message stringlengths 2 28.7k | message_type stringclasses 2 values | message_id int64 0 1 | conversation_id int64 21 109k | cluster float64 7 7 | __index_level_0__ int64 42 217k |
|---|---|---|---|---|---|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a grid of squares with H horizontal rows and W vertical columns, where each square is painted white or black. HW characters from A_{11} to A_{HW} represent the colors of the squares. A_{ij} is `#` if the square at the i-th row from the top and the j-th column from the left is black, and A_{ij} is `.` if that square is white.
We will repeatedly perform the following operation until all the squares are black:
* Every white square that shares a side with a black square, becomes black.
Find the number of operations that will be performed. The initial grid has at least one black square.
Constraints
* 1 \leq H,W \leq 1000
* A_{ij} is `#` or `.`.
* The given grid has at least one black square.
Input
Input is given from Standard Input in the following format:
H W
A_{11}A_{12}...A_{1W}
:
A_{H1}A_{H2}...A_{HW}
Output
Print the number of operations that will be performed.
Examples
Input
3 3
...
.#.
...
Output
2
Input
6 6
..#..#
......
..#..
......
.#....
....#.
Output
3
Submitted Solution:
```
from collections import deque
H,W = map(int, input().split())
A = 0
dp = [None] * H
mat = deque()
for h in range(H):
s = list(input())
for w, x in zip(range(W), s):
if x == "#":
mat.append((h, w, 0))
s[w] = 1
else:
s[w] = 0
dp[h] = s
while len(mat):
h, w, cnt = mat.popleft()
for t, h, w, cnt in ((h > 0, h - 1, w, cnt),(h < H-1, h + 1, w, cnt),(w > 0, h , w - 1, cnt),(w < W-1, h , w + 1, cnt)):
if t and dp[h][w] == 0:
mat.append((h, w, cnt+1))
dp[h][w] = 1
print(cnt)
``` | instruction | 0 | 3,144 | 7 | 6,288 |
Yes | output | 1 | 3,144 | 7 | 6,289 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a grid of squares with H horizontal rows and W vertical columns, where each square is painted white or black. HW characters from A_{11} to A_{HW} represent the colors of the squares. A_{ij} is `#` if the square at the i-th row from the top and the j-th column from the left is black, and A_{ij} is `.` if that square is white.
We will repeatedly perform the following operation until all the squares are black:
* Every white square that shares a side with a black square, becomes black.
Find the number of operations that will be performed. The initial grid has at least one black square.
Constraints
* 1 \leq H,W \leq 1000
* A_{ij} is `#` or `.`.
* The given grid has at least one black square.
Input
Input is given from Standard Input in the following format:
H W
A_{11}A_{12}...A_{1W}
:
A_{H1}A_{H2}...A_{HW}
Output
Print the number of operations that will be performed.
Examples
Input
3 3
...
.#.
...
Output
2
Input
6 6
..#..#
......
..#..
......
.#....
....#.
Output
3
Submitted Solution:
```
from collections import deque
h,w=map(int,input().split())
l=[]
l.append('@'*(w+2))
for i in range(h):
l.append('@'+input()+'@')
l.append('@'*(w+2))
q=deque()
p=[[-1 for i in j] for j in l]
for i in range(1,h+1):
for j in range(1,w+1):
if l[i][j]=='#':
q.append([i,j])
p[i][j]=0
while len(q)>0:
x,y=q.popleft()
for i,j in [[1,0],[-1,0],[0,-1],[0,1]]:
if l[x+i][y+j]=='.' and p[x+i][y+j]==-1:
q.append([x+i,y+j])
p[x+i][y+j]=p[x][y]+1
a=0
for i in p:
for j in i:
a=max(a,j)
print(a)
#print(*p,sep='\n')
``` | instruction | 0 | 3,145 | 7 | 6,290 |
Yes | output | 1 | 3,145 | 7 | 6,291 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a grid of squares with H horizontal rows and W vertical columns, where each square is painted white or black. HW characters from A_{11} to A_{HW} represent the colors of the squares. A_{ij} is `#` if the square at the i-th row from the top and the j-th column from the left is black, and A_{ij} is `.` if that square is white.
We will repeatedly perform the following operation until all the squares are black:
* Every white square that shares a side with a black square, becomes black.
Find the number of operations that will be performed. The initial grid has at least one black square.
Constraints
* 1 \leq H,W \leq 1000
* A_{ij} is `#` or `.`.
* The given grid has at least one black square.
Input
Input is given from Standard Input in the following format:
H W
A_{11}A_{12}...A_{1W}
:
A_{H1}A_{H2}...A_{HW}
Output
Print the number of operations that will be performed.
Examples
Input
3 3
...
.#.
...
Output
2
Input
6 6
..#..#
......
..#..
......
.#....
....#.
Output
3
Submitted Solution:
```
import numpy as np
H, W = map(int, input().split())
arr = [0]*H
for i in range(H):
arr[i] = [j for j in input()]
a, b = np.where(np.array(arr)=='#')
abb = np.array([list(a), list(b)])
res = 0
x = np.array(range(H))
y = np.array(range(W))
xx, yy = np.meshgrid(x,y)
z = np.c_[xx.ravel(),yy.ravel()]
for zz in z:
res1 = 3000
res1 = min(np.amin(np.sum(abs(zz.reshape(2, -1) - abb), axis=0)), res1)
res = max(res1, res)
print(res)
``` | instruction | 0 | 3,146 | 7 | 6,292 |
No | output | 1 | 3,146 | 7 | 6,293 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a grid of squares with H horizontal rows and W vertical columns, where each square is painted white or black. HW characters from A_{11} to A_{HW} represent the colors of the squares. A_{ij} is `#` if the square at the i-th row from the top and the j-th column from the left is black, and A_{ij} is `.` if that square is white.
We will repeatedly perform the following operation until all the squares are black:
* Every white square that shares a side with a black square, becomes black.
Find the number of operations that will be performed. The initial grid has at least one black square.
Constraints
* 1 \leq H,W \leq 1000
* A_{ij} is `#` or `.`.
* The given grid has at least one black square.
Input
Input is given from Standard Input in the following format:
H W
A_{11}A_{12}...A_{1W}
:
A_{H1}A_{H2}...A_{HW}
Output
Print the number of operations that will be performed.
Examples
Input
3 3
...
.#.
...
Output
2
Input
6 6
..#..#
......
..#..
......
.#....
....#.
Output
3
Submitted Solution:
```
h, w = map(int, input().split())
a = [["."]*w]*h
for i in range(h):
a[i] = list(input())
s=0
e=0
k=0
while s==0:
for i in range(h):
for j in range(w):
print(i,j)
if a[i][j]=='#':
for x in range(-1,2):
for y in range(-1,2):
if x+i >= 0 and x+i < h and y+j >= 0 and y+j < w and a[i+x][j+y] == '.':
a[i+x][j+y]='#'
else:
e=1
if e==0:
s=1
k+=1
e=0
print(k)
``` | instruction | 0 | 3,147 | 7 | 6,294 |
No | output | 1 | 3,147 | 7 | 6,295 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a grid of squares with H horizontal rows and W vertical columns, where each square is painted white or black. HW characters from A_{11} to A_{HW} represent the colors of the squares. A_{ij} is `#` if the square at the i-th row from the top and the j-th column from the left is black, and A_{ij} is `.` if that square is white.
We will repeatedly perform the following operation until all the squares are black:
* Every white square that shares a side with a black square, becomes black.
Find the number of operations that will be performed. The initial grid has at least one black square.
Constraints
* 1 \leq H,W \leq 1000
* A_{ij} is `#` or `.`.
* The given grid has at least one black square.
Input
Input is given from Standard Input in the following format:
H W
A_{11}A_{12}...A_{1W}
:
A_{H1}A_{H2}...A_{HW}
Output
Print the number of operations that will be performed.
Examples
Input
3 3
...
.#.
...
Output
2
Input
6 6
..#..#
......
..#..
......
.#....
....#.
Output
3
Submitted Solution:
```
from collections import*
import sys
input=sys.stdin.readline
h,w=map(int,input().split())
grid=defaultdict(str)
for i in range(w+2):
grid[(0,i)]="#"
grid[(h+1,i)]="#"
for i in range(h):
for k,v in enumerate(input()):
grid[(i+1,k+1)]=v
for i in range(1,h+1):
grid[(i,0)]="#"
grid[(i,w+1)]="#"
#for i in range(h+2):
# print(*[grid[(i,j)]for j in range(w+2)])
stack=deque()
for y in range(1,h+1):
for x in range(1,w+1):
if grid[(y,x)]=="#":
grid[(y,x)]=str(0)
stack.append((y,x))
for i in range(h*w):
tmp=deque()
while stack:
#print(stack)
my,mx=stack.pop()
for y,x in ([0,1],[1,0],[-1,0],[0,-1]):
if grid[(my+y,mx+x)]==".":
grid[(my+y,mx+x)]=str(i+1)
tmp.append((my+y,mx+x))
stack=tmp
if len(tmp)==0:
print(i)
break
``` | instruction | 0 | 3,148 | 7 | 6,296 |
No | output | 1 | 3,148 | 7 | 6,297 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a grid of squares with H horizontal rows and W vertical columns, where each square is painted white or black. HW characters from A_{11} to A_{HW} represent the colors of the squares. A_{ij} is `#` if the square at the i-th row from the top and the j-th column from the left is black, and A_{ij} is `.` if that square is white.
We will repeatedly perform the following operation until all the squares are black:
* Every white square that shares a side with a black square, becomes black.
Find the number of operations that will be performed. The initial grid has at least one black square.
Constraints
* 1 \leq H,W \leq 1000
* A_{ij} is `#` or `.`.
* The given grid has at least one black square.
Input
Input is given from Standard Input in the following format:
H W
A_{11}A_{12}...A_{1W}
:
A_{H1}A_{H2}...A_{HW}
Output
Print the number of operations that will be performed.
Examples
Input
3 3
...
.#.
...
Output
2
Input
6 6
..#..#
......
..#..
......
.#....
....#.
Output
3
Submitted Solution:
```
from collections import deque
h,w = map(int, input().split())
m = [list(input()) for _ in range(h)]
def get_start_points(m):
start_points = []
for y in range(h):
for x in range(w):
if m[y][x] == '#':
start_points.append((y,x))
return start_points
def bfs(start_points):
que = deque(start_points)
step = 0
count_flag = False
while len(que):
for q in range(len(que)):
y,x = que.popleft()
for dy,dx in [(1,0),(-1,0),(0,1),(0,-1)]:
ny = y + dy
nx = x + dx
if not (0<=ny<h and 0<=nx<w) or m[ny][nx] != '.':
continue
m[ny][nx] = '#'
count_flag = True
que.append((ny,nx))
if count_flag:
step += 1
count_flag = False
else:
break
return step
print(bfs(get_start_points(m)))
``` | instruction | 0 | 3,149 | 7 | 6,298 |
No | output | 1 | 3,149 | 7 | 6,299 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Catherine has a deck of n cards, each of which is either red, green, or blue. As long as there are at least two cards left, she can do one of two actions:
* take any two (not necessarily adjacent) cards with different colors and exchange them for a new card of the third color;
* take any two (not necessarily adjacent) cards with the same color and exchange them for a new card with that color.
She repeats this process until there is only one card left. What are the possible colors for the final card?
Input
The first line of the input contains a single integer n (1 ≤ n ≤ 200) — the total number of cards.
The next line contains a string s of length n — the colors of the cards. s contains only the characters 'B', 'G', and 'R', representing blue, green, and red, respectively.
Output
Print a single string of up to three characters — the possible colors of the final card (using the same symbols as the input) in alphabetical order.
Examples
Input
2
RB
Output
G
Input
3
GRG
Output
BR
Input
5
BBBBB
Output
B
Note
In the first sample, Catherine has one red card and one blue card, which she must exchange for a green card.
In the second sample, Catherine has two green cards and one red card. She has two options: she can exchange the two green cards for a green card, then exchange the new green card and the red card for a blue card. Alternatively, she can exchange a green and a red card for a blue card, then exchange the blue card and remaining green card for a red card.
In the third sample, Catherine only has blue cards, so she can only exchange them for more blue cards. | instruction | 0 | 3,780 | 7 | 7,560 |
Tags: constructive algorithms, dp, math
Correct Solution:
```
n = int(input())
s = input()
a,b,c = sorted((s.count(x),x) for x in 'RGB')
# print(a,b,c)
if a[0] or b[0]>1:
print ('BGR')
elif b[0] and c[0]>1:
print (''.join(sorted(a[1]+b[1])))
elif b[0]:
print (a[1])
elif c[0]:
print (c[1])
``` | output | 1 | 3,780 | 7 | 7,561 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Catherine has a deck of n cards, each of which is either red, green, or blue. As long as there are at least two cards left, she can do one of two actions:
* take any two (not necessarily adjacent) cards with different colors and exchange them for a new card of the third color;
* take any two (not necessarily adjacent) cards with the same color and exchange them for a new card with that color.
She repeats this process until there is only one card left. What are the possible colors for the final card?
Input
The first line of the input contains a single integer n (1 ≤ n ≤ 200) — the total number of cards.
The next line contains a string s of length n — the colors of the cards. s contains only the characters 'B', 'G', and 'R', representing blue, green, and red, respectively.
Output
Print a single string of up to three characters — the possible colors of the final card (using the same symbols as the input) in alphabetical order.
Examples
Input
2
RB
Output
G
Input
3
GRG
Output
BR
Input
5
BBBBB
Output
B
Note
In the first sample, Catherine has one red card and one blue card, which she must exchange for a green card.
In the second sample, Catherine has two green cards and one red card. She has two options: she can exchange the two green cards for a green card, then exchange the new green card and the red card for a blue card. Alternatively, she can exchange a green and a red card for a blue card, then exchange the blue card and remaining green card for a red card.
In the third sample, Catherine only has blue cards, so she can only exchange them for more blue cards. | instruction | 0 | 3,781 | 7 | 7,562 |
Tags: constructive algorithms, dp, math
Correct Solution:
```
def main():
n = int(input())
s = input()
b, g, r = [s.count(i) for i in "BGR"]
if min(b, g, r) > 0:
print("BGR")
return
if max(b, g, r) == n:
if b == n: print("B")
if g == n: print("G")
if r == n: print("R")
return
if max(b, g, r) == 1:
if b == 0: print("B")
if g == 0: print("G")
if r == 0: print("R")
return
if max(b, g, r) == n - 1:
if b == n - 1: print("GR")
if g == n - 1: print("BR")
if r == n - 1: print("BG")
return
print("BGR")
main()
``` | output | 1 | 3,781 | 7 | 7,563 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Catherine has a deck of n cards, each of which is either red, green, or blue. As long as there are at least two cards left, she can do one of two actions:
* take any two (not necessarily adjacent) cards with different colors and exchange them for a new card of the third color;
* take any two (not necessarily adjacent) cards with the same color and exchange them for a new card with that color.
She repeats this process until there is only one card left. What are the possible colors for the final card?
Input
The first line of the input contains a single integer n (1 ≤ n ≤ 200) — the total number of cards.
The next line contains a string s of length n — the colors of the cards. s contains only the characters 'B', 'G', and 'R', representing blue, green, and red, respectively.
Output
Print a single string of up to three characters — the possible colors of the final card (using the same symbols as the input) in alphabetical order.
Examples
Input
2
RB
Output
G
Input
3
GRG
Output
BR
Input
5
BBBBB
Output
B
Note
In the first sample, Catherine has one red card and one blue card, which she must exchange for a green card.
In the second sample, Catherine has two green cards and one red card. She has two options: she can exchange the two green cards for a green card, then exchange the new green card and the red card for a blue card. Alternatively, she can exchange a green and a red card for a blue card, then exchange the blue card and remaining green card for a red card.
