description
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This is the hard version of the problem. The difference between the versions is that in the easy version all prices $a_i$ are different. You can make hacks if and only if you solved both versions of the problem.
Today is Sage's birthday, and she will go shopping to buy ice spheres. All $n$ ice spheres are placed in a row and they are numbered from $1$ to $n$ from left to right. Each ice sphere has a positive integer price. In this version, some prices can be equal.
An ice sphere is cheap if it costs strictly less than two neighboring ice spheres: the nearest to the left and the nearest to the right. The leftmost and the rightmost ice spheres are not cheap. Sage will choose all cheap ice spheres and then buy only them.
You can visit the shop before Sage and reorder the ice spheres as you wish. Find out the maximum number of ice spheres that Sage can buy, and show how the ice spheres should be reordered.
-----Input-----
The first line contains a single integer $n$ $(1 \le n \le 10^5)$ — the number of ice spheres in the shop.
The second line contains $n$ integers $a_1, a_2, \dots, a_n$ $(1 \le a_i \le 10^9)$ — the prices of ice spheres.
-----Output-----
In the first line print the maximum number of ice spheres that Sage can buy.
In the second line print the prices of ice spheres in the optimal order. If there are several correct answers, you can print any of them.
-----Example-----
Input
7
1 3 2 2 4 5 4
Output
3
3 1 4 2 4 2 5
-----Note-----
In the sample it's not possible to place the ice spheres in any order so that Sage would buy $4$ of them. If the spheres are placed in the order $(3, 1, 4, 2, 4, 2, 5)$, then Sage will buy one sphere for $1$ and two spheres for $2$ each.
|
def winner(nums, n):
nums = sorted(nums)
ans = []
i = 0
j = n // 2
count = 0
while j < n:
ans.append(nums[j])
if i < n // 2:
ans.append(nums[i])
if j != n - 1 and nums[i] != nums[j]:
count += 1
j += 1
i += 1
return ans, count
n = int(input())
nums, count = winner([int(i) for i in input().split()], n)
print(count)
for num in nums:
print(num, end=" ")
|
FUNC_DEF ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR LIST ASSIGN VAR NUMBER ASSIGN VAR BIN_OP VAR NUMBER ASSIGN VAR NUMBER WHILE VAR VAR EXPR FUNC_CALL VAR VAR VAR IF VAR BIN_OP VAR NUMBER EXPR FUNC_CALL VAR VAR VAR IF VAR BIN_OP VAR NUMBER VAR VAR VAR VAR VAR NUMBER VAR NUMBER VAR NUMBER RETURN VAR VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR VAR FUNC_CALL VAR FUNC_CALL VAR VAR VAR FUNC_CALL FUNC_CALL VAR VAR EXPR FUNC_CALL VAR VAR FOR VAR VAR EXPR FUNC_CALL VAR VAR STRING
|
This is the hard version of the problem. The difference between the versions is that in the easy version all prices $a_i$ are different. You can make hacks if and only if you solved both versions of the problem.
Today is Sage's birthday, and she will go shopping to buy ice spheres. All $n$ ice spheres are placed in a row and they are numbered from $1$ to $n$ from left to right. Each ice sphere has a positive integer price. In this version, some prices can be equal.
An ice sphere is cheap if it costs strictly less than two neighboring ice spheres: the nearest to the left and the nearest to the right. The leftmost and the rightmost ice spheres are not cheap. Sage will choose all cheap ice spheres and then buy only them.
You can visit the shop before Sage and reorder the ice spheres as you wish. Find out the maximum number of ice spheres that Sage can buy, and show how the ice spheres should be reordered.
-----Input-----
The first line contains a single integer $n$ $(1 \le n \le 10^5)$ — the number of ice spheres in the shop.
The second line contains $n$ integers $a_1, a_2, \dots, a_n$ $(1 \le a_i \le 10^9)$ — the prices of ice spheres.
-----Output-----
In the first line print the maximum number of ice spheres that Sage can buy.
In the second line print the prices of ice spheres in the optimal order. If there are several correct answers, you can print any of them.
-----Example-----
Input
7
1 3 2 2 4 5 4
Output
3
3 1 4 2 4 2 5
-----Note-----
In the sample it's not possible to place the ice spheres in any order so that Sage would buy $4$ of them. If the spheres are placed in the order $(3, 1, 4, 2, 4, 2, 5)$, then Sage will buy one sphere for $1$ and two spheres for $2$ each.
|
import sys
input = sys.stdin.readline
n = int(input())
A = sorted(map(int, input().split()))
ans = [0] * n
j = n - 1
for i in range(0, n, 2):
ans[i] = A[j]
j -= 1
rem = []
cnt = 0
pre = -1
for i in range(1, n - 1, 2):
if j < 0:
break
while j >= 0 and A[j] >= ans[i + 1]:
rem.append(A[j])
j -= 1
cnt += 1
ans[i] = A[j]
pre = i
j -= 1
rem += A[: j + 1]
j = 0
if rem:
for k in range(pre + 2, n, 2):
ans[k] = rem[j]
j += 1
print(cnt)
print(*ans)
|
IMPORT ASSIGN VAR VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR BIN_OP LIST NUMBER VAR ASSIGN VAR BIN_OP VAR NUMBER FOR VAR FUNC_CALL VAR NUMBER VAR NUMBER ASSIGN VAR VAR VAR VAR VAR NUMBER ASSIGN VAR LIST ASSIGN VAR NUMBER ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR NUMBER BIN_OP VAR NUMBER NUMBER IF VAR NUMBER WHILE VAR NUMBER VAR VAR VAR BIN_OP VAR NUMBER EXPR FUNC_CALL VAR VAR VAR VAR NUMBER VAR NUMBER ASSIGN VAR VAR VAR VAR ASSIGN VAR VAR VAR NUMBER VAR VAR BIN_OP VAR NUMBER ASSIGN VAR NUMBER IF VAR FOR VAR FUNC_CALL VAR BIN_OP VAR NUMBER VAR NUMBER ASSIGN VAR VAR VAR VAR VAR NUMBER EXPR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR VAR
|
This is the hard version of the problem. The difference between the versions is that in the easy version all prices $a_i$ are different. You can make hacks if and only if you solved both versions of the problem.
Today is Sage's birthday, and she will go shopping to buy ice spheres. All $n$ ice spheres are placed in a row and they are numbered from $1$ to $n$ from left to right. Each ice sphere has a positive integer price. In this version, some prices can be equal.
An ice sphere is cheap if it costs strictly less than two neighboring ice spheres: the nearest to the left and the nearest to the right. The leftmost and the rightmost ice spheres are not cheap. Sage will choose all cheap ice spheres and then buy only them.
You can visit the shop before Sage and reorder the ice spheres as you wish. Find out the maximum number of ice spheres that Sage can buy, and show how the ice spheres should be reordered.
-----Input-----
The first line contains a single integer $n$ $(1 \le n \le 10^5)$ — the number of ice spheres in the shop.
The second line contains $n$ integers $a_1, a_2, \dots, a_n$ $(1 \le a_i \le 10^9)$ — the prices of ice spheres.
-----Output-----
In the first line print the maximum number of ice spheres that Sage can buy.
In the second line print the prices of ice spheres in the optimal order. If there are several correct answers, you can print any of them.
-----Example-----
Input
7
1 3 2 2 4 5 4
Output
3
3 1 4 2 4 2 5
-----Note-----
In the sample it's not possible to place the ice spheres in any order so that Sage would buy $4$ of them. If the spheres are placed in the order $(3, 1, 4, 2, 4, 2, 5)$, then Sage will buy one sphere for $1$ and two spheres for $2$ each.
|
n = int(input())
a = list(map(int, input().split()))
if n <= 2:
print(0)
print(" ".join(map(str, a)))
else:
a.sort()
b = [0] * n
j = 0
for i in range(1, n, 2):
b[i] = a[j]
j += 1
for i in range(0, n, 2):
b[i] = a[j]
j += 1
c = 0
for i in range(1, n - 1, 2):
if b[i - 1] > b[i] < b[i + 1]:
c += 1
print(c)
print(" ".join(map(str, b)))
|
ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR IF VAR NUMBER EXPR FUNC_CALL VAR NUMBER EXPR FUNC_CALL VAR FUNC_CALL STRING FUNC_CALL VAR VAR VAR EXPR FUNC_CALL VAR ASSIGN VAR BIN_OP LIST NUMBER VAR ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR NUMBER VAR NUMBER ASSIGN VAR VAR VAR VAR VAR NUMBER FOR VAR FUNC_CALL VAR NUMBER VAR NUMBER ASSIGN VAR VAR VAR VAR VAR NUMBER ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR NUMBER BIN_OP VAR NUMBER NUMBER IF VAR BIN_OP VAR NUMBER VAR VAR VAR BIN_OP VAR NUMBER VAR NUMBER EXPR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR FUNC_CALL STRING FUNC_CALL VAR VAR VAR
|
This is the hard version of the problem. The difference between the versions is that in the easy version all prices $a_i$ are different. You can make hacks if and only if you solved both versions of the problem.
Today is Sage's birthday, and she will go shopping to buy ice spheres. All $n$ ice spheres are placed in a row and they are numbered from $1$ to $n$ from left to right. Each ice sphere has a positive integer price. In this version, some prices can be equal.
An ice sphere is cheap if it costs strictly less than two neighboring ice spheres: the nearest to the left and the nearest to the right. The leftmost and the rightmost ice spheres are not cheap. Sage will choose all cheap ice spheres and then buy only them.
You can visit the shop before Sage and reorder the ice spheres as you wish. Find out the maximum number of ice spheres that Sage can buy, and show how the ice spheres should be reordered.
-----Input-----
The first line contains a single integer $n$ $(1 \le n \le 10^5)$ — the number of ice spheres in the shop.
The second line contains $n$ integers $a_1, a_2, \dots, a_n$ $(1 \le a_i \le 10^9)$ — the prices of ice spheres.
-----Output-----
In the first line print the maximum number of ice spheres that Sage can buy.
In the second line print the prices of ice spheres in the optimal order. If there are several correct answers, you can print any of them.
-----Example-----
Input
7
1 3 2 2 4 5 4
Output
3
3 1 4 2 4 2 5
-----Note-----
In the sample it's not possible to place the ice spheres in any order so that Sage would buy $4$ of them. If the spheres are placed in the order $(3, 1, 4, 2, 4, 2, 5)$, then Sage will buy one sphere for $1$ and two spheres for $2$ each.
|
t = 1
for _ in range(t):
n = int(input())
arr = [int(x) for x in input().split()]
arr = sorted(arr)
temp = [(0) for x in range(n)]
c = 0
for i in range(1, n, 2):
temp[i] = arr[c]
c += 1
for i in range(0, n, 2):
temp[i] = arr[c]
c += 1
ans = 0
val = n
if n % 2 == 0:
val -= 1
for i in range(1, val, 2):
if temp[i] < temp[i - 1] and temp[i] < temp[i + 1]:
ans += 1
print(ans)
print(*temp)
|
ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR NUMBER VAR FUNC_CALL VAR VAR ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR NUMBER VAR NUMBER ASSIGN VAR VAR VAR VAR VAR NUMBER FOR VAR FUNC_CALL VAR NUMBER VAR NUMBER ASSIGN VAR VAR VAR VAR VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR VAR IF BIN_OP VAR NUMBER NUMBER VAR NUMBER FOR VAR FUNC_CALL VAR NUMBER VAR NUMBER IF VAR VAR VAR BIN_OP VAR NUMBER VAR VAR VAR BIN_OP VAR NUMBER VAR NUMBER EXPR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR VAR
|
This is the hard version of the problem. The difference between the versions is that in the easy version all prices $a_i$ are different. You can make hacks if and only if you solved both versions of the problem.
Today is Sage's birthday, and she will go shopping to buy ice spheres. All $n$ ice spheres are placed in a row and they are numbered from $1$ to $n$ from left to right. Each ice sphere has a positive integer price. In this version, some prices can be equal.
An ice sphere is cheap if it costs strictly less than two neighboring ice spheres: the nearest to the left and the nearest to the right. The leftmost and the rightmost ice spheres are not cheap. Sage will choose all cheap ice spheres and then buy only them.
You can visit the shop before Sage and reorder the ice spheres as you wish. Find out the maximum number of ice spheres that Sage can buy, and show how the ice spheres should be reordered.
-----Input-----
The first line contains a single integer $n$ $(1 \le n \le 10^5)$ — the number of ice spheres in the shop.
The second line contains $n$ integers $a_1, a_2, \dots, a_n$ $(1 \le a_i \le 10^9)$ — the prices of ice spheres.
-----Output-----
In the first line print the maximum number of ice spheres that Sage can buy.
In the second line print the prices of ice spheres in the optimal order. If there are several correct answers, you can print any of them.
-----Example-----
Input
7
1 3 2 2 4 5 4
Output
3
3 1 4 2 4 2 5
-----Note-----
In the sample it's not possible to place the ice spheres in any order so that Sage would buy $4$ of them. If the spheres are placed in the order $(3, 1, 4, 2, 4, 2, 5)$, then Sage will buy one sphere for $1$ and two spheres for $2$ each.
|
n = int(input())
ls = list(map(int, input().split()))
ls.sort()
ReturnLs = []
i = 0
for j in range(n // 2, n):
ReturnLs.append(ls[j])
if i != n // 2:
ReturnLs.append(ls[i])
i += 1
count = 0
for i in range(1, n - 1, 2):
if ReturnLs[i] < ReturnLs[i + 1] and ReturnLs[i - 1] > ReturnLs[i]:
count += 1
print(count)
print(*ReturnLs)
|
ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR EXPR FUNC_CALL VAR ASSIGN VAR LIST ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR BIN_OP VAR NUMBER VAR EXPR FUNC_CALL VAR VAR VAR IF VAR BIN_OP VAR NUMBER EXPR FUNC_CALL VAR VAR VAR VAR NUMBER ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR NUMBER BIN_OP VAR NUMBER NUMBER IF VAR VAR VAR BIN_OP VAR NUMBER VAR BIN_OP VAR NUMBER VAR VAR VAR NUMBER EXPR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR VAR
|
This is the hard version of the problem. The difference between the versions is that in the easy version all prices $a_i$ are different. You can make hacks if and only if you solved both versions of the problem.
Today is Sage's birthday, and she will go shopping to buy ice spheres. All $n$ ice spheres are placed in a row and they are numbered from $1$ to $n$ from left to right. Each ice sphere has a positive integer price. In this version, some prices can be equal.
An ice sphere is cheap if it costs strictly less than two neighboring ice spheres: the nearest to the left and the nearest to the right. The leftmost and the rightmost ice spheres are not cheap. Sage will choose all cheap ice spheres and then buy only them.
You can visit the shop before Sage and reorder the ice spheres as you wish. Find out the maximum number of ice spheres that Sage can buy, and show how the ice spheres should be reordered.
-----Input-----
The first line contains a single integer $n$ $(1 \le n \le 10^5)$ — the number of ice spheres in the shop.
The second line contains $n$ integers $a_1, a_2, \dots, a_n$ $(1 \le a_i \le 10^9)$ — the prices of ice spheres.
-----Output-----
In the first line print the maximum number of ice spheres that Sage can buy.
In the second line print the prices of ice spheres in the optimal order. If there are several correct answers, you can print any of them.
-----Example-----
Input
7
1 3 2 2 4 5 4
Output
3
3 1 4 2 4 2 5
-----Note-----
In the sample it's not possible to place the ice spheres in any order so that Sage would buy $4$ of them. If the spheres are placed in the order $(3, 1, 4, 2, 4, 2, 5)$, then Sage will buy one sphere for $1$ and two spheres for $2$ each.
|
def inp():
return int(input())
def inlt():
return list(map(int, input().split()))
def insr():
s = input()
return list(s[: len(s) - 1])
def invr():
return map(int, input().split())
n, l = inp(), inlt()
res = []
l.sort()
for i in range(n // 2):
if i + n // 2 < n:
res.append(l[i + n // 2])
res.append(l[i])
if n % 2:
res.append(l[-1])
cnt = 0
for i in range(1, len(res) - 1):
if res[i - 1] > res[i] < res[i + 1]:
cnt += 1
print(cnt)
print(" ".join([str(x) for x in res]))
|
FUNC_DEF RETURN FUNC_CALL VAR FUNC_CALL VAR FUNC_DEF RETURN FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR FUNC_DEF ASSIGN VAR FUNC_CALL VAR RETURN FUNC_CALL VAR VAR BIN_OP FUNC_CALL VAR VAR NUMBER FUNC_DEF RETURN FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR LIST EXPR FUNC_CALL VAR FOR VAR FUNC_CALL VAR BIN_OP VAR NUMBER IF BIN_OP VAR BIN_OP VAR NUMBER VAR EXPR FUNC_CALL VAR VAR BIN_OP VAR BIN_OP VAR NUMBER EXPR FUNC_CALL VAR VAR VAR IF BIN_OP VAR NUMBER EXPR FUNC_CALL VAR VAR NUMBER ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR NUMBER BIN_OP FUNC_CALL VAR VAR NUMBER IF VAR BIN_OP VAR NUMBER VAR VAR VAR BIN_OP VAR NUMBER VAR NUMBER EXPR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR FUNC_CALL STRING FUNC_CALL VAR VAR VAR VAR
|
This is the hard version of the problem. The difference between the versions is that in the easy version all prices $a_i$ are different. You can make hacks if and only if you solved both versions of the problem.
Today is Sage's birthday, and she will go shopping to buy ice spheres. All $n$ ice spheres are placed in a row and they are numbered from $1$ to $n$ from left to right. Each ice sphere has a positive integer price. In this version, some prices can be equal.
An ice sphere is cheap if it costs strictly less than two neighboring ice spheres: the nearest to the left and the nearest to the right. The leftmost and the rightmost ice spheres are not cheap. Sage will choose all cheap ice spheres and then buy only them.
You can visit the shop before Sage and reorder the ice spheres as you wish. Find out the maximum number of ice spheres that Sage can buy, and show how the ice spheres should be reordered.
-----Input-----
The first line contains a single integer $n$ $(1 \le n \le 10^5)$ — the number of ice spheres in the shop.
The second line contains $n$ integers $a_1, a_2, \dots, a_n$ $(1 \le a_i \le 10^9)$ — the prices of ice spheres.
-----Output-----
In the first line print the maximum number of ice spheres that Sage can buy.
In the second line print the prices of ice spheres in the optimal order. If there are several correct answers, you can print any of them.
-----Example-----
Input
7
1 3 2 2 4 5 4
Output
3
3 1 4 2 4 2 5
-----Note-----
In the sample it's not possible to place the ice spheres in any order so that Sage would buy $4$ of them. If the spheres are placed in the order $(3, 1, 4, 2, 4, 2, 5)$, then Sage will buy one sphere for $1$ and two spheres for $2$ each.
|
import sys
def inp():
return sys.stdin.readline().rstrip("\n").encode("utf8")
def mpint():
return map(int, sys.stdin.readline().split(" "))
def itg():
return int(sys.stdin.readline())
n = itg()
arr = sorted(mpint())
a1 = arr[n >> 1 :]
a2 = arr[: n >> 1]
ans = []
for i in range(n):
if i & 1:
ans.append(a2.pop())
else:
ans.append(a1.pop())
for i in range(n - 2):
if ans[i] == ans[i + 1]:
ans[i + 1], ans[i + 2] = ans[i + 2], ans[i + 1]
ans1 = 0
for i in range(1, n - 1):
if ans[i] < ans[i - 1] and ans[i] < ans[i + 1]:
ans1 += 1
print(ans1)
print(*ans)
|
IMPORT FUNC_DEF RETURN FUNC_CALL FUNC_CALL FUNC_CALL VAR STRING STRING FUNC_DEF RETURN FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR STRING FUNC_DEF RETURN FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR VAR BIN_OP VAR NUMBER ASSIGN VAR VAR BIN_OP VAR NUMBER ASSIGN VAR LIST FOR VAR FUNC_CALL VAR VAR IF BIN_OP VAR NUMBER EXPR FUNC_CALL VAR FUNC_CALL VAR EXPR FUNC_CALL VAR FUNC_CALL VAR FOR VAR FUNC_CALL VAR BIN_OP VAR NUMBER IF VAR VAR VAR BIN_OP VAR NUMBER ASSIGN VAR BIN_OP VAR NUMBER VAR BIN_OP VAR NUMBER VAR BIN_OP VAR NUMBER VAR BIN_OP VAR NUMBER ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR NUMBER BIN_OP VAR NUMBER IF VAR VAR VAR BIN_OP VAR NUMBER VAR VAR VAR BIN_OP VAR NUMBER VAR NUMBER EXPR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR VAR
|
This is the hard version of the problem. The difference between the versions is that in the easy version all prices $a_i$ are different. You can make hacks if and only if you solved both versions of the problem.
Today is Sage's birthday, and she will go shopping to buy ice spheres. All $n$ ice spheres are placed in a row and they are numbered from $1$ to $n$ from left to right. Each ice sphere has a positive integer price. In this version, some prices can be equal.
An ice sphere is cheap if it costs strictly less than two neighboring ice spheres: the nearest to the left and the nearest to the right. The leftmost and the rightmost ice spheres are not cheap. Sage will choose all cheap ice spheres and then buy only them.
You can visit the shop before Sage and reorder the ice spheres as you wish. Find out the maximum number of ice spheres that Sage can buy, and show how the ice spheres should be reordered.
-----Input-----
The first line contains a single integer $n$ $(1 \le n \le 10^5)$ — the number of ice spheres in the shop.
The second line contains $n$ integers $a_1, a_2, \dots, a_n$ $(1 \le a_i \le 10^9)$ — the prices of ice spheres.
-----Output-----
In the first line print the maximum number of ice spheres that Sage can buy.
In the second line print the prices of ice spheres in the optimal order. If there are several correct answers, you can print any of them.
-----Example-----
Input
7
1 3 2 2 4 5 4
Output
3
3 1 4 2 4 2 5
-----Note-----
In the sample it's not possible to place the ice spheres in any order so that Sage would buy $4$ of them. If the spheres are placed in the order $(3, 1, 4, 2, 4, 2, 5)$, then Sage will buy one sphere for $1$ and two spheres for $2$ each.
|
t = 1
for _ in range(t):
n = int(input())
l = [int(j) for j in input().split()]
l.sort()
c, cnt = 0, 0
k = [0] * n
for i in range(n // 2):
k[c] = l[n // 2 + i]
c += 1
k[c] = l[i]
c += 1
if n % 2 != 0:
k[n - 1] = l[n - 1]
for i in range(1, n - 1):
if k[i - 1] > k[i] and k[i + 1] > k[i]:
cnt += 1
print(cnt)
print(*k)
|
ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR VAR VAR FUNC_CALL FUNC_CALL VAR EXPR FUNC_CALL VAR ASSIGN VAR VAR NUMBER NUMBER ASSIGN VAR BIN_OP LIST NUMBER VAR FOR VAR FUNC_CALL VAR BIN_OP VAR NUMBER ASSIGN VAR VAR VAR BIN_OP BIN_OP VAR NUMBER VAR VAR NUMBER ASSIGN VAR VAR VAR VAR VAR NUMBER IF BIN_OP VAR NUMBER NUMBER ASSIGN VAR BIN_OP VAR NUMBER VAR BIN_OP VAR NUMBER FOR VAR FUNC_CALL VAR NUMBER BIN_OP VAR NUMBER IF VAR BIN_OP VAR NUMBER VAR VAR VAR BIN_OP VAR NUMBER VAR VAR VAR NUMBER EXPR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR VAR
|
This is the hard version of the problem. The difference between the versions is that in the easy version all prices $a_i$ are different. You can make hacks if and only if you solved both versions of the problem.
Today is Sage's birthday, and she will go shopping to buy ice spheres. All $n$ ice spheres are placed in a row and they are numbered from $1$ to $n$ from left to right. Each ice sphere has a positive integer price. In this version, some prices can be equal.
An ice sphere is cheap if it costs strictly less than two neighboring ice spheres: the nearest to the left and the nearest to the right. The leftmost and the rightmost ice spheres are not cheap. Sage will choose all cheap ice spheres and then buy only them.
You can visit the shop before Sage and reorder the ice spheres as you wish. Find out the maximum number of ice spheres that Sage can buy, and show how the ice spheres should be reordered.
-----Input-----
The first line contains a single integer $n$ $(1 \le n \le 10^5)$ — the number of ice spheres in the shop.
The second line contains $n$ integers $a_1, a_2, \dots, a_n$ $(1 \le a_i \le 10^9)$ — the prices of ice spheres.
-----Output-----
In the first line print the maximum number of ice spheres that Sage can buy.
In the second line print the prices of ice spheres in the optimal order. If there are several correct answers, you can print any of them.
-----Example-----
Input
7
1 3 2 2 4 5 4
Output
3
3 1 4 2 4 2 5
-----Note-----
In the sample it's not possible to place the ice spheres in any order so that Sage would buy $4$ of them. If the spheres are placed in the order $(3, 1, 4, 2, 4, 2, 5)$, then Sage will buy one sphere for $1$ and two spheres for $2$ each.
|
a = int(input())
b = [int(x) for x in input().split()]
b.sort()
c = []
d = []
result = [0] * a
ans = 0
k = -1
l = 0
m = 0
if a == 1:
print(0)
print(*b)
else:
for i in range(a // 2):
c.append(b[i])
for j in range(i + 1, a):
d.append(b[j])
while k + 1 != a:
k += 1
if (k + 1) % 2 != 0:
result[k] = d[l]
l += 1
else:
result[k] = c[m]
m += 1
for n in range(1, a - 1):
if result[n] < result[n - 1] and result[n] < result[n + 1]:
ans += 1
print(ans)
print(*result)
|
ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR VAR VAR FUNC_CALL FUNC_CALL VAR EXPR FUNC_CALL VAR ASSIGN VAR LIST ASSIGN VAR LIST ASSIGN VAR BIN_OP LIST NUMBER VAR ASSIGN VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR NUMBER IF VAR NUMBER EXPR FUNC_CALL VAR NUMBER EXPR FUNC_CALL VAR VAR FOR VAR FUNC_CALL VAR BIN_OP VAR NUMBER EXPR FUNC_CALL VAR VAR VAR FOR VAR FUNC_CALL VAR BIN_OP VAR NUMBER VAR EXPR FUNC_CALL VAR VAR VAR WHILE BIN_OP VAR NUMBER VAR VAR NUMBER IF BIN_OP BIN_OP VAR NUMBER NUMBER NUMBER ASSIGN VAR VAR VAR VAR VAR NUMBER ASSIGN VAR VAR VAR VAR VAR NUMBER FOR VAR FUNC_CALL VAR NUMBER BIN_OP VAR NUMBER IF VAR VAR VAR BIN_OP VAR NUMBER VAR VAR VAR BIN_OP VAR NUMBER VAR NUMBER EXPR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR VAR
|
This is the hard version of the problem. The difference between the versions is that in the easy version all prices $a_i$ are different. You can make hacks if and only if you solved both versions of the problem.
Today is Sage's birthday, and she will go shopping to buy ice spheres. All $n$ ice spheres are placed in a row and they are numbered from $1$ to $n$ from left to right. Each ice sphere has a positive integer price. In this version, some prices can be equal.
An ice sphere is cheap if it costs strictly less than two neighboring ice spheres: the nearest to the left and the nearest to the right. The leftmost and the rightmost ice spheres are not cheap. Sage will choose all cheap ice spheres and then buy only them.
You can visit the shop before Sage and reorder the ice spheres as you wish. Find out the maximum number of ice spheres that Sage can buy, and show how the ice spheres should be reordered.
-----Input-----
The first line contains a single integer $n$ $(1 \le n \le 10^5)$ — the number of ice spheres in the shop.
The second line contains $n$ integers $a_1, a_2, \dots, a_n$ $(1 \le a_i \le 10^9)$ — the prices of ice spheres.
-----Output-----
In the first line print the maximum number of ice spheres that Sage can buy.
In the second line print the prices of ice spheres in the optimal order. If there are several correct answers, you can print any of them.
-----Example-----
Input
7
1 3 2 2 4 5 4
Output
3
3 1 4 2 4 2 5
-----Note-----
In the sample it's not possible to place the ice spheres in any order so that Sage would buy $4$ of them. If the spheres are placed in the order $(3, 1, 4, 2, 4, 2, 5)$, then Sage will buy one sphere for $1$ and two spheres for $2$ each.
|
import sys
try:
sys.stdin = open("input.txt", "r")
except FileNotFoundError:
pass
def next_array():
return [int(i) for i in input().split()]
def prefix(lst):
assert len(lst)
pre = [lst[0]]
for i in range(1, len(lst)):
pre.append(lst[i] + pre[i - 1])
return pre
def prefix_max(lst):
assert len(lst)
pre = [lst[0]]
for i in range(1, len(lst)):
pre.append(max(lst[i], pre[i - 1]))
return pre
def solve():
n = int(input())
lst = next_array()
lst.sort()
out = [None] * len(lst)
for i in range(len(lst) // 2, len(lst)):
out[(i - len(lst) // 2) * 2] = lst[i]
for i in range(0, len(lst) // 2):
out[i * 2 + 1] = lst[i]
c = 0
for i in range(1, len(lst) - 1, 2):
if out[i] < out[i - 1] and out[i] < out[i + 1]:
c += 1
print(c)
print(" ".join(str(i) for i in out))
solve()
|
IMPORT ASSIGN VAR FUNC_CALL VAR STRING STRING VAR FUNC_DEF RETURN FUNC_CALL VAR VAR VAR FUNC_CALL FUNC_CALL VAR FUNC_DEF FUNC_CALL VAR VAR ASSIGN VAR LIST VAR NUMBER FOR VAR FUNC_CALL VAR NUMBER FUNC_CALL VAR VAR EXPR FUNC_CALL VAR BIN_OP VAR VAR VAR BIN_OP VAR NUMBER RETURN VAR FUNC_DEF FUNC_CALL VAR VAR ASSIGN VAR LIST VAR NUMBER FOR VAR FUNC_CALL VAR NUMBER FUNC_CALL VAR VAR EXPR FUNC_CALL VAR FUNC_CALL VAR VAR VAR VAR BIN_OP VAR NUMBER RETURN VAR FUNC_DEF ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR EXPR FUNC_CALL VAR ASSIGN VAR BIN_OP LIST NONE FUNC_CALL VAR VAR FOR VAR FUNC_CALL VAR BIN_OP FUNC_CALL VAR VAR NUMBER FUNC_CALL VAR VAR ASSIGN VAR BIN_OP BIN_OP VAR BIN_OP FUNC_CALL VAR VAR NUMBER NUMBER VAR VAR FOR VAR FUNC_CALL VAR NUMBER BIN_OP FUNC_CALL VAR VAR NUMBER ASSIGN VAR BIN_OP BIN_OP VAR NUMBER NUMBER VAR VAR ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR NUMBER BIN_OP FUNC_CALL VAR VAR NUMBER NUMBER IF VAR VAR VAR BIN_OP VAR NUMBER VAR VAR VAR BIN_OP VAR NUMBER VAR NUMBER EXPR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR FUNC_CALL STRING FUNC_CALL VAR VAR VAR VAR EXPR FUNC_CALL VAR
|
This is the hard version of the problem. The difference between the versions is that in the easy version all prices $a_i$ are different. You can make hacks if and only if you solved both versions of the problem.
Today is Sage's birthday, and she will go shopping to buy ice spheres. All $n$ ice spheres are placed in a row and they are numbered from $1$ to $n$ from left to right. Each ice sphere has a positive integer price. In this version, some prices can be equal.
An ice sphere is cheap if it costs strictly less than two neighboring ice spheres: the nearest to the left and the nearest to the right. The leftmost and the rightmost ice spheres are not cheap. Sage will choose all cheap ice spheres and then buy only them.
You can visit the shop before Sage and reorder the ice spheres as you wish. Find out the maximum number of ice spheres that Sage can buy, and show how the ice spheres should be reordered.
-----Input-----
The first line contains a single integer $n$ $(1 \le n \le 10^5)$ — the number of ice spheres in the shop.
The second line contains $n$ integers $a_1, a_2, \dots, a_n$ $(1 \le a_i \le 10^9)$ — the prices of ice spheres.
-----Output-----
In the first line print the maximum number of ice spheres that Sage can buy.
In the second line print the prices of ice spheres in the optimal order. If there are several correct answers, you can print any of them.
-----Example-----
Input
7
1 3 2 2 4 5 4
Output
3
3 1 4 2 4 2 5
-----Note-----
In the sample it's not possible to place the ice spheres in any order so that Sage would buy $4$ of them. If the spheres are placed in the order $(3, 1, 4, 2, 4, 2, 5)$, then Sage will buy one sphere for $1$ and two spheres for $2$ each.
|
n = int(input())
a = sorted(list(map(int, input().split())))
val = 0
b, c = a[: n // 2], a[n // 2 : -1]
sol = [a[-1]]
d = []
b_ptr, c_ptr = len(b) - 1, len(c) - 1
while True:
if b_ptr < 0 or c_ptr < 0:
break
cc = c[c_ptr]
while b_ptr >= 0:
if b[b_ptr] >= cc:
d.append(b[b_ptr])
b_ptr -= 1
else:
break
if b_ptr < 0:
break
else:
sol.append(b[b_ptr])
b_ptr -= 1
sol.append(c[c_ptr])
c_ptr -= 1
val += 1
sol += b[: max(b_ptr + 1, 0)] + c[: max(c_ptr + 1, 0)] + d
print(val)
print(" ".join([str(i) for i in sol]))
|
ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR NUMBER ASSIGN VAR VAR VAR BIN_OP VAR NUMBER VAR BIN_OP VAR NUMBER NUMBER ASSIGN VAR LIST VAR NUMBER ASSIGN VAR LIST ASSIGN VAR VAR BIN_OP FUNC_CALL VAR VAR NUMBER BIN_OP FUNC_CALL VAR VAR NUMBER WHILE NUMBER IF VAR NUMBER VAR NUMBER ASSIGN VAR VAR VAR WHILE VAR NUMBER IF VAR VAR VAR EXPR FUNC_CALL VAR VAR VAR VAR NUMBER IF VAR NUMBER EXPR FUNC_CALL VAR VAR VAR VAR NUMBER EXPR FUNC_CALL VAR VAR VAR VAR NUMBER VAR NUMBER VAR BIN_OP BIN_OP VAR FUNC_CALL VAR BIN_OP VAR NUMBER NUMBER VAR FUNC_CALL VAR BIN_OP VAR NUMBER NUMBER VAR EXPR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR FUNC_CALL STRING FUNC_CALL VAR VAR VAR VAR
|
This is the hard version of the problem. The difference between the versions is that in the easy version all prices $a_i$ are different. You can make hacks if and only if you solved both versions of the problem.
Today is Sage's birthday, and she will go shopping to buy ice spheres. All $n$ ice spheres are placed in a row and they are numbered from $1$ to $n$ from left to right. Each ice sphere has a positive integer price. In this version, some prices can be equal.
An ice sphere is cheap if it costs strictly less than two neighboring ice spheres: the nearest to the left and the nearest to the right. The leftmost and the rightmost ice spheres are not cheap. Sage will choose all cheap ice spheres and then buy only them.
You can visit the shop before Sage and reorder the ice spheres as you wish. Find out the maximum number of ice spheres that Sage can buy, and show how the ice spheres should be reordered.
-----Input-----
The first line contains a single integer $n$ $(1 \le n \le 10^5)$ — the number of ice spheres in the shop.
The second line contains $n$ integers $a_1, a_2, \dots, a_n$ $(1 \le a_i \le 10^9)$ — the prices of ice spheres.
-----Output-----
In the first line print the maximum number of ice spheres that Sage can buy.
In the second line print the prices of ice spheres in the optimal order. If there are several correct answers, you can print any of them.
-----Example-----
Input
7
1 3 2 2 4 5 4
Output
3
3 1 4 2 4 2 5
-----Note-----
In the sample it's not possible to place the ice spheres in any order so that Sage would buy $4$ of them. If the spheres are placed in the order $(3, 1, 4, 2, 4, 2, 5)$, then Sage will buy one sphere for $1$ and two spheres for $2$ each.
|
n = int(input())
a = [int(x) for x in input().split()]
a.sort()
ans = 0
ansarr = []
if n < 3:
print(0)
print(" ".join([str(x) for x in a]))
elif n % 2 == 0:
for i in range(n // 2):
ansarr.append(a[n // 2 + i])
ansarr.append(a[i])
for i in range(1, len(a) - 1, 2):
if ansarr[i] < ansarr[i - 1] and ansarr[i] < ansarr[i + 1]:
ans += 1
print(ans)
print(" ".join([str(x) for x in ansarr]))
else:
for i in range(n // 2):
ansarr.append(a[n // 2 + i])
ansarr.append(a[i])
ansarr.append(a[len(a) - 1])
for i in range(1, len(a), 2):
if ansarr[i] < ansarr[i - 1] and ansarr[i] < ansarr[i + 1]:
ans += 1
print(ans)
print(" ".join([str(x) for x in ansarr]))
|
ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR VAR VAR FUNC_CALL FUNC_CALL VAR EXPR FUNC_CALL VAR ASSIGN VAR NUMBER ASSIGN VAR LIST IF VAR NUMBER EXPR FUNC_CALL VAR NUMBER EXPR FUNC_CALL VAR FUNC_CALL STRING FUNC_CALL VAR VAR VAR VAR IF BIN_OP VAR NUMBER NUMBER FOR VAR FUNC_CALL VAR BIN_OP VAR NUMBER EXPR FUNC_CALL VAR VAR BIN_OP BIN_OP VAR NUMBER VAR EXPR FUNC_CALL VAR VAR VAR FOR VAR FUNC_CALL VAR NUMBER BIN_OP FUNC_CALL VAR VAR NUMBER NUMBER IF VAR VAR VAR BIN_OP VAR NUMBER VAR VAR VAR BIN_OP VAR NUMBER VAR NUMBER EXPR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR FUNC_CALL STRING FUNC_CALL VAR VAR VAR VAR FOR VAR FUNC_CALL VAR BIN_OP VAR NUMBER EXPR FUNC_CALL VAR VAR BIN_OP BIN_OP VAR NUMBER VAR EXPR FUNC_CALL VAR VAR VAR EXPR FUNC_CALL VAR VAR BIN_OP FUNC_CALL VAR VAR NUMBER FOR VAR FUNC_CALL VAR NUMBER FUNC_CALL VAR VAR NUMBER IF VAR VAR VAR BIN_OP VAR NUMBER VAR VAR VAR BIN_OP VAR NUMBER VAR NUMBER EXPR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR FUNC_CALL STRING FUNC_CALL VAR VAR VAR VAR
|
This is the hard version of the problem. The difference between the versions is that in the easy version all prices $a_i$ are different. You can make hacks if and only if you solved both versions of the problem.
