Problem:
 Score : 800 points    Problem Statement  Niwango-kun, an employee of Dwango Co., Ltd., likes Niconico TV-chan, so he collected a lot of soft toys of her and spread them on the floor.  Niwango-kun has N black rare soft toys of Niconico TV-chan and they are spread together with ordinary ones. He wanted these black rare soft toys to be close together, so he decided to rearrange them.  In an infinitely large two-dimensional plane, every lattice point has a soft toy on it. The coordinates (x_i,y_i) of N black rare soft toys are given. All soft toys are considered to be points (without a length, area, or volume).  He may perform the following operation arbitrarily many times:   Put an axis-aligned square with side length D , rotate the square by 90 degrees with four soft toys on the four corners of the square. More specifically, if the left bottom corner's coordinate is (x, y) , rotate four points (x,y) \rightarrow (x+D,y) \rightarrow (x+D,y+D) \rightarrow (x,y+D) \rightarrow (x,y) in this order. Each of the four corners of the square must be on a lattice point.   Let's define the scatteredness of an arrangement by the minimum side length of an axis-aligned square enclosing all black rare soft toys. Black rare soft toys on the edges or the vertices of a square are considered to be enclosed by the square.  Find the minimum scatteredness after he performs arbitrarily many operations.      Constraints   2 \leq N \leq 10^5  1 \leq D \leq 1000  0 \leq x_i, y_i \leq 10^9  Given coordinates are pairwise distinct  All numbers given in input are integers       Partial Scores   500 points will be awarded for passing the test set satisfying 1 \leq D \leq 30 .         Input  Input is given from Standard Input in the following format:  N  D  x_1  y_1  :  x_N  y_N       Output  Print the answer.        Sample Input 1  3 1
0 0
1 0
2 0      Sample Output 1  1       Sample Input 2  19 2
1 3
2 3
0 1
1 1
2 1
3 1
4 4
5 4
6 4
7 4
8 4
8 3
8 2
8 1
8 0
7 0
6 0
5 0
4 0      Sample Output 2  4       Sample Input 3  8 3
0 0
0 3
3 0
3 3
2 2
2 5
5 2
5 5      Sample Output 3  4

Solution (Python):
n,d = map(int,input().split())
num = [[0]*d for i in range(d)]
a = 0
for i in range(n):
  x,y = map(int,input().split())
  x %=d
  y%=d
  num[x][y] += 1
  a = max(a,num[x][y])
x=1
while x*x<a:
  x += 1
r = (x-1)*d
a = x-1
dai = d-1
syo = 0
anum = [[[0]*d for i in range(d)]for i in range(3)]
rui = [[[0]*(2*d+1) for i in range(2*d+1)]for i in range(3)]
for i in range(d):
  for j in range(d):
    z = num[i][j]
    if z > (a+1)**2:
      anum[0][i][j]=1
      anum[1][i][j]=1
      anum[2][i][j]=1
    elif z> a*(a+1):
      anum[1][i][j]=1
      anum[2][i][j]=1
    elif z > a*a:
      anum[1][i][j]=1
for x in range(3):
  for i in range(1,2*d+1):
    for j in range(1,2*d+1):
      rui[x][i][j]+=rui[x][i-1][j]+rui[x][i][j-1]-rui[x][i-1][j-1]+anum[x][(i-1)%d][(j-1)%d]
def cheak(kon):
  for i in range(1,d+1):
    for j in range(1,d+1):
      if rui[0][i+kon][j+kon] - rui[0][i - 1][j+kon] - rui[0][i+kon][j - 1] + rui[0][i - 1][j - 1]:
        continue
      if rui[1][i+d-1][j+d-1] - rui[1][i+kon+1 - 1][j+d-1] - rui[1][i+d-1][j+kon+1 - 1] + rui[1][i+kon+1 - 1][j+kon+1 - 1]:
        continue
      if rui[2][i+d-1][j+kon] - rui[2][i+kon+1 - 1][j+kon] - rui[2][i+d-1][j - 1] + rui[2][i+kon+1 - 1][j - 1]:
        continue
      if rui[2][i+kon][j+d-1] - rui[2][i - 1][j+d-1] - rui[2][i+kon][j+kon+1 - 1] + rui[2][i - 1][j+kon+1 - 1]:
        continue
      return 1
  return 0 
kon = (dai+syo)//2
while True:
  if dai-syo <= 1:
    if cheak(syo) == 1:
       dai = syo
    break
  kon = (dai+syo)//2
  c = cheak(kon)
  if c == 1:
    dai = kon
  else:
    syo = kon
print(dai+r)

