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Provide a correct Python 3 solution for this coding contest problem.
C: Skewering
problem
One day, when Homura was playing with blocks, Tempura came. Homura decided to play with blocks with Tempura.
There is a rectangular parallelepiped of A \ times B \ times C, which is made by stacking A \ times B \ times C blocks of cubic blocks with a side length of 1 without any gaps. Each side of all cubes and rectangular parallelepipeds is parallel to the x-axis, y-axis, or z-axis.
Homura-chan and Tempura-kun alternately repeat the following operations.
* Select a row of blocks of building blocks lined up in a row from a rectangular parallelepiped in any of the vertical, horizontal, and depth directions, and paint all the blocks included in the row in red. However, you cannot select columns that contain blocks that are already painted red.
More precisely
* Select one of the blocks contained in the rectangular parallelepiped and one of the three directions x, y, and z.
* When the selected block is moved in the selected direction by an integer distance, all the blocks that completely overlap are painted in red (think of moving the distance of 0 or a negative integer). However, this operation cannot be performed if there is at least one block that meets the conditions and has already been painted.
Homura-chan is the first player to lose the game if he can't operate it first.
Also, initially all cubes are uncolored.
Determine which one wins when the two act optimally.
Input format
A B C
Constraint
* 1 \ leq A, B, C \ leq 100
* All inputs are integers.
Output format
When the two act optimally, if Homura wins, `Hom` is output, and if Tempura wins,` Tem` is output on one line.
Input example 1
1 1 10
Output example 1
Hom
* The first time Homura can paint all the blocks red.
Input example 2
4 3 5
Output example 2
Hom
Input example 3
6 4 10
Output example 3
Tem
Example
Input
1 1 10
Output
Hom
"Correct Solution:
```
A, B, C = map(int, input().split())
count = 0
if A%2 == 1:
count += 1
if B%2 == 1:
count += 1
if C%2 == 1:
count += 1
if count >= 2:
print ('Hom')
else:
print ('Tem')
```
| 100,300 |
Provide a correct Python 3 solution for this coding contest problem.
C: Skewering
problem
One day, when Homura was playing with blocks, Tempura came. Homura decided to play with blocks with Tempura.
There is a rectangular parallelepiped of A \ times B \ times C, which is made by stacking A \ times B \ times C blocks of cubic blocks with a side length of 1 without any gaps. Each side of all cubes and rectangular parallelepipeds is parallel to the x-axis, y-axis, or z-axis.
Homura-chan and Tempura-kun alternately repeat the following operations.
* Select a row of blocks of building blocks lined up in a row from a rectangular parallelepiped in any of the vertical, horizontal, and depth directions, and paint all the blocks included in the row in red. However, you cannot select columns that contain blocks that are already painted red.
More precisely
* Select one of the blocks contained in the rectangular parallelepiped and one of the three directions x, y, and z.
* When the selected block is moved in the selected direction by an integer distance, all the blocks that completely overlap are painted in red (think of moving the distance of 0 or a negative integer). However, this operation cannot be performed if there is at least one block that meets the conditions and has already been painted.
Homura-chan is the first player to lose the game if he can't operate it first.
Also, initially all cubes are uncolored.
Determine which one wins when the two act optimally.
Input format
A B C
Constraint
* 1 \ leq A, B, C \ leq 100
* All inputs are integers.
Output format
When the two act optimally, if Homura wins, `Hom` is output, and if Tempura wins,` Tem` is output on one line.
Input example 1
1 1 10
Output example 1
Hom
* The first time Homura can paint all the blocks red.
Input example 2
4 3 5
Output example 2
Hom
Input example 3
6 4 10
Output example 3
Tem
Example
Input
1 1 10
Output
Hom
"Correct Solution:
```
print('Hom' if len(list(i for i in map(int, input().split()) if i % 2 == 1)) >= 2 else 'Tem')
```
| 100,301 |
Provide a correct Python 3 solution for this coding contest problem.
C: Skewering
problem
One day, when Homura was playing with blocks, Tempura came. Homura decided to play with blocks with Tempura.
There is a rectangular parallelepiped of A \ times B \ times C, which is made by stacking A \ times B \ times C blocks of cubic blocks with a side length of 1 without any gaps. Each side of all cubes and rectangular parallelepipeds is parallel to the x-axis, y-axis, or z-axis.
Homura-chan and Tempura-kun alternately repeat the following operations.
* Select a row of blocks of building blocks lined up in a row from a rectangular parallelepiped in any of the vertical, horizontal, and depth directions, and paint all the blocks included in the row in red. However, you cannot select columns that contain blocks that are already painted red.
More precisely
* Select one of the blocks contained in the rectangular parallelepiped and one of the three directions x, y, and z.
* When the selected block is moved in the selected direction by an integer distance, all the blocks that completely overlap are painted in red (think of moving the distance of 0 or a negative integer). However, this operation cannot be performed if there is at least one block that meets the conditions and has already been painted.
Homura-chan is the first player to lose the game if he can't operate it first.
Also, initially all cubes are uncolored.
Determine which one wins when the two act optimally.
Input format
A B C
Constraint
* 1 \ leq A, B, C \ leq 100
* All inputs are integers.
Output format
When the two act optimally, if Homura wins, `Hom` is output, and if Tempura wins,` Tem` is output on one line.
Input example 1
1 1 10
Output example 1
Hom
* The first time Homura can paint all the blocks red.
Input example 2
4 3 5
Output example 2
Hom
Input example 3
6 4 10
Output example 3
Tem
Example
Input
1 1 10
Output
Hom
"Correct Solution:
```
a,b,c = map(int,input().split())
cnt = 0
if a %2 == 0:
cnt += 1
if b %2 == 0:
cnt += 1
if c %2 == 0:
cnt += 1
if cnt >= 2:
print('Tem')
else:
print('Hom')
```
| 100,302 |
Provide a correct Python 3 solution for this coding contest problem.
Write a program which manipulates a sequence A = {a0, a1, . . . , an−1} with the following operations:
* update(s, t, x): change as, as+1, ..., at to x.
* find(s, t): report the minimum element in as, as+1, ..., at.
Note that the initial values of ai (i = 0, 1, . . . , n−1) are 231-1.
Constraints
* 1 ≤ n ≤ 100000
* 1 ≤ q ≤ 100000
* 0 ≤ s ≤ t < n
* 0 ≤ x < 231−1
Input
n q
query1
query2
:
queryq
In the first line, n (the number of elements in A) and q (the number of queries) are given. Then, ith query queryi is given in the following format:
0 s t x
or
1 s t
The first digit represents the type of the query. '0' denotes update(s, t, x) and '1' denotes find(s, t).
Output
For each find operation, print the minimum value.
Examples
Input
3 5
0 0 1 1
0 1 2 3
0 2 2 2
1 0 2
1 1 2
Output
1
2
Input
1 3
1 0 0
0 0 0 5
1 0 0
Output
2147483647
5
"Correct Solution:
```
import sys
readline = sys.stdin.readline
write = sys.stdout.write
n, q = map(int, input().split())
INF = 2**31-1
LV = (n-1).bit_length()
N0 = 2**LV
data = [INF]*(2*N0)
lazy = [None]*(2*N0)
def gindex(l, r):
L = (l + N0) >> 1; R = (r + N0) >> 1
lc = 0 if l & 1 else (L & -L).bit_length()
rc = 0 if r & 1 else (R & -R).bit_length()
for i in range(LV):
if rc <= i:
yield R
if L < R and lc <= i:
yield L
L >>= 1; R >>= 1
def propagates(*ids):
for i in reversed(ids):
v = lazy[i-1]
if v is None:
continue
lazy[2*i-1] = data[2*i-1] = lazy[2*i] = data[2*i] = v
lazy[i-1] = None
def update(l, r, x):
*ids, = gindex(l, r)
propagates(*ids)
L = N0 + l; R = N0 + r
while L < R:
if R & 1:
R -= 1
lazy[R-1] = data[R-1] = x
if L & 1:
lazy[L-1] = data[L-1] = x
L += 1
L >>= 1; R >>= 1
for i in ids:
data[i-1] = min(data[2*i-1], data[2*i])
def query(l, r):
propagates(*gindex(l, r))
L = N0 + l; R = N0 + r
s = INF
while L < R:
if R & 1:
R -= 1
s = min(s, data[R-1])
if L & 1:
s = min(s, data[L-1])
L += 1
L >>= 1; R >>= 1
return s
ans = []
for i in range(q):
t, *cmd = map(int, readline().split())
if t:
s, t = cmd
ans.append(str(query(s, t+1)))
else:
s, t, x = cmd
update(s, t+1, x)
write("\n".join(ans))
write("\n")
```
| 100,303 |
Provide a correct Python 3 solution for this coding contest problem.
Write a program which manipulates a sequence A = {a0, a1, . . . , an−1} with the following operations:
* update(s, t, x): change as, as+1, ..., at to x.
* find(s, t): report the minimum element in as, as+1, ..., at.
Note that the initial values of ai (i = 0, 1, . . . , n−1) are 231-1.
Constraints
* 1 ≤ n ≤ 100000
* 1 ≤ q ≤ 100000
* 0 ≤ s ≤ t < n
* 0 ≤ x < 231−1
Input
n q
query1
query2
:
queryq
In the first line, n (the number of elements in A) and q (the number of queries) are given. Then, ith query queryi is given in the following format:
0 s t x
or
1 s t
The first digit represents the type of the query. '0' denotes update(s, t, x) and '1' denotes find(s, t).
Output
For each find operation, print the minimum value.
Examples
Input
3 5
0 0 1 1
0 1 2 3
0 2 2 2
1 0 2
1 1 2
Output
1
2
Input
1 3
1 0 0
0 0 0 5
1 0 0
Output
2147483647
5
"Correct Solution:
```
import math
class SegmentTree:
__slots__ = ["rank", "elem_size", "tree_size", "tree", "lazy", "default_value"]
def __init__(self, a: list, default: int):
self.default_value = default
real_size = len(a)
self.rank = math.ceil(math.log2(real_size))
self.elem_size = 1 << self.rank
self.tree_size = 2 * self.elem_size
self.tree = [default]*self.elem_size + a + [default]*(self.elem_size - real_size)
self.lazy = [None]*self.tree_size
# self.init_tree()
def init_tree(self):
tree = self.tree
for i in range(self.elem_size-1, 0, -1):
left, right = tree[i << 1], tree[(i << 1)+1]
# ===== change me =====
tree[i] = left if left < right else right
def process_query(self, l: int, r: int, value: int = None):
'''[x, y)'''
tree, lazy, elem_size, rank = self.tree, self.lazy, self.elem_size, self.rank-1
l, r, targets, p_l, p_r, l_rank, r_rank = l+elem_size, r+elem_size, [], 0, 0, 0, 0
t_ap = targets.append
while l < r:
if l & 1:
t_ap(l)
p_l = p_l or l >> 1
l_rank = l_rank or rank
l += 1
if r & 1:
r -= 1
t_ap(r)
p_r = p_r or r >> 1
r_rank = r_rank or rank
l >>= 1
r >>= 1
rank -= 1
deepest = (p_l, p_r)
paths = [[p_l >> n for n in range(l_rank-1, -1, -1)], [p_r >> n for n in range(r_rank-1, -1, -1)]]
for a in paths:
for i in a:
if lazy[i] is None:
continue
# ===== change me =====
tree[i] = lazy[i]
if i < elem_size:
lazy[i << 1] = lazy[i]
lazy[(i << 1)+1] = lazy[i]
lazy[i] = None
result = self.default_value
for i in targets:
v = value if value is not None else lazy[i]
# ===== change me =====
if v is not None:
if i < elem_size:
lazy[i << 1] = v
lazy[(i << 1)+1] = v
tree[i] = v
lazy[i] = None
if result > tree[i]:
result = tree[i]
self.update_tree(deepest)
return result
def update_tree(self, indexes: tuple):
''' ????????????lazy?????¨????????¬????????§???????????¨???????????¨???????????? '''
tree, lazy = self.tree, self.lazy
for k in indexes:
while k > 0:
left, right = k << 1, (k << 1)+1
# ===== change me =====
l_value = tree[left] if lazy[left] is None else lazy[left]
r_value = tree[right] if lazy[right] is None else lazy[right]
tree[k] = l_value if l_value < r_value else r_value
k >>= 1
n, q = map(int, input().split())
rmq = SegmentTree([2**31-1]*n, 2**31-1)
ans = []
append = ans.append
for _ in [0]*q:
l = list(map(int, input().split()))
if l[0] == 0:
rmq.process_query(l[1], l[2]+1, l[3])
else:
a = rmq.process_query(l[1], l[2]+1)
append(a)
print("\n".join([str(n) for n in ans]))
```
| 100,304 |
Provide a correct Python 3 solution for this coding contest problem.
Write a program which manipulates a sequence A = {a0, a1, . . . , an−1} with the following operations:
* update(s, t, x): change as, as+1, ..., at to x.
* find(s, t): report the minimum element in as, as+1, ..., at.
Note that the initial values of ai (i = 0, 1, . . . , n−1) are 231-1.
Constraints
* 1 ≤ n ≤ 100000
* 1 ≤ q ≤ 100000
* 0 ≤ s ≤ t < n
* 0 ≤ x < 231−1
Input
n q
query1
query2
:
queryq
In the first line, n (the number of elements in A) and q (the number of queries) are given. Then, ith query queryi is given in the following format:
0 s t x
or
1 s t
The first digit represents the type of the query. '0' denotes update(s, t, x) and '1' denotes find(s, t).
Output
For each find operation, print the minimum value.
Examples
Input
3 5
0 0 1 1
0 1 2 3
0 2 2 2
1 0 2
1 1 2
Output
1
2
Input
1 3
1 0 0
0 0 0 5
1 0 0
Output
2147483647
5
"Correct Solution:
```
import sys
readline = sys.stdin.readline
write = sys.stdout.write
N, Q = map(int, input().split())
INF = 2**31-1
LV = (N-1).bit_length()
N0 = 2**LV
data = [INF]*(2*N0)
lazy = [None]*(2*N0)
def gindex(l, r):
L = l + N0; R = r + N0
lm = (L // (L & -L)) >> 1
rm = (R // (R & -R)) >> 1
while L < R:
if R <= rm:
yield R
if L <= lm:
yield L
L >>= 1; R >>= 1
while L:
yield L
L >>= 1
def propagates(*ids):
for i in reversed(ids):
v = lazy[i-1]
if v is None:
continue
lazy[2*i-1] = data[2*i-1] = lazy[2*i] = data[2*i] = v
lazy[i-1] = None
def update(l, r, x):
*ids, = gindex(l, r)
propagates(*ids)
L = N0 + l; R = N0 + r
while L < R:
if R & 1:
R -= 1
lazy[R-1] = data[R-1] = x
if L & 1:
lazy[L-1] = data[L-1] = x
L += 1
L >>= 1; R >>= 1
for i in ids:
data[i-1] = min(data[2*i-1], data[2*i])
def query(l, r):
propagates(*gindex(l, r))
L = N0 + l; R = N0 + r
s = INF
while L < R:
if R & 1:
R -= 1
s = min(s, data[R-1])
if L & 1:
s = min(s, data[L-1])
L += 1
L >>= 1; R >>= 1
return s
ans = []
for q in range(Q):
t, *cmd = map(int, readline().split())
if t:
s, t = cmd
ans.append(str(query(s, t+1)))
else:
s, t, x = cmd
update(s, t+1, x)
write("\n".join(ans))
write("\n")
```
| 100,305 |
Provide a correct Python 3 solution for this coding contest problem.
Write a program which manipulates a sequence A = {a0, a1, . . . , an−1} with the following operations:
* update(s, t, x): change as, as+1, ..., at to x.
* find(s, t): report the minimum element in as, as+1, ..., at.
Note that the initial values of ai (i = 0, 1, . . . , n−1) are 231-1.
Constraints
* 1 ≤ n ≤ 100000
* 1 ≤ q ≤ 100000
* 0 ≤ s ≤ t < n
* 0 ≤ x < 231−1
Input
n q
query1
query2
:
queryq
In the first line, n (the number of elements in A) and q (the number of queries) are given. Then, ith query queryi is given in the following format:
0 s t x
or
1 s t
The first digit represents the type of the query. '0' denotes update(s, t, x) and '1' denotes find(s, t).
Output
For each find operation, print the minimum value.
Examples
Input
3 5
0 0 1 1
0 1 2 3
0 2 2 2
1 0 2
1 1 2
Output
1
2
Input
1 3
1 0 0
0 0 0 5
1 0 0
Output
2147483647
5
"Correct Solution:
```
import sys
input=sys.stdin.readline
def gindex(l,r):
L,R=l+N0,r+N0
lm=L//(L&-L)>>1
rm=R//(R&-R)>>1
while L<R:
if R<=rm:
yield R
if L<=lm:
yield L
L>>=1
R>>=1
while L:
yield L
L>>=1
def propagates(*ids):
for i in reversed(ids):
v=lazy[i-1]
if v==None:
continue
lazy[2*i-1]=v
lazy[2*i]=v
data[2*i-1]=v
data[2*i]=v
lazy[i-1]=None
def update(l,r,x):#1-index [s,t)
L,R=N0+l,N0+r
*ids,=gindex(l,r)
propagates(*ids)
while L<R:#上から
if R&1:
R-=1
lazy[R-1]=x
data[R-1]=x
if L&1:
lazy[L-1]=x
data[L-1]=x
L+=1
L>>=1
R>>=1
for i in ids:#下から
data[i-1]=min(data[2*i-1],data[2*i])
def query(l,r):
propagates(*gindex(l,r))
L,R=N0+l,N0+r
s=INF
while L<R:
if R&1:
R-=1
s=min(s,data[R-1])
if L&1:
s=min(s,data[L-1])
L+=1
L>>=1
R>>=1
return s
n,q=map(int,input().split())
LV=n.bit_length()
N0=2**LV
INF=2**40
data=[2**31-1]*2*N0
lazy=[None]*2*N0
for i in range(q):
l=list(map(int,input().split()))
if l[0]==0:
s,t,x=l[1:]
update(s,t+1,x)
else:
s,t=l[1:]
print(query(s,t+1))
```
| 100,306 |
Provide a correct Python 3 solution for this coding contest problem.
Write a program which manipulates a sequence A = {a0, a1, . . . , an−1} with the following operations:
* update(s, t, x): change as, as+1, ..., at to x.
* find(s, t): report the minimum element in as, as+1, ..., at.
Note that the initial values of ai (i = 0, 1, . . . , n−1) are 231-1.
Constraints
* 1 ≤ n ≤ 100000
* 1 ≤ q ≤ 100000
* 0 ≤ s ≤ t < n
* 0 ≤ x < 231−1
Input
n q
query1
query2
:
queryq
In the first line, n (the number of elements in A) and q (the number of queries) are given. Then, ith query queryi is given in the following format:
0 s t x
or
1 s t
The first digit represents the type of the query. '0' denotes update(s, t, x) and '1' denotes find(s, t).
Output
For each find operation, print the minimum value.
Examples
Input
3 5
0 0 1 1
0 1 2 3
0 2 2 2
1 0 2
1 1 2
Output
1
2
Input
1 3
1 0 0
0 0 0 5
1 0 0
Output
2147483647
5
"Correct Solution:
```
import math
from collections import deque
class SegmentTree:
__slots__ = ["rank", "elem_size", "tree_size", "tree", "lazy", "default_value"]
def __init__(self, a: list, default: int):
self.default_value = default
real_size = len(a)
self.rank = math.ceil(math.log2(real_size))
self.elem_size = 1 << self.rank
self.tree_size = 2 * self.elem_size
self.tree = [default]*self.elem_size + a + [default]*(self.elem_size - real_size)
self.lazy = [None]*self.tree_size
self.init_tree()
def init_tree(self):
tree = self.tree
for i in range(self.elem_size-1, 0, -1):
left, right = tree[i << 1], tree[(i << 1)+1]
# ===== change me =====
tree[i] = left if left < right else right
def propagate(self, l: int, r: int, value: int = None):
'''[x, y)'''
tree, lazy, elem_size, rank = self.tree, self.lazy, self.elem_size, self.rank-1
l, r, targets, p_l, p_r, l_rank, r_rank = l+elem_size, r+elem_size, deque(), 0, 0, 0, 0
t_ap = targets.append
while l < r:
if l & 1:
t_ap(l)
p_l = p_l or l >> 1
l_rank = l_rank or rank
l += 1
if r & 1:
r -= 1
t_ap(r)
p_r = p_r or r >> 1
r_rank = r_rank or rank
l >>= 1
r >>= 1
rank -= 1
deepest = (p_l, p_r)
paths = [[p_l >> n for n in range(l_rank-1, -1, -1)], [p_r >> n for n in range(r_rank-1, -1, -1)]]
for a in paths:
for i in a:
if lazy[i] is None:
continue
# ===== change me =====
tree[i] = lazy[i]
if i < elem_size:
lazy[i << 1] = lazy[i]
lazy[(i << 1)+1] = lazy[i]
lazy[i] = None
result = self.default_value
if value is None:
for i in targets:
if lazy[i] is not None:
tree[i] = lazy[i]
if i < elem_size:
lazy[i << 1] = lazy[i]
lazy[(i << 1)+1] = lazy[i]
lazy[i] = None
# ===== change me =====
if result > tree[i]:
result = tree[i]
else:
for i in targets:
# ===== change me =====
if i < elem_size:
lazy[i << 1] = value
lazy[(i << 1)+1] = value
tree[i] = value
lazy[i] = None
self.update_tree(deepest)
return result
def update_lazy(self, l, r, value):
self.propagate(l, r, value)
def get_value(self, l: int, r: int):
tree = self.tree
targets = self.propagate(l, r)
# ===== change me =====
return min(tree[n] for n in targets)
def update_tree(self, indexes: tuple):
''' ????????????lazy?????¨????????¬????????§???????????¨???????????¨???????????? '''
tree, lazy = self.tree, self.lazy
for k in indexes:
while k > 0:
left, right = k << 1, (k << 1)+1
# ===== change me =====
l_value = tree[left] if lazy[left] is None else lazy[left]
r_value = tree[right] if lazy[right] is None else lazy[right]
tree[k] = l_value if l_value < r_value else r_value
k >>= 1
n, q = map(int, input().split())
rmq = SegmentTree([2**31-1]*n, 2**31-1)
ans = []
append = ans.append
for _ in [0]*q:
l = list(map(int, input().split()))
if l[0] == 0:
rmq.propagate(l[1], l[2]+1, l[3])
else:
a = rmq.propagate(l[1], l[2]+1)
append(a)
print("\n".join([str(n) for n in ans]))
```
| 100,307 |
Provide a correct Python 3 solution for this coding contest problem.
Write a program which manipulates a sequence A = {a0, a1, . . . , an−1} with the following operations:
* update(s, t, x): change as, as+1, ..., at to x.
* find(s, t): report the minimum element in as, as+1, ..., at.
Note that the initial values of ai (i = 0, 1, . . . , n−1) are 231-1.
Constraints
* 1 ≤ n ≤ 100000
* 1 ≤ q ≤ 100000
* 0 ≤ s ≤ t < n
* 0 ≤ x < 231−1
Input
n q
query1
query2
:
queryq
In the first line, n (the number of elements in A) and q (the number of queries) are given. Then, ith query queryi is given in the following format:
0 s t x
or
1 s t
The first digit represents the type of the query. '0' denotes update(s, t, x) and '1' denotes find(s, t).
Output
For each find operation, print the minimum value.
Examples
Input
3 5
0 0 1 1
0 1 2 3
0 2 2 2
1 0 2
1 1 2
Output
1
2
Input
1 3
1 0 0
0 0 0 5
1 0 0
Output
2147483647
5
"Correct Solution:
```
import sys
readline = sys.stdin.readline
write = sys.stdout.write
N, Q = map(int, input().split())
INF = 2**31-1
LV = (N-1).bit_length()
N0 = 2**LV
data = [INF]*(2*N0)
lazy = [None]*(2*N0)
def gindex(l, r):
L = (l + N0) >> 1; R = (r + N0) >> 1
lc = 0 if l & 1 else (L & -L).bit_length()
rc = 0 if r & 1 else (R & -R).bit_length()
for i in range(LV):
if rc <= i:
yield R
if L < R and lc <= i:
yield L
L >>= 1; R >>= 1
def propagates(*ids):
for i in reversed(ids):
v = lazy[i-1]
if v is None:
continue
lazy[2*i-1] = data[2*i-1] = lazy[2*i] = data[2*i] = v
lazy[i-1] = None
def update(l, r, x):
*ids, = gindex(l, r)
propagates(*ids)
L = N0 + l; R = N0 + r
while L < R:
if R & 1:
R -= 1
lazy[R-1] = data[R-1] = x
if L & 1:
lazy[L-1] = data[L-1] = x
L += 1
L >>= 1; R >>= 1
for i in ids:
data[i-1] = min(data[2*i-1], data[2*i])
def query(l, r):
propagates(*gindex(l, r))
L = N0 + l; R = N0 + r
s = INF
while L < R:
if R & 1:
R -= 1
s = min(s, data[R-1])
if L & 1:
s = min(s, data[L-1])
L += 1
L >>= 1; R >>= 1
return s
answer = []
for q in range(Q):
t, *cmd = map(int, readline().split())
if t:
s, t = cmd
answer.append(str(query(s, t+1)))
else:
s, t, x = cmd
update(s, t+1, x)
write("\n".join(answer))
write("\n")
```
| 100,308 |
Provide a correct Python 3 solution for this coding contest problem.
Write a program which manipulates a sequence A = {a0, a1, . . . , an−1} with the following operations:
* update(s, t, x): change as, as+1, ..., at to x.
* find(s, t): report the minimum element in as, as+1, ..., at.
Note that the initial values of ai (i = 0, 1, . . . , n−1) are 231-1.
Constraints
* 1 ≤ n ≤ 100000
* 1 ≤ q ≤ 100000
* 0 ≤ s ≤ t < n
* 0 ≤ x < 231−1
Input
n q
query1
query2
:
queryq
In the first line, n (the number of elements in A) and q (the number of queries) are given. Then, ith query queryi is given in the following format:
0 s t x
or
1 s t
The first digit represents the type of the query. '0' denotes update(s, t, x) and '1' denotes find(s, t).
Output
For each find operation, print the minimum value.
Examples
Input
3 5
0 0 1 1
0 1 2 3
0 2 2 2
1 0 2
1 1 2
Output
1
2
Input
1 3
1 0 0
0 0 0 5
1 0 0
Output
2147483647
5
"Correct Solution:
```
import sys
input = sys.stdin.readline
N,Q=map(int,input().split())
seg_el=1<<(N.bit_length()) # Segment treeの台の要素数
SEG=[(1<<31)-1]*(2*seg_el) # 1-indexedなので、要素数2*seg_el.Segment treeの初期値で初期化
LAZY=[None]*(2*seg_el) # 1-indexedなので、要素数2*seg_el.Segment treeの初期値で初期化
def indexes(L,R):
INDLIST=[]
R-=1
L>>=1
R>>=1
while L!=R:
if L>R:
INDLIST.append(L)
L>>=1
else:
INDLIST.append(R)
R>>=1
while L!=0:
INDLIST.append(L)
L>>=1
return INDLIST
def updates(l,r,x): # 区間[l,r)をxに更新
L=l+seg_el
R=r+seg_el
L//=(L & (-L))
R//=(R & (-R))
UPIND=indexes(L,R)
for ind in UPIND[::-1]:
if LAZY[ind]!=None:
LAZY[ind<<1]=LAZY[1+(ind<<1)]=SEG[ind<<1]=SEG[1+(ind<<1)]=LAZY[ind]
LAZY[ind]=None
while L!=R:
if L > R:
SEG[L]=x
LAZY[L]=x
L+=1
L//=(L & (-L))
else:
R-=1
SEG[R]=x
LAZY[R]=x
R//=(R & (-R))
for ind in UPIND:
SEG[ind]=min(SEG[ind<<1],SEG[1+(ind<<1)])
def getvalues(l,r): # 区間[l,r)に関するminを調べる
L=l+seg_el
R=r+seg_el
L//=(L & (-L))
R//=(R & (-R))
UPIND=indexes(L,R)
for ind in UPIND[::-1]:
if LAZY[ind]!=None:
LAZY[ind<<1]=LAZY[1+(ind<<1)]=SEG[ind<<1]=SEG[1+(ind<<1)]=LAZY[ind]
LAZY[ind]=None
ANS=1<<31
while L!=R:
if L > R:
ANS=min(ANS , SEG[L])
L+=1
L//=(L & (-L))
else:
R-=1
ANS=min(ANS , SEG[R])
R//=(R & (-R))
return ANS
ANS=[]
for _ in range(Q):
query=list(map(int,input().split()))
if query[0]==0:
updates(query[1],query[2]+1,query[3])
else:
ANS.append(getvalues(query[1],query[2]+1))
print("\n".join([str(ans) for ans in ANS]))
```
| 100,309 |
Provide a correct Python 3 solution for this coding contest problem.
Write a program which manipulates a sequence A = {a0, a1, . . . , an−1} with the following operations:
* update(s, t, x): change as, as+1, ..., at to x.
* find(s, t): report the minimum element in as, as+1, ..., at.
Note that the initial values of ai (i = 0, 1, . . . , n−1) are 231-1.
Constraints
* 1 ≤ n ≤ 100000
* 1 ≤ q ≤ 100000
* 0 ≤ s ≤ t < n
* 0 ≤ x < 231−1
Input
n q
query1
query2
:
queryq
In the first line, n (the number of elements in A) and q (the number of queries) are given. Then, ith query queryi is given in the following format:
0 s t x
or
1 s t
The first digit represents the type of the query. '0' denotes update(s, t, x) and '1' denotes find(s, t).
Output
For each find operation, print the minimum value.
Examples
Input
3 5
0 0 1 1
0 1 2 3
0 2 2 2
1 0 2
1 1 2
Output
1
2
Input
1 3
1 0 0
0 0 0 5
1 0 0
Output
2147483647
5
"Correct Solution:
```
# -*- coding: utf-8 -*-
import sys
def input(): return sys.stdin.readline().strip()
def list2d(a, b, c): return [[c] * b for i in range(a)]
def list3d(a, b, c, d): return [[[d] * c for j in range(b)] for i in range(a)]
def list4d(a, b, c, d, e): return [[[[e] * d for j in range(c)] for j in range(b)] for i in range(a)]
def ceil(x, y=1): return int(-(-x // y))
def INT(): return int(input())
def MAP(): return map(int, input().split())
def LIST(N=None): return list(MAP()) if N is None else [INT() for i in range(N)]
def Yes(): print('Yes')
def No(): print('No')
def YES(): print('YES')
def NO(): print('NO')
sys.setrecursionlimit(10 ** 9)
INF = 2 ** 31 - 1
MOD = 10 ** 9 + 7
class SegTreeLazy:
""" 遅延評価セグメント木 """
def __init__(self, N, func, intv):
self.intv = intv
self.func = func
LV = (N-1).bit_length()
self.N0 = 2**LV
self.data = [intv]*(2*self.N0)
self.lazy = [None]*(2*self.N0)
# 伝搬される区間のインデックス(1-indexed)を全て列挙するgenerator
def gindex(self, l, r):
L = l + self.N0; R = r + self.N0
lm = (L // (L & -L)) >> 1
rm = (R // (R & -R)) >> 1
while L < R:
if R <= rm:
yield R
if L <= lm:
yield L
L >>= 1; R >>= 1
while L:
yield L
L >>= 1
# 遅延評価の伝搬処理
def propagates(self, *ids):
# 1-indexedで単調増加のインデックスリスト
for i in reversed(ids):
v = self.lazy[i-1]
if v is None:
continue
self.lazy[2*i-1] = self.data[2*i-1] = self.lazy[2*i] = self.data[2*i] = v
self.lazy[i-1] = None
def update(self, l, r, x):
""" 区間[l,r)の値をxに更新 """
*ids, = self.gindex(l, r)
# 1. トップダウンにlazyの値を伝搬
self.propagates(*ids)
# 2. 区間[l,r)のdata, lazyの値を更新
L = self.N0 + l; R = self.N0 + r
while L < R:
if R & 1:
R -= 1
self.lazy[R-1] = self.data[R-1] = x
if L & 1:
self.lazy[L-1] = self.data[L-1] = x
L += 1
L >>= 1; R >>= 1
# 3. 伝搬させた区間について、ボトムアップにdataの値を伝搬する
for i in ids:
self.data[i-1] = self.func(self.data[2*i-1], self.data[2*i])
def query(self, l, r):
""" 区間[l,r)の最小値を取得 """
# 1. トップダウンにlazyの値を伝搬
self.propagates(*self.gindex(l, r))
L = self.N0 + l; R = self.N0 + r
# 2. 区間[l, r)の最小値を求める
s = self.intv
while L < R:
if R & 1:
R -= 1
s = self.func(s, self.data[R-1])
if L & 1:
s = self.func(s, self.data[L-1])
L += 1
L >>= 1; R >>= 1
return s
N, Q = MAP()
stl = SegTreeLazy(N+1, min, INF)
ans = []
for i in range(Q):
cmd, *arg = MAP()
if cmd == 0:
s, t, x = arg
stl.update(s, t+1, x)
else:
s, t = arg
ans.append(str(stl.query(s, t+1)))
print('\n'.join(ans))
```
| 100,310 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Write a program which manipulates a sequence A = {a0, a1, . . . , an−1} with the following operations:
* update(s, t, x): change as, as+1, ..., at to x.
* find(s, t): report the minimum element in as, as+1, ..., at.