In the third sample, Catherine only has blue cards, so she can only exchange them for more blue cards. | instruction | 0 | 3,782 | 7 | 7,564 |
Tags: constructive algorithms, dp, math
Correct Solution:
```
from collections import Counter, deque
from sys import stdin
class UniqueDeque(deque):
def appendleft(self, *args, **kwargs):
if args[0] not in self:
return super().appendleft(*args, **kwargs)
def solve(s: str):
c = Counter(s)
t = (min(c['R'], 3), min(c['G'], 3), min(c['B'],3))
d = UniqueDeque()
d.append(t)
while any(filter(lambda x: x > 1, map(sum, d))):
i = d.pop()
if i[0] > 1:
d.appendleft((i[0] - 1, i[1], i[2]))
if i[1] > 1:
d.appendleft((i[0], i[1] - 1, i[2]))
if i[2] > 1:
d.appendleft((i[0], i[1], i[2] - 1))
if i[0] and i[1]:
d.appendleft((i[0] - 1, i[1] - 1, i[2] + 1))
if i[0] and i[2]:
d.appendleft((i[0] - 1, i[1] + 1, i[2] - 1))
if i[1] and i[2]:
d.appendleft((i[0] + 1, i[1] - 1, i[2] - 1))
res = []
if (1,0,0) in d:
res.append('R')
if (0,1,0) in d:
res.append('G')
if (0,0,1) in d:
res.append('B')
return ''.join(sorted(res))
if __name__ == '__main__':
_in = iter(stdin.readlines())
n = int(next(_in))
s = str(next(_in))[:-1]
print(solve(s))
``` | output | 1 | 3,782 | 7 | 7,565 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Catherine has a deck of n cards, each of which is either red, green, or blue. As long as there are at least two cards left, she can do one of two actions:
* take any two (not necessarily adjacent) cards with different colors and exchange them for a new card of the third color;
* take any two (not necessarily adjacent) cards with the same color and exchange them for a new card with that color.
She repeats this process until there is only one card left. What are the possible colors for the final card?
Input
The first line of the input contains a single integer n (1 ≤ n ≤ 200) — the total number of cards.
The next line contains a string s of length n — the colors of the cards. s contains only the characters 'B', 'G', and 'R', representing blue, green, and red, respectively.
Output
Print a single string of up to three characters — the possible colors of the final card (using the same symbols as the input) in alphabetical order.
Examples
Input
2
RB
Output
G
Input
3
GRG
Output
BR
Input
5
BBBBB
Output
B
Note
In the first sample, Catherine has one red card and one blue card, which she must exchange for a green card.
In the second sample, Catherine has two green cards and one red card. She has two options: she can exchange the two green cards for a green card, then exchange the new green card and the red card for a blue card. Alternatively, she can exchange a green and a red card for a blue card, then exchange the blue card and remaining green card for a red card.
In the third sample, Catherine only has blue cards, so she can only exchange them for more blue cards. | instruction | 0 | 3,783 | 7 | 7,566 |
Tags: constructive algorithms, dp, math
Correct Solution:
```
n= int(input())
l=[x for x in input()]
s=set(l)
sl = len(s)
alpa = ['B', 'G', 'R']
if sl==1:
print(l[0])
elif sl==3:
print("BGR")
else:
l2=[]
for x in s:
l2.append(x)
c1=l.count(l2[0])
c2=l.count(l2[1])
if c1>1 and c2>1:
print("BGR")
elif c1>1:
x=l2[0]
for i in alpa:
if i!=x:
print(i, end="")
print()
elif c2>1:
x = l2[1]
for i in alpa:
if i != x:
print(i, end="")
print()
else:
x = l2[1]
y=l2[0]
for i in alpa:
if i != x and i!=y:
print(i, end="")
print()
``` | output | 1 | 3,783 | 7 | 7,567 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Catherine has a deck of n cards, each of which is either red, green, or blue. As long as there are at least two cards left, she can do one of two actions:
* take any two (not necessarily adjacent) cards with different colors and exchange them for a new card of the third color;
* take any two (not necessarily adjacent) cards with the same color and exchange them for a new card with that color.
She repeats this process until there is only one card left. What are the possible colors for the final card?
Input
The first line of the input contains a single integer n (1 ≤ n ≤ 200) — the total number of cards.
The next line contains a string s of length n — the colors of the cards. s contains only the characters 'B', 'G', and 'R', representing blue, green, and red, respectively.
Output
Print a single string of up to three characters — the possible colors of the final card (using the same symbols as the input) in alphabetical order.
Examples
Input
2
RB
Output
G
Input
3
GRG
Output
BR
Input
5
BBBBB
Output
B
Note
In the first sample, Catherine has one red card and one blue card, which she must exchange for a green card.
In the second sample, Catherine has two green cards and one red card. She has two options: she can exchange the two green cards for a green card, then exchange the new green card and the red card for a blue card. Alternatively, she can exchange a green and a red card for a blue card, then exchange the blue card and remaining green card for a red card.
In the third sample, Catherine only has blue cards, so she can only exchange them for more blue cards. | instruction | 0 | 3,784 | 7 | 7,568 |
Tags: constructive algorithms, dp, math
Correct Solution:
```
NumberOfCards = int(input())
InputString = input()
NumbersOfColors = {'B':0,'G':0,'R':0}
for Letter in InputString:
NumbersOfColors[Letter] += 1 # = NumberOfColors.get(Letter,0) + 1
NumberOfColors = 0
for value in list(NumbersOfColors.values()):
if value > 0:
NumberOfColors += 1
if NumberOfColors == 1:
for key,value in NumbersOfColors.items():
if value > 0:
print(key)
quit()
if NumberOfColors == 2:
if NumberOfCards == 2:
for key,value in NumbersOfColors.items():
if value == 0:
print(key)
quit()
if 1 in NumbersOfColors.values():
s = ""
for key,value in NumbersOfColors.items():
if value <= 1:
s += str(key)
print("".join(sorted(s)))
quit()
print("BGR")
#print(NumberOfColors)
``` | output | 1 | 3,784 | 7 | 7,569 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Catherine has a deck of n cards, each of which is either red, green, or blue. As long as there are at least two cards left, she can do one of two actions:
* take any two (not necessarily adjacent) cards with different colors and exchange them for a new card of the third color;
* take any two (not necessarily adjacent) cards with the same color and exchange them for a new card with that color.
She repeats this process until there is only one card left. What are the possible colors for the final card?
Input
The first line of the input contains a single integer n (1 ≤ n ≤ 200) — the total number of cards.
The next line contains a string s of length n — the colors of the cards. s contains only the characters 'B', 'G', and 'R', representing blue, green, and red, respectively.
Output
Print a single string of up to three characters — the possible colors of the final card (using the same symbols as the input) in alphabetical order.
Examples
Input
2
RB
Output
G
Input
3
GRG
Output
BR
Input
5
BBBBB
Output
B
Note
In the first sample, Catherine has one red card and one blue card, which she must exchange for a green card.
In the second sample, Catherine has two green cards and one red card. She has two options: she can exchange the two green cards for a green card, then exchange the new green card and the red card for a blue card. Alternatively, she can exchange a green and a red card for a blue card, then exchange the blue card and remaining green card for a red card.
In the third sample, Catherine only has blue cards, so she can only exchange them for more blue cards. | instruction | 0 | 3,785 | 7 | 7,570 |
Tags: constructive algorithms, dp, math
Correct Solution:
```
a = int(input())
s = input()
arr = [0, 0, 0]
def sum(arr):
c = 0
for i in arr:
c += i
return c
def determine(arr, poss, seen):
if seen[arr[0]][arr[1]][arr[2]]:
return poss
elif sum(arr) == 1:
if arr[0] == 1:
seen[1][0][0] = True
poss[0] = True
elif arr[1] == 1:
seen[0][1][0] = True
poss[1] = True
else:
seen[0][0][1] = True
poss[2] = True
return poss
else:
seen[arr[0]][arr[1]][arr[2]] = True
if arr[0] >= 2:
a = [arr[0] - 1, arr[1], arr[2]]
p1 = determine(a, poss, seen)
for i in range(3):
if p1[i] == True:
poss[i] = True
if arr[1] >= 2:
a = [arr[0], arr[1]-1, arr[2]]
p1 = determine(a, poss, seen)
for i in range(3):
if p1[i] == True:
poss[i] = True
if arr[2] >= 2:
a = [arr[0], arr[1], arr[2]-1]
p1 = determine(a, poss, seen)
for i in range(3):
if p1[i] == True:
poss[i] = True
if arr[0] >= 1 and arr[1] >= 1:
a = [arr[0] - 1, arr[1]-1, arr[2]+1]
p1 = determine(a, poss, seen)
for i in range(3):
if p1[i] == True:
poss[i] = True
if arr[0] >= 1 and arr[2] >= 1:
a = [arr[0] - 1, arr[1]+1, arr[2]-1]
p1 = determine(a, poss, seen)
for i in range(3):
if p1[i] == True:
poss[i] = True
if arr[1] >= 1 and arr[2] >= 1:
a = [arr[0] + 1, arr[1] - 1, arr[2] - 1]
p1 = determine(a, poss, seen)
for i in range(3):
if p1[i] == True:
poss[i] = True
return poss
for i in range(len(s)):
if s[i] == "B":
arr[0] += 1
elif s[i] == "G":
arr[1] += 1
else:
arr[2] += 1
b = arr[0]
g = arr[1]
r = arr[2]
if b > 0 and g > 0 and r > 0:
a = [True, True, True]
elif b == 0 and g == 0:
a = [False, False, True]
elif b == 0 and r == 0:
a = [False, True, False]
elif g == 0 and r == 0:
a = [True, False, False]
elif b == 0 and g > 1 and r > 1:
a = [True, True, True]
elif g == 0 and b > 1 and r > 1:
a = [True, True, True]
elif r == 0 and b > 1 and g > 1:
a = [True, True, True]
else:
seen = [[[False for i in range(r+g+b+1)] for i in range(r+g+b+1)] for i in range(r+g+b+1)]
seen[0][0][0] = True
poss = [False, False, False]
a = determine(arr, poss, seen)
t = ""
if a[0] == True:
t += "B"
if a[1] == True:
t += "G"
if a[2] == True:
t += "R"
print(t)
``` | output | 1 | 3,785 | 7 | 7,571 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Catherine has a deck of n cards, each of which is either red, green, or blue. As long as there are at least two cards left, she can do one of two actions:
* take any two (not necessarily adjacent) cards with different colors and exchange them for a new card of the third color;
* take any two (not necessarily adjacent) cards with the same color and exchange them for a new card with that color.
She repeats this process until there is only one card left. What are the possible colors for the final card?
Input
The first line of the input contains a single integer n (1 ≤ n ≤ 200) — the total number of cards.
The next line contains a string s of length n — the colors of the cards. s contains only the characters 'B', 'G', and 'R', representing blue, green, and red, respectively.
Output
Print a single string of up to three characters — the possible colors of the final card (using the same symbols as the input) in alphabetical order.
Examples
Input
2
RB
Output
G
Input
3
GRG
Output
BR
Input
5
BBBBB
Output
B
Note
In the first sample, Catherine has one red card and one blue card, which she must exchange for a green card.
In the second sample, Catherine has two green cards and one red card. She has two options: she can exchange the two green cards for a green card, then exchange the new green card and the red card for a blue card. Alternatively, she can exchange a green and a red card for a blue card, then exchange the blue card and remaining green card for a red card.
In the third sample, Catherine only has blue cards, so she can only exchange them for more blue cards. | instruction | 0 | 3,786 | 7 | 7,572 |
Tags: constructive algorithms, dp, math
Correct Solution:
```
from collections import defaultdict
def mp(): return map(int,input().split())
def lt(): return list(map(int,input().split()))
def pt(x): print(x)
def ip(): return input()
def it(): return int(input())
def sl(x): return [t for t in x]
def spl(x): return x.split()
def aj(liste, item): liste.append(item)
def bin(x): return "{0:b}".format(x)
def printlist(l): print(''.join([str(x) for x in l]))
def listring(l): return ''.join([str(x) for x in l])
n = it()
s = [x for x in input()]
t = "RGB"
c = [s.count(t[i]) for i in range(3)]
res = []
for i in range(3):
x,y,z = c[i],c[(i+1)%3],c[(i+2)%3]
if (x >= 1 and y == z == 0) or (y >= 1 and z >= 1) or (x >= 1 and (y >= 2 or z >= 2)):
res += t[i]
res.sort()
printlist(res)
``` | output | 1 | 3,786 | 7 | 7,573 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Catherine has a deck of n cards, each of which is either red, green, or blue. As long as there are at least two cards left, she can do one of two actions:
* take any two (not necessarily adjacent) cards with different colors and exchange them for a new card of the third color;
* take any two (not necessarily adjacent) cards with the same color and exchange them for a new card with that color.
She repeats this process until there is only one card left. What are the possible colors for the final card?
Input
The first line of the input contains a single integer n (1 ≤ n ≤ 200) — the total number of cards.
The next line contains a string s of length n — the colors of the cards. s contains only the characters 'B', 'G', and 'R', representing blue, green, and red, respectively.
Output
Print a single string of up to three characters — the possible colors of the final card (using the same symbols as the input) in alphabetical order.
Examples
Input
2
RB
Output
G
Input
3
GRG
Output
BR
Input
5
BBBBB
Output
B
Note
In the first sample, Catherine has one red card and one blue card, which she must exchange for a green card.
In the second sample, Catherine has two green cards and one red card. She has two options: she can exchange the two green cards for a green card, then exchange the new green card and the red card for a blue card. Alternatively, she can exchange a green and a red card for a blue card, then exchange the blue card and remaining green card for a red card.
In the third sample, Catherine only has blue cards, so she can only exchange them for more blue cards. | instruction | 0 | 3,787 | 7 | 7,574 |
Tags: constructive algorithms, dp, math
Correct Solution:
```
n = int(input())
s = input()
r = s.count('R')
g = s.count('G')
b = s.count('B')
if r>=1 and g>=1 and b>=1:
print('BGR')
elif r+g==0:
print('B')
elif b+g==0:
print('R')
elif b+r==0:
print('G')
elif r==0:
if b+g==2:
print('R')
elif b==1 and g>1:
print('BR')
elif g==1 and b>1:
print('GR')
else:
print('BGR')
elif b==0:
if r+g==2:
print('B')
elif r==1 and g>1:
print('BR')
elif g==1 and r>1:
print('BG')
else:
print('BGR')
elif g==0:
if b+r==2:
print('G')
elif b==1 and r>1:
print('BG')
elif r==1 and b>1:
print('GR')
else:
print('BGR')
``` | output | 1 | 3,787 | 7 | 7,575 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Catherine has a deck of n cards, each of which is either red, green, or blue. As long as there are at least two cards left, she can do one of two actions:
* take any two (not necessarily adjacent) cards with different colors and exchange them for a new card of the third color;
* take any two (not necessarily adjacent) cards with the same color and exchange them for a new card with that color.
She repeats this process until there is only one card left. What are the possible colors for the final card?
Input
The first line of the input contains a single integer n (1 ≤ n ≤ 200) — the total number of cards.
The next line contains a string s of length n — the colors of the cards. s contains only the characters 'B', 'G', and 'R', representing blue, green, and red, respectively.
Output
Print a single string of up to three characters — the possible colors of the final card (using the same symbols as the input) in alphabetical order.
Examples
Input
2
RB
Output
G
Input
3
GRG
Output
BR
Input
5
BBBBB
Output
B
Note
In the first sample, Catherine has one red card and one blue card, which she must exchange for a green card.
In the second sample, Catherine has two green cards and one red card. She has two options: she can exchange the two green cards for a green card, then exchange the new green card and the red card for a blue card. Alternatively, she can exchange a green and a red card for a blue card, then exchange the blue card and remaining green card for a red card.
In the third sample, Catherine only has blue cards, so she can only exchange them for more blue cards.
Submitted Solution:
```
n = int(input())
cards = input()
rc = ('R', cards.count('R'))
gc = ('G', cards.count('G'))
bc = ('B', cards.count('B'))
r = "RGB"
a = [rc, gc, bc]
a.sort(key=lambda item: item[1])
if a[0][1] == 0:
if a[1][1] == 0:
r = a[2][0]
elif a[1][1] != a[2][1] and a[1][1] == 1:
r = a[1][0]+a[0][0]
elif a[1][1] == 1:
r = a[0][0]
print(''.join(sorted(r)))
``` | instruction | 0 | 3,788 | 7 | 7,576 |
Yes | output | 1 | 3,788 | 7 | 7,577 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Catherine has a deck of n cards, each of which is either red, green, or blue. As long as there are at least two cards left, she can do one of two actions:
* take any two (not necessarily adjacent) cards with different colors and exchange them for a new card of the third color;
* take any two (not necessarily adjacent) cards with the same color and exchange them for a new card with that color.
