description stringlengths 171 4k | code stringlengths 94 3.98k | normalized_code stringlengths 57 4.99k |
|---|---|---|
You are given an array $a$ consisting of $n$ integers. Indices of the array start from zero (i. e. the first element is $a_0$, the second one is $a_1$, and so on).
You can reverse at most one subarray (continuous subsegment) of this array. Recall that the subarray of $a$ with borders $l$ and $r$ is $a[l; r] = a_l, a_{l + 1}, \dots, a_{r}$.
Your task is to reverse such a subarray that the sum of elements on even positions of the resulting array is maximized (i. e. the sum of elements $a_0, a_2, \dots, a_{2k}$ for integer $k = \lfloor\frac{n-1}{2}\rfloor$ should be maximum possible).
You have to answer $t$ independent test cases.
-----Input-----
The first line of the input contains one integer $t$ ($1 \le t \le 2 \cdot 10^4$) — the number of test cases. Then $t$ test cases follow.
The first line of the test case contains one integer $n$ ($1 \le n \le 2 \cdot 10^5$) — the length of $a$. The second line of the test case contains $n$ integers $a_0, a_1, \dots, a_{n-1}$ ($1 \le a_i \le 10^9$), where $a_i$ is the $i$-th element of $a$.
It is guaranteed that the sum of $n$ does not exceed $2 \cdot 10^5$ ($\sum n \le 2 \cdot 10^5$).
-----Output-----
For each test case, print the answer on the separate line — the maximum possible sum of elements on even positions after reversing at most one subarray (continuous subsegment) of $a$.
-----Example-----
Input
4
8
1 7 3 4 7 6 2 9
5
1 2 1 2 1
10
7 8 4 5 7 6 8 9 7 3
4
3 1 2 1
Output
26
5
37
5 | for _ in range(int(input())):
n = int(input())
a = list(map(int, input().split()))
odd = [0] * n
even = [0] * n
for i in range(n):
if i % 2 == 0:
odd[i] = odd[i - 1]
even[i] = even[i - 1] + a[i]
else:
odd[i] = odd[i - 1] + a[i]
even[i] = even[i - 1]
ODD = [(odd[i] - even[i]) for i in range(n) if i % 2 == 1]
EVEN = [(odd[i] - even[i]) for i in range(n) if i % 2 == 0]
for i in range(len(ODD) - 2, -1, -1):
ODD[i] = max(ODD[i], ODD[i + 1])
for i in range(len(EVEN) - 2, -1, -1):
EVEN[i] = max(EVEN[i], EVEN[i + 1])
ans = -float("inf")
for i in range(n - 1):
if i != 0:
test = even[i - 1] - odd[i - 1]
else:
test = 0
if i % 2 == 0:
k = i // 2
test += ODD[k]
else:
k = i // 2
test += EVEN[k + 1]
ans = max(ans, test)
S = sum(a[i] for i in range(n) if i % 2 == 0)
print(max(S, S + ans)) | FOR VAR FUNC_CALL VAR FUNC_CALL VAR FUNC_CALL 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 LIST NUMBER VAR FOR VAR FUNC_CALL VAR VAR IF BIN_OP VAR NUMBER NUMBER ASSIGN VAR VAR VAR BIN_OP VAR NUMBER ASSIGN VAR VAR BIN_OP VAR BIN_OP VAR NUMBER VAR VAR ASSIGN VAR VAR BIN_OP VAR BIN_OP VAR NUMBER VAR VAR ASSIGN VAR VAR VAR BIN_OP VAR NUMBER ASSIGN VAR BIN_OP VAR VAR VAR VAR VAR FUNC_CALL VAR VAR BIN_OP VAR NUMBER NUMBER ASSIGN VAR BIN_OP VAR VAR VAR VAR VAR FUNC_CALL VAR VAR BIN_OP VAR NUMBER NUMBER FOR VAR FUNC_CALL VAR BIN_OP FUNC_CALL VAR VAR NUMBER NUMBER NUMBER ASSIGN VAR VAR FUNC_CALL VAR VAR VAR VAR BIN_OP VAR NUMBER FOR VAR FUNC_CALL VAR BIN_OP FUNC_CALL VAR VAR NUMBER NUMBER NUMBER ASSIGN VAR VAR FUNC_CALL VAR VAR VAR VAR BIN_OP VAR NUMBER ASSIGN VAR FUNC_CALL VAR STRING FOR VAR FUNC_CALL VAR BIN_OP VAR NUMBER IF VAR NUMBER ASSIGN VAR BIN_OP VAR BIN_OP VAR NUMBER VAR BIN_OP VAR NUMBER ASSIGN VAR NUMBER IF BIN_OP VAR NUMBER NUMBER ASSIGN VAR BIN_OP VAR NUMBER VAR VAR VAR ASSIGN VAR BIN_OP VAR NUMBER VAR VAR BIN_OP VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR VAR ASSIGN VAR FUNC_CALL VAR VAR VAR VAR FUNC_CALL VAR VAR BIN_OP VAR NUMBER NUMBER EXPR FUNC_CALL VAR FUNC_CALL VAR VAR BIN_OP VAR VAR |
You are given an array $a$ consisting of $n$ integers. Indices of the array start from zero (i. e. the first element is $a_0$, the second one is $a_1$, and so on).
You can reverse at most one subarray (continuous subsegment) of this array. Recall that the subarray of $a$ with borders $l$ and $r$ is $a[l; r] = a_l, a_{l + 1}, \dots, a_{r}$.
Your task is to reverse such a subarray that the sum of elements on even positions of the resulting array is maximized (i. e. the sum of elements $a_0, a_2, \dots, a_{2k}$ for integer $k = \lfloor\frac{n-1}{2}\rfloor$ should be maximum possible).
You have to answer $t$ independent test cases.
-----Input-----
The first line of the input contains one integer $t$ ($1 \le t \le 2 \cdot 10^4$) — the number of test cases. Then $t$ test cases follow.
The first line of the test case contains one integer $n$ ($1 \le n \le 2 \cdot 10^5$) — the length of $a$. The second line of the test case contains $n$ integers $a_0, a_1, \dots, a_{n-1}$ ($1 \le a_i \le 10^9$), where $a_i$ is the $i$-th element of $a$.
It is guaranteed that the sum of $n$ does not exceed $2 \cdot 10^5$ ($\sum n \le 2 \cdot 10^5$).
-----Output-----
For each test case, print the answer on the separate line — the maximum possible sum of elements on even positions after reversing at most one subarray (continuous subsegment) of $a$.
-----Example-----
Input
4
8
1 7 3 4 7 6 2 9
5
1 2 1 2 1
10
7 8 4 5 7 6 8 9 7 3
4
3 1 2 1
Output
26
5
37
5 | import sys
reader = (s.rstrip() for s in sys.stdin)
input = reader.__next__
def maxSubArraySum(a, size):
max_so_far = -float("inf")
max_ending_here = 0
for i in range(0, size):
max_ending_here = max_ending_here + a[i]
if max_so_far < max_ending_here:
max_so_far = max_ending_here
if max_ending_here < 0:
max_ending_here = 0
return max_so_far
def gift():
for _ in range(t):
n = int(input())
arry = list(map(int, input().split()))
sum1 = 0
for i in range(n // 2 + n % 2):
sum1 += arry[2 * i]
arra = []
arrb = []
for i in range(n // 2):
arra.append(arry[2 * i + 1] - arry[2 * i])
for i in range(n // 2 - 1 + n % 2):
arrb.append(arry[2 * i + 1] - arry[2 * (i + 1)])
diua = maxSubArraySum(arra, len(arra))
diub = maxSubArraySum(arrb, len(arrb))
yield sum1 + max(diua, diub, 0)
t = int(input())
ans = gift()
print(*ans, sep="\n") | IMPORT ASSIGN VAR FUNC_CALL VAR VAR VAR ASSIGN VAR VAR FUNC_DEF ASSIGN VAR FUNC_CALL VAR STRING ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR NUMBER VAR ASSIGN VAR BIN_OP VAR VAR VAR IF VAR VAR ASSIGN VAR VAR IF VAR NUMBER ASSIGN VAR NUMBER RETURN VAR FUNC_DEF FOR VAR FUNC_CALL 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 NUMBER FOR VAR FUNC_CALL VAR BIN_OP BIN_OP VAR NUMBER BIN_OP VAR NUMBER VAR VAR BIN_OP NUMBER VAR ASSIGN VAR LIST ASSIGN VAR LIST FOR VAR FUNC_CALL VAR BIN_OP VAR NUMBER EXPR FUNC_CALL VAR BIN_OP VAR BIN_OP BIN_OP NUMBER VAR NUMBER VAR BIN_OP NUMBER VAR FOR VAR FUNC_CALL VAR BIN_OP BIN_OP BIN_OP VAR NUMBER NUMBER BIN_OP VAR NUMBER EXPR FUNC_CALL VAR BIN_OP VAR BIN_OP BIN_OP NUMBER VAR NUMBER VAR BIN_OP NUMBER BIN_OP VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR VAR FUNC_CALL VAR VAR EXPR BIN_OP VAR FUNC_CALL VAR VAR VAR NUMBER ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR EXPR FUNC_CALL VAR VAR STRING |
You are given an array $a$ consisting of $n$ integers. Indices of the array start from zero (i. e. the first element is $a_0$, the second one is $a_1$, and so on).
You can reverse at most one subarray (continuous subsegment) of this array. Recall that the subarray of $a$ with borders $l$ and $r$ is $a[l; r] = a_l, a_{l + 1}, \dots, a_{r}$.
Your task is to reverse such a subarray that the sum of elements on even positions of the resulting array is maximized (i. e. the sum of elements $a_0, a_2, \dots, a_{2k}$ for integer $k = \lfloor\frac{n-1}{2}\rfloor$ should be maximum possible).
You have to answer $t$ independent test cases.
-----Input-----
The first line of the input contains one integer $t$ ($1 \le t \le 2 \cdot 10^4$) — the number of test cases. Then $t$ test cases follow.
The first line of the test case contains one integer $n$ ($1 \le n \le 2 \cdot 10^5$) — the length of $a$. The second line of the test case contains $n$ integers $a_0, a_1, \dots, a_{n-1}$ ($1 \le a_i \le 10^9$), where $a_i$ is the $i$-th element of $a$.
It is guaranteed that the sum of $n$ does not exceed $2 \cdot 10^5$ ($\sum n \le 2 \cdot 10^5$).
-----Output-----
For each test case, print the answer on the separate line — the maximum possible sum of elements on even positions after reversing at most one subarray (continuous subsegment) of $a$.
-----Example-----
Input
4
8
1 7 3 4 7 6 2 9
5
1 2 1 2 1
10
7 8 4 5 7 6 8 9 7 3
4
3 1 2 1
Output
26
5
37
5 | z = int(input())
for h in range(z):
n = int(input())
l = list(map(int, input().split()))
a = []
if n == 1:
print(l[0])
else:
x = l[1] - l[0]
if n > 2:
y = l[1] - l[2]
if y < 0:
y = 0
a.append(y)
if x < 0:
x = 0
a.append(x)
i = 3
while i < n:
if x < 0:
x = 0
if y < 0:
y = 0
x += l[i] - l[i - 1]
if i < n - 1:
y += l[i] - l[i + 1]
a.append(x)
a.append(y)
i += 2
b = []
i = 0
while i < n:
b.append(l[i])
i += 2
print(sum(b) + max(a)) | ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR FOR VAR FUNC_CALL 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 LIST IF VAR NUMBER EXPR FUNC_CALL VAR VAR NUMBER ASSIGN VAR BIN_OP VAR NUMBER VAR NUMBER IF VAR NUMBER ASSIGN VAR BIN_OP VAR NUMBER VAR NUMBER IF VAR NUMBER ASSIGN VAR NUMBER EXPR FUNC_CALL VAR VAR IF VAR NUMBER ASSIGN VAR NUMBER EXPR FUNC_CALL VAR VAR ASSIGN VAR NUMBER WHILE VAR VAR IF VAR NUMBER ASSIGN VAR NUMBER IF VAR NUMBER ASSIGN VAR NUMBER VAR BIN_OP VAR VAR VAR BIN_OP VAR NUMBER IF VAR BIN_OP VAR NUMBER VAR BIN_OP VAR VAR VAR BIN_OP VAR NUMBER EXPR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR VAR VAR NUMBER ASSIGN VAR LIST ASSIGN VAR NUMBER WHILE VAR VAR EXPR FUNC_CALL VAR VAR VAR VAR NUMBER EXPR FUNC_CALL VAR BIN_OP FUNC_CALL VAR VAR FUNC_CALL VAR VAR |
You are given an array $a$ consisting of $n$ integers. Indices of the array start from zero (i. e. the first element is $a_0$, the second one is $a_1$, and so on).
You can reverse at most one subarray (continuous subsegment) of this array. Recall that the subarray of $a$ with borders $l$ and $r$ is $a[l; r] = a_l, a_{l + 1}, \dots, a_{r}$.
Your task is to reverse such a subarray that the sum of elements on even positions of the resulting array is maximized (i. e. the sum of elements $a_0, a_2, \dots, a_{2k}$ for integer $k = \lfloor\frac{n-1}{2}\rfloor$ should be maximum possible).
You have to answer $t$ independent test cases.
-----Input-----
The first line of the input contains one integer $t$ ($1 \le t \le 2 \cdot 10^4$) — the number of test cases. Then $t$ test cases follow.
The first line of the test case contains one integer $n$ ($1 \le n \le 2 \cdot 10^5$) — the length of $a$. The second line of the test case contains $n$ integers $a_0, a_1, \dots, a_{n-1}$ ($1 \le a_i \le 10^9$), where $a_i$ is the $i$-th element of $a$.
It is guaranteed that the sum of $n$ does not exceed $2 \cdot 10^5$ ($\sum n \le 2 \cdot 10^5$).
-----Output-----
For each test case, print the answer on the separate line — the maximum possible sum of elements on even positions after reversing at most one subarray (continuous subsegment) of $a$.
-----Example-----
Input
4
8
1 7 3 4 7 6 2 9
5
1 2 1 2 1
10
7 8 4 5 7 6 8 9 7 3
4
3 1 2 1
Output
26
5
37
5 | from sys import stdin
for _ in range(int(stdin.readline().rstrip())):
n = int(stdin.readline().rstrip())
l = list(map(int, stdin.readline().rstrip().split(" ")))
a = [0] * (n // 2)
x = [0] * ((n - 1) // 2)
t = 0
j = 0
if n % 2:
t = l[-1]
for i in range(0, n - 1, 2):
t += l[i]
a[j] = l[i + 1] - l[i]
j += 1
j = 0
for i in range(1, n - 1, 2):
x[j] = l[i] - l[i + 1]
j += 1
s = b = 0
for i in range(len(a)):
s = max(a[i], s + a[i])
b = max(s, b)
s = 0
for i in range(len(x)):
s = max(x[i], s + x[i])
b = max(s, b)
print(t + b) | FOR VAR FUNC_CALL VAR FUNC_CALL VAR FUNC_CALL FUNC_CALL VAR 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 STRING ASSIGN VAR BIN_OP LIST NUMBER BIN_OP VAR NUMBER ASSIGN VAR BIN_OP LIST NUMBER BIN_OP BIN_OP VAR NUMBER NUMBER ASSIGN VAR NUMBER ASSIGN VAR NUMBER IF BIN_OP VAR NUMBER ASSIGN VAR VAR NUMBER FOR VAR FUNC_CALL VAR NUMBER BIN_OP VAR NUMBER NUMBER VAR VAR VAR ASSIGN VAR VAR BIN_OP VAR BIN_OP VAR NUMBER VAR VAR VAR NUMBER ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR NUMBER BIN_OP VAR NUMBER NUMBER ASSIGN VAR VAR BIN_OP VAR VAR VAR BIN_OP VAR NUMBER VAR NUMBER ASSIGN VAR VAR NUMBER FOR VAR FUNC_CALL VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR VAR VAR BIN_OP VAR VAR VAR ASSIGN VAR FUNC_CALL VAR VAR VAR ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR VAR VAR BIN_OP VAR VAR VAR ASSIGN VAR FUNC_CALL VAR VAR VAR EXPR FUNC_CALL VAR BIN_OP VAR VAR |
You are given an array $a$ consisting of $n$ integers. Indices of the array start from zero (i. e. the first element is $a_0$, the second one is $a_1$, and so on).
You can reverse at most one subarray (continuous subsegment) of this array. Recall that the subarray of $a$ with borders $l$ and $r$ is $a[l; r] = a_l, a_{l + 1}, \dots, a_{r}$.
Your task is to reverse such a subarray that the sum of elements on even positions of the resulting array is maximized (i. e. the sum of elements $a_0, a_2, \dots, a_{2k}$ for integer $k = \lfloor\frac{n-1}{2}\rfloor$ should be maximum possible).
You have to answer $t$ independent test cases.
-----Input-----
The first line of the input contains one integer $t$ ($1 \le t \le 2 \cdot 10^4$) — the number of test cases. Then $t$ test cases follow.
The first line of the test case contains one integer $n$ ($1 \le n \le 2 \cdot 10^5$) — the length of $a$. The second line of the test case contains $n$ integers $a_0, a_1, \dots, a_{n-1}$ ($1 \le a_i \le 10^9$), where $a_i$ is the $i$-th element of $a$.
It is guaranteed that the sum of $n$ does not exceed $2 \cdot 10^5$ ($\sum n \le 2 \cdot 10^5$).
-----Output-----
For each test case, print the answer on the separate line — the maximum possible sum of elements on even positions after reversing at most one subarray (continuous subsegment) of $a$.
-----Example-----
Input
4
8
1 7 3 4 7 6 2 9
5
1 2 1 2 1
10
7 8 4 5 7 6 8 9 7 3
4
3 1 2 1
Output
26
5
37
5 | import sys
input = sys.stdin.readline
T = int(input())
for _ in range(T):
N = int(input())
A = list(map(int, input().split()))
ev = A[::2]
od = A[1::2]
ans = sum(ev)
mn = cum = gain = 0
for a, b in zip(ev, od):
cum += b - a
mn = min(mn, cum)
gain = max(gain, cum - mn)
mn = cum = 0
for a, b in zip(ev[1:], od):
cum += b - a
mn = min(mn, cum)
gain = max(gain, cum - mn)
print(ans + gain) | IMPORT ASSIGN VAR VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR FOR VAR FUNC_CALL 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 VAR NUMBER ASSIGN VAR VAR NUMBER NUMBER ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR VAR VAR NUMBER FOR VAR VAR FUNC_CALL VAR VAR VAR VAR BIN_OP VAR VAR ASSIGN VAR FUNC_CALL VAR VAR VAR ASSIGN VAR FUNC_CALL VAR VAR BIN_OP VAR VAR ASSIGN VAR VAR NUMBER FOR VAR VAR FUNC_CALL VAR VAR NUMBER VAR VAR BIN_OP VAR VAR ASSIGN VAR FUNC_CALL VAR VAR VAR ASSIGN VAR FUNC_CALL VAR VAR BIN_OP VAR VAR EXPR FUNC_CALL VAR BIN_OP VAR VAR |
You are given an array $a$ consisting of $n$ integers. Indices of the array start from zero (i. e. the first element is $a_0$, the second one is $a_1$, and so on).
You can reverse at most one subarray (continuous subsegment) of this array. Recall that the subarray of $a$ with borders $l$ and $r$ is $a[l; r] = a_l, a_{l + 1}, \dots, a_{r}$.
Your task is to reverse such a subarray that the sum of elements on even positions of the resulting array is maximized (i. e. the sum of elements $a_0, a_2, \dots, a_{2k}$ for integer $k = \lfloor\frac{n-1}{2}\rfloor$ should be maximum possible).
You have to answer $t$ independent test cases.
-----Input-----
The first line of the input contains one integer $t$ ($1 \le t \le 2 \cdot 10^4$) — the number of test cases. Then $t$ test cases follow.
The first line of the test case contains one integer $n$ ($1 \le n \le 2 \cdot 10^5$) — the length of $a$. The second line of the test case contains $n$ integers $a_0, a_1, \dots, a_{n-1}$ ($1 \le a_i \le 10^9$), where $a_i$ is the $i$-th element of $a$.
It is guaranteed that the sum of $n$ does not exceed $2 \cdot 10^5$ ($\sum n \le 2 \cdot 10^5$).
-----Output-----
For each test case, print the answer on the separate line — the maximum possible sum of elements on even positions after reversing at most one subarray (continuous subsegment) of $a$.
-----Example-----
Input
4
8
1 7 3 4 7 6 2 9
5
1 2 1 2 1
10
7 8 4 5 7 6 8 9 7 3
4
3 1 2 1
Output
26
5
37
5 | def kadene(A):
max_g = 0
max_c = 0
for i in range(len(A)):
max_c = max(A[i], max_c + A[i])
if max_c > max_g:
max_g = max_c
return max_g
def answer(n, A):
if n == 1:
return A[0]
dp = [0] * (n // 2)
count = 0
s = 0
for i in range(1, n, 2):
dp[count] = A[i] - A[i - 1]
count += 1
s += A[i - 1]
dp1 = [0] * (n // 2)
count = 0
for i in range(1, n, 2):
if i + 1 < n:
dp1[count] = A[i] - A[i + 1]
count += 1
k1 = kadene(dp)
k2 = kadene(dp1)
if n % 2 != 0:
s += A[-1]
return s + max(k1, k2)
t = int(input())
for i in range(t):
n = int(input())
A = list(map(int, input().split()))
print(answer(n, A)) | FUNC_DEF ASSIGN VAR NUMBER ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR VAR VAR BIN_OP VAR VAR VAR IF VAR VAR ASSIGN VAR VAR RETURN VAR FUNC_DEF IF VAR NUMBER RETURN VAR NUMBER ASSIGN VAR BIN_OP LIST NUMBER BIN_OP VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR NUMBER VAR NUMBER ASSIGN VAR VAR BIN_OP VAR VAR VAR BIN_OP VAR NUMBER VAR NUMBER VAR VAR BIN_OP VAR NUMBER ASSIGN VAR BIN_OP LIST NUMBER BIN_OP VAR NUMBER ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR NUMBER VAR NUMBER IF BIN_OP VAR NUMBER VAR ASSIGN VAR VAR BIN_OP VAR VAR VAR BIN_OP VAR NUMBER VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR VAR IF BIN_OP VAR NUMBER NUMBER VAR VAR NUMBER RETURN BIN_OP VAR FUNC_CALL VAR VAR VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR FOR VAR FUNC_CALL 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 FUNC_CALL VAR VAR VAR |
You are given an array $a$ consisting of $n$ integers. Indices of the array start from zero (i. e. the first element is $a_0$, the second one is $a_1$, and so on).
You can reverse at most one subarray (continuous subsegment) of this array. Recall that the subarray of $a$ with borders $l$ and $r$ is $a[l; r] = a_l, a_{l + 1}, \dots, a_{r}$.
Your task is to reverse such a subarray that the sum of elements on even positions of the resulting array is maximized (i. e. the sum of elements $a_0, a_2, \dots, a_{2k}$ for integer $k = \lfloor\frac{n-1}{2}\rfloor$ should be maximum possible).
You have to answer $t$ independent test cases.
-----Input-----
The first line of the input contains one integer $t$ ($1 \le t \le 2 \cdot 10^4$) — the number of test cases. Then $t$ test cases follow.
The first line of the test case contains one integer $n$ ($1 \le n \le 2 \cdot 10^5$) — the length of $a$. The second line of the test case contains $n$ integers $a_0, a_1, \dots, a_{n-1}$ ($1 \le a_i \le 10^9$), where $a_i$ is the $i$-th element of $a$.
It is guaranteed that the sum of $n$ does not exceed $2 \cdot 10^5$ ($\sum n \le 2 \cdot 10^5$).
-----Output-----
For each test case, print the answer on the separate line — the maximum possible sum of elements on even positions after reversing at most one subarray (continuous subsegment) of $a$.
-----Example-----
Input
4
8
1 7 3 4 7 6 2 9
5
1 2 1 2 1
10
7 8 4 5 7 6 8 9 7 3
4
3 1 2 1
Output
26
5
37
5 | def maxSubArraySum(a, size):
max_so_far = 0
max_ending_here = 0
for i in range(0, size):
max_ending_here = max_ending_here + a[i]
if max_ending_here < 0:
max_ending_here = 0
elif max_so_far < max_ending_here:
max_so_far = max_ending_here
return max_so_far
for _ in range(int(input())):
n = int(input())
l = [int(i) for i in input().split()]
sev = 0
for i in range(len(l)):
if i % 2 == 0:
sev += l[i]
i = 0
a = []
while i + 1 < len(l):
a.append(l[i + 1] - l[i])
i += 2
i = 1
b = []
while i + 1 < len(l):
b.append(l[i] - l[i + 1])
i += 2
y = max(maxSubArraySum(a, len(a)), maxSubArraySum(b, len(b)))
print(sev + y) | FUNC_DEF ASSIGN VAR NUMBER ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR NUMBER VAR ASSIGN VAR BIN_OP VAR VAR VAR IF VAR NUMBER ASSIGN VAR NUMBER IF VAR VAR ASSIGN VAR VAR RETURN VAR FOR VAR FUNC_CALL VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR FUNC_CALL VAR VAR IF BIN_OP VAR NUMBER NUMBER VAR VAR VAR ASSIGN VAR NUMBER ASSIGN VAR LIST WHILE BIN_OP VAR NUMBER FUNC_CALL VAR VAR EXPR FUNC_CALL VAR BIN_OP VAR BIN_OP VAR NUMBER VAR VAR VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR LIST WHILE BIN_OP VAR NUMBER FUNC_CALL VAR VAR EXPR FUNC_CALL VAR BIN_OP VAR VAR VAR BIN_OP VAR NUMBER VAR NUMBER ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL VAR VAR FUNC_CALL VAR VAR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR BIN_OP VAR VAR |
You are given an array $a$ consisting of $n$ integers. Indices of the array start from zero (i. e. the first element is $a_0$, the second one is $a_1$, and so on).
You can reverse at most one subarray (continuous subsegment) of this array. Recall that the subarray of $a$ with borders $l$ and $r$ is $a[l; r] = a_l, a_{l + 1}, \dots, a_{r}$.
Your task is to reverse such a subarray that the sum of elements on even positions of the resulting array is maximized (i. e. the sum of elements $a_0, a_2, \dots, a_{2k}$ for integer $k = \lfloor\frac{n-1}{2}\rfloor$ should be maximum possible).
You have to answer $t$ independent test cases.
-----Input-----
The first line of the input contains one integer $t$ ($1 \le t \le 2 \cdot 10^4$) — the number of test cases. Then $t$ test cases follow.
The first line of the test case contains one integer $n$ ($1 \le n \le 2 \cdot 10^5$) — the length of $a$. The second line of the test case contains $n$ integers $a_0, a_1, \dots, a_{n-1}$ ($1 \le a_i \le 10^9$), where $a_i$ is the $i$-th element of $a$.
It is guaranteed that the sum of $n$ does not exceed $2 \cdot 10^5$ ($\sum n \le 2 \cdot 10^5$).
-----Output-----
For each test case, print the answer on the separate line — the maximum possible sum of elements on even positions after reversing at most one subarray (continuous subsegment) of $a$.
-----Example-----
Input
4
8
1 7 3 4 7 6 2 9
5
1 2 1 2 1
10
7 8 4 5 7 6 8 9 7 3
4
3 1 2 1
Output
26
5
37
5 | import sys
reader = (s.rstrip() for s in sys.stdin)
input = reader.__next__
def kadane(A):
t = -100000000000
s = 0
for x in A:
t = max(x, t + x)
s = max(s, t)
return s
t = int(input())
for _ in range(t):
n = int(input())
arr = [int(x) for x in input().split()]
diff1 = [(arr[x] - arr[x - 1]) for x in range(1, n, 2)]
diff2 = [(arr[x - 1] - arr[x]) for x in range(2, n, 2)]
ans = sum([arr[x] for x in range(0, n, 2)])
temp1 = kadane(diff1)
temp2 = kadane(diff2)
temp = max(temp1, temp2)
if temp > 0:
ans += temp
print(ans) | IMPORT ASSIGN VAR FUNC_CALL VAR VAR VAR ASSIGN VAR VAR FUNC_DEF ASSIGN VAR NUMBER ASSIGN VAR NUMBER FOR VAR VAR ASSIGN VAR FUNC_CALL VAR VAR BIN_OP VAR VAR ASSIGN VAR FUNC_CALL VAR VAR VAR RETURN VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR 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 BIN_OP VAR VAR VAR BIN_OP VAR NUMBER VAR FUNC_CALL VAR NUMBER VAR NUMBER ASSIGN VAR BIN_OP VAR BIN_OP VAR NUMBER VAR VAR VAR FUNC_CALL VAR NUMBER VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR VAR VAR FUNC_CALL VAR NUMBER VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR VAR VAR IF VAR NUMBER VAR VAR EXPR FUNC_CALL VAR VAR |
You are given an array $a$ consisting of $n$ integers. Indices of the array start from zero (i. e. the first element is $a_0$, the second one is $a_1$, and so on).
You can reverse at most one subarray (continuous subsegment) of this array. Recall that the subarray of $a$ with borders $l$ and $r$ is $a[l; r] = a_l, a_{l + 1}, \dots, a_{r}$.
Your task is to reverse such a subarray that the sum of elements on even positions of the resulting array is maximized (i. e. the sum of elements $a_0, a_2, \dots, a_{2k}$ for integer $k = \lfloor\frac{n-1}{2}\rfloor$ should be maximum possible).
You have to answer $t$ independent test cases.
-----Input-----
The first line of the input contains one integer $t$ ($1 \le t \le 2 \cdot 10^4$) — the number of test cases. Then $t$ test cases follow.
The first line of the test case contains one integer $n$ ($1 \le n \le 2 \cdot 10^5$) — the length of $a$. The second line of the test case contains $n$ integers $a_0, a_1, \dots, a_{n-1}$ ($1 \le a_i \le 10^9$), where $a_i$ is the $i$-th element of $a$.
It is guaranteed that the sum of $n$ does not exceed $2 \cdot 10^5$ ($\sum n \le 2 \cdot 10^5$).
-----Output-----
For each test case, print the answer on the separate line — the maximum possible sum of elements on even positions after reversing at most one subarray (continuous subsegment) of $a$.
-----Example-----
Input
4
8
1 7 3 4 7 6 2 9
5
1 2 1 2 1
10
7 8 4 5 7 6 8 9 7 3
4
3 1 2 1
Output
26
5
37
5 | def kadane_sum(lista):
max_local_np = max_local_p = max_local_np2 = max_local_p2 = 0
best, best2 = 0, 0
for y in range(1, len(lista), 2):
max_local_p += lista[y - 1]
max_local_np += lista[y]
best = max(best, max_local_np - max_local_p)
if max_local_p > max_local_np:
best = max(best, max_local_np - max_local_p)
max_local_np = 0
max_local_p = 0
best = max(best, max_local_np - max_local_p)
for y in range(2, len(lista), 2):
max_local_p2 += lista[y]
max_local_np2 += lista[y - 1]
best2 = max(best2, max_local_np2 - max_local_p2)
if max_local_p2 > max_local_np2:
best2 = max(best2, max_local_np2 - max_local_p2)
max_local_np2 = 0
max_local_p2 = 0
best2 = max(best2, max_local_np2 - max_local_p2)
best = max(best, best2)
return best
for _ in range(int(input())):
ilosc = input()
lista = [int(x) for x in input().split()]
best = kadane_sum(lista)
print(sum(lista[::2]) + best) | FUNC_DEF ASSIGN VAR VAR VAR VAR NUMBER ASSIGN VAR VAR NUMBER NUMBER FOR VAR FUNC_CALL VAR NUMBER FUNC_CALL VAR VAR NUMBER VAR VAR BIN_OP VAR NUMBER VAR VAR VAR ASSIGN VAR FUNC_CALL VAR VAR BIN_OP VAR VAR IF VAR VAR ASSIGN VAR FUNC_CALL VAR VAR BIN_OP VAR VAR ASSIGN VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR BIN_OP VAR VAR FOR VAR FUNC_CALL VAR NUMBER FUNC_CALL VAR VAR NUMBER VAR VAR VAR VAR VAR BIN_OP VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR BIN_OP VAR VAR IF VAR VAR ASSIGN VAR FUNC_CALL VAR VAR BIN_OP VAR VAR ASSIGN VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR BIN_OP VAR VAR ASSIGN VAR FUNC_CALL VAR VAR VAR RETURN VAR FOR VAR FUNC_CALL VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR BIN_OP FUNC_CALL VAR VAR NUMBER VAR |
You are given an array $a$ consisting of $n$ integers. Indices of the array start from zero (i. e. the first element is $a_0$, the second one is $a_1$, and so on).
You can reverse at most one subarray (continuous subsegment) of this array. Recall that the subarray of $a$ with borders $l$ and $r$ is $a[l; r] = a_l, a_{l + 1}, \dots, a_{r}$.
Your task is to reverse such a subarray that the sum of elements on even positions of the resulting array is maximized (i. e. the sum of elements $a_0, a_2, \dots, a_{2k}$ for integer $k = \lfloor\frac{n-1}{2}\rfloor$ should be maximum possible).
You have to answer $t$ independent test cases.
-----Input-----
The first line of the input contains one integer $t$ ($1 \le t \le 2 \cdot 10^4$) — the number of test cases. Then $t$ test cases follow.
The first line of the test case contains one integer $n$ ($1 \le n \le 2 \cdot 10^5$) — the length of $a$. The second line of the test case contains $n$ integers $a_0, a_1, \dots, a_{n-1}$ ($1 \le a_i \le 10^9$), where $a_i$ is the $i$-th element of $a$.
It is guaranteed that the sum of $n$ does not exceed $2 \cdot 10^5$ ($\sum n \le 2 \cdot 10^5$).
-----Output-----
For each test case, print the answer on the separate line — the maximum possible sum of elements on even positions after reversing at most one subarray (continuous subsegment) of $a$.
-----Example-----
Input
4
8
1 7 3 4 7 6 2 9
5
1 2 1 2 1
10
7 8 4 5 7 6 8 9 7 3
4
3 1 2 1
Output
26
5
37
5 | def qwerty(asa):
ans = 0
if bool(asa):
ans = asa[0]
summ = 0
min_sum = 0
for r in range(len(asa)):
summ += asa[r]
ans = max(ans, summ - min_sum)
min_sum = min(min_sum, summ)
return ans
for t in range(int(input())):
n = int(input())
data = [int(x) for x in input().split()]
ans = 0
if n == 1:
print(data[0])
continue
for i in range(0, n, 2):
ans += data[i]
sp1 = []
sp2 = []
for i in range(0, n - 1, 2):
sp1 += [data[i + 1] - data[i]]
if i + 2 != n:
sp2 += [data[i + 1] - data[i + 2]]
print(max(ans, ans + qwerty(sp1), ans + qwerty(sp2))) | FUNC_DEF ASSIGN VAR NUMBER IF FUNC_CALL VAR VAR ASSIGN VAR VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR FUNC_CALL VAR VAR VAR VAR VAR ASSIGN VAR FUNC_CALL VAR VAR BIN_OP VAR VAR ASSIGN VAR FUNC_CALL VAR VAR VAR RETURN VAR FOR VAR FUNC_CALL VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR NUMBER IF VAR NUMBER EXPR FUNC_CALL VAR VAR NUMBER FOR VAR FUNC_CALL VAR NUMBER VAR NUMBER VAR VAR VAR ASSIGN VAR LIST ASSIGN VAR LIST FOR VAR FUNC_CALL VAR NUMBER BIN_OP VAR NUMBER NUMBER VAR LIST BIN_OP VAR BIN_OP VAR NUMBER VAR VAR IF BIN_OP VAR NUMBER VAR VAR LIST BIN_OP VAR BIN_OP VAR NUMBER VAR BIN_OP VAR NUMBER EXPR FUNC_CALL VAR FUNC_CALL VAR VAR BIN_OP VAR FUNC_CALL VAR VAR BIN_OP VAR FUNC_CALL VAR VAR |
You are given an array $a$ consisting of $n$ integers. Indices of the array start from zero (i. e. the first element is $a_0$, the second one is $a_1$, and so on).
You can reverse at most one subarray (continuous subsegment) of this array. Recall that the subarray of $a$ with borders $l$ and $r$ is $a[l; r] = a_l, a_{l + 1}, \dots, a_{r}$.
Your task is to reverse such a subarray that the sum of elements on even positions of the resulting array is maximized (i. e. the sum of elements $a_0, a_2, \dots, a_{2k}$ for integer $k = \lfloor\frac{n-1}{2}\rfloor$ should be maximum possible).
You have to answer $t$ independent test cases.
-----Input-----
The first line of the input contains one integer $t$ ($1 \le t \le 2 \cdot 10^4$) — the number of test cases. Then $t$ test cases follow.
The first line of the test case contains one integer $n$ ($1 \le n \le 2 \cdot 10^5$) — the length of $a$. The second line of the test case contains $n$ integers $a_0, a_1, \dots, a_{n-1}$ ($1 \le a_i \le 10^9$), where $a_i$ is the $i$-th element of $a$.
It is guaranteed that the sum of $n$ does not exceed $2 \cdot 10^5$ ($\sum n \le 2 \cdot 10^5$).
-----Output-----
For each test case, print the answer on the separate line — the maximum possible sum of elements on even positions after reversing at most one subarray (continuous subsegment) of $a$.
