ≤ n ≤ 4⋅10^5). The second line contains n integers (1 ≤ a_i ≤ 10^9).
Output
In the first line print x (1 ≤ x ≤ n) — the total number of cells of the required maximum beautiful rectangle. In the second line print p and q (p ⋅ q=x): its sizes. In the next p lines print the required rectangle itself. If there are several answers, print any.
Examples
Input
12
3 1 4 1 5 9 2 6 5 3 5 8
Output
12
3 4
1 2 3 5
3 1 5 4
5 6 8 9
Input
5
1 1 1 1 1
Output
1
1 1
1 | instruction | 0 | 2,593 | 23 | 5,186 |
Tags: brute force, combinatorics, constructive algorithms, data structures, greedy, math
Correct Solution:
```
from collections import Counter
from itertools import accumulate
from math import sqrt
from operator import itemgetter
import sys
n = int(input())
cnt = Counter(map(int, input().split()))
nums, counts = zip(*sorted(cnt.items(), key=itemgetter(1)))
acc = [0] + list(accumulate(counts))
area = 1
h, w = 1, 1
i = len(counts)
for y in range(int(sqrt(n)), 0, -1):
while i and counts[i-1] > y:
i -= 1
total = acc[i] + (len(counts) - i) * y
x = total // y
if y <= x and area < x * y:
h, w, area = y, x, x*y
ans = [[0]*w for _ in range(h)]
i = len(counts)-1
num, count = nums[i], min(h, counts[i])
for x in range(w):
for y in range(h):
ans[y][(x + y) % w] = num
count -= 1
if count == 0:
i -= 1
num, count = nums[i], h if h < counts[i] else counts[i]
print(area)
print(h, w)
for y in range(h):
sys.stdout.write(' '.join(map(str, ans[y])) + '\n')
``` | output | 1 | 2,593 | 23 | 5,187 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given n integers. You need to choose a subset and put the chosen numbers in a beautiful rectangle (rectangular matrix). Each chosen number should occupy one of its rectangle cells, each cell must be filled with exactly one chosen number. Some of the n numbers may not be chosen.
A rectangle (rectangular matrix) is called beautiful if in each row and in each column all values are different.
What is the largest (by the total number of cells) beautiful rectangle you can construct? Print the rectangle itself.
Input
The first line contains n (1 ≤ n ≤ 4⋅10^5). The second line contains n integers (1 ≤ a_i ≤ 10^9).
Output
In the first line print x (1 ≤ x ≤ n) — the total number of cells of the required maximum beautiful rectangle. In the second line print p and q (p ⋅ q=x): its sizes. In the next p lines print the required rectangle itself. If there are several answers, print any.
Examples
Input
12
3 1 4 1 5 9 2 6 5 3 5 8
Output
12
3 4
1 2 3 5
3 1 5 4
5 6 8 9
Input
5
1 1 1 1 1
Output
1
1 1
1 | instruction | 0 | 2,594 | 23 | 5,188 |
Tags: brute force, combinatorics, constructive algorithms, data structures, greedy, math
Correct Solution:
```
n = int(input())
arr = list(map(int, input().split()))
d = {}
for i in arr:
d[i] = d.get(i, 0) + 1
d2 = {}
for k, v in d.items():
d2.setdefault(v, []).append(k)
s = n
prev = 0
ansp = ansq = anss = 0
for p in range(n, 0, -1):
q = s // p
if p <= q and q * p > anss:
anss = q * p
ansq = q
ansp = p
prev += len(d2.get(p, []))
s -= prev
def get_ans():
cur_i = 0
cur_j = 0
cur = 0
for k, v in d3:
for val in v:
f = min(k, anss - cur, ansp)
cur += f
for i in range(f):
cur_i = (cur_i + 1) % ansp
cur_j = (cur_j + 1) % ansq
if ans[cur_i][cur_j]:
cur_i = (cur_i + 1) % ansp
ans[cur_i][cur_j] = val
print(anss)
print(ansp, ansq)
d3 = sorted(d2.items(), reverse=True)
ans = [[0] * ansq for i in range(ansp)]
get_ans()
for i in range(ansp):
print(*ans[i])
``` | output | 1 | 2,594 | 23 | 5,189 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given n integers. You need to choose a subset and put the chosen numbers in a beautiful rectangle (rectangular matrix). Each chosen number should occupy one of its rectangle cells, each cell must be filled with exactly one chosen number. Some of the n numbers may not be chosen.
A rectangle (rectangular matrix) is called beautiful if in each row and in each column all values are different.
What is the largest (by the total number of cells) beautiful rectangle you can construct? Print the rectangle itself.
Input
The first line contains n (1 ≤ n ≤ 4⋅10^5). The second line contains n integers (1 ≤ a_i ≤ 10^9).
Output
In the first line print x (1 ≤ x ≤ n) — the total number of cells of the required maximum beautiful rectangle. In the second line print p and q (p ⋅ q=x): its sizes. In the next p lines print the required rectangle itself. If there are several answers, print any.
Examples
Input
12
3 1 4 1 5 9 2 6 5 3 5 8
Output
12
3 4
1 2 3 5
3 1 5 4
5 6 8 9
Input
5
1 1 1 1 1
Output
1
1 1
1 | instruction | 0 | 2,596 | 23 | 5,192 |
Tags: brute force, combinatorics, constructive algorithms, data structures, greedy, math
Correct Solution:
```
n = int(input())
l = list(map(int,input().split()))
if n == 1:
print(1)
print(1,1)
print(l[0])
else:
d = {}
for i in l:
d[i] = 0
for i in l:
d[i] += 1
equal = [0] * (n + 1)
for i in d:
equal[d[i]] += 1
atmost = [0] * (n + 1)
atmost[0] = equal[0]
for i in range(1, n+1):
atmost[i] = atmost[i-1] + equal[i]
sumka = 0
best_iloczyn = 0
best_a = 0
best_b = 0
for a in range(1, n):
if a**2 > n:
break
sumka += (len(d) - atmost[a-1])
b_cand = sumka//a
if b_cand < a:
continue
if a * b_cand > best_iloczyn:
best_iloczyn = a * b_cand
best_a = a
best_b = b_cand
print(best_iloczyn)
print(best_a, best_b)
li = []
for i in d:
if d[i] >= best_a:
li += [i]*min(best_a, d[i])
for i in d:
if d[i] < best_a:
li += [i]*min(best_a, d[i])
#print(li)
mat = [[0] * best_b for i in range(best_a)]
for dd in range(1, best_a + 1):
if best_a%dd==0 and best_b%dd==0:
du = dd
i = 0
for st in range(du):
for j in range(best_iloczyn//du):
mat[i%best_a][(st+i)%best_b] = li[i]
i += 1
for i in range(best_a):
print(*mat[i])
``` | output | 1 | 2,596 | 23 | 5,193 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given n integers. You need to choose a subset and put the chosen numbers in a beautiful rectangle (rectangular matrix). Each chosen number should occupy one of its rectangle cells, each cell must be filled with exactly one chosen number. Some of the n numbers may not be chosen.