Today is Sage's birthday, and she will go shopping to buy ice spheres. All $n$ ice spheres are placed in a row and they are numbered from $1$ to $n$ from left to right. Each ice sphere has a positive integer price. In this version, some prices can be equal.
An ice sphere is cheap if it costs strictly less than two neighboring ice spheres: the nearest to the left and the nearest to the right. The leftmost and the rightmost ice spheres are not cheap. Sage will choose all cheap ice spheres and then buy only them.
You can visit the shop before Sage and reorder the ice spheres as you wish. Find out the maximum number of ice spheres that Sage can buy, and show how the ice spheres should be reordered.
-----Input-----
The first line contains a single integer $n$ $(1 \le n \le 10^5)$ — the number of ice spheres in the shop.
The second line contains $n$ integers $a_1, a_2, \dots, a_n$ $(1 \le a_i \le 10^9)$ — the prices of ice spheres.
-----Output-----
In the first line print the maximum number of ice spheres that Sage can buy.
In the second line print the prices of ice spheres in the optimal order. If there are several correct answers, you can print any of them.
-----Example-----
Input
7
1 3 2 2 4 5 4
Output
3
3 1 4 2 4 2 5
-----Note-----
In the sample it's not possible to place the ice spheres in any order so that Sage would buy $4$ of them. If the spheres are placed in the order $(3, 1, 4, 2, 4, 2, 5)$, then Sage will buy one sphere for $1$ and two spheres for $2$ each.
|
import sys
readline = sys.stdin.readline
N = int(readline())
A = list(map(int, readline().split()))
A.sort()
Ans = [None] * N
A1 = A[: (N - 1) // 2][::-1]
A2 = A[(N - 1) // 2 :]
for i in range((N - 1) // 2):
Ans[2 * i + 1] = A1[i]
for i in range(N):
if Ans[i] is None:
Ans[i] = A2.pop()
ans = 0
for i in range(1, N - 1):
if Ans[i - 1] > Ans[i] and Ans[i] < Ans[i + 1]:
ans += 1
print(ans)
print(*Ans)
|
IMPORT ASSIGN VAR VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR EXPR FUNC_CALL VAR ASSIGN VAR BIN_OP LIST NONE VAR ASSIGN VAR VAR BIN_OP BIN_OP VAR NUMBER NUMBER NUMBER ASSIGN VAR VAR BIN_OP BIN_OP VAR NUMBER NUMBER FOR VAR FUNC_CALL VAR BIN_OP BIN_OP VAR NUMBER NUMBER ASSIGN VAR BIN_OP BIN_OP NUMBER VAR NUMBER VAR VAR FOR VAR FUNC_CALL VAR VAR IF VAR VAR NONE ASSIGN VAR VAR FUNC_CALL VAR ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR NUMBER BIN_OP VAR NUMBER IF VAR BIN_OP VAR NUMBER VAR VAR VAR VAR VAR BIN_OP VAR NUMBER VAR NUMBER EXPR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR VAR
|
This is the hard version of the problem. The difference between the versions is that in the easy version all prices $a_i$ are different. You can make hacks if and only if you solved both versions of the problem.
Today is Sage's birthday, and she will go shopping to buy ice spheres. All $n$ ice spheres are placed in a row and they are numbered from $1$ to $n$ from left to right. Each ice sphere has a positive integer price. In this version, some prices can be equal.
An ice sphere is cheap if it costs strictly less than two neighboring ice spheres: the nearest to the left and the nearest to the right. The leftmost and the rightmost ice spheres are not cheap. Sage will choose all cheap ice spheres and then buy only them.
You can visit the shop before Sage and reorder the ice spheres as you wish. Find out the maximum number of ice spheres that Sage can buy, and show how the ice spheres should be reordered.
-----Input-----
The first line contains a single integer $n$ $(1 \le n \le 10^5)$ — the number of ice spheres in the shop.
The second line contains $n$ integers $a_1, a_2, \dots, a_n$ $(1 \le a_i \le 10^9)$ — the prices of ice spheres.
-----Output-----
In the first line print the maximum number of ice spheres that Sage can buy.
In the second line print the prices of ice spheres in the optimal order. If there are several correct answers, you can print any of them.
-----Example-----
Input
7
1 3 2 2 4 5 4
Output
3
3 1 4 2 4 2 5
-----Note-----
In the sample it's not possible to place the ice spheres in any order so that Sage would buy $4$ of them. If the spheres are placed in the order $(3, 1, 4, 2, 4, 2, 5)$, then Sage will buy one sphere for $1$ and two spheres for $2$ each.
|
n = int(input())
s = input()
L = s.split()
for i in range(len(L)):
L[i] = int(L[i])
myLen = len(L)
if myLen <= 2:
print(0)
print(s)
quit()
L.sort()
gr = myLen // 2
cnt = gr if myLen % 2 else gr - 1
v = L[0]
for i in range(gr, myLen - 1):
if L[i] > v < L[i + 1]:
k = i
break
else:
print(0)
print(s)
quit()
shift = k - gr
L1 = []
for i in range(gr, gr + shift):
L1.append(L[i])
nn = cnt - shift
for i in range(nn):
L1.append(L[k + i])
L1.append(L[i])
L1.append(L[-1])
for i in range(shift):
L1.append(L[i + nn])
if not myLen % 2:
L1.append(L[cnt])
nSales = 0
for i in range(1, len(L1), 2):
if i + 1 < len(L1) and L1[i - 1] > L1[i] < L1[i + 1]:
nSales += 1
print(nSales)
L1 = [str(e) for e in L1]
print(" ".join(L1))
|
ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FOR VAR FUNC_CALL VAR FUNC_CALL VAR VAR ASSIGN VAR VAR FUNC_CALL VAR VAR VAR ASSIGN VAR FUNC_CALL VAR VAR IF VAR NUMBER EXPR FUNC_CALL VAR NUMBER EXPR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR EXPR FUNC_CALL VAR ASSIGN VAR BIN_OP VAR NUMBER ASSIGN VAR BIN_OP VAR NUMBER VAR BIN_OP VAR NUMBER ASSIGN VAR VAR NUMBER FOR VAR FUNC_CALL VAR VAR BIN_OP VAR NUMBER IF VAR VAR VAR VAR BIN_OP VAR NUMBER ASSIGN VAR VAR EXPR FUNC_CALL VAR NUMBER EXPR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR ASSIGN VAR BIN_OP VAR VAR ASSIGN VAR LIST FOR VAR FUNC_CALL VAR VAR BIN_OP VAR VAR EXPR FUNC_CALL VAR VAR VAR ASSIGN VAR BIN_OP VAR VAR FOR VAR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR VAR BIN_OP VAR VAR EXPR FUNC_CALL VAR VAR VAR EXPR FUNC_CALL VAR VAR NUMBER FOR VAR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR VAR BIN_OP VAR VAR IF BIN_OP VAR NUMBER EXPR FUNC_CALL VAR VAR VAR ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR NUMBER FUNC_CALL VAR VAR NUMBER IF BIN_OP VAR NUMBER FUNC_CALL VAR VAR VAR BIN_OP VAR NUMBER VAR VAR VAR BIN_OP VAR NUMBER VAR NUMBER EXPR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR VAR VAR VAR EXPR FUNC_CALL VAR FUNC_CALL STRING VAR
|
This is the hard version of the problem. The difference between the versions is that in the easy version all prices $a_i$ are different. You can make hacks if and only if you solved both versions of the problem.
Today is Sage's birthday, and she will go shopping to buy ice spheres. All $n$ ice spheres are placed in a row and they are numbered from $1$ to $n$ from left to right. Each ice sphere has a positive integer price. In this version, some prices can be equal.
An ice sphere is cheap if it costs strictly less than two neighboring ice spheres: the nearest to the left and the nearest to the right. The leftmost and the rightmost ice spheres are not cheap. Sage will choose all cheap ice spheres and then buy only them.
You can visit the shop before Sage and reorder the ice spheres as you wish. Find out the maximum number of ice spheres that Sage can buy, and show how the ice spheres should be reordered.
-----Input-----
The first line contains a single integer $n$ $(1 \le n \le 10^5)$ — the number of ice spheres in the shop.
The second line contains $n$ integers $a_1, a_2, \dots, a_n$ $(1 \le a_i \le 10^9)$ — the prices of ice spheres.
-----Output-----
In the first line print the maximum number of ice spheres that Sage can buy.
In the second line print the prices of ice spheres in the optimal order. If there are several correct answers, you can print any of them.
-----Example-----
Input
7
1 3 2 2 4 5 4
Output
3
3 1 4 2 4 2 5
-----Note-----
In the sample it's not possible to place the ice spheres in any order so that Sage would buy $4$ of them. If the spheres are placed in the order $(3, 1, 4, 2, 4, 2, 5)$, then Sage will buy one sphere for $1$ and two spheres for $2$ each.
|
n = int(input())
a = [int(x) for x in input().split()]
asort = sorted(a)
j = 0
ans = [0] * len(a)
for i in range(1, len(a), 2):
ans[i] = asort[j]
j += 1
for k in range(0, len(a), 2):
ans[k] = asort[j]
j += 1
total = 0
for num in range(1, len(ans) - 1, 2):
if ans[num] < ans[num - 1] and ans[num] < ans[num + 1]:
total += 1
print(total)
print(*ans)
|
ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR NUMBER ASSIGN VAR BIN_OP LIST NUMBER FUNC_CALL VAR VAR FOR VAR FUNC_CALL VAR NUMBER FUNC_CALL VAR VAR NUMBER ASSIGN VAR VAR VAR VAR VAR NUMBER FOR VAR FUNC_CALL VAR NUMBER FUNC_CALL VAR VAR NUMBER ASSIGN VAR VAR VAR VAR VAR NUMBER ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR NUMBER BIN_OP FUNC_CALL VAR VAR NUMBER NUMBER IF VAR VAR VAR BIN_OP VAR NUMBER VAR VAR VAR BIN_OP VAR NUMBER VAR NUMBER EXPR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR VAR
|
This is the hard version of the problem. The difference between the versions is that in the easy version all prices $a_i$ are different. You can make hacks if and only if you solved both versions of the problem.
Today is Sage's birthday, and she will go shopping to buy ice spheres. All $n$ ice spheres are placed in a row and they are numbered from $1$ to $n$ from left to right. Each ice sphere has a positive integer price. In this version, some prices can be equal.
An ice sphere is cheap if it costs strictly less than two neighboring ice spheres: the nearest to the left and the nearest to the right. The leftmost and the rightmost ice spheres are not cheap. Sage will choose all cheap ice spheres and then buy only them.
You can visit the shop before Sage and reorder the ice spheres as you wish. Find out the maximum number of ice spheres that Sage can buy, and show how the ice spheres should be reordered.
-----Input-----
The first line contains a single integer $n$ $(1 \le n \le 10^5)$ — the number of ice spheres in the shop.
The second line contains $n$ integers $a_1, a_2, \dots, a_n$ $(1 \le a_i \le 10^9)$ — the prices of ice spheres.
-----Output-----
In the first line print the maximum number of ice spheres that Sage can buy.
In the second line print the prices of ice spheres in the optimal order. If there are several correct answers, you can print any of them.
-----Example-----
Input
7
1 3 2 2 4 5 4
Output
3
3 1 4 2 4 2 5
-----Note-----
In the sample it's not possible to place the ice spheres in any order so that Sage would buy $4$ of them. If the spheres are placed in the order $(3, 1, 4, 2, 4, 2, 5)$, then Sage will buy one sphere for $1$ and two spheres for $2$ each.
|
n = int(input())
a = list(map(int, input().split()))
a.sort()
ans = [None for _ in range(n)]
i = 0
j = 1
while j < n:
ans[j] = a[i]
i += 1
j += 2
k = i
j = 0
while k < n and j < n:
ans[j] = a[k]
k += 1
j += 2
count = 0
for i in range(1, n - 1):
if ans[i] < ans[i + 1] and ans[i] < ans[i - 1]:
count += 1
print(count)
print(" ".join(map(str, ans)))
|
ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR EXPR FUNC_CALL VAR ASSIGN VAR NONE VAR FUNC_CALL VAR VAR ASSIGN VAR NUMBER ASSIGN VAR NUMBER WHILE VAR VAR ASSIGN VAR VAR VAR VAR VAR NUMBER VAR NUMBER ASSIGN VAR VAR ASSIGN VAR NUMBER WHILE VAR VAR VAR VAR ASSIGN VAR VAR VAR VAR VAR NUMBER VAR NUMBER ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR NUMBER BIN_OP VAR NUMBER IF VAR VAR VAR BIN_OP VAR NUMBER VAR VAR VAR BIN_OP VAR NUMBER VAR NUMBER EXPR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR FUNC_CALL STRING FUNC_CALL VAR VAR VAR
|
This is the hard version of the problem. The difference between the versions is that in the easy version all prices $a_i$ are different. You can make hacks if and only if you solved both versions of the problem.
Today is Sage's birthday, and she will go shopping to buy ice spheres. All $n$ ice spheres are placed in a row and they are numbered from $1$ to $n$ from left to right. Each ice sphere has a positive integer price. In this version, some prices can be equal.
An ice sphere is cheap if it costs strictly less than two neighboring ice spheres: the nearest to the left and the nearest to the right. The leftmost and the rightmost ice spheres are not cheap. Sage will choose all cheap ice spheres and then buy only them.
You can visit the shop before Sage and reorder the ice spheres as you wish. Find out the maximum number of ice spheres that Sage can buy, and show how the ice spheres should be reordered.
-----Input-----
The first line contains a single integer $n$ $(1 \le n \le 10^5)$ — the number of ice spheres in the shop.
The second line contains $n$ integers $a_1, a_2, \dots, a_n$ $(1 \le a_i \le 10^9)$ — the prices of ice spheres.
-----Output-----
In the first line print the maximum number of ice spheres that Sage can buy.
In the second line print the prices of ice spheres in the optimal order. If there are several correct answers, you can print any of them.
-----Example-----
Input
7
1 3 2 2 4 5 4
Output
3
3 1 4 2 4 2 5
-----Note-----
In the sample it's not possible to place the ice spheres in any order so that Sage would buy $4$ of them. If the spheres are placed in the order $(3, 1, 4, 2, 4, 2, 5)$, then Sage will buy one sphere for $1$ and two spheres for $2$ each.
|
n = int(input())
arr = list(map(int, input().split(" ")))
arr = list(sorted(arr))
p = n // 2
resultarr = []
i = 0
while p < n:
resultarr.append(arr[p])
if i < n // 2:
resultarr.append(arr[i])
i += 1
p += 1
result = 0
for i in range(1, len(arr) - 1):
if resultarr[i - 1] > resultarr[i] and resultarr[i + 1] > resultarr[i]:
result += 1
print(result)
print(" ".join(list(map(str, resultarr))))
|
ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR STRING ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR VAR ASSIGN VAR BIN_OP VAR NUMBER ASSIGN VAR LIST ASSIGN VAR NUMBER WHILE VAR VAR EXPR FUNC_CALL VAR VAR VAR IF VAR BIN_OP VAR NUMBER EXPR FUNC_CALL VAR VAR VAR VAR NUMBER VAR NUMBER ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR NUMBER BIN_OP FUNC_CALL VAR VAR NUMBER IF VAR BIN_OP VAR NUMBER VAR VAR VAR BIN_OP VAR NUMBER VAR VAR VAR NUMBER EXPR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR FUNC_CALL STRING FUNC_CALL VAR FUNC_CALL VAR VAR VAR
|
This is the hard version of the problem. The difference between the versions is that in the easy version all prices $a_i$ are different. You can make hacks if and only if you solved both versions of the problem.
Today is Sage's birthday, and she will go shopping to buy ice spheres. All $n$ ice spheres are placed in a row and they are numbered from $1$ to $n$ from left to right. Each ice sphere has a positive integer price. In this version, some prices can be equal.
An ice sphere is cheap if it costs strictly less than two neighboring ice spheres: the nearest to the left and the nearest to the right. The leftmost and the rightmost ice spheres are not cheap. Sage will choose all cheap ice spheres and then buy only them.
You can visit the shop before Sage and reorder the ice spheres as you wish. Find out the maximum number of ice spheres that Sage can buy, and show how the ice spheres should be reordered.
-----Input-----
The first line contains a single integer $n$ $(1 \le n \le 10^5)$ — the number of ice spheres in the shop.
The second line contains $n$ integers $a_1, a_2, \dots, a_n$ $(1 \le a_i \le 10^9)$ — the prices of ice spheres.
-----Output-----
In the first line print the maximum number of ice spheres that Sage can buy.
In the second line print the prices of ice spheres in the optimal order. If there are several correct answers, you can print any of them.
-----Example-----
Input
7
1 3 2 2 4 5 4
Output
3
3 1 4 2 4 2 5
-----Note-----
In the sample it's not possible to place the ice spheres in any order so that Sage would buy $4$ of them. If the spheres are placed in the order $(3, 1, 4, 2, 4, 2, 5)$, then Sage will buy one sphere for $1$ and two spheres for $2$ each.
|
n = int(input())
l = list(map(int, input().split()))
l.sort(reverse=True)
ans = [0] * n
for i in range(1, n, 2):
ans[i] = l.pop()
for i in range(0, n, 2):
ans[i] = l.pop()
count = 0
for i in range(1, n - 1):
if ans[i - 1] > ans[i] and ans[i] < ans[i + 1]:
count += 1
print(count)
print(*ans)
|
ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR EXPR FUNC_CALL VAR NUMBER ASSIGN VAR BIN_OP LIST NUMBER VAR FOR VAR FUNC_CALL VAR NUMBER VAR NUMBER ASSIGN VAR VAR FUNC_CALL VAR FOR VAR FUNC_CALL VAR NUMBER VAR NUMBER ASSIGN VAR VAR FUNC_CALL VAR ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR NUMBER BIN_OP VAR NUMBER IF VAR BIN_OP VAR NUMBER VAR VAR VAR VAR VAR BIN_OP VAR NUMBER VAR NUMBER EXPR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR VAR
|
This is the hard version of the problem. The difference between the versions is that in the easy version all prices $a_i$ are different. You can make hacks if and only if you solved both versions of the problem.
Today is Sage's birthday, and she will go shopping to buy ice spheres. All $n$ ice spheres are placed in a row and they are numbered from $1$ to $n$ from left to right. Each ice sphere has a positive integer price. In this version, some prices can be equal.
An ice sphere is cheap if it costs strictly less than two neighboring ice spheres: the nearest to the left and the nearest to the right. The leftmost and the rightmost ice spheres are not cheap. Sage will choose all cheap ice spheres and then buy only them.
You can visit the shop before Sage and reorder the ice spheres as you wish. Find out the maximum number of ice spheres that Sage can buy, and show how the ice spheres should be reordered.
-----Input-----
The first line contains a single integer $n$ $(1 \le n \le 10^5)$ — the number of ice spheres in the shop.
The second line contains $n$ integers $a_1, a_2, \dots, a_n$ $(1 \le a_i \le 10^9)$ — the prices of ice spheres.
-----Output-----
In the first line print the maximum number of ice spheres that Sage can buy.
In the second line print the prices of ice spheres in the optimal order. If there are several correct answers, you can print any of them.
-----Example-----
Input
7
1 3 2 2 4 5 4
Output
3
3 1 4 2 4 2 5
-----Note-----
In the sample it's not possible to place the ice spheres in any order so that Sage would buy $4$ of them. If the spheres are placed in the order $(3, 1, 4, 2, 4, 2, 5)$, then Sage will buy one sphere for $1$ and two spheres for $2$ each.
|
from sys import *
input = stdin.readline
for _ in range(1):
n = int(input())
a = list(map(int, input().split()))
a.sort()
b = [0] * n
k = n // 2
j = k
for i in range(0, n, 2):
b[i] = a[j]
j += 1
j = 0
for i in range(1, n, 2):
b[i] = a[j]
j += 1
ans = 0
for i in range(1, n - 1):
if b[i] < b[i - 1] and b[i] < b[i + 1]:
ans += 1
stdout.write(str(ans) + "\n")
print(*b)
|
ASSIGN VAR VAR FOR VAR FUNC_CALL VAR NUMBER ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR EXPR FUNC_CALL VAR ASSIGN VAR BIN_OP LIST NUMBER VAR ASSIGN VAR BIN_OP VAR NUMBER ASSIGN VAR VAR FOR VAR FUNC_CALL VAR NUMBER VAR NUMBER ASSIGN VAR VAR VAR VAR VAR NUMBER ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR NUMBER VAR NUMBER ASSIGN VAR VAR VAR VAR VAR NUMBER ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR NUMBER BIN_OP VAR NUMBER IF VAR VAR VAR BIN_OP VAR NUMBER VAR VAR VAR BIN_OP VAR NUMBER VAR NUMBER EXPR FUNC_CALL VAR BIN_OP FUNC_CALL VAR VAR STRING EXPR FUNC_CALL VAR VAR
|
This is the hard version of the problem. The difference between the versions is that in the easy version all prices $a_i$ are different. You can make hacks if and only if you solved both versions of the problem.
Today is Sage's birthday, and she will go shopping to buy ice spheres. All $n$ ice spheres are placed in a row and they are numbered from $1$ to $n$ from left to right. Each ice sphere has a positive integer price. In this version, some prices can be equal.
An ice sphere is cheap if it costs strictly less than two neighboring ice spheres: the nearest to the left and the nearest to the right. The leftmost and the rightmost ice spheres are not cheap. Sage will choose all cheap ice spheres and then buy only them.
You can visit the shop before Sage and reorder the ice spheres as you wish. Find out the maximum number of ice spheres that Sage can buy, and show how the ice spheres should be reordered.
-----Input-----
The first line contains a single integer $n$ $(1 \le n \le 10^5)$ — the number of ice spheres in the shop.
The second line contains $n$ integers $a_1, a_2, \dots, a_n$ $(1 \le a_i \le 10^9)$ — the prices of ice spheres.
-----Output-----
In the first line print the maximum number of ice spheres that Sage can buy.
In the second line print the prices of ice spheres in the optimal order. If there are several correct answers, you can print any of them.
-----Example-----
Input
7
1 3 2 2 4 5 4
Output
3
3 1 4 2 4 2 5
-----Note-----
In the sample it's not possible to place the ice spheres in any order so that Sage would buy $4$ of them. If the spheres are placed in the order $(3, 1, 4, 2, 4, 2, 5)$, then Sage will buy one sphere for $1$ and two spheres for $2$ each.
|
n = int(input())
arr = [int(num) for num in input().split()]
arr.sort()
arr0 = arr[: n // 2]
arr1 = arr[n // 2 :]
p0 = 0
p1 = 0
ans = []
while p0 < len(arr0) and p1 < len(arr1):
ans.append(arr1[p1])
ans.append(arr0[p0])
p0 += 1
p1 += 1
if n % 2:
ans.append(arr1[-1])
ans_num = 0
for ind in range(1, n - 1):
if ans[ind - 1] > ans[ind] < ans[ind + 1]:
ans_num += 1
print(ans_num)
print(*ans)
|
ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR VAR VAR FUNC_CALL FUNC_CALL VAR EXPR FUNC_CALL VAR ASSIGN VAR VAR BIN_OP VAR NUMBER ASSIGN VAR VAR BIN_OP VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR LIST WHILE VAR FUNC_CALL VAR VAR VAR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR VAR VAR EXPR FUNC_CALL VAR VAR VAR VAR NUMBER VAR NUMBER IF BIN_OP VAR NUMBER EXPR FUNC_CALL VAR VAR NUMBER ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR NUMBER BIN_OP VAR NUMBER IF VAR BIN_OP VAR NUMBER VAR VAR VAR BIN_OP VAR NUMBER VAR NUMBER EXPR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR VAR
|
This is the hard version of the problem. The difference between the versions is that in the easy version all prices $a_i$ are different. You can make hacks if and only if you solved both versions of the problem.
Today is Sage's birthday, and she will go shopping to buy ice spheres. All $n$ ice spheres are placed in a row and they are numbered from $1$ to $n$ from left to right. Each ice sphere has a positive integer price. In this version, some prices can be equal.
An ice sphere is cheap if it costs strictly less than two neighboring ice spheres: the nearest to the left and the nearest to the right. The leftmost and the rightmost ice spheres are not cheap. Sage will choose all cheap ice spheres and then buy only them.
You can visit the shop before Sage and reorder the ice spheres as you wish. Find out the maximum number of ice spheres that Sage can buy, and show how the ice spheres should be reordered.
-----Input-----
The first line contains a single integer $n$ $(1 \le n \le 10^5)$ — the number of ice spheres in the shop.
The second line contains $n$ integers $a_1, a_2, \dots, a_n$ $(1 \le a_i \le 10^9)$ — the prices of ice spheres.
-----Output-----
In the first line print the maximum number of ice spheres that Sage can buy.
In the second line print the prices of ice spheres in the optimal order. If there are several correct answers, you can print any of them.
-----Example-----
Input
7
1 3 2 2 4 5 4
Output
3
3 1 4 2 4 2 5
-----Note-----
In the sample it's not possible to place the ice spheres in any order so that Sage would buy $4$ of them. If the spheres are placed in the order $(3, 1, 4, 2, 4, 2, 5)$, then Sage will buy one sphere for $1$ and two spheres for $2$ each.
|
n = int(input())
a = list(map(int, input().split()))
a.sort()
b = [0] * n
for i in range((n + 1) // 2 - 1):
b[2 * i + 1] = a[i]
if n % 2 == 0:
b[n - 1] = a[n // 2 - 1]
for i in range((n + 1) // 2 - n % 2, n):
b[2 * i - 2 * ((n + 1) // 2 - n % 2)] = a[i]
cnt = 0
for i in range(n):
if i > 0 and i < n - 1 and b[i] < b[i - 1] and b[i] < b[i + 1]:
cnt += 1
print(cnt)
print(" ".join(map(str, b)))
|
ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR EXPR FUNC_CALL VAR ASSIGN VAR BIN_OP LIST NUMBER VAR FOR VAR FUNC_CALL VAR BIN_OP BIN_OP BIN_OP VAR NUMBER NUMBER NUMBER ASSIGN VAR BIN_OP BIN_OP NUMBER VAR NUMBER VAR VAR IF BIN_OP VAR NUMBER NUMBER ASSIGN VAR BIN_OP VAR NUMBER VAR BIN_OP BIN_OP VAR NUMBER NUMBER FOR VAR FUNC_CALL VAR BIN_OP BIN_OP BIN_OP VAR NUMBER NUMBER BIN_OP VAR NUMBER VAR ASSIGN VAR BIN_OP BIN_OP NUMBER VAR BIN_OP NUMBER BIN_OP BIN_OP BIN_OP VAR NUMBER NUMBER BIN_OP VAR NUMBER VAR VAR ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR VAR IF VAR NUMBER VAR BIN_OP VAR NUMBER VAR VAR VAR BIN_OP VAR NUMBER VAR VAR VAR BIN_OP VAR NUMBER VAR NUMBER EXPR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR FUNC_CALL STRING FUNC_CALL VAR VAR VAR
|
This is the hard version of the problem. The difference between the versions is that in the easy version all prices $a_i$ are different. You can make hacks if and only if you solved both versions of the problem.
Today is Sage's birthday, and she will go shopping to buy ice spheres. All $n$ ice spheres are placed in a row and they are numbered from $1$ to $n$ from left to right. Each ice sphere has a positive integer price. In this version, some prices can be equal.
An ice sphere is cheap if it costs strictly less than two neighboring ice spheres: the nearest to the left and the nearest to the right. The leftmost and the rightmost ice spheres are not cheap. Sage will choose all cheap ice spheres and then buy only them.
You can visit the shop before Sage and reorder the ice spheres as you wish. Find out the maximum number of ice spheres that Sage can buy, and show how the ice spheres should be reordered.
-----Input-----
The first line contains a single integer $n$ $(1 \le n \le 10^5)$ — the number of ice spheres in the shop.
The second line contains $n$ integers $a_1, a_2, \dots, a_n$ $(1 \le a_i \le 10^9)$ — the prices of ice spheres.
-----Output-----
In the first line print the maximum number of ice spheres that Sage can buy.
In the second line print the prices of ice spheres in the optimal order. If there are several correct answers, you can print any of them.
-----Example-----
Input
7
1 3 2 2 4 5 4
Output
3
3 1 4 2 4 2 5
-----Note-----
In the sample it's not possible to place the ice spheres in any order so that Sage would buy $4$ of them. If the spheres are placed in the order $(3, 1, 4, 2, 4, 2, 5)$, then Sage will buy one sphere for $1$ and two spheres for $2$ each.
|
def solve(n, A):
A.sort()
v = [0] * n
j = 0
for i in range(1, n, 2):
v[i] = A[j]
j += 1
for i in range(0, n, 2):
v[i] = A[j]
j += 1
count = 0
for i in range(1, n - 1):
pre = i - 1
cur = i
next = i + 1
if v[pre] > v[cur] and v[next] > v[cur]:
count += 1
print(count)
return v
res = []
n = int(input())
in_ = map(int, input().split(" "))
for c in in_:
res.append(int(c))
print(*solve(n, res))
|
FUNC_DEF EXPR FUNC_CALL VAR ASSIGN VAR BIN_OP LIST NUMBER VAR ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR NUMBER VAR NUMBER ASSIGN VAR VAR VAR VAR VAR NUMBER FOR VAR FUNC_CALL VAR NUMBER VAR NUMBER ASSIGN VAR VAR VAR VAR VAR NUMBER ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR NUMBER BIN_OP VAR NUMBER ASSIGN VAR BIN_OP VAR NUMBER ASSIGN VAR VAR ASSIGN VAR BIN_OP VAR NUMBER IF VAR VAR VAR VAR VAR VAR VAR VAR VAR NUMBER EXPR FUNC_CALL VAR VAR RETURN VAR ASSIGN VAR LIST ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR STRING FOR VAR VAR EXPR FUNC_CALL VAR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR FUNC_CALL VAR VAR VAR
|
This is the hard version of the problem. The difference between the versions is that in the easy version all prices $a_i$ are different. You can make hacks if and only if you solved both versions of the problem.
Today is Sage's birthday, and she will go shopping to buy ice spheres. All $n$ ice spheres are placed in a row and they are numbered from $1$ to $n$ from left to right. Each ice sphere has a positive integer price. In this version, some prices can be equal.
An ice sphere is cheap if it costs strictly less than two neighboring ice spheres: the nearest to the left and the nearest to the right. The leftmost and the rightmost ice spheres are not cheap. Sage will choose all cheap ice spheres and then buy only them.
You can visit the shop before Sage and reorder the ice spheres as you wish. Find out the maximum number of ice spheres that Sage can buy, and show how the ice spheres should be reordered.
-----Input-----
The first line contains a single integer $n$ $(1 \le n \le 10^5)$ — the number of ice spheres in the shop.
The second line contains $n$ integers $a_1, a_2, \dots, a_n$ $(1 \le a_i \le 10^9)$ — the prices of ice spheres.
-----Output-----
In the first line print the maximum number of ice spheres that Sage can buy.
In the second line print the prices of ice spheres in the optimal order. If there are several correct answers, you can print any of them.
-----Example-----
Input
7
1 3 2 2 4 5 4
Output
3
3 1 4 2 4 2 5
-----Note-----
In the sample it's not possible to place the ice spheres in any order so that Sage would buy $4$ of them. If the spheres are placed in the order $(3, 1, 4, 2, 4, 2, 5)$, then Sage will buy one sphere for $1$ and two spheres for $2$ each.
|
n = int(input())
a = sorted(list(map(int, input().split())))
arr = [0] * n
i = 0
for j in range(n // 2, n):
arr[i] = a[j]
i += 2
i = 1
for j in range(n // 2):
arr[i] = a[j]
i += 2
cnt = 0
for i in range(1, n - 1):
if arr[i - 1] > arr[i] < arr[i + 1]:
cnt += 1
print(cnt)
print(*arr)
|
ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR BIN_OP LIST NUMBER VAR ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR BIN_OP VAR NUMBER VAR ASSIGN VAR VAR VAR VAR VAR NUMBER ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR BIN_OP VAR NUMBER ASSIGN VAR VAR VAR VAR VAR NUMBER ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR NUMBER BIN_OP VAR NUMBER IF VAR BIN_OP VAR NUMBER VAR VAR VAR BIN_OP VAR NUMBER VAR NUMBER EXPR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR VAR
|
This is the hard version of the problem. The difference between the versions is that in the easy version all prices $a_i$ are different. You can make hacks if and only if you solved both versions of the problem.
Today is Sage's birthday, and she will go shopping to buy ice spheres. All $n$ ice spheres are placed in a row and they are numbered from $1$ to $n$ from left to right. Each ice sphere has a positive integer price. In this version, some prices can be equal.
An ice sphere is cheap if it costs strictly less than two neighboring ice spheres: the nearest to the left and the nearest to the right. The leftmost and the rightmost ice spheres are not cheap. Sage will choose all cheap ice spheres and then buy only them.
You can visit the shop before Sage and reorder the ice spheres as you wish. Find out the maximum number of ice spheres that Sage can buy, and show how the ice spheres should be reordered.
-----Input-----
The first line contains a single integer $n$ $(1 \le n \le 10^5)$ — the number of ice spheres in the shop.
The second line contains $n$ integers $a_1, a_2, \dots, a_n$ $(1 \le a_i \le 10^9)$ — the prices of ice spheres.
-----Output-----
In the first line print the maximum number of ice spheres that Sage can buy.
In the second line print the prices of ice spheres in the optimal order. If there are several correct answers, you can print any of them.
-----Example-----
Input
7
1 3 2 2 4 5 4
Output
3
3 1 4 2 4 2 5
-----Note-----
In the sample it's not possible to place the ice spheres in any order so that Sage would buy $4$ of them. If the spheres are placed in the order $(3, 1, 4, 2, 4, 2, 5)$, then Sage will buy one sphere for $1$ and two spheres for $2$ each.
|
def main():
n = int(input())
lst = list(map(int, input().split()))
lst = sorted(lst)
new_lst = []
for i in range(n // 2):
new_lst.append(lst[n // 2 + i])
new_lst.append(lst[i])
if n % 2 == 1:
new_lst.append(lst[n - 1])
cnt = 0
for i in range(1, n - 1):
if new_lst[i] < new_lst[i - 1] and new_lst[i] < new_lst[i + 1]:
cnt += 1
print(cnt)
line = ""
for i in new_lst:
line += str(i) + " "
print(line)
t = 1
for i in range(t):
main()
|
FUNC_DEF ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR LIST FOR VAR FUNC_CALL VAR BIN_OP VAR NUMBER EXPR FUNC_CALL VAR VAR BIN_OP BIN_OP VAR NUMBER VAR EXPR FUNC_CALL VAR VAR VAR IF BIN_OP VAR NUMBER NUMBER EXPR FUNC_CALL VAR VAR BIN_OP VAR NUMBER ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR NUMBER BIN_OP VAR NUMBER IF VAR VAR VAR BIN_OP VAR NUMBER VAR VAR VAR BIN_OP VAR NUMBER VAR NUMBER EXPR FUNC_CALL VAR VAR ASSIGN VAR STRING FOR VAR VAR VAR BIN_OP FUNC_CALL VAR VAR STRING EXPR FUNC_CALL VAR VAR ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR
|
This is the hard version of the problem. The difference between the versions is that in the easy version all prices $a_i$ are different. You can make hacks if and only if you solved both versions of the problem.
Today is Sage's birthday, and she will go shopping to buy ice spheres. All $n$ ice spheres are placed in a row and they are numbered from $1$ to $n$ from left to right. Each ice sphere has a positive integer price. In this version, some prices can be equal.
An ice sphere is cheap if it costs strictly less than two neighboring ice spheres: the nearest to the left and the nearest to the right. The leftmost and the rightmost ice spheres are not cheap. Sage will choose all cheap ice spheres and then buy only them.
You can visit the shop before Sage and reorder the ice spheres as you wish. Find out the maximum number of ice spheres that Sage can buy, and show how the ice spheres should be reordered.
-----Input-----
The first line contains a single integer $n$ $(1 \le n \le 10^5)$ — the number of ice spheres in the shop.
The second line contains $n$ integers $a_1, a_2, \dots, a_n$ $(1 \le a_i \le 10^9)$ — the prices of ice spheres.
-----Output-----
In the first line print the maximum number of ice spheres that Sage can buy.
In the second line print the prices of ice spheres in the optimal order. If there are several correct answers, you can print any of them.
-----Example-----
Input
7
1 3 2 2 4 5 4
Output
3
3 1 4 2 4 2 5
-----Note-----
In the sample it's not possible to place the ice spheres in any order so that Sage would buy $4$ of them. If the spheres are placed in the order $(3, 1, 4, 2, 4, 2, 5)$, then Sage will buy one sphere for $1$ and two spheres for $2$ each.
|
n = int(input())
l = list(map(int, input().split()))
ans = 0
l.sort()
la = l[-1]
k = []
if n < 3:
print(0)
print(*l)
elif n % 2 != 0:
ans = 0
l1 = l[0 : n // 2]
l2 = l[n // 2 :]
k.append(l[n // 2])
for i in range(n // 2, n - 1):
ma1 = l[i]
ma2 = l[i + 1]
if l[i - n // 2] < ma1 and l[i - n // 2] < ma2:
k.append(l[i - n // 2])
k.append(ma2)
ans += 1
else:
k.append(l[i - n // 2])
k.append(ma2)
print(ans)
print(*k)
else:
ans = 0
l1 = l[0 : n // 2 - 1]
l2 = l[n // 2 - 1 :]
k.append(l[n // 2])
for i in range(n // 2, n - 1):
ma1 = l[i]
ma2 = l[i + 1]
if l[i - n // 2] < ma1 and l[i - n // 2] < ma2:
k.append(l[i - n // 2])
k.append(ma2)
ans += 1
else:
k.append(l[i - n // 2])
k.append(ma2)
k.append(l[n // 2 - 1])
print(ans)
print(*k)
|
ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR NUMBER EXPR FUNC_CALL VAR ASSIGN VAR VAR NUMBER ASSIGN VAR LIST IF VAR NUMBER EXPR FUNC_CALL VAR NUMBER EXPR FUNC_CALL VAR VAR IF BIN_OP VAR NUMBER NUMBER ASSIGN VAR NUMBER ASSIGN VAR VAR NUMBER BIN_OP VAR NUMBER ASSIGN VAR VAR BIN_OP VAR NUMBER EXPR FUNC_CALL VAR VAR BIN_OP VAR NUMBER FOR VAR FUNC_CALL VAR BIN_OP VAR NUMBER BIN_OP VAR NUMBER ASSIGN VAR VAR VAR ASSIGN VAR VAR BIN_OP VAR NUMBER IF VAR BIN_OP VAR BIN_OP VAR NUMBER VAR VAR BIN_OP VAR BIN_OP VAR NUMBER VAR EXPR FUNC_CALL VAR VAR BIN_OP VAR BIN_OP VAR NUMBER EXPR FUNC_CALL VAR VAR VAR NUMBER EXPR FUNC_CALL VAR VAR BIN_OP VAR BIN_OP VAR NUMBER EXPR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR VAR ASSIGN VAR NUMBER ASSIGN VAR VAR NUMBER BIN_OP BIN_OP VAR NUMBER NUMBER ASSIGN VAR VAR BIN_OP BIN_OP VAR NUMBER NUMBER EXPR FUNC_CALL VAR VAR BIN_OP VAR NUMBER FOR VAR FUNC_CALL VAR BIN_OP VAR NUMBER BIN_OP VAR NUMBER ASSIGN VAR VAR VAR ASSIGN VAR VAR BIN_OP VAR NUMBER IF VAR BIN_OP VAR BIN_OP VAR NUMBER VAR VAR BIN_OP VAR BIN_OP VAR NUMBER VAR EXPR FUNC_CALL VAR VAR BIN_OP VAR BIN_OP VAR NUMBER EXPR FUNC_CALL VAR VAR VAR NUMBER EXPR FUNC_CALL VAR VAR BIN_OP VAR BIN_OP VAR NUMBER EXPR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR VAR BIN_OP BIN_OP VAR NUMBER NUMBER EXPR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR VAR
|
This is the hard version of the problem. The difference between the versions is that in the easy version all prices $a_i$ are different. You can make hacks if and only if you solved both versions of the problem.
