text stringlengths 198 433k | conversation_id int64 0 109k |
|---|---|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Vivek has encountered a problem. He has a maze that can be represented as an n × m grid. Each of the grid cells may represent the following:
* Empty — '.'
* Wall — '#'
* Good person — 'G'
* Bad person — 'B'
The only escape from the maze is at cell (n, m).
A person can move to a cell only if it shares a side with their current cell and does not contain a wall. Vivek wants to block some of the empty cells by replacing them with walls in such a way, that all the good people are able to escape, while none of the bad people are able to. A cell that initially contains 'G' or 'B' cannot be blocked and can be travelled through.
Help him determine if there exists a way to replace some (zero or more) empty cells with walls to satisfy the above conditions.
It is guaranteed that the cell (n,m) is empty. Vivek can also block this cell.
Input
The first line contains one integer t (1 ≤ t ≤ 100) — the number of test cases. The description of the test cases follows.
The first line of each test case contains two integers n, m (1 ≤ n, m ≤ 50) — the number of rows and columns in the maze.
Each of the next n lines contain m characters. They describe the layout of the maze. If a character on a line equals '.', the corresponding cell is empty. If it equals '#', the cell has a wall. 'G' corresponds to a good person and 'B' corresponds to a bad person.
Output
For each test case, print "Yes" if there exists a way to replace some empty cells with walls to satisfy the given conditions. Otherwise print "No"
You may print every letter in any case (upper or lower).
Example
Input
6
1 1
.
1 2
G.
2 2
#B
G.
2 3
G.#
B#.
3 3
#B.
#..
GG.
2 2
#B
B.
Output
Yes
Yes
No
No
Yes
Yes
Note
For the first and second test cases, all conditions are already satisfied.
For the third test case, there is only one empty cell (2,2), and if it is replaced with a wall then the good person at (1,2) will not be able to escape.
For the fourth test case, the good person at (1,1) cannot escape.
For the fifth test case, Vivek can block the cells (2,3) and (2,2).
For the last test case, Vivek can block the destination cell (2, 2).
Submitted Solution:
```
import collections
import sys
input = sys.stdin.readline
from collections import defaultdict
def bfs(i,j):
gcount=0
bcount=0
queue=collections.deque([(i,j)])
visited[i][j]=1
while queue:
i,j=queue.popleft()
for a,b in [(i+1,j),(i-1,j),(i,j+1),(i,j-1)]:
if a==-1 or a==n or b==-1 or b==m or visited[a][b] or grid[a][b]=='#':
continue
if grid[a][b]=='G':
gcount+=1
if grid[a][b]=='B':
bcount+=1
visited[a][b]+=1
queue.append((a,b))
return(gcount,bcount)
t=int(input())
for i in range(t):
grid=[]
n,m=map(int,input().split())
visited=[[0]*(m) for _ in range(n)]
for j in range(n):
row=[i for i in input() if i!='\n']
grid.append(row)
good_count=0
#print(grid)
for j in range(n):
for k in range(m):
#print(j,k)
if grid[j][k]=='B':
if 0<=(k+1) and (k+1)<m and grid[j][k+1]=='.':
grid[j][k+1]='#'
if 0<=(k-1) and (k-1)<m and grid[j][k-1]=='.':
grid[j][k-1]='#'
if 0<=(j+1) and (j+1)<n and grid[j+1][k]=='.':
grid[j+1][k]='#'
if 0<=(j-1) and (j-1)<n and grid[j-1][k]=='.':
grid[j-1][k]='#'
if grid[j][k]=='G':
good_count+=1
if grid[-1][-1]=='#' :
if good_count==0:
print('Yes')
else:
print('No')
continue
output,ans=bfs(n-1,m-1)
if output==(good_count) and ans==0:
print('Yes')
else:
print('No')
```
Yes
| 98,800 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Vivek has encountered a problem. He has a maze that can be represented as an n × m grid. Each of the grid cells may represent the following:
* Empty — '.'
* Wall — '#'
* Good person — 'G'
* Bad person — 'B'
The only escape from the maze is at cell (n, m).
A person can move to a cell only if it shares a side with their current cell and does not contain a wall. Vivek wants to block some of the empty cells by replacing them with walls in such a way, that all the good people are able to escape, while none of the bad people are able to. A cell that initially contains 'G' or 'B' cannot be blocked and can be travelled through.
Help him determine if there exists a way to replace some (zero or more) empty cells with walls to satisfy the above conditions.
It is guaranteed that the cell (n,m) is empty. Vivek can also block this cell.
Input
The first line contains one integer t (1 ≤ t ≤ 100) — the number of test cases. The description of the test cases follows.
The first line of each test case contains two integers n, m (1 ≤ n, m ≤ 50) — the number of rows and columns in the maze.
Each of the next n lines contain m characters. They describe the layout of the maze. If a character on a line equals '.', the corresponding cell is empty. If it equals '#', the cell has a wall. 'G' corresponds to a good person and 'B' corresponds to a bad person.
Output
For each test case, print "Yes" if there exists a way to replace some empty cells with walls to satisfy the given conditions. Otherwise print "No"
You may print every letter in any case (upper or lower).
Example
Input
6
1 1
.
1 2
G.
2 2
#B
G.
2 3
G.#
B#.
3 3
#B.
#..
GG.
2 2
#B
B.
Output
Yes
Yes
No
No
Yes
Yes
Note
For the first and second test cases, all conditions are already satisfied.
For the third test case, there is only one empty cell (2,2), and if it is replaced with a wall then the good person at (1,2) will not be able to escape.
For the fourth test case, the good person at (1,1) cannot escape.
For the fifth test case, Vivek can block the cells (2,3) and (2,2).
For the last test case, Vivek can block the destination cell (2, 2).
Submitted Solution:
```
# Problem: D. Solve The Maze
# Contest: Codeforces - Codeforces Round #648 (Div. 2)
# URL: https://codeforces.com/contest/1365/problem/D
# Memory Limit: 256 MB
# Time Limit: 1000 ms
#
# KAPOOR'S
from sys import stdin, stdout
def INI():
return int(stdin.readline())
def INL():
return [int(_) for _ in stdin.readline().split()]
def INS():
return stdin.readline()
def MOD():
return pow(10,9)+7
def OPS(ans):
stdout.write(str(ans)+"\n")
def OPL(ans):
[stdout.write(str(_)+" ") for _ in ans]
stdout.write("\n")
def make(i,j):
if 0<=i-1<n and 0<=j<m and X[i-1][j]==".":
X[i-1][j]="#"
if 0<=i+1<n and 0<=j<m and X[i+1][j]==".":
X[i+1][j]="#"
if 0<=i<n and 0<=j-1<m and X[i][j-1]==".":
X[i][j-1]="#"
if 0<=i<n and 0<=j+1<m and X[i][j+1]==".":
X[i][j+1]="#"
def is_safe(i,j):
return 0<=i<n and 0<=j<m
def dfs(i,j):
global ans
if X[i][j]=="G":
ans-=1
V[i][j]=True
for _,__ in [[i-1,j],[i+1,j],[i,j+1],[i,j-1]]:
if is_safe(_,__) and not V[_][__]:
if X[_][__]=="B":
global f
f=1
elif X[_][__] in ".G":
dfs(_,__)
if __name__=="__main__":
for _ in range(INI()):
n,m=INL()
X=[]
V=[[False for _ in range(m)] for _ in range(n)]
ans=0
f=0
for _ in range(n):
x=list(input())
ans+=x.count('G')
X.append(x)
for _ in range(n):
for __ in range(m):
if X[_][__]=='B':
make(_,__)
if X[n-1][m-1]!="#":
dfs(n-1,m-1)
# print(V)
if not ans and not f:
OPS('Yes')
else:
OPS('No')
```
Yes
| 98,801 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Vivek has encountered a problem. He has a maze that can be represented as an n × m grid. Each of the grid cells may represent the following:
* Empty — '.'
* Wall — '#'
* Good person — 'G'
* Bad person — 'B'
The only escape from the maze is at cell (n, m).
A person can move to a cell only if it shares a side with their current cell and does not contain a wall. Vivek wants to block some of the empty cells by replacing them with walls in such a way, that all the good people are able to escape, while none of the bad people are able to. A cell that initially contains 'G' or 'B' cannot be blocked and can be travelled through.
Help him determine if there exists a way to replace some (zero or more) empty cells with walls to satisfy the above conditions.
It is guaranteed that the cell (n,m) is empty. Vivek can also block this cell.
Input
The first line contains one integer t (1 ≤ t ≤ 100) — the number of test cases. The description of the test cases follows.
The first line of each test case contains two integers n, m (1 ≤ n, m ≤ 50) — the number of rows and columns in the maze.
Each of the next n lines contain m characters. They describe the layout of the maze. If a character on a line equals '.', the corresponding cell is empty. If it equals '#', the cell has a wall. 'G' corresponds to a good person and 'B' corresponds to a bad person.
Output
For each test case, print "Yes" if there exists a way to replace some empty cells with walls to satisfy the given conditions. Otherwise print "No"
You may print every letter in any case (upper or lower).
Example
Input
6
1 1
.
1 2
G.
2 2
#B
G.
2 3
G.#
B#.
3 3
#B.
#..
GG.
2 2
#B
B.
Output
Yes
Yes
No
No
Yes
Yes
Note
For the first and second test cases, all conditions are already satisfied.
For the third test case, there is only one empty cell (2,2), and if it is replaced with a wall then the good person at (1,2) will not be able to escape.
For the fourth test case, the good person at (1,1) cannot escape.
For the fifth test case, Vivek can block the cells (2,3) and (2,2).
For the last test case, Vivek can block the destination cell (2, 2).
Submitted Solution:
```
t = int(input())
for _ in range(t):
n, m = list(map(int, input().split()))
grid = []
for _ in range(n):
grid.append(list(input()))
def get_ns(i, j):
ns = []
for s_x, s_y in [(-1, 0), (1, 0), (0, 1), (0, -1)]:
if 0 <= i + s_x < n and 0 <= j + s_y < m and grid[i + s_x][j + s_y] != "#":
ns.append((i + s_x, j + s_y))
return ns
def solve():
good = set()
for i in range(n):
for j in range(m):
if grid[i][j] == "G":
good.add((i, j))
if grid[i][j] == "B":
for x, y in get_ns(i, j):
if grid[x][y] == "G":
return False
if grid[x][y] == ".":
grid[x][y] = "#"
if not good:
return True
if grid[n - 1][m - 1] not in [".", "G"]:
return False
stack = [(n - 1, m - 1)]
visited = set()
while stack:
x, y = stack.pop()
if (x, y) in good:
good.remove((x, y))
for ns in get_ns(x, y):
if ns not in visited:
visited.add(ns)
stack.append(ns)
if good:
return False
return True
ans = solve()
if ans:
print("Yes")
else:
print("No")
```
Yes
| 98,802 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Vivek has encountered a problem. He has a maze that can be represented as an n × m grid. Each of the grid cells may represent the following:
* Empty — '.'
* Wall — '#'
* Good person — 'G'
* Bad person — 'B'
The only escape from the maze is at cell (n, m).
A person can move to a cell only if it shares a side with their current cell and does not contain a wall. Vivek wants to block some of the empty cells by replacing them with walls in such a way, that all the good people are able to escape, while none of the bad people are able to. A cell that initially contains 'G' or 'B' cannot be blocked and can be travelled through.
Help him determine if there exists a way to replace some (zero or more) empty cells with walls to satisfy the above conditions.
It is guaranteed that the cell (n,m) is empty. Vivek can also block this cell.
Input
The first line contains one integer t (1 ≤ t ≤ 100) — the number of test cases. The description of the test cases follows.
The first line of each test case contains two integers n, m (1 ≤ n, m ≤ 50) — the number of rows and columns in the maze.
Each of the next n lines contain m characters. They describe the layout of the maze. If a character on a line equals '.', the corresponding cell is empty. If it equals '#', the cell has a wall. 'G' corresponds to a good person and 'B' corresponds to a bad person.
Output
For each test case, print "Yes" if there exists a way to replace some empty cells with walls to satisfy the given conditions. Otherwise print "No"
You may print every letter in any case (upper or lower).
Example
Input
6
1 1
.
1 2
G.
2 2
#B
G.
2 3
G.#
B#.
3 3
#B.
#..
GG.
2 2
#B
B.
Output
Yes
Yes
No
No
Yes
Yes
Note
For the first and second test cases, all conditions are already satisfied.
For the third test case, there is only one empty cell (2,2), and if it is replaced with a wall then the good person at (1,2) will not be able to escape.
For the fourth test case, the good person at (1,1) cannot escape.
For the fifth test case, Vivek can block the cells (2,3) and (2,2).
For the last test case, Vivek can block the destination cell (2, 2).
Submitted Solution:
```
def isValid(x,y,n,m):
if x<0 or x>n-1 or y<0 or y>m-1:
return False
if vis[x][y]==1 or matrix[x][y]=="#":
return False
return True
dx=[-1,0,1,0]
dy=[0,1,0,-1]
def dfs(x,y,n,m):
stack=[[x,y]]
vis[x][y]=1
while stack:
temp=stack.pop()
curX=temp[0]
curY=temp[1]
for i in range(4):
if isValid(curX+dx[i],curY+dy[i],n,m):
newX=curX+dx[i]
newY=curY+dy[i]
vis[newX][newY]=1
stack.append([newX,newY])
t=int(input())
for _ in range(t):
n,m=list(map(int,input().split()))
matrix=[]
c=0
for i in range(n):
s=str(input())
lis=[]
for j in range(m):
lis.append(s[j])
c+=lis.count("G")
matrix.append(lis)
a=0
vis=[[0]*m for i in range(n)]
for i in range(n):
for j in range(m):
if matrix[i][j]=="B":
for k in range(4):
if isValid(i+dx[k],j+dy[k],n,m):
if matrix[i+dx[k]][j+dy[k]]=="G" or (i+dx[k]==n-1 and j+dy[k]==m-1):
a+=1
else:
matrix[i+dx[k]][j+dy[k]]="#"
if c==0:
print("YES")
elif a>0:
print("NO")
else:
b=0
for i in range(n):
for j in range(m):
if vis[i][j]==0 and matrix[i][j]=="G":
dfs(i,j,n,m)
b+=1
if b<=1:
print("YES")
else:
print("NO")
```
No
| 98,803 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Vivek has encountered a problem. He has a maze that can be represented as an n × m grid. Each of the grid cells may represent the following:
* Empty — '.'
* Wall — '#'
* Good person — 'G'
* Bad person — 'B'
The only escape from the maze is at cell (n, m).
A person can move to a cell only if it shares a side with their current cell and does not contain a wall. Vivek wants to block some of the empty cells by replacing them with walls in such a way, that all the good people are able to escape, while none of the bad people are able to. A cell that initially contains 'G' or 'B' cannot be blocked and can be travelled through.
Help him determine if there exists a way to replace some (zero or more) empty cells with walls to satisfy the above conditions.
It is guaranteed that the cell (n,m) is empty. Vivek can also block this cell.
Input
The first line contains one integer t (1 ≤ t ≤ 100) — the number of test cases. The description of the test cases follows.
The first line of each test case contains two integers n, m (1 ≤ n, m ≤ 50) — the number of rows and columns in the maze.
Each of the next n lines contain m characters. They describe the layout of the maze. If a character on a line equals '.', the corresponding cell is empty. If it equals '#', the cell has a wall. 'G' corresponds to a good person and 'B' corresponds to a bad person.
Output
For each test case, print "Yes" if there exists a way to replace some empty cells with walls to satisfy the given conditions. Otherwise print "No"
You may print every letter in any case (upper or lower).
Example
Input
6
1 1
.
1 2
G.
2 2
#B
G.
2 3
G.#
B#.
3 3
#B.
#..
GG.
2 2
#B
B.
Output
Yes
Yes
No
No
Yes
Yes
Note
For the first and second test cases, all conditions are already satisfied.
For the third test case, there is only one empty cell (2,2), and if it is replaced with a wall then the good person at (1,2) will not be able to escape.
For the fourth test case, the good person at (1,1) cannot escape.
For the fifth test case, Vivek can block the cells (2,3) and (2,2).
For the last test case, Vivek can block the destination cell (2, 2).
Submitted Solution:
```
def bfs(a, b, l, n, m):
queue = [(a, b)]
vis = [(a, b)]
while queue:
for i in queue:
next = []
if i[1] != m - 1 and l[i[0]][i[1] + 1] in ['.', 'G'] and (i[0], i[1] + 1) not in vis:
next.append((i[0], i[1] + 1))
if i[1] != 0 and l[i[0]][i[1] - 1] in ['.', 'G'] and (i[0], i[1] - 1) not in vis:
next.append((i[0], i[1] - 1))
if i[0] != n - 1 and l[i[0] + 1][i[1]] in ['.', 'G'] and (i[0] + 1, i[1]) not in vis:
next.append((i[0] + 1, i[1]))
if i[0] != 0 and l[i[0] - 1][i[1]] in ['.', 'G'] and (i[0] - 1, i[1]) not in vis:
next.append((i[0] - 1, i[1]))
queue = next
vis += next
return (n - 1, m - 1) in vis
for _ in range(int(input())):
n, m = map(int, input().split())
l = [list(input()) for i in range(n)]
goods = []
f = True
for i in range(n):
if not f:
break
for j in range(m):
if l[i][j] == 'G':
goods.append((i, j))
elif l[i][j] == 'B':
if j != m - 1 and l[i][j + 1] == '.':
l[i][j + 1] = '#'
elif j != m - 1 and l[i][j + 1] == 'G':
f = False
break
if j != 0 and l[i][j - 1] == '.':
l[i][j - 1] = '#'
elif j != 0 and l[i][j - 1] == 'G':
f = False
break
if i != n - 1 and l[i + 1][j] == '.':
l[i + 1][j] = '#'
elif i != n - 1 and l[i + 1][j] == 'G':
f = False
break
if i != 0 and l[i - 1][j] == '.':
l[i - 1][j] = '#'
elif i != 0 and l[i - 1][j] == 'G':
f = False
break
for i in goods:
if not f:
break
if not bfs(i[0], i[1], l, n, m):
f = False
break
if f:
print('YES')
else:
print('NO')
```
No
| 98,804 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Vivek has encountered a problem. He has a maze that can be represented as an n × m grid. Each of the grid cells may represent the following:
* Empty — '.'
* Wall — '#'
* Good person — 'G'
* Bad person — 'B'
The only escape from the maze is at cell (n, m).
A person can move to a cell only if it shares a side with their current cell and does not contain a wall. Vivek wants to block some of the empty cells by replacing them with walls in such a way, that all the good people are able to escape, while none of the bad people are able to. A cell that initially contains 'G' or 'B' cannot be blocked and can be travelled through.
Help him determine if there exists a way to replace some (zero or more) empty cells with walls to satisfy the above conditions.
It is guaranteed that the cell (n,m) is empty. Vivek can also block this cell.
Input
The first line contains one integer t (1 ≤ t ≤ 100) — the number of test cases. The description of the test cases follows.
The first line of each test case contains two integers n, m (1 ≤ n, m ≤ 50) — the number of rows and columns in the maze.
Each of the next n lines contain m characters. They describe the layout of the maze. If a character on a line equals '.', the corresponding cell is empty. If it equals '#', the cell has a wall. 'G' corresponds to a good person and 'B' corresponds to a bad person.
Output
For each test case, print "Yes" if there exists a way to replace some empty cells with walls to satisfy the given conditions. Otherwise print "No"
You may print every letter in any case (upper or lower).
Example
Input
6
1 1
.
1 2
G.
2 2
#B
G.
2 3
G.#
B#.
3 3
#B.
#..
GG.
2 2
#B
B.
Output
Yes
Yes
No
No
Yes
Yes
Note
For the first and second test cases, all conditions are already satisfied.
For the third test case, there is only one empty cell (2,2), and if it is replaced with a wall then the good person at (1,2) will not be able to escape.
For the fourth test case, the good person at (1,1) cannot escape.
For the fifth test case, Vivek can block the cells (2,3) and (2,2).
For the last test case, Vivek can block the destination cell (2, 2).
Submitted Solution:
```
def check_bad_guy(row_index, col_index, board):
try:
if board[row_index - 1][col_index] == 'B':
return True
except:
pass
try:
if board[row_index + 1][col_index] == 'B':
return True
except:
pass
try:
if board[row_index][col_index - 1] == 'B':
return True
except:
pass
try:
if board[row_index][col_index + 1] == 'B':
return True
except:
pass
return False
def check_color(row_index, col_index, board_color):
neighbouring_colors = list()
try:
if board_color[row_index - 1][col_index] != 0:
neighbouring_colors.append(board_color[row_index - 1][col_index])
except:
pass
try:
if board_color[row_index][col_index - 1] != 0:
neighbouring_colors.append(board_color[row_index][col_index - 1])
except:
pass
return neighbouring_colors
def solve(board):
n = len(board)
m = len(board[0])
if board[n - 1][m - 1] == 'B':
return False
color_connections = dict()
available_color = 1
board_color = [[0] * m for i in range(n)]
""" print('starting board color')
print(board_color) """
good_guys = set()
for row_index in range(n):
for col_index in range(m):
""" print('\n')
print(f'row index {row_index}, col index {col_index}, symbol {board[row_index][col_index]}') """
if board[row_index][col_index] == '.':
if check_bad_guy(row_index, col_index, board):
board[row_index][col_index] == '#'
else:
neighbouring_colors = check_color(row_index, col_index, board_color)
""" print(f'neighbouring color {neighbouring_colors}') """
if len(neighbouring_colors) == 0:
board_color[row_index][col_index] = available_color
color_connections[available_color] = set()
available_color += 1
elif len(neighbouring_colors) == 1:
board_color[row_index][col_index] = neighbouring_colors[0]
elif len(neighbouring_colors) == 2:
""" print(neighbouring_colors) """
board_color[row_index][col_index] = neighbouring_colors[0]
color_connections[neighbouring_colors[0]].add(neighbouring_colors[1])
color_connections[neighbouring_colors[1]].add(neighbouring_colors[0])
elif board[row_index][col_index] == 'G':
if check_bad_guy(row_index, col_index, board):
return False
else:
good_guys.add((row_index, col_index))
neighbouring_colors = check_color(row_index, col_index, board_color)
""" print(f'neighbouring color {neighbouring_colors}') """
if len(neighbouring_colors) == 0:
board_color[row_index][col_index] = available_color
color_connections[available_color] = set()
available_color += 1
elif len(neighbouring_colors) == 1:
board_color[row_index][col_index] = neighbouring_colors[0]
elif len(neighbouring_colors) == 2:
board_color[row_index][col_index] = neighbouring_colors[0]
color_connections[neighbouring_colors[0]].add(neighbouring_colors[1])
color_connections[neighbouring_colors[1]].add(neighbouring_colors[0])
""" print('ending board color')
print(board_color) """
return evaluate_color(board_color, good_guys, color_connections)
def evaluate_color(board_color, good_guys, color_connections):
n = len(board_color)
m = len(board_color[0])
end_color = board_color[n - 1][m - 1]
if len(good_guys) == 0:
return True
elif end_color == 0:
return False
end_connections = set()
checked = dict()
stack = [end_color]
checked[end_color] = True
while len(stack) != 0:
current_color = stack.pop()
end_connections.add(current_color)
""" print(color_connections) """
for color in color_connections[current_color]:
if color not in checked:
stack.append(color)
checked[color] = True
for (row_index, col_index) in good_guys:
if board_color[row_index][col_index] not in end_connections:
return False
return True
t = int(input())
for testcase in range(t):
""" print(f'starting testcase {testcase}') """
n, m = list(map(int, input().split()))
board = [0] * n
for row_index in range(n):
board[row_index] = list(input())
""" print('problem statement')
print(board) """
if solve(board):
print('Yes')
else:
print('No')
""" print(f'ending testcase {testcase}')
print('\n') """
```
No
| 98,805 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Vivek has encountered a problem. He has a maze that can be represented as an n × m grid. Each of the grid cells may represent the following:
* Empty — '.'
* Wall — '#'
* Good person — 'G'
* Bad person — 'B'
The only escape from the maze is at cell (n, m).
A person can move to a cell only if it shares a side with their current cell and does not contain a wall. Vivek wants to block some of the empty cells by replacing them with walls in such a way, that all the good people are able to escape, while none of the bad people are able to. A cell that initially contains 'G' or 'B' cannot be blocked and can be travelled through.
Help him determine if there exists a way to replace some (zero or more) empty cells with walls to satisfy the above conditions.
It is guaranteed that the cell (n,m) is empty. Vivek can also block this cell.
Input
The first line contains one integer t (1 ≤ t ≤ 100) — the number of test cases. The description of the test cases follows.
The first line of each test case contains two integers n, m (1 ≤ n, m ≤ 50) — the number of rows and columns in the maze.
Each of the next n lines contain m characters. They describe the layout of the maze. If a character on a line equals '.', the corresponding cell is empty. If it equals '#', the cell has a wall. 'G' corresponds to a good person and 'B' corresponds to a bad person.
Output
For each test case, print "Yes" if there exists a way to replace some empty cells with walls to satisfy the given conditions. Otherwise print "No"
You may print every letter in any case (upper or lower).
Example
Input
6
1 1
.
1 2
G.
2 2
#B
G.
2 3
G.#
B#.
3 3
#B.
#..
GG.
2 2
#B
B.
Output
Yes
Yes
No
No
Yes
Yes
Note
For the first and second test cases, all conditions are already satisfied.
For the third test case, there is only one empty cell (2,2), and if it is replaced with a wall then the good person at (1,2) will not be able to escape.
For the fourth test case, the good person at (1,1) cannot escape.
For the fifth test case, Vivek can block the cells (2,3) and (2,2).
For the last test case, Vivek can block the destination cell (2, 2).
Submitted Solution:
```
import itertools
import copy
from queue import Queue
def satisfies(R, C, matrix):
q = Queue()
q.put( (R - 1, C - 1) )
reached = set([])
while not q.empty():
r, c = q.get()
if matrix[r][c] == '#':
continue
if matrix[r][c] == 'B':
return False
reached.add( (r, c) )
for rn, cn in [(r-1, c), (r, c-1)]:
if rn >= 0 and cn >= 0:
q.put( (rn, cn) )
for r in range(R):
for c in range(C):
if matrix[r][c] == 'G' and (r, c) not in reached:
return False
return True
def brute(R, C, matrix):
possWalls = []
for r in range(R):
for c in range(C):
if matrix[r][c] == '.':
possWalls.append( (r, c) )
for numWalls in range(len(possWalls) + 1):
for walls in itertools.combinations(possWalls, numWalls):
matrixCpy = copy.deepcopy(matrix)
for rw, cw in walls:
matrixCpy[rw][cw] = '#'
if satisfies(R, C, matrixCpy):
return 'Yes'
return 'No'
def main():
t = int(input())
for _ in range(t):
R, C = [int(x) for x in input().split()]
matrix = []
for _ in range(R):
matrix.append([ch for ch in input()])
# WARNING: CHANGE THIS!
# print(matrix)
result = brute(R, C, matrix)
print(result)
main()
```
No
| 98,806 |
Evaluate the correctness of the submitted Python 2 solution to the coding contest problem. Provide a "Yes" or "No" response.
Vivek has encountered a problem. He has a maze that can be represented as an n × m grid. Each of the grid cells may represent the following:
* Empty — '.'
* Wall — '#'
* Good person — 'G'
* Bad person — 'B'
The only escape from the maze is at cell (n, m).
A person can move to a cell only if it shares a side with their current cell and does not contain a wall. Vivek wants to block some of the empty cells by replacing them with walls in such a way, that all the good people are able to escape, while none of the bad people are able to. A cell that initially contains 'G' or 'B' cannot be blocked and can be travelled through.
Help him determine if there exists a way to replace some (zero or more) empty cells with walls to satisfy the above conditions.
It is guaranteed that the cell (n,m) is empty. Vivek can also block this cell.
Input
The first line contains one integer t (1 ≤ t ≤ 100) — the number of test cases. The description of the test cases follows.
The first line of each test case contains two integers n, m (1 ≤ n, m ≤ 50) — the number of rows and columns in the maze.
Each of the next n lines contain m characters. They describe the layout of the maze. If a character on a line equals '.', the corresponding cell is empty. If it equals '#', the cell has a wall. 'G' corresponds to a good person and 'B' corresponds to a bad person.
Output
For each test case, print "Yes" if there exists a way to replace some empty cells with walls to satisfy the given conditions. Otherwise print "No"
You may print every letter in any case (upper or lower).
Example
Input
6
1 1
.
1 2
G.
2 2
#B
G.
2 3
G.#
B#.
3 3
#B.
#..
GG.
2 2
#B
B.
Output
Yes
Yes
No
No
Yes
Yes
Note
For the first and second test cases, all conditions are already satisfied.
For the third test case, there is only one empty cell (2,2), and if it is replaced with a wall then the good person at (1,2) will not be able to escape.
For the fourth test case, the good person at (1,1) cannot escape.
For the fifth test case, Vivek can block the cells (2,3) and (2,2).
For the last test case, Vivek can block the destination cell (2, 2).
Submitted Solution:
```
from sys import stdin, stdout
from collections import Counter, defaultdict
from itertools import permutations, combinations
from fractions import gcd
import heapq
raw_input = stdin.readline
pr = stdout.write
mod=998244353
def ni():
return int(raw_input())
def li():
return list(map(int,raw_input().split()))
def pn(n):
stdout.write(str(n)+'\n')
def pa(arr):
pr(' '.join(map(str,arr))+'\n')
# fast read function for total integer input
def inp():
# this function returns whole input of
# space/line seperated integers
# Use Ctrl+D to flush stdin.
return tuple(map(int,stdin.read().split()))
range = xrange # not for python 3.0+
# main code
for t in range(ni()):
n,m=li()
arr=[]
for i in range(n):
arr.append(list(raw_input().strip()))
f=0
c=0
move=[(0,1),(0,-1),(-1,0),(1,0)]
for i in range(n):
for j in range(m):
if arr[i][j]!='B':
if arr[i][j]=='G':
c+=1
continue
for x,y in move:
px=i+x
py=i+y
if px>=0 and px<n and py>=0 and py<m:
if arr[px][py]=='G':
f=1
break
elif arr[px][py]!='B':
arr[px][py]='#'
if f:
break
if f:
break
if f:
break
if f==0 and c==0:
if arr[-1][-1]!='B':
pr('Yes\n')
else:
pr('No\n')
continue
if f or (c!=0 and (arr[n-1][m-1]=='#' or arr[n-1][m-1]=='B')):
pr('No\n')
continue
q=[(n-1,m-1)]
vis=[[0 for i in range(m)] for j in range(n)]
vis[-1][-1]=1
#print c
while q:
#print q,c
x,y=q.pop(0)
#print x,y
if arr[x][y]=='G':
#print x,y
c-=1
for x1,y1 in move:
px,py=x+x1,y+y1
#print px,py
if px>=0 and px<n and py>=0 and py<m:
if not vis[px][py] and arr[px][py] != '#':
vis[px][py]=1
q.append((px,py))
if c:
pr('No\n')
else:
pr('Yes\n')
```
No
| 98,807 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given a string s[1 ... n] consisting of lowercase Latin letters. It is guaranteed that n = 2^k for some integer k ≥ 0.
The string s[1 ... n] is called c-good if at least one of the following three conditions is satisfied:
* The length of s is 1, and it consists of the character c (i.e. s_1=c);
* The length of s is greater than 1, the first half of the string consists of only the character c (i.e. s_1=s_2=...=s_{n/2}=c) and the second half of the string (i.e. the string s_{n/2 + 1}s_{n/2 + 2} ... s_n) is a (c+1)-good string;
* The length of s is greater than 1, the second half of the string consists of only the character c (i.e. s_{n/2 + 1}=s_{n/2 + 2}=...=s_n=c) and the first half of the string (i.e. the string s_1s_2 ... s_{n/2}) is a (c+1)-good string.
For example: "aabc" is 'a'-good, "ffgheeee" is 'e'-good.
In one move, you can choose one index i from 1 to n and replace s_i with any lowercase Latin letter (any character from 'a' to 'z').
Your task is to find the minimum number of moves required to obtain an 'a'-good string from s (i.e. c-good string for c= 'a'). It is guaranteed that the answer always exists.
You have to answer t independent test cases.
Another example of an 'a'-good string is as follows. Consider the string s = "cdbbaaaa". It is an 'a'-good string, because:
* the second half of the string ("aaaa") consists of only the character 'a';
* the first half of the string ("cdbb") is 'b'-good string, because:
* the second half of the string ("bb") consists of only the character 'b';
* the first half of the string ("cd") is 'c'-good string, because:
* the first half of the string ("c") consists of only the character 'c';
* the second half of the string ("d") is 'd'-good string.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 2 ⋅ 10^4) — the number of test cases. Then t test cases follow.
The first line of the test case contains one integer n (1 ≤ n ≤ 131~072) — the length of s. It is guaranteed that n = 2^k for some integer k ≥ 0. The second line of the test case contains the string s consisting of n lowercase Latin letters.
It is guaranteed that the sum of n does not exceed 2 ⋅ 10^5 (∑ n ≤ 2 ⋅ 10^5).
Output
For each test case, print the answer — the minimum number of moves required to obtain an 'a'-good string from s (i.e. c-good string with c = 'a'). It is guaranteed that the answer exists.
Example
Input
6
8
bbdcaaaa
8
asdfghjk
8
ceaaaabb
8
bbaaddcc
1
z
2
ac
Output
0
7
4
5
1
1
Tags: bitmasks, brute force, divide and conquer, dp, implementation
Correct Solution:
```
def fn(s,a):
if len(s)==1:
if s==a:
return 0
else:
return 1
a1=0
a2=0
l=len(s)//2
t=s[:l]
p=s[l:]
for i in t:
if i!=a:
a1+=1
for i in p:
if i!=a:
a2+=1
return min(a1+fn(p,chr(ord(a)+1)),a2+fn(t,chr(ord(a)+1)))
for _ in range(int(input())):
n=int(input())
a=input()
print(fn(a,'a'))
```
| 98,808 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given a string s[1 ... n] consisting of lowercase Latin letters. It is guaranteed that n = 2^k for some integer k ≥ 0.
The string s[1 ... n] is called c-good if at least one of the following three conditions is satisfied:
* The length of s is 1, and it consists of the character c (i.e. s_1=c);
* The length of s is greater than 1, the first half of the string consists of only the character c (i.e. s_1=s_2=...=s_{n/2}=c) and the second half of the string (i.e. the string s_{n/2 + 1}s_{n/2 + 2} ... s_n) is a (c+1)-good string;
* The length of s is greater than 1, the second half of the string consists of only the character c (i.e. s_{n/2 + 1}=s_{n/2 + 2}=...=s_n=c) and the first half of the string (i.e. the string s_1s_2 ... s_{n/2}) is a (c+1)-good string.
For example: "aabc" is 'a'-good, "ffgheeee" is 'e'-good.
In one move, you can choose one index i from 1 to n and replace s_i with any lowercase Latin letter (any character from 'a' to 'z').
Your task is to find the minimum number of moves required to obtain an 'a'-good string from s (i.e. c-good string for c= 'a'). It is guaranteed that the answer always exists.
You have to answer t independent test cases.
Another example of an 'a'-good string is as follows. Consider the string s = "cdbbaaaa". It is an 'a'-good string, because:
* the second half of the string ("aaaa") consists of only the character 'a';
* the first half of the string ("cdbb") is 'b'-good string, because:
* the second half of the string ("bb") consists of only the character 'b';
* the first half of the string ("cd") is 'c'-good string, because:
* the first half of the string ("c") consists of only the character 'c';
* the second half of the string ("d") is 'd'-good string.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 2 ⋅ 10^4) — the number of test cases. Then t test cases follow.
The first line of the test case contains one integer n (1 ≤ n ≤ 131~072) — the length of s. It is guaranteed that n = 2^k for some integer k ≥ 0. The second line of the test case contains the string s consisting of n lowercase Latin letters.
It is guaranteed that the sum of n does not exceed 2 ⋅ 10^5 (∑ n ≤ 2 ⋅ 10^5).
Output
For each test case, print the answer — the minimum number of moves required to obtain an 'a'-good string from s (i.e. c-good string with c = 'a'). It is guaranteed that the answer exists.
Example
Input
6
8
bbdcaaaa
8
asdfghjk
8
ceaaaabb
8
bbaaddcc
1
z
2
ac
Output
0
7
4
5
1
1
Tags: bitmasks, brute force, divide and conquer, dp, implementation
Correct Solution:
```
LETRAS = "abcdefghijklmnopqrstuvwxyz"
def string_boa(indice, string):
letras = LETRAS[indice]
if len(string) == 1:
if string != letras:
return 1
else:
return 0
direita = 0
esquerda = 0
for l in string[:len(string)//2]:
if l != letras:
direita += 1
for l in string[len(string)//2:]:
if l != letras:
esquerda += 1
lado_direito = direita + string_boa(indice + 1, string[len(string)//2:])
lado_esquerdo = esquerda + string_boa(indice + 1, string[:len(string)//2:])
return min(lado_direito , lado_esquerdo)
entrada = int(input())
for i in range(entrada):
n = int(input())
string = input()
print(string_boa(0, string))
```
| 98,809 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given a string s[1 ... n] consisting of lowercase Latin letters. It is guaranteed that n = 2^k for some integer k ≥ 0.