-----Example-----
Input
4
8
1 7 3 4 7 6 2 9
5
1 2 1 2 1
10
7 8 4 5 7 6 8 9 7 3
4
3 1 2 1
Output
26
5
37
5 | import sys
input = sys.stdin.readline
for _ in range(int(input())):
n = int(input())
ar = list(map(int, input().split()))
li = []
for i in range(1, n, 2):
li.append(ar[i] - ar[i - 1])
sm = [0]
counter = 0
for i in li:
counter += i
if counter < 0:
counter = 0
else:
sm.append(counter)
ans1 = max(sm)
ans = 0
li = []
if n > 2:
for i in range(2, n, 2):
ans += ar[i]
li.append(ar[i - 1] - ar[i])
sm = [0]
counter = 0
for i in li:
counter += i
if counter < 0:
counter = 0
else:
sm.append(counter)
ans2 = max(sm)
ans += ar[0]
print(ans + max(ans1, ans2)) | IMPORT ASSIGN VAR VAR FOR VAR FUNC_CALL VAR FUNC_CALL VAR FUNC_CALL 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 LIST FOR VAR FUNC_CALL VAR NUMBER VAR NUMBER EXPR FUNC_CALL VAR BIN_OP VAR VAR VAR BIN_OP VAR NUMBER ASSIGN VAR LIST NUMBER ASSIGN VAR NUMBER FOR VAR VAR VAR VAR IF VAR NUMBER ASSIGN VAR NUMBER EXPR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR NUMBER ASSIGN VAR LIST IF VAR NUMBER FOR VAR FUNC_CALL VAR NUMBER VAR NUMBER VAR VAR VAR EXPR FUNC_CALL VAR BIN_OP VAR BIN_OP VAR NUMBER VAR VAR ASSIGN VAR LIST NUMBER ASSIGN VAR NUMBER FOR VAR VAR VAR VAR IF VAR NUMBER ASSIGN VAR NUMBER EXPR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR VAR VAR VAR NUMBER EXPR FUNC_CALL VAR BIN_OP VAR FUNC_CALL VAR VAR VAR |
You are given an array $a$ consisting of $n$ integers. Indices of the array start from zero (i. e. the first element is $a_0$, the second one is $a_1$, and so on).
You can reverse at most one subarray (continuous subsegment) of this array. Recall that the subarray of $a$ with borders $l$ and $r$ is $a[l; r] = a_l, a_{l + 1}, \dots, a_{r}$.
Your task is to reverse such a subarray that the sum of elements on even positions of the resulting array is maximized (i. e. the sum of elements $a_0, a_2, \dots, a_{2k}$ for integer $k = \lfloor\frac{n-1}{2}\rfloor$ should be maximum possible).
You have to answer $t$ independent test cases.
-----Input-----
The first line of the input contains one integer $t$ ($1 \le t \le 2 \cdot 10^4$) — the number of test cases. Then $t$ test cases follow.
The first line of the test case contains one integer $n$ ($1 \le n \le 2 \cdot 10^5$) — the length of $a$. The second line of the test case contains $n$ integers $a_0, a_1, \dots, a_{n-1}$ ($1 \le a_i \le 10^9$), where $a_i$ is the $i$-th element of $a$.
It is guaranteed that the sum of $n$ does not exceed $2 \cdot 10^5$ ($\sum n \le 2 \cdot 10^5$).
-----Output-----
For each test case, print the answer on the separate line — the maximum possible sum of elements on even positions after reversing at most one subarray (continuous subsegment) of $a$.
-----Example-----
Input
4
8
1 7 3 4 7 6 2 9
5
1 2 1 2 1
10
7 8 4 5 7 6 8 9 7 3
4
3 1 2 1
Output
26
5
37
5 | def solve(arr, n, ans):
odd = 0
even = 0
dp = []
even_indices = [(float("inf"), -1)]
odd_indices = [(float("inf"), -1)]
for i in range(n):
if i % 2 == 0:
even += arr[i]
else:
odd += arr[i]
dp.append(odd - even)
if i % 2 == 0:
odd_indices.append(odd_indices[-1])
if dp[-1] <= even_indices[-1][0]:
even_indices.append((dp[-1], i))
else:
even_indices.append(even_indices[-1])
else:
even_indices.append(even_indices[-1])
if dp[-1] <= odd_indices[-1][0]:
odd_indices.append((dp[-1], i))
else:
odd_indices.append(odd_indices[-1])
even_indices.pop(0)
odd_indices.pop(0)
max_val = 0
l = -1
r = -1
for i in range(1, n):
if i % 2 != 0:
diff = dp[i] - odd_indices[i - 1][0]
if diff > max_val:
max_val = diff
l = odd_indices[i - 1][1] + 1
r = i
if dp[i] > max_val:
max_val = dp[i]
l = 0
r = i
else:
diff = dp[i] - even_indices[i - 1][0]
if diff > max_val:
max_val = diff
l = even_indices[i - 1][1] + 1
r = i
if l != -1:
reverse(arr, l, r, n)
total = find(arr, n)
ans.append(total)
def reverse(arr, l, r, n):
for i in range((r - l + 1) // 2):
arr[l + i], arr[r - i] = arr[r - i], arr[l + i]
def find(arr, n):
total = 0
for i in range(n):
if i % 2 == 0:
total += arr[i]
return total
def brute(arr, n, ans):
ans = 0
li = -1
ri = -1
for l in range(n):
for r in range(l, n):
reverse(arr, l, r, n)
total = find(arr, n)
if total > ans:
ans = total
li = l
ri = r
reverse(arr, l, r, n)
def main():
t = int(input())
ans = []
for i in range(t):
n = int(input())
arr = list(map(int, input().split()))
solve(arr, n, ans)
for i in ans:
print(i)
main() | FUNC_DEF ASSIGN VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR LIST ASSIGN VAR LIST FUNC_CALL VAR STRING NUMBER ASSIGN VAR LIST FUNC_CALL VAR STRING NUMBER FOR VAR FUNC_CALL VAR VAR IF BIN_OP VAR NUMBER NUMBER VAR VAR VAR VAR VAR VAR EXPR FUNC_CALL VAR BIN_OP VAR VAR IF BIN_OP VAR NUMBER NUMBER EXPR FUNC_CALL VAR VAR NUMBER IF VAR NUMBER VAR NUMBER NUMBER EXPR FUNC_CALL VAR VAR NUMBER VAR EXPR FUNC_CALL VAR VAR NUMBER EXPR FUNC_CALL VAR VAR NUMBER IF VAR NUMBER VAR NUMBER NUMBER EXPR FUNC_CALL VAR VAR NUMBER VAR EXPR FUNC_CALL VAR VAR NUMBER EXPR FUNC_CALL VAR NUMBER EXPR FUNC_CALL VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR NUMBER VAR IF BIN_OP VAR NUMBER NUMBER ASSIGN VAR BIN_OP VAR VAR VAR BIN_OP VAR NUMBER NUMBER IF VAR VAR ASSIGN VAR VAR ASSIGN VAR BIN_OP VAR BIN_OP VAR NUMBER NUMBER NUMBER ASSIGN VAR VAR IF VAR VAR VAR ASSIGN VAR VAR VAR ASSIGN VAR NUMBER ASSIGN VAR VAR ASSIGN VAR BIN_OP VAR VAR VAR BIN_OP VAR NUMBER NUMBER IF VAR VAR ASSIGN VAR VAR ASSIGN VAR BIN_OP VAR BIN_OP VAR NUMBER NUMBER NUMBER ASSIGN VAR VAR IF VAR NUMBER EXPR FUNC_CALL VAR VAR VAR VAR VAR ASSIGN VAR FUNC_CALL VAR VAR VAR EXPR FUNC_CALL VAR VAR FUNC_DEF FOR VAR FUNC_CALL VAR BIN_OP BIN_OP BIN_OP VAR VAR NUMBER NUMBER ASSIGN VAR BIN_OP VAR VAR VAR BIN_OP VAR VAR VAR BIN_OP VAR VAR VAR BIN_OP VAR VAR FUNC_DEF ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR VAR IF BIN_OP VAR NUMBER NUMBER VAR VAR VAR RETURN VAR FUNC_DEF ASSIGN VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR VAR FOR VAR FUNC_CALL VAR VAR VAR EXPR FUNC_CALL VAR VAR VAR VAR VAR ASSIGN VAR FUNC_CALL VAR VAR VAR IF VAR VAR ASSIGN VAR VAR ASSIGN VAR VAR ASSIGN VAR VAR EXPR FUNC_CALL VAR VAR VAR VAR VAR FUNC_DEF ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR LIST FOR VAR FUNC_CALL 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 VAR FOR VAR VAR EXPR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR |
You are given an array $a$ consisting of $n$ integers. Indices of the array start from zero (i. e. the first element is $a_0$, the second one is $a_1$, and so on).
You can reverse at most one subarray (continuous subsegment) of this array. Recall that the subarray of $a$ with borders $l$ and $r$ is $a[l; r] = a_l, a_{l + 1}, \dots, a_{r}$.
Your task is to reverse such a subarray that the sum of elements on even positions of the resulting array is maximized (i. e. the sum of elements $a_0, a_2, \dots, a_{2k}$ for integer $k = \lfloor\frac{n-1}{2}\rfloor$ should be maximum possible).
You have to answer $t$ independent test cases.
-----Input-----
The first line of the input contains one integer $t$ ($1 \le t \le 2 \cdot 10^4$) — the number of test cases. Then $t$ test cases follow.
The first line of the test case contains one integer $n$ ($1 \le n \le 2 \cdot 10^5$) — the length of $a$. The second line of the test case contains $n$ integers $a_0, a_1, \dots, a_{n-1}$ ($1 \le a_i \le 10^9$), where $a_i$ is the $i$-th element of $a$.
It is guaranteed that the sum of $n$ does not exceed $2 \cdot 10^5$ ($\sum n \le 2 \cdot 10^5$).
-----Output-----
For each test case, print the answer on the separate line — the maximum possible sum of elements on even positions after reversing at most one subarray (continuous subsegment) of $a$.
-----Example-----
Input
4
8
1 7 3 4 7 6 2 9
5
1 2 1 2 1
10
7 8 4 5 7 6 8 9 7 3
4
3 1 2 1
Output
26
5
37
5 | def solh(arr, neg=False):
sm = 0
mxend = 0
i = 1
while i < len(arr):
sm += (-1 if neg else 1) * (arr[i] - arr[i - 1])
mxend = max(mxend, sm)
if sm < 0:
sm = 0
i += 2
return mxend
def solve():
n = int(input())
arr = [int(v) for v in input().split()]
i = 0
sm = 0
while i < n:
sm += arr[i]
i += 2
print(max(solh(arr), solh(arr[1:], True)) + sm)
t = int(input())
for _ in range(t):
solve() | FUNC_DEF NUMBER ASSIGN VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR NUMBER WHILE VAR FUNC_CALL VAR VAR VAR BIN_OP VAR NUMBER NUMBER BIN_OP VAR VAR VAR BIN_OP VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR VAR IF VAR NUMBER ASSIGN VAR NUMBER VAR NUMBER RETURN VAR FUNC_DEF ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR NUMBER ASSIGN VAR NUMBER WHILE VAR VAR VAR VAR VAR VAR NUMBER EXPR FUNC_CALL VAR BIN_OP FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL VAR VAR NUMBER NUMBER VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR FOR VAR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR |
You are given an array $a$ consisting of $n$ integers. Indices of the array start from zero (i. e. the first element is $a_0$, the second one is $a_1$, and so on).
You can reverse at most one subarray (continuous subsegment) of this array. Recall that the subarray of $a$ with borders $l$ and $r$ is $a[l; r] = a_l, a_{l + 1}, \dots, a_{r}$.
Your task is to reverse such a subarray that the sum of elements on even positions of the resulting array is maximized (i. e. the sum of elements $a_0, a_2, \dots, a_{2k}$ for integer $k = \lfloor\frac{n-1}{2}\rfloor$ should be maximum possible).
You have to answer $t$ independent test cases.
-----Input-----
The first line of the input contains one integer $t$ ($1 \le t \le 2 \cdot 10^4$) — the number of test cases. Then $t$ test cases follow.
The first line of the test case contains one integer $n$ ($1 \le n \le 2 \cdot 10^5$) — the length of $a$. The second line of the test case contains $n$ integers $a_0, a_1, \dots, a_{n-1}$ ($1 \le a_i \le 10^9$), where $a_i$ is the $i$-th element of $a$.
It is guaranteed that the sum of $n$ does not exceed $2 \cdot 10^5$ ($\sum n \le 2 \cdot 10^5$).
-----Output-----
For each test case, print the answer on the separate line — the maximum possible sum of elements on even positions after reversing at most one subarray (continuous subsegment) of $a$.
-----Example-----
Input
4
8
1 7 3 4 7 6 2 9
5
1 2 1 2 1
10
7 8 4 5 7 6 8 9 7 3
4
3 1 2 1
Output
26
5
37
5 | def maxSubArraySum(a, size):
max_so_far = a[0]
curr_max = a[0]
for i in range(1, size):
curr_max = max(a[i], curr_max + a[i])
max_so_far = max(max_so_far, curr_max, 0)
return max(max_so_far, 0)
for _ in range(int(input())):
N = int(input())
A = list(map(int, input().split()))
temp1 = []
temp2 = []
if N == 1:
print(A[0])
elif N % 2 == 0:
suma = 0
for i in range(0, N, 2):
temp1.append(A[i + 1] - A[i])
suma += A[i]
for i in range(1, N - 1, 2):
temp2.append(A[i] - A[i + 1])
t = maxSubArraySum(temp1, len(temp1))
if len(temp2) == 0:
t2 = 0
else:
t2 = maxSubArraySum(temp2, len(temp2))
r = max(t, t2)
print(r + suma)
else:
suma = 0
for i in range(0, N - 1, 2):
temp1.append(A[i + 1] - A[i])
suma += A[i]
suma += A[N - 1]
for i in range(1, N, 2):
temp2.append(A[i] - A[i + 1])
t1 = maxSubArraySum(temp1, len(temp1))
t2 = maxSubArraySum(temp2, len(temp2))
r = max(t1, t2)
print(suma + r) | FUNC_DEF ASSIGN VAR VAR NUMBER ASSIGN VAR VAR NUMBER FOR VAR FUNC_CALL VAR NUMBER VAR ASSIGN VAR FUNC_CALL VAR VAR VAR BIN_OP VAR VAR VAR ASSIGN VAR FUNC_CALL VAR VAR VAR NUMBER RETURN FUNC_CALL VAR VAR NUMBER FOR VAR FUNC_CALL VAR FUNC_CALL VAR FUNC_CALL 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 LIST ASSIGN VAR LIST IF VAR NUMBER EXPR FUNC_CALL VAR VAR NUMBER IF BIN_OP VAR NUMBER NUMBER ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR NUMBER VAR NUMBER EXPR FUNC_CALL VAR BIN_OP VAR BIN_OP VAR NUMBER VAR VAR VAR VAR VAR FOR VAR FUNC_CALL VAR NUMBER BIN_OP VAR NUMBER NUMBER EXPR FUNC_CALL VAR BIN_OP VAR VAR VAR BIN_OP VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR FUNC_CALL VAR VAR IF FUNC_CALL VAR VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR VAR VAR EXPR FUNC_CALL VAR BIN_OP VAR VAR ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR NUMBER BIN_OP VAR NUMBER NUMBER EXPR FUNC_CALL VAR BIN_OP VAR BIN_OP VAR NUMBER VAR VAR VAR VAR VAR VAR VAR BIN_OP VAR NUMBER FOR VAR FUNC_CALL VAR NUMBER VAR NUMBER EXPR FUNC_CALL VAR BIN_OP VAR VAR VAR BIN_OP VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR VAR VAR EXPR FUNC_CALL VAR BIN_OP VAR VAR |
You are given an array $a$ consisting of $n$ integers. Indices of the array start from zero (i. e. the first element is $a_0$, the second one is $a_1$, and so on).
You can reverse at most one subarray (continuous subsegment) of this array. Recall that the subarray of $a$ with borders $l$ and $r$ is $a[l; r] = a_l, a_{l + 1}, \dots, a_{r}$.
Your task is to reverse such a subarray that the sum of elements on even positions of the resulting array is maximized (i. e. the sum of elements $a_0, a_2, \dots, a_{2k}$ for integer $k = \lfloor\frac{n-1}{2}\rfloor$ should be maximum possible).
You have to answer $t$ independent test cases.
-----Input-----
The first line of the input contains one integer $t$ ($1 \le t \le 2 \cdot 10^4$) — the number of test cases. Then $t$ test cases follow.
The first line of the test case contains one integer $n$ ($1 \le n \le 2 \cdot 10^5$) — the length of $a$. The second line of the test case contains $n$ integers $a_0, a_1, \dots, a_{n-1}$ ($1 \le a_i \le 10^9$), where $a_i$ is the $i$-th element of $a$.
It is guaranteed that the sum of $n$ does not exceed $2 \cdot 10^5$ ($\sum n \le 2 \cdot 10^5$).
-----Output-----
For each test case, print the answer on the separate line — the maximum possible sum of elements on even positions after reversing at most one subarray (continuous subsegment) of $a$.
-----Example-----
Input
4
8
1 7 3 4 7 6 2 9
5
1 2 1 2 1
10
7 8 4 5 7 6 8 9 7 3
4
3 1 2 1
Output
26
5
37
5 | def kadane(l):
length = len(l)
maxsub = globalmax = l[0]
for i in range(1, length):
maxsub = max(l[i], maxsub + l[i])
globalmax = max(globalmax, maxsub)
return globalmax
t = int(input(""))
while t > 0:
n = int(input(""))
a = list(map(int, input().split()))
neighbours = []
oddsum = evensum = 0
l1 = [0]
l2 = [0]
if n == 1:
ans = a[0]
else:
for i in range(0, len(a), 2):
if i != len(a) - 1:
l1.append(a[i + 1] - a[i])
evensum += a[i]
ans = evensum
for j in range(1, len(a), 2):
if j != len(a) - 1:
l2.append(a[j] - a[j + 1])
change = max(kadane(l1), kadane(l2))
if change > 0:
ans += change
print(ans)
t -= 1 | FUNC_DEF ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR VAR VAR NUMBER FOR VAR FUNC_CALL VAR NUMBER VAR ASSIGN VAR FUNC_CALL VAR VAR VAR BIN_OP VAR VAR VAR ASSIGN VAR FUNC_CALL VAR VAR VAR RETURN VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR STRING WHILE VAR NUMBER ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR STRING ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR LIST ASSIGN VAR VAR NUMBER ASSIGN VAR LIST NUMBER ASSIGN VAR LIST NUMBER IF VAR NUMBER ASSIGN VAR VAR NUMBER FOR VAR FUNC_CALL VAR NUMBER FUNC_CALL VAR VAR NUMBER IF VAR BIN_OP FUNC_CALL VAR VAR NUMBER EXPR FUNC_CALL VAR BIN_OP VAR BIN_OP VAR NUMBER VAR VAR VAR VAR VAR ASSIGN VAR VAR FOR VAR FUNC_CALL VAR NUMBER FUNC_CALL VAR VAR NUMBER IF VAR BIN_OP FUNC_CALL VAR VAR NUMBER EXPR FUNC_CALL VAR BIN_OP VAR VAR VAR BIN_OP VAR NUMBER ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL VAR VAR IF VAR NUMBER VAR VAR EXPR FUNC_CALL VAR VAR VAR NUMBER |
You are given an array $a$ consisting of $n$ integers. Indices of the array start from zero (i. e. the first element is $a_0$, the second one is $a_1$, and so on).
You can reverse at most one subarray (continuous subsegment) of this array. Recall that the subarray of $a$ with borders $l$ and $r$ is $a[l; r] = a_l, a_{l + 1}, \dots, a_{r}$.
Your task is to reverse such a subarray that the sum of elements on even positions of the resulting array is maximized (i. e. the sum of elements $a_0, a_2, \dots, a_{2k}$ for integer $k = \lfloor\frac{n-1}{2}\rfloor$ should be maximum possible).
You have to answer $t$ independent test cases.
-----Input-----
The first line of the input contains one integer $t$ ($1 \le t \le 2 \cdot 10^4$) — the number of test cases. Then $t$ test cases follow.
The first line of the test case contains one integer $n$ ($1 \le n \le 2 \cdot 10^5$) — the length of $a$. The second line of the test case contains $n$ integers $a_0, a_1, \dots, a_{n-1}$ ($1 \le a_i \le 10^9$), where $a_i$ is the $i$-th element of $a$.
It is guaranteed that the sum of $n$ does not exceed $2 \cdot 10^5$ ($\sum n \le 2 \cdot 10^5$).
-----Output-----
For each test case, print the answer on the separate line — the maximum possible sum of elements on even positions after reversing at most one subarray (continuous subsegment) of $a$.
-----Example-----
Input
4
8
1 7 3 4 7 6 2 9
5
1 2 1 2 1
10
7 8 4 5 7 6 8 9 7 3
4
3 1 2 1
Output
26
5
37
5 | t = int(input())
for _ in range(t):
n = int(input())
arr = [int(i) for i in input().split()]
diff1 = []
diff2 = []
evensum = 0
for i in range(1, n, 2):
diff1.append(arr[i] - arr[i - 1])
for i in range(1, n - 1, 2):
diff2.append(arr[i] - arr[i + 1])
for i in range(0, n, 2):
evensum += arr[i]
m1 = 0
m2 = 0
l = 0
r = 0
s = 0
for i in range(len(diff1)):
s += diff1[i]
if s < 0:
s = 0
if m1 < s:
m1 = s
s = 0
for i in range(len(diff2)):
s += diff2[i]
if s < 0:
s = 0
if m2 < s:
m2 = s
print(evensum + max(m1, m2)) | ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR 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 LIST ASSIGN VAR LIST ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR NUMBER VAR NUMBER EXPR FUNC_CALL VAR BIN_OP VAR VAR VAR BIN_OP VAR NUMBER FOR VAR FUNC_CALL VAR NUMBER BIN_OP VAR NUMBER NUMBER EXPR FUNC_CALL VAR BIN_OP VAR VAR VAR BIN_OP VAR NUMBER FOR VAR FUNC_CALL VAR NUMBER VAR NUMBER VAR VAR VAR ASSIGN VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR FUNC_CALL VAR VAR VAR VAR VAR IF VAR NUMBER ASSIGN VAR NUMBER IF VAR VAR ASSIGN VAR VAR ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR FUNC_CALL VAR VAR VAR VAR VAR IF VAR NUMBER ASSIGN VAR NUMBER IF VAR VAR ASSIGN VAR VAR EXPR FUNC_CALL VAR BIN_OP VAR FUNC_CALL VAR VAR VAR |
You are given an array $a$ consisting of $n$ integers. Indices of the array start from zero (i. e. the first element is $a_0$, the second one is $a_1$, and so on).
You can reverse at most one subarray (continuous subsegment) of this array. Recall that the subarray of $a$ with borders $l$ and $r$ is $a[l; r] = a_l, a_{l + 1}, \dots, a_{r}$.
Your task is to reverse such a subarray that the sum of elements on even positions of the resulting array is maximized (i. e. the sum of elements $a_0, a_2, \dots, a_{2k}$ for integer $k = \lfloor\frac{n-1}{2}\rfloor$ should be maximum possible).
You have to answer $t$ independent test cases.
-----Input-----
The first line of the input contains one integer $t$ ($1 \le t \le 2 \cdot 10^4$) — the number of test cases. Then $t$ test cases follow.
The first line of the test case contains one integer $n$ ($1 \le n \le 2 \cdot 10^5$) — the length of $a$. The second line of the test case contains $n$ integers $a_0, a_1, \dots, a_{n-1}$ ($1 \le a_i \le 10^9$), where $a_i$ is the $i$-th element of $a$.
It is guaranteed that the sum of $n$ does not exceed $2 \cdot 10^5$ ($\sum n \le 2 \cdot 10^5$).
-----Output-----
For each test case, print the answer on the separate line — the maximum possible sum of elements on even positions after reversing at most one subarray (continuous subsegment) of $a$.
-----Example-----
Input
4
8
1 7 3 4 7 6 2 9
5
1 2 1 2 1
10
7 8 4 5 7 6 8 9 7 3
4
3 1 2 1
Output
26
5
37
5 | for tt in range(int(input())):
n = int(input())
a = list(map(int, input().split()))
if n == 1 or n == 2:
print(max(a))
continue
ans = 0
for s in range(2):
suff = (s == 0 and n % 2) * a[-1]
for i in range(s, n - (n - s) % 2, 2):
suff += a[i + s]
pref = s * a[0]
ans = max(ans, pref + suff)
cur = 0
for i in range(s, n - (n - s) % 2, 2):
suff -= a[i + s]
cur = max(cur, pref + a[i + (s ^ 1)], cur + a[i + (s ^ 1)])
pref += a[i + s]
ans = max(ans, max(pref, cur) + suff)
print(ans) | FOR VAR FUNC_CALL VAR FUNC_CALL VAR FUNC_CALL VAR 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 VAR NUMBER EXPR FUNC_CALL VAR FUNC_CALL VAR VAR ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR NUMBER ASSIGN VAR BIN_OP VAR NUMBER BIN_OP VAR NUMBER VAR NUMBER FOR VAR FUNC_CALL VAR VAR BIN_OP VAR BIN_OP BIN_OP VAR VAR NUMBER NUMBER VAR VAR BIN_OP VAR VAR ASSIGN VAR BIN_OP VAR VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR BIN_OP VAR VAR ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR VAR BIN_OP VAR BIN_OP BIN_OP VAR VAR NUMBER NUMBER VAR VAR BIN_OP VAR VAR ASSIGN VAR FUNC_CALL VAR VAR BIN_OP VAR VAR BIN_OP VAR BIN_OP VAR NUMBER BIN_OP VAR VAR BIN_OP VAR BIN_OP VAR NUMBER VAR VAR BIN_OP VAR VAR ASSIGN VAR FUNC_CALL VAR VAR BIN_OP FUNC_CALL VAR VAR VAR VAR EXPR FUNC_CALL VAR VAR |
You are given an array $a$ consisting of $n$ integers. Indices of the array start from zero (i. e. the first element is $a_0$, the second one is $a_1$, and so on).
You can reverse at most one subarray (continuous subsegment) of this array. Recall that the subarray of $a$ with borders $l$ and $r$ is $a[l; r] = a_l, a_{l + 1}, \dots, a_{r}$.
Your task is to reverse such a subarray that the sum of elements on even positions of the resulting array is maximized (i. e. the sum of elements $a_0, a_2, \dots, a_{2k}$ for integer $k = \lfloor\frac{n-1}{2}\rfloor$ should be maximum possible).
You have to answer $t$ independent test cases.
-----Input-----
The first line of the input contains one integer $t$ ($1 \le t \le 2 \cdot 10^4$) — the number of test cases. Then $t$ test cases follow.
The first line of the test case contains one integer $n$ ($1 \le n \le 2 \cdot 10^5$) — the length of $a$. The second line of the test case contains $n$ integers $a_0, a_1, \dots, a_{n-1}$ ($1 \le a_i \le 10^9$), where $a_i$ is the $i$-th element of $a$.
It is guaranteed that the sum of $n$ does not exceed $2 \cdot 10^5$ ($\sum n \le 2 \cdot 10^5$).
-----Output-----
For each test case, print the answer on the separate line — the maximum possible sum of elements on even positions after reversing at most one subarray (continuous subsegment) of $a$.
-----Example-----
Input
4
8
1 7 3 4 7 6 2 9
5
1 2 1 2 1
10
7 8 4 5 7 6 8 9 7 3
4
3 1 2 1
Output
26
5
37
5 | for t in range(int(input())):
n = int(input())
lst = list(map(int, input().split()))
r = s = rs = 0
x = n // 2
for i in range(x):
s = s + (lst[2 * i + 1] - lst[2 * i])
rs = rs + lst[2 * i]
s = max(s, 0)
if s > r:
r = s
if n % 2:
rs += lst[-1]
s = 0
y = (n + 1) // 2
for i in range(1, y):
s += lst[2 * i - 1] - lst[2 * i]
s = max(s, 0)
if s > r:
r = s
print(rs + r) | FOR VAR FUNC_CALL VAR FUNC_CALL VAR FUNC_CALL 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 VAR VAR NUMBER ASSIGN VAR BIN_OP VAR NUMBER FOR VAR FUNC_CALL VAR VAR ASSIGN VAR BIN_OP VAR BIN_OP VAR BIN_OP BIN_OP NUMBER VAR NUMBER VAR BIN_OP NUMBER VAR ASSIGN VAR BIN_OP VAR VAR BIN_OP NUMBER VAR ASSIGN VAR FUNC_CALL VAR VAR NUMBER IF VAR VAR ASSIGN VAR VAR IF BIN_OP VAR NUMBER VAR VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR BIN_OP BIN_OP VAR NUMBER NUMBER FOR VAR FUNC_CALL VAR NUMBER VAR VAR BIN_OP VAR BIN_OP BIN_OP NUMBER VAR NUMBER VAR BIN_OP NUMBER VAR ASSIGN VAR FUNC_CALL VAR VAR NUMBER IF VAR VAR ASSIGN VAR VAR EXPR FUNC_CALL VAR BIN_OP VAR VAR |
You are given an array $a$ consisting of $n$ integers. Indices of the array start from zero (i. e. the first element is $a_0$, the second one is $a_1$, and so on).
You can reverse at most one subarray (continuous subsegment) of this array. Recall that the subarray of $a$ with borders $l$ and $r$ is $a[l; r] = a_l, a_{l + 1}, \dots, a_{r}$.
Your task is to reverse such a subarray that the sum of elements on even positions of the resulting array is maximized (i. e. the sum of elements $a_0, a_2, \dots, a_{2k}$ for integer $k = \lfloor\frac{n-1}{2}\rfloor$ should be maximum possible).
You have to answer $t$ independent test cases.
-----Input-----
The first line of the input contains one integer $t$ ($1 \le t \le 2 \cdot 10^4$) — the number of test cases. Then $t$ test cases follow.
The first line of the test case contains one integer $n$ ($1 \le n \le 2 \cdot 10^5$) — the length of $a$. The second line of the test case contains $n$ integers $a_0, a_1, \dots, a_{n-1}$ ($1 \le a_i \le 10^9$), where $a_i$ is the $i$-th element of $a$.
It is guaranteed that the sum of $n$ does not exceed $2 \cdot 10^5$ ($\sum n \le 2 \cdot 10^5$).
-----Output-----
For each test case, print the answer on the separate line — the maximum possible sum of elements on even positions after reversing at most one subarray (continuous subsegment) of $a$.
-----Example-----
Input
4
8
1 7 3 4 7 6 2 9
5
1 2 1 2 1
10
7 8 4 5 7 6 8 9 7 3
4
3 1 2 1
Output
26
5
37
5 | def main():
n = int(input())
a = list(map(int, input().split()))
best = [0, 0, 0]
st = 0
end = 0
res = 0
for i in range(n // 2):
res += -a[i * 2] + a[i * 2 + 1]
if res < 0:
st = -1
end = -1
res = 0
else:
if st != -1:
end = i * 2 + 1
else:
st = i * 2
end = i * 2 + 1
if res > best[0]:
best[0] = res
best[1] = st
best[2] = end
res = 0
end = 0
st = 0
for i in range(n // 2 - 1 + n % 2):
res += a[i * 2 + 1] - a[i * 2 + 2]
if res < 0:
st = -1
end = -1
res = 0
else:
if st != -1:
end = i * 2 + 1
else:
st = i * 2
end = i * 2 + 1
if res > best[0]:
best[0] = res
best[1] = st
best[2] = end
sm = 0
for i in range(0, n, 2):
sm += a[i]
print(sm + best[0])
t = int(input())
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 LIST NUMBER NUMBER NUMBER ASSIGN VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR BIN_OP VAR NUMBER VAR BIN_OP VAR BIN_OP VAR NUMBER VAR BIN_OP BIN_OP VAR NUMBER NUMBER IF VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR NUMBER IF VAR NUMBER ASSIGN VAR BIN_OP BIN_OP VAR NUMBER NUMBER ASSIGN VAR BIN_OP VAR NUMBER ASSIGN VAR BIN_OP BIN_OP VAR NUMBER NUMBER IF VAR VAR NUMBER ASSIGN VAR NUMBER VAR ASSIGN VAR NUMBER VAR ASSIGN VAR NUMBER VAR ASSIGN VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR BIN_OP BIN_OP BIN_OP VAR NUMBER NUMBER BIN_OP VAR NUMBER VAR BIN_OP VAR BIN_OP BIN_OP VAR NUMBER NUMBER VAR BIN_OP BIN_OP VAR NUMBER NUMBER IF VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR NUMBER IF VAR NUMBER ASSIGN VAR BIN_OP BIN_OP VAR NUMBER NUMBER ASSIGN VAR BIN_OP VAR NUMBER ASSIGN VAR BIN_OP BIN_OP VAR NUMBER NUMBER IF VAR VAR NUMBER ASSIGN VAR NUMBER VAR ASSIGN VAR NUMBER VAR ASSIGN VAR NUMBER VAR ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR NUMBER VAR NUMBER VAR VAR VAR EXPR FUNC_CALL VAR BIN_OP VAR VAR NUMBER ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR FOR VAR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR |
You are given an array $a$ consisting of $n$ integers. Indices of the array start from zero (i. e. the first element is $a_0$, the second one is $a_1$, and so on).
You can reverse at most one subarray (continuous subsegment) of this array. Recall that the subarray of $a$ with borders $l$ and $r$ is $a[l; r] = a_l, a_{l + 1}, \dots, a_{r}$.
Your task is to reverse such a subarray that the sum of elements on even positions of the resulting array is maximized (i. e. the sum of elements $a_0, a_2, \dots, a_{2k}$ for integer $k = \lfloor\frac{n-1}{2}\rfloor$ should be maximum possible).
You have to answer $t$ independent test cases.
-----Input-----
The first line of the input contains one integer $t$ ($1 \le t \le 2 \cdot 10^4$) — the number of test cases. Then $t$ test cases follow.
The first line of the test case contains one integer $n$ ($1 \le n \le 2 \cdot 10^5$) — the length of $a$. The second line of the test case contains $n$ integers $a_0, a_1, \dots, a_{n-1}$ ($1 \le a_i \le 10^9$), where $a_i$ is the $i$-th element of $a$.
It is guaranteed that the sum of $n$ does not exceed $2 \cdot 10^5$ ($\sum n \le 2 \cdot 10^5$).
-----Output-----
For each test case, print the answer on the separate line — the maximum possible sum of elements on even positions after reversing at most one subarray (continuous subsegment) of $a$.
-----Example-----
Input
4
8
1 7 3 4 7 6 2 9
5
1 2 1 2 1
10
7 8 4 5 7 6 8 9 7 3
4
3 1 2 1
Output
26
5
37
5 | def ok(A):
g = 0
l = 0
for i in range(len(A)):
l = max(A[i], l + A[i])
g = max(g, l)
return g
for _ in range(int(input())):
n = int(input())
a = list(map(int, input().split()))
s = 0
for i in range(n):
if i % 2 == 0:
s += a[i]
odd = []
even = []
for i in range(0, n - 1, 2):
even.append(a[i + 1] - a[i])
for i in range(1, n - 1, 2):
odd.append(a[i] - a[i + 1])
b = ok(odd)
c = ok(even)
print(s + max(b, c)) | FUNC_DEF ASSIGN VAR NUMBER ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR VAR VAR BIN_OP VAR VAR VAR ASSIGN VAR FUNC_CALL VAR VAR VAR RETURN VAR FOR VAR FUNC_CALL VAR FUNC_CALL VAR FUNC_CALL 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 NUMBER FOR VAR FUNC_CALL VAR VAR IF BIN_OP VAR NUMBER NUMBER VAR VAR VAR ASSIGN VAR LIST ASSIGN VAR LIST FOR VAR FUNC_CALL VAR NUMBER BIN_OP VAR NUMBER NUMBER EXPR FUNC_CALL VAR BIN_OP VAR BIN_OP VAR NUMBER VAR VAR FOR VAR FUNC_CALL VAR NUMBER BIN_OP VAR NUMBER NUMBER EXPR FUNC_CALL VAR BIN_OP VAR VAR VAR BIN_OP VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR BIN_OP VAR FUNC_CALL VAR VAR VAR |
You are given an array $a$ consisting of $n$ integers. Indices of the array start from zero (i. e. the first element is $a_0$, the second one is $a_1$, and so on).
You can reverse at most one subarray (continuous subsegment) of this array. Recall that the subarray of $a$ with borders $l$ and $r$ is $a[l; r] = a_l, a_{l + 1}, \dots, a_{r}$.
Your task is to reverse such a subarray that the sum of elements on even positions of the resulting array is maximized (i. e. the sum of elements $a_0, a_2, \dots, a_{2k}$ for integer $k = \lfloor\frac{n-1}{2}\rfloor$ should be maximum possible).
You have to answer $t$ independent test cases.
-----Input-----
The first line of the input contains one integer $t$ ($1 \le t \le 2 \cdot 10^4$) — the number of test cases. Then $t$ test cases follow.
The first line of the test case contains one integer $n$ ($1 \le n \le 2 \cdot 10^5$) — the length of $a$. The second line of the test case contains $n$ integers $a_0, a_1, \dots, a_{n-1}$ ($1 \le a_i \le 10^9$), where $a_i$ is the $i$-th element of $a$.
It is guaranteed that the sum of $n$ does not exceed $2 \cdot 10^5$ ($\sum n \le 2 \cdot 10^5$).
-----Output-----
For each test case, print the answer on the separate line — the maximum possible sum of elements on even positions after reversing at most one subarray (continuous subsegment) of $a$.