Today is Sage's birthday, and she will go shopping to buy ice spheres. All $n$ ice spheres are placed in a row and they are numbered from $1$ to $n$ from left to right. Each ice sphere has a positive integer price. In this version, some prices can be equal.
An ice sphere is cheap if it costs strictly less than two neighboring ice spheres: the nearest to the left and the nearest to the right. The leftmost and the rightmost ice spheres are not cheap. Sage will choose all cheap ice spheres and then buy only them.
You can visit the shop before Sage and reorder the ice spheres as you wish. Find out the maximum number of ice spheres that Sage can buy, and show how the ice spheres should be reordered.
-----Input-----
The first line contains a single integer $n$ $(1 \le n \le 10^5)$ — the number of ice spheres in the shop.
The second line contains $n$ integers $a_1, a_2, \dots, a_n$ $(1 \le a_i \le 10^9)$ — the prices of ice spheres.
-----Output-----
In the first line print the maximum number of ice spheres that Sage can buy.
In the second line print the prices of ice spheres in the optimal order. If there are several correct answers, you can print any of them.
-----Example-----
Input
7
1 3 2 2 4 5 4
Output
3
3 1 4 2 4 2 5
-----Note-----
In the sample it's not possible to place the ice spheres in any order so that Sage would buy $4$ of them. If the spheres are placed in the order $(3, 1, 4, 2, 4, 2, 5)$, then Sage will buy one sphere for $1$ and two spheres for $2$ each.
|
n = int(input())
arr = [int(x) for x in input().split()]
arr.sort()
arr[0::2], arr[1::2] = arr[n // 2 :], arr[: n // 2]
ans = 0
for i in range(1, n - 1):
if arr[i] < arr[i - 1] and arr[i] < arr[i + 1]:
ans += 1
print(ans)
print(*arr)
|
ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR VAR VAR FUNC_CALL FUNC_CALL VAR EXPR FUNC_CALL VAR ASSIGN VAR NUMBER NUMBER VAR NUMBER NUMBER VAR BIN_OP VAR NUMBER VAR BIN_OP VAR NUMBER ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR NUMBER BIN_OP VAR NUMBER IF VAR VAR VAR BIN_OP VAR NUMBER VAR VAR VAR BIN_OP VAR NUMBER VAR NUMBER EXPR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR VAR
|
This is the hard version of the problem. The difference between the versions is that in the easy version all prices $a_i$ are different. You can make hacks if and only if you solved both versions of the problem.
Today is Sage's birthday, and she will go shopping to buy ice spheres. All $n$ ice spheres are placed in a row and they are numbered from $1$ to $n$ from left to right. Each ice sphere has a positive integer price. In this version, some prices can be equal.
An ice sphere is cheap if it costs strictly less than two neighboring ice spheres: the nearest to the left and the nearest to the right. The leftmost and the rightmost ice spheres are not cheap. Sage will choose all cheap ice spheres and then buy only them.
You can visit the shop before Sage and reorder the ice spheres as you wish. Find out the maximum number of ice spheres that Sage can buy, and show how the ice spheres should be reordered.
-----Input-----
The first line contains a single integer $n$ $(1 \le n \le 10^5)$ — the number of ice spheres in the shop.
The second line contains $n$ integers $a_1, a_2, \dots, a_n$ $(1 \le a_i \le 10^9)$ — the prices of ice spheres.
-----Output-----
In the first line print the maximum number of ice spheres that Sage can buy.
In the second line print the prices of ice spheres in the optimal order. If there are several correct answers, you can print any of them.
-----Example-----
Input
7
1 3 2 2 4 5 4
Output
3
3 1 4 2 4 2 5
-----Note-----
In the sample it's not possible to place the ice spheres in any order so that Sage would buy $4$ of them. If the spheres are placed in the order $(3, 1, 4, 2, 4, 2, 5)$, then Sage will buy one sphere for $1$ and two spheres for $2$ each.
|
def main():
n = int(input())
a = list(map(int, input().split()))
a.sort()
l = []
if n < 3:
print(0)
print(*a)
return
half = n // 2
for i in range(half):
l.append(a[half + i])
l.append(a[i])
if n % 2 != 0:
l.append(a[-1])
total = 0
for j in range(1, n - 1):
if l[j] < l[j - 1] and l[j] < l[j + 1]:
total += 1
print(total)
print(*l)
main()
|
FUNC_DEF ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR EXPR FUNC_CALL VAR ASSIGN VAR LIST IF VAR NUMBER EXPR FUNC_CALL VAR NUMBER EXPR FUNC_CALL VAR VAR RETURN ASSIGN VAR BIN_OP VAR NUMBER FOR VAR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR VAR BIN_OP VAR VAR EXPR FUNC_CALL VAR VAR VAR IF BIN_OP VAR NUMBER NUMBER EXPR FUNC_CALL VAR VAR NUMBER ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR NUMBER BIN_OP VAR NUMBER IF VAR VAR VAR BIN_OP VAR NUMBER VAR VAR VAR BIN_OP VAR NUMBER VAR NUMBER EXPR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR
|
This is the hard version of the problem. The difference between the versions is that in the easy version all prices $a_i$ are different. You can make hacks if and only if you solved both versions of the problem.
Today is Sage's birthday, and she will go shopping to buy ice spheres. All $n$ ice spheres are placed in a row and they are numbered from $1$ to $n$ from left to right. Each ice sphere has a positive integer price. In this version, some prices can be equal.
An ice sphere is cheap if it costs strictly less than two neighboring ice spheres: the nearest to the left and the nearest to the right. The leftmost and the rightmost ice spheres are not cheap. Sage will choose all cheap ice spheres and then buy only them.
You can visit the shop before Sage and reorder the ice spheres as you wish. Find out the maximum number of ice spheres that Sage can buy, and show how the ice spheres should be reordered.
-----Input-----
The first line contains a single integer $n$ $(1 \le n \le 10^5)$ — the number of ice spheres in the shop.
The second line contains $n$ integers $a_1, a_2, \dots, a_n$ $(1 \le a_i \le 10^9)$ — the prices of ice spheres.
-----Output-----
In the first line print the maximum number of ice spheres that Sage can buy.
In the second line print the prices of ice spheres in the optimal order. If there are several correct answers, you can print any of them.
-----Example-----
Input
7
1 3 2 2 4 5 4
Output
3
3 1 4 2 4 2 5
-----Note-----
In the sample it's not possible to place the ice spheres in any order so that Sage would buy $4$ of them. If the spheres are placed in the order $(3, 1, 4, 2, 4, 2, 5)$, then Sage will buy one sphere for $1$ and two spheres for $2$ each.
|
def solve():
n = int(input())
p = list(map(int, input().split()))
if n <= 2:
print(0)
print(*p)
return
p.sort()
a = [0] * n
l = 0
k = n
if n % 2 == 0:
k -= 1
for i in range(1, k, 2):
a[i] = p[l]
l += 1
l = n - 1
for i in range(k - 1, -1, -2):
a[i] = p[l]
l -= 1
if n % 2 == 0:
a[-1] = p[l]
cnt = 0
for i in range(1, n - 1):
if a[i] < a[i - 1] and a[i] < a[i + 1]:
cnt += 1
print(cnt)
print(*a)
return
def main():
t = 1
for _ in range(t):
solve()
main()
|
FUNC_DEF ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR IF VAR NUMBER EXPR FUNC_CALL VAR NUMBER EXPR FUNC_CALL VAR VAR RETURN EXPR FUNC_CALL VAR ASSIGN VAR BIN_OP LIST NUMBER VAR ASSIGN VAR NUMBER ASSIGN VAR VAR IF BIN_OP VAR NUMBER NUMBER VAR NUMBER FOR VAR FUNC_CALL VAR NUMBER VAR NUMBER ASSIGN VAR VAR VAR VAR VAR NUMBER ASSIGN VAR BIN_OP VAR NUMBER FOR VAR FUNC_CALL VAR BIN_OP VAR NUMBER NUMBER NUMBER ASSIGN VAR VAR VAR VAR VAR NUMBER IF BIN_OP VAR NUMBER NUMBER ASSIGN VAR NUMBER VAR VAR ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR NUMBER BIN_OP VAR NUMBER IF VAR VAR VAR BIN_OP VAR NUMBER VAR VAR VAR BIN_OP VAR NUMBER VAR NUMBER EXPR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR VAR RETURN FUNC_DEF ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR EXPR FUNC_CALL VAR
|
This is the hard version of the problem. The difference between the versions is that in the easy version all prices $a_i$ are different. You can make hacks if and only if you solved both versions of the problem.
Today is Sage's birthday, and she will go shopping to buy ice spheres. All $n$ ice spheres are placed in a row and they are numbered from $1$ to $n$ from left to right. Each ice sphere has a positive integer price. In this version, some prices can be equal.
An ice sphere is cheap if it costs strictly less than two neighboring ice spheres: the nearest to the left and the nearest to the right. The leftmost and the rightmost ice spheres are not cheap. Sage will choose all cheap ice spheres and then buy only them.
You can visit the shop before Sage and reorder the ice spheres as you wish. Find out the maximum number of ice spheres that Sage can buy, and show how the ice spheres should be reordered.
-----Input-----
The first line contains a single integer $n$ $(1 \le n \le 10^5)$ — the number of ice spheres in the shop.
The second line contains $n$ integers $a_1, a_2, \dots, a_n$ $(1 \le a_i \le 10^9)$ — the prices of ice spheres.
-----Output-----
In the first line print the maximum number of ice spheres that Sage can buy.
In the second line print the prices of ice spheres in the optimal order. If there are several correct answers, you can print any of them.
-----Example-----
Input
7
1 3 2 2 4 5 4
Output
3
3 1 4 2 4 2 5
-----Note-----
In the sample it's not possible to place the ice spheres in any order so that Sage would buy $4$ of them. If the spheres are placed in the order $(3, 1, 4, 2, 4, 2, 5)$, then Sage will buy one sphere for $1$ and two spheres for $2$ each.
|
def solve(n, ices):
ices = sorted(ices)
former, latter = ices[: n // 2], ices[n // 2 :]
res = 0
res_ice = []
for i in range((n - 1) // 2):
res_ice.append(latter[i])
res_ice.append(former[i])
if former[i] < latter[i]:
res += 1
res_ice.append(latter[-1])
if n % 2 == 0:
res_ice.append(former[-1])
return res, " ".join(list(map(str, res_ice)))
def main():
n = int(input())
ices = list(map(int, input().split()))
res1, res2 = solve(n, ices)
print(res1)
print(res2)
main()
|
FUNC_DEF ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR VAR VAR BIN_OP VAR NUMBER VAR BIN_OP VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR LIST FOR VAR FUNC_CALL VAR BIN_OP BIN_OP VAR NUMBER NUMBER EXPR FUNC_CALL VAR VAR VAR EXPR FUNC_CALL VAR VAR VAR IF VAR VAR VAR VAR VAR NUMBER EXPR FUNC_CALL VAR VAR NUMBER IF BIN_OP VAR NUMBER NUMBER EXPR FUNC_CALL VAR VAR NUMBER RETURN VAR FUNC_CALL STRING FUNC_CALL VAR FUNC_CALL VAR VAR VAR FUNC_DEF ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR VAR FUNC_CALL VAR VAR VAR EXPR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR
|
This is the hard version of the problem. The difference between the versions is that in the easy version all prices $a_i$ are different. You can make hacks if and only if you solved both versions of the problem.
Today is Sage's birthday, and she will go shopping to buy ice spheres. All $n$ ice spheres are placed in a row and they are numbered from $1$ to $n$ from left to right. Each ice sphere has a positive integer price. In this version, some prices can be equal.
An ice sphere is cheap if it costs strictly less than two neighboring ice spheres: the nearest to the left and the nearest to the right. The leftmost and the rightmost ice spheres are not cheap. Sage will choose all cheap ice spheres and then buy only them.
You can visit the shop before Sage and reorder the ice spheres as you wish. Find out the maximum number of ice spheres that Sage can buy, and show how the ice spheres should be reordered.
-----Input-----
The first line contains a single integer $n$ $(1 \le n \le 10^5)$ — the number of ice spheres in the shop.
The second line contains $n$ integers $a_1, a_2, \dots, a_n$ $(1 \le a_i \le 10^9)$ — the prices of ice spheres.
-----Output-----
In the first line print the maximum number of ice spheres that Sage can buy.
In the second line print the prices of ice spheres in the optimal order. If there are several correct answers, you can print any of them.
-----Example-----
Input
7
1 3 2 2 4 5 4
Output
3
3 1 4 2 4 2 5
-----Note-----
In the sample it's not possible to place the ice spheres in any order so that Sage would buy $4$ of them. If the spheres are placed in the order $(3, 1, 4, 2, 4, 2, 5)$, then Sage will buy one sphere for $1$ and two spheres for $2$ each.
|
t = int(input())
ice = list(map(int, input().split()))
new_ice = []
count = 0
ice.sort()
if len(ice) % 2 != 0:
for i in range(0, len(ice) // 2 + 1):
new_ice.append(ice[len(ice) // 2 + i])
new_ice.append(ice[i])
del new_ice[-1]
elif len(ice) <= 2:
for i in range(len(ice)):
new_ice.append(ice[i])
else:
for i in range(0, len(ice) // 2):
new_ice.append(ice[len(ice) // 2 + i])
new_ice.append(ice[i])
for i in range(1, len(new_ice) - 1, 1):
if new_ice[i] < new_ice[i - 1] and new_ice[i] < new_ice[i + 1]:
count += 1
print(count)
print(" ".join(map(str, new_ice)))
|
ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR LIST ASSIGN VAR NUMBER EXPR FUNC_CALL VAR IF BIN_OP FUNC_CALL VAR VAR NUMBER NUMBER FOR VAR FUNC_CALL VAR NUMBER BIN_OP BIN_OP FUNC_CALL VAR VAR NUMBER NUMBER EXPR FUNC_CALL VAR VAR BIN_OP BIN_OP FUNC_CALL VAR VAR NUMBER VAR EXPR FUNC_CALL VAR VAR VAR VAR NUMBER IF FUNC_CALL VAR VAR NUMBER FOR VAR FUNC_CALL VAR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR VAR VAR FOR VAR FUNC_CALL VAR NUMBER BIN_OP FUNC_CALL VAR VAR NUMBER EXPR FUNC_CALL VAR VAR BIN_OP BIN_OP FUNC_CALL VAR VAR NUMBER VAR EXPR FUNC_CALL VAR VAR VAR FOR VAR FUNC_CALL VAR NUMBER BIN_OP FUNC_CALL VAR VAR NUMBER NUMBER IF VAR VAR VAR BIN_OP VAR NUMBER VAR VAR VAR BIN_OP VAR NUMBER VAR NUMBER EXPR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR FUNC_CALL STRING FUNC_CALL VAR VAR VAR
|
This is the hard version of the problem. The difference between the versions is that in the easy version all prices $a_i$ are different. You can make hacks if and only if you solved both versions of the problem.
Today is Sage's birthday, and she will go shopping to buy ice spheres. All $n$ ice spheres are placed in a row and they are numbered from $1$ to $n$ from left to right. Each ice sphere has a positive integer price. In this version, some prices can be equal.
An ice sphere is cheap if it costs strictly less than two neighboring ice spheres: the nearest to the left and the nearest to the right. The leftmost and the rightmost ice spheres are not cheap. Sage will choose all cheap ice spheres and then buy only them.
You can visit the shop before Sage and reorder the ice spheres as you wish. Find out the maximum number of ice spheres that Sage can buy, and show how the ice spheres should be reordered.
-----Input-----
The first line contains a single integer $n$ $(1 \le n \le 10^5)$ — the number of ice spheres in the shop.
The second line contains $n$ integers $a_1, a_2, \dots, a_n$ $(1 \le a_i \le 10^9)$ — the prices of ice spheres.
-----Output-----
In the first line print the maximum number of ice spheres that Sage can buy.
In the second line print the prices of ice spheres in the optimal order. If there are several correct answers, you can print any of them.
-----Example-----
Input
7
1 3 2 2 4 5 4
Output
3
3 1 4 2 4 2 5
-----Note-----
In the sample it's not possible to place the ice spheres in any order so that Sage would buy $4$ of them. If the spheres are placed in the order $(3, 1, 4, 2, 4, 2, 5)$, then Sage will buy one sphere for $1$ and two spheres for $2$ each.
|
n = int(input())
a = list(sorted(list(map(int, input().split()))))
k = n // 2
b = []
j = k
l = 0
for i in range(n):
if i % 2 == 0:
b.append(a[j])
j += 1
else:
b.append(a[l])
l += 1
c = 0
for i in range(1, n - 1, 2):
if b[i] < b[i + 1] and b[i] < b[i - 1]:
c += 1
print(c)
print(*b, sep=" ")
|
ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR BIN_OP VAR NUMBER ASSIGN VAR LIST ASSIGN VAR VAR ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR VAR IF BIN_OP VAR NUMBER NUMBER EXPR FUNC_CALL VAR VAR VAR VAR NUMBER EXPR FUNC_CALL VAR VAR VAR VAR NUMBER ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR NUMBER BIN_OP VAR NUMBER NUMBER IF VAR VAR VAR BIN_OP VAR NUMBER VAR VAR VAR BIN_OP VAR NUMBER VAR NUMBER EXPR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR VAR STRING
|
This is the hard version of the problem. The difference between the versions is that in the easy version all prices $a_i$ are different. You can make hacks if and only if you solved both versions of the problem.
Today is Sage's birthday, and she will go shopping to buy ice spheres. All $n$ ice spheres are placed in a row and they are numbered from $1$ to $n$ from left to right. Each ice sphere has a positive integer price. In this version, some prices can be equal.
An ice sphere is cheap if it costs strictly less than two neighboring ice spheres: the nearest to the left and the nearest to the right. The leftmost and the rightmost ice spheres are not cheap. Sage will choose all cheap ice spheres and then buy only them.
You can visit the shop before Sage and reorder the ice spheres as you wish. Find out the maximum number of ice spheres that Sage can buy, and show how the ice spheres should be reordered.
-----Input-----
The first line contains a single integer $n$ $(1 \le n \le 10^5)$ — the number of ice spheres in the shop.
The second line contains $n$ integers $a_1, a_2, \dots, a_n$ $(1 \le a_i \le 10^9)$ — the prices of ice spheres.
-----Output-----
In the first line print the maximum number of ice spheres that Sage can buy.
In the second line print the prices of ice spheres in the optimal order. If there are several correct answers, you can print any of them.
-----Example-----
Input
7
1 3 2 2 4 5 4
Output
3
3 1 4 2 4 2 5
-----Note-----
In the sample it's not possible to place the ice spheres in any order so that Sage would buy $4$ of them. If the spheres are placed in the order $(3, 1, 4, 2, 4, 2, 5)$, then Sage will buy one sphere for $1$ and two spheres for $2$ each.
|
n = int(input())
a = sorted(map(int, input().split()))
mas = [0] * n
j = 1
for i in range(n):
if j >= n:
j = 0
mas[j] = a[i]
j += 2
ans = 0
for i in range(1, n - 1):
ans += mas[i - 1] > mas[i] < mas[i + 1]
print(ans)
print(*mas)
|
ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR BIN_OP LIST NUMBER VAR ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR VAR IF VAR VAR ASSIGN VAR NUMBER ASSIGN VAR VAR VAR VAR VAR NUMBER ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR NUMBER BIN_OP VAR NUMBER VAR VAR BIN_OP VAR NUMBER VAR VAR VAR BIN_OP VAR NUMBER EXPR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR VAR
|
This is the hard version of the problem. The difference between the versions is that in the easy version all prices $a_i$ are different. You can make hacks if and only if you solved both versions of the problem.
Today is Sage's birthday, and she will go shopping to buy ice spheres. All $n$ ice spheres are placed in a row and they are numbered from $1$ to $n$ from left to right. Each ice sphere has a positive integer price. In this version, some prices can be equal.
An ice sphere is cheap if it costs strictly less than two neighboring ice spheres: the nearest to the left and the nearest to the right. The leftmost and the rightmost ice spheres are not cheap. Sage will choose all cheap ice spheres and then buy only them.
You can visit the shop before Sage and reorder the ice spheres as you wish. Find out the maximum number of ice spheres that Sage can buy, and show how the ice spheres should be reordered.
-----Input-----
The first line contains a single integer $n$ $(1 \le n \le 10^5)$ — the number of ice spheres in the shop.
The second line contains $n$ integers $a_1, a_2, \dots, a_n$ $(1 \le a_i \le 10^9)$ — the prices of ice spheres.
-----Output-----
In the first line print the maximum number of ice spheres that Sage can buy.
In the second line print the prices of ice spheres in the optimal order. If there are several correct answers, you can print any of them.
-----Example-----
Input
7
1 3 2 2 4 5 4
Output
3
3 1 4 2 4 2 5
-----Note-----
In the sample it's not possible to place the ice spheres in any order so that Sage would buy $4$ of them. If the spheres are placed in the order $(3, 1, 4, 2, 4, 2, 5)$, then Sage will buy one sphere for $1$ and two spheres for $2$ each.
|
import sys
n = int(sys.stdin.readline())
pricelist = [int(v) for v in sys.stdin.readline().strip().split(" ")]
pricelist.sort()
newlist = [(0) for i in range(0, n)]
start = int(n / 2)
for i in range(0, start):
newlist[i * 2] = pricelist[i + start]
newlist[i * 2 + 1] = pricelist[i]
if n % 2 == 1:
newlist[n - 1] = pricelist[n - 1]
total = 0
for i in range(1, n - 1, 2):
if newlist[i] < newlist[i - 1] and newlist[i] < newlist[i + 1]:
total += 1
print(total)
print(" ".join(str(v) for v in newlist))
|
IMPORT ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR VAR VAR FUNC_CALL FUNC_CALL FUNC_CALL VAR STRING EXPR FUNC_CALL VAR ASSIGN VAR NUMBER VAR FUNC_CALL VAR NUMBER VAR ASSIGN VAR FUNC_CALL VAR BIN_OP VAR NUMBER FOR VAR FUNC_CALL VAR NUMBER VAR ASSIGN VAR BIN_OP VAR NUMBER VAR BIN_OP VAR VAR ASSIGN VAR BIN_OP BIN_OP VAR NUMBER NUMBER VAR VAR IF BIN_OP VAR NUMBER NUMBER ASSIGN VAR BIN_OP VAR NUMBER VAR BIN_OP VAR NUMBER ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR NUMBER BIN_OP VAR NUMBER NUMBER IF VAR VAR VAR BIN_OP VAR NUMBER VAR VAR VAR BIN_OP VAR NUMBER VAR NUMBER EXPR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR FUNC_CALL STRING FUNC_CALL VAR VAR VAR VAR
|
This is the hard version of the problem. The difference between the versions is that in the easy version all prices $a_i$ are different. You can make hacks if and only if you solved both versions of the problem.
Today is Sage's birthday, and she will go shopping to buy ice spheres. All $n$ ice spheres are placed in a row and they are numbered from $1$ to $n$ from left to right. Each ice sphere has a positive integer price. In this version, some prices can be equal.
An ice sphere is cheap if it costs strictly less than two neighboring ice spheres: the nearest to the left and the nearest to the right. The leftmost and the rightmost ice spheres are not cheap. Sage will choose all cheap ice spheres and then buy only them.
You can visit the shop before Sage and reorder the ice spheres as you wish. Find out the maximum number of ice spheres that Sage can buy, and show how the ice spheres should be reordered.
-----Input-----
The first line contains a single integer $n$ $(1 \le n \le 10^5)$ — the number of ice spheres in the shop.
The second line contains $n$ integers $a_1, a_2, \dots, a_n$ $(1 \le a_i \le 10^9)$ — the prices of ice spheres.
-----Output-----
In the first line print the maximum number of ice spheres that Sage can buy.
In the second line print the prices of ice spheres in the optimal order. If there are several correct answers, you can print any of them.
-----Example-----
Input
7
1 3 2 2 4 5 4
Output
3
3 1 4 2 4 2 5
-----Note-----
In the sample it's not possible to place the ice spheres in any order so that Sage would buy $4$ of them. If the spheres are placed in the order $(3, 1, 4, 2, 4, 2, 5)$, then Sage will buy one sphere for $1$ and two spheres for $2$ each.
|
n = int(input())
array = list(map(int, input().split(" ")))
array.sort()
menores = array[: int(n / 2)]
maiores = array[int(n / 2) :]
index = 0
solucao = []
qtd = 0
flag_left = False
for i in range(n):
if i % 2 != 0:
solucao.append(menores[index])
if menores[index] < solucao[i - 1]:
flag_left = True
else:
flag_left = False
index += 1
else:
solucao.append(maiores[int(i / 2)])
if flag_left and maiores[int(i / 2)] > solucao[i - 1]:
qtd += 1
flag_left = False
print(qtd)
solucao = [str(int) for int in solucao]
print(" ".join(solucao))
|
ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR STRING EXPR FUNC_CALL VAR ASSIGN VAR VAR FUNC_CALL VAR BIN_OP VAR NUMBER ASSIGN VAR VAR FUNC_CALL VAR BIN_OP VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR LIST ASSIGN VAR NUMBER ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR VAR IF BIN_OP VAR NUMBER NUMBER EXPR FUNC_CALL VAR VAR VAR IF VAR VAR VAR BIN_OP VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR NUMBER VAR NUMBER EXPR FUNC_CALL VAR VAR FUNC_CALL VAR BIN_OP VAR NUMBER IF VAR VAR FUNC_CALL VAR BIN_OP VAR NUMBER VAR BIN_OP VAR NUMBER VAR NUMBER ASSIGN VAR NUMBER EXPR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR VAR VAR VAR EXPR FUNC_CALL VAR FUNC_CALL STRING VAR
|
This is the hard version of the problem. The difference between the versions is that in the easy version all prices $a_i$ are different. You can make hacks if and only if you solved both versions of the problem.
Today is Sage's birthday, and she will go shopping to buy ice spheres. All $n$ ice spheres are placed in a row and they are numbered from $1$ to $n$ from left to right. Each ice sphere has a positive integer price. In this version, some prices can be equal.
An ice sphere is cheap if it costs strictly less than two neighboring ice spheres: the nearest to the left and the nearest to the right. The leftmost and the rightmost ice spheres are not cheap. Sage will choose all cheap ice spheres and then buy only them.
You can visit the shop before Sage and reorder the ice spheres as you wish. Find out the maximum number of ice spheres that Sage can buy, and show how the ice spheres should be reordered.
-----Input-----
The first line contains a single integer $n$ $(1 \le n \le 10^5)$ — the number of ice spheres in the shop.
The second line contains $n$ integers $a_1, a_2, \dots, a_n$ $(1 \le a_i \le 10^9)$ — the prices of ice spheres.
-----Output-----
In the first line print the maximum number of ice spheres that Sage can buy.
In the second line print the prices of ice spheres in the optimal order. If there are several correct answers, you can print any of them.
-----Example-----
Input
7
1 3 2 2 4 5 4
Output
3
3 1 4 2 4 2 5
-----Note-----
In the sample it's not possible to place the ice spheres in any order so that Sage would buy $4$ of them. If the spheres are placed in the order $(3, 1, 4, 2, 4, 2, 5)$, then Sage will buy one sphere for $1$ and two spheres for $2$ each.
|
n = int(input())
list1 = list(map(int, input().split()))
list1 = sorted(list1)
res = []
first = list1[: n // 2]
second = list1[n // 2 :]
c = 0
if n % 2 == 0:
for i in range(len(second)):
res.append(second[i])
res.append(first[i])
else:
for i in range(len(second) - 1):
res.append(second[i])
res.append(first[i])
res.append(second[-1])
for i in range(n - 1):
if res[i] < res[i - 1] and res[i] < res[i + 1]:
c += 1
print(c)
for num in res:
print(num, end=" ")
|
ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR LIST ASSIGN VAR VAR BIN_OP VAR NUMBER ASSIGN VAR VAR BIN_OP VAR NUMBER ASSIGN VAR NUMBER IF BIN_OP VAR NUMBER NUMBER FOR VAR FUNC_CALL VAR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR VAR VAR EXPR FUNC_CALL VAR VAR VAR FOR VAR FUNC_CALL VAR BIN_OP FUNC_CALL VAR VAR NUMBER EXPR FUNC_CALL VAR VAR VAR EXPR FUNC_CALL VAR VAR VAR EXPR FUNC_CALL VAR VAR NUMBER FOR VAR FUNC_CALL VAR BIN_OP VAR NUMBER IF VAR VAR VAR BIN_OP VAR NUMBER VAR VAR VAR BIN_OP VAR NUMBER VAR NUMBER EXPR FUNC_CALL VAR VAR FOR VAR VAR EXPR FUNC_CALL VAR VAR STRING
|
This is the hard version of the problem. The difference between the versions is that in the easy version all prices $a_i$ are different. You can make hacks if and only if you solved both versions of the problem.
Today is Sage's birthday, and she will go shopping to buy ice spheres. All $n$ ice spheres are placed in a row and they are numbered from $1$ to $n$ from left to right. Each ice sphere has a positive integer price. In this version, some prices can be equal.
An ice sphere is cheap if it costs strictly less than two neighboring ice spheres: the nearest to the left and the nearest to the right. The leftmost and the rightmost ice spheres are not cheap. Sage will choose all cheap ice spheres and then buy only them.
You can visit the shop before Sage and reorder the ice spheres as you wish. Find out the maximum number of ice spheres that Sage can buy, and show how the ice spheres should be reordered.
-----Input-----
The first line contains a single integer $n$ $(1 \le n \le 10^5)$ — the number of ice spheres in the shop.
The second line contains $n$ integers $a_1, a_2, \dots, a_n$ $(1 \le a_i \le 10^9)$ — the prices of ice spheres.
-----Output-----
In the first line print the maximum number of ice spheres that Sage can buy.
In the second line print the prices of ice spheres in the optimal order. If there are several correct answers, you can print any of them.
-----Example-----
Input
7
1 3 2 2 4 5 4
Output
3
3 1 4 2 4 2 5
-----Note-----
In the sample it's not possible to place the ice spheres in any order so that Sage would buy $4$ of them. If the spheres are placed in the order $(3, 1, 4, 2, 4, 2, 5)$, then Sage will buy one sphere for $1$ and two spheres for $2$ each.
|
n = int(input())
l = list(map(int, input().split()))[:n]
f = [0] * n
l.sort(reverse=True)
cnt = 0
for i in range(n):
if i % 2 == 1:
f[i] = l.pop()
for i in range(n):
if f[i] == 0:
f[i] = l.pop()
if n % 2 == 0:
for i in range(1, n - 2, 2):
if f[i] < f[i - 1] and f[i] < f[i + 1]:
cnt += 1
else:
for i in range(1, n - 1, 2):
if f[i] < f[i - 1] and f[i] < f[i + 1]:
cnt += 1
print(cnt)
print(*f)
|
ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR VAR ASSIGN VAR BIN_OP LIST NUMBER VAR EXPR FUNC_CALL VAR NUMBER ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR VAR IF BIN_OP VAR NUMBER NUMBER ASSIGN VAR VAR FUNC_CALL VAR FOR VAR FUNC_CALL VAR VAR IF VAR VAR NUMBER ASSIGN VAR VAR FUNC_CALL VAR IF BIN_OP VAR NUMBER NUMBER FOR VAR FUNC_CALL VAR NUMBER BIN_OP VAR NUMBER NUMBER IF VAR VAR VAR BIN_OP VAR NUMBER VAR VAR VAR BIN_OP VAR NUMBER VAR NUMBER FOR VAR FUNC_CALL VAR NUMBER BIN_OP VAR NUMBER NUMBER IF VAR VAR VAR BIN_OP VAR NUMBER VAR VAR VAR BIN_OP VAR NUMBER VAR NUMBER EXPR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR VAR
|
This is the hard version of the problem. The difference between the versions is that in the easy version all prices $a_i$ are different. You can make hacks if and only if you solved both versions of the problem.
Today is Sage's birthday, and she will go shopping to buy ice spheres. All $n$ ice spheres are placed in a row and they are numbered from $1$ to $n$ from left to right. Each ice sphere has a positive integer price. In this version, some prices can be equal.
An ice sphere is cheap if it costs strictly less than two neighboring ice spheres: the nearest to the left and the nearest to the right. The leftmost and the rightmost ice spheres are not cheap. Sage will choose all cheap ice spheres and then buy only them.
You can visit the shop before Sage and reorder the ice spheres as you wish. Find out the maximum number of ice spheres that Sage can buy, and show how the ice spheres should be reordered.
-----Input-----
The first line contains a single integer $n$ $(1 \le n \le 10^5)$ — the number of ice spheres in the shop.
The second line contains $n$ integers $a_1, a_2, \dots, a_n$ $(1 \le a_i \le 10^9)$ — the prices of ice spheres.
-----Output-----
In the first line print the maximum number of ice spheres that Sage can buy.
In the second line print the prices of ice spheres in the optimal order. If there are several correct answers, you can print any of them.
-----Example-----
Input
7
1 3 2 2 4 5 4
Output
3
3 1 4 2 4 2 5
-----Note-----
In the sample it's not possible to place the ice spheres in any order so that Sage would buy $4$ of them. If the spheres are placed in the order $(3, 1, 4, 2, 4, 2, 5)$, then Sage will buy one sphere for $1$ and two spheres for $2$ each.
|
n = int(input())
bolas = []
valores = input().split()
cont = 0
for e in range(n):
bolas.append(int(valores[e]))
bolas.sort()
v1 = bolas[0 : n // 2]
v2 = bolas[n // 2 :]
v3 = []
for e in range(n // 2):
v3.append(v2[e])
v3.append(v1[e])
if n % 2 != 0:
v3.append(v2[-1])
for e in range(1, n - 1):
if v3[e - 1] > v3[e] and v3[e] < v3[e + 1]:
cont += 1
print(cont)
for e in v3:
print(e, end=" ")
|
ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR LIST ASSIGN VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR FUNC_CALL VAR VAR VAR EXPR FUNC_CALL VAR ASSIGN VAR VAR NUMBER BIN_OP VAR NUMBER ASSIGN VAR VAR BIN_OP VAR NUMBER ASSIGN VAR LIST FOR VAR FUNC_CALL VAR BIN_OP VAR NUMBER EXPR FUNC_CALL VAR VAR VAR EXPR FUNC_CALL VAR VAR VAR IF BIN_OP VAR NUMBER NUMBER EXPR FUNC_CALL VAR VAR NUMBER FOR VAR FUNC_CALL VAR NUMBER BIN_OP VAR NUMBER IF VAR BIN_OP VAR NUMBER VAR VAR VAR VAR VAR BIN_OP VAR NUMBER VAR NUMBER EXPR FUNC_CALL VAR VAR FOR VAR VAR EXPR FUNC_CALL VAR VAR STRING
|
This is the hard version of the problem. The difference between the versions is that in the easy version all prices $a_i$ are different. You can make hacks if and only if you solved both versions of the problem.
Today is Sage's birthday, and she will go shopping to buy ice spheres. All $n$ ice spheres are placed in a row and they are numbered from $1$ to $n$ from left to right. Each ice sphere has a positive integer price. In this version, some prices can be equal.
An ice sphere is cheap if it costs strictly less than two neighboring ice spheres: the nearest to the left and the nearest to the right. The leftmost and the rightmost ice spheres are not cheap. Sage will choose all cheap ice spheres and then buy only them.
You can visit the shop before Sage and reorder the ice spheres as you wish. Find out the maximum number of ice spheres that Sage can buy, and show how the ice spheres should be reordered.
-----Input-----
The first line contains a single integer $n$ $(1 \le n \le 10^5)$ — the number of ice spheres in the shop.
The second line contains $n$ integers $a_1, a_2, \dots, a_n$ $(1 \le a_i \le 10^9)$ — the prices of ice spheres.
-----Output-----
In the first line print the maximum number of ice spheres that Sage can buy.
In the second line print the prices of ice spheres in the optimal order. If there are several correct answers, you can print any of them.
-----Example-----
Input
7
1 3 2 2 4 5 4
Output
3
3 1 4 2 4 2 5
-----Note-----
In the sample it's not possible to place the ice spheres in any order so that Sage would buy $4$ of them. If the spheres are placed in the order $(3, 1, 4, 2, 4, 2, 5)$, then Sage will buy one sphere for $1$ and two spheres for $2$ each.
|
n = input()
p = [int(x) for x in input().strip().split(" ")]
if len(p) <= 2:
print(0)
print(" ".join([str(x) for x in p]))
exit()
p.sort()
res_list = []
if len(p) % 2 == 0:
index = int(len(p) / 2)
if p[index - 1] == p[index - 2]:
res_list.append(p[index - 1])
del p[index - 1]
else:
res_list.append(p[index])
del p[index]
res = 0
gap = int((len(p) + 1) / 2)
res_list.append(p[gap - 1])
for i in range(gap - 1):
if p[i] < res_list[-1] and p[i] < p[i + gap]:
res += 1
res_list.append(p[i])
res_list.append(p[i + gap])
print(res)
print(" ".join([str(x) for x in res_list]))
|
ASSIGN VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR VAR VAR FUNC_CALL FUNC_CALL FUNC_CALL VAR STRING IF FUNC_CALL VAR VAR NUMBER EXPR FUNC_CALL VAR NUMBER EXPR FUNC_CALL VAR FUNC_CALL STRING FUNC_CALL VAR VAR VAR VAR EXPR FUNC_CALL VAR EXPR FUNC_CALL VAR ASSIGN VAR LIST IF BIN_OP FUNC_CALL VAR VAR NUMBER NUMBER ASSIGN VAR FUNC_CALL VAR BIN_OP FUNC_CALL VAR VAR NUMBER IF VAR BIN_OP VAR NUMBER VAR BIN_OP VAR NUMBER EXPR FUNC_CALL VAR VAR BIN_OP VAR NUMBER VAR BIN_OP VAR NUMBER EXPR FUNC_CALL VAR VAR VAR VAR VAR ASSIGN VAR NUMBER ASSIGN VAR FUNC_CALL VAR BIN_OP BIN_OP FUNC_CALL VAR VAR NUMBER NUMBER EXPR FUNC_CALL VAR VAR BIN_OP VAR NUMBER FOR VAR FUNC_CALL VAR BIN_OP VAR NUMBER IF VAR VAR VAR NUMBER VAR VAR VAR BIN_OP VAR VAR VAR NUMBER EXPR FUNC_CALL VAR VAR VAR EXPR FUNC_CALL VAR VAR BIN_OP VAR VAR EXPR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR FUNC_CALL STRING FUNC_CALL VAR VAR VAR VAR
|
This is the hard version of the problem. The difference between the versions is that in the easy version all prices $a_i$ are different. You can make hacks if and only if you solved both versions of the problem.