A rectangle (rectangular matrix) is called beautiful if in each row and in each column all values are different.
What is the largest (by the total number of cells) beautiful rectangle you can construct? Print the rectangle itself.
Input
The first line contains n (1 ≤ n ≤ 4⋅10^5). The second line contains n integers (1 ≤ a_i ≤ 10^9).
Output
In the first line print x (1 ≤ x ≤ n) — the total number of cells of the required maximum beautiful rectangle. In the second line print p and q (p ⋅ q=x): its sizes. In the next p lines print the required rectangle itself. If there are several answers, print any.
Examples
Input
12
3 1 4 1 5 9 2 6 5 3 5 8
Output
12
3 4
1 2 3 5
3 1 5 4
5 6 8 9
Input
5
1 1 1 1 1
Output
1
1 1
1
Submitted Solution:
```
n = int(input())
l = list(map(int,input().split()))
if n == 1:
print(1)
print(1,1)
print(l[0])
else:
d = {}
for i in l:
d[i] = 0
for i in l:
d[i] += 1
equal = [0] * (n + 1)
for i in d:
equal[d[i]] += 1
atmost = [0] * (n + 1)
atmost[0] = equal[0]
for i in range(1, n+1):
atmost[i] = atmost[i-1] + equal[i]
sumka = 0
best_iloczyn = 0
best_a = 0
best_b = 0
for a in range(1, n):
if a**2 > n:
break
sumka += (len(d) - atmost[a-1])
b_cand = sumka//a
if b_cand < a:
continue
if a * b_cand > best_iloczyn:
best_iloczyn = a * b_cand
best_a = a
best_b = b_cand
print(best_iloczyn)
print(best_a, best_b)
li = []
for i in d:
if d[i] == best_a:
li += [i]*min(best_a, d[i])
for i in d:
if d[i] != best_a:
li += [i]*min(best_a, d[i])
mat = [[0] * best_b for i in range(best_a)]
for dd in range(1, best_a + 1):
if best_a%dd==0 and best_b%dd==0:
du = dd
i = 0
for st in range(du):
for j in range(best_iloczyn//du):
mat[i%best_a][(st+i)%best_b] = li[i]
i += 1
for i in range(best_a):
print(*mat[i])
``` | instruction | 0 | 2,597 | 23 | 5,194 |
No | output | 1 | 2,597 | 23 | 5,195 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given n integers. You need to choose a subset and put the chosen numbers in a beautiful rectangle (rectangular matrix). Each chosen number should occupy one of its rectangle cells, each cell must be filled with exactly one chosen number. Some of the n numbers may not be chosen.
A rectangle (rectangular matrix) is called beautiful if in each row and in each column all values are different.
What is the largest (by the total number of cells) beautiful rectangle you can construct? Print the rectangle itself.
Input
The first line contains n (1 ≤ n ≤ 4⋅10^5). The second line contains n integers (1 ≤ a_i ≤ 10^9).
Output
In the first line print x (1 ≤ x ≤ n) — the total number of cells of the required maximum beautiful rectangle. In the second line print p and q (p ⋅ q=x): its sizes. In the next p lines print the required rectangle itself. If there are several answers, print any.
Examples
Input
12
3 1 4 1 5 9 2 6 5 3 5 8
Output
12
3 4
1 2 3 5
3 1 5 4
5 6 8 9
Input
5
1 1 1 1 1
Output
1
1 1
1
Submitted Solution:
```
from collections import Counter
from math import sqrt
def main():
n = int(input())
aa = map(int,input().split())
cc = list(Counter(aa).most_common())
val_by_count = [[] for _ in range(n+1)]
for v, c in cc:
val_by_count[c].append(v)
geq = [len(val_by_count[n])]
for v in reversed(val_by_count[:n]):
geq.append(geq[-1] + len(v))
geq.reverse()
tot = 0
best = 0
for a in range(1,n+1):
tot += geq[a]
b = tot//a
if a <= b and best < a * b:
best = a * b
h = a
w = b
mat = [[0] * w for _ in range(h) ]
i = 0
j = 0
for v, c in cc:
for k in range(c):
if i+k < h:
mat[i+k][(j+k) % w ] = v
while i < h and mat[i][j] != 0:
j+=1
if j == w:
i+=1
j=0
print(h*w)
print(h, w)
for i in range(h):
print(' '.join(map(str, mat[i])))
if __name__ == "__main__":
main()
``` | instruction | 0 | 2,598 | 23 | 5,196 |
No | output | 1 | 2,598 | 23 | 5,197 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given n integers. You need to choose a subset and put the chosen numbers in a beautiful rectangle (rectangular matrix). Each chosen number should occupy one of its rectangle cells, each cell must be filled with exactly one chosen number. Some of the n numbers may not be chosen.
A rectangle (rectangular matrix) is called beautiful if in each row and in each column all values are different.
What is the largest (by the total number of cells) beautiful rectangle you can construct? Print the rectangle itself.
Input
The first line contains n (1 ≤ n ≤ 4⋅10^5). The second line contains n integers (1 ≤ a_i ≤ 10^9).
Output
In the first line print x (1 ≤ x ≤ n) — the total number of cells of the required maximum beautiful rectangle. In the second line print p and q (p ⋅ q=x): its sizes. In the next p lines print the required rectangle itself. If there are several answers, print any.
Examples
Input
12
3 1 4 1 5 9 2 6 5 3 5 8
Output
12
3 4
1 2 3 5
3 1 5 4
5 6 8 9
Input
5
1 1 1 1 1
Output
1
1 1
1
Submitted Solution:
```
n = int(input())
l = list(map(int,input().split()))
if n == 1:
print(1)
print(1,1)
print(l[0])
else:
d = {}
for i in l:
d[i] = 0
for i in l:
d[i] += 1
equal = [0] * (n + 1)
for i in d:
equal[d[i]] += 1
atmost = [0] * (n + 1)
atmost[0] = equal[0]
for i in range(1, n+1):
atmost[i] = atmost[i-1] + equal[i]
sumka = 0
best_iloczyn = 0
best_a = 0
best_b = 0
for a in range(1, n):
if a**2 > n:
break
sumka += (len(d) - atmost[a-1])
b_cand = sumka//a
if b_cand < a:
continue
if a * b_cand > best_iloczyn:
best_iloczyn = a * b_cand
best_a = a
best_b = b_cand
print(best_iloczyn)
print(best_a, best_b)
li = []
for i in d:
li += [i]*min(best_a, d[i])
mat = [[0] * best_b for i in range(best_a)]
for dd in range(1, best_a + 1):
if best_a%dd==0 and best_b%dd==0:
du = dd
i = 0
for st in range(du):
for j in range(best_iloczyn//du):
mat[i%best_a][(st+i)%best_b] = li[i]
i += 1
for i in range(best_a):
print(*mat[i])
``` | instruction | 0 | 2,599 | 23 | 5,198 |
No | output | 1 | 2,599 | 23 | 5,199 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given n integers. You need to choose a subset and put the chosen numbers in a beautiful rectangle (rectangular matrix). Each chosen number should occupy one of its rectangle cells, each cell must be filled with exactly one chosen number. Some of the n numbers may not be chosen.