-----Example-----
Input
4
8
1 7 3 4 7 6 2 9
5
1 2 1 2 1
10
7 8 4 5 7 6 8 9 7 3
4
3 1 2 1
Output
26
5
37
5 | for _ in range(int(input())):
n = int(input())
a = list(map(int, input().split()))
a1 = []
a2 = []
for i in range(0, n - 1, 2):
a1.append(a[i + 1] - a[i])
for i in range(1, n - 1, 2):
a2.append(a[i] - a[i + 1])
ans = 0
tot = 0
for i in range(len(a1)):
tot += a1[i]
if tot < 0:
tot = 0
ans = max(ans, tot)
tot = 0
for i in range(len(a2)):
tot += a2[i]
if tot < 0:
tot = 0
ans = max(ans, tot)
tot = 0
for i in range(0, n, 2):
tot += a[i]
print(tot + ans) | FOR VAR FUNC_CALL VAR FUNC_CALL VAR FUNC_CALL 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 LIST ASSIGN VAR LIST FOR VAR FUNC_CALL VAR NUMBER BIN_OP VAR NUMBER NUMBER EXPR FUNC_CALL VAR BIN_OP VAR BIN_OP VAR NUMBER VAR VAR FOR VAR FUNC_CALL VAR NUMBER BIN_OP VAR NUMBER NUMBER EXPR FUNC_CALL VAR BIN_OP VAR VAR VAR BIN_OP VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR FUNC_CALL VAR VAR VAR VAR VAR IF VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR VAR ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR FUNC_CALL VAR VAR VAR VAR VAR IF VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR VAR ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR NUMBER VAR NUMBER VAR VAR VAR EXPR FUNC_CALL VAR BIN_OP VAR VAR |
You are given an array $a$ consisting of $n$ integers. Indices of the array start from zero (i. e. the first element is $a_0$, the second one is $a_1$, and so on).
You can reverse at most one subarray (continuous subsegment) of this array. Recall that the subarray of $a$ with borders $l$ and $r$ is $a[l; r] = a_l, a_{l + 1}, \dots, a_{r}$.
Your task is to reverse such a subarray that the sum of elements on even positions of the resulting array is maximized (i. e. the sum of elements $a_0, a_2, \dots, a_{2k}$ for integer $k = \lfloor\frac{n-1}{2}\rfloor$ should be maximum possible).
You have to answer $t$ independent test cases.
-----Input-----
The first line of the input contains one integer $t$ ($1 \le t \le 2 \cdot 10^4$) — the number of test cases. Then $t$ test cases follow.
The first line of the test case contains one integer $n$ ($1 \le n \le 2 \cdot 10^5$) — the length of $a$. The second line of the test case contains $n$ integers $a_0, a_1, \dots, a_{n-1}$ ($1 \le a_i \le 10^9$), where $a_i$ is the $i$-th element of $a$.
It is guaranteed that the sum of $n$ does not exceed $2 \cdot 10^5$ ($\sum n \le 2 \cdot 10^5$).
-----Output-----
For each test case, print the answer on the separate line — the maximum possible sum of elements on even positions after reversing at most one subarray (continuous subsegment) of $a$.
-----Example-----
Input
4
8
1 7 3 4 7 6 2 9
5
1 2 1 2 1
10
7 8 4 5 7 6 8 9 7 3
4
3 1 2 1
Output
26
5
37
5 | for w in range(int(input())):
n = int(input())
a = list(map(int, input().split()))
ans = 0
for i in range(0, n, 2):
ans += a[i]
l = []
for i in range(1, n, 2):
l.append(a[i] - a[i - 1])
temp = 0
kadane = -1
for i in range(len(l)):
temp += l[i]
if temp < 0:
temp = 0
kadane = max(kadane, temp)
kadane_endgame = kadane
kadane = -1
temp = 0
l = []
for i in range(1, n, 2):
if i + 1 < n:
l.append(a[i] - a[i + 1])
for i in range(len(l)):
temp += l[i]
if temp < 0:
temp = 0
kadane = max(kadane, temp)
kadane_endgame = max(kadane, kadane_endgame)
ans = max(ans, ans + kadane_endgame)
print(ans) | FOR VAR FUNC_CALL VAR FUNC_CALL VAR FUNC_CALL 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 NUMBER FOR VAR FUNC_CALL VAR NUMBER VAR NUMBER VAR VAR VAR ASSIGN VAR LIST FOR VAR FUNC_CALL VAR NUMBER VAR NUMBER EXPR FUNC_CALL VAR BIN_OP VAR VAR VAR BIN_OP VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR FUNC_CALL VAR VAR VAR VAR VAR IF VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR VAR ASSIGN VAR VAR ASSIGN VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR LIST FOR VAR FUNC_CALL VAR NUMBER VAR NUMBER IF BIN_OP VAR NUMBER VAR EXPR FUNC_CALL VAR BIN_OP VAR VAR VAR BIN_OP VAR NUMBER FOR VAR FUNC_CALL VAR FUNC_CALL VAR VAR VAR VAR VAR IF VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR VAR ASSIGN VAR FUNC_CALL VAR VAR VAR ASSIGN VAR FUNC_CALL VAR VAR BIN_OP VAR VAR EXPR FUNC_CALL VAR VAR |
You are given an array $a$ consisting of $n$ integers. Indices of the array start from zero (i. e. the first element is $a_0$, the second one is $a_1$, and so on).
You can reverse at most one subarray (continuous subsegment) of this array. Recall that the subarray of $a$ with borders $l$ and $r$ is $a[l; r] = a_l, a_{l + 1}, \dots, a_{r}$.
Your task is to reverse such a subarray that the sum of elements on even positions of the resulting array is maximized (i. e. the sum of elements $a_0, a_2, \dots, a_{2k}$ for integer $k = \lfloor\frac{n-1}{2}\rfloor$ should be maximum possible).
You have to answer $t$ independent test cases.
-----Input-----
The first line of the input contains one integer $t$ ($1 \le t \le 2 \cdot 10^4$) — the number of test cases. Then $t$ test cases follow.
The first line of the test case contains one integer $n$ ($1 \le n \le 2 \cdot 10^5$) — the length of $a$. The second line of the test case contains $n$ integers $a_0, a_1, \dots, a_{n-1}$ ($1 \le a_i \le 10^9$), where $a_i$ is the $i$-th element of $a$.
It is guaranteed that the sum of $n$ does not exceed $2 \cdot 10^5$ ($\sum n \le 2 \cdot 10^5$).
-----Output-----
For each test case, print the answer on the separate line — the maximum possible sum of elements on even positions after reversing at most one subarray (continuous subsegment) of $a$.
-----Example-----
Input
4
8
1 7 3 4 7 6 2 9
5
1 2 1 2 1
10
7 8 4 5 7 6 8 9 7 3
4
3 1 2 1
Output
26
5
37
5 | t = int(input())
for _ in range(t):
n = int(input())
a = list(map(int, input().split()))
se = [0]
for i in range(n):
if i % 2 == 0:
se.append(se[-1] + a[i])
ans = se[-1]
tmp = 0
A = 0
for i in range(0, n - 1, 2):
tmp += a[i + 1] - a[i]
A = max(A, tmp)
if tmp < 0:
tmp = 0
tmp = 0
B = 0
for i in range(1, n - 1, 2):
tmp += a[i] - a[i + 1]
B = max(B, tmp)
if tmp < 0:
tmp = 0
ans += max(A, B, 0)
print(ans) | ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR FOR VAR FUNC_CALL 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 LIST NUMBER FOR VAR FUNC_CALL VAR VAR IF BIN_OP VAR NUMBER NUMBER EXPR FUNC_CALL VAR BIN_OP VAR NUMBER VAR VAR ASSIGN VAR VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR NUMBER BIN_OP VAR NUMBER NUMBER VAR BIN_OP VAR BIN_OP VAR NUMBER VAR VAR ASSIGN VAR FUNC_CALL VAR VAR VAR IF VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR NUMBER BIN_OP VAR NUMBER NUMBER VAR BIN_OP VAR VAR VAR BIN_OP VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR VAR IF VAR NUMBER ASSIGN VAR NUMBER VAR FUNC_CALL VAR VAR VAR NUMBER EXPR FUNC_CALL VAR VAR |
You are given an array $a$ consisting of $n$ integers. Indices of the array start from zero (i. e. the first element is $a_0$, the second one is $a_1$, and so on).
You can reverse at most one subarray (continuous subsegment) of this array. Recall that the subarray of $a$ with borders $l$ and $r$ is $a[l; r] = a_l, a_{l + 1}, \dots, a_{r}$.
Your task is to reverse such a subarray that the sum of elements on even positions of the resulting array is maximized (i. e. the sum of elements $a_0, a_2, \dots, a_{2k}$ for integer $k = \lfloor\frac{n-1}{2}\rfloor$ should be maximum possible).
You have to answer $t$ independent test cases.
-----Input-----
The first line of the input contains one integer $t$ ($1 \le t \le 2 \cdot 10^4$) — the number of test cases. Then $t$ test cases follow.
The first line of the test case contains one integer $n$ ($1 \le n \le 2 \cdot 10^5$) — the length of $a$. The second line of the test case contains $n$ integers $a_0, a_1, \dots, a_{n-1}$ ($1 \le a_i \le 10^9$), where $a_i$ is the $i$-th element of $a$.
It is guaranteed that the sum of $n$ does not exceed $2 \cdot 10^5$ ($\sum n \le 2 \cdot 10^5$).
-----Output-----
For each test case, print the answer on the separate line — the maximum possible sum of elements on even positions after reversing at most one subarray (continuous subsegment) of $a$.
-----Example-----
Input
4
8
1 7 3 4 7 6 2 9
5
1 2 1 2 1
10
7 8 4 5 7 6 8 9 7 3
4
3 1 2 1
Output
26
5
37
5 | t = int(input())
for _ in range(t):
n = int(input())
line = input()
ar = line.split()
ar = [int(i) for i in ar]
ans1 = list()
ans2 = [0, 0]
ans3 = list()
ans1.append(0)
ans3.append(0)
for i in range(n):
if i % 2 == 0:
ans1.append(ans1[i] + ar[i])
else:
ans1.append(ans1[i] + 0)
if i % 2 == 0 and i + 2 <= n:
ans2.append(ar[i + 1] + max(ans1[i], ans2[i]))
elif i + 2 <= n:
ans2.append(ar[i] + max(ans1[i], ans2[i]))
if i % 2 == 0:
ans3.append(max(ans1[i], ans2[i], ans3[i]) + ar[i])
else:
ans3.append(max(ans1[i], ans2[i], ans3[i]) + 0)
print(max(max(ans1), max(ans2), max(ans3))) | ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR FOR VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR VAR VAR VAR ASSIGN VAR FUNC_CALL VAR ASSIGN VAR LIST NUMBER NUMBER ASSIGN VAR FUNC_CALL VAR EXPR FUNC_CALL VAR NUMBER EXPR FUNC_CALL VAR NUMBER FOR VAR FUNC_CALL VAR VAR IF BIN_OP VAR NUMBER NUMBER EXPR FUNC_CALL VAR BIN_OP VAR VAR VAR VAR EXPR FUNC_CALL VAR BIN_OP VAR VAR NUMBER IF BIN_OP VAR NUMBER NUMBER BIN_OP VAR NUMBER VAR EXPR FUNC_CALL VAR BIN_OP VAR BIN_OP VAR NUMBER FUNC_CALL VAR VAR VAR VAR VAR IF BIN_OP VAR NUMBER VAR EXPR FUNC_CALL VAR BIN_OP VAR VAR FUNC_CALL VAR VAR VAR VAR VAR IF BIN_OP VAR NUMBER NUMBER EXPR FUNC_CALL VAR BIN_OP FUNC_CALL VAR VAR VAR VAR VAR VAR VAR VAR VAR EXPR FUNC_CALL VAR BIN_OP FUNC_CALL VAR VAR VAR VAR VAR VAR VAR NUMBER EXPR FUNC_CALL VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL VAR VAR FUNC_CALL VAR VAR |
You are given an array $a$ consisting of $n$ integers. Indices of the array start from zero (i. e. the first element is $a_0$, the second one is $a_1$, and so on).
You can reverse at most one subarray (continuous subsegment) of this array. Recall that the subarray of $a$ with borders $l$ and $r$ is $a[l; r] = a_l, a_{l + 1}, \dots, a_{r}$.
Your task is to reverse such a subarray that the sum of elements on even positions of the resulting array is maximized (i. e. the sum of elements $a_0, a_2, \dots, a_{2k}$ for integer $k = \lfloor\frac{n-1}{2}\rfloor$ should be maximum possible).
You have to answer $t$ independent test cases.
-----Input-----
The first line of the input contains one integer $t$ ($1 \le t \le 2 \cdot 10^4$) — the number of test cases. Then $t$ test cases follow.
The first line of the test case contains one integer $n$ ($1 \le n \le 2 \cdot 10^5$) — the length of $a$. The second line of the test case contains $n$ integers $a_0, a_1, \dots, a_{n-1}$ ($1 \le a_i \le 10^9$), where $a_i$ is the $i$-th element of $a$.
It is guaranteed that the sum of $n$ does not exceed $2 \cdot 10^5$ ($\sum n \le 2 \cdot 10^5$).
-----Output-----
For each test case, print the answer on the separate line — the maximum possible sum of elements on even positions after reversing at most one subarray (continuous subsegment) of $a$.
-----Example-----
Input
4
8
1 7 3 4 7 6 2 9
5
1 2 1 2 1
10
7 8 4 5 7 6 8 9 7 3
4
3 1 2 1
Output
26
5
37
5 | for _ in range(int(input())):
n = int(input())
arr = list(map(int, input().split()))
eveSum = 0
for i in range(n):
if i % 2 == 0:
eveSum += arr[i]
preSum = 0
maxDiffSum = 0
for i in range(1, n, 2):
preSum = max(preSum + arr[i] - arr[i - 1], 0)
maxDiffSum = max(maxDiffSum, preSum)
preSum = 0
for i in range(1, n - 1, 2):
preSum = max(preSum + arr[i] - arr[i + 1], 0)
maxDiffSum = max(maxDiffSum, preSum)
print(eveSum + maxDiffSum) | FOR VAR FUNC_CALL VAR FUNC_CALL VAR FUNC_CALL 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 NUMBER FOR VAR FUNC_CALL VAR VAR IF BIN_OP VAR NUMBER NUMBER VAR VAR VAR ASSIGN VAR NUMBER ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR NUMBER VAR NUMBER ASSIGN VAR FUNC_CALL VAR BIN_OP BIN_OP VAR VAR VAR VAR BIN_OP VAR NUMBER NUMBER ASSIGN VAR FUNC_CALL VAR VAR VAR ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR NUMBER BIN_OP VAR NUMBER NUMBER ASSIGN VAR FUNC_CALL VAR BIN_OP BIN_OP VAR VAR VAR VAR BIN_OP VAR NUMBER NUMBER ASSIGN VAR FUNC_CALL VAR VAR VAR EXPR FUNC_CALL VAR BIN_OP VAR VAR |
You are given an array $a$ consisting of $n$ integers. Indices of the array start from zero (i. e. the first element is $a_0$, the second one is $a_1$, and so on).
You can reverse at most one subarray (continuous subsegment) of this array. Recall that the subarray of $a$ with borders $l$ and $r$ is $a[l; r] = a_l, a_{l + 1}, \dots, a_{r}$.
Your task is to reverse such a subarray that the sum of elements on even positions of the resulting array is maximized (i. e. the sum of elements $a_0, a_2, \dots, a_{2k}$ for integer $k = \lfloor\frac{n-1}{2}\rfloor$ should be maximum possible).
You have to answer $t$ independent test cases.
-----Input-----
The first line of the input contains one integer $t$ ($1 \le t \le 2 \cdot 10^4$) — the number of test cases. Then $t$ test cases follow.
The first line of the test case contains one integer $n$ ($1 \le n \le 2 \cdot 10^5$) — the length of $a$. The second line of the test case contains $n$ integers $a_0, a_1, \dots, a_{n-1}$ ($1 \le a_i \le 10^9$), where $a_i$ is the $i$-th element of $a$.
It is guaranteed that the sum of $n$ does not exceed $2 \cdot 10^5$ ($\sum n \le 2 \cdot 10^5$).
-----Output-----
For each test case, print the answer on the separate line — the maximum possible sum of elements on even positions after reversing at most one subarray (continuous subsegment) of $a$.
-----Example-----
Input
4
8
1 7 3 4 7 6 2 9
5
1 2 1 2 1
10
7 8 4 5 7 6 8 9 7 3
4
3 1 2 1
Output
26
5
37
5 | from sys import stdin
input = stdin.readline
def answer():
dp = [[0, 0, 0, 0] for i in range(n + 1)]
for i in range(1, n + 1):
dp[i][0] = dp[i - 1][0]
dp[i][1] = dp[i - 1][1]
dp[i][2] = dp[i - 1][2]
dp[i][3] = dp[i - 1][3]
if i & 1:
dp[i][0] += a[i - 1]
dp[i][3] += a[i - 1]
dp[i][1] = max(dp[i][1], dp[i - 1][0])
dp[i][3] = max(dp[i][3], dp[i][2], dp[i][0])
else:
dp[i][1] += a[i - 1]
dp[i][2] = max(dp[i][2], dp[i - 1][0]) + a[i - 1]
dp[i][3] = max(dp[i][3], dp[i][1])
return dp[n][3]
for T in range(int(input())):
n = int(input())
a = list(map(int, input().split()))
print(answer()) | ASSIGN VAR VAR FUNC_DEF ASSIGN VAR LIST NUMBER NUMBER NUMBER NUMBER VAR FUNC_CALL VAR BIN_OP VAR NUMBER FOR VAR FUNC_CALL VAR NUMBER BIN_OP VAR NUMBER ASSIGN VAR VAR NUMBER VAR BIN_OP VAR NUMBER NUMBER ASSIGN VAR VAR NUMBER VAR BIN_OP VAR NUMBER NUMBER ASSIGN VAR VAR NUMBER VAR BIN_OP VAR NUMBER NUMBER ASSIGN VAR VAR NUMBER VAR BIN_OP VAR NUMBER NUMBER IF BIN_OP VAR NUMBER VAR VAR NUMBER VAR BIN_OP VAR NUMBER VAR VAR NUMBER VAR BIN_OP VAR NUMBER ASSIGN VAR VAR NUMBER FUNC_CALL VAR VAR VAR NUMBER VAR BIN_OP VAR NUMBER NUMBER ASSIGN VAR VAR NUMBER FUNC_CALL VAR VAR VAR NUMBER VAR VAR NUMBER VAR VAR NUMBER VAR VAR NUMBER VAR BIN_OP VAR NUMBER ASSIGN VAR VAR NUMBER BIN_OP FUNC_CALL VAR VAR VAR NUMBER VAR BIN_OP VAR NUMBER NUMBER VAR BIN_OP VAR NUMBER ASSIGN VAR VAR NUMBER FUNC_CALL VAR VAR VAR NUMBER VAR VAR NUMBER RETURN VAR VAR NUMBER FOR VAR FUNC_CALL VAR FUNC_CALL VAR FUNC_CALL 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 FUNC_CALL VAR |
You are given an array $a$ consisting of $n$ integers. Indices of the array start from zero (i. e. the first element is $a_0$, the second one is $a_1$, and so on).
You can reverse at most one subarray (continuous subsegment) of this array. Recall that the subarray of $a$ with borders $l$ and $r$ is $a[l; r] = a_l, a_{l + 1}, \dots, a_{r}$.
Your task is to reverse such a subarray that the sum of elements on even positions of the resulting array is maximized (i. e. the sum of elements $a_0, a_2, \dots, a_{2k}$ for integer $k = \lfloor\frac{n-1}{2}\rfloor$ should be maximum possible).
You have to answer $t$ independent test cases.
-----Input-----
The first line of the input contains one integer $t$ ($1 \le t \le 2 \cdot 10^4$) — the number of test cases. Then $t$ test cases follow.
The first line of the test case contains one integer $n$ ($1 \le n \le 2 \cdot 10^5$) — the length of $a$. The second line of the test case contains $n$ integers $a_0, a_1, \dots, a_{n-1}$ ($1 \le a_i \le 10^9$), where $a_i$ is the $i$-th element of $a$.
It is guaranteed that the sum of $n$ does not exceed $2 \cdot 10^5$ ($\sum n \le 2 \cdot 10^5$).
-----Output-----
For each test case, print the answer on the separate line — the maximum possible sum of elements on even positions after reversing at most one subarray (continuous subsegment) of $a$.
-----Example-----
Input
4
8
1 7 3 4 7 6 2 9
5
1 2 1 2 1
10
7 8 4 5 7 6 8 9 7 3
4
3 1 2 1
Output
26
5
37
5 | def max_subarray(numbers):
best_sum = 0
best_start = best_end = 0
current_sum = 0
for current_end, x in enumerate(numbers):
if current_sum <= 0:
current_start = current_end
current_sum = x
else:
current_sum += x
if current_sum > best_sum:
best_sum = current_sum
best_start = current_start
best_end = current_end + 1
return best_sum, best_start, best_end
def solve():
n = int(input())
a = [int(x) for x in input().split()]
b = []
c = []
s = 0
for i in range(n):
if i % 2 == 0:
if i + 1 < n:
b.append(a[i + 1] - a[i])
s += a[i]
elif i + 1 < n:
c.append(a[i] - a[i + 1])
p, q, r = max_subarray(b)
p1, q1, r1 = max_subarray(c)
print(s + max(p, p1))
t = int(input())
for i in range(t):
solve() | FUNC_DEF ASSIGN VAR NUMBER ASSIGN VAR VAR NUMBER ASSIGN VAR NUMBER FOR VAR VAR FUNC_CALL VAR VAR IF VAR NUMBER ASSIGN VAR VAR ASSIGN VAR VAR VAR VAR IF VAR VAR ASSIGN VAR VAR ASSIGN VAR VAR ASSIGN VAR BIN_OP VAR NUMBER RETURN VAR VAR VAR FUNC_DEF ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR LIST ASSIGN VAR LIST ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR VAR IF BIN_OP VAR NUMBER NUMBER IF BIN_OP VAR NUMBER VAR EXPR FUNC_CALL VAR BIN_OP VAR BIN_OP VAR NUMBER VAR VAR VAR VAR VAR IF BIN_OP VAR NUMBER VAR EXPR FUNC_CALL VAR BIN_OP VAR VAR VAR BIN_OP VAR NUMBER ASSIGN VAR VAR VAR FUNC_CALL VAR VAR ASSIGN VAR VAR VAR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR BIN_OP VAR FUNC_CALL VAR VAR VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR FOR VAR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR |
You are given an array $a$ consisting of $n$ integers. Indices of the array start from zero (i. e. the first element is $a_0$, the second one is $a_1$, and so on).
You can reverse at most one subarray (continuous subsegment) of this array. Recall that the subarray of $a$ with borders $l$ and $r$ is $a[l; r] = a_l, a_{l + 1}, \dots, a_{r}$.
Your task is to reverse such a subarray that the sum of elements on even positions of the resulting array is maximized (i. e. the sum of elements $a_0, a_2, \dots, a_{2k}$ for integer $k = \lfloor\frac{n-1}{2}\rfloor$ should be maximum possible).
You have to answer $t$ independent test cases.
-----Input-----
The first line of the input contains one integer $t$ ($1 \le t \le 2 \cdot 10^4$) — the number of test cases. Then $t$ test cases follow.
The first line of the test case contains one integer $n$ ($1 \le n \le 2 \cdot 10^5$) — the length of $a$. The second line of the test case contains $n$ integers $a_0, a_1, \dots, a_{n-1}$ ($1 \le a_i \le 10^9$), where $a_i$ is the $i$-th element of $a$.
It is guaranteed that the sum of $n$ does not exceed $2 \cdot 10^5$ ($\sum n \le 2 \cdot 10^5$).
-----Output-----
For each test case, print the answer on the separate line — the maximum possible sum of elements on even positions after reversing at most one subarray (continuous subsegment) of $a$.
-----Example-----
Input
4
8
1 7 3 4 7 6 2 9
5
1 2 1 2 1
10
7 8 4 5 7 6 8 9 7 3
4
3 1 2 1
Output
26
5
37
5 | def subsolve(a):
c = 0
m = 0
for i in a:
c += i
if c < 0:
c = 0
m = max(m, c)
return m
def solve():
n = int(input())
a = list(map(int, input().split()))
s = 0
for i in range(0, n, 2):
s += a[i]
u = subsolve([(a[i] - a[i - 1]) for i in range(1, n, 2)])
v = subsolve([(a[i - 1] - a[i]) for i in range(2, n, 2)])
print(max(u, v) + s)
t = int(input())
for _ in range(t):
solve() | FUNC_DEF ASSIGN VAR NUMBER ASSIGN VAR NUMBER FOR VAR VAR VAR VAR IF VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR VAR RETURN 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 NUMBER FOR VAR FUNC_CALL VAR NUMBER VAR NUMBER VAR VAR VAR ASSIGN VAR FUNC_CALL VAR BIN_OP VAR VAR VAR BIN_OP VAR NUMBER VAR FUNC_CALL VAR NUMBER VAR NUMBER ASSIGN VAR FUNC_CALL VAR BIN_OP VAR BIN_OP VAR NUMBER VAR VAR VAR FUNC_CALL VAR NUMBER VAR NUMBER EXPR FUNC_CALL VAR BIN_OP FUNC_CALL VAR VAR VAR VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR FOR VAR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR |
You are given an array $a$ consisting of $n$ integers. Indices of the array start from zero (i. e. the first element is $a_0$, the second one is $a_1$, and so on).
You can reverse at most one subarray (continuous subsegment) of this array. Recall that the subarray of $a$ with borders $l$ and $r$ is $a[l; r] = a_l, a_{l + 1}, \dots, a_{r}$.
Your task is to reverse such a subarray that the sum of elements on even positions of the resulting array is maximized (i. e. the sum of elements $a_0, a_2, \dots, a_{2k}$ for integer $k = \lfloor\frac{n-1}{2}\rfloor$ should be maximum possible).
You have to answer $t$ independent test cases.
-----Input-----
The first line of the input contains one integer $t$ ($1 \le t \le 2 \cdot 10^4$) — the number of test cases. Then $t$ test cases follow.
The first line of the test case contains one integer $n$ ($1 \le n \le 2 \cdot 10^5$) — the length of $a$. The second line of the test case contains $n$ integers $a_0, a_1, \dots, a_{n-1}$ ($1 \le a_i \le 10^9$), where $a_i$ is the $i$-th element of $a$.
It is guaranteed that the sum of $n$ does not exceed $2 \cdot 10^5$ ($\sum n \le 2 \cdot 10^5$).
-----Output-----
For each test case, print the answer on the separate line — the maximum possible sum of elements on even positions after reversing at most one subarray (continuous subsegment) of $a$.
-----Example-----
Input
4
8
1 7 3 4 7 6 2 9
5
1 2 1 2 1
10
7 8 4 5 7 6 8 9 7 3
4
3 1 2 1
Output
26
5
37
5 | cases = int(input())
for i in range(cases):
n = int(input())
l2 = input().split(" ")
arr = []
sum1 = 0
for a in l2:
arr.append(int(a))
diff1 = []
for j in range(0, len(arr), 2):
if j + 1 >= len(arr):
break
ele1 = arr[j]
ele2 = arr[j + 1]
diff1.append(ele2 - ele1)
diff2 = []
for k in range(1, len(arr), 2):
if k + 1 >= len(arr):
break
ele1 = arr[k]
ele2 = arr[k + 1]
diff2.append(ele1 - ele2)
for p in range(len(arr)):
if p % 2 == 0:
sum1 += arr[p]
s1 = s2 = -1
for a in range(len(diff1)):
start = diff1[a]
if start > 0:
s1 = a
break
for b in range(len(diff2)):
start = diff2[b]
if start > 0:
s2 = b
break
e1 = e2 = -1
for c in range(len(diff1) - 1, -1, -1):
end = diff1[c]
if end > 0:
e1 = c
break
for d in range(len(diff2) - 1, -1, -1):
end = diff2[d]
if end > 0:
e2 = d
break
if s1 == e1 == s2 == e2 == -1:
print(sum1)
continue
sum2 = sum3 = 0
t1 = t2 = 0
if e1 >= s1 and e1 != -1:
for w in range(s1, e1 + 1):
num = diff1[w]
sum2 += num
if sum2 > t1:
t1 = sum2
if sum2 < 0:
sum2 = 0
if e2 >= s2 and e2 != -1:
for x in range(s2, e2 + 1):
num = diff2[x]
sum3 += num
if sum3 > t2:
t2 = sum3
if sum3 < 0:
sum3 = 0
m = max(t1, t2)
print(sum1 + m) | ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR FOR VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL FUNC_CALL VAR STRING ASSIGN VAR LIST ASSIGN VAR NUMBER FOR VAR VAR EXPR FUNC_CALL VAR FUNC_CALL VAR VAR ASSIGN VAR LIST FOR VAR FUNC_CALL VAR NUMBER FUNC_CALL VAR VAR NUMBER IF BIN_OP VAR NUMBER FUNC_CALL VAR VAR ASSIGN VAR VAR VAR ASSIGN VAR VAR BIN_OP VAR NUMBER EXPR FUNC_CALL VAR BIN_OP VAR VAR ASSIGN VAR LIST FOR VAR FUNC_CALL VAR NUMBER FUNC_CALL VAR VAR NUMBER IF BIN_OP VAR NUMBER FUNC_CALL VAR VAR ASSIGN VAR VAR VAR ASSIGN VAR VAR BIN_OP VAR NUMBER EXPR FUNC_CALL VAR BIN_OP VAR VAR FOR VAR FUNC_CALL VAR FUNC_CALL VAR VAR IF BIN_OP VAR NUMBER NUMBER VAR VAR VAR ASSIGN VAR VAR NUMBER FOR VAR FUNC_CALL VAR FUNC_CALL VAR VAR ASSIGN VAR VAR VAR IF VAR NUMBER ASSIGN VAR VAR FOR VAR FUNC_CALL VAR FUNC_CALL VAR VAR ASSIGN VAR VAR VAR IF VAR NUMBER ASSIGN VAR VAR ASSIGN VAR VAR NUMBER FOR VAR FUNC_CALL VAR BIN_OP FUNC_CALL VAR VAR NUMBER NUMBER NUMBER ASSIGN VAR VAR VAR IF VAR NUMBER ASSIGN VAR VAR FOR VAR FUNC_CALL VAR BIN_OP FUNC_CALL VAR VAR NUMBER NUMBER NUMBER ASSIGN VAR VAR VAR IF VAR NUMBER ASSIGN VAR VAR IF VAR VAR VAR VAR NUMBER EXPR FUNC_CALL VAR VAR ASSIGN VAR VAR NUMBER ASSIGN VAR VAR NUMBER IF VAR VAR VAR NUMBER FOR VAR FUNC_CALL VAR VAR BIN_OP VAR NUMBER ASSIGN VAR VAR VAR VAR VAR IF VAR VAR ASSIGN VAR VAR IF VAR NUMBER ASSIGN VAR NUMBER IF VAR VAR VAR NUMBER FOR VAR FUNC_CALL VAR VAR BIN_OP VAR NUMBER ASSIGN VAR VAR VAR VAR VAR IF VAR VAR ASSIGN VAR VAR IF VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR VAR EXPR FUNC_CALL VAR BIN_OP VAR VAR |
You are given an array $a$ consisting of $n$ integers. Indices of the array start from zero (i. e. the first element is $a_0$, the second one is $a_1$, and so on).
You can reverse at most one subarray (continuous subsegment) of this array. Recall that the subarray of $a$ with borders $l$ and $r$ is $a[l; r] = a_l, a_{l + 1}, \dots, a_{r}$.
Your task is to reverse such a subarray that the sum of elements on even positions of the resulting array is maximized (i. e. the sum of elements $a_0, a_2, \dots, a_{2k}$ for integer $k = \lfloor\frac{n-1}{2}\rfloor$ should be maximum possible).
You have to answer $t$ independent test cases.
-----Input-----
The first line of the input contains one integer $t$ ($1 \le t \le 2 \cdot 10^4$) — the number of test cases. Then $t$ test cases follow.
The first line of the test case contains one integer $n$ ($1 \le n \le 2 \cdot 10^5$) — the length of $a$. The second line of the test case contains $n$ integers $a_0, a_1, \dots, a_{n-1}$ ($1 \le a_i \le 10^9$), where $a_i$ is the $i$-th element of $a$.
It is guaranteed that the sum of $n$ does not exceed $2 \cdot 10^5$ ($\sum n \le 2 \cdot 10^5$).
-----Output-----
For each test case, print the answer on the separate line — the maximum possible sum of elements on even positions after reversing at most one subarray (continuous subsegment) of $a$.
-----Example-----
Input
4
8
1 7 3 4 7 6 2 9
5
1 2 1 2 1
10
7 8 4 5 7 6 8 9 7 3
4
3 1 2 1
Output
26
5
37
5 | t = int(input())
for T in range(t):
n = int(input())
a = [int(x) for x in input().split()]
ans, s, mx = 0, 0, 0
for i in range(0, n, 2):
ans += a[i]
l1, l2 = [], []
for i in range(0, n - 1, 2):
s += a[i + 1] - a[i]
l1.append(s)
mx = max(mx, s)
s = max(s, 0)
s = 0
for i in range(1, n - 1, 2):
s += a[i] - a[i + 1]
l2.append(s)
mx = max(mx, s)
s = max(s, 0)
print(mx + ans) | ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR 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 VAR VAR NUMBER NUMBER NUMBER FOR VAR FUNC_CALL VAR NUMBER VAR NUMBER VAR VAR VAR ASSIGN VAR VAR LIST LIST FOR VAR FUNC_CALL VAR NUMBER BIN_OP VAR NUMBER NUMBER VAR BIN_OP VAR BIN_OP VAR NUMBER VAR VAR EXPR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR VAR VAR ASSIGN VAR FUNC_CALL VAR VAR NUMBER ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR NUMBER BIN_OP VAR NUMBER NUMBER VAR BIN_OP VAR VAR VAR BIN_OP VAR NUMBER EXPR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR VAR VAR ASSIGN VAR FUNC_CALL VAR VAR NUMBER EXPR FUNC_CALL VAR BIN_OP VAR VAR |
You are given an array $a$ consisting of $n$ integers. Indices of the array start from zero (i. e. the first element is $a_0$, the second one is $a_1$, and so on).
You can reverse at most one subarray (continuous subsegment) of this array. Recall that the subarray of $a$ with borders $l$ and $r$ is $a[l; r] = a_l, a_{l + 1}, \dots, a_{r}$.
Your task is to reverse such a subarray that the sum of elements on even positions of the resulting array is maximized (i. e. the sum of elements $a_0, a_2, \dots, a_{2k}$ for integer $k = \lfloor\frac{n-1}{2}\rfloor$ should be maximum possible).
You have to answer $t$ independent test cases.
-----Input-----
The first line of the input contains one integer $t$ ($1 \le t \le 2 \cdot 10^4$) — the number of test cases. Then $t$ test cases follow.
The first line of the test case contains one integer $n$ ($1 \le n \le 2 \cdot 10^5$) — the length of $a$. The second line of the test case contains $n$ integers $a_0, a_1, \dots, a_{n-1}$ ($1 \le a_i \le 10^9$), where $a_i$ is the $i$-th element of $a$.
It is guaranteed that the sum of $n$ does not exceed $2 \cdot 10^5$ ($\sum n \le 2 \cdot 10^5$).
-----Output-----
For each test case, print the answer on the separate line — the maximum possible sum of elements on even positions after reversing at most one subarray (continuous subsegment) of $a$.
-----Example-----
Input
4
8
1 7 3 4 7 6 2 9
5
1 2 1 2 1
10
7 8 4 5 7 6 8 9 7 3
4
3 1 2 1
Output
26
5
37
5 | for _ in range(int(input())):
n = int(input())
arr = list(map(int, input().split()))
ev = 0
l = []
m = []
for i in range(0, n, 2):
ev += arr[i]
for i in range(1, n, 2):
l.append(arr[i] - arr[i - 1])
if i + 1 < n:
m.append(arr[i] - arr[i + 1])
k = len(l)
ma = -10000000
cur = 0
for i in range(k):
cur = max(l[i], cur + l[i])
ma = max(ma, cur)
k = len(m)
ma2 = -10000000
cur = 0
for i in range(k):
cur = max(m[i], cur + m[i])
ma2 = max(ma2, cur)
print(max(ev, ev + max(ma, ma2))) | FOR VAR FUNC_CALL VAR FUNC_CALL VAR FUNC_CALL 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 NUMBER ASSIGN VAR LIST ASSIGN VAR LIST FOR VAR FUNC_CALL VAR NUMBER VAR NUMBER VAR VAR VAR FOR VAR FUNC_CALL VAR NUMBER VAR NUMBER EXPR FUNC_CALL VAR BIN_OP VAR VAR VAR BIN_OP VAR NUMBER IF BIN_OP VAR NUMBER VAR EXPR FUNC_CALL VAR BIN_OP VAR VAR VAR BIN_OP VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR NUMBER ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR VAR VAR BIN_OP VAR VAR VAR ASSIGN VAR FUNC_CALL VAR VAR VAR ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR NUMBER ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR VAR VAR BIN_OP VAR VAR VAR ASSIGN VAR FUNC_CALL VAR VAR VAR EXPR FUNC_CALL VAR FUNC_CALL VAR VAR BIN_OP VAR FUNC_CALL VAR VAR VAR |
You are given an array $a$ consisting of $n$ integers. Indices of the array start from zero (i. e. the first element is $a_0$, the second one is $a_1$, and so on).
You can reverse at most one subarray (continuous subsegment) of this array. Recall that the subarray of $a$ with borders $l$ and $r$ is $a[l; r] = a_l, a_{l + 1}, \dots, a_{r}$.
Your task is to reverse such a subarray that the sum of elements on even positions of the resulting array is maximized (i. e. the sum of elements $a_0, a_2, \dots, a_{2k}$ for integer $k = \lfloor\frac{n-1}{2}\rfloor$ should be maximum possible).
You have to answer $t$ independent test cases.