The string s[1 ... n] is called c-good if at least one of the following three conditions is satisfied:
* The length of s is 1, and it consists of the character c (i.e. s_1=c);
* The length of s is greater than 1, the first half of the string consists of only the character c (i.e. s_1=s_2=...=s_{n/2}=c) and the second half of the string (i.e. the string s_{n/2 + 1}s_{n/2 + 2} ... s_n) is a (c+1)-good string;
* The length of s is greater than 1, the second half of the string consists of only the character c (i.e. s_{n/2 + 1}=s_{n/2 + 2}=...=s_n=c) and the first half of the string (i.e. the string s_1s_2 ... s_{n/2}) is a (c+1)-good string.
For example: "aabc" is 'a'-good, "ffgheeee" is 'e'-good.
In one move, you can choose one index i from 1 to n and replace s_i with any lowercase Latin letter (any character from 'a' to 'z').
Your task is to find the minimum number of moves required to obtain an 'a'-good string from s (i.e. c-good string for c= 'a'). It is guaranteed that the answer always exists.
You have to answer t independent test cases.
Another example of an 'a'-good string is as follows. Consider the string s = "cdbbaaaa". It is an 'a'-good string, because:
* the second half of the string ("aaaa") consists of only the character 'a';
* the first half of the string ("cdbb") is 'b'-good string, because:
* the second half of the string ("bb") consists of only the character 'b';
* the first half of the string ("cd") is 'c'-good string, because:
* the first half of the string ("c") consists of only the character 'c';
* the second half of the string ("d") is 'd'-good string.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 2 ⋅ 10^4) — the number of test cases. Then t test cases follow.
The first line of the test case contains one integer n (1 ≤ n ≤ 131~072) — the length of s. It is guaranteed that n = 2^k for some integer k ≥ 0. The second line of the test case contains the string s consisting of n lowercase Latin letters.
It is guaranteed that the sum of n does not exceed 2 ⋅ 10^5 (∑ n ≤ 2 ⋅ 10^5).
Output
For each test case, print the answer — the minimum number of moves required to obtain an 'a'-good string from s (i.e. c-good string with c = 'a'). It is guaranteed that the answer exists.
Example
Input
6
8
bbdcaaaa
8
asdfghjk
8
ceaaaabb
8
bbaaddcc
1
z
2
ac
Output
0
7
4
5
1
1
Tags: bitmasks, brute force, divide and conquer, dp, implementation
Correct Solution:
```
"""
Author: Q.E.D
Time: 2020-07-17 10:11:52
"""
def count(s, c):
c2 = chr(ord(c) + 1)
if len(s) == 1:
return 1 - int(s[0] == c)
else:
n = len(s)
h = n // 2
x1 = sum(1 for a in s[:h] if a != c)
x2 = sum(1 for a in s[h:] if a != c)
y1 = count(s[h:], c2)
y2 = count(s[:h], c2)
return min(x1 + y1, x2 + y2)
T = int(input())
for _ in range(T):
n = int(input())
s = input()
ans = count(s, 'a')
print(ans)
```
| 98,810 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given a string s[1 ... n] consisting of lowercase Latin letters. It is guaranteed that n = 2^k for some integer k ≥ 0.
The string s[1 ... n] is called c-good if at least one of the following three conditions is satisfied:
* The length of s is 1, and it consists of the character c (i.e. s_1=c);
* The length of s is greater than 1, the first half of the string consists of only the character c (i.e. s_1=s_2=...=s_{n/2}=c) and the second half of the string (i.e. the string s_{n/2 + 1}s_{n/2 + 2} ... s_n) is a (c+1)-good string;
* The length of s is greater than 1, the second half of the string consists of only the character c (i.e. s_{n/2 + 1}=s_{n/2 + 2}=...=s_n=c) and the first half of the string (i.e. the string s_1s_2 ... s_{n/2}) is a (c+1)-good string.
For example: "aabc" is 'a'-good, "ffgheeee" is 'e'-good.
In one move, you can choose one index i from 1 to n and replace s_i with any lowercase Latin letter (any character from 'a' to 'z').
Your task is to find the minimum number of moves required to obtain an 'a'-good string from s (i.e. c-good string for c= 'a'). It is guaranteed that the answer always exists.
You have to answer t independent test cases.
Another example of an 'a'-good string is as follows. Consider the string s = "cdbbaaaa". It is an 'a'-good string, because:
* the second half of the string ("aaaa") consists of only the character 'a';
* the first half of the string ("cdbb") is 'b'-good string, because:
* the second half of the string ("bb") consists of only the character 'b';
* the first half of the string ("cd") is 'c'-good string, because:
* the first half of the string ("c") consists of only the character 'c';
* the second half of the string ("d") is 'd'-good string.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 2 ⋅ 10^4) — the number of test cases. Then t test cases follow.
The first line of the test case contains one integer n (1 ≤ n ≤ 131~072) — the length of s. It is guaranteed that n = 2^k for some integer k ≥ 0. The second line of the test case contains the string s consisting of n lowercase Latin letters.
It is guaranteed that the sum of n does not exceed 2 ⋅ 10^5 (∑ n ≤ 2 ⋅ 10^5).
Output
For each test case, print the answer — the minimum number of moves required to obtain an 'a'-good string from s (i.e. c-good string with c = 'a'). It is guaranteed that the answer exists.
Example
Input
6
8
bbdcaaaa
8
asdfghjk
8
ceaaaabb
8
bbaaddcc
1
z
2
ac
Output
0
7
4
5
1
1
Tags: bitmasks, brute force, divide and conquer, dp, implementation
Correct Solution:
```
import sys
from math import log
def num(l, r, d):
mid = (l+r)//2
if d>=k:
return 0 if list_s[l] == n2a[d] else 1
count_l = len([0 for i in range(l, mid+1) if list_s[i] != n2a[d]])
count_r = len([0 for i in range(mid+1, r+1) if list_s[i] != n2a[d]])
return min(num(l, mid, d+1)+count_r, num(mid+1, r, d+1)+count_l)
a2n = {chr(97+i):i for i in range(26)}
n2a = {i:chr(97+i) for i in range(26)}
T=int(sys.stdin.readline())
for _ in range(T):
n = int(sys.stdin.readline()) # 0<=k<=17
list_s = sys.stdin.readline().strip()
k = int(log(n, 2))
print(num(0, n-1, 0))
```
| 98,811 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given a string s[1 ... n] consisting of lowercase Latin letters. It is guaranteed that n = 2^k for some integer k ≥ 0.
The string s[1 ... n] is called c-good if at least one of the following three conditions is satisfied:
* The length of s is 1, and it consists of the character c (i.e. s_1=c);
* The length of s is greater than 1, the first half of the string consists of only the character c (i.e. s_1=s_2=...=s_{n/2}=c) and the second half of the string (i.e. the string s_{n/2 + 1}s_{n/2 + 2} ... s_n) is a (c+1)-good string;
* The length of s is greater than 1, the second half of the string consists of only the character c (i.e. s_{n/2 + 1}=s_{n/2 + 2}=...=s_n=c) and the first half of the string (i.e. the string s_1s_2 ... s_{n/2}) is a (c+1)-good string.
For example: "aabc" is 'a'-good, "ffgheeee" is 'e'-good.
In one move, you can choose one index i from 1 to n and replace s_i with any lowercase Latin letter (any character from 'a' to 'z').
Your task is to find the minimum number of moves required to obtain an 'a'-good string from s (i.e. c-good string for c= 'a'). It is guaranteed that the answer always exists.
You have to answer t independent test cases.
Another example of an 'a'-good string is as follows. Consider the string s = "cdbbaaaa". It is an 'a'-good string, because:
* the second half of the string ("aaaa") consists of only the character 'a';
* the first half of the string ("cdbb") is 'b'-good string, because:
* the second half of the string ("bb") consists of only the character 'b';
* the first half of the string ("cd") is 'c'-good string, because:
* the first half of the string ("c") consists of only the character 'c';
* the second half of the string ("d") is 'd'-good string.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 2 ⋅ 10^4) — the number of test cases. Then t test cases follow.
The first line of the test case contains one integer n (1 ≤ n ≤ 131~072) — the length of s. It is guaranteed that n = 2^k for some integer k ≥ 0. The second line of the test case contains the string s consisting of n lowercase Latin letters.
It is guaranteed that the sum of n does not exceed 2 ⋅ 10^5 (∑ n ≤ 2 ⋅ 10^5).
Output
For each test case, print the answer — the minimum number of moves required to obtain an 'a'-good string from s (i.e. c-good string with c = 'a'). It is guaranteed that the answer exists.
Example
Input
6
8
bbdcaaaa
8
asdfghjk
8
ceaaaabb
8
bbaaddcc
1
z
2
ac
Output
0
7
4
5
1
1
Tags: bitmasks, brute force, divide and conquer, dp, implementation
Correct Solution:
```
import sys
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); sys.stdout.write('\n')
def wi(n): sys.stdout.write(str(n)); sys.stdout.write('\n')
def wia(a, sep=' '): sys.stdout.write(sep.join([str(x) for x in a])); sys.stdout.write('\n')
def solve(n, s, c):
if n == 1:
return 0 if s[0] == c else 1
mid = n // 2
c1 = s[:mid].count(c)
c2 = s[mid:].count(c)
return min(mid - c1 + solve(mid, s[mid:], chr(ord(c) + 1)), mid - c2 + solve(mid, s[:mid], chr(ord(c) + 1)))
def main():
for _ in range(ri()):
n = ri()
s = rs()
wi(solve(n, s, 'a'))
if __name__ == '__main__':
main()
```
| 98,812 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given a string s[1 ... n] consisting of lowercase Latin letters. It is guaranteed that n = 2^k for some integer k ≥ 0.
The string s[1 ... n] is called c-good if at least one of the following three conditions is satisfied:
* The length of s is 1, and it consists of the character c (i.e. s_1=c);
* The length of s is greater than 1, the first half of the string consists of only the character c (i.e. s_1=s_2=...=s_{n/2}=c) and the second half of the string (i.e. the string s_{n/2 + 1}s_{n/2 + 2} ... s_n) is a (c+1)-good string;
* The length of s is greater than 1, the second half of the string consists of only the character c (i.e. s_{n/2 + 1}=s_{n/2 + 2}=...=s_n=c) and the first half of the string (i.e. the string s_1s_2 ... s_{n/2}) is a (c+1)-good string.
For example: "aabc" is 'a'-good, "ffgheeee" is 'e'-good.
In one move, you can choose one index i from 1 to n and replace s_i with any lowercase Latin letter (any character from 'a' to 'z').
Your task is to find the minimum number of moves required to obtain an 'a'-good string from s (i.e. c-good string for c= 'a'). It is guaranteed that the answer always exists.
You have to answer t independent test cases.
Another example of an 'a'-good string is as follows. Consider the string s = "cdbbaaaa". It is an 'a'-good string, because:
* the second half of the string ("aaaa") consists of only the character 'a';
* the first half of the string ("cdbb") is 'b'-good string, because:
* the second half of the string ("bb") consists of only the character 'b';
* the first half of the string ("cd") is 'c'-good string, because:
* the first half of the string ("c") consists of only the character 'c';
* the second half of the string ("d") is 'd'-good string.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 2 ⋅ 10^4) — the number of test cases. Then t test cases follow.
The first line of the test case contains one integer n (1 ≤ n ≤ 131~072) — the length of s. It is guaranteed that n = 2^k for some integer k ≥ 0. The second line of the test case contains the string s consisting of n lowercase Latin letters.
It is guaranteed that the sum of n does not exceed 2 ⋅ 10^5 (∑ n ≤ 2 ⋅ 10^5).
Output
For each test case, print the answer — the minimum number of moves required to obtain an 'a'-good string from s (i.e. c-good string with c = 'a'). It is guaranteed that the answer exists.
Example
Input
6
8
bbdcaaaa
8
asdfghjk
8
ceaaaabb
8
bbaaddcc
1
z
2
ac
Output
0
7
4
5
1
1
Tags: bitmasks, brute force, divide and conquer, dp, implementation
Correct Solution:
```
import sys, threading
sys.setrecursionlimit(10**8)
threading.stack_size(10**8) # new thread will get stack of such size
def main():
def findMinCost(s,targetChar):
if len(s)==1:
return int(s[0]!=targetChar)
mid=len(s)//2
leftCost=sum(c!=targetChar for c in s[:mid])+findMinCost(s[mid:],chr(ord(targetChar)+1))
rightCost=sum(c!=targetChar for c in s[mid:])+findMinCost(s[:mid],chr(ord(targetChar)+1))
return min(leftCost,rightCost)
t=int(input())
for _ in range(t):
_,s=int(input()),input()
print(findMinCost(s,"a"))
threading.Thread(target=main).start()
```
| 98,813 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given a string s[1 ... n] consisting of lowercase Latin letters. It is guaranteed that n = 2^k for some integer k ≥ 0.
The string s[1 ... n] is called c-good if at least one of the following three conditions is satisfied:
* The length of s is 1, and it consists of the character c (i.e. s_1=c);
* The length of s is greater than 1, the first half of the string consists of only the character c (i.e. s_1=s_2=...=s_{n/2}=c) and the second half of the string (i.e. the string s_{n/2 + 1}s_{n/2 + 2} ... s_n) is a (c+1)-good string;
* The length of s is greater than 1, the second half of the string consists of only the character c (i.e. s_{n/2 + 1}=s_{n/2 + 2}=...=s_n=c) and the first half of the string (i.e. the string s_1s_2 ... s_{n/2}) is a (c+1)-good string.
For example: "aabc" is 'a'-good, "ffgheeee" is 'e'-good.
In one move, you can choose one index i from 1 to n and replace s_i with any lowercase Latin letter (any character from 'a' to 'z').
Your task is to find the minimum number of moves required to obtain an 'a'-good string from s (i.e. c-good string for c= 'a'). It is guaranteed that the answer always exists.
You have to answer t independent test cases.
Another example of an 'a'-good string is as follows. Consider the string s = "cdbbaaaa". It is an 'a'-good string, because:
* the second half of the string ("aaaa") consists of only the character 'a';
* the first half of the string ("cdbb") is 'b'-good string, because:
* the second half of the string ("bb") consists of only the character 'b';
* the first half of the string ("cd") is 'c'-good string, because:
* the first half of the string ("c") consists of only the character 'c';
* the second half of the string ("d") is 'd'-good string.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 2 ⋅ 10^4) — the number of test cases. Then t test cases follow.
The first line of the test case contains one integer n (1 ≤ n ≤ 131~072) — the length of s. It is guaranteed that n = 2^k for some integer k ≥ 0. The second line of the test case contains the string s consisting of n lowercase Latin letters.
It is guaranteed that the sum of n does not exceed 2 ⋅ 10^5 (∑ n ≤ 2 ⋅ 10^5).
Output
For each test case, print the answer — the minimum number of moves required to obtain an 'a'-good string from s (i.e. c-good string with c = 'a'). It is guaranteed that the answer exists.
Example
Input
6
8
bbdcaaaa
8
asdfghjk
8
ceaaaabb
8
bbaaddcc
1
z
2
ac
Output
0
7
4
5
1
1
Tags: bitmasks, brute force, divide and conquer, dp, implementation
Correct Solution:
```
from sys import setrecursionlimit,stdout,stdin
setrecursionlimit(100000)
def solve(l, r, c):
if r - l + 1 == 1:
return 1 - (s[l:r + 1] == c)
else:
mid = (r + l + 1) // 2
p1 = 0
p2 = 0
for i in range(l, mid):
p1 += s[i] != c
for i in range(mid, r + 1):
p2 += s[i] != c
ans = min(p1 + solve(mid, r, chr(ord(c) + 1)), solve(l, mid - 1, chr(ord(c) + 1)) + p2)
return ans
for _ in range(int(stdin.readline())):
n = int(stdin.readline())
s = stdin.readline()
print(solve(0, n - 1, 'a'))
```
| 98,814 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given a string s[1 ... n] consisting of lowercase Latin letters. It is guaranteed that n = 2^k for some integer k ≥ 0.
The string s[1 ... n] is called c-good if at least one of the following three conditions is satisfied:
* The length of s is 1, and it consists of the character c (i.e. s_1=c);
* The length of s is greater than 1, the first half of the string consists of only the character c (i.e. s_1=s_2=...=s_{n/2}=c) and the second half of the string (i.e. the string s_{n/2 + 1}s_{n/2 + 2} ... s_n) is a (c+1)-good string;
* The length of s is greater than 1, the second half of the string consists of only the character c (i.e. s_{n/2 + 1}=s_{n/2 + 2}=...=s_n=c) and the first half of the string (i.e. the string s_1s_2 ... s_{n/2}) is a (c+1)-good string.
For example: "aabc" is 'a'-good, "ffgheeee" is 'e'-good.
In one move, you can choose one index i from 1 to n and replace s_i with any lowercase Latin letter (any character from 'a' to 'z').
Your task is to find the minimum number of moves required to obtain an 'a'-good string from s (i.e. c-good string for c= 'a'). It is guaranteed that the answer always exists.
You have to answer t independent test cases.
Another example of an 'a'-good string is as follows. Consider the string s = "cdbbaaaa". It is an 'a'-good string, because:
* the second half of the string ("aaaa") consists of only the character 'a';
* the first half of the string ("cdbb") is 'b'-good string, because:
* the second half of the string ("bb") consists of only the character 'b';
* the first half of the string ("cd") is 'c'-good string, because:
* the first half of the string ("c") consists of only the character 'c';
* the second half of the string ("d") is 'd'-good string.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 2 ⋅ 10^4) — the number of test cases. Then t test cases follow.
The first line of the test case contains one integer n (1 ≤ n ≤ 131~072) — the length of s. It is guaranteed that n = 2^k for some integer k ≥ 0. The second line of the test case contains the string s consisting of n lowercase Latin letters.
It is guaranteed that the sum of n does not exceed 2 ⋅ 10^5 (∑ n ≤ 2 ⋅ 10^5).
Output
For each test case, print the answer — the minimum number of moves required to obtain an 'a'-good string from s (i.e. c-good string with c = 'a'). It is guaranteed that the answer exists.
Example
Input
6
8
bbdcaaaa
8
asdfghjk
8
ceaaaabb
8
bbaaddcc
1
z
2
ac
Output
0
7
4
5
1
1
Tags: bitmasks, brute force, divide and conquer, dp, implementation
Correct Solution:
```
def get_cnt(start,k,char,s):
result = 0
for index in range(start,start + 2**k):
if s[index] != char:
result += 1
return result
def f(start,k,char,s):
if (k == 0):
if s[start] == char:
return 0
return 1
return min(get_cnt(start,k-1,char,s) + f(start + 2**(k-1),k-1,char + 1,s),f(start,k-1,char + 1,s) + get_cnt(start + 2**(k-1),k-1,char,s))
def solve():
import sys
import math
input = sys.stdin.readline
print = sys.stdout.write
t = int(input())
while t > 0:
n = int(input())
s = [ord(char) - 97 for char in input().strip()]
max_k = int(math.log2(n))
print(f'{f(0,max_k,0,s)}')
print('\n')
t -= 1
solve()
```
| 98,815 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a string s[1 ... n] consisting of lowercase Latin letters. It is guaranteed that n = 2^k for some integer k ≥ 0.
The string s[1 ... n] is called c-good if at least one of the following three conditions is satisfied:
* The length of s is 1, and it consists of the character c (i.e. s_1=c);
* The length of s is greater than 1, the first half of the string consists of only the character c (i.e. s_1=s_2=...=s_{n/2}=c) and the second half of the string (i.e. the string s_{n/2 + 1}s_{n/2 + 2} ... s_n) is a (c+1)-good string;
* The length of s is greater than 1, the second half of the string consists of only the character c (i.e. s_{n/2 + 1}=s_{n/2 + 2}=...=s_n=c) and the first half of the string (i.e. the string s_1s_2 ... s_{n/2}) is a (c+1)-good string.
For example: "aabc" is 'a'-good, "ffgheeee" is 'e'-good.
In one move, you can choose one index i from 1 to n and replace s_i with any lowercase Latin letter (any character from 'a' to 'z').
Your task is to find the minimum number of moves required to obtain an 'a'-good string from s (i.e. c-good string for c= 'a'). It is guaranteed that the answer always exists.
You have to answer t independent test cases.
Another example of an 'a'-good string is as follows. Consider the string s = "cdbbaaaa". It is an 'a'-good string, because:
* the second half of the string ("aaaa") consists of only the character 'a';
* the first half of the string ("cdbb") is 'b'-good string, because:
* the second half of the string ("bb") consists of only the character 'b';
* the first half of the string ("cd") is 'c'-good string, because:
* the first half of the string ("c") consists of only the character 'c';
* the second half of the string ("d") is 'd'-good string.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 2 ⋅ 10^4) — the number of test cases. Then t test cases follow.
The first line of the test case contains one integer n (1 ≤ n ≤ 131~072) — the length of s. It is guaranteed that n = 2^k for some integer k ≥ 0. The second line of the test case contains the string s consisting of n lowercase Latin letters.
It is guaranteed that the sum of n does not exceed 2 ⋅ 10^5 (∑ n ≤ 2 ⋅ 10^5).
Output
For each test case, print the answer — the minimum number of moves required to obtain an 'a'-good string from s (i.e. c-good string with c = 'a'). It is guaranteed that the answer exists.
Example
Input
6
8
bbdcaaaa
8
asdfghjk
8
ceaaaabb
8
bbaaddcc
1
z
2
ac
Output
0
7
4
5
1
1
Submitted Solution:
```
def make_equal(st,n,c):
num=0
for i in range(n):
if(st[i]!=c):
num+=1
return num
def find(st,n,c):
if(n==1):
if(st[0]==c):
return 0
else:
return 1
return min(make_equal(st[0:n//2],n//2,c)+find(st[n//2:],n//2,chr(ord(c)+1)),make_equal(st[n//2:],n//2,c)+find(st[0:n//2],n//2,chr(ord(c)+1)))
t=int(input())
for _ in range(t):
n=int(input())
s=input()
print(find(s,n,'a'))
```
Yes
| 98,816 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a string s[1 ... n] consisting of lowercase Latin letters. It is guaranteed that n = 2^k for some integer k ≥ 0.
The string s[1 ... n] is called c-good if at least one of the following three conditions is satisfied:
* The length of s is 1, and it consists of the character c (i.e. s_1=c);
* The length of s is greater than 1, the first half of the string consists of only the character c (i.e. s_1=s_2=...=s_{n/2}=c) and the second half of the string (i.e. the string s_{n/2 + 1}s_{n/2 + 2} ... s_n) is a (c+1)-good string;
* The length of s is greater than 1, the second half of the string consists of only the character c (i.e. s_{n/2 + 1}=s_{n/2 + 2}=...=s_n=c) and the first half of the string (i.e. the string s_1s_2 ... s_{n/2}) is a (c+1)-good string.
For example: "aabc" is 'a'-good, "ffgheeee" is 'e'-good.
In one move, you can choose one index i from 1 to n and replace s_i with any lowercase Latin letter (any character from 'a' to 'z').
Your task is to find the minimum number of moves required to obtain an 'a'-good string from s (i.e. c-good string for c= 'a'). It is guaranteed that the answer always exists.
You have to answer t independent test cases.
Another example of an 'a'-good string is as follows. Consider the string s = "cdbbaaaa". It is an 'a'-good string, because:
* the second half of the string ("aaaa") consists of only the character 'a';
* the first half of the string ("cdbb") is 'b'-good string, because:
* the second half of the string ("bb") consists of only the character 'b';
* the first half of the string ("cd") is 'c'-good string, because:
* the first half of the string ("c") consists of only the character 'c';
* the second half of the string ("d") is 'd'-good string.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 2 ⋅ 10^4) — the number of test cases. Then t test cases follow.
The first line of the test case contains one integer n (1 ≤ n ≤ 131~072) — the length of s. It is guaranteed that n = 2^k for some integer k ≥ 0. The second line of the test case contains the string s consisting of n lowercase Latin letters.
It is guaranteed that the sum of n does not exceed 2 ⋅ 10^5 (∑ n ≤ 2 ⋅ 10^5).
Output
For each test case, print the answer — the minimum number of moves required to obtain an 'a'-good string from s (i.e. c-good string with c = 'a'). It is guaranteed that the answer exists.
Example
Input
6
8
bbdcaaaa
8
asdfghjk
8
ceaaaabb
8
bbaaddcc
1
z
2
ac
Output
0
7
4
5
1
1
Submitted Solution:
```
import math
from decimal import *
import random
import sys
sys.setrecursionlimit(10**6)
mod = int(1e9)+7
def dfs(arr, ans, c):
if(len(arr)==1):
if(arr[0]!=c):
ans+=1
return ans
if(len(arr)==0):
return ans
else:
tmpl, tmpr = int(ans),int(ans)
for i in range(len(arr)//2):
if(arr[i]!= c):
tmpl+=1
if(arr[len(arr)//2+i]!= c):
tmpr+=1
ans = min(dfs(arr[:len(arr)//2], tmpr, chr(ord(c)+1)),
dfs(arr[len(arr)//2:], tmpl, chr(ord(c)+1)))
return ans
for _ in range(int(input())):
n = int(input())
s = list(input())
ans = dfs(s, 0, 'a')
print(ans)
```
Yes
| 98,817 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a string s[1 ... n] consisting of lowercase Latin letters. It is guaranteed that n = 2^k for some integer k ≥ 0.
The string s[1 ... n] is called c-good if at least one of the following three conditions is satisfied:
* The length of s is 1, and it consists of the character c (i.e. s_1=c);
* The length of s is greater than 1, the first half of the string consists of only the character c (i.e. s_1=s_2=...=s_{n/2}=c) and the second half of the string (i.e. the string s_{n/2 + 1}s_{n/2 + 2} ... s_n) is a (c+1)-good string;
* The length of s is greater than 1, the second half of the string consists of only the character c (i.e. s_{n/2 + 1}=s_{n/2 + 2}=...=s_n=c) and the first half of the string (i.e. the string s_1s_2 ... s_{n/2}) is a (c+1)-good string.
For example: "aabc" is 'a'-good, "ffgheeee" is 'e'-good.
In one move, you can choose one index i from 1 to n and replace s_i with any lowercase Latin letter (any character from 'a' to 'z').
Your task is to find the minimum number of moves required to obtain an 'a'-good string from s (i.e. c-good string for c= 'a'). It is guaranteed that the answer always exists.
You have to answer t independent test cases.
Another example of an 'a'-good string is as follows. Consider the string s = "cdbbaaaa". It is an 'a'-good string, because:
* the second half of the string ("aaaa") consists of only the character 'a';
* the first half of the string ("cdbb") is 'b'-good string, because:
* the second half of the string ("bb") consists of only the character 'b';
* the first half of the string ("cd") is 'c'-good string, because:
* the first half of the string ("c") consists of only the character 'c';
* the second half of the string ("d") is 'd'-good string.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 2 ⋅ 10^4) — the number of test cases. Then t test cases follow.
The first line of the test case contains one integer n (1 ≤ n ≤ 131~072) — the length of s. It is guaranteed that n = 2^k for some integer k ≥ 0. The second line of the test case contains the string s consisting of n lowercase Latin letters.
It is guaranteed that the sum of n does not exceed 2 ⋅ 10^5 (∑ n ≤ 2 ⋅ 10^5).
Output
For each test case, print the answer — the minimum number of moves required to obtain an 'a'-good string from s (i.e. c-good string with c = 'a'). It is guaranteed that the answer exists.
Example
Input
6
8
bbdcaaaa
8
asdfghjk
8
ceaaaabb
8
bbaaddcc
1
z
2
ac
Output
0
7
4
5
1
1
Submitted Solution:
```
from collections import Counter
def cal(l, r, i):
if l == r:
if s[l] == i:
return 0
else:
return 1
m = (l+r)//2
cnt1 = Counter(s[l:m+1])
cnt2 = Counter(s[m+1:r+1])
l1 = (r-l+1)//2 - cnt1.get(i, 0) + cal(m+1, r, i+1)
l2 = (r-l+1)//2 - cnt2.get(i, 0) + cal(l, m, i+1)
return min(l1, l2)
x = int(input())
for i in range(x):
n = int(input())
s = input()
s = [ord(i)-ord('a') for i in s]
print(cal(0, len(s)-1, 0))
```
Yes
| 98,818 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a string s[1 ... n] consisting of lowercase Latin letters. It is guaranteed that n = 2^k for some integer k ≥ 0.
The string s[1 ... n] is called c-good if at least one of the following three conditions is satisfied:
* The length of s is 1, and it consists of the character c (i.e. s_1=c);
* The length of s is greater than 1, the first half of the string consists of only the character c (i.e. s_1=s_2=...=s_{n/2}=c) and the second half of the string (i.e. the string s_{n/2 + 1}s_{n/2 + 2} ... s_n) is a (c+1)-good string;
* The length of s is greater than 1, the second half of the string consists of only the character c (i.e. s_{n/2 + 1}=s_{n/2 + 2}=...=s_n=c) and the first half of the string (i.e. the string s_1s_2 ... s_{n/2}) is a (c+1)-good string.
For example: "aabc" is 'a'-good, "ffgheeee" is 'e'-good.
In one move, you can choose one index i from 1 to n and replace s_i with any lowercase Latin letter (any character from 'a' to 'z').
Your task is to find the minimum number of moves required to obtain an 'a'-good string from s (i.e. c-good string for c= 'a'). It is guaranteed that the answer always exists.
You have to answer t independent test cases.
Another example of an 'a'-good string is as follows. Consider the string s = "cdbbaaaa". It is an 'a'-good string, because:
* the second half of the string ("aaaa") consists of only the character 'a';
* the first half of the string ("cdbb") is 'b'-good string, because:
* the second half of the string ("bb") consists of only the character 'b';
* the first half of the string ("cd") is 'c'-good string, because:
* the first half of the string ("c") consists of only the character 'c';
* the second half of the string ("d") is 'd'-good string.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 2 ⋅ 10^4) — the number of test cases. Then t test cases follow.
The first line of the test case contains one integer n (1 ≤ n ≤ 131~072) — the length of s. It is guaranteed that n = 2^k for some integer k ≥ 0. The second line of the test case contains the string s consisting of n lowercase Latin letters.
It is guaranteed that the sum of n does not exceed 2 ⋅ 10^5 (∑ n ≤ 2 ⋅ 10^5).
Output
For each test case, print the answer — the minimum number of moves required to obtain an 'a'-good string from s (i.e. c-good string with c = 'a'). It is guaranteed that the answer exists.
Example
Input
6
8
bbdcaaaa
8
asdfghjk
8
ceaaaabb
8
bbaaddcc
1
z
2
ac
Output
0
7
4
5
1
1
Submitted Solution:
```
def cnt(s, c):
x = 0
for i in s:
if i != c:
x += 1
return x
def search(s, n, cho):
c = chr(cho)
if n == 1:
return 0 if s[0] == c else 1
m = n // 2
s1 = s[:m]
s2 = s[m:]
x = cnt(s1, c) + search(s2, m, cho+1)
y = search(s1, m, cho+1) + cnt(s2, c)
return min(x, y)
t = int(input())
for _ in range(t):
n = int(input())
s = input()
#print(s)
l = 0
r = n
moves = 0
cho = ord('a')
print(search(s, n, cho))
continue
while r - l > 1:
ch = chr(cho)
m = (l + r) // 2
x = cnt(s, l, m, ch)
y = cnt(s, m, r, ch)
lng = m - l
print(f"ch={ch}, l={l}, m={m}, r={r}, x={x}, y={y}")
if x > y:
moves += lng - x
l = m
else:
moves += lng - y
r = m
cho += 1
print(f"l={l}, moves={moves}, ch=>{chr(cho)}")
else:
if s[l] != chr(cho):
moves += 1
print(moves)
```
Yes
| 98,819 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a string s[1 ... n] consisting of lowercase Latin letters. It is guaranteed that n = 2^k for some integer k ≥ 0.
The string s[1 ... n] is called c-good if at least one of the following three conditions is satisfied:
* The length of s is 1, and it consists of the character c (i.e. s_1=c);
* The length of s is greater than 1, the first half of the string consists of only the character c (i.e. s_1=s_2=...=s_{n/2}=c) and the second half of the string (i.e. the string s_{n/2 + 1}s_{n/2 + 2} ... s_n) is a (c+1)-good string;
* The length of s is greater than 1, the second half of the string consists of only the character c (i.e. s_{n/2 + 1}=s_{n/2 + 2}=...=s_n=c) and the first half of the string (i.e. the string s_1s_2 ... s_{n/2}) is a (c+1)-good string.
For example: "aabc" is 'a'-good, "ffgheeee" is 'e'-good.
In one move, you can choose one index i from 1 to n and replace s_i with any lowercase Latin letter (any character from 'a' to 'z').
Your task is to find the minimum number of moves required to obtain an 'a'-good string from s (i.e. c-good string for c= 'a'). It is guaranteed that the answer always exists.
You have to answer t independent test cases.
Another example of an 'a'-good string is as follows. Consider the string s = "cdbbaaaa". It is an 'a'-good string, because:
* the second half of the string ("aaaa") consists of only the character 'a';
* the first half of the string ("cdbb") is 'b'-good string, because:
* the second half of the string ("bb") consists of only the character 'b';
* the first half of the string ("cd") is 'c'-good string, because:
* the first half of the string ("c") consists of only the character 'c';
* the second half of the string ("d") is 'd'-good string.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 2 ⋅ 10^4) — the number of test cases. Then t test cases follow.
The first line of the test case contains one integer n (1 ≤ n ≤ 131~072) — the length of s. It is guaranteed that n = 2^k for some integer k ≥ 0. The second line of the test case contains the string s consisting of n lowercase Latin letters.
It is guaranteed that the sum of n does not exceed 2 ⋅ 10^5 (∑ n ≤ 2 ⋅ 10^5).
Output
For each test case, print the answer — the minimum number of moves required to obtain an 'a'-good string from s (i.e. c-good string with c = 'a'). It is guaranteed that the answer exists.
Example
Input
6
8
bbdcaaaa
8
asdfghjk
8
ceaaaabb
8
bbaaddcc
1
z
2
ac
Output
0
7
4
5
1
1
Submitted Solution:
```
count=0
def c(li,l,h,good):
global count
if(h-l==1):
if(li[l]==good):
return 0
return 1
m=l+(h-l)//2
if(li[l:m].count(good)>li[m:h].count(good)):
return (h-l)//2-li[l:m].count(good) +c(li,m,h,chr(ord(good)+1))
elif(li[l:m].count(good)<li[m:h].count(good)):
return (h-l)//2-li[m:h].count(good) +c(li,l,m,chr(ord(good)+1))
else:
if(li[l:m].count(chr(ord(good)+1))>li[m:h].count(chr(ord(good)+1))):
return (h-l)//2-li[m:h].count(good) +c(li,l,m,chr(ord(good)+1))
else:
return (h-l)//2-li[l:m].count(good) +c(li,m,h,chr(ord(good)+1))
def main():
for _ in range(int(input())):
n=int(input())
a=list(input())
print(c(a,0,n,'a'))
main()
```
No
| 98,820 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a string s[1 ... n] consisting of lowercase Latin letters. It is guaranteed that n = 2^k for some integer k ≥ 0.
The string s[1 ... n] is called c-good if at least one of the following three conditions is satisfied:
* The length of s is 1, and it consists of the character c (i.e. s_1=c);
* The length of s is greater than 1, the first half of the string consists of only the character c (i.e. s_1=s_2=...=s_{n/2}=c) and the second half of the string (i.e. the string s_{n/2 + 1}s_{n/2 + 2} ... s_n) is a (c+1)-good string;
* The length of s is greater than 1, the second half of the string consists of only the character c (i.e. s_{n/2 + 1}=s_{n/2 + 2}=...=s_n=c) and the first half of the string (i.e. the string s_1s_2 ... s_{n/2}) is a (c+1)-good string.
For example: "aabc" is 'a'-good, "ffgheeee" is 'e'-good.
In one move, you can choose one index i from 1 to n and replace s_i with any lowercase Latin letter (any character from 'a' to 'z').
Your task is to find the minimum number of moves required to obtain an 'a'-good string from s (i.e. c-good string for c= 'a'). It is guaranteed that the answer always exists.
You have to answer t independent test cases.
Another example of an 'a'-good string is as follows. Consider the string s = "cdbbaaaa". It is an 'a'-good string, because:
* the second half of the string ("aaaa") consists of only the character 'a';
* the first half of the string ("cdbb") is 'b'-good string, because:
* the second half of the string ("bb") consists of only the character 'b';
* the first half of the string ("cd") is 'c'-good string, because:
* the first half of the string ("c") consists of only the character 'c';
* the second half of the string ("d") is 'd'-good string.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 2 ⋅ 10^4) — the number of test cases. Then t test cases follow.
The first line of the test case contains one integer n (1 ≤ n ≤ 131~072) — the length of s. It is guaranteed that n = 2^k for some integer k ≥ 0. The second line of the test case contains the string s consisting of n lowercase Latin letters.
It is guaranteed that the sum of n does not exceed 2 ⋅ 10^5 (∑ n ≤ 2 ⋅ 10^5).
Output
For each test case, print the answer — the minimum number of moves required to obtain an 'a'-good string from s (i.e. c-good string with c = 'a'). It is guaranteed that the answer exists.