She repeats this process until there is only one card left. What are the possible colors for the final card?
Input
The first line of the input contains a single integer n (1 ≤ n ≤ 200) — the total number of cards.
The next line contains a string s of length n — the colors of the cards. s contains only the characters 'B', 'G', and 'R', representing blue, green, and red, respectively.
Output
Print a single string of up to three characters — the possible colors of the final card (using the same symbols as the input) in alphabetical order.
Examples
Input
2
RB
Output
G
Input
3
GRG
Output
BR
Input
5
BBBBB
Output
B
Note
In the first sample, Catherine has one red card and one blue card, which she must exchange for a green card.
In the second sample, Catherine has two green cards and one red card. She has two options: she can exchange the two green cards for a green card, then exchange the new green card and the red card for a blue card. Alternatively, she can exchange a green and a red card for a blue card, then exchange the blue card and remaining green card for a red card.
In the third sample, Catherine only has blue cards, so she can only exchange them for more blue cards.
Submitted Solution:
```
from sys import stdin, stdout
from math import floor, gcd, fabs, factorial, fmod, sqrt, inf, log
from collections import defaultdict as dd, deque
from heapq import merge, heapify, heappop, heappush, nsmallest
from bisect import bisect_left as bl, bisect_right as br, bisect
mod = pow(10, 9) + 7
mod2 = 998244353
def inp(): return stdin.readline().strip()
def iinp(): return int(inp())
def out(var, end="\n"): stdout.write(str(var)+"\n")
def outa(*var, end="\n"): stdout.write(' '.join(map(str, var)) + end)
def lmp(): return list(mp())
def mp(): return map(int, inp().split())
def l1d(n, val=0): return [val for i in range(n)]
def l2d(n, m, val=0): return [l1d(m, val) for j in range(n)]
def ceil(a,b): return (a+b-1)//b
S1 = 'abcdefghijklmnopqrstuvwxyz'
S2 = 'ABCDEFGHIJKLMNOPQRSTUVWXYZ'
def isprime(x):
if x<=1: return False
if x in (2, 3): return True
if x%2 == 0: return False
for i in range(3, int(sqrt(x))+1, 2):
if x%i == 0: return False
return True
n = iinp()
r = g = b = 0
s = inp()
cnt = l1d(3)
t = 'BGR'
for c in s:
cnt[t.index(c)] += 1
cnt = [(cnt[i], t[i]) for i in range(3)]
cnt.sort()
if not cnt[1][0]:
print(cnt[2][1])
elif cnt[0][0] >= 1 or cnt[1][0] >= 2:
print('BGR')
elif cnt[0][0]==cnt[1][0]==1:
print(cnt[2][1])
elif cnt[1][0]==cnt[2][0]==1:
print(cnt[0][1])
else:
print(''.join(sorted([cnt[0][1], cnt[1][1]])))
``` | instruction | 0 | 3,789 | 7 | 7,578 |
Yes | output | 1 | 3,789 | 7 | 7,579 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Catherine has a deck of n cards, each of which is either red, green, or blue. As long as there are at least two cards left, she can do one of two actions:
* take any two (not necessarily adjacent) cards with different colors and exchange them for a new card of the third color;
* take any two (not necessarily adjacent) cards with the same color and exchange them for a new card with that color.
She repeats this process until there is only one card left. What are the possible colors for the final card?
Input
The first line of the input contains a single integer n (1 ≤ n ≤ 200) — the total number of cards.
The next line contains a string s of length n — the colors of the cards. s contains only the characters 'B', 'G', and 'R', representing blue, green, and red, respectively.
Output
Print a single string of up to three characters — the possible colors of the final card (using the same symbols as the input) in alphabetical order.
Examples
Input
2
RB
Output
G
Input
3
GRG
Output
BR
Input
5
BBBBB
Output
B
Note
In the first sample, Catherine has one red card and one blue card, which she must exchange for a green card.
In the second sample, Catherine has two green cards and one red card. She has two options: she can exchange the two green cards for a green card, then exchange the new green card and the red card for a blue card. Alternatively, she can exchange a green and a red card for a blue card, then exchange the blue card and remaining green card for a red card.
In the third sample, Catherine only has blue cards, so she can only exchange them for more blue cards.
Submitted Solution:
```
def main():
from collections import Counter
n = int(input())
if n == 1:
print(input())
return
cnt = Counter({"R": 0, "G": 0, "B": 0})
cnt.update(input())
r, g, b = sorted("RGB", key=cnt.get)
print(''.join(sorted("RGB" if cnt[r] else (r if n == 2 else r + g) if cnt[g] == 1 else "RGB" if cnt[g] else b)))
if __name__ == '__main__':
main()
``` | instruction | 0 | 3,790 | 7 | 7,580 |
Yes | output | 1 | 3,790 | 7 | 7,581 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Catherine has a deck of n cards, each of which is either red, green, or blue. As long as there are at least two cards left, she can do one of two actions:
* take any two (not necessarily adjacent) cards with different colors and exchange them for a new card of the third color;
* take any two (not necessarily adjacent) cards with the same color and exchange them for a new card with that color.
She repeats this process until there is only one card left. What are the possible colors for the final card?
Input
The first line of the input contains a single integer n (1 ≤ n ≤ 200) — the total number of cards.
The next line contains a string s of length n — the colors of the cards. s contains only the characters 'B', 'G', and 'R', representing blue, green, and red, respectively.
Output
Print a single string of up to three characters — the possible colors of the final card (using the same symbols as the input) in alphabetical order.
Examples
Input
2
RB
Output
G
Input
3
GRG
Output
BR
Input
5
BBBBB
Output
B
Note
In the first sample, Catherine has one red card and one blue card, which she must exchange for a green card.
In the second sample, Catherine has two green cards and one red card. She has two options: she can exchange the two green cards for a green card, then exchange the new green card and the red card for a blue card. Alternatively, she can exchange a green and a red card for a blue card, then exchange the blue card and remaining green card for a red card.
In the third sample, Catherine only has blue cards, so she can only exchange them for more blue cards.
Submitted Solution:
```
n = int(input())
s = input()
g = 0
r = 0
b = 0
for i in s:
if i == 'R':
r = r+1
elif i == 'G':
g = g+1
else:
b= b+1
if b == n:
print('B')
elif g == n:
print("G")
elif r == n:
print("R")
elif b >= 1 and g>=1 and r>=1:
print("BGR")
elif b>1 and (g == 1 or r == 1):
print("GR")
elif g>1 and (b==1 or r ==1):
print("BR")
elif r>1 and (g == 1 or b==1):
print("BG")
elif (b>1 and (g>1 or r>1)) or (g>1 and(b>1 or r>1)) or (r>1 and(b>1 or g>1 )):
print("BGR")
elif b == 1 and r == 1:
print("G")
elif b == 1 and g == 1:
print("R")
elif r == 1 and g == 1:
print("B")
``` | instruction | 0 | 3,791 | 7 | 7,582 |
Yes | output | 1 | 3,791 | 7 | 7,583 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Catherine has a deck of n cards, each of which is either red, green, or blue. As long as there are at least two cards left, she can do one of two actions:
* take any two (not necessarily adjacent) cards with different colors and exchange them for a new card of the third color;
* take any two (not necessarily adjacent) cards with the same color and exchange them for a new card with that color.
She repeats this process until there is only one card left. What are the possible colors for the final card?
Input
The first line of the input contains a single integer n (1 ≤ n ≤ 200) — the total number of cards.
The next line contains a string s of length n — the colors of the cards. s contains only the characters 'B', 'G', and 'R', representing blue, green, and red, respectively.
Output
Print a single string of up to three characters — the possible colors of the final card (using the same symbols as the input) in alphabetical order.
Examples
Input
2
RB
Output
G
Input
3
GRG
Output
BR
Input
5
BBBBB
Output
B
Note
In the first sample, Catherine has one red card and one blue card, which she must exchange for a green card.
In the second sample, Catherine has two green cards and one red card. She has two options: she can exchange the two green cards for a green card, then exchange the new green card and the red card for a blue card. Alternatively, she can exchange a green and a red card for a blue card, then exchange the blue card and remaining green card for a red card.
In the third sample, Catherine only has blue cards, so she can only exchange them for more blue cards.
Submitted Solution:
```
N = 205
memo = [[[0]*205 for i in range(N)] for j in range(N)]
n = int(input())
s = input()
ans = [0,0,0]
def dp(r,g,b):
global ans,memo
if memo[r][g][b]: return
memo[r][g][b] = 1
if max(r,g,b) == 1:
ans[0 if r else 1 if g else 2] = 1
else:
if r >= 2:
dp(r-1,g,b)
if g >= 2:
dp(r,g-1,b)
if b >= 2:
dp(r,g,b-1)
if r and g:
dp(r-1,g-1,b+1)
if g and b:
dp(r+1,g-1,b-1)
if b and r:
dp(r-1,g+1,b-1)
dp(s.count("R"),s.count("G"),s.count("B"))
ass = "RGB"
print(*[ass[i] for i in range(3) if ans[i]],sep="")
``` | instruction | 0 | 3,792 | 7 | 7,584 |
No | output | 1 | 3,792 | 7 | 7,585 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Catherine has a deck of n cards, each of which is either red, green, or blue. As long as there are at least two cards left, she can do one of two actions:
* take any two (not necessarily adjacent) cards with different colors and exchange them for a new card of the third color;
* take any two (not necessarily adjacent) cards with the same color and exchange them for a new card with that color.
She repeats this process until there is only one card left. What are the possible colors for the final card?
Input
The first line of the input contains a single integer n (1 ≤ n ≤ 200) — the total number of cards.
The next line contains a string s of length n — the colors of the cards. s contains only the characters 'B', 'G', and 'R', representing blue, green, and red, respectively.
Output
Print a single string of up to three characters — the possible colors of the final card (using the same symbols as the input) in alphabetical order.
Examples
Input
2
RB
Output
G
Input
3
GRG
Output
BR
Input
5
BBBBB
Output
B
Note
In the first sample, Catherine has one red card and one blue card, which she must exchange for a green card.
In the second sample, Catherine has two green cards and one red card. She has two options: she can exchange the two green cards for a green card, then exchange the new green card and the red card for a blue card. Alternatively, she can exchange a green and a red card for a blue card, then exchange the blue card and remaining green card for a red card.
In the third sample, Catherine only has blue cards, so she can only exchange them for more blue cards.
Submitted Solution:
```
n, s = int(input()), input()
r, g = s.count('R'), s.count('G')
b = n - r - g
print('%s%s%s' % ('B' if g and r or b and g + r > 1 or b > 1 and g + r < 1 else '', 'G' if r and b or g and b + r > 1 or g > 1 and b + r < 1 else '', 'R' if g and b or r and g + b > 1 or r > 1 and g + b < 1 else ''))
``` | instruction | 0 | 3,793 | 7 | 7,586 |
No | output | 1 | 3,793 | 7 | 7,587 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Catherine has a deck of n cards, each of which is either red, green, or blue. As long as there are at least two cards left, she can do one of two actions:
* take any two (not necessarily adjacent) cards with different colors and exchange them for a new card of the third color;
* take any two (not necessarily adjacent) cards with the same color and exchange them for a new card with that color.
She repeats this process until there is only one card left. What are the possible colors for the final card?
Input
The first line of the input contains a single integer n (1 ≤ n ≤ 200) — the total number of cards.
The next line contains a string s of length n — the colors of the cards. s contains only the characters 'B', 'G', and 'R', representing blue, green, and red, respectively.
Output
Print a single string of up to three characters — the possible colors of the final card (using the same symbols as the input) in alphabetical order.
Examples
Input
2
RB
Output
G
Input
3
GRG
Output
BR
Input
5
BBBBB
Output
B
Note
In the first sample, Catherine has one red card and one blue card, which she must exchange for a green card.
In the second sample, Catherine has two green cards and one red card. She has two options: she can exchange the two green cards for a green card, then exchange the new green card and the red card for a blue card. Alternatively, she can exchange a green and a red card for a blue card, then exchange the blue card and remaining green card for a red card.
In the third sample, Catherine only has blue cards, so she can only exchange them for more blue cards.
Submitted Solution:
```
n = int(input())
s = input()
r,g,b = 0,0,0
for i in range(n):
if s[i] == "R":
r+=1
if s[i] == "G":
g+=1
if s[i] == "B":
b+=1
if r != 0 and g == 0 and b == 0:
print("R")
elif r == 0 and g != 0 and b == 0:
print("G")
elif r == 0 and g == 0 and b != 0:
print("B")
elif r > 0 and g > 0 and b > 0:
print("RGB")
elif (r > 1 and b > 1 and b == 0):
print("RGB")
elif (r > 1 and b == 0 and b > 1):
print("RGB")
elif (r == 0 and b > 1 and b > 1):
print("RGB")
elif r == 1 and g == 1 and b == 0:
print("B")
elif r == 1 and g == 0 and b == 1:
print("G")
elif r == 0 and g == 1 and b == 1:
print("R")
else:
if r > 1 and b + g == 1:
print("BG")
elif g > 1 and b + r == 1:
print("BR")
elif b > 1 and r + g == 1:
print("RG")
``` | instruction | 0 | 3,794 | 7 | 7,588 |
No | output | 1 | 3,794 | 7 | 7,589 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Catherine has a deck of n cards, each of which is either red, green, or blue. As long as there are at least two cards left, she can do one of two actions:
* take any two (not necessarily adjacent) cards with different colors and exchange them for a new card of the third color;
* take any two (not necessarily adjacent) cards with the same color and exchange them for a new card with that color.
She repeats this process until there is only one card left. What are the possible colors for the final card?
Input
The first line of the input contains a single integer n (1 ≤ n ≤ 200) — the total number of cards.
The next line contains a string s of length n — the colors of the cards. s contains only the characters 'B', 'G', and 'R', representing blue, green, and red, respectively.
Output
Print a single string of up to three characters — the possible colors of the final card (using the same symbols as the input) in alphabetical order.
Examples
Input
2
RB
Output
G
Input
3
GRG
Output
BR
Input
5
BBBBB
Output
B
Note
In the first sample, Catherine has one red card and one blue card, which she must exchange for a green card.
In the second sample, Catherine has two green cards and one red card. She has two options: she can exchange the two green cards for a green card, then exchange the new green card and the red card for a blue card. Alternatively, she can exchange a green and a red card for a blue card, then exchange the blue card and remaining green card for a red card.
In the third sample, Catherine only has blue cards, so she can only exchange them for more blue cards.
Submitted Solution:
```
n=int(input())
s=input()
i,e=0,0
while i<n:
k=i+1
while k<=n:
c,d=0,0
for j in range(i,k):
if s[j]=='U':
c+=1
elif s[j]=='D':
c-=1
elif s[j]=='R':
d+=1
else:
d-=1
if c==0 and c==0:
e+=1
k+=1
i+=1
print(e)
``` | instruction | 0 | 3,795 | 7 | 7,590 |
No | output | 1 | 3,795 | 7 | 7,591 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Professor Ibrahim has prepared the final homework for his algorithm’s class. He asked his students to implement the Posterization Image Filter.
Their algorithm will be tested on an array of integers, where the i-th integer represents the color of the i-th pixel in the image. The image is in black and white, therefore the color of each pixel will be an integer between 0 and 255 (inclusive).
To implement the filter, students are required to divide the black and white color range [0, 255] into groups of consecutive colors, and select one color in each group to be the group’s key. In order to preserve image details, the size of a group must not be greater than k, and each color should belong to exactly one group.
Finally, the students will replace the color of each pixel in the array with that color’s assigned group key.
To better understand the effect, here is an image of a basking turtle where the Posterization Filter was applied with increasing k to the right.
<image>
To make the process of checking the final answer easier, Professor Ibrahim wants students to divide the groups and assign the keys in a way that produces the lexicographically smallest possible array.
Input
The first line of input contains two integers n and k (1 ≤ n ≤ 10^5, 1 ≤ k ≤ 256), the number of pixels in the image, and the maximum size of a group, respectively.
The second line contains n integers p_1, p_2, ..., p_n (0 ≤ p_i ≤ 255), where p_i is the color of the i-th pixel.
Output
Print n space-separated integers; the lexicographically smallest possible array that represents the image after applying the Posterization filter.