-----Input-----
The first line of the input contains one integer $t$ ($1 \le t \le 2 \cdot 10^4$) — the number of test cases. Then $t$ test cases follow.
The first line of the test case contains one integer $n$ ($1 \le n \le 2 \cdot 10^5$) — the length of $a$. The second line of the test case contains $n$ integers $a_0, a_1, \dots, a_{n-1}$ ($1 \le a_i \le 10^9$), where $a_i$ is the $i$-th element of $a$.
It is guaranteed that the sum of $n$ does not exceed $2 \cdot 10^5$ ($\sum n \le 2 \cdot 10^5$).
-----Output-----
For each test case, print the answer on the separate line — the maximum possible sum of elements on even positions after reversing at most one subarray (continuous subsegment) of $a$.
-----Example-----
Input
4
8
1 7 3 4 7 6 2 9
5
1 2 1 2 1
10
7 8 4 5 7 6 8 9 7 3
4
3 1 2 1
Output
26
5
37
5 | t = int(input())
def forward_pass(x):
n_a = [(x[i + 1] - x[i]) for i in range(0, len(x) - 1, 2)]
m = 0
c = 0
for n in n_a:
c += n
if c < 0:
c = 0
m = max(m, c)
return m
for i in range(t):
n = int(input())
a = [int(x) for x in input().split(" ")]
ans = sum([x for i, x in enumerate(a) if i % 2 == 0])
if len(a) % 2 == 0:
rev = a[:-1]
else:
rev = a
ans += max(forward_pass(a), forward_pass(rev[::-1]))
print(ans) | ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR FUNC_DEF ASSIGN VAR BIN_OP VAR BIN_OP VAR NUMBER VAR VAR VAR FUNC_CALL VAR NUMBER BIN_OP FUNC_CALL VAR VAR NUMBER NUMBER ASSIGN VAR NUMBER ASSIGN VAR NUMBER FOR VAR VAR VAR VAR IF VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR VAR RETURN VAR 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 STRING ASSIGN VAR FUNC_CALL VAR VAR VAR VAR FUNC_CALL VAR VAR BIN_OP VAR NUMBER NUMBER IF BIN_OP FUNC_CALL VAR VAR NUMBER NUMBER ASSIGN VAR VAR NUMBER ASSIGN VAR VAR VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL VAR VAR NUMBER EXPR FUNC_CALL VAR VAR |
You are given an array $a$ consisting of $n$ integers. Indices of the array start from zero (i. e. the first element is $a_0$, the second one is $a_1$, and so on).
You can reverse at most one subarray (continuous subsegment) of this array. Recall that the subarray of $a$ with borders $l$ and $r$ is $a[l; r] = a_l, a_{l + 1}, \dots, a_{r}$.
Your task is to reverse such a subarray that the sum of elements on even positions of the resulting array is maximized (i. e. the sum of elements $a_0, a_2, \dots, a_{2k}$ for integer $k = \lfloor\frac{n-1}{2}\rfloor$ should be maximum possible).
You have to answer $t$ independent test cases.
-----Input-----
The first line of the input contains one integer $t$ ($1 \le t \le 2 \cdot 10^4$) — the number of test cases. Then $t$ test cases follow.
The first line of the test case contains one integer $n$ ($1 \le n \le 2 \cdot 10^5$) — the length of $a$. The second line of the test case contains $n$ integers $a_0, a_1, \dots, a_{n-1}$ ($1 \le a_i \le 10^9$), where $a_i$ is the $i$-th element of $a$.
It is guaranteed that the sum of $n$ does not exceed $2 \cdot 10^5$ ($\sum n \le 2 \cdot 10^5$).
-----Output-----
For each test case, print the answer on the separate line — the maximum possible sum of elements on even positions after reversing at most one subarray (continuous subsegment) of $a$.
-----Example-----
Input
4
8
1 7 3 4 7 6 2 9
5
1 2 1 2 1
10
7 8 4 5 7 6 8 9 7 3
4
3 1 2 1
Output
26
5
37
5 | t = int(input())
for _ in range(t):
n = int(input())
l = list(map(int, input().split()))
if n % 2:
l.append(0)
dp = [([0] * 4) for i in range((n + 1) // 2 + 1)]
for i in range((n + 1) // 2):
dp[i + 1][0] = dp[i][0] + l[i * 2]
dp[i + 1][1] = max(dp[i][1], dp[i][0])
if i >= 1:
dp[i + 1][1] += l[i * 2 - 1]
dp[i + 1][2] = max(dp[i][0], dp[i][2]) + l[i * 2 + 1]
dp[i + 1][3] = max(dp[i][3], dp[i][1], dp[i][2]) + l[i * 2]
print(max(dp[-1])) | ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR FOR VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR IF BIN_OP VAR NUMBER EXPR FUNC_CALL VAR NUMBER ASSIGN VAR BIN_OP LIST NUMBER NUMBER VAR FUNC_CALL VAR BIN_OP BIN_OP BIN_OP VAR NUMBER NUMBER NUMBER FOR VAR FUNC_CALL VAR BIN_OP BIN_OP VAR NUMBER NUMBER ASSIGN VAR BIN_OP VAR NUMBER NUMBER BIN_OP VAR VAR NUMBER VAR BIN_OP VAR NUMBER ASSIGN VAR BIN_OP VAR NUMBER NUMBER FUNC_CALL VAR VAR VAR NUMBER VAR VAR NUMBER IF VAR NUMBER VAR BIN_OP VAR NUMBER NUMBER VAR BIN_OP BIN_OP VAR NUMBER NUMBER ASSIGN VAR BIN_OP VAR NUMBER NUMBER BIN_OP FUNC_CALL VAR VAR VAR NUMBER VAR VAR NUMBER VAR BIN_OP BIN_OP VAR NUMBER NUMBER ASSIGN VAR BIN_OP VAR NUMBER NUMBER BIN_OP FUNC_CALL VAR VAR VAR NUMBER VAR VAR NUMBER VAR VAR NUMBER VAR BIN_OP VAR NUMBER EXPR FUNC_CALL VAR FUNC_CALL VAR VAR NUMBER |
You are given an array $a$ consisting of $n$ integers. Indices of the array start from zero (i. e. the first element is $a_0$, the second one is $a_1$, and so on).
You can reverse at most one subarray (continuous subsegment) of this array. Recall that the subarray of $a$ with borders $l$ and $r$ is $a[l; r] = a_l, a_{l + 1}, \dots, a_{r}$.
Your task is to reverse such a subarray that the sum of elements on even positions of the resulting array is maximized (i. e. the sum of elements $a_0, a_2, \dots, a_{2k}$ for integer $k = \lfloor\frac{n-1}{2}\rfloor$ should be maximum possible).
You have to answer $t$ independent test cases.
-----Input-----
The first line of the input contains one integer $t$ ($1 \le t \le 2 \cdot 10^4$) — the number of test cases. Then $t$ test cases follow.
The first line of the test case contains one integer $n$ ($1 \le n \le 2 \cdot 10^5$) — the length of $a$. The second line of the test case contains $n$ integers $a_0, a_1, \dots, a_{n-1}$ ($1 \le a_i \le 10^9$), where $a_i$ is the $i$-th element of $a$.
It is guaranteed that the sum of $n$ does not exceed $2 \cdot 10^5$ ($\sum n \le 2 \cdot 10^5$).
-----Output-----
For each test case, print the answer on the separate line — the maximum possible sum of elements on even positions after reversing at most one subarray (continuous subsegment) of $a$.
-----Example-----
Input
4
8
1 7 3 4 7 6 2 9
5
1 2 1 2 1
10
7 8 4 5 7 6 8 9 7 3
4
3 1 2 1
Output
26
5
37
5 | t = int(input())
for _ in range(t):
n = int(input())
l = list(map(int, input().split()))
even_sum = sum(l[::2])
v1 = []
v2 = []
for i in range(1, n - 1, 2):
v1.append(l[i] - l[i + 1])
for i in range(0, n - 1, 2):
v2.append(l[i + 1] - l[i])
c = 0
m = 0
for i in range(len(v1)):
c += v1[i]
m = max(m, c)
if c < 0:
c = 0
c = 0
for i in range(len(v2)):
c += v2[i]
m = max(m, c)
if c < 0:
c = 0
print(even_sum + m) | ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR FOR VAR FUNC_CALL 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 FUNC_CALL VAR VAR NUMBER ASSIGN VAR LIST ASSIGN VAR LIST FOR VAR FUNC_CALL VAR NUMBER BIN_OP VAR NUMBER NUMBER EXPR FUNC_CALL VAR BIN_OP VAR VAR VAR BIN_OP VAR NUMBER FOR VAR FUNC_CALL VAR NUMBER BIN_OP VAR NUMBER NUMBER EXPR FUNC_CALL VAR BIN_OP VAR BIN_OP VAR NUMBER VAR VAR ASSIGN VAR NUMBER ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR FUNC_CALL VAR VAR VAR VAR VAR ASSIGN VAR FUNC_CALL VAR VAR VAR IF VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR FUNC_CALL VAR VAR VAR VAR VAR ASSIGN VAR FUNC_CALL VAR VAR VAR IF VAR NUMBER ASSIGN VAR NUMBER EXPR FUNC_CALL VAR BIN_OP VAR VAR |
You are given an array $a$ consisting of $n$ integers. Indices of the array start from zero (i. e. the first element is $a_0$, the second one is $a_1$, and so on).
You can reverse at most one subarray (continuous subsegment) of this array. Recall that the subarray of $a$ with borders $l$ and $r$ is $a[l; r] = a_l, a_{l + 1}, \dots, a_{r}$.
Your task is to reverse such a subarray that the sum of elements on even positions of the resulting array is maximized (i. e. the sum of elements $a_0, a_2, \dots, a_{2k}$ for integer $k = \lfloor\frac{n-1}{2}\rfloor$ should be maximum possible).
You have to answer $t$ independent test cases.
-----Input-----
The first line of the input contains one integer $t$ ($1 \le t \le 2 \cdot 10^4$) — the number of test cases. Then $t$ test cases follow.
The first line of the test case contains one integer $n$ ($1 \le n \le 2 \cdot 10^5$) — the length of $a$. The second line of the test case contains $n$ integers $a_0, a_1, \dots, a_{n-1}$ ($1 \le a_i \le 10^9$), where $a_i$ is the $i$-th element of $a$.
It is guaranteed that the sum of $n$ does not exceed $2 \cdot 10^5$ ($\sum n \le 2 \cdot 10^5$).
-----Output-----
For each test case, print the answer on the separate line — the maximum possible sum of elements on even positions after reversing at most one subarray (continuous subsegment) of $a$.
-----Example-----
Input
4
8
1 7 3 4 7 6 2 9
5
1 2 1 2 1
10
7 8 4 5 7 6 8 9 7 3
4
3 1 2 1
Output
26
5
37
5 | import sys
input = sys.stdin.buffer.readline
t = int(input())
for ___ in range(t):
n = int(input())
numbers = [int(x) for x in input().split()]
if n == 1:
print(numbers[0])
continue
dpEven = [[(-1) for _ in range(len(numbers))] for __ in range(3)]
dpEven[0][0] = numbers[0]
for i in range(2, len(numbers), 2):
dpEven[0][i] = dpEven[0][i - 2] + numbers[i]
dpEven[1][0] = numbers[1]
for i in range(2, len(numbers) // 2 * 2, 2):
dpEven[1][i] = max(dpEven[1][i - 2], dpEven[0][i - 2]) + numbers[i + 1]
dpEven[2][0] = numbers[0]
for i in range(2, len(numbers), 2):
dpEven[2][i] = max(dpEven[1][i - 2], dpEven[2][i - 2]) + numbers[i]
if len(numbers) % 2 == 0:
evenMax = max(dpEven[2][-2], dpEven[1][-2], dpEven[0][-1])
else:
evenMax = max(dpEven[2][-1], dpEven[1][-1], dpEven[0][-1])
dpOdd = [[(-1) for _ in range(len(numbers))] for __ in range(3)]
dpOdd[0][0] = numbers[0]
for i in range(2, len(numbers), 2):
dpOdd[0][i] = dpOdd[0][i - 2] + numbers[i]
dpOdd[1][0] = numbers[0]
for i in range(2, len(numbers), 2):
dpOdd[1][i] = max(dpOdd[0][i - 2], dpOdd[1][i - 2]) + numbers[i - 1]
dpOdd[2][0] = numbers[0]
for i in range(2, len(numbers), 2):
dpOdd[2][i] = max(dpOdd[1][i - 2], dpOdd[2][i - 2]) + numbers[i]
if len(numbers) % 2 == 0:
oddMax = max(dpOdd[0][-2], dpOdd[1][-2], dpOdd[2][-2])
else:
oddMax = max(dpOdd[0][-1], dpOdd[1][-1], dpOdd[2][-1])
print(max(oddMax, evenMax)) | IMPORT ASSIGN VAR VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR 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 IF VAR NUMBER EXPR FUNC_CALL VAR VAR NUMBER ASSIGN VAR NUMBER VAR FUNC_CALL VAR FUNC_CALL VAR VAR VAR FUNC_CALL VAR NUMBER ASSIGN VAR NUMBER NUMBER VAR NUMBER FOR VAR FUNC_CALL VAR NUMBER FUNC_CALL VAR VAR NUMBER ASSIGN VAR NUMBER VAR BIN_OP VAR NUMBER BIN_OP VAR NUMBER VAR VAR ASSIGN VAR NUMBER NUMBER VAR NUMBER FOR VAR FUNC_CALL VAR NUMBER BIN_OP BIN_OP FUNC_CALL VAR VAR NUMBER NUMBER NUMBER ASSIGN VAR NUMBER VAR BIN_OP FUNC_CALL VAR VAR NUMBER BIN_OP VAR NUMBER VAR NUMBER BIN_OP VAR NUMBER VAR BIN_OP VAR NUMBER ASSIGN VAR NUMBER NUMBER VAR NUMBER FOR VAR FUNC_CALL VAR NUMBER FUNC_CALL VAR VAR NUMBER ASSIGN VAR NUMBER VAR BIN_OP FUNC_CALL VAR VAR NUMBER BIN_OP VAR NUMBER VAR NUMBER BIN_OP VAR NUMBER VAR VAR IF BIN_OP FUNC_CALL VAR VAR NUMBER NUMBER ASSIGN VAR FUNC_CALL VAR VAR NUMBER NUMBER VAR NUMBER NUMBER VAR NUMBER NUMBER ASSIGN VAR FUNC_CALL VAR VAR NUMBER NUMBER VAR NUMBER NUMBER VAR NUMBER NUMBER ASSIGN VAR NUMBER VAR FUNC_CALL VAR FUNC_CALL VAR VAR VAR FUNC_CALL VAR NUMBER ASSIGN VAR NUMBER NUMBER VAR NUMBER FOR VAR FUNC_CALL VAR NUMBER FUNC_CALL VAR VAR NUMBER ASSIGN VAR NUMBER VAR BIN_OP VAR NUMBER BIN_OP VAR NUMBER VAR VAR ASSIGN VAR NUMBER NUMBER VAR NUMBER FOR VAR FUNC_CALL VAR NUMBER FUNC_CALL VAR VAR NUMBER ASSIGN VAR NUMBER VAR BIN_OP FUNC_CALL VAR VAR NUMBER BIN_OP VAR NUMBER VAR NUMBER BIN_OP VAR NUMBER VAR BIN_OP VAR NUMBER ASSIGN VAR NUMBER NUMBER VAR NUMBER FOR VAR FUNC_CALL VAR NUMBER FUNC_CALL VAR VAR NUMBER ASSIGN VAR NUMBER VAR BIN_OP FUNC_CALL VAR VAR NUMBER BIN_OP VAR NUMBER VAR NUMBER BIN_OP VAR NUMBER VAR VAR IF BIN_OP FUNC_CALL VAR VAR NUMBER NUMBER ASSIGN VAR FUNC_CALL VAR VAR NUMBER NUMBER VAR NUMBER NUMBER VAR NUMBER NUMBER ASSIGN VAR FUNC_CALL VAR VAR NUMBER NUMBER VAR NUMBER NUMBER VAR NUMBER NUMBER EXPR FUNC_CALL VAR FUNC_CALL VAR VAR VAR |
You are given an array $a$ consisting of $n$ integers. Indices of the array start from zero (i. e. the first element is $a_0$, the second one is $a_1$, and so on).
You can reverse at most one subarray (continuous subsegment) of this array. Recall that the subarray of $a$ with borders $l$ and $r$ is $a[l; r] = a_l, a_{l + 1}, \dots, a_{r}$.
Your task is to reverse such a subarray that the sum of elements on even positions of the resulting array is maximized (i. e. the sum of elements $a_0, a_2, \dots, a_{2k}$ for integer $k = \lfloor\frac{n-1}{2}\rfloor$ should be maximum possible).
You have to answer $t$ independent test cases.
-----Input-----
The first line of the input contains one integer $t$ ($1 \le t \le 2 \cdot 10^4$) — the number of test cases. Then $t$ test cases follow.
The first line of the test case contains one integer $n$ ($1 \le n \le 2 \cdot 10^5$) — the length of $a$. The second line of the test case contains $n$ integers $a_0, a_1, \dots, a_{n-1}$ ($1 \le a_i \le 10^9$), where $a_i$ is the $i$-th element of $a$.
It is guaranteed that the sum of $n$ does not exceed $2 \cdot 10^5$ ($\sum n \le 2 \cdot 10^5$).
-----Output-----
For each test case, print the answer on the separate line — the maximum possible sum of elements on even positions after reversing at most one subarray (continuous subsegment) of $a$.
-----Example-----
Input
4
8
1 7 3 4 7 6 2 9
5
1 2 1 2 1
10
7 8 4 5 7 6 8 9 7 3
4
3 1 2 1
Output
26
5
37
5 | from sys import stdin
input = stdin.readline
for _ in range(int(input())):
n = int(input())
a = [*map(int, input().split())]
v1, v2, v3, v4 = 0, 0, 0, 0
for i, j in enumerate(a):
if i >= 1:
if i % 2 == 1:
v3 = max(a[i] - a[i - 1], a[i] - a[i - 1] + v1, v3)
else:
v3 = max(a[i - 1] - a[i], a[i - 1] - a[i] + v1, v3)
v4 = max(v4, v3)
v1, v2, v3 = v2, v3, 0
print(sum(a[i] for i in range(0, n, 2)) + v4) | ASSIGN VAR VAR FOR VAR FUNC_CALL VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR LIST FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR VAR VAR VAR NUMBER NUMBER NUMBER NUMBER FOR VAR VAR FUNC_CALL VAR VAR IF VAR NUMBER IF BIN_OP VAR NUMBER NUMBER ASSIGN VAR FUNC_CALL VAR BIN_OP VAR VAR VAR BIN_OP VAR NUMBER BIN_OP BIN_OP VAR VAR VAR BIN_OP VAR NUMBER VAR VAR ASSIGN VAR FUNC_CALL VAR BIN_OP VAR BIN_OP VAR NUMBER VAR VAR BIN_OP BIN_OP VAR BIN_OP VAR NUMBER VAR VAR VAR VAR ASSIGN VAR FUNC_CALL VAR VAR VAR ASSIGN VAR VAR VAR VAR VAR NUMBER EXPR FUNC_CALL VAR BIN_OP FUNC_CALL VAR VAR VAR VAR FUNC_CALL VAR NUMBER VAR NUMBER VAR |
You are given an array $a$ consisting of $n$ integers. Indices of the array start from zero (i. e. the first element is $a_0$, the second one is $a_1$, and so on).
You can reverse at most one subarray (continuous subsegment) of this array. Recall that the subarray of $a$ with borders $l$ and $r$ is $a[l; r] = a_l, a_{l + 1}, \dots, a_{r}$.
Your task is to reverse such a subarray that the sum of elements on even positions of the resulting array is maximized (i. e. the sum of elements $a_0, a_2, \dots, a_{2k}$ for integer $k = \lfloor\frac{n-1}{2}\rfloor$ should be maximum possible).
You have to answer $t$ independent test cases.
-----Input-----
The first line of the input contains one integer $t$ ($1 \le t \le 2 \cdot 10^4$) — the number of test cases. Then $t$ test cases follow.
The first line of the test case contains one integer $n$ ($1 \le n \le 2 \cdot 10^5$) — the length of $a$. The second line of the test case contains $n$ integers $a_0, a_1, \dots, a_{n-1}$ ($1 \le a_i \le 10^9$), where $a_i$ is the $i$-th element of $a$.
It is guaranteed that the sum of $n$ does not exceed $2 \cdot 10^5$ ($\sum n \le 2 \cdot 10^5$).
-----Output-----
For each test case, print the answer on the separate line — the maximum possible sum of elements on even positions after reversing at most one subarray (continuous subsegment) of $a$.
-----Example-----
Input
4
8
1 7 3 4 7 6 2 9
5
1 2 1 2 1
10
7 8 4 5 7 6 8 9 7 3
4
3 1 2 1
Output
26
5
37
5 | def main():
t = int(input())
for g in range(t):
n = int(input())
values = list(map(int, input().split()))
sums = [(0) for i in range(n)]
minimum_odd = [(10**9) for i in range(n)]
minimum_even = [(10**9) for i in range(n)]
sums[0] = -values[0]
minimum_odd[0] = 0
minimum_even[0] = -values[0]
add = 0
s = values[0]
for i in range(1, n):
if i % 2 == 0:
sums[i] = sums[i - 1] - values[i]
s += values[i]
add = max(add, sums[i] - minimum_even[i - 1])
minimum_even[i] = min(minimum_even[i - 1], sums[i])
minimum_odd[i] = minimum_odd[i - 1]
else:
sums[i] = sums[i - 1] + values[i]
add = max(add, sums[i] - minimum_odd[i - 1])
minimum_odd[i] = min(minimum_odd[i - 1], sums[i])
minimum_even[i] = minimum_even[i - 1]
print(s + add)
main() | FUNC_DEF ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR FOR VAR FUNC_CALL 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 NUMBER VAR FUNC_CALL VAR VAR ASSIGN VAR BIN_OP NUMBER NUMBER VAR FUNC_CALL VAR VAR ASSIGN VAR BIN_OP NUMBER NUMBER VAR FUNC_CALL VAR VAR ASSIGN VAR NUMBER VAR NUMBER ASSIGN VAR NUMBER NUMBER ASSIGN VAR NUMBER VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR VAR NUMBER FOR VAR FUNC_CALL VAR NUMBER VAR IF BIN_OP VAR NUMBER NUMBER ASSIGN VAR VAR BIN_OP VAR BIN_OP VAR NUMBER VAR VAR VAR VAR VAR ASSIGN VAR FUNC_CALL VAR VAR BIN_OP VAR VAR VAR BIN_OP VAR NUMBER ASSIGN VAR VAR FUNC_CALL VAR VAR BIN_OP VAR NUMBER VAR VAR ASSIGN VAR VAR VAR BIN_OP VAR NUMBER ASSIGN VAR VAR BIN_OP VAR BIN_OP VAR NUMBER VAR VAR ASSIGN VAR FUNC_CALL VAR VAR BIN_OP VAR VAR VAR BIN_OP VAR NUMBER ASSIGN VAR VAR FUNC_CALL VAR VAR BIN_OP VAR NUMBER VAR VAR ASSIGN VAR VAR VAR BIN_OP VAR NUMBER EXPR FUNC_CALL VAR BIN_OP VAR VAR EXPR FUNC_CALL VAR |
You are given an array $a$ consisting of $n$ integers. Indices of the array start from zero (i. e. the first element is $a_0$, the second one is $a_1$, and so on).
You can reverse at most one subarray (continuous subsegment) of this array. Recall that the subarray of $a$ with borders $l$ and $r$ is $a[l; r] = a_l, a_{l + 1}, \dots, a_{r}$.
Your task is to reverse such a subarray that the sum of elements on even positions of the resulting array is maximized (i. e. the sum of elements $a_0, a_2, \dots, a_{2k}$ for integer $k = \lfloor\frac{n-1}{2}\rfloor$ should be maximum possible).
You have to answer $t$ independent test cases.
-----Input-----
The first line of the input contains one integer $t$ ($1 \le t \le 2 \cdot 10^4$) — the number of test cases. Then $t$ test cases follow.
The first line of the test case contains one integer $n$ ($1 \le n \le 2 \cdot 10^5$) — the length of $a$. The second line of the test case contains $n$ integers $a_0, a_1, \dots, a_{n-1}$ ($1 \le a_i \le 10^9$), where $a_i$ is the $i$-th element of $a$.
It is guaranteed that the sum of $n$ does not exceed $2 \cdot 10^5$ ($\sum n \le 2 \cdot 10^5$).
-----Output-----
For each test case, print the answer on the separate line — the maximum possible sum of elements on even positions after reversing at most one subarray (continuous subsegment) of $a$.
-----Example-----
Input
4
8
1 7 3 4 7 6 2 9
5
1 2 1 2 1
10
7 8 4 5 7 6 8 9 7 3
4
3 1 2 1
Output
26
5
37
5 | import sys
input = lambda: sys.stdin.readline().strip()
t = int(input())
def mss(b):
ans = 0
msf = 0
for i in b:
msf = max(msf + i, 0)
ans = max(ans, msf)
return ans
while t:
t -= 1
n = int(input())
a = list(map(int, input().split()))
s = 0
for i in range(0, n, 2):
s += a[i]
b1 = []
b2 = []
for i in range(0, n - 1, 2):
b1.append(a[i + 1] - a[i])
for i in range(1, n - 1, 2):
b2.append(a[i] - a[i + 1])
print(s + max(0, mss(b1), mss(b2))) | IMPORT ASSIGN VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR FUNC_DEF ASSIGN VAR NUMBER ASSIGN VAR NUMBER FOR VAR VAR ASSIGN VAR FUNC_CALL VAR BIN_OP VAR VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR VAR RETURN VAR WHILE VAR VAR NUMBER 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 FOR VAR FUNC_CALL VAR NUMBER VAR NUMBER VAR VAR VAR ASSIGN VAR LIST ASSIGN VAR LIST FOR VAR FUNC_CALL VAR NUMBER BIN_OP VAR NUMBER NUMBER EXPR FUNC_CALL VAR BIN_OP VAR BIN_OP VAR NUMBER VAR VAR FOR VAR FUNC_CALL VAR NUMBER BIN_OP VAR NUMBER NUMBER EXPR FUNC_CALL VAR BIN_OP VAR VAR VAR BIN_OP VAR NUMBER EXPR FUNC_CALL VAR BIN_OP VAR FUNC_CALL VAR NUMBER FUNC_CALL VAR VAR FUNC_CALL VAR VAR |
You are given an array $a$ consisting of $n$ integers. Indices of the array start from zero (i. e. the first element is $a_0$, the second one is $a_1$, and so on).
You can reverse at most one subarray (continuous subsegment) of this array. Recall that the subarray of $a$ with borders $l$ and $r$ is $a[l; r] = a_l, a_{l + 1}, \dots, a_{r}$.
Your task is to reverse such a subarray that the sum of elements on even positions of the resulting array is maximized (i. e. the sum of elements $a_0, a_2, \dots, a_{2k}$ for integer $k = \lfloor\frac{n-1}{2}\rfloor$ should be maximum possible).
You have to answer $t$ independent test cases.
-----Input-----
The first line of the input contains one integer $t$ ($1 \le t \le 2 \cdot 10^4$) — the number of test cases. Then $t$ test cases follow.
The first line of the test case contains one integer $n$ ($1 \le n \le 2 \cdot 10^5$) — the length of $a$. The second line of the test case contains $n$ integers $a_0, a_1, \dots, a_{n-1}$ ($1 \le a_i \le 10^9$), where $a_i$ is the $i$-th element of $a$.
It is guaranteed that the sum of $n$ does not exceed $2 \cdot 10^5$ ($\sum n \le 2 \cdot 10^5$).
-----Output-----
For each test case, print the answer on the separate line — the maximum possible sum of elements on even positions after reversing at most one subarray (continuous subsegment) of $a$.
-----Example-----
Input
4
8
1 7 3 4 7 6 2 9
5
1 2 1 2 1
10
7 8 4 5 7 6 8 9 7 3
4
3 1 2 1
Output
26
5
37
5 | def lis(a, n):
i = 0
sum = 0
max = 0
k = -1
j = [0, 0]
while i < n:
sum += a[i]
if sum <= 0:
k = i
sum = 0
if sum > max:
j[0] = k + 2
j[1] = i + 1
max = sum
i += 1
return max
for _ in range(int(input())):
n = int(input())
a = list(map(int, input().split()))
s = sum(a[::2])
if n & 1:
a.append(0)
n += 1
i = 0
b = []
while i < n:
b.append(a[i + 1] - a[i])
i += 2
m = lis(b, len(b))
i = 2
b = []
while i < n:
b.append(a[i - 1] - a[i])
i += 2
n = lis(b, len(b))
print(s + max(m, n)) | FUNC_DEF ASSIGN VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR LIST NUMBER NUMBER WHILE VAR VAR VAR VAR VAR IF VAR NUMBER ASSIGN VAR VAR ASSIGN VAR NUMBER IF VAR VAR ASSIGN VAR NUMBER BIN_OP VAR NUMBER ASSIGN VAR NUMBER BIN_OP VAR NUMBER ASSIGN VAR VAR VAR NUMBER RETURN VAR FOR VAR FUNC_CALL VAR FUNC_CALL VAR FUNC_CALL 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 FUNC_CALL VAR VAR NUMBER IF BIN_OP VAR NUMBER EXPR FUNC_CALL VAR NUMBER VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR LIST WHILE VAR VAR EXPR FUNC_CALL VAR BIN_OP VAR BIN_OP VAR NUMBER VAR VAR VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR FUNC_CALL VAR VAR ASSIGN VAR NUMBER ASSIGN VAR LIST WHILE VAR VAR EXPR FUNC_CALL VAR BIN_OP VAR BIN_OP VAR NUMBER VAR VAR VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR BIN_OP VAR FUNC_CALL VAR VAR VAR |
You are given an array $a$ consisting of $n$ integers. Indices of the array start from zero (i. e. the first element is $a_0$, the second one is $a_1$, and so on).
You can reverse at most one subarray (continuous subsegment) of this array. Recall that the subarray of $a$ with borders $l$ and $r$ is $a[l; r] = a_l, a_{l + 1}, \dots, a_{r}$.
Your task is to reverse such a subarray that the sum of elements on even positions of the resulting array is maximized (i. e. the sum of elements $a_0, a_2, \dots, a_{2k}$ for integer $k = \lfloor\frac{n-1}{2}\rfloor$ should be maximum possible).
You have to answer $t$ independent test cases.
-----Input-----
The first line of the input contains one integer $t$ ($1 \le t \le 2 \cdot 10^4$) — the number of test cases. Then $t$ test cases follow.
The first line of the test case contains one integer $n$ ($1 \le n \le 2 \cdot 10^5$) — the length of $a$. The second line of the test case contains $n$ integers $a_0, a_1, \dots, a_{n-1}$ ($1 \le a_i \le 10^9$), where $a_i$ is the $i$-th element of $a$.
It is guaranteed that the sum of $n$ does not exceed $2 \cdot 10^5$ ($\sum n \le 2 \cdot 10^5$).
-----Output-----
For each test case, print the answer on the separate line — the maximum possible sum of elements on even positions after reversing at most one subarray (continuous subsegment) of $a$.
-----Example-----
Input
4
8
1 7 3 4 7 6 2 9
5
1 2 1 2 1
10
7 8 4 5 7 6 8 9 7 3
4
3 1 2 1
Output
26
5
37
5 | def maxeve(arr):
sm = sum(arr[::2])
ms, cs = 0, 0
for i in range(0, len(arr) - 1, 2):
cs = max(cs + arr[i + 1] - arr[i], 0)
ms = max(cs, ms)
cs = 0
for i in range(1, len(arr) - 1, 2):
cs = max(0, cs + arr[i] - arr[i + 1])
ms = max(cs, ms)
return ms + sm
for i in range(int(input())):
a = input()
lst = list(map(int, input().strip().split()))
print(maxeve(lst)) | FUNC_DEF ASSIGN VAR FUNC_CALL VAR VAR NUMBER ASSIGN VAR VAR NUMBER NUMBER FOR VAR FUNC_CALL VAR NUMBER BIN_OP FUNC_CALL VAR VAR NUMBER NUMBER ASSIGN VAR FUNC_CALL VAR BIN_OP BIN_OP VAR VAR BIN_OP VAR NUMBER VAR VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR VAR ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR NUMBER BIN_OP FUNC_CALL VAR VAR NUMBER NUMBER ASSIGN VAR FUNC_CALL VAR NUMBER BIN_OP BIN_OP VAR VAR VAR VAR BIN_OP VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR VAR RETURN BIN_OP VAR VAR FOR VAR FUNC_CALL VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL FUNC_CALL VAR EXPR FUNC_CALL VAR FUNC_CALL VAR VAR |
You are given an array $a$ consisting of $n$ integers. Indices of the array start from zero (i. e. the first element is $a_0$, the second one is $a_1$, and so on).
You can reverse at most one subarray (continuous subsegment) of this array. Recall that the subarray of $a$ with borders $l$ and $r$ is $a[l; r] = a_l, a_{l + 1}, \dots, a_{r}$.
Your task is to reverse such a subarray that the sum of elements on even positions of the resulting array is maximized (i. e. the sum of elements $a_0, a_2, \dots, a_{2k}$ for integer $k = \lfloor\frac{n-1}{2}\rfloor$ should be maximum possible).
You have to answer $t$ independent test cases.
-----Input-----
The first line of the input contains one integer $t$ ($1 \le t \le 2 \cdot 10^4$) — the number of test cases. Then $t$ test cases follow.
The first line of the test case contains one integer $n$ ($1 \le n \le 2 \cdot 10^5$) — the length of $a$. The second line of the test case contains $n$ integers $a_0, a_1, \dots, a_{n-1}$ ($1 \le a_i \le 10^9$), where $a_i$ is the $i$-th element of $a$.
It is guaranteed that the sum of $n$ does not exceed $2 \cdot 10^5$ ($\sum n \le 2 \cdot 10^5$).
-----Output-----
For each test case, print the answer on the separate line — the maximum possible sum of elements on even positions after reversing at most one subarray (continuous subsegment) of $a$.
-----Example-----
Input
4
8
1 7 3 4 7 6 2 9
5
1 2 1 2 1
10
7 8 4 5 7 6 8 9 7 3
4
3 1 2 1
Output
26
5
37
5 | t = int(input())
def f(a):
sums = [0]
for i in range(len(a)):
sums.append(a[i] + sums[-1])
minn = [sums[0]]
for i in range(1, len(sums)):
minn.append(min(minn[-1], sums[i]))
maxx = [sums[-1]]
for i in range(1, len(sums)):
maxx.append(max(maxx[-1], sums[len(sums) - i - 1]))
maxx.reverse()
diff = [0]
for i in range(1, len(sums)):
diff.append(maxx[i] - minn[i])
return max(diff)
for tt in range(t):
n = int(input())
a = [int(x) for x in input().split(" ")]
ch = []
for i in range(1, n):
if i % 2 == 1:
ch.append(a[i] - a[i - 1])
else:
ch.append(a[i - 1] - a[i])
l = []
r = []
for i in range(len(ch)):
if i % 2 == 0:
l.append(ch[i])
else:
r.append(ch[i])
sum_org = 0
for i in range(n):
if i % 2 == 0:
sum_org += a[i]
summ = sum_org + max(f(l), f(r), 0)
print(summ) | ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR FUNC_DEF ASSIGN VAR LIST NUMBER FOR VAR FUNC_CALL VAR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR BIN_OP VAR VAR VAR NUMBER ASSIGN VAR LIST VAR NUMBER FOR VAR FUNC_CALL VAR NUMBER FUNC_CALL VAR VAR EXPR FUNC_CALL VAR FUNC_CALL VAR VAR NUMBER 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 NUMBER VAR BIN_OP BIN_OP FUNC_CALL VAR VAR VAR NUMBER EXPR FUNC_CALL VAR ASSIGN VAR LIST NUMBER FOR VAR FUNC_CALL VAR NUMBER FUNC_CALL VAR VAR EXPR FUNC_CALL VAR BIN_OP VAR VAR VAR VAR RETURN FUNC_CALL VAR VAR 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 STRING ASSIGN VAR LIST FOR VAR FUNC_CALL VAR NUMBER VAR IF BIN_OP VAR NUMBER NUMBER EXPR FUNC_CALL VAR BIN_OP VAR VAR VAR BIN_OP VAR NUMBER EXPR FUNC_CALL VAR BIN_OP VAR BIN_OP VAR NUMBER VAR VAR ASSIGN VAR LIST ASSIGN VAR LIST FOR VAR FUNC_CALL VAR FUNC_CALL VAR VAR IF BIN_OP VAR NUMBER NUMBER EXPR FUNC_CALL VAR VAR VAR EXPR FUNC_CALL VAR VAR VAR ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR VAR IF BIN_OP VAR NUMBER NUMBER VAR VAR VAR ASSIGN VAR BIN_OP VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL VAR VAR NUMBER EXPR FUNC_CALL VAR VAR |
You are given an array $a$ consisting of $n$ integers. Indices of the array start from zero (i. e. the first element is $a_0$, the second one is $a_1$, and so on).
You can reverse at most one subarray (continuous subsegment) of this array. Recall that the subarray of $a$ with borders $l$ and $r$ is $a[l; r] = a_l, a_{l + 1}, \dots, a_{r}$.
Your task is to reverse such a subarray that the sum of elements on even positions of the resulting array is maximized (i. e. the sum of elements $a_0, a_2, \dots, a_{2k}$ for integer $k = \lfloor\frac{n-1}{2}\rfloor$ should be maximum possible).
You have to answer $t$ independent test cases.
-----Input-----
The first line of the input contains one integer $t$ ($1 \le t \le 2 \cdot 10^4$) — the number of test cases. Then $t$ test cases follow.