Solution (Python):
import sys
read = sys.stdin.buffer.read
readline = sys.stdin.buffer.readline
readlines = sys.stdin.buffer.readlines
sys.setrecursionlimit(10 ** 7)

import numpy as np

N,D = map(int,readline().split())
XY = np.array(list(map(int,read().split())))

A = np.zeros((D,D),np.int64)
X = XY[::2]; Y = XY[1::2]
np.add.at(A,(X%D,Y%D),1)

maxA = A.max()
n = int((maxA-1)**.5)


AA = np.tile(A,(2,2))

gr1 = (AA > n*n).cumsum(axis=0).cumsum(axis=1)
gr2 = (AA > n*(n+1)).cumsum(axis=0).cumsum(axis=1)

def test(x):
    gr1_rd = gr1[D:D+D,D:D+D] + gr1[x:x+D,x:x+D] - gr1[x:x+D,D:D+D] - gr1[D:D+D,x:x+D]
    condition_rd = (gr1_rd == 0)
    gr2_ru = gr2[x:x+D,D:D+D] + gr2[0:D,x:x+D] - gr2[x:x+D,x:x+D] - gr2[0:D,D:D+D]
    gr2_ld = gr2[D:D+D,x:x+D] + gr2[x:x+D,0:D] - gr2[x:x+D,x:x+D] - gr2[D:D+D,0:D]
    condition_ru = (gr2_ru == 0)
    condition_ld = (gr2_ld == 0)
    return (condition_rd&condition_ru&condition_ld).any()

left = 0 
right = D 
while left+1 < right:
    mid = (left+right)//2
    if test(mid):
        right = mid
    else:
        left = mid

answer = n*D + right - 1
print(answer)

Solution (Python):
n,d = map(int,input().split())
num = [[0]*d for i in range(d)]
a = 0
for i in range(n):
  x,y = map(int,input().split())
  x %=d
  y%=d
  num[x][y] += 1
  a = max(a,num[x][y])
x=1
while x*x<a:
  x += 1
r = (x-1)*d
a = x-1
dai = d-1
syo = 0
anum = [[[0]*d for i in range(d)]for i in range(3)]
rui = [[[0]*(2*d+1) for i in range(2*d+1)]for i in range(3)]
for i in range(d):
  for j in range(d):
    z = num[i][j]
    if z > (a+1)**2:
      anum[0][i][j]=1
      anum[1][i][j]=1
      anum[2][i][j]=1
    elif z> a*(a+1):
      anum[1][i][j]=1
      anum[2][i][j]=1
    elif z > a*a:
      anum[1][i][j]=1
for x in range(3):
  for i in range(1,2*d+1):
    for j in range(1,2*d+1):
      rui[x][i][j]+=rui[x][i-1][j]+rui[x][i][j-1]-rui[x][i-1][j-1]+anum[x][(i-1)%d][(j-1)%d]
def cheak(kon):
  for i in range(1,d+1):
    for j in range(1,d+1):
      if rui[0][i+kon][j+kon] - rui[0][i - 1][j+kon] - rui[0][i+kon][j - 1] + rui[0][i - 1][j - 1]:
        continue
      if rui[1][i+d-1][j+d-1] - rui[1][i+kon+1 - 1][j+d-1] - rui[1][i+d-1][j+kon+1 - 1] + rui[1][i+kon+1 - 1][j+kon+1 - 1]:
        continue
      if rui[2][i+d-1][j+kon] - rui[2][i+kon+1 - 1][j+kon] - rui[2][i+d-1][j - 1] + rui[2][i+kon+1 - 1][j - 1]:
        continue
      if rui[2][i+kon][j+d-1] - rui[2][i - 1][j+d-1] - rui[2][i+kon][j+kon+1 - 1] + rui[2][i - 1][j+kon+1 - 1]:
        continue
      return 1
  return 0 
kon = (dai+syo)//2
while True:
  if dai-syo <= 1:
    if cheak(syo) == 1:
       dai = syo
    break
  kon = (dai+syo)//2
  c = cheak(kon)
  if c == 1:
    dai = kon
  else:
    syo = kon
print(dai+r)