Note that the initial values of ai (i = 0, 1, . . . , n−1) are 231-1.
Constraints
* 1 ≤ n ≤ 100000
* 1 ≤ q ≤ 100000
* 0 ≤ s ≤ t < n
* 0 ≤ x < 231−1
Input
n q
query1
query2
:
queryq
In the first line, n (the number of elements in A) and q (the number of queries) are given. Then, ith query queryi is given in the following format:
0 s t x
or
1 s t
The first digit represents the type of the query. '0' denotes update(s, t, x) and '1' denotes find(s, t).
Output
For each find operation, print the minimum value.
Examples
Input
3 5
0 0 1 1
0 1 2 3
0 2 2 2
1 0 2
1 1 2
Output
1
2
Input
1 3
1 0 0
0 0 0 5
1 0 0
Output
2147483647
5
Submitted Solution:
```
import sys
readline = sys.stdin.readline
write = sys.stdout.write
N, Q = map(int, input().split())
INF = 2**31-1
LV = (N-1).bit_length()
N0 = 2**LV
data = [INF]*(2*N0)
lazy = [None]*(2*N0)
def gindex(l, r):
L = (l + N0) >> 1; R = (r + N0) >> 1
lc = 0 if l & 1 else (L & -L).bit_length()
rc = 0 if r & 1 else (R & -R).bit_length()
for i in range(LV):
if rc <= i:
yield R
if L < R and lc <= i:
yield L
L >>= 1; R >>= 1
def propagates(*ids):
for i in reversed(ids):
v = lazy[i-1]
if v is None:
continue
lazy[2*i-1] = data[2*i-1] = lazy[2*i] = data[2*i] = v
lazy[i-1] = None
def update(l, r, x):
*ids, = gindex(l, r)
propagates(*ids)
L = N0 + l; R = N0 + r
while L < R:
if R & 1:
R -= 1
lazy[R-1] = data[R-1] = x
if L & 1:
lazy[L-1] = data[L-1] = x
L += 1
L >>= 1; R >>= 1
for i in ids:
data[i-1] = min(data[2*i-1], data[2*i])
def query(l, r):
propagates(*gindex(l, r))
L = N0 + l; R = N0 + r
s = INF
while L < R:
if R & 1:
R -= 1
s = min(s, data[R-1])
if L & 1:
s = min(s, data[L-1])
L += 1
L >>= 1; R >>= 1
return s
ans = []
for q in range(Q):
t, *cmd = map(int, readline().split())
if t:
s, t = cmd
ans.append(str(query(s, t+1)))
else:
s, t, x = cmd
update(s, t+1, x)
write("\n".join(ans))
write("\n")
```
Yes
| 100,311 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Write a program which manipulates a sequence A = {a0, a1, . . . , an−1} with the following operations:
* update(s, t, x): change as, as+1, ..., at to x.
* find(s, t): report the minimum element in as, as+1, ..., at.
Note that the initial values of ai (i = 0, 1, . . . , n−1) are 231-1.
Constraints
* 1 ≤ n ≤ 100000
* 1 ≤ q ≤ 100000
* 0 ≤ s ≤ t < n
* 0 ≤ x < 231−1
Input
n q
query1
query2
:
queryq
In the first line, n (the number of elements in A) and q (the number of queries) are given. Then, ith query queryi is given in the following format:
0 s t x
or
1 s t
The first digit represents the type of the query. '0' denotes update(s, t, x) and '1' denotes find(s, t).
Output
For each find operation, print the minimum value.
Examples
Input
3 5
0 0 1 1
0 1 2 3
0 2 2 2
1 0 2
1 1 2
Output
1
2
Input
1 3
1 0 0
0 0 0 5
1 0 0
Output
2147483647
5
Submitted Solution:
```
class LazySegTree: # Non Recursion, RmQ, RUQ
def __init__(self, N):
self.N = N
self._N = 1<<((N-1).bit_length())
self.INF = (1<<31) - 1
self.node = [self.INF]*2*self._N
self.lazy = [None]*2*self._N
def build(self, lis):
for i in range(self.N):
self.node[i+self._N-1] = lis[i]
for i in range(self._N-2,-1,-1):
self.node[i] = self.node[i*2+1] + self.node[i*2+2]
def _gindex(self, l, r):
left = l + self._N; right = r + self._N
lm = (left // (left & -left)) >> 1
rm = (right // (right & -right)) >> 1
while left < right:
if right <= rm: yield right
if left <= lm: yield left
left >>= 1; right >>= 1
while left:
yield left
left >>= 1
def propagates(self, *ids): # ids: 1-indexded
for i in reversed(ids):
v = self.lazy[i-1]
if v is None: continue
self.lazy[2*i-1] = self.node[2*i-1] = v
self.lazy[2*i] = self.node[2*i] = v
self.lazy[i-1] = None
return
def update(self, l, r, x): # change all[left, right) to x
*ids, = self._gindex(l, r)
self.propagates(*ids)
left = l + self._N; right = r + self._N
while left < right:
if left & 1:
self.lazy[left-1] = self.node[left-1] = x
left += 1
if right & 1:
right -= 1
self.lazy[right-1] = self.node[right-1] = x
left >>= 1; right >>= 1
for i in ids:
self.node[i-1] = min(self.node[2*i-1], self.node[2*i])
def query(self, l, r):
self.propagates(*self._gindex(l, r))
left = l + self._N; right = r + self._N
ret = self.INF
while left < right:
if right & 1:
right -= 1
ret = min(ret, self.node[right-1])
if left & 1:
ret = min(ret, self.node[left-1])
left += 1
left >>= 1; right >>= 1
return ret
nm = lambda: map(int, input().split())
n,q = nm()
S = LazySegTree(n)
for _ in range(q):
v = list(nm())
if v[0]:
# print(S.node)
# print(S.lazy)
print(S.query(v[1], v[2]+1))
else:
S.update(v[1], v[2]+1, v[3])
```
Yes
| 100,312 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Write a program which manipulates a sequence A = {a0, a1, . . . , an−1} with the following operations:
* update(s, t, x): change as, as+1, ..., at to x.
* find(s, t): report the minimum element in as, as+1, ..., at.
Note that the initial values of ai (i = 0, 1, . . . , n−1) are 231-1.
Constraints
* 1 ≤ n ≤ 100000
* 1 ≤ q ≤ 100000
* 0 ≤ s ≤ t < n
* 0 ≤ x < 231−1
Input
n q
query1
query2
:
queryq
In the first line, n (the number of elements in A) and q (the number of queries) are given. Then, ith query queryi is given in the following format:
0 s t x
or
1 s t
The first digit represents the type of the query. '0' denotes update(s, t, x) and '1' denotes find(s, t).
Output
For each find operation, print the minimum value.
Examples
Input
3 5
0 0 1 1
0 1 2 3
0 2 2 2
1 0 2
1 1 2
Output
1
2
Input
1 3
1 0 0
0 0 0 5
1 0 0
Output
2147483647
5
Submitted Solution:
```
class LazyPropSegmentTree:
def __init__(self, lst, op, apply, comp, e, identity):
self.n = len(lst)
self.depth = (self.n - 1).bit_length()
self.N = 1 << self.depth
self.op = op # binary operation of elements
self.apply = apply # function to apply to an element
self.comp = comp # composition of functions
self.e = e # identity element w.r.t. op
self.identity = identity # identity element w.r.t. comp
self.v = self._build(lst) # self.v is set to be 1-indexed for simplicity
self.lazy = [self.identity] * (2 * self.N)
def __getitem__(self, i):
return self.fold(i, i+1)
def _build(self, lst):
# construction of a tree
# total 2 * self.N elements (tree[0] is not used)
tree = [self.e] * (self.N) + lst + [self.e] * (self.N - self.n)
for i in range(self.N - 1, 0, -1): tree[i] = self.op(tree[i << 1], tree[(i << 1)|1])
return tree
def _indices(self, l, r):
left = l + self.N; right = r + self.N
left //= (left & (-left)); right //= (right & (-right))
left >>= 1; right >>= 1
while left != right:
if left > right: yield left; left >>= 1
else: yield right; right >>= 1
while left > 0: yield left; left >>= 1
# propagate self.lazy and self.v in a top-down manner
def _propagate_topdown(self, *indices):
identity, v, lazy, apply, comp = self.identity, self.v, self.lazy, self.apply, self.comp
for k in reversed(indices):
x = lazy[k]
if x == identity: continue
lazy[k << 1] = comp(lazy[k << 1], x)
lazy[(k << 1)|1] = comp(lazy[(k << 1)|1], x)
v[k << 1] = apply(v[k << 1], x)
v[(k << 1)|1] = apply(v[(k << 1)|1], x)
lazy[k] = identity # propagated
# propagate self.v in a bottom-up manner
def _propagate_bottomup(self, indices):
v, op = self.v, self.op
for k in indices: v[k] = op(v[k << 1], v[(k << 1)|1])
# update for the query interval [l, r) with function x
def update(self, l, r, x):
*indices, = self._indices(l, r)
self._propagate_topdown(*indices)
N, v, lazy, apply, comp = self.N, self.v, self.lazy, self.apply, self.comp
# update self.v and self.lazy for the query interval [l, r)
left = l + N; right = r + N
if left & 1: v[left] = apply(v[left], x); left += 1
if right & 1: right -= 1; v[right] = apply(v[right], x)
left >>= 1; right >>= 1
while left < right:
if left & 1:
lazy[left] = comp(lazy[left], x)
v[left] = apply(v[left], x)
left += 1
if right & 1:
right -= 1
lazy[right] = comp(lazy[right], x)
v[right] = apply(v[right], x)
left >>= 1; right >>= 1
self._propagate_bottomup(indices)
# returns answer for the query interval [l, r)
def fold(self, l, r):
self._propagate_topdown(*self._indices(l, r))
e, N, v, op = self.e, self.N, self.v, self.op
# calculate the answer for the query interval [l, r)
left = l + N; right = r + N
L = R = e
while left < right:
if left & 1: # self.v[left] is the right child
L = op(L, v[left])
left += 1
if right & 1: # self.v[right-1] is the left child
right -= 1
R = op(v[right], R)
left >>= 1; right >>= 1
return op(L, R)
N, Q = map(int, input().split())
op = min
apply = lambda x, f: f
comp = lambda f, g: g
e = 2**31 - 1
identity = None
A = [e] * N
lpsg = LazyPropSegmentTree(A, op, apply, comp, e, identity)
ans = []
for _ in range(Q):
t, *arg, = map(int, input().split())
if t == 0:
s, t, x = arg
lpsg.update(s, t+1, x)
else:
s, t = arg
ans.append(lpsg.fold(s, t+1))
print('\n'.join(map(str, ans)))
```
Yes
| 100,313 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Write a program which manipulates a sequence A = {a0, a1, . . . , an−1} with the following operations:
* update(s, t, x): change as, as+1, ..., at to x.
* find(s, t): report the minimum element in as, as+1, ..., at.
Note that the initial values of ai (i = 0, 1, . . . , n−1) are 231-1.
Constraints
* 1 ≤ n ≤ 100000
* 1 ≤ q ≤ 100000
* 0 ≤ s ≤ t < n
* 0 ≤ x < 231−1
Input
n q
query1
query2
:
queryq
In the first line, n (the number of elements in A) and q (the number of queries) are given. Then, ith query queryi is given in the following format:
0 s t x
or
1 s t
The first digit represents the type of the query. '0' denotes update(s, t, x) and '1' denotes find(s, t).
Output
For each find operation, print the minimum value.
Examples
Input
3 5
0 0 1 1
0 1 2 3
0 2 2 2
1 0 2
1 1 2
Output
1
2
Input
1 3
1 0 0
0 0 0 5
1 0 0
Output
2147483647
5
Submitted Solution:
```
import sys
input = sys.stdin.readline
n, q = map(int, input().split())
INF = 2**31-1
Len = (n-1).bit_length()
size = 2**Len
tree = [INF]*(2*size)
lazy = [None]*(2*size)
def gindex(l,r):
L = (l+size)>>1;R=(r+size)>>1
lc = 0 if l & 1 else (L&-L).bit_length()
rc = 0 if r & 1 else (R&-R).bit_length()
for i in range(Len):
if rc <= i:
yield R
if L < R and lc <= i:
yield L
L >>= 1; R >>= 1
def propagates(ids):
#上から更新
for i in reversed(ids):
v = lazy[i]
if v is None:
continue
lazy[2*i]=tree[2*i]=lazy[2*i+1]=tree[2*i+1]=v
lazy[i]=None
#[l,r)の探索
def update(l,r,x):
*ids, = gindex(l, r)
propagates(ids)
L = size+l
R = size+r
while L<R:
if R&1:
R -= 1
lazy[R]=tree[R]=x
if L&1:
lazy[L]=tree[L]=x
L+=1
L>>=1;R>>=1
#値を更新し終わったら、最小値を下から更新
for i in ids:
if 2*i+1<size*2:
tree[i]=min(tree[i*2],tree[i*2+1])
def query(l, r):
*ids, = gindex(l, r)
propagates(ids)
L = size + l
R = size + r
s = INF
while L<R:
if R&1:
R-=1
s = min(s,tree[R])
if L&1:
s = min(s,tree[L])
L+=1
L>>=1;R>>=1
return s
ans = []
for i in range(q):
a, *b = map(int, input().split())
if a:
ans.append(query(b[0],b[1]+1))
else:
update(b[0],b[1]+1,b[2])
print('\n'.join(map(str,ans)))
```
Yes
| 100,314 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Write a program which manipulates a sequence A = {a0, a1, . . . , an−1} with the following operations:
* update(s, t, x): change as, as+1, ..., at to x.
* find(s, t): report the minimum element in as, as+1, ..., at.
Note that the initial values of ai (i = 0, 1, . . . , n−1) are 231-1.
Constraints
* 1 ≤ n ≤ 100000
* 1 ≤ q ≤ 100000
* 0 ≤ s ≤ t < n
* 0 ≤ x < 231−1
Input
n q
query1
query2
:
queryq
In the first line, n (the number of elements in A) and q (the number of queries) are given. Then, ith query queryi is given in the following format:
0 s t x
or
1 s t
The first digit represents the type of the query. '0' denotes update(s, t, x) and '1' denotes find(s, t).
Output
For each find operation, print the minimum value.
Examples
Input
3 5
0 0 1 1
0 1 2 3
0 2 2 2
1 0 2
1 1 2
Output
1
2
Input
1 3
1 0 0
0 0 0 5
1 0 0
Output
2147483647
5
Submitted Solution:
```
import math
lazy_default = None
def update_tree(k: int):
if k >= l:
k >>= l
while k > 0:
left, right = tree[k << 1], tree[(k << 1) + 1]
# ===== change me =====
tree[k] = left if left < right else right
k >>= 1
def merge(node_index, value, update=False):
lazy_value = lazy[node_index]
# ===== change me =====
value = lazy_value if value is None else value
lazy[node_index] = None
# ===== change me =====
if value is not lazy_default:
tree[node_index] = value
if node_index < l and value != lazy_default:
# ===== change me =====
if node_index < l // 2:
tree[node_index << 1] = value
tree[(node_index << 1) + 1] = value
else:
lazy[node_index << 1] = value
lazy[(node_index << 1) + 1] = value
def hoge(s, e, value):
s, e, d, l_active, r_active = s+l, e+l, rank, 1, 1
values = []
append = values.append
while l_active or r_active:
l_node, r_node = s >> d, e >> d
l_leaf = l_node << d, (l_node + 1 << d) - 1
r_leaf = r_node << d, (r_node + 1 << d) - 1
lazy_left, lazy_right = lazy[l_node], lazy[r_node]
if l_node == r_node:
if l_leaf[0] == s and r_leaf[1] == e:
# match
merge(l_node, value, True)
append(tree[l_node])
break
else:
if l_active:
if l_leaf[0] == s:
# match
merge(l_node, value, True)
append(tree[l_node])
l_active = 0
# ??????????????????????????????????????????????¶?????????????????????´???
else:
if lazy_left != None:
merge(l_node, None)
if l_node >> 1 << 1 == l_node and l_node+1 < r_node:
# match
merge(l_node+1, value)
append(tree[l_node+1])
if r_active:
if r_leaf[1] == e:
# match
merge(r_node, value)
update_tree(r_node)
append(tree[r_node])
r_active = 0
else:
if lazy_right != None:
merge(r_node, None)
if r_node >> 1 << 1 == r_node-1 and l_node+1 < r_node:
# match
merge(r_node-1, value)
append(tree[r_node-1])
d -= 1
return values
def get_value(i):
i += l
v = 0
for j in range(rank, -1, -1):
v += lazy[i>>j]
return v + tree[i]
n, q = map(int,input().split())
l = 1 << math.ceil(math.log2(n))
tree = [2**31-1]*(2*l)
lazy = [None]*(2*l)
rank = int(math.log2(l))
ans = []
ap = ans.append
for _ in [None]*q:
query = list(map(int, input().split()))
if query[0] == 0:
hoge(query[1], query[2], query[3])
else:
a = hoge(query[1], query[2], None)
ap(min(a))
print("\n".join((str(n) for n in ans)))
```
No
| 100,315 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Write a program which manipulates a sequence A = {a0, a1, . . . , an−1} with the following operations:
* update(s, t, x): change as, as+1, ..., at to x.
* find(s, t): report the minimum element in as, as+1, ..., at.
Note that the initial values of ai (i = 0, 1, . . . , n−1) are 231-1.
Constraints
* 1 ≤ n ≤ 100000
* 1 ≤ q ≤ 100000
* 0 ≤ s ≤ t < n
* 0 ≤ x < 231−1
Input
n q
query1
query2
:
queryq
In the first line, n (the number of elements in A) and q (the number of queries) are given. Then, ith query queryi is given in the following format:
0 s t x
or
1 s t
The first digit represents the type of the query. '0' denotes update(s, t, x) and '1' denotes find(s, t).
Output
For each find operation, print the minimum value.
Examples
Input
3 5
0 0 1 1
0 1 2 3
0 2 2 2
1 0 2
1 1 2
Output
1
2
Input
1 3
1 0 0
0 0 0 5
1 0 0
Output
2147483647
5
Submitted Solution:
```
from math import log2, ceil
class SegmentTree:
SENTINEL = 2 ** 31 - 1
def __init__(self, n):
tn = 2 ** ceil(log2(n))
self.a = [2 ** 31 - 1] * (tn * 2)
def find(self, c, l, r, s, t):
if l == s and r == t:
return self.a[c]
mid = (l + r) // 2
if t <= mid:
return self.find(c * 2, l, mid, s, t)
elif s > mid:
return self.find(c * 2 + 1, mid + 1, r, s, t)
else:
return min(
self.find(c * 2, l, mid, s, mid),
self.find(c * 2 + 1, mid + 1, r, mid + 1, t))
def update(self, c, l, r, s, t, x):
if l == s and r == t:
self.a[c] = x
return x
mid = (l + r) // 2
if t <= mid:
if self.a[c * 2 + 1] == self.SENTINEL:
self.a[c * 2 + 1] = self.a[c]
u = min(self.update(c * 2, l, mid, s, t, x), self.a[c * 2 + 1])
elif s > mid:
if self.a[c * 2] == self.SENTINEL:
self.a[c * 2] = self.a[c]
u = min(self.a[c * 2], self.update(c * 2 + 1, mid + 1, r, s, t, x))
else:
u = min(
self.update(c * 2, l, mid, s, mid, x),
self.update(c * 2 + 1, mid + 1, r, mid + 1, t, x))
self.a[c] = u
return u
n, q = map(int, input().split())
st = SegmentTree(n)
for _ in range(q):
query = input().split()
if query[0] == '0':
s, t, x = map(int, query[1:])
st.update(1, 0, n - 1, s, t, x)
else:
s, t = map(int, query[1:])
print(st.find(1, 0, n - 1, s, t))
```
No
| 100,316 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Write a program which manipulates a sequence A = {a0, a1, . . . , an−1} with the following operations:
* update(s, t, x): change as, as+1, ..., at to x.
* find(s, t): report the minimum element in as, as+1, ..., at.
Note that the initial values of ai (i = 0, 1, . . . , n−1) are 231-1.
Constraints
* 1 ≤ n ≤ 100000
* 1 ≤ q ≤ 100000
* 0 ≤ s ≤ t < n
* 0 ≤ x < 231−1
Input
n q
query1
query2
:
queryq
In the first line, n (the number of elements in A) and q (the number of queries) are given. Then, ith query queryi is given in the following format:
0 s t x
or
1 s t
The first digit represents the type of the query. '0' denotes update(s, t, x) and '1' denotes find(s, t).
Output
For each find operation, print the minimum value.
Examples
Input
3 5
0 0 1 1
0 1 2 3
0 2 2 2
1 0 2
1 1 2
Output
1
2
Input
1 3
1 0 0
0 0 0 5
1 0 0
Output
2147483647
5
Submitted Solution:
```
INT_MAX = 1<<31 - 1
_N,query_num = map(int, input().split())
queries = [list(map(int, input().split())) for i in range(query_num)]
N = 1<<17
arr_size = 2*N - 1
A = [INT_MAX]*(arr_size)
lazy_A = [None]*(arr_size)
def tree_propagate(k):
if(lazy_A[k] is None):
return
if(k < N // 2):
A[k*2+1] = A[k*2+2] = lazy_A[k*2+1] = lazy_A[k*2+2] = lazy_A[k]
lazy_A[k] = None
return
if(k < N-1):
A[k*2+1] = A[k*2+2] = lazy_A[k]
lazy_A[k] = None
def update(s,t,x,k,l,r):
if (r <= s or t <= l):
return
if (s <= l and r <= t):
A[k] = x
if(k < N-1):
lazy_A[k] = x
return
tree_propagate(k)
center = (l+r)//2
update(s, t, x, k*2 + 1, l, center)
update(s, t, x, k*2 + 2, center, r)
A[k] = min(A[k*2+1], A[k*2+2])
def find(s,t,k,l,r):
if (r <= s or t <= l):
return INT_MAX
if(s <= l and r <= t):
return A[k]
tree_propagate(k)
center = (l+r)//2
v_left = find(s, t, k*2 + 1, l, center)
v_right = find(s, t, k*2 + 2, center, r)
return min(v_left,v_right)
answers = []
for query in queries:
# update
if(query[0]==0):
update(query[1],query[2]+1,query[3],0,0,N)
# find
else:
answers.append(find(query[1],query[2]+1,0,0,N))
print("\n".join([str(ans) for ans in answers]))
```
No
| 100,317 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Write a program which manipulates a sequence A = {a0, a1, . . . , an−1} with the following operations:
* update(s, t, x): change as, as+1, ..., at to x.
* find(s, t): report the minimum element in as, as+1, ..., at.
Note that the initial values of ai (i = 0, 1, . . . , n−1) are 231-1.
Constraints
* 1 ≤ n ≤ 100000
* 1 ≤ q ≤ 100000
* 0 ≤ s ≤ t < n
* 0 ≤ x < 231−1
Input
n q
query1
query2
:
queryq
In the first line, n (the number of elements in A) and q (the number of queries) are given. Then, ith query queryi is given in the following format:
0 s t x
or
1 s t
The first digit represents the type of the query. '0' denotes update(s, t, x) and '1' denotes find(s, t).
Output
For each find operation, print the minimum value.
Examples
Input
3 5
0 0 1 1
0 1 2 3
0 2 2 2
1 0 2
1 1 2
Output
1
2
Input
1 3
1 0 0
0 0 0 5
1 0 0
Output
2147483647
5
Submitted Solution:
```
import math
class SegmentTree:
__slots__ = ["elem_size", "tree_size", "tree", "lazy"]
def __init__(self, a: list, default: int):
real_size = len(a)
self.elem_size = 1 << math.ceil(math.log2(real_size))
self.tree_size = 2 * self.elem_size
self.tree = [default]*self.elem_size + a + [default]*(self.elem_size - real_size)
self.lazy = [None]*self.tree_size
self.init_tree()
def init_tree(self):
tree = self.tree
for i in range(self.elem_size-1, 0, -1):
left, right = tree[i << 1], tree[(i << 1)+1]
# ===== change me =====
tree[i] = left if left < right else right
def get_indexes(self, l: int, r: int):
'''[x, y)'''
l, r, indexes = l+self.elem_size, r+self.elem_size, []
append = indexes.append
while l < r:
if l & 1:
append(l)
l += 1
if r & 1:
r -= 1
append(r)
l, r = l >> 1, r >> 1
return indexes
def get_indexes_with_propagation(self, l: int, r: int, current_node: int, l_end: int, r_end: int):
# print(l,r,current_node,l_end,r_end,self.lazy[current_node])
indexes = []
tree, lazy = self.tree, self.lazy
lazy_value = lazy[current_node]
left_child, right_child = current_node << 1, (current_node << 1) + 1
if lazy_value is not None:
self.lazy[current_node] = None
tree[current_node] = lazy_value
if left_child < self.tree_size:
if left_child < self.elem_size:
lazy[left_child] = lazy[right_child] = lazy_value
else:
tree[left_child] = tree[right_child] = lazy_value
if l == l_end and r == r_end:
return [current_node]
mid = (l_end + r_end) // 2
if l < mid:
l_r = r if r < mid else mid
indexes += self.get_indexes_with_propagation(l, l_r, left_child, l_end, mid)
if r > mid:
r_l = l if mid < l else mid
indexes += self.get_indexes_with_propagation(r_l, r, right_child, mid, r_end)
return indexes
def update_lazy(self, l, r, value):
indexes = sorted(self.get_indexes_with_propagation(l, r, 1, 0, self.elem_size))
l = self.elem_size
lazy, tree = self.lazy, self.tree
for n in indexes:
lazy[n] = None
if n < l//2:
lazy[n << 1] = value
lazy[(n << 1)+1] = value
elif n < l:
tree[n << 1] = value
tree[(n << 1)+1] = value
else:
tree[n] = value
self.update_tree(n)
def get_value(self, l: int, r: int):
tree = self.tree
index_list = self.get_indexes_with_propagation(l, r, 1, 0, self.elem_size)
# ===== change me =====
return min(tree[n] for n in index_list)
def update_tree(self, k: int):
tree, lazy = self.tree, self.lazy
if k >= self.elem_size:
k >>= 1
while k > 1:
left, right = k << 1, (k << 1)+1
left_value = lazy[left] or tree[left]
right_value = lazy[right] or tree[right]
# ===== change me =====
tree[k] = left_value if left_value < right_value else right_value
k >>= 1
def set_value(self, i: int, value: int, op: str):
k = self.elem_size + i
if op == "=":
self.tree[k] = value
elif op == "+":
self.tree[k] += value
self.update_tree(k)
n, q = map(int, input().split())
rmq = SegmentTree([2**31-1]*n, 2**31-1)
ans = []
for _ in [0]*q:
l = list(map(int, input().split()))
if l[0] == 0:
rmq.update_lazy(l[1], l[2]+1, l[3])
else:
ans.append(rmq.get_value(l[1], l[2]+1))
print("\n".join([str(n) for n in ans]))
```
No
| 100,318 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Now Vasya is taking an exam in mathematics. In order to get a good mark, Vasya needs to guess the matrix that the teacher has constructed!
Vasya knows that the matrix consists of n rows and m columns. For each row, he knows the xor (bitwise excluding or) of the elements in this row. The sequence a1, a2, ..., an denotes the xor of elements in rows with indices 1, 2, ..., n, respectively. Similarly, for each column, he knows the xor of the elements in this column. The sequence b1, b2, ..., bm denotes the xor of elements in columns with indices 1, 2, ..., m, respectively.
Help Vasya! Find a matrix satisfying the given constraints or tell him that there is no suitable matrix.
Input
The first line contains two numbers n and m (2 ≤ n, m ≤ 100) — the dimensions of the matrix.
The second line contains n numbers a1, a2, ..., an (0 ≤ ai ≤ 109), where ai is the xor of all elements in row i.
The third line contains m numbers b1, b2, ..., bm (0 ≤ bi ≤ 109), where bi is the xor of all elements in column i.
Output
If there is no matrix satisfying the given constraints in the first line, output "NO".
Otherwise, on the first line output "YES", and then n rows of m numbers in each ci1, ci2, ... , cim (0 ≤ cij ≤ 2·109) — the description of the matrix.
If there are several suitable matrices, it is allowed to print any of them.
Examples
Input
2 3
2 9
5 3 13
Output
YES
3 4 5
6 7 8
Input
3 3
1 7 6
2 15 12
Output
NO
Tags: constructive algorithms, flows, math
Correct Solution:
```
USE_STDIO = False
if not USE_STDIO:
try: import mypc
except: pass
def getXor(a):
ans = 0
for x in a:
ans ^= x
return ans
def main():
n, m = map(int, input().split(' '))
a = [x for x in map(int, input().split(' '))]
b = [x for x in map(int, input().split(' '))]
rxor = getXor(a)
cxor = getXor(b)
if rxor != cxor:
print('NO')
return
x = [[0] * m for _ in range(n)]
if n >= m:
x[0][0] = a[0]
for i in range(1, m):
x[i][i-1] = b[i-1] ^ x[i-1][i-1]
x[i][i] = a[i] ^ x[i][i-1]
for i in range(m, n):
x[i][-1] = a[i]
else:
x[0][0] = b[0]
for i in range(1, n):
x[i-1][i] = a[i-1] ^ x[i-1][i-1]
x[i][i] = b[i] ^ x[i-1][i]
for i in range(n, m):
x[-1][i] = b[i]
print('YES')
for i in range(n):
print(' '.join(map(str, x[i])))
if __name__ == '__main__':
main()
```
| 100,319 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Now Vasya is taking an exam in mathematics. In order to get a good mark, Vasya needs to guess the matrix that the teacher has constructed!
Vasya knows that the matrix consists of n rows and m columns. For each row, he knows the xor (bitwise excluding or) of the elements in this row. The sequence a1, a2, ..., an denotes the xor of elements in rows with indices 1, 2, ..., n, respectively. Similarly, for each column, he knows the xor of the elements in this column. The sequence b1, b2, ..., bm denotes the xor of elements in columns with indices 1, 2, ..., m, respectively.
Help Vasya! Find a matrix satisfying the given constraints or tell him that there is no suitable matrix.
Input
The first line contains two numbers n and m (2 ≤ n, m ≤ 100) — the dimensions of the matrix.
The second line contains n numbers a1, a2, ..., an (0 ≤ ai ≤ 109), where ai is the xor of all elements in row i.
The third line contains m numbers b1, b2, ..., bm (0 ≤ bi ≤ 109), where bi is the xor of all elements in column i.
Output
If there is no matrix satisfying the given constraints in the first line, output "NO".
Otherwise, on the first line output "YES", and then n rows of m numbers in each ci1, ci2, ... , cim (0 ≤ cij ≤ 2·109) — the description of the matrix.
If there are several suitable matrices, it is allowed to print any of them.
Examples
Input
2 3
2 9
5 3 13
Output
YES
3 4 5
6 7 8
Input
3 3
1 7 6
2 15 12
Output
NO
Tags: constructive algorithms, flows, math
Correct Solution:
```
n,m = map(int,input().split())
r = list(map(int,input().split()))
c = list(map(int,input().split()))
mat = [[0]*m for i in range(n)]
fail = False
for p in range(60):
totr = 0
needr = []
totc = 0
needc = []
for i in range(n):
if (r[i]>>p)&1:
needr.append(i)
totr += 1
for j in range(m):
if (c[j]>>p)&1:
needc.append(j)
totc += 1
if (totr-totc) % 2 != 0:
fail = True
break
else:
mn = min(len(needr),len(needc))
for k in range(mn):
mat[needr[k]][needc[k]] += (1 << p)
if len(needr) > len(needc):
em = 0
for i in range(m):
if i not in needc:
em = i
break
for i in range(mn,len(needr)):
mat[needr[i]][em] += (1 << p)
else:
em = 0
for i in range(n):
if i not in needr:
em = i
break
for i in range(mn, len(needc)):
mat[em][needc[i]] += (1 << p)
if fail:
print("NO")
else:
print("YES")
for i in mat:
print(*i)
```
| 100,320 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Now Vasya is taking an exam in mathematics. In order to get a good mark, Vasya needs to guess the matrix that the teacher has constructed!
Vasya knows that the matrix consists of n rows and m columns. For each row, he knows the xor (bitwise excluding or) of the elements in this row. The sequence a1, a2, ..., an denotes the xor of elements in rows with indices 1, 2, ..., n, respectively. Similarly, for each column, he knows the xor of the elements in this column. The sequence b1, b2, ..., bm denotes the xor of elements in columns with indices 1, 2, ..., m, respectively.
Help Vasya! Find a matrix satisfying the given constraints or tell him that there is no suitable matrix.
Input
The first line contains two numbers n and m (2 ≤ n, m ≤ 100) — the dimensions of the matrix.
The second line contains n numbers a1, a2, ..., an (0 ≤ ai ≤ 109), where ai is the xor of all elements in row i.
The third line contains m numbers b1, b2, ..., bm (0 ≤ bi ≤ 109), where bi is the xor of all elements in column i.
Output
If there is no matrix satisfying the given constraints in the first line, output "NO".
Otherwise, on the first line output "YES", and then n rows of m numbers in each ci1, ci2, ... , cim (0 ≤ cij ≤ 2·109) — the description of the matrix.
If there are several suitable matrices, it is allowed to print any of them.