Today is Sage's birthday, and she will go shopping to buy ice spheres. All $n$ ice spheres are placed in a row and they are numbered from $1$ to $n$ from left to right. Each ice sphere has a positive integer price. In this version, some prices can be equal.
An ice sphere is cheap if it costs strictly less than two neighboring ice spheres: the nearest to the left and the nearest to the right. The leftmost and the rightmost ice spheres are not cheap. Sage will choose all cheap ice spheres and then buy only them.
You can visit the shop before Sage and reorder the ice spheres as you wish. Find out the maximum number of ice spheres that Sage can buy, and show how the ice spheres should be reordered.
-----Input-----
The first line contains a single integer $n$ $(1 \le n \le 10^5)$ — the number of ice spheres in the shop.
The second line contains $n$ integers $a_1, a_2, \dots, a_n$ $(1 \le a_i \le 10^9)$ — the prices of ice spheres.
-----Output-----
In the first line print the maximum number of ice spheres that Sage can buy.
In the second line print the prices of ice spheres in the optimal order. If there are several correct answers, you can print any of them.
-----Example-----
Input
7
1 3 2 2 4 5 4
Output
3
3 1 4 2 4 2 5
-----Note-----
In the sample it's not possible to place the ice spheres in any order so that Sage would buy $4$ of them. If the spheres are placed in the order $(3, 1, 4, 2, 4, 2, 5)$, then Sage will buy one sphere for $1$ and two spheres for $2$ each.
|
n = int(input())
a = [int(x) for x in input().split()]
b = [0] * n
a.sort()
k = 0
c = 0
for i in range(1, n, 2):
b[i] = a[k]
k += 1
for i in range(0, n, 2):
b[i] = a[k]
k += 1
for i in range(1, n - 1):
if b[i] < b[i - 1] and b[i] < b[i + 1]:
c += 1
print(c)
print(*b)
|
ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR BIN_OP LIST NUMBER VAR EXPR FUNC_CALL VAR ASSIGN VAR NUMBER ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR NUMBER VAR NUMBER ASSIGN VAR VAR VAR VAR VAR NUMBER FOR VAR FUNC_CALL VAR NUMBER VAR NUMBER ASSIGN VAR VAR VAR VAR VAR NUMBER FOR VAR FUNC_CALL VAR NUMBER BIN_OP VAR NUMBER IF VAR VAR VAR BIN_OP VAR NUMBER VAR VAR VAR BIN_OP VAR NUMBER VAR NUMBER EXPR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR VAR
|
This is the hard version of the problem. The difference between the versions is that in the easy version all prices $a_i$ are different. You can make hacks if and only if you solved both versions of the problem.
Today is Sage's birthday, and she will go shopping to buy ice spheres. All $n$ ice spheres are placed in a row and they are numbered from $1$ to $n$ from left to right. Each ice sphere has a positive integer price. In this version, some prices can be equal.
An ice sphere is cheap if it costs strictly less than two neighboring ice spheres: the nearest to the left and the nearest to the right. The leftmost and the rightmost ice spheres are not cheap. Sage will choose all cheap ice spheres and then buy only them.
You can visit the shop before Sage and reorder the ice spheres as you wish. Find out the maximum number of ice spheres that Sage can buy, and show how the ice spheres should be reordered.
-----Input-----
The first line contains a single integer $n$ $(1 \le n \le 10^5)$ — the number of ice spheres in the shop.
The second line contains $n$ integers $a_1, a_2, \dots, a_n$ $(1 \le a_i \le 10^9)$ — the prices of ice spheres.
-----Output-----
In the first line print the maximum number of ice spheres that Sage can buy.
In the second line print the prices of ice spheres in the optimal order. If there are several correct answers, you can print any of them.
-----Example-----
Input
7
1 3 2 2 4 5 4
Output
3
3 1 4 2 4 2 5
-----Note-----
In the sample it's not possible to place the ice spheres in any order so that Sage would buy $4$ of them. If the spheres are placed in the order $(3, 1, 4, 2, 4, 2, 5)$, then Sage will buy one sphere for $1$ and two spheres for $2$ each.
|
n = int(input())
arr = list(map(int, input().split()))
arr.sort()
res = []
for i in range(n // 2):
res.append(arr[n // 2 + i])
res.append(arr[i])
if n % 2:
res.append(arr[-1])
res2 = 0
for i in range(1, n - 1):
if res[i] < res[i - 1] and res[i] < res[i + 1]:
res2 += 1
print(res2)
print(" ".join([str(i) for i in res]))
|
ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR EXPR FUNC_CALL VAR ASSIGN VAR LIST FOR VAR FUNC_CALL VAR BIN_OP VAR NUMBER EXPR FUNC_CALL VAR VAR BIN_OP BIN_OP VAR NUMBER VAR EXPR FUNC_CALL VAR VAR VAR IF BIN_OP VAR NUMBER EXPR FUNC_CALL VAR VAR NUMBER ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR NUMBER BIN_OP VAR NUMBER IF VAR VAR VAR BIN_OP VAR NUMBER VAR VAR VAR BIN_OP VAR NUMBER VAR NUMBER EXPR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR FUNC_CALL STRING FUNC_CALL VAR VAR VAR VAR
|
This is the hard version of the problem. The difference between the versions is that in the easy version all prices $a_i$ are different. You can make hacks if and only if you solved both versions of the problem.
Today is Sage's birthday, and she will go shopping to buy ice spheres. All $n$ ice spheres are placed in a row and they are numbered from $1$ to $n$ from left to right. Each ice sphere has a positive integer price. In this version, some prices can be equal.
An ice sphere is cheap if it costs strictly less than two neighboring ice spheres: the nearest to the left and the nearest to the right. The leftmost and the rightmost ice spheres are not cheap. Sage will choose all cheap ice spheres and then buy only them.
You can visit the shop before Sage and reorder the ice spheres as you wish. Find out the maximum number of ice spheres that Sage can buy, and show how the ice spheres should be reordered.
-----Input-----
The first line contains a single integer $n$ $(1 \le n \le 10^5)$ — the number of ice spheres in the shop.
The second line contains $n$ integers $a_1, a_2, \dots, a_n$ $(1 \le a_i \le 10^9)$ — the prices of ice spheres.
-----Output-----
In the first line print the maximum number of ice spheres that Sage can buy.
In the second line print the prices of ice spheres in the optimal order. If there are several correct answers, you can print any of them.
-----Example-----
Input
7
1 3 2 2 4 5 4
Output
3
3 1 4 2 4 2 5
-----Note-----
In the sample it's not possible to place the ice spheres in any order so that Sage would buy $4$ of them. If the spheres are placed in the order $(3, 1, 4, 2, 4, 2, 5)$, then Sage will buy one sphere for $1$ and two spheres for $2$ each.
|
import sys
input = sys.stdin.readline
N = int(input())
A = [int(_) for _ in input().split()]
sA = sorted(A)
B = ["?"] * N
for i in range(N // 2):
B[2 * i + 1] = sA[i]
for i in range(N // 2, N):
B[2 * (i - N // 2)] = sA[i]
answer = 0
for i in range(N // 2):
if 2 * i + 2 < N:
if B[2 * i + 1] < B[2 * i] and B[2 * i + 1] < B[2 * i + 2]:
answer += 1
print(answer)
print(" ".join(map(str, B)))
|
IMPORT ASSIGN VAR VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR BIN_OP LIST STRING VAR FOR VAR FUNC_CALL VAR BIN_OP VAR NUMBER ASSIGN VAR BIN_OP BIN_OP NUMBER VAR NUMBER VAR VAR FOR VAR FUNC_CALL VAR BIN_OP VAR NUMBER VAR ASSIGN VAR BIN_OP NUMBER BIN_OP VAR BIN_OP VAR NUMBER VAR VAR ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR BIN_OP VAR NUMBER IF BIN_OP BIN_OP NUMBER VAR NUMBER VAR IF VAR BIN_OP BIN_OP NUMBER VAR NUMBER VAR BIN_OP NUMBER VAR VAR BIN_OP BIN_OP NUMBER VAR NUMBER VAR BIN_OP BIN_OP NUMBER VAR NUMBER VAR NUMBER EXPR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR FUNC_CALL STRING FUNC_CALL VAR VAR VAR
|
This is the hard version of the problem. The difference between the versions is that in the easy version all prices $a_i$ are different. You can make hacks if and only if you solved both versions of the problem.
Today is Sage's birthday, and she will go shopping to buy ice spheres. All $n$ ice spheres are placed in a row and they are numbered from $1$ to $n$ from left to right. Each ice sphere has a positive integer price. In this version, some prices can be equal.
An ice sphere is cheap if it costs strictly less than two neighboring ice spheres: the nearest to the left and the nearest to the right. The leftmost and the rightmost ice spheres are not cheap. Sage will choose all cheap ice spheres and then buy only them.
You can visit the shop before Sage and reorder the ice spheres as you wish. Find out the maximum number of ice spheres that Sage can buy, and show how the ice spheres should be reordered.
-----Input-----
The first line contains a single integer $n$ $(1 \le n \le 10^5)$ — the number of ice spheres in the shop.
The second line contains $n$ integers $a_1, a_2, \dots, a_n$ $(1 \le a_i \le 10^9)$ — the prices of ice spheres.
-----Output-----
In the first line print the maximum number of ice spheres that Sage can buy.
In the second line print the prices of ice spheres in the optimal order. If there are several correct answers, you can print any of them.
-----Example-----
Input
7
1 3 2 2 4 5 4
Output
3
3 1 4 2 4 2 5
-----Note-----
In the sample it's not possible to place the ice spheres in any order so that Sage would buy $4$ of them. If the spheres are placed in the order $(3, 1, 4, 2, 4, 2, 5)$, then Sage will buy one sphere for $1$ and two spheres for $2$ each.
|
n = int(input())
a = list(map(int, input().split()))
a.sort()
cur = 0
i = 1
res = [0] * n
while i < n:
res[i] = a[cur]
i += 2
cur += 1
i = 0
while i < n:
res[i] = a[cur]
i += 2
cur += 1
ans = 0
for i in range(1, n - 1):
if res[i - 1] > res[i] and res[i] < res[i + 1]:
ans += 1
print(ans)
print(*res)
|
ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR EXPR FUNC_CALL VAR ASSIGN VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR BIN_OP LIST NUMBER VAR WHILE VAR VAR ASSIGN VAR VAR VAR VAR VAR NUMBER VAR NUMBER ASSIGN VAR NUMBER WHILE VAR VAR ASSIGN VAR VAR VAR VAR VAR NUMBER VAR NUMBER ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR NUMBER BIN_OP VAR NUMBER IF VAR BIN_OP VAR NUMBER VAR VAR VAR VAR VAR BIN_OP VAR NUMBER VAR NUMBER EXPR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR VAR
|
This is the hard version of the problem. The difference between the versions is that in the easy version all prices $a_i$ are different. You can make hacks if and only if you solved both versions of the problem.
Today is Sage's birthday, and she will go shopping to buy ice spheres. All $n$ ice spheres are placed in a row and they are numbered from $1$ to $n$ from left to right. Each ice sphere has a positive integer price. In this version, some prices can be equal.
An ice sphere is cheap if it costs strictly less than two neighboring ice spheres: the nearest to the left and the nearest to the right. The leftmost and the rightmost ice spheres are not cheap. Sage will choose all cheap ice spheres and then buy only them.
You can visit the shop before Sage and reorder the ice spheres as you wish. Find out the maximum number of ice spheres that Sage can buy, and show how the ice spheres should be reordered.
-----Input-----
The first line contains a single integer $n$ $(1 \le n \le 10^5)$ — the number of ice spheres in the shop.
The second line contains $n$ integers $a_1, a_2, \dots, a_n$ $(1 \le a_i \le 10^9)$ — the prices of ice spheres.
-----Output-----
In the first line print the maximum number of ice spheres that Sage can buy.
In the second line print the prices of ice spheres in the optimal order. If there are several correct answers, you can print any of them.
-----Example-----
Input
7
1 3 2 2 4 5 4
Output
3
3 1 4 2 4 2 5
-----Note-----
In the sample it's not possible to place the ice spheres in any order so that Sage would buy $4$ of them. If the spheres are placed in the order $(3, 1, 4, 2, 4, 2, 5)$, then Sage will buy one sphere for $1$ and two spheres for $2$ each.
|
import sys
n = int(input())
spheres = list(map(int, input().split()))
spheres.sort()
small = spheres[: n // 2]
big = spheres[n // 2 :]
res = []
for i in range(n):
if i % 2 == 0:
res.append(big[i // 2])
else:
res.append(small[i // 2])
c = 0
for i in range(n):
if i == 0 or i == n - 1:
continue
if res[i - 1] > res[i] < res[i + 1]:
c += 1
print(c)
print(*res)
|
IMPORT ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR EXPR FUNC_CALL VAR ASSIGN VAR VAR BIN_OP VAR NUMBER ASSIGN VAR VAR BIN_OP VAR NUMBER ASSIGN VAR LIST FOR VAR FUNC_CALL VAR VAR IF BIN_OP VAR NUMBER NUMBER EXPR FUNC_CALL VAR VAR BIN_OP VAR NUMBER EXPR FUNC_CALL VAR VAR BIN_OP VAR NUMBER ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR VAR IF VAR NUMBER VAR BIN_OP VAR NUMBER IF VAR BIN_OP VAR NUMBER VAR VAR VAR BIN_OP VAR NUMBER VAR NUMBER EXPR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR VAR
|
This is the hard version of the problem. The difference between the versions is that in the easy version all prices $a_i$ are different. You can make hacks if and only if you solved both versions of the problem.
Today is Sage's birthday, and she will go shopping to buy ice spheres. All $n$ ice spheres are placed in a row and they are numbered from $1$ to $n$ from left to right. Each ice sphere has a positive integer price. In this version, some prices can be equal.
An ice sphere is cheap if it costs strictly less than two neighboring ice spheres: the nearest to the left and the nearest to the right. The leftmost and the rightmost ice spheres are not cheap. Sage will choose all cheap ice spheres and then buy only them.
You can visit the shop before Sage and reorder the ice spheres as you wish. Find out the maximum number of ice spheres that Sage can buy, and show how the ice spheres should be reordered.
-----Input-----
The first line contains a single integer $n$ $(1 \le n \le 10^5)$ — the number of ice spheres in the shop.
The second line contains $n$ integers $a_1, a_2, \dots, a_n$ $(1 \le a_i \le 10^9)$ — the prices of ice spheres.
-----Output-----
In the first line print the maximum number of ice spheres that Sage can buy.
In the second line print the prices of ice spheres in the optimal order. If there are several correct answers, you can print any of them.
-----Example-----
Input
7
1 3 2 2 4 5 4
Output
3
3 1 4 2 4 2 5
-----Note-----
In the sample it's not possible to place the ice spheres in any order so that Sage would buy $4$ of them. If the spheres are placed in the order $(3, 1, 4, 2, 4, 2, 5)$, then Sage will buy one sphere for $1$ and two spheres for $2$ each.
|
from sys import *
input = stdin.readline
listInput = lambda: list(map(int, input().strip().split()))
lineInput = lambda: map(int, input().strip().split())
sJoin = lambda a, sep: f"{sep}".join(a)
arrJoin = lambda a, sep: f"{sep}".join(map(str, a))
def main():
n = int(input())
arr = sorted(listInput())
i, j = 0, n // 2
ans = []
while i < n // 2:
ans.append(arr[j])
ans.append(arr[i])
i += 1
j += 1
if n % 2 == 1:
ans.append(arr[j])
c = 0
j = 1
while j < n - 1:
if ans[j - 1] > ans[j] < ans[j + 1]:
c += 1
j += 2
print(c)
print(*ans)
main()
|
ASSIGN VAR VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR VAR VAR FUNC_DEF ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR VAR NUMBER BIN_OP VAR NUMBER ASSIGN VAR LIST WHILE VAR BIN_OP VAR NUMBER EXPR FUNC_CALL VAR VAR VAR EXPR FUNC_CALL VAR VAR VAR VAR NUMBER VAR NUMBER IF BIN_OP VAR NUMBER NUMBER EXPR FUNC_CALL VAR VAR VAR ASSIGN VAR NUMBER ASSIGN VAR NUMBER WHILE VAR BIN_OP VAR NUMBER IF VAR BIN_OP VAR NUMBER VAR VAR VAR BIN_OP VAR NUMBER VAR NUMBER VAR NUMBER EXPR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR
|
This is the hard version of the problem. The difference between the versions is that in the easy version all prices $a_i$ are different. You can make hacks if and only if you solved both versions of the problem.
Today is Sage's birthday, and she will go shopping to buy ice spheres. All $n$ ice spheres are placed in a row and they are numbered from $1$ to $n$ from left to right. Each ice sphere has a positive integer price. In this version, some prices can be equal.
An ice sphere is cheap if it costs strictly less than two neighboring ice spheres: the nearest to the left and the nearest to the right. The leftmost and the rightmost ice spheres are not cheap. Sage will choose all cheap ice spheres and then buy only them.
You can visit the shop before Sage and reorder the ice spheres as you wish. Find out the maximum number of ice spheres that Sage can buy, and show how the ice spheres should be reordered.
-----Input-----
The first line contains a single integer $n$ $(1 \le n \le 10^5)$ — the number of ice spheres in the shop.
The second line contains $n$ integers $a_1, a_2, \dots, a_n$ $(1 \le a_i \le 10^9)$ — the prices of ice spheres.
-----Output-----
In the first line print the maximum number of ice spheres that Sage can buy.
In the second line print the prices of ice spheres in the optimal order. If there are several correct answers, you can print any of them.
-----Example-----
Input
7
1 3 2 2 4 5 4
Output
3
3 1 4 2 4 2 5
-----Note-----
In the sample it's not possible to place the ice spheres in any order so that Sage would buy $4$ of them. If the spheres are placed in the order $(3, 1, 4, 2, 4, 2, 5)$, then Sage will buy one sphere for $1$ and two spheres for $2$ each.
|
def solve():
n = int(input().strip())
a = list(map(int, input().strip().split()))
a.sort()
a1 = a[: n // 2]
a2 = a[n // 2 :]
flag = True
cursor1 = 0
cursor2 = 0
prevPrice = -100
ans = 0
for i in range(len(a2)):
if i + 1 < len(a2):
ans += 1 if a2[i] > a1[i] and a1[i] < a2[i + 1] else 0
print(ans)
for i in range(n):
if not flag:
print(a1[cursor1], end=" ")
cursor1 += 1
else:
print(a2[cursor2], end=" ")
cursor2 += 1
flag = not flag
t = 1
for _ in range(t):
solve()
|
FUNC_DEF ASSIGN VAR FUNC_CALL VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL FUNC_CALL VAR EXPR FUNC_CALL VAR ASSIGN VAR VAR BIN_OP VAR NUMBER ASSIGN VAR VAR BIN_OP VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR FUNC_CALL VAR VAR IF BIN_OP VAR NUMBER FUNC_CALL VAR VAR VAR VAR VAR VAR VAR VAR VAR VAR BIN_OP VAR NUMBER NUMBER NUMBER EXPR FUNC_CALL VAR VAR FOR VAR FUNC_CALL VAR VAR IF VAR EXPR FUNC_CALL VAR VAR VAR STRING VAR NUMBER EXPR FUNC_CALL VAR VAR VAR STRING VAR NUMBER ASSIGN VAR VAR ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR
|
This is the hard version of the problem. The difference between the versions is that in the easy version all prices $a_i$ are different. You can make hacks if and only if you solved both versions of the problem.
Today is Sage's birthday, and she will go shopping to buy ice spheres. All $n$ ice spheres are placed in a row and they are numbered from $1$ to $n$ from left to right. Each ice sphere has a positive integer price. In this version, some prices can be equal.
An ice sphere is cheap if it costs strictly less than two neighboring ice spheres: the nearest to the left and the nearest to the right. The leftmost and the rightmost ice spheres are not cheap. Sage will choose all cheap ice spheres and then buy only them.
You can visit the shop before Sage and reorder the ice spheres as you wish. Find out the maximum number of ice spheres that Sage can buy, and show how the ice spheres should be reordered.
-----Input-----
The first line contains a single integer $n$ $(1 \le n \le 10^5)$ — the number of ice spheres in the shop.
The second line contains $n$ integers $a_1, a_2, \dots, a_n$ $(1 \le a_i \le 10^9)$ — the prices of ice spheres.
-----Output-----
In the first line print the maximum number of ice spheres that Sage can buy.
In the second line print the prices of ice spheres in the optimal order. If there are several correct answers, you can print any of them.
-----Example-----
Input
7
1 3 2 2 4 5 4
Output
3
3 1 4 2 4 2 5
-----Note-----
In the sample it's not possible to place the ice spheres in any order so that Sage would buy $4$ of them. If the spheres are placed in the order $(3, 1, 4, 2, 4, 2, 5)$, then Sage will buy one sphere for $1$ and two spheres for $2$ each.
|
n = int(input())
mas = list(map(int, input().split()))
mas = sorted(mas)
middle = n // 2
masans = []
ans = 0
if n > 1:
for i in range(n // 2):
if mas[i] < mas[i + middle]:
ans += 1
masans.append(mas[i + middle])
masans.append(mas[i])
if n % 2 != 0:
masans.append(mas[n - 1])
if n % 2 == 0:
if masans[-1] < masans[-2]:
ans -= 1
else:
ans = 0
masans = [mas[0]]
print(ans)
for x in masans:
print(x, end=" ")
|
ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR BIN_OP VAR NUMBER ASSIGN VAR LIST ASSIGN VAR NUMBER IF VAR NUMBER FOR VAR FUNC_CALL VAR BIN_OP VAR NUMBER IF VAR VAR VAR BIN_OP VAR VAR VAR NUMBER EXPR FUNC_CALL VAR VAR BIN_OP VAR VAR EXPR FUNC_CALL VAR VAR VAR IF BIN_OP VAR NUMBER NUMBER EXPR FUNC_CALL VAR VAR BIN_OP VAR NUMBER IF BIN_OP VAR NUMBER NUMBER IF VAR NUMBER VAR NUMBER VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR LIST VAR NUMBER EXPR FUNC_CALL VAR VAR FOR VAR VAR EXPR FUNC_CALL VAR VAR STRING
|
This is the hard version of the problem. The difference between the versions is that in the easy version all prices $a_i$ are different. You can make hacks if and only if you solved both versions of the problem.
Today is Sage's birthday, and she will go shopping to buy ice spheres. All $n$ ice spheres are placed in a row and they are numbered from $1$ to $n$ from left to right. Each ice sphere has a positive integer price. In this version, some prices can be equal.
An ice sphere is cheap if it costs strictly less than two neighboring ice spheres: the nearest to the left and the nearest to the right. The leftmost and the rightmost ice spheres are not cheap. Sage will choose all cheap ice spheres and then buy only them.
You can visit the shop before Sage and reorder the ice spheres as you wish. Find out the maximum number of ice spheres that Sage can buy, and show how the ice spheres should be reordered.
-----Input-----
The first line contains a single integer $n$ $(1 \le n \le 10^5)$ — the number of ice spheres in the shop.
The second line contains $n$ integers $a_1, a_2, \dots, a_n$ $(1 \le a_i \le 10^9)$ — the prices of ice spheres.
-----Output-----
In the first line print the maximum number of ice spheres that Sage can buy.
In the second line print the prices of ice spheres in the optimal order. If there are several correct answers, you can print any of them.
-----Example-----
Input
7
1 3 2 2 4 5 4
Output
3
3 1 4 2 4 2 5
-----Note-----
In the sample it's not possible to place the ice spheres in any order so that Sage would buy $4$ of them. If the spheres are placed in the order $(3, 1, 4, 2, 4, 2, 5)$, then Sage will buy one sphere for $1$ and two spheres for $2$ each.
|
def do(l, n):
o = l[0 : n // 2]
k = l[n // 2 : n]
nlist = []
for i in range(n // 2):
nlist.append(str(k[i]))
nlist.append(str(o[i]))
if n % 2 == 1:
nlist.append(str(k[n // 2]))
return nlist
num = int(input())
l = list(map(int, input().split()))
l.sort()
a = do(l, num)
cnt = 0
for i in range(1, num - 1):
if int(a[i - 1]) > int(a[i]) and int(a[i + 1]) > int(a[i]):
cnt += 1
print(cnt)
print(" ".join(a))
|
FUNC_DEF ASSIGN VAR VAR NUMBER BIN_OP VAR NUMBER ASSIGN VAR VAR BIN_OP VAR NUMBER VAR ASSIGN VAR LIST FOR VAR FUNC_CALL VAR BIN_OP VAR NUMBER EXPR FUNC_CALL VAR FUNC_CALL VAR VAR VAR EXPR FUNC_CALL VAR FUNC_CALL VAR VAR VAR IF BIN_OP VAR NUMBER NUMBER EXPR FUNC_CALL VAR FUNC_CALL VAR VAR BIN_OP VAR NUMBER RETURN VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR EXPR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR VAR VAR ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR NUMBER BIN_OP VAR NUMBER IF FUNC_CALL VAR VAR BIN_OP VAR NUMBER FUNC_CALL VAR VAR VAR FUNC_CALL VAR VAR BIN_OP VAR NUMBER FUNC_CALL VAR VAR VAR VAR NUMBER EXPR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR FUNC_CALL STRING VAR
|
This is the hard version of the problem. The difference between the versions is that in the easy version all prices $a_i$ are different. You can make hacks if and only if you solved both versions of the problem.
Today is Sage's birthday, and she will go shopping to buy ice spheres. All $n$ ice spheres are placed in a row and they are numbered from $1$ to $n$ from left to right. Each ice sphere has a positive integer price. In this version, some prices can be equal.
An ice sphere is cheap if it costs strictly less than two neighboring ice spheres: the nearest to the left and the nearest to the right. The leftmost and the rightmost ice spheres are not cheap. Sage will choose all cheap ice spheres and then buy only them.
You can visit the shop before Sage and reorder the ice spheres as you wish. Find out the maximum number of ice spheres that Sage can buy, and show how the ice spheres should be reordered.
-----Input-----
The first line contains a single integer $n$ $(1 \le n \le 10^5)$ — the number of ice spheres in the shop.
The second line contains $n$ integers $a_1, a_2, \dots, a_n$ $(1 \le a_i \le 10^9)$ — the prices of ice spheres.
-----Output-----
In the first line print the maximum number of ice spheres that Sage can buy.
In the second line print the prices of ice spheres in the optimal order. If there are several correct answers, you can print any of them.
-----Example-----
Input
7
1 3 2 2 4 5 4
Output
3
3 1 4 2 4 2 5
-----Note-----
In the sample it's not possible to place the ice spheres in any order so that Sage would buy $4$ of them. If the spheres are placed in the order $(3, 1, 4, 2, 4, 2, 5)$, then Sage will buy one sphere for $1$ and two spheres for $2$ each.
|
n = int(input())
a = [int(inp) for inp in input().split()]
a.sort()
i = 0
j = len(a) // 2
b = []
sw = 1
while j < len(a) or i < len(a) // 2:
if sw:
b.append(a[j])
j += 1
else:
b.append(a[i])
i += 1
sw = 1 - sw
res = 0
i = 1
while i < len(b) - 1:
if b[i] < b[i - 1] and b[i] < b[i + 1]:
res += 1
i += 1
i += 1
print(res)
print(*b, sep=" ")
|
ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR VAR VAR FUNC_CALL FUNC_CALL VAR EXPR FUNC_CALL VAR ASSIGN VAR NUMBER ASSIGN VAR BIN_OP FUNC_CALL VAR VAR NUMBER ASSIGN VAR LIST ASSIGN VAR NUMBER WHILE VAR FUNC_CALL VAR VAR VAR BIN_OP FUNC_CALL VAR VAR NUMBER IF VAR EXPR FUNC_CALL VAR VAR VAR VAR NUMBER EXPR FUNC_CALL VAR VAR VAR VAR NUMBER ASSIGN VAR BIN_OP NUMBER VAR ASSIGN VAR NUMBER ASSIGN VAR NUMBER WHILE VAR BIN_OP FUNC_CALL VAR VAR NUMBER IF VAR VAR VAR BIN_OP VAR NUMBER VAR VAR VAR BIN_OP VAR NUMBER VAR NUMBER VAR NUMBER VAR NUMBER EXPR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR VAR STRING
|
This is the hard version of the problem. The difference between the versions is that in the easy version all prices $a_i$ are different. You can make hacks if and only if you solved both versions of the problem.
Today is Sage's birthday, and she will go shopping to buy ice spheres. All $n$ ice spheres are placed in a row and they are numbered from $1$ to $n$ from left to right. Each ice sphere has a positive integer price. In this version, some prices can be equal.
An ice sphere is cheap if it costs strictly less than two neighboring ice spheres: the nearest to the left and the nearest to the right. The leftmost and the rightmost ice spheres are not cheap. Sage will choose all cheap ice spheres and then buy only them.
You can visit the shop before Sage and reorder the ice spheres as you wish. Find out the maximum number of ice spheres that Sage can buy, and show how the ice spheres should be reordered.
-----Input-----
The first line contains a single integer $n$ $(1 \le n \le 10^5)$ — the number of ice spheres in the shop.
The second line contains $n$ integers $a_1, a_2, \dots, a_n$ $(1 \le a_i \le 10^9)$ — the prices of ice spheres.
-----Output-----
In the first line print the maximum number of ice spheres that Sage can buy.
In the second line print the prices of ice spheres in the optimal order. If there are several correct answers, you can print any of them.
-----Example-----
Input
7
1 3 2 2 4 5 4
Output
3
3 1 4 2 4 2 5
-----Note-----
In the sample it's not possible to place the ice spheres in any order so that Sage would buy $4$ of them. If the spheres are placed in the order $(3, 1, 4, 2, 4, 2, 5)$, then Sage will buy one sphere for $1$ and two spheres for $2$ each.
|
n = int(input())
a = sorted(map(int, input().split()))
m = (n - 1) // 2
i, j = 0, m + 1
ans = 0
temp = a[m]
if n < 3:
print(ans)
for i in a:
print(i, end=" ")
else:
z = n * [0]
z[::2] = a[n // 2 :]
z[1::2] = a[: n // 2]
for i in range(m):
if z[2 * i + 1] < z[2 * i] and z[2 * i + 1] < z[2 * i + 2]:
ans += 1
print(ans)
for i in z:
print(i, end=" ")
|
ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR BIN_OP BIN_OP VAR NUMBER NUMBER ASSIGN VAR VAR NUMBER BIN_OP VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR VAR VAR IF VAR NUMBER EXPR FUNC_CALL VAR VAR FOR VAR VAR EXPR FUNC_CALL VAR VAR STRING ASSIGN VAR BIN_OP VAR LIST NUMBER ASSIGN VAR NUMBER VAR BIN_OP VAR NUMBER ASSIGN VAR NUMBER NUMBER VAR BIN_OP VAR NUMBER FOR VAR FUNC_CALL VAR VAR IF VAR BIN_OP BIN_OP NUMBER VAR NUMBER VAR BIN_OP NUMBER VAR VAR BIN_OP BIN_OP NUMBER VAR NUMBER VAR BIN_OP BIN_OP NUMBER VAR NUMBER VAR NUMBER EXPR FUNC_CALL VAR VAR FOR VAR VAR EXPR FUNC_CALL VAR VAR STRING
|
This is the hard version of the problem. The difference between the versions is that in the easy version all prices $a_i$ are different. You can make hacks if and only if you solved both versions of the problem.
Today is Sage's birthday, and she will go shopping to buy ice spheres. All $n$ ice spheres are placed in a row and they are numbered from $1$ to $n$ from left to right. Each ice sphere has a positive integer price. In this version, some prices can be equal.
An ice sphere is cheap if it costs strictly less than two neighboring ice spheres: the nearest to the left and the nearest to the right. The leftmost and the rightmost ice spheres are not cheap. Sage will choose all cheap ice spheres and then buy only them.
You can visit the shop before Sage and reorder the ice spheres as you wish. Find out the maximum number of ice spheres that Sage can buy, and show how the ice spheres should be reordered.
-----Input-----
The first line contains a single integer $n$ $(1 \le n \le 10^5)$ — the number of ice spheres in the shop.
The second line contains $n$ integers $a_1, a_2, \dots, a_n$ $(1 \le a_i \le 10^9)$ — the prices of ice spheres.
-----Output-----
In the first line print the maximum number of ice spheres that Sage can buy.
In the second line print the prices of ice spheres in the optimal order. If there are several correct answers, you can print any of them.
-----Example-----
Input
7
1 3 2 2 4 5 4
Output
3
3 1 4 2 4 2 5
-----Note-----
In the sample it's not possible to place the ice spheres in any order so that Sage would buy $4$ of them. If the spheres are placed in the order $(3, 1, 4, 2, 4, 2, 5)$, then Sage will buy one sphere for $1$ and two spheres for $2$ each.
|
n = int(input())
a = [int(i) for i in input().split()]
a.sort()
ans = []
p = 0
if n % 2:
for i in range(n // 2, n):
ans.append(a[i])
if i < n - 1:
ans.append(a[p])
p += 1
else:
for i in range(n // 2, n):
ans.append(a[i])
ans.append(a[p])
p += 1
cnt = 0
for i in range(1, n - 1):
if ans[i] < ans[i - 1] and ans[i] < ans[i + 1]:
cnt += 1
print(cnt)
print(*ans)
|
ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR VAR VAR FUNC_CALL FUNC_CALL VAR EXPR FUNC_CALL VAR ASSIGN VAR LIST ASSIGN VAR NUMBER IF BIN_OP VAR NUMBER FOR VAR FUNC_CALL VAR BIN_OP VAR NUMBER VAR EXPR FUNC_CALL VAR VAR VAR IF VAR BIN_OP VAR NUMBER EXPR FUNC_CALL VAR VAR VAR VAR NUMBER FOR VAR FUNC_CALL VAR BIN_OP VAR NUMBER VAR EXPR FUNC_CALL VAR VAR VAR EXPR FUNC_CALL VAR VAR VAR VAR NUMBER ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR NUMBER BIN_OP VAR NUMBER IF VAR VAR VAR BIN_OP VAR NUMBER VAR VAR VAR BIN_OP VAR NUMBER VAR NUMBER EXPR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR VAR
|
This is the hard version of the problem. The difference between the versions is that in the easy version all prices $a_i$ are different. You can make hacks if and only if you solved both versions of the problem.
Today is Sage's birthday, and she will go shopping to buy ice spheres. All $n$ ice spheres are placed in a row and they are numbered from $1$ to $n$ from left to right. Each ice sphere has a positive integer price. In this version, some prices can be equal.
An ice sphere is cheap if it costs strictly less than two neighboring ice spheres: the nearest to the left and the nearest to the right. The leftmost and the rightmost ice spheres are not cheap. Sage will choose all cheap ice spheres and then buy only them.
You can visit the shop before Sage and reorder the ice spheres as you wish. Find out the maximum number of ice spheres that Sage can buy, and show how the ice spheres should be reordered.
-----Input-----
The first line contains a single integer $n$ $(1 \le n \le 10^5)$ — the number of ice spheres in the shop.
The second line contains $n$ integers $a_1, a_2, \dots, a_n$ $(1 \le a_i \le 10^9)$ — the prices of ice spheres.
-----Output-----
In the first line print the maximum number of ice spheres that Sage can buy.
In the second line print the prices of ice spheres in the optimal order. If there are several correct answers, you can print any of them.
-----Example-----
Input
7
1 3 2 2 4 5 4
Output
3
3 1 4 2 4 2 5
-----Note-----
In the sample it's not possible to place the ice spheres in any order so that Sage would buy $4$ of them. If the spheres are placed in the order $(3, 1, 4, 2, 4, 2, 5)$, then Sage will buy one sphere for $1$ and two spheres for $2$ each.
|
n = int(input())
l = sorted(list(map(int, input().split())))
x = []
if n <= 2:
print(0)
print(*l)
elif n % 2 == 0:
a = l[: n // 2]
b = l[n // 2 :]
ans = 0
for i in range(n // 2):
x.append(b[i])
x.append(a[i])
for j in range(1, n - 1, 2):
if x[j] < x[j + 1] and x[j] < x[j - 1]:
ans += 1
print(ans)
print(*x)
else:
a = l[: n // 2]
b = l[n // 2 :]
ans = 0
for i in range(n // 2):
x.append(b[i])
x.append(a[i])
x.append(b[-1])
for j in range(1, n, 2):
if x[j] < x[j + 1] and x[j] < x[j - 1]:
ans += 1
print(ans)
print(*x)
|
ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR LIST IF VAR NUMBER EXPR FUNC_CALL VAR NUMBER EXPR FUNC_CALL VAR VAR IF BIN_OP VAR NUMBER NUMBER ASSIGN VAR VAR BIN_OP VAR NUMBER ASSIGN VAR VAR BIN_OP VAR NUMBER ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR BIN_OP VAR NUMBER EXPR FUNC_CALL VAR VAR VAR EXPR FUNC_CALL VAR VAR VAR FOR VAR FUNC_CALL VAR NUMBER BIN_OP VAR NUMBER NUMBER IF VAR VAR VAR BIN_OP VAR NUMBER VAR VAR VAR BIN_OP VAR NUMBER VAR NUMBER EXPR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR VAR ASSIGN VAR VAR BIN_OP VAR NUMBER ASSIGN VAR VAR BIN_OP VAR NUMBER ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR BIN_OP VAR NUMBER EXPR FUNC_CALL VAR VAR VAR EXPR FUNC_CALL VAR VAR VAR EXPR FUNC_CALL VAR VAR NUMBER FOR VAR FUNC_CALL VAR NUMBER VAR NUMBER IF VAR VAR VAR BIN_OP VAR NUMBER VAR VAR VAR BIN_OP VAR NUMBER VAR NUMBER EXPR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR VAR
|
This is the hard version of the problem. The difference between the versions is that in the easy version all prices $a_i$ are different. You can make hacks if and only if you solved both versions of the problem.