A rectangle (rectangular matrix) is called beautiful if in each row and in each column all values are different.
What is the largest (by the total number of cells) beautiful rectangle you can construct? Print the rectangle itself.
Input
The first line contains n (1 ≤ n ≤ 4⋅10^5). The second line contains n integers (1 ≤ a_i ≤ 10^9).
Output
In the first line print x (1 ≤ x ≤ n) — the total number of cells of the required maximum beautiful rectangle. In the second line print p and q (p ⋅ q=x): its sizes. In the next p lines print the required rectangle itself. If there are several answers, print any.
Examples
Input
12
3 1 4 1 5 9 2 6 5 3 5 8
Output
12
3 4
1 2 3 5
3 1 5 4
5 6 8 9
Input
5
1 1 1 1 1
Output
1
1 1
1
Submitted Solution:
```
def play(arr):
n = len(arr)
number2Count = {}
for p in arr:
number2Count[p] = number2Count.get(p, 0) + 1
count2Numbers = {}
for num in number2Count:
cnt = number2Count[num]
if not cnt in count2Numbers:
count2Numbers[cnt] = []
count2Numbers[cnt].append(num)
numRepeats = [0] * (n + 1)
numRepeats[n] = len(count2Numbers.get(n, []))
for i in range(n - 1, 0, -1):
numRepeats[i] = numRepeats[i + 1] + len(count2Numbers.get(i, []))
a_ideal = 0
b_ideal = 0
square = 0
square_ideal = 0
for a in range(1, n + 1):
square += numRepeats[a]
b = int(square / a)
if a <= b:
if square_ideal < a * b:
square_ideal = a * b
a_ideal = a
b_ideal = b
print(a_ideal * b_ideal)
print(str(a_ideal) + ' ' + str(b_ideal))
matrix = [[0] * b_ideal for p in range(0, a_ideal)]
x = 0
y = 0
for cnt in count2Numbers:
for num in count2Numbers[cnt]:
for i in range(0, min(cnt, a_ideal)):
if matrix[x][y] > 0:
x = (x + 1) % a_ideal
if matrix[x][y] == 0:
matrix[x][y] = num
x = (x + 1) % a_ideal
y = (y + 1) % b_ideal
for i in range(0, a_ideal):
print(*matrix[i])
def main():
input()
arr = list(map(int, input().split()))
play(arr)
main()
#print(play([3, 1, 4, 1, 5, 9, 2, 6, 5, 3, 5, 8]))
# play(['010', '101', '0'])
# play(['00000', '00001'])
# play(['01', '001', '0001', '00001'])
``` | instruction | 0 | 2,600 | 23 | 5,200 |
No | output | 1 | 2,600 | 23 | 5,201 |
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 on a plane.
Please find the minimum sum of areas of two axis-aligned rectangles, such that each point is contained in at least one of these rectangles.
Note that the chosen rectangles can be degenerate. Rectangle contains all the points that lie inside it or on its boundary.
Input
The first line contains one integer t (1 ≤ t ≤ 2 ⋅ 10^5) — the number of test cases.
The first line of each test case contains a single integer n (1 ≤ n ≤ 2 ⋅ 10^5) — the number of points.
The following n lines contain the coordinates of the points x_i and y_i (0 ≤ x_i, y_i ≤ 10^9). It is guaranteed that the points are distinct.
It is guaranteed that the sum of values n over all test cases does not exceed 2⋅10^5.
Output
For each test case print one integer — the minimum sum of areas.
Example
Input
3
2
9 1
1 6
2
8 10
0 7
4
0 0
1 1
9 9
10 10
Output
0
0
2
Note
In the first two test cases the answer consists of 2 degenerate rectangles. In the third test case one of the possible answers consists of two rectangles 1 × 1 with bottom left corners (0,0) and (9,9).
Submitted Solution:
```
iterations=int(input('insert the number of tests that you want to run:'))
allpoints=allp=set()
puntiInterni = pI = set()
puntiEsterni = pE = set()
toPrint=''
for iteration in range(0,iterations):
pI = set()
pE = set()
allp=set()
setx=set()
sety=set()
numberOfPoints=np=int(input('insert the number of poins that you want to test :'))
for i in range(np):
point=p=input(f'insert point {i+1} :')
pointaslist=pl=p.split(' ')
setx.add(int(pl[0]))
sety.add(int(pl[1]))
allp.add((int(pl[0]),int(pl[1])))
maxx=max(setx)
minx=min(setx)
maxy=max(sety)
miny=min(sety)
confrontoIniziale=c1=(maxx-minx)*(maxy-miny)
vincitore=c1
for p in allp:
if p[0]<maxx and p[0]>minx and p[1]<maxy and p[1]>miny:
pI.add(p)
else:
pE.add(p)
if len(allp)<=2:
vincitore=0
else:
for pe in pE:
xset=set()
yset=set()
xset.add(pe[0])
yset.add(pe[1])
tallp0=allp.copy()
tallp0.discard(pe)
newarea = A =(max(tallp0, key=lambda x: x[1])[1] - min(tallp0, key=lambda x: x[1])[1]) * (max(tallp0, key=lambda x: x[0])[0] - min(tallp0, key=lambda x: x[0])[0])
if newarea<vincitore:
vincitore=newarea
for p1 in tallp0:
tallp1=tallp0.copy()
tallp1.discard(p1)
xset1 = xset.copy()
yset1 = yset.copy()
xset1.add(p1[0])
yset1.add(p1[1])
newarea=A=(max(xset1)-min(xset1))*(max(yset1)-min(yset1))+(max(tallp1,key= lambda x: x[1])[1]-min(tallp1,key= lambda x: x[1])[1])*(max(tallp1,key= lambda x: x[0])[0]-min(tallp1,key= lambda x: x[0])[0])
if newarea<vincitore:
vincitore=newarea
if len(allp)>3:
for p2 in tallp1:
tallp2 = tallp1.copy()
tallp2.discard(p2)
xset2 = xset1.copy()
yset2 = yset1.copy()
xset2.add(p2[0])
yset2.add(p2[1])
tmaxx=max(xset2)
tminx=min(xset2)
tmaxy=max(yset2)
tminy=min(yset2)
newarea=(tmaxx-tminx)*(tmaxy-tminy)+(max(tallp2,key= lambda x: x[1])[1]-min(tallp2,key= lambda x: x[1])[1])*(max(tallp2,key= lambda x: x[0])[0]-min(tallp2,key= lambda x: x[0])[0])
if newarea<vincitore:
vincitore=newarea
for p3 in tallp2:
tallp3 = tallp2.copy()
tallp2.discard(p2)
xset3 = xset2.copy()
yset3 = yset2.copy()
xset3.add(p3[0])
yset3.add(p3[1])
tmaxx=max(xset3)
tminx=min(xset3)
tmaxy=max(yset3)
tminy=min(yset3)
newarea=(tmaxx-tminx)*(tmaxy-tminy)+(max(tallp3,key= lambda x: x[1])[1]-min(tallp3,key= lambda x: x[1])[1])*(max(tallp3,key= lambda x: x[0])[0]-min(tallp3,key= lambda x: x[0])[0])
if newarea < vincitore:
vincitore = newarea
else: pass
toPrint += str(vincitore)
if iteration != (iterations - 1):
toPrint +='\n'
print(toPrint)
``` | instruction | 0 | 2,651 | 23 | 5,302 |
No | output | 1 | 2,651 | 23 | 5,303 |
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 on a plane.