Examples
Input
2 3
2 9
5 3 13
Output
YES
3 4 5
6 7 8
Input
3 3
1 7 6
2 15 12
Output
NO
Tags: constructive algorithms, flows, math
Correct Solution:
```
i=lambda:[*map(int,input().split())]
n,m=i()
a,b=i(),i()
def s(a):
r=0
for i in a: r^=i
return r
if s(a)==s(b):
print("YES")
b[0]^=s(a[1:])
print(*b)
for i in a[1:]:print(str(i)+' 0'*(m-1))
else:
print("NO")
```
| 100,321 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Now Vasya is taking an exam in mathematics. In order to get a good mark, Vasya needs to guess the matrix that the teacher has constructed!
Vasya knows that the matrix consists of n rows and m columns. For each row, he knows the xor (bitwise excluding or) of the elements in this row. The sequence a1, a2, ..., an denotes the xor of elements in rows with indices 1, 2, ..., n, respectively. Similarly, for each column, he knows the xor of the elements in this column. The sequence b1, b2, ..., bm denotes the xor of elements in columns with indices 1, 2, ..., m, respectively.
Help Vasya! Find a matrix satisfying the given constraints or tell him that there is no suitable matrix.
Input
The first line contains two numbers n and m (2 ≤ n, m ≤ 100) — the dimensions of the matrix.
The second line contains n numbers a1, a2, ..., an (0 ≤ ai ≤ 109), where ai is the xor of all elements in row i.
The third line contains m numbers b1, b2, ..., bm (0 ≤ bi ≤ 109), where bi is the xor of all elements in column i.
Output
If there is no matrix satisfying the given constraints in the first line, output "NO".
Otherwise, on the first line output "YES", and then n rows of m numbers in each ci1, ci2, ... , cim (0 ≤ cij ≤ 2·109) — the description of the matrix.
If there are several suitable matrices, it is allowed to print any of them.
Examples
Input
2 3
2 9
5 3 13
Output
YES
3 4 5
6 7 8
Input
3 3
1 7 6
2 15 12
Output
NO
Tags: constructive algorithms, flows, math
Correct Solution:
```
from functools import reduce
n, m = map(int, input().split())
a = list(map(int, input().split()))
b = list(map(int, input().split()))
if reduce( (lambda x, y: x^y), a) != reduce( (lambda x, y: x^y), b):
print('NO')
else:
print('YES')
for row in range(n-1):
for col in range(m - 1):
print(0, end=' ')
print(a[row])
for col in range(m - 1):
print(b[col], end=' ')
print(reduce( (lambda x, y: x^y), b[:-1]) ^ a[n-1])
# print(3 ^ 0)
```
| 100,322 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Now Vasya is taking an exam in mathematics. In order to get a good mark, Vasya needs to guess the matrix that the teacher has constructed!
Vasya knows that the matrix consists of n rows and m columns. For each row, he knows the xor (bitwise excluding or) of the elements in this row. The sequence a1, a2, ..., an denotes the xor of elements in rows with indices 1, 2, ..., n, respectively. Similarly, for each column, he knows the xor of the elements in this column. The sequence b1, b2, ..., bm denotes the xor of elements in columns with indices 1, 2, ..., m, respectively.
Help Vasya! Find a matrix satisfying the given constraints or tell him that there is no suitable matrix.
Input
The first line contains two numbers n and m (2 ≤ n, m ≤ 100) — the dimensions of the matrix.
The second line contains n numbers a1, a2, ..., an (0 ≤ ai ≤ 109), where ai is the xor of all elements in row i.
The third line contains m numbers b1, b2, ..., bm (0 ≤ bi ≤ 109), where bi is the xor of all elements in column i.
Output
If there is no matrix satisfying the given constraints in the first line, output "NO".
Otherwise, on the first line output "YES", and then n rows of m numbers in each ci1, ci2, ... , cim (0 ≤ cij ≤ 2·109) — the description of the matrix.
If there are several suitable matrices, it is allowed to print any of them.
Examples
Input
2 3
2 9
5 3 13
Output
YES
3 4 5
6 7 8
Input
3 3
1 7 6
2 15 12
Output
NO
Tags: constructive algorithms, flows, math
Correct Solution:
```
n, m = map(int, input().split())
xor_sum = []
a = []
for i in range(2):
a.append(list(map(int, input().split())))
xor_sum.append(0)
for x in a[-1]:
xor_sum[-1] ^= x
if xor_sum[0] != xor_sum[1]:
print('NO')
exit(0)
print('YES')
for i in range(n):
for j in range(m):
if (i < n - 1) and (j < m - 1):
print(0, end = ' ')
elif (i == n - 1) and (j < m - 1):
print(a[1][j], end = ' ')
elif (i < n - 1) and (j == m - 1):
print(a[0][i], end = ' ')
else:
t = a[0][-1]
for k in range(0, m - 1):
t ^= a[1][k]
print(t, end = ' ')
print()
```
| 100,323 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Now Vasya is taking an exam in mathematics. In order to get a good mark, Vasya needs to guess the matrix that the teacher has constructed!
Vasya knows that the matrix consists of n rows and m columns. For each row, he knows the xor (bitwise excluding or) of the elements in this row. The sequence a1, a2, ..., an denotes the xor of elements in rows with indices 1, 2, ..., n, respectively. Similarly, for each column, he knows the xor of the elements in this column. The sequence b1, b2, ..., bm denotes the xor of elements in columns with indices 1, 2, ..., m, respectively.
Help Vasya! Find a matrix satisfying the given constraints or tell him that there is no suitable matrix.
Input
The first line contains two numbers n and m (2 ≤ n, m ≤ 100) — the dimensions of the matrix.
The second line contains n numbers a1, a2, ..., an (0 ≤ ai ≤ 109), where ai is the xor of all elements in row i.
The third line contains m numbers b1, b2, ..., bm (0 ≤ bi ≤ 109), where bi is the xor of all elements in column i.
Output
If there is no matrix satisfying the given constraints in the first line, output "NO".
Otherwise, on the first line output "YES", and then n rows of m numbers in each ci1, ci2, ... , cim (0 ≤ cij ≤ 2·109) — the description of the matrix.
If there are several suitable matrices, it is allowed to print any of them.
Examples
Input
2 3
2 9
5 3 13
Output
YES
3 4 5
6 7 8
Input
3 3
1 7 6
2 15 12
Output
NO
Tags: constructive algorithms, flows, math
Correct Solution:
```
n,m=map(int,input().split())
a=list(map(int,input().split()))
b=list(map(int,input().split()))
x=a[-1]
y=b[-1]
for j in range(m-1):
x=x^b[j]
for j in range(n-1):
y=y^a[j]
if x!=y:
print("NO")
else:
print("YES")
c=[]
for i in range(n):
c.append([])
for i in range(n):
for j in range(m):
if i!=(n-1) and j!=(m-1):
c[i].append(0)
elif i!=(n-1) and j==(m-1):
c[i].append(a[i])
elif i==(n-1) and j!=(m-1):
c[i].append(b[j])
else:
c[i].append(x)
for j in c:
print(*j)
```
| 100,324 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Now Vasya is taking an exam in mathematics. In order to get a good mark, Vasya needs to guess the matrix that the teacher has constructed!
Vasya knows that the matrix consists of n rows and m columns. For each row, he knows the xor (bitwise excluding or) of the elements in this row. The sequence a1, a2, ..., an denotes the xor of elements in rows with indices 1, 2, ..., n, respectively. Similarly, for each column, he knows the xor of the elements in this column. The sequence b1, b2, ..., bm denotes the xor of elements in columns with indices 1, 2, ..., m, respectively.
Help Vasya! Find a matrix satisfying the given constraints or tell him that there is no suitable matrix.
Input
The first line contains two numbers n and m (2 ≤ n, m ≤ 100) — the dimensions of the matrix.
The second line contains n numbers a1, a2, ..., an (0 ≤ ai ≤ 109), where ai is the xor of all elements in row i.
The third line contains m numbers b1, b2, ..., bm (0 ≤ bi ≤ 109), where bi is the xor of all elements in column i.
Output
If there is no matrix satisfying the given constraints in the first line, output "NO".
Otherwise, on the first line output "YES", and then n rows of m numbers in each ci1, ci2, ... , cim (0 ≤ cij ≤ 2·109) — the description of the matrix.
If there are several suitable matrices, it is allowed to print any of them.
Examples
Input
2 3
2 9
5 3 13
Output
YES
3 4 5
6 7 8
Input
3 3
1 7 6
2 15 12
Output
NO
Tags: constructive algorithms, flows, math
Correct Solution:
```
n,m=map(int,input().split())
a=list(map(int,input().split()))
b=list(map(int,input().split()))
xorall=0
def xor(a):
x=0
for i in a:
x^=i
return x
if xor(a)!=xor(b):
print("NO")
else:
print("YES")
a[0]^=xor(b[1:])
print(a[0],*b[1:])
for i in a[1:]:
print(str(i)+' 0'*(m-1))
```
| 100,325 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Now Vasya is taking an exam in mathematics. In order to get a good mark, Vasya needs to guess the matrix that the teacher has constructed!
Vasya knows that the matrix consists of n rows and m columns. For each row, he knows the xor (bitwise excluding or) of the elements in this row. The sequence a1, a2, ..., an denotes the xor of elements in rows with indices 1, 2, ..., n, respectively. Similarly, for each column, he knows the xor of the elements in this column. The sequence b1, b2, ..., bm denotes the xor of elements in columns with indices 1, 2, ..., m, respectively.
Help Vasya! Find a matrix satisfying the given constraints or tell him that there is no suitable matrix.
Input
The first line contains two numbers n and m (2 ≤ n, m ≤ 100) — the dimensions of the matrix.
The second line contains n numbers a1, a2, ..., an (0 ≤ ai ≤ 109), where ai is the xor of all elements in row i.
The third line contains m numbers b1, b2, ..., bm (0 ≤ bi ≤ 109), where bi is the xor of all elements in column i.
Output
If there is no matrix satisfying the given constraints in the first line, output "NO".
Otherwise, on the first line output "YES", and then n rows of m numbers in each ci1, ci2, ... , cim (0 ≤ cij ≤ 2·109) — the description of the matrix.
If there are several suitable matrices, it is allowed to print any of them.
Examples
Input
2 3
2 9
5 3 13
Output
YES
3 4 5
6 7 8
Input
3 3
1 7 6
2 15 12
Output
NO
Tags: constructive algorithms, flows, math
Correct Solution:
```
###### ### ####### ####### ## # ##### ### #####
# # # # # # # # # # # # # ###
# # # # # # # # # # # # # ###
###### ######### # # # # # # ######### #
###### ######### # # # # # # ######### #
# # # # # # # # # # #### # # #
# # # # # # # ## # # # # #
###### # # ####### ####### # # ##### # # # #
from __future__ import print_function # for PyPy2
# from itertools import permutations as perm
# from functools import cmp_to_key # for adding custom comparator
# from fractions import Fraction
# from collections import *
from sys import stdin
# from bisect import *
# from heapq import *
# from math import *
g = lambda : stdin.readline().strip()
gl = lambda : g().split()
gil = lambda : [int(var) for var in gl()]
gfl = lambda : [float(var) for var in gl()]
gcl = lambda : list(g())
gbs = lambda : [int(var) for var in g()]
mod = int(1e9)+7
inf = float("inf")
n, m = gil()
r, c = gil(), gil()
rv, cv = 0, 0
for v in r:
rv ^= v
for v in c:
cv ^= v
if rv == cv:
print('YES')
x = r[0]^cv^c[0]
ans = [[0 for _ in range(m)] for _ in range(n)]
for j in range(1, m):
ans[0][j] = c[j]
for i in range(1, n):
ans[i][0] = r[i]
ans[0][0] = x
for r in ans:
print(*r)
else:
print('NO')
```
| 100,326 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Now Vasya is taking an exam in mathematics. In order to get a good mark, Vasya needs to guess the matrix that the teacher has constructed!
Vasya knows that the matrix consists of n rows and m columns. For each row, he knows the xor (bitwise excluding or) of the elements in this row. The sequence a1, a2, ..., an denotes the xor of elements in rows with indices 1, 2, ..., n, respectively. Similarly, for each column, he knows the xor of the elements in this column. The sequence b1, b2, ..., bm denotes the xor of elements in columns with indices 1, 2, ..., m, respectively.
Help Vasya! Find a matrix satisfying the given constraints or tell him that there is no suitable matrix.
Input
The first line contains two numbers n and m (2 ≤ n, m ≤ 100) — the dimensions of the matrix.
The second line contains n numbers a1, a2, ..., an (0 ≤ ai ≤ 109), where ai is the xor of all elements in row i.
The third line contains m numbers b1, b2, ..., bm (0 ≤ bi ≤ 109), where bi is the xor of all elements in column i.
Output
If there is no matrix satisfying the given constraints in the first line, output "NO".
Otherwise, on the first line output "YES", and then n rows of m numbers in each ci1, ci2, ... , cim (0 ≤ cij ≤ 2·109) — the description of the matrix.
If there are several suitable matrices, it is allowed to print any of them.
Examples
Input
2 3
2 9
5 3 13
Output
YES
3 4 5
6 7 8
Input
3 3
1 7 6
2 15 12
Output
NO
Submitted Solution:
```
n,m = list(map(int,input().split(' ')))
a = list(map(int,input().split(' ')))
b = list(map(int,input().split(' ')))
aa = 0
bb = 0
cc = 0
for i in range(max(n,m)):
if i == n-1:
cc = aa
if i<n:
aa ^= a[i]
if i<m:
bb ^= b[i]
if aa != bb:
print('NO')
else:
print('YES')
for i in range(n-1):
for j in range(m):
if j < m-1:
print('0',end=' ')
else:
print(a[i])
for j in range(m-1):
print(b[j], end=' ')
print(cc^b[-1])
```
Yes
| 100,327 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Now Vasya is taking an exam in mathematics. In order to get a good mark, Vasya needs to guess the matrix that the teacher has constructed!
Vasya knows that the matrix consists of n rows and m columns. For each row, he knows the xor (bitwise excluding or) of the elements in this row. The sequence a1, a2, ..., an denotes the xor of elements in rows with indices 1, 2, ..., n, respectively. Similarly, for each column, he knows the xor of the elements in this column. The sequence b1, b2, ..., bm denotes the xor of elements in columns with indices 1, 2, ..., m, respectively.
Help Vasya! Find a matrix satisfying the given constraints or tell him that there is no suitable matrix.
Input
The first line contains two numbers n and m (2 ≤ n, m ≤ 100) — the dimensions of the matrix.
The second line contains n numbers a1, a2, ..., an (0 ≤ ai ≤ 109), where ai is the xor of all elements in row i.
The third line contains m numbers b1, b2, ..., bm (0 ≤ bi ≤ 109), where bi is the xor of all elements in column i.
Output
If there is no matrix satisfying the given constraints in the first line, output "NO".
Otherwise, on the first line output "YES", and then n rows of m numbers in each ci1, ci2, ... , cim (0 ≤ cij ≤ 2·109) — the description of the matrix.
If there are several suitable matrices, it is allowed to print any of them.
Examples
Input
2 3
2 9
5 3 13
Output
YES
3 4 5
6 7 8
Input
3 3
1 7 6
2 15 12
Output
NO
Submitted Solution:
```
n, m = map(int, input().split())
a = [int(i) for i in input().split()]
b = [int(i) for i in input().split()]
ans = [[int(0)for i in range(m)] for j in range(n)]
for i in range(n):
ans[i][0] = a[i]
for i in range(m):
ans[0][i] = b[i]
ans[0][0] = 0
kekw = b[0]
for i in range(1,n):
kekw ^= a[i]
kekw1 = a[0]
for j in range(1,m):
kekw1 ^= b[j]
if kekw != kekw1:
print("NO")
else:
ans[0][0] = kekw
print("YES")
for i in range(n):
print(*ans[i])
```
Yes
| 100,328 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Now Vasya is taking an exam in mathematics. In order to get a good mark, Vasya needs to guess the matrix that the teacher has constructed!
Vasya knows that the matrix consists of n rows and m columns. For each row, he knows the xor (bitwise excluding or) of the elements in this row. The sequence a1, a2, ..., an denotes the xor of elements in rows with indices 1, 2, ..., n, respectively. Similarly, for each column, he knows the xor of the elements in this column. The sequence b1, b2, ..., bm denotes the xor of elements in columns with indices 1, 2, ..., m, respectively.
Help Vasya! Find a matrix satisfying the given constraints or tell him that there is no suitable matrix.
Input
The first line contains two numbers n and m (2 ≤ n, m ≤ 100) — the dimensions of the matrix.
The second line contains n numbers a1, a2, ..., an (0 ≤ ai ≤ 109), where ai is the xor of all elements in row i.
The third line contains m numbers b1, b2, ..., bm (0 ≤ bi ≤ 109), where bi is the xor of all elements in column i.
Output
If there is no matrix satisfying the given constraints in the first line, output "NO".
Otherwise, on the first line output "YES", and then n rows of m numbers in each ci1, ci2, ... , cim (0 ≤ cij ≤ 2·109) — the description of the matrix.
If there are several suitable matrices, it is allowed to print any of them.
Examples
Input
2 3
2 9
5 3 13
Output
YES
3 4 5
6 7 8
Input
3 3
1 7 6
2 15 12
Output
NO
Submitted Solution:
```
I=input
P=print
R=range
I()
a=[int(i)for i in I().split()]
b=[int(i)for i in I().split()]
n=len(a)
m=len(b)
x=b[-1]
for i in R(n-1):x^=a[i]
y=a[-1]
for i in R(m-1):y^=b[i]
if x!=y:P("NO");exit()
P("YES")
for i in R(n-1):P("0 "*(m-1)+str(a[i]))
for i in R(m-1):P(b[i],end=" ")
P(x)
```
Yes
| 100,329 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Now Vasya is taking an exam in mathematics. In order to get a good mark, Vasya needs to guess the matrix that the teacher has constructed!
Vasya knows that the matrix consists of n rows and m columns. For each row, he knows the xor (bitwise excluding or) of the elements in this row. The sequence a1, a2, ..., an denotes the xor of elements in rows with indices 1, 2, ..., n, respectively. Similarly, for each column, he knows the xor of the elements in this column. The sequence b1, b2, ..., bm denotes the xor of elements in columns with indices 1, 2, ..., m, respectively.
Help Vasya! Find a matrix satisfying the given constraints or tell him that there is no suitable matrix.
Input
The first line contains two numbers n and m (2 ≤ n, m ≤ 100) — the dimensions of the matrix.
The second line contains n numbers a1, a2, ..., an (0 ≤ ai ≤ 109), where ai is the xor of all elements in row i.
The third line contains m numbers b1, b2, ..., bm (0 ≤ bi ≤ 109), where bi is the xor of all elements in column i.
Output
If there is no matrix satisfying the given constraints in the first line, output "NO".
Otherwise, on the first line output "YES", and then n rows of m numbers in each ci1, ci2, ... , cim (0 ≤ cij ≤ 2·109) — the description of the matrix.
If there are several suitable matrices, it is allowed to print any of them.
Examples
Input
2 3
2 9
5 3 13
Output
YES
3 4 5
6 7 8
Input
3 3
1 7 6
2 15 12
Output
NO
Submitted Solution:
```
n,m = map(int,input().split())
a = list(map(int,input().split()))
b = list(map(int,input().split()))
x1,x2 = a[0],b[0]
for i in range(1,n):
x1 ^= a[i]
for i in range(1,m):
x2 ^= b[i]
if x1!=x2:
print ("NO")
exit()
ans = [[0 for i in range(m)] for j in range(n)]
ans[0][0] = a[0]^x2^b[0]
for i in range(1,m):
ans[0][i] = b[i]
for i in range(1,n):
ans[i][0] = a[i]
print ("YES")
for i in range(n):
print (*ans[i])
```
Yes
| 100,330 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Now Vasya is taking an exam in mathematics. In order to get a good mark, Vasya needs to guess the matrix that the teacher has constructed!
Vasya knows that the matrix consists of n rows and m columns. For each row, he knows the xor (bitwise excluding or) of the elements in this row. The sequence a1, a2, ..., an denotes the xor of elements in rows with indices 1, 2, ..., n, respectively. Similarly, for each column, he knows the xor of the elements in this column. The sequence b1, b2, ..., bm denotes the xor of elements in columns with indices 1, 2, ..., m, respectively.
Help Vasya! Find a matrix satisfying the given constraints or tell him that there is no suitable matrix.
Input
The first line contains two numbers n and m (2 ≤ n, m ≤ 100) — the dimensions of the matrix.
The second line contains n numbers a1, a2, ..., an (0 ≤ ai ≤ 109), where ai is the xor of all elements in row i.
The third line contains m numbers b1, b2, ..., bm (0 ≤ bi ≤ 109), where bi is the xor of all elements in column i.
Output
If there is no matrix satisfying the given constraints in the first line, output "NO".
Otherwise, on the first line output "YES", and then n rows of m numbers in each ci1, ci2, ... , cim (0 ≤ cij ≤ 2·109) — the description of the matrix.
If there are several suitable matrices, it is allowed to print any of them.
Examples
Input
2 3
2 9
5 3 13
Output
YES
3 4 5
6 7 8
Input
3 3
1 7 6
2 15 12
Output
NO
Submitted Solution:
```
def getXor(a):
ans = 0
for x in a:
ans |= x
return ans
def main():
n, m = map(int, input().split(' '))
a = [x for x in map(int, input().split(' '))]
b = [x for x in map(int, input().split(' '))]
rxor = getXor(a)
cxor = getXor(b)
if rxor != cxor:
print('NO')
return
print('YES')
for i in range(n-1):
print(a[i], end='')
for j in range(1, m):
print(' 0', end='')
print()
print(b[0] | getXor(a[:-1]), end='')
for j in range(1, m):
print(' %d' % b[i], end ='')
print()
if __name__ == '__main__':
main()
```
No
| 100,331 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Now Vasya is taking an exam in mathematics. In order to get a good mark, Vasya needs to guess the matrix that the teacher has constructed!
Vasya knows that the matrix consists of n rows and m columns. For each row, he knows the xor (bitwise excluding or) of the elements in this row. The sequence a1, a2, ..., an denotes the xor of elements in rows with indices 1, 2, ..., n, respectively. Similarly, for each column, he knows the xor of the elements in this column. The sequence b1, b2, ..., bm denotes the xor of elements in columns with indices 1, 2, ..., m, respectively.
Help Vasya! Find a matrix satisfying the given constraints or tell him that there is no suitable matrix.
Input
The first line contains two numbers n and m (2 ≤ n, m ≤ 100) — the dimensions of the matrix.
The second line contains n numbers a1, a2, ..., an (0 ≤ ai ≤ 109), where ai is the xor of all elements in row i.
The third line contains m numbers b1, b2, ..., bm (0 ≤ bi ≤ 109), where bi is the xor of all elements in column i.
Output
If there is no matrix satisfying the given constraints in the first line, output "NO".
Otherwise, on the first line output "YES", and then n rows of m numbers in each ci1, ci2, ... , cim (0 ≤ cij ≤ 2·109) — the description of the matrix.
If there are several suitable matrices, it is allowed to print any of them.
Examples
Input
2 3
2 9
5 3 13
Output
YES
3 4 5
6 7 8
Input
3 3
1 7 6
2 15 12
Output
NO
Submitted Solution:
```
def getXor(a):
ans = 0
for x in a:
ans ^= x
return ans
def main():
n, m = map(int, input().split(' '))
a = [x for x in map(int, input().split(' '))]
b = [x for x in map(int, input().split(' '))]
rxor = getXor(a)
cxor = getXor(b)
if rxor != cxor:
print('NO')
return
print('YES')
for i in range(n-1):
print(a[i], end='')
for j in range(1, m):
print(' 0', end='')
print()
print(b[0] ^ getXor(a[:-1]), end='')
for j in range(1, m):
print(' %d' % b[i], end ='')
print()
if __name__ == '__main__':
main()
```
No
| 100,332 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Now Vasya is taking an exam in mathematics. In order to get a good mark, Vasya needs to guess the matrix that the teacher has constructed!
Vasya knows that the matrix consists of n rows and m columns. For each row, he knows the xor (bitwise excluding or) of the elements in this row. The sequence a1, a2, ..., an denotes the xor of elements in rows with indices 1, 2, ..., n, respectively. Similarly, for each column, he knows the xor of the elements in this column. The sequence b1, b2, ..., bm denotes the xor of elements in columns with indices 1, 2, ..., m, respectively.
Help Vasya! Find a matrix satisfying the given constraints or tell him that there is no suitable matrix.
Input
The first line contains two numbers n and m (2 ≤ n, m ≤ 100) — the dimensions of the matrix.
The second line contains n numbers a1, a2, ..., an (0 ≤ ai ≤ 109), where ai is the xor of all elements in row i.
The third line contains m numbers b1, b2, ..., bm (0 ≤ bi ≤ 109), where bi is the xor of all elements in column i.
Output
If there is no matrix satisfying the given constraints in the first line, output "NO".
Otherwise, on the first line output "YES", and then n rows of m numbers in each ci1, ci2, ... , cim (0 ≤ cij ≤ 2·109) — the description of the matrix.
If there are several suitable matrices, it is allowed to print any of them.
Examples
Input
2 3
2 9
5 3 13
Output
YES
3 4 5
6 7 8
Input
3 3
1 7 6
2 15 12
Output
NO
Submitted Solution:
```
def getXor(a):
ans = 0
for x in a:
ans ^= x
return ans
def main():
n, m = map(int, input().split(' '))
a = [x for x in map(int, input().split(' '))]
b = [x for x in map(int, input().split(' '))]
rxor = getXor(a)
cxor = getXor(b)
if rxor != cxor:
print('NO')
return
x = [[0] * m for _ in range(n)]
x[0][0] = a[0]
for i in range(1, min(n, m)):
x[i][i-1] = b[i-1] ^ x[i-1][i-1]
x[i][i] = a[i] ^ x[i][i-1]
for i in range(m-1, n):
x[i][-1] = a[i]
for j in range(n-1, m):
x[-1][j] = b[j]
for i in range(n):
print(' '.join(map(str, x[i])))
if __name__ == '__main__':
main()
```
No
| 100,333 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Now Vasya is taking an exam in mathematics. In order to get a good mark, Vasya needs to guess the matrix that the teacher has constructed!
Vasya knows that the matrix consists of n rows and m columns. For each row, he knows the xor (bitwise excluding or) of the elements in this row. The sequence a1, a2, ..., an denotes the xor of elements in rows with indices 1, 2, ..., n, respectively. Similarly, for each column, he knows the xor of the elements in this column. The sequence b1, b2, ..., bm denotes the xor of elements in columns with indices 1, 2, ..., m, respectively.
Help Vasya! Find a matrix satisfying the given constraints or tell him that there is no suitable matrix.
Input
The first line contains two numbers n and m (2 ≤ n, m ≤ 100) — the dimensions of the matrix.
The second line contains n numbers a1, a2, ..., an (0 ≤ ai ≤ 109), where ai is the xor of all elements in row i.
The third line contains m numbers b1, b2, ..., bm (0 ≤ bi ≤ 109), where bi is the xor of all elements in column i.
Output
If there is no matrix satisfying the given constraints in the first line, output "NO".
Otherwise, on the first line output "YES", and then n rows of m numbers in each ci1, ci2, ... , cim (0 ≤ cij ≤ 2·109) — the description of the matrix.
If there are several suitable matrices, it is allowed to print any of them.
Examples
Input
2 3
2 9
5 3 13
Output
YES
3 4 5
6 7 8
Input
3 3
1 7 6
2 15 12
Output
NO
Submitted Solution:
```
USE_STDIO = False
if not USE_STDIO:
try: import mypc
except: pass
def getXor(a):
ans = 0
for x in a:
ans ^= x
return ans
def main():
n, m = map(int, input().split(' '))
a = [x for x in map(int, input().split(' '))]
b = [x for x in map(int, input().split(' '))]
rxor = getXor(a)
cxor = getXor(b)
if rxor != cxor:
print('NO')
return
x = [[0] * m for _ in range(n)]
if n >= m:
x[0][0] = a[0]
for i in range(1, m):
x[i][i-1] = b[i-1] ^ x[i-1][i-1]
x[i][i] = a[i] ^ x[i][i-1]
for i in range(m, n):
x[i][-1] = a[i]
else:
x[0][0] = b[0]
for i in range(1, n):
x[i-1][i] = a[i-1] ^ x[i-1][i-1]
x[i][i] = b[i] ^ x[i-1][i]
for i in range(m, n):
x[-1][i] = b[i]
print('YES')
for i in range(n):
print(' '.join(map(str, x[i])))
if __name__ == '__main__':
main()
```
No
| 100,334 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
As you know, the most intelligent beings on the Earth are, of course, cows. This conclusion was reached long ago by the Martian aliens, as well as a number of other intelligent civilizations from outer space.
Sometimes cows gather into cowavans. This seems to be seasonal. But at this time the cows become passive and react poorly to external stimuli. A cowavan is a perfect target for the Martian scientific saucer, it's time for large-scale abductions, or, as the Martians say, raids. Simply put, a cowavan is a set of cows in a row.
If we number all cows in the cowavan with positive integers from 1 to n, then we can formalize the popular model of abduction, known as the (a, b)-Cowavan Raid: first they steal a cow number a, then number a + b, then — number a + 2·b, and so on, until the number of an abducted cow exceeds n. During one raid the cows are not renumbered.
The aliens would be happy to place all the cows on board of their hospitable ship, but unfortunately, the amount of cargo space is very, very limited. The researchers, knowing the mass of each cow in the cowavan, made p scenarios of the (a, b)-raid. Now they want to identify the following thing for each scenario individually: what total mass of pure beef will get on board of the ship. All the scenarios are independent, in the process of performing the calculations the cows are not being stolen.
<image>
Input
The first line contains the only positive integer n (1 ≤ n ≤ 3·105) — the number of cows in the cowavan.
The second number contains n positive integer wi, separated by spaces, where the i-th number describes the mass of the i-th cow in the cowavan (1 ≤ wi ≤ 109).
The third line contains the only positive integer p — the number of scenarios of (a, b)-raids (1 ≤ p ≤ 3·105).
Each following line contains integer parameters a and b of the corresponding scenario (1 ≤ a, b ≤ n).
Output
Print for each scenario of the (a, b)-raid the total mass of cows, that can be stolen using only this scenario.
Please, do not use the %lld specificator to read or write 64-bit integers in С++. It is recommended to use the cin, cout streams of the %I64d specificator.
Examples
Input
3
1 2 3
2
1 1
1 2
Output
6
4
Input
4
2 3 5 7
3
1 3
2 3
2 2
Output
9
3
10
Submitted Solution:
```
#!/usr/bin/env python3
from math import ceil, sqrt
from operator import itemgetter
n = int(input().rstrip())
arr = list(map(int, input().rstrip().split()))
p = int(input().rstrip())
data = [[0, 0, 0, 0] for i in range(p)]
for i in range(p):
a, b = map(int, input().rstrip().split())
data[i][0], data[i][1], data[i][2] = a, b, i
data = sorted(data, key=itemgetter(1))
threshold = ceil(sqrt(n))
dp = [None for i in range(n)]
prev_b = -1
for i in range(p):
if data[i][1] >= threshold:
ans = 0
for j in range(a-1, n, b):
ans += arr[j]
data[i][3] = ans
else:
if data[i][1] != prev_b:
prev_b = data[i][1]
for j in range(n-1, -1, -1):
dp[j] = (0 if j + prev_b >= n else dp[j+prev_b]) + arr[j]
data[i][3] = dp[data[i][0]-1]
data = sorted(data, key=itemgetter(2))
for i in range(p):
print(data[i][3])
```
No
| 100,335 |
Provide tags and a correct Python 3 solution for this coding contest problem.
The Fair Nut is going to travel to the Tree Country, in which there are n cities. Most of the land of this country is covered by forest. Furthermore, the local road system forms a tree (connected graph without cycles). Nut wants to rent a car in the city u and go by a simple path to city v. He hasn't determined the path, so it's time to do it. Note that chosen path can consist of only one vertex.
A filling station is located in every city. Because of strange law, Nut can buy only w_i liters of gasoline in the i-th city. We can assume, that he has infinite money. Each road has a length, and as soon as Nut drives through this road, the amount of gasoline decreases by length. Of course, Nut can't choose a path, which consists of roads, where he runs out of gasoline. He can buy gasoline in every visited city, even in the first and the last.
He also wants to find the maximum amount of gasoline that he can have at the end of the path. Help him: count it.
Input
The first line contains a single integer n (1 ≤ n ≤ 3 ⋅ 10^5) — the number of cities.
The second line contains n integers w_1, w_2, …, w_n (0 ≤ w_{i} ≤ 10^9) — the maximum amounts of liters of gasoline that Nut can buy in cities.
Each of the next n - 1 lines describes road and contains three integers u, v, c (1 ≤ u, v ≤ n, 1 ≤ c ≤ 10^9, u ≠ v), where u and v — cities that are connected by this road and c — its length.
It is guaranteed that graph of road connectivity is a tree.
Output
Print one number — the maximum amount of gasoline that he can have at the end of the path.
Examples
Input
3
1 3 3
1 2 2
1 3 2
Output
3
Input
5
6 3 2 5 0
1 2 10
2 3 3
2 4 1
1 5 1
Output
7
Note
The optimal way in the first example is 2 → 1 → 3.
<image>
The optimal way in the second example is 2 → 4.