Example
Input
6
8
bbdcaaaa
8
asdfghjk
8
ceaaaabb
8
bbaaddcc
1
z
2
ac
Output
0
7
4
5
1
1
Submitted Solution:
```
from sys import stdin, stdout, setrecursionlimit
from collections import deque, defaultdict, Counter
from functools import lru_cache
from heapq import heappush, heappop
import math
setrecursionlimit(100000)
rl = lambda: stdin.readline()
rll = lambda: stdin.readline().split()
rli = lambda: map(int, stdin.readline().split())
rlf = lambda: map(float, stdin.readline().split())
INF, NINF = float('inf'), float('-inf')
tonum = lambda c: ord(c) - 97
tochr = lambda num: chr(num + 97)
def solv(s, i, j, need):
if i > j: return 0
if i == j: return 0 if s[i] == tochr(need) else 1
mid = (i + j)//2
cnt1, cnt2 = 0, 0
cneed = tochr(need)
for x in range(i, mid+1):
if s[x] != cneed: cnt1 += 1
for x in range(mid+1, j+1):
if s[x] != cneed: cnt2 += 1
# print(f"i, j, need, mid = {i, j, cneed, mid}. fix_a = {cnt1}, fix_b = {cnt2}")
if cnt1 < cnt2:
return cnt1 + solv(s, mid+1, j, need+1)
elif cnt2 > cnt1:
return cnt2 + solv(s, i, mid, need+1)
else:
return min(cnt1+solv(s,mid+1,j,need+1), cnt2+solv(s,i,mid,need+1))
def main():
T = int(rl())
for _ in range(T):
n = int(rl())
s = rll()[0]
x = solv(s, 0, len(s)-1, 0)
print(x)
stdout.close()
if __name__ == "__main__":
main()
```
No
| 98,821 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a string s[1 ... n] consisting of lowercase Latin letters. It is guaranteed that n = 2^k for some integer k ≥ 0.
The string s[1 ... n] is called c-good if at least one of the following three conditions is satisfied:
* The length of s is 1, and it consists of the character c (i.e. s_1=c);
* The length of s is greater than 1, the first half of the string consists of only the character c (i.e. s_1=s_2=...=s_{n/2}=c) and the second half of the string (i.e. the string s_{n/2 + 1}s_{n/2 + 2} ... s_n) is a (c+1)-good string;
* The length of s is greater than 1, the second half of the string consists of only the character c (i.e. s_{n/2 + 1}=s_{n/2 + 2}=...=s_n=c) and the first half of the string (i.e. the string s_1s_2 ... s_{n/2}) is a (c+1)-good string.
For example: "aabc" is 'a'-good, "ffgheeee" is 'e'-good.
In one move, you can choose one index i from 1 to n and replace s_i with any lowercase Latin letter (any character from 'a' to 'z').
Your task is to find the minimum number of moves required to obtain an 'a'-good string from s (i.e. c-good string for c= 'a'). It is guaranteed that the answer always exists.
You have to answer t independent test cases.
Another example of an 'a'-good string is as follows. Consider the string s = "cdbbaaaa". It is an 'a'-good string, because:
* the second half of the string ("aaaa") consists of only the character 'a';
* the first half of the string ("cdbb") is 'b'-good string, because:
* the second half of the string ("bb") consists of only the character 'b';
* the first half of the string ("cd") is 'c'-good string, because:
* the first half of the string ("c") consists of only the character 'c';
* the second half of the string ("d") is 'd'-good string.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 2 ⋅ 10^4) — the number of test cases. Then t test cases follow.
The first line of the test case contains one integer n (1 ≤ n ≤ 131~072) — the length of s. It is guaranteed that n = 2^k for some integer k ≥ 0. The second line of the test case contains the string s consisting of n lowercase Latin letters.
It is guaranteed that the sum of n does not exceed 2 ⋅ 10^5 (∑ n ≤ 2 ⋅ 10^5).
Output
For each test case, print the answer — the minimum number of moves required to obtain an 'a'-good string from s (i.e. c-good string with c = 'a'). It is guaranteed that the answer exists.
Example
Input
6
8
bbdcaaaa
8
asdfghjk
8
ceaaaabb
8
bbaaddcc
1
z
2
ac
Output
0
7
4
5
1
1
Submitted Solution:
```
from sys import stdin
input = lambda: stdin.readline().rstrip("\r\n")
from collections import deque as que, defaultdict as vector
from heapq import*
inin = lambda: int(input())
inar = lambda: list(map(int,input().split()))
from math import log2
def solve(c,st):
cnt=0
for i in st:
if i!=c:
cnt+=1
return cnt
alp='abcdefghijklmnopqrstuvwxyz'
Testcase=inin()
for i in range(Testcase):
n=inin()
s=input()
sc=s
k=int(log2(n)+1)
copy=k
#print(k)
ans=0
while(k):
c=alp[copy-k]
k-=1
#print(c)
n=len(s)
if len(s)==1:
ans+= solve(c,s)
break
l=solve(c,s[0:2**(k-1)])
r=solve(c,s[2**(k-1):n])
#print(s[0:2**(k-1)],s[2**(k-1):n])
#print(l,r)
if l<r:
s=s[2**(k-1):n]
ans+=l
else:
s=s[0:2**(k-1)]
#print(s[0:2**(k-1)])
ans+=r
ans1=0
k=copy
s=sc
while(k):
c=alp[copy-k]
k-=1
#print(c)
n=len(s)
if len(s)==1:
ans1+= solve(c,s)
break
l=solve(c,s[0:2**(k-1)])
r=solve(c,s[2**(k-1):n])
#print(s[0:2**(k-1)],s[2**(k-1):n])
#print(l,r)
if l<=r:
s=s[2**(k-1):n]
ans1+=l
else:
s=s[0:2**(k-1)]
#print(s[0:2**(k-1)])
ans1+=r
print(min(ans,ans1))
```
No
| 98,822 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a string s[1 ... n] consisting of lowercase Latin letters. It is guaranteed that n = 2^k for some integer k ≥ 0.
The string s[1 ... n] is called c-good if at least one of the following three conditions is satisfied:
* The length of s is 1, and it consists of the character c (i.e. s_1=c);
* The length of s is greater than 1, the first half of the string consists of only the character c (i.e. s_1=s_2=...=s_{n/2}=c) and the second half of the string (i.e. the string s_{n/2 + 1}s_{n/2 + 2} ... s_n) is a (c+1)-good string;
* The length of s is greater than 1, the second half of the string consists of only the character c (i.e. s_{n/2 + 1}=s_{n/2 + 2}=...=s_n=c) and the first half of the string (i.e. the string s_1s_2 ... s_{n/2}) is a (c+1)-good string.
For example: "aabc" is 'a'-good, "ffgheeee" is 'e'-good.
In one move, you can choose one index i from 1 to n and replace s_i with any lowercase Latin letter (any character from 'a' to 'z').
Your task is to find the minimum number of moves required to obtain an 'a'-good string from s (i.e. c-good string for c= 'a'). It is guaranteed that the answer always exists.
You have to answer t independent test cases.
Another example of an 'a'-good string is as follows. Consider the string s = "cdbbaaaa". It is an 'a'-good string, because:
* the second half of the string ("aaaa") consists of only the character 'a';
* the first half of the string ("cdbb") is 'b'-good string, because:
* the second half of the string ("bb") consists of only the character 'b';
* the first half of the string ("cd") is 'c'-good string, because:
* the first half of the string ("c") consists of only the character 'c';
* the second half of the string ("d") is 'd'-good string.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 2 ⋅ 10^4) — the number of test cases. Then t test cases follow.
The first line of the test case contains one integer n (1 ≤ n ≤ 131~072) — the length of s. It is guaranteed that n = 2^k for some integer k ≥ 0. The second line of the test case contains the string s consisting of n lowercase Latin letters.
It is guaranteed that the sum of n does not exceed 2 ⋅ 10^5 (∑ n ≤ 2 ⋅ 10^5).
Output
For each test case, print the answer — the minimum number of moves required to obtain an 'a'-good string from s (i.e. c-good string with c = 'a'). It is guaranteed that the answer exists.
Example
Input
6
8
bbdcaaaa
8
asdfghjk
8
ceaaaabb
8
bbaaddcc
1
z
2
ac
Output
0
7
4
5
1
1
Submitted Solution:
```
def rec(s,c):
n=len(s)
if n==2:
if s[0]==c and s[1]==chr(ord(c)+1):
return 0
elif s[1]==c and s[0]==chr(ord(c)+1):
return 0
elif s[0]==c:
return 1
elif s[1]==c:
return 1
else:
return 2
a1,a2=0,0
for i in range(n//2):
if s[i]!=c:
a1+=1
for i in range(n//2,n):
if s[i]!=c:
a2+=1
an1=rec(s[n//2:],chr(ord(c)+1))+a1
an2=rec(s[:n//2],chr(ord(c)+1))+a2
return min(an1,an2)
t=int(input())
for q in range(t):
n=int(input())
s=input()
if n==1:
if s[0]=='a':
print(0)
else:
print(1)
else:
print(rec(s,'a'))
```
No
| 98,823 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Alexander is a well-known programmer. Today he decided to finally go out and play football, but with the first hit he left a dent on the new Rolls-Royce of the wealthy businessman Big Vova. Vladimir has recently opened a store on the popular online marketplace "Zmey-Gorynych", and offers Alex a job: if he shows his programming skills by solving a task, he'll work as a cybersecurity specialist. Otherwise, he'll be delivering some doubtful products for the next two years.
You're given n positive integers a_1, a_2, ..., a_n. Using each of them exactly at once, you're to make such sequence b_1, b_2, ..., b_n that sequence c_1, c_2, ..., c_n is lexicographically maximal, where c_i=GCD(b_1,...,b_i) - the greatest common divisor of the first i elements of b.
Alexander is really afraid of the conditions of this simple task, so he asks you to solve it.
A sequence a is lexicographically smaller than a sequence b if and only if one of the following holds:
* a is a prefix of b, but a ≠ b;
* in the first position where a and b differ, the sequence a has a smaller element than the corresponding element in b.
Input
Each test contains multiple test cases. The first line contains the number of test cases t (1 ≤ t ≤ 10^3). Description of the test cases follows.
The first line of each test case contains a single integer n (1 ≤ n ≤ 10^3) — the length of the sequence a.
The second line of each test case contains n integers a_1,...,a_n (1 ≤ a_i ≤ 10^3) — the sequence a.
It is guaranteed that the sum of n over all test cases does not exceed 10^3.
Output
For each test case output the answer in a single line — the desired sequence b. If there are multiple answers, print any.
Example
Input
7
2
2 5
4
1 8 2 3
3
3 8 9
5
64 25 75 100 50
1
42
6
96 128 88 80 52 7
5
2 4 8 16 17
Output
5 2
8 2 1 3
9 3 8
100 50 25 75 64
42
128 96 80 88 52 7
17 2 4 8 16
Note
In the first test case of the example, there are only two possible permutations b — [2, 5] and [5, 2]: for the first one c=[2, 1], for the second one c=[5, 1].
In the third test case of the example, number 9 should be the first in b, and GCD(9, 3)=3, GCD(9, 8)=1, so the second number of b should be 3.
In the seventh test case of the example, first four numbers pairwise have a common divisor (a power of two), but none of them can be the first in the optimal permutation b.
Tags: brute force, greedy, math, number theory
Correct Solution:
```
t=int(input())
import math
while t:
t-=1
n=int(input())
a=[int(i) for i in input().split()]
a.sort()
ans=[]
ans.append(a[n-1])
del a[n-1]
curg=ans[0]
while len(a):
ma=0
inl=0
flag=0
mi=1
for i in range(len(a)):
if math.gcd(curg,a[i])>mi:
ma=a[i]
inl=i
mi=math.gcd(curg,a[i])
flag+=1
if flag==0:
break
else:
ans.append(ma)
del a[inl]
curg=math.gcd(curg,ans[-1])
ans+=a
print(*ans,sep=" ")
```
| 98,824 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Alexander is a well-known programmer. Today he decided to finally go out and play football, but with the first hit he left a dent on the new Rolls-Royce of the wealthy businessman Big Vova. Vladimir has recently opened a store on the popular online marketplace "Zmey-Gorynych", and offers Alex a job: if he shows his programming skills by solving a task, he'll work as a cybersecurity specialist. Otherwise, he'll be delivering some doubtful products for the next two years.
You're given n positive integers a_1, a_2, ..., a_n. Using each of them exactly at once, you're to make such sequence b_1, b_2, ..., b_n that sequence c_1, c_2, ..., c_n is lexicographically maximal, where c_i=GCD(b_1,...,b_i) - the greatest common divisor of the first i elements of b.
Alexander is really afraid of the conditions of this simple task, so he asks you to solve it.
A sequence a is lexicographically smaller than a sequence b if and only if one of the following holds:
* a is a prefix of b, but a ≠ b;
* in the first position where a and b differ, the sequence a has a smaller element than the corresponding element in b.
Input
Each test contains multiple test cases. The first line contains the number of test cases t (1 ≤ t ≤ 10^3). Description of the test cases follows.
The first line of each test case contains a single integer n (1 ≤ n ≤ 10^3) — the length of the sequence a.
The second line of each test case contains n integers a_1,...,a_n (1 ≤ a_i ≤ 10^3) — the sequence a.
It is guaranteed that the sum of n over all test cases does not exceed 10^3.
Output
For each test case output the answer in a single line — the desired sequence b. If there are multiple answers, print any.
Example
Input
7
2
2 5
4
1 8 2 3
3
3 8 9
5
64 25 75 100 50
1
42
6
96 128 88 80 52 7
5
2 4 8 16 17
Output
5 2
8 2 1 3
9 3 8
100 50 25 75 64
42
128 96 80 88 52 7
17 2 4 8 16
Note
In the first test case of the example, there are only two possible permutations b — [2, 5] and [5, 2]: for the first one c=[2, 1], for the second one c=[5, 1].
In the third test case of the example, number 9 should be the first in b, and GCD(9, 3)=3, GCD(9, 8)=1, so the second number of b should be 3.
In the seventh test case of the example, first four numbers pairwise have a common divisor (a power of two), but none of them can be the first in the optimal permutation b.
Tags: brute force, greedy, math, number theory
Correct Solution:
```
from sys import stdin
def inp():
return stdin.buffer.readline().rstrip().decode('utf8')
def itg():
return int(stdin.buffer.readline())
def mpint():
return map(int, stdin.buffer.readline().split())
# ############################## import
def gcd(x, y):
"""greatest common divisor of x and y"""
while y:
x, y = y, x % y
return x
# ############################## main
from collections import Counter
for __ in range(itg()):
n = itg()
arr = list(mpint())
arr.sort()
gg = arr.pop()
ans = [gg]
counter = Counter(arr)
while counter:
best_key, best_value = None, 0
for key in counter.keys():
g = gcd(key, gg)
if g > best_value:
best_key, best_value = key, g
elif g == best_value:
best_key = max(best_key, key)
ans.append(best_key)
gg = gcd(gg, best_key)
counter[best_key] -= 1
if counter[best_key] == 0:
del counter[best_key]
print(*ans)
# Please check!
```
| 98,825 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Alexander is a well-known programmer. Today he decided to finally go out and play football, but with the first hit he left a dent on the new Rolls-Royce of the wealthy businessman Big Vova. Vladimir has recently opened a store on the popular online marketplace "Zmey-Gorynych", and offers Alex a job: if he shows his programming skills by solving a task, he'll work as a cybersecurity specialist. Otherwise, he'll be delivering some doubtful products for the next two years.
You're given n positive integers a_1, a_2, ..., a_n. Using each of them exactly at once, you're to make such sequence b_1, b_2, ..., b_n that sequence c_1, c_2, ..., c_n is lexicographically maximal, where c_i=GCD(b_1,...,b_i) - the greatest common divisor of the first i elements of b.
Alexander is really afraid of the conditions of this simple task, so he asks you to solve it.
A sequence a is lexicographically smaller than a sequence b if and only if one of the following holds:
* a is a prefix of b, but a ≠ b;
* in the first position where a and b differ, the sequence a has a smaller element than the corresponding element in b.
Input
Each test contains multiple test cases. The first line contains the number of test cases t (1 ≤ t ≤ 10^3). Description of the test cases follows.
The first line of each test case contains a single integer n (1 ≤ n ≤ 10^3) — the length of the sequence a.
The second line of each test case contains n integers a_1,...,a_n (1 ≤ a_i ≤ 10^3) — the sequence a.
It is guaranteed that the sum of n over all test cases does not exceed 10^3.
Output
For each test case output the answer in a single line — the desired sequence b. If there are multiple answers, print any.
Example
Input
7
2
2 5
4
1 8 2 3
3
3 8 9
5
64 25 75 100 50
1
42
6
96 128 88 80 52 7
5
2 4 8 16 17
Output
5 2
8 2 1 3
9 3 8
100 50 25 75 64
42
128 96 80 88 52 7
17 2 4 8 16
Note
In the first test case of the example, there are only two possible permutations b — [2, 5] and [5, 2]: for the first one c=[2, 1], for the second one c=[5, 1].
In the third test case of the example, number 9 should be the first in b, and GCD(9, 3)=3, GCD(9, 8)=1, so the second number of b should be 3.
In the seventh test case of the example, first four numbers pairwise have a common divisor (a power of two), but none of them can be the first in the optimal permutation b.
Tags: brute force, greedy, math, number theory
Correct Solution:
```
import sys
sys.setrecursionlimit(10**6)
from collections import Counter,defaultdict as dft
def mp():return map(int,input().split())
def ml():return list(map(int,input().split()))
def solve():
n=int(input())
arr=ml()
vis=[0]*(n+5)
#print(fac)
arr.sort(reverse=True)
res=[]
for p in range(10**3+1,0,-1):
case=0
for i in range(len(res)):
if res[i]%p!=0:
case=1
break
if case:
continue
for i in range(n):
if vis[i]==0 and arr[i]%p==0:
res.append(arr[i])
vis[i]=1
print(*res)
t=int(input())
for _ in range(t):
solve()
#pass
```
| 98,826 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Alexander is a well-known programmer. Today he decided to finally go out and play football, but with the first hit he left a dent on the new Rolls-Royce of the wealthy businessman Big Vova. Vladimir has recently opened a store on the popular online marketplace "Zmey-Gorynych", and offers Alex a job: if he shows his programming skills by solving a task, he'll work as a cybersecurity specialist. Otherwise, he'll be delivering some doubtful products for the next two years.
You're given n positive integers a_1, a_2, ..., a_n. Using each of them exactly at once, you're to make such sequence b_1, b_2, ..., b_n that sequence c_1, c_2, ..., c_n is lexicographically maximal, where c_i=GCD(b_1,...,b_i) - the greatest common divisor of the first i elements of b.
Alexander is really afraid of the conditions of this simple task, so he asks you to solve it.
A sequence a is lexicographically smaller than a sequence b if and only if one of the following holds:
* a is a prefix of b, but a ≠ b;
* in the first position where a and b differ, the sequence a has a smaller element than the corresponding element in b.
Input
Each test contains multiple test cases. The first line contains the number of test cases t (1 ≤ t ≤ 10^3). Description of the test cases follows.
The first line of each test case contains a single integer n (1 ≤ n ≤ 10^3) — the length of the sequence a.
The second line of each test case contains n integers a_1,...,a_n (1 ≤ a_i ≤ 10^3) — the sequence a.
It is guaranteed that the sum of n over all test cases does not exceed 10^3.
Output
For each test case output the answer in a single line — the desired sequence b. If there are multiple answers, print any.
Example
Input
7
2
2 5
4
1 8 2 3
3
3 8 9
5
64 25 75 100 50
1
42
6
96 128 88 80 52 7
5
2 4 8 16 17
Output
5 2
8 2 1 3
9 3 8
100 50 25 75 64
42
128 96 80 88 52 7
17 2 4 8 16
Note
In the first test case of the example, there are only two possible permutations b — [2, 5] and [5, 2]: for the first one c=[2, 1], for the second one c=[5, 1].
In the third test case of the example, number 9 should be the first in b, and GCD(9, 3)=3, GCD(9, 8)=1, so the second number of b should be 3.
In the seventh test case of the example, first four numbers pairwise have a common divisor (a power of two), but none of them can be the first in the optimal permutation b.
Tags: brute force, greedy, math, number theory
Correct Solution:
```
def bigVova(arr):
res = [None for _ in arr]
visited = [False for _ in arr]
d = 0
for i in range(len(arr)):
maxItem = 0
index = 0
for j in range(len(arr)):
currGcd = gcd(d, arr[j])
if visited[j] == False and currGcd > maxItem:
maxItem = currGcd
index = j
d = maxItem
res[i] = arr[index]
visited[index] = True
return res
def gcd(a, b):
while b > 0:
rem = a % b
a = b
b = rem
return a
for i in range(int(input())):
N = int(input())
arr = [int(val) for val in input().split(" ")]
print(" ".join(str(x) for x in bigVova(arr)))
```
| 98,827 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Alexander is a well-known programmer. Today he decided to finally go out and play football, but with the first hit he left a dent on the new Rolls-Royce of the wealthy businessman Big Vova. Vladimir has recently opened a store on the popular online marketplace "Zmey-Gorynych", and offers Alex a job: if he shows his programming skills by solving a task, he'll work as a cybersecurity specialist. Otherwise, he'll be delivering some doubtful products for the next two years.
You're given n positive integers a_1, a_2, ..., a_n. Using each of them exactly at once, you're to make such sequence b_1, b_2, ..., b_n that sequence c_1, c_2, ..., c_n is lexicographically maximal, where c_i=GCD(b_1,...,b_i) - the greatest common divisor of the first i elements of b.
Alexander is really afraid of the conditions of this simple task, so he asks you to solve it.
A sequence a is lexicographically smaller than a sequence b if and only if one of the following holds:
* a is a prefix of b, but a ≠ b;
* in the first position where a and b differ, the sequence a has a smaller element than the corresponding element in b.
Input
Each test contains multiple test cases. The first line contains the number of test cases t (1 ≤ t ≤ 10^3). Description of the test cases follows.
The first line of each test case contains a single integer n (1 ≤ n ≤ 10^3) — the length of the sequence a.
The second line of each test case contains n integers a_1,...,a_n (1 ≤ a_i ≤ 10^3) — the sequence a.
It is guaranteed that the sum of n over all test cases does not exceed 10^3.
Output
For each test case output the answer in a single line — the desired sequence b. If there are multiple answers, print any.
Example
Input
7
2
2 5
4
1 8 2 3
3
3 8 9
5
64 25 75 100 50
1
42
6
96 128 88 80 52 7
5
2 4 8 16 17
Output
5 2
8 2 1 3
9 3 8
100 50 25 75 64
42
128 96 80 88 52 7
17 2 4 8 16
Note
In the first test case of the example, there are only two possible permutations b — [2, 5] and [5, 2]: for the first one c=[2, 1], for the second one c=[5, 1].
In the third test case of the example, number 9 should be the first in b, and GCD(9, 3)=3, GCD(9, 8)=1, so the second number of b should be 3.
In the seventh test case of the example, first four numbers pairwise have a common divisor (a power of two), but none of them can be the first in the optimal permutation b.
Tags: brute force, greedy, math, number theory
Correct Solution:
```
from math import gcd
t = int(input())
for _ in range(t):
n = int(input())
arr = list(map(int,input().strip().split()))[:n]
#arr = sorted(arr,reverse=True)
maxi_index = None
maxi = -float('inf')
for i in range(n):
if arr[i] > maxi:
maxi = arr[i]
maxi_index = i
(arr[0],arr[maxi_index]) = (arr[maxi_index],arr[0])
curr_gcd = arr[0]
for i in range(n-1):
max_gcd = 0
max_index = None
for j in range(i+1,n):
index_gcd = gcd(curr_gcd,arr[j])
#print(index_gcd,'index gcd deb')
if index_gcd > max_gcd:
max_index = j
max_gcd = index_gcd
#print(max_gcd,'max gcd deb')
curr_gcd = max_gcd
(arr[i+1],arr[max_index]) = (arr[max_index],arr[i+1])
for i in arr:
print(i,end=' ')
print()
```
| 98,828 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Alexander is a well-known programmer. Today he decided to finally go out and play football, but with the first hit he left a dent on the new Rolls-Royce of the wealthy businessman Big Vova. Vladimir has recently opened a store on the popular online marketplace "Zmey-Gorynych", and offers Alex a job: if he shows his programming skills by solving a task, he'll work as a cybersecurity specialist. Otherwise, he'll be delivering some doubtful products for the next two years.
You're given n positive integers a_1, a_2, ..., a_n. Using each of them exactly at once, you're to make such sequence b_1, b_2, ..., b_n that sequence c_1, c_2, ..., c_n is lexicographically maximal, where c_i=GCD(b_1,...,b_i) - the greatest common divisor of the first i elements of b.
Alexander is really afraid of the conditions of this simple task, so he asks you to solve it.
A sequence a is lexicographically smaller than a sequence b if and only if one of the following holds:
* a is a prefix of b, but a ≠ b;
* in the first position where a and b differ, the sequence a has a smaller element than the corresponding element in b.
Input
Each test contains multiple test cases. The first line contains the number of test cases t (1 ≤ t ≤ 10^3). Description of the test cases follows.
The first line of each test case contains a single integer n (1 ≤ n ≤ 10^3) — the length of the sequence a.
The second line of each test case contains n integers a_1,...,a_n (1 ≤ a_i ≤ 10^3) — the sequence a.
It is guaranteed that the sum of n over all test cases does not exceed 10^3.
Output
For each test case output the answer in a single line — the desired sequence b. If there are multiple answers, print any.
Example
Input
7
2
2 5
4
1 8 2 3
3
3 8 9
5
64 25 75 100 50
1
42
6
96 128 88 80 52 7
5
2 4 8 16 17
Output
5 2
8 2 1 3
9 3 8
100 50 25 75 64
42
128 96 80 88 52 7
17 2 4 8 16
Note
In the first test case of the example, there are only two possible permutations b — [2, 5] and [5, 2]: for the first one c=[2, 1], for the second one c=[5, 1].
In the third test case of the example, number 9 should be the first in b, and GCD(9, 3)=3, GCD(9, 8)=1, so the second number of b should be 3.
In the seventh test case of the example, first four numbers pairwise have a common divisor (a power of two), but none of them can be the first in the optimal permutation b.
Tags: brute force, greedy, math, number theory
Correct Solution:
```
def gcd(a,b):
# Everything divides 0
if (b == 0):
return a
return gcd(b, a%b)
for _ in range(int(input())):
n=int(input())
arr=list(map(int,input().split()))
x=max(arr)
b=[x]
arr.remove(x)
for i in range(n-1):
temp=-1
k=0
m=0
for j in range(n-1):
if arr[j]!=-1:
if gcd(x,arr[j])>temp:
k=j
m=arr[j]
temp=gcd(x,arr[j])
b.append(m)
arr[k]=-1
x=temp
b=list(map(str,b))
s=" ".join(b)
print(s)
```
| 98,829 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Alexander is a well-known programmer. Today he decided to finally go out and play football, but with the first hit he left a dent on the new Rolls-Royce of the wealthy businessman Big Vova. Vladimir has recently opened a store on the popular online marketplace "Zmey-Gorynych", and offers Alex a job: if he shows his programming skills by solving a task, he'll work as a cybersecurity specialist. Otherwise, he'll be delivering some doubtful products for the next two years.
You're given n positive integers a_1, a_2, ..., a_n. Using each of them exactly at once, you're to make such sequence b_1, b_2, ..., b_n that sequence c_1, c_2, ..., c_n is lexicographically maximal, where c_i=GCD(b_1,...,b_i) - the greatest common divisor of the first i elements of b.
Alexander is really afraid of the conditions of this simple task, so he asks you to solve it.
A sequence a is lexicographically smaller than a sequence b if and only if one of the following holds:
* a is a prefix of b, but a ≠ b;
* in the first position where a and b differ, the sequence a has a smaller element than the corresponding element in b.
Input
Each test contains multiple test cases. The first line contains the number of test cases t (1 ≤ t ≤ 10^3). Description of the test cases follows.
The first line of each test case contains a single integer n (1 ≤ n ≤ 10^3) — the length of the sequence a.
The second line of each test case contains n integers a_1,...,a_n (1 ≤ a_i ≤ 10^3) — the sequence a.
It is guaranteed that the sum of n over all test cases does not exceed 10^3.
Output
For each test case output the answer in a single line — the desired sequence b. If there are multiple answers, print any.
Example
Input
7
2
2 5
4
1 8 2 3
3
3 8 9
5
64 25 75 100 50
1
42
6
96 128 88 80 52 7
5
2 4 8 16 17
Output
5 2
8 2 1 3
9 3 8
100 50 25 75 64
42
128 96 80 88 52 7
17 2 4 8 16
Note
In the first test case of the example, there are only two possible permutations b — [2, 5] and [5, 2]: for the first one c=[2, 1], for the second one c=[5, 1].
In the third test case of the example, number 9 should be the first in b, and GCD(9, 3)=3, GCD(9, 8)=1, so the second number of b should be 3.
In the seventh test case of the example, first four numbers pairwise have a common divisor (a power of two), but none of them can be the first in the optimal permutation b.
Tags: brute force, greedy, math, number theory
Correct Solution:
```
import math
t = int(input())
for _ in range(t):
n = int(input())
a = list(map(int,input().split(" ")))
s = dict()
for i in a:
s[i] = 0
for i in a:
s[i] += 1
a = list(set(a))
max_a = max(a)
ans = [max_a]*s[max_a]
visited = [False]*len(a)
visited[a.index(max_a)] = True
temp = max_a
for i in range(len(a)):
max_ = 1
x = ans[-1]
for j in range(len(a)):
if not visited[j] and max_ < math.gcd(temp,a[j]):
max_ = math.gcd(temp,a[j])
y = a[j]
idx = j
if max_ != 1:
ans += [y]*s[y]
visited[idx] = True
temp = max_
else:
break
for i in range(len(a)):
if not visited[i]:
ans += [a[i]]*s[a[i]]
for i in ans:
print(i,end=" ")
print()
```
| 98,830 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Alexander is a well-known programmer. Today he decided to finally go out and play football, but with the first hit he left a dent on the new Rolls-Royce of the wealthy businessman Big Vova. Vladimir has recently opened a store on the popular online marketplace "Zmey-Gorynych", and offers Alex a job: if he shows his programming skills by solving a task, he'll work as a cybersecurity specialist. Otherwise, he'll be delivering some doubtful products for the next two years.
You're given n positive integers a_1, a_2, ..., a_n. Using each of them exactly at once, you're to make such sequence b_1, b_2, ..., b_n that sequence c_1, c_2, ..., c_n is lexicographically maximal, where c_i=GCD(b_1,...,b_i) - the greatest common divisor of the first i elements of b.
Alexander is really afraid of the conditions of this simple task, so he asks you to solve it.
A sequence a is lexicographically smaller than a sequence b if and only if one of the following holds:
* a is a prefix of b, but a ≠ b;
* in the first position where a and b differ, the sequence a has a smaller element than the corresponding element in b.
Input
Each test contains multiple test cases. The first line contains the number of test cases t (1 ≤ t ≤ 10^3). Description of the test cases follows.
The first line of each test case contains a single integer n (1 ≤ n ≤ 10^3) — the length of the sequence a.
The second line of each test case contains n integers a_1,...,a_n (1 ≤ a_i ≤ 10^3) — the sequence a.
It is guaranteed that the sum of n over all test cases does not exceed 10^3.
Output
For each test case output the answer in a single line — the desired sequence b. If there are multiple answers, print any.
Example
Input
7
2
2 5
4
1 8 2 3
3
3 8 9
5
64 25 75 100 50
1
42
6
96 128 88 80 52 7
5
2 4 8 16 17
Output
5 2
8 2 1 3
9 3 8
100 50 25 75 64
42
128 96 80 88 52 7
17 2 4 8 16
Note
In the first test case of the example, there are only two possible permutations b — [2, 5] and [5, 2]: for the first one c=[2, 1], for the second one c=[5, 1].
In the third test case of the example, number 9 should be the first in b, and GCD(9, 3)=3, GCD(9, 8)=1, so the second number of b should be 3.
In the seventh test case of the example, first four numbers pairwise have a common divisor (a power of two), but none of them can be the first in the optimal permutation b.
Tags: brute force, greedy, math, number theory
Correct Solution:
```
from collections import defaultdict
import math
t = int(input())
while t:
t -= 1
n = int(input())
a = list(map(int, input().split()))
freq = defaultdict(int)
maxi = float("-inf")
for x in a:
freq[x] += 1
maxi = max(maxi, x)
ans = [maxi]
i = 1
freq[maxi] -= 1
while i < n:
curr_max = 0
next = None
for num in a:
if freq[num]:
if math.gcd(maxi, num) > curr_max:
curr_max = math.gcd(maxi, num)
next = num
freq[next]-=1
ans.append(next)
maxi = curr_max
i+=1
print(*ans)
```
| 98,831 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Alexander is a well-known programmer. Today he decided to finally go out and play football, but with the first hit he left a dent on the new Rolls-Royce of the wealthy businessman Big Vova. Vladimir has recently opened a store on the popular online marketplace "Zmey-Gorynych", and offers Alex a job: if he shows his programming skills by solving a task, he'll work as a cybersecurity specialist. Otherwise, he'll be delivering some doubtful products for the next two years.
You're given n positive integers a_1, a_2, ..., a_n. Using each of them exactly at once, you're to make such sequence b_1, b_2, ..., b_n that sequence c_1, c_2, ..., c_n is lexicographically maximal, where c_i=GCD(b_1,...,b_i) - the greatest common divisor of the first i elements of b.
Alexander is really afraid of the conditions of this simple task, so he asks you to solve it.
A sequence a is lexicographically smaller than a sequence b if and only if one of the following holds:
* a is a prefix of b, but a ≠ b;
* in the first position where a and b differ, the sequence a has a smaller element than the corresponding element in b.
Input
Each test contains multiple test cases. The first line contains the number of test cases t (1 ≤ t ≤ 10^3). Description of the test cases follows.
The first line of each test case contains a single integer n (1 ≤ n ≤ 10^3) — the length of the sequence a.
The second line of each test case contains n integers a_1,...,a_n (1 ≤ a_i ≤ 10^3) — the sequence a.
It is guaranteed that the sum of n over all test cases does not exceed 10^3.
Output
For each test case output the answer in a single line — the desired sequence b. If there are multiple answers, print any.
Example
Input
7
2
2 5
4
1 8 2 3
3
3 8 9
5
64 25 75 100 50
1
42
6
96 128 88 80 52 7
5
2 4 8 16 17
Output
5 2
8 2 1 3
9 3 8
100 50 25 75 64
42
128 96 80 88 52 7
17 2 4 8 16
Note
In the first test case of the example, there are only two possible permutations b — [2, 5] and [5, 2]: for the first one c=[2, 1], for the second one c=[5, 1].
In the third test case of the example, number 9 should be the first in b, and GCD(9, 3)=3, GCD(9, 8)=1, so the second number of b should be 3.
In the seventh test case of the example, first four numbers pairwise have a common divisor (a power of two), but none of them can be the first in the optimal permutation b.
Submitted Solution:
```
def gcd(a,b):
while a % b != 0:
a,b = b, a % b
return b
def bigGCD(a):
maxGcd = max(a)
b=[]
while maxGcd > 1:
if maxGcd in a:
b.append(maxGcd)
a.remove(maxGcd)
maxNow = 1
best = 0
for i in a:
if gcd(maxGcd,i) > maxNow:
maxNow = gcd(maxGcd,i)
best = i
if best != 0:
b.append(best)
a.remove(best)
maxGcd = maxNow
while a:
b.append(a.pop(0))
return b
if __name__ == '__main__':
cases = int(input())
for i in range(cases):
n = int(input())
a = list(map(int,input().strip().split()))
ans = bigGCD(a)
print(f"{' '.join(map(str,ans))}")
```
Yes
| 98,832 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Alexander is a well-known programmer. Today he decided to finally go out and play football, but with the first hit he left a dent on the new Rolls-Royce of the wealthy businessman Big Vova. Vladimir has recently opened a store on the popular online marketplace "Zmey-Gorynych", and offers Alex a job: if he shows his programming skills by solving a task, he'll work as a cybersecurity specialist. Otherwise, he'll be delivering some doubtful products for the next two years.
You're given n positive integers a_1, a_2, ..., a_n. Using each of them exactly at once, you're to make such sequence b_1, b_2, ..., b_n that sequence c_1, c_2, ..., c_n is lexicographically maximal, where c_i=GCD(b_1,...,b_i) - the greatest common divisor of the first i elements of b.
Alexander is really afraid of the conditions of this simple task, so he asks you to solve it.
A sequence a is lexicographically smaller than a sequence b if and only if one of the following holds:
* a is a prefix of b, but a ≠ b;
* in the first position where a and b differ, the sequence a has a smaller element than the corresponding element in b.
Input
Each test contains multiple test cases. The first line contains the number of test cases t (1 ≤ t ≤ 10^3). Description of the test cases follows.
The first line of each test case contains a single integer n (1 ≤ n ≤ 10^3) — the length of the sequence a.
The second line of each test case contains n integers a_1,...,a_n (1 ≤ a_i ≤ 10^3) — the sequence a.
It is guaranteed that the sum of n over all test cases does not exceed 10^3.
Output
For each test case output the answer in a single line — the desired sequence b. If there are multiple answers, print any.
Example
Input
7
2
2 5
4
1 8 2 3
3
3 8 9
5
64 25 75 100 50
1
42
6
96 128 88 80 52 7
5
2 4 8 16 17
Output
5 2
8 2 1 3
9 3 8
100 50 25 75 64
42
128 96 80 88 52 7
17 2 4 8 16
Note
In the first test case of the example, there are only two possible permutations b — [2, 5] and [5, 2]: for the first one c=[2, 1], for the second one c=[5, 1].