Please find the minimum sum of areas of two axis-aligned rectangles, such that each point is contained in at least one of these rectangles.
Note that the chosen rectangles can be degenerate. Rectangle contains all the points that lie inside it or on its boundary.
Input
The first line contains one integer t (1 ≤ t ≤ 2 ⋅ 10^5) — the number of test cases.
The first line of each test case contains a single integer n (1 ≤ n ≤ 2 ⋅ 10^5) — the number of points.
The following n lines contain the coordinates of the points x_i and y_i (0 ≤ x_i, y_i ≤ 10^9). It is guaranteed that the points are distinct.
It is guaranteed that the sum of values n over all test cases does not exceed 2⋅10^5.
Output
For each test case print one integer — the minimum sum of areas.
Example
Input
3
2
9 1
1 6
2
8 10
0 7
4
0 0
1 1
9 9
10 10
Output
0
0
2
Note
In the first two test cases the answer consists of 2 degenerate rectangles. In the third test case one of the possible answers consists of two rectangles 1 × 1 with bottom left corners (0,0) and (9,9).
Submitted Solution:
```
def StrangeCovering():
iterations=int(input('insert the number of tests that you want to run:'))
allpoints=allp=set()
puntiInterni = pI = set()
puntiEsterni = pE = set()
for iteration in range(0,iterations):
pI = set()
pE = set()
allp=set()
setx=set()
sety=set()
numberOfPoints=np=int(input('insert the number of poins that you want to test :'))
for i in range(np):
point=p=input(f'insert point {i+1} :')
pointaslist=pl=p.split(' ')
setx.add(int(pl[0]))
sety.add(int(pl[1]))
allp.add((int(pl[0]),int(pl[1])))
maxx=max(setx)
minx=min(setx)
maxy=max(sety)
miny=min(sety)
confrontoIniziale=c1=(maxx-minx)*(maxy-miny)
vincitore=c1
for p in allp:
if p[0]<maxx and p[0]>minx and p[1]<maxy and p[1]>miny:
pI.add(p)
else:
pE.add(p)
if len(allp)<=2:
vincitore=0
else:
for pe in pE:
xset=set()
yset=set()
xset.add(pe[0])
yset.add(pe[1])
tallp0=allp.copy()
tallp0.discard(pe)
newarea = A =(max(tallp0, key=lambda x: x[1])[1] - min(tallp0, key=lambda x: x[1])[1]) * (max(tallp0, key=lambda x: x[0])[0] - min(tallp0, key=lambda x: x[0])[0])
if newarea<vincitore:
vincitore=newarea
for p1 in tallp0:
tallp1=tallp0.copy()
tallp1.discard(p1)
xset1 = xset.copy()
yset1 = yset.copy()
xset1.add(p1[0])
yset1.add(p1[1])
newarea=A=(max(xset1)-min(xset1))*(max(yset1)-min(yset1))+(max(tallp1,key= lambda x: x[1])[1]-min(tallp1,key= lambda x: x[1])[1])*(max(tallp1,key= lambda x: x[0])[0]-min(tallp1,key= lambda x: x[0])[0])
if newarea<vincitore:
vincitore=newarea
if len(allp)>3:
for p2 in tallp1:
tallp2 = tallp1.copy()
tallp2.discard(p2)
xset2 = xset1.copy()
yset2 = yset1.copy()
xset2.add(p2[0])
yset2.add(p2[1])
tmaxx=max(xset2)
tminx=min(xset2)
tmaxy=max(yset2)
tminy=min(yset2)
newarea=(tmaxx-tminx)*(tmaxy-tminy)+(max(tallp2,key= lambda x: x[1])[1]-min(tallp2,key= lambda x: x[1])[1])*(max(tallp2,key= lambda x: x[0])[0]-min(tallp2,key= lambda x: x[0])[0])
if newarea<vincitore:
vincitore=newarea
for p3 in tallp2:
tallp3 = tallp2.copy()
tallp2.discard(p2)
xset3 = xset2.copy()
yset3 = yset2.copy()
xset3.add(p3[0])
yset3.add(p3[1])
tmaxx=max(xset3)
tminx=min(xset3)
tmaxy=max(yset3)
tminy=min(yset3)
newarea=(tmaxx-tminx)*(tmaxy-tminy)+(max(tallp3,key= lambda x: x[1])[1]-min(tallp3,key= lambda x: x[1])[1])*(max(tallp3,key= lambda x: x[0])[0]-min(tallp3,key= lambda x: x[0])[0])
if newarea < vincitore:
vincitore = newarea
else: pass
print(vincitore)
StrangeCovering()
``` | instruction | 0 | 2,652 | 23 | 5,304 |
No | output | 1 | 2,652 | 23 | 5,305 |
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 on a plane.
Please find the minimum sum of areas of two axis-aligned rectangles, such that each point is contained in at least one of these rectangles.
Note that the chosen rectangles can be degenerate. Rectangle contains all the points that lie inside it or on its boundary.
Input
The first line contains one integer t (1 ≤ t ≤ 2 ⋅ 10^5) — the number of test cases.
The first line of each test case contains a single integer n (1 ≤ n ≤ 2 ⋅ 10^5) — the number of points.
The following n lines contain the coordinates of the points x_i and y_i (0 ≤ x_i, y_i ≤ 10^9). It is guaranteed that the points are distinct.