Examples
Input
4 3
2 14 3 4
Output
0 12 3 3
Input
5 2
0 2 1 255 254
Output
0 1 1 254 254
Note
One possible way to group colors and assign keys for the first sample:
Color 2 belongs to the group [0,2], with group key 0.
Color 14 belongs to the group [12,14], with group key 12.
Colors 3 and 4 belong to group [3, 5], with group key 3.
Other groups won't affect the result so they are not listed here. | instruction | 0 | 3,916 | 7 | 7,832 |
Tags: games, greedy
Correct Solution:
```
n, k = map(int, input().strip().split())
data = map(int, input().strip().split())
sol = []
mapping = [(-1,1000)]*256
for x in data:
if mapping[x][0] == -1:
for i in range(max(x-k+1,0), x+1):
if mapping[i][0] == -1:
if i > 0 and mapping[i-1][1]+(x-i+1) <= k:
p = mapping[i-1][1]+1
for j in range(i, x+1):
mapping[j] = (mapping[i-1][0], p)
p += 1
else:
p = 1
for j in range(i, x+1):
mapping[j] = (i, p)
p += 1
break
sol.append(mapping[x][0])
print(' '.join(map(str, sol)))
``` | output | 1 | 3,916 | 7 | 7,833 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Professor Ibrahim has prepared the final homework for his algorithm’s class. He asked his students to implement the Posterization Image Filter.
Their algorithm will be tested on an array of integers, where the i-th integer represents the color of the i-th pixel in the image. The image is in black and white, therefore the color of each pixel will be an integer between 0 and 255 (inclusive).
To implement the filter, students are required to divide the black and white color range [0, 255] into groups of consecutive colors, and select one color in each group to be the group’s key. In order to preserve image details, the size of a group must not be greater than k, and each color should belong to exactly one group.
Finally, the students will replace the color of each pixel in the array with that color’s assigned group key.
To better understand the effect, here is an image of a basking turtle where the Posterization Filter was applied with increasing k to the right.
<image>
To make the process of checking the final answer easier, Professor Ibrahim wants students to divide the groups and assign the keys in a way that produces the lexicographically smallest possible array.
Input
The first line of input contains two integers n and k (1 ≤ n ≤ 10^5, 1 ≤ k ≤ 256), the number of pixels in the image, and the maximum size of a group, respectively.
The second line contains n integers p_1, p_2, ..., p_n (0 ≤ p_i ≤ 255), where p_i is the color of the i-th pixel.
Output
Print n space-separated integers; the lexicographically smallest possible array that represents the image after applying the Posterization filter.
Examples
Input
4 3
2 14 3 4
Output
0 12 3 3
Input
5 2
0 2 1 255 254
Output
0 1 1 254 254
Note
One possible way to group colors and assign keys for the first sample:
Color 2 belongs to the group [0,2], with group key 0.
Color 14 belongs to the group [12,14], with group key 12.
Colors 3 and 4 belong to group [3, 5], with group key 3.
Other groups won't affect the result so they are not listed here. | instruction | 0 | 3,917 | 7 | 7,834 |
Tags: games, greedy
Correct Solution:
```
R = lambda: map(int, input().split())
n, k = R()
a = list(range(0, 257)); v = [1]*257
# print(a)
# print(v)
for p in R():
if v[p]:
t = p
# print(t)
while t >= 0 and p-a[t]<=k-1:
# print(t)
t -= 1
t += 1
for i in range(t, p+1): a[i] = a[t]; v[i] = 0
print(a[p], end=' ')
``` | output | 1 | 3,917 | 7 | 7,835 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Professor Ibrahim has prepared the final homework for his algorithm’s class. He asked his students to implement the Posterization Image Filter.
Their algorithm will be tested on an array of integers, where the i-th integer represents the color of the i-th pixel in the image. The image is in black and white, therefore the color of each pixel will be an integer between 0 and 255 (inclusive).
To implement the filter, students are required to divide the black and white color range [0, 255] into groups of consecutive colors, and select one color in each group to be the group’s key. In order to preserve image details, the size of a group must not be greater than k, and each color should belong to exactly one group.
Finally, the students will replace the color of each pixel in the array with that color’s assigned group key.
To better understand the effect, here is an image of a basking turtle where the Posterization Filter was applied with increasing k to the right.
<image>
To make the process of checking the final answer easier, Professor Ibrahim wants students to divide the groups and assign the keys in a way that produces the lexicographically smallest possible array.
Input
The first line of input contains two integers n and k (1 ≤ n ≤ 10^5, 1 ≤ k ≤ 256), the number of pixels in the image, and the maximum size of a group, respectively.
The second line contains n integers p_1, p_2, ..., p_n (0 ≤ p_i ≤ 255), where p_i is the color of the i-th pixel.
Output
Print n space-separated integers; the lexicographically smallest possible array that represents the image after applying the Posterization filter.
Examples
Input
4 3
2 14 3 4
Output
0 12 3 3
Input
5 2
0 2 1 255 254
Output
0 1 1 254 254
Note
One possible way to group colors and assign keys for the first sample:
Color 2 belongs to the group [0,2], with group key 0.
Color 14 belongs to the group [12,14], with group key 12.
Colors 3 and 4 belong to group [3, 5], with group key 3.
Other groups won't affect the result so they are not listed here. | instruction | 0 | 3,918 | 7 | 7,836 |
Tags: games, greedy
Correct Solution:
```
import atexit
import io
import sys
# Buffering IO
_INPUT_LINES = sys.stdin.read().splitlines()
input = iter(_INPUT_LINES).__next__
_OUTPUT_BUFFER = io.StringIO()
sys.stdout = _OUTPUT_BUFFER
@atexit.register
def write():
sys.__stdout__.write(_OUTPUT_BUFFER.getvalue())
def main():
n, k = [int(x) for x in input().split()]
p = [int(x) for x in input().split()]
t = []
g={}
for x in p:
if x in g:
t.append(g[x])
continue
kk = x - 1
while True:
if kk in g:
if x - g[kk] < k:
ttt = g[kk]
else:
ttt= kk + 1
for i in range(kk +1 , x + 1):
g[i] = ttt
t.append(g[x])
break
elif kk<0 or x - kk == k:
for i in range(kk +1 , x + 1):
g[i] = kk + 1
t.append(g[x])
break
kk -= 1
print(' '.join((str(x) for x in t)))
if __name__ == '__main__':
main()
``` | output | 1 | 3,918 | 7 | 7,837 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Professor Ibrahim has prepared the final homework for his algorithm’s class. He asked his students to implement the Posterization Image Filter.
Their algorithm will be tested on an array of integers, where the i-th integer represents the color of the i-th pixel in the image. The image is in black and white, therefore the color of each pixel will be an integer between 0 and 255 (inclusive).
To implement the filter, students are required to divide the black and white color range [0, 255] into groups of consecutive colors, and select one color in each group to be the group’s key. In order to preserve image details, the size of a group must not be greater than k, and each color should belong to exactly one group.
Finally, the students will replace the color of each pixel in the array with that color’s assigned group key.
To better understand the effect, here is an image of a basking turtle where the Posterization Filter was applied with increasing k to the right.
<image>
To make the process of checking the final answer easier, Professor Ibrahim wants students to divide the groups and assign the keys in a way that produces the lexicographically smallest possible array.
Input
The first line of input contains two integers n and k (1 ≤ n ≤ 10^5, 1 ≤ k ≤ 256), the number of pixels in the image, and the maximum size of a group, respectively.
The second line contains n integers p_1, p_2, ..., p_n (0 ≤ p_i ≤ 255), where p_i is the color of the i-th pixel.
Output
Print n space-separated integers; the lexicographically smallest possible array that represents the image after applying the Posterization filter.
Examples
Input
4 3
2 14 3 4
Output
0 12 3 3
Input
5 2
0 2 1 255 254
Output
0 1 1 254 254
Note
One possible way to group colors and assign keys for the first sample:
Color 2 belongs to the group [0,2], with group key 0.
Color 14 belongs to the group [12,14], with group key 12.
Colors 3 and 4 belong to group [3, 5], with group key 3.
Other groups won't affect the result so they are not listed here. | instruction | 0 | 3,919 | 7 | 7,838 |
Tags: games, greedy
Correct Solution:
```
N, K = input().split()
N, K = int(N), int(K)
P = [int(x) for x in input().split()]
A = [None]*256
A[0] = 0
for i in range(N):
pn = P[i]
if A[pn] is None:
for j in range(K-1, -1, -1):
if pn < j: continue
if A[pn-j] is None:
A[pn-j] = pn-j
break
else:
if A[pn-j] + K - 1 >= pn:
break
for jj in range(j, -1, -1):
A[pn-jj] = A[pn-j]
print(*[A[P[i]] for i in range(N)])
``` | output | 1 | 3,919 | 7 | 7,839 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Professor Ibrahim has prepared the final homework for his algorithm’s class. He asked his students to implement the Posterization Image Filter.
Their algorithm will be tested on an array of integers, where the i-th integer represents the color of the i-th pixel in the image. The image is in black and white, therefore the color of each pixel will be an integer between 0 and 255 (inclusive).
To implement the filter, students are required to divide the black and white color range [0, 255] into groups of consecutive colors, and select one color in each group to be the group’s key. In order to preserve image details, the size of a group must not be greater than k, and each color should belong to exactly one group.
Finally, the students will replace the color of each pixel in the array with that color’s assigned group key.
To better understand the effect, here is an image of a basking turtle where the Posterization Filter was applied with increasing k to the right.
<image>
To make the process of checking the final answer easier, Professor Ibrahim wants students to divide the groups and assign the keys in a way that produces the lexicographically smallest possible array.
Input
The first line of input contains two integers n and k (1 ≤ n ≤ 10^5, 1 ≤ k ≤ 256), the number of pixels in the image, and the maximum size of a group, respectively.
The second line contains n integers p_1, p_2, ..., p_n (0 ≤ p_i ≤ 255), where p_i is the color of the i-th pixel.
Output
Print n space-separated integers; the lexicographically smallest possible array that represents the image after applying the Posterization filter.
Examples
Input
4 3
2 14 3 4
Output
0 12 3 3
Input
5 2
0 2 1 255 254
Output
0 1 1 254 254
Note
One possible way to group colors and assign keys for the first sample:
Color 2 belongs to the group [0,2], with group key 0.
Color 14 belongs to the group [12,14], with group key 12.
Colors 3 and 4 belong to group [3, 5], with group key 3.
Other groups won't affect the result so they are not listed here. | instruction | 0 | 3,920 | 7 | 7,840 |
Tags: games, greedy
Correct Solution:
```
n, k = map(int, input().split())
a = map(int, input().split())
ans = [0]*n
f = [-1]*256
i = -1
for x in a:
i += 1
if f[x] < 0:
l = x
while l > 0 > f[l - 1] and x - l < k - 1:
l -= 1
if l > 0 and x - f[l - 1] < k and f[l - 1] >= 0:
f[x] = f[l - 1]
else:
f[x] = l
for j in range(l, x):
f[j] = f[x]
ans[i] = f[x]
for x in ans:
print(x, end=' ')
``` | output | 1 | 3,920 | 7 | 7,841 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Professor Ibrahim has prepared the final homework for his algorithm’s class. He asked his students to implement the Posterization Image Filter.
Their algorithm will be tested on an array of integers, where the i-th integer represents the color of the i-th pixel in the image. The image is in black and white, therefore the color of each pixel will be an integer between 0 and 255 (inclusive).
To implement the filter, students are required to divide the black and white color range [0, 255] into groups of consecutive colors, and select one color in each group to be the group’s key. In order to preserve image details, the size of a group must not be greater than k, and each color should belong to exactly one group.
Finally, the students will replace the color of each pixel in the array with that color’s assigned group key.
To better understand the effect, here is an image of a basking turtle where the Posterization Filter was applied with increasing k to the right.
<image>
To make the process of checking the final answer easier, Professor Ibrahim wants students to divide the groups and assign the keys in a way that produces the lexicographically smallest possible array.
Input
The first line of input contains two integers n and k (1 ≤ n ≤ 10^5, 1 ≤ k ≤ 256), the number of pixels in the image, and the maximum size of a group, respectively.
The second line contains n integers p_1, p_2, ..., p_n (0 ≤ p_i ≤ 255), where p_i is the color of the i-th pixel.
Output
Print n space-separated integers; the lexicographically smallest possible array that represents the image after applying the Posterization filter.
Examples
Input
4 3
2 14 3 4
Output
0 12 3 3
Input
5 2
0 2 1 255 254
Output
0 1 1 254 254
Note
One possible way to group colors and assign keys for the first sample:
Color 2 belongs to the group [0,2], with group key 0.
Color 14 belongs to the group [12,14], with group key 12.
Colors 3 and 4 belong to group [3, 5], with group key 3.
Other groups won't affect the result so they are not listed here. | instruction | 0 | 3,921 | 7 | 7,842 |
Tags: games, greedy
Correct Solution:
```
n, k = list(map(int, input().split()))
p = list(map(int, input().split()))
processed = set()
color = {}
length = {}
ans = []
def exists(p, elt, d):
for e in p:
if e > elt:
if e <= elt + d:
return True
elif e - d <= elt + d:
return False
return False
def exists2(p, elt, d):
for e in p:
if e > elt:
if e <= elt + d:
return False
elif e - d <= elt + d:
return [True, e - d]
return False
for i in range(n):
elt = p[i]
if elt in processed:
ans.append(color[elt])
else:
processed.add(elt)
new = 1
run = True
for j in range(1, k):
if elt - j < 0:
break
elif (elt - j) not in processed:
processed.add(elt - j)
new += 1
elif length[elt - j] + new <= k:
for i2 in range(length[elt - j] + new):
color[elt - i2] = color[elt - j]
length[elt] = length[elt - j] + new
run = False
break
else:
break
if run:
for j in range(new):
color[elt - j] = elt - new + 1
length[elt] = new
s = str(color[p[0]])
for elt in p[1:]:
s += ' ' + str(color[elt])
print(s)
``` | output | 1 | 3,921 | 7 | 7,843 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Professor Ibrahim has prepared the final homework for his algorithm’s class. He asked his students to implement the Posterization Image Filter.
Their algorithm will be tested on an array of integers, where the i-th integer represents the color of the i-th pixel in the image. The image is in black and white, therefore the color of each pixel will be an integer between 0 and 255 (inclusive).
To implement the filter, students are required to divide the black and white color range [0, 255] into groups of consecutive colors, and select one color in each group to be the group’s key. In order to preserve image details, the size of a group must not be greater than k, and each color should belong to exactly one group.
Finally, the students will replace the color of each pixel in the array with that color’s assigned group key.
To better understand the effect, here is an image of a basking turtle where the Posterization Filter was applied with increasing k to the right.
<image>
To make the process of checking the final answer easier, Professor Ibrahim wants students to divide the groups and assign the keys in a way that produces the lexicographically smallest possible array.
Input
The first line of input contains two integers n and k (1 ≤ n ≤ 10^5, 1 ≤ k ≤ 256), the number of pixels in the image, and the maximum size of a group, respectively.
The second line contains n integers p_1, p_2, ..., p_n (0 ≤ p_i ≤ 255), where p_i is the color of the i-th pixel.
Output
Print n space-separated integers; the lexicographically smallest possible array that represents the image after applying the Posterization filter.
Examples
Input
4 3
2 14 3 4
Output
0 12 3 3
Input
5 2
0 2 1 255 254
Output
0 1 1 254 254
Note
One possible way to group colors and assign keys for the first sample:
Color 2 belongs to the group [0,2], with group key 0.
Color 14 belongs to the group [12,14], with group key 12.
Colors 3 and 4 belong to group [3, 5], with group key 3.
Other groups won't affect the result so they are not listed here. | instruction | 0 | 3,922 | 7 | 7,844 |
Tags: games, greedy
Correct Solution:
```
R = lambda: map(int, input().split())
n, k = R()
a = list(range(0, 257)); v = [1]*257
for p in R():
if v[p]:
t = p
while t >= 0 and p-a[t]<=k-1: t -= 1
t += 1
for i in range(t, p+1): a[i] = a[t]; v[i] = 0
print(a[p], end=' ')
``` | output | 1 | 3,922 | 7 | 7,845 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Professor Ibrahim has prepared the final homework for his algorithm’s class. He asked his students to implement the Posterization Image Filter.