In the third test case of the example, number 9 should be the first in b, and GCD(9, 3)=3, GCD(9, 8)=1, so the second number of b should be 3.
In the seventh test case of the example, first four numbers pairwise have a common divisor (a power of two), but none of them can be the first in the optimal permutation b.
Submitted Solution:
```
import math
def main():
n = int(input())
lst = list(map(int, input().split()))
cur = max(lst)
res = [lst.index(cur)]
st = set()
st.add(lst.index(cur))
for i in range(n - 1):
el = -1
mx = 0
for k in range(n):
if k in st:
pass
elif math.gcd(cur, lst[k]) > mx:
mx= math.gcd(cur, lst[k])
el = k
st.add(el)
res.append(el)
cur = mx
for i in res:
print(lst[i], end=" ")
print()
if __name__ == '__main__':
t = int(input())
for i in range(t):
main()
"""
60, 61
"""
"""
"""
```
Yes
| 98,833 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Alexander is a well-known programmer. Today he decided to finally go out and play football, but with the first hit he left a dent on the new Rolls-Royce of the wealthy businessman Big Vova. Vladimir has recently opened a store on the popular online marketplace "Zmey-Gorynych", and offers Alex a job: if he shows his programming skills by solving a task, he'll work as a cybersecurity specialist. Otherwise, he'll be delivering some doubtful products for the next two years.
You're given n positive integers a_1, a_2, ..., a_n. Using each of them exactly at once, you're to make such sequence b_1, b_2, ..., b_n that sequence c_1, c_2, ..., c_n is lexicographically maximal, where c_i=GCD(b_1,...,b_i) - the greatest common divisor of the first i elements of b.
Alexander is really afraid of the conditions of this simple task, so he asks you to solve it.
A sequence a is lexicographically smaller than a sequence b if and only if one of the following holds:
* a is a prefix of b, but a ≠ b;
* in the first position where a and b differ, the sequence a has a smaller element than the corresponding element in b.
Input
Each test contains multiple test cases. The first line contains the number of test cases t (1 ≤ t ≤ 10^3). Description of the test cases follows.
The first line of each test case contains a single integer n (1 ≤ n ≤ 10^3) — the length of the sequence a.
The second line of each test case contains n integers a_1,...,a_n (1 ≤ a_i ≤ 10^3) — the sequence a.
It is guaranteed that the sum of n over all test cases does not exceed 10^3.
Output
For each test case output the answer in a single line — the desired sequence b. If there are multiple answers, print any.
Example
Input
7
2
2 5
4
1 8 2 3
3
3 8 9
5
64 25 75 100 50
1
42
6
96 128 88 80 52 7
5
2 4 8 16 17
Output
5 2
8 2 1 3
9 3 8
100 50 25 75 64
42
128 96 80 88 52 7
17 2 4 8 16
Note
In the first test case of the example, there are only two possible permutations b — [2, 5] and [5, 2]: for the first one c=[2, 1], for the second one c=[5, 1].
In the third test case of the example, number 9 should be the first in b, and GCD(9, 3)=3, GCD(9, 8)=1, so the second number of b should be 3.
In the seventh test case of the example, first four numbers pairwise have a common divisor (a power of two), but none of them can be the first in the optimal permutation b.
Submitted Solution:
```
#!/usr/bin/env python3
# encoding: utf-8
#----------
# Constants
#----------
#----------
# Functions
#----------
def gcd(a, b):
a, b = max(a, b), min(a, b)
while b > 0:
a, b = b, a % b
return a
# The function that solves the task
def calc(a):
a.sort()
max_val = a[-1]
del a[-1]
res = [ max_val ]
res_gcd = max_val
while a:
val, pos = 0, 0
for i, item in enumerate(a):
v = gcd(res_gcd, item)
if v > val:
val = v
pos = i
res.append(a[pos])
del a[pos]
res_gcd = val
return res
# Reads a string from stdin, splits it by space chars, converts each
# substring to int, adds it to a list and returns the list as a result.
def get_ints():
return [ int(n) for n in input().split() ]
# Reads a string from stdin, splits it by space chars, converts each substring
# to floating point number, adds it to a list and returns the list as a result.
def get_floats():
return [ float(n) for n in input().split() ]
# Converts a sequence to the space separated string
def seq2str(seq):
return ' '.join(str(item) for item in seq)
#----------
# Execution start point
#----------
if __name__ == "__main__":
zzz = get_ints()
assert len(zzz) == 1
t = zzz[0]
for i in range(t):
zzz = get_ints()
assert len(zzz) == 1
n = zzz[0]
zzz = get_ints()
assert len(zzz) == n
a = zzz
res = calc(a)
print(seq2str(res))
```
Yes
| 98,834 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Alexander is a well-known programmer. Today he decided to finally go out and play football, but with the first hit he left a dent on the new Rolls-Royce of the wealthy businessman Big Vova. Vladimir has recently opened a store on the popular online marketplace "Zmey-Gorynych", and offers Alex a job: if he shows his programming skills by solving a task, he'll work as a cybersecurity specialist. Otherwise, he'll be delivering some doubtful products for the next two years.
You're given n positive integers a_1, a_2, ..., a_n. Using each of them exactly at once, you're to make such sequence b_1, b_2, ..., b_n that sequence c_1, c_2, ..., c_n is lexicographically maximal, where c_i=GCD(b_1,...,b_i) - the greatest common divisor of the first i elements of b.
Alexander is really afraid of the conditions of this simple task, so he asks you to solve it.
A sequence a is lexicographically smaller than a sequence b if and only if one of the following holds:
* a is a prefix of b, but a ≠ b;
* in the first position where a and b differ, the sequence a has a smaller element than the corresponding element in b.
Input
Each test contains multiple test cases. The first line contains the number of test cases t (1 ≤ t ≤ 10^3). Description of the test cases follows.
The first line of each test case contains a single integer n (1 ≤ n ≤ 10^3) — the length of the sequence a.
The second line of each test case contains n integers a_1,...,a_n (1 ≤ a_i ≤ 10^3) — the sequence a.
It is guaranteed that the sum of n over all test cases does not exceed 10^3.
Output
For each test case output the answer in a single line — the desired sequence b. If there are multiple answers, print any.
Example
Input
7
2
2 5
4
1 8 2 3
3
3 8 9
5
64 25 75 100 50
1
42
6
96 128 88 80 52 7
5
2 4 8 16 17
Output
5 2
8 2 1 3
9 3 8
100 50 25 75 64
42
128 96 80 88 52 7
17 2 4 8 16
Note
In the first test case of the example, there are only two possible permutations b — [2, 5] and [5, 2]: for the first one c=[2, 1], for the second one c=[5, 1].
In the third test case of the example, number 9 should be the first in b, and GCD(9, 3)=3, GCD(9, 8)=1, so the second number of b should be 3.
In the seventh test case of the example, first four numbers pairwise have a common divisor (a power of two), but none of them can be the first in the optimal permutation b.
Submitted Solution:
```
from sys import stdin
from math import gcd
input = stdin.readline
t = int(input())
for _ in range(t):
n = int(input())
a = [int(x) for x in input().split()]
used = [False]*n
cur = 0
b = []
for j in range(n):
best = 0
for i in range(n):
if used[i]:
continue
g = gcd(cur, a[i])
if g > best:
best = g
best_idx = i
used[best_idx] = True
cur = best
b.append(a[best_idx])
print(*b)
```
Yes
| 98,835 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Alexander is a well-known programmer. Today he decided to finally go out and play football, but with the first hit he left a dent on the new Rolls-Royce of the wealthy businessman Big Vova. Vladimir has recently opened a store on the popular online marketplace "Zmey-Gorynych", and offers Alex a job: if he shows his programming skills by solving a task, he'll work as a cybersecurity specialist. Otherwise, he'll be delivering some doubtful products for the next two years.
You're given n positive integers a_1, a_2, ..., a_n. Using each of them exactly at once, you're to make such sequence b_1, b_2, ..., b_n that sequence c_1, c_2, ..., c_n is lexicographically maximal, where c_i=GCD(b_1,...,b_i) - the greatest common divisor of the first i elements of b.
Alexander is really afraid of the conditions of this simple task, so he asks you to solve it.
A sequence a is lexicographically smaller than a sequence b if and only if one of the following holds:
* a is a prefix of b, but a ≠ b;
* in the first position where a and b differ, the sequence a has a smaller element than the corresponding element in b.
Input
Each test contains multiple test cases. The first line contains the number of test cases t (1 ≤ t ≤ 10^3). Description of the test cases follows.
The first line of each test case contains a single integer n (1 ≤ n ≤ 10^3) — the length of the sequence a.
The second line of each test case contains n integers a_1,...,a_n (1 ≤ a_i ≤ 10^3) — the sequence a.
It is guaranteed that the sum of n over all test cases does not exceed 10^3.
Output
For each test case output the answer in a single line — the desired sequence b. If there are multiple answers, print any.
Example
Input
7
2
2 5
4
1 8 2 3
3
3 8 9
5
64 25 75 100 50
1
42
6
96 128 88 80 52 7
5
2 4 8 16 17
Output
5 2
8 2 1 3
9 3 8
100 50 25 75 64
42
128 96 80 88 52 7
17 2 4 8 16
Note
In the first test case of the example, there are only two possible permutations b — [2, 5] and [5, 2]: for the first one c=[2, 1], for the second one c=[5, 1].
In the third test case of the example, number 9 should be the first in b, and GCD(9, 3)=3, GCD(9, 8)=1, so the second number of b should be 3.
In the seventh test case of the example, first four numbers pairwise have a common divisor (a power of two), but none of them can be the first in the optimal permutation b.
Submitted Solution:
```
t=int(input())
import math
while t:
t-=1
n=int(input())
a=[int(i) for i in input().split()]
a.sort()
ans=[]
ans.append(a[n-1])
del a[n-1]
curg=ans[0]
while len(a):
ma=0
inl=0
flag=0
for i in range(len(a)):
if curg%a[i]==0:
ma=a[i]
inl=i
flag+=1
if flag==0:
break
else:
ans.append(ma)
del a[inl]
curg=math.gcd(curg,ans[-1])
ans+=a
print(*ans,sep=" ")
```
No
| 98,836 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Alexander is a well-known programmer. Today he decided to finally go out and play football, but with the first hit he left a dent on the new Rolls-Royce of the wealthy businessman Big Vova. Vladimir has recently opened a store on the popular online marketplace "Zmey-Gorynych", and offers Alex a job: if he shows his programming skills by solving a task, he'll work as a cybersecurity specialist. Otherwise, he'll be delivering some doubtful products for the next two years.
You're given n positive integers a_1, a_2, ..., a_n. Using each of them exactly at once, you're to make such sequence b_1, b_2, ..., b_n that sequence c_1, c_2, ..., c_n is lexicographically maximal, where c_i=GCD(b_1,...,b_i) - the greatest common divisor of the first i elements of b.
Alexander is really afraid of the conditions of this simple task, so he asks you to solve it.
A sequence a is lexicographically smaller than a sequence b if and only if one of the following holds:
* a is a prefix of b, but a ≠ b;
* in the first position where a and b differ, the sequence a has a smaller element than the corresponding element in b.
Input
Each test contains multiple test cases. The first line contains the number of test cases t (1 ≤ t ≤ 10^3). Description of the test cases follows.
The first line of each test case contains a single integer n (1 ≤ n ≤ 10^3) — the length of the sequence a.
The second line of each test case contains n integers a_1,...,a_n (1 ≤ a_i ≤ 10^3) — the sequence a.
It is guaranteed that the sum of n over all test cases does not exceed 10^3.
Output
For each test case output the answer in a single line — the desired sequence b. If there are multiple answers, print any.
Example
Input
7
2
2 5
4
1 8 2 3
3
3 8 9
5
64 25 75 100 50
1
42
6
96 128 88 80 52 7
5
2 4 8 16 17
Output
5 2
8 2 1 3
9 3 8
100 50 25 75 64
42
128 96 80 88 52 7
17 2 4 8 16
Note
In the first test case of the example, there are only two possible permutations b — [2, 5] and [5, 2]: for the first one c=[2, 1], for the second one c=[5, 1].
In the third test case of the example, number 9 should be the first in b, and GCD(9, 3)=3, GCD(9, 8)=1, so the second number of b should be 3.
In the seventh test case of the example, first four numbers pairwise have a common divisor (a power of two), but none of them can be the first in the optimal permutation b.
Submitted Solution:
```
t=int(input())
for _ in range(t):
n=int(input())
l1=list(map(int,input().split()))
l1.sort(reverse=True)
l2,l3=[],[]
l2.append(l1[0])
m=l1[0]
for i in range(1,n):
if m%l1[i]==0:
l2.append(l1[i])
else:
l3.append(l1[i])
l3.sort()
l2+=l3
for i in range(n-1):
print(l2[i],end=" ")
print(l2[n-1])
```
No
| 98,837 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Alexander is a well-known programmer. Today he decided to finally go out and play football, but with the first hit he left a dent on the new Rolls-Royce of the wealthy businessman Big Vova. Vladimir has recently opened a store on the popular online marketplace "Zmey-Gorynych", and offers Alex a job: if he shows his programming skills by solving a task, he'll work as a cybersecurity specialist. Otherwise, he'll be delivering some doubtful products for the next two years.
You're given n positive integers a_1, a_2, ..., a_n. Using each of them exactly at once, you're to make such sequence b_1, b_2, ..., b_n that sequence c_1, c_2, ..., c_n is lexicographically maximal, where c_i=GCD(b_1,...,b_i) - the greatest common divisor of the first i elements of b.
Alexander is really afraid of the conditions of this simple task, so he asks you to solve it.
A sequence a is lexicographically smaller than a sequence b if and only if one of the following holds:
* a is a prefix of b, but a ≠ b;
* in the first position where a and b differ, the sequence a has a smaller element than the corresponding element in b.
Input
Each test contains multiple test cases. The first line contains the number of test cases t (1 ≤ t ≤ 10^3). Description of the test cases follows.
The first line of each test case contains a single integer n (1 ≤ n ≤ 10^3) — the length of the sequence a.
The second line of each test case contains n integers a_1,...,a_n (1 ≤ a_i ≤ 10^3) — the sequence a.
It is guaranteed that the sum of n over all test cases does not exceed 10^3.
Output
For each test case output the answer in a single line — the desired sequence b. If there are multiple answers, print any.
Example
Input
7
2
2 5
4
1 8 2 3
3
3 8 9
5
64 25 75 100 50
1
42
6
96 128 88 80 52 7
5
2 4 8 16 17
Output
5 2
8 2 1 3
9 3 8
100 50 25 75 64
42
128 96 80 88 52 7
17 2 4 8 16
Note
In the first test case of the example, there are only two possible permutations b — [2, 5] and [5, 2]: for the first one c=[2, 1], for the second one c=[5, 1].
In the third test case of the example, number 9 should be the first in b, and GCD(9, 3)=3, GCD(9, 8)=1, so the second number of b should be 3.
In the seventh test case of the example, first four numbers pairwise have a common divisor (a power of two), but none of them can be the first in the optimal permutation b.
Submitted Solution:
```
import math
import collections
for t in range(int(input())):
n=int(input())
num=list(map(int,input().split()))
val=max(num)
k=collections.defaultdict(list)
for i in num:
v=math.gcd(val,i)
k[v].append(i)
out=[]
for i in sorted(k.keys(),reverse=True):
out+=k[i]
print(*out)
```
No
| 98,838 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Alexander is a well-known programmer. Today he decided to finally go out and play football, but with the first hit he left a dent on the new Rolls-Royce of the wealthy businessman Big Vova. Vladimir has recently opened a store on the popular online marketplace "Zmey-Gorynych", and offers Alex a job: if he shows his programming skills by solving a task, he'll work as a cybersecurity specialist. Otherwise, he'll be delivering some doubtful products for the next two years.
You're given n positive integers a_1, a_2, ..., a_n. Using each of them exactly at once, you're to make such sequence b_1, b_2, ..., b_n that sequence c_1, c_2, ..., c_n is lexicographically maximal, where c_i=GCD(b_1,...,b_i) - the greatest common divisor of the first i elements of b.
Alexander is really afraid of the conditions of this simple task, so he asks you to solve it.
A sequence a is lexicographically smaller than a sequence b if and only if one of the following holds:
* a is a prefix of b, but a ≠ b;
* in the first position where a and b differ, the sequence a has a smaller element than the corresponding element in b.
Input
Each test contains multiple test cases. The first line contains the number of test cases t (1 ≤ t ≤ 10^3). Description of the test cases follows.
The first line of each test case contains a single integer n (1 ≤ n ≤ 10^3) — the length of the sequence a.
The second line of each test case contains n integers a_1,...,a_n (1 ≤ a_i ≤ 10^3) — the sequence a.
It is guaranteed that the sum of n over all test cases does not exceed 10^3.
Output
For each test case output the answer in a single line — the desired sequence b. If there are multiple answers, print any.
Example
Input
7
2
2 5
4
1 8 2 3
3
3 8 9
5
64 25 75 100 50
1
42
6
96 128 88 80 52 7
5
2 4 8 16 17
Output
5 2
8 2 1 3
9 3 8
100 50 25 75 64
42
128 96 80 88 52 7
17 2 4 8 16
Note
In the first test case of the example, there are only two possible permutations b — [2, 5] and [5, 2]: for the first one c=[2, 1], for the second one c=[5, 1].
In the third test case of the example, number 9 should be the first in b, and GCD(9, 3)=3, GCD(9, 8)=1, so the second number of b should be 3.
In the seventh test case of the example, first four numbers pairwise have a common divisor (a power of two), but none of them can be the first in the optimal permutation b.
Submitted Solution:
```
import math
import collections
def solution(n, a):
max_ = max(a)
max_i = a.index(max_)
x = [max_]
del a[max_i]
while a != []:
out = collections.defaultdict(list)
for e in a:
out[math.gcd(x[-1], e)].append(e)
temp = out.pop(max(list(out.keys())))
x.extend(temp)
for e in temp:
a.remove(e)
return x
if __name__ == "__main__":
t = int(input())
for _ in range(t):
n = int(input())
a = list(map(int, input().split(' ')))
print(*solution(n, a))
```
No
| 98,839 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A prisoner wants to escape from a prison. The prison is represented by the interior of the convex polygon with vertices P_1, P_2, P_3, …, P_{n+1}, P_{n+2}, P_{n+3}. It holds P_1=(0,0), P_{n+1}=(0, h), P_{n+2}=(-10^{18}, h) and P_{n+3}=(-10^{18}, 0).
<image>
The prison walls P_{n+1}P_{n+2}, P_{n+2}P_{n+3} and P_{n+3}P_1 are very high and the prisoner is not able to climb them. Hence his only chance is to reach a point on one of the walls P_1P_2, P_2P_3,..., P_{n}P_{n+1} and escape from there. On the perimeter of the prison, there are two guards. The prisoner moves at speed 1 while the guards move, remaining always on the perimeter of the prison, with speed v.
If the prisoner reaches a point of the perimeter where there is a guard, the guard kills the prisoner. If the prisoner reaches a point of the part of the perimeter he is able to climb and there is no guard there, he escapes immediately. Initially the prisoner is at the point (-10^{17}, h/2) and the guards are at P_1.
Find the minimum speed v such that the guards can guarantee that the prisoner will not escape (assuming that both the prisoner and the guards move optimally).
Notes:
* At any moment, the guards and the prisoner can see each other.
* The "climbing part" of the escape takes no time.
* You may assume that both the prisoner and the guards can change direction and velocity instantly and that they both have perfect reflexes (so they can react instantly to whatever the other one is doing).
* The two guards can plan ahead how to react to the prisoner movements.
Input
The first line of the input contains n (1 ≤ n ≤ 50).
The following n+1 lines describe P_1, P_2,..., P_{n+1}. The i-th of such lines contain two integers x_i, y_i (0≤ x_i, y_i≤ 1,000) — the coordinates of P_i=(x_i, y_i).
It is guaranteed that P_1=(0,0) and x_{n+1}=0. The polygon with vertices P_1,P_2,..., P_{n+1}, P_{n+2}, P_{n+3} (where P_{n+2}, P_{n+3} shall be constructed as described in the statement) is guaranteed to be convex and such that there is no line containing three of its vertices.
Output
Print a single real number, the minimum speed v that allows the guards to guarantee that the prisoner will not escape. Your answer will be considered correct if its relative or absolute error does not exceed 10^{-6}.
Examples
Input
2
0 0
223 464
0 749
Output
1
Input
3
0 0
2 2
2 4
0 6
Output
1.0823922
Input
4
0 0
7 3
7 4
5 7
0 8
Output
1.130309669
Input
5
0 0
562 248
460 610
281 702
206 723
0 746
Output
1.148649561
Input
7
0 0
412 36
745 180
747 184
746 268
611 359
213 441
0 450
Output
1.134745994
Submitted Solution:
```
#!/usr/local/bin/python3
```
No
| 98,840 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A prisoner wants to escape from a prison. The prison is represented by the interior of the convex polygon with vertices P_1, P_2, P_3, …, P_{n+1}, P_{n+2}, P_{n+3}. It holds P_1=(0,0), P_{n+1}=(0, h), P_{n+2}=(-10^{18}, h) and P_{n+3}=(-10^{18}, 0).
<image>
The prison walls P_{n+1}P_{n+2}, P_{n+2}P_{n+3} and P_{n+3}P_1 are very high and the prisoner is not able to climb them. Hence his only chance is to reach a point on one of the walls P_1P_2, P_2P_3,..., P_{n}P_{n+1} and escape from there. On the perimeter of the prison, there are two guards. The prisoner moves at speed 1 while the guards move, remaining always on the perimeter of the prison, with speed v.
If the prisoner reaches a point of the perimeter where there is a guard, the guard kills the prisoner. If the prisoner reaches a point of the part of the perimeter he is able to climb and there is no guard there, he escapes immediately. Initially the prisoner is at the point (-10^{17}, h/2) and the guards are at P_1.
Find the minimum speed v such that the guards can guarantee that the prisoner will not escape (assuming that both the prisoner and the guards move optimally).
Notes:
* At any moment, the guards and the prisoner can see each other.
* The "climbing part" of the escape takes no time.
* You may assume that both the prisoner and the guards can change direction and velocity instantly and that they both have perfect reflexes (so they can react instantly to whatever the other one is doing).
* The two guards can plan ahead how to react to the prisoner movements.
Input
The first line of the input contains n (1 ≤ n ≤ 50).
The following n+1 lines describe P_1, P_2,..., P_{n+1}. The i-th of such lines contain two integers x_i, y_i (0≤ x_i, y_i≤ 1,000) — the coordinates of P_i=(x_i, y_i).
It is guaranteed that P_1=(0,0) and x_{n+1}=0. The polygon with vertices P_1,P_2,..., P_{n+1}, P_{n+2}, P_{n+3} (where P_{n+2}, P_{n+3} shall be constructed as described in the statement) is guaranteed to be convex and such that there is no line containing three of its vertices.
Output
Print a single real number, the minimum speed v that allows the guards to guarantee that the prisoner will not escape. Your answer will be considered correct if its relative or absolute error does not exceed 10^{-6}.
Examples
Input
2
0 0
223 464
0 749
Output
1
Input
3
0 0
2 2
2 4
0 6
Output
1.0823922
Input
4
0 0
7 3
7 4
5 7
0 8
Output
1.130309669
Input
5
0 0
562 248
460 610
281 702
206 723
0 746
Output
1.148649561
Input
7
0 0
412 36
745 180
747 184
746 268
611 359
213 441
0 450
Output
1.134745994
Submitted Solution:
```
print(1)
```
No
| 98,841 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A prisoner wants to escape from a prison. The prison is represented by the interior of the convex polygon with vertices P_1, P_2, P_3, …, P_{n+1}, P_{n+2}, P_{n+3}. It holds P_1=(0,0), P_{n+1}=(0, h), P_{n+2}=(-10^{18}, h) and P_{n+3}=(-10^{18}, 0).
<image>
The prison walls P_{n+1}P_{n+2}, P_{n+2}P_{n+3} and P_{n+3}P_1 are very high and the prisoner is not able to climb them. Hence his only chance is to reach a point on one of the walls P_1P_2, P_2P_3,..., P_{n}P_{n+1} and escape from there. On the perimeter of the prison, there are two guards. The prisoner moves at speed 1 while the guards move, remaining always on the perimeter of the prison, with speed v.
If the prisoner reaches a point of the perimeter where there is a guard, the guard kills the prisoner. If the prisoner reaches a point of the part of the perimeter he is able to climb and there is no guard there, he escapes immediately. Initially the prisoner is at the point (-10^{17}, h/2) and the guards are at P_1.
Find the minimum speed v such that the guards can guarantee that the prisoner will not escape (assuming that both the prisoner and the guards move optimally).
Notes:
* At any moment, the guards and the prisoner can see each other.
* The "climbing part" of the escape takes no time.
* You may assume that both the prisoner and the guards can change direction and velocity instantly and that they both have perfect reflexes (so they can react instantly to whatever the other one is doing).
* The two guards can plan ahead how to react to the prisoner movements.
Input
The first line of the input contains n (1 ≤ n ≤ 50).
The following n+1 lines describe P_1, P_2,..., P_{n+1}. The i-th of such lines contain two integers x_i, y_i (0≤ x_i, y_i≤ 1,000) — the coordinates of P_i=(x_i, y_i).
It is guaranteed that P_1=(0,0) and x_{n+1}=0. The polygon with vertices P_1,P_2,..., P_{n+1}, P_{n+2}, P_{n+3} (where P_{n+2}, P_{n+3} shall be constructed as described in the statement) is guaranteed to be convex and such that there is no line containing three of its vertices.
Output
Print a single real number, the minimum speed v that allows the guards to guarantee that the prisoner will not escape. Your answer will be considered correct if its relative or absolute error does not exceed 10^{-6}.
Examples
Input
2
0 0
223 464
0 749
Output
1
Input
3
0 0
2 2
2 4
0 6
Output
1.0823922
Input
4
0 0
7 3
7 4
5 7
0 8
Output
1.130309669
Input
5
0 0
562 248
460 610
281 702
206 723
0 746
Output
1.148649561
Input
7
0 0
412 36
745 180
747 184
746 268
611 359
213 441
0 450
Output
1.134745994
Submitted Solution:
```
print("hello , worldgchgfhgfkhoifyufouyouyfolyfvfuvhfdgffgcgfgcgc ng nfn gv ")
```
No
| 98,842 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Igor had a sequence d_1, d_2, ..., d_n of integers. When Igor entered the classroom there was an integer x written on the blackboard.
Igor generated sequence p using the following algorithm:
1. initially, p = [x];
2. for each 1 ≤ i ≤ n he did the following operation |d_i| times:
* if d_i ≥ 0, then he looked at the last element of p (let it be y) and appended y + 1 to the end of p;
* if d_i < 0, then he looked at the last element of p (let it be y) and appended y - 1 to the end of p.
For example, if x = 3, and d = [1, -1, 2], p will be equal [3, 4, 3, 4, 5].
Igor decided to calculate the length of the longest increasing subsequence of p and the number of them.
A sequence a is a subsequence of a sequence b if a can be obtained from b by deletion of several (possibly, zero or all) elements.
A sequence a is an increasing sequence if each element of a (except the first one) is strictly greater than the previous element.
For p = [3, 4, 3, 4, 5], the length of longest increasing subsequence is 3 and there are 3 of them: [\underline{3}, \underline{4}, 3, 4, \underline{5}], [\underline{3}, 4, 3, \underline{4}, \underline{5}], [3, 4, \underline{3}, \underline{4}, \underline{5}].
Input
The first line contains a single integer n (1 ≤ n ≤ 50) — the length of the sequence d.
The second line contains a single integer x (-10^9 ≤ x ≤ 10^9) — the integer on the blackboard.
The third line contains n integers d_1, d_2, …, d_n (-10^9 ≤ d_i ≤ 10^9).
Output
Print two integers:
* the first integer should be equal to the length of the longest increasing subsequence of p;
* the second should be equal to the number of them modulo 998244353.
You should print only the second number modulo 998244353.
Examples
Input
3
3
1 -1 2
Output
3 3
Input
3
100
5 -3 6
Output
9 7
Input
3
1
999999999 0 1000000000
Output
2000000000 1
Input
5
34
1337 -146 42 -69 228
Output
1393 3876
Note
The first test case was explained in the statement.
In the second test case p = [100, 101, 102, 103, 104, 105, 104, 103, 102, 103, 104, 105, 106, 107, 108].
In the third test case p = [1, 2, …, 2000000000].
Tags: dp, math, matrices
Correct Solution:
```
import sys, io, os
if os.environ['USERNAME']=='kissz':
inp=open('in55.txt','r').readline
def debug(*args):
print(*args,file=sys.stderr)
else:
inp=io.BytesIO(os.read(0,os.fstat(0).st_size)).readline
def debug(*args):
pass
def mexp(size,power):
A=[]
for i in range(size):
A.append([int(j<=(i//2)*2) for j in range(size)])
powers={1: A}
p=1
while p*2<=power:
powers[p*2]=mmmult(powers[p],powers[p])
p*=2
A=powers[p]
power-=p
while power>0:
p=p//2
if p<=power:
A=mmmult(A,powers[p])
power-=p
return A
def mvmult(A,V):
res=[]
for i in range(len(A)):
res.append(sum(a*v for a,v in zip(A[i],V)) % 998244353 )
return res
def mmmult(A,B):
res=[]
for i in range(len(A)):
res.append([sum(a*B[j][k] for j,a in enumerate(A[i])) for k in range(len(B[0]))])
return res
def get_rep(corners):
corners[0]-=1
corners[-1]+=1
bps=sorted(list(set(corners)))
m=len(bps)
X=[1]
active=[1]
for i in range(1,m):
x,y=bps[i-1],bps[i]
d=y-x
A=mexp(len(active),d)
X=mvmult(A,X)
#debug(active,X)
#debug(A)
if i<m-1:
for j,c in enumerate(corners):
if c==y:
if j%2: # top: j and j+1 in active
idx=active.index(j)
X[idx+2]+=X[idx]
active.pop(idx)
active.pop(idx)
X.pop(idx)
X.pop(idx)
else: # bottom
active+=[j,j+1]
active.sort()
idx=active.index(j)
X=X[:idx]+[0,X[idx-1]]+X[idx:]
else:
return X[0]
n=int(inp())
inp()
d,*D=map(int,inp().split())
if d==0 and all(dd==0 for dd in D):
print(1,1)
else:
while d==0: d,*D=D
up=(d>=0)
corners=[0,d]
for d in D:
x=corners[-1]+d
if up==(d>=0):
corners[-1]=x
if up!=(d>=0):
up=(d>=0)
corners.append(x)
debug(corners)
cands=[(-1,0,0)]
low=(0,0)
maxdiff=(0,0,0)
for i,corner in enumerate(corners):
if corner<low[0]: low=(corner,i)
if corner-low[0]>=cands[0][0]:
if corner-low[0]==cands[0][0] and low[1]>cands[0][1]:
cands+=[(corner-low[0],low[1],i)]
else:
cands=[(corner-low[0],low[1],i)]
L=cands[0][0]+1
if L>1:
X=0
debug(cands)
for _, starti, endi in cands:
#debug(corners[starti:endi+1])
X+=get_rep(corners[starti:endi+1])
else:
X=1-corners[-1]
print(L,X % 998244353)
```
| 98,843 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Igor had a sequence d_1, d_2, ..., d_n of integers. When Igor entered the classroom there was an integer x written on the blackboard.
Igor generated sequence p using the following algorithm:
1. initially, p = [x];
2. for each 1 ≤ i ≤ n he did the following operation |d_i| times:
* if d_i ≥ 0, then he looked at the last element of p (let it be y) and appended y + 1 to the end of p;
* if d_i < 0, then he looked at the last element of p (let it be y) and appended y - 1 to the end of p.
For example, if x = 3, and d = [1, -1, 2], p will be equal [3, 4, 3, 4, 5].
Igor decided to calculate the length of the longest increasing subsequence of p and the number of them.
A sequence a is a subsequence of a sequence b if a can be obtained from b by deletion of several (possibly, zero or all) elements.
A sequence a is an increasing sequence if each element of a (except the first one) is strictly greater than the previous element.
For p = [3, 4, 3, 4, 5], the length of longest increasing subsequence is 3 and there are 3 of them: [\underline{3}, \underline{4}, 3, 4, \underline{5}], [\underline{3}, 4, 3, \underline{4}, \underline{5}], [3, 4, \underline{3}, \underline{4}, \underline{5}].
Input
The first line contains a single integer n (1 ≤ n ≤ 50) — the length of the sequence d.
The second line contains a single integer x (-10^9 ≤ x ≤ 10^9) — the integer on the blackboard.
The third line contains n integers d_1, d_2, …, d_n (-10^9 ≤ d_i ≤ 10^9).
Output
Print two integers:
* the first integer should be equal to the length of the longest increasing subsequence of p;
* the second should be equal to the number of them modulo 998244353.
You should print only the second number modulo 998244353.
Examples
Input
3
3
1 -1 2
Output
3 3
Input
3
100
5 -3 6
Output
9 7
Input
3
1
999999999 0 1000000000
Output
2000000000 1
Input
5
34
1337 -146 42 -69 228
Output
1393 3876
Note
The first test case was explained in the statement.
In the second test case p = [100, 101, 102, 103, 104, 105, 104, 103, 102, 103, 104, 105, 106, 107, 108].
In the third test case p = [1, 2, …, 2000000000].
Submitted Solution:
```
print("atulpandey")
```
No
| 98,844 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Igor had a sequence d_1, d_2, ..., d_n of integers. When Igor entered the classroom there was an integer x written on the blackboard.
Igor generated sequence p using the following algorithm:
1. initially, p = [x];
2. for each 1 ≤ i ≤ n he did the following operation |d_i| times:
* if d_i ≥ 0, then he looked at the last element of p (let it be y) and appended y + 1 to the end of p;
* if d_i < 0, then he looked at the last element of p (let it be y) and appended y - 1 to the end of p.
For example, if x = 3, and d = [1, -1, 2], p will be equal [3, 4, 3, 4, 5].
Igor decided to calculate the length of the longest increasing subsequence of p and the number of them.
A sequence a is a subsequence of a sequence b if a can be obtained from b by deletion of several (possibly, zero or all) elements.
A sequence a is an increasing sequence if each element of a (except the first one) is strictly greater than the previous element.
For p = [3, 4, 3, 4, 5], the length of longest increasing subsequence is 3 and there are 3 of them: [\underline{3}, \underline{4}, 3, 4, \underline{5}], [\underline{3}, 4, 3, \underline{4}, \underline{5}], [3, 4, \underline{3}, \underline{4}, \underline{5}].
Input
The first line contains a single integer n (1 ≤ n ≤ 50) — the length of the sequence d.
The second line contains a single integer x (-10^9 ≤ x ≤ 10^9) — the integer on the blackboard.
The third line contains n integers d_1, d_2, …, d_n (-10^9 ≤ d_i ≤ 10^9).
Output
Print two integers:
* the first integer should be equal to the length of the longest increasing subsequence of p;
* the second should be equal to the number of them modulo 998244353.
You should print only the second number modulo 998244353.
Examples
Input
3
3
1 -1 2
Output
3 3
Input
3
100
5 -3 6
Output
9 7
Input
3
1
999999999 0 1000000000
Output
2000000000 1
Input
5
34
1337 -146 42 -69 228
Output
1393 3876
Note
The first test case was explained in the statement.
In the second test case p = [100, 101, 102, 103, 104, 105, 104, 103, 102, 103, 104, 105, 106, 107, 108].
In the third test case p = [1, 2, …, 2000000000].
Submitted Solution:
```
import sys, io, os
if os.environ['USERNAME']=='kissz':
inp=open('in4.txt','r').readline
def debug(*args):
print(*args,file=sys.stderr)
else:
inp=io.BytesIO(os.read(0,os.fstat(0).st_size)).readline
def debug(*args):
pass
def mexp(size,power):
A=[]
for i in range(size):
A.append([int(j<=(i//2)*2) for j in range(size)])
powers={1: A}
p=1
while p*2<=power:
powers[p*2]=mmmult(powers[p],powers[p])
p*=2
A=powers[p]
power-=p
while power>0:
p=p//2
if p<=power:
A=mmmult(A,powers[p])
power-=p
return A
def mvmult(A,V):
res=[]
for i in range(len(A)):
res.append(sum(a*v for a,v in zip(A[i],V)))
return res
def mmmult(A,B):
res=[]
for i in range(len(A)):
res.append([sum(a*B[j][k] for j,a in enumerate(A[i])) for k in range(len(B[0]))])
return res
n=int(inp())
inp()
d,*D=map(int,inp().split())
while d==0: d,*D=D
up=(d>=0)
corners=[0,d]
for d in D:
x=corners[-1]+d
if up==(d>=0):
corners[-1]=x
if up!=(d>=0):
up=(d>=0)
corners.append(x)
low=(0,0)
maxdiff=(0,0,0)
for i,corner in enumerate(corners):
if corner<low[0]: low=(corner,i)
if corner-low[0]>=maxdiff[0]:
maxdiff=(corner-low[0],low[1],i)
debug(maxdiff)
corners=corners[maxdiff[1]:maxdiff[2]+1]
debug(corners)
L=maxdiff[0]+1
corners[0]-=1
corners[-1]+=1
bps=sorted(list(set(corners)))
m=len(bps)
X=[1]
active=[1]
for i in range(1,m):
x,y=bps[i-1],bps[i]
d=y-x
A=mexp(len(active),d)
X=mvmult(A,X)
debug(active,X)
debug(A)
if i<m-1:
for j,c in enumerate(corners):
if c==y:
if j%2: # top: j and j+1 in active
idx=active.index(j)
X[idx+2]+=X[idx]
active.pop(idx)
active.pop(idx)
X.pop(idx)
X.pop(idx)
else: # bottom
active+=[j,j+1]
active.sort()
idx=active.index(j)
X=X[:idx]+[0,X[idx-1]]+X[idx:]
else:
print(L,X[0])
```
No
| 98,845 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Igor had a sequence d_1, d_2, ..., d_n of integers. When Igor entered the classroom there was an integer x written on the blackboard.