<image>
Tags: data structures, dp, trees
Correct Solution:
```
n=int(input())
w=list(map(int,input().split()))
edges=[[] for i in range(n)]
for i in range(n-1):
a,b,c=map(int,input().split())
edges[a-1].append((b-1,c))
edges[b-1].append((a-1,c))
parents=[None]*n
layers=[{0}]
while True:
newlayer=[]
for guy in layers[-1]:
for neigh in edges[guy]:
if neigh[0]!=parents[guy]:
newlayer.append(neigh[0])
parents[neigh[0]]=guy
if newlayer!=[]:
layers.append(set(newlayer))
else:
break
b1=[None]*n
b2=[None]*n
for layer in layers[::-1]:
for guy in layer:
goods=[]
for guy1 in edges[guy]:
if guy1[0]!=parents[guy]:
goods.append(b1[guy1[0]]-guy1[1])
big1=0
big2=0
for boi in goods:
if boi>big1:
big1,big2=boi,big1
elif boi>big2:
big2=boi
b1[guy]=w[guy]+big1
b2[guy]=w[guy]+big1+big2
print(max(b2))
```
| 100,336 |
Provide tags and a correct Python 3 solution for this coding contest problem.
The Fair Nut is going to travel to the Tree Country, in which there are n cities. Most of the land of this country is covered by forest. Furthermore, the local road system forms a tree (connected graph without cycles). Nut wants to rent a car in the city u and go by a simple path to city v. He hasn't determined the path, so it's time to do it. Note that chosen path can consist of only one vertex.
A filling station is located in every city. Because of strange law, Nut can buy only w_i liters of gasoline in the i-th city. We can assume, that he has infinite money. Each road has a length, and as soon as Nut drives through this road, the amount of gasoline decreases by length. Of course, Nut can't choose a path, which consists of roads, where he runs out of gasoline. He can buy gasoline in every visited city, even in the first and the last.
He also wants to find the maximum amount of gasoline that he can have at the end of the path. Help him: count it.
Input
The first line contains a single integer n (1 ≤ n ≤ 3 ⋅ 10^5) — the number of cities.
The second line contains n integers w_1, w_2, …, w_n (0 ≤ w_{i} ≤ 10^9) — the maximum amounts of liters of gasoline that Nut can buy in cities.
Each of the next n - 1 lines describes road and contains three integers u, v, c (1 ≤ u, v ≤ n, 1 ≤ c ≤ 10^9, u ≠ v), where u and v — cities that are connected by this road and c — its length.
It is guaranteed that graph of road connectivity is a tree.
Output
Print one number — the maximum amount of gasoline that he can have at the end of the path.
Examples
Input
3
1 3 3
1 2 2
1 3 2
Output
3
Input
5
6 3 2 5 0
1 2 10
2 3 3
2 4 1
1 5 1
Output
7
Note
The optimal way in the first example is 2 → 1 → 3.
<image>
The optimal way in the second example is 2 → 4.
<image>
Tags: data structures, dp, trees
Correct Solution:
```
import sys
from collections import deque
sys.setrecursionlimit(1000000)
input = sys.stdin.readline
n=int(input())
W=list(map(int,input().split()))
EDGE=[list(map(int,input().split())) for i in range(n-1)]
#n=30000
#W=[10 for i in range(n)]
#EDGE=[[i+1,i+2,1] for i in range(n-1)]
EDGELIST=[[] for i in range(n+1)]
for i,j,c in EDGE:
EDGELIST[i].append([j,c])
EDGELIST[j].append([i,c])
QUE = deque([1])
COST=[[None] for i in range(n+1)]
COST[1]=W[0]
USED=[0]*(n+1)
USED[1]=1
HLIST=[]
EDGELIST2=[[] for i in range(n+1)]
while QUE:
x=QUE.pop()
for to ,length in EDGELIST[x]:
if USED[to]==0:
COST[to]=COST[x]-length+W[to-1]
QUE.append(to)
EDGELIST2[x].append(to)
USED[x]=1
HLIST.append(x)
MAXCOST=dict()
def best(i,j):#i→jから繋がるsubtreeで最大のcost
if MAXCOST.get((i,j))!=None:
return MAXCOST[(i,j)]
else:
ANS=COST[j]
for k,_ in EDGELIST[j]:
if k==i:
continue
if ANS<best(j,k):
ANS=best(j,k)
MAXCOST[(i,j)]=ANS
return ANS
ANS=max(W)
for i in HLIST[::-1]:
for j in EDGELIST2[i]:
if ANS<best(i,j)-COST[i]+W[i-1]:
ANS=best(i,j)-COST[i]+W[i-1]
CLIST=[]
for j in EDGELIST2[i]:
CLIST.append(best(i,j))
if len(CLIST)<=1:
continue
else:
CLIST.sort(reverse=True)
if ANS<CLIST[0]+CLIST[1]-COST[i]*2+W[i-1]:
ANS=CLIST[0]+CLIST[1]-COST[i]*2+W[i-1]
print(ANS)
```
| 100,337 |
Provide tags and a correct Python 3 solution for this coding contest problem.
The Fair Nut is going to travel to the Tree Country, in which there are n cities. Most of the land of this country is covered by forest. Furthermore, the local road system forms a tree (connected graph without cycles). Nut wants to rent a car in the city u and go by a simple path to city v. He hasn't determined the path, so it's time to do it. Note that chosen path can consist of only one vertex.
A filling station is located in every city. Because of strange law, Nut can buy only w_i liters of gasoline in the i-th city. We can assume, that he has infinite money. Each road has a length, and as soon as Nut drives through this road, the amount of gasoline decreases by length. Of course, Nut can't choose a path, which consists of roads, where he runs out of gasoline. He can buy gasoline in every visited city, even in the first and the last.
He also wants to find the maximum amount of gasoline that he can have at the end of the path. Help him: count it.
Input
The first line contains a single integer n (1 ≤ n ≤ 3 ⋅ 10^5) — the number of cities.
The second line contains n integers w_1, w_2, …, w_n (0 ≤ w_{i} ≤ 10^9) — the maximum amounts of liters of gasoline that Nut can buy in cities.
Each of the next n - 1 lines describes road and contains three integers u, v, c (1 ≤ u, v ≤ n, 1 ≤ c ≤ 10^9, u ≠ v), where u and v — cities that are connected by this road and c — its length.
It is guaranteed that graph of road connectivity is a tree.
Output
Print one number — the maximum amount of gasoline that he can have at the end of the path.
Examples
Input
3
1 3 3
1 2 2
1 3 2
Output
3
Input
5
6 3 2 5 0
1 2 10
2 3 3
2 4 1
1 5 1
Output
7
Note
The optimal way in the first example is 2 → 1 → 3.
<image>
The optimal way in the second example is 2 → 4.
<image>
Tags: data structures, dp, trees
Correct Solution:
```
import os,io
input=io.BytesIO(os.read(0,os.fstat(0).st_size)).readline
n=int(input())
w=list(map(int,input().split()))
ans=max(w)
graph=[]
for i in range(n):
graph.append([])
for i in range(n-1):
u,v,c=map(int,input().split())
graph[u-1].append((v-1,c))
graph[v-1].append((u-1,c))
children=[]
parents=[-1]*n
for i in range(n):
children.append([])
stack=[(0,-1)]
while stack:
curr,parent=stack.pop()
for i,j in graph[curr]:
if i!=parent:
children[curr].append((i,j))
parents[i]=curr
stack.append((i,curr))
donechildren=[0]*n
value=[0]*n
stack=[]
for i in range(n):
if not children[i]:
stack.append(i)
while stack:
curr=stack.pop()
k=[]
for child,cost in children[curr]:
k.append(value[child]-cost)
k.sort()
if len(children[curr])>=2:
ans=max(ans,k[-1]+k[-2]+w[curr])
if children[curr]:
value[curr]=max(w[curr],w[curr]+k[-1])
else:
value[curr]=w[curr]
ans=max(ans,value[curr])
if curr==0:
break
parent=parents[curr]
donechildren[parent]+=1
if donechildren[parent]==len(children[parent]):
stack.append(parent)
print(ans)
```
| 100,338 |
Provide tags and a correct Python 3 solution for this coding contest problem.
The Fair Nut is going to travel to the Tree Country, in which there are n cities. Most of the land of this country is covered by forest. Furthermore, the local road system forms a tree (connected graph without cycles). Nut wants to rent a car in the city u and go by a simple path to city v. He hasn't determined the path, so it's time to do it. Note that chosen path can consist of only one vertex.
A filling station is located in every city. Because of strange law, Nut can buy only w_i liters of gasoline in the i-th city. We can assume, that he has infinite money. Each road has a length, and as soon as Nut drives through this road, the amount of gasoline decreases by length. Of course, Nut can't choose a path, which consists of roads, where he runs out of gasoline. He can buy gasoline in every visited city, even in the first and the last.
He also wants to find the maximum amount of gasoline that he can have at the end of the path. Help him: count it.
Input
The first line contains a single integer n (1 ≤ n ≤ 3 ⋅ 10^5) — the number of cities.
The second line contains n integers w_1, w_2, …, w_n (0 ≤ w_{i} ≤ 10^9) — the maximum amounts of liters of gasoline that Nut can buy in cities.
Each of the next n - 1 lines describes road and contains three integers u, v, c (1 ≤ u, v ≤ n, 1 ≤ c ≤ 10^9, u ≠ v), where u and v — cities that are connected by this road and c — its length.
It is guaranteed that graph of road connectivity is a tree.
Output
Print one number — the maximum amount of gasoline that he can have at the end of the path.
Examples
Input
3
1 3 3
1 2 2
1 3 2
Output
3
Input
5
6 3 2 5 0
1 2 10
2 3 3
2 4 1
1 5 1
Output
7
Note
The optimal way in the first example is 2 → 1 → 3.
<image>
The optimal way in the second example is 2 → 4.
<image>
Tags: data structures, dp, trees
Correct Solution:
```
import sys
readline = sys.stdin.readline
from collections import Counter
def parorder(Edge, p):
N = len(Edge)
par = [0]*N
par[p] = -1
stack = [p]
visited = set([p])
order = []
while stack:
vn = stack.pop()
order.append(vn)
for vf in Edge[vn]:
if vf in visited:
continue
visited.add(vf)
par[vf] = vn
stack.append(vf)
return par, order
def getcld(p):
res = [[] for _ in range(len(p))]
for i, v in enumerate(p[1:], 1):
res[v].append(i)
return res
N = int(readline())
We = list(map(int, readline().split()))
Edge = [[] for _ in range(N)]
Cost = Counter()
geta = N+1
for _ in range(N-1):
a, b, c = map(int, readline().split())
a -= 1
b -= 1
Edge[a].append(b)
Edge[b].append(a)
Cost[b*geta+a] = c
Cost[a*geta+b] = c
P, L = parorder(Edge, 0)
#C = getcld(P)
dp = [0]*N
candi = [[0, 0] for _ in range(N)]
ans = 0
for l in L[::-1][:-1]:
dp[l] += We[l]
p = P[l]
k = dp[l] - Cost[l*geta + p]
if k > 0:
dp[p] = max(dp[p], k)
candi[p].append(k)
res = max(candi[l])
candi[l].remove(res)
ans = max(ans, We[l] + res + max(candi[l]))
res = max(candi[0])
candi[0].remove(res)
ans = max(ans, We[0] + res + max(candi[0]))
print(ans)
```
| 100,339 |
Provide tags and a correct Python 3 solution for this coding contest problem.
The Fair Nut is going to travel to the Tree Country, in which there are n cities. Most of the land of this country is covered by forest. Furthermore, the local road system forms a tree (connected graph without cycles). Nut wants to rent a car in the city u and go by a simple path to city v. He hasn't determined the path, so it's time to do it. Note that chosen path can consist of only one vertex.
A filling station is located in every city. Because of strange law, Nut can buy only w_i liters of gasoline in the i-th city. We can assume, that he has infinite money. Each road has a length, and as soon as Nut drives through this road, the amount of gasoline decreases by length. Of course, Nut can't choose a path, which consists of roads, where he runs out of gasoline. He can buy gasoline in every visited city, even in the first and the last.
He also wants to find the maximum amount of gasoline that he can have at the end of the path. Help him: count it.
Input
The first line contains a single integer n (1 ≤ n ≤ 3 ⋅ 10^5) — the number of cities.
The second line contains n integers w_1, w_2, …, w_n (0 ≤ w_{i} ≤ 10^9) — the maximum amounts of liters of gasoline that Nut can buy in cities.
Each of the next n - 1 lines describes road and contains three integers u, v, c (1 ≤ u, v ≤ n, 1 ≤ c ≤ 10^9, u ≠ v), where u and v — cities that are connected by this road and c — its length.
It is guaranteed that graph of road connectivity is a tree.
Output
Print one number — the maximum amount of gasoline that he can have at the end of the path.
Examples
Input
3
1 3 3
1 2 2
1 3 2
Output
3
Input
5
6 3 2 5 0
1 2 10
2 3 3
2 4 1
1 5 1
Output
7
Note
The optimal way in the first example is 2 → 1 → 3.
<image>
The optimal way in the second example is 2 → 4.
<image>
Tags: data structures, dp, trees
Correct Solution:
```
from sys import stdin
input=lambda : stdin.readline().strip()
from math import ceil,sqrt,factorial,gcd
from collections import deque
n=int(input())
l=list(map(int,input().split()))
visited=set()
graph={i:set() for i in range(1,n+1)}
d={}
papa=[0 for i in range(n+1)]
level=[[] for i in range(n+1)]
z=[[0] for i in range(n+1)]
for i in range(n-1):
a,b,c=map(int,input().split())
graph[a].add(b)
graph[b].add(a)
d[(a,b)]=c
stack=deque()
# print(graph)
for i in graph:
if len(graph[i])==1:
stack.append([i,0])
m=0
while stack:
# print(stack)
x,y=stack.popleft()
if len(graph[x])>=1:
for i in graph[x]:
t=i
break
if (t,x) in d:
q=d[(t,x)]
else:
q=d[(x,t)]
z[t].append(y+l[x-1]-q)
graph[t].remove(x)
if len(graph[t])==1:
stack.append([t,max(z[t])])
for i in range(1,n+1):
z[i].sort()
if len(z[i])>=3:
m=max(m,l[i-1]+z[i][-2]+z[i][-1])
m=max(m,z[i][-1]+l[i-1])
print(m)
```
| 100,340 |
Provide tags and a correct Python 3 solution for this coding contest problem.
The Fair Nut is going to travel to the Tree Country, in which there are n cities. Most of the land of this country is covered by forest. Furthermore, the local road system forms a tree (connected graph without cycles). Nut wants to rent a car in the city u and go by a simple path to city v. He hasn't determined the path, so it's time to do it. Note that chosen path can consist of only one vertex.
A filling station is located in every city. Because of strange law, Nut can buy only w_i liters of gasoline in the i-th city. We can assume, that he has infinite money. Each road has a length, and as soon as Nut drives through this road, the amount of gasoline decreases by length. Of course, Nut can't choose a path, which consists of roads, where he runs out of gasoline. He can buy gasoline in every visited city, even in the first and the last.
He also wants to find the maximum amount of gasoline that he can have at the end of the path. Help him: count it.
Input
The first line contains a single integer n (1 ≤ n ≤ 3 ⋅ 10^5) — the number of cities.
The second line contains n integers w_1, w_2, …, w_n (0 ≤ w_{i} ≤ 10^9) — the maximum amounts of liters of gasoline that Nut can buy in cities.
Each of the next n - 1 lines describes road and contains three integers u, v, c (1 ≤ u, v ≤ n, 1 ≤ c ≤ 10^9, u ≠ v), where u and v — cities that are connected by this road and c — its length.
It is guaranteed that graph of road connectivity is a tree.
Output
Print one number — the maximum amount of gasoline that he can have at the end of the path.
Examples
Input
3
1 3 3
1 2 2
1 3 2
Output
3
Input
5
6 3 2 5 0
1 2 10
2 3 3
2 4 1
1 5 1
Output
7
Note
The optimal way in the first example is 2 → 1 → 3.
<image>
The optimal way in the second example is 2 → 4.
<image>
Tags: data structures, dp, trees
Correct Solution:
```
from sys import stdin, setrecursionlimit
import threading
n = int(stdin.readline())
w = [int(x) for x in stdin.readline().split()]
graph = [{} for x in range(n)]
for road in range(n-1):
u,v,c = [int(x) for x in stdin.readline().split()]
u -= 1
v -= 1
if v in graph[u]:
graph[u][v] = min(graph[u][v], c)
else:
graph[u][v] = c
if u in graph[v]:
graph[v][u] = min(graph[v][u], c)
else:
graph[v][u] = c
gas = [{} for x in range(n)]
highs = [[0,0] for x in range(n)]
path = [(0,0)]
ind = 0
while ind < len(path):
cur, par = path[ind]
edges = graph[cur]
for x in edges:
if x != par:
path.append((x,cur))
ind += 1
def mostGas(node,parent):
edges = graph[node]
high = w[node]
high2 = w[node]
for x in edges:
if x != parent:
gas[node][x] = highs[x][0] + w[node] - edges[x]
if gas[node][x] > high:
high,high2 = gas[node][x], high
elif gas[node][x] > high2:
high2 = gas[node][x]
highs[node] = [high,high2]
return high
'''def fillIn(node,parent):
edges = graph[node]
high,high2 = highs[node]
for x in edges:
if x != parent:
if gas[node][x] == high:
gas[x][node] = high2 - edges[x]
else:
gas[x][node] = high - edges[x]
if gas[x][node] > highs[x][0]:
highs[x] = [gas[x][node], highs[x][0]]
elif gas[x][node] > highs[x][1]:
highs[x][1] = gas[x][node]
fillIn(x,node)'''
for x,y in path[::-1]:
mostGas(x,y)
#fillIn(0,0)
high = 0
for x in range(n):
high = max(high, highs[x][0]+highs[x][1]-w[x])
print(high)
```
| 100,341 |
Provide tags and a correct Python 3 solution for this coding contest problem.
The Fair Nut is going to travel to the Tree Country, in which there are n cities. Most of the land of this country is covered by forest. Furthermore, the local road system forms a tree (connected graph without cycles). Nut wants to rent a car in the city u and go by a simple path to city v. He hasn't determined the path, so it's time to do it. Note that chosen path can consist of only one vertex.
A filling station is located in every city. Because of strange law, Nut can buy only w_i liters of gasoline in the i-th city. We can assume, that he has infinite money. Each road has a length, and as soon as Nut drives through this road, the amount of gasoline decreases by length. Of course, Nut can't choose a path, which consists of roads, where he runs out of gasoline. He can buy gasoline in every visited city, even in the first and the last.
He also wants to find the maximum amount of gasoline that he can have at the end of the path. Help him: count it.
Input
The first line contains a single integer n (1 ≤ n ≤ 3 ⋅ 10^5) — the number of cities.
The second line contains n integers w_1, w_2, …, w_n (0 ≤ w_{i} ≤ 10^9) — the maximum amounts of liters of gasoline that Nut can buy in cities.
Each of the next n - 1 lines describes road and contains three integers u, v, c (1 ≤ u, v ≤ n, 1 ≤ c ≤ 10^9, u ≠ v), where u and v — cities that are connected by this road and c — its length.
It is guaranteed that graph of road connectivity is a tree.
Output
Print one number — the maximum amount of gasoline that he can have at the end of the path.
Examples
Input
3
1 3 3
1 2 2
1 3 2
Output
3
Input
5
6 3 2 5 0
1 2 10
2 3 3
2 4 1
1 5 1
Output
7
Note
The optimal way in the first example is 2 → 1 → 3.
<image>
The optimal way in the second example is 2 → 4.
<image>
Tags: data structures, dp, trees
Correct Solution:
```
import sys
readline = sys.stdin.readline
from collections import Counter
def parorder(Edge, p):
N = len(Edge)
par = [0]*N
par[p] = -1
stack = [p]
order = []
visited = set([p])
ast = stack.append
apo = order.append
while stack:
vn = stack.pop()
apo(vn)
for vf in Edge[vn]:
if vf in visited:
continue
visited.add(vf)
par[vf] = vn
ast(vf)
return par, order
def getcld(p):
res = [[] for _ in range(len(p))]
for i, v in enumerate(p[1:], 1):
res[v].append(i)
return res
N = int(readline())
We = list(map(int, readline().split()))
Edge = [[] for _ in range(N)]
Cost = Counter()
geta = N+1
for _ in range(N-1):
a, b, c = map(int, readline().split())
a -= 1
b -= 1
Edge[a].append(b)
Edge[b].append(a)
Cost[b*geta+a] = c
Cost[a*geta+b] = c
P, L = parorder(Edge, 0)
#C = getcld(P)
dp = [0]*N
candi = [[0, 0] for _ in range(N)]
ans = 0
for l in L[::-1][:-1]:
dp[l] += We[l]
p = P[l]
k = dp[l] - Cost[l*geta + p]
if k > 0:
dp[p] = max(dp[p], k)
candi[p].append(k)
res = max(candi[l])
candi[l].remove(res)
ans = max(ans, We[l] + res + max(candi[l]))
res = max(candi[0])
candi[0].remove(res)
ans = max(ans, We[0] + res + max(candi[0]))
print(ans)
```
| 100,342 |
Provide tags and a correct Python 3 solution for this coding contest problem.
The Fair Nut is going to travel to the Tree Country, in which there are n cities. Most of the land of this country is covered by forest. Furthermore, the local road system forms a tree (connected graph without cycles). Nut wants to rent a car in the city u and go by a simple path to city v. He hasn't determined the path, so it's time to do it. Note that chosen path can consist of only one vertex.
A filling station is located in every city. Because of strange law, Nut can buy only w_i liters of gasoline in the i-th city. We can assume, that he has infinite money. Each road has a length, and as soon as Nut drives through this road, the amount of gasoline decreases by length. Of course, Nut can't choose a path, which consists of roads, where he runs out of gasoline. He can buy gasoline in every visited city, even in the first and the last.
He also wants to find the maximum amount of gasoline that he can have at the end of the path. Help him: count it.
Input
The first line contains a single integer n (1 ≤ n ≤ 3 ⋅ 10^5) — the number of cities.
The second line contains n integers w_1, w_2, …, w_n (0 ≤ w_{i} ≤ 10^9) — the maximum amounts of liters of gasoline that Nut can buy in cities.
Each of the next n - 1 lines describes road and contains three integers u, v, c (1 ≤ u, v ≤ n, 1 ≤ c ≤ 10^9, u ≠ v), where u and v — cities that are connected by this road and c — its length.
It is guaranteed that graph of road connectivity is a tree.
Output
Print one number — the maximum amount of gasoline that he can have at the end of the path.
Examples
Input
3
1 3 3
1 2 2
1 3 2
Output
3
Input
5
6 3 2 5 0
1 2 10
2 3 3
2 4 1
1 5 1
Output
7
Note
The optimal way in the first example is 2 → 1 → 3.
<image>
The optimal way in the second example is 2 → 4.
<image>
Tags: data structures, dp, trees
Correct Solution:
```
import sys
import io, os
#input = sys.stdin.buffer.readline
input = io.BytesIO(os.read(0,os.fstat(0).st_size)).readline
n = int(input())
W = list(map(int, input().split()))
g = [[] for i in range(n)]
edge = []
for i in range(n-1):
u, v, c = map(int, input().split())
u, v = u-1, v-1
g[u].append((c, v))
g[v].append((c, u))
edge.append((u, v, c))
s = []
s.append(0)
par = [-1]*n
order = []
while s:
v = s.pop()
order.append(v)
for c, u in g[v]:
if u == par[v]:
continue
s.append(u)
par[u] = v
C = [0]*n
for u, v, c in edge:
if u == par[v]:
C[v] = c
else:
C[u] = c
order.reverse()
import heapq
hq = []
for i in range(n):
q = []
heapq.heapify(q)
hq.append(q)
ans = [0]*n
dp = [0]*n
for v in order:
if not hq[v]:
ans[v] = W[v]
dp[v] = W[v]
elif len(hq[v]) == 1:
temp = W[v]
x = heapq.heappop(hq[v])
x = -x
ans[v] = W[v]+x
dp[v] = W[v]+x
else:
temp = W[v]
x = heapq.heappop(hq[v])
x = -x
ans[v] = W[v]+x
dp[v] = W[v]+x
x = heapq.heappop(hq[v])
x = -x
ans[v] += x
if par[v] != -1:
heapq.heappush(hq[par[v]], -max(0, dp[v]-C[v]))
print(max(ans))
```
| 100,343 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
The Fair Nut is going to travel to the Tree Country, in which there are n cities. Most of the land of this country is covered by forest. Furthermore, the local road system forms a tree (connected graph without cycles). Nut wants to rent a car in the city u and go by a simple path to city v. He hasn't determined the path, so it's time to do it. Note that chosen path can consist of only one vertex.
A filling station is located in every city. Because of strange law, Nut can buy only w_i liters of gasoline in the i-th city. We can assume, that he has infinite money. Each road has a length, and as soon as Nut drives through this road, the amount of gasoline decreases by length. Of course, Nut can't choose a path, which consists of roads, where he runs out of gasoline. He can buy gasoline in every visited city, even in the first and the last.
He also wants to find the maximum amount of gasoline that he can have at the end of the path. Help him: count it.
Input
The first line contains a single integer n (1 ≤ n ≤ 3 ⋅ 10^5) — the number of cities.
The second line contains n integers w_1, w_2, …, w_n (0 ≤ w_{i} ≤ 10^9) — the maximum amounts of liters of gasoline that Nut can buy in cities.
Each of the next n - 1 lines describes road and contains three integers u, v, c (1 ≤ u, v ≤ n, 1 ≤ c ≤ 10^9, u ≠ v), where u and v — cities that are connected by this road and c — its length.
It is guaranteed that graph of road connectivity is a tree.
Output
Print one number — the maximum amount of gasoline that he can have at the end of the path.
Examples
Input
3
1 3 3
1 2 2
1 3 2
Output
3
Input
5
6 3 2 5 0
1 2 10
2 3 3
2 4 1
1 5 1
Output
7
Note
The optimal way in the first example is 2 → 1 → 3.
<image>
The optimal way in the second example is 2 → 4.
<image>
Submitted Solution:
```
import sys
input = sys.stdin.readline
n = int(input())
a = list(map(int, input().split()))
adj = [[] for i in range(n)]
for i in range(n-1):
u, v, w = map(int, input().split())
u -= 1
v -= 1
adj[u].append((v, w))
adj[v].append((u, w))
best = [0] * n
ans = 0
def dfs(u):
stack = list()
visit = [False] * n
stack.append((u, -1))
while stack:
u, par = stack[-1]
if not visit[u]:
visit[u] = True
for v, w in adj[u]:
if v != par:
stack.append((v, u))
else:
cand = []
for v, w in adj[u]:
if v != par:
cand.append(best[v] + a[v] - w)
cand.sort(reverse=True)
cur = a[u]
for i in range(2):
if i < len(cand) and cand[i] > 0:
cur += cand[i]
global ans
ans = max(ans, cur)
best[u] = cand[0] if len(cand) > 0 and cand[0] > 0 else 0
stack.pop()
dfs(0)
print(ans)
```
Yes
| 100,344 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
The Fair Nut is going to travel to the Tree Country, in which there are n cities. Most of the land of this country is covered by forest. Furthermore, the local road system forms a tree (connected graph without cycles). Nut wants to rent a car in the city u and go by a simple path to city v. He hasn't determined the path, so it's time to do it. Note that chosen path can consist of only one vertex.
A filling station is located in every city. Because of strange law, Nut can buy only w_i liters of gasoline in the i-th city. We can assume, that he has infinite money. Each road has a length, and as soon as Nut drives through this road, the amount of gasoline decreases by length. Of course, Nut can't choose a path, which consists of roads, where he runs out of gasoline. He can buy gasoline in every visited city, even in the first and the last.
He also wants to find the maximum amount of gasoline that he can have at the end of the path. Help him: count it.
Input
The first line contains a single integer n (1 ≤ n ≤ 3 ⋅ 10^5) — the number of cities.
The second line contains n integers w_1, w_2, …, w_n (0 ≤ w_{i} ≤ 10^9) — the maximum amounts of liters of gasoline that Nut can buy in cities.
Each of the next n - 1 lines describes road and contains three integers u, v, c (1 ≤ u, v ≤ n, 1 ≤ c ≤ 10^9, u ≠ v), where u and v — cities that are connected by this road and c — its length.
It is guaranteed that graph of road connectivity is a tree.
Output
Print one number — the maximum amount of gasoline that he can have at the end of the path.
Examples
Input
3
1 3 3
1 2 2
1 3 2
Output
3
Input
5
6 3 2 5 0
1 2 10
2 3 3
2 4 1
1 5 1
Output
7
Note
The optimal way in the first example is 2 → 1 → 3.
<image>
The optimal way in the second example is 2 → 4.
<image>
Submitted Solution:
```
# if there >=2 branches u have the option to take the best in ur path
# also u need to change the node good shit with the best 1 or
from collections import defaultdict
from sys import stdin,setrecursionlimit
input=stdin.readline
#first dfs i can do for best 1 ? this works i think
ans=0
from types import GeneratorType
def bootstrap(f, stack=[]):
def wrappedfunc(*args, **kwargs):
if stack:
return f(*args, **kwargs)
else:
to = f(*args, **kwargs)
while True:
if type(to) is GeneratorType:
stack.append(to)
to = next(to)
else:
stack.pop()
if not stack:
break
to = stack[-1].send(to)
return to
return wrappedfunc
# *************************************************************************************#
@bootstrap
def dfs(node,parent,wt,g,dp):
global ans
l=wt[node]
t=[]
for i,w in g[node]:
if i!=parent:
yield dfs(i,node,wt,g,dp)
l=max(dp[i]+wt[node]-w,l)
t.append(dp[i]+wt[node]-w)
dp[node]=l
t=sorted(t,reverse=True)
ans=max(ans,l)
if len(t)>=2:
ans=max(ans,t[0]+t[1]-wt[node])
yield l
def main():
global ans
n=int(input())
wt=list(map(int,input().strip().split()))
g=defaultdict(list)
for _ in range(n-1):
x,y,w=map(int,input().strip().split())
g[x-1].append((y-1,w))
g[y-1].append((x-1,w))
dp=[0]*(n)
dfs(0,-1,wt,g,dp)
print(ans)
main()
```
Yes
| 100,345 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
The Fair Nut is going to travel to the Tree Country, in which there are n cities. Most of the land of this country is covered by forest. Furthermore, the local road system forms a tree (connected graph without cycles). Nut wants to rent a car in the city u and go by a simple path to city v. He hasn't determined the path, so it's time to do it. Note that chosen path can consist of only one vertex.
A filling station is located in every city. Because of strange law, Nut can buy only w_i liters of gasoline in the i-th city. We can assume, that he has infinite money. Each road has a length, and as soon as Nut drives through this road, the amount of gasoline decreases by length. Of course, Nut can't choose a path, which consists of roads, where he runs out of gasoline. He can buy gasoline in every visited city, even in the first and the last.
He also wants to find the maximum amount of gasoline that he can have at the end of the path. Help him: count it.
Input
The first line contains a single integer n (1 ≤ n ≤ 3 ⋅ 10^5) — the number of cities.
The second line contains n integers w_1, w_2, …, w_n (0 ≤ w_{i} ≤ 10^9) — the maximum amounts of liters of gasoline that Nut can buy in cities.
Each of the next n - 1 lines describes road and contains three integers u, v, c (1 ≤ u, v ≤ n, 1 ≤ c ≤ 10^9, u ≠ v), where u and v — cities that are connected by this road and c — its length.
It is guaranteed that graph of road connectivity is a tree.
Output
Print one number — the maximum amount of gasoline that he can have at the end of the path.
Examples
Input
3
1 3 3
1 2 2
1 3 2
Output
3
Input
5
6 3 2 5 0
1 2 10
2 3 3
2 4 1
1 5 1
Output
7
Note
The optimal way in the first example is 2 → 1 → 3.
<image>
The optimal way in the second example is 2 → 4.
<image>
Submitted Solution:
```
from collections import deque
def recurse(x,d,w,parent,v,vb):
best = 0
bestt = 0
ans = 0
for t in d[x]:
node = t[0]
if node == parent:
continue
weight = int(w[node-1])-t[1]
ans = max(ans,v[node])
tot = weight+vb[node]
if tot > best:
bestt = best
best = tot
elif tot > bestt:
bestt = tot
ans = max(ans,best+bestt+int(w[x-1]))
v[x] = ans
vb[x] = best
return (ans,best)
n = int(input())
w = input().split()
dic = {}
for i in range(1,n+1):
dic[i] = []
for i in range(n-1):
u,v,c = map(int,input().split())
dic[u].append((v,c))
dic[v].append((u,c))
dq = deque()
dq.append(1)
visit = set()
l = []
l.append((1,0))
while len(dq) > 0:
cur = dq.pop()
visit.add(cur)
for t in dic[cur]:
node = t[0]
if node not in visit:
l.append((node,cur))
dq.append(node)
val = [0]*(n+1)
valb = [0]*(n+1)
for i in range(len(l)-1,-1,-1):
recurse(l[i][0],dic,w,l[i][1],val,valb)
print(val[1])
```
Yes
| 100,346 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
The Fair Nut is going to travel to the Tree Country, in which there are n cities. Most of the land of this country is covered by forest. Furthermore, the local road system forms a tree (connected graph without cycles). Nut wants to rent a car in the city u and go by a simple path to city v. He hasn't determined the path, so it's time to do it. Note that chosen path can consist of only one vertex.
A filling station is located in every city. Because of strange law, Nut can buy only w_i liters of gasoline in the i-th city. We can assume, that he has infinite money. Each road has a length, and as soon as Nut drives through this road, the amount of gasoline decreases by length. Of course, Nut can't choose a path, which consists of roads, where he runs out of gasoline. He can buy gasoline in every visited city, even in the first and the last.
He also wants to find the maximum amount of gasoline that he can have at the end of the path. Help him: count it.