Today is Sage's birthday, and she will go shopping to buy ice spheres. All $n$ ice spheres are placed in a row and they are numbered from $1$ to $n$ from left to right. Each ice sphere has a positive integer price. In this version, some prices can be equal.
An ice sphere is cheap if it costs strictly less than two neighboring ice spheres: the nearest to the left and the nearest to the right. The leftmost and the rightmost ice spheres are not cheap. Sage will choose all cheap ice spheres and then buy only them.
You can visit the shop before Sage and reorder the ice spheres as you wish. Find out the maximum number of ice spheres that Sage can buy, and show how the ice spheres should be reordered.
-----Input-----
The first line contains a single integer $n$ $(1 \le n \le 10^5)$ — the number of ice spheres in the shop.
The second line contains $n$ integers $a_1, a_2, \dots, a_n$ $(1 \le a_i \le 10^9)$ — the prices of ice spheres.
-----Output-----
In the first line print the maximum number of ice spheres that Sage can buy.
In the second line print the prices of ice spheres in the optimal order. If there are several correct answers, you can print any of them.
-----Example-----
Input
7
1 3 2 2 4 5 4
Output
3
3 1 4 2 4 2 5
-----Note-----
In the sample it's not possible to place the ice spheres in any order so that Sage would buy $4$ of them. If the spheres are placed in the order $(3, 1, 4, 2, 4, 2, 5)$, then Sage will buy one sphere for $1$ and two spheres for $2$ each.
|
import sys
n = int(sys.stdin.readline())
pricelist = [int(v) for v in sys.stdin.readline().strip().split(" ")]
pricelist.sort(reverse=True)
newlist = [(0) for i in range(0, n)]
start = int((n + 1) / 2)
for i in range(0, start):
newlist[i * 2] = pricelist[i]
waitlist = []
pt = 1
total = 0
for i in range(start, n):
if pt < n - 1 and pricelist[i] < newlist[pt - 1] and pricelist[i] < newlist[pt + 1]:
newlist[pt] = pricelist[i]
total += 1
pt += 2
else:
waitlist.append(pricelist[i])
for v in waitlist:
newlist[pt] = v
pt += 2
print(total)
print(" ".join(str(v) for v in newlist))
|
IMPORT ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR VAR VAR FUNC_CALL FUNC_CALL FUNC_CALL VAR STRING EXPR FUNC_CALL VAR NUMBER ASSIGN VAR NUMBER VAR FUNC_CALL VAR NUMBER VAR ASSIGN VAR FUNC_CALL VAR BIN_OP BIN_OP VAR NUMBER NUMBER FOR VAR FUNC_CALL VAR NUMBER VAR ASSIGN VAR BIN_OP VAR NUMBER VAR VAR ASSIGN VAR LIST ASSIGN VAR NUMBER ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR VAR VAR IF VAR BIN_OP VAR NUMBER VAR VAR VAR BIN_OP VAR NUMBER VAR VAR VAR BIN_OP VAR NUMBER ASSIGN VAR VAR VAR VAR VAR NUMBER VAR NUMBER EXPR FUNC_CALL VAR VAR VAR FOR VAR VAR ASSIGN VAR VAR VAR VAR NUMBER EXPR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR FUNC_CALL STRING FUNC_CALL VAR VAR VAR VAR
|
This is the hard version of the problem. The difference between the versions is that in the easy version all prices $a_i$ are different. You can make hacks if and only if you solved both versions of the problem.
Today is Sage's birthday, and she will go shopping to buy ice spheres. All $n$ ice spheres are placed in a row and they are numbered from $1$ to $n$ from left to right. Each ice sphere has a positive integer price. In this version, some prices can be equal.
An ice sphere is cheap if it costs strictly less than two neighboring ice spheres: the nearest to the left and the nearest to the right. The leftmost and the rightmost ice spheres are not cheap. Sage will choose all cheap ice spheres and then buy only them.
You can visit the shop before Sage and reorder the ice spheres as you wish. Find out the maximum number of ice spheres that Sage can buy, and show how the ice spheres should be reordered.
-----Input-----
The first line contains a single integer $n$ $(1 \le n \le 10^5)$ — the number of ice spheres in the shop.
The second line contains $n$ integers $a_1, a_2, \dots, a_n$ $(1 \le a_i \le 10^9)$ — the prices of ice spheres.
-----Output-----
In the first line print the maximum number of ice spheres that Sage can buy.
In the second line print the prices of ice spheres in the optimal order. If there are several correct answers, you can print any of them.
-----Example-----
Input
7
1 3 2 2 4 5 4
Output
3
3 1 4 2 4 2 5
-----Note-----
In the sample it's not possible to place the ice spheres in any order so that Sage would buy $4$ of them. If the spheres are placed in the order $(3, 1, 4, 2, 4, 2, 5)$, then Sage will buy one sphere for $1$ and two spheres for $2$ each.
|
import sys
input = sys.stdin.readline
n = int(input())
a = [int(i) for i in input().split() if i != "\n"]
a.sort()
ans = [0] * n
j = 0
for i in range(1, n, 2):
ans[i] = a[j]
j += 1
for i in range(n):
if ans[i] == 0:
ans[i] = a[j]
j += 1
count = 0
for i in range(1, n - 1):
if ans[i - 1] > ans[i] and ans[i + 1] > ans[i]:
count += 1
print(count)
print(*ans)
|
IMPORT ASSIGN VAR VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR VAR VAR FUNC_CALL FUNC_CALL VAR VAR STRING EXPR FUNC_CALL VAR ASSIGN VAR BIN_OP LIST NUMBER VAR ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR NUMBER VAR NUMBER ASSIGN VAR VAR VAR VAR VAR NUMBER FOR VAR FUNC_CALL VAR VAR IF VAR VAR NUMBER ASSIGN VAR VAR VAR VAR VAR NUMBER ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR NUMBER BIN_OP VAR NUMBER IF VAR BIN_OP VAR NUMBER VAR VAR VAR BIN_OP VAR NUMBER VAR VAR VAR NUMBER EXPR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR VAR
|
This is the hard version of the problem. The difference between the versions is that in the easy version all prices $a_i$ are different. You can make hacks if and only if you solved both versions of the problem.
Today is Sage's birthday, and she will go shopping to buy ice spheres. All $n$ ice spheres are placed in a row and they are numbered from $1$ to $n$ from left to right. Each ice sphere has a positive integer price. In this version, some prices can be equal.
An ice sphere is cheap if it costs strictly less than two neighboring ice spheres: the nearest to the left and the nearest to the right. The leftmost and the rightmost ice spheres are not cheap. Sage will choose all cheap ice spheres and then buy only them.
You can visit the shop before Sage and reorder the ice spheres as you wish. Find out the maximum number of ice spheres that Sage can buy, and show how the ice spheres should be reordered.
-----Input-----
The first line contains a single integer $n$ $(1 \le n \le 10^5)$ — the number of ice spheres in the shop.
The second line contains $n$ integers $a_1, a_2, \dots, a_n$ $(1 \le a_i \le 10^9)$ — the prices of ice spheres.
-----Output-----
In the first line print the maximum number of ice spheres that Sage can buy.
In the second line print the prices of ice spheres in the optimal order. If there are several correct answers, you can print any of them.
-----Example-----
Input
7
1 3 2 2 4 5 4
Output
3
3 1 4 2 4 2 5
-----Note-----
In the sample it's not possible to place the ice spheres in any order so that Sage would buy $4$ of them. If the spheres are placed in the order $(3, 1, 4, 2, 4, 2, 5)$, then Sage will buy one sphere for $1$ and two spheres for $2$ each.
|
n = int(input())
esferas = list(map(int, input().split(" ")))
esferas.sort()
limite = int(n / 2)
menores = esferas[:limite]
maiores = esferas[limite:]
indice = 0
reordenadas = []
compras = 0
esquerda = False
for i in range(n):
if i % 2 == 0:
reordenadas.append(maiores[int(i / 2)])
if esquerda and maiores[int(i / 2)] > reordenadas[i - 1]:
compras += 1
esquerda = False
else:
reordenadas.append(menores[indice])
if menores[indice] < reordenadas[i - 1]:
esquerda = True
else:
esquerda = False
indice += 1
print(compras)
reordenadas = [str(int) for int in reordenadas]
print(" ".join(reordenadas))
|
ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR STRING EXPR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR BIN_OP VAR NUMBER ASSIGN VAR VAR VAR ASSIGN VAR VAR VAR ASSIGN VAR NUMBER ASSIGN VAR LIST ASSIGN VAR NUMBER ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR VAR IF BIN_OP VAR NUMBER NUMBER EXPR FUNC_CALL VAR VAR FUNC_CALL VAR BIN_OP VAR NUMBER IF VAR VAR FUNC_CALL VAR BIN_OP VAR NUMBER VAR BIN_OP VAR NUMBER VAR NUMBER ASSIGN VAR NUMBER EXPR FUNC_CALL VAR VAR VAR IF VAR VAR VAR BIN_OP VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR NUMBER VAR NUMBER EXPR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR VAR VAR VAR EXPR FUNC_CALL VAR FUNC_CALL STRING VAR
|
This is the hard version of the problem. The difference between the versions is that in the easy version all prices $a_i$ are different. You can make hacks if and only if you solved both versions of the problem.
Today is Sage's birthday, and she will go shopping to buy ice spheres. All $n$ ice spheres are placed in a row and they are numbered from $1$ to $n$ from left to right. Each ice sphere has a positive integer price. In this version, some prices can be equal.
An ice sphere is cheap if it costs strictly less than two neighboring ice spheres: the nearest to the left and the nearest to the right. The leftmost and the rightmost ice spheres are not cheap. Sage will choose all cheap ice spheres and then buy only them.
You can visit the shop before Sage and reorder the ice spheres as you wish. Find out the maximum number of ice spheres that Sage can buy, and show how the ice spheres should be reordered.
-----Input-----
The first line contains a single integer $n$ $(1 \le n \le 10^5)$ — the number of ice spheres in the shop.
The second line contains $n$ integers $a_1, a_2, \dots, a_n$ $(1 \le a_i \le 10^9)$ — the prices of ice spheres.
-----Output-----
In the first line print the maximum number of ice spheres that Sage can buy.
In the second line print the prices of ice spheres in the optimal order. If there are several correct answers, you can print any of them.
-----Example-----
Input
7
1 3 2 2 4 5 4
Output
3
3 1 4 2 4 2 5
-----Note-----
In the sample it's not possible to place the ice spheres in any order so that Sage would buy $4$ of them. If the spheres are placed in the order $(3, 1, 4, 2, 4, 2, 5)$, then Sage will buy one sphere for $1$ and two spheres for $2$ each.
|
n = int(input())
l = list(map(int, input().split()))
lena = len(l)
l.sort()
left = l[: lena // 2]
right = l[lena // 2 :]
left.sort()
right.sort()
answer = []
count = 0
for i in range(len(right)):
answer.append(right[i])
if i < len(left):
answer.append(left[i])
if answer[-1] < answer[-2] and i < len(right) - 1:
count += 1
print(count)
print(*answer)
|
ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR ASSIGN VAR VAR BIN_OP VAR NUMBER ASSIGN VAR VAR BIN_OP VAR NUMBER EXPR FUNC_CALL VAR EXPR FUNC_CALL VAR ASSIGN VAR LIST ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR VAR VAR IF VAR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR VAR VAR IF VAR NUMBER VAR NUMBER VAR BIN_OP FUNC_CALL VAR VAR NUMBER VAR NUMBER EXPR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR VAR
|
This is the hard version of the problem. The difference between the versions is that in the easy version all prices $a_i$ are different. You can make hacks if and only if you solved both versions of the problem.
Today is Sage's birthday, and she will go shopping to buy ice spheres. All $n$ ice spheres are placed in a row and they are numbered from $1$ to $n$ from left to right. Each ice sphere has a positive integer price. In this version, some prices can be equal.
An ice sphere is cheap if it costs strictly less than two neighboring ice spheres: the nearest to the left and the nearest to the right. The leftmost and the rightmost ice spheres are not cheap. Sage will choose all cheap ice spheres and then buy only them.
You can visit the shop before Sage and reorder the ice spheres as you wish. Find out the maximum number of ice spheres that Sage can buy, and show how the ice spheres should be reordered.
-----Input-----
The first line contains a single integer $n$ $(1 \le n \le 10^5)$ — the number of ice spheres in the shop.
The second line contains $n$ integers $a_1, a_2, \dots, a_n$ $(1 \le a_i \le 10^9)$ — the prices of ice spheres.
-----Output-----
In the first line print the maximum number of ice spheres that Sage can buy.
In the second line print the prices of ice spheres in the optimal order. If there are several correct answers, you can print any of them.
-----Example-----
Input
7
1 3 2 2 4 5 4
Output
3
3 1 4 2 4 2 5
-----Note-----
In the sample it's not possible to place the ice spheres in any order so that Sage would buy $4$ of them. If the spheres are placed in the order $(3, 1, 4, 2, 4, 2, 5)$, then Sage will buy one sphere for $1$ and two spheres for $2$ each.
|
n = int(input())
F = [int(i) for i in input().split()]
F.sort()
if n % 2 != 0:
A = F[: int((n - 1) / 2)]
B = F[int((n - 1) / 2) :]
for i in range(int((n + 1) / 2)):
F[2 * i] = B[i]
for i in range(int((n - 1) / 2)):
F[2 * i + 1] = A[i]
else:
A1 = F[: int(n / 2)]
B1 = F[int(n / 2) :]
for i in range(int(n / 2)):
F[2 * i] = B1[i]
for i in range(int(n / 2)):
F[2 * i + 1] = A1[i]
count = 0
for i in range(1, n - 1):
if F[i] < F[i - 1] and F[i] < F[i + 1]:
count += 1
print(count)
print(*F)
|
ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR VAR VAR FUNC_CALL FUNC_CALL VAR EXPR FUNC_CALL VAR IF BIN_OP VAR NUMBER NUMBER ASSIGN VAR VAR FUNC_CALL VAR BIN_OP BIN_OP VAR NUMBER NUMBER ASSIGN VAR VAR FUNC_CALL VAR BIN_OP BIN_OP VAR NUMBER NUMBER FOR VAR FUNC_CALL VAR FUNC_CALL VAR BIN_OP BIN_OP VAR NUMBER NUMBER ASSIGN VAR BIN_OP NUMBER VAR VAR VAR FOR VAR FUNC_CALL VAR FUNC_CALL VAR BIN_OP BIN_OP VAR NUMBER NUMBER ASSIGN VAR BIN_OP BIN_OP NUMBER VAR NUMBER VAR VAR ASSIGN VAR VAR FUNC_CALL VAR BIN_OP VAR NUMBER ASSIGN VAR VAR FUNC_CALL VAR BIN_OP VAR NUMBER FOR VAR FUNC_CALL VAR FUNC_CALL VAR BIN_OP VAR NUMBER ASSIGN VAR BIN_OP NUMBER VAR VAR VAR FOR VAR FUNC_CALL VAR FUNC_CALL VAR BIN_OP VAR NUMBER ASSIGN VAR BIN_OP BIN_OP NUMBER VAR NUMBER VAR VAR ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR NUMBER BIN_OP VAR NUMBER IF VAR VAR VAR BIN_OP VAR NUMBER VAR VAR VAR BIN_OP VAR NUMBER VAR NUMBER EXPR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR VAR
|
This is the hard version of the problem. The difference between the versions is that in the easy version all prices $a_i$ are different. You can make hacks if and only if you solved both versions of the problem.
Today is Sage's birthday, and she will go shopping to buy ice spheres. All $n$ ice spheres are placed in a row and they are numbered from $1$ to $n$ from left to right. Each ice sphere has a positive integer price. In this version, some prices can be equal.
An ice sphere is cheap if it costs strictly less than two neighboring ice spheres: the nearest to the left and the nearest to the right. The leftmost and the rightmost ice spheres are not cheap. Sage will choose all cheap ice spheres and then buy only them.
You can visit the shop before Sage and reorder the ice spheres as you wish. Find out the maximum number of ice spheres that Sage can buy, and show how the ice spheres should be reordered.
-----Input-----
The first line contains a single integer $n$ $(1 \le n \le 10^5)$ — the number of ice spheres in the shop.
The second line contains $n$ integers $a_1, a_2, \dots, a_n$ $(1 \le a_i \le 10^9)$ — the prices of ice spheres.
-----Output-----
In the first line print the maximum number of ice spheres that Sage can buy.
In the second line print the prices of ice spheres in the optimal order. If there are several correct answers, you can print any of them.
-----Example-----
Input
7
1 3 2 2 4 5 4
Output
3
3 1 4 2 4 2 5
-----Note-----
In the sample it's not possible to place the ice spheres in any order so that Sage would buy $4$ of them. If the spheres are placed in the order $(3, 1, 4, 2, 4, 2, 5)$, then Sage will buy one sphere for $1$ and two spheres for $2$ each.
|
n = int(input())
a = sorted(map(int, input().split()))
b = [0] * n
b[1::2], b[::2] = a[: n // 2], a[n // 2 :]
print(sum(b[i - 1] > b[i] < b[i + 1] for i in range(1, n - 1)))
print(*b)
|
ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR BIN_OP LIST NUMBER VAR ASSIGN VAR NUMBER NUMBER VAR NUMBER VAR BIN_OP VAR NUMBER VAR BIN_OP VAR NUMBER EXPR FUNC_CALL VAR FUNC_CALL VAR VAR BIN_OP VAR NUMBER VAR VAR VAR BIN_OP VAR NUMBER VAR FUNC_CALL VAR NUMBER BIN_OP VAR NUMBER EXPR FUNC_CALL VAR VAR
|
This is the hard version of the problem. The difference between the versions is that in the easy version all prices $a_i$ are different. You can make hacks if and only if you solved both versions of the problem.
Today is Sage's birthday, and she will go shopping to buy ice spheres. All $n$ ice spheres are placed in a row and they are numbered from $1$ to $n$ from left to right. Each ice sphere has a positive integer price. In this version, some prices can be equal.
An ice sphere is cheap if it costs strictly less than two neighboring ice spheres: the nearest to the left and the nearest to the right. The leftmost and the rightmost ice spheres are not cheap. Sage will choose all cheap ice spheres and then buy only them.
You can visit the shop before Sage and reorder the ice spheres as you wish. Find out the maximum number of ice spheres that Sage can buy, and show how the ice spheres should be reordered.
-----Input-----
The first line contains a single integer $n$ $(1 \le n \le 10^5)$ — the number of ice spheres in the shop.
The second line contains $n$ integers $a_1, a_2, \dots, a_n$ $(1 \le a_i \le 10^9)$ — the prices of ice spheres.
-----Output-----
In the first line print the maximum number of ice spheres that Sage can buy.
In the second line print the prices of ice spheres in the optimal order. If there are several correct answers, you can print any of them.
-----Example-----
Input
7
1 3 2 2 4 5 4
Output
3
3 1 4 2 4 2 5
-----Note-----
In the sample it's not possible to place the ice spheres in any order so that Sage would buy $4$ of them. If the spheres are placed in the order $(3, 1, 4, 2, 4, 2, 5)$, then Sage will buy one sphere for $1$ and two spheres for $2$ each.
|
from sys import stdin
n = int(stdin.readline())
a = sorted(map(int, input().split()))
m = (n - 1) // 2
c = 0
b = []
for i in range(n):
b.append(a[(1 - i % 2) * (n // 2) + i // 2])
for i in range(1, m + 1):
if b[i * 2 - 1] < min(b[i * 2 - 2], b[i * 2]):
c += 1
print(c)
for i in range(n):
print(b[i], end=" ")
|
ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR BIN_OP BIN_OP VAR NUMBER NUMBER ASSIGN VAR NUMBER ASSIGN VAR LIST FOR VAR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR VAR BIN_OP BIN_OP BIN_OP NUMBER BIN_OP VAR NUMBER BIN_OP VAR NUMBER BIN_OP VAR NUMBER FOR VAR FUNC_CALL VAR NUMBER BIN_OP VAR NUMBER IF VAR BIN_OP BIN_OP VAR NUMBER NUMBER FUNC_CALL VAR VAR BIN_OP BIN_OP VAR NUMBER NUMBER VAR BIN_OP VAR NUMBER VAR NUMBER EXPR FUNC_CALL VAR VAR FOR VAR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR VAR VAR STRING
|
This is the hard version of the problem. The difference between the versions is that in the easy version all prices $a_i$ are different. You can make hacks if and only if you solved both versions of the problem.
Today is Sage's birthday, and she will go shopping to buy ice spheres. All $n$ ice spheres are placed in a row and they are numbered from $1$ to $n$ from left to right. Each ice sphere has a positive integer price. In this version, some prices can be equal.
An ice sphere is cheap if it costs strictly less than two neighboring ice spheres: the nearest to the left and the nearest to the right. The leftmost and the rightmost ice spheres are not cheap. Sage will choose all cheap ice spheres and then buy only them.
You can visit the shop before Sage and reorder the ice spheres as you wish. Find out the maximum number of ice spheres that Sage can buy, and show how the ice spheres should be reordered.
-----Input-----
The first line contains a single integer $n$ $(1 \le n \le 10^5)$ — the number of ice spheres in the shop.
The second line contains $n$ integers $a_1, a_2, \dots, a_n$ $(1 \le a_i \le 10^9)$ — the prices of ice spheres.
-----Output-----
In the first line print the maximum number of ice spheres that Sage can buy.
In the second line print the prices of ice spheres in the optimal order. If there are several correct answers, you can print any of them.
-----Example-----
Input
7
1 3 2 2 4 5 4
Output
3
3 1 4 2 4 2 5
-----Note-----
In the sample it's not possible to place the ice spheres in any order so that Sage would buy $4$ of them. If the spheres are placed in the order $(3, 1, 4, 2, 4, 2, 5)$, then Sage will buy one sphere for $1$ and two spheres for $2$ each.
|
n = int(input())
a = list(map(int, input().split()))
a.sort()
i = 0
j = 0
a1 = a[: n // 2]
a2 = a[n // 2 :]
ans = []
while i < len(a1) and j < len(a2):
ans.append(a2[j])
j += 1
ans.append(a1[i])
i += 1
while i < len(a1):
ans.append(a1[i])
i += 1
while j < len(a2):
ans.append(a2[j])
j += 1
c = 0
for i in range(1, len(ans) - 1):
if ans[i] < ans[i - 1] and ans[i] < ans[i + 1]:
c += 1
print(c)
print(*ans)
|
ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR EXPR FUNC_CALL VAR ASSIGN VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR VAR BIN_OP VAR NUMBER ASSIGN VAR VAR BIN_OP VAR NUMBER ASSIGN VAR LIST WHILE VAR FUNC_CALL VAR VAR VAR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR VAR VAR VAR NUMBER EXPR FUNC_CALL VAR VAR VAR VAR NUMBER WHILE VAR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR VAR VAR VAR NUMBER WHILE VAR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR VAR VAR VAR NUMBER ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR NUMBER BIN_OP FUNC_CALL VAR VAR NUMBER IF VAR VAR VAR BIN_OP VAR NUMBER VAR VAR VAR BIN_OP VAR NUMBER VAR NUMBER EXPR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR VAR
|
This is the hard version of the problem. The difference between the versions is that in the easy version all prices $a_i$ are different. You can make hacks if and only if you solved both versions of the problem.
Today is Sage's birthday, and she will go shopping to buy ice spheres. All $n$ ice spheres are placed in a row and they are numbered from $1$ to $n$ from left to right. Each ice sphere has a positive integer price. In this version, some prices can be equal.
An ice sphere is cheap if it costs strictly less than two neighboring ice spheres: the nearest to the left and the nearest to the right. The leftmost and the rightmost ice spheres are not cheap. Sage will choose all cheap ice spheres and then buy only them.
You can visit the shop before Sage and reorder the ice spheres as you wish. Find out the maximum number of ice spheres that Sage can buy, and show how the ice spheres should be reordered.
-----Input-----
The first line contains a single integer $n$ $(1 \le n \le 10^5)$ — the number of ice spheres in the shop.
The second line contains $n$ integers $a_1, a_2, \dots, a_n$ $(1 \le a_i \le 10^9)$ — the prices of ice spheres.
-----Output-----
In the first line print the maximum number of ice spheres that Sage can buy.
In the second line print the prices of ice spheres in the optimal order. If there are several correct answers, you can print any of them.
-----Example-----
Input
7
1 3 2 2 4 5 4
Output
3
3 1 4 2 4 2 5
-----Note-----
In the sample it's not possible to place the ice spheres in any order so that Sage would buy $4$ of them. If the spheres are placed in the order $(3, 1, 4, 2, 4, 2, 5)$, then Sage will buy one sphere for $1$ and two spheres for $2$ each.
|
n = int(input())
p = [int(x) for x in input().split()]
p.sort()
a = [None] * n
k = 0
if n % 2 == 1:
for i in range(1, n, 2):
a[i] = p[k]
k += 1
for i in range(0, n, 2):
a[i] = p[k]
k += 1
else:
for i in range(1, n, 2):
a[i] = p[k]
k += 1
for i in range(0, n, 2):
a[i] = p[k]
k += 1
ct = 0
for i in range(1, n - 1):
if a[i] < a[i - 1] and a[i] < a[i + 1]:
ct += 1
print(ct)
for e in a:
print(e, end=" ")
print()
|
ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR VAR VAR FUNC_CALL FUNC_CALL VAR EXPR FUNC_CALL VAR ASSIGN VAR BIN_OP LIST NONE VAR ASSIGN VAR NUMBER IF BIN_OP VAR NUMBER NUMBER FOR VAR FUNC_CALL VAR NUMBER VAR NUMBER ASSIGN VAR VAR VAR VAR VAR NUMBER FOR VAR FUNC_CALL VAR NUMBER VAR NUMBER ASSIGN VAR VAR VAR VAR VAR NUMBER FOR VAR FUNC_CALL VAR NUMBER VAR NUMBER ASSIGN VAR VAR VAR VAR VAR NUMBER FOR VAR FUNC_CALL VAR NUMBER VAR NUMBER ASSIGN VAR VAR VAR VAR VAR NUMBER ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR NUMBER BIN_OP VAR NUMBER IF VAR VAR VAR BIN_OP VAR NUMBER VAR VAR VAR BIN_OP VAR NUMBER VAR NUMBER EXPR FUNC_CALL VAR VAR FOR VAR VAR EXPR FUNC_CALL VAR VAR STRING EXPR FUNC_CALL VAR
|
This is the hard version of the problem. The difference between the versions is that in the easy version all prices $a_i$ are different. You can make hacks if and only if you solved both versions of the problem.
Today is Sage's birthday, and she will go shopping to buy ice spheres. All $n$ ice spheres are placed in a row and they are numbered from $1$ to $n$ from left to right. Each ice sphere has a positive integer price. In this version, some prices can be equal.
An ice sphere is cheap if it costs strictly less than two neighboring ice spheres: the nearest to the left and the nearest to the right. The leftmost and the rightmost ice spheres are not cheap. Sage will choose all cheap ice spheres and then buy only them.
You can visit the shop before Sage and reorder the ice spheres as you wish. Find out the maximum number of ice spheres that Sage can buy, and show how the ice spheres should be reordered.
-----Input-----
The first line contains a single integer $n$ $(1 \le n \le 10^5)$ — the number of ice spheres in the shop.
The second line contains $n$ integers $a_1, a_2, \dots, a_n$ $(1 \le a_i \le 10^9)$ — the prices of ice spheres.
-----Output-----
In the first line print the maximum number of ice spheres that Sage can buy.
In the second line print the prices of ice spheres in the optimal order. If there are several correct answers, you can print any of them.
-----Example-----
Input
7
1 3 2 2 4 5 4
Output
3
3 1 4 2 4 2 5
-----Note-----
In the sample it's not possible to place the ice spheres in any order so that Sage would buy $4$ of them. If the spheres are placed in the order $(3, 1, 4, 2, 4, 2, 5)$, then Sage will buy one sphere for $1$ and two spheres for $2$ each.
|
n = int(input())
arr = list(map(int, input().split()))
arr.sort()
small = arr[: len(arr) // 2]
large = arr[len(arr) // 2 :]
i = 0
while i < len(arr):
if i % 2 == 0:
arr[i] = large[i // 2]
else:
arr[i] = small[i // 2]
i += 1
ans = 0
for i in range(1, len(arr) - 1):
if arr[i] < arr[i + 1] and arr[i] < arr[i - 1]:
ans += 1
print(ans)
for i in arr:
print(i, end=" ")
|
ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR EXPR FUNC_CALL VAR ASSIGN VAR VAR BIN_OP FUNC_CALL VAR VAR NUMBER ASSIGN VAR VAR BIN_OP FUNC_CALL VAR VAR NUMBER ASSIGN VAR NUMBER WHILE VAR FUNC_CALL VAR VAR IF BIN_OP VAR NUMBER NUMBER ASSIGN VAR VAR VAR BIN_OP VAR NUMBER ASSIGN VAR VAR VAR BIN_OP VAR NUMBER VAR NUMBER ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR NUMBER BIN_OP FUNC_CALL VAR VAR NUMBER IF VAR VAR VAR BIN_OP VAR NUMBER VAR VAR VAR BIN_OP VAR NUMBER VAR NUMBER EXPR FUNC_CALL VAR VAR FOR VAR VAR EXPR FUNC_CALL VAR VAR STRING
|
This is the hard version of the problem. The difference between the versions is that in the easy version all prices $a_i$ are different. You can make hacks if and only if you solved both versions of the problem.
Today is Sage's birthday, and she will go shopping to buy ice spheres. All $n$ ice spheres are placed in a row and they are numbered from $1$ to $n$ from left to right. Each ice sphere has a positive integer price. In this version, some prices can be equal.
An ice sphere is cheap if it costs strictly less than two neighboring ice spheres: the nearest to the left and the nearest to the right. The leftmost and the rightmost ice spheres are not cheap. Sage will choose all cheap ice spheres and then buy only them.
You can visit the shop before Sage and reorder the ice spheres as you wish. Find out the maximum number of ice spheres that Sage can buy, and show how the ice spheres should be reordered.
-----Input-----
The first line contains a single integer $n$ $(1 \le n \le 10^5)$ — the number of ice spheres in the shop.
The second line contains $n$ integers $a_1, a_2, \dots, a_n$ $(1 \le a_i \le 10^9)$ — the prices of ice spheres.
-----Output-----
In the first line print the maximum number of ice spheres that Sage can buy.
In the second line print the prices of ice spheres in the optimal order. If there are several correct answers, you can print any of them.
-----Example-----
Input
7
1 3 2 2 4 5 4
Output
3
3 1 4 2 4 2 5
-----Note-----
In the sample it's not possible to place the ice spheres in any order so that Sage would buy $4$ of them. If the spheres are placed in the order $(3, 1, 4, 2, 4, 2, 5)$, then Sage will buy one sphere for $1$ and two spheres for $2$ each.
|
n = int(input())
a = list(map(int, input().split()))
a.sort()
b = n // 2
c = []
i = 0
while len(c) < n:
c.append(a[b])
if len(c) == n:
break
c.append(a[i])
i += 1
b += 1
ans = 0
i = 0
while i < n - 1:
if c[i] > c[i + 1]:
ans += 1
i += 2
if n % 2 == 0 and c[-1] != c[-2]:
ans -= 1
print(ans)
print(" ".join(str(x) for x in c))
|
ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR EXPR FUNC_CALL VAR ASSIGN VAR BIN_OP VAR NUMBER ASSIGN VAR LIST ASSIGN VAR NUMBER WHILE FUNC_CALL VAR VAR VAR EXPR FUNC_CALL VAR VAR VAR IF FUNC_CALL VAR VAR VAR EXPR FUNC_CALL VAR VAR VAR VAR NUMBER VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR NUMBER WHILE VAR BIN_OP VAR NUMBER IF VAR VAR VAR BIN_OP VAR NUMBER VAR NUMBER VAR NUMBER IF BIN_OP VAR NUMBER NUMBER VAR NUMBER VAR NUMBER VAR NUMBER EXPR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR FUNC_CALL STRING FUNC_CALL VAR VAR VAR VAR
|
This is the hard version of the problem. The difference between the versions is that in the easy version all prices $a_i$ are different. You can make hacks if and only if you solved both versions of the problem.
Today is Sage's birthday, and she will go shopping to buy ice spheres. All $n$ ice spheres are placed in a row and they are numbered from $1$ to $n$ from left to right. Each ice sphere has a positive integer price. In this version, some prices can be equal.
An ice sphere is cheap if it costs strictly less than two neighboring ice spheres: the nearest to the left and the nearest to the right. The leftmost and the rightmost ice spheres are not cheap. Sage will choose all cheap ice spheres and then buy only them.
You can visit the shop before Sage and reorder the ice spheres as you wish. Find out the maximum number of ice spheres that Sage can buy, and show how the ice spheres should be reordered.
-----Input-----
The first line contains a single integer $n$ $(1 \le n \le 10^5)$ — the number of ice spheres in the shop.
The second line contains $n$ integers $a_1, a_2, \dots, a_n$ $(1 \le a_i \le 10^9)$ — the prices of ice spheres.
-----Output-----
In the first line print the maximum number of ice spheres that Sage can buy.
In the second line print the prices of ice spheres in the optimal order. If there are several correct answers, you can print any of them.
-----Example-----
Input
7
1 3 2 2 4 5 4
Output
3
3 1 4 2 4 2 5
-----Note-----
In the sample it's not possible to place the ice spheres in any order so that Sage would buy $4$ of them. If the spheres are placed in the order $(3, 1, 4, 2, 4, 2, 5)$, then Sage will buy one sphere for $1$ and two spheres for $2$ each.
|
import sys
input = sys.stdin.readline
n = int(input())
x = sorted(list(map(int, input().split())))
a, b = x[: n // 2], x[n // 2 :]
tmp = []
for i in range(n // 2):
tmp.append(b[i])
tmp.append(a[i])
if n % 2:
tmp.append(b[-1])
cnt = 0
for i in range(1, n - 1, 2):
if tmp[i + 1] > tmp[i] < tmp[i - 1]:
cnt += 1
print(cnt)
print(*tmp)
|
IMPORT ASSIGN VAR VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR VAR VAR BIN_OP VAR NUMBER VAR BIN_OP VAR NUMBER ASSIGN VAR LIST FOR VAR FUNC_CALL VAR BIN_OP VAR NUMBER EXPR FUNC_CALL VAR VAR VAR EXPR FUNC_CALL VAR VAR VAR IF BIN_OP VAR NUMBER EXPR FUNC_CALL VAR VAR NUMBER ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR NUMBER BIN_OP VAR NUMBER NUMBER IF VAR BIN_OP VAR NUMBER VAR VAR VAR BIN_OP VAR NUMBER VAR NUMBER EXPR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR VAR
|
This is the hard version of the problem. The difference between the versions is that in the easy version all prices $a_i$ are different. You can make hacks if and only if you solved both versions of the problem.
Today is Sage's birthday, and she will go shopping to buy ice spheres. All $n$ ice spheres are placed in a row and they are numbered from $1$ to $n$ from left to right. Each ice sphere has a positive integer price. In this version, some prices can be equal.
An ice sphere is cheap if it costs strictly less than two neighboring ice spheres: the nearest to the left and the nearest to the right. The leftmost and the rightmost ice spheres are not cheap. Sage will choose all cheap ice spheres and then buy only them.
You can visit the shop before Sage and reorder the ice spheres as you wish. Find out the maximum number of ice spheres that Sage can buy, and show how the ice spheres should be reordered.
-----Input-----
The first line contains a single integer $n$ $(1 \le n \le 10^5)$ — the number of ice spheres in the shop.
The second line contains $n$ integers $a_1, a_2, \dots, a_n$ $(1 \le a_i \le 10^9)$ — the prices of ice spheres.
-----Output-----
In the first line print the maximum number of ice spheres that Sage can buy.
In the second line print the prices of ice spheres in the optimal order. If there are several correct answers, you can print any of them.
-----Example-----
Input
7
1 3 2 2 4 5 4
Output
3
3 1 4 2 4 2 5
-----Note-----
In the sample it's not possible to place the ice spheres in any order so that Sage would buy $4$ of them. If the spheres are placed in the order $(3, 1, 4, 2, 4, 2, 5)$, then Sage will buy one sphere for $1$ and two spheres for $2$ each.
|
n = int(input())
b = list(map(int, input().split()))
if n < 3:
print(0)
print(*b)
else:
out = 0
ans = []
b.sort()
k = b[: n // 2]
l = b[n // 2 :]
if n % 2:
for i in range(n // 2):
ans.append(l.pop())
ans.append(k.pop())
ans.append(l.pop())
for i in range(1, n, 2):
if ans[i] < ans[i - 1] and ans[i] < ans[i + 1]:
out += 1
print(out)
print(*ans)
else:
for i in range(len(k)):
ans.append(k.pop())
ans.append(l.pop())
for i in range(2, n, 2):
if ans[i] < ans[i - 1] and ans[i] < ans[i + 1]:
out += 1
print(out)
print(*ans)
|
ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR IF VAR NUMBER EXPR FUNC_CALL VAR NUMBER EXPR FUNC_CALL VAR VAR ASSIGN VAR NUMBER ASSIGN VAR LIST EXPR FUNC_CALL VAR ASSIGN VAR VAR BIN_OP VAR NUMBER ASSIGN VAR VAR BIN_OP VAR NUMBER IF BIN_OP VAR NUMBER FOR VAR FUNC_CALL VAR BIN_OP VAR NUMBER EXPR FUNC_CALL VAR FUNC_CALL VAR EXPR FUNC_CALL VAR FUNC_CALL VAR EXPR FUNC_CALL VAR FUNC_CALL VAR FOR VAR FUNC_CALL VAR NUMBER VAR NUMBER IF VAR VAR VAR BIN_OP VAR NUMBER VAR VAR VAR BIN_OP VAR NUMBER VAR NUMBER EXPR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR VAR FOR VAR FUNC_CALL VAR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR FUNC_CALL VAR EXPR FUNC_CALL VAR FUNC_CALL VAR FOR VAR FUNC_CALL VAR NUMBER VAR NUMBER IF VAR VAR VAR BIN_OP VAR NUMBER VAR VAR VAR BIN_OP VAR NUMBER VAR NUMBER EXPR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR VAR
|
This is the hard version of the problem. The difference between the versions is that in the easy version all prices $a_i$ are different. You can make hacks if and only if you solved both versions of the problem.
Today is Sage's birthday, and she will go shopping to buy ice spheres. All $n$ ice spheres are placed in a row and they are numbered from $1$ to $n$ from left to right. Each ice sphere has a positive integer price. In this version, some prices can be equal.
An ice sphere is cheap if it costs strictly less than two neighboring ice spheres: the nearest to the left and the nearest to the right. The leftmost and the rightmost ice spheres are not cheap. Sage will choose all cheap ice spheres and then buy only them.