It is guaranteed that the sum of values n over all test cases does not exceed 2⋅10^5.
Output
For each test case print one integer — the minimum sum of areas.
Example
Input
3
2
9 1
1 6
2
8 10
0 7
4
0 0
1 1
9 9
10 10
Output
0
0
2
Note
In the first two test cases the answer consists of 2 degenerate rectangles. In the third test case one of the possible answers consists of two rectangles 1 × 1 with bottom left corners (0,0) and (9,9).
Submitted Solution:
```
for _ in range(int(input())):
n = int(input())
a = []
for i in range(n):
x,y = input().split()
a.append((int(x),int(y)))
xa = sorted(a,key=lambda x:x[0])
ya = sorted(a,key=lambda x:x[1])
dp_asc_x = [((-1,-1),(-1,-1),-1) for i in range(n)]
dp_asc_y = [((-1,-1),(-1,-1),-1) for i in range(n)]
dp_desc_x = [((-1,-1),(-1,-1),-1) for i in range(n)]
dp_desc_y = [((-1,-1),(-1,-1),-1) for i in range(n)]
#print(xa)
dp_asc_x[0] = (xa[0],xa[0], 0)
for i in range(1, n):
p_mx_x,p_mx_y = dp_asc_x[i-1][0]
p_mn_x,p_mn_y = dp_asc_x[i-1][1]
mx_x, mn_x = max(xa[i][0], p_mx_x), min(xa[i][0], p_mn_x)
mx_y, mn_y = max(xa[i][1], p_mx_y), min(xa[i][1], p_mn_y)
#print((mx_x, mx_y), (mn_x, mn_y))
dp_asc_x[i] = ((mx_x, mx_y), (mn_x, mn_y), (mx_x-mn_x)*(mx_y-mn_y))
dp_desc_x[n-1] = (xa[n-1],xa[n-1], 0)
for i in range(n-2, -1, -1):
p_mx_x,p_mx_y = dp_desc_x[i+1][0]
p_mn_x,p_mn_y = dp_desc_x[i+1][1]
mx_x, mn_x = max(xa[i][0], p_mx_x), min(xa[i][0], p_mn_x)
mx_y, mn_y = max(xa[i][1], p_mx_y), min(xa[i][1], p_mn_y)
#print((mx_x, mx_y), (mn_x, mn_y))
dp_desc_x[i] = ((mx_x, mx_y), (mn_x, mn_y), (mx_x-mn_x)*(mx_y-mn_y))
dp_asc_y[0] = (ya[0],ya[0], 0)
for i in range(1, n):
p_mx_x,p_mx_y = dp_asc_y[i-1][0]
p_mn_x,p_mn_y = dp_asc_y[i-1][1]
mx_x, mn_x = max(ya[i][0], p_mx_x), min(ya[i][0], p_mn_x)
mx_y, mn_y = max(ya[i][1], p_mx_y), min(ya[i][1], p_mn_y)
#print((mx_x, mx_y), (mn_x, mn_y))
dp_asc_y[i] = ((mx_x, mx_y), (mn_x, mn_y), (mx_x-mn_x)*(mx_y-mn_y))
dp_desc_y[n-1] = (ya[n-1],ya[n-1], 0)
for i in range(n-2, -1, -1):
p_mx_x,p_mx_y = dp_desc_y[i+1][0]
p_mn_x,p_mn_y = dp_desc_y[i+1][1]
mx_x, mn_x = max(ya[i][0], p_mx_x), min(ya[i][0], p_mn_x)
mx_y, mn_y = max(ya[i][1], p_mx_y), min(ya[i][1], p_mn_y)
#print((mx_x, mx_y), (mn_x, mn_y))
dp_desc_y[i] = ((mx_x, mx_y), (mn_x, mn_y), (mx_x-mn_x)*(mx_y-mn_y))
#print(dp_asc_x)
#print(dp_desc_x)
#print(dp_asc_y)
#print(dp_desc_y)
mn = 10**20
for i in range(n-1):
if dp_asc_x[i][2]+dp_desc_x[i+1][2] < mn:
mn = dp_asc_x[i][2]+dp_desc_x[i+1][2]
for i in range(n-1):
if dp_asc_y[i][2]+dp_desc_y[i+1][2] < mn:
mn = dp_asc_y[i][2]+dp_desc_y[i+1][2]
print(mn)
``` | instruction | 0 | 2,653 | 23 | 5,306 |
No | output | 1 | 2,653 | 23 | 5,307 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Little Chris is a huge fan of linear algebra. This time he has been given a homework about the unusual square of a square matrix.
The dot product of two integer number vectors x and y of size n is the sum of the products of the corresponding components of the vectors. The unusual square of an n × n square matrix A is defined as the sum of n dot products. The i-th of them is the dot product of the i-th row vector and the i-th column vector in the matrix A.
Fortunately for Chris, he has to work only in GF(2)! This means that all operations (addition, multiplication) are calculated modulo 2. In fact, the matrix A is binary: each element of A is either 0 or 1. For example, consider the following matrix A:
<image>
The unusual square of A is equal to (1·1 + 1·0 + 1·1) + (0·1 + 1·1 + 1·0) + (1·1 + 0·1 + 0·0) = 0 + 1 + 1 = 0.
However, there is much more to the homework. Chris has to process q queries; each query can be one of the following:
1. given a row index i, flip all the values in the i-th row in A;
2. given a column index i, flip all the values in the i-th column in A;
3. find the unusual square of A.
To flip a bit value w means to change it to 1 - w, i.e., 1 changes to 0 and 0 changes to 1.
Given the initial matrix A, output the answers for each query of the third type! Can you solve Chris's homework?
Input
The first line of input contains an integer n (1 ≤ n ≤ 1000), the number of rows and the number of columns in the matrix A. The next n lines describe the matrix: the i-th line contains n space-separated bits and describes the i-th row of A. The j-th number of the i-th line aij (0 ≤ aij ≤ 1) is the element on the intersection of the i-th row and the j-th column of A.
The next line of input contains an integer q (1 ≤ q ≤ 106), the number of queries. Each of the next q lines describes a single query, which can be one of the following:
* 1 i — flip the values of the i-th row;
* 2 i — flip the values of the i-th column;
* 3 — output the unusual square of A.
Note: since the size of the input and output could be very large, don't use slow output techniques in your language. For example, do not use input and output streams (cin, cout) in C++.
Output
Let the number of the 3rd type queries in the input be m. Output a single string s of length m, where the i-th symbol of s is the value of the unusual square of A for the i-th query of the 3rd type as it appears in the input.