The first line of the test case contains one integer $n$ ($1 \le n \le 2 \cdot 10^5$) — the length of $a$. The second line of the test case contains $n$ integers $a_0, a_1, \dots, a_{n-1}$ ($1 \le a_i \le 10^9$), where $a_i$ is the $i$-th element of $a$.
It is guaranteed that the sum of $n$ does not exceed $2 \cdot 10^5$ ($\sum n \le 2 \cdot 10^5$).
-----Output-----
For each test case, print the answer on the separate line — the maximum possible sum of elements on even positions after reversing at most one subarray (continuous subsegment) of $a$.
-----Example-----
Input
4
8
1 7 3 4 7 6 2 9
5
1 2 1 2 1
10
7 8 4 5 7 6 8 9 7 3
4
3 1 2 1
Output
26
5
37
5 | t = int(input())
for _ in range(t):
n = int(input())
A = [int(a) for a in input().split()]
ans = 0
num = 0
m = 0
for i in range(n):
if i % 2 == 0:
ans += A[i]
num += A[i]
else:
num -= A[i]
num = min(0, num)
m = min(m, num)
num = 0
for i in range(1, n):
if i % 2 == 0:
num += A[i]
num = min(0, num)
m = min(m, num)
else:
num -= A[i]
print(ans - m) | ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR 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 NUMBER ASSIGN VAR NUMBER ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR VAR IF BIN_OP VAR NUMBER NUMBER VAR VAR VAR VAR VAR VAR VAR VAR VAR ASSIGN VAR FUNC_CALL VAR NUMBER VAR ASSIGN VAR FUNC_CALL VAR VAR VAR ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR NUMBER VAR IF BIN_OP VAR NUMBER NUMBER VAR VAR VAR ASSIGN VAR FUNC_CALL VAR NUMBER VAR ASSIGN VAR FUNC_CALL VAR VAR VAR VAR VAR VAR EXPR FUNC_CALL VAR BIN_OP VAR VAR |
You are given an array $a$ consisting of $n$ integers. Indices of the array start from zero (i. e. the first element is $a_0$, the second one is $a_1$, and so on).
You can reverse at most one subarray (continuous subsegment) of this array. Recall that the subarray of $a$ with borders $l$ and $r$ is $a[l; r] = a_l, a_{l + 1}, \dots, a_{r}$.
Your task is to reverse such a subarray that the sum of elements on even positions of the resulting array is maximized (i. e. the sum of elements $a_0, a_2, \dots, a_{2k}$ for integer $k = \lfloor\frac{n-1}{2}\rfloor$ should be maximum possible).
You have to answer $t$ independent test cases.
-----Input-----
The first line of the input contains one integer $t$ ($1 \le t \le 2 \cdot 10^4$) — the number of test cases. Then $t$ test cases follow.
The first line of the test case contains one integer $n$ ($1 \le n \le 2 \cdot 10^5$) — the length of $a$. The second line of the test case contains $n$ integers $a_0, a_1, \dots, a_{n-1}$ ($1 \le a_i \le 10^9$), where $a_i$ is the $i$-th element of $a$.
It is guaranteed that the sum of $n$ does not exceed $2 \cdot 10^5$ ($\sum n \le 2 \cdot 10^5$).
-----Output-----
For each test case, print the answer on the separate line — the maximum possible sum of elements on even positions after reversing at most one subarray (continuous subsegment) of $a$.
-----Example-----
Input
4
8
1 7 3 4 7 6 2 9
5
1 2 1 2 1
10
7 8 4 5 7 6 8 9 7 3
4
3 1 2 1
Output
26
5
37
5 | for _ in range(int(input())):
n = int(input())
arr = list(map(int, input().split()))
Sum = sum(arr[::2])
ms, cs = 0, 0
for i in range(0, n - n % 2, 2):
cs = max(0, cs + arr[i + 1] - arr[i])
ms = max(cs, ms)
cs = 0
for i in range(1, n - int(n % 2 == 0), 2):
cs = max(0, cs + arr[i] - arr[i + 1])
ms = max(cs, ms)
print(ms + Sum) | FOR VAR FUNC_CALL VAR FUNC_CALL VAR FUNC_CALL 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 FUNC_CALL VAR VAR NUMBER ASSIGN VAR VAR NUMBER NUMBER FOR VAR FUNC_CALL VAR NUMBER BIN_OP VAR BIN_OP VAR NUMBER NUMBER ASSIGN VAR FUNC_CALL VAR NUMBER BIN_OP BIN_OP VAR VAR BIN_OP VAR NUMBER VAR VAR ASSIGN VAR FUNC_CALL VAR VAR VAR ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR NUMBER BIN_OP VAR FUNC_CALL VAR BIN_OP VAR NUMBER NUMBER NUMBER ASSIGN VAR FUNC_CALL VAR NUMBER BIN_OP BIN_OP VAR VAR VAR VAR BIN_OP VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR VAR EXPR FUNC_CALL VAR BIN_OP VAR VAR |
You are given an array $a$ consisting of $n$ integers. Indices of the array start from zero (i. e. the first element is $a_0$, the second one is $a_1$, and so on).
You can reverse at most one subarray (continuous subsegment) of this array. Recall that the subarray of $a$ with borders $l$ and $r$ is $a[l; r] = a_l, a_{l + 1}, \dots, a_{r}$.
Your task is to reverse such a subarray that the sum of elements on even positions of the resulting array is maximized (i. e. the sum of elements $a_0, a_2, \dots, a_{2k}$ for integer $k = \lfloor\frac{n-1}{2}\rfloor$ should be maximum possible).
You have to answer $t$ independent test cases.
-----Input-----
The first line of the input contains one integer $t$ ($1 \le t \le 2 \cdot 10^4$) — the number of test cases. Then $t$ test cases follow.
The first line of the test case contains one integer $n$ ($1 \le n \le 2 \cdot 10^5$) — the length of $a$. The second line of the test case contains $n$ integers $a_0, a_1, \dots, a_{n-1}$ ($1 \le a_i \le 10^9$), where $a_i$ is the $i$-th element of $a$.
It is guaranteed that the sum of $n$ does not exceed $2 \cdot 10^5$ ($\sum n \le 2 \cdot 10^5$).
-----Output-----
For each test case, print the answer on the separate line — the maximum possible sum of elements on even positions after reversing at most one subarray (continuous subsegment) of $a$.
-----Example-----
Input
4
8
1 7 3 4 7 6 2 9
5
1 2 1 2 1
10
7 8 4 5 7 6 8 9 7 3
4
3 1 2 1
Output
26
5
37
5 | for t in range(int(input())):
n = int(input())
(*a,) = map(int, input().split())
dp = [([0] * 3) for _ in range(n + 2)]
for i in range(n):
dp[i + 1][0] = dp[i][0] + (a[i] if i % 2 == 0 else 0)
if i + 1 < n:
dp[i + 2][1] = max(dp[i][0], dp[i][1]) + (a[i + 1] if i % 2 == 0 else a[i])
dp[i + 1][2] = max(dp[i][1], dp[i][2]) + (a[i] if i % 2 == 0 else 0)
print(max(dp[n])) | FOR VAR FUNC_CALL VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR BIN_OP LIST NUMBER NUMBER VAR FUNC_CALL VAR BIN_OP VAR NUMBER FOR VAR FUNC_CALL VAR VAR ASSIGN VAR BIN_OP VAR NUMBER NUMBER BIN_OP VAR VAR NUMBER BIN_OP VAR NUMBER NUMBER VAR VAR NUMBER IF BIN_OP VAR NUMBER VAR ASSIGN VAR BIN_OP VAR NUMBER NUMBER BIN_OP FUNC_CALL VAR VAR VAR NUMBER VAR VAR NUMBER BIN_OP VAR NUMBER NUMBER VAR BIN_OP VAR NUMBER VAR VAR ASSIGN VAR BIN_OP VAR NUMBER NUMBER BIN_OP FUNC_CALL VAR VAR VAR NUMBER VAR VAR NUMBER BIN_OP VAR NUMBER NUMBER VAR VAR NUMBER EXPR FUNC_CALL VAR FUNC_CALL VAR VAR VAR |
You are given an array $a$ consisting of $n$ integers. Indices of the array start from zero (i. e. the first element is $a_0$, the second one is $a_1$, and so on).
You can reverse at most one subarray (continuous subsegment) of this array. Recall that the subarray of $a$ with borders $l$ and $r$ is $a[l; r] = a_l, a_{l + 1}, \dots, a_{r}$.
Your task is to reverse such a subarray that the sum of elements on even positions of the resulting array is maximized (i. e. the sum of elements $a_0, a_2, \dots, a_{2k}$ for integer $k = \lfloor\frac{n-1}{2}\rfloor$ should be maximum possible).
You have to answer $t$ independent test cases.
-----Input-----
The first line of the input contains one integer $t$ ($1 \le t \le 2 \cdot 10^4$) — the number of test cases. Then $t$ test cases follow.
The first line of the test case contains one integer $n$ ($1 \le n \le 2 \cdot 10^5$) — the length of $a$. The second line of the test case contains $n$ integers $a_0, a_1, \dots, a_{n-1}$ ($1 \le a_i \le 10^9$), where $a_i$ is the $i$-th element of $a$.
It is guaranteed that the sum of $n$ does not exceed $2 \cdot 10^5$ ($\sum n \le 2 \cdot 10^5$).
-----Output-----
For each test case, print the answer on the separate line — the maximum possible sum of elements on even positions after reversing at most one subarray (continuous subsegment) of $a$.
-----Example-----
Input
4
8
1 7 3 4 7 6 2 9
5
1 2 1 2 1
10
7 8 4 5 7 6 8 9 7 3
4
3 1 2 1
Output
26
5
37
5 | import sys
input = sys.stdin.readline
def swaparr(arr, a, b):
temp = arr[a]
arr[a] = arr[b]
arr[b] = temp
def gcd(a, b):
if b == 0:
return a
return gcd(b, a % b)
def nCr(n, k):
if k > n - k:
k = n - k
res = 1
for i in range(k):
res = res * (n - i)
res = res / (i + 1)
return int(res)
def upper_bound(a, x, lo=0):
hi = len(a)
while lo < hi:
mid = (lo + hi) // 2
if a[mid] < x:
lo = mid + 1
else:
hi = mid
return lo
def primefs(n):
primes = {}
while n % 2 == 0 and n > 0:
primes[2] = primes.get(2, 0) + 1
n = n // 2
for i in range(3, int(n**0.5) + 2, 2):
while n % i == 0 and n > 0:
primes[i] = primes.get(i, 0) + 1
n = n // i
if n > 2:
primes[n] = primes.get(n, 0) + 1
return primes
def power(x, y, p):
res = 1
x = x % p
if x == 0:
return 0
while y > 0:
if y & 1 == 1:
res = res * x % p
y = y >> 1
x = x * x % p
return res
def swap(a, b):
temp = a
a = b
b = temp
return a, b
def find(x, link):
p = x
while p != link[p]:
p = link[p]
while x != p:
nex = link[x]
link[x] = p
x = nex
return p
def union(x, y, link, size):
x = find(x, link)
y = find(y, link)
if size[x] < size[y]:
x, y = swap(x, y)
if x != y:
size[x] += size[y]
link[y] = x
def sieve(n):
prime = [(True) for i in range(n + 1)]
p = 2
while p * p <= n:
if prime[p] == True:
for i in range(p * p, n + 1, p):
prime[i] = False
p += 1
return prime
MAXN = int(10000000.0 + 5)
def spf_sieve():
spf[1] = 1
for i in range(2, MAXN):
spf[i] = i
for i in range(4, MAXN, 2):
spf[i] = 2
for i in range(3, ceil(MAXN**0.5), 2):
if spf[i] == i:
for j in range(i * i, MAXN, i):
if spf[j] == j:
spf[j] = i
def factoriazation(x):
ret = {}
while x != 1:
ret[spf[x]] = ret.get(spf[x], 0) + 1
x = x // spf[x]
return ret
def int_array():
return list(map(int, input().strip().split()))
def str_array():
return input().strip().split()
MOD = int(1000000000.0) + 7
CMOD = 998244353
INF = float("inf")
NINF = -float("inf")
def maxEvenLenSum(a, n):
ans1 = 0
this = 0
l = 0
for i in a:
this += i
l += 1
if this < 0 and l % 2 == 0:
this = 0
l = 0
if l % 2 == 0:
ans1 = max(ans1, this)
ans2 = 0
this = 0
l = 0
for i in a:
this += i
l += 1
if this < 0:
this = 0
l = 0
if l % 2 == 0:
ans2 = max(ans2, this)
return max(ans1, ans2)
for _ in range(int(input())):
n = int(input())
a = int_array()
ans = sum([a[i] for i in range(0, n, 2)])
for i in range(0, n, 2):
a[i] *= -1
ans = ans + maxEvenLenSum(a, n)
print(ans) | IMPORT ASSIGN VAR VAR FUNC_DEF ASSIGN VAR VAR VAR ASSIGN VAR VAR VAR VAR ASSIGN VAR VAR VAR FUNC_DEF IF VAR NUMBER RETURN VAR RETURN FUNC_CALL VAR VAR BIN_OP VAR VAR FUNC_DEF IF VAR BIN_OP VAR VAR ASSIGN VAR BIN_OP VAR VAR ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR VAR ASSIGN VAR BIN_OP VAR BIN_OP VAR VAR ASSIGN VAR BIN_OP VAR BIN_OP VAR NUMBER RETURN FUNC_CALL VAR VAR FUNC_DEF NUMBER ASSIGN VAR FUNC_CALL VAR VAR WHILE VAR VAR ASSIGN VAR BIN_OP BIN_OP VAR VAR NUMBER IF VAR VAR VAR ASSIGN VAR BIN_OP VAR NUMBER ASSIGN VAR VAR RETURN VAR FUNC_DEF ASSIGN VAR DICT WHILE BIN_OP VAR NUMBER NUMBER VAR NUMBER ASSIGN VAR NUMBER BIN_OP FUNC_CALL VAR NUMBER NUMBER NUMBER ASSIGN VAR BIN_OP VAR NUMBER FOR VAR FUNC_CALL VAR NUMBER BIN_OP FUNC_CALL VAR BIN_OP VAR NUMBER NUMBER NUMBER WHILE BIN_OP VAR VAR NUMBER VAR NUMBER ASSIGN VAR VAR BIN_OP FUNC_CALL VAR VAR NUMBER NUMBER ASSIGN VAR BIN_OP VAR VAR IF VAR NUMBER ASSIGN VAR VAR BIN_OP FUNC_CALL VAR VAR NUMBER NUMBER RETURN VAR FUNC_DEF ASSIGN VAR NUMBER ASSIGN VAR BIN_OP VAR VAR IF VAR NUMBER RETURN NUMBER WHILE VAR NUMBER IF BIN_OP VAR NUMBER NUMBER ASSIGN VAR BIN_OP BIN_OP VAR VAR VAR ASSIGN VAR BIN_OP VAR NUMBER ASSIGN VAR BIN_OP BIN_OP VAR VAR VAR RETURN VAR FUNC_DEF ASSIGN VAR VAR ASSIGN VAR VAR ASSIGN VAR VAR RETURN VAR VAR FUNC_DEF ASSIGN VAR VAR WHILE VAR VAR VAR ASSIGN VAR VAR VAR WHILE VAR VAR ASSIGN VAR VAR VAR ASSIGN VAR VAR VAR ASSIGN VAR VAR RETURN VAR FUNC_DEF ASSIGN VAR FUNC_CALL VAR VAR VAR ASSIGN VAR FUNC_CALL VAR VAR VAR IF VAR VAR VAR VAR ASSIGN VAR VAR FUNC_CALL VAR VAR VAR IF VAR VAR VAR VAR VAR VAR ASSIGN VAR VAR VAR FUNC_DEF ASSIGN VAR NUMBER VAR FUNC_CALL VAR BIN_OP VAR NUMBER ASSIGN VAR NUMBER WHILE BIN_OP VAR VAR VAR IF VAR VAR NUMBER FOR VAR FUNC_CALL VAR BIN_OP VAR VAR BIN_OP VAR NUMBER VAR ASSIGN VAR VAR NUMBER VAR NUMBER RETURN VAR ASSIGN VAR FUNC_CALL VAR BIN_OP NUMBER NUMBER FUNC_DEF ASSIGN VAR NUMBER NUMBER FOR VAR FUNC_CALL VAR NUMBER VAR ASSIGN VAR VAR VAR FOR VAR FUNC_CALL VAR NUMBER VAR NUMBER ASSIGN VAR VAR NUMBER FOR VAR FUNC_CALL VAR NUMBER FUNC_CALL VAR BIN_OP VAR NUMBER NUMBER IF VAR VAR VAR FOR VAR FUNC_CALL VAR BIN_OP VAR VAR VAR VAR IF VAR VAR VAR ASSIGN VAR VAR VAR FUNC_DEF ASSIGN VAR DICT WHILE VAR NUMBER ASSIGN VAR VAR VAR BIN_OP FUNC_CALL VAR VAR VAR NUMBER NUMBER ASSIGN VAR BIN_OP VAR VAR VAR RETURN VAR FUNC_DEF RETURN FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL FUNC_CALL VAR FUNC_DEF RETURN FUNC_CALL FUNC_CALL FUNC_CALL VAR ASSIGN VAR BIN_OP FUNC_CALL VAR NUMBER NUMBER ASSIGN VAR NUMBER ASSIGN VAR FUNC_CALL VAR STRING ASSIGN VAR FUNC_CALL VAR STRING FUNC_DEF ASSIGN VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR NUMBER FOR VAR VAR VAR VAR VAR NUMBER IF VAR NUMBER BIN_OP VAR NUMBER NUMBER ASSIGN VAR NUMBER ASSIGN VAR NUMBER IF BIN_OP VAR NUMBER NUMBER ASSIGN VAR FUNC_CALL VAR VAR VAR ASSIGN VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR NUMBER FOR VAR VAR VAR VAR VAR NUMBER IF VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR NUMBER IF BIN_OP VAR NUMBER NUMBER ASSIGN VAR FUNC_CALL VAR VAR VAR RETURN FUNC_CALL VAR VAR VAR FOR VAR FUNC_CALL VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR VAR VAR VAR FUNC_CALL VAR NUMBER VAR NUMBER FOR VAR FUNC_CALL VAR NUMBER VAR NUMBER VAR VAR NUMBER ASSIGN VAR BIN_OP VAR FUNC_CALL VAR VAR VAR EXPR FUNC_CALL VAR VAR |
You are given an array $a$ consisting of $n$ integers. Indices of the array start from zero (i. e. the first element is $a_0$, the second one is $a_1$, and so on).
You can reverse at most one subarray (continuous subsegment) of this array. Recall that the subarray of $a$ with borders $l$ and $r$ is $a[l; r] = a_l, a_{l + 1}, \dots, a_{r}$.
Your task is to reverse such a subarray that the sum of elements on even positions of the resulting array is maximized (i. e. the sum of elements $a_0, a_2, \dots, a_{2k}$ for integer $k = \lfloor\frac{n-1}{2}\rfloor$ should be maximum possible).
You have to answer $t$ independent test cases.
-----Input-----
The first line of the input contains one integer $t$ ($1 \le t \le 2 \cdot 10^4$) — the number of test cases. Then $t$ test cases follow.
The first line of the test case contains one integer $n$ ($1 \le n \le 2 \cdot 10^5$) — the length of $a$. The second line of the test case contains $n$ integers $a_0, a_1, \dots, a_{n-1}$ ($1 \le a_i \le 10^9$), where $a_i$ is the $i$-th element of $a$.
It is guaranteed that the sum of $n$ does not exceed $2 \cdot 10^5$ ($\sum n \le 2 \cdot 10^5$).
-----Output-----
For each test case, print the answer on the separate line — the maximum possible sum of elements on even positions after reversing at most one subarray (continuous subsegment) of $a$.
-----Example-----
Input
4
8
1 7 3 4 7 6 2 9
5
1 2 1 2 1
10
7 8 4 5 7 6 8 9 7 3
4
3 1 2 1
Output
26
5
37
5 | T = int(input())
for _ in range(T):
n = int(input())
a = list(map(int, input().split()))
if n % 2 == 0 or n % 2 == 1:
xx = [0] * (n - 1)
for i in range(n - 1):
if i % 2 == 0:
xx[i] = a[i + 1] - a[i]
else:
xx[i] = a[i] - a[i + 1]
a1 = []
a2 = []
x, y = 0, 0
xma, yma = 0, 0
ans = 0
for i in range(n - 1):
if i % 2 == 0:
x += xx[i]
xma = max(xma, x)
if x < 0:
x = 0
else:
y += xx[i]
yma = max(yma, y)
if y < 0:
y = 0
ans = 0
for i in range((n + 1) // 2):
ans += a[i * 2]
print(ans + max(0, xma, yma)) | ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR FOR VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR IF BIN_OP VAR NUMBER NUMBER BIN_OP VAR NUMBER NUMBER ASSIGN VAR BIN_OP LIST NUMBER BIN_OP VAR NUMBER FOR VAR FUNC_CALL VAR BIN_OP VAR NUMBER IF BIN_OP VAR NUMBER NUMBER ASSIGN VAR VAR BIN_OP VAR BIN_OP VAR NUMBER VAR VAR ASSIGN VAR VAR BIN_OP VAR VAR VAR BIN_OP VAR NUMBER ASSIGN VAR LIST ASSIGN VAR LIST ASSIGN VAR VAR NUMBER NUMBER ASSIGN VAR VAR NUMBER NUMBER ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR BIN_OP VAR NUMBER IF BIN_OP VAR NUMBER NUMBER VAR VAR VAR ASSIGN VAR FUNC_CALL VAR VAR VAR IF VAR NUMBER ASSIGN VAR NUMBER VAR VAR VAR ASSIGN VAR FUNC_CALL VAR VAR VAR IF VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR BIN_OP BIN_OP VAR NUMBER NUMBER VAR VAR BIN_OP VAR NUMBER EXPR FUNC_CALL VAR BIN_OP VAR FUNC_CALL VAR NUMBER VAR VAR |
You are given an array $a$ consisting of $n$ integers. Indices of the array start from zero (i. e. the first element is $a_0$, the second one is $a_1$, and so on).
You can reverse at most one subarray (continuous subsegment) of this array. Recall that the subarray of $a$ with borders $l$ and $r$ is $a[l; r] = a_l, a_{l + 1}, \dots, a_{r}$.
Your task is to reverse such a subarray that the sum of elements on even positions of the resulting array is maximized (i. e. the sum of elements $a_0, a_2, \dots, a_{2k}$ for integer $k = \lfloor\frac{n-1}{2}\rfloor$ should be maximum possible).
You have to answer $t$ independent test cases.
-----Input-----
The first line of the input contains one integer $t$ ($1 \le t \le 2 \cdot 10^4$) — the number of test cases. Then $t$ test cases follow.
The first line of the test case contains one integer $n$ ($1 \le n \le 2 \cdot 10^5$) — the length of $a$. The second line of the test case contains $n$ integers $a_0, a_1, \dots, a_{n-1}$ ($1 \le a_i \le 10^9$), where $a_i$ is the $i$-th element of $a$.
It is guaranteed that the sum of $n$ does not exceed $2 \cdot 10^5$ ($\sum n \le 2 \cdot 10^5$).
-----Output-----
For each test case, print the answer on the separate line — the maximum possible sum of elements on even positions after reversing at most one subarray (continuous subsegment) of $a$.
-----Example-----
Input
4
8
1 7 3 4 7 6 2 9
5
1 2 1 2 1
10
7 8 4 5 7 6 8 9 7 3
4
3 1 2 1
Output
26
5
37
5 | import sys
t = int(input())
for _ in range(t):
n = int(input())
lis = list(map(int, input().split()))
s = sum(lis[::2])
arr = [0] * (n // 2)
for i in range(1, n, 2):
arr[i // 2] = lis[i] - lis[i - 1]
m1 = 0
l = 0
for i in range(len(arr)):
l += arr[i]
m1 = max(l, m1)
l = max(l, 0)
arr = [0] * ((n - 1) // 2)
for i in range(2, n, 2):
arr[(i - 1) // 2] = lis[i - 1] - lis[i]
m2 = 0
l = 0
for i in range(len(arr)):
l += arr[i]
m2 = max(l, m2)
l = max(l, 0)
print(s + max(m1, m2)) | IMPORT ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR FOR VAR FUNC_CALL 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 FUNC_CALL VAR VAR NUMBER ASSIGN VAR BIN_OP LIST NUMBER BIN_OP VAR NUMBER FOR VAR FUNC_CALL VAR NUMBER VAR NUMBER ASSIGN VAR BIN_OP VAR NUMBER BIN_OP VAR VAR VAR BIN_OP VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR FUNC_CALL VAR VAR VAR VAR VAR ASSIGN VAR FUNC_CALL VAR VAR VAR ASSIGN VAR FUNC_CALL VAR VAR NUMBER ASSIGN VAR BIN_OP LIST NUMBER BIN_OP BIN_OP VAR NUMBER NUMBER FOR VAR FUNC_CALL VAR NUMBER VAR NUMBER ASSIGN VAR BIN_OP BIN_OP VAR NUMBER NUMBER BIN_OP VAR BIN_OP VAR NUMBER VAR VAR ASSIGN VAR NUMBER ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR FUNC_CALL VAR VAR VAR VAR VAR ASSIGN VAR FUNC_CALL VAR VAR VAR ASSIGN VAR FUNC_CALL VAR VAR NUMBER EXPR FUNC_CALL VAR BIN_OP VAR FUNC_CALL VAR VAR VAR |
You are given an array $a$ consisting of $n$ integers. Indices of the array start from zero (i. e. the first element is $a_0$, the second one is $a_1$, and so on).
You can reverse at most one subarray (continuous subsegment) of this array. Recall that the subarray of $a$ with borders $l$ and $r$ is $a[l; r] = a_l, a_{l + 1}, \dots, a_{r}$.
Your task is to reverse such a subarray that the sum of elements on even positions of the resulting array is maximized (i. e. the sum of elements $a_0, a_2, \dots, a_{2k}$ for integer $k = \lfloor\frac{n-1}{2}\rfloor$ should be maximum possible).
You have to answer $t$ independent test cases.
-----Input-----
The first line of the input contains one integer $t$ ($1 \le t \le 2 \cdot 10^4$) — the number of test cases. Then $t$ test cases follow.
The first line of the test case contains one integer $n$ ($1 \le n \le 2 \cdot 10^5$) — the length of $a$. The second line of the test case contains $n$ integers $a_0, a_1, \dots, a_{n-1}$ ($1 \le a_i \le 10^9$), where $a_i$ is the $i$-th element of $a$.
It is guaranteed that the sum of $n$ does not exceed $2 \cdot 10^5$ ($\sum n \le 2 \cdot 10^5$).
-----Output-----
For each test case, print the answer on the separate line — the maximum possible sum of elements on even positions after reversing at most one subarray (continuous subsegment) of $a$.
-----Example-----
Input
4
8
1 7 3 4 7 6 2 9
5
1 2 1 2 1
10
7 8 4 5 7 6 8 9 7 3
4
3 1 2 1
Output
26
5
37
5 | import sys
input = sys.stdin.readline
I = lambda: list(map(int, input().split()))
(t,) = I()
def ma(a, size):
max_so_far = a[0]
curr_max = a[0]
for i in range(1, size):
curr_max = max(a[i], curr_max + a[i])
max_so_far = max(max_so_far, curr_max)
return max_so_far
for _ in range(t):
(n,) = I()
l = I()
pre = [0] * n
suf = [0] * n
an = 0
if n <= 2:
print(max(l))
continue
d = []
for i in range(n):
if i > 0:
d.append(l[i - 1] - l[i])
if i % 2 == 0:
pre[i] = pre[max(0, i - 2)] + l[i]
if (n - i - 1) % 2:
suf[n - i - 1] = suf[min(n - 1, n - i + 1)] + l[n - i - 1]
x = [d[i] for i in range(n - 1) if i % 2]
y = [(-d[i]) for i in range(n - 1) if i % 2 == 0]
r = ma(x, len(x))
p = ma(y, len(y))
r = max(r, p)
if r < 0:
r = 0
an = max(max(pre[n - 2], pre[n - 1]) + r, suf[1])
for i in range(0, n - 3, 2):
an = max(an, pre[i] + suf[i + 3])
print(an) | IMPORT ASSIGN VAR VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FUNC_DEF ASSIGN VAR VAR NUMBER ASSIGN VAR VAR NUMBER FOR VAR FUNC_CALL VAR NUMBER VAR ASSIGN VAR FUNC_CALL VAR VAR VAR BIN_OP VAR VAR VAR ASSIGN VAR FUNC_CALL VAR VAR VAR RETURN VAR FOR VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR ASSIGN VAR BIN_OP LIST NUMBER VAR ASSIGN VAR BIN_OP LIST NUMBER VAR ASSIGN VAR NUMBER IF VAR NUMBER EXPR FUNC_CALL VAR FUNC_CALL VAR VAR ASSIGN VAR LIST FOR VAR FUNC_CALL VAR VAR IF VAR NUMBER EXPR FUNC_CALL VAR BIN_OP VAR BIN_OP VAR NUMBER VAR VAR IF BIN_OP VAR NUMBER NUMBER ASSIGN VAR VAR BIN_OP VAR FUNC_CALL VAR NUMBER BIN_OP VAR NUMBER VAR VAR IF BIN_OP BIN_OP BIN_OP VAR VAR NUMBER NUMBER ASSIGN VAR BIN_OP BIN_OP VAR VAR NUMBER BIN_OP VAR FUNC_CALL VAR BIN_OP VAR NUMBER BIN_OP BIN_OP VAR VAR NUMBER VAR BIN_OP BIN_OP VAR VAR NUMBER ASSIGN VAR VAR VAR VAR FUNC_CALL VAR BIN_OP VAR NUMBER BIN_OP VAR NUMBER ASSIGN VAR VAR VAR VAR FUNC_CALL VAR BIN_OP VAR NUMBER BIN_OP VAR NUMBER NUMBER ASSIGN VAR FUNC_CALL VAR VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR VAR VAR IF VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR FUNC_CALL VAR BIN_OP FUNC_CALL VAR VAR BIN_OP VAR NUMBER VAR BIN_OP VAR NUMBER VAR VAR NUMBER FOR VAR FUNC_CALL VAR NUMBER BIN_OP VAR NUMBER NUMBER ASSIGN VAR FUNC_CALL VAR VAR BIN_OP VAR VAR VAR BIN_OP VAR NUMBER EXPR FUNC_CALL VAR VAR |
You are given an array $a$ consisting of $n$ integers. Indices of the array start from zero (i. e. the first element is $a_0$, the second one is $a_1$, and so on).
You can reverse at most one subarray (continuous subsegment) of this array. Recall that the subarray of $a$ with borders $l$ and $r$ is $a[l; r] = a_l, a_{l + 1}, \dots, a_{r}$.
Your task is to reverse such a subarray that the sum of elements on even positions of the resulting array is maximized (i. e. the sum of elements $a_0, a_2, \dots, a_{2k}$ for integer $k = \lfloor\frac{n-1}{2}\rfloor$ should be maximum possible).
You have to answer $t$ independent test cases.
-----Input-----
The first line of the input contains one integer $t$ ($1 \le t \le 2 \cdot 10^4$) — the number of test cases. Then $t$ test cases follow.
The first line of the test case contains one integer $n$ ($1 \le n \le 2 \cdot 10^5$) — the length of $a$. The second line of the test case contains $n$ integers $a_0, a_1, \dots, a_{n-1}$ ($1 \le a_i \le 10^9$), where $a_i$ is the $i$-th element of $a$.
It is guaranteed that the sum of $n$ does not exceed $2 \cdot 10^5$ ($\sum n \le 2 \cdot 10^5$).
-----Output-----
For each test case, print the answer on the separate line — the maximum possible sum of elements on even positions after reversing at most one subarray (continuous subsegment) of $a$.
-----Example-----
Input
4
8
1 7 3 4 7 6 2 9
5
1 2 1 2 1
10
7 8 4 5 7 6 8 9 7 3
4
3 1 2 1
Output
26
5
37
5 | import sys
input = sys.stdin.readline
t = int(input())
for _ in range(t):
n = int(input())
a = [int(item) for item in input().split()]
even_sum = sum(a[::2])
max_total = 0
diff = []
for a1, a2 in zip(a[::2], a[1::2]):
diff.append(a2 - a1)
total = 0
for item in diff:
total += item
if item > 0:
max_total = max(max_total, total)
elif total < 0:
total = 0
diff = []
for a1, a2 in zip(a[1::2], a[2::2]):
diff.append(a1 - a2)
total = 0
for item in diff:
total += item
if item > 0:
max_total = max(max_total, total)
elif total < 0:
total = 0
print(max_total + even_sum) | IMPORT ASSIGN VAR VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR 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 NUMBER ASSIGN VAR NUMBER ASSIGN VAR LIST FOR VAR VAR FUNC_CALL VAR VAR NUMBER VAR NUMBER NUMBER EXPR FUNC_CALL VAR BIN_OP VAR VAR ASSIGN VAR NUMBER FOR VAR VAR VAR VAR IF VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR VAR IF VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR LIST FOR VAR VAR FUNC_CALL VAR VAR NUMBER NUMBER VAR NUMBER NUMBER EXPR FUNC_CALL VAR BIN_OP VAR VAR ASSIGN VAR NUMBER FOR VAR VAR VAR VAR IF VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR VAR IF VAR NUMBER ASSIGN VAR NUMBER EXPR FUNC_CALL VAR BIN_OP VAR VAR |
You are given an array $a$ consisting of $n$ integers. Indices of the array start from zero (i. e. the first element is $a_0$, the second one is $a_1$, and so on).
You can reverse at most one subarray (continuous subsegment) of this array. Recall that the subarray of $a$ with borders $l$ and $r$ is $a[l; r] = a_l, a_{l + 1}, \dots, a_{r}$.
Your task is to reverse such a subarray that the sum of elements on even positions of the resulting array is maximized (i. e. the sum of elements $a_0, a_2, \dots, a_{2k}$ for integer $k = \lfloor\frac{n-1}{2}\rfloor$ should be maximum possible).
You have to answer $t$ independent test cases.
-----Input-----
The first line of the input contains one integer $t$ ($1 \le t \le 2 \cdot 10^4$) — the number of test cases. Then $t$ test cases follow.
The first line of the test case contains one integer $n$ ($1 \le n \le 2 \cdot 10^5$) — the length of $a$. The second line of the test case contains $n$ integers $a_0, a_1, \dots, a_{n-1}$ ($1 \le a_i \le 10^9$), where $a_i$ is the $i$-th element of $a$.
It is guaranteed that the sum of $n$ does not exceed $2 \cdot 10^5$ ($\sum n \le 2 \cdot 10^5$).
-----Output-----
For each test case, print the answer on the separate line — the maximum possible sum of elements on even positions after reversing at most one subarray (continuous subsegment) of $a$.
-----Example-----
Input
4
8
1 7 3 4 7 6 2 9
5
1 2 1 2 1
10
7 8 4 5 7 6 8 9 7 3
4
3 1 2 1
Output
26
5
37
5 | for lo in range(int(input())):
n = int(input())
ls = [int(x) for x in input().split()]
mxd = 0
currmxd = 0
se = 0
so = 0
m = n
if n % 2 != 0:
m = n - 1
for i in range(0, m, 2):
so += ls[i + 1]
se += ls[i]
if so >= se:
currmxd = so - se
if mxd <= currmxd:
mxd = currmxd
else:
so = 0
se = 0
currmxd = 0
m = n
if n % 2 == 0:
m = n - 1
se = 0
so = 0
currmxd = 0
for i in range(1, m, 2):
so += ls[i]
se += ls[i + 1]
if so >= se:
currmxd = so - se
if mxd <= currmxd:
mxd = currmxd
else:
so = 0
se = 0
currmxd = 0
res = 0
for i in range(0, n, 2):
res += ls[i]
print(res + mxd) | FOR VAR FUNC_CALL VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR VAR IF BIN_OP VAR NUMBER NUMBER ASSIGN VAR BIN_OP VAR NUMBER FOR VAR FUNC_CALL VAR NUMBER VAR NUMBER VAR VAR BIN_OP VAR NUMBER VAR VAR VAR IF VAR VAR ASSIGN VAR BIN_OP VAR VAR IF VAR VAR ASSIGN VAR VAR ASSIGN VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR VAR IF BIN_OP VAR NUMBER NUMBER ASSIGN VAR BIN_OP VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR NUMBER VAR NUMBER VAR VAR VAR VAR VAR BIN_OP VAR NUMBER IF VAR VAR ASSIGN VAR BIN_OP VAR VAR IF VAR VAR ASSIGN VAR VAR ASSIGN VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR NUMBER VAR NUMBER VAR VAR VAR EXPR FUNC_CALL VAR BIN_OP VAR VAR |
You are given an array $a$ consisting of $n$ integers. Indices of the array start from zero (i. e. the first element is $a_0$, the second one is $a_1$, and so on).
You can reverse at most one subarray (continuous subsegment) of this array. Recall that the subarray of $a$ with borders $l$ and $r$ is $a[l; r] = a_l, a_{l + 1}, \dots, a_{r}$.
Your task is to reverse such a subarray that the sum of elements on even positions of the resulting array is maximized (i. e. the sum of elements $a_0, a_2, \dots, a_{2k}$ for integer $k = \lfloor\frac{n-1}{2}\rfloor$ should be maximum possible).
You have to answer $t$ independent test cases.
-----Input-----
The first line of the input contains one integer $t$ ($1 \le t \le 2 \cdot 10^4$) — the number of test cases. Then $t$ test cases follow.