Input
The first line contains a single integer n (1 ≤ n ≤ 3 ⋅ 10^5) — the number of cities.
The second line contains n integers w_1, w_2, …, w_n (0 ≤ w_{i} ≤ 10^9) — the maximum amounts of liters of gasoline that Nut can buy in cities.
Each of the next n - 1 lines describes road and contains three integers u, v, c (1 ≤ u, v ≤ n, 1 ≤ c ≤ 10^9, u ≠ v), where u and v — cities that are connected by this road and c — its length.
It is guaranteed that graph of road connectivity is a tree.
Output
Print one number — the maximum amount of gasoline that he can have at the end of the path.
Examples
Input
3
1 3 3
1 2 2
1 3 2
Output
3
Input
5
6 3 2 5 0
1 2 10
2 3 3
2 4 1
1 5 1
Output
7
Note
The optimal way in the first example is 2 → 1 → 3.
<image>
The optimal way in the second example is 2 → 4.
<image>
Submitted Solution:
```
import sys
readline = sys.stdin.readline
from collections import Counter
def getpar(Edge, p):
N = len(Edge)
par = [0]*N
par[0] = -1
par[p] -1
stack = [p]
visited = set([p])
while stack:
vn = stack.pop()
for vf in Edge[vn]:
if vf in visited:
continue
visited.add(vf)
par[vf] = vn
stack.append(vf)
return par
def topological_sort_tree(E, r):
Q = [r]
L = []
visited = set([r])
while Q:
vn = Q.pop()
L.append(vn)
for vf in E[vn]:
if vf not in visited:
visited.add(vf)
Q.append(vf)
return L
def getcld(p):
res = [[] for _ in range(len(p))]
for i, v in enumerate(p[1:], 1):
res[v].append(i)
return res
N = int(readline())
We = list(map(int, readline().split()))
Edge = [[] for _ in range(N)]
Cost = Counter()
geta = N+1
for _ in range(N-1):
a, b, c = map(int, readline().split())
a -= 1
b -= 1
Edge[a].append(b)
Edge[b].append(a)
Cost[b*geta+a] = c
Cost[a*geta+b] = c
P = getpar(Edge, 0)
L = topological_sort_tree(Edge, 0)
C = getcld(P)
dp = [0]*N
candi = [[0, 0] for _ in range(N)]
ans = 0
for l in L[::-1][:-1]:
dp[l] += We[l]
p = P[l]
k = dp[l] - Cost[l*geta + p]
if k > 0:
dp[p] = max(dp[p], k)
candi[p].append(k)
res = max(candi[l])
candi[l].remove(res)
ans = max(ans, We[l] + res + max(candi[l]))
res = max(candi[0])
candi[0].remove(res)
ans = max(ans, We[0] + res + max(candi[0]))
print(ans)
```
Yes
| 100,347 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
The Fair Nut is going to travel to the Tree Country, in which there are n cities. Most of the land of this country is covered by forest. Furthermore, the local road system forms a tree (connected graph without cycles). Nut wants to rent a car in the city u and go by a simple path to city v. He hasn't determined the path, so it's time to do it. Note that chosen path can consist of only one vertex.
A filling station is located in every city. Because of strange law, Nut can buy only w_i liters of gasoline in the i-th city. We can assume, that he has infinite money. Each road has a length, and as soon as Nut drives through this road, the amount of gasoline decreases by length. Of course, Nut can't choose a path, which consists of roads, where he runs out of gasoline. He can buy gasoline in every visited city, even in the first and the last.
He also wants to find the maximum amount of gasoline that he can have at the end of the path. Help him: count it.
Input
The first line contains a single integer n (1 ≤ n ≤ 3 ⋅ 10^5) — the number of cities.
The second line contains n integers w_1, w_2, …, w_n (0 ≤ w_{i} ≤ 10^9) — the maximum amounts of liters of gasoline that Nut can buy in cities.
Each of the next n - 1 lines describes road and contains three integers u, v, c (1 ≤ u, v ≤ n, 1 ≤ c ≤ 10^9, u ≠ v), where u and v — cities that are connected by this road and c — its length.
It is guaranteed that graph of road connectivity is a tree.
Output
Print one number — the maximum amount of gasoline that he can have at the end of the path.
Examples
Input
3
1 3 3
1 2 2
1 3 2
Output
3
Input
5
6 3 2 5 0
1 2 10
2 3 3
2 4 1
1 5 1
Output
7
Note
The optimal way in the first example is 2 → 1 → 3.
<image>
The optimal way in the second example is 2 → 4.
<image>
Submitted Solution:
```
def ii():
return int(input())
def mi():
return map(int, input().split())
def li():
return list(mi())
import sys
sys.setrecursionlimit(10 ** 7)
# A. The Fair Nut and the Best Path
n = ii()
a = [None] + li()
g = [[] for i in range(n + 1)]
for i in range(n - 1):
u, v, c = mi()
g[u].append((v, c))
g[v].append((u, c))
ans = -1
def dfs(u, p):
global ans
difs = []
for v, w in g[u]:
if v == p: continue
dif = dfs(v, u)
difs.append(dif - w)
if len(difs) >= 2:
ans = max(ans, difs[-1] + difs[-2] + a[u])
mx = 0
if len(difs) >= 1:
ans = max(ans, difs[-1] + a[u])
mx = max(mx, difs[-1])
return mx + a[u]
dfs(1, 1)
print(max(0, ans))
```
No
| 100,348 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
The Fair Nut is going to travel to the Tree Country, in which there are n cities. Most of the land of this country is covered by forest. Furthermore, the local road system forms a tree (connected graph without cycles). Nut wants to rent a car in the city u and go by a simple path to city v. He hasn't determined the path, so it's time to do it. Note that chosen path can consist of only one vertex.
A filling station is located in every city. Because of strange law, Nut can buy only w_i liters of gasoline in the i-th city. We can assume, that he has infinite money. Each road has a length, and as soon as Nut drives through this road, the amount of gasoline decreases by length. Of course, Nut can't choose a path, which consists of roads, where he runs out of gasoline. He can buy gasoline in every visited city, even in the first and the last.
He also wants to find the maximum amount of gasoline that he can have at the end of the path. Help him: count it.
Input
The first line contains a single integer n (1 ≤ n ≤ 3 ⋅ 10^5) — the number of cities.
The second line contains n integers w_1, w_2, …, w_n (0 ≤ w_{i} ≤ 10^9) — the maximum amounts of liters of gasoline that Nut can buy in cities.
Each of the next n - 1 lines describes road and contains three integers u, v, c (1 ≤ u, v ≤ n, 1 ≤ c ≤ 10^9, u ≠ v), where u and v — cities that are connected by this road and c — its length.
It is guaranteed that graph of road connectivity is a tree.
Output
Print one number — the maximum amount of gasoline that he can have at the end of the path.
Examples
Input
3
1 3 3
1 2 2
1 3 2
Output
3
Input
5
6 3 2 5 0
1 2 10
2 3 3
2 4 1
1 5 1
Output
7
Note
The optimal way in the first example is 2 → 1 → 3.
<image>
The optimal way in the second example is 2 → 4.
<image>
Submitted Solution:
```
n=int(input())
w=list(map(int,input().split()))
def findshit(nweights,tree,memo,vertex):
if vertex in memo:
return memo[vertex]
childs=[]
for child in tree[vertex]:
childs.append(findshit(nweights,tree,memo,child[0])-child[1])
childs.sort()
childs+=[0,0]
memo[vertex]=nweights[vertex]+max((childs[0],0))+max((childs[1],0))
return nweights[vertex]+max((childs[0],0))+max((childs[1],0))
nweights={}
tree={}
memo={}
for i in range(n):
tree[i+1]=[]
for i in range(n):
nweights[i+1]=w[i]
for i in range(n-1):
a,b,c=map(int,input().split())
tree[a].append((b,c))
tree[b].append((a,c))
visited=[1]
lastlayer=[1]
while len(visited)<n:
newlayer=[]
for vert in lastlayer:
for guy in tree[vert]:
tree[guy[0]].remove((vert,guy[1]))
newlayer.append(guy[0])
lastlayer=newlayer
visited+=newlayer
findshit(nweights,tree,memo,1)
best=0
for guy in memo:
best=max((memo[guy],best))
print(best)
```
No
| 100,349 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
The Fair Nut is going to travel to the Tree Country, in which there are n cities. Most of the land of this country is covered by forest. Furthermore, the local road system forms a tree (connected graph without cycles). Nut wants to rent a car in the city u and go by a simple path to city v. He hasn't determined the path, so it's time to do it. Note that chosen path can consist of only one vertex.
A filling station is located in every city. Because of strange law, Nut can buy only w_i liters of gasoline in the i-th city. We can assume, that he has infinite money. Each road has a length, and as soon as Nut drives through this road, the amount of gasoline decreases by length. Of course, Nut can't choose a path, which consists of roads, where he runs out of gasoline. He can buy gasoline in every visited city, even in the first and the last.
He also wants to find the maximum amount of gasoline that he can have at the end of the path. Help him: count it.
Input
The first line contains a single integer n (1 ≤ n ≤ 3 ⋅ 10^5) — the number of cities.
The second line contains n integers w_1, w_2, …, w_n (0 ≤ w_{i} ≤ 10^9) — the maximum amounts of liters of gasoline that Nut can buy in cities.
Each of the next n - 1 lines describes road and contains three integers u, v, c (1 ≤ u, v ≤ n, 1 ≤ c ≤ 10^9, u ≠ v), where u and v — cities that are connected by this road and c — its length.
It is guaranteed that graph of road connectivity is a tree.
Output
Print one number — the maximum amount of gasoline that he can have at the end of the path.
Examples
Input
3
1 3 3
1 2 2
1 3 2
Output
3
Input
5
6 3 2 5 0
1 2 10
2 3 3
2 4 1
1 5 1
Output
7
Note
The optimal way in the first example is 2 → 1 → 3.
<image>
The optimal way in the second example is 2 → 4.
<image>
Submitted Solution:
```
import sys
import math
from collections import defaultdict,deque
tree=defaultdict(list)
n=int(sys.stdin.readline())
gas=list(map(int,sys.stdin.readline().split()))
dist=defaultdict(int)
for i in range(n-1):
u,v,w=map(int,sys.stdin.readline().split())
tree[u].append([v,w])
tree[v].append([u,w])
dist[(min(u,v),max(u,v))]=w
leaves=deque()
dp=[0 for _ in range(n+1)]
for i in tree:
if len(tree[i])==1:
leaves.append([i,gas[i-1]])
dp[i]=gas[i-1]
#print(leaves,'leaves')
vis=defaultdict(int)
for i in range(1,n+1):
dp[i]=gas[i-1]
while leaves:
a,g=leaves.popleft()
for j,w in tree[a]:
if vis[(min(j,a),max(j,a))]==0:
vis[(min(j,a),max(j,a))]=1
if g>=w:
val=g-w+gas[j-1]
dp[j]=max(dp[j]+g-w,g-w+gas[j-1],gas[j-1])
leaves.append([j,val])
#print(dp,'dp')
#print(vis,'vis')
if n==1:
print(gas[0])
sys.exit()
print(max(dp))
```
No
| 100,350 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
The Fair Nut is going to travel to the Tree Country, in which there are n cities. Most of the land of this country is covered by forest. Furthermore, the local road system forms a tree (connected graph without cycles). Nut wants to rent a car in the city u and go by a simple path to city v. He hasn't determined the path, so it's time to do it. Note that chosen path can consist of only one vertex.
A filling station is located in every city. Because of strange law, Nut can buy only w_i liters of gasoline in the i-th city. We can assume, that he has infinite money. Each road has a length, and as soon as Nut drives through this road, the amount of gasoline decreases by length. Of course, Nut can't choose a path, which consists of roads, where he runs out of gasoline. He can buy gasoline in every visited city, even in the first and the last.
He also wants to find the maximum amount of gasoline that he can have at the end of the path. Help him: count it.
Input
The first line contains a single integer n (1 ≤ n ≤ 3 ⋅ 10^5) — the number of cities.
The second line contains n integers w_1, w_2, …, w_n (0 ≤ w_{i} ≤ 10^9) — the maximum amounts of liters of gasoline that Nut can buy in cities.
Each of the next n - 1 lines describes road and contains three integers u, v, c (1 ≤ u, v ≤ n, 1 ≤ c ≤ 10^9, u ≠ v), where u and v — cities that are connected by this road and c — its length.
It is guaranteed that graph of road connectivity is a tree.
Output
Print one number — the maximum amount of gasoline that he can have at the end of the path.
Examples
Input
3
1 3 3
1 2 2
1 3 2
Output
3
Input
5
6 3 2 5 0
1 2 10
2 3 3
2 4 1
1 5 1
Output
7
Note
The optimal way in the first example is 2 → 1 → 3.
<image>
The optimal way in the second example is 2 → 4.
<image>
Submitted Solution:
```
o = []
oGraph = []
#print("Enter Number of oil station")
n=int(input())
#print("Enter Number of roads")
oGraph = [[-1]*n for i in range(n)]
o=[0]*n
oil = []
a=[]
b=[]
dist = []
maxO=[]
track = []
#print("Enter amount of oil in each station")
o=[int(x) for x in input().split()]
# for i in range(n):
# oil = int(input())
# o[i]=oil
#print("Enter roads distance with connected roads")
for j in range(n-1):
# a=int(input())
# b=int(input())
# dist = int(input())
gg=[int(x) for x in input().split()]
if((o[gg[0]-1]>(o[gg[0]-1]+o[gg[1]-1])-gg[2]) or (o[gg[1]-1]>(o[gg[0]-1]+o[gg[1]-1])-gg[2])):
oGraph[gg[0]-1][gg[1]-1]=-1
else:
oGraph[gg[0]-1][gg[1]-1]=gg[2]
stk = []
visited = [0]*n
amax=-1
for e in range(n):
if(visited[e]==0):
stk.append(e)
visited[e]=1
maxO=o[e]
track.append(e)
while stk:
v = stk.pop()
visited[v]=2
for j in range(n):
if ((oGraph[v][j]>-1) and (visited[j]==0)):
if(maxO<((maxO+o[j])-oGraph[v][j])):
maxO=((maxO+o[j])-oGraph[v][j])
track.append(j)
if(o[j]>maxO):
#maxO=o[j]
track = []
track.append(j)
#print("----------------")
visited[j]=1
stk.append(j)
#print(maxO)
if(maxO>amax):
amax=maxO
print(amax)
#print(track)
```
No
| 100,351 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Vasya has his favourite number n. He wants to split it to some non-zero digits. It means, that he wants to choose some digits d_1, d_2, …, d_k, such that 1 ≤ d_i ≤ 9 for all i and d_1 + d_2 + … + d_k = n.
Vasya likes beauty in everything, so he wants to find any solution with the minimal possible number of different digits among d_1, d_2, …, d_k. Help him!
Input
The first line contains a single integer n — the number that Vasya wants to split (1 ≤ n ≤ 1000).
Output
In the first line print one integer k — the number of digits in the partition. Note that k must satisfy the inequality 1 ≤ k ≤ n. In the next line print k digits d_1, d_2, …, d_k separated by spaces. All digits must satisfy the inequalities 1 ≤ d_i ≤ 9.
You should find a partition of n in which the number of different digits among d_1, d_2, …, d_k will be minimal possible among all partitions of n into non-zero digits. Among such partitions, it is allowed to find any. It is guaranteed that there exists at least one partition of the number n into digits.
Examples
Input
1
Output
1
1
Input
4
Output
2
2 2
Input
27
Output
3
9 9 9
Note
In the first test, the number 1 can be divided into 1 digit equal to 1.
In the second test, there are 3 partitions of the number 4 into digits in which the number of different digits is 1. This partitions are [1, 1, 1, 1], [2, 2] and [4]. Any of these partitions can be found. And, for example, dividing the number 4 to the digits [1, 1, 2] isn't an answer, because it has 2 different digits, that isn't the minimum possible number.
Tags: constructive algorithms, implementation, math
Correct Solution:
```
n=int(input())
re=0
for i in range(9,1,-1):
if(n%i==0):
re=i
break
if(re==0):
print(n)
for i in range(n):
if(i==n-1):
print(1)
else:
print(1,end=' ')
else:
print(n//re)
res=n//re
for i in range(res):
if(i==res-1):
print(re)
else:
print(re,end=' ')
n-=re
```
| 100,352 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Vasya has his favourite number n. He wants to split it to some non-zero digits. It means, that he wants to choose some digits d_1, d_2, …, d_k, such that 1 ≤ d_i ≤ 9 for all i and d_1 + d_2 + … + d_k = n.
Vasya likes beauty in everything, so he wants to find any solution with the minimal possible number of different digits among d_1, d_2, …, d_k. Help him!
Input
The first line contains a single integer n — the number that Vasya wants to split (1 ≤ n ≤ 1000).
Output
In the first line print one integer k — the number of digits in the partition. Note that k must satisfy the inequality 1 ≤ k ≤ n. In the next line print k digits d_1, d_2, …, d_k separated by spaces. All digits must satisfy the inequalities 1 ≤ d_i ≤ 9.
You should find a partition of n in which the number of different digits among d_1, d_2, …, d_k will be minimal possible among all partitions of n into non-zero digits. Among such partitions, it is allowed to find any. It is guaranteed that there exists at least one partition of the number n into digits.
Examples
Input
1
Output
1
1
Input
4
Output
2
2 2
Input
27
Output
3
9 9 9
Note
In the first test, the number 1 can be divided into 1 digit equal to 1.
In the second test, there are 3 partitions of the number 4 into digits in which the number of different digits is 1. This partitions are [1, 1, 1, 1], [2, 2] and [4]. Any of these partitions can be found. And, for example, dividing the number 4 to the digits [1, 1, 2] isn't an answer, because it has 2 different digits, that isn't the minimum possible number.
Tags: constructive algorithms, implementation, math
Correct Solution:
```
import sys
number = int(input())
if(number <= 2):
print(int(number))
print((str(1) + ' ') * number)
else:
for i in range(9,1,-1):
if(number % i == 0 and i != number):
print(int(number/i))
print((str(i) + ' ') * int(number/i))
sys.exit()
print(int(number))
print((str(1) + ' ') * number)
```
| 100,353 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Vasya has his favourite number n. He wants to split it to some non-zero digits. It means, that he wants to choose some digits d_1, d_2, …, d_k, such that 1 ≤ d_i ≤ 9 for all i and d_1 + d_2 + … + d_k = n.
Vasya likes beauty in everything, so he wants to find any solution with the minimal possible number of different digits among d_1, d_2, …, d_k. Help him!
Input
The first line contains a single integer n — the number that Vasya wants to split (1 ≤ n ≤ 1000).
Output
In the first line print one integer k — the number of digits in the partition. Note that k must satisfy the inequality 1 ≤ k ≤ n. In the next line print k digits d_1, d_2, …, d_k separated by spaces. All digits must satisfy the inequalities 1 ≤ d_i ≤ 9.
You should find a partition of n in which the number of different digits among d_1, d_2, …, d_k will be minimal possible among all partitions of n into non-zero digits. Among such partitions, it is allowed to find any. It is guaranteed that there exists at least one partition of the number n into digits.
Examples
Input
1
Output
1
1
Input
4
Output
2
2 2
Input
27
Output
3
9 9 9
Note
In the first test, the number 1 can be divided into 1 digit equal to 1.
In the second test, there are 3 partitions of the number 4 into digits in which the number of different digits is 1. This partitions are [1, 1, 1, 1], [2, 2] and [4]. Any of these partitions can be found. And, for example, dividing the number 4 to the digits [1, 1, 2] isn't an answer, because it has 2 different digits, that isn't the minimum possible number.
Tags: constructive algorithms, implementation, math
Correct Solution:
```
x = int(input())
z = []
if x == 1:
print(1)
print(1)
else:
for i in range(1,10):
if x%i == 0:
z.append(i)
r = int((x/max(z)))
print(int(x/max(z)))
k = str(max(z))
print((k+' ')*r)
```
| 100,354 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Vasya has his favourite number n. He wants to split it to some non-zero digits. It means, that he wants to choose some digits d_1, d_2, …, d_k, such that 1 ≤ d_i ≤ 9 for all i and d_1 + d_2 + … + d_k = n.
Vasya likes beauty in everything, so he wants to find any solution with the minimal possible number of different digits among d_1, d_2, …, d_k. Help him!
Input
The first line contains a single integer n — the number that Vasya wants to split (1 ≤ n ≤ 1000).
Output
In the first line print one integer k — the number of digits in the partition. Note that k must satisfy the inequality 1 ≤ k ≤ n. In the next line print k digits d_1, d_2, …, d_k separated by spaces. All digits must satisfy the inequalities 1 ≤ d_i ≤ 9.
You should find a partition of n in which the number of different digits among d_1, d_2, …, d_k will be minimal possible among all partitions of n into non-zero digits. Among such partitions, it is allowed to find any. It is guaranteed that there exists at least one partition of the number n into digits.
Examples
Input
1
Output
1
1
Input
4
Output
2
2 2
Input
27
Output
3
9 9 9
Note
In the first test, the number 1 can be divided into 1 digit equal to 1.
In the second test, there are 3 partitions of the number 4 into digits in which the number of different digits is 1. This partitions are [1, 1, 1, 1], [2, 2] and [4]. Any of these partitions can be found. And, for example, dividing the number 4 to the digits [1, 1, 2] isn't an answer, because it has 2 different digits, that isn't the minimum possible number.
Tags: constructive algorithms, implementation, math
Correct Solution:
```
n = int(input())
a = [1] * n
print(n)
print(*a)
```
| 100,355 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Vasya has his favourite number n. He wants to split it to some non-zero digits. It means, that he wants to choose some digits d_1, d_2, …, d_k, such that 1 ≤ d_i ≤ 9 for all i and d_1 + d_2 + … + d_k = n.
Vasya likes beauty in everything, so he wants to find any solution with the minimal possible number of different digits among d_1, d_2, …, d_k. Help him!
Input
The first line contains a single integer n — the number that Vasya wants to split (1 ≤ n ≤ 1000).
Output
In the first line print one integer k — the number of digits in the partition. Note that k must satisfy the inequality 1 ≤ k ≤ n. In the next line print k digits d_1, d_2, …, d_k separated by spaces. All digits must satisfy the inequalities 1 ≤ d_i ≤ 9.
You should find a partition of n in which the number of different digits among d_1, d_2, …, d_k will be minimal possible among all partitions of n into non-zero digits. Among such partitions, it is allowed to find any. It is guaranteed that there exists at least one partition of the number n into digits.
Examples
Input
1
Output
1
1
Input
4
Output
2
2 2
Input
27
Output
3
9 9 9
Note
In the first test, the number 1 can be divided into 1 digit equal to 1.
In the second test, there are 3 partitions of the number 4 into digits in which the number of different digits is 1. This partitions are [1, 1, 1, 1], [2, 2] and [4]. Any of these partitions can be found. And, for example, dividing the number 4 to the digits [1, 1, 2] isn't an answer, because it has 2 different digits, that isn't the minimum possible number.
Tags: constructive algorithms, implementation, math
Correct Solution:
```
# cook your dish here
n=int(input())
print(n)
for i in range(n):
print("1",end=" ")
```
| 100,356 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Vasya has his favourite number n. He wants to split it to some non-zero digits. It means, that he wants to choose some digits d_1, d_2, …, d_k, such that 1 ≤ d_i ≤ 9 for all i and d_1 + d_2 + … + d_k = n.
Vasya likes beauty in everything, so he wants to find any solution with the minimal possible number of different digits among d_1, d_2, …, d_k. Help him!
Input
The first line contains a single integer n — the number that Vasya wants to split (1 ≤ n ≤ 1000).
Output
In the first line print one integer k — the number of digits in the partition. Note that k must satisfy the inequality 1 ≤ k ≤ n. In the next line print k digits d_1, d_2, …, d_k separated by spaces. All digits must satisfy the inequalities 1 ≤ d_i ≤ 9.
You should find a partition of n in which the number of different digits among d_1, d_2, …, d_k will be minimal possible among all partitions of n into non-zero digits. Among such partitions, it is allowed to find any. It is guaranteed that there exists at least one partition of the number n into digits.
Examples
Input
1
Output
1
1
Input
4
Output
2
2 2
Input
27
Output
3
9 9 9
Note
In the first test, the number 1 can be divided into 1 digit equal to 1.
In the second test, there are 3 partitions of the number 4 into digits in which the number of different digits is 1. This partitions are [1, 1, 1, 1], [2, 2] and [4]. Any of these partitions can be found. And, for example, dividing the number 4 to the digits [1, 1, 2] isn't an answer, because it has 2 different digits, that isn't the minimum possible number.
Tags: constructive algorithms, implementation, math
Correct Solution:
```
n = int(input())
c = 1
j = n
for i in range(1,10):
if n // i > c and n % i == 0 and n//i != n and n//i <= 9:
c = n//i
j = i
s = ""
print(j)
for k in range(j):
print(c, end=" ")
```
| 100,357 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Vasya has his favourite number n. He wants to split it to some non-zero digits. It means, that he wants to choose some digits d_1, d_2, …, d_k, such that 1 ≤ d_i ≤ 9 for all i and d_1 + d_2 + … + d_k = n.
Vasya likes beauty in everything, so he wants to find any solution with the minimal possible number of different digits among d_1, d_2, …, d_k. Help him!
Input
The first line contains a single integer n — the number that Vasya wants to split (1 ≤ n ≤ 1000).
Output
In the first line print one integer k — the number of digits in the partition. Note that k must satisfy the inequality 1 ≤ k ≤ n. In the next line print k digits d_1, d_2, …, d_k separated by spaces. All digits must satisfy the inequalities 1 ≤ d_i ≤ 9.
You should find a partition of n in which the number of different digits among d_1, d_2, …, d_k will be minimal possible among all partitions of n into non-zero digits. Among such partitions, it is allowed to find any. It is guaranteed that there exists at least one partition of the number n into digits.
Examples
Input
1
Output
1
1
Input
4
Output
2
2 2
Input
27
Output
3
9 9 9
Note
In the first test, the number 1 can be divided into 1 digit equal to 1.
In the second test, there are 3 partitions of the number 4 into digits in which the number of different digits is 1. This partitions are [1, 1, 1, 1], [2, 2] and [4]. Any of these partitions can be found. And, for example, dividing the number 4 to the digits [1, 1, 2] isn't an answer, because it has 2 different digits, that isn't the minimum possible number.
Tags: constructive algorithms, implementation, math
Correct Solution:
```
n = int(input())
number = 1
for i in range(9,1,-1):
if(n%i == 0):
number = i
break
times = n//number
print(times)
for i in range(times):
print(number, end=' ')
```
| 100,358 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Vasya has his favourite number n. He wants to split it to some non-zero digits. It means, that he wants to choose some digits d_1, d_2, …, d_k, such that 1 ≤ d_i ≤ 9 for all i and d_1 + d_2 + … + d_k = n.
Vasya likes beauty in everything, so he wants to find any solution with the minimal possible number of different digits among d_1, d_2, …, d_k. Help him!
Input
The first line contains a single integer n — the number that Vasya wants to split (1 ≤ n ≤ 1000).
Output
In the first line print one integer k — the number of digits in the partition. Note that k must satisfy the inequality 1 ≤ k ≤ n. In the next line print k digits d_1, d_2, …, d_k separated by spaces. All digits must satisfy the inequalities 1 ≤ d_i ≤ 9.
You should find a partition of n in which the number of different digits among d_1, d_2, …, d_k will be minimal possible among all partitions of n into non-zero digits. Among such partitions, it is allowed to find any. It is guaranteed that there exists at least one partition of the number n into digits.
Examples
Input
1
Output
1
1
Input
4
Output
2
2 2
Input
27
Output
3
9 9 9
Note
In the first test, the number 1 can be divided into 1 digit equal to 1.
In the second test, there are 3 partitions of the number 4 into digits in which the number of different digits is 1. This partitions are [1, 1, 1, 1], [2, 2] and [4]. Any of these partitions can be found. And, for example, dividing the number 4 to the digits [1, 1, 2] isn't an answer, because it has 2 different digits, that isn't the minimum possible number.
Tags: constructive algorithms, implementation, math
Correct Solution:
```
if __name__ == '__main__':
n = int(input())
print(n)
print("1",end = "")
print(" 1"*(n-1))
```
| 100,359 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Vasya has his favourite number n. He wants to split it to some non-zero digits. It means, that he wants to choose some digits d_1, d_2, …, d_k, such that 1 ≤ d_i ≤ 9 for all i and d_1 + d_2 + … + d_k = n.
Vasya likes beauty in everything, so he wants to find any solution with the minimal possible number of different digits among d_1, d_2, …, d_k. Help him!
Input
The first line contains a single integer n — the number that Vasya wants to split (1 ≤ n ≤ 1000).
Output
In the first line print one integer k — the number of digits in the partition. Note that k must satisfy the inequality 1 ≤ k ≤ n. In the next line print k digits d_1, d_2, …, d_k separated by spaces. All digits must satisfy the inequalities 1 ≤ d_i ≤ 9.
You should find a partition of n in which the number of different digits among d_1, d_2, …, d_k will be minimal possible among all partitions of n into non-zero digits. Among such partitions, it is allowed to find any. It is guaranteed that there exists at least one partition of the number n into digits.
Examples
Input
1
Output
1
1
Input
4
Output
2
2 2
Input
27
Output
3
9 9 9
Note
In the first test, the number 1 can be divided into 1 digit equal to 1.
In the second test, there are 3 partitions of the number 4 into digits in which the number of different digits is 1. This partitions are [1, 1, 1, 1], [2, 2] and [4]. Any of these partitions can be found. And, for example, dividing the number 4 to the digits [1, 1, 2] isn't an answer, because it has 2 different digits, that isn't the minimum possible number.
Submitted Solution:
```
n=int(input())
if(n<9):
print(1)
print(n)
else:
l=[1]*n
print(n)
print(*l)
```
Yes
| 100,360 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Vasya has his favourite number n. He wants to split it to some non-zero digits. It means, that he wants to choose some digits d_1, d_2, …, d_k, such that 1 ≤ d_i ≤ 9 for all i and d_1 + d_2 + … + d_k = n.
Vasya likes beauty in everything, so he wants to find any solution with the minimal possible number of different digits among d_1, d_2, …, d_k. Help him!
Input
The first line contains a single integer n — the number that Vasya wants to split (1 ≤ n ≤ 1000).
Output
In the first line print one integer k — the number of digits in the partition. Note that k must satisfy the inequality 1 ≤ k ≤ n. In the next line print k digits d_1, d_2, …, d_k separated by spaces. All digits must satisfy the inequalities 1 ≤ d_i ≤ 9.
You should find a partition of n in which the number of different digits among d_1, d_2, …, d_k will be minimal possible among all partitions of n into non-zero digits. Among such partitions, it is allowed to find any. It is guaranteed that there exists at least one partition of the number n into digits.
Examples
Input
1
Output
1
1
Input
4
Output
2
2 2
Input
27
Output
3
9 9 9
Note
In the first test, the number 1 can be divided into 1 digit equal to 1.
In the second test, there are 3 partitions of the number 4 into digits in which the number of different digits is 1. This partitions are [1, 1, 1, 1], [2, 2] and [4]. Any of these partitions can be found. And, for example, dividing the number 4 to the digits [1, 1, 2] isn't an answer, because it has 2 different digits, that isn't the minimum possible number.
Submitted Solution:
```
import sys
input = sys.stdin.readline
N=int(input())
L=[1]*N
print(N)
print(' '.join(map(str,L)))
```
Yes
| 100,361 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Vasya has his favourite number n. He wants to split it to some non-zero digits. It means, that he wants to choose some digits d_1, d_2, …, d_k, such that 1 ≤ d_i ≤ 9 for all i and d_1 + d_2 + … + d_k = n.
Vasya likes beauty in everything, so he wants to find any solution with the minimal possible number of different digits among d_1, d_2, …, d_k. Help him!
Input
The first line contains a single integer n — the number that Vasya wants to split (1 ≤ n ≤ 1000).
Output
In the first line print one integer k — the number of digits in the partition. Note that k must satisfy the inequality 1 ≤ k ≤ n. In the next line print k digits d_1, d_2, …, d_k separated by spaces. All digits must satisfy the inequalities 1 ≤ d_i ≤ 9.
You should find a partition of n in which the number of different digits among d_1, d_2, …, d_k will be minimal possible among all partitions of n into non-zero digits. Among such partitions, it is allowed to find any. It is guaranteed that there exists at least one partition of the number n into digits.
Examples
Input
1
Output
1
1
Input
4
Output
2
2 2
Input
27
Output
3
9 9 9
Note
In the first test, the number 1 can be divided into 1 digit equal to 1.
In the second test, there are 3 partitions of the number 4 into digits in which the number of different digits is 1. This partitions are [1, 1, 1, 1], [2, 2] and [4]. Any of these partitions can be found. And, for example, dividing the number 4 to the digits [1, 1, 2] isn't an answer, because it has 2 different digits, that isn't the minimum possible number.
Submitted Solution:
```
n=int(input())
if n<=9:
print(1)
print(n)
if n>9:
print(n)
print("1 "*n)
```
Yes
| 100,362 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Vasya has his favourite number n. He wants to split it to some non-zero digits. It means, that he wants to choose some digits d_1, d_2, …, d_k, such that 1 ≤ d_i ≤ 9 for all i and d_1 + d_2 + … + d_k = n.
Vasya likes beauty in everything, so he wants to find any solution with the minimal possible number of different digits among d_1, d_2, …, d_k. Help him!