Examples
Input
3
1 1 1
0 1 1
1 0 0
12
3
2 3
3
2 2
2 2
1 3
3
3
1 2
2 1
1 1
3
Output
01001 | instruction | 0 | 2,811 | 23 | 5,622 |
Tags: implementation, math
Correct Solution:
```
from sys import stdin
test = stdin.readlines()
n = int(test[0])
matrix = [test[i+1].split() for i in range(n)]
dot = 0
for i in range(n):
dot ^= matrix[i][i] == '1'
out = []
for q in range(int(test[n + 1])):
query = test[n + q + 2].split()
if len(query) == 1:
out.append(dot)
else:
dot ^= 1
print(''.join(map(str, out)))
``` | output | 1 | 2,811 | 23 | 5,623 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Little Chris is a huge fan of linear algebra. This time he has been given a homework about the unusual square of a square matrix.
The dot product of two integer number vectors x and y of size n is the sum of the products of the corresponding components of the vectors. The unusual square of an n × n square matrix A is defined as the sum of n dot products. The i-th of them is the dot product of the i-th row vector and the i-th column vector in the matrix A.
Fortunately for Chris, he has to work only in GF(2)! This means that all operations (addition, multiplication) are calculated modulo 2. In fact, the matrix A is binary: each element of A is either 0 or 1. For example, consider the following matrix A:
<image>
The unusual square of A is equal to (1·1 + 1·0 + 1·1) + (0·1 + 1·1 + 1·0) + (1·1 + 0·1 + 0·0) = 0 + 1 + 1 = 0.
However, there is much more to the homework. Chris has to process q queries; each query can be one of the following:
1. given a row index i, flip all the values in the i-th row in A;
2. given a column index i, flip all the values in the i-th column in A;
3. find the unusual square of A.
To flip a bit value w means to change it to 1 - w, i.e., 1 changes to 0 and 0 changes to 1.
Given the initial matrix A, output the answers for each query of the third type! Can you solve Chris's homework?
Input
The first line of input contains an integer n (1 ≤ n ≤ 1000), the number of rows and the number of columns in the matrix A. The next n lines describe the matrix: the i-th line contains n space-separated bits and describes the i-th row of A. The j-th number of the i-th line aij (0 ≤ aij ≤ 1) is the element on the intersection of the i-th row and the j-th column of A.
The next line of input contains an integer q (1 ≤ q ≤ 106), the number of queries. Each of the next q lines describes a single query, which can be one of the following:
* 1 i — flip the values of the i-th row;
* 2 i — flip the values of the i-th column;
* 3 — output the unusual square of A.
Note: since the size of the input and output could be very large, don't use slow output techniques in your language. For example, do not use input and output streams (cin, cout) in C++.
Output
Let the number of the 3rd type queries in the input be m. Output a single string s of length m, where the i-th symbol of s is the value of the unusual square of A for the i-th query of the 3rd type as it appears in the input.
Examples
Input
3
1 1 1
0 1 1
1 0 0
12
3
2 3
3
2 2
2 2
1 3
3
3
1 2
2 1
1 1
3
Output
01001 | instruction | 0 | 2,812 | 23 | 5,624 |
Tags: implementation, math
Correct Solution:
```
#Code by Sounak, IIESTS
#------------------------------warmup----------------------------
import os
import sys
import math
from io import BytesIO, IOBase
BUFSIZE = 8192
class FastIO(IOBase):
newlines = 0
def __init__(self, file):
self._fd = file.fileno()
self.buffer = BytesIO()
self.writable = "x" in file.mode or "r" not in file.mode
self.write = self.buffer.write if self.writable else None
def read(self):
while True:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
if not b:
break
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines = 0
return self.buffer.read()
def readline(self):
while self.newlines == 0:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
self.newlines = b.count(b"\n") + (not b)
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines -= 1
return self.buffer.readline()
def flush(self):
if self.writable:
os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
self.write = lambda s: self.buffer.write(s.encode("ascii"))
self.read = lambda: self.buffer.read().decode("ascii")
self.readline = lambda: self.buffer.readline().decode("ascii")
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")
#-------------------game starts now-----------------------------------------------------
n = int(input())
lis=list(list(map(int, input().split())) for _ in range(n))
u=0
for i in range(n):
for j in range(n):
if i==j:
u^=lis[i][j]
ans =[]
k = int(input())
for i in range(k):
s = input()
if s[0]=='3':
ans.append(str(u))
else:
u^=1
print(''.join(ans))
``` | output | 1 | 2,812 | 23 | 5,625 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Little Chris is a huge fan of linear algebra. This time he has been given a homework about the unusual square of a square matrix.
The dot product of two integer number vectors x and y of size n is the sum of the products of the corresponding components of the vectors. The unusual square of an n × n square matrix A is defined as the sum of n dot products. The i-th of them is the dot product of the i-th row vector and the i-th column vector in the matrix A.
Fortunately for Chris, he has to work only in GF(2)! This means that all operations (addition, multiplication) are calculated modulo 2. In fact, the matrix A is binary: each element of A is either 0 or 1. For example, consider the following matrix A:
<image>
The unusual square of A is equal to (1·1 + 1·0 + 1·1) + (0·1 + 1·1 + 1·0) + (1·1 + 0·1 + 0·0) = 0 + 1 + 1 = 0.
However, there is much more to the homework. Chris has to process q queries; each query can be one of the following:
1. given a row index i, flip all the values in the i-th row in A;
2. given a column index i, flip all the values in the i-th column in A;
3. find the unusual square of A.
To flip a bit value w means to change it to 1 - w, i.e., 1 changes to 0 and 0 changes to 1.
Given the initial matrix A, output the answers for each query of the third type! Can you solve Chris's homework?
Input
The first line of input contains an integer n (1 ≤ n ≤ 1000), the number of rows and the number of columns in the matrix A. The next n lines describe the matrix: the i-th line contains n space-separated bits and describes the i-th row of A. The j-th number of the i-th line aij (0 ≤ aij ≤ 1) is the element on the intersection of the i-th row and the j-th column of A.
The next line of input contains an integer q (1 ≤ q ≤ 106), the number of queries. Each of the next q lines describes a single query, which can be one of the following:
* 1 i — flip the values of the i-th row;
* 2 i — flip the values of the i-th column;
* 3 — output the unusual square of A.
Note: since the size of the input and output could be very large, don't use slow output techniques in your language. For example, do not use input and output streams (cin, cout) in C++.
Output
Let the number of the 3rd type queries in the input be m. Output a single string s of length m, where the i-th symbol of s is the value of the unusual square of A for the i-th query of the 3rd type as it appears in the input.