Their algorithm will be tested on an array of integers, where the i-th integer represents the color of the i-th pixel in the image. The image is in black and white, therefore the color of each pixel will be an integer between 0 and 255 (inclusive).
To implement the filter, students are required to divide the black and white color range [0, 255] into groups of consecutive colors, and select one color in each group to be the group’s key. In order to preserve image details, the size of a group must not be greater than k, and each color should belong to exactly one group.
Finally, the students will replace the color of each pixel in the array with that color’s assigned group key.
To better understand the effect, here is an image of a basking turtle where the Posterization Filter was applied with increasing k to the right.
<image>
To make the process of checking the final answer easier, Professor Ibrahim wants students to divide the groups and assign the keys in a way that produces the lexicographically smallest possible array.
Input
The first line of input contains two integers n and k (1 ≤ n ≤ 10^5, 1 ≤ k ≤ 256), the number of pixels in the image, and the maximum size of a group, respectively.
The second line contains n integers p_1, p_2, ..., p_n (0 ≤ p_i ≤ 255), where p_i is the color of the i-th pixel.
Output
Print n space-separated integers; the lexicographically smallest possible array that represents the image after applying the Posterization filter.
Examples
Input
4 3
2 14 3 4
Output
0 12 3 3
Input
5 2
0 2 1 255 254
Output
0 1 1 254 254
Note
One possible way to group colors and assign keys for the first sample:
Color 2 belongs to the group [0,2], with group key 0.
Color 14 belongs to the group [12,14], with group key 12.
Colors 3 and 4 belong to group [3, 5], with group key 3.
Other groups won't affect the result so they are not listed here. | instruction | 0 | 3,923 | 7 | 7,846 |
Tags: games, greedy
Correct Solution:
```
n,k = list(map(int,input().split()))
arr = list(map(int,input().split()))
val = [-1]*256
val[0] = 0
for i in range(0,n):
if(val[arr[i]] > -1):
print(val[arr[i]],end = " ")
else:
flag = 0
for j in range(1,k):
if( val[arr[i]-j] > -1 ):
if(arr[i] - val[arr[i]-j] < k):
for p in range(arr[i] - j + 1,arr[i]+1):
val[p] = val[arr[i]-j]
else:
for t in range(1,j+1):
val[arr[i]- t + 1] = arr[i] - j + 1
# print(val[arr[i]],"OOO")
#print(val[:i],i)
print(val[arr[i]],end = " ")
flag = 1
break
if(flag == 0):
for j in range(0,k):
val[arr[i] - j] = arr[i] -k + 1
print(val[arr[i]],end = " ")
``` | output | 1 | 3,923 | 7 | 7,847 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Professor Ibrahim has prepared the final homework for his algorithm’s class. He asked his students to implement the Posterization Image Filter.
Their algorithm will be tested on an array of integers, where the i-th integer represents the color of the i-th pixel in the image. The image is in black and white, therefore the color of each pixel will be an integer between 0 and 255 (inclusive).
To implement the filter, students are required to divide the black and white color range [0, 255] into groups of consecutive colors, and select one color in each group to be the group’s key. In order to preserve image details, the size of a group must not be greater than k, and each color should belong to exactly one group.
Finally, the students will replace the color of each pixel in the array with that color’s assigned group key.
To better understand the effect, here is an image of a basking turtle where the Posterization Filter was applied with increasing k to the right.
<image>
To make the process of checking the final answer easier, Professor Ibrahim wants students to divide the groups and assign the keys in a way that produces the lexicographically smallest possible array.
Input
The first line of input contains two integers n and k (1 ≤ n ≤ 10^5, 1 ≤ k ≤ 256), the number of pixels in the image, and the maximum size of a group, respectively.
The second line contains n integers p_1, p_2, ..., p_n (0 ≤ p_i ≤ 255), where p_i is the color of the i-th pixel.
Output
Print n space-separated integers; the lexicographically smallest possible array that represents the image after applying the Posterization filter.
Examples
Input
4 3
2 14 3 4
Output
0 12 3 3
Input
5 2
0 2 1 255 254
Output
0 1 1 254 254
Note
One possible way to group colors and assign keys for the first sample:
Color 2 belongs to the group [0,2], with group key 0.
Color 14 belongs to the group [12,14], with group key 12.
Colors 3 and 4 belong to group [3, 5], with group key 3.
Other groups won't affect the result so they are not listed here.
Submitted Solution:
```
n,k = [int(s) for s in input().split()]
p = [int(s) for s in input().split()]
map = {}
res = []
for pi in p:
if map.get(pi) is None:
key = pi
for j in range(pi, pi-k, -1):
if j < 0: break
if map.get(j) is None:
key = j
else:
if map[j] >= pi-k+1: key = map[j]
break
for j in range(pi, key-1, -1):
if map.get(j):
break
map[j] = key
res.append(map[pi])
print(*res, sep=" ")
``` | instruction | 0 | 3,924 | 7 | 7,848 |
Yes | output | 1 | 3,924 | 7 | 7,849 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Professor Ibrahim has prepared the final homework for his algorithm’s class. He asked his students to implement the Posterization Image Filter.
Their algorithm will be tested on an array of integers, where the i-th integer represents the color of the i-th pixel in the image. The image is in black and white, therefore the color of each pixel will be an integer between 0 and 255 (inclusive).
To implement the filter, students are required to divide the black and white color range [0, 255] into groups of consecutive colors, and select one color in each group to be the group’s key. In order to preserve image details, the size of a group must not be greater than k, and each color should belong to exactly one group.
Finally, the students will replace the color of each pixel in the array with that color’s assigned group key.
To better understand the effect, here is an image of a basking turtle where the Posterization Filter was applied with increasing k to the right.
<image>
To make the process of checking the final answer easier, Professor Ibrahim wants students to divide the groups and assign the keys in a way that produces the lexicographically smallest possible array.
Input
The first line of input contains two integers n and k (1 ≤ n ≤ 10^5, 1 ≤ k ≤ 256), the number of pixels in the image, and the maximum size of a group, respectively.
The second line contains n integers p_1, p_2, ..., p_n (0 ≤ p_i ≤ 255), where p_i is the color of the i-th pixel.
Output
Print n space-separated integers; the lexicographically smallest possible array that represents the image after applying the Posterization filter.
Examples
Input
4 3
2 14 3 4
Output
0 12 3 3
Input
5 2
0 2 1 255 254
Output
0 1 1 254 254
Note
One possible way to group colors and assign keys for the first sample:
Color 2 belongs to the group [0,2], with group key 0.
Color 14 belongs to the group [12,14], with group key 12.
Colors 3 and 4 belong to group [3, 5], with group key 3.
Other groups won't affect the result so they are not listed here.
Submitted Solution:
```
read=lambda:map(int,input().split())
n,k=read()
p=list(read())
r=[-1]*256
for i in p:
for u in range(i,max(i-k,-1),-1):
if r[u]>=0: u+=1;break
if u<=i:
if u>0 and r[u-1]+k>i: r[i]=r[u-1]
else:
for j in range(u,i+1): r[j]=u
print(r[i])
``` | instruction | 0 | 3,925 | 7 | 7,850 |
Yes | output | 1 | 3,925 | 7 | 7,851 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Professor Ibrahim has prepared the final homework for his algorithm’s class. He asked his students to implement the Posterization Image Filter.
Their algorithm will be tested on an array of integers, where the i-th integer represents the color of the i-th pixel in the image. The image is in black and white, therefore the color of each pixel will be an integer between 0 and 255 (inclusive).
To implement the filter, students are required to divide the black and white color range [0, 255] into groups of consecutive colors, and select one color in each group to be the group’s key. In order to preserve image details, the size of a group must not be greater than k, and each color should belong to exactly one group.
Finally, the students will replace the color of each pixel in the array with that color’s assigned group key.
To better understand the effect, here is an image of a basking turtle where the Posterization Filter was applied with increasing k to the right.
<image>
To make the process of checking the final answer easier, Professor Ibrahim wants students to divide the groups and assign the keys in a way that produces the lexicographically smallest possible array.
Input
The first line of input contains two integers n and k (1 ≤ n ≤ 10^5, 1 ≤ k ≤ 256), the number of pixels in the image, and the maximum size of a group, respectively.
The second line contains n integers p_1, p_2, ..., p_n (0 ≤ p_i ≤ 255), where p_i is the color of the i-th pixel.
Output
Print n space-separated integers; the lexicographically smallest possible array that represents the image after applying the Posterization filter.
Examples
Input
4 3
2 14 3 4
Output
0 12 3 3
Input
5 2
0 2 1 255 254
Output
0 1 1 254 254
Note
One possible way to group colors and assign keys for the first sample:
Color 2 belongs to the group [0,2], with group key 0.
Color 14 belongs to the group [12,14], with group key 12.
Colors 3 and 4 belong to group [3, 5], with group key 3.
Other groups won't affect the result so they are not listed here.
Submitted Solution:
```
n,k=map(int,input().split())
arr=list(map(int,input().split()))
par=[i for i in range(260)]
path=[-1 for i in range(260)]
for i in range(n):
j=arr[i]
if path[j] >=0:
par[j] =par[path[j]]
continue
jump=1
while j>0 and path[j] ==-1 and jump <k:
path[j] =arr[i]
j-=1
jump +=1
if arr[i] -par[j] +1 <=k:
par[arr[i]] =par[j]
path[j] =arr[i]
else:
par[arr[i]] =par[j+1]
for i in range(n):
print(par[arr[i]],end=' ')
print()
``` | instruction | 0 | 3,926 | 7 | 7,852 |
Yes | output | 1 | 3,926 | 7 | 7,853 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Professor Ibrahim has prepared the final homework for his algorithm’s class. He asked his students to implement the Posterization Image Filter.
Their algorithm will be tested on an array of integers, where the i-th integer represents the color of the i-th pixel in the image. The image is in black and white, therefore the color of each pixel will be an integer between 0 and 255 (inclusive).
To implement the filter, students are required to divide the black and white color range [0, 255] into groups of consecutive colors, and select one color in each group to be the group’s key. In order to preserve image details, the size of a group must not be greater than k, and each color should belong to exactly one group.
Finally, the students will replace the color of each pixel in the array with that color’s assigned group key.
To better understand the effect, here is an image of a basking turtle where the Posterization Filter was applied with increasing k to the right.
<image>
To make the process of checking the final answer easier, Professor Ibrahim wants students to divide the groups and assign the keys in a way that produces the lexicographically smallest possible array.
Input
The first line of input contains two integers n and k (1 ≤ n ≤ 10^5, 1 ≤ k ≤ 256), the number of pixels in the image, and the maximum size of a group, respectively.
The second line contains n integers p_1, p_2, ..., p_n (0 ≤ p_i ≤ 255), where p_i is the color of the i-th pixel.
Output
Print n space-separated integers; the lexicographically smallest possible array that represents the image after applying the Posterization filter.
Examples
Input
4 3
2 14 3 4
Output
0 12 3 3
Input
5 2
0 2 1 255 254
Output
0 1 1 254 254
Note
One possible way to group colors and assign keys for the first sample:
Color 2 belongs to the group [0,2], with group key 0.
Color 14 belongs to the group [12,14], with group key 12.
Colors 3 and 4 belong to group [3, 5], with group key 3.
Other groups won't affect the result so they are not listed here.
Submitted Solution:
```
def main():
n, k = map(int, input().split())
p = list(map(int, input().split()))
solve(n, k, p)
def solve(n, k, p):
group = 256 * [None]
r = p[:]
for i, pi in enumerate(p):
# print([(i, gi) for i, gi in enumerate(group)if gi is not None])
if group[pi] is not None:
r[i] = group[pi][0]
else:
lo = pi
while lo >= 0 and pi - lo < k and group[lo] is None:
lo -= 1
if lo < 0 or pi - lo == k:
lo += 1
hi = pi + 1
else: # group[lo] is not None
if pi - group[lo][0] < k:
lo = group[lo][0]
hi = pi + 1
else:
lo += 1
hi = pi + 1
lohi = (lo, hi)
for j in range(lo, hi):
group[j] = lohi
r[i] = group[pi][0]
print(" ".join(map(str, r)))
main()
``` | instruction | 0 | 3,927 | 7 | 7,854 |
Yes | output | 1 | 3,927 | 7 | 7,855 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Professor Ibrahim has prepared the final homework for his algorithm’s class. He asked his students to implement the Posterization Image Filter.
Their algorithm will be tested on an array of integers, where the i-th integer represents the color of the i-th pixel in the image. The image is in black and white, therefore the color of each pixel will be an integer between 0 and 255 (inclusive).
To implement the filter, students are required to divide the black and white color range [0, 255] into groups of consecutive colors, and select one color in each group to be the group’s key. In order to preserve image details, the size of a group must not be greater than k, and each color should belong to exactly one group.
Finally, the students will replace the color of each pixel in the array with that color’s assigned group key.
To better understand the effect, here is an image of a basking turtle where the Posterization Filter was applied with increasing k to the right.
<image>
To make the process of checking the final answer easier, Professor Ibrahim wants students to divide the groups and assign the keys in a way that produces the lexicographically smallest possible array.
Input
The first line of input contains two integers n and k (1 ≤ n ≤ 10^5, 1 ≤ k ≤ 256), the number of pixels in the image, and the maximum size of a group, respectively.
The second line contains n integers p_1, p_2, ..., p_n (0 ≤ p_i ≤ 255), where p_i is the color of the i-th pixel.
Output
Print n space-separated integers; the lexicographically smallest possible array that represents the image after applying the Posterization filter.
Examples
Input
4 3
2 14 3 4
Output
0 12 3 3
Input
5 2
0 2 1 255 254
Output
0 1 1 254 254
Note
One possible way to group colors and assign keys for the first sample:
Color 2 belongs to the group [0,2], with group key 0.
Color 14 belongs to the group [12,14], with group key 12.
Colors 3 and 4 belong to group [3, 5], with group key 3.
Other groups won't affect the result so they are not listed here.
Submitted Solution:
```
from sys import stdin, stdout
a = stdin.readline()
p=stdin.readline()
n, k = [int(x) for x in a.split()]
b=p.split()
# print(b)
m=int(256/k)
out=[]
for i in b :
j = 0
for j in range(0, int(256/k)):
if int(i)>=(k*j) and int(i)<= ((k*j)+k-1):
out.append(str(k * j))
new_numbers = [];
for n in out:
new_numbers.append(int(n));
out = new_numbers;
print(out)
``` | instruction | 0 | 3,928 | 7 | 7,856 |
No | output | 1 | 3,928 | 7 | 7,857 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Professor Ibrahim has prepared the final homework for his algorithm’s class. He asked his students to implement the Posterization Image Filter.
Their algorithm will be tested on an array of integers, where the i-th integer represents the color of the i-th pixel in the image. The image is in black and white, therefore the color of each pixel will be an integer between 0 and 255 (inclusive).
To implement the filter, students are required to divide the black and white color range [0, 255] into groups of consecutive colors, and select one color in each group to be the group’s key. In order to preserve image details, the size of a group must not be greater than k, and each color should belong to exactly one group.
Finally, the students will replace the color of each pixel in the array with that color’s assigned group key.
To better understand the effect, here is an image of a basking turtle where the Posterization Filter was applied with increasing k to the right.
<image>
To make the process of checking the final answer easier, Professor Ibrahim wants students to divide the groups and assign the keys in a way that produces the lexicographically smallest possible array.
Input
The first line of input contains two integers n and k (1 ≤ n ≤ 10^5, 1 ≤ k ≤ 256), the number of pixels in the image, and the maximum size of a group, respectively.
The second line contains n integers p_1, p_2, ..., p_n (0 ≤ p_i ≤ 255), where p_i is the color of the i-th pixel.
Output
Print n space-separated integers; the lexicographically smallest possible array that represents the image after applying the Posterization filter.
Examples
Input
4 3
2 14 3 4
Output
0 12 3 3
Input
5 2
0 2 1 255 254
Output
0 1 1 254 254
Note
One possible way to group colors and assign keys for the first sample:
Color 2 belongs to the group [0,2], with group key 0.
Color 14 belongs to the group [12,14], with group key 12.
Colors 3 and 4 belong to group [3, 5], with group key 3.
Other groups won't affect the result so they are not listed here.