The first line of the test case contains one integer $n$ ($1 \le n \le 2 \cdot 10^5$) — the length of $a$. The second line of the test case contains $n$ integers $a_0, a_1, \dots, a_{n-1}$ ($1 \le a_i \le 10^9$), where $a_i$ is the $i$-th element of $a$.
It is guaranteed that the sum of $n$ does not exceed $2 \cdot 10^5$ ($\sum n \le 2 \cdot 10^5$).
-----Output-----
For each test case, print the answer on the separate line — the maximum possible sum of elements on even positions after reversing at most one subarray (continuous subsegment) of $a$.
-----Example-----
Input
4
8
1 7 3 4 7 6 2 9
5
1 2 1 2 1
10
7 8 4 5 7 6 8 9 7 3
4
3 1 2 1
Output
26
5
37
5 | for _ in range(int(input())):
n = int(input())
arr = list(map(int, input().split()))
sm = 0
for i, x in enumerate(arr):
if not i % 2:
sm += x
dp1 = [0]
dp2 = [0]
for i in range(n - 1):
if i % 2:
dp2.append(arr[i] - arr[i + 1])
else:
dp1.append(arr[i + 1] - arr[i])
for i in range(1, len(dp1)):
dp1[i] = max(dp1[i] + dp1[i - 1], 0)
for i in range(1, len(dp2)):
dp2[i] = max(dp2[i] + dp2[i - 1], 0)
print(sm + max(max(dp1), max(dp2))) | FOR VAR FUNC_CALL VAR FUNC_CALL VAR FUNC_CALL 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 NUMBER FOR VAR VAR FUNC_CALL VAR VAR IF BIN_OP VAR NUMBER VAR VAR ASSIGN VAR LIST NUMBER ASSIGN VAR LIST NUMBER FOR VAR FUNC_CALL VAR BIN_OP VAR NUMBER IF BIN_OP VAR NUMBER EXPR FUNC_CALL VAR BIN_OP VAR VAR VAR BIN_OP VAR NUMBER EXPR FUNC_CALL VAR BIN_OP VAR BIN_OP VAR NUMBER VAR VAR FOR VAR FUNC_CALL VAR NUMBER FUNC_CALL VAR VAR ASSIGN VAR VAR FUNC_CALL VAR BIN_OP VAR VAR VAR BIN_OP VAR NUMBER NUMBER FOR VAR FUNC_CALL VAR NUMBER FUNC_CALL VAR VAR ASSIGN VAR VAR FUNC_CALL VAR BIN_OP VAR VAR VAR BIN_OP VAR NUMBER NUMBER EXPR FUNC_CALL VAR BIN_OP VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL VAR VAR |
You are given an array $a$ consisting of $n$ integers. Indices of the array start from zero (i. e. the first element is $a_0$, the second one is $a_1$, and so on).
You can reverse at most one subarray (continuous subsegment) of this array. Recall that the subarray of $a$ with borders $l$ and $r$ is $a[l; r] = a_l, a_{l + 1}, \dots, a_{r}$.
Your task is to reverse such a subarray that the sum of elements on even positions of the resulting array is maximized (i. e. the sum of elements $a_0, a_2, \dots, a_{2k}$ for integer $k = \lfloor\frac{n-1}{2}\rfloor$ should be maximum possible).
You have to answer $t$ independent test cases.
-----Input-----
The first line of the input contains one integer $t$ ($1 \le t \le 2 \cdot 10^4$) — the number of test cases. Then $t$ test cases follow.
The first line of the test case contains one integer $n$ ($1 \le n \le 2 \cdot 10^5$) — the length of $a$. The second line of the test case contains $n$ integers $a_0, a_1, \dots, a_{n-1}$ ($1 \le a_i \le 10^9$), where $a_i$ is the $i$-th element of $a$.
It is guaranteed that the sum of $n$ does not exceed $2 \cdot 10^5$ ($\sum n \le 2 \cdot 10^5$).
-----Output-----
For each test case, print the answer on the separate line — the maximum possible sum of elements on even positions after reversing at most one subarray (continuous subsegment) of $a$.
-----Example-----
Input
4
8
1 7 3 4 7 6 2 9
5
1 2 1 2 1
10
7 8 4 5 7 6 8 9 7 3
4
3 1 2 1
Output
26
5
37
5 | t = int(input())
for you in range(t):
n = int(input())
l = input().split()
li = [int(i) for i in l]
z = 0
for i in range(0, n, 2):
z += li[i]
dp = [(0) for i in range(n)]
if n > 1:
dp[1] = max(0, li[1] - li[0])
if n > 2:
dp[2] = max(0, li[1] - li[2])
for i in range(3, n):
if i % 2 == 0:
dp[i] = max(0, dp[i - 2] + li[i - 1] - li[i])
else:
dp[i] = max(0, dp[i - 2] + li[i] - li[i - 1])
ok = 0
for i in dp:
ok = max(ok, i)
print(ok + z) | ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR FOR VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR VAR VAR VAR ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR NUMBER VAR NUMBER VAR VAR VAR ASSIGN VAR NUMBER VAR FUNC_CALL VAR VAR IF VAR NUMBER ASSIGN VAR NUMBER FUNC_CALL VAR NUMBER BIN_OP VAR NUMBER VAR NUMBER IF VAR NUMBER ASSIGN VAR NUMBER FUNC_CALL VAR NUMBER BIN_OP VAR NUMBER VAR NUMBER FOR VAR FUNC_CALL VAR NUMBER VAR IF BIN_OP VAR NUMBER NUMBER ASSIGN VAR VAR FUNC_CALL VAR NUMBER BIN_OP BIN_OP VAR BIN_OP VAR NUMBER VAR BIN_OP VAR NUMBER VAR VAR ASSIGN VAR VAR FUNC_CALL VAR NUMBER BIN_OP BIN_OP VAR BIN_OP VAR NUMBER VAR VAR VAR BIN_OP VAR NUMBER ASSIGN VAR NUMBER FOR VAR VAR ASSIGN VAR FUNC_CALL VAR VAR VAR EXPR FUNC_CALL VAR BIN_OP VAR VAR |
You are given an array $a$ consisting of $n$ integers. Indices of the array start from zero (i. e. the first element is $a_0$, the second one is $a_1$, and so on).
You can reverse at most one subarray (continuous subsegment) of this array. Recall that the subarray of $a$ with borders $l$ and $r$ is $a[l; r] = a_l, a_{l + 1}, \dots, a_{r}$.
Your task is to reverse such a subarray that the sum of elements on even positions of the resulting array is maximized (i. e. the sum of elements $a_0, a_2, \dots, a_{2k}$ for integer $k = \lfloor\frac{n-1}{2}\rfloor$ should be maximum possible).
You have to answer $t$ independent test cases.
-----Input-----
The first line of the input contains one integer $t$ ($1 \le t \le 2 \cdot 10^4$) — the number of test cases. Then $t$ test cases follow.
The first line of the test case contains one integer $n$ ($1 \le n \le 2 \cdot 10^5$) — the length of $a$. The second line of the test case contains $n$ integers $a_0, a_1, \dots, a_{n-1}$ ($1 \le a_i \le 10^9$), where $a_i$ is the $i$-th element of $a$.
It is guaranteed that the sum of $n$ does not exceed $2 \cdot 10^5$ ($\sum n \le 2 \cdot 10^5$).
-----Output-----
For each test case, print the answer on the separate line — the maximum possible sum of elements on even positions after reversing at most one subarray (continuous subsegment) of $a$.
-----Example-----
Input
4
8
1 7 3 4 7 6 2 9
5
1 2 1 2 1
10
7 8 4 5 7 6 8 9 7 3
4
3 1 2 1
Output
26
5
37
5 | for i in range(int(input())):
n = int(input())
l = list(map(int, input().split()))
ans = 0
ans1 = 0
i = 0
ans2 = 0
ans3 = 0
while i <= len(l) - 2:
if ans + (l[i + 1] - l[i]) > 0 and i % 2 == 0:
ans1 = max(ans, ans1)
ans += l[i + 1]
ans -= l[i]
ans1 = max(ans, ans1)
elif i % 2 != 1:
ans = 0
if ans2 + (l[i] - l[i + 1]) > 0 and i % 2 != 0:
ans3 = max(ans2, ans3)
ans2 += l[i]
ans2 -= l[i + 1]
ans3 = max(ans2, ans3)
elif i % 2 != 0:
ans2 = 0
i += 1
ans1 = max(ans1, ans3)
for i in range(len(l)):
if i % 2 == 0:
ans1 += l[i]
print(ans1) | FOR VAR FUNC_CALL VAR FUNC_CALL VAR FUNC_CALL 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 NUMBER ASSIGN VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR NUMBER WHILE VAR BIN_OP FUNC_CALL VAR VAR NUMBER IF BIN_OP VAR BIN_OP VAR BIN_OP VAR NUMBER VAR VAR NUMBER BIN_OP VAR NUMBER NUMBER ASSIGN VAR FUNC_CALL VAR VAR VAR VAR VAR BIN_OP VAR NUMBER VAR VAR VAR ASSIGN VAR FUNC_CALL VAR VAR VAR IF BIN_OP VAR NUMBER NUMBER ASSIGN VAR NUMBER IF BIN_OP VAR BIN_OP VAR VAR VAR BIN_OP VAR NUMBER NUMBER BIN_OP VAR NUMBER NUMBER ASSIGN VAR FUNC_CALL VAR VAR VAR VAR VAR VAR VAR VAR BIN_OP VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR VAR IF BIN_OP VAR NUMBER NUMBER ASSIGN VAR NUMBER VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR VAR FOR VAR FUNC_CALL VAR FUNC_CALL VAR VAR IF BIN_OP VAR NUMBER NUMBER VAR VAR VAR EXPR FUNC_CALL VAR VAR |
You are given an array $a$ consisting of $n$ integers. Indices of the array start from zero (i. e. the first element is $a_0$, the second one is $a_1$, and so on).
You can reverse at most one subarray (continuous subsegment) of this array. Recall that the subarray of $a$ with borders $l$ and $r$ is $a[l; r] = a_l, a_{l + 1}, \dots, a_{r}$.
Your task is to reverse such a subarray that the sum of elements on even positions of the resulting array is maximized (i. e. the sum of elements $a_0, a_2, \dots, a_{2k}$ for integer $k = \lfloor\frac{n-1}{2}\rfloor$ should be maximum possible).
You have to answer $t$ independent test cases.
-----Input-----
The first line of the input contains one integer $t$ ($1 \le t \le 2 \cdot 10^4$) — the number of test cases. Then $t$ test cases follow.
The first line of the test case contains one integer $n$ ($1 \le n \le 2 \cdot 10^5$) — the length of $a$. The second line of the test case contains $n$ integers $a_0, a_1, \dots, a_{n-1}$ ($1 \le a_i \le 10^9$), where $a_i$ is the $i$-th element of $a$.
It is guaranteed that the sum of $n$ does not exceed $2 \cdot 10^5$ ($\sum n \le 2 \cdot 10^5$).
-----Output-----
For each test case, print the answer on the separate line — the maximum possible sum of elements on even positions after reversing at most one subarray (continuous subsegment) of $a$.
-----Example-----
Input
4
8
1 7 3 4 7 6 2 9
5
1 2 1 2 1
10
7 8 4 5 7 6 8 9 7 3
4
3 1 2 1
Output
26
5
37
5 | def main():
for case in range(int(input())):
input()
vs = [int(t) for t in input().strip().split()]
max_sum = 0
cum_sum = 0
for i in range(len(vs) // 2):
cum_sum = max(0, cum_sum + vs[2 * i + 1] - vs[2 * i])
max_sum = max(max_sum, cum_sum)
cum_sum = 0
for i in range((len(vs) - 1) // 2):
cum_sum = max(0, cum_sum + vs[2 * i + 1] - vs[2 * i + 2])
max_sum = max(max_sum, cum_sum)
print(sum(v for i, v in enumerate(vs) if i % 2 == 0) + max_sum)
main() | FUNC_DEF FOR VAR FUNC_CALL VAR FUNC_CALL VAR FUNC_CALL VAR EXPR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR VAR VAR FUNC_CALL FUNC_CALL FUNC_CALL VAR ASSIGN VAR NUMBER ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR BIN_OP FUNC_CALL VAR VAR NUMBER ASSIGN VAR FUNC_CALL VAR NUMBER BIN_OP BIN_OP VAR VAR BIN_OP BIN_OP NUMBER VAR NUMBER VAR BIN_OP NUMBER VAR ASSIGN VAR FUNC_CALL VAR VAR VAR ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR BIN_OP BIN_OP FUNC_CALL VAR VAR NUMBER NUMBER ASSIGN VAR FUNC_CALL VAR NUMBER BIN_OP BIN_OP VAR VAR BIN_OP BIN_OP NUMBER VAR NUMBER VAR BIN_OP BIN_OP NUMBER VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR VAR EXPR FUNC_CALL VAR BIN_OP FUNC_CALL VAR VAR VAR VAR FUNC_CALL VAR VAR BIN_OP VAR NUMBER NUMBER VAR EXPR FUNC_CALL VAR |
You are given an array $a$ consisting of $n$ integers. Indices of the array start from zero (i. e. the first element is $a_0$, the second one is $a_1$, and so on).
You can reverse at most one subarray (continuous subsegment) of this array. Recall that the subarray of $a$ with borders $l$ and $r$ is $a[l; r] = a_l, a_{l + 1}, \dots, a_{r}$.
Your task is to reverse such a subarray that the sum of elements on even positions of the resulting array is maximized (i. e. the sum of elements $a_0, a_2, \dots, a_{2k}$ for integer $k = \lfloor\frac{n-1}{2}\rfloor$ should be maximum possible).
You have to answer $t$ independent test cases.
-----Input-----
The first line of the input contains one integer $t$ ($1 \le t \le 2 \cdot 10^4$) — the number of test cases. Then $t$ test cases follow.
The first line of the test case contains one integer $n$ ($1 \le n \le 2 \cdot 10^5$) — the length of $a$. The second line of the test case contains $n$ integers $a_0, a_1, \dots, a_{n-1}$ ($1 \le a_i \le 10^9$), where $a_i$ is the $i$-th element of $a$.
It is guaranteed that the sum of $n$ does not exceed $2 \cdot 10^5$ ($\sum n \le 2 \cdot 10^5$).
-----Output-----
For each test case, print the answer on the separate line — the maximum possible sum of elements on even positions after reversing at most one subarray (continuous subsegment) of $a$.
-----Example-----
Input
4
8
1 7 3 4 7 6 2 9
5
1 2 1 2 1
10
7 8 4 5 7 6 8 9 7 3
4
3 1 2 1
Output
26
5
37
5 | for it in range(int(input())):
n = int(input())
A = list(map(int, input().split()))
sum = 0
for i in range(n):
if i % 2 == 0:
sum += A[i]
B = []
C = []
for i in range(0, n - 1, 2):
B.append(A[i + 1] - A[i])
for i in range(2, n, 2):
C.append(A[i - 1] - A[i])
r1 = -1 * 1000000007
r2 = -1 * 1000000007
sum1 = 0
sum2 = 0
for i in range(len(B)):
sum1 += B[i]
r1 = max(r1, sum1)
if sum1 < 0:
sum1 = 0
for i in range(len(C)):
sum2 += C[i]
r2 = max(r2, sum2)
if sum2 < 0:
sum2 = 0
sum += max(0, r1, r2)
print(sum) | FOR VAR FUNC_CALL VAR FUNC_CALL VAR FUNC_CALL 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 NUMBER FOR VAR FUNC_CALL VAR VAR IF BIN_OP VAR NUMBER NUMBER VAR VAR VAR ASSIGN VAR LIST ASSIGN VAR LIST FOR VAR FUNC_CALL VAR NUMBER BIN_OP VAR NUMBER NUMBER EXPR FUNC_CALL VAR BIN_OP VAR BIN_OP VAR NUMBER VAR VAR FOR VAR FUNC_CALL VAR NUMBER VAR NUMBER EXPR FUNC_CALL VAR BIN_OP VAR BIN_OP VAR NUMBER VAR VAR ASSIGN VAR BIN_OP NUMBER NUMBER ASSIGN VAR BIN_OP NUMBER NUMBER ASSIGN VAR NUMBER ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR FUNC_CALL VAR VAR VAR VAR VAR ASSIGN VAR FUNC_CALL VAR VAR VAR IF VAR NUMBER ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR FUNC_CALL VAR VAR VAR VAR VAR ASSIGN VAR FUNC_CALL VAR VAR VAR IF VAR NUMBER ASSIGN VAR NUMBER VAR FUNC_CALL VAR NUMBER VAR VAR EXPR FUNC_CALL VAR VAR |
You are given an array $a$ consisting of $n$ integers. Indices of the array start from zero (i. e. the first element is $a_0$, the second one is $a_1$, and so on).
You can reverse at most one subarray (continuous subsegment) of this array. Recall that the subarray of $a$ with borders $l$ and $r$ is $a[l; r] = a_l, a_{l + 1}, \dots, a_{r}$.
Your task is to reverse such a subarray that the sum of elements on even positions of the resulting array is maximized (i. e. the sum of elements $a_0, a_2, \dots, a_{2k}$ for integer $k = \lfloor\frac{n-1}{2}\rfloor$ should be maximum possible).
You have to answer $t$ independent test cases.
-----Input-----
The first line of the input contains one integer $t$ ($1 \le t \le 2 \cdot 10^4$) — the number of test cases. Then $t$ test cases follow.
The first line of the test case contains one integer $n$ ($1 \le n \le 2 \cdot 10^5$) — the length of $a$. The second line of the test case contains $n$ integers $a_0, a_1, \dots, a_{n-1}$ ($1 \le a_i \le 10^9$), where $a_i$ is the $i$-th element of $a$.
It is guaranteed that the sum of $n$ does not exceed $2 \cdot 10^5$ ($\sum n \le 2 \cdot 10^5$).
-----Output-----
For each test case, print the answer on the separate line — the maximum possible sum of elements on even positions after reversing at most one subarray (continuous subsegment) of $a$.
-----Example-----
Input
4
8
1 7 3 4 7 6 2 9
5
1 2 1 2 1
10
7 8 4 5 7 6 8 9 7 3
4
3 1 2 1
Output
26
5
37
5 | def maiorLucro(arr):
maior = soma = 0
for x in arr:
soma += x
if soma < 0:
soma = 0
if soma > maior:
maior = soma
return maior
def solve():
input()
num = list(map(int, input().split()))
prefImp = [(j - i) for i, j in zip(num[0::2], num[1::2])]
prefPar = [(i - j) for i, j in zip(num[1::2], num[2::2])]
maxImp = maiorLucro(prefImp)
maxPar = maiorLucro(prefPar)
adicional = max(maxImp, maxPar)
print(sum(num[::2]) + (adicional if adicional > 0 else 0))
for _ in range(int(input())):
solve() | FUNC_DEF ASSIGN VAR VAR NUMBER FOR VAR VAR VAR VAR IF VAR NUMBER ASSIGN VAR NUMBER IF VAR VAR ASSIGN VAR VAR RETURN VAR FUNC_DEF EXPR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR BIN_OP VAR VAR VAR VAR FUNC_CALL VAR VAR NUMBER NUMBER VAR NUMBER NUMBER ASSIGN VAR BIN_OP VAR VAR VAR VAR FUNC_CALL VAR VAR NUMBER NUMBER VAR NUMBER NUMBER ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR VAR VAR EXPR FUNC_CALL VAR BIN_OP FUNC_CALL VAR VAR NUMBER VAR NUMBER VAR NUMBER FOR VAR FUNC_CALL VAR FUNC_CALL VAR FUNC_CALL VAR EXPR FUNC_CALL VAR |
You are given an array $a$ consisting of $n$ integers. Indices of the array start from zero (i. e. the first element is $a_0$, the second one is $a_1$, and so on).
You can reverse at most one subarray (continuous subsegment) of this array. Recall that the subarray of $a$ with borders $l$ and $r$ is $a[l; r] = a_l, a_{l + 1}, \dots, a_{r}$.
Your task is to reverse such a subarray that the sum of elements on even positions of the resulting array is maximized (i. e. the sum of elements $a_0, a_2, \dots, a_{2k}$ for integer $k = \lfloor\frac{n-1}{2}\rfloor$ should be maximum possible).
You have to answer $t$ independent test cases.
-----Input-----
The first line of the input contains one integer $t$ ($1 \le t \le 2 \cdot 10^4$) — the number of test cases. Then $t$ test cases follow.
The first line of the test case contains one integer $n$ ($1 \le n \le 2 \cdot 10^5$) — the length of $a$. The second line of the test case contains $n$ integers $a_0, a_1, \dots, a_{n-1}$ ($1 \le a_i \le 10^9$), where $a_i$ is the $i$-th element of $a$.
It is guaranteed that the sum of $n$ does not exceed $2 \cdot 10^5$ ($\sum n \le 2 \cdot 10^5$).
-----Output-----
For each test case, print the answer on the separate line — the maximum possible sum of elements on even positions after reversing at most one subarray (continuous subsegment) of $a$.
-----Example-----
Input
4
8
1 7 3 4 7 6 2 9
5
1 2 1 2 1
10
7 8 4 5 7 6 8 9 7 3
4
3 1 2 1
Output
26
5
37
5 | for _ in range(int(input())):
n = int(input())
a = list(map(int, input().split()))
dp0 = [[0, 0, 0] for i in range((n - 1) // 2 + 1)]
for i in range((n - 1) // 2 + 1):
if i == 0:
dp0[i][0] = a[2 * i]
continue
dp0[i][0] = dp0[i - 1][0] + a[2 * i]
dp0[i][1] = max(dp0[i - 1][0], dp0[i - 1][1]) + a[2 * i - 1]
dp0[i][2] = max(dp0[i - 1][1], dp0[i - 1][2]) + a[2 * i]
dp1 = [[0, 0, 0] for i in range((n - 1) // 2 + 1)]
for i in range((n - 1) // 2 + 1):
if i == 0:
dp1[i][0] = a[2 * i]
if 2 * i + 1 < n:
dp1[i][1] = a[2 * i + 1]
continue
dp1[i][0] = dp1[i - 1][0] + a[2 * i]
if 2 * i + 1 < n:
dp1[i][1] = max(dp1[i - 1][0], dp1[i - 1][1]) + a[2 * i + 1]
if 2 * i < n:
dp1[i][2] = max(dp1[i - 1][1], dp1[i - 1][2]) + a[2 * i]
print(max(dp0[(n - 1) // 2] + dp1[(n - 1) // 2])) | FOR VAR FUNC_CALL VAR FUNC_CALL VAR FUNC_CALL 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 LIST NUMBER NUMBER NUMBER VAR FUNC_CALL VAR BIN_OP BIN_OP BIN_OP VAR NUMBER NUMBER NUMBER FOR VAR FUNC_CALL VAR BIN_OP BIN_OP BIN_OP VAR NUMBER NUMBER NUMBER IF VAR NUMBER ASSIGN VAR VAR NUMBER VAR BIN_OP NUMBER VAR ASSIGN VAR VAR NUMBER BIN_OP VAR BIN_OP VAR NUMBER NUMBER VAR BIN_OP NUMBER VAR ASSIGN VAR VAR NUMBER BIN_OP FUNC_CALL VAR VAR BIN_OP VAR NUMBER NUMBER VAR BIN_OP VAR NUMBER NUMBER VAR BIN_OP BIN_OP NUMBER VAR NUMBER ASSIGN VAR VAR NUMBER BIN_OP FUNC_CALL VAR VAR BIN_OP VAR NUMBER NUMBER VAR BIN_OP VAR NUMBER NUMBER VAR BIN_OP NUMBER VAR ASSIGN VAR LIST NUMBER NUMBER NUMBER VAR FUNC_CALL VAR BIN_OP BIN_OP BIN_OP VAR NUMBER NUMBER NUMBER FOR VAR FUNC_CALL VAR BIN_OP BIN_OP BIN_OP VAR NUMBER NUMBER NUMBER IF VAR NUMBER ASSIGN VAR VAR NUMBER VAR BIN_OP NUMBER VAR IF BIN_OP BIN_OP NUMBER VAR NUMBER VAR ASSIGN VAR VAR NUMBER VAR BIN_OP BIN_OP NUMBER VAR NUMBER ASSIGN VAR VAR NUMBER BIN_OP VAR BIN_OP VAR NUMBER NUMBER VAR BIN_OP NUMBER VAR IF BIN_OP BIN_OP NUMBER VAR NUMBER VAR ASSIGN VAR VAR NUMBER BIN_OP FUNC_CALL VAR VAR BIN_OP VAR NUMBER NUMBER VAR BIN_OP VAR NUMBER NUMBER VAR BIN_OP BIN_OP NUMBER VAR NUMBER IF BIN_OP NUMBER VAR VAR ASSIGN VAR VAR NUMBER BIN_OP FUNC_CALL VAR VAR BIN_OP VAR NUMBER NUMBER VAR BIN_OP VAR NUMBER NUMBER VAR BIN_OP NUMBER VAR EXPR FUNC_CALL VAR FUNC_CALL VAR BIN_OP VAR BIN_OP BIN_OP VAR NUMBER NUMBER VAR BIN_OP BIN_OP VAR NUMBER NUMBER |
You are given an array $a$ consisting of $n$ integers. Indices of the array start from zero (i. e. the first element is $a_0$, the second one is $a_1$, and so on).
You can reverse at most one subarray (continuous subsegment) of this array. Recall that the subarray of $a$ with borders $l$ and $r$ is $a[l; r] = a_l, a_{l + 1}, \dots, a_{r}$.
Your task is to reverse such a subarray that the sum of elements on even positions of the resulting array is maximized (i. e. the sum of elements $a_0, a_2, \dots, a_{2k}$ for integer $k = \lfloor\frac{n-1}{2}\rfloor$ should be maximum possible).
You have to answer $t$ independent test cases.
-----Input-----
The first line of the input contains one integer $t$ ($1 \le t \le 2 \cdot 10^4$) — the number of test cases. Then $t$ test cases follow.
The first line of the test case contains one integer $n$ ($1 \le n \le 2 \cdot 10^5$) — the length of $a$. The second line of the test case contains $n$ integers $a_0, a_1, \dots, a_{n-1}$ ($1 \le a_i \le 10^9$), where $a_i$ is the $i$-th element of $a$.
It is guaranteed that the sum of $n$ does not exceed $2 \cdot 10^5$ ($\sum n \le 2 \cdot 10^5$).
-----Output-----
For each test case, print the answer on the separate line — the maximum possible sum of elements on even positions after reversing at most one subarray (continuous subsegment) of $a$.
-----Example-----
Input
4
8
1 7 3 4 7 6 2 9
5
1 2 1 2 1
10
7 8 4 5 7 6 8 9 7 3
4
3 1 2 1
Output
26
5
37
5 | import sys
readline = sys.stdin.readline
def solve():
N = int(readline())
A = list(map(int, readline().split()))
def _sub(d):
gain0 = [0] * (N // 2)
j = 0
for i in range(0, N, 2):
if not 0 <= i + d < N:
continue
gain0[j] = A[i + d] - A[i]
j += 1
for i in range(len(gain0)):
gain0[i] = gain0[0] if i == 0 else gain0[i] + gain0[i - 1]
m = 0
gain = 0
for v in gain0:
gain = max(gain, v - m)
m = min(m, v)
return gain
a = sum(A[0:N:2])
a0 = _sub(-1) + a
a1 = _sub(1) + a
print(max(a0, a1))
T = int(readline())
for i in range(T):
solve() | IMPORT ASSIGN 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 FUNC_DEF ASSIGN VAR BIN_OP LIST NUMBER BIN_OP VAR NUMBER ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR NUMBER VAR NUMBER IF NUMBER BIN_OP VAR VAR VAR ASSIGN VAR VAR BIN_OP VAR BIN_OP VAR VAR VAR VAR VAR NUMBER FOR VAR FUNC_CALL VAR FUNC_CALL VAR VAR ASSIGN VAR VAR VAR NUMBER VAR NUMBER BIN_OP VAR VAR VAR BIN_OP VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR NUMBER FOR VAR VAR ASSIGN VAR FUNC_CALL VAR VAR BIN_OP VAR VAR ASSIGN VAR FUNC_CALL VAR VAR VAR RETURN VAR ASSIGN VAR FUNC_CALL VAR VAR NUMBER VAR NUMBER ASSIGN VAR BIN_OP FUNC_CALL VAR NUMBER VAR ASSIGN VAR BIN_OP FUNC_CALL VAR NUMBER VAR EXPR FUNC_CALL VAR FUNC_CALL VAR VAR VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR FOR VAR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR |
You are given an array $a$ consisting of $n$ integers. Indices of the array start from zero (i. e. the first element is $a_0$, the second one is $a_1$, and so on).
You can reverse at most one subarray (continuous subsegment) of this array. Recall that the subarray of $a$ with borders $l$ and $r$ is $a[l; r] = a_l, a_{l + 1}, \dots, a_{r}$.
Your task is to reverse such a subarray that the sum of elements on even positions of the resulting array is maximized (i. e. the sum of elements $a_0, a_2, \dots, a_{2k}$ for integer $k = \lfloor\frac{n-1}{2}\rfloor$ should be maximum possible).
You have to answer $t$ independent test cases.
-----Input-----
The first line of the input contains one integer $t$ ($1 \le t \le 2 \cdot 10^4$) — the number of test cases. Then $t$ test cases follow.
The first line of the test case contains one integer $n$ ($1 \le n \le 2 \cdot 10^5$) — the length of $a$. The second line of the test case contains $n$ integers $a_0, a_1, \dots, a_{n-1}$ ($1 \le a_i \le 10^9$), where $a_i$ is the $i$-th element of $a$.
It is guaranteed that the sum of $n$ does not exceed $2 \cdot 10^5$ ($\sum n \le 2 \cdot 10^5$).
-----Output-----
For each test case, print the answer on the separate line — the maximum possible sum of elements on even positions after reversing at most one subarray (continuous subsegment) of $a$.
-----Example-----
Input
4
8
1 7 3 4 7 6 2 9
5
1 2 1 2 1
10
7 8 4 5 7 6 8 9 7 3
4
3 1 2 1
Output
26
5
37
5 | def solve():
n = int(input())
a = [int(x) for x in input().split(" ")]
ans = 0
for i in range(n):
if i % 2 == 0:
ans += a[i]
v = list()
for i in range(n - 1):
if i % 2 == 0:
v.append(a[i + 1] - a[i])
max_sum = 0
cur = 0
for i in v:
cur += i
max_sum = max(max_sum, cur)
cur = max(cur, 0)
tmp = max_sum
v.clear()
for i in range(n - 1):
if i % 2 == 1:
v.append(a[i] - a[i + 1])
max_sum = 0
cur = 0
for i in v:
cur += i
max_sum = max(max_sum, cur)
cur = max(cur, 0)
ans += max(max_sum, tmp)
print(ans)
t = int(input())
while t > 0:
t -= 1
solve() | FUNC_DEF ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR VAR VAR FUNC_CALL FUNC_CALL VAR STRING ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR VAR IF BIN_OP VAR NUMBER NUMBER VAR VAR VAR ASSIGN VAR FUNC_CALL VAR FOR VAR FUNC_CALL VAR BIN_OP VAR NUMBER IF BIN_OP VAR NUMBER NUMBER EXPR FUNC_CALL VAR BIN_OP VAR BIN_OP VAR NUMBER VAR VAR ASSIGN VAR NUMBER ASSIGN VAR NUMBER FOR VAR VAR VAR VAR ASSIGN VAR FUNC_CALL VAR VAR VAR ASSIGN VAR FUNC_CALL VAR VAR NUMBER ASSIGN VAR VAR EXPR FUNC_CALL VAR FOR VAR FUNC_CALL VAR BIN_OP VAR NUMBER IF BIN_OP VAR NUMBER NUMBER EXPR FUNC_CALL VAR BIN_OP VAR VAR VAR BIN_OP VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR NUMBER FOR VAR VAR VAR VAR ASSIGN VAR FUNC_CALL VAR VAR VAR ASSIGN VAR FUNC_CALL VAR VAR NUMBER VAR FUNC_CALL VAR VAR VAR EXPR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR WHILE VAR NUMBER VAR NUMBER EXPR FUNC_CALL VAR |
You are given an array $a$ consisting of $n$ integers. Indices of the array start from zero (i. e. the first element is $a_0$, the second one is $a_1$, and so on).
You can reverse at most one subarray (continuous subsegment) of this array. Recall that the subarray of $a$ with borders $l$ and $r$ is $a[l; r] = a_l, a_{l + 1}, \dots, a_{r}$.
Your task is to reverse such a subarray that the sum of elements on even positions of the resulting array is maximized (i. e. the sum of elements $a_0, a_2, \dots, a_{2k}$ for integer $k = \lfloor\frac{n-1}{2}\rfloor$ should be maximum possible).
You have to answer $t$ independent test cases.
-----Input-----
The first line of the input contains one integer $t$ ($1 \le t \le 2 \cdot 10^4$) — the number of test cases. Then $t$ test cases follow.
The first line of the test case contains one integer $n$ ($1 \le n \le 2 \cdot 10^5$) — the length of $a$. The second line of the test case contains $n$ integers $a_0, a_1, \dots, a_{n-1}$ ($1 \le a_i \le 10^9$), where $a_i$ is the $i$-th element of $a$.
It is guaranteed that the sum of $n$ does not exceed $2 \cdot 10^5$ ($\sum n \le 2 \cdot 10^5$).
-----Output-----
For each test case, print the answer on the separate line — the maximum possible sum of elements on even positions after reversing at most one subarray (continuous subsegment) of $a$.
-----Example-----
Input
4
8
1 7 3 4 7 6 2 9
5
1 2 1 2 1
10
7 8 4 5 7 6 8 9 7 3
4
3 1 2 1
Output
26
5
37
5 | import sys
input = sys.stdin.readline
def max_sub_array_sum(a):
s = temp = 0
for i in a:
temp += i
if temp < 0:
temp = 0
s = max(temp, s)
return s
for _ in range(int(input())):
n = int(input())
arr = list(map(int, input().split()))
tot = sum([arr[i] for i in range(0, n, 2)])
even = [(arr[i + 1] - arr[i]) for i in range(0, n - 1, 2)]
odd = [(arr[i] - arr[i + 1]) for i in range(1, n - 1, 2)]
print(max(tot, tot + max_sub_array_sum(even), tot + max_sub_array_sum(odd))) | IMPORT ASSIGN VAR VAR FUNC_DEF ASSIGN VAR VAR NUMBER FOR VAR VAR VAR VAR IF VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR VAR RETURN VAR FOR VAR FUNC_CALL VAR FUNC_CALL VAR FUNC_CALL 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 FUNC_CALL VAR VAR VAR VAR FUNC_CALL VAR NUMBER VAR NUMBER ASSIGN VAR BIN_OP VAR BIN_OP VAR NUMBER VAR VAR VAR FUNC_CALL VAR NUMBER BIN_OP VAR NUMBER NUMBER ASSIGN VAR BIN_OP VAR VAR VAR BIN_OP VAR NUMBER VAR FUNC_CALL VAR NUMBER BIN_OP VAR NUMBER NUMBER EXPR FUNC_CALL VAR FUNC_CALL VAR VAR BIN_OP VAR FUNC_CALL VAR VAR BIN_OP VAR FUNC_CALL VAR VAR |
You are given an array $a$ consisting of $n$ integers. Indices of the array start from zero (i. e. the first element is $a_0$, the second one is $a_1$, and so on).
You can reverse at most one subarray (continuous subsegment) of this array. Recall that the subarray of $a$ with borders $l$ and $r$ is $a[l; r] = a_l, a_{l + 1}, \dots, a_{r}$.
Your task is to reverse such a subarray that the sum of elements on even positions of the resulting array is maximized (i. e. the sum of elements $a_0, a_2, \dots, a_{2k}$ for integer $k = \lfloor\frac{n-1}{2}\rfloor$ should be maximum possible).
You have to answer $t$ independent test cases.
-----Input-----
The first line of the input contains one integer $t$ ($1 \le t \le 2 \cdot 10^4$) — the number of test cases. Then $t$ test cases follow.
The first line of the test case contains one integer $n$ ($1 \le n \le 2 \cdot 10^5$) — the length of $a$. The second line of the test case contains $n$ integers $a_0, a_1, \dots, a_{n-1}$ ($1 \le a_i \le 10^9$), where $a_i$ is the $i$-th element of $a$.
It is guaranteed that the sum of $n$ does not exceed $2 \cdot 10^5$ ($\sum n \le 2 \cdot 10^5$).
-----Output-----
For each test case, print the answer on the separate line — the maximum possible sum of elements on even positions after reversing at most one subarray (continuous subsegment) of $a$.