Igor generated sequence p using the following algorithm:
1. initially, p = [x];
2. for each 1 ≤ i ≤ n he did the following operation |d_i| times:
* if d_i ≥ 0, then he looked at the last element of p (let it be y) and appended y + 1 to the end of p;
* if d_i < 0, then he looked at the last element of p (let it be y) and appended y - 1 to the end of p.
For example, if x = 3, and d = [1, -1, 2], p will be equal [3, 4, 3, 4, 5].
Igor decided to calculate the length of the longest increasing subsequence of p and the number of them.
A sequence a is a subsequence of a sequence b if a can be obtained from b by deletion of several (possibly, zero or all) elements.
A sequence a is an increasing sequence if each element of a (except the first one) is strictly greater than the previous element.
For p = [3, 4, 3, 4, 5], the length of longest increasing subsequence is 3 and there are 3 of them: [\underline{3}, \underline{4}, 3, 4, \underline{5}], [\underline{3}, 4, 3, \underline{4}, \underline{5}], [3, 4, \underline{3}, \underline{4}, \underline{5}].
Input
The first line contains a single integer n (1 ≤ n ≤ 50) — the length of the sequence d.
The second line contains a single integer x (-10^9 ≤ x ≤ 10^9) — the integer on the blackboard.
The third line contains n integers d_1, d_2, …, d_n (-10^9 ≤ d_i ≤ 10^9).
Output
Print two integers:
* the first integer should be equal to the length of the longest increasing subsequence of p;
* the second should be equal to the number of them modulo 998244353.
You should print only the second number modulo 998244353.
Examples
Input
3
3
1 -1 2
Output
3 3
Input
3
100
5 -3 6
Output
9 7
Input
3
1
999999999 0 1000000000
Output
2000000000 1
Input
5
34
1337 -146 42 -69 228
Output
1393 3876
Note
The first test case was explained in the statement.
In the second test case p = [100, 101, 102, 103, 104, 105, 104, 103, 102, 103, 104, 105, 106, 107, 108].
In the third test case p = [1, 2, …, 2000000000].
Submitted Solution:
```
import sys, io, os
if os.environ['USERNAME']=='kissz':
inp=open('in4.txt','r').readline
def debug(*args):
print(*args,file=sys.stderr)
else:
inp=io.BytesIO(os.read(0,os.fstat(0).st_size)).readline
def debug(*args):
pass
def mexp(size,power):
A=[]
for i in range(size):
A.append([int(j<=(i//2)*2) for j in range(size)])
powers={1: A}
p=1
while p*2<=power:
powers[p*2]=mmmult(powers[p],powers[p])
p*=2
A=powers[p]
power-=p
while power>0:
p=p//2
if p<=power:
A=mmmult(A,powers[p])
power-=p
return A
def mvmult(A,V):
res=[]
for i in range(len(A)):
res.append(sum(a*v for a,v in zip(A[i],V)) % 998244353 )
return res
def mmmult(A,B):
res=[]
for i in range(len(A)):
res.append([sum(a*B[j][k] for j,a in enumerate(A[i])) for k in range(len(B[0]))])
return res
def get_rep(corners):
corners[0]-=1
corners[-1]+=1
bps=sorted(list(set(corners)))
m=len(bps)
X=[1]
active=[1]
for i in range(1,m):
x,y=bps[i-1],bps[i]
d=y-x
A=mexp(len(active),d)
X=mvmult(A,X)
debug(active,X)
debug(A)
if i<m-1:
for j,c in enumerate(corners):
if c==y:
if j%2: # top: j and j+1 in active
idx=active.index(j)
X[idx+2]+=X[idx]
active.pop(idx)
active.pop(idx)
X.pop(idx)
X.pop(idx)
else: # bottom
active+=[j,j+1]
active.sort()
idx=active.index(j)
X=X[:idx]+[0,X[idx-1]]+X[idx:]
else:
return X[0]
n=int(inp())
inp()
d,*D=map(int,inp().split())
while d==0: d,*D=D
up=(d>=0)
corners=[0,d]
for d in D:
x=corners[-1]+d
if up==(d>=0):
corners[-1]=x
if up!=(d>=0):
up=(d>=0)
corners.append(x)
debug(corners)
cands=[(-1,0,0)]
low=(0,0)
maxdiff=(0,0,0)
for i,corner in enumerate(corners):
if corner<low[0]: low=(corner,i)
if corner-low[0]>cands[0][0]:
cands=[(corner-low[0],low[1],i)]
elif corner-low[0]==cands[0][0]:
cands+=[(corner-low[0],low[1],i)]
L=cands[0][0]+1
if L>1:
X=0
debug(cands)
for _, starti, endi in cands:
debug(corners[starti:endi+1])
X+=get_rep(corners[starti:endi+1])
else:
X=1-corners[-1]
print(L,X)
```
No
| 98,846 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Igor had a sequence d_1, d_2, ..., d_n of integers. When Igor entered the classroom there was an integer x written on the blackboard.
Igor generated sequence p using the following algorithm:
1. initially, p = [x];
2. for each 1 ≤ i ≤ n he did the following operation |d_i| times:
* if d_i ≥ 0, then he looked at the last element of p (let it be y) and appended y + 1 to the end of p;
* if d_i < 0, then he looked at the last element of p (let it be y) and appended y - 1 to the end of p.
For example, if x = 3, and d = [1, -1, 2], p will be equal [3, 4, 3, 4, 5].
Igor decided to calculate the length of the longest increasing subsequence of p and the number of them.
A sequence a is a subsequence of a sequence b if a can be obtained from b by deletion of several (possibly, zero or all) elements.
A sequence a is an increasing sequence if each element of a (except the first one) is strictly greater than the previous element.
For p = [3, 4, 3, 4, 5], the length of longest increasing subsequence is 3 and there are 3 of them: [\underline{3}, \underline{4}, 3, 4, \underline{5}], [\underline{3}, 4, 3, \underline{4}, \underline{5}], [3, 4, \underline{3}, \underline{4}, \underline{5}].
Input
The first line contains a single integer n (1 ≤ n ≤ 50) — the length of the sequence d.
The second line contains a single integer x (-10^9 ≤ x ≤ 10^9) — the integer on the blackboard.
The third line contains n integers d_1, d_2, …, d_n (-10^9 ≤ d_i ≤ 10^9).
Output
Print two integers:
* the first integer should be equal to the length of the longest increasing subsequence of p;
* the second should be equal to the number of them modulo 998244353.
You should print only the second number modulo 998244353.
Examples
Input
3
3
1 -1 2
Output
3 3
Input
3
100
5 -3 6
Output
9 7
Input
3
1
999999999 0 1000000000
Output
2000000000 1
Input
5
34
1337 -146 42 -69 228
Output
1393 3876
Note
The first test case was explained in the statement.
In the second test case p = [100, 101, 102, 103, 104, 105, 104, 103, 102, 103, 104, 105, 106, 107, 108].
In the third test case p = [1, 2, …, 2000000000].
Submitted Solution:
```
import sys, io, os
if os.environ['USERNAME']=='kissz':
inp=open('in3.txt','r').readline
def debug(*args):
print(*args,file=sys.stderr)
else:
inp=io.BytesIO(os.read(0,os.fstat(0).st_size)).readline
def debug(*args):
pass
# SCRIPT STARTS
def polyadd(poly1,poly2):
poly=[0]*max(len(poly1),len(poly2))
for i,p in enumerate(poly1):
poly[i]+=p
for i,p in enumerate(poly2):
poly[i]+=p
return poly
def polysub(poly1,poly2):
poly=[0]*max(len(poly1),len(poly2))
for i,p in enumerate(poly1):
poly[i]+=p
for i,p in enumerate(poly2):
poly[i]-=p
return poly
def polymult(poly1,poly2):
poly=[0]*(len(poly1)+len(poly2)-1)
for i,pi in enumerate(poly1):
for j,pj in enumerate(poly2):
poly[i+j]+=pi*pj % 998244353
return poly
ex=[1]
N=[ex]
ndiv=[1]
div=1
for i in range(50):
div*=(i+1)
ex=polymult(ex,[i,1])
ndiv.append(div)
N.append(ex)
class multipoly():
def __init__(self,min_element,max_element):
self.ranges=[min_element-1,min_element,min_element+1] # keep sorted!
self.poly={min_element-1: [1], min_element:[1], min_element+1:[0]}
self.pos=min_element
def set_poly(self,rmin,poly,caller_name=None):
self.poly[rmin]=poly
if rmin not in self.ranges:
self.ranges.append(rmin)
self.ranges.sort()
if caller_name:
debug(caller_name,rmin,poly)
def find_range(self,x): # brute
for rmin in reversed(self.ranges):
if rmin<=x: return rmin
def rebase_poly(self,poly,rmin,x):
if x>rmin:
rebase=x-rmin
ex=[1]
new_poly=[0]*len(poly)
for p in poly:
for i,e in enumerate(ex):
new_poly[i]+=e*p
ex=polymult(ex, [rebase,1])
return new_poly
else:
return poly
def get_poly(self,x):
rmin=self.find_range(x)
poly=self.poly[rmin]
return self.rebase_poly(poly,rmin,x)
def convert_poly(self,poly):
series=[0]*len(poly)
while any(p!=0 for p in poly):
while not poly[-1]: poly.pop()
n=len(poly)-1
serie=N[n]
d=poly[-1]*ndiv[n]//serie[-1]
series[n]=d
poly=polysub(poly,polymult(serie,[d//ndiv[n]]))
return series
def convert_series(self,series):
poly=[0]
for i,s in enumerate(series):
poly=polyadd(poly,polymult(N[i],[s//ndiv[i]]))
return poly
def integrate_series(self,series):
return [0]+series
def x_eval(self,x):
rmin=self.find_range(x)
poly=self.poly[rmin]
return sum(p*((x-rmin+1)**i) for i,p in enumerate(poly)) % 998244353
def integrate(self,low):
poly=self.get_poly(low)
series=self.convert_poly(poly)
series=self.integrate_series(series)
new_poly=self.convert_series(series)
base=self.x_eval(low-1)
new_poly[0]+=base
self.set_poly(low,new_poly)
def up(self,low,high):
self.set_poly(high+1,self.get_poly(high+1))
self.integrate(low)
for rmin in self.ranges:
if low<rmin<=high:
self.integrate(rmin)
def downsum(self,low):
poly1=self.get_poly(low)
poly2=self.get_poly(low-1)
poly=polyadd(poly1,poly2)
self.set_poly(low,poly)
def down(self,high,low):
self.set_poly(high+1,self.get_poly(high+1))
checked=[]
for rmin in reversed(self.ranges):
if low<=rmin<=high:
if rmin<high and rmin+1 not in checked:
self.downsum(rmin+1)
checked.append(rmin+1)
if rmin not in checked:
self.downsum(rmin)
checked.append(rmin)
if low not in checked:
self.downsum(low)
def move(self,to):
if to>self.pos:
self.up(self.pos+1,to)
else:
self.down(self.pos-1,to)
self.pos=to
debug('arrive',to)
n=int(inp())
inp()
d,*D=map(int,inp().split())
while d==0: d,*D=D
up=(d>=0)
corners=[0,d]
for d in D:
x=corners[-1]+d
if up==(d>=0):
corners[-1]=x
if up!=(d>=0):
up=(d>=0)
corners.append(x)
low=(0,0)
maxdiff=(0,0,0)
for i,corner in enumerate(corners):
if corner<low[0]: low=(corner,i)
if corner-low[0]>=maxdiff[0]:
maxdiff=(corner-low[0],low[1],i)
corners=corners[maxdiff[1]:maxdiff[2]+1]
MP=multipoly(corners[0],corners[-1])
for target in corners[1:]:
MP.move(target)
print(maxdiff[0]+1,MP.x_eval(corners[-1]))
```
No
| 98,847 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Several months later Alex finally got his brother Bob's creation by post. And now, in his turn, Alex wants to boast about something to his brother. He thought for a while, and came to the conclusion that he has no ready creations, and decided to write a program for rectangles detection. According to his plan, the program detects if the four given segments form a rectangle of a positive area and with sides parallel to coordinate axes. As Alex does badly at school and can't write this program by himself, he asks you to help him.
Input
The input data contain four lines. Each of these lines contains four integers x1, y1, x2, y2 ( - 109 ≤ x1, y1, x2, y2 ≤ 109) — coordinates of segment's beginning and end positions. The given segments can degenerate into points.
Output
Output the word «YES», if the given four segments form the required rectangle, otherwise output «NO».
Examples
Input
1 1 6 1
1 0 6 0
6 0 6 1
1 1 1 0
Output
YES
Input
0 0 0 3
2 0 0 0
2 2 2 0
0 2 2 2
Output
NO
Tags: brute force, constructive algorithms, geometry, implementation, math
Correct Solution:
```
from collections import Counter
puntos = Counter()
segmentos = Counter()
for _ in range(4):
px, py, qx, qy = list(map(int, input().split()))
p = (px, py)
q = (qx, qy)
puntos[p] += 1
puntos[q] += 1
if (q, p) not in segmentos.keys():
segmentos[(p, q)] += 1
else:
segmentos[(q, p)] += 1
def f(puntos, segmentos):
if len(puntos) == 4 and len(segmentos) == 4:
v, h = 0, 0
for L in segmentos:
p, q = L[0], L[1]
if p == q:
return('NO') # si un segmento es un punto retorna no
else:
px, py = p[0], p[1]
qx, qy = q[0], q[1]
if qx - px == 0:
v += 1
elif qy - py == 0:
h += 1
else:
return('NO') # si no es h ni v retorna no
if v == 2 and h == 2:
return('YES') # si hay 2 h y 2 v retorna si
else:
return('NO') # si no hay 2 h y no hay 2 v retorna no
else:
return('NO')
print(f(puntos, segmentos))
```
| 98,848 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Several months later Alex finally got his brother Bob's creation by post. And now, in his turn, Alex wants to boast about something to his brother. He thought for a while, and came to the conclusion that he has no ready creations, and decided to write a program for rectangles detection. According to his plan, the program detects if the four given segments form a rectangle of a positive area and with sides parallel to coordinate axes. As Alex does badly at school and can't write this program by himself, he asks you to help him.
Input
The input data contain four lines. Each of these lines contains four integers x1, y1, x2, y2 ( - 109 ≤ x1, y1, x2, y2 ≤ 109) — coordinates of segment's beginning and end positions. The given segments can degenerate into points.
Output
Output the word «YES», if the given four segments form the required rectangle, otherwise output «NO».
Examples
Input
1 1 6 1
1 0 6 0
6 0 6 1
1 1 1 0
Output
YES
Input
0 0 0 3
2 0 0 0
2 2 2 0
0 2 2 2
Output
NO
Tags: brute force, constructive algorithms, geometry, implementation, math
Correct Solution:
```
def main():
line_1 = input().strip().split()
line_1 = list(map(int, line_1))
line_2 = input().strip().split()
line_2 = list(map(int, line_2))
line_3 = input().strip().split()
line_3 = list(map(int, line_3))
line_4 = input().strip().split()
line_4 = list(map(int, line_4))
x_line_count = 0
y_line_count = 0
points = set()
if line_1 == line_2 or line_1 == line_3 or line_1 == line_4 or line_2 == line_3 or line_2 == line_4 or line_3 == line_4:
return 'NO'
for line in (line_1, line_2, line_3, line_4):
if line[0] == line[2] and line[1] != line[3]:
x_line_count += 1
elif line[1] == line[3] and line[0] != line[2]:
y_line_count += 1
points.add((line[0], line[1]))
points.add((line[2], line[3]))
if x_line_count != 2 or y_line_count != 2:
return 'NO'
if len(points) != 4:
return 'NO'
return 'YES'
print(main())
```
| 98,849 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Several months later Alex finally got his brother Bob's creation by post. And now, in his turn, Alex wants to boast about something to his brother. He thought for a while, and came to the conclusion that he has no ready creations, and decided to write a program for rectangles detection. According to his plan, the program detects if the four given segments form a rectangle of a positive area and with sides parallel to coordinate axes. As Alex does badly at school and can't write this program by himself, he asks you to help him.
Input
The input data contain four lines. Each of these lines contains four integers x1, y1, x2, y2 ( - 109 ≤ x1, y1, x2, y2 ≤ 109) — coordinates of segment's beginning and end positions. The given segments can degenerate into points.
Output
Output the word «YES», if the given four segments form the required rectangle, otherwise output «NO».
Examples
Input
1 1 6 1
1 0 6 0
6 0 6 1
1 1 1 0
Output
YES
Input
0 0 0 3
2 0 0 0
2 2 2 0
0 2 2 2
Output
NO
Tags: brute force, constructive algorithms, geometry, implementation, math
Correct Solution:
```
def main():
line_1 = input().strip().split()
line_1 = list(map(int, line_1))
line_2 = input().strip().split()
line_2 = list(map(int, line_2))
line_3 = input().strip().split()
line_3 = list(map(int, line_3))
line_4 = input().strip().split()
line_4 = list(map(int, line_4))
x_line_count = 0
y_line_count = 0
points = {}
if line_1 == line_2 or line_1 == line_3 or line_1 == line_4 or line_2 == line_3 or line_2 == line_4 or line_3 == line_4:
return 'NO'
for line in (line_1, line_2, line_3, line_4):
if line[0] == line[2] and line[1] != line[3]:
x_line_count += 1
elif line[1] == line[3] and line[0] != line[2]:
y_line_count += 1
if (line[0], line[1]) in points:
points[(line[0], line[1])] += 1
else:
points[(line[0], line[1])] = 1
if (line[2], line[3]) in points:
points[(line[2], line[3])] += 1
else:
points[(line[2], line[3])] = 1
if x_line_count != 2 or y_line_count != 2:
return 'NO'
if len(points) != 4:
return 'NO'
return 'YES'
print(main())
```
| 98,850 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Several months later Alex finally got his brother Bob's creation by post. And now, in his turn, Alex wants to boast about something to his brother. He thought for a while, and came to the conclusion that he has no ready creations, and decided to write a program for rectangles detection. According to his plan, the program detects if the four given segments form a rectangle of a positive area and with sides parallel to coordinate axes. As Alex does badly at school and can't write this program by himself, he asks you to help him.
Input
The input data contain four lines. Each of these lines contains four integers x1, y1, x2, y2 ( - 109 ≤ x1, y1, x2, y2 ≤ 109) — coordinates of segment's beginning and end positions. The given segments can degenerate into points.
Output
Output the word «YES», if the given four segments form the required rectangle, otherwise output «NO».
Examples
Input
1 1 6 1
1 0 6 0
6 0 6 1
1 1 1 0
Output
YES
Input
0 0 0 3
2 0 0 0
2 2 2 0
0 2 2 2
Output
NO
Tags: brute force, constructive algorithms, geometry, implementation, math
Correct Solution:
```
# Try to do in 30 minutes
# 4 segments are given
def read_points():
hSegs = []
vSegs = []
for i in range(4):
rawline = input().split()
line = {"p1":[int(rawline[0]),int(rawline[1])],"p2":[int(rawline[2]),int(rawline[3])]}
if line["p1"][0] == line["p2"][0]: # X values the same, Vertical line
if line["p1"][1] > line["p2"][1]: # Swap order so p1 is smaller
line["p2"][1], line["p1"][1] = line["p1"][1], line["p2"][1]
vSegs.append(line)
elif line["p1"][1] == line["p2"][1]: # Y values the same, horizontal line
if line["p1"][0] > line["p2"][0]: # Swap order so p1 is smaller
line["p2"][0], line["p1"][0] = line["p1"][0], line["p2"][0]
hSegs.append(line)
else:
return False
if line["p1"][1] == line["p2"][1] and line["p1"][0] == line["p2"][0]: # line segments length = 0
return False
if len(hSegs) != 2 or len(vSegs) != 2:
return False
hSegs.sort(key=lambda a : a["p1"][1])
vSegs.sort(key=lambda a : a["p1"][0])
if hSegs[0]['p1'][1] == hSegs[1]['p1'][1] or vSegs[0]['p1'][0] == vSegs[1]['p1'][0]: # no area
return False
for h in hSegs:
if h['p1'][0] != vSegs[0]['p1'][0]: #line does not touch 1st vert
return False
if h['p2'][0] != vSegs[1]['p1'][0]: #line does not touch 2nd vert
return False
for v in vSegs:
if v['p1'][1] != hSegs[0]['p1'][1]: #line does not touch 1st horizontal
return False
if v['p2'][1] != hSegs[1]['p1'][1]: #line does not touch 2nd horizontal
return False
return True
print("YES" if read_points() else "NO")
```
| 98,851 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Several months later Alex finally got his brother Bob's creation by post. And now, in his turn, Alex wants to boast about something to his brother. He thought for a while, and came to the conclusion that he has no ready creations, and decided to write a program for rectangles detection. According to his plan, the program detects if the four given segments form a rectangle of a positive area and with sides parallel to coordinate axes. As Alex does badly at school and can't write this program by himself, he asks you to help him.
Input
The input data contain four lines. Each of these lines contains four integers x1, y1, x2, y2 ( - 109 ≤ x1, y1, x2, y2 ≤ 109) — coordinates of segment's beginning and end positions. The given segments can degenerate into points.
Output
Output the word «YES», if the given four segments form the required rectangle, otherwise output «NO».
Examples
Input
1 1 6 1
1 0 6 0
6 0 6 1
1 1 1 0
Output
YES
Input
0 0 0 3
2 0 0 0
2 2 2 0
0 2 2 2
Output
NO
Tags: brute force, constructive algorithms, geometry, implementation, math
Correct Solution:
```
def is_rect(es):
v = set([])
for e in es:
if not ((e[0] == e[2]) or (e[1] == e[3])):
return False
v.add((e[0], e[1]))
v.add((e[2], e[3]))
if len(v) != 4:
return False
xs = set([])
ys = set([])
for vi in v:
xs.add(vi[0])
ys.add(vi[1])
if len(xs) != 2 or len(ys) != 2:
return False
t = b = r = l = False
for e in es:
t = t or (e[1] == e[3] and e[0] != e[2] and e[1] == max(ys))
b = b or (e[1] == e[3] and e[0] != e[2] and e[1] == min(ys))
r = r or (e[0] == e[2] and e[1] != e[3] and e[0] == max(xs))
l = l or (e[0] == e[2] and e[1] != e[3] and e[0] == min(xs))
return t and b and l and r
e1 = list(map(int, input().split(' ')))
e2 = list(map(int, input().split(' ')))
e3 = list(map(int, input().split(' ')))
e4 = list(map(int, input().split(' ')))
if is_rect((e1, e2, e3, e4)):
print("YES")
else:
print("NO")
```
| 98,852 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Several months later Alex finally got his brother Bob's creation by post. And now, in his turn, Alex wants to boast about something to his brother. He thought for a while, and came to the conclusion that he has no ready creations, and decided to write a program for rectangles detection. According to his plan, the program detects if the four given segments form a rectangle of a positive area and with sides parallel to coordinate axes. As Alex does badly at school and can't write this program by himself, he asks you to help him.
Input
The input data contain four lines. Each of these lines contains four integers x1, y1, x2, y2 ( - 109 ≤ x1, y1, x2, y2 ≤ 109) — coordinates of segment's beginning and end positions. The given segments can degenerate into points.
Output
Output the word «YES», if the given four segments form the required rectangle, otherwise output «NO».
Examples
Input
1 1 6 1
1 0 6 0
6 0 6 1
1 1 1 0
Output
YES
Input
0 0 0 3
2 0 0 0
2 2 2 0
0 2 2 2
Output
NO
Tags: brute force, constructive algorithms, geometry, implementation, math
Correct Solution:
```
def is_rect():
def seg():
[x0, y0, x1, y1] = input().split(' ')
return int(x0), int(y0), int(x1), int(y1)
def parallel(x0, x1, y0, y1, x2, x3, y2, y3):
return y0 == y1 and y2 == y3 and y0 != y2 and ((x2 == x0 and x3 == x1) or (x3 == x0 and x2 == x1))
def shared(sx0, sy0, x1, y1, x2, y2):
return sx0, y1, x2, y1, x2, sy0, x2, y1
x0, y0, x1, y1 = seg()
x2, y2, x3, y3 = seg()
x4, y4, x5, y5 = seg()
x6, y6, x7, y7 = seg()
def find_corner(x0, y0, x1, y1, x2, y2, x3, y3):
if x0 == x2 and y0 == y2:
return x0, y0
elif x0 == x3 and y0 == y3:
return x0, y0
elif x1 == x2 and y1 == y2:
return x1, y1
elif x1 == x3 and y1 == y3:
return x1, y1
else:
return None, None
def flip_seg(cx, cy, x0, y0, x1, y1):
if cx == x0 and cy == y0:
return x0, y0, x1, y1
elif cx == x1 and cy == y1:
return x1, y1, x0, y0
cx0, cy0 = find_corner(x0, y0, x1, y1, x2, y2, x3, y3)
if cx0 == None:
# Segments are parallel
cx0, cy0 = find_corner(x0, y0, x1, y1, x4, y4, x5, y5)
if cx0 == None:
cx0, cy0 = find_corner(x0, y0, x1, y1, x6, y6, x7, y7)
if cx0 == None:
return False
else:
x2, y2, x3, y3, x6, y6, x7, y7 = x6, y6, x7, y7, x2, y2, x3, y3
else:
x2, y2, x3, y3, x4, y4, x5, y5 = x4, y4, x5, y5, x2, y2, x3, y3
cx1, cy1 = find_corner(x4, y4, x5, y5, x6, y6, x7, y7)
if cx1 == None:
return False
if cx0 == cx1 or cy0 == cy1:
return False
x0, y0, x1, y1 = flip_seg(cx0, cy0, x0, y0, x1, y1)
x2, y2, x3, y3 = flip_seg(cx0, cy0, x2, y2, x3, y3)
x4, y4, x5, y5 = flip_seg(cx1, cy1, x4, y4, x5, y5)
x6, y6, x7, y7 = flip_seg(cx1, cy1, x6, y6, x7, y7)
if x1 == x0:
pass
elif y1 == y0:
x0, y0, x1, y1, x2, y2, x3, y3 = x2, y2, x3, y3, x0, y0, x1, y1
else:
return False
if x0 != x1 or y2 != y3:
return False
if x5 == x4:
pass
elif y5 == y4:
x4, y4, x5, y5, x6, y6, x7, y7 = x6, y6, x7, y7, x4, y4, x5, y5
else:
return False
if x4 != x5 or y6 != y7:
return False
if y1 == cy1 and x3 == cx1 and y5 == cy0 and x7 == cx0:
return True
else:
return False
if is_rect():
print("YES")
else:
print("NO")
```
| 98,853 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Several months later Alex finally got his brother Bob's creation by post. And now, in his turn, Alex wants to boast about something to his brother. He thought for a while, and came to the conclusion that he has no ready creations, and decided to write a program for rectangles detection. According to his plan, the program detects if the four given segments form a rectangle of a positive area and with sides parallel to coordinate axes. As Alex does badly at school and can't write this program by himself, he asks you to help him.
Input
The input data contain four lines. Each of these lines contains four integers x1, y1, x2, y2 ( - 109 ≤ x1, y1, x2, y2 ≤ 109) — coordinates of segment's beginning and end positions. The given segments can degenerate into points.
Output
Output the word «YES», if the given four segments form the required rectangle, otherwise output «NO».
Examples
Input
1 1 6 1
1 0 6 0
6 0 6 1
1 1 1 0
Output
YES
Input
0 0 0 3
2 0 0 0
2 2 2 0
0 2 2 2
Output
NO
Tags: brute force, constructive algorithms, geometry, implementation, math
Correct Solution:
```
accepted = True
lines = []
dic = {}
for i in range(4):
line = list(map(int,input().split(" ")))
if line[0] == line[2] and line[1] == line[3]: accepted = False
lines.append(((line[0], line[1]),(line[2],line[3])))
if not (line[0] == line[2] or line[1] == line[3]): accepted = False
if lines[-1][0] not in dic.keys():
dic[lines[-1][0]] = [(i, 0)]
else: dic[lines[-1][0]].append((i,0))
if lines[-1][1] not in dic.keys():
dic[lines[-1][1]] = [(i, 1)]
else: dic[lines[-1][1]].append((i,1))
zero, one = set(), set()
for key in dic.keys():
if len(set(dic[key])) != 2: accepted = False
zero.add(key[0])
one.add(key[1])
j = 0
for entry1 in lines[j:]:
for entry2 in lines[j+1:]:
if entry1[0] == entry2[1] and entry1[1] == entry2[0]: accepted = False
if entry1 == entry2: accepted = False
j+=1
if len(zero) == 1 or len(one) == 1: accepted = False
print("YES") if accepted else print("NO")
#http://codeforces.com/problemset/problem/14/C
```
| 98,854 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Several months later Alex finally got his brother Bob's creation by post. And now, in his turn, Alex wants to boast about something to his brother. He thought for a while, and came to the conclusion that he has no ready creations, and decided to write a program for rectangles detection. According to his plan, the program detects if the four given segments form a rectangle of a positive area and with sides parallel to coordinate axes. As Alex does badly at school and can't write this program by himself, he asks you to help him.
Input
The input data contain four lines. Each of these lines contains four integers x1, y1, x2, y2 ( - 109 ≤ x1, y1, x2, y2 ≤ 109) — coordinates of segment's beginning and end positions. The given segments can degenerate into points.
Output
Output the word «YES», if the given four segments form the required rectangle, otherwise output «NO».
Examples
Input
1 1 6 1
1 0 6 0
6 0 6 1
1 1 1 0
Output
YES
Input
0 0 0 3
2 0 0 0
2 2 2 0
0 2 2 2
Output
NO
Tags: brute force, constructive algorithms, geometry, implementation, math
Correct Solution:
```
liste=[]
i=0
i2=0
distance=[]
answer='YES'
for _ in range(4):
x1,y1,x2,y2=map(int,input().split(" "))
liste.append([(x1,y1),(x2,y2)])
def perpendiculaire(vect1,vect2):
return vect1[0]*vect2[0]+vect1[1]*vect2[1]
def search(liste,point,excluded_index):
for i in range(4):
if point in liste[i] and i!=excluded_index:
return i,liste[i].index(point)
return -1,-1
index,i=0,0
for _ in range(4):
element=liste[index]
(x1,y1),(x2,y2)=element[i],element[1-i]
vector1=(x2-x1,y2-y1)
distance.append(abs(x2-x1)+abs(y2-y1))
index,i=search(liste,(x2,y2),index)
if index==-1:
answer='NO'
break
element=liste[index]
(x1,y1),(x2,y2)=element[i],element[1-i]
vector2=(x2-x1,y2-y1)
if perpendiculaire(vector1,vector2)!=0 or 0 not in vector1 or 0 not in vector2 or vector1==(0,0) or vector2==(0,0):
answer='NO'
break
if answer=='YES':
if distance[0]!=distance[2] or distance[1]!=distance[3]:
answer='NO'
print(answer)
```
| 98,855 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Several months later Alex finally got his brother Bob's creation by post. And now, in his turn, Alex wants to boast about something to his brother. He thought for a while, and came to the conclusion that he has no ready creations, and decided to write a program for rectangles detection. According to his plan, the program detects if the four given segments form a rectangle of a positive area and with sides parallel to coordinate axes. As Alex does badly at school and can't write this program by himself, he asks you to help him.
Input
The input data contain four lines. Each of these lines contains four integers x1, y1, x2, y2 ( - 109 ≤ x1, y1, x2, y2 ≤ 109) — coordinates of segment's beginning and end positions. The given segments can degenerate into points.
Output
Output the word «YES», if the given four segments form the required rectangle, otherwise output «NO».
Examples
Input
1 1 6 1
1 0 6 0
6 0 6 1
1 1 1 0
Output
YES
Input
0 0 0 3
2 0 0 0
2 2 2 0
0 2 2 2
Output
NO
Submitted Solution:
```
a={}
ox,oy=0,0
for i in range(4):
x1,y1,x2,y2=map(int,input().split())
a[(x1,y1)]=a.get((x1,y1),0)+1
a[(x2,y2)]=a.get((x2,y2),0)+1
if x1==x2 and y1!=y2: ox+=1
if y1==y2 and x1!=x2: oy+=1
ans="YES"
for p in a.values():
if p!=2:
ans="NO"
break
if ox!=2 or oy!=2:
ans="NO"
print(ans)
```
Yes
| 98,856 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Several months later Alex finally got his brother Bob's creation by post. And now, in his turn, Alex wants to boast about something to his brother. He thought for a while, and came to the conclusion that he has no ready creations, and decided to write a program for rectangles detection. According to his plan, the program detects if the four given segments form a rectangle of a positive area and with sides parallel to coordinate axes. As Alex does badly at school and can't write this program by himself, he asks you to help him.
Input
The input data contain four lines. Each of these lines contains four integers x1, y1, x2, y2 ( - 109 ≤ x1, y1, x2, y2 ≤ 109) — coordinates of segment's beginning and end positions. The given segments can degenerate into points.
Output
Output the word «YES», if the given four segments form the required rectangle, otherwise output «NO».
Examples
Input
1 1 6 1
1 0 6 0
6 0 6 1
1 1 1 0
Output
YES
Input
0 0 0 3
2 0 0 0
2 2 2 0
0 2 2 2
Output
NO
Submitted Solution:
```
Lines = 4*[None]
for i in range(4):
Lines[i] = list(map(int, input().split()))
if Lines[i][0] == Lines[i][2] and Lines[i][1] == Lines[i][3]:
print("NO")
exit(0)
if Lines[i][0] == Lines[i][2]:
Lines[i][1], Lines[i][3] = min(
Lines[i][1], Lines[i][3]), max(Lines[i][1], Lines[i][3])
elif Lines[i][1] == Lines[i][3]:
Lines[i][0], Lines[i][2] = min(
Lines[i][0], Lines[i][2]), max(Lines[i][0], Lines[i][2])
else:
print("NO")
exit(0)
Lines.sort()
if Lines[0] == Lines[1] or Lines[1] == Lines[2] or Lines[2] == Lines[3]:
print("NO")
exit(0)
if Lines[0][:2] == Lines[1][:2] and Lines[0][2:] == Lines[2][:2] and Lines[1][2:] == Lines[3][:2] and Lines[2][2:] == Lines[3][2:]:
print("YES")
else:
print("NO")
```
Yes
| 98,857 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Several months later Alex finally got his brother Bob's creation by post. And now, in his turn, Alex wants to boast about something to his brother. He thought for a while, and came to the conclusion that he has no ready creations, and decided to write a program for rectangles detection. According to his plan, the program detects if the four given segments form a rectangle of a positive area and with sides parallel to coordinate axes. As Alex does badly at school and can't write this program by himself, he asks you to help him.
Input
The input data contain four lines. Each of these lines contains four integers x1, y1, x2, y2 ( - 109 ≤ x1, y1, x2, y2 ≤ 109) — coordinates of segment's beginning and end positions. The given segments can degenerate into points.
Output
Output the word «YES», if the given four segments form the required rectangle, otherwise output «NO».
Examples
Input
1 1 6 1
1 0 6 0
6 0 6 1
1 1 1 0
Output
YES
Input
0 0 0 3
2 0 0 0
2 2 2 0
0 2 2 2
Output
NO
Submitted Solution:
```
puntos={0:0,1:0}
for i in range(4):
linea=list(map(int,input().split()))
p1=(linea[0],linea[1])
p2=(linea[2],linea[3])
if p1[0]==p2[0] and p1[1]!=p2[1]: puntos[0]+=1
elif p1[0]!=p2[0] and p1[1]==p2[1]: puntos[1]+=1
for punto in [p1,p2]:
puntos[punto]=puntos.get(punto,0)+1
if all( i==2 for i in puntos.values()):
print("YES")
else:
print("NO")
```
Yes
| 98,858 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Several months later Alex finally got his brother Bob's creation by post. And now, in his turn, Alex wants to boast about something to his brother. He thought for a while, and came to the conclusion that he has no ready creations, and decided to write a program for rectangles detection. According to his plan, the program detects if the four given segments form a rectangle of a positive area and with sides parallel to coordinate axes. As Alex does badly at school and can't write this program by himself, he asks you to help him.
Input
The input data contain four lines. Each of these lines contains four integers x1, y1, x2, y2 ( - 109 ≤ x1, y1, x2, y2 ≤ 109) — coordinates of segment's beginning and end positions. The given segments can degenerate into points.
Output
Output the word «YES», if the given four segments form the required rectangle, otherwise output «NO».