Input
The first line contains a single integer n — the number that Vasya wants to split (1 ≤ n ≤ 1000).
Output
In the first line print one integer k — the number of digits in the partition. Note that k must satisfy the inequality 1 ≤ k ≤ n. In the next line print k digits d_1, d_2, …, d_k separated by spaces. All digits must satisfy the inequalities 1 ≤ d_i ≤ 9.
You should find a partition of n in which the number of different digits among d_1, d_2, …, d_k will be minimal possible among all partitions of n into non-zero digits. Among such partitions, it is allowed to find any. It is guaranteed that there exists at least one partition of the number n into digits.
Examples
Input
1
Output
1
1
Input
4
Output
2
2 2
Input
27
Output
3
9 9 9
Note
In the first test, the number 1 can be divided into 1 digit equal to 1.
In the second test, there are 3 partitions of the number 4 into digits in which the number of different digits is 1. This partitions are [1, 1, 1, 1], [2, 2] and [4]. Any of these partitions can be found. And, for example, dividing the number 4 to the digits [1, 1, 2] isn't an answer, because it has 2 different digits, that isn't the minimum possible number.
Submitted Solution:
```
N = int(input())
Max = 0
for i in range(N, 0, -1):
if N % i == 0:
Max = i
break
print(N)
print("1 "*N)
```
Yes
| 100,363 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Vasya has his favourite number n. He wants to split it to some non-zero digits. It means, that he wants to choose some digits d_1, d_2, …, d_k, such that 1 ≤ d_i ≤ 9 for all i and d_1 + d_2 + … + d_k = n.
Vasya likes beauty in everything, so he wants to find any solution with the minimal possible number of different digits among d_1, d_2, …, d_k. Help him!
Input
The first line contains a single integer n — the number that Vasya wants to split (1 ≤ n ≤ 1000).
Output
In the first line print one integer k — the number of digits in the partition. Note that k must satisfy the inequality 1 ≤ k ≤ n. In the next line print k digits d_1, d_2, …, d_k separated by spaces. All digits must satisfy the inequalities 1 ≤ d_i ≤ 9.
You should find a partition of n in which the number of different digits among d_1, d_2, …, d_k will be minimal possible among all partitions of n into non-zero digits. Among such partitions, it is allowed to find any. It is guaranteed that there exists at least one partition of the number n into digits.
Examples
Input
1
Output
1
1
Input
4
Output
2
2 2
Input
27
Output
3
9 9 9
Note
In the first test, the number 1 can be divided into 1 digit equal to 1.
In the second test, there are 3 partitions of the number 4 into digits in which the number of different digits is 1. This partitions are [1, 1, 1, 1], [2, 2] and [4]. Any of these partitions can be found. And, for example, dividing the number 4 to the digits [1, 1, 2] isn't an answer, because it has 2 different digits, that isn't the minimum possible number.
Submitted Solution:
```
n=int(input())
flag=0
for i in [2,3,5,7]:
if(n%i==0):
print(n//i)
for j in range(n//i):
print(i,end=' ')
flag=1
break
if(flag==0):
print((n//2)+1)
for i in range(n//2):
print(2,end=' ')
print(1)
```
No
| 100,364 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Vasya has his favourite number n. He wants to split it to some non-zero digits. It means, that he wants to choose some digits d_1, d_2, …, d_k, such that 1 ≤ d_i ≤ 9 for all i and d_1 + d_2 + … + d_k = n.
Vasya likes beauty in everything, so he wants to find any solution with the minimal possible number of different digits among d_1, d_2, …, d_k. Help him!
Input
The first line contains a single integer n — the number that Vasya wants to split (1 ≤ n ≤ 1000).
Output
In the first line print one integer k — the number of digits in the partition. Note that k must satisfy the inequality 1 ≤ k ≤ n. In the next line print k digits d_1, d_2, …, d_k separated by spaces. All digits must satisfy the inequalities 1 ≤ d_i ≤ 9.
You should find a partition of n in which the number of different digits among d_1, d_2, …, d_k will be minimal possible among all partitions of n into non-zero digits. Among such partitions, it is allowed to find any. It is guaranteed that there exists at least one partition of the number n into digits.
Examples
Input
1
Output
1
1
Input
4
Output
2
2 2
Input
27
Output
3
9 9 9
Note
In the first test, the number 1 can be divided into 1 digit equal to 1.
In the second test, there are 3 partitions of the number 4 into digits in which the number of different digits is 1. This partitions are [1, 1, 1, 1], [2, 2] and [4]. Any of these partitions can be found. And, for example, dividing the number 4 to the digits [1, 1, 2] isn't an answer, because it has 2 different digits, that isn't the minimum possible number.
Submitted Solution:
```
n = 1
first_digit = None
include_digits = [9, 8, 7, 6, 5, 4, 3, 2, 1]
for digit in include_digits:
if n % digit == 0:
first_digit = digit
break
times = int(n / first_digit)
print(times)
for t in range(times):
print(first_digit, end=" ")
```
No
| 100,365 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Vasya has his favourite number n. He wants to split it to some non-zero digits. It means, that he wants to choose some digits d_1, d_2, …, d_k, such that 1 ≤ d_i ≤ 9 for all i and d_1 + d_2 + … + d_k = n.
Vasya likes beauty in everything, so he wants to find any solution with the minimal possible number of different digits among d_1, d_2, …, d_k. Help him!
Input
The first line contains a single integer n — the number that Vasya wants to split (1 ≤ n ≤ 1000).
Output
In the first line print one integer k — the number of digits in the partition. Note that k must satisfy the inequality 1 ≤ k ≤ n. In the next line print k digits d_1, d_2, …, d_k separated by spaces. All digits must satisfy the inequalities 1 ≤ d_i ≤ 9.
You should find a partition of n in which the number of different digits among d_1, d_2, …, d_k will be minimal possible among all partitions of n into non-zero digits. Among such partitions, it is allowed to find any. It is guaranteed that there exists at least one partition of the number n into digits.
Examples
Input
1
Output
1
1
Input
4
Output
2
2 2
Input
27
Output
3
9 9 9
Note
In the first test, the number 1 can be divided into 1 digit equal to 1.
In the second test, there are 3 partitions of the number 4 into digits in which the number of different digits is 1. This partitions are [1, 1, 1, 1], [2, 2] and [4]. Any of these partitions can be found. And, for example, dividing the number 4 to the digits [1, 1, 2] isn't an answer, because it has 2 different digits, that isn't the minimum possible number.
Submitted Solution:
```
n = int(input())
a = []
while n > 9:
a.append(9)
n -= 9
if n > 0:
a.append(n)
print(" ".join(map(str, a)))
```
No
| 100,366 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Vasya has his favourite number n. He wants to split it to some non-zero digits. It means, that he wants to choose some digits d_1, d_2, …, d_k, such that 1 ≤ d_i ≤ 9 for all i and d_1 + d_2 + … + d_k = n.
Vasya likes beauty in everything, so he wants to find any solution with the minimal possible number of different digits among d_1, d_2, …, d_k. Help him!
Input
The first line contains a single integer n — the number that Vasya wants to split (1 ≤ n ≤ 1000).
Output
In the first line print one integer k — the number of digits in the partition. Note that k must satisfy the inequality 1 ≤ k ≤ n. In the next line print k digits d_1, d_2, …, d_k separated by spaces. All digits must satisfy the inequalities 1 ≤ d_i ≤ 9.
You should find a partition of n in which the number of different digits among d_1, d_2, …, d_k will be minimal possible among all partitions of n into non-zero digits. Among such partitions, it is allowed to find any. It is guaranteed that there exists at least one partition of the number n into digits.
Examples
Input
1
Output
1
1
Input
4
Output
2
2 2
Input
27
Output
3
9 9 9
Note
In the first test, the number 1 can be divided into 1 digit equal to 1.
In the second test, there are 3 partitions of the number 4 into digits in which the number of different digits is 1. This partitions are [1, 1, 1, 1], [2, 2] and [4]. Any of these partitions can be found. And, for example, dividing the number 4 to the digits [1, 1, 2] isn't an answer, because it has 2 different digits, that isn't the minimum possible number.
Submitted Solution:
```
import math
mod=1000000007
#c=[[0 for x in range(1001)] for y in range(1001)]
n=int(input())
if n<10:
print("1")
print(n)
else:
m=n//9
x=n-(m*9)
if x>0:
print(m+1)
else:
print(m)
for i in range(m):
print("9",end=" ")
if(x>0):
print(x)
```
No
| 100,367 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Asya loves animals very much. Recently, she purchased n kittens, enumerated them from 1 and n and then put them into the cage. The cage consists of one row of n cells, enumerated with integers from 1 to n from left to right. Adjacent cells had a partially transparent partition wall between them, hence there were n - 1 partitions originally. Initially, each cell contained exactly one kitten with some number.
Observing the kittens, Asya noticed, that they are very friendly and often a pair of kittens in neighboring cells wants to play together. So Asya started to remove partitions between neighboring cells. In particular, on the day i, Asya:
* Noticed, that the kittens x_i and y_i, located in neighboring cells want to play together.
* Removed the partition between these two cells, efficiently creating a single cell, having all kittens from two original cells.
Since Asya has never putted partitions back, after n - 1 days the cage contained a single cell, having all kittens.
For every day, Asya remembers numbers of kittens x_i and y_i, who wanted to play together, however she doesn't remember how she placed kittens in the cage in the beginning. Please help her and find any possible initial arrangement of the kittens into n cells.
Input
The first line contains a single integer n (2 ≤ n ≤ 150 000) — the number of kittens.
Each of the following n - 1 lines contains integers x_i and y_i (1 ≤ x_i, y_i ≤ n, x_i ≠ y_i) — indices of kittens, which got together due to the border removal on the corresponding day.
It's guaranteed, that the kittens x_i and y_i were in the different cells before this day.
Output
For every cell from 1 to n print a single integer — the index of the kitten from 1 to n, who was originally in it.
All printed integers must be distinct.
It's guaranteed, that there is at least one answer possible. In case there are multiple possible answers, print any of them.
Example
Input
5
1 4
2 5
3 1
4 5
Output
3 1 4 2 5
Note
The answer for the example contains one of several possible initial arrangements of the kittens.
The picture below shows how the cells were united for this initial arrangement. Note, that the kittens who wanted to play together on each day were indeed in adjacent cells.
<image>
Tags: constructive algorithms, dsu
Correct Solution:
```
from math import *
from collections import *
import sys
sys.setrecursionlimit(10**9)
def find(x):
if (p[x] == x):
return x
p[x] = find(p[x])
return p[x]
def merge(x,y):
a = find(x)
b = find(y)
#print(a,b)
if(a != b):
if(sz[a] > sz[b]): a,b = b,a
p[a] = b
sz[b] += sz[a]
arr[b].extend(arr[a])
n = int(input())
p = [i for i in range(n+1)]
sz = [1 for i in range(n+1)]
arr = [[i] for i in range(n+1)]
for i in range(n-1):
u,v = map(int,input().split())
merge(u,v)
#print(p)
for i in arr:
if len(i) == n:
for j in i:
print(j,end = " ")
print()
break
```
| 100,368 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Asya loves animals very much. Recently, she purchased n kittens, enumerated them from 1 and n and then put them into the cage. The cage consists of one row of n cells, enumerated with integers from 1 to n from left to right. Adjacent cells had a partially transparent partition wall between them, hence there were n - 1 partitions originally. Initially, each cell contained exactly one kitten with some number.
Observing the kittens, Asya noticed, that they are very friendly and often a pair of kittens in neighboring cells wants to play together. So Asya started to remove partitions between neighboring cells. In particular, on the day i, Asya:
* Noticed, that the kittens x_i and y_i, located in neighboring cells want to play together.
* Removed the partition between these two cells, efficiently creating a single cell, having all kittens from two original cells.
Since Asya has never putted partitions back, after n - 1 days the cage contained a single cell, having all kittens.
For every day, Asya remembers numbers of kittens x_i and y_i, who wanted to play together, however she doesn't remember how she placed kittens in the cage in the beginning. Please help her and find any possible initial arrangement of the kittens into n cells.
Input
The first line contains a single integer n (2 ≤ n ≤ 150 000) — the number of kittens.
Each of the following n - 1 lines contains integers x_i and y_i (1 ≤ x_i, y_i ≤ n, x_i ≠ y_i) — indices of kittens, which got together due to the border removal on the corresponding day.
It's guaranteed, that the kittens x_i and y_i were in the different cells before this day.
Output
For every cell from 1 to n print a single integer — the index of the kitten from 1 to n, who was originally in it.
All printed integers must be distinct.
It's guaranteed, that there is at least one answer possible. In case there are multiple possible answers, print any of them.
Example
Input
5
1 4
2 5
3 1
4 5
Output
3 1 4 2 5
Note
The answer for the example contains one of several possible initial arrangements of the kittens.
The picture below shows how the cells were united for this initial arrangement. Note, that the kittens who wanted to play together on each day were indeed in adjacent cells.
<image>
Tags: constructive algorithms, dsu
Correct Solution:
```
n=int(input())
d={}
for i in range(n-1):
a,b=map(int,input().split())
if (a not in d.keys()) and (b not in d.keys()):
d[a]=[a,b]
d[b]=d[a]
elif (a in d.keys()) and (b not in d.keys()):
d[a]+=[b]
d[b]=d[a]
elif (a not in d.keys()) and (b in d.keys()):
d[b]+=[a]
d[a]=d[b]
else:
d[a]+=d[b]
for j in d[b]:
d[j]=d[a]
print(*d[1])
```
| 100,369 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Asya loves animals very much. Recently, she purchased n kittens, enumerated them from 1 and n and then put them into the cage. The cage consists of one row of n cells, enumerated with integers from 1 to n from left to right. Adjacent cells had a partially transparent partition wall between them, hence there were n - 1 partitions originally. Initially, each cell contained exactly one kitten with some number.
Observing the kittens, Asya noticed, that they are very friendly and often a pair of kittens in neighboring cells wants to play together. So Asya started to remove partitions between neighboring cells. In particular, on the day i, Asya:
* Noticed, that the kittens x_i and y_i, located in neighboring cells want to play together.
* Removed the partition between these two cells, efficiently creating a single cell, having all kittens from two original cells.
Since Asya has never putted partitions back, after n - 1 days the cage contained a single cell, having all kittens.
For every day, Asya remembers numbers of kittens x_i and y_i, who wanted to play together, however she doesn't remember how she placed kittens in the cage in the beginning. Please help her and find any possible initial arrangement of the kittens into n cells.
Input
The first line contains a single integer n (2 ≤ n ≤ 150 000) — the number of kittens.
Each of the following n - 1 lines contains integers x_i and y_i (1 ≤ x_i, y_i ≤ n, x_i ≠ y_i) — indices of kittens, which got together due to the border removal on the corresponding day.
It's guaranteed, that the kittens x_i and y_i were in the different cells before this day.
Output
For every cell from 1 to n print a single integer — the index of the kitten from 1 to n, who was originally in it.
All printed integers must be distinct.
It's guaranteed, that there is at least one answer possible. In case there are multiple possible answers, print any of them.
Example
Input
5
1 4
2 5
3 1
4 5
Output
3 1 4 2 5
Note
The answer for the example contains one of several possible initial arrangements of the kittens.
The picture below shows how the cells were united for this initial arrangement. Note, that the kittens who wanted to play together on each day were indeed in adjacent cells.
<image>
Tags: constructive algorithms, dsu
Correct Solution:
```
n, = map(int, input().split())
p = [i for i in range(0,n+1)]
def find(i):
if p[i] == i:
return i
par = find(p[i])
p[i] = par
return par
def join(a,b):
p[find(a)] = find(b)
N = [-1 for i in range(0,n+1)]
class List:
def __init__(self, val):
self.front = self
self.value = val
self.rear = self
self.next = None
def add(self, other):
self.rear.next = other
self.rear = other.rear
other.front = self
m = {i: List(i) for i in range(1,n+1)}
def printList(i):
if i is None:
return
list = []
while i!=None:
list.append(i.value)
i = i.next
print(str(list).replace(',','').replace('[','').replace(']',''))
ans = 1
for i in range(1,n):
a,b = map(int, input().split())
m[find(a)].add(m[find(b)])
temp = m[find(a)]
del m[find(a)]
del m[find(b)]
join(a,b)
key = find(a)
ans = key
m[key] = temp.front
printList(m[ans].front)
```
| 100,370 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Asya loves animals very much. Recently, she purchased n kittens, enumerated them from 1 and n and then put them into the cage. The cage consists of one row of n cells, enumerated with integers from 1 to n from left to right. Adjacent cells had a partially transparent partition wall between them, hence there were n - 1 partitions originally. Initially, each cell contained exactly one kitten with some number.
Observing the kittens, Asya noticed, that they are very friendly and often a pair of kittens in neighboring cells wants to play together. So Asya started to remove partitions between neighboring cells. In particular, on the day i, Asya:
* Noticed, that the kittens x_i and y_i, located in neighboring cells want to play together.
* Removed the partition between these two cells, efficiently creating a single cell, having all kittens from two original cells.
Since Asya has never putted partitions back, after n - 1 days the cage contained a single cell, having all kittens.
For every day, Asya remembers numbers of kittens x_i and y_i, who wanted to play together, however she doesn't remember how she placed kittens in the cage in the beginning. Please help her and find any possible initial arrangement of the kittens into n cells.
Input
The first line contains a single integer n (2 ≤ n ≤ 150 000) — the number of kittens.
Each of the following n - 1 lines contains integers x_i and y_i (1 ≤ x_i, y_i ≤ n, x_i ≠ y_i) — indices of kittens, which got together due to the border removal on the corresponding day.
It's guaranteed, that the kittens x_i and y_i were in the different cells before this day.
Output
For every cell from 1 to n print a single integer — the index of the kitten from 1 to n, who was originally in it.
All printed integers must be distinct.
It's guaranteed, that there is at least one answer possible. In case there are multiple possible answers, print any of them.
Example
Input
5
1 4
2 5
3 1
4 5
Output
3 1 4 2 5
Note
The answer for the example contains one of several possible initial arrangements of the kittens.
The picture below shows how the cells were united for this initial arrangement. Note, that the kittens who wanted to play together on each day were indeed in adjacent cells.
<image>
Tags: constructive algorithms, dsu
Correct Solution:
```
n = int(input())
parent = [i for i in range(n+1)]
group = [[i] for i in range(n+1)]
for i in range(n-1):
x, y = map(int, input().split())
if len(group[parent[x]]) < len(group[parent[y]]):
x, y = y, x
group[parent[x]] += group[parent[y]]
for j in group[parent[y]]:
parent[j] = parent[x]
print(*group[parent[1]])
```
| 100,371 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Asya loves animals very much. Recently, she purchased n kittens, enumerated them from 1 and n and then put them into the cage. The cage consists of one row of n cells, enumerated with integers from 1 to n from left to right. Adjacent cells had a partially transparent partition wall between them, hence there were n - 1 partitions originally. Initially, each cell contained exactly one kitten with some number.
Observing the kittens, Asya noticed, that they are very friendly and often a pair of kittens in neighboring cells wants to play together. So Asya started to remove partitions between neighboring cells. In particular, on the day i, Asya:
* Noticed, that the kittens x_i and y_i, located in neighboring cells want to play together.
* Removed the partition between these two cells, efficiently creating a single cell, having all kittens from two original cells.
Since Asya has never putted partitions back, after n - 1 days the cage contained a single cell, having all kittens.
For every day, Asya remembers numbers of kittens x_i and y_i, who wanted to play together, however she doesn't remember how she placed kittens in the cage in the beginning. Please help her and find any possible initial arrangement of the kittens into n cells.
Input
The first line contains a single integer n (2 ≤ n ≤ 150 000) — the number of kittens.
Each of the following n - 1 lines contains integers x_i and y_i (1 ≤ x_i, y_i ≤ n, x_i ≠ y_i) — indices of kittens, which got together due to the border removal on the corresponding day.
It's guaranteed, that the kittens x_i and y_i were in the different cells before this day.
Output
For every cell from 1 to n print a single integer — the index of the kitten from 1 to n, who was originally in it.
All printed integers must be distinct.
It's guaranteed, that there is at least one answer possible. In case there are multiple possible answers, print any of them.
Example
Input
5
1 4
2 5
3 1
4 5
Output
3 1 4 2 5
Note
The answer for the example contains one of several possible initial arrangements of the kittens.
The picture below shows how the cells were united for this initial arrangement. Note, that the kittens who wanted to play together on each day were indeed in adjacent cells.
<image>
Tags: constructive algorithms, dsu
Correct Solution:
```
n = int(input())
ks = [[0]]
for i in range(n):
ks.append([i + 1])
for _ in range(n - 1):
a, b = map(int, input().split())
while type(ks[a]) == int:
a = ks[a]
while type(ks[b]) == int:
b = ks[b]
if len(ks[a]) >= len(ks[b]):
ks[a] += ks[b]
ks[b] = a
else:
ks[b] += ks[a]
ks[a] = b
if _ == n - 2:
while type(ks[a]) == int:
a = ks[a]
print(*ks[a])
```
| 100,372 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Asya loves animals very much. Recently, she purchased n kittens, enumerated them from 1 and n and then put them into the cage. The cage consists of one row of n cells, enumerated with integers from 1 to n from left to right. Adjacent cells had a partially transparent partition wall between them, hence there were n - 1 partitions originally. Initially, each cell contained exactly one kitten with some number.
Observing the kittens, Asya noticed, that they are very friendly and often a pair of kittens in neighboring cells wants to play together. So Asya started to remove partitions between neighboring cells. In particular, on the day i, Asya:
* Noticed, that the kittens x_i and y_i, located in neighboring cells want to play together.
* Removed the partition between these two cells, efficiently creating a single cell, having all kittens from two original cells.
Since Asya has never putted partitions back, after n - 1 days the cage contained a single cell, having all kittens.
For every day, Asya remembers numbers of kittens x_i and y_i, who wanted to play together, however she doesn't remember how she placed kittens in the cage in the beginning. Please help her and find any possible initial arrangement of the kittens into n cells.
Input
The first line contains a single integer n (2 ≤ n ≤ 150 000) — the number of kittens.
Each of the following n - 1 lines contains integers x_i and y_i (1 ≤ x_i, y_i ≤ n, x_i ≠ y_i) — indices of kittens, which got together due to the border removal on the corresponding day.
It's guaranteed, that the kittens x_i and y_i were in the different cells before this day.
Output
For every cell from 1 to n print a single integer — the index of the kitten from 1 to n, who was originally in it.
All printed integers must be distinct.
It's guaranteed, that there is at least one answer possible. In case there are multiple possible answers, print any of them.
Example
Input
5
1 4
2 5
3 1
4 5
Output
3 1 4 2 5
Note
The answer for the example contains one of several possible initial arrangements of the kittens.
The picture below shows how the cells were united for this initial arrangement. Note, that the kittens who wanted to play together on each day were indeed in adjacent cells.
<image>
Tags: constructive algorithms, dsu
Correct Solution:
```
import sys
input=sys.stdin.readline
n=int(input())
p=[list(map(int,input().split())) for i in range(n-1)]
g=[i for i in range(n+1)]
def find(x):
while(g[x]!=x):
x=g[x]
return x
def union(x,y):
if find(x)!=find(y):
g[find(x)]=g[find(y)]=min(find(x),find(y))
ans=[[i] for i in range(n+1)]
for i,j in p:
a=find(i)
b=find(j)
if b<a:
ans[b]+=ans[a]
else:
ans[a]+=ans[b]
union(i,j)
print(*ans[1])
```
| 100,373 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Asya loves animals very much. Recently, she purchased n kittens, enumerated them from 1 and n and then put them into the cage. The cage consists of one row of n cells, enumerated with integers from 1 to n from left to right. Adjacent cells had a partially transparent partition wall between them, hence there were n - 1 partitions originally. Initially, each cell contained exactly one kitten with some number.
Observing the kittens, Asya noticed, that they are very friendly and often a pair of kittens in neighboring cells wants to play together. So Asya started to remove partitions between neighboring cells. In particular, on the day i, Asya:
* Noticed, that the kittens x_i and y_i, located in neighboring cells want to play together.
* Removed the partition between these two cells, efficiently creating a single cell, having all kittens from two original cells.
Since Asya has never putted partitions back, after n - 1 days the cage contained a single cell, having all kittens.
For every day, Asya remembers numbers of kittens x_i and y_i, who wanted to play together, however she doesn't remember how she placed kittens in the cage in the beginning. Please help her and find any possible initial arrangement of the kittens into n cells.
Input
The first line contains a single integer n (2 ≤ n ≤ 150 000) — the number of kittens.
Each of the following n - 1 lines contains integers x_i and y_i (1 ≤ x_i, y_i ≤ n, x_i ≠ y_i) — indices of kittens, which got together due to the border removal on the corresponding day.
It's guaranteed, that the kittens x_i and y_i were in the different cells before this day.
Output
For every cell from 1 to n print a single integer — the index of the kitten from 1 to n, who was originally in it.
All printed integers must be distinct.
It's guaranteed, that there is at least one answer possible. In case there are multiple possible answers, print any of them.
Example
Input
5
1 4
2 5
3 1
4 5
Output
3 1 4 2 5
Note
The answer for the example contains one of several possible initial arrangements of the kittens.
The picture below shows how the cells were united for this initial arrangement. Note, that the kittens who wanted to play together on each day were indeed in adjacent cells.
<image>
Tags: constructive algorithms, dsu
Correct Solution:
```
n = int(input())
class Element:
def __init__(self, x):
self.things = [x]
self.rep = x
class DisjSet:
def __init__(self, n):
# Constructor to create and
# initialize sets of n items
self.rank = [1] * n
self.elements = [Element(x) for x in range(n)]
# Finds set of given item x
def find(self, x):
# Finds the representative of the set
# that x is an element of
if (self.elements[x].rep != x):
# if x is not the parent of itself
# Then x is not the representative of
# its set,
self.elements[x].rep = self.find(self.elements[x].rep)
# so we recursively call Find on its parent
# and move i's node directly under the
# representative of this set
return self.elements[x].rep
# Do union of two sets represented
# by x and y.
def Union(self, x, y):
# Find current sets of x and y
xset = self.find(x)
yset = self.find(y)
# If they are already in same set
if xset == yset:
return
# Put smaller ranked item under
# bigger ranked item if ranks are
# different
if self.rank[xset] < self.rank[yset]:
self.elements[xset].rep = yset
self.elements[yset].things.extend(self.elements[xset].things)
elif self.rank[xset] > self.rank[yset]:
self.elements[yset].rep = xset
self.elements[xset].things.extend(self.elements[yset].things)
# If ranks are same, then move y under
# x (doesn't matter which one goes where)
# and increment rank of x's tree
else:
self.elements[yset].rep = xset
self.rank[xset] = self.rank[xset] + 1
self.elements[xset].things.extend(self.elements[yset].things)
obj = DisjSet(n)
for _ in range(n - 1):
a, b = list(map(lambda x: int(x), input().split()))
obj.Union(a-1,b-1)
for ele in obj.elements:
if len(ele.things)==n:
print(*[x+1 for x in ele.things])
```
| 100,374 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Asya loves animals very much. Recently, she purchased n kittens, enumerated them from 1 and n and then put them into the cage. The cage consists of one row of n cells, enumerated with integers from 1 to n from left to right. Adjacent cells had a partially transparent partition wall between them, hence there were n - 1 partitions originally. Initially, each cell contained exactly one kitten with some number.
Observing the kittens, Asya noticed, that they are very friendly and often a pair of kittens in neighboring cells wants to play together. So Asya started to remove partitions between neighboring cells. In particular, on the day i, Asya:
* Noticed, that the kittens x_i and y_i, located in neighboring cells want to play together.
* Removed the partition between these two cells, efficiently creating a single cell, having all kittens from two original cells.
Since Asya has never putted partitions back, after n - 1 days the cage contained a single cell, having all kittens.
For every day, Asya remembers numbers of kittens x_i and y_i, who wanted to play together, however she doesn't remember how she placed kittens in the cage in the beginning. Please help her and find any possible initial arrangement of the kittens into n cells.
Input
The first line contains a single integer n (2 ≤ n ≤ 150 000) — the number of kittens.
Each of the following n - 1 lines contains integers x_i and y_i (1 ≤ x_i, y_i ≤ n, x_i ≠ y_i) — indices of kittens, which got together due to the border removal on the corresponding day.
It's guaranteed, that the kittens x_i and y_i were in the different cells before this day.
Output
For every cell from 1 to n print a single integer — the index of the kitten from 1 to n, who was originally in it.
All printed integers must be distinct.
It's guaranteed, that there is at least one answer possible. In case there are multiple possible answers, print any of them.
Example
Input
5
1 4
2 5
3 1
4 5
Output
3 1 4 2 5
Note
The answer for the example contains one of several possible initial arrangements of the kittens.
The picture below shows how the cells were united for this initial arrangement. Note, that the kittens who wanted to play together on each day were indeed in adjacent cells.
<image>
Tags: constructive algorithms, dsu
Correct Solution:
```
import sys
input = sys.stdin.readline
class Unionfind:
def __init__(self, n):
self.par = [-1]*n
self.left = [-1]*n
self.right = [-1]*n
self.most_right = [i for i in range(n)]
def root(self, x):
p = x
while not self.par[p]<0:
p = self.par[p]
while x!=p:
tmp = x
x = self.par[x]
self.par[tmp] = p
return p
def unite(self, x, y):
rx, ry = self.root(x), self.root(y)
self.par[ry] = rx
self.left[ry] = self.most_right[rx]
self.right[self.most_right[rx]] = ry
self.most_right[rx] = self.most_right[ry]
n = int(input())
uf = Unionfind(n)
for _ in range(n-1):
x, y = map(int, input().split())
uf.unite(x-1, y-1)
ans = []
for i in range(n):
if uf.left[i]==-1:
ans.append(i)
break
for _ in range(n-1):
ans.append(uf.right[ans[-1]])
ans = list(map(lambda x: x+1, ans))
print(*ans)
```
| 100,375 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Asya loves animals very much. Recently, she purchased n kittens, enumerated them from 1 and n and then put them into the cage. The cage consists of one row of n cells, enumerated with integers from 1 to n from left to right. Adjacent cells had a partially transparent partition wall between them, hence there were n - 1 partitions originally. Initially, each cell contained exactly one kitten with some number.
Observing the kittens, Asya noticed, that they are very friendly and often a pair of kittens in neighboring cells wants to play together. So Asya started to remove partitions between neighboring cells. In particular, on the day i, Asya:
* Noticed, that the kittens x_i and y_i, located in neighboring cells want to play together.
* Removed the partition between these two cells, efficiently creating a single cell, having all kittens from two original cells.
Since Asya has never putted partitions back, after n - 1 days the cage contained a single cell, having all kittens.
For every day, Asya remembers numbers of kittens x_i and y_i, who wanted to play together, however she doesn't remember how she placed kittens in the cage in the beginning. Please help her and find any possible initial arrangement of the kittens into n cells.
Input
The first line contains a single integer n (2 ≤ n ≤ 150 000) — the number of kittens.
Each of the following n - 1 lines contains integers x_i and y_i (1 ≤ x_i, y_i ≤ n, x_i ≠ y_i) — indices of kittens, which got together due to the border removal on the corresponding day.
It's guaranteed, that the kittens x_i and y_i were in the different cells before this day.
Output
For every cell from 1 to n print a single integer — the index of the kitten from 1 to n, who was originally in it.
All printed integers must be distinct.
It's guaranteed, that there is at least one answer possible. In case there are multiple possible answers, print any of them.
Example
Input
5
1 4
2 5
3 1
4 5
Output
3 1 4 2 5
Note
The answer for the example contains one of several possible initial arrangements of the kittens.
The picture below shows how the cells were united for this initial arrangement. Note, that the kittens who wanted to play together on each day were indeed in adjacent cells.
<image>
Submitted Solution:
```
class Disjoint_set:
class node:
def __init__(self,a):
self.value=a
self.p=self
self.rank=0
self.ans=[a]
def __init__(self,a):
self.data={i:self.make_set(i) for i in range(1,a+1)}
def make_set(self,val):
return self.node(val)
def union(self,x,y):
self.link(self.find_set(self.data[x]),self.find_set(self.data[y]))
def link(self,val1,val2):
if val1.rank>val2.rank:
val2.p=val1
val1.ans.extend(val2.ans)
val1.rank+=1
else:
val1.p=val2
val2.ans.extend(val1.ans)
val2.rank+=1
def find_set(self,val):
if val.p.value!=val.value:
val.p=self.find_set(val.p)
return val.p
def sol(self):
x=1
for i in self.data:
x=i
break
print(*(self.find_set(self.data[x])).ans)
n=int(input())
a=[list(map(int,input().split())) for i in range(n-1)]
sa=Disjoint_set(n)
for i,j in a:
sa.union(i,j)
sa.sol()
```
Yes
| 100,376 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Asya loves animals very much. Recently, she purchased n kittens, enumerated them from 1 and n and then put them into the cage. The cage consists of one row of n cells, enumerated with integers from 1 to n from left to right. Adjacent cells had a partially transparent partition wall between them, hence there were n - 1 partitions originally. Initially, each cell contained exactly one kitten with some number.
Observing the kittens, Asya noticed, that they are very friendly and often a pair of kittens in neighboring cells wants to play together. So Asya started to remove partitions between neighboring cells. In particular, on the day i, Asya:
* Noticed, that the kittens x_i and y_i, located in neighboring cells want to play together.
* Removed the partition between these two cells, efficiently creating a single cell, having all kittens from two original cells.
Since Asya has never putted partitions back, after n - 1 days the cage contained a single cell, having all kittens.