Submitted Solution:
```
n, k = map(int, input().split())
p = list(map(int, input().split()))
ranged = [None] * 256
for i in range(256):
ranged[i] = [0, None]
result = ""
for i, pi in enumerate(p):
if ranged[pi][0] == 2:
idx = pi
while ranged[idx][0] == 2:
ranged[idx][0] = 1
idx -= 1
elif ranged[pi][0] == 0:
# find smallest lexicographically possible value for pi
idx = pi
count = 0
_min = pi
while idx > 0 and ranged[idx - 1][0] != 1 and count < k - 1:
idx -= 1
count += 1
if str(idx) < str(_min):
_min = idx
count = 0
lowest = _min
# print("Min for {} is {}".format(pi, _min))
while count < k and idx < 256:
if idx <= pi:
ranged[idx][0] = 1
else:
ranged[idx][0] = 2
ranged[idx][1] = lowest
idx += 1
count += 1
# print(count, k - 1)
# print(ranged)
result += str(ranged[pi][1]) + " "
print(result[:-1])
``` | instruction | 0 | 3,929 | 7 | 7,858 |
No | output | 1 | 3,929 | 7 | 7,859 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Professor Ibrahim has prepared the final homework for his algorithm’s class. He asked his students to implement the Posterization Image Filter.
Their algorithm will be tested on an array of integers, where the i-th integer represents the color of the i-th pixel in the image. The image is in black and white, therefore the color of each pixel will be an integer between 0 and 255 (inclusive).
To implement the filter, students are required to divide the black and white color range [0, 255] into groups of consecutive colors, and select one color in each group to be the group’s key. In order to preserve image details, the size of a group must not be greater than k, and each color should belong to exactly one group.
Finally, the students will replace the color of each pixel in the array with that color’s assigned group key.
To better understand the effect, here is an image of a basking turtle where the Posterization Filter was applied with increasing k to the right.
<image>
To make the process of checking the final answer easier, Professor Ibrahim wants students to divide the groups and assign the keys in a way that produces the lexicographically smallest possible array.
Input
The first line of input contains two integers n and k (1 ≤ n ≤ 10^5, 1 ≤ k ≤ 256), the number of pixels in the image, and the maximum size of a group, respectively.
The second line contains n integers p_1, p_2, ..., p_n (0 ≤ p_i ≤ 255), where p_i is the color of the i-th pixel.
Output
Print n space-separated integers; the lexicographically smallest possible array that represents the image after applying the Posterization filter.
Examples
Input
4 3
2 14 3 4
Output
0 12 3 3
Input
5 2
0 2 1 255 254
Output
0 1 1 254 254
Note
One possible way to group colors and assign keys for the first sample:
Color 2 belongs to the group [0,2], with group key 0.
Color 14 belongs to the group [12,14], with group key 12.
Colors 3 and 4 belong to group [3, 5], with group key 3.
Other groups won't affect the result so they are not listed here.
Submitted Solution:
```
n,k=map(int,input().split())
ar=list(map(int,input().split()))
tmp=""
for j in ar:
if (j-k+1)>=0:
print(j-k+1,end=' ')
else:
print(0,end=' ')
``` | instruction | 0 | 3,930 | 7 | 7,860 |
No | output | 1 | 3,930 | 7 | 7,861 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Professor Ibrahim has prepared the final homework for his algorithm’s class. He asked his students to implement the Posterization Image Filter.
Their algorithm will be tested on an array of integers, where the i-th integer represents the color of the i-th pixel in the image. The image is in black and white, therefore the color of each pixel will be an integer between 0 and 255 (inclusive).
To implement the filter, students are required to divide the black and white color range [0, 255] into groups of consecutive colors, and select one color in each group to be the group’s key. In order to preserve image details, the size of a group must not be greater than k, and each color should belong to exactly one group.
Finally, the students will replace the color of each pixel in the array with that color’s assigned group key.
To better understand the effect, here is an image of a basking turtle where the Posterization Filter was applied with increasing k to the right.
<image>
To make the process of checking the final answer easier, Professor Ibrahim wants students to divide the groups and assign the keys in a way that produces the lexicographically smallest possible array.
Input
The first line of input contains two integers n and k (1 ≤ n ≤ 10^5, 1 ≤ k ≤ 256), the number of pixels in the image, and the maximum size of a group, respectively.
The second line contains n integers p_1, p_2, ..., p_n (0 ≤ p_i ≤ 255), where p_i is the color of the i-th pixel.
Output
Print n space-separated integers; the lexicographically smallest possible array that represents the image after applying the Posterization filter.
Examples
Input
4 3
2 14 3 4
Output
0 12 3 3
Input
5 2
0 2 1 255 254
Output
0 1 1 254 254
Note
One possible way to group colors and assign keys for the first sample:
Color 2 belongs to the group [0,2], with group key 0.
Color 14 belongs to the group [12,14], with group key 12.
Colors 3 and 4 belong to group [3, 5], with group key 3.
Other groups won't affect the result so they are not listed here.
Submitted Solution:
```
n,k=map(int,input().split())
par=[-1 for i in range(512)]
ls=map(int,input().split())
for v in ls:
v+=256
ans=v
if par[v]==-1:
for j in range(v-k+1,v+1):
if par[j]==-1:
ans=j
break
par[ans]=ans
#print('ans:%d'%(ans))
for j in range(ans+1,min(511,ans+k)):
if par[j]==-1:
par[j]=ans
else:
break
else:
ans=par[v]
print(max(0,ans-256),end=' ')
``` | instruction | 0 | 3,931 | 7 | 7,862 |
No | output | 1 | 3,931 | 7 | 7,863 |
Provide a correct Python 3 solution for this coding contest problem.
A young boy John is playing with eight triangular panels. These panels are all regular triangles of the same size, each painted in a single color; John is forming various octahedra with them.
While he enjoys his playing, his father is wondering how many octahedra can be made of these panels since he is a pseudo-mathematician. Your task is to help his father: write a program that reports the number of possible octahedra for given panels. Here, a pair of octahedra should be considered identical when they have the same combination of the colors allowing rotation.
Input
The input consists of multiple datasets. Each dataset has the following format:
Color1 Color2 ... Color8
Each Colori (1 ≤ i ≤ 8) is a string of up to 20 lowercase alphabets and represents the color of the i-th triangular panel.
The input ends with EOF.
Output
For each dataset, output the number of different octahedra that can be made of given panels.
Example
Input
blue blue blue blue blue blue blue blue
red blue blue blue blue blue blue blue
red red blue blue blue blue blue blue
Output
1
1
3 | instruction | 0 | 4,127 | 7 | 8,254 |
"Correct Solution:
```
import math,string,itertools,fractions,heapq,collections,re,array,bisect,sys,random,time,copy,functools
sys.setrecursionlimit(10**7)
inf = 10**20
eps = 1.0 / 10**13
mod = 10**9+7
dd = [(-1,0),(0,1),(1,0),(0,-1)]
ddn = [(-1,0),(-1,1),(0,1),(1,1),(1,0),(1,-1),(0,-1),(-1,-1)]
def LI(): return [int(x) for x in sys.stdin.readline().split()]
def LI_(): return [int(x)-1 for x in sys.stdin.readline().split()]
def LF(): return [float(x) for x in sys.stdin.readline().split()]
def LS(): return sys.stdin.readline().split()
def I(): return int(sys.stdin.readline())
def F(): return float(sys.stdin.readline())
def S(): return input()
def pf(s): return print(s, flush=True)
# aa
# 1 2 4 5
# 2 3 1 6
# 3 4 2 7
# 4 1 3 8
# 5 8 6 1
# 6 5 7 2
# 7 6 8 3
# 8 7 5 4
#
def main():
rr = []
k = [
[2, 4, 5],
[3, 1, 6],
[4, 2, 7],
[1, 3, 8],
[8, 6, 1],
[5, 7, 2],
[6, 8, 3],
[7, 5, 4]
]
for i in range(8):
for j in range(3):
k[i][j] -= 1
ptn = []
for i in range(8):
for j in range(3):
t = [-1] * 8
t[0] = i
t[1] = k[i][j]
t[3] = k[i][(j+1)%3]
for l in range(8):
if l == i:
continue
if t[1] in k[l] and t[3] in k[l]:
t[2] = l
break
for l in range(4):
kl = k[t[l]]
for m in range(3):
if kl[m] not in t:
t[l+4] = kl[m]
break
ptn.append(t)
def f(a):
s = set()
r = 0
tc = 0
for t in itertools.permutations(a):
tc += 1
if t in s:
continue
r += 1
for p in ptn:
s.add(tuple(t[p[i]] for i in range(8)))
return r
while True:
a = LS()
if len(a) == 0:
break
rr.append(f(a))
return '\n'.join(map(str,rr))
print(main())
``` | output | 1 | 4,127 | 7 | 8,255 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Pasha is a young technician, nevertheless, he has already got a huge goal: to assemble a PC. The first task he has to become familiar with is to assemble an electric scheme.
The scheme Pasha assembled yesterday consists of several wires. Each wire is a segment that connects two points on a plane with integer coordinates within the segment [1, 10^9].
There are wires of two colors in the scheme:
* red wires: these wires are horizontal segments, i.e. if such a wire connects two points (x_1, y_1) and (x_2, y_2), then y_1 = y_2;
* blue wires: these wires are vertical segments, i.e. if such a wire connects two points (x_1, y_1) and (x_2, y_2), then x_1 = x_2.
Note that if a wire connects a point to itself, it may be blue, and it can be red. Also, in Pasha's scheme no two wires of the same color intersect, i.e. there are no two wires of same color that have common points.
The imperfection of Pasha's scheme was that the wires were not isolated, so in the points where two wires of different colors intersect, Pasha saw sparks. Pasha wrote down all the points where he saw sparks and obtained a set of n distinct points. After that he disassembled the scheme.
Next morning Pasha looked at the set of n points where he had seen sparks and wondered how many wires had he used. Unfortunately, he does not remember that, so he wonders now what is the smallest number of wires he might have used in the scheme. Help him to determine this number and place the wires in such a way that in the resulting scheme the sparks occur in the same places.
Input
The first line contains a single integer n (1 ≤ n ≤ 1000) — the number of points where Pasha saw sparks.
Each of the next n lines contain two integers x and y (1 ≤ x, y ≤ 10^9) — the coordinates of a point with sparks. It is guaranteed that all points are distinct.
Output
Print the description of the scheme in the following format.
In the first line print h — the number of horizontal red wires (0 ≤ h). In each of the next h lines print 4 integers x_1, y_1, x_2, y_2 — the coordinates of two points (x_1, y_1) and (x_2, y_2) that are connected with this red wire. The segmenst must be horizontal, i.e. y_1 = y_2 must hold. Also, the constraint 1 ≤ x_1, y_1, x_2, y_2 ≤ 10^9 must hold.
After that print v — the number of vertical blue wires (0 ≤ v). In each of the next v lines print 4 integers x_1, y_1, x_2, y_2 — the coordinates of two points (x_1, y_1) and (x_2, y_2) that are connected with this blue wire. The segmenst must be vertical, i.e. x_1 = x_2 shomustuld hold. Also, the constraint 1 ≤ x_1, y_1, x_2, y_2 ≤ 10^9 must hold.
No two segments of the same color should have common points. The set of points where sparks appear should be the same as given in the input.
The number of segments (h + v) should be minimum possible. It's easy to see that the answer always exists. If there are multiple possible answers, print any.
Examples
Input
4
2 2
2 4
4 2
4 4
Output
2
5 2 1 2
1 4 5 4
2
2 1 2 5
4 5 4 1
Input
4
2 1
3 2
2 3
1 2
Output
4
2 1 2 1
3 2 3 2
1 2 1 2
2 3 2 3
3
1 2 1 2
3 2 3 2
2 3 2 1
Note
In the first example Pasha could have assembled the following scheme:
<image>
In this scheme there are 2 wires of each color: red ones connecting (5, 2) with (1, 2) and (1, 4) with (5, 4), blue ones connecting (2, 1) with (2, 5) and (4, 5) with (4, 1). Note that the sparks appear in the points that are described in the input, they are shown in yellow on the picture. For example, Pasha will see the spark in the point (2, 4) because the second red wire and the first blue wire intersect there. It is possible to show that we can't have less than 4 wires to produce a scheme with same sparks positions.
Submitted Solution:
```
n=int(input())
x=[]
y=[]
X={}
Y={}
for i in range(n):
a,b=map(int,input().split())
x.append(a)
y.append(b)
s1=list(set(x))
s2=list(set(x))
s1.sort()
s2.sort()
for i in s1:
X[i]=[]
for i in s2:
Y[i]=[]
for i in range(n):
X[x[i]].append(y[i])
Y[y[i]].append(x[i])
for i in s1:
X[i].sort()
for i in s2:
Y[i].sort()
print(len(s2))
for i in s2:
k=len(Y[i])
print(Y[i][0],i,Y[i][k-1],i)
print(len(s1))
for i in s1:
h=len(X[i])
print(i,X[i][0],i,X[i][h-1])
``` | instruction | 0 | 4,177 | 7 | 8,354 |
No | output | 1 | 4,177 | 7 | 8,355 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Pasha is a young technician, nevertheless, he has already got a huge goal: to assemble a PC. The first task he has to become familiar with is to assemble an electric scheme.
The scheme Pasha assembled yesterday consists of several wires. Each wire is a segment that connects two points on a plane with integer coordinates within the segment [1, 10^9].
There are wires of two colors in the scheme:
* red wires: these wires are horizontal segments, i.e. if such a wire connects two points (x_1, y_1) and (x_2, y_2), then y_1 = y_2;
* blue wires: these wires are vertical segments, i.e. if such a wire connects two points (x_1, y_1) and (x_2, y_2), then x_1 = x_2.
Note that if a wire connects a point to itself, it may be blue, and it can be red. Also, in Pasha's scheme no two wires of the same color intersect, i.e. there are no two wires of same color that have common points.
The imperfection of Pasha's scheme was that the wires were not isolated, so in the points where two wires of different colors intersect, Pasha saw sparks. Pasha wrote down all the points where he saw sparks and obtained a set of n distinct points. After that he disassembled the scheme.
Next morning Pasha looked at the set of n points where he had seen sparks and wondered how many wires had he used. Unfortunately, he does not remember that, so he wonders now what is the smallest number of wires he might have used in the scheme. Help him to determine this number and place the wires in such a way that in the resulting scheme the sparks occur in the same places.
Input
The first line contains a single integer n (1 ≤ n ≤ 1000) — the number of points where Pasha saw sparks.
Each of the next n lines contain two integers x and y (1 ≤ x, y ≤ 10^9) — the coordinates of a point with sparks. It is guaranteed that all points are distinct.
Output
Print the description of the scheme in the following format.
In the first line print h — the number of horizontal red wires (0 ≤ h). In each of the next h lines print 4 integers x_1, y_1, x_2, y_2 — the coordinates of two points (x_1, y_1) and (x_2, y_2) that are connected with this red wire. The segmenst must be horizontal, i.e. y_1 = y_2 must hold. Also, the constraint 1 ≤ x_1, y_1, x_2, y_2 ≤ 10^9 must hold.
After that print v — the number of vertical blue wires (0 ≤ v). In each of the next v lines print 4 integers x_1, y_1, x_2, y_2 — the coordinates of two points (x_1, y_1) and (x_2, y_2) that are connected with this blue wire. The segmenst must be vertical, i.e. x_1 = x_2 shomustuld hold. Also, the constraint 1 ≤ x_1, y_1, x_2, y_2 ≤ 10^9 must hold.
No two segments of the same color should have common points. The set of points where sparks appear should be the same as given in the input.
The number of segments (h + v) should be minimum possible. It's easy to see that the answer always exists. If there are multiple possible answers, print any.
Examples
Input
4
2 2
2 4
4 2
4 4
Output
2
5 2 1 2
1 4 5 4
2
2 1 2 5
4 5 4 1
Input
4
2 1
3 2
2 3
1 2
Output
4
2 1 2 1
3 2 3 2
1 2 1 2
2 3 2 3
3
1 2 1 2
3 2 3 2
2 3 2 1
Note
In the first example Pasha could have assembled the following scheme:
<image>
In this scheme there are 2 wires of each color: red ones connecting (5, 2) with (1, 2) and (1, 4) with (5, 4), blue ones connecting (2, 1) with (2, 5) and (4, 5) with (4, 1). Note that the sparks appear in the points that are described in the input, they are shown in yellow on the picture. For example, Pasha will see the spark in the point (2, 4) because the second red wire and the first blue wire intersect there. It is possible to show that we can't have less than 4 wires to produce a scheme with same sparks positions.