-----Example-----
Input
4
8
1 7 3 4 7 6 2 9
5
1 2 1 2 1
10
7 8 4 5 7 6 8 9 7 3
4
3 1 2 1
Output
26
5
37
5 | def max_list(l):
if len(l) == 1:
return l[0]
elif len(l) <= 2:
return max(l[0], l[1])
evens = [0]
odds = [0]
for p in range(len(l)):
if p % 2 == 0:
evens.append(evens[-1] + l[p])
odds.append(odds[-1])
else:
odds.append(odds[-1] + l[p])
evens.append(evens[-1])
combined = [(evens[a] - odds[a]) for a in range(len(evens))]
min_combined_even = [(2 * ((len(combined) - 1) // 2)) for i in range(len(combined))]
min_combined_odd = [
(2 * (len(combined) // 2 - 1) + 1) for i in range(len(combined))
]
for i in range(len(combined) - 2, -1, -1):
if i % 2 == 0:
if combined[i] <= combined[min_combined_even[i + 1]]:
min_combined_even[i] = i
else:
min_combined_even[i] = min_combined_even[i + 1]
min_combined_odd[i] = min_combined_odd[i + 1]
else:
if combined[i] <= combined[min_combined_odd[i + 1]]:
min_combined_odd[i] = i
else:
min_combined_odd[i] = min_combined_odd[i + 1]
min_combined_even[i] = min_combined_even[i + 1]
max_so_far = evens[-1]
for i in range(len(combined)):
if i % 2 == 0:
max_so_far = max(
max_so_far, combined[i] - combined[min_combined_even[i]] + evens[-1]
)
else:
max_so_far = max(
max_so_far, combined[i] - combined[min_combined_odd[i]] + evens[-1]
)
return max_so_far
t = int(input())
for i in range(t):
n = int(input())
print(max_list(list(map(int, input().split())))) | FUNC_DEF IF FUNC_CALL VAR VAR NUMBER RETURN VAR NUMBER IF FUNC_CALL VAR VAR NUMBER RETURN FUNC_CALL VAR VAR NUMBER VAR NUMBER ASSIGN VAR LIST NUMBER ASSIGN VAR LIST NUMBER FOR VAR FUNC_CALL VAR FUNC_CALL VAR VAR IF BIN_OP VAR NUMBER NUMBER EXPR FUNC_CALL VAR BIN_OP VAR NUMBER VAR VAR EXPR FUNC_CALL VAR VAR NUMBER EXPR FUNC_CALL VAR BIN_OP VAR NUMBER VAR VAR EXPR FUNC_CALL VAR VAR NUMBER ASSIGN VAR BIN_OP VAR VAR VAR VAR VAR FUNC_CALL VAR FUNC_CALL VAR VAR ASSIGN VAR BIN_OP NUMBER BIN_OP BIN_OP FUNC_CALL VAR VAR NUMBER NUMBER VAR FUNC_CALL VAR FUNC_CALL VAR VAR ASSIGN VAR BIN_OP BIN_OP NUMBER BIN_OP BIN_OP FUNC_CALL VAR VAR NUMBER NUMBER NUMBER VAR FUNC_CALL VAR FUNC_CALL VAR VAR FOR VAR FUNC_CALL VAR BIN_OP FUNC_CALL VAR VAR NUMBER NUMBER NUMBER IF BIN_OP VAR NUMBER NUMBER IF VAR VAR VAR VAR BIN_OP VAR NUMBER ASSIGN VAR VAR VAR ASSIGN VAR VAR VAR BIN_OP VAR NUMBER ASSIGN VAR VAR VAR BIN_OP VAR NUMBER IF VAR VAR VAR VAR BIN_OP VAR NUMBER ASSIGN VAR VAR VAR ASSIGN VAR VAR VAR BIN_OP VAR NUMBER ASSIGN VAR VAR VAR BIN_OP VAR NUMBER ASSIGN VAR VAR NUMBER FOR VAR FUNC_CALL VAR FUNC_CALL VAR VAR IF BIN_OP VAR NUMBER NUMBER ASSIGN VAR FUNC_CALL VAR VAR BIN_OP BIN_OP VAR VAR VAR VAR VAR VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR BIN_OP BIN_OP VAR VAR VAR VAR VAR VAR NUMBER RETURN VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR FOR VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR EXPR FUNC_CALL VAR FUNC_CALL VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR |
You are given an array $a$ consisting of $n$ integers. Indices of the array start from zero (i. e. the first element is $a_0$, the second one is $a_1$, and so on).
You can reverse at most one subarray (continuous subsegment) of this array. Recall that the subarray of $a$ with borders $l$ and $r$ is $a[l; r] = a_l, a_{l + 1}, \dots, a_{r}$.
Your task is to reverse such a subarray that the sum of elements on even positions of the resulting array is maximized (i. e. the sum of elements $a_0, a_2, \dots, a_{2k}$ for integer $k = \lfloor\frac{n-1}{2}\rfloor$ should be maximum possible).
You have to answer $t$ independent test cases.
-----Input-----
The first line of the input contains one integer $t$ ($1 \le t \le 2 \cdot 10^4$) — the number of test cases. Then $t$ test cases follow.
The first line of the test case contains one integer $n$ ($1 \le n \le 2 \cdot 10^5$) — the length of $a$. The second line of the test case contains $n$ integers $a_0, a_1, \dots, a_{n-1}$ ($1 \le a_i \le 10^9$), where $a_i$ is the $i$-th element of $a$.
It is guaranteed that the sum of $n$ does not exceed $2 \cdot 10^5$ ($\sum n \le 2 \cdot 10^5$).
-----Output-----
For each test case, print the answer on the separate line — the maximum possible sum of elements on even positions after reversing at most one subarray (continuous subsegment) of $a$.
-----Example-----
Input
4
8
1 7 3 4 7 6 2 9
5
1 2 1 2 1
10
7 8 4 5 7 6 8 9 7 3
4
3 1 2 1
Output
26
5
37
5 | def msa(a):
if a == []:
return 0
ans, cur = 0, 0
for x in a:
if x < 0:
ans = max(ans, cur)
cur += x
if cur < 0:
ans = max(ans, cur - x)
cur = 0
ans = max(cur, ans)
return ans if ans > 0 else max(a)
for _ in range(int(input())):
n = int(input())
a = list(map(int, input().split()))
a1, a2 = [], []
initsum = 0
for i in range(0, n, 2):
initsum += a[i]
if i + 1 >= n:
break
a1.append(a[i + 1] - a[i])
for i in range(1, n, 2):
if i + 1 >= n:
break
a2.append(a[i] - a[i + 1])
addition = max(msa(a1), msa(a2))
print(initsum + (addition if addition > 0 else 0)) | FUNC_DEF IF VAR LIST RETURN NUMBER ASSIGN VAR VAR NUMBER NUMBER FOR VAR VAR IF VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR VAR VAR VAR IF VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR BIN_OP VAR VAR ASSIGN VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR VAR RETURN VAR NUMBER VAR FUNC_CALL VAR VAR FOR VAR FUNC_CALL VAR FUNC_CALL VAR FUNC_CALL 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 VAR LIST LIST ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR NUMBER VAR NUMBER VAR VAR VAR IF BIN_OP VAR NUMBER VAR EXPR FUNC_CALL VAR BIN_OP VAR BIN_OP VAR NUMBER VAR VAR FOR VAR FUNC_CALL VAR NUMBER VAR NUMBER IF BIN_OP VAR NUMBER VAR EXPR FUNC_CALL VAR BIN_OP VAR VAR VAR BIN_OP VAR NUMBER ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR BIN_OP VAR VAR NUMBER VAR NUMBER |
You are given an array $a$ consisting of $n$ integers. Indices of the array start from zero (i. e. the first element is $a_0$, the second one is $a_1$, and so on).
You can reverse at most one subarray (continuous subsegment) of this array. Recall that the subarray of $a$ with borders $l$ and $r$ is $a[l; r] = a_l, a_{l + 1}, \dots, a_{r}$.
Your task is to reverse such a subarray that the sum of elements on even positions of the resulting array is maximized (i. e. the sum of elements $a_0, a_2, \dots, a_{2k}$ for integer $k = \lfloor\frac{n-1}{2}\rfloor$ should be maximum possible).
You have to answer $t$ independent test cases.
-----Input-----
The first line of the input contains one integer $t$ ($1 \le t \le 2 \cdot 10^4$) — the number of test cases. Then $t$ test cases follow.
The first line of the test case contains one integer $n$ ($1 \le n \le 2 \cdot 10^5$) — the length of $a$. The second line of the test case contains $n$ integers $a_0, a_1, \dots, a_{n-1}$ ($1 \le a_i \le 10^9$), where $a_i$ is the $i$-th element of $a$.
It is guaranteed that the sum of $n$ does not exceed $2 \cdot 10^5$ ($\sum n \le 2 \cdot 10^5$).
-----Output-----
For each test case, print the answer on the separate line — the maximum possible sum of elements on even positions after reversing at most one subarray (continuous subsegment) of $a$.
-----Example-----
Input
4
8
1 7 3 4 7 6 2 9
5
1 2 1 2 1
10
7 8 4 5 7 6 8 9 7 3
4
3 1 2 1
Output
26
5
37
5 | def inp():
return int(input())
def inlt():
return list(map(int, input().split()))
def insr():
s = input()
return list(s[: len(s)])
def invr():
return map(int, input().split())
entries = inp()
for i in range(entries):
n = inp()
l = inlt()
if len(l) == 1 or len(l) == 2:
print(max(l))
else:
two_string = []
i = 0
while i < len(l):
if i + 1 < len(l):
two_string.append(l[i + 1] - l[i])
i += 2
cur = -10000000
max_val = -100000000
cur_ind_start = 0
ind_start = 0
ind_end = 0
for i in range(len(two_string)):
x = two_string[i]
if x > cur + x:
cur_ind_start = i
cur = x
else:
cur += x
if cur > max_val:
ind_start = cur_ind_start
ind_end = i
max_val = cur
two_string2 = []
i = 1
while i < len(l):
if i + 1 < len(l):
two_string2.append(l[i] - l[i + 1])
i += 2
cur2 = -100000000
max_val2 = -1000000000
cur_ind_start2 = 0
ind_start2 = -1
ind_end2 = -1
for i in range(len(two_string2)):
x = two_string2[i]
if x > cur2 + x:
cur_ind_start2 = i
cur2 = x
else:
cur2 += x
if cur2 > max_val2:
ind_start2 = cur_ind_start2
ind_end2 = i
max_val2 = cur2
ind_start *= 2
ind_end *= 2
ind_start2 *= 2
ind_end2 *= 2
res = 0
k = 0
if max_val <= 0 and max_val2 <= 0:
while k < len(l):
res += l[k]
k += 2
elif max_val > max_val2:
while k < len(l):
if k >= ind_start and k <= ind_end:
res += l[k + 1]
else:
res += l[k]
k += 2
else:
k = 0
while k < len(l):
if k >= ind_start2 + 2 and k <= ind_end2 + 2:
res += l[k - 1]
else:
res += l[k]
k += 2
print(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 FUNC_CALL VAR VAR FUNC_DEF RETURN FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FOR VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR IF FUNC_CALL VAR VAR NUMBER FUNC_CALL VAR VAR NUMBER EXPR FUNC_CALL VAR FUNC_CALL VAR VAR ASSIGN VAR LIST ASSIGN VAR NUMBER WHILE VAR FUNC_CALL VAR VAR IF BIN_OP VAR NUMBER FUNC_CALL VAR VAR EXPR FUNC_CALL VAR BIN_OP VAR BIN_OP VAR NUMBER VAR VAR 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 ASSIGN VAR VAR VAR IF VAR BIN_OP VAR VAR ASSIGN VAR VAR ASSIGN VAR VAR VAR VAR IF VAR VAR ASSIGN VAR VAR ASSIGN VAR VAR ASSIGN VAR VAR ASSIGN VAR LIST ASSIGN VAR NUMBER WHILE VAR FUNC_CALL VAR VAR IF BIN_OP VAR NUMBER FUNC_CALL VAR VAR EXPR FUNC_CALL VAR BIN_OP VAR VAR VAR BIN_OP VAR NUMBER 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 ASSIGN VAR VAR VAR IF VAR BIN_OP VAR VAR ASSIGN VAR VAR ASSIGN VAR VAR VAR VAR IF VAR VAR ASSIGN VAR VAR ASSIGN VAR VAR ASSIGN VAR VAR VAR NUMBER VAR NUMBER VAR NUMBER VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR NUMBER IF VAR NUMBER VAR NUMBER WHILE VAR FUNC_CALL VAR VAR VAR VAR VAR VAR NUMBER IF VAR VAR WHILE VAR FUNC_CALL VAR VAR IF VAR VAR VAR VAR VAR VAR BIN_OP VAR NUMBER VAR VAR VAR VAR NUMBER ASSIGN VAR NUMBER WHILE VAR FUNC_CALL VAR VAR IF VAR BIN_OP VAR NUMBER VAR BIN_OP VAR NUMBER VAR VAR BIN_OP VAR NUMBER VAR VAR VAR VAR NUMBER EXPR FUNC_CALL VAR VAR |
You are given an array $a$ consisting of $n$ integers. Indices of the array start from zero (i. e. the first element is $a_0$, the second one is $a_1$, and so on).
You can reverse at most one subarray (continuous subsegment) of this array. Recall that the subarray of $a$ with borders $l$ and $r$ is $a[l; r] = a_l, a_{l + 1}, \dots, a_{r}$.
Your task is to reverse such a subarray that the sum of elements on even positions of the resulting array is maximized (i. e. the sum of elements $a_0, a_2, \dots, a_{2k}$ for integer $k = \lfloor\frac{n-1}{2}\rfloor$ should be maximum possible).
You have to answer $t$ independent test cases.
-----Input-----
The first line of the input contains one integer $t$ ($1 \le t \le 2 \cdot 10^4$) — the number of test cases. Then $t$ test cases follow.
The first line of the test case contains one integer $n$ ($1 \le n \le 2 \cdot 10^5$) — the length of $a$. The second line of the test case contains $n$ integers $a_0, a_1, \dots, a_{n-1}$ ($1 \le a_i \le 10^9$), where $a_i$ is the $i$-th element of $a$.
It is guaranteed that the sum of $n$ does not exceed $2 \cdot 10^5$ ($\sum n \le 2 \cdot 10^5$).
-----Output-----
For each test case, print the answer on the separate line — the maximum possible sum of elements on even positions after reversing at most one subarray (continuous subsegment) of $a$.
-----Example-----
Input
4
8
1 7 3 4 7 6 2 9
5
1 2 1 2 1
10
7 8 4 5 7 6 8 9 7 3
4
3 1 2 1
Output
26
5
37
5 | q = int(input())
for i in range(q):
n = int(input())
lst = [int(x) for x in input().split()]
if n == 1:
print(lst[0])
continue
se = so = mx = mn = 0
k = 1
while k < n:
if se >= 0:
se += lst[k] - lst[k - 1]
else:
se = lst[k] - lst[k - 1]
k += 2
mx = max(mx, se)
k = 2
while k < n:
if so <= 0:
so += lst[k] - lst[k - 1]
else:
so = lst[k] - lst[k - 1]
mn = min(mn, so)
k += 2
ret = 0
k = 0
mx = max(abs(mn), mx)
while k < n:
ret += lst[k]
k += 2
if mx > 0:
print(ret + mx)
else:
print(ret) | ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR 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 IF VAR NUMBER EXPR FUNC_CALL VAR VAR NUMBER ASSIGN VAR VAR VAR VAR NUMBER ASSIGN VAR NUMBER WHILE VAR VAR IF VAR NUMBER VAR BIN_OP VAR VAR VAR BIN_OP VAR NUMBER ASSIGN VAR BIN_OP VAR VAR VAR BIN_OP VAR NUMBER VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR VAR ASSIGN VAR NUMBER WHILE VAR VAR IF VAR NUMBER VAR BIN_OP VAR VAR VAR BIN_OP VAR NUMBER ASSIGN VAR BIN_OP VAR VAR VAR BIN_OP VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR VAR VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR VAR VAR WHILE VAR VAR VAR VAR VAR VAR NUMBER IF VAR NUMBER EXPR FUNC_CALL VAR BIN_OP VAR VAR EXPR FUNC_CALL VAR VAR |
You are given an array $a$ consisting of $n$ integers. Indices of the array start from zero (i. e. the first element is $a_0$, the second one is $a_1$, and so on).
You can reverse at most one subarray (continuous subsegment) of this array. Recall that the subarray of $a$ with borders $l$ and $r$ is $a[l; r] = a_l, a_{l + 1}, \dots, a_{r}$.
Your task is to reverse such a subarray that the sum of elements on even positions of the resulting array is maximized (i. e. the sum of elements $a_0, a_2, \dots, a_{2k}$ for integer $k = \lfloor\frac{n-1}{2}\rfloor$ should be maximum possible).
You have to answer $t$ independent test cases.
-----Input-----
The first line of the input contains one integer $t$ ($1 \le t \le 2 \cdot 10^4$) — the number of test cases. Then $t$ test cases follow.
The first line of the test case contains one integer $n$ ($1 \le n \le 2 \cdot 10^5$) — the length of $a$. The second line of the test case contains $n$ integers $a_0, a_1, \dots, a_{n-1}$ ($1 \le a_i \le 10^9$), where $a_i$ is the $i$-th element of $a$.
It is guaranteed that the sum of $n$ does not exceed $2 \cdot 10^5$ ($\sum n \le 2 \cdot 10^5$).
-----Output-----
For each test case, print the answer on the separate line — the maximum possible sum of elements on even positions after reversing at most one subarray (continuous subsegment) of $a$.
-----Example-----
Input
4
8
1 7 3 4 7 6 2 9
5
1 2 1 2 1
10
7 8 4 5 7 6 8 9 7 3
4
3 1 2 1
Output
26
5
37
5 | import sys
input = lambda: sys.stdin.readline().rstrip()
t = int(input())
v = []
for i in range(t):
n = int(input())
x = list(map(int, input().split()))
v.append(x)
inf = -(10**18)
for i in range(t):
dpx = [inf] * len(v[i])
dpy = [inf] * len(v[i])
ans = 0
for j in range(len(v[i]) - 1):
if j % 2 == 0:
cnt = v[i][j + 1] - v[i][j]
dpx[j // 2 + 1] = max(dpx[j // 2] + cnt, cnt)
else:
cnt = -v[i][j + 1] + v[i][j]
dpy[j // 2 + 1] = max(dpy[j // 2] + cnt, cnt)
for j in range(len(v[i])):
if j % 2 == 0:
ans += v[i][j]
ans += max(max(dpx), max(dpy), 0)
print(ans) | IMPORT ASSIGN VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR LIST FOR VAR FUNC_CALL 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 ASSIGN VAR BIN_OP NUMBER NUMBER FOR VAR FUNC_CALL VAR VAR ASSIGN VAR BIN_OP LIST VAR FUNC_CALL VAR VAR VAR ASSIGN VAR BIN_OP LIST VAR FUNC_CALL VAR VAR VAR ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR BIN_OP FUNC_CALL VAR VAR VAR NUMBER IF BIN_OP VAR NUMBER NUMBER ASSIGN VAR BIN_OP VAR VAR BIN_OP VAR NUMBER VAR VAR VAR ASSIGN VAR BIN_OP BIN_OP VAR NUMBER NUMBER FUNC_CALL VAR BIN_OP VAR BIN_OP VAR NUMBER VAR VAR ASSIGN VAR BIN_OP VAR VAR BIN_OP VAR NUMBER VAR VAR VAR ASSIGN VAR BIN_OP BIN_OP VAR NUMBER NUMBER FUNC_CALL VAR BIN_OP VAR BIN_OP VAR NUMBER VAR VAR FOR VAR FUNC_CALL VAR FUNC_CALL VAR VAR VAR IF BIN_OP VAR NUMBER NUMBER VAR VAR VAR VAR VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL VAR VAR NUMBER EXPR FUNC_CALL VAR VAR |
You are given an array $a$ consisting of $n$ integers. Indices of the array start from zero (i. e. the first element is $a_0$, the second one is $a_1$, and so on).
You can reverse at most one subarray (continuous subsegment) of this array. Recall that the subarray of $a$ with borders $l$ and $r$ is $a[l; r] = a_l, a_{l + 1}, \dots, a_{r}$.
Your task is to reverse such a subarray that the sum of elements on even positions of the resulting array is maximized (i. e. the sum of elements $a_0, a_2, \dots, a_{2k}$ for integer $k = \lfloor\frac{n-1}{2}\rfloor$ should be maximum possible).
You have to answer $t$ independent test cases.
-----Input-----
The first line of the input contains one integer $t$ ($1 \le t \le 2 \cdot 10^4$) — the number of test cases. Then $t$ test cases follow.
The first line of the test case contains one integer $n$ ($1 \le n \le 2 \cdot 10^5$) — the length of $a$. The second line of the test case contains $n$ integers $a_0, a_1, \dots, a_{n-1}$ ($1 \le a_i \le 10^9$), where $a_i$ is the $i$-th element of $a$.
It is guaranteed that the sum of $n$ does not exceed $2 \cdot 10^5$ ($\sum n \le 2 \cdot 10^5$).
-----Output-----
For each test case, print the answer on the separate line — the maximum possible sum of elements on even positions after reversing at most one subarray (continuous subsegment) of $a$.
-----Example-----
Input
4
8
1 7 3 4 7 6 2 9
5
1 2 1 2 1
10
7 8 4 5 7 6 8 9 7 3
4
3 1 2 1
Output
26
5
37
5 | t = int(input())
for _ in range(t):
n = int(input())
a = list(map(int, input().split()))
ans = total_even = sum(a[::2])
min_diff = diff = 0
for i in range(0, n - 1, 2):
diff += a[i + 1] - a[i]
ans = max(ans, total_even + diff - min_diff)
min_diff = min(min_diff, diff)
min_diff = diff = 0
for i in range(1, n - 1, 2):
diff += a[i] - a[i + 1]
ans = max(ans, total_even + diff - min_diff)
min_diff = min(min_diff, diff)
print(ans) | ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR FOR VAR FUNC_CALL 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 VAR FUNC_CALL VAR VAR NUMBER ASSIGN VAR VAR NUMBER FOR VAR FUNC_CALL VAR NUMBER BIN_OP VAR NUMBER NUMBER VAR BIN_OP VAR BIN_OP VAR NUMBER VAR VAR ASSIGN VAR FUNC_CALL VAR VAR BIN_OP BIN_OP VAR VAR VAR ASSIGN VAR FUNC_CALL VAR VAR VAR ASSIGN VAR VAR NUMBER FOR VAR FUNC_CALL VAR NUMBER BIN_OP VAR NUMBER NUMBER VAR BIN_OP VAR VAR VAR BIN_OP VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR BIN_OP BIN_OP VAR VAR VAR ASSIGN VAR FUNC_CALL VAR VAR VAR EXPR FUNC_CALL VAR VAR |
You are given an array $a$ consisting of $n$ integers. Indices of the array start from zero (i. e. the first element is $a_0$, the second one is $a_1$, and so on).
You can reverse at most one subarray (continuous subsegment) of this array. Recall that the subarray of $a$ with borders $l$ and $r$ is $a[l; r] = a_l, a_{l + 1}, \dots, a_{r}$.
Your task is to reverse such a subarray that the sum of elements on even positions of the resulting array is maximized (i. e. the sum of elements $a_0, a_2, \dots, a_{2k}$ for integer $k = \lfloor\frac{n-1}{2}\rfloor$ should be maximum possible).
You have to answer $t$ independent test cases.
-----Input-----
The first line of the input contains one integer $t$ ($1 \le t \le 2 \cdot 10^4$) — the number of test cases. Then $t$ test cases follow.
The first line of the test case contains one integer $n$ ($1 \le n \le 2 \cdot 10^5$) — the length of $a$. The second line of the test case contains $n$ integers $a_0, a_1, \dots, a_{n-1}$ ($1 \le a_i \le 10^9$), where $a_i$ is the $i$-th element of $a$.
It is guaranteed that the sum of $n$ does not exceed $2 \cdot 10^5$ ($\sum n \le 2 \cdot 10^5$).
-----Output-----
For each test case, print the answer on the separate line — the maximum possible sum of elements on even positions after reversing at most one subarray (continuous subsegment) of $a$.
-----Example-----
Input
4
8
1 7 3 4 7 6 2 9
5
1 2 1 2 1
10
7 8 4 5 7 6 8 9 7 3
4
3 1 2 1
Output
26
5
37
5 | def maxi_sum_sub(list1):
maxi = 0
sum1 = 0
i = 0
while i < len(list1):
if list1[i] >= 0:
while i < len(list1) and list1[i] >= 0:
sum1 += list1[i]
i += 1
maxi = max(maxi, sum1)
else:
while i < len(list1) and list1[i] < 0 and sum1 >= 0:
sum1 += list1[i]
i += 1
if sum1 <= 0:
sum1 = 0
return maxi
t = int(input())
while t != 0:
n = int(input())
list1 = list(map(int, input().split()))
evs = 0
for i in range(0, n, 2):
if i >= n:
break
evs += list1[i]
list2 = list()
list3 = list()
for i in range(1, n, 2):
if i >= n:
break
list2.append(list1[i] - list1[i - 1])
for i in range(1, n, 2):
if i >= n - 1:
break
list3.append(list1[i] - list1[i + 1])
p = maxi_sum_sub(list2)
q = maxi_sum_sub(list3)
print(max(evs, evs + p, evs + q))
t -= 1 | FUNC_DEF ASSIGN VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR NUMBER WHILE VAR FUNC_CALL VAR VAR IF VAR VAR NUMBER WHILE VAR FUNC_CALL VAR VAR VAR VAR NUMBER VAR VAR VAR VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR VAR WHILE VAR FUNC_CALL VAR VAR VAR VAR NUMBER VAR NUMBER VAR VAR VAR VAR NUMBER IF VAR NUMBER ASSIGN VAR NUMBER RETURN VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR WHILE VAR NUMBER 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 FOR VAR FUNC_CALL VAR NUMBER VAR NUMBER IF VAR VAR VAR VAR VAR ASSIGN VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FOR VAR FUNC_CALL VAR NUMBER VAR NUMBER IF VAR VAR EXPR FUNC_CALL VAR BIN_OP VAR VAR VAR BIN_OP VAR NUMBER FOR VAR FUNC_CALL VAR NUMBER VAR NUMBER IF VAR BIN_OP VAR NUMBER EXPR FUNC_CALL VAR BIN_OP VAR VAR VAR BIN_OP VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR FUNC_CALL VAR VAR BIN_OP VAR VAR BIN_OP VAR VAR VAR NUMBER |
You are given an array $a$ consisting of $n$ integers. Indices of the array start from zero (i. e. the first element is $a_0$, the second one is $a_1$, and so on).
You can reverse at most one subarray (continuous subsegment) of this array. Recall that the subarray of $a$ with borders $l$ and $r$ is $a[l; r] = a_l, a_{l + 1}, \dots, a_{r}$.
Your task is to reverse such a subarray that the sum of elements on even positions of the resulting array is maximized (i. e. the sum of elements $a_0, a_2, \dots, a_{2k}$ for integer $k = \lfloor\frac{n-1}{2}\rfloor$ should be maximum possible).
You have to answer $t$ independent test cases.
-----Input-----
The first line of the input contains one integer $t$ ($1 \le t \le 2 \cdot 10^4$) — the number of test cases. Then $t$ test cases follow.
The first line of the test case contains one integer $n$ ($1 \le n \le 2 \cdot 10^5$) — the length of $a$. The second line of the test case contains $n$ integers $a_0, a_1, \dots, a_{n-1}$ ($1 \le a_i \le 10^9$), where $a_i$ is the $i$-th element of $a$.
It is guaranteed that the sum of $n$ does not exceed $2 \cdot 10^5$ ($\sum n \le 2 \cdot 10^5$).
-----Output-----
For each test case, print the answer on the separate line — the maximum possible sum of elements on even positions after reversing at most one subarray (continuous subsegment) of $a$.
-----Example-----
Input
4
8
1 7 3 4 7 6 2 9
5
1 2 1 2 1
10
7 8 4 5 7 6 8 9 7 3
4
3 1 2 1
Output
26
5
37
5 | line = input()
t = int(line)
for _ in range(t):
line = input()
n = int(line)
line = input()
nums = [int(i) for i in line.split(" ")]
evensum, oddsum = [0] * (n + 1), [0] * (n + 1)
for i in range(n):
evensum[i + 1] = evensum[i] + (nums[i] if i % 2 == 0 else 0)
oddsum[i + 1] = oddsum[i] + (nums[i] if i % 2 == 1 else 0)
res = evensum[-1]
minD = 9999999999
for i in range(0, n + 1, 2):
res = max(res, evensum[-1] + (oddsum[i] - evensum[i]) - minD)
minD = min(minD, oddsum[i] - evensum[i])
minD = 9999999999
for i in range(1, n + 1, 2):
res = max(res, evensum[-1] + (oddsum[i] - evensum[i]) - minD)
minD = min(minD, oddsum[i] - evensum[i])
print(res) | ASSIGN VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR VAR FOR VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR VAR VAR FUNC_CALL VAR STRING ASSIGN VAR VAR BIN_OP LIST NUMBER BIN_OP VAR NUMBER BIN_OP LIST NUMBER BIN_OP VAR NUMBER FOR VAR FUNC_CALL VAR VAR ASSIGN VAR BIN_OP VAR NUMBER BIN_OP VAR VAR BIN_OP VAR NUMBER NUMBER VAR VAR NUMBER ASSIGN VAR BIN_OP VAR NUMBER BIN_OP VAR VAR BIN_OP VAR NUMBER NUMBER VAR VAR NUMBER ASSIGN VAR VAR NUMBER ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR NUMBER BIN_OP VAR NUMBER NUMBER ASSIGN VAR FUNC_CALL VAR VAR BIN_OP BIN_OP VAR NUMBER BIN_OP VAR VAR VAR VAR VAR ASSIGN VAR FUNC_CALL VAR VAR BIN_OP VAR VAR VAR VAR ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR NUMBER BIN_OP VAR NUMBER NUMBER ASSIGN VAR FUNC_CALL VAR VAR BIN_OP BIN_OP VAR NUMBER BIN_OP VAR VAR VAR VAR VAR ASSIGN VAR FUNC_CALL VAR VAR BIN_OP VAR VAR VAR VAR EXPR FUNC_CALL VAR VAR |
You are given an array $a$ consisting of $n$ integers. Indices of the array start from zero (i. e. the first element is $a_0$, the second one is $a_1$, and so on).
You can reverse at most one subarray (continuous subsegment) of this array. Recall that the subarray of $a$ with borders $l$ and $r$ is $a[l; r] = a_l, a_{l + 1}, \dots, a_{r}$.
Your task is to reverse such a subarray that the sum of elements on even positions of the resulting array is maximized (i. e. the sum of elements $a_0, a_2, \dots, a_{2k}$ for integer $k = \lfloor\frac{n-1}{2}\rfloor$ should be maximum possible).
You have to answer $t$ independent test cases.
-----Input-----
The first line of the input contains one integer $t$ ($1 \le t \le 2 \cdot 10^4$) — the number of test cases. Then $t$ test cases follow.
The first line of the test case contains one integer $n$ ($1 \le n \le 2 \cdot 10^5$) — the length of $a$. The second line of the test case contains $n$ integers $a_0, a_1, \dots, a_{n-1}$ ($1 \le a_i \le 10^9$), where $a_i$ is the $i$-th element of $a$.
It is guaranteed that the sum of $n$ does not exceed $2 \cdot 10^5$ ($\sum n \le 2 \cdot 10^5$).
-----Output-----
For each test case, print the answer on the separate line — the maximum possible sum of elements on even positions after reversing at most one subarray (continuous subsegment) of $a$.
-----Example-----
Input
4
8
1 7 3 4 7 6 2 9
5
1 2 1 2 1
10
7 8 4 5 7 6 8 9 7 3
4
3 1 2 1
Output
26
5
37
5 | def mii():
return map(int, input().split())
def maxSubArraySum(a, size):
max_so_far = -10000000000.0 - 1
max_ending_here = 0
for i in range(0, size):
max_ending_here = max_ending_here + a[i]
if max_so_far < max_ending_here:
max_so_far = max_ending_here
if max_ending_here < 0:
max_ending_here = 0
return max_so_far
for _ in range(int(input())):
n = int(input())
a = list(mii())
lol = []
lol1 = []
s = 0
for i in range(1, n, 2):
s += a[i - 1]
lol.append(a[i] - a[i - 1])
if i > 1:
lol1.append(a[i - 2] - a[i - 1])
if n % 2:
s += a[-1]
if n > 1:
lol1.append(a[-2] - a[-1])
print(
s
+ max(
max(0, maxSubArraySum(lol, len(lol))),
max(0, maxSubArraySum(lol1, len(lol1))),
)
) | FUNC_DEF RETURN FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR FUNC_DEF ASSIGN VAR BIN_OP NUMBER NUMBER ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR NUMBER VAR ASSIGN VAR BIN_OP VAR VAR VAR IF VAR VAR ASSIGN VAR VAR IF VAR NUMBER ASSIGN VAR NUMBER RETURN VAR FOR VAR FUNC_CALL VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR LIST ASSIGN VAR LIST ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR NUMBER VAR NUMBER VAR VAR BIN_OP VAR NUMBER EXPR FUNC_CALL VAR BIN_OP VAR VAR VAR BIN_OP VAR NUMBER IF VAR NUMBER EXPR FUNC_CALL VAR BIN_OP VAR BIN_OP VAR NUMBER VAR BIN_OP VAR NUMBER IF BIN_OP VAR NUMBER VAR VAR NUMBER IF VAR NUMBER EXPR FUNC_CALL VAR BIN_OP VAR NUMBER VAR NUMBER EXPR FUNC_CALL VAR BIN_OP VAR FUNC_CALL VAR FUNC_CALL VAR NUMBER FUNC_CALL VAR VAR FUNC_CALL VAR VAR FUNC_CALL VAR NUMBER FUNC_CALL VAR VAR FUNC_CALL VAR VAR |
You are given an array $a$ consisting of $n$ integers. Indices of the array start from zero (i. e. the first element is $a_0$, the second one is $a_1$, and so on).
You can reverse at most one subarray (continuous subsegment) of this array. Recall that the subarray of $a$ with borders $l$ and $r$ is $a[l; r] = a_l, a_{l + 1}, \dots, a_{r}$.
Your task is to reverse such a subarray that the sum of elements on even positions of the resulting array is maximized (i. e. the sum of elements $a_0, a_2, \dots, a_{2k}$ for integer $k = \lfloor\frac{n-1}{2}\rfloor$ should be maximum possible).
You have to answer $t$ independent test cases.
-----Input-----
The first line of the input contains one integer $t$ ($1 \le t \le 2 \cdot 10^4$) — the number of test cases. Then $t$ test cases follow.
The first line of the test case contains one integer $n$ ($1 \le n \le 2 \cdot 10^5$) — the length of $a$. The second line of the test case contains $n$ integers $a_0, a_1, \dots, a_{n-1}$ ($1 \le a_i \le 10^9$), where $a_i$ is the $i$-th element of $a$.
It is guaranteed that the sum of $n$ does not exceed $2 \cdot 10^5$ ($\sum n \le 2 \cdot 10^5$).
-----Output-----
For each test case, print the answer on the separate line — the maximum possible sum of elements on even positions after reversing at most one subarray (continuous subsegment) of $a$.
-----Example-----
Input
4
8
1 7 3 4 7 6 2 9
5
1 2 1 2 1
10
7 8 4 5 7 6 8 9 7 3
4
3 1 2 1
Output
26
5
37
5 | import sys
input = sys.stdin.readline
def kadane(l):
max_so_far = 0
max_ending_here = 0
for i in range(len(l)):
max_ending_here = max_ending_here + l[i]
if max_ending_here < 0:
max_ending_here = 0
elif max_so_far < max_ending_here:
max_so_far = max_ending_here
return max_so_far
t = int(input())
for ii in range(t):
n = int(input())
a = [int(i) for i in input().split()]
if n == 1:
print(a[0])
elif n == 2:
print(max(a[1], a[0]))
else:
evensum = 0
for i in range(0, n, 2):
evensum += a[i]
freq1, freq2 = [], []
freq1.append(max(0, a[1] - a[0]))
freq2.append(max(0, a[1] - a[2]))
for i in range(3, n, 2):
freq1.append(a[i] - a[i - 1])
if not n & 1:
n = n - 1
for i in range(3, n, 2):
freq2.append(a[i] - a[i + 1])
p, q = kadane(freq1), kadane(freq2)
print(evensum + max(p, q)) | IMPORT ASSIGN VAR VAR FUNC_DEF ASSIGN VAR NUMBER ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR FUNC_CALL VAR VAR ASSIGN VAR BIN_OP VAR VAR VAR IF VAR NUMBER ASSIGN VAR NUMBER IF VAR VAR ASSIGN VAR VAR RETURN VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR 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 IF VAR NUMBER EXPR FUNC_CALL VAR VAR NUMBER IF VAR NUMBER EXPR FUNC_CALL VAR FUNC_CALL VAR VAR NUMBER VAR NUMBER ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR NUMBER VAR NUMBER VAR VAR VAR ASSIGN VAR VAR LIST LIST EXPR FUNC_CALL VAR FUNC_CALL VAR NUMBER BIN_OP VAR NUMBER VAR NUMBER EXPR FUNC_CALL VAR FUNC_CALL VAR NUMBER BIN_OP VAR NUMBER VAR NUMBER FOR VAR FUNC_CALL VAR NUMBER VAR NUMBER EXPR FUNC_CALL VAR BIN_OP VAR VAR VAR BIN_OP VAR NUMBER IF BIN_OP VAR NUMBER ASSIGN VAR BIN_OP VAR NUMBER FOR VAR FUNC_CALL VAR NUMBER VAR NUMBER EXPR FUNC_CALL VAR BIN_OP VAR VAR VAR BIN_OP VAR NUMBER ASSIGN VAR VAR FUNC_CALL VAR VAR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR BIN_OP VAR FUNC_CALL VAR VAR VAR |
You are given an array $a$ consisting of $n$ integers. Indices of the array start from zero (i. e. the first element is $a_0$, the second one is $a_1$, and so on).
You can reverse at most one subarray (continuous subsegment) of this array. Recall that the subarray of $a$ with borders $l$ and $r$ is $a[l; r] = a_l, a_{l + 1}, \dots, a_{r}$.
Your task is to reverse such a subarray that the sum of elements on even positions of the resulting array is maximized (i. e. the sum of elements $a_0, a_2, \dots, a_{2k}$ for integer $k = \lfloor\frac{n-1}{2}\rfloor$ should be maximum possible).
You have to answer $t$ independent test cases.
-----Input-----
The first line of the input contains one integer $t$ ($1 \le t \le 2 \cdot 10^4$) — the number of test cases. Then $t$ test cases follow.