For every day, Asya remembers numbers of kittens x_i and y_i, who wanted to play together, however she doesn't remember how she placed kittens in the cage in the beginning. Please help her and find any possible initial arrangement of the kittens into n cells.
Input
The first line contains a single integer n (2 ≤ n ≤ 150 000) — the number of kittens.
Each of the following n - 1 lines contains integers x_i and y_i (1 ≤ x_i, y_i ≤ n, x_i ≠ y_i) — indices of kittens, which got together due to the border removal on the corresponding day.
It's guaranteed, that the kittens x_i and y_i were in the different cells before this day.
Output
For every cell from 1 to n print a single integer — the index of the kitten from 1 to n, who was originally in it.
All printed integers must be distinct.
It's guaranteed, that there is at least one answer possible. In case there are multiple possible answers, print any of them.
Example
Input
5
1 4
2 5
3 1
4 5
Output
3 1 4 2 5
Note
The answer for the example contains one of several possible initial arrangements of the kittens.
The picture below shows how the cells were united for this initial arrangement. Note, that the kittens who wanted to play together on each day were indeed in adjacent cells.
<image>
Submitted Solution:
```
nxt= []
last = []
rt = []
def dsu(n):
for i in range(0,n+1):
rt.append(-1)
last.append(i)
nxt.append(-1)
def root(x):
if(rt[x]<0):
return x
else:
rt[x] = root(rt[x])
return rt[x]
def union(x,y):
x = root(x)
y = root(y)
if(x!=y):
rt[x] = rt[x]+rt[y]
rt[y] = x
nxt[last[x]] = y
last[x] = last[y]
n = int(input())
dsu(n)
for z in range(n-1):
x,y = [int(i) for i in input().split()]
union(x,y)
i = root(1)
while(i!=-1):
print(i,end= " ")
i = nxt[i]
```
Yes
| 100,377 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Asya loves animals very much. Recently, she purchased n kittens, enumerated them from 1 and n and then put them into the cage. The cage consists of one row of n cells, enumerated with integers from 1 to n from left to right. Adjacent cells had a partially transparent partition wall between them, hence there were n - 1 partitions originally. Initially, each cell contained exactly one kitten with some number.
Observing the kittens, Asya noticed, that they are very friendly and often a pair of kittens in neighboring cells wants to play together. So Asya started to remove partitions between neighboring cells. In particular, on the day i, Asya:
* Noticed, that the kittens x_i and y_i, located in neighboring cells want to play together.
* Removed the partition between these two cells, efficiently creating a single cell, having all kittens from two original cells.
Since Asya has never putted partitions back, after n - 1 days the cage contained a single cell, having all kittens.
For every day, Asya remembers numbers of kittens x_i and y_i, who wanted to play together, however she doesn't remember how she placed kittens in the cage in the beginning. Please help her and find any possible initial arrangement of the kittens into n cells.
Input
The first line contains a single integer n (2 ≤ n ≤ 150 000) — the number of kittens.
Each of the following n - 1 lines contains integers x_i and y_i (1 ≤ x_i, y_i ≤ n, x_i ≠ y_i) — indices of kittens, which got together due to the border removal on the corresponding day.
It's guaranteed, that the kittens x_i and y_i were in the different cells before this day.
Output
For every cell from 1 to n print a single integer — the index of the kitten from 1 to n, who was originally in it.
All printed integers must be distinct.
It's guaranteed, that there is at least one answer possible. In case there are multiple possible answers, print any of them.
Example
Input
5
1 4
2 5
3 1
4 5
Output
3 1 4 2 5
Note
The answer for the example contains one of several possible initial arrangements of the kittens.
The picture below shows how the cells were united for this initial arrangement. Note, that the kittens who wanted to play together on each day were indeed in adjacent cells.
<image>
Submitted Solution:
```
from collections import defaultdict
import sys
import bisect
input=sys.stdin.readline
def find(x):
while p[x]!=x:
x=p[p[x]]
return(x)
def union(a,b):
if (sz[a]>sz[b]):
p[b]=a
ans[a]+=ans[b]
elif (sz[a]<sz[b]):
p[a]=b
ans[b]+=ans[a]
else:
p[b]=a
ans[a]+=ans[b]
sz[a]+=1
n=int(input())
p=[i for i in range(n)]
ans=[[i] for i in range(n)]
sz=[0]*(n)
for ii in range(n-1):
u,v=map(int,input().split())
u-=1
v-=1
a=find(u)
b=find(v)
union(a,b)
leng,out=0,[]
for i in ans:
if len(i)>leng:
leng=len(i)
out=i
out=[i+1 for i in out]
print(*out)
```
Yes
| 100,378 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Asya loves animals very much. Recently, she purchased n kittens, enumerated them from 1 and n and then put them into the cage. The cage consists of one row of n cells, enumerated with integers from 1 to n from left to right. Adjacent cells had a partially transparent partition wall between them, hence there were n - 1 partitions originally. Initially, each cell contained exactly one kitten with some number.
Observing the kittens, Asya noticed, that they are very friendly and often a pair of kittens in neighboring cells wants to play together. So Asya started to remove partitions between neighboring cells. In particular, on the day i, Asya:
* Noticed, that the kittens x_i and y_i, located in neighboring cells want to play together.
* Removed the partition between these two cells, efficiently creating a single cell, having all kittens from two original cells.
Since Asya has never putted partitions back, after n - 1 days the cage contained a single cell, having all kittens.
For every day, Asya remembers numbers of kittens x_i and y_i, who wanted to play together, however she doesn't remember how she placed kittens in the cage in the beginning. Please help her and find any possible initial arrangement of the kittens into n cells.
Input
The first line contains a single integer n (2 ≤ n ≤ 150 000) — the number of kittens.
Each of the following n - 1 lines contains integers x_i and y_i (1 ≤ x_i, y_i ≤ n, x_i ≠ y_i) — indices of kittens, which got together due to the border removal on the corresponding day.
It's guaranteed, that the kittens x_i and y_i were in the different cells before this day.
Output
For every cell from 1 to n print a single integer — the index of the kitten from 1 to n, who was originally in it.
All printed integers must be distinct.
It's guaranteed, that there is at least one answer possible. In case there are multiple possible answers, print any of them.
Example
Input
5
1 4
2 5
3 1
4 5
Output
3 1 4 2 5
Note
The answer for the example contains one of several possible initial arrangements of the kittens.
The picture below shows how the cells were united for this initial arrangement. Note, that the kittens who wanted to play together on each day were indeed in adjacent cells.
<image>
Submitted Solution:
```
class DisjointSets:
parent = {}
rank = {}
ls = {}
def __init__(self, n):
for i in range(1, n + 1):
self.makeSet(i)
def makeSet(self, node):
self.node = node
self.parent[node] = node
self.rank[node] = 1
self.ls[node] = [node]
def find(self, node):
parent = self.parent
if parent[node] != node:
parent[node] = self.find(parent[node])
return parent[node]
def union(self, node, other):
node = self.find(node)
other = self.find(other)
if self.rank[node] > self.rank[other]:
self.parent[other] = node
self.ls[node].extend(self.ls[other])
elif self.rank[other] > self.rank[node]:
self.parent[node] = other
self.ls[other].extend(self.ls[node])
else:
self.parent[other] = node
self.ls[node] += self.ls[other]
self.rank[node] += 1
n = int(input())
ds = DisjointSets(n)
for i in range(n - 1):
a, b = map(int, input().split())
ds.union(a, b)
print(" ".join(map(str, ds.ls[ds.find(1)])))
```
Yes
| 100,379 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Asya loves animals very much. Recently, she purchased n kittens, enumerated them from 1 and n and then put them into the cage. The cage consists of one row of n cells, enumerated with integers from 1 to n from left to right. Adjacent cells had a partially transparent partition wall between them, hence there were n - 1 partitions originally. Initially, each cell contained exactly one kitten with some number.
Observing the kittens, Asya noticed, that they are very friendly and often a pair of kittens in neighboring cells wants to play together. So Asya started to remove partitions between neighboring cells. In particular, on the day i, Asya:
* Noticed, that the kittens x_i and y_i, located in neighboring cells want to play together.
* Removed the partition between these two cells, efficiently creating a single cell, having all kittens from two original cells.
Since Asya has never putted partitions back, after n - 1 days the cage contained a single cell, having all kittens.
For every day, Asya remembers numbers of kittens x_i and y_i, who wanted to play together, however she doesn't remember how she placed kittens in the cage in the beginning. Please help her and find any possible initial arrangement of the kittens into n cells.
Input
The first line contains a single integer n (2 ≤ n ≤ 150 000) — the number of kittens.
Each of the following n - 1 lines contains integers x_i and y_i (1 ≤ x_i, y_i ≤ n, x_i ≠ y_i) — indices of kittens, which got together due to the border removal on the corresponding day.
It's guaranteed, that the kittens x_i and y_i were in the different cells before this day.
Output
For every cell from 1 to n print a single integer — the index of the kitten from 1 to n, who was originally in it.
All printed integers must be distinct.
It's guaranteed, that there is at least one answer possible. In case there are multiple possible answers, print any of them.
Example
Input
5
1 4
2 5
3 1
4 5
Output
3 1 4 2 5
Note
The answer for the example contains one of several possible initial arrangements of the kittens.
The picture below shows how the cells were united for this initial arrangement. Note, that the kittens who wanted to play together on each day were indeed in adjacent cells.
<image>
Submitted Solution:
```
n=int(input())
dct={}
l=1
for i in range(n-1):
u,v=map(int,input().split())
u,v=min(u,v),max(u,v)
if u in dct and v in dct:
dct[u]=dct[u]+dct[v]
dct[v]=dct[u]
elif u in dct:
dct[u]+=[v]
dct[v]=dct[u]
elif v in dct:
dct[v]+=[u]
dct[u]=dct[v]
else:
dct[u]=[u,v]
dct[v]=dct[u]
for i in dct[u]:
print(i,end=" ")
```
No
| 100,380 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Asya loves animals very much. Recently, she purchased n kittens, enumerated them from 1 and n and then put them into the cage. The cage consists of one row of n cells, enumerated with integers from 1 to n from left to right. Adjacent cells had a partially transparent partition wall between them, hence there were n - 1 partitions originally. Initially, each cell contained exactly one kitten with some number.
Observing the kittens, Asya noticed, that they are very friendly and often a pair of kittens in neighboring cells wants to play together. So Asya started to remove partitions between neighboring cells. In particular, on the day i, Asya:
* Noticed, that the kittens x_i and y_i, located in neighboring cells want to play together.
* Removed the partition between these two cells, efficiently creating a single cell, having all kittens from two original cells.
Since Asya has never putted partitions back, after n - 1 days the cage contained a single cell, having all kittens.
For every day, Asya remembers numbers of kittens x_i and y_i, who wanted to play together, however she doesn't remember how she placed kittens in the cage in the beginning. Please help her and find any possible initial arrangement of the kittens into n cells.
Input
The first line contains a single integer n (2 ≤ n ≤ 150 000) — the number of kittens.
Each of the following n - 1 lines contains integers x_i and y_i (1 ≤ x_i, y_i ≤ n, x_i ≠ y_i) — indices of kittens, which got together due to the border removal on the corresponding day.
It's guaranteed, that the kittens x_i and y_i were in the different cells before this day.
Output
For every cell from 1 to n print a single integer — the index of the kitten from 1 to n, who was originally in it.
All printed integers must be distinct.
It's guaranteed, that there is at least one answer possible. In case there are multiple possible answers, print any of them.
Example
Input
5
1 4
2 5
3 1
4 5
Output
3 1 4 2 5
Note
The answer for the example contains one of several possible initial arrangements of the kittens.
The picture below shows how the cells were united for this initial arrangement. Note, that the kittens who wanted to play together on each day were indeed in adjacent cells.
<image>
Submitted Solution:
```
from collections import defaultdict
# def dfs(visited, vertex, g):
# visited[vertex] = True
# print(vertex + 1,end=" ")
# for i in g[vertex]:
# if visited[i] is False:
# dfs(visited, i, g)
n = int(input())
g = defaultdict(list)
visited = [False] * n
for i in range(n-1):
u, v = map(int, input().split())
g[u-1].append(v-1)
g[v-1].append(u-1)
for i in range(n):
q = [i]
if visited[i] is False:
while len(q):
vertex = q.pop(0)
visited[vertex] = True
print(vertex + 1, end= " ")
for i in g[vertex]:
if visited[i] is False:
q.append(i)
```
No
| 100,381 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Asya loves animals very much. Recently, she purchased n kittens, enumerated them from 1 and n and then put them into the cage. The cage consists of one row of n cells, enumerated with integers from 1 to n from left to right. Adjacent cells had a partially transparent partition wall between them, hence there were n - 1 partitions originally. Initially, each cell contained exactly one kitten with some number.
Observing the kittens, Asya noticed, that they are very friendly and often a pair of kittens in neighboring cells wants to play together. So Asya started to remove partitions between neighboring cells. In particular, on the day i, Asya:
* Noticed, that the kittens x_i and y_i, located in neighboring cells want to play together.
* Removed the partition between these two cells, efficiently creating a single cell, having all kittens from two original cells.
Since Asya has never putted partitions back, after n - 1 days the cage contained a single cell, having all kittens.
For every day, Asya remembers numbers of kittens x_i and y_i, who wanted to play together, however she doesn't remember how she placed kittens in the cage in the beginning. Please help her and find any possible initial arrangement of the kittens into n cells.
Input
The first line contains a single integer n (2 ≤ n ≤ 150 000) — the number of kittens.
Each of the following n - 1 lines contains integers x_i and y_i (1 ≤ x_i, y_i ≤ n, x_i ≠ y_i) — indices of kittens, which got together due to the border removal on the corresponding day.
It's guaranteed, that the kittens x_i and y_i were in the different cells before this day.
Output
For every cell from 1 to n print a single integer — the index of the kitten from 1 to n, who was originally in it.
All printed integers must be distinct.
It's guaranteed, that there is at least one answer possible. In case there are multiple possible answers, print any of them.
Example
Input
5
1 4
2 5
3 1
4 5
Output
3 1 4 2 5
Note
The answer for the example contains one of several possible initial arrangements of the kittens.
The picture below shows how the cells were united for this initial arrangement. Note, that the kittens who wanted to play together on each day were indeed in adjacent cells.
<image>
Submitted Solution:
```
"""
5
1 4
2 5
3 1
4 5
"""
n = int(input())
d = dict()
l = []
for _ in range(n-1):
l += [list(map(int, input().split()))]
for pair in l:
print(pair)
p0 = pair[0] in d.keys()
p1 = pair[1] in d.keys()
if not (p0 or p1):
d[pair[0]] = [pair[0], pair[1]]
d[pair[1]] = [pair[0], pair[1]]
elif p0 and not p1:
d[pair[0]] = d[pair[0]] + [pair[1]]
d[pair[1]] = d[pair[0]]
elif p1 and not p0:
d[pair[1]] = d[pair[1]] + [pair[0]]
else:
d[pair[0]] = d[pair[0]] + d[pair[1]]
d[pair[1]] = d[pair[0]]
for i in d[pair[1]]:
d[i] = d[pair[1]]
print(' '.join(map(str,d[1])))
```
No
| 100,382 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Asya loves animals very much. Recently, she purchased n kittens, enumerated them from 1 and n and then put them into the cage. The cage consists of one row of n cells, enumerated with integers from 1 to n from left to right. Adjacent cells had a partially transparent partition wall between them, hence there were n - 1 partitions originally. Initially, each cell contained exactly one kitten with some number.
Observing the kittens, Asya noticed, that they are very friendly and often a pair of kittens in neighboring cells wants to play together. So Asya started to remove partitions between neighboring cells. In particular, on the day i, Asya:
* Noticed, that the kittens x_i and y_i, located in neighboring cells want to play together.
* Removed the partition between these two cells, efficiently creating a single cell, having all kittens from two original cells.
Since Asya has never putted partitions back, after n - 1 days the cage contained a single cell, having all kittens.
For every day, Asya remembers numbers of kittens x_i and y_i, who wanted to play together, however she doesn't remember how she placed kittens in the cage in the beginning. Please help her and find any possible initial arrangement of the kittens into n cells.
Input
The first line contains a single integer n (2 ≤ n ≤ 150 000) — the number of kittens.
Each of the following n - 1 lines contains integers x_i and y_i (1 ≤ x_i, y_i ≤ n, x_i ≠ y_i) — indices of kittens, which got together due to the border removal on the corresponding day.
It's guaranteed, that the kittens x_i and y_i were in the different cells before this day.
Output
For every cell from 1 to n print a single integer — the index of the kitten from 1 to n, who was originally in it.
All printed integers must be distinct.
It's guaranteed, that there is at least one answer possible. In case there are multiple possible answers, print any of them.
Example
Input
5
1 4
2 5
3 1
4 5
Output
3 1 4 2 5
Note
The answer for the example contains one of several possible initial arrangements of the kittens.
The picture below shows how the cells were united for this initial arrangement. Note, that the kittens who wanted to play together on each day were indeed in adjacent cells.
<image>
Submitted Solution:
```
n = int(input())
uni = {}
sol = []
for _ in range(n-1):
x, y = map(int, input().split())
if x in uni:
if y in uni:
uni[x] += uni[y]
else:
uni[x].append(y)
uni[y] = uni[x]
sol = uni[x].copy()
elif y in uni:
if x in uni:
uni[y] += uni[x]
else:
uni[y].append(x)
uni[x] = uni[y]
sol = uni[y].copy()
else:
uni[x] = [x, y]
uni[y] = uni[x]
sol = uni[x].copy()
# print(uni)
print(*sol)
```
No
| 100,383 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Cat Furrier Transform is a popular algorithm among cat programmers to create longcats. As one of the greatest cat programmers ever exist, Neko wants to utilize this algorithm to create the perfect longcat.
Assume that we have a cat with a number x. A perfect longcat is a cat with a number equal 2^m - 1 for some non-negative integer m. For example, the numbers 0, 1, 3, 7, 15 and so on are suitable for the perfect longcats.
In the Cat Furrier Transform, the following operations can be performed on x:
* (Operation A): you select any non-negative integer n and replace x with x ⊕ (2^n - 1), with ⊕ being a [bitwise XOR operator](https://en.wikipedia.org/wiki/Bitwise_operation#XOR).
* (Operation B): replace x with x + 1.
The first applied operation must be of type A, the second of type B, the third of type A again, and so on. Formally, if we number operations from one in the order they are executed, then odd-numbered operations must be of type A and the even-numbered operations must be of type B.
Neko wants to produce perfect longcats at industrial scale, thus for each cat Neko only wants to perform at most 40 operations. Can you help Neko writing a transformation plan?
Note that it is not required to minimize the number of operations. You just need to use no more than 40 operations.
Input
The only line contains a single integer x (1 ≤ x ≤ 10^6).
Output
The first line should contain a single integer t (0 ≤ t ≤ 40) — the number of operations to apply.
Then for each odd-numbered operation print the corresponding number n_i in it. That is, print ⌈ t/2 ⌉ integers n_i (0 ≤ n_i ≤ 30), denoting the replacement x with x ⊕ (2^{n_i} - 1) in the corresponding step.
If there are multiple possible answers, you can print any of them. It is possible to show, that there is at least one answer in the constraints of this problem.
Examples
Input
39
Output
4
5 3
Input
1
Output
0
Input
7
Output
0
Note
In the first test, one of the transforms might be as follows: 39 → 56 → 57 → 62 → 63. Or more precisely:
1. Pick n = 5. x is transformed into 39 ⊕ 31, or 56.
2. Increase x by 1, changing its value to 57.
3. Pick n = 3. x is transformed into 57 ⊕ 7, or 62.
4. Increase x by 1, changing its value to 63 = 2^6 - 1.
In the second and third test, the number already satisfies the goal requirement.
Tags: bitmasks, constructive algorithms, dfs and similar, math
Correct Solution:
```
i = int(input())
def getZ(n):
s = "{0:0b}".format(n)
return len(s) - s.find("0")
def check(n):
return "{0:0b}".format(n).find("0") == -1
steps = 0
ans = []
while not check(i):
m = getZ(i)
i = i ^ ((2 ** m) - 1)
steps +=1
ans.append(m)
if check(i):
print(steps)
print(" ".join(map(str, ans)))
exit(0)
i += 1
steps += 1
print(steps)
print(" ".join(map(str, ans)))
exit(0)
```
| 100,384 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Cat Furrier Transform is a popular algorithm among cat programmers to create longcats. As one of the greatest cat programmers ever exist, Neko wants to utilize this algorithm to create the perfect longcat.
Assume that we have a cat with a number x. A perfect longcat is a cat with a number equal 2^m - 1 for some non-negative integer m. For example, the numbers 0, 1, 3, 7, 15 and so on are suitable for the perfect longcats.
In the Cat Furrier Transform, the following operations can be performed on x:
* (Operation A): you select any non-negative integer n and replace x with x ⊕ (2^n - 1), with ⊕ being a [bitwise XOR operator](https://en.wikipedia.org/wiki/Bitwise_operation#XOR).
* (Operation B): replace x with x + 1.
The first applied operation must be of type A, the second of type B, the third of type A again, and so on. Formally, if we number operations from one in the order they are executed, then odd-numbered operations must be of type A and the even-numbered operations must be of type B.
Neko wants to produce perfect longcats at industrial scale, thus for each cat Neko only wants to perform at most 40 operations. Can you help Neko writing a transformation plan?
Note that it is not required to minimize the number of operations. You just need to use no more than 40 operations.
Input
The only line contains a single integer x (1 ≤ x ≤ 10^6).
Output
The first line should contain a single integer t (0 ≤ t ≤ 40) — the number of operations to apply.
Then for each odd-numbered operation print the corresponding number n_i in it. That is, print ⌈ t/2 ⌉ integers n_i (0 ≤ n_i ≤ 30), denoting the replacement x with x ⊕ (2^{n_i} - 1) in the corresponding step.
If there are multiple possible answers, you can print any of them. It is possible to show, that there is at least one answer in the constraints of this problem.
Examples
Input
39
Output
4
5 3
Input
1
Output
0
Input
7
Output
0
Note
In the first test, one of the transforms might be as follows: 39 → 56 → 57 → 62 → 63. Or more precisely:
1. Pick n = 5. x is transformed into 39 ⊕ 31, or 56.
2. Increase x by 1, changing its value to 57.
3. Pick n = 3. x is transformed into 57 ⊕ 7, or 62.
4. Increase x by 1, changing its value to 63 = 2^6 - 1.
In the second and third test, the number already satisfies the goal requirement.
Tags: bitmasks, constructive algorithms, dfs and similar, math
Correct Solution:
```
import bisect
import decimal
from decimal import Decimal
import os
from collections import Counter
import bisect
from collections import defaultdict
import math
import random
import heapq
from math import sqrt
import sys
from functools import reduce, cmp_to_key
from collections import deque
import threading
from itertools import combinations
from io import BytesIO, IOBase
from itertools import accumulate
# sys.setrecursionlimit(200000)
# mod = 10**9+7
# mod = 998244353
decimal.getcontext().prec = 46
def primeFactors(n):
prime = set()
while n % 2 == 0:
prime.add(2)
n = n//2
for i in range(3,int(math.sqrt(n))+1,2):
while n % i== 0:
prime.add(i)
n = n//i
if n > 2:
prime.add(n)
return list(prime)
def getFactors(n) :
factors = []
i = 1
while i <= math.sqrt(n):
if (n % i == 0) :
if (n // i == i) :
factors.append(i)
else :
factors.append(i)
factors.append(n//i)
i = i + 1
return factors
def modefiedSieve():
mx=10**7+1
sieve=[-1]*mx
for i in range(2,mx):
if sieve[i]==-1:
sieve[i]=i
for j in range(i*i,mx,i):
if sieve[j]==-1:
sieve[j]=i
return sieve
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
num = []
for p in range(2, n+1):
if prime[p]:
num.append(p)
return num
def lcm(a,b):
return (a*b)//math.gcd(a,b)
def sort_dict(key_value):
return sorted(key_value.items(), key = lambda kv:(kv[1], kv[0]), reverse=True)
def list_input():
return list(map(int,input().split()))
def num_input():
return map(int,input().split())
def string_list():
return list(input())
def decimalToBinary(n):
return bin(n).replace("0b", "")
def binaryToDecimal(n):
return int(n,2)
def DFS(n,s,adj):
visited = [False for i in range(n+1)]
stack = []
stack.append(s)
while (len(stack)):
s = stack[-1]
stack.pop()
if (not visited[s]):
visited[s] = True
for node in adj[s]:
if (not visited[node]):
stack.append(node)
def maxSubArraySum(a,size):
max_so_far = -sys.maxsize - 1
max_ending_here = 0
start = 0
end = 0
s = 0
for i in range(0,size):
max_ending_here += a[i]
if max_so_far < max_ending_here:
max_so_far = max_ending_here
start = s
end = i
if max_ending_here < 0:
max_ending_here = 0
s = i+1
return max_so_far,start,end
def lis(arr):
n = len(arr)
lis = [1]*n
for i in range (1 , n):
for j in range(0 , i):
if arr[i] >= arr[j] and lis[i]< lis[j] + 1 :
lis[i] = lis[j]+1
maximum = 0
for i in range(n):
maximum = max(maximum , lis[i])
return maximum
def solve():
n = int(input())
ans = []
x = math.ceil(math.log(n,2))
num = 2**x-1
while n != num:
n = n^num
ans.append(x)
x = math.ceil(math.log(n,2))
num = 2**x-1
if n == num:
break
n += 1
ans.append(n)
x = math.ceil(math.log(n,2))
num = 2**x-1
print(len(ans))
for i in range(0,len(ans),2):
print(ans[i],end=' ')
t = 1
#t = int(input())
for _ in range(t):
solve()
```
| 100,385 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Cat Furrier Transform is a popular algorithm among cat programmers to create longcats. As one of the greatest cat programmers ever exist, Neko wants to utilize this algorithm to create the perfect longcat.
Assume that we have a cat with a number x. A perfect longcat is a cat with a number equal 2^m - 1 for some non-negative integer m. For example, the numbers 0, 1, 3, 7, 15 and so on are suitable for the perfect longcats.
In the Cat Furrier Transform, the following operations can be performed on x:
* (Operation A): you select any non-negative integer n and replace x with x ⊕ (2^n - 1), with ⊕ being a [bitwise XOR operator](https://en.wikipedia.org/wiki/Bitwise_operation#XOR).
* (Operation B): replace x with x + 1.
The first applied operation must be of type A, the second of type B, the third of type A again, and so on. Formally, if we number operations from one in the order they are executed, then odd-numbered operations must be of type A and the even-numbered operations must be of type B.
Neko wants to produce perfect longcats at industrial scale, thus for each cat Neko only wants to perform at most 40 operations. Can you help Neko writing a transformation plan?
Note that it is not required to minimize the number of operations. You just need to use no more than 40 operations.
Input
The only line contains a single integer x (1 ≤ x ≤ 10^6).
Output
The first line should contain a single integer t (0 ≤ t ≤ 40) — the number of operations to apply.
Then for each odd-numbered operation print the corresponding number n_i in it. That is, print ⌈ t/2 ⌉ integers n_i (0 ≤ n_i ≤ 30), denoting the replacement x with x ⊕ (2^{n_i} - 1) in the corresponding step.
If there are multiple possible answers, you can print any of them. It is possible to show, that there is at least one answer in the constraints of this problem.
Examples
Input
39
Output
4
5 3
Input
1
Output
0
Input
7
Output
0
Note
In the first test, one of the transforms might be as follows: 39 → 56 → 57 → 62 → 63. Or more precisely:
1. Pick n = 5. x is transformed into 39 ⊕ 31, or 56.
2. Increase x by 1, changing its value to 57.
3. Pick n = 3. x is transformed into 57 ⊕ 7, or 62.
4. Increase x by 1, changing its value to 63 = 2^6 - 1.
In the second and third test, the number already satisfies the goal requirement.
Tags: bitmasks, constructive algorithms, dfs and similar, math
Correct Solution:
```
from sys import *
from math import *
from bisect import *
n=int(stdin.readline())
x=n.bit_length()
ans=[]
f=0
for i in range(x,0,-1):
y=(bin(1<<(i-1)))
if (1<<(i-1) & n)==0:
n=n^(2**i-1)
k=bin(n)
ans.append(i)
if n==2**(x)-1:
f=1
break
n+=1
if n==(2**x)-1:
break
if f==0:
print(len(ans)*2)
else:
print(len(ans)*2-1)
if len(ans)!=0:
print(*ans)
```
| 100,386 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Cat Furrier Transform is a popular algorithm among cat programmers to create longcats. As one of the greatest cat programmers ever exist, Neko wants to utilize this algorithm to create the perfect longcat.
Assume that we have a cat with a number x. A perfect longcat is a cat with a number equal 2^m - 1 for some non-negative integer m. For example, the numbers 0, 1, 3, 7, 15 and so on are suitable for the perfect longcats.
In the Cat Furrier Transform, the following operations can be performed on x:
* (Operation A): you select any non-negative integer n and replace x with x ⊕ (2^n - 1), with ⊕ being a [bitwise XOR operator](https://en.wikipedia.org/wiki/Bitwise_operation#XOR).
* (Operation B): replace x with x + 1.
The first applied operation must be of type A, the second of type B, the third of type A again, and so on. Formally, if we number operations from one in the order they are executed, then odd-numbered operations must be of type A and the even-numbered operations must be of type B.
Neko wants to produce perfect longcats at industrial scale, thus for each cat Neko only wants to perform at most 40 operations. Can you help Neko writing a transformation plan?
Note that it is not required to minimize the number of operations. You just need to use no more than 40 operations.
Input
The only line contains a single integer x (1 ≤ x ≤ 10^6).
Output
The first line should contain a single integer t (0 ≤ t ≤ 40) — the number of operations to apply.
Then for each odd-numbered operation print the corresponding number n_i in it. That is, print ⌈ t/2 ⌉ integers n_i (0 ≤ n_i ≤ 30), denoting the replacement x with x ⊕ (2^{n_i} - 1) in the corresponding step.
If there are multiple possible answers, you can print any of them. It is possible to show, that there is at least one answer in the constraints of this problem.
Examples
Input
39
Output
4
5 3
Input
1
Output
0
Input
7
Output
0
Note
In the first test, one of the transforms might be as follows: 39 → 56 → 57 → 62 → 63. Or more precisely:
1. Pick n = 5. x is transformed into 39 ⊕ 31, or 56.
2. Increase x by 1, changing its value to 57.
3. Pick n = 3. x is transformed into 57 ⊕ 7, or 62.
4. Increase x by 1, changing its value to 63 = 2^6 - 1.
In the second and third test, the number already satisfies the goal requirement.
Tags: bitmasks, constructive algorithms, dfs and similar, math
Correct Solution:
```
x = int(input())
b = len(bin(x)) - 2
ans = []
n = 0
while x ^ ((1 << b) - 1):
bn = bin(x)[2:]
for i in range(b):
if bn[i] == '0':
x ^= ((1 << (b-i)) - 1)
ans.append(b-i)
break
if x ^ ((1 << b) - 1) == 0:
n = len(ans) * 2 - 1
break
x += 1
b = len(bin(x)) - 2
bn = bin(x)[2:]
print(n if n != 0 else len(ans) * 2)
print(*ans)
```
| 100,387 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Cat Furrier Transform is a popular algorithm among cat programmers to create longcats. As one of the greatest cat programmers ever exist, Neko wants to utilize this algorithm to create the perfect longcat.
Assume that we have a cat with a number x. A perfect longcat is a cat with a number equal 2^m - 1 for some non-negative integer m. For example, the numbers 0, 1, 3, 7, 15 and so on are suitable for the perfect longcats.
In the Cat Furrier Transform, the following operations can be performed on x:
* (Operation A): you select any non-negative integer n and replace x with x ⊕ (2^n - 1), with ⊕ being a [bitwise XOR operator](https://en.wikipedia.org/wiki/Bitwise_operation#XOR).
* (Operation B): replace x with x + 1.
The first applied operation must be of type A, the second of type B, the third of type A again, and so on. Formally, if we number operations from one in the order they are executed, then odd-numbered operations must be of type A and the even-numbered operations must be of type B.
Neko wants to produce perfect longcats at industrial scale, thus for each cat Neko only wants to perform at most 40 operations. Can you help Neko writing a transformation plan?
Note that it is not required to minimize the number of operations. You just need to use no more than 40 operations.
Input
The only line contains a single integer x (1 ≤ x ≤ 10^6).
Output
The first line should contain a single integer t (0 ≤ t ≤ 40) — the number of operations to apply.
Then for each odd-numbered operation print the corresponding number n_i in it. That is, print ⌈ t/2 ⌉ integers n_i (0 ≤ n_i ≤ 30), denoting the replacement x with x ⊕ (2^{n_i} - 1) in the corresponding step.
If there are multiple possible answers, you can print any of them. It is possible to show, that there is at least one answer in the constraints of this problem.
Examples
Input
39
Output
4
5 3
Input
1
Output
0
Input
7
Output
0
Note
In the first test, one of the transforms might be as follows: 39 → 56 → 57 → 62 → 63. Or more precisely:
1. Pick n = 5. x is transformed into 39 ⊕ 31, or 56.
2. Increase x by 1, changing its value to 57.
3. Pick n = 3. x is transformed into 57 ⊕ 7, or 62.
4. Increase x by 1, changing its value to 63 = 2^6 - 1.
In the second and third test, the number already satisfies the goal requirement.