Examples
Input
1 1 6 1
1 0 6 0
6 0 6 1
1 1 1 0
Output
YES
Input
0 0 0 3
2 0 0 0
2 2 2 0
0 2 2 2
Output
NO
Submitted Solution:
```
def segmentos_iguales(l1, l2):
return (l1[:2] == l2[2:4] and l1[2:4] == l2[:2]) or (l1 == l2)
def paralela(linea):
return linea[0] == linea[2] or linea[1] == linea[3]
def degenerada(linea):
return linea[0] == linea[2] and linea[1] == linea[3]
lineas = []
for _ in range(4):
lineas.append(input().split())
x, y = set(), set()
for i in range(4):
x.add(lineas[i][0])
y.add(lineas[i][1])
x.add(lineas[i][2])
y.add(lineas[i][3])
if len(x) != 2 or len(y) != 2:
print("NO")
else:
for i in range(4):
if not paralela(lineas[i]) or degenerada(lineas[i]):
print("NO")
break
else:
flag = False
for i in range(3):
for j in range(i+1, 4):
if segmentos_iguales(lineas[i], lineas[j]):
print("NO")
flag = True
break
if flag:
break
else:
print("YES")
```
Yes
| 98,859 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Several months later Alex finally got his brother Bob's creation by post. And now, in his turn, Alex wants to boast about something to his brother. He thought for a while, and came to the conclusion that he has no ready creations, and decided to write a program for rectangles detection. According to his plan, the program detects if the four given segments form a rectangle of a positive area and with sides parallel to coordinate axes. As Alex does badly at school and can't write this program by himself, he asks you to help him.
Input
The input data contain four lines. Each of these lines contains four integers x1, y1, x2, y2 ( - 109 ≤ x1, y1, x2, y2 ≤ 109) — coordinates of segment's beginning and end positions. The given segments can degenerate into points.
Output
Output the word «YES», if the given four segments form the required rectangle, otherwise output «NO».
Examples
Input
1 1 6 1
1 0 6 0
6 0 6 1
1 1 1 0
Output
YES
Input
0 0 0 3
2 0 0 0
2 2 2 0
0 2 2 2
Output
NO
Submitted Solution:
```
def distance(a, b, c, d):
return (d - b) ** 2 + (c - a) ** 2
def slope(a, b, c, d):
if a == c:
return float('inf')
else:
return (d - b) / (c - a)
def CF14C():
infinity = 0
zero = 0
counter = {} # Sides counter
points = set()
for i in range(4):
a, b, c, d = list(map(int, input().split()))
points.add((a, b))
points.add((c, d))
s = slope(a, b, c, d)
if s == float('inf'):
infinity += 1
elif s == 0:
zero += 1
side = distance(a, b, c, d)
counter[side] = counter.get(side, 0) + 1
# There should be two sides parallel to x-axis and two sides parallel to y-axis
if zero != 2 and infinity != 2: return False
# print("Slopes ma problem vayena")
# There should be only two values and both of them should have two elements
if len(counter) != 2: return False
counts = list(counter.values())
if counts[1] != counts[0]: return False
# print("Sides pani duita equal cha")
# Lets do the pythagoras test now. Final test.
points = sorted(list(points), key = lambda x: (x[0], x[1]))
a, b = points[0]
c, d = points[1]
e, f = points[2]
sides = list(counter.keys())
d1 = distance(a, b, c, d)
d2 = distance(b, c, d, e)
d3 = distance(a, b, e, f)
if sides[0] + sides[1] != max(d1, d2, d3): return False
# For everything else.
return True
res = CF14C()
print("YES" if res else "NO")
print(res)
```
No
| 98,860 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Several months later Alex finally got his brother Bob's creation by post. And now, in his turn, Alex wants to boast about something to his brother. He thought for a while, and came to the conclusion that he has no ready creations, and decided to write a program for rectangles detection. According to his plan, the program detects if the four given segments form a rectangle of a positive area and with sides parallel to coordinate axes. As Alex does badly at school and can't write this program by himself, he asks you to help him.
Input
The input data contain four lines. Each of these lines contains four integers x1, y1, x2, y2 ( - 109 ≤ x1, y1, x2, y2 ≤ 109) — coordinates of segment's beginning and end positions. The given segments can degenerate into points.
Output
Output the word «YES», if the given four segments form the required rectangle, otherwise output «NO».
Examples
Input
1 1 6 1
1 0 6 0
6 0 6 1
1 1 1 0
Output
YES
Input
0 0 0 3
2 0 0 0
2 2 2 0
0 2 2 2
Output
NO
Submitted Solution:
```
from sys import stdin
class my_vector:
def __init__(self, x1, y1, x2, y2):
self.i_coord = (x1, y1)
self.end_coord = (x2, y2)
self.my_vector = (y2 - y1, x2 - x1)
def __mul__(self, other):
return self.my_vector[0] * other.my_vector[0] + self.my_vector[1] * other.my_vector[1]
def problem(lines):
v_dict = {}
number = 0
for row in lines:
number += 1
x1, y1, x2, y2 = row.strip().split()
v_dict[number] = my_vector(int(x1), int(y1), int(x2), int(y2))
# Check if the first my_vector has connected i and end:
A = 1
B = []
C = []
D = []
possible_candidates = {2, 3, 4}
for i in range(2, 5):
if v_dict[A].i_coord == v_dict[i].i_coord:
if v_dict[A].end_coord == v_dict[i].end_coord:
return False
possible_candidates.remove(i)
B.append(i)
elif v_dict[A].i_coord == v_dict[i].end_coord:
if v_dict[A].end_coord == v_dict[i].i_coord:
return False
possible_candidates.remove(i)
B.append(i)
elif v_dict[A].end_coord == v_dict[i].end_coord:
if v_dict[A].i_coord == v_dict[i].i_coord:
return False
possible_candidates.remove(i)
C.append(i)
elif v_dict[A].end_coord == v_dict[i].i_coord:
if v_dict[A].i_coord == v_dict[i].end_coord:
return False
possible_candidates.remove(i)
C.append(i)
if len(B) == 1 and len(C) == 1:
B = B.pop()
C = C.pop()
D = possible_candidates.pop()
if v_dict[D].i_coord == v_dict[B].i_coord or v_dict[D].i_coord == v_dict[B].end_coord:
if v_dict[D].end_coord != v_dict[C].i_coord and v_dict[D].end_coord != v_dict[C].end_coord:
return False
elif v_dict[D].end_coord == v_dict[B].i_coord or v_dict[D].end_coord == v_dict[B].end_coord:
if v_dict[D].i_coord != v_dict[C].i_coord and v_dict[D].i_coord != v_dict[C].end_coord:
return False
else:
return False
else:
return False
x_coord = my_vector(0,0,0,1)
y_coord = my_vector(0,0,1,0)
for i in range(1, 5):
if x_coord * v_dict[i] != 0 and y_coord * v_dict[i] != 0:
return False
min_x = min(v_dict[A].i_coord[0], v_dict[B].i_coord[0],v_dict[C].i_coord[0])
max_x = max(v_dict[A].i_coord[0], v_dict[B].i_coord[0],v_dict[C].i_coord[0])
min_y = min(v_dict[A].i_coord[1], v_dict[B].i_coord[1],v_dict[C].i_coord[1])
max_y = max(v_dict[A].i_coord[1], v_dict[B].i_coord[1],v_dict[C].i_coord[1])
if min_y < 0:
area = max_y * (max_x - min_x) + min_y * (max_x - min_x)
else:
area = max_y * (max_x - min_x) - min_y * (max_x - min_x)
if area <= 0:
return False
return True
def main():
if problem(stdin.readlines()):
print("YES")
else:
print("NO")
main()
```
No
| 98,861 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Several months later Alex finally got his brother Bob's creation by post. And now, in his turn, Alex wants to boast about something to his brother. He thought for a while, and came to the conclusion that he has no ready creations, and decided to write a program for rectangles detection. According to his plan, the program detects if the four given segments form a rectangle of a positive area and with sides parallel to coordinate axes. As Alex does badly at school and can't write this program by himself, he asks you to help him.
Input
The input data contain four lines. Each of these lines contains four integers x1, y1, x2, y2 ( - 109 ≤ x1, y1, x2, y2 ≤ 109) — coordinates of segment's beginning and end positions. The given segments can degenerate into points.
Output
Output the word «YES», if the given four segments form the required rectangle, otherwise output «NO».
Examples
Input
1 1 6 1
1 0 6 0
6 0 6 1
1 1 1 0
Output
YES
Input
0 0 0 3
2 0 0 0
2 2 2 0
0 2 2 2
Output
NO
Submitted Solution:
```
from collections import Counter
def main():
segs = [tuple(map(int, input().split())) for _ in range(4)]
points = []
for x1, y1, x2, y2 in segs:
if (x1, y1) == (x2, y2):
print('NO')
return
points.append((x1, y1))
points.append((x2, y2))
c = Counter(points)
xs = Counter(p[0] for p in points)
ys = Counter(p[1] for p in points)
if any(c[p] != 2 for p in c):
print('NO')
return
if any(xs[x] != 4 for x in xs):
print('NO')
return
if any(ys[y] != 4 for y in ys):
print('NO')
return
if len(set(points)) != 4:
print('NO')
return
print('YES')
main()
```
No
| 98,862 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Several months later Alex finally got his brother Bob's creation by post. And now, in his turn, Alex wants to boast about something to his brother. He thought for a while, and came to the conclusion that he has no ready creations, and decided to write a program for rectangles detection. According to his plan, the program detects if the four given segments form a rectangle of a positive area and with sides parallel to coordinate axes. As Alex does badly at school and can't write this program by himself, he asks you to help him.
Input
The input data contain four lines. Each of these lines contains four integers x1, y1, x2, y2 ( - 109 ≤ x1, y1, x2, y2 ≤ 109) — coordinates of segment's beginning and end positions. The given segments can degenerate into points.
Output
Output the word «YES», if the given four segments form the required rectangle, otherwise output «NO».
Examples
Input
1 1 6 1
1 0 6 0
6 0 6 1
1 1 1 0
Output
YES
Input
0 0 0 3
2 0 0 0
2 2 2 0
0 2 2 2
Output
NO
Submitted Solution:
```
def Prod_punto(v1,v2):
return v1[0]*v2[0]+v1[1]*v2[1]
puntos = []
rect = True
for i in range(4):
a = tuple(input().split(" "))
for k,j in enumerate(a):
if k%2 == 0:
b = [j]
else:
b.append(j)
puntos.append(tuple(b))
for j,i in enumerate(puntos):
if puntos.count(i) != 2:
rect = False
break
if j%2 == 0 and j+1 <= 7:
if i == puntos[j+1]:
rect = False
break
segmentos = []
for i in range(0,len(puntos),2):
punto = (int(puntos[i][0])-int(puntos[i+1][0])),(int(puntos[i][1])-int(puntos[i+1][1]))
segmentos.append(punto)
largos = []
for i in segmentos:
largos.append(Prod_punto(i,i))
for i in largos:
if largos.count(i) != 2:
rect = False
if rect:
for i,j in zip(segmentos,largos):
suma = 0
for k in segmentos:
suma += abs(Prod_punto(i,k))
if suma != 2*j:
rect = False
if rect:
print("YES")
else:
print("NO")
```
No
| 98,863 |
Provide tags and a correct Python 3 solution for this coding contest problem.
As Sherlock Holmes was investigating a crime, he identified n suspects. He knows for sure that exactly one of them committed the crime. To find out which one did it, the detective lines up the suspects and numbered them from 1 to n. After that, he asked each one: "Which one committed the crime?". Suspect number i answered either "The crime was committed by suspect number ai", or "Suspect number ai didn't commit the crime". Also, the suspect could say so about himself (ai = i).
Sherlock Holmes understood for sure that exactly m answers were the truth and all other answers were a lie. Now help him understand this: which suspect lied and which one told the truth?
Input
The first line contains two integers n and m (1 ≤ n ≤ 105, 0 ≤ m ≤ n) — the total number of suspects and the number of suspects who told the truth. Next n lines contain the suspects' answers. The i-th line contains either "+ai" (without the quotes), if the suspect number i says that the crime was committed by suspect number ai, or "-ai" (without the quotes), if the suspect number i says that the suspect number ai didn't commit the crime (ai is an integer, 1 ≤ ai ≤ n).
It is guaranteed that at least one suspect exists, such that if he committed the crime, then exactly m people told the truth.
Output
Print n lines. Line number i should contain "Truth" if suspect number i has told the truth for sure. Print "Lie" if the suspect number i lied for sure and print "Not defined" if he could lie and could tell the truth, too, depending on who committed the crime.
Examples
Input
1 1
+1
Output
Truth
Input
3 2
-1
-2
-3
Output
Not defined
Not defined
Not defined
Input
4 1
+2
-3
+4
-1
Output
Lie
Not defined
Lie
Not defined
Note
The first sample has the single person and he confesses to the crime, and Sherlock Holmes knows that one person is telling the truth. That means that this person is telling the truth.
In the second sample there are three suspects and each one denies his guilt. Sherlock Holmes knows that only two of them are telling the truth. Any one of them can be the criminal, so we don't know for any of them, whether this person is telling the truth or not.
In the third sample the second and the fourth suspect defend the first and the third one. But only one is telling the truth, thus, the first or the third one is the criminal. Both of them can be criminals, so the second and the fourth one can either be lying or telling the truth. The first and the third one are lying for sure as they are blaming the second and the fourth one.
Tags: constructive algorithms, data structures, implementation
Correct Solution:
```
# SHRi GANESHA author: Kunal Verma #
import os
import sys
from bisect import bisect_right
from collections import Counter, defaultdict, deque
from heapq import *
from io import BytesIO, IOBase
from math import gcd, inf, sqrt, ceil
def lcm(a, b):
return (a * b) // gcd(a, b)
'''
mod = 10 ** 9 + 7
fac = [1]
for i in range(1, 2 * 10 ** 5 + 1):
fac.append((fac[-1] * i) % mod)
fac_in = [pow(fac[-1], mod - 2, mod)]
for i in range(2 * 10 ** 5, 0, -1):
fac_in.append((fac_in[-1] * i) % mod)
fac_in.reverse()
def comb(a, b):
if a < b:
return 0
return (fac[a] * fac_in[b] * fac_in[a - b]) % mod
'''
#MAXN = 10000004
# spf = [0 for i in range(MAXN)]
# adj = [[] for i in range(MAXN)]
def sieve():
global spf, adj, MAXN
spf[1] = 1
for i in range(2, MAXN):
spf[i] = i
for i in range(2, MAXN):
if i * i > MAXN:
break
if (spf[i] == i):
for j in range(i * i, MAXN, i):
if (spf[j] == j):
spf[j] = i
def getdistinctFactorization(n):
global adj, spf, MAXN
for i in range(1, n + 1):
index = 1
x = i
if (x != 1):
adj[i].append(spf[x])
x = x // spf[x]
while (x != 1):
if (adj[i][index - 1] != spf[x]):
adj[i].append(spf[x])
index += 1
x = x // spf[x]
def printDivisors(n):
i = 2
z = [1, n]
while i <= sqrt(n):
if (n % i == 0):
if (n / i == i):
z.append(i)
else:
z.append(i)
z.append(n // i)
i = i + 1
return z
def create(n, x, f):
pq = len(bin(n)[2:])
if f == 0:
tt = min
else:
tt = max
dp = [[inf] * n for _ in range(pq)]
dp[0] = x
for i in range(1, pq):
for j in range(n - (1 << i) + 1):
dp[i][j] = tt(dp[i - 1][j], dp[i - 1][j + (1 << (i - 1))])
return dp
def enquiry(l, r, dp, f):
if l > r:
return inf if not f else -inf
if f == 1:
tt = max
else:
tt = min
pq1 = len(bin(r - l + 1)[2:]) - 1
return tt(dp[pq1][l], dp[pq1][r - (1 << pq1) + 1])
def SieveOfEratosthenes(n):
prime = [True for i in range(n + 1)]
p = 2
while (p * p <= n):
if (prime[p] == True):
for i in range(p * p, n + 1, p):
prime[i] = False
p += 1
x = []
for i in range(2, n + 1):
if prime[i]:
x.append(i)
return x
def main():
n,m=map(int,input().split())
z=[[0,0]for i in range(n+1) ]
xx=[]
for i in range(n):
p=int(input())
if p>0:
z[p][1]+=1
else:
z[abs(p)][0]+=1
xx.append(p)
s=0
for i in range(1,n+1):
s+=z[i][1]
an=[]
# print(z)
for i in range(1,n+1):
# print(n - z[i][0] - s + z[i][1],s)
if n-z[i][0]-s+z[i][1]==m:
an.append(i)
z=Counter(an)
m=[0]*n
# print(an)
for j in range(n):
if xx[j]>0:
if z[xx[j]]==len(an):
m[j]=1
elif z[xx[j]]==0:
m[j]=0
else:
m[j]=0.7
else:
if z[abs(xx[j])]==len(an):
m[j]=0
elif z[abs(xx[j])]>0:
m[j]=0.7
else:
m[j]=1
#print(an)
for i in range(n):
if m[i]==1:
m[i]="Truth"
elif m[i]==0:
m[i]= "Lie"
else:
m[i]="Not defined"
print(*m,sep='\n')
# Fast IO Region
BUFSIZE = 8192
class FastIO(IOBase):
newlines = 0
def __init__(self, file):
self._fd = file.fileno()
self.buffer = BytesIO()
self.writable = "x" in file.mode or "r" not in file.mode
self.write = self.buffer.write if self.writable else None
def read(self):
while True:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
if not b:
break
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines = 0
return self.buffer.read()
def readline(self):
while self.newlines == 0:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
self.newlines = b.count(b"\n") + (not b)
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines -= 1
return self.buffer.readline()
def flush(self):
if self.writable:
os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
self.write = lambda s: self.buffer.write(s.encode("ascii"))
self.read = lambda: self.buffer.read().decode("ascii")
self.readline = lambda: self.buffer.readline().decode("ascii")
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")
if __name__ == '__main__':
main()
```
| 98,864 |
Provide tags and a correct Python 3 solution for this coding contest problem.
As Sherlock Holmes was investigating a crime, he identified n suspects. He knows for sure that exactly one of them committed the crime. To find out which one did it, the detective lines up the suspects and numbered them from 1 to n. After that, he asked each one: "Which one committed the crime?". Suspect number i answered either "The crime was committed by suspect number ai", or "Suspect number ai didn't commit the crime". Also, the suspect could say so about himself (ai = i).
Sherlock Holmes understood for sure that exactly m answers were the truth and all other answers were a lie. Now help him understand this: which suspect lied and which one told the truth?
Input
The first line contains two integers n and m (1 ≤ n ≤ 105, 0 ≤ m ≤ n) — the total number of suspects and the number of suspects who told the truth. Next n lines contain the suspects' answers. The i-th line contains either "+ai" (without the quotes), if the suspect number i says that the crime was committed by suspect number ai, or "-ai" (without the quotes), if the suspect number i says that the suspect number ai didn't commit the crime (ai is an integer, 1 ≤ ai ≤ n).
It is guaranteed that at least one suspect exists, such that if he committed the crime, then exactly m people told the truth.
Output
Print n lines. Line number i should contain "Truth" if suspect number i has told the truth for sure. Print "Lie" if the suspect number i lied for sure and print "Not defined" if he could lie and could tell the truth, too, depending on who committed the crime.
Examples
Input
1 1
+1
Output
Truth
Input
3 2
-1
-2
-3
Output
Not defined
Not defined
Not defined
Input
4 1
+2
-3
+4
-1
Output
Lie
Not defined
Lie
Not defined
Note
The first sample has the single person and he confesses to the crime, and Sherlock Holmes knows that one person is telling the truth. That means that this person is telling the truth.
In the second sample there are three suspects and each one denies his guilt. Sherlock Holmes knows that only two of them are telling the truth. Any one of them can be the criminal, so we don't know for any of them, whether this person is telling the truth or not.
In the third sample the second and the fourth suspect defend the first and the third one. But only one is telling the truth, thus, the first or the third one is the criminal. Both of them can be criminals, so the second and the fourth one can either be lying or telling the truth. The first and the third one are lying for sure as they are blaming the second and the fourth one.
Tags: constructive algorithms, data structures, implementation
Correct Solution:
```
a,b=map(int,input().split())
r=[0]*a
c1=[0]*(a+1)
c2=[0]*(a+1)
f=0
for i in range(a):
n=int(input())
r[i]=n
if n>0:c1[n]+=1
else:
f+=1
c2[-n]+=1
possible=[False]*(a+1)
# print(c1,c2)
np=0#number of suspects
for i in range(1,a+1):
if c1[i]+f-c2[i]==b:
#f-c2[i] is truth
possible[i]=True
np+=1
# print(possible)
for i in range(a):
if r[i]>0:#he said +
if possible[r[i]] and np==1:#unique
print("Truth")
elif not possible[r[i]]:#Lie
print("Lie")
else:
print("Not defined")
else:#claims r[i] is not
r[i]*=-1
if possible[r[i]] and np==1:print("Lie")
elif not possible[r[i]]:print("Truth")
else:print("Not defined")
# print (possible)
```
| 98,865 |
Provide tags and a correct Python 3 solution for this coding contest problem.
As Sherlock Holmes was investigating a crime, he identified n suspects. He knows for sure that exactly one of them committed the crime. To find out which one did it, the detective lines up the suspects and numbered them from 1 to n. After that, he asked each one: "Which one committed the crime?". Suspect number i answered either "The crime was committed by suspect number ai", or "Suspect number ai didn't commit the crime". Also, the suspect could say so about himself (ai = i).
Sherlock Holmes understood for sure that exactly m answers were the truth and all other answers were a lie. Now help him understand this: which suspect lied and which one told the truth?
Input
The first line contains two integers n and m (1 ≤ n ≤ 105, 0 ≤ m ≤ n) — the total number of suspects and the number of suspects who told the truth. Next n lines contain the suspects' answers. The i-th line contains either "+ai" (without the quotes), if the suspect number i says that the crime was committed by suspect number ai, or "-ai" (without the quotes), if the suspect number i says that the suspect number ai didn't commit the crime (ai is an integer, 1 ≤ ai ≤ n).
It is guaranteed that at least one suspect exists, such that if he committed the crime, then exactly m people told the truth.
Output
Print n lines. Line number i should contain "Truth" if suspect number i has told the truth for sure. Print "Lie" if the suspect number i lied for sure and print "Not defined" if he could lie and could tell the truth, too, depending on who committed the crime.
Examples
Input
1 1
+1
Output
Truth
Input
3 2
-1
-2
-3
Output
Not defined
Not defined
Not defined
Input
4 1
+2
-3
+4
-1
Output
Lie
Not defined
Lie
Not defined
Note
The first sample has the single person and he confesses to the crime, and Sherlock Holmes knows that one person is telling the truth. That means that this person is telling the truth.
In the second sample there are three suspects and each one denies his guilt. Sherlock Holmes knows that only two of them are telling the truth. Any one of them can be the criminal, so we don't know for any of them, whether this person is telling the truth or not.
In the third sample the second and the fourth suspect defend the first and the third one. But only one is telling the truth, thus, the first or the third one is the criminal. Both of them can be criminals, so the second and the fourth one can either be lying or telling the truth. The first and the third one are lying for sure as they are blaming the second and the fourth one.
Tags: constructive algorithms, data structures, implementation
Correct Solution:
```
import sys
input=lambda:sys.stdin.readline().strip()
print=lambda s:sys.stdout.write(str(s)+"\n")
a,b=map(int,input().split())
r=[0]*a
c1=[0]*(a+1)
c2=[0]*(a+1)
f=0
for i in range(a):
n=int(input())
r[i]=n
if n>0:c1[n]+=1
else:
f+=1
c2[-n]+=1
possible=[False]*(a+1)
# print(c1,c2)
np=0#number of suspects
for i in range(1,a+1):
if c1[i]+f-c2[i]==b:
#f-c2[i] is truth
possible[i]=True
np+=1
# print(possible)
for i in range(a):
if r[i]>0:#he said +
if possible[r[i]] and np==1:#unique
print("Truth")
elif not possible[r[i]]:#Lie
print("Lie")
else:
print("Not defined")
else:#claims r[i] is not
r[i]*=-1
if possible[r[i]] and np==1:print("Lie")
elif not possible[r[i]]:print("Truth")
else:print("Not defined")
# print (possible)
```
| 98,866 |
Provide tags and a correct Python 3 solution for this coding contest problem.
As Sherlock Holmes was investigating a crime, he identified n suspects. He knows for sure that exactly one of them committed the crime. To find out which one did it, the detective lines up the suspects and numbered them from 1 to n. After that, he asked each one: "Which one committed the crime?". Suspect number i answered either "The crime was committed by suspect number ai", or "Suspect number ai didn't commit the crime". Also, the suspect could say so about himself (ai = i).
Sherlock Holmes understood for sure that exactly m answers were the truth and all other answers were a lie. Now help him understand this: which suspect lied and which one told the truth?
Input
The first line contains two integers n and m (1 ≤ n ≤ 105, 0 ≤ m ≤ n) — the total number of suspects and the number of suspects who told the truth. Next n lines contain the suspects' answers. The i-th line contains either "+ai" (without the quotes), if the suspect number i says that the crime was committed by suspect number ai, or "-ai" (without the quotes), if the suspect number i says that the suspect number ai didn't commit the crime (ai is an integer, 1 ≤ ai ≤ n).
It is guaranteed that at least one suspect exists, such that if he committed the crime, then exactly m people told the truth.
Output
Print n lines. Line number i should contain "Truth" if suspect number i has told the truth for sure. Print "Lie" if the suspect number i lied for sure and print "Not defined" if he could lie and could tell the truth, too, depending on who committed the crime.
Examples
Input
1 1
+1
Output
Truth
Input
3 2
-1
-2
-3
Output
Not defined
Not defined
Not defined
Input
4 1
+2
-3
+4
-1
Output
Lie
Not defined
Lie
Not defined
Note
The first sample has the single person and he confesses to the crime, and Sherlock Holmes knows that one person is telling the truth. That means that this person is telling the truth.
In the second sample there are three suspects and each one denies his guilt. Sherlock Holmes knows that only two of them are telling the truth. Any one of them can be the criminal, so we don't know for any of them, whether this person is telling the truth or not.
In the third sample the second and the fourth suspect defend the first and the third one. But only one is telling the truth, thus, the first or the third one is the criminal. Both of them can be criminals, so the second and the fourth one can either be lying or telling the truth. The first and the third one are lying for sure as they are blaming the second and the fourth one.
Tags: constructive algorithms, data structures, implementation
Correct Solution:
```
from collections import Counter
c = Counter()
n, m = map(int, input().split())
sumdc = 0
mark = []
suspect = set()
for _ in range(n):
k = int(input())
mark.append(k)
if k < 0: sumdc += 1
c = Counter(mark)
k = 0
for i in range(1, n + 1):
if c[i] + sumdc - c[-1 * i] == m:
suspect.add(i)
k += 1
if k==0:print("Not defined\n"*n)
elif k==1:
j=suspect.pop()
for x in mark:
if x==j or(x<0 and abs(x)!=j):print("Truth")
else: print("Lie")
else:
suspect.update({-x for x in suspect})
for x in mark:
if x in suspect:print("Not defined")
elif x<0:print("Truth")
else:print("Lie")
```
| 98,867 |
Provide tags and a correct Python 3 solution for this coding contest problem.
As Sherlock Holmes was investigating a crime, he identified n suspects. He knows for sure that exactly one of them committed the crime. To find out which one did it, the detective lines up the suspects and numbered them from 1 to n. After that, he asked each one: "Which one committed the crime?". Suspect number i answered either "The crime was committed by suspect number ai", or "Suspect number ai didn't commit the crime". Also, the suspect could say so about himself (ai = i).
Sherlock Holmes understood for sure that exactly m answers were the truth and all other answers were a lie. Now help him understand this: which suspect lied and which one told the truth?
Input
The first line contains two integers n and m (1 ≤ n ≤ 105, 0 ≤ m ≤ n) — the total number of suspects and the number of suspects who told the truth. Next n lines contain the suspects' answers. The i-th line contains either "+ai" (without the quotes), if the suspect number i says that the crime was committed by suspect number ai, or "-ai" (without the quotes), if the suspect number i says that the suspect number ai didn't commit the crime (ai is an integer, 1 ≤ ai ≤ n).
It is guaranteed that at least one suspect exists, such that if he committed the crime, then exactly m people told the truth.
Output
Print n lines. Line number i should contain "Truth" if suspect number i has told the truth for sure. Print "Lie" if the suspect number i lied for sure and print "Not defined" if he could lie and could tell the truth, too, depending on who committed the crime.
Examples
Input
1 1
+1
Output
Truth
Input
3 2
-1
-2
-3
Output
Not defined
Not defined
Not defined
Input
4 1
+2
-3
+4
-1
Output
Lie
Not defined
Lie
Not defined
Note
The first sample has the single person and he confesses to the crime, and Sherlock Holmes knows that one person is telling the truth. That means that this person is telling the truth.
In the second sample there are three suspects and each one denies his guilt. Sherlock Holmes knows that only two of them are telling the truth. Any one of them can be the criminal, so we don't know for any of them, whether this person is telling the truth or not.
In the third sample the second and the fourth suspect defend the first and the third one. But only one is telling the truth, thus, the first or the third one is the criminal. Both of them can be criminals, so the second and the fourth one can either be lying or telling the truth. The first and the third one are lying for sure as they are blaming the second and the fourth one.
Tags: constructive algorithms, data structures, implementation
Correct Solution:
```
n, m = map(int, input().split())
t = [int(input()) for i in range(n)]
s, p = 0, [0] * (n + 1)
for i in t:
if i < 0:
m -= 1
p[-i] -= 1
else: p[i] += 1
q = {i for i in range(1, n + 1) if p[i] == m}
if len(q) == 0: print('Not defined\n' * n)
elif len(q) == 1:
j = q.pop()
print('\n'.join(['Truth' if i == j or (i < 0 and i + j) else 'Lie' for i in t]))
else:
q.update({-i for i in q})
print('\n'.join(['Not defined' if i in q else ('Truth' if i < 0 else 'Lie') for i in t]))
```
| 98,868 |
Provide tags and a correct Python 3 solution for this coding contest problem.
As Sherlock Holmes was investigating a crime, he identified n suspects. He knows for sure that exactly one of them committed the crime. To find out which one did it, the detective lines up the suspects and numbered them from 1 to n. After that, he asked each one: "Which one committed the crime?". Suspect number i answered either "The crime was committed by suspect number ai", or "Suspect number ai didn't commit the crime". Also, the suspect could say so about himself (ai = i).
Sherlock Holmes understood for sure that exactly m answers were the truth and all other answers were a lie. Now help him understand this: which suspect lied and which one told the truth?
Input
The first line contains two integers n and m (1 ≤ n ≤ 105, 0 ≤ m ≤ n) — the total number of suspects and the number of suspects who told the truth. Next n lines contain the suspects' answers. The i-th line contains either "+ai" (without the quotes), if the suspect number i says that the crime was committed by suspect number ai, or "-ai" (without the quotes), if the suspect number i says that the suspect number ai didn't commit the crime (ai is an integer, 1 ≤ ai ≤ n).
It is guaranteed that at least one suspect exists, such that if he committed the crime, then exactly m people told the truth.
Output
Print n lines. Line number i should contain "Truth" if suspect number i has told the truth for sure. Print "Lie" if the suspect number i lied for sure and print "Not defined" if he could lie and could tell the truth, too, depending on who committed the crime.
Examples
Input
1 1
+1
Output
Truth
Input
3 2
-1
-2
-3
Output
Not defined
Not defined
Not defined
Input
4 1
+2
-3
+4
-1
Output
Lie
Not defined
Lie
Not defined
Note
The first sample has the single person and he confesses to the crime, and Sherlock Holmes knows that one person is telling the truth. That means that this person is telling the truth.
In the second sample there are three suspects and each one denies his guilt. Sherlock Holmes knows that only two of them are telling the truth. Any one of them can be the criminal, so we don't know for any of them, whether this person is telling the truth or not.
In the third sample the second and the fourth suspect defend the first and the third one. But only one is telling the truth, thus, the first or the third one is the criminal. Both of them can be criminals, so the second and the fourth one can either be lying or telling the truth. The first and the third one are lying for sure as they are blaming the second and the fourth one.
Tags: constructive algorithms, data structures, implementation
Correct Solution:
```
n,m=list(map(int,input().split()))
a=[[0,0] for i in range(n)]
b=[]
e=0
f=0
for i in range(n):
c=input()
d=int(c[1:])
if c[0]=='+':
a[d-1][0]+=1
e+=1
else:
a[d-1][1]+=1
f+=1
b.append([c[0],d])
g=[a[i][0]+f-a[i][1] for i in range(n)]
h=g.count(m)
for i in range(n):
d=b[i][1]
if b[i][0]=='+':
if g[d-1]==m:
if h>1:
print('Not defined')
else:
print('Truth')
else:
print('Lie')
else:
if h>1 or h==1 and g[d-1]!=m:
if g[d-1]==m:
print('Not defined')
else:
print('Truth')
else:
print('Lie')
```
| 98,869 |
Provide tags and a correct Python 3 solution for this coding contest problem.
As Sherlock Holmes was investigating a crime, he identified n suspects. He knows for sure that exactly one of them committed the crime. To find out which one did it, the detective lines up the suspects and numbered them from 1 to n. After that, he asked each one: "Which one committed the crime?". Suspect number i answered either "The crime was committed by suspect number ai", or "Suspect number ai didn't commit the crime". Also, the suspect could say so about himself (ai = i).
Sherlock Holmes understood for sure that exactly m answers were the truth and all other answers were a lie. Now help him understand this: which suspect lied and which one told the truth?
Input
The first line contains two integers n and m (1 ≤ n ≤ 105, 0 ≤ m ≤ n) — the total number of suspects and the number of suspects who told the truth. Next n lines contain the suspects' answers. The i-th line contains either "+ai" (without the quotes), if the suspect number i says that the crime was committed by suspect number ai, or "-ai" (without the quotes), if the suspect number i says that the suspect number ai didn't commit the crime (ai is an integer, 1 ≤ ai ≤ n).
It is guaranteed that at least one suspect exists, such that if he committed the crime, then exactly m people told the truth.
Output
Print n lines. Line number i should contain "Truth" if suspect number i has told the truth for sure. Print "Lie" if the suspect number i lied for sure and print "Not defined" if he could lie and could tell the truth, too, depending on who committed the crime.
Examples
Input
1 1
+1
Output
Truth
Input
3 2
-1
-2
-3
Output
Not defined
Not defined
Not defined
Input
4 1
+2
-3
+4
-1
Output
Lie
Not defined
Lie
Not defined
Note
The first sample has the single person and he confesses to the crime, and Sherlock Holmes knows that one person is telling the truth. That means that this person is telling the truth.
In the second sample there are three suspects and each one denies his guilt. Sherlock Holmes knows that only two of them are telling the truth. Any one of them can be the criminal, so we don't know for any of them, whether this person is telling the truth or not.
In the third sample the second and the fourth suspect defend the first and the third one. But only one is telling the truth, thus, the first or the third one is the criminal. Both of them can be criminals, so the second and the fourth one can either be lying or telling the truth. The first and the third one are lying for sure as they are blaming the second and the fourth one.
Tags: constructive algorithms, data structures, implementation
Correct Solution:
```
from sys import stdin, stdout
n, m = map(int, stdin.readline().split())
cntf = [0 for i in range(n + 1)]
cnts = [0 for i in range(n + 1)]
challengers = []
for i in range(n):
s = stdin.readline().strip()
challengers.append(s)
if s[0] == '+':
cntf[int(s[1:])] += 1
else:
cnts[int(s[1:])] += 1
first = sum(cntf)
second = sum(cnts)
ans = set()
for i in range(1, n + 1):
if cntf[i] + second - cnts[i] == m:
ans.add(i)
for i in range(n):
s = challengers[i]
if s[0] == '+':
if int(s[1:]) in ans:
if len(ans) > 1:
stdout.write('Not defined\n')
else:
stdout.write('Truth\n')
else:
stdout.write('Lie\n')
else:
if int(s[1:]) in ans:
if len(ans) > 1:
stdout.write('Not defined\n')
else:
stdout.write('Lie\n')
else:
stdout.write('Truth\n')
```
| 98,870 |
Provide tags and a correct Python 3 solution for this coding contest problem.
As Sherlock Holmes was investigating a crime, he identified n suspects. He knows for sure that exactly one of them committed the crime. To find out which one did it, the detective lines up the suspects and numbered them from 1 to n. After that, he asked each one: "Which one committed the crime?". Suspect number i answered either "The crime was committed by suspect number ai", or "Suspect number ai didn't commit the crime". Also, the suspect could say so about himself (ai = i).
Sherlock Holmes understood for sure that exactly m answers were the truth and all other answers were a lie. Now help him understand this: which suspect lied and which one told the truth?
Input
The first line contains two integers n and m (1 ≤ n ≤ 105, 0 ≤ m ≤ n) — the total number of suspects and the number of suspects who told the truth. Next n lines contain the suspects' answers. The i-th line contains either "+ai" (without the quotes), if the suspect number i says that the crime was committed by suspect number ai, or "-ai" (without the quotes), if the suspect number i says that the suspect number ai didn't commit the crime (ai is an integer, 1 ≤ ai ≤ n).
It is guaranteed that at least one suspect exists, such that if he committed the crime, then exactly m people told the truth.
Output
Print n lines. Line number i should contain "Truth" if suspect number i has told the truth for sure. Print "Lie" if the suspect number i lied for sure and print "Not defined" if he could lie and could tell the truth, too, depending on who committed the crime.
Examples
Input
1 1
+1
Output
Truth
Input
3 2
-1
-2
-3
Output
Not defined
Not defined
Not defined
Input
4 1
+2
-3
+4
-1
Output
Lie
Not defined
Lie
Not defined
Note
The first sample has the single person and he confesses to the crime, and Sherlock Holmes knows that one person is telling the truth. That means that this person is telling the truth.