Examples
Input
3
1 1 1
0 1 1
1 0 0
12
3
2 3
3
2 2
2 2
1 3
3
3
1 2
2 1
1 1
3
Output
01001 | instruction | 0 | 2,813 | 23 | 5,626 |
Tags: implementation, math
Correct Solution:
```
def main():
from sys import stdin, stdout
from functools import reduce
from operator import xor
n = int(int(input()))
mat = [list(map(int, input().split())) for _ in range(n)]
ans, q, a = reduce(xor, [mat[i][i] for i in range(n)]), int(input()), []
queries = [stdin.readline() for i in range(q)]
for query in queries:
if query[0] == '3':
a.append(str(ans))
else:
ans ^= 1
print(''.join(a))
if __name__ == '__main__':
main()
``` | output | 1 | 2,813 | 23 | 5,627 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Little Chris is a huge fan of linear algebra. This time he has been given a homework about the unusual square of a square matrix.
The dot product of two integer number vectors x and y of size n is the sum of the products of the corresponding components of the vectors. The unusual square of an n × n square matrix A is defined as the sum of n dot products. The i-th of them is the dot product of the i-th row vector and the i-th column vector in the matrix A.
Fortunately for Chris, he has to work only in GF(2)! This means that all operations (addition, multiplication) are calculated modulo 2. In fact, the matrix A is binary: each element of A is either 0 or 1. For example, consider the following matrix A:
<image>
The unusual square of A is equal to (1·1 + 1·0 + 1·1) + (0·1 + 1·1 + 1·0) + (1·1 + 0·1 + 0·0) = 0 + 1 + 1 = 0.
However, there is much more to the homework. Chris has to process q queries; each query can be one of the following:
1. given a row index i, flip all the values in the i-th row in A;
2. given a column index i, flip all the values in the i-th column in A;
3. find the unusual square of A.
To flip a bit value w means to change it to 1 - w, i.e., 1 changes to 0 and 0 changes to 1.
Given the initial matrix A, output the answers for each query of the third type! Can you solve Chris's homework?
Input
The first line of input contains an integer n (1 ≤ n ≤ 1000), the number of rows and the number of columns in the matrix A. The next n lines describe the matrix: the i-th line contains n space-separated bits and describes the i-th row of A. The j-th number of the i-th line aij (0 ≤ aij ≤ 1) is the element on the intersection of the i-th row and the j-th column of A.
The next line of input contains an integer q (1 ≤ q ≤ 106), the number of queries. Each of the next q lines describes a single query, which can be one of the following:
* 1 i — flip the values of the i-th row;
* 2 i — flip the values of the i-th column;
* 3 — output the unusual square of A.
Note: since the size of the input and output could be very large, don't use slow output techniques in your language. For example, do not use input and output streams (cin, cout) in C++.
Output
Let the number of the 3rd type queries in the input be m. Output a single string s of length m, where the i-th symbol of s is the value of the unusual square of A for the i-th query of the 3rd type as it appears in the input.
Examples
Input
3
1 1 1
0 1 1
1 0 0
12
3
2 3
3
2 2
2 2
1 3
3
3
1 2
2 1
1 1
3
Output
01001 | instruction | 0 | 2,814 | 23 | 5,628 |
Tags: implementation, math
Correct Solution:
```
from sys import stdin, stdout
from functools import reduce
from operator import xor
def main():
n = int(int(input()))
mat = [list(map(int, input().split())) for _ in range(n)]
ans, q, a = reduce(xor, [mat[i][i] for i in range(n)]), int(input()), []
queries = [stdin.readline() for i in range(q)]
for query in queries:
if query[0] == '3':
a.append(str(ans))
else:
ans ^= 1
print(''.join(a))
if __name__ == '__main__':
main()
``` | output | 1 | 2,814 | 23 | 5,629 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Little Chris is a huge fan of linear algebra. This time he has been given a homework about the unusual square of a square matrix.
The dot product of two integer number vectors x and y of size n is the sum of the products of the corresponding components of the vectors. The unusual square of an n × n square matrix A is defined as the sum of n dot products. The i-th of them is the dot product of the i-th row vector and the i-th column vector in the matrix A.
Fortunately for Chris, he has to work only in GF(2)! This means that all operations (addition, multiplication) are calculated modulo 2. In fact, the matrix A is binary: each element of A is either 0 or 1. For example, consider the following matrix A:
<image>
The unusual square of A is equal to (1·1 + 1·0 + 1·1) + (0·1 + 1·1 + 1·0) + (1·1 + 0·1 + 0·0) = 0 + 1 + 1 = 0.
However, there is much more to the homework. Chris has to process q queries; each query can be one of the following:
1. given a row index i, flip all the values in the i-th row in A;
2. given a column index i, flip all the values in the i-th column in A;
3. find the unusual square of A.
To flip a bit value w means to change it to 1 - w, i.e., 1 changes to 0 and 0 changes to 1.
Given the initial matrix A, output the answers for each query of the third type! Can you solve Chris's homework?
Input
The first line of input contains an integer n (1 ≤ n ≤ 1000), the number of rows and the number of columns in the matrix A. The next n lines describe the matrix: the i-th line contains n space-separated bits and describes the i-th row of A. The j-th number of the i-th line aij (0 ≤ aij ≤ 1) is the element on the intersection of the i-th row and the j-th column of A.
The next line of input contains an integer q (1 ≤ q ≤ 106), the number of queries. Each of the next q lines describes a single query, which can be one of the following:
* 1 i — flip the values of the i-th row;
* 2 i — flip the values of the i-th column;
* 3 — output the unusual square of A.
Note: since the size of the input and output could be very large, don't use slow output techniques in your language. For example, do not use input and output streams (cin, cout) in C++.
Output
Let the number of the 3rd type queries in the input be m. Output a single string s of length m, where the i-th symbol of s is the value of the unusual square of A for the i-th query of the 3rd type as it appears in the input.
Examples
Input
3
1 1 1
0 1 1
1 0 0
12
3
2 3
3
2 2
2 2
1 3
3
3
1 2
2 1
1 1
3
Output
01001 | instruction | 0 | 2,815 | 23 | 5,630 |
Tags: implementation, math
Correct Solution:
```
from sys import stdin
test = stdin.readlines()
n = int(test[0])
dot = 0
j = 0
for i in range(n):
if test[i+1][j] == '1':
dot ^= 1
j += 2
out = []
for q in range(int(test[n + 1])):
query = test[n + q + 2].split()
if len(query) == 1:
out.append(dot)
else:
dot ^= 1
print(''.join(map(str, out)))
``` | output | 1 | 2,815 | 23 | 5,631 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Little Chris is a huge fan of linear algebra. This time he has been given a homework about the unusual square of a square matrix.
The dot product of two integer number vectors x and y of size n is the sum of the products of the corresponding components of the vectors. The unusual square of an n × n square matrix A is defined as the sum of n dot products. The i-th of them is the dot product of the i-th row vector and the i-th column vector in the matrix A.