Submitted Solution:
```
def intersect(x, y, x_line, y_line):
return (x_line[0] <= y <= x_line[1]) and (y_line[0] <= x <= y_line[1])
n = int(input())
scs = []
for i in range(n):
l = input().split()
scs.append((int(l[0]), int(l[1])))
red_wires = {} # horizontal
blue_wires = {} # vertical
for sc in scs:
if not sc[0] in blue_wires.keys():
blue_wires[sc[0]] = []
blue_wires[sc[0]].append(sc[1])
if not sc[1] in red_wires.keys():
red_wires[sc[1]] = []
red_wires[sc[1]].append(sc[0])
for y in red_wires.keys():
red_wires[y].sort()
min_x = min(red_wires[y])
max_x = max(red_wires[y])
red_wires[y].clear()
red_wires[y].append((min_x, max_x))
for x in blue_wires.keys():
blue_wires[x].sort()
min_y = min(blue_wires[x])
max_y = max(blue_wires[x])
blue_wires[x].clear()
blue_wires[x].append((min_y, max_y))
for y in red_wires.keys():
for x in red_wires.keys():
# if any lines intersect
for red_line in red_wires[y]:
for blue_line in blue_wires[x]:
if intersect(x, y, blue_line, red_line) and not (x, y) in scs:
red_wires[y].append((x + 1, red_line[1]))
red_wires[y].append((red_line[0], x - 1))
red_wires[y].remove(red_line)
rsize = 0
bsize = 0
for y in red_wires.keys():
rsize += len(red_wires[y])
for x in red_wires.keys():
bsize += len(blue_wires[x])
print(rsize)
for y in red_wires.keys():
for line in red_wires[y]:
print(str(line[0]) + str(y) + str(line[1]) + str(y))
print(bsize)
for x in blue_wires.keys():
for line in blue_wires[x]:
print(str(x) + str(line[0]) + str(x) + str(line[1]))
``` | instruction | 0 | 4,178 | 7 | 8,356 |
No | output | 1 | 4,178 | 7 | 8,357 |
Provide tags and a correct Python 3 solution for this coding contest problem.
RedDreamer has an array a consisting of n non-negative integers, and an unlucky integer T.
Let's denote the misfortune of array b having length m as f(b) — the number of pairs of integers (i, j) such that 1 ≤ i < j ≤ m and b_i + b_j = T. RedDreamer has to paint each element of a into one of two colors, white and black (for each element, the color is chosen independently), and then create two arrays c and d so that all white elements belong to c, and all black elements belong to d (it is possible that one of these two arrays becomes empty). RedDreamer wants to paint the elements in such a way that f(c) + f(d) is minimum possible.
For example:
* if n = 6, T = 7 and a = [1, 2, 3, 4, 5, 6], it is possible to paint the 1-st, the 4-th and the 5-th elements white, and all other elements black. So c = [1, 4, 5], d = [2, 3, 6], and f(c) + f(d) = 0 + 0 = 0;
* if n = 3, T = 6 and a = [3, 3, 3], it is possible to paint the 1-st element white, and all other elements black. So c = [3], d = [3, 3], and f(c) + f(d) = 0 + 1 = 1.
Help RedDreamer to paint the array optimally!
Input
The first line contains one integer t (1 ≤ t ≤ 1000) — the number of test cases. Then t test cases follow.
The first line of each test case contains two integers n and T (1 ≤ n ≤ 10^5, 0 ≤ T ≤ 10^9) — the number of elements in the array and the unlucky integer, respectively.
The second line contains n integers a_1, a_2, ..., a_n (0 ≤ a_i ≤ 10^9) — the elements of the array.
The sum of n over all test cases does not exceed 10^5.
Output
For each test case print n integers: p_1, p_2, ..., p_n (each p_i is either 0 or 1) denoting the colors. If p_i is 0, then a_i is white and belongs to the array c, otherwise it is black and belongs to the array d.
If there are multiple answers that minimize the value of f(c) + f(d), print any of them.
Example
Input
2
6 7
1 2 3 4 5 6
3 6
3 3 3
Output
1 0 0 1 1 0
1 0 0 | instruction | 0 | 4,383 | 7 | 8,766 |
Tags: greedy, math, sortings
Correct Solution:
```
# cook your dish here
#_________________ Mukul Mohan Varshney _______________#
#Template
import sys
import os
import math
import copy
from math import gcd
from bisect import bisect
from io import BytesIO, IOBase
from math import sqrt,floor,factorial,gcd,log,ceil
from collections import deque,Counter,defaultdict
from itertools import permutations, combinations
from itertools import accumulate
#define function
def Int(): return int(sys.stdin.readline())
def Mint(): return map(int,sys.stdin.readline().split())
def Lstr(): return list(sys.stdin.readline().strip())
def Str(): return sys.stdin.readline().strip()
def Mstr(): return map(str,sys.stdin.readline().strip().split())
def List(): return list(map(int,sys.stdin.readline().split()))
def Hash(): return dict()
def Mod(): return 1000000007
def Ncr(n,r,p): return ((fact[n])*((ifact[r]*ifact[n-r])%p))%p
def Most_frequent(list): return max(set(list), key = list.count)
def Mat2x2(n): return [List() for _ in range(n)]
def Lcm(x,y): return (x*y)//gcd(x,y)
def dtob(n): return bin(n).replace("0b","")
def btod(n): return int(n,2)
def common(l1, l2):
return set(l1).intersection(l2)
# Driver Code
def solution():
for _ in range(Int()):
n,t=Mint()
a=List()
ans=[]
f=0
for i in range(n):
if(2*a[i]<t):
ans.append(1)
elif(2*a[i]==t):
if(f==0):
ans.append(0)
f=1
else:
ans.append(1)
f=0
else:
ans.append(0)
print(*ans)
#Call the solve function
if __name__ == "__main__":
solution()
``` | output | 1 | 4,383 | 7 | 8,767 |
Provide tags and a correct Python 3 solution for this coding contest problem.
RedDreamer has an array a consisting of n non-negative integers, and an unlucky integer T.
Let's denote the misfortune of array b having length m as f(b) — the number of pairs of integers (i, j) such that 1 ≤ i < j ≤ m and b_i + b_j = T. RedDreamer has to paint each element of a into one of two colors, white and black (for each element, the color is chosen independently), and then create two arrays c and d so that all white elements belong to c, and all black elements belong to d (it is possible that one of these two arrays becomes empty). RedDreamer wants to paint the elements in such a way that f(c) + f(d) is minimum possible.
For example:
* if n = 6, T = 7 and a = [1, 2, 3, 4, 5, 6], it is possible to paint the 1-st, the 4-th and the 5-th elements white, and all other elements black. So c = [1, 4, 5], d = [2, 3, 6], and f(c) + f(d) = 0 + 0 = 0;
* if n = 3, T = 6 and a = [3, 3, 3], it is possible to paint the 1-st element white, and all other elements black. So c = [3], d = [3, 3], and f(c) + f(d) = 0 + 1 = 1.
Help RedDreamer to paint the array optimally!
Input
The first line contains one integer t (1 ≤ t ≤ 1000) — the number of test cases. Then t test cases follow.
The first line of each test case contains two integers n and T (1 ≤ n ≤ 10^5, 0 ≤ T ≤ 10^9) — the number of elements in the array and the unlucky integer, respectively.
The second line contains n integers a_1, a_2, ..., a_n (0 ≤ a_i ≤ 10^9) — the elements of the array.
The sum of n over all test cases does not exceed 10^5.
Output
For each test case print n integers: p_1, p_2, ..., p_n (each p_i is either 0 or 1) denoting the colors. If p_i is 0, then a_i is white and belongs to the array c, otherwise it is black and belongs to the array d.
If there are multiple answers that minimize the value of f(c) + f(d), print any of them.
Example
Input
2
6 7
1 2 3 4 5 6
3 6
3 3 3
Output
1 0 0 1 1 0
1 0 0 | instruction | 0 | 4,384 | 7 | 8,768 |
Tags: greedy, math, sortings
Correct Solution:
```
import sys, os, io
def rs(): return sys.stdin.readline().rstrip()
def ri(): return int(sys.stdin.readline())
def ria(): return list(map(int, sys.stdin.readline().split()))
def ws(s): sys.stdout.write(s + '\n')
def wi(n): sys.stdout.write(str(n) + '\n')
def wia(a): sys.stdout.write(' '.join([str(x) for x in a]) + '\n')
import math,datetime,functools,itertools,operator,bisect,fractions
from collections import deque,defaultdict,OrderedDict,Counter
from fractions import Fraction
from decimal import Decimal
N = 10005
# array to store inverse of 1 to N
factorialNumInverse = [None] * (N + 1)
# array to precompute inverse of 1! to N!
naturalNumInverse = factorialNumInverse.copy()
# array to store factorial of
# first N numbers
fact = factorialNumInverse.copy()
# Function to precompute inverse of numbers
def InverseofNumber(p):
naturalNumInverse[0] = naturalNumInverse[1] = 1
for i in range(2, N + 1, 1):
naturalNumInverse[i] = (naturalNumInverse[p % i] *
(p - int(p / i)) % p)
# Function to precompute inverse
# of factorials
def InverseofFactorial(p):
factorialNumInverse[0] = factorialNumInverse[1] = 1
# precompute inverse of natural numbers
for i in range(2, N + 1, 1):
factorialNumInverse[i] = (naturalNumInverse[i] *
factorialNumInverse[i - 1]) % p
# Function to calculate factorial of 1 to N
def factorial(p):
fact[0] = 1
# precompute factorials
for i in range(1, N + 1):
fact[i] = (fact[i - 1] * i) % p
# Function to return nCr % p in O(1) time
def Binomial(N, R, p):
# n C r = n!*inverse(r!)*inverse((n-r)!)
ans = ((fact[N] * factorialNumInverse[R])% p *
factorialNumInverse[N - R])% p
return ans
def main():
starttime=datetime.datetime.now()
if(os.path.exists('input.txt')):
sys.stdin = open("input.txt","r")
sys.stdout = open("output.txt","w")
for _ in range(ri()):
N,n=ria()
a=ria()
g1={}
g0={}
t=0
if n%2!=0:
for i in a:
if i<=(n//2):
if i in g0:
g0[i]+=1
else:
g0[i]=1
if i>(n//2):
if i in g1:
g1[i]+=1
else:
g1[i]=1
else:
for i in a:
if i<(n//2):
if i in g0:
g0[i]+=1
else:
g0[i]=1
if i>(n//2):
if i in g1:
g1[i]+=1
else:
g1[i]=1
if i==(n//2):
if t==0:
if i in g0:
g0[i]+=1
else:
g0[i]=1
t=1
else:
if i in g1:
g1[i]+=1
else:
g1[i]=1
t=0
ans=[]
for i in a:
if i in g1 and g1[i]>0:
ans.append(1)
g1[i]-=1
else:
if i in g0 and g0[i]>0:
ans.append(0)
g0[i]-=1
print(*ans)
#<--Solving Area Ends
endtime=datetime.datetime.now()
time=(endtime-starttime).total_seconds()*1000
if(os.path.exists('input.txt')):
print("Time:",time,"ms")
class FastReader(io.IOBase):
newlines = 0
def __init__(self, fd, chunk_size=1024 * 8):
self._fd = fd
self._chunk_size = chunk_size
self.buffer = io.BytesIO()
def read(self):
while True:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, self._chunk_size))
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, size=-1):
while self.newlines == 0:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, self._chunk_size if size == -1 else size))
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()
class FastWriter(io.IOBase):
def __init__(self, fd):
self._fd = fd
self.buffer = io.BytesIO()
self.write = self.buffer.write
def flush(self):
os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class FastStdin(io.IOBase):
def __init__(self, fd=0):
self.buffer = FastReader(fd)
self.read = lambda: self.buffer.read().decode("ascii")
self.readline = lambda: self.buffer.readline().decode("ascii")
class FastStdout(io.IOBase):
def __init__(self, fd=1):
self.buffer = FastWriter(fd)
self.write = lambda s: self.buffer.write(s.encode("ascii"))
self.flush = self.buffer.flush
if __name__ == '__main__':
sys.stdin = FastStdin()
sys.stdout = FastStdout()
main()
``` | output | 1 | 4,384 | 7 | 8,769 |
Provide tags and a correct Python 3 solution for this coding contest problem.
RedDreamer has an array a consisting of n non-negative integers, and an unlucky integer T.
Let's denote the misfortune of array b having length m as f(b) — the number of pairs of integers (i, j) such that 1 ≤ i < j ≤ m and b_i + b_j = T. RedDreamer has to paint each element of a into one of two colors, white and black (for each element, the color is chosen independently), and then create two arrays c and d so that all white elements belong to c, and all black elements belong to d (it is possible that one of these two arrays becomes empty). RedDreamer wants to paint the elements in such a way that f(c) + f(d) is minimum possible.
For example:
* if n = 6, T = 7 and a = [1, 2, 3, 4, 5, 6], it is possible to paint the 1-st, the 4-th and the 5-th elements white, and all other elements black. So c = [1, 4, 5], d = [2, 3, 6], and f(c) + f(d) = 0 + 0 = 0;
* if n = 3, T = 6 and a = [3, 3, 3], it is possible to paint the 1-st element white, and all other elements black. So c = [3], d = [3, 3], and f(c) + f(d) = 0 + 1 = 1.
Help RedDreamer to paint the array optimally!
Input
The first line contains one integer t (1 ≤ t ≤ 1000) — the number of test cases. Then t test cases follow.
The first line of each test case contains two integers n and T (1 ≤ n ≤ 10^5, 0 ≤ T ≤ 10^9) — the number of elements in the array and the unlucky integer, respectively.
The second line contains n integers a_1, a_2, ..., a_n (0 ≤ a_i ≤ 10^9) — the elements of the array.
The sum of n over all test cases does not exceed 10^5.
Output
For each test case print n integers: p_1, p_2, ..., p_n (each p_i is either 0 or 1) denoting the colors. If p_i is 0, then a_i is white and belongs to the array c, otherwise it is black and belongs to the array d.
If there are multiple answers that minimize the value of f(c) + f(d), print any of them.
Example
Input
2
6 7
1 2 3 4 5 6
3 6
3 3 3
Output
1 0 0 1 1 0
1 0 0 | instruction | 0 | 4,385 | 7 | 8,770 |
Tags: greedy, math, sortings
Correct Solution:
```
import sys
from collections import defaultdict as dd
from collections import Counter as cc
from queue import Queue
import math
import itertools
try:
sys.stdin = open('input.txt', 'r')
sys.stdout = open('output.txt', 'w')
except:
pass
input = lambda: sys.stdin.buffer.readline().rstrip()
for _ in range(int(input())):
n,k=map(int,input().split())
a=list(map(int,input().split()))
r=1
for i in range(n):
if a[i]==k/2:
a[i]=r
r=abs(r-1)
elif a[i]<k/2:
a[i]=1
else:
a[i]=0
print(*a)
``` | output | 1 | 4,385 | 7 | 8,771 |
Provide tags and a correct Python 3 solution for this coding contest problem.
RedDreamer has an array a consisting of n non-negative integers, and an unlucky integer T.
Let's denote the misfortune of array b having length m as f(b) — the number of pairs of integers (i, j) such that 1 ≤ i < j ≤ m and b_i + b_j = T. RedDreamer has to paint each element of a into one of two colors, white and black (for each element, the color is chosen independently), and then create two arrays c and d so that all white elements belong to c, and all black elements belong to d (it is possible that one of these two arrays becomes empty). RedDreamer wants to paint the elements in such a way that f(c) + f(d) is minimum possible.
For example:
* if n = 6, T = 7 and a = [1, 2, 3, 4, 5, 6], it is possible to paint the 1-st, the 4-th and the 5-th elements white, and all other elements black. So c = [1, 4, 5], d = [2, 3, 6], and f(c) + f(d) = 0 + 0 = 0;
* if n = 3, T = 6 and a = [3, 3, 3], it is possible to paint the 1-st element white, and all other elements black. So c = [3], d = [3, 3], and f(c) + f(d) = 0 + 1 = 1.