You can visit the shop before Sage and reorder the ice spheres as you wish. Find out the maximum number of ice spheres that Sage can buy, and show how the ice spheres should be reordered.
-----Input-----
The first line contains a single integer $n$ $(1 \le n \le 10^5)$ — the number of ice spheres in the shop.
The second line contains $n$ integers $a_1, a_2, \dots, a_n$ $(1 \le a_i \le 10^9)$ — the prices of ice spheres.
-----Output-----
In the first line print the maximum number of ice spheres that Sage can buy.
In the second line print the prices of ice spheres in the optimal order. If there are several correct answers, you can print any of them.
-----Example-----
Input
7
1 3 2 2 4 5 4
Output
3
3 1 4 2 4 2 5
-----Note-----
In the sample it's not possible to place the ice spheres in any order so that Sage would buy $4$ of them. If the spheres are placed in the order $(3, 1, 4, 2, 4, 2, 5)$, then Sage will buy one sphere for $1$ and two spheres for $2$ each.
|
n = int(input())
l = list(map(int, input().split()))
l.sort()
ans = []
if n % 2 == 1:
ans.append(l[n // 2])
for i in range(n // 2):
ans.append(l[i])
ans.append(l[(n + 1) // 2 + i])
else:
for i in range(n // 2):
ans.append(l[(n + 1) // 2 + i])
ans.append(l[i])
num = 0
for i in range(1, n - 1):
if ans[i] < min(ans[i - 1], ans[i + 1]):
num += 1
print(num)
print(" ".join(map(str, ans)))
|
ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR EXPR FUNC_CALL VAR ASSIGN VAR LIST IF BIN_OP VAR NUMBER NUMBER EXPR FUNC_CALL VAR VAR BIN_OP VAR NUMBER FOR VAR FUNC_CALL VAR BIN_OP VAR NUMBER EXPR FUNC_CALL VAR VAR VAR EXPR FUNC_CALL VAR VAR BIN_OP BIN_OP BIN_OP VAR NUMBER NUMBER VAR FOR VAR FUNC_CALL VAR BIN_OP VAR NUMBER EXPR FUNC_CALL VAR VAR BIN_OP BIN_OP BIN_OP VAR NUMBER NUMBER VAR EXPR FUNC_CALL VAR VAR VAR ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR NUMBER BIN_OP VAR NUMBER IF VAR VAR FUNC_CALL VAR VAR BIN_OP VAR NUMBER VAR BIN_OP VAR NUMBER VAR NUMBER EXPR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR FUNC_CALL STRING FUNC_CALL VAR VAR VAR
|
This is the hard version of the problem. The difference between the versions is that in the easy version all prices $a_i$ are different. You can make hacks if and only if you solved both versions of the problem.
Today is Sage's birthday, and she will go shopping to buy ice spheres. All $n$ ice spheres are placed in a row and they are numbered from $1$ to $n$ from left to right. Each ice sphere has a positive integer price. In this version, some prices can be equal.
An ice sphere is cheap if it costs strictly less than two neighboring ice spheres: the nearest to the left and the nearest to the right. The leftmost and the rightmost ice spheres are not cheap. Sage will choose all cheap ice spheres and then buy only them.
You can visit the shop before Sage and reorder the ice spheres as you wish. Find out the maximum number of ice spheres that Sage can buy, and show how the ice spheres should be reordered.
-----Input-----
The first line contains a single integer $n$ $(1 \le n \le 10^5)$ — the number of ice spheres in the shop.
The second line contains $n$ integers $a_1, a_2, \dots, a_n$ $(1 \le a_i \le 10^9)$ — the prices of ice spheres.
-----Output-----
In the first line print the maximum number of ice spheres that Sage can buy.
In the second line print the prices of ice spheres in the optimal order. If there are several correct answers, you can print any of them.
-----Example-----
Input
7
1 3 2 2 4 5 4
Output
3
3 1 4 2 4 2 5
-----Note-----
In the sample it's not possible to place the ice spheres in any order so that Sage would buy $4$ of them. If the spheres are placed in the order $(3, 1, 4, 2, 4, 2, 5)$, then Sage will buy one sphere for $1$ and two spheres for $2$ each.
|
n = int(input())
arr = list(map(int, input().split()))
arr.sort()
ans = []
f = 0
less = arr[: n // 2]
gre = arr[n // 2 :]
l = 0
while l != n // 2:
ans.append(gre[l])
ans.append(less[l])
l += 1
if n % 2 != 0:
ans.append(gre[-1])
for i in range(1, n - 1):
if ans[i - 1] > ans[i] and ans[i] < ans[i + 1]:
f += 1
print(f)
print(*ans)
|
ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR EXPR FUNC_CALL VAR ASSIGN VAR LIST ASSIGN VAR NUMBER ASSIGN VAR VAR BIN_OP VAR NUMBER ASSIGN VAR VAR BIN_OP VAR NUMBER ASSIGN VAR NUMBER WHILE VAR BIN_OP VAR NUMBER EXPR FUNC_CALL VAR VAR VAR EXPR FUNC_CALL VAR VAR VAR VAR NUMBER IF BIN_OP VAR NUMBER NUMBER EXPR FUNC_CALL VAR VAR NUMBER FOR VAR FUNC_CALL VAR NUMBER BIN_OP VAR NUMBER IF VAR BIN_OP VAR NUMBER VAR VAR VAR VAR VAR BIN_OP VAR NUMBER VAR NUMBER EXPR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR VAR
|
This is the hard version of the problem. The difference between the versions is that in the easy version all prices $a_i$ are different. You can make hacks if and only if you solved both versions of the problem.
Today is Sage's birthday, and she will go shopping to buy ice spheres. All $n$ ice spheres are placed in a row and they are numbered from $1$ to $n$ from left to right. Each ice sphere has a positive integer price. In this version, some prices can be equal.
An ice sphere is cheap if it costs strictly less than two neighboring ice spheres: the nearest to the left and the nearest to the right. The leftmost and the rightmost ice spheres are not cheap. Sage will choose all cheap ice spheres and then buy only them.
You can visit the shop before Sage and reorder the ice spheres as you wish. Find out the maximum number of ice spheres that Sage can buy, and show how the ice spheres should be reordered.
-----Input-----
The first line contains a single integer $n$ $(1 \le n \le 10^5)$ — the number of ice spheres in the shop.
The second line contains $n$ integers $a_1, a_2, \dots, a_n$ $(1 \le a_i \le 10^9)$ — the prices of ice spheres.
-----Output-----
In the first line print the maximum number of ice spheres that Sage can buy.
In the second line print the prices of ice spheres in the optimal order. If there are several correct answers, you can print any of them.
-----Example-----
Input
7
1 3 2 2 4 5 4
Output
3
3 1 4 2 4 2 5
-----Note-----
In the sample it's not possible to place the ice spheres in any order so that Sage would buy $4$ of them. If the spheres are placed in the order $(3, 1, 4, 2, 4, 2, 5)$, then Sage will buy one sphere for $1$ and two spheres for $2$ each.
|
n = int(input())
data = list(map(int, input().split()))
def cal(data, n):
data.sort()
res = [0] * n
m = n // 2
res[0::2] = data[m:]
res[1::2] = data[:m]
ans = 0
for i in range(1, n, 2):
if i == n - 1:
break
if res[i - 1] > res[i] and res[i] < res[i + 1]:
ans += 1
print(ans)
s = " ".join(list(map(str, res)))
print(s)
cal(data, n)
|
ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR FUNC_DEF EXPR FUNC_CALL VAR ASSIGN VAR BIN_OP LIST NUMBER VAR ASSIGN VAR BIN_OP VAR NUMBER ASSIGN VAR NUMBER NUMBER VAR VAR ASSIGN VAR NUMBER NUMBER VAR VAR ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR NUMBER VAR NUMBER IF VAR BIN_OP VAR NUMBER IF VAR BIN_OP VAR NUMBER VAR VAR VAR VAR VAR BIN_OP VAR NUMBER VAR NUMBER EXPR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL STRING FUNC_CALL VAR FUNC_CALL VAR VAR VAR EXPR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR VAR VAR
|
This is the hard version of the problem. The difference between the versions is that in the easy version all prices $a_i$ are different. You can make hacks if and only if you solved both versions of the problem.
Today is Sage's birthday, and she will go shopping to buy ice spheres. All $n$ ice spheres are placed in a row and they are numbered from $1$ to $n$ from left to right. Each ice sphere has a positive integer price. In this version, some prices can be equal.
An ice sphere is cheap if it costs strictly less than two neighboring ice spheres: the nearest to the left and the nearest to the right. The leftmost and the rightmost ice spheres are not cheap. Sage will choose all cheap ice spheres and then buy only them.
You can visit the shop before Sage and reorder the ice spheres as you wish. Find out the maximum number of ice spheres that Sage can buy, and show how the ice spheres should be reordered.
-----Input-----
The first line contains a single integer $n$ $(1 \le n \le 10^5)$ — the number of ice spheres in the shop.
The second line contains $n$ integers $a_1, a_2, \dots, a_n$ $(1 \le a_i \le 10^9)$ — the prices of ice spheres.
-----Output-----
In the first line print the maximum number of ice spheres that Sage can buy.
In the second line print the prices of ice spheres in the optimal order. If there are several correct answers, you can print any of them.
-----Example-----
Input
7
1 3 2 2 4 5 4
Output
3
3 1 4 2 4 2 5
-----Note-----
In the sample it's not possible to place the ice spheres in any order so that Sage would buy $4$ of them. If the spheres are placed in the order $(3, 1, 4, 2, 4, 2, 5)$, then Sage will buy one sphere for $1$ and two spheres for $2$ each.
|
l = []
if 1 == 1:
n = int(input())
a = list(map(int, input().split()))
a.sort()
s = []
if n % 2 == 0:
if a[n // 2 - 1] != a[n // 2]:
print(n // 2 - 1)
for i in range(n):
if i % 2 == 0:
s.append(str(a[i // 2 + n // 2]))
else:
s.append(str(a[i // 2]))
print(" ".join(s))
else:
c1 = a[: n // 2].count(a[n // 2])
c2 = a[n // 2 :].count(a[n // 2])
if n // 2 - c2 - c1 >= 1:
print(n // 2 - 1)
for i in range(n):
if i % 2 == 0:
s.append(str(a[n - 1 - i // 2]))
else:
s.append(str(a[n // 2 - 1 - i // 2]))
print(" ".join(s))
else:
print((n // 2 - c2 + 1) // 2 + n // 2 - c1)
for i in range(n):
if i % 2 == 0:
s.append(str(a[n // 2 + i // 2]))
else:
s.append(str(a[i // 2]))
print(" ".join(s))
elif a[n // 2] != a[n // 2 - 1]:
print(n // 2)
for i in range(n):
if i % 2 == 0:
s.append(str(a[n // 2 + i // 2]))
else:
s.append(str(a[i // 2]))
print(" ".join(s))
else:
c1 = a[: n // 2].count(a[n // 2])
c2 = a[n // 2 :].count(a[n // 2])
if n // 2 - c2 - c1 + 1 >= 1:
print(n // 2)
else:
print((n // 2 + 1 - c2 + 1) // 2 + n // 2 - c1)
for i in range(n):
if i % 2 == 0:
s.append(str(a[n // 2 + i // 2]))
else:
s.append(str(a[i // 2]))
print(" ".join(s))
|
ASSIGN VAR LIST IF NUMBER NUMBER ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR EXPR FUNC_CALL VAR ASSIGN VAR LIST IF BIN_OP VAR NUMBER NUMBER IF VAR BIN_OP BIN_OP VAR NUMBER NUMBER VAR BIN_OP VAR NUMBER EXPR FUNC_CALL VAR BIN_OP BIN_OP VAR NUMBER NUMBER FOR VAR FUNC_CALL VAR VAR IF BIN_OP VAR NUMBER NUMBER EXPR FUNC_CALL VAR FUNC_CALL VAR VAR BIN_OP BIN_OP VAR NUMBER BIN_OP VAR NUMBER EXPR FUNC_CALL VAR FUNC_CALL VAR VAR BIN_OP VAR NUMBER EXPR FUNC_CALL VAR FUNC_CALL STRING VAR ASSIGN VAR FUNC_CALL VAR BIN_OP VAR NUMBER VAR BIN_OP VAR NUMBER ASSIGN VAR FUNC_CALL VAR BIN_OP VAR NUMBER VAR BIN_OP VAR NUMBER IF BIN_OP BIN_OP BIN_OP VAR NUMBER VAR VAR NUMBER EXPR FUNC_CALL VAR BIN_OP BIN_OP VAR NUMBER NUMBER FOR VAR FUNC_CALL VAR VAR IF BIN_OP VAR NUMBER NUMBER EXPR FUNC_CALL VAR FUNC_CALL VAR VAR BIN_OP BIN_OP VAR NUMBER BIN_OP VAR NUMBER EXPR FUNC_CALL VAR FUNC_CALL VAR VAR BIN_OP BIN_OP BIN_OP VAR NUMBER NUMBER BIN_OP VAR NUMBER EXPR FUNC_CALL VAR FUNC_CALL STRING VAR EXPR FUNC_CALL VAR BIN_OP BIN_OP BIN_OP BIN_OP BIN_OP BIN_OP VAR NUMBER VAR NUMBER NUMBER BIN_OP VAR NUMBER VAR FOR VAR FUNC_CALL VAR VAR IF BIN_OP VAR NUMBER NUMBER EXPR FUNC_CALL VAR FUNC_CALL VAR VAR BIN_OP BIN_OP VAR NUMBER BIN_OP VAR NUMBER EXPR FUNC_CALL VAR FUNC_CALL VAR VAR BIN_OP VAR NUMBER EXPR FUNC_CALL VAR FUNC_CALL STRING VAR IF VAR BIN_OP VAR NUMBER VAR BIN_OP BIN_OP VAR NUMBER NUMBER EXPR FUNC_CALL VAR BIN_OP VAR NUMBER FOR VAR FUNC_CALL VAR VAR IF BIN_OP VAR NUMBER NUMBER EXPR FUNC_CALL VAR FUNC_CALL VAR VAR BIN_OP BIN_OP VAR NUMBER BIN_OP VAR NUMBER EXPR FUNC_CALL VAR FUNC_CALL VAR VAR BIN_OP VAR NUMBER EXPR FUNC_CALL VAR FUNC_CALL STRING VAR ASSIGN VAR FUNC_CALL VAR BIN_OP VAR NUMBER VAR BIN_OP VAR NUMBER ASSIGN VAR FUNC_CALL VAR BIN_OP VAR NUMBER VAR BIN_OP VAR NUMBER IF BIN_OP BIN_OP BIN_OP BIN_OP VAR NUMBER VAR VAR NUMBER NUMBER EXPR FUNC_CALL VAR BIN_OP VAR NUMBER EXPR FUNC_CALL VAR BIN_OP BIN_OP BIN_OP BIN_OP BIN_OP BIN_OP BIN_OP VAR NUMBER NUMBER VAR NUMBER NUMBER BIN_OP VAR NUMBER VAR FOR VAR FUNC_CALL VAR VAR IF BIN_OP VAR NUMBER NUMBER EXPR FUNC_CALL VAR FUNC_CALL VAR VAR BIN_OP BIN_OP VAR NUMBER BIN_OP VAR NUMBER EXPR FUNC_CALL VAR FUNC_CALL VAR VAR BIN_OP VAR NUMBER EXPR FUNC_CALL VAR FUNC_CALL STRING VAR
|
This is the hard version of the problem. The difference between the versions is that in the easy version all prices $a_i$ are different. You can make hacks if and only if you solved both versions of the problem.
Today is Sage's birthday, and she will go shopping to buy ice spheres. All $n$ ice spheres are placed in a row and they are numbered from $1$ to $n$ from left to right. Each ice sphere has a positive integer price. In this version, some prices can be equal.
An ice sphere is cheap if it costs strictly less than two neighboring ice spheres: the nearest to the left and the nearest to the right. The leftmost and the rightmost ice spheres are not cheap. Sage will choose all cheap ice spheres and then buy only them.
You can visit the shop before Sage and reorder the ice spheres as you wish. Find out the maximum number of ice spheres that Sage can buy, and show how the ice spheres should be reordered.
-----Input-----
The first line contains a single integer $n$ $(1 \le n \le 10^5)$ — the number of ice spheres in the shop.
The second line contains $n$ integers $a_1, a_2, \dots, a_n$ $(1 \le a_i \le 10^9)$ — the prices of ice spheres.
-----Output-----
In the first line print the maximum number of ice spheres that Sage can buy.
In the second line print the prices of ice spheres in the optimal order. If there are several correct answers, you can print any of them.
-----Example-----
Input
7
1 3 2 2 4 5 4
Output
3
3 1 4 2 4 2 5
-----Note-----
In the sample it's not possible to place the ice spheres in any order so that Sage would buy $4$ of them. If the spheres are placed in the order $(3, 1, 4, 2, 4, 2, 5)$, then Sage will buy one sphere for $1$ and two spheres for $2$ each.
|
def solve(n, ar):
ar = sorted(ar)
l = 0
r = n + 1
while r - l > 1:
m = (r + l) // 2
f = True
if 2 * m + 1 > n:
f = False
else:
b = []
pos_a = 0
pos_b = n - (m + 1)
for i in range(2 * m + 1):
if i % 2 == 0:
b.append(ar[pos_b])
pos_b += 1
else:
b.append(ar[pos_a])
pos_a += 1
for i in range(1, 2 * m + 1, 2):
if b[i - 1] <= b[i] or b[i + 1] <= b[i]:
f = False
if f:
l = m
else:
r = m
print(l)
pos_a = 0
pos_b = n - (l + 1)
b = []
for i in range(2 * l + 1):
if i % 2 == 0:
b.append(ar[pos_b])
pos_b += 1
else:
b.append(ar[pos_a])
pos_a += 1
for i in range(pos_a, n - (l + 1)):
b.append(ar[i])
print(" ".join(map(str, b)))
n = int(input())
ar = list(map(int, input().split()))
solve(n, ar)
|
FUNC_DEF ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR NUMBER ASSIGN VAR BIN_OP VAR NUMBER WHILE BIN_OP VAR VAR NUMBER ASSIGN VAR BIN_OP BIN_OP VAR VAR NUMBER ASSIGN VAR NUMBER IF BIN_OP BIN_OP NUMBER VAR NUMBER VAR ASSIGN VAR NUMBER ASSIGN VAR LIST ASSIGN VAR NUMBER ASSIGN VAR BIN_OP VAR BIN_OP VAR NUMBER FOR VAR FUNC_CALL VAR BIN_OP BIN_OP NUMBER VAR NUMBER IF BIN_OP VAR NUMBER NUMBER EXPR FUNC_CALL VAR VAR VAR VAR NUMBER EXPR FUNC_CALL VAR VAR VAR VAR NUMBER FOR VAR FUNC_CALL VAR NUMBER BIN_OP BIN_OP NUMBER VAR NUMBER NUMBER IF VAR BIN_OP VAR NUMBER VAR VAR VAR BIN_OP VAR NUMBER VAR VAR ASSIGN VAR NUMBER IF VAR ASSIGN VAR VAR ASSIGN VAR VAR EXPR FUNC_CALL VAR VAR ASSIGN VAR NUMBER ASSIGN VAR BIN_OP VAR BIN_OP VAR NUMBER ASSIGN VAR LIST FOR VAR FUNC_CALL VAR BIN_OP BIN_OP NUMBER VAR NUMBER IF BIN_OP VAR NUMBER NUMBER EXPR FUNC_CALL VAR VAR VAR VAR NUMBER EXPR FUNC_CALL VAR VAR VAR VAR NUMBER FOR VAR FUNC_CALL VAR VAR BIN_OP VAR BIN_OP VAR NUMBER EXPR FUNC_CALL VAR VAR VAR EXPR FUNC_CALL VAR FUNC_CALL STRING FUNC_CALL VAR VAR VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR EXPR FUNC_CALL VAR VAR VAR
|
This is the hard version of the problem. The difference between the versions is that in the easy version all prices $a_i$ are different. You can make hacks if and only if you solved both versions of the problem.
Today is Sage's birthday, and she will go shopping to buy ice spheres. All $n$ ice spheres are placed in a row and they are numbered from $1$ to $n$ from left to right. Each ice sphere has a positive integer price. In this version, some prices can be equal.
An ice sphere is cheap if it costs strictly less than two neighboring ice spheres: the nearest to the left and the nearest to the right. The leftmost and the rightmost ice spheres are not cheap. Sage will choose all cheap ice spheres and then buy only them.
You can visit the shop before Sage and reorder the ice spheres as you wish. Find out the maximum number of ice spheres that Sage can buy, and show how the ice spheres should be reordered.
-----Input-----
The first line contains a single integer $n$ $(1 \le n \le 10^5)$ — the number of ice spheres in the shop.
The second line contains $n$ integers $a_1, a_2, \dots, a_n$ $(1 \le a_i \le 10^9)$ — the prices of ice spheres.
-----Output-----
In the first line print the maximum number of ice spheres that Sage can buy.
In the second line print the prices of ice spheres in the optimal order. If there are several correct answers, you can print any of them.
-----Example-----
Input
7
1 3 2 2 4 5 4
Output
3
3 1 4 2 4 2 5
-----Note-----
In the sample it's not possible to place the ice spheres in any order so that Sage would buy $4$ of them. If the spheres are placed in the order $(3, 1, 4, 2, 4, 2, 5)$, then Sage will buy one sphere for $1$ and two spheres for $2$ each.
|
n = int(input())
arr = sorted(list(map(int, input().split())), reverse=True)
t = n - (n - 1) // 2
ans = [(0) for _ in range(n)]
for i in range(t):
x = i * 2
if x == n:
x -= 1
ans[x] = arr[i]
cur = 1
cur2 = n - 2
if n % 2 == 0:
cur2 -= 1
for i in range(t, n):
if ans[cur - 1] > arr[i] < ans[cur + 1]:
ans[cur] = arr[i]
cur += 2
else:
ans[cur2] = arr[i]
cur2 -= 2
answer = sum(ans[i - 1] > ans[i] < ans[i + 1] for i in range(1, n - 1))
print(answer)
print(*ans)
|
ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR NUMBER ASSIGN VAR BIN_OP VAR BIN_OP BIN_OP VAR NUMBER NUMBER ASSIGN VAR NUMBER VAR FUNC_CALL VAR VAR FOR VAR FUNC_CALL VAR VAR ASSIGN VAR BIN_OP VAR NUMBER IF VAR VAR VAR NUMBER ASSIGN VAR VAR VAR VAR ASSIGN VAR NUMBER ASSIGN VAR BIN_OP VAR NUMBER IF BIN_OP VAR NUMBER NUMBER VAR NUMBER FOR VAR FUNC_CALL VAR VAR VAR IF VAR BIN_OP VAR NUMBER VAR VAR VAR BIN_OP VAR NUMBER ASSIGN VAR VAR VAR VAR VAR NUMBER ASSIGN VAR VAR VAR VAR VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR BIN_OP VAR NUMBER VAR VAR VAR BIN_OP VAR NUMBER VAR FUNC_CALL VAR NUMBER BIN_OP VAR NUMBER EXPR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR VAR
|
This is the hard version of the problem. The difference between the versions is that in the easy version all prices $a_i$ are different. You can make hacks if and only if you solved both versions of the problem.
Today is Sage's birthday, and she will go shopping to buy ice spheres. All $n$ ice spheres are placed in a row and they are numbered from $1$ to $n$ from left to right. Each ice sphere has a positive integer price. In this version, some prices can be equal.
An ice sphere is cheap if it costs strictly less than two neighboring ice spheres: the nearest to the left and the nearest to the right. The leftmost and the rightmost ice spheres are not cheap. Sage will choose all cheap ice spheres and then buy only them.
You can visit the shop before Sage and reorder the ice spheres as you wish. Find out the maximum number of ice spheres that Sage can buy, and show how the ice spheres should be reordered.
-----Input-----
The first line contains a single integer $n$ $(1 \le n \le 10^5)$ — the number of ice spheres in the shop.
The second line contains $n$ integers $a_1, a_2, \dots, a_n$ $(1 \le a_i \le 10^9)$ — the prices of ice spheres.
-----Output-----
In the first line print the maximum number of ice spheres that Sage can buy.
In the second line print the prices of ice spheres in the optimal order. If there are several correct answers, you can print any of them.
-----Example-----
Input
7
1 3 2 2 4 5 4
Output
3
3 1 4 2 4 2 5
-----Note-----
In the sample it's not possible to place the ice spheres in any order so that Sage would buy $4$ of them. If the spheres are placed in the order $(3, 1, 4, 2, 4, 2, 5)$, then Sage will buy one sphere for $1$ and two spheres for $2$ each.
|
n = int(input())
array = list(map(int, input().split()))
array.sort()
output = []
array1 = array[0 : n // 2]
array2 = array[n // 2 :]
i = 0
j = 0
count = 0
while i < n // 2 and j < (n - 1) // 2:
b = array2[j]
a = array1[i]
output.append(b)
output.append(a)
if b > a:
count += 1
i += 1
j += 1
if n % 2 != 0:
output.append(array2[-1])
else:
output.append(array2[-1])
output.append(array1[-1])
print(count)
print(" ".join(map(str, output)))
|
ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR EXPR FUNC_CALL VAR ASSIGN VAR LIST ASSIGN VAR VAR NUMBER BIN_OP VAR NUMBER ASSIGN VAR VAR BIN_OP VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR NUMBER WHILE VAR BIN_OP VAR NUMBER VAR BIN_OP BIN_OP VAR NUMBER NUMBER ASSIGN VAR VAR VAR ASSIGN VAR VAR VAR EXPR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR VAR IF VAR VAR VAR NUMBER VAR NUMBER VAR NUMBER IF BIN_OP VAR NUMBER NUMBER EXPR FUNC_CALL VAR VAR NUMBER EXPR FUNC_CALL VAR VAR NUMBER EXPR FUNC_CALL VAR VAR NUMBER EXPR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR FUNC_CALL STRING FUNC_CALL VAR VAR VAR
|
This is the hard version of the problem. The difference between the versions is that in the easy version all prices $a_i$ are different. You can make hacks if and only if you solved both versions of the problem.
Today is Sage's birthday, and she will go shopping to buy ice spheres. All $n$ ice spheres are placed in a row and they are numbered from $1$ to $n$ from left to right. Each ice sphere has a positive integer price. In this version, some prices can be equal.
An ice sphere is cheap if it costs strictly less than two neighboring ice spheres: the nearest to the left and the nearest to the right. The leftmost and the rightmost ice spheres are not cheap. Sage will choose all cheap ice spheres and then buy only them.
You can visit the shop before Sage and reorder the ice spheres as you wish. Find out the maximum number of ice spheres that Sage can buy, and show how the ice spheres should be reordered.
-----Input-----
The first line contains a single integer $n$ $(1 \le n \le 10^5)$ — the number of ice spheres in the shop.
The second line contains $n$ integers $a_1, a_2, \dots, a_n$ $(1 \le a_i \le 10^9)$ — the prices of ice spheres.
-----Output-----
In the first line print the maximum number of ice spheres that Sage can buy.
In the second line print the prices of ice spheres in the optimal order. If there are several correct answers, you can print any of them.
-----Example-----
Input
7
1 3 2 2 4 5 4
Output
3
3 1 4 2 4 2 5
-----Note-----
In the sample it's not possible to place the ice spheres in any order so that Sage would buy $4$ of them. If the spheres are placed in the order $(3, 1, 4, 2, 4, 2, 5)$, then Sage will buy one sphere for $1$ and two spheres for $2$ each.
|
n = int(input())
a = sorted(map(int, input().split()), reverse=True)
i = n - n // 2
order = [a[0]]
skipped = []
j = 1
while i < n:
if len(order) % 2:
while a[i] == a[j]:
skipped.append(a[i])
i += 1
if i == n:
break
else:
order.append(a[i])
i += 1
else:
order.append(a[j])
j += 1
skipped.extend(a[j : j + n - len(order) - len(skipped)])
purchases = len(order) // 2
if not skipped:
purchases = max(purchases - 1, 0)
print(purchases)
print(*(order + skipped))
|
ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR NUMBER ASSIGN VAR BIN_OP VAR BIN_OP VAR NUMBER ASSIGN VAR LIST VAR NUMBER ASSIGN VAR LIST ASSIGN VAR NUMBER WHILE VAR VAR IF BIN_OP FUNC_CALL VAR VAR NUMBER WHILE VAR VAR VAR VAR EXPR FUNC_CALL VAR VAR VAR VAR NUMBER IF VAR VAR EXPR FUNC_CALL VAR VAR VAR VAR NUMBER EXPR FUNC_CALL VAR VAR VAR VAR NUMBER EXPR FUNC_CALL VAR VAR VAR BIN_OP BIN_OP BIN_OP VAR VAR FUNC_CALL VAR VAR FUNC_CALL VAR VAR ASSIGN VAR BIN_OP FUNC_CALL VAR VAR NUMBER IF VAR ASSIGN VAR FUNC_CALL VAR BIN_OP VAR NUMBER NUMBER EXPR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR BIN_OP VAR VAR
|
This is the hard version of the problem. The difference between the versions is that in the easy version all prices $a_i$ are different. You can make hacks if and only if you solved both versions of the problem.
Today is Sage's birthday, and she will go shopping to buy ice spheres. All $n$ ice spheres are placed in a row and they are numbered from $1$ to $n$ from left to right. Each ice sphere has a positive integer price. In this version, some prices can be equal.
An ice sphere is cheap if it costs strictly less than two neighboring ice spheres: the nearest to the left and the nearest to the right. The leftmost and the rightmost ice spheres are not cheap. Sage will choose all cheap ice spheres and then buy only them.
You can visit the shop before Sage and reorder the ice spheres as you wish. Find out the maximum number of ice spheres that Sage can buy, and show how the ice spheres should be reordered.
-----Input-----
The first line contains a single integer $n$ $(1 \le n \le 10^5)$ — the number of ice spheres in the shop.
The second line contains $n$ integers $a_1, a_2, \dots, a_n$ $(1 \le a_i \le 10^9)$ — the prices of ice spheres.
-----Output-----
In the first line print the maximum number of ice spheres that Sage can buy.
In the second line print the prices of ice spheres in the optimal order. If there are several correct answers, you can print any of them.
-----Example-----
Input
7
1 3 2 2 4 5 4
Output
3
3 1 4 2 4 2 5
-----Note-----
In the sample it's not possible to place the ice spheres in any order so that Sage would buy $4$ of them. If the spheres are placed in the order $(3, 1, 4, 2, 4, 2, 5)$, then Sage will buy one sphere for $1$ and two spheres for $2$ each.
|
n = int(input())
spheres = list(map(int, input().split(" ")))
spheres = sorted(spheres)
ret = [(-1) for _ in range(n)]
l, r = 0, n - 1
cnt = 0
block_n = n // 2 + n % 2
inner_n = n - block_n
inners = spheres[:inner_n]
blocks = spheres[inner_n:n]
st = 0
if n % 2 == 0:
ret[0] = inners.pop()
st = 1
for i in range(st, n):
if i % 2 == st:
ret[i] = blocks.pop()
else:
ret[i] = inners.pop()
if i > 1 and ret[i - 1] < ret[i - 2] and ret[i - 1] < ret[i]:
cnt += 1
print(cnt)
print(" ".join(list(map(str, ret))))
|
ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR STRING ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR NUMBER VAR FUNC_CALL VAR VAR ASSIGN VAR VAR NUMBER BIN_OP VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR BIN_OP BIN_OP VAR NUMBER BIN_OP VAR NUMBER ASSIGN VAR BIN_OP VAR VAR ASSIGN VAR VAR VAR ASSIGN VAR VAR VAR VAR ASSIGN VAR NUMBER IF BIN_OP VAR NUMBER NUMBER ASSIGN VAR NUMBER FUNC_CALL VAR ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR VAR VAR IF BIN_OP VAR NUMBER VAR ASSIGN VAR VAR FUNC_CALL VAR ASSIGN VAR VAR FUNC_CALL VAR IF VAR NUMBER VAR BIN_OP VAR NUMBER VAR BIN_OP VAR NUMBER VAR BIN_OP VAR NUMBER VAR VAR VAR NUMBER EXPR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR FUNC_CALL STRING FUNC_CALL VAR FUNC_CALL VAR VAR VAR
|
This is the hard version of the problem. The difference between the versions is that in the easy version all prices $a_i$ are different. You can make hacks if and only if you solved both versions of the problem.
Today is Sage's birthday, and she will go shopping to buy ice spheres. All $n$ ice spheres are placed in a row and they are numbered from $1$ to $n$ from left to right. Each ice sphere has a positive integer price. In this version, some prices can be equal.
An ice sphere is cheap if it costs strictly less than two neighboring ice spheres: the nearest to the left and the nearest to the right. The leftmost and the rightmost ice spheres are not cheap. Sage will choose all cheap ice spheres and then buy only them.
You can visit the shop before Sage and reorder the ice spheres as you wish. Find out the maximum number of ice spheres that Sage can buy, and show how the ice spheres should be reordered.
-----Input-----
The first line contains a single integer $n$ $(1 \le n \le 10^5)$ — the number of ice spheres in the shop.
The second line contains $n$ integers $a_1, a_2, \dots, a_n$ $(1 \le a_i \le 10^9)$ — the prices of ice spheres.
-----Output-----
In the first line print the maximum number of ice spheres that Sage can buy.
In the second line print the prices of ice spheres in the optimal order. If there are several correct answers, you can print any of them.
-----Example-----
Input
7
1 3 2 2 4 5 4
Output
3
3 1 4 2 4 2 5
-----Note-----
In the sample it's not possible to place the ice spheres in any order so that Sage would buy $4$ of them. If the spheres are placed in the order $(3, 1, 4, 2, 4, 2, 5)$, then Sage will buy one sphere for $1$ and two spheres for $2$ each.
|
def f():
n = int(input())
arr = list(map(int, input().split()))
arr.sort()
newarr = []
for i in range(n // 2):
newarr.append(arr[i + n // 2])
newarr.append(arr[i])
if n % 2:
newarr.append(arr[n - 1])
total = 0
i = 1
while i + 1 < n:
if newarr[i - 1] > newarr[i] and newarr[i] < newarr[i + 1]:
total += 1
i += 2
newarr = list(map(str, newarr))
newarr = " ".join(newarr)
print(total)
print(newarr)
f()
|
FUNC_DEF ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR EXPR FUNC_CALL VAR ASSIGN VAR LIST FOR VAR FUNC_CALL VAR BIN_OP VAR NUMBER EXPR FUNC_CALL VAR VAR BIN_OP VAR BIN_OP VAR NUMBER EXPR FUNC_CALL VAR VAR VAR IF BIN_OP VAR NUMBER EXPR FUNC_CALL VAR VAR BIN_OP VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR NUMBER WHILE BIN_OP VAR NUMBER VAR IF VAR BIN_OP VAR NUMBER VAR VAR VAR VAR VAR BIN_OP VAR NUMBER VAR NUMBER VAR NUMBER ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR VAR VAR ASSIGN VAR FUNC_CALL STRING VAR EXPR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR
|
This is the hard version of the problem. The difference between the versions is that in the easy version all prices $a_i$ are different. You can make hacks if and only if you solved both versions of the problem.
Today is Sage's birthday, and she will go shopping to buy ice spheres. All $n$ ice spheres are placed in a row and they are numbered from $1$ to $n$ from left to right. Each ice sphere has a positive integer price. In this version, some prices can be equal.
An ice sphere is cheap if it costs strictly less than two neighboring ice spheres: the nearest to the left and the nearest to the right. The leftmost and the rightmost ice spheres are not cheap. Sage will choose all cheap ice spheres and then buy only them.
You can visit the shop before Sage and reorder the ice spheres as you wish. Find out the maximum number of ice spheres that Sage can buy, and show how the ice spheres should be reordered.
-----Input-----
The first line contains a single integer $n$ $(1 \le n \le 10^5)$ — the number of ice spheres in the shop.
The second line contains $n$ integers $a_1, a_2, \dots, a_n$ $(1 \le a_i \le 10^9)$ — the prices of ice spheres.
-----Output-----
In the first line print the maximum number of ice spheres that Sage can buy.
In the second line print the prices of ice spheres in the optimal order. If there are several correct answers, you can print any of them.
-----Example-----
Input
7
1 3 2 2 4 5 4
Output
3
3 1 4 2 4 2 5
-----Note-----
In the sample it's not possible to place the ice spheres in any order so that Sage would buy $4$ of them. If the spheres are placed in the order $(3, 1, 4, 2, 4, 2, 5)$, then Sage will buy one sphere for $1$ and two spheres for $2$ each.
|
n = int(input())
a = list(map(int, input().split()))
a.sort()
lo = 0
hi = (n + 1) // 2
while hi - lo > 1:
test = (hi + lo) // 2
smol = a[:test]
large = a[-test - 1 :]
rest = a[test : -test - 1]
works = True
for i in range(test):
if smol[i] == large[i] or smol[i] == large[i + 1]:
works = False
if works:
lo = test
else:
hi = test
smol = a[:lo]
large = a[-lo - 1 :]
rest = a[lo : -lo - 1]
o = []
while smol:
o += [large.pop()]
o += [smol.pop()]
o += large
o += rest
print(lo)
print(" ".join(map(str, o)))
|
ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR EXPR FUNC_CALL VAR ASSIGN VAR NUMBER ASSIGN VAR BIN_OP BIN_OP VAR NUMBER NUMBER WHILE BIN_OP VAR VAR NUMBER ASSIGN VAR BIN_OP BIN_OP VAR VAR NUMBER ASSIGN VAR VAR VAR ASSIGN VAR VAR BIN_OP VAR NUMBER ASSIGN VAR VAR VAR BIN_OP VAR NUMBER ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR VAR IF VAR VAR VAR VAR VAR VAR VAR BIN_OP VAR NUMBER ASSIGN VAR NUMBER IF VAR ASSIGN VAR VAR ASSIGN VAR VAR ASSIGN VAR VAR VAR ASSIGN VAR VAR BIN_OP VAR NUMBER ASSIGN VAR VAR VAR BIN_OP VAR NUMBER ASSIGN VAR LIST WHILE VAR VAR LIST FUNC_CALL VAR VAR LIST FUNC_CALL VAR VAR VAR VAR VAR EXPR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR FUNC_CALL STRING FUNC_CALL VAR VAR VAR
|
This is the hard version of the problem. The difference between the versions is that in the easy version all prices $a_i$ are different. You can make hacks if and only if you solved both versions of the problem.
Today is Sage's birthday, and she will go shopping to buy ice spheres. All $n$ ice spheres are placed in a row and they are numbered from $1$ to $n$ from left to right. Each ice sphere has a positive integer price. In this version, some prices can be equal.
An ice sphere is cheap if it costs strictly less than two neighboring ice spheres: the nearest to the left and the nearest to the right. The leftmost and the rightmost ice spheres are not cheap. Sage will choose all cheap ice spheres and then buy only them.