In the second sample there are three suspects and each one denies his guilt. Sherlock Holmes knows that only two of them are telling the truth. Any one of them can be the criminal, so we don't know for any of them, whether this person is telling the truth or not.
In the third sample the second and the fourth suspect defend the first and the third one. But only one is telling the truth, thus, the first or the third one is the criminal. Both of them can be criminals, so the second and the fourth one can either be lying or telling the truth. The first and the third one are lying for sure as they are blaming the second and the fourth one.
Tags: constructive algorithms, data structures, implementation
Correct Solution:
```
from collections import Counter
n, m = map(int, input().split())
pank = 0
buben = []
bober = set()
for _ in range(n):
k = int(input())
buben.append(k)
if k < 0:
pank += 1
cnt = Counter(buben)
k = 0
for i in range(1, n + 1):
if cnt[i] + pank - cnt[-1 * i] == m:
bober.add(i)
k += 1
if k == 0:
print('Not defined\n' * n)
elif k == 1:
j = bober.pop()
for x in buben:
if x == j or (x < 0 and abs(x) != j):
print('Truth')
else:
print('Lie')
else:
bober.update({-x for x in bober})
for x in buben:
if x in bober:
print('Not defined')
elif x < 0:
print('Truth')
else:
print('Lie')
```
| 98,871 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
As Sherlock Holmes was investigating a crime, he identified n suspects. He knows for sure that exactly one of them committed the crime. To find out which one did it, the detective lines up the suspects and numbered them from 1 to n. After that, he asked each one: "Which one committed the crime?". Suspect number i answered either "The crime was committed by suspect number ai", or "Suspect number ai didn't commit the crime". Also, the suspect could say so about himself (ai = i).
Sherlock Holmes understood for sure that exactly m answers were the truth and all other answers were a lie. Now help him understand this: which suspect lied and which one told the truth?
Input
The first line contains two integers n and m (1 ≤ n ≤ 105, 0 ≤ m ≤ n) — the total number of suspects and the number of suspects who told the truth. Next n lines contain the suspects' answers. The i-th line contains either "+ai" (without the quotes), if the suspect number i says that the crime was committed by suspect number ai, or "-ai" (without the quotes), if the suspect number i says that the suspect number ai didn't commit the crime (ai is an integer, 1 ≤ ai ≤ n).
It is guaranteed that at least one suspect exists, such that if he committed the crime, then exactly m people told the truth.
Output
Print n lines. Line number i should contain "Truth" if suspect number i has told the truth for sure. Print "Lie" if the suspect number i lied for sure and print "Not defined" if he could lie and could tell the truth, too, depending on who committed the crime.
Examples
Input
1 1
+1
Output
Truth
Input
3 2
-1
-2
-3
Output
Not defined
Not defined
Not defined
Input
4 1
+2
-3
+4
-1
Output
Lie
Not defined
Lie
Not defined
Note
The first sample has the single person and he confesses to the crime, and Sherlock Holmes knows that one person is telling the truth. That means that this person is telling the truth.
In the second sample there are three suspects and each one denies his guilt. Sherlock Holmes knows that only two of them are telling the truth. Any one of them can be the criminal, so we don't know for any of them, whether this person is telling the truth or not.
In the third sample the second and the fourth suspect defend the first and the third one. But only one is telling the truth, thus, the first or the third one is the criminal. Both of them can be criminals, so the second and the fourth one can either be lying or telling the truth. The first and the third one are lying for sure as they are blaming the second and the fourth one.
Submitted Solution:
```
"""
Brandt Smith, Lemuel Gorion and Peter Haddad
codeforces.com
Problem 156B
"""
import sys
n, m = map(int, input().split(' '))
inp = []
guess = [0] * (n + 1)
for i in range(n):
temp = int(input())
inp.append(temp)
if temp < 0:
m -= 1
guess[-temp] -= 1
else:
guess[temp] += 1
dic = {temp for temp in range(1, n + 1) if guess[temp] == m}
if len(dic) == 0:
for i in range(n):
print('Not defined')
elif len(dic) == 1:
temp = dic.pop()
for i in inp:
if i == temp or (i < 0 and i + temp):
print('Truth')
else:
print('Lie')
else:
temp = dic.update({-i for i in dic})
for i in inp:
if i in dic:
print('Not defined')
else:
if i < 0:
print('Truth')
else:
print('Lie')
```
Yes
| 98,872 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
As Sherlock Holmes was investigating a crime, he identified n suspects. He knows for sure that exactly one of them committed the crime. To find out which one did it, the detective lines up the suspects and numbered them from 1 to n. After that, he asked each one: "Which one committed the crime?". Suspect number i answered either "The crime was committed by suspect number ai", or "Suspect number ai didn't commit the crime". Also, the suspect could say so about himself (ai = i).
Sherlock Holmes understood for sure that exactly m answers were the truth and all other answers were a lie. Now help him understand this: which suspect lied and which one told the truth?
Input
The first line contains two integers n and m (1 ≤ n ≤ 105, 0 ≤ m ≤ n) — the total number of suspects and the number of suspects who told the truth. Next n lines contain the suspects' answers. The i-th line contains either "+ai" (without the quotes), if the suspect number i says that the crime was committed by suspect number ai, or "-ai" (without the quotes), if the suspect number i says that the suspect number ai didn't commit the crime (ai is an integer, 1 ≤ ai ≤ n).
It is guaranteed that at least one suspect exists, such that if he committed the crime, then exactly m people told the truth.
Output
Print n lines. Line number i should contain "Truth" if suspect number i has told the truth for sure. Print "Lie" if the suspect number i lied for sure and print "Not defined" if he could lie and could tell the truth, too, depending on who committed the crime.
Examples
Input
1 1
+1
Output
Truth
Input
3 2
-1
-2
-3
Output
Not defined
Not defined
Not defined
Input
4 1
+2
-3
+4
-1
Output
Lie
Not defined
Lie
Not defined
Note
The first sample has the single person and he confesses to the crime, and Sherlock Holmes knows that one person is telling the truth. That means that this person is telling the truth.
In the second sample there are three suspects and each one denies his guilt. Sherlock Holmes knows that only two of them are telling the truth. Any one of them can be the criminal, so we don't know for any of them, whether this person is telling the truth or not.
In the third sample the second and the fourth suspect defend the first and the third one. But only one is telling the truth, thus, the first or the third one is the criminal. Both of them can be criminals, so the second and the fourth one can either be lying or telling the truth. The first and the third one are lying for sure as they are blaming the second and the fourth one.
Submitted Solution:
```
from collections import defaultdict
n, m = map(int, input().split())
a = []
for i in range(n):
a.append(int(input()))
d = defaultdict(int)
pos, neg = 0, 0
for x in a:
d[x] += 1
if x > 0:
pos += 1
else:
neg += 1
possible = [False] * n
for i in range(1, n + 1):
t = d[i] + neg - d[-i]
if t == m:
possible[i - 1] = True
cnt = sum(possible)
for i in range(n):
if cnt == 0:
print('Lie')
continue
if a[i] > 0:
if possible[a[i] - 1]:
print('Truth' if cnt == 1 else 'Not defined')
else:
print('Lie')
else:
if not possible[-a[i] - 1]:
print('Truth')
else:
print('Lie' if cnt == 1 else 'Not defined')
```
Yes
| 98,873 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
As Sherlock Holmes was investigating a crime, he identified n suspects. He knows for sure that exactly one of them committed the crime. To find out which one did it, the detective lines up the suspects and numbered them from 1 to n. After that, he asked each one: "Which one committed the crime?". Suspect number i answered either "The crime was committed by suspect number ai", or "Suspect number ai didn't commit the crime". Also, the suspect could say so about himself (ai = i).
Sherlock Holmes understood for sure that exactly m answers were the truth and all other answers were a lie. Now help him understand this: which suspect lied and which one told the truth?
Input
The first line contains two integers n and m (1 ≤ n ≤ 105, 0 ≤ m ≤ n) — the total number of suspects and the number of suspects who told the truth. Next n lines contain the suspects' answers. The i-th line contains either "+ai" (without the quotes), if the suspect number i says that the crime was committed by suspect number ai, or "-ai" (without the quotes), if the suspect number i says that the suspect number ai didn't commit the crime (ai is an integer, 1 ≤ ai ≤ n).
It is guaranteed that at least one suspect exists, such that if he committed the crime, then exactly m people told the truth.
Output
Print n lines. Line number i should contain "Truth" if suspect number i has told the truth for sure. Print "Lie" if the suspect number i lied for sure and print "Not defined" if he could lie and could tell the truth, too, depending on who committed the crime.
Examples
Input
1 1
+1
Output
Truth
Input
3 2
-1
-2
-3
Output
Not defined
Not defined
Not defined
Input
4 1
+2
-3
+4
-1
Output
Lie
Not defined
Lie
Not defined
Note
The first sample has the single person and he confesses to the crime, and Sherlock Holmes knows that one person is telling the truth. That means that this person is telling the truth.
In the second sample there are three suspects and each one denies his guilt. Sherlock Holmes knows that only two of them are telling the truth. Any one of them can be the criminal, so we don't know for any of them, whether this person is telling the truth or not.
In the third sample the second and the fourth suspect defend the first and the third one. But only one is telling the truth, thus, the first or the third one is the criminal. Both of them can be criminals, so the second and the fourth one can either be lying or telling the truth. The first and the third one are lying for sure as they are blaming the second and the fourth one.
Submitted Solution:
```
"""
Brandt Smith, Lemuel Gorion and Peter Haddad
codeforces.com
Problem 156B
"""
import sys
n, m = map(int, input().split(' '))
inp = []
guess = [0] * (n + 1)
for i in range(n):
temp = int(input())
inp.append(temp)
if temp < 0:
m -= 1
guess[-temp] -= 1
else:
guess[temp] += 1
dic = {temp for temp in range(1, n + 1) if guess[temp] == m}
if len(dic) == 0:
for i in range(n):
print('Not defined')
elif len(dic) == 1:
temp = dic.pop()
for i in inp:
if i == temp or (i < 0 and i + temp):
print('Truth')
else:
print('Lie')
else:
temp = dic.update({-i for i in dic})
for i in inp:
if i in dic:
print('Not defined')
else:
if i < 0:
print('Truth')
else:
print('Lie')
# Made By Mostafa_Khaled
```
Yes
| 98,874 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
As Sherlock Holmes was investigating a crime, he identified n suspects. He knows for sure that exactly one of them committed the crime. To find out which one did it, the detective lines up the suspects and numbered them from 1 to n. After that, he asked each one: "Which one committed the crime?". Suspect number i answered either "The crime was committed by suspect number ai", or "Suspect number ai didn't commit the crime". Also, the suspect could say so about himself (ai = i).
Sherlock Holmes understood for sure that exactly m answers were the truth and all other answers were a lie. Now help him understand this: which suspect lied and which one told the truth?
Input
The first line contains two integers n and m (1 ≤ n ≤ 105, 0 ≤ m ≤ n) — the total number of suspects and the number of suspects who told the truth. Next n lines contain the suspects' answers. The i-th line contains either "+ai" (without the quotes), if the suspect number i says that the crime was committed by suspect number ai, or "-ai" (without the quotes), if the suspect number i says that the suspect number ai didn't commit the crime (ai is an integer, 1 ≤ ai ≤ n).
It is guaranteed that at least one suspect exists, such that if he committed the crime, then exactly m people told the truth.
Output
Print n lines. Line number i should contain "Truth" if suspect number i has told the truth for sure. Print "Lie" if the suspect number i lied for sure and print "Not defined" if he could lie and could tell the truth, too, depending on who committed the crime.
Examples
Input
1 1
+1
Output
Truth
Input
3 2
-1
-2
-3
Output
Not defined
Not defined
Not defined
Input
4 1
+2
-3
+4
-1
Output
Lie
Not defined
Lie
Not defined
Note
The first sample has the single person and he confesses to the crime, and Sherlock Holmes knows that one person is telling the truth. That means that this person is telling the truth.
In the second sample there are three suspects and each one denies his guilt. Sherlock Holmes knows that only two of them are telling the truth. Any one of them can be the criminal, so we don't know for any of them, whether this person is telling the truth or not.
In the third sample the second and the fourth suspect defend the first and the third one. But only one is telling the truth, thus, the first or the third one is the criminal. Both of them can be criminals, so the second and the fourth one can either be lying or telling the truth. The first and the third one are lying for sure as they are blaming the second and the fourth one.
Submitted Solution:
```
n,m=list(map(int,input().split()))
a=[[0,0] for i in range(n)]
b=[]
e=0
f=0
for i in range(n):
c=input()
d=int(c[1:])
if c[0]=='+':
a[d-1][0]+=1
e+=1
else:
a[d-1][1]+=1
f+=1
b.append([c[0],d])
g=[a[int(b[i][1])-1][0]+f-a[int(b[i][1])-1][1] for i in range(n)]
h=g.count(m)
for i in range(n):
d=b[i][1]
if b[i][0]=='+':
if g[i]==m:
if h>1:
print('Not defined')
else:
print('Truth')
else:
print('Lie')
else:
if h>1 or h==1 and g[i]!=m:
if g[i]==m:
print('Not defined')
else:
print('Truth')
else:
print('Lie')
```
No
| 98,875 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
As Sherlock Holmes was investigating a crime, he identified n suspects. He knows for sure that exactly one of them committed the crime. To find out which one did it, the detective lines up the suspects and numbered them from 1 to n. After that, he asked each one: "Which one committed the crime?". Suspect number i answered either "The crime was committed by suspect number ai", or "Suspect number ai didn't commit the crime". Also, the suspect could say so about himself (ai = i).
Sherlock Holmes understood for sure that exactly m answers were the truth and all other answers were a lie. Now help him understand this: which suspect lied and which one told the truth?
Input
The first line contains two integers n and m (1 ≤ n ≤ 105, 0 ≤ m ≤ n) — the total number of suspects and the number of suspects who told the truth. Next n lines contain the suspects' answers. The i-th line contains either "+ai" (without the quotes), if the suspect number i says that the crime was committed by suspect number ai, or "-ai" (without the quotes), if the suspect number i says that the suspect number ai didn't commit the crime (ai is an integer, 1 ≤ ai ≤ n).
It is guaranteed that at least one suspect exists, such that if he committed the crime, then exactly m people told the truth.
Output
Print n lines. Line number i should contain "Truth" if suspect number i has told the truth for sure. Print "Lie" if the suspect number i lied for sure and print "Not defined" if he could lie and could tell the truth, too, depending on who committed the crime.
Examples
Input
1 1
+1
Output
Truth
Input
3 2
-1
-2
-3
Output
Not defined
Not defined
Not defined
Input
4 1
+2
-3
+4
-1
Output
Lie
Not defined
Lie
Not defined
Note
The first sample has the single person and he confesses to the crime, and Sherlock Holmes knows that one person is telling the truth. That means that this person is telling the truth.
In the second sample there are three suspects and each one denies his guilt. Sherlock Holmes knows that only two of them are telling the truth. Any one of them can be the criminal, so we don't know for any of them, whether this person is telling the truth or not.
In the third sample the second and the fourth suspect defend the first and the third one. But only one is telling the truth, thus, the first or the third one is the criminal. Both of them can be criminals, so the second and the fourth one can either be lying or telling the truth. The first and the third one are lying for sure as they are blaming the second and the fourth one.
Submitted Solution:
```
a,b=map(int,input().split())
r=[0]*a
c1=[0]*(a+1)
c2=[0]*(a+1)
f=0
for i in range(a):
n=int(input())
r[i]=n
if n>0:c1[n]+=1
else:
f+=1
c2[n]+=1
possible=[False]*(a+1)
np=0#number of suspects
for i in range(1,a+1):
if c1[i]+f-c2[i]==b:
#f-c2[i] is truth
possible[i]=True
np+=1
for i in range(a):
if r[i]>0:#he said +
if possible[r[i]] and np==1:#unique
print("Truth")
elif not possible[r[i]]:#Lie
print("Lie")
else:
print("Not defined")
else:#claims r[i] is not
r[i]*=-1
if possible[r[i]] and np==1:print("Lie")
elif not possible[r[i]]:print("Truth")
else:print("Not defined")
# print (possible)
```
No
| 98,876 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
As Sherlock Holmes was investigating a crime, he identified n suspects. He knows for sure that exactly one of them committed the crime. To find out which one did it, the detective lines up the suspects and numbered them from 1 to n. After that, he asked each one: "Which one committed the crime?". Suspect number i answered either "The crime was committed by suspect number ai", or "Suspect number ai didn't commit the crime". Also, the suspect could say so about himself (ai = i).
Sherlock Holmes understood for sure that exactly m answers were the truth and all other answers were a lie. Now help him understand this: which suspect lied and which one told the truth?
Input
The first line contains two integers n and m (1 ≤ n ≤ 105, 0 ≤ m ≤ n) — the total number of suspects and the number of suspects who told the truth. Next n lines contain the suspects' answers. The i-th line contains either "+ai" (without the quotes), if the suspect number i says that the crime was committed by suspect number ai, or "-ai" (without the quotes), if the suspect number i says that the suspect number ai didn't commit the crime (ai is an integer, 1 ≤ ai ≤ n).
It is guaranteed that at least one suspect exists, such that if he committed the crime, then exactly m people told the truth.
Output
Print n lines. Line number i should contain "Truth" if suspect number i has told the truth for sure. Print "Lie" if the suspect number i lied for sure and print "Not defined" if he could lie and could tell the truth, too, depending on who committed the crime.
Examples
Input
1 1
+1
Output
Truth
Input
3 2
-1
-2
-3
Output
Not defined
Not defined
Not defined
Input
4 1
+2
-3
+4
-1
Output
Lie
Not defined
Lie
Not defined
Note
The first sample has the single person and he confesses to the crime, and Sherlock Holmes knows that one person is telling the truth. That means that this person is telling the truth.
In the second sample there are three suspects and each one denies his guilt. Sherlock Holmes knows that only two of them are telling the truth. Any one of them can be the criminal, so we don't know for any of them, whether this person is telling the truth or not.
In the third sample the second and the fourth suspect defend the first and the third one. But only one is telling the truth, thus, the first or the third one is the criminal. Both of them can be criminals, so the second and the fourth one can either be lying or telling the truth. The first and the third one are lying for sure as they are blaming the second and the fourth one.
Submitted Solution:
```
n,m=map(int,input().split())
c=[0]*(n+1)
dc=[0]*(n+1)
sumdc=0
sus=0
mark=[]
suspect=[]
for _ in range(n):
k=int(input())
mark.append(k)
if k>0:
c[k]+=1
else:
dc[abs(k)]+=1
sumdc+=1
for i in range(1,n+1): #假定每一个I 为罪犯
if c[i]+sumdc-dc[i]==m:
suspect.append(i)
if len(suspect)==1:
for x in mark:
print ("Truth") if x==suspect[0] else print("Lie")
else:
for x in mark:
if x>0:
print("Not defined") if suspect.count(x)>0 else print("Lie")
else:
if suspect.count(abs(x))>0:
print("Not defined")
elif suspect.count(abs(x))==0:
print("Truth")
```
No
| 98,877 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
As Sherlock Holmes was investigating a crime, he identified n suspects. He knows for sure that exactly one of them committed the crime. To find out which one did it, the detective lines up the suspects and numbered them from 1 to n. After that, he asked each one: "Which one committed the crime?". Suspect number i answered either "The crime was committed by suspect number ai", or "Suspect number ai didn't commit the crime". Also, the suspect could say so about himself (ai = i).
Sherlock Holmes understood for sure that exactly m answers were the truth and all other answers were a lie. Now help him understand this: which suspect lied and which one told the truth?
Input
The first line contains two integers n and m (1 ≤ n ≤ 105, 0 ≤ m ≤ n) — the total number of suspects and the number of suspects who told the truth. Next n lines contain the suspects' answers. The i-th line contains either "+ai" (without the quotes), if the suspect number i says that the crime was committed by suspect number ai, or "-ai" (without the quotes), if the suspect number i says that the suspect number ai didn't commit the crime (ai is an integer, 1 ≤ ai ≤ n).
It is guaranteed that at least one suspect exists, such that if he committed the crime, then exactly m people told the truth.
Output
Print n lines. Line number i should contain "Truth" if suspect number i has told the truth for sure. Print "Lie" if the suspect number i lied for sure and print "Not defined" if he could lie and could tell the truth, too, depending on who committed the crime.
Examples
Input
1 1
+1
Output
Truth
Input
3 2
-1
-2
-3
Output
Not defined
Not defined
Not defined
Input
4 1
+2
-3
+4
-1
Output
Lie
Not defined
Lie
Not defined
Note
The first sample has the single person and he confesses to the crime, and Sherlock Holmes knows that one person is telling the truth. That means that this person is telling the truth.
In the second sample there are three suspects and each one denies his guilt. Sherlock Holmes knows that only two of them are telling the truth. Any one of them can be the criminal, so we don't know for any of them, whether this person is telling the truth or not.
In the third sample the second and the fourth suspect defend the first and the third one. But only one is telling the truth, thus, the first or the third one is the criminal. Both of them can be criminals, so the second and the fourth one can either be lying or telling the truth. The first and the third one are lying for sure as they are blaming the second and the fourth one.
Submitted Solution:
```
n,m=map(int,input().split())
c=[0]*(n+1)
dc=[0]*(n+1)
sumdc=0
sus=0
mark=[]
suspect=[]
for _ in range(n):
k=int(input())
mark.append(k)
if k>0:
c[k]+=1
else:
dc[abs(k)]+=1
sumdc+=1
for i in range(1,n+1): #假定每一个I 为罪犯
if c[i]+sumdc-dc[i]==m:
suspect.append(i)
#print(suspect)
if len(suspect)==1:
for x in mark:
if x>0:
print ("Truth") if x==suspect[0] else print("Lie")
else:
print ("Lie") if abs(x)==suspect[0] else print("True")
else:
for x in mark:
if x>0:
print("Not defined") if suspect.count(x)>0 else print("Lie")
else:
if suspect.count(abs(x))>0:
print("Not defined")
elif suspect.count(abs(x))==0:
print("Truth")
```
No
| 98,878 |
Provide a correct Python 3 solution for this coding contest problem.
The Smart Beaver from ABBYY was offered a job of a screenwriter for the ongoing TV series. In particular, he needs to automate the hard decision: which main characters will get married by the end of the series.
There are n single men and n single women among the main characters. An opinion poll showed that viewers like several couples, and a marriage of any of them will make the audience happy. The Smart Beaver formalized this fact as k triples of numbers (h, w, r), where h is the index of the man, w is the index of the woman, and r is the measure of the audience's delight in case of the marriage of this couple. The same poll showed that the marriage of any other couple will leave the audience indifferent, so the screenwriters decided not to include any such marriages in the plot.
The script allows you to arrange several marriages between the heroes or not to arrange marriages at all. A subset of some of the k marriages is considered acceptable if each man and each woman is involved in at most one marriage of the subset (the series won't allow any divorces). The value of the acceptable set of marriages is the total delight the spectators will get from the marriages included in this set.
Obviously, there is a finite number of acceptable sets, and they all describe some variants of the script. The screenwriters do not want to choose a set with maximum value — it would make the plot too predictable. So the Smart Beaver offers the following option: sort all the acceptable sets in increasing order of value and choose the t-th set from the sorted list. Thus, t = 1 corresponds to a plot without marriages, t = 2 — to a single marriage resulting in minimal delight for the audience, and so on.
Help the Beaver to implement the algorithm for selecting the desired set.
Input
The first input line contains integers n, k and t (1 ≤ k ≤ min(100, n2), 1 ≤ t ≤ 2·105), separated by single spaces. Next k lines contain triples of integers (h, w, r) (1 ≤ h, w ≤ n; 1 ≤ r ≤ 1000), separated by single spaces, which describe the possible marriages. It is guaranteed that the input data is correct: t doesn't exceed the total number of acceptable sets, and each pair (h, w) is present in at most one triple.
The input limitations for getting 30 points are:
* 1 ≤ n ≤ 5
The input limitations for getting 100 points are:
* 1 ≤ n ≤ 20
Output
Print a single number — the value of the t-th acceptable variant.
Examples
Input
2 4 3
1 1 1
1 2 2
2 1 3
2 2 7
Output
2
Input
2 4 7
1 1 1
1 2 2
2 1 3
2 2 7
Output
8
Note
The figure shows 7 acceptable sets of marriages that exist in the first sample.
<image>
"Correct Solution:
```
n, k, t = map(int, input().split())
l = (1 << (n + 1)) - 1
p = [tuple(map(int, input().split())) for i in range(k)]
p.sort(key = lambda i: i[2])
a, b = -1, sum(i[2] for i in p)
def g(j, s, u, v):
global k, p, t
k += 1
if k > t: return
for i, (h, w, r) in enumerate(p[j:], j + 1):
if r > s: break
if ((u >> h) & 1) and ((v >> w) & 1):
g(i, s - r, u - (1 << h), v - (1 << w))
def f(s):
global k, l
k = 0
g(0, s, l, l)
return k
while b - a > 1:
c = (a + b) // 2
if f(c) < t: a = c
else: b = c
print(b)
```
| 98,879 |
Provide a correct Python 3 solution for this coding contest problem.
The Smart Beaver from ABBYY was offered a job of a screenwriter for the ongoing TV series. In particular, he needs to automate the hard decision: which main characters will get married by the end of the series.
There are n single men and n single women among the main characters. An opinion poll showed that viewers like several couples, and a marriage of any of them will make the audience happy. The Smart Beaver formalized this fact as k triples of numbers (h, w, r), where h is the index of the man, w is the index of the woman, and r is the measure of the audience's delight in case of the marriage of this couple. The same poll showed that the marriage of any other couple will leave the audience indifferent, so the screenwriters decided not to include any such marriages in the plot.
The script allows you to arrange several marriages between the heroes or not to arrange marriages at all. A subset of some of the k marriages is considered acceptable if each man and each woman is involved in at most one marriage of the subset (the series won't allow any divorces). The value of the acceptable set of marriages is the total delight the spectators will get from the marriages included in this set.
Obviously, there is a finite number of acceptable sets, and they all describe some variants of the script. The screenwriters do not want to choose a set with maximum value — it would make the plot too predictable. So the Smart Beaver offers the following option: sort all the acceptable sets in increasing order of value and choose the t-th set from the sorted list. Thus, t = 1 corresponds to a plot without marriages, t = 2 — to a single marriage resulting in minimal delight for the audience, and so on.
Help the Beaver to implement the algorithm for selecting the desired set.
Input
The first input line contains integers n, k and t (1 ≤ k ≤ min(100, n2), 1 ≤ t ≤ 2·105), separated by single spaces. Next k lines contain triples of integers (h, w, r) (1 ≤ h, w ≤ n; 1 ≤ r ≤ 1000), separated by single spaces, which describe the possible marriages. It is guaranteed that the input data is correct: t doesn't exceed the total number of acceptable sets, and each pair (h, w) is present in at most one triple.
The input limitations for getting 30 points are:
* 1 ≤ n ≤ 5
The input limitations for getting 100 points are:
* 1 ≤ n ≤ 20
Output
Print a single number — the value of the t-th acceptable variant.
Examples
Input
2 4 3
1 1 1
1 2 2
2 1 3
2 2 7
Output
2
Input
2 4 7
1 1 1
1 2 2
2 1 3
2 2 7
Output
8
Note
The figure shows 7 acceptable sets of marriages that exist in the first sample.
<image>
"Correct Solution:
```
def g(t, d):
j = len(d) - 1
i = len(t) - 1
while j >= 0:
while i >= 0 and t[i] > d[j]: i -= 1
t.insert(i + 1, d[j])
j -= 1
n, k, t = map(int, input().split())
p = [0] * k
for i in range(k):
h, w, r = map(int, input().split())
p[i] = (r, w, h)
p.sort()
q, a, b = [0], [0] * (n + 1), [0] * (n + 1)
def f(i, s):
global p, q, a, b, k, t
r = [(p[j][1], p[j][2], s + p[j][0], j + 1) for j in range(i, k) if a[p[j][1]] == b[p[j][2]] == 0]
g(q, [s[2] for s in r])
if len(q) > t: q = q[:t]
for h, w, s, j in r:
if len(q) == t and s >= q[t - 1]: break
a[h], b[w] = 1, 1
f(j, s)
a[h], b[w] = 0, 0
f(0, 0)
print(q[t - 1])
```
| 98,880 |
Provide a correct Python 3 solution for this coding contest problem.
The Smart Beaver from ABBYY was offered a job of a screenwriter for the ongoing TV series. In particular, he needs to automate the hard decision: which main characters will get married by the end of the series.
There are n single men and n single women among the main characters. An opinion poll showed that viewers like several couples, and a marriage of any of them will make the audience happy. The Smart Beaver formalized this fact as k triples of numbers (h, w, r), where h is the index of the man, w is the index of the woman, and r is the measure of the audience's delight in case of the marriage of this couple. The same poll showed that the marriage of any other couple will leave the audience indifferent, so the screenwriters decided not to include any such marriages in the plot.
The script allows you to arrange several marriages between the heroes or not to arrange marriages at all. A subset of some of the k marriages is considered acceptable if each man and each woman is involved in at most one marriage of the subset (the series won't allow any divorces). The value of the acceptable set of marriages is the total delight the spectators will get from the marriages included in this set.
Obviously, there is a finite number of acceptable sets, and they all describe some variants of the script. The screenwriters do not want to choose a set with maximum value — it would make the plot too predictable. So the Smart Beaver offers the following option: sort all the acceptable sets in increasing order of value and choose the t-th set from the sorted list. Thus, t = 1 corresponds to a plot without marriages, t = 2 — to a single marriage resulting in minimal delight for the audience, and so on.
Help the Beaver to implement the algorithm for selecting the desired set.
Input
The first input line contains integers n, k and t (1 ≤ k ≤ min(100, n2), 1 ≤ t ≤ 2·105), separated by single spaces. Next k lines contain triples of integers (h, w, r) (1 ≤ h, w ≤ n; 1 ≤ r ≤ 1000), separated by single spaces, which describe the possible marriages. It is guaranteed that the input data is correct: t doesn't exceed the total number of acceptable sets, and each pair (h, w) is present in at most one triple.
The input limitations for getting 30 points are:
* 1 ≤ n ≤ 5
The input limitations for getting 100 points are:
* 1 ≤ n ≤ 20
Output
Print a single number — the value of the t-th acceptable variant.
Examples
Input
2 4 3
1 1 1
1 2 2
2 1 3
2 2 7
Output
2
Input
2 4 7
1 1 1
1 2 2
2 1 3
2 2 7
Output
8
Note
The figure shows 7 acceptable sets of marriages that exist in the first sample.
<image>
"Correct Solution:
```
n, k, t = map(int, input().split())
p = [0] * k
for i in range(k):
h, w, r = map(int, input().split())
p[i] = (r, w, h)
p.sort()
q, a, b = [], [0] * (n + 1), [0] * (n + 1)
def f(i, s):
global p, q, a, b, k
q.append(s)
while i < k:
h, w = p[i][1], p[i][2]
if a[h] == b[w] == 0:
a[h], b[w] = 1, 1
f(i + 1, s + p[i][0])
a[h], b[w] = 0, 0
i += 1
f(0, 0)
q.sort()
print(q[t - 1])
```
| 98,881 |
Provide a correct Python 3 solution for this coding contest problem.
The Smart Beaver from ABBYY was offered a job of a screenwriter for the ongoing TV series. In particular, he needs to automate the hard decision: which main characters will get married by the end of the series.
There are n single men and n single women among the main characters. An opinion poll showed that viewers like several couples, and a marriage of any of them will make the audience happy. The Smart Beaver formalized this fact as k triples of numbers (h, w, r), where h is the index of the man, w is the index of the woman, and r is the measure of the audience's delight in case of the marriage of this couple. The same poll showed that the marriage of any other couple will leave the audience indifferent, so the screenwriters decided not to include any such marriages in the plot.
The script allows you to arrange several marriages between the heroes or not to arrange marriages at all. A subset of some of the k marriages is considered acceptable if each man and each woman is involved in at most one marriage of the subset (the series won't allow any divorces). The value of the acceptable set of marriages is the total delight the spectators will get from the marriages included in this set.
Obviously, there is a finite number of acceptable sets, and they all describe some variants of the script. The screenwriters do not want to choose a set with maximum value — it would make the plot too predictable. So the Smart Beaver offers the following option: sort all the acceptable sets in increasing order of value and choose the t-th set from the sorted list. Thus, t = 1 corresponds to a plot without marriages, t = 2 — to a single marriage resulting in minimal delight for the audience, and so on.
Help the Beaver to implement the algorithm for selecting the desired set.
Input
The first input line contains integers n, k and t (1 ≤ k ≤ min(100, n2), 1 ≤ t ≤ 2·105), separated by single spaces. Next k lines contain triples of integers (h, w, r) (1 ≤ h, w ≤ n; 1 ≤ r ≤ 1000), separated by single spaces, which describe the possible marriages. It is guaranteed that the input data is correct: t doesn't exceed the total number of acceptable sets, and each pair (h, w) is present in at most one triple.
The input limitations for getting 30 points are:
* 1 ≤ n ≤ 5
The input limitations for getting 100 points are:
* 1 ≤ n ≤ 20
Output
Print a single number — the value of the t-th acceptable variant.
Examples
Input
2 4 3
1 1 1
1 2 2
2 1 3
2 2 7
Output
2
Input
2 4 7
1 1 1
1 2 2
2 1 3
2 2 7
Output
8
Note
The figure shows 7 acceptable sets of marriages that exist in the first sample.
<image>
"Correct Solution:
```
I=lambda:list(map(int,input().split()))
n,k,T=I()
t=[I()for _ in '0'*k]
def b(h,w,r,a):
if h>n:a+=[r]
else:
b(h+1,w,r,a)
for f,s,v in t:
if f==h and s in w:b(h+1,w-set([s]),r+v,a)
return a
print(sorted(b(1,set(range(1,n+1)), 0,[]))[T-1])
```
| 98,882 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
The Smart Beaver from ABBYY was offered a job of a screenwriter for the ongoing TV series. In particular, he needs to automate the hard decision: which main characters will get married by the end of the series.
There are n single men and n single women among the main characters. An opinion poll showed that viewers like several couples, and a marriage of any of them will make the audience happy. The Smart Beaver formalized this fact as k triples of numbers (h, w, r), where h is the index of the man, w is the index of the woman, and r is the measure of the audience's delight in case of the marriage of this couple. The same poll showed that the marriage of any other couple will leave the audience indifferent, so the screenwriters decided not to include any such marriages in the plot.
The script allows you to arrange several marriages between the heroes or not to arrange marriages at all. A subset of some of the k marriages is considered acceptable if each man and each woman is involved in at most one marriage of the subset (the series won't allow any divorces). The value of the acceptable set of marriages is the total delight the spectators will get from the marriages included in this set.
Obviously, there is a finite number of acceptable sets, and they all describe some variants of the script. The screenwriters do not want to choose a set with maximum value — it would make the plot too predictable. So the Smart Beaver offers the following option: sort all the acceptable sets in increasing order of value and choose the t-th set from the sorted list. Thus, t = 1 corresponds to a plot without marriages, t = 2 — to a single marriage resulting in minimal delight for the audience, and so on.
Help the Beaver to implement the algorithm for selecting the desired set.
Input
The first input line contains integers n, k and t (1 ≤ k ≤ min(100, n2), 1 ≤ t ≤ 2·105), separated by single spaces. Next k lines contain triples of integers (h, w, r) (1 ≤ h, w ≤ n; 1 ≤ r ≤ 1000), separated by single spaces, which describe the possible marriages. It is guaranteed that the input data is correct: t doesn't exceed the total number of acceptable sets, and each pair (h, w) is present in at most one triple.
The input limitations for getting 30 points are:
* 1 ≤ n ≤ 5
The input limitations for getting 100 points are:
* 1 ≤ n ≤ 20
Output
Print a single number — the value of the t-th acceptable variant.
Examples
Input
2 4 3
1 1 1
1 2 2
2 1 3
2 2 7
Output
2
Input
2 4 7
1 1 1
1 2 2
2 1 3
2 2 7
Output
8
Note
The figure shows 7 acceptable sets of marriages that exist in the first sample.
<image>
Submitted Solution:
```
n, k, t = map(int, input().split())
p = [0] * k
for i in range(k):
h, w, r = map(int, input().split())
p[i] = (r, w, h)
p.sort()
q, a, b = [0], [0] * (n + 1), [0] * (n + 1)
def f(i, s):
global p, q, a, b, k, t
r = [(p[j][1], p[j][2], s + p[j][0], j + 1) for j in range(i, k) if a[p[j][1]] == b[p[j][2]] == 0]
print(r)
q += [s[2] for s in r]
if len(q) > t:
q.sort()
q = q[:t]
for h, w, s, j in r:
if len(q) == t and s >= q[t - 1]: break
a[h], b[w] = 1, 1
f(j, s)
a[h], b[w] = 0, 0
f(0, 0)
q.sort()
print(q[t - 1])
```
No
| 98,883 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
The Smart Beaver from ABBYY was offered a job of a screenwriter for the ongoing TV series. In particular, he needs to automate the hard decision: which main characters will get married by the end of the series.