Help RedDreamer to paint the array optimally!
Input
The first line contains one integer t (1 ≤ t ≤ 1000) — the number of test cases. Then t test cases follow.
The first line of each test case contains two integers n and T (1 ≤ n ≤ 10^5, 0 ≤ T ≤ 10^9) — the number of elements in the array and the unlucky integer, respectively.
The second line contains n integers a_1, a_2, ..., a_n (0 ≤ a_i ≤ 10^9) — the elements of the array.
The sum of n over all test cases does not exceed 10^5.
Output
For each test case print n integers: p_1, p_2, ..., p_n (each p_i is either 0 or 1) denoting the colors. If p_i is 0, then a_i is white and belongs to the array c, otherwise it is black and belongs to the array d.
If there are multiple answers that minimize the value of f(c) + f(d), print any of them.
Example
Input
2
6 7
1 2 3 4 5 6
3 6
3 3 3
Output
1 0 0 1 1 0
1 0 0 | instruction | 0 | 4,386 | 7 | 8,772 |
Tags: greedy, math, sortings
Correct Solution:
```
from sys import stdin
tt = int(stdin.readline())
for loop in range(tt):
n,T = map(int,stdin.readline().split())
a = list(map(int,stdin.readline().split()))
cnt = 0
p = [None] * n
for i in range(n):
if a[i]*2 < T:
p[i] = 0
elif a[i]*2 > T:
p[i] = 1
else:
p[i] = cnt
cnt ^= 1
print (*p)
``` | output | 1 | 4,386 | 7 | 8,773 |
Provide tags and a correct Python 3 solution for this coding contest problem.
RedDreamer has an array a consisting of n non-negative integers, and an unlucky integer T.
Let's denote the misfortune of array b having length m as f(b) — the number of pairs of integers (i, j) such that 1 ≤ i < j ≤ m and b_i + b_j = T. RedDreamer has to paint each element of a into one of two colors, white and black (for each element, the color is chosen independently), and then create two arrays c and d so that all white elements belong to c, and all black elements belong to d (it is possible that one of these two arrays becomes empty). RedDreamer wants to paint the elements in such a way that f(c) + f(d) is minimum possible.
For example:
* if n = 6, T = 7 and a = [1, 2, 3, 4, 5, 6], it is possible to paint the 1-st, the 4-th and the 5-th elements white, and all other elements black. So c = [1, 4, 5], d = [2, 3, 6], and f(c) + f(d) = 0 + 0 = 0;
* if n = 3, T = 6 and a = [3, 3, 3], it is possible to paint the 1-st element white, and all other elements black. So c = [3], d = [3, 3], and f(c) + f(d) = 0 + 1 = 1.
Help RedDreamer to paint the array optimally!
Input
The first line contains one integer t (1 ≤ t ≤ 1000) — the number of test cases. Then t test cases follow.
The first line of each test case contains two integers n and T (1 ≤ n ≤ 10^5, 0 ≤ T ≤ 10^9) — the number of elements in the array and the unlucky integer, respectively.
The second line contains n integers a_1, a_2, ..., a_n (0 ≤ a_i ≤ 10^9) — the elements of the array.
The sum of n over all test cases does not exceed 10^5.
Output
For each test case print n integers: p_1, p_2, ..., p_n (each p_i is either 0 or 1) denoting the colors. If p_i is 0, then a_i is white and belongs to the array c, otherwise it is black and belongs to the array d.
If there are multiple answers that minimize the value of f(c) + f(d), print any of them.
Example
Input
2
6 7
1 2 3 4 5 6
3 6
3 3 3
Output
1 0 0 1 1 0
1 0 0 | instruction | 0 | 4,387 | 7 | 8,774 |
Tags: greedy, math, sortings
Correct Solution:
```
for T in range(int(input())):
n,t=map(int,input().split())
a=list(map(int,input().split()))
arr,d=[],{}
for i in range(len(a)):
arr.append([a[i],i])
d[a[i],i]=0
arr.sort(key=lambda x : x[0])
i,j=0,n-1
while(i<j):
if(arr[i][0]+arr[j][0]<=t):
if(arr[i][0]+arr[j][0]==t):
d[arr[i][0],arr[i][1]]=1
if(arr[i][0]==arr[j][0]):
j-=1
i+=1
else:
j-=1
for i in range(len(a)):
print(d[a[i],i],end=' ')
print()
``` | output | 1 | 4,387 | 7 | 8,775 |
Provide tags and a correct Python 3 solution for this coding contest problem.
RedDreamer has an array a consisting of n non-negative integers, and an unlucky integer T.
Let's denote the misfortune of array b having length m as f(b) — the number of pairs of integers (i, j) such that 1 ≤ i < j ≤ m and b_i + b_j = T. RedDreamer has to paint each element of a into one of two colors, white and black (for each element, the color is chosen independently), and then create two arrays c and d so that all white elements belong to c, and all black elements belong to d (it is possible that one of these two arrays becomes empty). RedDreamer wants to paint the elements in such a way that f(c) + f(d) is minimum possible.
For example:
* if n = 6, T = 7 and a = [1, 2, 3, 4, 5, 6], it is possible to paint the 1-st, the 4-th and the 5-th elements white, and all other elements black. So c = [1, 4, 5], d = [2, 3, 6], and f(c) + f(d) = 0 + 0 = 0;
* if n = 3, T = 6 and a = [3, 3, 3], it is possible to paint the 1-st element white, and all other elements black. So c = [3], d = [3, 3], and f(c) + f(d) = 0 + 1 = 1.
Help RedDreamer to paint the array optimally!
Input
The first line contains one integer t (1 ≤ t ≤ 1000) — the number of test cases. Then t test cases follow.
The first line of each test case contains two integers n and T (1 ≤ n ≤ 10^5, 0 ≤ T ≤ 10^9) — the number of elements in the array and the unlucky integer, respectively.
The second line contains n integers a_1, a_2, ..., a_n (0 ≤ a_i ≤ 10^9) — the elements of the array.
The sum of n over all test cases does not exceed 10^5.
Output
For each test case print n integers: p_1, p_2, ..., p_n (each p_i is either 0 or 1) denoting the colors. If p_i is 0, then a_i is white and belongs to the array c, otherwise it is black and belongs to the array d.
If there are multiple answers that minimize the value of f(c) + f(d), print any of them.
Example
Input
2
6 7
1 2 3 4 5 6
3 6
3 3 3
Output
1 0 0 1 1 0
1 0 0 | instruction | 0 | 4,388 | 7 | 8,776 |
Tags: greedy, math, sortings
Correct Solution:
```
for _ in range(int(input())):
n,k = map(int,input().split())
w=[];b=[];sd={};bd={};sw=set();sb=set()
lis = list(map(int,input().split()))
ans=[0]*(n)
c=0
for i in lis:
if k-i not in sw:
sw.add(i)
sd[i]=1
#w.append(i)
elif k-i not in sb:
sb.add(i)
bd[i]=1
ans[c]=1
#b.append(i)
elif k-i in sw and k-i in sb:
if bd[k-i]<=sd[k-i]:
ans[c]=1
# b.append(i)
bd[i]+=1
sb.add(i)
else:
# w.append(i)
sd[i]+=1
sw.add(i)
elif k-i in sb:
#w.append(i)
sd[i]+=1
sw.add(i)
elif k-i in sw:
ans[c]=1
#b.append(i)
bd[i]+=1
sb.add(i)
c+=1
print(*ans)
``` | output | 1 | 4,388 | 7 | 8,777 |
Provide tags and a correct Python 3 solution for this coding contest problem.
RedDreamer has an array a consisting of n non-negative integers, and an unlucky integer T.
Let's denote the misfortune of array b having length m as f(b) — the number of pairs of integers (i, j) such that 1 ≤ i < j ≤ m and b_i + b_j = T. RedDreamer has to paint each element of a into one of two colors, white and black (for each element, the color is chosen independently), and then create two arrays c and d so that all white elements belong to c, and all black elements belong to d (it is possible that one of these two arrays becomes empty). RedDreamer wants to paint the elements in such a way that f(c) + f(d) is minimum possible.
For example:
* if n = 6, T = 7 and a = [1, 2, 3, 4, 5, 6], it is possible to paint the 1-st, the 4-th and the 5-th elements white, and all other elements black. So c = [1, 4, 5], d = [2, 3, 6], and f(c) + f(d) = 0 + 0 = 0;
* if n = 3, T = 6 and a = [3, 3, 3], it is possible to paint the 1-st element white, and all other elements black. So c = [3], d = [3, 3], and f(c) + f(d) = 0 + 1 = 1.
Help RedDreamer to paint the array optimally!
Input
The first line contains one integer t (1 ≤ t ≤ 1000) — the number of test cases. Then t test cases follow.
The first line of each test case contains two integers n and T (1 ≤ n ≤ 10^5, 0 ≤ T ≤ 10^9) — the number of elements in the array and the unlucky integer, respectively.
The second line contains n integers a_1, a_2, ..., a_n (0 ≤ a_i ≤ 10^9) — the elements of the array.
The sum of n over all test cases does not exceed 10^5.
Output
For each test case print n integers: p_1, p_2, ..., p_n (each p_i is either 0 or 1) denoting the colors. If p_i is 0, then a_i is white and belongs to the array c, otherwise it is black and belongs to the array d.
If there are multiple answers that minimize the value of f(c) + f(d), print any of them.
Example
Input
2
6 7
1 2 3 4 5 6
3 6
3 3 3
Output
1 0 0 1 1 0
1 0 0 | instruction | 0 | 4,389 | 7 | 8,778 |
Tags: greedy, math, sortings
Correct Solution:
```
import random
def gcd(a, b):
if a == 0:
return b
return gcd(b % a, a)
def lcm(a, b):
return (a * b) / gcd(a, b)
for _ in range(int(input())):
#n = int(input())
n,t= map(int, input().split())
a = list(map(int, input().split()))
d={}
for i in range(n):
if a[i] in d:
d[a[i]].append(i)
else:
d[a[i]]=[i]
ans=[-1]*n
for i in d.keys():
if ans[d[i][0]]==-1:
if i==t//2:
for j in range(len(d[i])//2):
ans[d[i][j]]=0
for j in range(len(d[i])//2,len(d[i])):
ans[d[i][j]] = 1
else:
for j in range(len(d[i])):
ans[d[i][j]]=0
if t-i in d:
for j in range(len(d[t-i])):
ans[d[t-i][j]]=1
for i in ans:
print(i,end=' ')
print('')
``` | output | 1 | 4,389 | 7 | 8,779 |
Provide tags and a correct Python 3 solution for this coding contest problem.
RedDreamer has an array a consisting of n non-negative integers, and an unlucky integer T.
Let's denote the misfortune of array b having length m as f(b) — the number of pairs of integers (i, j) such that 1 ≤ i < j ≤ m and b_i + b_j = T. RedDreamer has to paint each element of a into one of two colors, white and black (for each element, the color is chosen independently), and then create two arrays c and d so that all white elements belong to c, and all black elements belong to d (it is possible that one of these two arrays becomes empty). RedDreamer wants to paint the elements in such a way that f(c) + f(d) is minimum possible.
For example:
* if n = 6, T = 7 and a = [1, 2, 3, 4, 5, 6], it is possible to paint the 1-st, the 4-th and the 5-th elements white, and all other elements black. So c = [1, 4, 5], d = [2, 3, 6], and f(c) + f(d) = 0 + 0 = 0;
* if n = 3, T = 6 and a = [3, 3, 3], it is possible to paint the 1-st element white, and all other elements black. So c = [3], d = [3, 3], and f(c) + f(d) = 0 + 1 = 1.
Help RedDreamer to paint the array optimally!
Input
The first line contains one integer t (1 ≤ t ≤ 1000) — the number of test cases. Then t test cases follow.
The first line of each test case contains two integers n and T (1 ≤ n ≤ 10^5, 0 ≤ T ≤ 10^9) — the number of elements in the array and the unlucky integer, respectively.
The second line contains n integers a_1, a_2, ..., a_n (0 ≤ a_i ≤ 10^9) — the elements of the array.
The sum of n over all test cases does not exceed 10^5.
Output
For each test case print n integers: p_1, p_2, ..., p_n (each p_i is either 0 or 1) denoting the colors. If p_i is 0, then a_i is white and belongs to the array c, otherwise it is black and belongs to the array d.
If there are multiple answers that minimize the value of f(c) + f(d), print any of them.
Example
Input
2
6 7
1 2 3 4 5 6
3 6
3 3 3
Output
1 0 0 1 1 0
1 0 0 | instruction | 0 | 4,390 | 7 | 8,780 |
Tags: greedy, math, sortings
Correct Solution:
```
from sys import stdin, stdout
import math,sys
from itertools import permutations, combinations
from collections import defaultdict,deque,OrderedDict,Counter
from os import path
from bisect import bisect_left
import heapq
mod=10**9+7
def yes():print('YES')
def no():print('NO')
if (path.exists('input.txt')):
#------------------Sublime--------------------------------------#
sys.stdin=open('input.txt','r');sys.stdout=open('output.txt','w');
def inp():return (int(input()))
def minp():return(map(int,input().split()))
else:
#------------------PYPY FAst I/o--------------------------------#
def inp():return (int(stdin.readline()))
def minp():return(map(int,stdin.readline().split()))
####### ---- Start Your Program From Here ---- #######
for _ in range(inp()):
n,t=minp()
a=list(minp())
for i in range(n):
if a[i]>t/2:
a[i]=0
elif a[i]<t/2:
a[i]=1
else:
a[i]=2
equal=a.count(2)
for i in range(n):
if a[i]==2:
if equal&1:
a[i]=0
else:
a[i]=1
equal-=1
print(*a)
``` | output | 1 | 4,390 | 7 | 8,781 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
RedDreamer has an array a consisting of n non-negative integers, and an unlucky integer T.
Let's denote the misfortune of array b having length m as f(b) — the number of pairs of integers (i, j) such that 1 ≤ i < j ≤ m and b_i + b_j = T. RedDreamer has to paint each element of a into one of two colors, white and black (for each element, the color is chosen independently), and then create two arrays c and d so that all white elements belong to c, and all black elements belong to d (it is possible that one of these two arrays becomes empty). RedDreamer wants to paint the elements in such a way that f(c) + f(d) is minimum possible.
For example:
* if n = 6, T = 7 and a = [1, 2, 3, 4, 5, 6], it is possible to paint the 1-st, the 4-th and the 5-th elements white, and all other elements black. So c = [1, 4, 5], d = [2, 3, 6], and f(c) + f(d) = 0 + 0 = 0;
* if n = 3, T = 6 and a = [3, 3, 3], it is possible to paint the 1-st element white, and all other elements black. So c = [3], d = [3, 3], and f(c) + f(d) = 0 + 1 = 1.
Help RedDreamer to paint the array optimally!
Input
The first line contains one integer t (1 ≤ t ≤ 1000) — the number of test cases. Then t test cases follow.
The first line of each test case contains two integers n and T (1 ≤ n ≤ 10^5, 0 ≤ T ≤ 10^9) — the number of elements in the array and the unlucky integer, respectively.
The second line contains n integers a_1, a_2, ..., a_n (0 ≤ a_i ≤ 10^9) — the elements of the array.
The sum of n over all test cases does not exceed 10^5.
Output
For each test case print n integers: p_1, p_2, ..., p_n (each p_i is either 0 or 1) denoting the colors. If p_i is 0, then a_i is white and belongs to the array c, otherwise it is black and belongs to the array d.
If there are multiple answers that minimize the value of f(c) + f(d), print any of them.
Example
Input
2
6 7
1 2 3 4 5 6
3 6
3 3 3
Output
1 0 0 1 1 0
1 0 0
Submitted Solution:
```
for _ in range(int(input())):
n,t = map(int,input().split())
a = list(map(int,input().split()))
x = t/2
sol = []
y = True
for i in a:
if i<x:
sol.append(0)
elif i>x:
sol.append(1)
else:
sol.append(1 if y else 0)
y = not y
print(' '.join([str(i) for i in sol]))
``` | instruction | 0 | 4,391 | 7 | 8,782 |
Yes | output | 1 | 4,391 | 7 | 8,783 |
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