You can visit the shop before Sage and reorder the ice spheres as you wish. Find out the maximum number of ice spheres that Sage can buy, and show how the ice spheres should be reordered.
-----Input-----
The first line contains a single integer $n$ $(1 \le n \le 10^5)$ — the number of ice spheres in the shop.
The second line contains $n$ integers $a_1, a_2, \dots, a_n$ $(1 \le a_i \le 10^9)$ — the prices of ice spheres.
-----Output-----
In the first line print the maximum number of ice spheres that Sage can buy.
In the second line print the prices of ice spheres in the optimal order. If there are several correct answers, you can print any of them.
-----Example-----
Input
7
1 3 2 2 4 5 4
Output
3
3 1 4 2 4 2 5
-----Note-----
In the sample it's not possible to place the ice spheres in any order so that Sage would buy $4$ of them. If the spheres are placed in the order $(3, 1, 4, 2, 4, 2, 5)$, then Sage will buy one sphere for $1$ and two spheres for $2$ each.
|
n = int(input())
a = list(map(int, input().split()))
a.sort()
ind = n // 2
res = []
i = 0
f = True
for j in range(n):
if f == True:
res.append(a[ind])
ind += 1
f = False
else:
res.append(a[i])
i += 1
f = True
ans = 0
for i in range(1, n - 1):
if res[i - 1] > res[i] and res[i] < res[i + 1]:
ans += 1
print(ans)
print(*res)
|
ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR EXPR FUNC_CALL VAR ASSIGN VAR BIN_OP VAR NUMBER ASSIGN VAR LIST ASSIGN VAR NUMBER ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR VAR IF VAR NUMBER EXPR FUNC_CALL VAR VAR VAR VAR NUMBER ASSIGN VAR NUMBER EXPR FUNC_CALL VAR VAR VAR VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR NUMBER BIN_OP VAR NUMBER IF VAR BIN_OP VAR NUMBER VAR VAR VAR VAR VAR BIN_OP VAR NUMBER VAR NUMBER EXPR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR VAR
|
This is the hard version of the problem. The difference between the versions is that in the easy version all prices $a_i$ are different. You can make hacks if and only if you solved both versions of the problem.
Today is Sage's birthday, and she will go shopping to buy ice spheres. All $n$ ice spheres are placed in a row and they are numbered from $1$ to $n$ from left to right. Each ice sphere has a positive integer price. In this version, some prices can be equal.
An ice sphere is cheap if it costs strictly less than two neighboring ice spheres: the nearest to the left and the nearest to the right. The leftmost and the rightmost ice spheres are not cheap. Sage will choose all cheap ice spheres and then buy only them.
You can visit the shop before Sage and reorder the ice spheres as you wish. Find out the maximum number of ice spheres that Sage can buy, and show how the ice spheres should be reordered.
-----Input-----
The first line contains a single integer $n$ $(1 \le n \le 10^5)$ — the number of ice spheres in the shop.
The second line contains $n$ integers $a_1, a_2, \dots, a_n$ $(1 \le a_i \le 10^9)$ — the prices of ice spheres.
-----Output-----
In the first line print the maximum number of ice spheres that Sage can buy.
In the second line print the prices of ice spheres in the optimal order. If there are several correct answers, you can print any of them.
-----Example-----
Input
7
1 3 2 2 4 5 4
Output
3
3 1 4 2 4 2 5
-----Note-----
In the sample it's not possible to place the ice spheres in any order so that Sage would buy $4$ of them. If the spheres are placed in the order $(3, 1, 4, 2, 4, 2, 5)$, then Sage will buy one sphere for $1$ and two spheres for $2$ each.
|
n = int(input())
d = [int(i) for i in input().split()]
d.sort()
if n < 3 or d.count(d[0]) >= n - 1:
print(0)
print(" ".join(map(str, d)))
exit()
mxd = [min(d[i], d[i + 1]) for i in range(n - 1)]
hi = (n - 3) // 2
lo = 0
while lo < hi:
mid = (lo + hi + 1) // 2
kk = n - mid - 2
if sum(mxd[kk + i] > d[i] for i in range(mid + 1)) == mid + 1:
lo = mid
else:
hi = mid - 1
print(lo + 1)
ans = []
for i in range(lo + 1):
ans.append(d[n - lo - 2 + i])
ans.append(d[i])
ans.append(d[-1])
ans.extend(d[lo + 1 : n - lo - 2])
print(" ".join(map(str, ans)))
|
ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR VAR VAR FUNC_CALL FUNC_CALL VAR EXPR FUNC_CALL VAR IF VAR NUMBER FUNC_CALL VAR VAR NUMBER BIN_OP VAR NUMBER EXPR FUNC_CALL VAR NUMBER EXPR FUNC_CALL VAR FUNC_CALL STRING FUNC_CALL VAR VAR VAR EXPR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR VAR VAR VAR BIN_OP VAR NUMBER VAR FUNC_CALL VAR BIN_OP VAR NUMBER ASSIGN VAR BIN_OP BIN_OP VAR NUMBER NUMBER ASSIGN VAR NUMBER WHILE VAR VAR ASSIGN VAR BIN_OP BIN_OP BIN_OP VAR VAR NUMBER NUMBER ASSIGN VAR BIN_OP BIN_OP VAR VAR NUMBER IF FUNC_CALL VAR VAR BIN_OP VAR VAR VAR VAR VAR FUNC_CALL VAR BIN_OP VAR NUMBER BIN_OP VAR NUMBER ASSIGN VAR VAR ASSIGN VAR BIN_OP VAR NUMBER EXPR FUNC_CALL VAR BIN_OP VAR NUMBER ASSIGN VAR LIST FOR VAR FUNC_CALL VAR BIN_OP VAR NUMBER EXPR FUNC_CALL VAR VAR BIN_OP BIN_OP BIN_OP VAR VAR NUMBER VAR EXPR FUNC_CALL VAR VAR VAR EXPR FUNC_CALL VAR VAR NUMBER EXPR FUNC_CALL VAR VAR BIN_OP VAR NUMBER BIN_OP BIN_OP VAR VAR NUMBER EXPR FUNC_CALL VAR FUNC_CALL STRING FUNC_CALL VAR VAR VAR
|
This is the hard version of the problem. The difference between the versions is that in the easy version all prices $a_i$ are different. You can make hacks if and only if you solved both versions of the problem.
Today is Sage's birthday, and she will go shopping to buy ice spheres. All $n$ ice spheres are placed in a row and they are numbered from $1$ to $n$ from left to right. Each ice sphere has a positive integer price. In this version, some prices can be equal.
An ice sphere is cheap if it costs strictly less than two neighboring ice spheres: the nearest to the left and the nearest to the right. The leftmost and the rightmost ice spheres are not cheap. Sage will choose all cheap ice spheres and then buy only them.
You can visit the shop before Sage and reorder the ice spheres as you wish. Find out the maximum number of ice spheres that Sage can buy, and show how the ice spheres should be reordered.
-----Input-----
The first line contains a single integer $n$ $(1 \le n \le 10^5)$ — the number of ice spheres in the shop.
The second line contains $n$ integers $a_1, a_2, \dots, a_n$ $(1 \le a_i \le 10^9)$ — the prices of ice spheres.
-----Output-----
In the first line print the maximum number of ice spheres that Sage can buy.
In the second line print the prices of ice spheres in the optimal order. If there are several correct answers, you can print any of them.
-----Example-----
Input
7
1 3 2 2 4 5 4
Output
3
3 1 4 2 4 2 5
-----Note-----
In the sample it's not possible to place the ice spheres in any order so that Sage would buy $4$ of them. If the spheres are placed in the order $(3, 1, 4, 2, 4, 2, 5)$, then Sage will buy one sphere for $1$ and two spheres for $2$ each.
|
n = int(input())
a = list(map(int, input().split()))
a.sort()
l = len(a) // 2
f = a[:l]
s = a[l:]
t = [0] * len(a)
ind = 0
for i in range(len(s)):
t[ind] = s[i]
ind += 2
ind = 1
for i in range(len(f)):
t[ind] = f[i]
ind += 2
x = 0
for i in range(1, len(t) - 1):
if t[i] < t[i + 1] and t[i] < t[i - 1]:
x += 1
print(x)
print(" ".join([str(i) for i in t]))
|
ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR EXPR FUNC_CALL VAR ASSIGN VAR BIN_OP FUNC_CALL VAR VAR NUMBER ASSIGN VAR VAR VAR ASSIGN VAR VAR VAR ASSIGN VAR BIN_OP LIST NUMBER FUNC_CALL VAR VAR ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR FUNC_CALL VAR VAR ASSIGN VAR VAR VAR VAR VAR NUMBER ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR FUNC_CALL VAR VAR ASSIGN VAR VAR VAR VAR VAR NUMBER ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR NUMBER BIN_OP FUNC_CALL VAR VAR NUMBER IF VAR VAR VAR BIN_OP VAR NUMBER VAR VAR VAR BIN_OP VAR NUMBER VAR NUMBER EXPR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR FUNC_CALL STRING FUNC_CALL VAR VAR VAR VAR
|
This is the hard version of the problem. The difference between the versions is that in the easy version all prices $a_i$ are different. You can make hacks if and only if you solved both versions of the problem.
Today is Sage's birthday, and she will go shopping to buy ice spheres. All $n$ ice spheres are placed in a row and they are numbered from $1$ to $n$ from left to right. Each ice sphere has a positive integer price. In this version, some prices can be equal.
An ice sphere is cheap if it costs strictly less than two neighboring ice spheres: the nearest to the left and the nearest to the right. The leftmost and the rightmost ice spheres are not cheap. Sage will choose all cheap ice spheres and then buy only them.
You can visit the shop before Sage and reorder the ice spheres as you wish. Find out the maximum number of ice spheres that Sage can buy, and show how the ice spheres should be reordered.
-----Input-----
The first line contains a single integer $n$ $(1 \le n \le 10^5)$ — the number of ice spheres in the shop.
The second line contains $n$ integers $a_1, a_2, \dots, a_n$ $(1 \le a_i \le 10^9)$ — the prices of ice spheres.
-----Output-----
In the first line print the maximum number of ice spheres that Sage can buy.
In the second line print the prices of ice spheres in the optimal order. If there are several correct answers, you can print any of them.
-----Example-----
Input
7
1 3 2 2 4 5 4
Output
3
3 1 4 2 4 2 5
-----Note-----
In the sample it's not possible to place the ice spheres in any order so that Sage would buy $4$ of them. If the spheres are placed in the order $(3, 1, 4, 2, 4, 2, 5)$, then Sage will buy one sphere for $1$ and two spheres for $2$ each.
|
n = int(input())
arr = list(map(int, input().strip().split()))
if n % 2 == 0:
ans = n // 2 - 1
else:
ans = n // 2
arr.sort()
low = arr[: n // 2]
hi = arr[n // 2 :]
arr[::2] = hi
arr[1::2] = low
for i in range(1, n - 1, 2):
if arr[i] == arr[i - 1] or arr[i] == arr[i + 1]:
ans -= 1
print(ans)
print(*arr, sep=" ")
|
ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL FUNC_CALL VAR IF BIN_OP VAR NUMBER NUMBER ASSIGN VAR BIN_OP BIN_OP VAR NUMBER NUMBER ASSIGN VAR BIN_OP VAR NUMBER EXPR FUNC_CALL VAR ASSIGN VAR VAR BIN_OP VAR NUMBER ASSIGN VAR VAR BIN_OP VAR NUMBER ASSIGN VAR NUMBER VAR ASSIGN VAR NUMBER NUMBER VAR FOR VAR FUNC_CALL VAR NUMBER BIN_OP VAR NUMBER NUMBER IF VAR VAR VAR BIN_OP VAR NUMBER VAR VAR VAR BIN_OP VAR NUMBER VAR NUMBER EXPR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR VAR STRING
|
This is the hard version of the problem. The difference between the versions is that in the easy version all prices $a_i$ are different. You can make hacks if and only if you solved both versions of the problem.
Today is Sage's birthday, and she will go shopping to buy ice spheres. All $n$ ice spheres are placed in a row and they are numbered from $1$ to $n$ from left to right. Each ice sphere has a positive integer price. In this version, some prices can be equal.
An ice sphere is cheap if it costs strictly less than two neighboring ice spheres: the nearest to the left and the nearest to the right. The leftmost and the rightmost ice spheres are not cheap. Sage will choose all cheap ice spheres and then buy only them.
You can visit the shop before Sage and reorder the ice spheres as you wish. Find out the maximum number of ice spheres that Sage can buy, and show how the ice spheres should be reordered.
-----Input-----
The first line contains a single integer $n$ $(1 \le n \le 10^5)$ — the number of ice spheres in the shop.
The second line contains $n$ integers $a_1, a_2, \dots, a_n$ $(1 \le a_i \le 10^9)$ — the prices of ice spheres.
-----Output-----
In the first line print the maximum number of ice spheres that Sage can buy.
In the second line print the prices of ice spheres in the optimal order. If there are several correct answers, you can print any of them.
-----Example-----
Input
7
1 3 2 2 4 5 4
Output
3
3 1 4 2 4 2 5
-----Note-----
In the sample it's not possible to place the ice spheres in any order so that Sage would buy $4$ of them. If the spheres are placed in the order $(3, 1, 4, 2, 4, 2, 5)$, then Sage will buy one sphere for $1$ and two spheres for $2$ each.
|
n = int(input())
a = list(map(int, input().split()))
a.sort()
g = []
h = []
for j in range(n // 2):
g.append(a[j])
for j in range(n // 2, n):
h.append(a[j])
gx = len(g)
hx = len(h)
gy = 0
hy = 0
c = 0
k = []
while gy < gx or hy < hx:
if gy < gx and hy < hx:
if g[gy] < h[hy]:
k.append(h[hy])
k.append(g[gy])
hy = hy + 1
gy = gy + 1
if hy < hx:
c = c + 1
elif g[gy] == h[hy]:
k.append(h[hy])
hy = hy + 1
elif gy == gx:
k.append(h[hy])
hy = hy + 1
else:
k.append(g[gy])
gy = gy + 1
print(c)
print(" ".join(map(str, k)))
|
ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR EXPR FUNC_CALL VAR ASSIGN VAR LIST ASSIGN VAR LIST FOR VAR FUNC_CALL VAR BIN_OP VAR NUMBER EXPR FUNC_CALL VAR VAR VAR FOR VAR FUNC_CALL VAR BIN_OP VAR NUMBER VAR EXPR FUNC_CALL VAR VAR VAR ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR LIST WHILE VAR VAR VAR VAR IF VAR VAR VAR VAR IF VAR VAR VAR VAR EXPR FUNC_CALL VAR VAR VAR EXPR FUNC_CALL VAR VAR VAR ASSIGN VAR BIN_OP VAR NUMBER ASSIGN VAR BIN_OP VAR NUMBER IF VAR VAR ASSIGN VAR BIN_OP VAR NUMBER IF VAR VAR VAR VAR EXPR FUNC_CALL VAR VAR VAR ASSIGN VAR BIN_OP VAR NUMBER IF VAR VAR EXPR FUNC_CALL VAR VAR VAR ASSIGN VAR BIN_OP VAR NUMBER EXPR FUNC_CALL VAR VAR VAR ASSIGN VAR BIN_OP VAR NUMBER EXPR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR FUNC_CALL STRING FUNC_CALL VAR VAR VAR
|
This is the hard version of the problem. The difference between the versions is that in the easy version all prices $a_i$ are different. You can make hacks if and only if you solved both versions of the problem.
Today is Sage's birthday, and she will go shopping to buy ice spheres. All $n$ ice spheres are placed in a row and they are numbered from $1$ to $n$ from left to right. Each ice sphere has a positive integer price. In this version, some prices can be equal.
An ice sphere is cheap if it costs strictly less than two neighboring ice spheres: the nearest to the left and the nearest to the right. The leftmost and the rightmost ice spheres are not cheap. Sage will choose all cheap ice spheres and then buy only them.
You can visit the shop before Sage and reorder the ice spheres as you wish. Find out the maximum number of ice spheres that Sage can buy, and show how the ice spheres should be reordered.
-----Input-----
The first line contains a single integer $n$ $(1 \le n \le 10^5)$ — the number of ice spheres in the shop.
The second line contains $n$ integers $a_1, a_2, \dots, a_n$ $(1 \le a_i \le 10^9)$ — the prices of ice spheres.
-----Output-----
In the first line print the maximum number of ice spheres that Sage can buy.
In the second line print the prices of ice spheres in the optimal order. If there are several correct answers, you can print any of them.
-----Example-----
Input
7
1 3 2 2 4 5 4
Output
3
3 1 4 2 4 2 5
-----Note-----
In the sample it's not possible to place the ice spheres in any order so that Sage would buy $4$ of them. If the spheres are placed in the order $(3, 1, 4, 2, 4, 2, 5)$, then Sage will buy one sphere for $1$ and two spheres for $2$ each.
|
n = int(input())
arr = list(map(int, input().split()))
flag = 1
if n <= 2:
print(0)
print(*arr)
flag = 0
arr.sort(reverse=True)
final = ["0"]
pair = 0
if flag == 1:
k = 0
j = 1
if n % 2 != 0:
temp1 = arr[: n // 2 + 1]
temp2 = arr[n // 2 + 1 :]
else:
temp1 = arr[: n // 2]
temp2 = arr[n // 2 :]
rem_val = temp2.pop(0)
for i in range(0, len(temp2)):
if temp1[k] == temp2[i] or temp1[j] == temp2[i]:
pass
else:
pair += 1
final.pop(-1)
final.append(temp1[k])
final.append(temp2[i])
final.append(temp1[j])
k += 1
j += 1
if n % 2 == 0:
final.append(rem_val)
if flag == 1:
print(pair)
print(*final)
|
ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR NUMBER IF VAR NUMBER EXPR FUNC_CALL VAR NUMBER EXPR FUNC_CALL VAR VAR ASSIGN VAR NUMBER EXPR FUNC_CALL VAR NUMBER ASSIGN VAR LIST STRING ASSIGN VAR NUMBER IF VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR NUMBER IF BIN_OP VAR NUMBER NUMBER ASSIGN VAR VAR BIN_OP BIN_OP VAR NUMBER NUMBER ASSIGN VAR VAR BIN_OP BIN_OP VAR NUMBER NUMBER ASSIGN VAR VAR BIN_OP VAR NUMBER ASSIGN VAR VAR BIN_OP VAR NUMBER ASSIGN VAR FUNC_CALL VAR NUMBER FOR VAR FUNC_CALL VAR NUMBER FUNC_CALL VAR VAR IF VAR VAR VAR VAR VAR VAR VAR VAR VAR NUMBER EXPR FUNC_CALL VAR NUMBER EXPR FUNC_CALL VAR VAR VAR EXPR FUNC_CALL VAR VAR VAR EXPR FUNC_CALL VAR VAR VAR VAR NUMBER VAR NUMBER IF BIN_OP VAR NUMBER NUMBER EXPR FUNC_CALL VAR VAR IF VAR NUMBER EXPR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR VAR
|
This is the hard version of the problem. The difference between the versions is that in the easy version all prices $a_i$ are different. You can make hacks if and only if you solved both versions of the problem.
Today is Sage's birthday, and she will go shopping to buy ice spheres. All $n$ ice spheres are placed in a row and they are numbered from $1$ to $n$ from left to right. Each ice sphere has a positive integer price. In this version, some prices can be equal.
An ice sphere is cheap if it costs strictly less than two neighboring ice spheres: the nearest to the left and the nearest to the right. The leftmost and the rightmost ice spheres are not cheap. Sage will choose all cheap ice spheres and then buy only them.
You can visit the shop before Sage and reorder the ice spheres as you wish. Find out the maximum number of ice spheres that Sage can buy, and show how the ice spheres should be reordered.
-----Input-----
The first line contains a single integer $n$ $(1 \le n \le 10^5)$ — the number of ice spheres in the shop.
The second line contains $n$ integers $a_1, a_2, \dots, a_n$ $(1 \le a_i \le 10^9)$ — the prices of ice spheres.
-----Output-----
In the first line print the maximum number of ice spheres that Sage can buy.
In the second line print the prices of ice spheres in the optimal order. If there are several correct answers, you can print any of them.
-----Example-----
Input
7
1 3 2 2 4 5 4
Output
3
3 1 4 2 4 2 5
-----Note-----
In the sample it's not possible to place the ice spheres in any order so that Sage would buy $4$ of them. If the spheres are placed in the order $(3, 1, 4, 2, 4, 2, 5)$, then Sage will buy one sphere for $1$ and two spheres for $2$ each.
|
n = int(input())
lst = list(map(int, input().split()))
lst.sort()
result = lst[n // 2 :]
for i in range(n // 2):
result.insert(2 * i + 1, lst[i])
cnt = 0
for i in range(1, n - 1):
if result[i - 1] > result[i] and result[i] < result[i + 1]:
cnt += 1
print(cnt)
print(" ".join(map(str, result)))
|
ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR EXPR FUNC_CALL VAR ASSIGN VAR VAR BIN_OP VAR NUMBER FOR VAR FUNC_CALL VAR BIN_OP VAR NUMBER EXPR FUNC_CALL VAR BIN_OP BIN_OP NUMBER VAR NUMBER VAR VAR ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR NUMBER BIN_OP VAR NUMBER IF VAR BIN_OP VAR NUMBER VAR VAR VAR VAR VAR BIN_OP VAR NUMBER VAR NUMBER EXPR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR FUNC_CALL STRING FUNC_CALL VAR VAR VAR
|
This is the hard version of the problem. The difference between the versions is that in the easy version all prices $a_i$ are different. You can make hacks if and only if you solved both versions of the problem.
Today is Sage's birthday, and she will go shopping to buy ice spheres. All $n$ ice spheres are placed in a row and they are numbered from $1$ to $n$ from left to right. Each ice sphere has a positive integer price. In this version, some prices can be equal.
An ice sphere is cheap if it costs strictly less than two neighboring ice spheres: the nearest to the left and the nearest to the right. The leftmost and the rightmost ice spheres are not cheap. Sage will choose all cheap ice spheres and then buy only them.
You can visit the shop before Sage and reorder the ice spheres as you wish. Find out the maximum number of ice spheres that Sage can buy, and show how the ice spheres should be reordered.
-----Input-----
The first line contains a single integer $n$ $(1 \le n \le 10^5)$ — the number of ice spheres in the shop.
The second line contains $n$ integers $a_1, a_2, \dots, a_n$ $(1 \le a_i \le 10^9)$ — the prices of ice spheres.
-----Output-----
In the first line print the maximum number of ice spheres that Sage can buy.
In the second line print the prices of ice spheres in the optimal order. If there are several correct answers, you can print any of them.
-----Example-----
Input
7
1 3 2 2 4 5 4
Output
3
3 1 4 2 4 2 5
-----Note-----
In the sample it's not possible to place the ice spheres in any order so that Sage would buy $4$ of them. If the spheres are placed in the order $(3, 1, 4, 2, 4, 2, 5)$, then Sage will buy one sphere for $1$ and two spheres for $2$ each.
|
import sys
input = sys.stdin.readline
n = int(input())
a = [int(item) for item in input().split()]
a.sort()
ans = [-1] * n
num = 0
if n % 2 == 1:
for i in range(0, n, 2):
ans[i] = a.pop()
for i in range(1, n, 2):
if ans[i - 1] > a[-1] < ans[i + 1]:
num += 1
ans[i] = a.pop()
print(num)
print(*ans)
else:
for i in range(0, n, 2):
ans[i] = a.pop()
ans[-1] = a.pop()
for i in range(1, n - 1, 2):
if ans[i - 1] > a[-1] < ans[i + 1]:
num += 1
ans[i] = a.pop()
print(num)
print(*ans)
|
IMPORT ASSIGN VAR VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR VAR VAR FUNC_CALL FUNC_CALL VAR EXPR FUNC_CALL VAR ASSIGN VAR BIN_OP LIST NUMBER VAR ASSIGN VAR NUMBER IF BIN_OP VAR NUMBER NUMBER FOR VAR FUNC_CALL VAR NUMBER VAR NUMBER ASSIGN VAR VAR FUNC_CALL VAR FOR VAR FUNC_CALL VAR NUMBER VAR NUMBER IF VAR BIN_OP VAR NUMBER VAR NUMBER VAR BIN_OP VAR NUMBER VAR NUMBER ASSIGN VAR VAR FUNC_CALL VAR EXPR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR VAR FOR VAR FUNC_CALL VAR NUMBER VAR NUMBER ASSIGN VAR VAR FUNC_CALL VAR ASSIGN VAR NUMBER FUNC_CALL VAR FOR VAR FUNC_CALL VAR NUMBER BIN_OP VAR NUMBER NUMBER IF VAR BIN_OP VAR NUMBER VAR NUMBER VAR BIN_OP VAR NUMBER VAR NUMBER ASSIGN VAR VAR FUNC_CALL VAR EXPR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR VAR
|
This is the hard version of the problem. The difference between the versions is that in the easy version all prices $a_i$ are different. You can make hacks if and only if you solved both versions of the problem.
Today is Sage's birthday, and she will go shopping to buy ice spheres. All $n$ ice spheres are placed in a row and they are numbered from $1$ to $n$ from left to right. Each ice sphere has a positive integer price. In this version, some prices can be equal.
An ice sphere is cheap if it costs strictly less than two neighboring ice spheres: the nearest to the left and the nearest to the right. The leftmost and the rightmost ice spheres are not cheap. Sage will choose all cheap ice spheres and then buy only them.
You can visit the shop before Sage and reorder the ice spheres as you wish. Find out the maximum number of ice spheres that Sage can buy, and show how the ice spheres should be reordered.
-----Input-----
The first line contains a single integer $n$ $(1 \le n \le 10^5)$ — the number of ice spheres in the shop.
The second line contains $n$ integers $a_1, a_2, \dots, a_n$ $(1 \le a_i \le 10^9)$ — the prices of ice spheres.
-----Output-----
In the first line print the maximum number of ice spheres that Sage can buy.
In the second line print the prices of ice spheres in the optimal order. If there are several correct answers, you can print any of them.
-----Example-----
Input
7
1 3 2 2 4 5 4
Output
3
3 1 4 2 4 2 5
-----Note-----
In the sample it's not possible to place the ice spheres in any order so that Sage would buy $4$ of them. If the spheres are placed in the order $(3, 1, 4, 2, 4, 2, 5)$, then Sage will buy one sphere for $1$ and two spheres for $2$ each.
|
n = int(input())
arr = list(map(int, input().split()))
res = [0] * n
arr.sort()
for i in range(n):
if i < n // 2:
res[i * 2 + 1] = arr[i]
else:
res[(i - n // 2) * 2] = arr[i]
x = 0
for i in range(1, n - 1):
if res[i - 1] > res[i] < res[i + 1]:
x += 1
print(x)
print(*res)
|
ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR BIN_OP LIST NUMBER VAR EXPR FUNC_CALL VAR FOR VAR FUNC_CALL VAR VAR IF VAR BIN_OP VAR NUMBER ASSIGN VAR BIN_OP BIN_OP VAR NUMBER NUMBER VAR VAR ASSIGN VAR BIN_OP BIN_OP VAR BIN_OP VAR NUMBER NUMBER VAR VAR ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR NUMBER BIN_OP VAR NUMBER IF VAR BIN_OP VAR NUMBER VAR VAR VAR BIN_OP VAR NUMBER VAR NUMBER EXPR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR VAR
|
This is the hard version of the problem. The difference between the versions is that in the easy version all prices $a_i$ are different. You can make hacks if and only if you solved both versions of the problem.
Today is Sage's birthday, and she will go shopping to buy ice spheres. All $n$ ice spheres are placed in a row and they are numbered from $1$ to $n$ from left to right. Each ice sphere has a positive integer price. In this version, some prices can be equal.
An ice sphere is cheap if it costs strictly less than two neighboring ice spheres: the nearest to the left and the nearest to the right. The leftmost and the rightmost ice spheres are not cheap. Sage will choose all cheap ice spheres and then buy only them.
You can visit the shop before Sage and reorder the ice spheres as you wish. Find out the maximum number of ice spheres that Sage can buy, and show how the ice spheres should be reordered.
-----Input-----
The first line contains a single integer $n$ $(1 \le n \le 10^5)$ — the number of ice spheres in the shop.
The second line contains $n$ integers $a_1, a_2, \dots, a_n$ $(1 \le a_i \le 10^9)$ — the prices of ice spheres.
-----Output-----
In the first line print the maximum number of ice spheres that Sage can buy.
In the second line print the prices of ice spheres in the optimal order. If there are several correct answers, you can print any of them.
-----Example-----
Input
7
1 3 2 2 4 5 4
Output
3
3 1 4 2 4 2 5
-----Note-----
In the sample it's not possible to place the ice spheres in any order so that Sage would buy $4$ of them. If the spheres are placed in the order $(3, 1, 4, 2, 4, 2, 5)$, then Sage will buy one sphere for $1$ and two spheres for $2$ each.
|
n = int(input())
arr = list(map(int, input().split()))
arr.sort()
ans = [-1] * n
k = 0
c = 0
for i in range(1, n, 2):
ans[i] = arr[k]
k += 1
for i in range(n):
if ans[i] == -1:
ans[i] = arr[k]
k += 1
for i in range(1, n - 1, 2):
if ans[i] < ans[i - 1] and ans[i] < ans[i + 1]:
c += 1
print(c)
for i in range(n):
print(ans[i], end=" ")
|
ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR EXPR FUNC_CALL VAR ASSIGN VAR BIN_OP LIST NUMBER VAR ASSIGN VAR NUMBER ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR NUMBER VAR NUMBER ASSIGN VAR VAR VAR VAR VAR NUMBER FOR VAR FUNC_CALL VAR VAR IF VAR VAR NUMBER ASSIGN VAR VAR VAR VAR VAR NUMBER FOR VAR FUNC_CALL VAR NUMBER BIN_OP VAR NUMBER NUMBER IF VAR VAR VAR BIN_OP VAR NUMBER VAR VAR VAR BIN_OP VAR NUMBER VAR NUMBER EXPR FUNC_CALL VAR VAR FOR VAR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR VAR VAR STRING
|
This is the hard version of the problem. The difference between the versions is that in the easy version all prices $a_i$ are different. You can make hacks if and only if you solved both versions of the problem.
Today is Sage's birthday, and she will go shopping to buy ice spheres. All $n$ ice spheres are placed in a row and they are numbered from $1$ to $n$ from left to right. Each ice sphere has a positive integer price. In this version, some prices can be equal.
An ice sphere is cheap if it costs strictly less than two neighboring ice spheres: the nearest to the left and the nearest to the right. The leftmost and the rightmost ice spheres are not cheap. Sage will choose all cheap ice spheres and then buy only them.
You can visit the shop before Sage and reorder the ice spheres as you wish. Find out the maximum number of ice spheres that Sage can buy, and show how the ice spheres should be reordered.
-----Input-----
The first line contains a single integer $n$ $(1 \le n \le 10^5)$ — the number of ice spheres in the shop.
The second line contains $n$ integers $a_1, a_2, \dots, a_n$ $(1 \le a_i \le 10^9)$ — the prices of ice spheres.
-----Output-----
In the first line print the maximum number of ice spheres that Sage can buy.
In the second line print the prices of ice spheres in the optimal order. If there are several correct answers, you can print any of them.
-----Example-----
Input
7
1 3 2 2 4 5 4
Output
3
3 1 4 2 4 2 5
-----Note-----
In the sample it's not possible to place the ice spheres in any order so that Sage would buy $4$ of them. If the spheres are placed in the order $(3, 1, 4, 2, 4, 2, 5)$, then Sage will buy one sphere for $1$ and two spheres for $2$ each.
|
n = int(input())
l = list(map(int, input().split()))
l.sort()
j = 0
mid = len(l) // 2
a = [0] * (n + 2)
for i in range(mid):
a[i + j], a[i + 1 + j] = l[mid + i], l[i]
j += 1
if n % 2 != 0:
a[n - 1] = l[n - 1]
p = 0
for i in range(1, n):
if a[i - 1] > a[i] and a[i] < a[i + 1]:
p += 1
print(p)
a = a[:n]
s = " ".join(map(str, a))
print(s)
|
ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR EXPR FUNC_CALL VAR ASSIGN VAR NUMBER ASSIGN VAR BIN_OP FUNC_CALL VAR VAR NUMBER ASSIGN VAR BIN_OP LIST NUMBER BIN_OP VAR NUMBER FOR VAR FUNC_CALL VAR VAR ASSIGN VAR BIN_OP VAR VAR VAR BIN_OP BIN_OP VAR NUMBER VAR VAR BIN_OP VAR VAR VAR VAR VAR NUMBER IF BIN_OP VAR NUMBER NUMBER ASSIGN VAR BIN_OP VAR NUMBER VAR BIN_OP VAR NUMBER ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR NUMBER VAR IF VAR BIN_OP VAR NUMBER VAR VAR VAR VAR VAR BIN_OP VAR NUMBER VAR NUMBER EXPR FUNC_CALL VAR VAR ASSIGN VAR VAR VAR ASSIGN VAR FUNC_CALL STRING FUNC_CALL VAR VAR VAR EXPR FUNC_CALL VAR VAR
|
This is the hard version of the problem. The difference between the versions is that in the easy version all prices $a_i$ are different. You can make hacks if and only if you solved both versions of the problem.
Today is Sage's birthday, and she will go shopping to buy ice spheres. All $n$ ice spheres are placed in a row and they are numbered from $1$ to $n$ from left to right. Each ice sphere has a positive integer price. In this version, some prices can be equal.
An ice sphere is cheap if it costs strictly less than two neighboring ice spheres: the nearest to the left and the nearest to the right. The leftmost and the rightmost ice spheres are not cheap. Sage will choose all cheap ice spheres and then buy only them.
You can visit the shop before Sage and reorder the ice spheres as you wish. Find out the maximum number of ice spheres that Sage can buy, and show how the ice spheres should be reordered.
-----Input-----
The first line contains a single integer $n$ $(1 \le n \le 10^5)$ — the number of ice spheres in the shop.
The second line contains $n$ integers $a_1, a_2, \dots, a_n$ $(1 \le a_i \le 10^9)$ — the prices of ice spheres.
-----Output-----
In the first line print the maximum number of ice spheres that Sage can buy.
In the second line print the prices of ice spheres in the optimal order. If there are several correct answers, you can print any of them.
-----Example-----
Input
7
1 3 2 2 4 5 4
Output
3
3 1 4 2 4 2 5
-----Note-----
In the sample it's not possible to place the ice spheres in any order so that Sage would buy $4$ of them. If the spheres are placed in the order $(3, 1, 4, 2, 4, 2, 5)$, then Sage will buy one sphere for $1$ and two spheres for $2$ each.
|
R = lambda: map(int, input().split())
n = int(input())
arr = sorted(R(), reverse=True)
vst = [0] * n
res = []
j = 0
k = (n + 1) // 2
cnt = 0
for i in range(n):
if not vst[i]:
res.append(arr[i])
j = max(j, i + 1)
while j < n and vst[j]:
j += 1
k = max(k, j + 1)
while k < n and (vst[k] or arr[k] >= arr[j]):
k += 1
if k < n:
res.append(arr[k])
vst[k] = 1
cnt += 1
print(cnt)
print(" ".join(map(str, res)))
|
ASSIGN VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR NUMBER ASSIGN VAR BIN_OP LIST NUMBER VAR ASSIGN VAR LIST ASSIGN VAR NUMBER ASSIGN VAR BIN_OP BIN_OP VAR NUMBER NUMBER ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR VAR IF VAR VAR EXPR FUNC_CALL VAR VAR VAR ASSIGN VAR FUNC_CALL VAR VAR BIN_OP VAR NUMBER WHILE VAR VAR VAR VAR VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR BIN_OP VAR NUMBER WHILE VAR VAR VAR VAR VAR VAR VAR VAR VAR NUMBER IF VAR VAR EXPR FUNC_CALL VAR VAR VAR ASSIGN VAR VAR NUMBER VAR NUMBER EXPR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR FUNC_CALL STRING FUNC_CALL VAR VAR VAR
|
This is the hard version of the problem. The difference between the versions is that in the easy version all prices $a_i$ are different. You can make hacks if and only if you solved both versions of the problem.
Today is Sage's birthday, and she will go shopping to buy ice spheres. All $n$ ice spheres are placed in a row and they are numbered from $1$ to $n$ from left to right. Each ice sphere has a positive integer price. In this version, some prices can be equal.
An ice sphere is cheap if it costs strictly less than two neighboring ice spheres: the nearest to the left and the nearest to the right. The leftmost and the rightmost ice spheres are not cheap. Sage will choose all cheap ice spheres and then buy only them.
You can visit the shop before Sage and reorder the ice spheres as you wish. Find out the maximum number of ice spheres that Sage can buy, and show how the ice spheres should be reordered.
-----Input-----
The first line contains a single integer $n$ $(1 \le n \le 10^5)$ — the number of ice spheres in the shop.
The second line contains $n$ integers $a_1, a_2, \dots, a_n$ $(1 \le a_i \le 10^9)$ — the prices of ice spheres.
-----Output-----
In the first line print the maximum number of ice spheres that Sage can buy.
In the second line print the prices of ice spheres in the optimal order. If there are several correct answers, you can print any of them.
-----Example-----
Input
7
1 3 2 2 4 5 4
Output
3
3 1 4 2 4 2 5
-----Note-----
In the sample it's not possible to place the ice spheres in any order so that Sage would buy $4$ of them. If the spheres are placed in the order $(3, 1, 4, 2, 4, 2, 5)$, then Sage will buy one sphere for $1$ and two spheres for $2$ each.
|
def main():
n = int(input())
a = map(int, input().split())
a = sorted(a)
result = [-1] * n
i = 0
j = n - 1 - 1
l = 0
r = n - 1
if n % 2 == 0:
i += 1
while j > 0:
result[i] = a[r]
result[j] = a[l]
i += 2
j -= 2
r -= 1
l += 1
if i < n:
result[i] = a[r]
if j >= 0:
result[j] = a[l]
cnt = 0
for r in range(1, len(result) - 1):
if result[r - 1] > result[r] and result[r] < result[r + 1]:
cnt += 1
print(cnt)
print(*result)
main()
|
FUNC_DEF ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR BIN_OP LIST NUMBER VAR ASSIGN VAR NUMBER ASSIGN VAR BIN_OP BIN_OP VAR NUMBER NUMBER ASSIGN VAR NUMBER ASSIGN VAR BIN_OP VAR NUMBER IF BIN_OP VAR NUMBER NUMBER VAR NUMBER WHILE VAR NUMBER ASSIGN VAR VAR VAR VAR ASSIGN VAR VAR VAR VAR VAR NUMBER VAR NUMBER VAR NUMBER VAR NUMBER IF VAR VAR ASSIGN VAR VAR VAR VAR IF VAR NUMBER ASSIGN VAR VAR VAR VAR ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR NUMBER BIN_OP FUNC_CALL VAR VAR NUMBER IF VAR BIN_OP VAR NUMBER VAR VAR VAR VAR VAR BIN_OP VAR NUMBER VAR NUMBER EXPR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR
|
This is the hard version of the problem. The difference between the versions is that in the easy version all prices $a_i$ are different. You can make hacks if and only if you solved both versions of the problem.