Tags: bitmasks, constructive algorithms, dfs and similar, math
Correct Solution:
```
#554_B
import sys
import math
num = int(sys.stdin.readline().rstrip())
nl = math.floor(math.log(num, 2))
a = []
op = 0
while True:
if num == 2 ** (nl + 1) - 1:
break
pw = math.floor(math.log((2 ** (nl + 1) - 1) - num, 2)) + 1
op += 1
num = num ^ ((2 ** pw) - 1)
a.append(pw)
if num == 2 ** (nl + 1) - 1:
break
op += 1
num += 1
if len(a) == 0:
print(0)
else:
print(op)
print(" ".join([str(i) for i in a]))
```
| 100,388 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Cat Furrier Transform is a popular algorithm among cat programmers to create longcats. As one of the greatest cat programmers ever exist, Neko wants to utilize this algorithm to create the perfect longcat.
Assume that we have a cat with a number x. A perfect longcat is a cat with a number equal 2^m - 1 for some non-negative integer m. For example, the numbers 0, 1, 3, 7, 15 and so on are suitable for the perfect longcats.
In the Cat Furrier Transform, the following operations can be performed on x:
* (Operation A): you select any non-negative integer n and replace x with x ⊕ (2^n - 1), with ⊕ being a [bitwise XOR operator](https://en.wikipedia.org/wiki/Bitwise_operation#XOR).
* (Operation B): replace x with x + 1.
The first applied operation must be of type A, the second of type B, the third of type A again, and so on. Formally, if we number operations from one in the order they are executed, then odd-numbered operations must be of type A and the even-numbered operations must be of type B.
Neko wants to produce perfect longcats at industrial scale, thus for each cat Neko only wants to perform at most 40 operations. Can you help Neko writing a transformation plan?
Note that it is not required to minimize the number of operations. You just need to use no more than 40 operations.
Input
The only line contains a single integer x (1 ≤ x ≤ 10^6).
Output
The first line should contain a single integer t (0 ≤ t ≤ 40) — the number of operations to apply.
Then for each odd-numbered operation print the corresponding number n_i in it. That is, print ⌈ t/2 ⌉ integers n_i (0 ≤ n_i ≤ 30), denoting the replacement x with x ⊕ (2^{n_i} - 1) in the corresponding step.
If there are multiple possible answers, you can print any of them. It is possible to show, that there is at least one answer in the constraints of this problem.
Examples
Input
39
Output
4
5 3
Input
1
Output
0
Input
7
Output
0
Note
In the first test, one of the transforms might be as follows: 39 → 56 → 57 → 62 → 63. Or more precisely:
1. Pick n = 5. x is transformed into 39 ⊕ 31, or 56.
2. Increase x by 1, changing its value to 57.
3. Pick n = 3. x is transformed into 57 ⊕ 7, or 62.
4. Increase x by 1, changing its value to 63 = 2^6 - 1.
In the second and third test, the number already satisfies the goal requirement.
Tags: bitmasks, constructive algorithms, dfs and similar, math
Correct Solution:
```
#--------------------
#WA
#--------------------
# from math import log
# def operation(b):
# flag = 0
# s = ""
# for i in range(len(b)-1,1,-1):
# if(b[i]=='0'):
# flag = 1
# if(flag==0 and b[i]=='1'):
# s=b[i]+s
# if(flag==1):
# if(b[i]=='0'):
# x = '1'
# else:
# x = '0'
# s = x + s
# return s
# # Driver program
# x = int(input())
# while(x&(x+1)!=0): # checking if it's of the form 2^n-1 or not
# p = int(operation(str(bin(x))),2)
# print(int(log(p,2))+1,end=" ")
# x = (x^p)+1
from math import log
x = int(input())
if(x<1 or x==1 or x&(x+1)==0):
print(0)
elif(x&(x-1)==0):
print(1)
print(int(log(x,2)))
else:
count = 0
a = []
while(x&(x+1)!=0 and count<40):
d = int(log(x,2))
p = 2**(d+1)-1
a.append(d+1)
#print(p,"XOR",x,"=",x^p)
x = (x^p)
count+=1
if(x&(x+1)==0):
break
#print(x&(x+1))
x+=1
#print("x = ",x)
count+=1
print(count)
for i in range(len(a)):
print(a[i],end=" ")
```
| 100,389 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Cat Furrier Transform is a popular algorithm among cat programmers to create longcats. As one of the greatest cat programmers ever exist, Neko wants to utilize this algorithm to create the perfect longcat.
Assume that we have a cat with a number x. A perfect longcat is a cat with a number equal 2^m - 1 for some non-negative integer m. For example, the numbers 0, 1, 3, 7, 15 and so on are suitable for the perfect longcats.
In the Cat Furrier Transform, the following operations can be performed on x:
* (Operation A): you select any non-negative integer n and replace x with x ⊕ (2^n - 1), with ⊕ being a [bitwise XOR operator](https://en.wikipedia.org/wiki/Bitwise_operation#XOR).
* (Operation B): replace x with x + 1.
The first applied operation must be of type A, the second of type B, the third of type A again, and so on. Formally, if we number operations from one in the order they are executed, then odd-numbered operations must be of type A and the even-numbered operations must be of type B.
Neko wants to produce perfect longcats at industrial scale, thus for each cat Neko only wants to perform at most 40 operations. Can you help Neko writing a transformation plan?
Note that it is not required to minimize the number of operations. You just need to use no more than 40 operations.
Input
The only line contains a single integer x (1 ≤ x ≤ 10^6).
Output
The first line should contain a single integer t (0 ≤ t ≤ 40) — the number of operations to apply.
Then for each odd-numbered operation print the corresponding number n_i in it. That is, print ⌈ t/2 ⌉ integers n_i (0 ≤ n_i ≤ 30), denoting the replacement x with x ⊕ (2^{n_i} - 1) in the corresponding step.
If there are multiple possible answers, you can print any of them. It is possible to show, that there is at least one answer in the constraints of this problem.
Examples
Input
39
Output
4
5 3
Input
1
Output
0
Input
7
Output
0
Note
In the first test, one of the transforms might be as follows: 39 → 56 → 57 → 62 → 63. Or more precisely:
1. Pick n = 5. x is transformed into 39 ⊕ 31, or 56.
2. Increase x by 1, changing its value to 57.
3. Pick n = 3. x is transformed into 57 ⊕ 7, or 62.
4. Increase x by 1, changing its value to 63 = 2^6 - 1.
In the second and third test, the number already satisfies the goal requirement.
Tags: bitmasks, constructive algorithms, dfs and similar, math
Correct Solution:
```
def check(t):
y = t
while y:
if y % 2 == 0:
return False
y = y >> 1
return True
x = int(input())
c = 0
res = []
def cnt(k):
y = k
t = 0
while y:
t += 1
y = y >> 1
return t
while not check(x):
c += 1
t = cnt(x)
res.append(t)
x = x ^ (2 ** t - 1)
if check(x):
break
x += 1
c += 1
print(c)
print(' '.join([str(x) for x in res]))
```
| 100,390 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Cat Furrier Transform is a popular algorithm among cat programmers to create longcats. As one of the greatest cat programmers ever exist, Neko wants to utilize this algorithm to create the perfect longcat.
Assume that we have a cat with a number x. A perfect longcat is a cat with a number equal 2^m - 1 for some non-negative integer m. For example, the numbers 0, 1, 3, 7, 15 and so on are suitable for the perfect longcats.
In the Cat Furrier Transform, the following operations can be performed on x:
* (Operation A): you select any non-negative integer n and replace x with x ⊕ (2^n - 1), with ⊕ being a [bitwise XOR operator](https://en.wikipedia.org/wiki/Bitwise_operation#XOR).
* (Operation B): replace x with x + 1.
The first applied operation must be of type A, the second of type B, the third of type A again, and so on. Formally, if we number operations from one in the order they are executed, then odd-numbered operations must be of type A and the even-numbered operations must be of type B.
Neko wants to produce perfect longcats at industrial scale, thus for each cat Neko only wants to perform at most 40 operations. Can you help Neko writing a transformation plan?
Note that it is not required to minimize the number of operations. You just need to use no more than 40 operations.
Input
The only line contains a single integer x (1 ≤ x ≤ 10^6).
Output
The first line should contain a single integer t (0 ≤ t ≤ 40) — the number of operations to apply.
Then for each odd-numbered operation print the corresponding number n_i in it. That is, print ⌈ t/2 ⌉ integers n_i (0 ≤ n_i ≤ 30), denoting the replacement x with x ⊕ (2^{n_i} - 1) in the corresponding step.
If there are multiple possible answers, you can print any of them. It is possible to show, that there is at least one answer in the constraints of this problem.
Examples
Input
39
Output
4
5 3
Input
1
Output
0
Input
7
Output
0
Note
In the first test, one of the transforms might be as follows: 39 → 56 → 57 → 62 → 63. Or more precisely:
1. Pick n = 5. x is transformed into 39 ⊕ 31, or 56.
2. Increase x by 1, changing its value to 57.
3. Pick n = 3. x is transformed into 57 ⊕ 7, or 62.
4. Increase x by 1, changing its value to 63 = 2^6 - 1.
In the second and third test, the number already satisfies the goal requirement.
Tags: bitmasks, constructive algorithms, dfs and similar, math
Correct Solution:
```
from bisect import bisect_right as br
from bisect import bisect_left as bl
from collections import *
from itertools import *
import functools
import sys
from math import *
MAX = sys.maxsize
MAXN = 10**5+10
MOD = 10**9+7
def isprime(n):
n = abs(int(n))
if n < 2:
return False
if n == 2:
return True
if not n & 1:
return False
for x in range(3, int(n**0.5) + 1, 2):
if n % x == 0:
return False
return True
def mhd(a,b,x,y):
return abs(a-x)+abs(b-y)
def numIN(x = " "):
return(map(int,sys.stdin.readline().strip().split(x)))
def charIN(x= ' '):
return(sys.stdin.readline().strip().split(x))
def arrIN():
return list(numIN())
def dis(x,y):
a = y[0]-x[0]
b = x[1]-y[1]
return (a*a+b*b)**0.5
def lgcd(a):
g = a[0]
for i in range(1,len(a)):
g = math.gcd(g,a[i])
return g
def ms(a):
msf = -MAX
meh = 0
st = en = be = 0
for i in range(len(a)):
meh+=a[i]
if msf<meh:
msf = meh
st = be
en = i
if meh<0:
meh = 0
be = i+1
return msf,st,en
def res(ans,t):
print('Case #{}: {}'.format(t,ans))
def chk(n):
x = bin(n)[2:]
if x.count('0')==0:
return 1
return 0
n = int(input())
x = bin(n)[2:]
if chk(n):
print(0)
else:
ans = []
op=0
while not chk(n):
x = bin(n)[2:]
idx = x.index('0')
ans.append(len(x)-idx)
n = n^(2**(ans[-1])-1)
op+=1
if chk(n):
break
n+=1
op+=1
print(op)
print(' '.join([str(i) for i in ans]))
```
| 100,391 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Cat Furrier Transform is a popular algorithm among cat programmers to create longcats. As one of the greatest cat programmers ever exist, Neko wants to utilize this algorithm to create the perfect longcat.
Assume that we have a cat with a number x. A perfect longcat is a cat with a number equal 2^m - 1 for some non-negative integer m. For example, the numbers 0, 1, 3, 7, 15 and so on are suitable for the perfect longcats.
In the Cat Furrier Transform, the following operations can be performed on x:
* (Operation A): you select any non-negative integer n and replace x with x ⊕ (2^n - 1), with ⊕ being a [bitwise XOR operator](https://en.wikipedia.org/wiki/Bitwise_operation#XOR).
* (Operation B): replace x with x + 1.
The first applied operation must be of type A, the second of type B, the third of type A again, and so on. Formally, if we number operations from one in the order they are executed, then odd-numbered operations must be of type A and the even-numbered operations must be of type B.
Neko wants to produce perfect longcats at industrial scale, thus for each cat Neko only wants to perform at most 40 operations. Can you help Neko writing a transformation plan?
Note that it is not required to minimize the number of operations. You just need to use no more than 40 operations.
Input
The only line contains a single integer x (1 ≤ x ≤ 10^6).
Output
The first line should contain a single integer t (0 ≤ t ≤ 40) — the number of operations to apply.
Then for each odd-numbered operation print the corresponding number n_i in it. That is, print ⌈ t/2 ⌉ integers n_i (0 ≤ n_i ≤ 30), denoting the replacement x with x ⊕ (2^{n_i} - 1) in the corresponding step.
If there are multiple possible answers, you can print any of them. It is possible to show, that there is at least one answer in the constraints of this problem.
Examples
Input
39
Output
4
5 3
Input
1
Output
0
Input
7
Output
0
Note
In the first test, one of the transforms might be as follows: 39 → 56 → 57 → 62 → 63. Or more precisely:
1. Pick n = 5. x is transformed into 39 ⊕ 31, or 56.
2. Increase x by 1, changing its value to 57.
3. Pick n = 3. x is transformed into 57 ⊕ 7, or 62.
4. Increase x by 1, changing its value to 63 = 2^6 - 1.
In the second and third test, the number already satisfies the goal requirement.
Submitted Solution:
```
import math
n=int(input())
b=bin(n)
b=b[2:]
l=[]
s=""
c=0
m=0
if len(set(b))==1:
print(0)
else:
while(len(set(b))!=1):
for i in range(len(b)):
if b[i]=="0":
c=i
break
s="1"*(len(b)-c)
k=int(s,2)
n=n^k
l.append(int(math.log((k+1),2)))
b=bin(n)
b=b[2:]
m=m+1
if len(set(b))==1:
break
else:
n=n+1
b=bin(n)
b=b[2:]
m=m+1
#print(n)
#print(n)
#print(bin(n))
print(m)
print(*l)
```
Yes
| 100,392 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Cat Furrier Transform is a popular algorithm among cat programmers to create longcats. As one of the greatest cat programmers ever exist, Neko wants to utilize this algorithm to create the perfect longcat.
Assume that we have a cat with a number x. A perfect longcat is a cat with a number equal 2^m - 1 for some non-negative integer m. For example, the numbers 0, 1, 3, 7, 15 and so on are suitable for the perfect longcats.
In the Cat Furrier Transform, the following operations can be performed on x:
* (Operation A): you select any non-negative integer n and replace x with x ⊕ (2^n - 1), with ⊕ being a [bitwise XOR operator](https://en.wikipedia.org/wiki/Bitwise_operation#XOR).
* (Operation B): replace x with x + 1.
The first applied operation must be of type A, the second of type B, the third of type A again, and so on. Formally, if we number operations from one in the order they are executed, then odd-numbered operations must be of type A and the even-numbered operations must be of type B.
Neko wants to produce perfect longcats at industrial scale, thus for each cat Neko only wants to perform at most 40 operations. Can you help Neko writing a transformation plan?
Note that it is not required to minimize the number of operations. You just need to use no more than 40 operations.
Input
The only line contains a single integer x (1 ≤ x ≤ 10^6).
Output
The first line should contain a single integer t (0 ≤ t ≤ 40) — the number of operations to apply.
Then for each odd-numbered operation print the corresponding number n_i in it. That is, print ⌈ t/2 ⌉ integers n_i (0 ≤ n_i ≤ 30), denoting the replacement x with x ⊕ (2^{n_i} - 1) in the corresponding step.
If there are multiple possible answers, you can print any of them. It is possible to show, that there is at least one answer in the constraints of this problem.
Examples
Input
39
Output
4
5 3
Input
1
Output
0
Input
7
Output
0
Note
In the first test, one of the transforms might be as follows: 39 → 56 → 57 → 62 → 63. Or more precisely:
1. Pick n = 5. x is transformed into 39 ⊕ 31, or 56.
2. Increase x by 1, changing its value to 57.
3. Pick n = 3. x is transformed into 57 ⊕ 7, or 62.
4. Increase x by 1, changing its value to 63 = 2^6 - 1.
In the second and third test, the number already satisfies the goal requirement.
Submitted Solution:
```
import math
n = int(input())
powers = []
x = []
temp = 1
for i in range(21):
temp = 2 ** i
powers.append(temp - 1)
if (temp - 1 > n):
break
if n in powers:
print(0)
exit()
for t in range(21):
diff = abs(powers[-1] - n)
l = math.log2(diff)
l = math.ceil(l)
if diff == 2:
l += 1
x.append(l)
l = 2 ** l - 1
n = n ^ l
if n == powers[-1]:
print(2*t+1)
print(*x)
break
n += 1
if n == powers[-1]:
print(2*t+2)
print(*x)
break
```
Yes
| 100,393 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Cat Furrier Transform is a popular algorithm among cat programmers to create longcats. As one of the greatest cat programmers ever exist, Neko wants to utilize this algorithm to create the perfect longcat.
Assume that we have a cat with a number x. A perfect longcat is a cat with a number equal 2^m - 1 for some non-negative integer m. For example, the numbers 0, 1, 3, 7, 15 and so on are suitable for the perfect longcats.
In the Cat Furrier Transform, the following operations can be performed on x:
* (Operation A): you select any non-negative integer n and replace x with x ⊕ (2^n - 1), with ⊕ being a [bitwise XOR operator](https://en.wikipedia.org/wiki/Bitwise_operation#XOR).
* (Operation B): replace x with x + 1.
The first applied operation must be of type A, the second of type B, the third of type A again, and so on. Formally, if we number operations from one in the order they are executed, then odd-numbered operations must be of type A and the even-numbered operations must be of type B.
Neko wants to produce perfect longcats at industrial scale, thus for each cat Neko only wants to perform at most 40 operations. Can you help Neko writing a transformation plan?
Note that it is not required to minimize the number of operations. You just need to use no more than 40 operations.
Input
The only line contains a single integer x (1 ≤ x ≤ 10^6).
Output
The first line should contain a single integer t (0 ≤ t ≤ 40) — the number of operations to apply.
Then for each odd-numbered operation print the corresponding number n_i in it. That is, print ⌈ t/2 ⌉ integers n_i (0 ≤ n_i ≤ 30), denoting the replacement x with x ⊕ (2^{n_i} - 1) in the corresponding step.
If there are multiple possible answers, you can print any of them. It is possible to show, that there is at least one answer in the constraints of this problem.
Examples
Input
39
Output
4
5 3
Input
1
Output
0
Input
7
Output
0
Note
In the first test, one of the transforms might be as follows: 39 → 56 → 57 → 62 → 63. Or more precisely:
1. Pick n = 5. x is transformed into 39 ⊕ 31, or 56.
2. Increase x by 1, changing its value to 57.
3. Pick n = 3. x is transformed into 57 ⊕ 7, or 62.
4. Increase x by 1, changing its value to 63 = 2^6 - 1.
In the second and third test, the number already satisfies the goal requirement.
Submitted Solution:
```
# AC
import sys
class Main:
def __init__(self):
self.buff = None
self.index = 0
def next(self):
if self.buff is None or self.index == len(self.buff):
self.buff = sys.stdin.readline().split()
self.index = 0
val = self.buff[self.index]
self.index += 1
return val
def next_int(self):
return int(self.next())
def solve(self):
n = self.next_int()
stp = 0
res = []
while not self.test(n):
d = self.get(n)
res.append(self.cal(d))
n ^= d
stp += 1
if not self.test(n):
n += 1
stp += 1
print(stp)
if stp > 0:
print(' '.join(str(x) for x in res))
def test(self, n):
d = n + 1
return d & -d == d
def get(self, n):
d = 1
while not self.test(n | d):
d = d * 2 + 1
return d
def cal(self, d):
x = 0
while d > 0:
x += 1
d >>= 1
return x
if __name__ == '__main__':
Main().solve()
```
Yes
| 100,394 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Cat Furrier Transform is a popular algorithm among cat programmers to create longcats. As one of the greatest cat programmers ever exist, Neko wants to utilize this algorithm to create the perfect longcat.
Assume that we have a cat with a number x. A perfect longcat is a cat with a number equal 2^m - 1 for some non-negative integer m. For example, the numbers 0, 1, 3, 7, 15 and so on are suitable for the perfect longcats.
In the Cat Furrier Transform, the following operations can be performed on x:
* (Operation A): you select any non-negative integer n and replace x with x ⊕ (2^n - 1), with ⊕ being a [bitwise XOR operator](https://en.wikipedia.org/wiki/Bitwise_operation#XOR).
* (Operation B): replace x with x + 1.
The first applied operation must be of type A, the second of type B, the third of type A again, and so on. Formally, if we number operations from one in the order they are executed, then odd-numbered operations must be of type A and the even-numbered operations must be of type B.
Neko wants to produce perfect longcats at industrial scale, thus for each cat Neko only wants to perform at most 40 operations. Can you help Neko writing a transformation plan?
Note that it is not required to minimize the number of operations. You just need to use no more than 40 operations.
Input
The only line contains a single integer x (1 ≤ x ≤ 10^6).
Output
The first line should contain a single integer t (0 ≤ t ≤ 40) — the number of operations to apply.
Then for each odd-numbered operation print the corresponding number n_i in it. That is, print ⌈ t/2 ⌉ integers n_i (0 ≤ n_i ≤ 30), denoting the replacement x with x ⊕ (2^{n_i} - 1) in the corresponding step.
If there are multiple possible answers, you can print any of them. It is possible to show, that there is at least one answer in the constraints of this problem.
Examples
Input
39
Output
4
5 3
Input
1
Output
0
Input
7
Output
0
Note
In the first test, one of the transforms might be as follows: 39 → 56 → 57 → 62 → 63. Or more precisely:
1. Pick n = 5. x is transformed into 39 ⊕ 31, or 56.
2. Increase x by 1, changing its value to 57.
3. Pick n = 3. x is transformed into 57 ⊕ 7, or 62.
4. Increase x by 1, changing its value to 63 = 2^6 - 1.
In the second and third test, the number already satisfies the goal requirement.
Submitted Solution:
```
x = int(input())
prop = {2**i - 1 for i in range(1, 40)}
ans = 0
Ans = []
while x not in prop:
j = x.bit_length()
ans += 1
Ans.append(j)
x ^= (1<<j) - 1
if x in prop:
break
x += 1
ans += 1
print(ans)
if ans:
print(*Ans)
```
Yes
| 100,395 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Cat Furrier Transform is a popular algorithm among cat programmers to create longcats. As one of the greatest cat programmers ever exist, Neko wants to utilize this algorithm to create the perfect longcat.
Assume that we have a cat with a number x. A perfect longcat is a cat with a number equal 2^m - 1 for some non-negative integer m. For example, the numbers 0, 1, 3, 7, 15 and so on are suitable for the perfect longcats.
In the Cat Furrier Transform, the following operations can be performed on x:
* (Operation A): you select any non-negative integer n and replace x with x ⊕ (2^n - 1), with ⊕ being a [bitwise XOR operator](https://en.wikipedia.org/wiki/Bitwise_operation#XOR).
* (Operation B): replace x with x + 1.
The first applied operation must be of type A, the second of type B, the third of type A again, and so on. Formally, if we number operations from one in the order they are executed, then odd-numbered operations must be of type A and the even-numbered operations must be of type B.
Neko wants to produce perfect longcats at industrial scale, thus for each cat Neko only wants to perform at most 40 operations. Can you help Neko writing a transformation plan?
Note that it is not required to minimize the number of operations. You just need to use no more than 40 operations.
Input
The only line contains a single integer x (1 ≤ x ≤ 10^6).
Output
The first line should contain a single integer t (0 ≤ t ≤ 40) — the number of operations to apply.
Then for each odd-numbered operation print the corresponding number n_i in it. That is, print ⌈ t/2 ⌉ integers n_i (0 ≤ n_i ≤ 30), denoting the replacement x with x ⊕ (2^{n_i} - 1) in the corresponding step.
If there are multiple possible answers, you can print any of them. It is possible to show, that there is at least one answer in the constraints of this problem.
Examples
Input
39
Output
4
5 3
Input
1
Output
0
Input
7
Output
0
Note
In the first test, one of the transforms might be as follows: 39 → 56 → 57 → 62 → 63. Or more precisely:
1. Pick n = 5. x is transformed into 39 ⊕ 31, or 56.
2. Increase x by 1, changing its value to 57.
3. Pick n = 3. x is transformed into 57 ⊕ 7, or 62.
4. Increase x by 1, changing its value to 63 = 2^6 - 1.
In the second and third test, the number already satisfies the goal requirement.
Submitted Solution:
```
x = int(input())
arr = []
k = 2 ** 30
s = 0
while k > 0:
if x >= k:
arr.append(1)
x -= k
s = 1
elif s == 1:
arr.append(0)
k = k // 2
#print(arr)
res = 0
op = 1
ooo = 0
oper = []
while res == 0:
res = 1
for el in arr:
if el == 0:
res = 0
if res == 1:
break
if op == 0:
ooo += 1
x += 1
arr = []
k = 2 ** 30
s = 0
while k > 0:
if x >= k:
arr.append(1)
x -= k
s = 1
elif s == 1:
arr.append(0)
k = k // 2
op = 1
#print(x)
#print(arr)
continue
a = []
for i in range(len(arr)):
if arr[i] == 0:
for j in range(i, len(arr)):
a.append(1)
break
#print('a =', a)
e = []
for i in range(len(arr) - len(a)):
e.append(arr[i])
d = len(arr) - len(a)
for i in range(d, len(arr)):
if (arr[i] != a[i - d]):
e.append(1)
else:
e.append(0)
arr = e.copy()
#print(arr)
summ = 0
k = 1
x = 0
for i in range(len(arr) - 1, -1, -1):
x += arr[i] * k
k *= 2
k = 1
for i in range(len(a) - 1, -1, -1):
summ += a[i] * k
k *= 2
#print(summ)
l = 30
while 2 ** l >= summ:
l -= 1
oper.append(l + 1)
ooo += 1
op = 0
if ooo > 40:
print('olololo')
break
print(ooo)
print(*oper)
```
No
| 100,396 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Cat Furrier Transform is a popular algorithm among cat programmers to create longcats. As one of the greatest cat programmers ever exist, Neko wants to utilize this algorithm to create the perfect longcat.
Assume that we have a cat with a number x. A perfect longcat is a cat with a number equal 2^m - 1 for some non-negative integer m. For example, the numbers 0, 1, 3, 7, 15 and so on are suitable for the perfect longcats.
In the Cat Furrier Transform, the following operations can be performed on x:
* (Operation A): you select any non-negative integer n and replace x with x ⊕ (2^n - 1), with ⊕ being a [bitwise XOR operator](https://en.wikipedia.org/wiki/Bitwise_operation#XOR).
* (Operation B): replace x with x + 1.
The first applied operation must be of type A, the second of type B, the third of type A again, and so on. Formally, if we number operations from one in the order they are executed, then odd-numbered operations must be of type A and the even-numbered operations must be of type B.
Neko wants to produce perfect longcats at industrial scale, thus for each cat Neko only wants to perform at most 40 operations. Can you help Neko writing a transformation plan?
Note that it is not required to minimize the number of operations. You just need to use no more than 40 operations.
Input
The only line contains a single integer x (1 ≤ x ≤ 10^6).
Output
The first line should contain a single integer t (0 ≤ t ≤ 40) — the number of operations to apply.
Then for each odd-numbered operation print the corresponding number n_i in it. That is, print ⌈ t/2 ⌉ integers n_i (0 ≤ n_i ≤ 30), denoting the replacement x with x ⊕ (2^{n_i} - 1) in the corresponding step.
If there are multiple possible answers, you can print any of them. It is possible to show, that there is at least one answer in the constraints of this problem.
Examples
Input
39
Output
4
5 3
Input
1
Output
0
Input
7
Output
0
Note
In the first test, one of the transforms might be as follows: 39 → 56 → 57 → 62 → 63. Or more precisely:
1. Pick n = 5. x is transformed into 39 ⊕ 31, or 56.
2. Increase x by 1, changing its value to 57.
3. Pick n = 3. x is transformed into 57 ⊕ 7, or 62.
4. Increase x by 1, changing its value to 63 = 2^6 - 1.
In the second and third test, the number already satisfies the goal requirement.
Submitted Solution:
```
import math
n=int(input())
if n==1:
print(0)
exit()
ans = []
r = 0
while True:
if math.log2(n + 1).is_integer():
break
else:
r+=1
s = bin(n)
s=s[2:]
pos=-1
for i in range(len(s)):
if s[i]=='0':
pos = len(s)-i
ans.append(pos)
break
n = n^(int(pow(2,pos)-1))
n+=1
print(r)
for i in ans:
print(i,end=' ')
```
No
| 100,397 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Cat Furrier Transform is a popular algorithm among cat programmers to create longcats. As one of the greatest cat programmers ever exist, Neko wants to utilize this algorithm to create the perfect longcat.
Assume that we have a cat with a number x. A perfect longcat is a cat with a number equal 2^m - 1 for some non-negative integer m. For example, the numbers 0, 1, 3, 7, 15 and so on are suitable for the perfect longcats.
In the Cat Furrier Transform, the following operations can be performed on x:
* (Operation A): you select any non-negative integer n and replace x with x ⊕ (2^n - 1), with ⊕ being a [bitwise XOR operator](https://en.wikipedia.org/wiki/Bitwise_operation#XOR).
* (Operation B): replace x with x + 1.
The first applied operation must be of type A, the second of type B, the third of type A again, and so on. Formally, if we number operations from one in the order they are executed, then odd-numbered operations must be of type A and the even-numbered operations must be of type B.
Neko wants to produce perfect longcats at industrial scale, thus for each cat Neko only wants to perform at most 40 operations. Can you help Neko writing a transformation plan?
Note that it is not required to minimize the number of operations. You just need to use no more than 40 operations.
Input
The only line contains a single integer x (1 ≤ x ≤ 10^6).
Output
The first line should contain a single integer t (0 ≤ t ≤ 40) — the number of operations to apply.
Then for each odd-numbered operation print the corresponding number n_i in it. That is, print ⌈ t/2 ⌉ integers n_i (0 ≤ n_i ≤ 30), denoting the replacement x with x ⊕ (2^{n_i} - 1) in the corresponding step.
If there are multiple possible answers, you can print any of them. It is possible to show, that there is at least one answer in the constraints of this problem.
Examples
Input
39
Output
4
5 3
Input
1
Output
0
Input
7
Output
0
Note
In the first test, one of the transforms might be as follows: 39 → 56 → 57 → 62 → 63. Or more precisely:
1. Pick n = 5. x is transformed into 39 ⊕ 31, or 56.
2. Increase x by 1, changing its value to 57.
3. Pick n = 3. x is transformed into 57 ⊕ 7, or 62.
4. Increase x by 1, changing its value to 63 = 2^6 - 1.
In the second and third test, the number already satisfies the goal requirement.
Submitted Solution:
```
if __name__ == '__main__':
x = int(input())
resultarr = []
prevbit = x & 1
x >>= 1
i = 1
while x != 0:
if x & 1 != prevbit:
resultarr.append(i)
i += 1
prevbit = x & 1
x >>= 1
print(len(resultarr) * 2 - len(resultarr) % 2)
for elem in resultarr:
print(elem,end=' ')
```
No
| 100,398 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Cat Furrier Transform is a popular algorithm among cat programmers to create longcats. As one of the greatest cat programmers ever exist, Neko wants to utilize this algorithm to create the perfect longcat.
Assume that we have a cat with a number x. A perfect longcat is a cat with a number equal 2^m - 1 for some non-negative integer m. For example, the numbers 0, 1, 3, 7, 15 and so on are suitable for the perfect longcats.
In the Cat Furrier Transform, the following operations can be performed on x:
* (Operation A): you select any non-negative integer n and replace x with x ⊕ (2^n - 1), with ⊕ being a [bitwise XOR operator](https://en.wikipedia.org/wiki/Bitwise_operation#XOR).
* (Operation B): replace x with x + 1.
The first applied operation must be of type A, the second of type B, the third of type A again, and so on. Formally, if we number operations from one in the order they are executed, then odd-numbered operations must be of type A and the even-numbered operations must be of type B.
Neko wants to produce perfect longcats at industrial scale, thus for each cat Neko only wants to perform at most 40 operations. Can you help Neko writing a transformation plan?
Note that it is not required to minimize the number of operations. You just need to use no more than 40 operations.
Input
The only line contains a single integer x (1 ≤ x ≤ 10^6).
Output
The first line should contain a single integer t (0 ≤ t ≤ 40) — the number of operations to apply.
Then for each odd-numbered operation print the corresponding number n_i in it. That is, print ⌈ t/2 ⌉ integers n_i (0 ≤ n_i ≤ 30), denoting the replacement x with x ⊕ (2^{n_i} - 1) in the corresponding step.
If there are multiple possible answers, you can print any of them. It is possible to show, that there is at least one answer in the constraints of this problem.
Examples
Input
39
Output
4
5 3
Input
1
Output
0
Input
7
Output
0
Note
In the first test, one of the transforms might be as follows: 39 → 56 → 57 → 62 → 63. Or more precisely:
1. Pick n = 5. x is transformed into 39 ⊕ 31, or 56.
2. Increase x by 1, changing its value to 57.
3. Pick n = 3. x is transformed into 57 ⊕ 7, or 62.
4. Increase x by 1, changing its value to 63 = 2^6 - 1.
In the second and third test, the number already satisfies the goal requirement.
Submitted Solution:
```
s = {2**i -1 for i in range(25)}
n = int(input())
tbp = []
c=0
while n not in s:
last = -1
for i in range(20):
if (n & (2**i)) == 0 and (2**i) < n:
last = i
if last == -1:
break # done
n ^= (-1+2**(last+1))
c+=1
tbp.append(str(last))
if n in s: break
c+=1
n+=1
print(c)
print(" ".join(tbp))
```
No
| 100,399 |
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