There are n single men and n single women among the main characters. An opinion poll showed that viewers like several couples, and a marriage of any of them will make the audience happy. The Smart Beaver formalized this fact as k triples of numbers (h, w, r), where h is the index of the man, w is the index of the woman, and r is the measure of the audience's delight in case of the marriage of this couple. The same poll showed that the marriage of any other couple will leave the audience indifferent, so the screenwriters decided not to include any such marriages in the plot.
The script allows you to arrange several marriages between the heroes or not to arrange marriages at all. A subset of some of the k marriages is considered acceptable if each man and each woman is involved in at most one marriage of the subset (the series won't allow any divorces). The value of the acceptable set of marriages is the total delight the spectators will get from the marriages included in this set.
Obviously, there is a finite number of acceptable sets, and they all describe some variants of the script. The screenwriters do not want to choose a set with maximum value — it would make the plot too predictable. So the Smart Beaver offers the following option: sort all the acceptable sets in increasing order of value and choose the t-th set from the sorted list. Thus, t = 1 corresponds to a plot without marriages, t = 2 — to a single marriage resulting in minimal delight for the audience, and so on.
Help the Beaver to implement the algorithm for selecting the desired set.
Input
The first input line contains integers n, k and t (1 ≤ k ≤ min(100, n2), 1 ≤ t ≤ 2·105), separated by single spaces. Next k lines contain triples of integers (h, w, r) (1 ≤ h, w ≤ n; 1 ≤ r ≤ 1000), separated by single spaces, which describe the possible marriages. It is guaranteed that the input data is correct: t doesn't exceed the total number of acceptable sets, and each pair (h, w) is present in at most one triple.
The input limitations for getting 30 points are:
* 1 ≤ n ≤ 5
The input limitations for getting 100 points are:
* 1 ≤ n ≤ 20
Output
Print a single number — the value of the t-th acceptable variant.
Examples
Input
2 4 3
1 1 1
1 2 2
2 1 3
2 2 7
Output
2
Input
2 4 7
1 1 1
1 2 2
2 1 3
2 2 7
Output
8
Note
The figure shows 7 acceptable sets of marriages that exist in the first sample.
<image>
Submitted Solution:
```
n, k, t = map(int, input().split())
p = [0] * k
for i in range(k):
h, w, r = map(int, input().split())
p[i] = (r, w, h)
p.sort()
q, g, a, b = [0], 10 ** 10, [0] * (n + 1), [0] * (n + 1)
def f(i, s):
global p, q, a, b, k, t, g
r = [(p[j][1], p[j][2], s + p[j][0]) for j in range(i, k) if a[p[j][1]] == b[p[j][2]] == 0]
q += [s[2] for s in r]
i += 1
if len(q) > t:
q.sort()
g = q[t - 1] - 1
for h, w, s in r:
if s > g: break
a[h], b[w] = 1, 1
f(i, s)
a[h], b[w] = 0, 0
f(0, 0)
print(g + 1)
```
No
| 98,884 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
The Smart Beaver from ABBYY was offered a job of a screenwriter for the ongoing TV series. In particular, he needs to automate the hard decision: which main characters will get married by the end of the series.
There are n single men and n single women among the main characters. An opinion poll showed that viewers like several couples, and a marriage of any of them will make the audience happy. The Smart Beaver formalized this fact as k triples of numbers (h, w, r), where h is the index of the man, w is the index of the woman, and r is the measure of the audience's delight in case of the marriage of this couple. The same poll showed that the marriage of any other couple will leave the audience indifferent, so the screenwriters decided not to include any such marriages in the plot.
The script allows you to arrange several marriages between the heroes or not to arrange marriages at all. A subset of some of the k marriages is considered acceptable if each man and each woman is involved in at most one marriage of the subset (the series won't allow any divorces). The value of the acceptable set of marriages is the total delight the spectators will get from the marriages included in this set.
Obviously, there is a finite number of acceptable sets, and they all describe some variants of the script. The screenwriters do not want to choose a set with maximum value — it would make the plot too predictable. So the Smart Beaver offers the following option: sort all the acceptable sets in increasing order of value and choose the t-th set from the sorted list. Thus, t = 1 corresponds to a plot without marriages, t = 2 — to a single marriage resulting in minimal delight for the audience, and so on.
Help the Beaver to implement the algorithm for selecting the desired set.
Input
The first input line contains integers n, k and t (1 ≤ k ≤ min(100, n2), 1 ≤ t ≤ 2·105), separated by single spaces. Next k lines contain triples of integers (h, w, r) (1 ≤ h, w ≤ n; 1 ≤ r ≤ 1000), separated by single spaces, which describe the possible marriages. It is guaranteed that the input data is correct: t doesn't exceed the total number of acceptable sets, and each pair (h, w) is present in at most one triple.
The input limitations for getting 30 points are:
* 1 ≤ n ≤ 5
The input limitations for getting 100 points are:
* 1 ≤ n ≤ 20
Output
Print a single number — the value of the t-th acceptable variant.
Examples
Input
2 4 3
1 1 1
1 2 2
2 1 3
2 2 7
Output
2
Input
2 4 7
1 1 1
1 2 2
2 1 3
2 2 7
Output
8
Note
The figure shows 7 acceptable sets of marriages that exist in the first sample.
<image>
Submitted Solution:
```
n, k, t = map(int, input().split())
l = (1 << (n + 1)) - 1
p = [tuple(map(int, input().split())) for i in range(k)]
p.sort(key = lambda i: i[2])
a, b = 0, sum(i[2] for i in p)
def g(j, s, u, v):
global k, p, t
k += 1
if k > t: return
for i, (h, w, r) in enumerate(p[j:], j + 1):
if r > s: break
if ((u >> h) & 1) and ((v >> w) & 1):
g(i, s - r, u - (1 << h), v - (1 << w))
def f(s):
global k, l
k = 0
g(0, s, l, l)
return k
while b - a > 1:
c = (a + b) // 2
if f(c) < t: a = c
else: b = c
print(b)
```
No
| 98,885 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Qwerty the Ranger took up a government job and arrived on planet Mars. He should stay in the secret lab and conduct some experiments on bacteria that have funny and abnormal properties. The job isn't difficult, but the salary is high.
At the beginning of the first experiment there is a single bacterium in the test tube. Every second each bacterium in the test tube divides itself into k bacteria. After that some abnormal effects create b more bacteria in the test tube. Thus, if at the beginning of some second the test tube had x bacteria, then at the end of the second it will have kx + b bacteria.
The experiment showed that after n seconds there were exactly z bacteria and the experiment ended at this point.
For the second experiment Qwerty is going to sterilize the test tube and put there t bacteria. He hasn't started the experiment yet but he already wonders, how many seconds he will need to grow at least z bacteria. The ranger thinks that the bacteria will divide by the same rule as in the first experiment.
Help Qwerty and find the minimum number of seconds needed to get a tube with at least z bacteria in the second experiment.
Input
The first line contains four space-separated integers k, b, n and t (1 ≤ k, b, n, t ≤ 106) — the parameters of bacterial growth, the time Qwerty needed to grow z bacteria in the first experiment and the initial number of bacteria in the second experiment, correspondingly.
Output
Print a single number — the minimum number of seconds Qwerty needs to grow at least z bacteria in the tube.
Examples
Input
3 1 3 5
Output
2
Input
1 4 4 7
Output
3
Input
2 2 4 100
Output
0
Tags: implementation, math
Correct Solution:
```
import math
kbnt=list(map(int,input().split()))
k=kbnt[0]
b=kbnt[1]
n=kbnt[2]
t=kbnt[3]
z=(k+b-1)/((k-1)*t+b)
if(k!=1):
new_n=n+math.log(z,k)
else:
new_n=n-(t-1)/b
print(max(0,math.ceil(new_n)))
```
| 98,886 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Qwerty the Ranger took up a government job and arrived on planet Mars. He should stay in the secret lab and conduct some experiments on bacteria that have funny and abnormal properties. The job isn't difficult, but the salary is high.
At the beginning of the first experiment there is a single bacterium in the test tube. Every second each bacterium in the test tube divides itself into k bacteria. After that some abnormal effects create b more bacteria in the test tube. Thus, if at the beginning of some second the test tube had x bacteria, then at the end of the second it will have kx + b bacteria.
The experiment showed that after n seconds there were exactly z bacteria and the experiment ended at this point.
For the second experiment Qwerty is going to sterilize the test tube and put there t bacteria. He hasn't started the experiment yet but he already wonders, how many seconds he will need to grow at least z bacteria. The ranger thinks that the bacteria will divide by the same rule as in the first experiment.
Help Qwerty and find the minimum number of seconds needed to get a tube with at least z bacteria in the second experiment.
Input
The first line contains four space-separated integers k, b, n and t (1 ≤ k, b, n, t ≤ 106) — the parameters of bacterial growth, the time Qwerty needed to grow z bacteria in the first experiment and the initial number of bacteria in the second experiment, correspondingly.
Output
Print a single number — the minimum number of seconds Qwerty needs to grow at least z bacteria in the tube.
Examples
Input
3 1 3 5
Output
2
Input
1 4 4 7
Output
3
Input
2 2 4 100
Output
0
Tags: implementation, math
Correct Solution:
```
#testing minimario's code in pypy
import math
k, b, n, t = map(int, input().split(' '))
if k == 1:
cur = n*b+1
print(max(math.ceil((-t + n*b + 1)/b), 0))
else:
god = (b+k-1)/(b+t*k-t)
m = 0
while not (k**(m-n) >= god):
m += 1
print(m)
```
| 98,887 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Qwerty the Ranger took up a government job and arrived on planet Mars. He should stay in the secret lab and conduct some experiments on bacteria that have funny and abnormal properties. The job isn't difficult, but the salary is high.
At the beginning of the first experiment there is a single bacterium in the test tube. Every second each bacterium in the test tube divides itself into k bacteria. After that some abnormal effects create b more bacteria in the test tube. Thus, if at the beginning of some second the test tube had x bacteria, then at the end of the second it will have kx + b bacteria.
The experiment showed that after n seconds there were exactly z bacteria and the experiment ended at this point.
For the second experiment Qwerty is going to sterilize the test tube and put there t bacteria. He hasn't started the experiment yet but he already wonders, how many seconds he will need to grow at least z bacteria. The ranger thinks that the bacteria will divide by the same rule as in the first experiment.
Help Qwerty and find the minimum number of seconds needed to get a tube with at least z bacteria in the second experiment.
Input
The first line contains four space-separated integers k, b, n and t (1 ≤ k, b, n, t ≤ 106) — the parameters of bacterial growth, the time Qwerty needed to grow z bacteria in the first experiment and the initial number of bacteria in the second experiment, correspondingly.
Output
Print a single number — the minimum number of seconds Qwerty needs to grow at least z bacteria in the tube.
Examples
Input
3 1 3 5
Output
2
Input
1 4 4 7
Output
3
Input
2 2 4 100
Output
0
Tags: implementation, math
Correct Solution:
```
#------------------------template--------------------------#
import os
import sys
from math import *
from collections import *
from fractions import *
from bisect import *
from heapq import*
from io import BytesIO, IOBase
def vsInput():
sys.stdin = open('input.txt', 'r')
sys.stdout = open('output.txt', 'w')
BUFSIZE = 8192
class FastIO(IOBase):
newlines = 0
def __init__(self, file):
self._fd = file.fileno()
self.buffer = BytesIO()
self.writable = "x" in file.mode or "r" not in file.mode
self.write = self.buffer.write if self.writable else None
def read(self):
while True:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
if not b:
break
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines = 0
return self.buffer.read()
def readline(self):
while self.newlines == 0:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
self.newlines = b.count(b"\n") + (not b)
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines -= 1
return self.buffer.readline()
def flush(self):
if self.writable:
os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
self.write = lambda s: self.buffer.write(s.encode("ascii"))
self.read = lambda: self.buffer.read().decode("ascii")
self.readline = lambda: self.buffer.readline().decode("ascii")
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")
ALPHA='abcdefghijklmnopqrstuvwxyz'
M=10**9+7
def value():return tuple(map(int,input().split()))
def array():return [int(i) for i in input().split()]
def Int():return int(input())
def Str():return input()
def arrayS():return [i for i in input().split()]
#-------------------------code---------------------------#
# vsInput()
def canMake(iter):
pass
k,b,n,t=value()
z=1
reduced=0
# for i in range(n):
# z=z*k+b
# print(z)
# print()
# z=t
# for i in range(n):
# z=z*k+b
# print(z)
while(z<=t):
reduced+=1
z=z*k+b
print(max(0,n-reduced+1))
```
| 98,888 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Qwerty the Ranger took up a government job and arrived on planet Mars. He should stay in the secret lab and conduct some experiments on bacteria that have funny and abnormal properties. The job isn't difficult, but the salary is high.
At the beginning of the first experiment there is a single bacterium in the test tube. Every second each bacterium in the test tube divides itself into k bacteria. After that some abnormal effects create b more bacteria in the test tube. Thus, if at the beginning of some second the test tube had x bacteria, then at the end of the second it will have kx + b bacteria.
The experiment showed that after n seconds there were exactly z bacteria and the experiment ended at this point.
For the second experiment Qwerty is going to sterilize the test tube and put there t bacteria. He hasn't started the experiment yet but he already wonders, how many seconds he will need to grow at least z bacteria. The ranger thinks that the bacteria will divide by the same rule as in the first experiment.
Help Qwerty and find the minimum number of seconds needed to get a tube with at least z bacteria in the second experiment.
Input
The first line contains four space-separated integers k, b, n and t (1 ≤ k, b, n, t ≤ 106) — the parameters of bacterial growth, the time Qwerty needed to grow z bacteria in the first experiment and the initial number of bacteria in the second experiment, correspondingly.
Output
Print a single number — the minimum number of seconds Qwerty needs to grow at least z bacteria in the tube.
Examples
Input
3 1 3 5
Output
2
Input
1 4 4 7
Output
3
Input
2 2 4 100
Output
0
Tags: implementation, math
Correct Solution:
```
k, b, n, t = map(int, input().split())
p = 1
while p <= t:
p = k * p + b
n -= 1
print(max(0, n+1))
```
| 98,889 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Qwerty the Ranger took up a government job and arrived on planet Mars. He should stay in the secret lab and conduct some experiments on bacteria that have funny and abnormal properties. The job isn't difficult, but the salary is high.
At the beginning of the first experiment there is a single bacterium in the test tube. Every second each bacterium in the test tube divides itself into k bacteria. After that some abnormal effects create b more bacteria in the test tube. Thus, if at the beginning of some second the test tube had x bacteria, then at the end of the second it will have kx + b bacteria.
The experiment showed that after n seconds there were exactly z bacteria and the experiment ended at this point.
For the second experiment Qwerty is going to sterilize the test tube and put there t bacteria. He hasn't started the experiment yet but he already wonders, how many seconds he will need to grow at least z bacteria. The ranger thinks that the bacteria will divide by the same rule as in the first experiment.
Help Qwerty and find the minimum number of seconds needed to get a tube with at least z bacteria in the second experiment.
Input
The first line contains four space-separated integers k, b, n and t (1 ≤ k, b, n, t ≤ 106) — the parameters of bacterial growth, the time Qwerty needed to grow z bacteria in the first experiment and the initial number of bacteria in the second experiment, correspondingly.
Output
Print a single number — the minimum number of seconds Qwerty needs to grow at least z bacteria in the tube.
Examples
Input
3 1 3 5
Output
2
Input
1 4 4 7
Output
3
Input
2 2 4 100
Output
0
Tags: implementation, math
Correct Solution:
```
k, b, n, t = map(int, input().split())
z = int(1)
res = int(n)
for i in range(n):
z = k*z + b
if z <= t:
res -= 1
else:
break
print(res)
# Made By Mostafa_Khaled
```
| 98,890 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Qwerty the Ranger took up a government job and arrived on planet Mars. He should stay in the secret lab and conduct some experiments on bacteria that have funny and abnormal properties. The job isn't difficult, but the salary is high.
At the beginning of the first experiment there is a single bacterium in the test tube. Every second each bacterium in the test tube divides itself into k bacteria. After that some abnormal effects create b more bacteria in the test tube. Thus, if at the beginning of some second the test tube had x bacteria, then at the end of the second it will have kx + b bacteria.
The experiment showed that after n seconds there were exactly z bacteria and the experiment ended at this point.
For the second experiment Qwerty is going to sterilize the test tube and put there t bacteria. He hasn't started the experiment yet but he already wonders, how many seconds he will need to grow at least z bacteria. The ranger thinks that the bacteria will divide by the same rule as in the first experiment.
Help Qwerty and find the minimum number of seconds needed to get a tube with at least z bacteria in the second experiment.
Input
The first line contains four space-separated integers k, b, n and t (1 ≤ k, b, n, t ≤ 106) — the parameters of bacterial growth, the time Qwerty needed to grow z bacteria in the first experiment and the initial number of bacteria in the second experiment, correspondingly.
Output
Print a single number — the minimum number of seconds Qwerty needs to grow at least z bacteria in the tube.
Examples
Input
3 1 3 5
Output
2
Input
1 4 4 7
Output
3
Input
2 2 4 100
Output
0
Tags: implementation, math
Correct Solution:
```
import math
k, b, n, t = map(int, input().split(' '))
if k == 1:
cur = n*b+1
print(max(math.ceil((-t + n*b + 1)/b), 0))
else:
god = (b+k-1)/(b+t*k-t)
m = 0
while not (k**(m-n) >= god):
m += 1
print(m)
```
| 98,891 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Qwerty the Ranger took up a government job and arrived on planet Mars. He should stay in the secret lab and conduct some experiments on bacteria that have funny and abnormal properties. The job isn't difficult, but the salary is high.
At the beginning of the first experiment there is a single bacterium in the test tube. Every second each bacterium in the test tube divides itself into k bacteria. After that some abnormal effects create b more bacteria in the test tube. Thus, if at the beginning of some second the test tube had x bacteria, then at the end of the second it will have kx + b bacteria.
The experiment showed that after n seconds there were exactly z bacteria and the experiment ended at this point.
For the second experiment Qwerty is going to sterilize the test tube and put there t bacteria. He hasn't started the experiment yet but he already wonders, how many seconds he will need to grow at least z bacteria. The ranger thinks that the bacteria will divide by the same rule as in the first experiment.
Help Qwerty and find the minimum number of seconds needed to get a tube with at least z bacteria in the second experiment.
Input
The first line contains four space-separated integers k, b, n and t (1 ≤ k, b, n, t ≤ 106) — the parameters of bacterial growth, the time Qwerty needed to grow z bacteria in the first experiment and the initial number of bacteria in the second experiment, correspondingly.
Output
Print a single number — the minimum number of seconds Qwerty needs to grow at least z bacteria in the tube.
Examples
Input
3 1 3 5
Output
2
Input
1 4 4 7
Output
3
Input
2 2 4 100
Output
0
Tags: implementation, math
Correct Solution:
```
#from sys import stdin, stdout
#import math
#k, b, n, t = list(map(int, input().split()))
#
#if k == 1000000 and n == 1000000 and t ==1:
# x = 1000000
#elif k == 185 and b == 58 and n == 579474 and t == 889969:
# x = 579472
#elif k == 9821 and b == 62 and n == 965712 and t == 703044:
# x = 965711
#elif k == 78993 and b == 99 and n == 646044 and t == 456226:
# x = 646043
#elif k == 193877 and b == 3 and n == 362586 and t == 6779:
# x = 362586
#elif k == 702841 and b == 39 and n == 622448 and t == 218727:
# x = 622448
#elif k == 987899 and b == 74 and n== 490126 and t == 87643:
# x = 490126
#elif k == 1000000 and b == 69 and n == 296123 and t == 144040:
# x = 296123
#elif k == t:
# x = n
#elif k != 1:
# if -1*(k+b-1) * (1-k**n) // (t*k+b-t) <= 0:
# x = 0
# else:
# x = math.ceil(math.log(-1*(k+b-1) * (1-k**n) // (t*k+b-t), k))
# if x == 0:
# x = 1
#else:
# if n == 1 and t >= b+1:
# x = 0
# elif n == 1:
# if t >= n:
# x = 1
# elif t < b:
# x = n
# elif b == 1047 and n == 230 and t == 1199:
# x = 229
# elif t > b and t != 1000000:
# xd = (t -1) // b
# x = n - xd
# else:
# x = math.ceil(((b * n + 1) - (t - k)) / n)
# if x < 0:
# x = 0
#stdout.write(str(x)+'\n')
k, b, n, t = map(int, input().split())
p = 1
while p <= t:
p = k * p + b
n -= 1
print(max(0, n+1))
```
| 98,892 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Qwerty the Ranger took up a government job and arrived on planet Mars. He should stay in the secret lab and conduct some experiments on bacteria that have funny and abnormal properties. The job isn't difficult, but the salary is high.
At the beginning of the first experiment there is a single bacterium in the test tube. Every second each bacterium in the test tube divides itself into k bacteria. After that some abnormal effects create b more bacteria in the test tube. Thus, if at the beginning of some second the test tube had x bacteria, then at the end of the second it will have kx + b bacteria.
The experiment showed that after n seconds there were exactly z bacteria and the experiment ended at this point.
For the second experiment Qwerty is going to sterilize the test tube and put there t bacteria. He hasn't started the experiment yet but he already wonders, how many seconds he will need to grow at least z bacteria. The ranger thinks that the bacteria will divide by the same rule as in the first experiment.
Help Qwerty and find the minimum number of seconds needed to get a tube with at least z bacteria in the second experiment.
Input
The first line contains four space-separated integers k, b, n and t (1 ≤ k, b, n, t ≤ 106) — the parameters of bacterial growth, the time Qwerty needed to grow z bacteria in the first experiment and the initial number of bacteria in the second experiment, correspondingly.
Output
Print a single number — the minimum number of seconds Qwerty needs to grow at least z bacteria in the tube.
Examples
Input
3 1 3 5
Output
2
Input
1 4 4 7
Output
3
Input
2 2 4 100
Output
0
Tags: implementation, math
Correct Solution:
```
from sys import stdin, stdout
k, b, n, t = map(int, stdin.readline().split())
cnt = 1
day = 0
while cnt < t:
cnt = cnt * k + b
day += 1
if day > n:
stdout.write('0')
elif cnt == t:
stdout.write(str(n - day))
else:
stdout.write(str(n - day + 1))
```
| 98,893 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Qwerty the Ranger took up a government job and arrived on planet Mars. He should stay in the secret lab and conduct some experiments on bacteria that have funny and abnormal properties. The job isn't difficult, but the salary is high.
At the beginning of the first experiment there is a single bacterium in the test tube. Every second each bacterium in the test tube divides itself into k bacteria. After that some abnormal effects create b more bacteria in the test tube. Thus, if at the beginning of some second the test tube had x bacteria, then at the end of the second it will have kx + b bacteria.
The experiment showed that after n seconds there were exactly z bacteria and the experiment ended at this point.
For the second experiment Qwerty is going to sterilize the test tube and put there t bacteria. He hasn't started the experiment yet but he already wonders, how many seconds he will need to grow at least z bacteria. The ranger thinks that the bacteria will divide by the same rule as in the first experiment.
Help Qwerty and find the minimum number of seconds needed to get a tube with at least z bacteria in the second experiment.
Input
The first line contains four space-separated integers k, b, n and t (1 ≤ k, b, n, t ≤ 106) — the parameters of bacterial growth, the time Qwerty needed to grow z bacteria in the first experiment and the initial number of bacteria in the second experiment, correspondingly.
Output
Print a single number — the minimum number of seconds Qwerty needs to grow at least z bacteria in the tube.
Examples
Input
3 1 3 5
Output
2
Input
1 4 4 7
Output
3
Input
2 2 4 100
Output
0
Submitted Solution:
```
k, b, n, t = list(map(int, input().split()))
p = 1
while p <= t:
p = k * p + b
n -= 1
print(max(0, n+1))
```
Yes
| 98,894 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Qwerty the Ranger took up a government job and arrived on planet Mars. He should stay in the secret lab and conduct some experiments on bacteria that have funny and abnormal properties. The job isn't difficult, but the salary is high.
At the beginning of the first experiment there is a single bacterium in the test tube. Every second each bacterium in the test tube divides itself into k bacteria. After that some abnormal effects create b more bacteria in the test tube. Thus, if at the beginning of some second the test tube had x bacteria, then at the end of the second it will have kx + b bacteria.
The experiment showed that after n seconds there were exactly z bacteria and the experiment ended at this point.
For the second experiment Qwerty is going to sterilize the test tube and put there t bacteria. He hasn't started the experiment yet but he already wonders, how many seconds he will need to grow at least z bacteria. The ranger thinks that the bacteria will divide by the same rule as in the first experiment.
Help Qwerty and find the minimum number of seconds needed to get a tube with at least z bacteria in the second experiment.
Input
The first line contains four space-separated integers k, b, n and t (1 ≤ k, b, n, t ≤ 106) — the parameters of bacterial growth, the time Qwerty needed to grow z bacteria in the first experiment and the initial number of bacteria in the second experiment, correspondingly.
Output
Print a single number — the minimum number of seconds Qwerty needs to grow at least z bacteria in the tube.
Examples
Input
3 1 3 5
Output
2
Input
1 4 4 7
Output
3
Input
2 2 4 100
Output
0
Submitted Solution:
```
from sys import stdin, stdout
import math
k, b, n, t = list(map(int, input().split()))
if k == n:
x = k // t
elif k == t:
x = n
elif k != 1:
if -1*(k+b-1) * (1-k**n) // (t*k+b-t) <= 0:
x = 0
else:
x = math.ceil(math.log(-1*(k+b-1) * (1-k**n) // (t*k+b-t), k))
else:
if n == 1 and t >= b+1:
x = 0
elif n == 1:
if t >= n:
x = 1
else:
x = math.ceil(((b * n + 1) - (t - k)) / n)
if x < 0:
x = 0
stdout.write(str(x)+'\n')
## k b n t
## 847 374 283 485756
#pierwsza = 1
#roznica = pierwsza*k + b - pierwsza
#if k != 1:
# pierwsza = roznica * ((1 - k**n) / (1 - k)) + 1
#else:
# pierwsza = 1+n*b
#
#sekundy = 0
#druga = t
#while druga < pierwsza:
# druga = druga*k + b
# sekundy = sekundy + 1
#stdout.write(str(sekundy))#+'\n')
#
#
#
#pierwsza = 1
#roznica = pierwsza*k + b - pierwsza
#if k != 1:
# pierwsza = roznica * ((1 - k**n) / (1 - k)) + 1
#else:
# pierwsza = 1+n*b
#roznica = t*k + b - t
#druga = roznica * ((1 - k**(n)) / (1 - k)) + t
#x = math.ceil(math.log(-1*(k+b-1)*(1-k**n) / (t*k+b-t), k))
#y = -1*(k+b-1)*(1-k**n) / (t*k+b-t) #log(z czego, o podstawie) log(8, 2) = 3
# k b n t
# 1 4 4 7
#print(x)
```
No
| 98,895 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Qwerty the Ranger took up a government job and arrived on planet Mars. He should stay in the secret lab and conduct some experiments on bacteria that have funny and abnormal properties. The job isn't difficult, but the salary is high.
At the beginning of the first experiment there is a single bacterium in the test tube. Every second each bacterium in the test tube divides itself into k bacteria. After that some abnormal effects create b more bacteria in the test tube. Thus, if at the beginning of some second the test tube had x bacteria, then at the end of the second it will have kx + b bacteria.
The experiment showed that after n seconds there were exactly z bacteria and the experiment ended at this point.
For the second experiment Qwerty is going to sterilize the test tube and put there t bacteria. He hasn't started the experiment yet but he already wonders, how many seconds he will need to grow at least z bacteria. The ranger thinks that the bacteria will divide by the same rule as in the first experiment.
Help Qwerty and find the minimum number of seconds needed to get a tube with at least z bacteria in the second experiment.
Input
The first line contains four space-separated integers k, b, n and t (1 ≤ k, b, n, t ≤ 106) — the parameters of bacterial growth, the time Qwerty needed to grow z bacteria in the first experiment and the initial number of bacteria in the second experiment, correspondingly.
Output
Print a single number — the minimum number of seconds Qwerty needs to grow at least z bacteria in the tube.
Examples
Input
3 1 3 5
Output
2
Input
1 4 4 7
Output
3
Input
2 2 4 100
Output
0
Submitted Solution:
```
from sys import stdin, stdout
import math
k, b, n, t = list(map(int, input().split()))
if k == 1000000 and n == 1000000 and t ==1:
x = 1000000
elif k == t:
x = n
elif k != 1:
if -1*(k+b-1) * (1-k**n) // (t*k+b-t) <= 0:
x = 0
else:
x = math.ceil(math.log(-1*(k+b-1) * (1-k**n) // (t*k+b-t), k))
else:
if n == 1 and t >= b+1:
x = 0
elif n == 1:
if t >= n:
x = 1
elif t < b:
x = n
elif t > b:
xd = t // b
x = n - xd
else:
x = math.ceil(((b * n + 1) - (t - k)) / n)
if x < 0:
x = 0
stdout.write(str(x)+'\n')
## k b n t
## 847 374 283 485756
#pierwsza = 1
#roznica = pierwsza*k + b - pierwsza
#if k != 1:
# pierwsza = roznica * ((1 - k**n) / (1 - k)) + 1
#else:
# pierwsza = 1+n*b
#
#sekundy = 0
#druga = t
#while druga < pierwsza:
# druga = druga*k + b
# sekundy = sekundy + 1
#stdout.write(str(sekundy))#+'\n')
#
#
#
#pierwsza = 1
#roznica = pierwsza*k + b - pierwsza
#if k != 1:
# pierwsza = roznica * ((1 - k**n) / (1 - k)) + 1
#else:
# pierwsza = 1+n*b
#roznica = t*k + b - t
#druga = roznica * ((1 - k**(n)) / (1 - k)) + t
#x = math.ceil(math.log(-1*(k+b-1)*(1-k**n) / (t*k+b-t), k))
#y = -1*(k+b-1)*(1-k**n) / (t*k+b-t) #log(z czego, o podstawie) log(8, 2) = 3
# k b n t
# 1 4 4 7
#print(x)
```
No
| 98,896 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Qwerty the Ranger took up a government job and arrived on planet Mars. He should stay in the secret lab and conduct some experiments on bacteria that have funny and abnormal properties. The job isn't difficult, but the salary is high.
At the beginning of the first experiment there is a single bacterium in the test tube. Every second each bacterium in the test tube divides itself into k bacteria. After that some abnormal effects create b more bacteria in the test tube. Thus, if at the beginning of some second the test tube had x bacteria, then at the end of the second it will have kx + b bacteria.
The experiment showed that after n seconds there were exactly z bacteria and the experiment ended at this point.
For the second experiment Qwerty is going to sterilize the test tube and put there t bacteria. He hasn't started the experiment yet but he already wonders, how many seconds he will need to grow at least z bacteria. The ranger thinks that the bacteria will divide by the same rule as in the first experiment.
Help Qwerty and find the minimum number of seconds needed to get a tube with at least z bacteria in the second experiment.
Input
The first line contains four space-separated integers k, b, n and t (1 ≤ k, b, n, t ≤ 106) — the parameters of bacterial growth, the time Qwerty needed to grow z bacteria in the first experiment and the initial number of bacteria in the second experiment, correspondingly.
Output
Print a single number — the minimum number of seconds Qwerty needs to grow at least z bacteria in the tube.
Examples
Input
3 1 3 5
Output
2
Input
1 4 4 7
Output
3
Input
2 2 4 100
Output
0
Submitted Solution:
```
from sys import stdin, stdout
import math
k, b, n, t = list(map(int, input().split()))
if k == 1000000 and n == 1000000 and t ==1:
x = 1000000
elif k == t:
x = n
elif k != 1:
if -1*(k+b-1) * (1-k**n) // (t*k+b-t) <= 0:
x = 0
else:
x = math.ceil(math.log(-1*(k+b-1) * (1-k**n) // (t*k+b-t), k))
else:
if n == 1 and t >= b+1:
x = 0
elif n == 1:
if t >= n:
x = 1
elif t < b:
x = n
else:
x = math.ceil(((b * n + 1) - (t - k)) / n)
if x < 0:
x = 0
stdout.write(str(x)+'\n')
## k b n t
## 847 374 283 485756
#pierwsza = 1
#roznica = pierwsza*k + b - pierwsza
#if k != 1:
# pierwsza = roznica * ((1 - k**n) / (1 - k)) + 1
#else:
# pierwsza = 1+n*b
#
#sekundy = 0
#druga = t
#while druga < pierwsza:
# druga = druga*k + b
# sekundy = sekundy + 1
#stdout.write(str(sekundy))#+'\n')
#
#
#
#pierwsza = 1
#roznica = pierwsza*k + b - pierwsza
#if k != 1:
# pierwsza = roznica * ((1 - k**n) / (1 - k)) + 1
#else:
# pierwsza = 1+n*b
#roznica = t*k + b - t
#druga = roznica * ((1 - k**(n)) / (1 - k)) + t
#x = math.ceil(math.log(-1*(k+b-1)*(1-k**n) / (t*k+b-t), k))
#y = -1*(k+b-1)*(1-k**n) / (t*k+b-t) #log(z czego, o podstawie) log(8, 2) = 3
# k b n t
# 1 4 4 7
#print(x)
```
No
| 98,897 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Qwerty the Ranger took up a government job and arrived on planet Mars. He should stay in the secret lab and conduct some experiments on bacteria that have funny and abnormal properties. The job isn't difficult, but the salary is high.
At the beginning of the first experiment there is a single bacterium in the test tube. Every second each bacterium in the test tube divides itself into k bacteria. After that some abnormal effects create b more bacteria in the test tube. Thus, if at the beginning of some second the test tube had x bacteria, then at the end of the second it will have kx + b bacteria.
The experiment showed that after n seconds there were exactly z bacteria and the experiment ended at this point.
For the second experiment Qwerty is going to sterilize the test tube and put there t bacteria. He hasn't started the experiment yet but he already wonders, how many seconds he will need to grow at least z bacteria. The ranger thinks that the bacteria will divide by the same rule as in the first experiment.
Help Qwerty and find the minimum number of seconds needed to get a tube with at least z bacteria in the second experiment.
Input
The first line contains four space-separated integers k, b, n and t (1 ≤ k, b, n, t ≤ 106) — the parameters of bacterial growth, the time Qwerty needed to grow z bacteria in the first experiment and the initial number of bacteria in the second experiment, correspondingly.
Output
Print a single number — the minimum number of seconds Qwerty needs to grow at least z bacteria in the tube.
Examples
Input
3 1 3 5
Output
2
Input
1 4 4 7
Output
3
Input
2 2 4 100
Output
0
Submitted Solution:
```
import math
kbnt=list(map(int,input().split()))
k=kbnt[0]
b=kbnt[1]
n=kbnt[2]
t=kbnt[3]
z=(k**n)*(k+b-1)/((k-1)*t+b)
if(k!=1):
new_n=math.log(z,k)
else:
new_n=n-(t-1)/b
print(max(0,math.ceil(new_n)))
```
No
| 98,898 |
Provide tags and a correct Python 3 solution for this coding contest problem.
A boy named Vasya has taken part in an Olympiad. His teacher knows that in total Vasya got at least x points for both tours of the Olympiad. The teacher has the results of the first and the second tour of the Olympiad but the problem is, the results have only points, no names. The teacher has to know Vasya's chances.
Help Vasya's teacher, find two numbers — the best and the worst place Vasya could have won. Note that the total results' table sorts the participants by the sum of points for both tours (the first place has the participant who has got the most points). If two or more participants have got the same number of points, it's up to the jury to assign places to them according to their choice. It is guaranteed that each participant of the Olympiad participated in both tours of the Olympiad.
Input
The first line contains two space-separated integers n, x (1 ≤ n ≤ 105; 0 ≤ x ≤ 2·105) — the number of Olympiad participants and the minimum number of points Vasya earned.
The second line contains n space-separated integers: a1, a2, ..., an (0 ≤ ai ≤ 105) — the participants' points in the first tour.
The third line contains n space-separated integers: b1, b2, ..., bn (0 ≤ bi ≤ 105) — the participants' points in the second tour.
The participants' points are given in the arbitrary order. It is guaranteed that Vasya was present in the Olympiad — there are two integers i, j (1 ≤ i, j ≤ n) such, that ai + bj ≥ x.
Output
Print two space-separated integers — the best and the worst place Vasya could have got on the Olympiad.
Examples
Input
5 2
1 1 1 1 1
1 1 1 1 1
Output
1 5
Input
6 7
4 3 5 6 4 4
8 6 0 4 3 4
Output
1 5
Note
In the first text sample all 5 participants earn 2 points each in any case. Depending on the jury's decision, Vasya can get the first (the best) as well as the last (the worst) fifth place.
In the second test sample in the best case scenario Vasya wins again: he can win 12 points and become the absolute winner if the total results' table looks like that — {4:8, 6:4, 3:6, 4:4, 4:3, 5:0}.
In this table all participants are sorted by decreasing points and we can see how much a participant earned in the first and in the second tour.
In the worst case scenario Vasya can get the fifth place if the table looks like that — {4:8, 4:6, 6:4, 5:4, 4:3, 3:0}, and he earned 4 and 3 points in the first and second tours, correspondingly.
Tags: binary search, greedy, sortings, two pointers
Correct Solution:
```
n, s = map(int, input().split())
a = sorted(map(int, input().split()), reverse = True)
b = sorted(map(int, input().split()))
i = j = 0
while i < n and j < n:
if a[i] + b[j] < s: j += 1
else:
i += 1
j += 1
print(1, i)
```
| 98,899 |
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