Today is Sage's birthday, and she will go shopping to buy ice spheres. All $n$ ice spheres are placed in a row and they are numbered from $1$ to $n$ from left to right. Each ice sphere has a positive integer price. In this version, some prices can be equal.
An ice sphere is cheap if it costs strictly less than two neighboring ice spheres: the nearest to the left and the nearest to the right. The leftmost and the rightmost ice spheres are not cheap. Sage will choose all cheap ice spheres and then buy only them.
You can visit the shop before Sage and reorder the ice spheres as you wish. Find out the maximum number of ice spheres that Sage can buy, and show how the ice spheres should be reordered.
-----Input-----
The first line contains a single integer $n$ $(1 \le n \le 10^5)$ — the number of ice spheres in the shop.
The second line contains $n$ integers $a_1, a_2, \dots, a_n$ $(1 \le a_i \le 10^9)$ — the prices of ice spheres.
-----Output-----
In the first line print the maximum number of ice spheres that Sage can buy.
In the second line print the prices of ice spheres in the optimal order. If there are several correct answers, you can print any of them.
-----Example-----
Input
7
1 3 2 2 4 5 4
Output
3
3 1 4 2 4 2 5
-----Note-----
In the sample it's not possible to place the ice spheres in any order so that Sage would buy $4$ of them. If the spheres are placed in the order $(3, 1, 4, 2, 4, 2, 5)$, then Sage will buy one sphere for $1$ and two spheres for $2$ each.
|
from itertools import chain, zip_longest
n = int(input())
a = [int(s) for s in input().split(" ")]
a.sort(reverse=True)
cnt = 0
ans = [x for x in chain(*zip_longest(a[: n // 2 + 1], a[n // 2 + 1 :])) if x]
for i in range(1, n - 1):
if min(ans[i - 1], ans[i + 1]) > ans[i]:
cnt += 1
print(cnt)
print(*ans)
|
ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR VAR VAR FUNC_CALL FUNC_CALL VAR STRING EXPR FUNC_CALL VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR VAR VAR FUNC_CALL VAR FUNC_CALL VAR VAR BIN_OP BIN_OP VAR NUMBER NUMBER VAR BIN_OP BIN_OP VAR NUMBER NUMBER VAR FOR VAR FUNC_CALL VAR NUMBER BIN_OP VAR NUMBER IF FUNC_CALL VAR VAR BIN_OP VAR NUMBER VAR BIN_OP VAR NUMBER VAR VAR VAR NUMBER EXPR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR VAR
|
This is the hard version of the problem. The difference between the versions is that in the easy version all prices $a_i$ are different. You can make hacks if and only if you solved both versions of the problem.
Today is Sage's birthday, and she will go shopping to buy ice spheres. All $n$ ice spheres are placed in a row and they are numbered from $1$ to $n$ from left to right. Each ice sphere has a positive integer price. In this version, some prices can be equal.
An ice sphere is cheap if it costs strictly less than two neighboring ice spheres: the nearest to the left and the nearest to the right. The leftmost and the rightmost ice spheres are not cheap. Sage will choose all cheap ice spheres and then buy only them.
You can visit the shop before Sage and reorder the ice spheres as you wish. Find out the maximum number of ice spheres that Sage can buy, and show how the ice spheres should be reordered.
-----Input-----
The first line contains a single integer $n$ $(1 \le n \le 10^5)$ — the number of ice spheres in the shop.
The second line contains $n$ integers $a_1, a_2, \dots, a_n$ $(1 \le a_i \le 10^9)$ — the prices of ice spheres.
-----Output-----
In the first line print the maximum number of ice spheres that Sage can buy.
In the second line print the prices of ice spheres in the optimal order. If there are several correct answers, you can print any of them.
-----Example-----
Input
7
1 3 2 2 4 5 4
Output
3
3 1 4 2 4 2 5
-----Note-----
In the sample it's not possible to place the ice spheres in any order so that Sage would buy $4$ of them. If the spheres are placed in the order $(3, 1, 4, 2, 4, 2, 5)$, then Sage will buy one sphere for $1$ and two spheres for $2$ each.
|
n = int(input())
a = list(map(int, input().split()))
a.sort()
b = a[: n // 2]
c = a[n // 2 :]
A = []
for i in c:
A.append(i)
index = 1
indB = 0
while index < n:
A.insert(index, b[indB])
indB += 1
index += 2
index = 1
ans = 0
while index < len(A) - 1:
if A[index - 1] > A[index] < A[index + 1]:
ans += 1
index += 2
print(ans)
answer = ""
for i in A:
answer += str(i) + " "
print(answer)
|
ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR EXPR FUNC_CALL VAR ASSIGN VAR VAR BIN_OP VAR NUMBER ASSIGN VAR VAR BIN_OP VAR NUMBER ASSIGN VAR LIST FOR VAR VAR EXPR FUNC_CALL VAR VAR ASSIGN VAR NUMBER ASSIGN VAR NUMBER WHILE VAR VAR EXPR FUNC_CALL VAR VAR VAR VAR VAR NUMBER VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR NUMBER WHILE VAR BIN_OP FUNC_CALL VAR VAR NUMBER IF VAR BIN_OP VAR NUMBER VAR VAR VAR BIN_OP VAR NUMBER VAR NUMBER VAR NUMBER EXPR FUNC_CALL VAR VAR ASSIGN VAR STRING FOR VAR VAR VAR BIN_OP FUNC_CALL VAR VAR STRING EXPR FUNC_CALL VAR VAR
|
This is the hard version of the problem. The difference between the versions is that in the easy version all prices $a_i$ are different. You can make hacks if and only if you solved both versions of the problem.
Today is Sage's birthday, and she will go shopping to buy ice spheres. All $n$ ice spheres are placed in a row and they are numbered from $1$ to $n$ from left to right. Each ice sphere has a positive integer price. In this version, some prices can be equal.
An ice sphere is cheap if it costs strictly less than two neighboring ice spheres: the nearest to the left and the nearest to the right. The leftmost and the rightmost ice spheres are not cheap. Sage will choose all cheap ice spheres and then buy only them.
You can visit the shop before Sage and reorder the ice spheres as you wish. Find out the maximum number of ice spheres that Sage can buy, and show how the ice spheres should be reordered.
-----Input-----
The first line contains a single integer $n$ $(1 \le n \le 10^5)$ — the number of ice spheres in the shop.
The second line contains $n$ integers $a_1, a_2, \dots, a_n$ $(1 \le a_i \le 10^9)$ — the prices of ice spheres.
-----Output-----
In the first line print the maximum number of ice spheres that Sage can buy.
In the second line print the prices of ice spheres in the optimal order. If there are several correct answers, you can print any of them.
-----Example-----
Input
7
1 3 2 2 4 5 4
Output
3
3 1 4 2 4 2 5
-----Note-----
In the sample it's not possible to place the ice spheres in any order so that Sage would buy $4$ of them. If the spheres are placed in the order $(3, 1, 4, 2, 4, 2, 5)$, then Sage will buy one sphere for $1$ and two spheres for $2$ each.
|
N = int(input())
A = list(map(int, input().split()))
A.sort()
B = [0] * N
for i in range(1, N, 2):
B[i] = A[i // 2]
Astart = N // 2
Bstart = N - 2 if N % 2 == 0 else N - 1
rem = []
while Astart < N and A[Astart] == B[Bstart - 1]:
rem.append(A[Astart])
Astart += 1
for i in range(Bstart, -1, -2):
if Astart >= N:
B[i] = rem.pop()
else:
B[i] = A[Astart]
Astart += 1
ans = 0
for i in range(1, N - 1, 2):
if B[i - 1] > B[i] < B[i + 1]:
ans += 1
print(ans)
print(*B)
|
ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR EXPR FUNC_CALL VAR ASSIGN VAR BIN_OP LIST NUMBER VAR FOR VAR FUNC_CALL VAR NUMBER VAR NUMBER ASSIGN VAR VAR VAR BIN_OP VAR NUMBER ASSIGN VAR BIN_OP VAR NUMBER ASSIGN VAR BIN_OP VAR NUMBER NUMBER BIN_OP VAR NUMBER BIN_OP VAR NUMBER ASSIGN VAR LIST WHILE VAR VAR VAR VAR VAR BIN_OP VAR NUMBER EXPR FUNC_CALL VAR VAR VAR VAR NUMBER FOR VAR FUNC_CALL VAR VAR NUMBER NUMBER IF VAR VAR ASSIGN VAR VAR FUNC_CALL VAR ASSIGN VAR VAR VAR VAR VAR NUMBER ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR NUMBER BIN_OP VAR NUMBER NUMBER IF VAR BIN_OP VAR NUMBER VAR VAR VAR BIN_OP VAR NUMBER VAR NUMBER EXPR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR VAR
|
This is the hard version of the problem. The difference between the versions is that in the easy version all prices $a_i$ are different. You can make hacks if and only if you solved both versions of the problem.
Today is Sage's birthday, and she will go shopping to buy ice spheres. All $n$ ice spheres are placed in a row and they are numbered from $1$ to $n$ from left to right. Each ice sphere has a positive integer price. In this version, some prices can be equal.
An ice sphere is cheap if it costs strictly less than two neighboring ice spheres: the nearest to the left and the nearest to the right. The leftmost and the rightmost ice spheres are not cheap. Sage will choose all cheap ice spheres and then buy only them.
You can visit the shop before Sage and reorder the ice spheres as you wish. Find out the maximum number of ice spheres that Sage can buy, and show how the ice spheres should be reordered.
-----Input-----
The first line contains a single integer $n$ $(1 \le n \le 10^5)$ — the number of ice spheres in the shop.
The second line contains $n$ integers $a_1, a_2, \dots, a_n$ $(1 \le a_i \le 10^9)$ — the prices of ice spheres.
-----Output-----
In the first line print the maximum number of ice spheres that Sage can buy.
In the second line print the prices of ice spheres in the optimal order. If there are several correct answers, you can print any of them.
-----Example-----
Input
7
1 3 2 2 4 5 4
Output
3
3 1 4 2 4 2 5
-----Note-----
In the sample it's not possible to place the ice spheres in any order so that Sage would buy $4$ of them. If the spheres are placed in the order $(3, 1, 4, 2, 4, 2, 5)$, then Sage will buy one sphere for $1$ and two spheres for $2$ each.
|
n = int(input())
a = sorted(map(int, input().split()))
l = [a[(1 - i % 2) * (n // 2) + i // 2] for i in range(n)]
c = 0
for i in range(1, n - 1):
if l[i - 1] > l[i] < l[i + 1]:
c += 1
print(c)
print(" ".join(map(str, l)))
|
ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR VAR BIN_OP BIN_OP BIN_OP NUMBER BIN_OP VAR NUMBER BIN_OP VAR NUMBER BIN_OP VAR NUMBER VAR FUNC_CALL VAR VAR ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR NUMBER BIN_OP VAR NUMBER IF VAR BIN_OP VAR NUMBER VAR VAR VAR BIN_OP VAR NUMBER VAR NUMBER EXPR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR FUNC_CALL STRING FUNC_CALL VAR VAR VAR
|
This is the hard version of the problem. The difference between the versions is that in the easy version all prices $a_i$ are different. You can make hacks if and only if you solved both versions of the problem.
Today is Sage's birthday, and she will go shopping to buy ice spheres. All $n$ ice spheres are placed in a row and they are numbered from $1$ to $n$ from left to right. Each ice sphere has a positive integer price. In this version, some prices can be equal.
An ice sphere is cheap if it costs strictly less than two neighboring ice spheres: the nearest to the left and the nearest to the right. The leftmost and the rightmost ice spheres are not cheap. Sage will choose all cheap ice spheres and then buy only them.
You can visit the shop before Sage and reorder the ice spheres as you wish. Find out the maximum number of ice spheres that Sage can buy, and show how the ice spheres should be reordered.
-----Input-----
The first line contains a single integer $n$ $(1 \le n \le 10^5)$ — the number of ice spheres in the shop.
The second line contains $n$ integers $a_1, a_2, \dots, a_n$ $(1 \le a_i \le 10^9)$ — the prices of ice spheres.
-----Output-----
In the first line print the maximum number of ice spheres that Sage can buy.
In the second line print the prices of ice spheres in the optimal order. If there are several correct answers, you can print any of them.
-----Example-----
Input
7
1 3 2 2 4 5 4
Output
3
3 1 4 2 4 2 5
-----Note-----
In the sample it's not possible to place the ice spheres in any order so that Sage would buy $4$ of them. If the spheres are placed in the order $(3, 1, 4, 2, 4, 2, 5)$, then Sage will buy one sphere for $1$ and two spheres for $2$ each.
|
n = int(input())
l = list(sorted(map(int, input().split())))
left = l[0 : n // 2]
right = l[n // 2 : n]
arr = [(-1) for x in range(n)]
count = 0
flag = True
j = 0
k = 0
for i in range(n):
if flag == True:
arr[i] = right[j]
j += 1
flag = False
elif flag == False:
arr[i] = left[k]
k += 1
flag = True
for a in range(1, n - 1):
if arr[a] < arr[a - 1] and arr[a] < arr[a + 1]:
count += 1
print(count)
print(*arr)
|
ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR VAR NUMBER BIN_OP VAR NUMBER ASSIGN VAR VAR BIN_OP VAR NUMBER VAR ASSIGN VAR NUMBER VAR FUNC_CALL VAR VAR ASSIGN VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR VAR IF VAR NUMBER ASSIGN VAR VAR VAR VAR VAR NUMBER ASSIGN VAR NUMBER IF VAR NUMBER ASSIGN VAR VAR VAR VAR VAR NUMBER ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR NUMBER BIN_OP VAR NUMBER IF VAR VAR VAR BIN_OP VAR NUMBER VAR VAR VAR BIN_OP VAR NUMBER VAR NUMBER EXPR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR VAR
|
This is the hard version of the problem. The difference between the versions is that in the easy version all prices $a_i$ are different. You can make hacks if and only if you solved both versions of the problem.
Today is Sage's birthday, and she will go shopping to buy ice spheres. All $n$ ice spheres are placed in a row and they are numbered from $1$ to $n$ from left to right. Each ice sphere has a positive integer price. In this version, some prices can be equal.
An ice sphere is cheap if it costs strictly less than two neighboring ice spheres: the nearest to the left and the nearest to the right. The leftmost and the rightmost ice spheres are not cheap. Sage will choose all cheap ice spheres and then buy only them.
You can visit the shop before Sage and reorder the ice spheres as you wish. Find out the maximum number of ice spheres that Sage can buy, and show how the ice spheres should be reordered.
-----Input-----
The first line contains a single integer $n$ $(1 \le n \le 10^5)$ — the number of ice spheres in the shop.
The second line contains $n$ integers $a_1, a_2, \dots, a_n$ $(1 \le a_i \le 10^9)$ — the prices of ice spheres.
-----Output-----
In the first line print the maximum number of ice spheres that Sage can buy.
In the second line print the prices of ice spheres in the optimal order. If there are several correct answers, you can print any of them.
-----Example-----
Input
7
1 3 2 2 4 5 4
Output
3
3 1 4 2 4 2 5
-----Note-----
In the sample it's not possible to place the ice spheres in any order so that Sage would buy $4$ of them. If the spheres are placed in the order $(3, 1, 4, 2, 4, 2, 5)$, then Sage will buy one sphere for $1$ and two spheres for $2$ each.
|
def check_cou(s, x, dx1, dx2):
if (n := len(s)) - x >= x + 1:
if s[x - 1] != s[n - x - 1]:
return True
elif x - dx1 >= dx2 and dx1 <= n - x + dx2:
return True
else:
return False
else:
return False
def cou(s):
x = int((len(s) - 1) / 2)
j = x - 1
while j > 0 and s[j] == s[x - 1]:
j -= 1
dx1 = x - 1 - j
j = n - x - 1
while j < n - 1 and s[j] == s[n - x - 1]:
j += 1
dx2 = j - n + x + 1
while not check_cou(s, x, dx1, dx2) and x > 0:
x -= 1
dx1 -= 1
dx2 -= 1
return x, n - x - 1
def fu():
return map(int, input().split(" "))
n = int(input())
s = sorted(list(fu()))
c = cou(s)
p = c[0]
d = c[1]
s1 = s[:p]
sost = s[p:d]
s2 = s[d:]
A = ""
for i in range(p):
A += str(s2[i]) + " " + str(s1[i]) + " "
for i in range(p, len(s2)):
A += str(s2[i]) + " "
for i in range(len(sost)):
A += str(sost[i]) + " "
print(p)
print(A[:-1])
|
FUNC_DEF IF BIN_OP VAR FUNC_CALL VAR VAR VAR BIN_OP VAR NUMBER IF VAR BIN_OP VAR NUMBER VAR BIN_OP BIN_OP VAR VAR NUMBER RETURN NUMBER IF BIN_OP VAR VAR VAR VAR BIN_OP BIN_OP VAR VAR VAR RETURN NUMBER RETURN NUMBER RETURN NUMBER FUNC_DEF ASSIGN VAR FUNC_CALL VAR BIN_OP BIN_OP FUNC_CALL VAR VAR NUMBER NUMBER ASSIGN VAR BIN_OP VAR NUMBER WHILE VAR NUMBER VAR VAR VAR BIN_OP VAR NUMBER VAR NUMBER ASSIGN VAR BIN_OP BIN_OP VAR NUMBER VAR ASSIGN VAR BIN_OP BIN_OP VAR VAR NUMBER WHILE VAR BIN_OP VAR NUMBER VAR VAR VAR BIN_OP BIN_OP VAR VAR NUMBER VAR NUMBER ASSIGN VAR BIN_OP BIN_OP BIN_OP VAR VAR VAR NUMBER WHILE FUNC_CALL VAR VAR VAR VAR VAR VAR NUMBER VAR NUMBER VAR NUMBER VAR NUMBER RETURN VAR BIN_OP BIN_OP VAR VAR NUMBER FUNC_DEF RETURN FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR STRING ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR VAR NUMBER ASSIGN VAR VAR NUMBER ASSIGN VAR VAR VAR ASSIGN VAR VAR VAR VAR ASSIGN VAR VAR VAR ASSIGN VAR STRING FOR VAR FUNC_CALL VAR VAR VAR BIN_OP BIN_OP BIN_OP FUNC_CALL VAR VAR VAR STRING FUNC_CALL VAR VAR VAR STRING FOR VAR FUNC_CALL VAR VAR FUNC_CALL VAR VAR VAR BIN_OP FUNC_CALL VAR VAR VAR STRING FOR VAR FUNC_CALL VAR FUNC_CALL VAR VAR VAR BIN_OP FUNC_CALL VAR VAR VAR STRING EXPR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR VAR NUMBER
|
This is the hard version of the problem. The difference between the versions is that in the easy version all prices $a_i$ are different. You can make hacks if and only if you solved both versions of the problem.
Today is Sage's birthday, and she will go shopping to buy ice spheres. All $n$ ice spheres are placed in a row and they are numbered from $1$ to $n$ from left to right. Each ice sphere has a positive integer price. In this version, some prices can be equal.
An ice sphere is cheap if it costs strictly less than two neighboring ice spheres: the nearest to the left and the nearest to the right. The leftmost and the rightmost ice spheres are not cheap. Sage will choose all cheap ice spheres and then buy only them.
You can visit the shop before Sage and reorder the ice spheres as you wish. Find out the maximum number of ice spheres that Sage can buy, and show how the ice spheres should be reordered.
-----Input-----
The first line contains a single integer $n$ $(1 \le n \le 10^5)$ — the number of ice spheres in the shop.
The second line contains $n$ integers $a_1, a_2, \dots, a_n$ $(1 \le a_i \le 10^9)$ — the prices of ice spheres.
-----Output-----
In the first line print the maximum number of ice spheres that Sage can buy.
In the second line print the prices of ice spheres in the optimal order. If there are several correct answers, you can print any of them.
-----Example-----
Input
7
1 3 2 2 4 5 4
Output
3
3 1 4 2 4 2 5
-----Note-----
In the sample it's not possible to place the ice spheres in any order so that Sage would buy $4$ of them. If the spheres are placed in the order $(3, 1, 4, 2, 4, 2, 5)$, then Sage will buy one sphere for $1$ and two spheres for $2$ each.
|
def check(arr):
n = len(arr)
ans = 0
for i in range(1, n - 1):
if arr[i] < arr[i - 1] and arr[i] < arr[i + 1]:
ans += 1
return ans
n = int(input())
a = list(map(int, input().split()))
arr = [None] * n
a.sort()
start = 0
for i in range(1, n, 2):
arr[i] = a[start]
start += 1
for i in range(n):
if arr[i] == None:
arr[i] = a[start]
start += 1
print(check(arr))
print(*arr)
|
FUNC_DEF ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR NUMBER BIN_OP VAR NUMBER IF VAR VAR VAR BIN_OP VAR NUMBER VAR VAR VAR BIN_OP VAR NUMBER VAR NUMBER RETURN VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR BIN_OP LIST NONE VAR EXPR FUNC_CALL VAR ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR NUMBER VAR NUMBER ASSIGN VAR VAR VAR VAR VAR NUMBER FOR VAR FUNC_CALL VAR VAR IF VAR VAR NONE ASSIGN VAR VAR VAR VAR VAR NUMBER EXPR FUNC_CALL VAR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR VAR
|
This is the hard version of the problem. The difference between the versions is that in the easy version all prices $a_i$ are different. You can make hacks if and only if you solved both versions of the problem.
Today is Sage's birthday, and she will go shopping to buy ice spheres. All $n$ ice spheres are placed in a row and they are numbered from $1$ to $n$ from left to right. Each ice sphere has a positive integer price. In this version, some prices can be equal.
An ice sphere is cheap if it costs strictly less than two neighboring ice spheres: the nearest to the left and the nearest to the right. The leftmost and the rightmost ice spheres are not cheap. Sage will choose all cheap ice spheres and then buy only them.
You can visit the shop before Sage and reorder the ice spheres as you wish. Find out the maximum number of ice spheres that Sage can buy, and show how the ice spheres should be reordered.
-----Input-----
The first line contains a single integer $n$ $(1 \le n \le 10^5)$ — the number of ice spheres in the shop.
The second line contains $n$ integers $a_1, a_2, \dots, a_n$ $(1 \le a_i \le 10^9)$ — the prices of ice spheres.
-----Output-----
In the first line print the maximum number of ice spheres that Sage can buy.
In the second line print the prices of ice spheres in the optimal order. If there are several correct answers, you can print any of them.
-----Example-----
Input
7
1 3 2 2 4 5 4
Output
3
3 1 4 2 4 2 5
-----Note-----
In the sample it's not possible to place the ice spheres in any order so that Sage would buy $4$ of them. If the spheres are placed in the order $(3, 1, 4, 2, 4, 2, 5)$, then Sage will buy one sphere for $1$ and two spheres for $2$ each.
|
n = int(input())
lis = sorted(map(int, input().split()))
ans = [0] * n
i = j = n - 1
while j >= 0 and i >= 0:
ans[j] = lis[i]
i -= 1
j -= 2
j = n - 2
tmp = []
while j > 0 and i >= 0:
while j > 0 and i >= 0 and (ans[j + 1] == lis[i] or ans[j - 1] == lis[i]):
tmp.append(lis[i])
i -= 1
if i >= 0:
ans[j] = lis[i]
i -= 1
j -= 2
j = 0
while i >= 0:
tmp.append(lis[i])
i -= 1
for i in range(n):
if ans[i] == 0:
ans[i] = tmp[j]
j += 1
tmp = 0
for i in range(1, n - 1):
if ans[i - 1] > ans[i] < ans[i + 1]:
tmp += 1
print(tmp)
print(*ans)
|
ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR BIN_OP LIST NUMBER VAR ASSIGN VAR VAR BIN_OP VAR NUMBER WHILE VAR NUMBER VAR NUMBER ASSIGN VAR VAR VAR VAR VAR NUMBER VAR NUMBER ASSIGN VAR BIN_OP VAR NUMBER ASSIGN VAR LIST WHILE VAR NUMBER VAR NUMBER WHILE VAR NUMBER VAR NUMBER VAR BIN_OP VAR NUMBER VAR VAR VAR BIN_OP VAR NUMBER VAR VAR EXPR FUNC_CALL VAR VAR VAR VAR NUMBER IF VAR NUMBER ASSIGN VAR VAR VAR VAR VAR NUMBER VAR NUMBER ASSIGN VAR NUMBER WHILE VAR NUMBER EXPR FUNC_CALL VAR VAR VAR VAR NUMBER FOR VAR FUNC_CALL VAR VAR IF VAR VAR NUMBER ASSIGN VAR VAR VAR VAR VAR NUMBER ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR NUMBER BIN_OP VAR NUMBER IF VAR BIN_OP VAR NUMBER VAR VAR VAR BIN_OP VAR NUMBER VAR NUMBER EXPR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR VAR
|
This is the hard version of the problem. The difference between the versions is that in the easy version all prices $a_i$ are different. You can make hacks if and only if you solved both versions of the problem.
Today is Sage's birthday, and she will go shopping to buy ice spheres. All $n$ ice spheres are placed in a row and they are numbered from $1$ to $n$ from left to right. Each ice sphere has a positive integer price. In this version, some prices can be equal.
An ice sphere is cheap if it costs strictly less than two neighboring ice spheres: the nearest to the left and the nearest to the right. The leftmost and the rightmost ice spheres are not cheap. Sage will choose all cheap ice spheres and then buy only them.
You can visit the shop before Sage and reorder the ice spheres as you wish. Find out the maximum number of ice spheres that Sage can buy, and show how the ice spheres should be reordered.
-----Input-----
The first line contains a single integer $n$ $(1 \le n \le 10^5)$ — the number of ice spheres in the shop.
The second line contains $n$ integers $a_1, a_2, \dots, a_n$ $(1 \le a_i \le 10^9)$ — the prices of ice spheres.
-----Output-----
In the first line print the maximum number of ice spheres that Sage can buy.
In the second line print the prices of ice spheres in the optimal order. If there are several correct answers, you can print any of them.
-----Example-----
Input
7
1 3 2 2 4 5 4
Output
3
3 1 4 2 4 2 5
-----Note-----
In the sample it's not possible to place the ice spheres in any order so that Sage would buy $4$ of them. If the spheres are placed in the order $(3, 1, 4, 2, 4, 2, 5)$, then Sage will buy one sphere for $1$ and two spheres for $2$ each.
|
import sys
input = sys.stdin.readline
n = int(input())
a = list(map(int, input().split()))
res = [0] * n
a = sorted(a, reverse=True)
ai = 0
for i in range(0, n, 2):
res[i] = a[ai]
ai += 1
if n % 2 == 0:
res[-1] = a[ai]
ai += 1
for i in range(1, n - 1, 2):
res[i] = a[ai]
ai += 1
ans = 0
for i in range(n):
if not (i - 1 >= 0 and i + 1 < n):
continue
if res[i] < res[i - 1] and res[i] < res[i + 1]:
ans += 1
print(ans)
print(*res)
|
IMPORT ASSIGN VAR VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR BIN_OP LIST NUMBER VAR ASSIGN VAR FUNC_CALL VAR VAR NUMBER ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR NUMBER VAR NUMBER ASSIGN VAR VAR VAR VAR VAR NUMBER IF BIN_OP VAR NUMBER NUMBER ASSIGN VAR NUMBER VAR VAR VAR NUMBER FOR VAR FUNC_CALL VAR NUMBER BIN_OP VAR NUMBER NUMBER ASSIGN VAR VAR VAR VAR VAR NUMBER ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR VAR IF BIN_OP VAR NUMBER NUMBER BIN_OP VAR NUMBER VAR IF VAR VAR VAR BIN_OP VAR NUMBER VAR VAR VAR BIN_OP VAR NUMBER VAR NUMBER EXPR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR VAR
|
This is the hard version of the problem. The difference between the versions is that in the easy version all prices $a_i$ are different. You can make hacks if and only if you solved both versions of the problem.
Today is Sage's birthday, and she will go shopping to buy ice spheres. All $n$ ice spheres are placed in a row and they are numbered from $1$ to $n$ from left to right. Each ice sphere has a positive integer price. In this version, some prices can be equal.
An ice sphere is cheap if it costs strictly less than two neighboring ice spheres: the nearest to the left and the nearest to the right. The leftmost and the rightmost ice spheres are not cheap. Sage will choose all cheap ice spheres and then buy only them.
You can visit the shop before Sage and reorder the ice spheres as you wish. Find out the maximum number of ice spheres that Sage can buy, and show how the ice spheres should be reordered.
-----Input-----
The first line contains a single integer $n$ $(1 \le n \le 10^5)$ — the number of ice spheres in the shop.
The second line contains $n$ integers $a_1, a_2, \dots, a_n$ $(1 \le a_i \le 10^9)$ — the prices of ice spheres.
-----Output-----
In the first line print the maximum number of ice spheres that Sage can buy.
In the second line print the prices of ice spheres in the optimal order. If there are several correct answers, you can print any of them.
-----Example-----
Input
7
1 3 2 2 4 5 4
Output
3
3 1 4 2 4 2 5
-----Note-----
In the sample it's not possible to place the ice spheres in any order so that Sage would buy $4$ of them. If the spheres are placed in the order $(3, 1, 4, 2, 4, 2, 5)$, then Sage will buy one sphere for $1$ and two spheres for $2$ each.
|
n = int(input())
a = list(map(int, input().split()))
a.sort(reverse=True)
ans = [-1] * n
j = 0
for i in range(0, n, 2):
ans[i] = a[j]
j += 1
ansl = 0
check = False
for i in range(1, n, 2):
ans[i] = a[j]
if i < n - 1 and ans[i - 1] > ans[i] < ans[i + 1]:
ansl += 1
elif i == n - 1:
check = True
j += 1
if check:
for i in range(1, n - 1, 2):
if not ans[i - 1] > ans[i] < ans[i + 1] and ans[i - 1] > ans[-1] < ans[i + 1]:
ansl += 1
ans[i], ans[-1] = ans[-1], ans[i]
break
print(ansl)
print(*ans)
|
ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR EXPR FUNC_CALL VAR NUMBER ASSIGN VAR BIN_OP LIST NUMBER VAR ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR NUMBER VAR NUMBER ASSIGN VAR VAR VAR VAR VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR NUMBER VAR NUMBER ASSIGN VAR VAR VAR VAR IF VAR BIN_OP VAR NUMBER VAR BIN_OP VAR NUMBER VAR VAR VAR BIN_OP VAR NUMBER VAR NUMBER IF VAR BIN_OP VAR NUMBER ASSIGN VAR NUMBER VAR NUMBER IF VAR FOR VAR FUNC_CALL VAR NUMBER BIN_OP VAR NUMBER NUMBER IF VAR BIN_OP VAR NUMBER VAR VAR VAR BIN_OP VAR NUMBER VAR BIN_OP VAR NUMBER VAR NUMBER VAR BIN_OP VAR NUMBER VAR NUMBER ASSIGN VAR VAR VAR NUMBER VAR NUMBER VAR VAR EXPR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR VAR
|
This is the hard version of the problem. The difference between the versions is that in the easy version all prices $a_i$ are different. You can make hacks if and only if you solved both versions of the problem.
Today is Sage's birthday, and she will go shopping to buy ice spheres. All $n$ ice spheres are placed in a row and they are numbered from $1$ to $n$ from left to right. Each ice sphere has a positive integer price. In this version, some prices can be equal.
An ice sphere is cheap if it costs strictly less than two neighboring ice spheres: the nearest to the left and the nearest to the right. The leftmost and the rightmost ice spheres are not cheap. Sage will choose all cheap ice spheres and then buy only them.
You can visit the shop before Sage and reorder the ice spheres as you wish. Find out the maximum number of ice spheres that Sage can buy, and show how the ice spheres should be reordered.
-----Input-----
The first line contains a single integer $n$ $(1 \le n \le 10^5)$ — the number of ice spheres in the shop.
The second line contains $n$ integers $a_1, a_2, \dots, a_n$ $(1 \le a_i \le 10^9)$ — the prices of ice spheres.
-----Output-----
In the first line print the maximum number of ice spheres that Sage can buy.
In the second line print the prices of ice spheres in the optimal order. If there are several correct answers, you can print any of them.
-----Example-----
Input
7
1 3 2 2 4 5 4
Output
3
3 1 4 2 4 2 5
-----Note-----
In the sample it's not possible to place the ice spheres in any order so that Sage would buy $4$ of them. If the spheres are placed in the order $(3, 1, 4, 2, 4, 2, 5)$, then Sage will buy one sphere for $1$ and two spheres for $2$ each.
|
n = int(input())
l = list(map(int, input().split()))
l = sorted(l)
ans = [(0) for _ in range(n)]
t = n - 1
x = 0
if n % 2 == 0:
x = 1
for i in range(n - 1 - x, -1, -2):
ans[i] = l[t]
t -= 1
t = 0
for i in range(1, n, 2):
ans[i] = l[t]
t += 1
t = 0
for i in range(1, n - 1):
if ans[i] < ans[i - 1] and ans[i] < ans[i + 1]:
t += 1
print(t)
print(*ans)
|
ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR NUMBER VAR FUNC_CALL VAR VAR ASSIGN VAR BIN_OP VAR NUMBER ASSIGN VAR NUMBER IF BIN_OP VAR NUMBER NUMBER ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR BIN_OP BIN_OP VAR NUMBER VAR NUMBER NUMBER ASSIGN VAR VAR VAR VAR VAR NUMBER ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR NUMBER VAR NUMBER ASSIGN VAR VAR VAR VAR VAR NUMBER ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR NUMBER BIN_OP VAR NUMBER IF VAR VAR VAR BIN_OP VAR NUMBER VAR VAR VAR BIN_OP VAR NUMBER VAR NUMBER EXPR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR VAR
|
This is the hard version of the problem. The difference between the versions is that in the easy version all prices $a_i$ are different. You can make hacks if and only if you solved both versions of the problem.
Today is Sage's birthday, and she will go shopping to buy ice spheres. All $n$ ice spheres are placed in a row and they are numbered from $1$ to $n$ from left to right. Each ice sphere has a positive integer price. In this version, some prices can be equal.
An ice sphere is cheap if it costs strictly less than two neighboring ice spheres: the nearest to the left and the nearest to the right. The leftmost and the rightmost ice spheres are not cheap. Sage will choose all cheap ice spheres and then buy only them.
You can visit the shop before Sage and reorder the ice spheres as you wish. Find out the maximum number of ice spheres that Sage can buy, and show how the ice spheres should be reordered.
-----Input-----
The first line contains a single integer $n$ $(1 \le n \le 10^5)$ — the number of ice spheres in the shop.
The second line contains $n$ integers $a_1, a_2, \dots, a_n$ $(1 \le a_i \le 10^9)$ — the prices of ice spheres.
-----Output-----
In the first line print the maximum number of ice spheres that Sage can buy.
In the second line print the prices of ice spheres in the optimal order. If there are several correct answers, you can print any of them.
-----Example-----
Input
7
1 3 2 2 4 5 4
Output
3
3 1 4 2 4 2 5
-----Note-----
In the sample it's not possible to place the ice spheres in any order so that Sage would buy $4$ of them. If the spheres are placed in the order $(3, 1, 4, 2, 4, 2, 5)$, then Sage will buy one sphere for $1$ and two spheres for $2$ each.
|
n = int(input())
a = list(map(int, input().split()))
a.sort(reverse=True)
x = n // 2
b = a[:x]
c = a[x:]
c.reverse()
b.reverse()
z = []
for i in range(x):
z.append(b[i])
z.append(c[i])
c[i] = 0
for i in range(len(c)):
if c[i] != 0:
z.append(c[i])
ans2 = 0
for i in range(1, n - 1, 2):
if z[i] < z[i - 1] and z[i] < z[i + 1]:
ans2 += 1
b = []
x = n // 2
pos = x
if 2 * x != n:
pos = x + 1
for i in range(x):
b.append(a[i])
b.append(a[i + pos])
if 2 * x != n:
b.append(a[x])
c = []
for i in range(x):
c.append(a[i])
c.append(a[-i])
if 2 * x != n:
c.append(a[x])
ans = 0
for i in range(1, n - 1, 2):
if b[i] < b[i - 1] and b[i] < b[i + 1]:
ans += 1
tmp = 0
for i in range(1, n - 1, 2):
if c[i] < c[i - 1] and c[i] < c[i + 1]:
tmp += 1
if max(ans, tmp, ans2) == ans:
print(ans)
print(*b)
elif max(ans, tmp, ans2) == tmp:
print(tmp)
print(*c)
else:
print(ans2)
print(*z)
|
ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR EXPR FUNC_CALL VAR NUMBER ASSIGN VAR BIN_OP VAR NUMBER ASSIGN VAR VAR VAR ASSIGN VAR VAR VAR EXPR FUNC_CALL VAR EXPR FUNC_CALL VAR ASSIGN VAR LIST FOR VAR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR VAR VAR EXPR FUNC_CALL VAR VAR VAR ASSIGN VAR VAR NUMBER FOR VAR FUNC_CALL VAR FUNC_CALL VAR VAR IF VAR VAR NUMBER EXPR FUNC_CALL VAR VAR VAR ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR NUMBER BIN_OP VAR NUMBER NUMBER IF VAR VAR VAR BIN_OP VAR NUMBER VAR VAR VAR BIN_OP VAR NUMBER VAR NUMBER ASSIGN VAR LIST ASSIGN VAR BIN_OP VAR NUMBER ASSIGN VAR VAR IF BIN_OP NUMBER VAR VAR ASSIGN VAR BIN_OP VAR NUMBER FOR VAR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR VAR VAR EXPR FUNC_CALL VAR VAR BIN_OP VAR VAR IF BIN_OP NUMBER VAR VAR EXPR FUNC_CALL VAR VAR VAR ASSIGN VAR LIST FOR VAR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR VAR VAR EXPR FUNC_CALL VAR VAR VAR IF BIN_OP NUMBER VAR VAR EXPR FUNC_CALL VAR VAR VAR ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR NUMBER BIN_OP VAR NUMBER NUMBER IF VAR VAR VAR BIN_OP VAR NUMBER VAR VAR VAR BIN_OP VAR NUMBER VAR NUMBER ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR NUMBER BIN_OP VAR NUMBER NUMBER IF VAR VAR VAR BIN_OP VAR NUMBER VAR VAR VAR BIN_OP VAR NUMBER VAR NUMBER IF FUNC_CALL VAR VAR VAR VAR VAR EXPR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR VAR IF FUNC_CALL VAR VAR VAR VAR VAR EXPR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR VAR
|
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