Fortunately for Chris, he has to work only in GF(2)! This means that all operations (addition, multiplication) are calculated modulo 2. In fact, the matrix A is binary: each element of A is either 0 or 1. For example, consider the following matrix A:
<image>
The unusual square of A is equal to (1·1 + 1·0 + 1·1) + (0·1 + 1·1 + 1·0) + (1·1 + 0·1 + 0·0) = 0 + 1 + 1 = 0.
However, there is much more to the homework. Chris has to process q queries; each query can be one of the following:
1. given a row index i, flip all the values in the i-th row in A;
2. given a column index i, flip all the values in the i-th column in A;
3. find the unusual square of A.
To flip a bit value w means to change it to 1 - w, i.e., 1 changes to 0 and 0 changes to 1.
Given the initial matrix A, output the answers for each query of the third type! Can you solve Chris's homework?
Input
The first line of input contains an integer n (1 ≤ n ≤ 1000), the number of rows and the number of columns in the matrix A. The next n lines describe the matrix: the i-th line contains n space-separated bits and describes the i-th row of A. The j-th number of the i-th line aij (0 ≤ aij ≤ 1) is the element on the intersection of the i-th row and the j-th column of A.
The next line of input contains an integer q (1 ≤ q ≤ 106), the number of queries. Each of the next q lines describes a single query, which can be one of the following:
* 1 i — flip the values of the i-th row;
* 2 i — flip the values of the i-th column;
* 3 — output the unusual square of A.
Note: since the size of the input and output could be very large, don't use slow output techniques in your language. For example, do not use input and output streams (cin, cout) in C++.
Output
Let the number of the 3rd type queries in the input be m. Output a single string s of length m, where the i-th symbol of s is the value of the unusual square of A for the i-th query of the 3rd type as it appears in the input.
Examples
Input
3
1 1 1
0 1 1
1 0 0
12
3
2 3
3
2 2
2 2
1 3
3
3
1 2
2 1
1 1
3
Output
01001 | instruction | 0 | 2,816 | 23 | 5,632 |
Tags: implementation, math
Correct Solution:
```
def main():
from sys import stdin
from operator import xor
from functools import reduce
x, res = reduce(xor, (input()[i] == '1' for i in range(0, int(input()) * 2, 2))), []
input()
for s in stdin.read().splitlines():
if s == '3':
res.append("01"[x])
else:
x ^= True
print(''.join(res))
if __name__ == "__main__":
main()
``` | output | 1 | 2,816 | 23 | 5,633 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Little Chris is a huge fan of linear algebra. This time he has been given a homework about the unusual square of a square matrix.
The dot product of two integer number vectors x and y of size n is the sum of the products of the corresponding components of the vectors. The unusual square of an n × n square matrix A is defined as the sum of n dot products. The i-th of them is the dot product of the i-th row vector and the i-th column vector in the matrix A.
Fortunately for Chris, he has to work only in GF(2)! This means that all operations (addition, multiplication) are calculated modulo 2. In fact, the matrix A is binary: each element of A is either 0 or 1. For example, consider the following matrix A:
<image>
The unusual square of A is equal to (1·1 + 1·0 + 1·1) + (0·1 + 1·1 + 1·0) + (1·1 + 0·1 + 0·0) = 0 + 1 + 1 = 0.
However, there is much more to the homework. Chris has to process q queries; each query can be one of the following:
1. given a row index i, flip all the values in the i-th row in A;
2. given a column index i, flip all the values in the i-th column in A;
3. find the unusual square of A.
To flip a bit value w means to change it to 1 - w, i.e., 1 changes to 0 and 0 changes to 1.
Given the initial matrix A, output the answers for each query of the third type! Can you solve Chris's homework?
Input
The first line of input contains an integer n (1 ≤ n ≤ 1000), the number of rows and the number of columns in the matrix A. The next n lines describe the matrix: the i-th line contains n space-separated bits and describes the i-th row of A. The j-th number of the i-th line aij (0 ≤ aij ≤ 1) is the element on the intersection of the i-th row and the j-th column of A.
The next line of input contains an integer q (1 ≤ q ≤ 106), the number of queries. Each of the next q lines describes a single query, which can be one of the following:
* 1 i — flip the values of the i-th row;
* 2 i — flip the values of the i-th column;
* 3 — output the unusual square of A.
Note: since the size of the input and output could be very large, don't use slow output techniques in your language. For example, do not use input and output streams (cin, cout) in C++.
Output
Let the number of the 3rd type queries in the input be m. Output a single string s of length m, where the i-th symbol of s is the value of the unusual square of A for the i-th query of the 3rd type as it appears in the input.
Examples
Input
3
1 1 1
0 1 1
1 0 0
12
3
2 3
3
2 2
2 2
1 3
3
3
1 2
2 1
1 1
3
Output
01001 | instruction | 0 | 2,817 | 23 | 5,634 |
Tags: implementation, math
Correct Solution:
```
from sys import stdin, stdout
from functools import reduce
from operator import xor
def arr_inp(n):
return [int(x) for x in input().split()]
class Matrix:
def __init__(self, r, c, mat=None):
self.r, self.c = r, c
if mat != None:
self.mat = mat
else:
self.mat = [[0 for i in range(c)] for j in range(r)]
def __add__(self, other):
mat0 = Matrix(self.r, self.c)
for i in range(self.r):
for j in range(self.c):
mat0.mat[i][j] = self.mat[i][j] + other.mat[i][j]
return mat0.mat
def __mul__(self, other):
mat0 = Matrix(self.r, other.c)
for i in range(self.r):
for j in range(other.c):
for k in range(self.c):
mat0.mat[i][j] += self.mat[i][k] * other.mat[k][j]
return mat0.mat
def trace(self):
res = 0
for i in range(self.r):
res += self.mat[i][i]
return res % 2
def dot_mul(self, other):
res = 0
for i in range(self.r):
for j in range(self.c):
res += self.mat[i][j] * other.mat[j][i]
return res % 2
def rotate(self):
mat0 = Matrix(self.c, self.r)
for i in range(self.r):
for j in range(self.c):
mat0.mat[j][self.r - (i + 1)] = self.mat[i][j]
self.mat, self.r, self.c = mat0.mat.copy(), self.c, self.r
return self.mat
def reflect(self):
mat0 = Matrix(self.r, self.c)
for i in range(self.r):
for j in range(self.c):
mat0.mat[i][self.c - (j + 1)] = self.mat[i][j]
self.mat = mat0.mat.copy()
return self.mat
n = int(int(input()))
mat = Matrix(n, n, [arr_inp(1) for _ in range(n)])
ans, q, a = mat.trace(), int(input()), []
queries = [stdin.readline() for i in range(q)]
for query in queries:
if query[0] == '3':
a.append(str(ans))
else:
ans ^= 1
print(''.join(a))
``` | output | 1 | 2,817 | 23 | 5,635 |
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