The first line of the test case contains one integer $n$ ($1 \le n \le 2 \cdot 10^5$) — the length of $a$. The second line of the test case contains $n$ integers $a_0, a_1, \dots, a_{n-1}$ ($1 \le a_i \le 10^9$), where $a_i$ is the $i$-th element of $a$.
It is guaranteed that the sum of $n$ does not exceed $2 \cdot 10^5$ ($\sum n \le 2 \cdot 10^5$).
-----Output-----
For each test case, print the answer on the separate line — the maximum possible sum of elements on even positions after reversing at most one subarray (continuous subsegment) of $a$.
-----Example-----
Input
4
8
1 7 3 4 7 6 2 9
5
1 2 1 2 1
10
7 8 4 5 7 6 8 9 7 3
4
3 1 2 1
Output
26
5
37
5 | def kadane(arr):
s = 0
maxi = 0
for i in arr:
s = max(0, s + i)
maxi = max(maxi, s)
return maxi
t = int(input())
for _ in range(0, t):
n = int(input())
aa = [int(i) for i in input().split()]
ss = sum([aa[i] for i in range(0, n, 2)])
diff1 = [(aa[i] - aa[i - 1]) for i in range(1, n, 2)]
diff2 = [(aa[i - 1] - aa[i]) for i in range(2, n, 2)]
print(ss + max(kadane(diff1), kadane(diff2))) | FUNC_DEF ASSIGN VAR NUMBER ASSIGN VAR NUMBER FOR VAR VAR ASSIGN VAR FUNC_CALL VAR NUMBER BIN_OP VAR VAR ASSIGN VAR FUNC_CALL VAR VAR VAR RETURN VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR FOR VAR FUNC_CALL VAR NUMBER 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 VAR VAR FUNC_CALL VAR NUMBER VAR NUMBER ASSIGN VAR BIN_OP VAR VAR VAR BIN_OP VAR NUMBER VAR FUNC_CALL VAR NUMBER VAR NUMBER ASSIGN VAR BIN_OP VAR BIN_OP VAR NUMBER VAR VAR VAR FUNC_CALL VAR NUMBER VAR NUMBER EXPR FUNC_CALL VAR BIN_OP VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL VAR VAR |
You are given an array $a$ consisting of $n$ integers. Indices of the array start from zero (i. e. the first element is $a_0$, the second one is $a_1$, and so on).
You can reverse at most one subarray (continuous subsegment) of this array. Recall that the subarray of $a$ with borders $l$ and $r$ is $a[l; r] = a_l, a_{l + 1}, \dots, a_{r}$.
Your task is to reverse such a subarray that the sum of elements on even positions of the resulting array is maximized (i. e. the sum of elements $a_0, a_2, \dots, a_{2k}$ for integer $k = \lfloor\frac{n-1}{2}\rfloor$ should be maximum possible).
You have to answer $t$ independent test cases.
-----Input-----
The first line of the input contains one integer $t$ ($1 \le t \le 2 \cdot 10^4$) — the number of test cases. Then $t$ test cases follow.
The first line of the test case contains one integer $n$ ($1 \le n \le 2 \cdot 10^5$) — the length of $a$. The second line of the test case contains $n$ integers $a_0, a_1, \dots, a_{n-1}$ ($1 \le a_i \le 10^9$), where $a_i$ is the $i$-th element of $a$.
It is guaranteed that the sum of $n$ does not exceed $2 \cdot 10^5$ ($\sum n \le 2 \cdot 10^5$).
-----Output-----
For each test case, print the answer on the separate line — the maximum possible sum of elements on even positions after reversing at most one subarray (continuous subsegment) of $a$.
-----Example-----
Input
4
8
1 7 3 4 7 6 2 9
5
1 2 1 2 1
10
7 8 4 5 7 6 8 9 7 3
4
3 1 2 1
Output
26
5
37
5 | from sys import stdin
def inp():
return stdin.buffer.readline().rstrip().decode("utf8")
def itg():
return int(stdin.buffer.readline())
def mpint():
return map(int, stdin.buffer.readline().split())
for case in range(itg()):
n = itg()
arr = tuple(mpint())
ans = sum(arr[::2])
c1 = c2 = 0
c = 0
for i in range(0, n - 1, 2):
c1 = max(c1, c)
c += arr[i + 1] - arr[i]
if c < 0:
c = 0
c1 = max(c1, c)
c = 0
for i in range(1, n - 1, 2):
c2 = max(c2, c)
c += arr[i] - arr[i + 1]
if c < 0:
c = 0
c2 = max(c2, c)
ans += max(c1, c2)
print(ans) | FUNC_DEF RETURN FUNC_CALL FUNC_CALL FUNC_CALL VAR STRING FUNC_DEF RETURN FUNC_CALL VAR FUNC_CALL VAR FUNC_DEF RETURN FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR FOR VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR VAR NUMBER ASSIGN VAR VAR NUMBER ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR NUMBER BIN_OP VAR NUMBER NUMBER ASSIGN VAR FUNC_CALL VAR VAR VAR VAR BIN_OP VAR BIN_OP VAR NUMBER VAR VAR IF VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR VAR ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR NUMBER BIN_OP VAR NUMBER NUMBER ASSIGN VAR FUNC_CALL VAR VAR VAR VAR BIN_OP VAR VAR VAR BIN_OP VAR NUMBER IF VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR VAR VAR FUNC_CALL VAR VAR VAR EXPR FUNC_CALL VAR VAR |
You are given an array $a$ consisting of $n$ integers. Indices of the array start from zero (i. e. the first element is $a_0$, the second one is $a_1$, and so on).
You can reverse at most one subarray (continuous subsegment) of this array. Recall that the subarray of $a$ with borders $l$ and $r$ is $a[l; r] = a_l, a_{l + 1}, \dots, a_{r}$.
Your task is to reverse such a subarray that the sum of elements on even positions of the resulting array is maximized (i. e. the sum of elements $a_0, a_2, \dots, a_{2k}$ for integer $k = \lfloor\frac{n-1}{2}\rfloor$ should be maximum possible).
You have to answer $t$ independent test cases.
-----Input-----
The first line of the input contains one integer $t$ ($1 \le t \le 2 \cdot 10^4$) — the number of test cases. Then $t$ test cases follow.
The first line of the test case contains one integer $n$ ($1 \le n \le 2 \cdot 10^5$) — the length of $a$. The second line of the test case contains $n$ integers $a_0, a_1, \dots, a_{n-1}$ ($1 \le a_i \le 10^9$), where $a_i$ is the $i$-th element of $a$.
It is guaranteed that the sum of $n$ does not exceed $2 \cdot 10^5$ ($\sum n \le 2 \cdot 10^5$).
-----Output-----
For each test case, print the answer on the separate line — the maximum possible sum of elements on even positions after reversing at most one subarray (continuous subsegment) of $a$.
-----Example-----
Input
4
8
1 7 3 4 7 6 2 9
5
1 2 1 2 1
10
7 8 4 5 7 6 8 9 7 3
4
3 1 2 1
Output
26
5
37
5 | import sys
input = sys.stdin.buffer.readline
Q = int(input())
Query = []
for _ in range(Q):
N = int(input())
A = list(map(int, input().split()))
Query.append((N, A))
def solve(A, ans):
nowmin = 0
w = 0
for p1 in A:
w += p1
nowmin = min(nowmin, w)
ans = max(ans, T + w - nowmin)
return ans
for N, A in Query:
T = 0
P1 = []
P2 = []
for i in range((N - 1) // 2 + 1):
T += A[2 * i]
if 2 * i + 1 <= N - 1:
d = A[2 * i + 1] - A[2 * i]
P1.append(d)
if 0 <= 2 * i - 1:
d = A[2 * i - 1] - A[2 * i]
P2.append(d)
ans = solve(P1, T)
ans = solve(P2, ans)
print(ans) | IMPORT ASSIGN VAR VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR LIST FOR VAR FUNC_CALL 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 FUNC_DEF ASSIGN VAR NUMBER ASSIGN VAR NUMBER FOR VAR VAR VAR VAR ASSIGN VAR FUNC_CALL VAR VAR VAR ASSIGN VAR FUNC_CALL VAR VAR BIN_OP BIN_OP VAR VAR VAR RETURN VAR FOR VAR VAR VAR ASSIGN VAR NUMBER ASSIGN VAR LIST ASSIGN VAR LIST FOR VAR FUNC_CALL VAR BIN_OP BIN_OP BIN_OP VAR NUMBER NUMBER NUMBER VAR VAR BIN_OP NUMBER VAR IF BIN_OP BIN_OP NUMBER VAR NUMBER BIN_OP VAR NUMBER ASSIGN VAR BIN_OP VAR BIN_OP BIN_OP NUMBER VAR NUMBER VAR BIN_OP NUMBER VAR EXPR FUNC_CALL VAR VAR IF NUMBER BIN_OP BIN_OP NUMBER VAR NUMBER ASSIGN VAR BIN_OP VAR BIN_OP BIN_OP NUMBER VAR NUMBER VAR BIN_OP NUMBER VAR EXPR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR VAR VAR ASSIGN VAR FUNC_CALL VAR VAR VAR EXPR FUNC_CALL VAR VAR |
You are given an array $a$ consisting of $n$ integers. Indices of the array start from zero (i. e. the first element is $a_0$, the second one is $a_1$, and so on).
You can reverse at most one subarray (continuous subsegment) of this array. Recall that the subarray of $a$ with borders $l$ and $r$ is $a[l; r] = a_l, a_{l + 1}, \dots, a_{r}$.
Your task is to reverse such a subarray that the sum of elements on even positions of the resulting array is maximized (i. e. the sum of elements $a_0, a_2, \dots, a_{2k}$ for integer $k = \lfloor\frac{n-1}{2}\rfloor$ should be maximum possible).
You have to answer $t$ independent test cases.
-----Input-----
The first line of the input contains one integer $t$ ($1 \le t \le 2 \cdot 10^4$) — the number of test cases. Then $t$ test cases follow.
The first line of the test case contains one integer $n$ ($1 \le n \le 2 \cdot 10^5$) — the length of $a$. The second line of the test case contains $n$ integers $a_0, a_1, \dots, a_{n-1}$ ($1 \le a_i \le 10^9$), where $a_i$ is the $i$-th element of $a$.
It is guaranteed that the sum of $n$ does not exceed $2 \cdot 10^5$ ($\sum n \le 2 \cdot 10^5$).
-----Output-----
For each test case, print the answer on the separate line — the maximum possible sum of elements on even positions after reversing at most one subarray (continuous subsegment) of $a$.
-----Example-----
Input
4
8
1 7 3 4 7 6 2 9
5
1 2 1 2 1
10
7 8 4 5 7 6 8 9 7 3
4
3 1 2 1
Output
26
5
37
5 | for t in range(int(input())):
n = int(input())
a = list(map(int, input().split()))
smallestOdd = 0
smallestEven = None
sumOdd = 0
sumEven = 0
largest = 0
for i, ai in enumerate(a):
if i % 2 == 0:
sumEven += ai
if smallestEven is not None:
largest = max(largest, sumOdd - sumEven - smallestEven)
smallestEven = min(smallestEven, sumOdd - sumEven)
else:
smallestEven = sumOdd - sumEven
else:
sumOdd += ai
largest = max(largest, sumOdd - sumEven - smallestOdd)
smallestOdd = min(smallestOdd, sumOdd - sumEven)
print(sumEven + largest) | FOR VAR FUNC_CALL VAR FUNC_CALL VAR FUNC_CALL 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 NUMBER ASSIGN VAR NONE ASSIGN VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR NUMBER FOR VAR VAR FUNC_CALL VAR VAR IF BIN_OP VAR NUMBER NUMBER VAR VAR IF VAR NONE ASSIGN VAR FUNC_CALL VAR VAR BIN_OP BIN_OP VAR VAR VAR ASSIGN VAR FUNC_CALL VAR VAR BIN_OP VAR VAR ASSIGN VAR BIN_OP VAR VAR VAR VAR ASSIGN VAR FUNC_CALL VAR VAR BIN_OP BIN_OP VAR VAR VAR ASSIGN VAR FUNC_CALL VAR VAR BIN_OP VAR VAR EXPR FUNC_CALL VAR BIN_OP VAR VAR |
You are given an array $a$ consisting of $n$ integers. Indices of the array start from zero (i. e. the first element is $a_0$, the second one is $a_1$, and so on).
You can reverse at most one subarray (continuous subsegment) of this array. Recall that the subarray of $a$ with borders $l$ and $r$ is $a[l; r] = a_l, a_{l + 1}, \dots, a_{r}$.
Your task is to reverse such a subarray that the sum of elements on even positions of the resulting array is maximized (i. e. the sum of elements $a_0, a_2, \dots, a_{2k}$ for integer $k = \lfloor\frac{n-1}{2}\rfloor$ should be maximum possible).
You have to answer $t$ independent test cases.
-----Input-----
The first line of the input contains one integer $t$ ($1 \le t \le 2 \cdot 10^4$) — the number of test cases. Then $t$ test cases follow.
The first line of the test case contains one integer $n$ ($1 \le n \le 2 \cdot 10^5$) — the length of $a$. The second line of the test case contains $n$ integers $a_0, a_1, \dots, a_{n-1}$ ($1 \le a_i \le 10^9$), where $a_i$ is the $i$-th element of $a$.
It is guaranteed that the sum of $n$ does not exceed $2 \cdot 10^5$ ($\sum n \le 2 \cdot 10^5$).
-----Output-----
For each test case, print the answer on the separate line — the maximum possible sum of elements on even positions after reversing at most one subarray (continuous subsegment) of $a$.
-----Example-----
Input
4
8
1 7 3 4 7 6 2 9
5
1 2 1 2 1
10
7 8 4 5 7 6 8 9 7 3
4
3 1 2 1
Output
26
5
37
5 | for t in range(int(input())):
n = int(input())
A = input().split(" ")
for j in range(n):
A[j] = int(A[j])
k = int((n - 1) / 2)
evfree = 0
evnon = 0
odfree = 0
odnon = 0
for i in range(k):
evnon = max(evnon - A[2 * i] + A[2 * i + 1], 0)
odnon = max(odnon - A[2 * i + 2] + A[2 * i + 1], 0)
evfree = max(evfree, evnon)
odfree = max(odfree, odnon)
if n % 2 == 0:
evnon = max(evnon - A[2 * k] + A[2 * k + 1], 0)
evfree = max(evfree, evnon)
ans = 0
for i in range(k + 1):
ans += A[2 * i]
print(ans + max(evfree, odfree)) | FOR VAR FUNC_CALL VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL FUNC_CALL VAR STRING FOR VAR FUNC_CALL VAR VAR ASSIGN VAR VAR FUNC_CALL VAR VAR VAR ASSIGN VAR FUNC_CALL VAR BIN_OP BIN_OP VAR NUMBER NUMBER ASSIGN VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR BIN_OP BIN_OP VAR VAR BIN_OP NUMBER VAR VAR BIN_OP BIN_OP NUMBER VAR NUMBER NUMBER ASSIGN VAR FUNC_CALL VAR BIN_OP BIN_OP VAR VAR BIN_OP BIN_OP NUMBER VAR NUMBER VAR BIN_OP BIN_OP NUMBER VAR NUMBER NUMBER ASSIGN VAR FUNC_CALL VAR VAR VAR ASSIGN VAR FUNC_CALL VAR VAR VAR IF BIN_OP VAR NUMBER NUMBER ASSIGN VAR FUNC_CALL VAR BIN_OP BIN_OP VAR VAR BIN_OP NUMBER VAR VAR BIN_OP BIN_OP NUMBER VAR NUMBER NUMBER ASSIGN VAR FUNC_CALL VAR VAR VAR ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR BIN_OP VAR NUMBER VAR VAR BIN_OP NUMBER VAR EXPR FUNC_CALL VAR BIN_OP VAR FUNC_CALL VAR VAR VAR |
You are given an array $a$ consisting of $n$ integers. Indices of the array start from zero (i. e. the first element is $a_0$, the second one is $a_1$, and so on).
You can reverse at most one subarray (continuous subsegment) of this array. Recall that the subarray of $a$ with borders $l$ and $r$ is $a[l; r] = a_l, a_{l + 1}, \dots, a_{r}$.
Your task is to reverse such a subarray that the sum of elements on even positions of the resulting array is maximized (i. e. the sum of elements $a_0, a_2, \dots, a_{2k}$ for integer $k = \lfloor\frac{n-1}{2}\rfloor$ should be maximum possible).
You have to answer $t$ independent test cases.
-----Input-----
The first line of the input contains one integer $t$ ($1 \le t \le 2 \cdot 10^4$) — the number of test cases. Then $t$ test cases follow.
The first line of the test case contains one integer $n$ ($1 \le n \le 2 \cdot 10^5$) — the length of $a$. The second line of the test case contains $n$ integers $a_0, a_1, \dots, a_{n-1}$ ($1 \le a_i \le 10^9$), where $a_i$ is the $i$-th element of $a$.
It is guaranteed that the sum of $n$ does not exceed $2 \cdot 10^5$ ($\sum n \le 2 \cdot 10^5$).
-----Output-----
For each test case, print the answer on the separate line — the maximum possible sum of elements on even positions after reversing at most one subarray (continuous subsegment) of $a$.
-----Example-----
Input
4
8
1 7 3 4 7 6 2 9
5
1 2 1 2 1
10
7 8 4 5 7 6 8 9 7 3
4
3 1 2 1
Output
26
5
37
5 | for _ in range(int(input())):
n = int(input())
a = list(map(int, input().split()))
o = 0
e = 0
oe = []
for i in range(n):
if i & 1:
o += a[i]
else:
e += a[i]
oe.append(o - e)
om = int(1000000000000000.0)
em = int(1000000000000000.0)
ma = -int(1000000000000000.0)
for i in range(n):
if i & 1:
ma = max(oe[i], oe[i] - om, ma)
om = min(oe[i], om)
else:
ma = max(oe[i] - em, ma)
em = min(oe[i], em)
print(max(e, e + ma)) | FOR VAR FUNC_CALL VAR FUNC_CALL VAR FUNC_CALL 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 NUMBER ASSIGN VAR NUMBER ASSIGN VAR LIST FOR VAR FUNC_CALL VAR VAR IF BIN_OP VAR NUMBER VAR VAR VAR VAR VAR VAR EXPR FUNC_CALL VAR BIN_OP VAR VAR ASSIGN VAR FUNC_CALL VAR NUMBER ASSIGN VAR FUNC_CALL VAR NUMBER ASSIGN VAR FUNC_CALL VAR NUMBER FOR VAR FUNC_CALL VAR VAR IF BIN_OP VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR VAR BIN_OP VAR VAR VAR VAR ASSIGN VAR FUNC_CALL VAR VAR VAR VAR ASSIGN VAR FUNC_CALL VAR BIN_OP VAR VAR VAR VAR ASSIGN VAR FUNC_CALL VAR VAR VAR VAR EXPR FUNC_CALL VAR FUNC_CALL VAR VAR BIN_OP VAR VAR |
You are given an array $a$ consisting of $n$ integers. Indices of the array start from zero (i. e. the first element is $a_0$, the second one is $a_1$, and so on).
You can reverse at most one subarray (continuous subsegment) of this array. Recall that the subarray of $a$ with borders $l$ and $r$ is $a[l; r] = a_l, a_{l + 1}, \dots, a_{r}$.
Your task is to reverse such a subarray that the sum of elements on even positions of the resulting array is maximized (i. e. the sum of elements $a_0, a_2, \dots, a_{2k}$ for integer $k = \lfloor\frac{n-1}{2}\rfloor$ should be maximum possible).
You have to answer $t$ independent test cases.
-----Input-----
The first line of the input contains one integer $t$ ($1 \le t \le 2 \cdot 10^4$) — the number of test cases. Then $t$ test cases follow.
The first line of the test case contains one integer $n$ ($1 \le n \le 2 \cdot 10^5$) — the length of $a$. The second line of the test case contains $n$ integers $a_0, a_1, \dots, a_{n-1}$ ($1 \le a_i \le 10^9$), where $a_i$ is the $i$-th element of $a$.
It is guaranteed that the sum of $n$ does not exceed $2 \cdot 10^5$ ($\sum n \le 2 \cdot 10^5$).
-----Output-----
For each test case, print the answer on the separate line — the maximum possible sum of elements on even positions after reversing at most one subarray (continuous subsegment) of $a$.
-----Example-----
Input
4
8
1 7 3 4 7 6 2 9
5
1 2 1 2 1
10
7 8 4 5 7 6 8 9 7 3
4
3 1 2 1
Output
26
5
37
5 | t = int(input())
for _ in range(t):
n = int(input())
inp = list(map(int, input().split()))
if n == 1:
print(inp[0])
else:
lef = [0]
ri = [max(0, inp[1] - inp[0])]
for i in range(1, len(inp)):
if i % 2 == 1:
lef.append(0)
ri.append(0)
continue
if i - 2 >= 0:
lef.append(max(0, lef[-2] + inp[i - 1] - inp[i]))
else:
lef.append(max(0, inp[i - 1] - inp[i]))
if i + 1 < len(inp):
if i - 2 >= 0:
ri.append(max(0, ri[-2] + inp[i + 1] - inp[i]))
else:
ri.append(max(0, inp[i + 1] - inp[i]))
ans = 0
for i in range(0, len(inp), 2):
ans += inp[i]
ans += max(max(ri), max(lef))
print(ans) | ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR FOR VAR FUNC_CALL VAR VAR 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 VAR NUMBER ASSIGN VAR LIST NUMBER ASSIGN VAR LIST FUNC_CALL VAR NUMBER BIN_OP VAR NUMBER VAR NUMBER FOR VAR FUNC_CALL VAR NUMBER FUNC_CALL VAR VAR IF BIN_OP VAR NUMBER NUMBER EXPR FUNC_CALL VAR NUMBER EXPR FUNC_CALL VAR NUMBER IF BIN_OP VAR NUMBER NUMBER EXPR FUNC_CALL VAR FUNC_CALL VAR NUMBER BIN_OP BIN_OP VAR NUMBER VAR BIN_OP VAR NUMBER VAR VAR EXPR FUNC_CALL VAR FUNC_CALL VAR NUMBER BIN_OP VAR BIN_OP VAR NUMBER VAR VAR IF BIN_OP VAR NUMBER FUNC_CALL VAR VAR IF BIN_OP VAR NUMBER NUMBER EXPR FUNC_CALL VAR FUNC_CALL VAR NUMBER BIN_OP BIN_OP VAR NUMBER VAR BIN_OP VAR NUMBER VAR VAR EXPR FUNC_CALL VAR FUNC_CALL VAR NUMBER BIN_OP VAR BIN_OP VAR NUMBER VAR VAR ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR NUMBER FUNC_CALL VAR VAR NUMBER VAR VAR VAR VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR VAR |
You are given an array $a$ consisting of $n$ integers. Indices of the array start from zero (i. e. the first element is $a_0$, the second one is $a_1$, and so on).
You can reverse at most one subarray (continuous subsegment) of this array. Recall that the subarray of $a$ with borders $l$ and $r$ is $a[l; r] = a_l, a_{l + 1}, \dots, a_{r}$.
Your task is to reverse such a subarray that the sum of elements on even positions of the resulting array is maximized (i. e. the sum of elements $a_0, a_2, \dots, a_{2k}$ for integer $k = \lfloor\frac{n-1}{2}\rfloor$ should be maximum possible).
You have to answer $t$ independent test cases.
-----Input-----
The first line of the input contains one integer $t$ ($1 \le t \le 2 \cdot 10^4$) — the number of test cases. Then $t$ test cases follow.
The first line of the test case contains one integer $n$ ($1 \le n \le 2 \cdot 10^5$) — the length of $a$. The second line of the test case contains $n$ integers $a_0, a_1, \dots, a_{n-1}$ ($1 \le a_i \le 10^9$), where $a_i$ is the $i$-th element of $a$.
It is guaranteed that the sum of $n$ does not exceed $2 \cdot 10^5$ ($\sum n \le 2 \cdot 10^5$).
-----Output-----
For each test case, print the answer on the separate line — the maximum possible sum of elements on even positions after reversing at most one subarray (continuous subsegment) of $a$.
-----Example-----
Input
4
8
1 7 3 4 7 6 2 9
5
1 2 1 2 1
10
7 8 4 5 7 6 8 9 7 3
4
3 1 2 1
Output
26
5
37
5 | from sys import *
input = stdin.readline
def maxSubArraySum(a, size):
max_so_far = -1e28 - 1
max_ending_here = 0
for i in range(0, size):
max_ending_here = max_ending_here + a[i]
if max_so_far < max_ending_here:
max_so_far = max_ending_here
if max_ending_here < 0:
max_ending_here = 0
return max_so_far
for _ in range(int(input())):
n = int(input())
k = n
a = list(map(int, input().split()))
b = []
c = []
for i in range(1, n, 2):
b.append(a[i] - a[i - 1])
mxb = maxSubArraySum(b, len(b))
mxc = 0
if n % 2 == 0:
n -= 1
for i in range(n - 2, -1, -2):
c.append(a[i] - a[i + 1])
mxc = maxSubArraySum(c, len(c))
mx = max(mxb, mxc, 0)
sn = 0
for i in range(0, k, 2):
sn += a[i]
sn += mx
stdout.write(str(sn) + "\n") | ASSIGN VAR VAR FUNC_DEF ASSIGN VAR BIN_OP NUMBER NUMBER ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR NUMBER VAR ASSIGN VAR BIN_OP VAR VAR VAR IF VAR VAR ASSIGN VAR VAR IF VAR NUMBER ASSIGN VAR NUMBER RETURN VAR FOR VAR FUNC_CALL VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR LIST ASSIGN VAR LIST FOR VAR FUNC_CALL VAR NUMBER VAR NUMBER EXPR FUNC_CALL VAR BIN_OP VAR VAR VAR BIN_OP VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR FUNC_CALL VAR VAR ASSIGN VAR NUMBER IF BIN_OP VAR NUMBER NUMBER VAR NUMBER FOR VAR FUNC_CALL VAR BIN_OP VAR NUMBER NUMBER NUMBER EXPR FUNC_CALL VAR BIN_OP VAR VAR VAR BIN_OP VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR VAR VAR NUMBER ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR NUMBER VAR NUMBER VAR VAR VAR VAR VAR EXPR FUNC_CALL VAR BIN_OP FUNC_CALL VAR VAR STRING |
You are given an array $a$ consisting of $n$ integers. Indices of the array start from zero (i. e. the first element is $a_0$, the second one is $a_1$, and so on).
You can reverse at most one subarray (continuous subsegment) of this array. Recall that the subarray of $a$ with borders $l$ and $r$ is $a[l; r] = a_l, a_{l + 1}, \dots, a_{r}$.
Your task is to reverse such a subarray that the sum of elements on even positions of the resulting array is maximized (i. e. the sum of elements $a_0, a_2, \dots, a_{2k}$ for integer $k = \lfloor\frac{n-1}{2}\rfloor$ should be maximum possible).
You have to answer $t$ independent test cases.
-----Input-----
The first line of the input contains one integer $t$ ($1 \le t \le 2 \cdot 10^4$) — the number of test cases. Then $t$ test cases follow.
The first line of the test case contains one integer $n$ ($1 \le n \le 2 \cdot 10^5$) — the length of $a$. The second line of the test case contains $n$ integers $a_0, a_1, \dots, a_{n-1}$ ($1 \le a_i \le 10^9$), where $a_i$ is the $i$-th element of $a$.
It is guaranteed that the sum of $n$ does not exceed $2 \cdot 10^5$ ($\sum n \le 2 \cdot 10^5$).
-----Output-----
For each test case, print the answer on the separate line — the maximum possible sum of elements on even positions after reversing at most one subarray (continuous subsegment) of $a$.
-----Example-----
Input
4
8
1 7 3 4 7 6 2 9
5
1 2 1 2 1
10
7 8 4 5 7 6 8 9 7 3
4
3 1 2 1
Output
26
5
37
5 | def maxSubArraySum(a, size):
max_so_far = -(10**9) - 1
max_ending_here = 0
for i in range(0, size):
max_ending_here = max_ending_here + a[i]
if max_so_far < max_ending_here:
max_so_far = max_ending_here
if max_ending_here < 0:
max_ending_here = 0
return max_so_far
for _ in range(int(input())):
n = int(input())
arr = list(map(int, input().split()))
x, y, ans = [], [], 0
p = n if n % 2 == 0 else n - 1
q = n if n % 2 == 1 else n - 1
for i in range(0, p, 2):
x.append(arr[i + 1] - arr[i])
ans += arr[i]
if n % 2 == 1:
ans += arr[n - 1]
for i in range(1, q, 2):
y.append(arr[i] - arr[i + 1])
k = max(maxSubArraySum(x, len(x)), maxSubArraySum(y, len(y)), 0)
print(ans + k) | FUNC_DEF ASSIGN VAR BIN_OP BIN_OP NUMBER NUMBER NUMBER ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR NUMBER VAR ASSIGN VAR BIN_OP VAR VAR VAR IF VAR VAR ASSIGN VAR VAR IF VAR NUMBER ASSIGN VAR NUMBER RETURN VAR FOR VAR FUNC_CALL VAR FUNC_CALL VAR FUNC_CALL 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 VAR VAR LIST LIST NUMBER ASSIGN VAR BIN_OP VAR NUMBER NUMBER VAR BIN_OP VAR NUMBER ASSIGN VAR BIN_OP VAR NUMBER NUMBER VAR BIN_OP VAR NUMBER FOR VAR FUNC_CALL VAR NUMBER VAR NUMBER EXPR FUNC_CALL VAR BIN_OP VAR BIN_OP VAR NUMBER VAR VAR VAR VAR VAR IF BIN_OP VAR NUMBER NUMBER VAR VAR BIN_OP VAR NUMBER FOR VAR FUNC_CALL VAR NUMBER VAR NUMBER EXPR FUNC_CALL VAR BIN_OP VAR VAR VAR BIN_OP VAR NUMBER ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL VAR VAR FUNC_CALL VAR VAR FUNC_CALL VAR VAR NUMBER EXPR FUNC_CALL VAR BIN_OP VAR VAR |
You are given an array $a$ consisting of $n$ integers. Indices of the array start from zero (i. e. the first element is $a_0$, the second one is $a_1$, and so on).
You can reverse at most one subarray (continuous subsegment) of this array. Recall that the subarray of $a$ with borders $l$ and $r$ is $a[l; r] = a_l, a_{l + 1}, \dots, a_{r}$.
Your task is to reverse such a subarray that the sum of elements on even positions of the resulting array is maximized (i. e. the sum of elements $a_0, a_2, \dots, a_{2k}$ for integer $k = \lfloor\frac{n-1}{2}\rfloor$ should be maximum possible).
You have to answer $t$ independent test cases.
-----Input-----
The first line of the input contains one integer $t$ ($1 \le t \le 2 \cdot 10^4$) — the number of test cases. Then $t$ test cases follow.
The first line of the test case contains one integer $n$ ($1 \le n \le 2 \cdot 10^5$) — the length of $a$. The second line of the test case contains $n$ integers $a_0, a_1, \dots, a_{n-1}$ ($1 \le a_i \le 10^9$), where $a_i$ is the $i$-th element of $a$.
It is guaranteed that the sum of $n$ does not exceed $2 \cdot 10^5$ ($\sum n \le 2 \cdot 10^5$).
-----Output-----
For each test case, print the answer on the separate line — the maximum possible sum of elements on even positions after reversing at most one subarray (continuous subsegment) of $a$.
-----Example-----
Input
4
8
1 7 3 4 7 6 2 9
5
1 2 1 2 1
10
7 8 4 5 7 6 8 9 7 3
4
3 1 2 1
Output
26
5
37
5 | import sys
ii = lambda: sys.stdin.readline().strip()
idata = lambda: [int(x) for x in ii().split()]
def ad(asa):
ans = 0
if bool(asa):
ans = asa[0]
summ = 0
min_sum = 0
for r in range(len(asa)):
summ += asa[r]
ans = max(ans, summ - min_sum)
min_sum = min(min_sum, summ)
return ans
def solve():
n = int(ii())
data = idata()
if n == 1:
print(data[0])
return
ans = 0
for i in range(0, n, 2):
ans += data[i]
sp1 = []
sp2 = []
for i in range(0, n - 1, 2):
sp1 += [data[i + 1] - data[i]]
if i + 2 != n:
sp2 += [data[i + 1] - data[i + 2]]
print(max(ans, ans + ad(sp1), ans + ad(sp2)))
return
for t in range(int(ii())):
solve() | IMPORT ASSIGN VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR VAR VAR FUNC_CALL FUNC_CALL VAR FUNC_DEF ASSIGN VAR NUMBER IF FUNC_CALL VAR VAR ASSIGN VAR VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR FUNC_CALL VAR VAR VAR VAR VAR ASSIGN VAR FUNC_CALL VAR VAR BIN_OP VAR VAR ASSIGN VAR FUNC_CALL VAR VAR VAR RETURN VAR FUNC_DEF ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR IF VAR NUMBER EXPR FUNC_CALL VAR VAR NUMBER RETURN ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR NUMBER VAR NUMBER VAR VAR VAR ASSIGN VAR LIST ASSIGN VAR LIST FOR VAR FUNC_CALL VAR NUMBER BIN_OP VAR NUMBER NUMBER VAR LIST BIN_OP VAR BIN_OP VAR NUMBER VAR VAR IF BIN_OP VAR NUMBER VAR VAR LIST BIN_OP VAR BIN_OP VAR NUMBER VAR BIN_OP VAR NUMBER EXPR FUNC_CALL VAR FUNC_CALL VAR VAR BIN_OP VAR FUNC_CALL VAR VAR BIN_OP VAR FUNC_CALL VAR VAR RETURN FOR VAR FUNC_CALL VAR FUNC_CALL VAR FUNC_CALL VAR EXPR FUNC_CALL VAR |
You are given an array $a$ consisting of $n$ integers. Indices of the array start from zero (i. e. the first element is $a_0$, the second one is $a_1$, and so on).
You can reverse at most one subarray (continuous subsegment) of this array. Recall that the subarray of $a$ with borders $l$ and $r$ is $a[l; r] = a_l, a_{l + 1}, \dots, a_{r}$.
Your task is to reverse such a subarray that the sum of elements on even positions of the resulting array is maximized (i. e. the sum of elements $a_0, a_2, \dots, a_{2k}$ for integer $k = \lfloor\frac{n-1}{2}\rfloor$ should be maximum possible).
You have to answer $t$ independent test cases.
-----Input-----
The first line of the input contains one integer $t$ ($1 \le t \le 2 \cdot 10^4$) — the number of test cases. Then $t$ test cases follow.
The first line of the test case contains one integer $n$ ($1 \le n \le 2 \cdot 10^5$) — the length of $a$. The second line of the test case contains $n$ integers $a_0, a_1, \dots, a_{n-1}$ ($1 \le a_i \le 10^9$), where $a_i$ is the $i$-th element of $a$.
It is guaranteed that the sum of $n$ does not exceed $2 \cdot 10^5$ ($\sum n \le 2 \cdot 10^5$).
-----Output-----
For each test case, print the answer on the separate line — the maximum possible sum of elements on even positions after reversing at most one subarray (continuous subsegment) of $a$.
-----Example-----
Input
4
8
1 7 3 4 7 6 2 9
5
1 2 1 2 1
10
7 8 4 5 7 6 8 9 7 3
4
3 1 2 1
Output
26
5
37
5 | import sys
input = sys.stdin.readline
t = int(input())
for _ in range(t):
n = int(input())
a = list(map(int, input().split()))
num_ini = 0
for i in range(0, n, 2):
num_ini += a[i]
dif1 = []
for i in range(0, n - 1, 2):
dif1.append(a[i + 1] - a[i])
dif2 = []
for i in range(1, n - 1, 2):
dif2.append(a[i] - a[i + 1])
dp1 = [0] * (len(dif1) + 1)
dp2 = [0] * (len(dif2) + 1)
for i in range(1, len(dp1)):
dp1[i] = max(dp1[i - 1] + dif1[i - 1], 0)
for j in range(1, len(dp2)):
dp2[j] = max(dp2[j - 1] + dif2[j - 1], 0)
print(max(max(dp1), max(dp2)) + num_ini) | IMPORT ASSIGN VAR VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR FOR VAR FUNC_CALL 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 NUMBER FOR VAR FUNC_CALL VAR NUMBER VAR NUMBER VAR VAR VAR ASSIGN VAR LIST FOR VAR FUNC_CALL VAR NUMBER BIN_OP VAR NUMBER NUMBER EXPR FUNC_CALL VAR BIN_OP VAR BIN_OP VAR NUMBER VAR VAR ASSIGN VAR LIST FOR VAR FUNC_CALL VAR NUMBER BIN_OP VAR NUMBER NUMBER EXPR FUNC_CALL VAR BIN_OP VAR VAR VAR BIN_OP VAR NUMBER ASSIGN VAR BIN_OP LIST NUMBER BIN_OP FUNC_CALL VAR VAR NUMBER ASSIGN VAR BIN_OP LIST NUMBER BIN_OP FUNC_CALL VAR VAR NUMBER FOR VAR FUNC_CALL VAR NUMBER FUNC_CALL VAR VAR ASSIGN VAR VAR FUNC_CALL VAR BIN_OP VAR BIN_OP VAR NUMBER VAR BIN_OP VAR NUMBER NUMBER FOR VAR FUNC_CALL VAR NUMBER FUNC_CALL VAR VAR ASSIGN VAR VAR FUNC_CALL VAR BIN_OP VAR BIN_OP VAR NUMBER VAR BIN_OP VAR NUMBER NUMBER EXPR FUNC_CALL VAR BIN_OP FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL VAR VAR VAR |
You are given an array $a$ consisting of $n$ integers. Indices of the array start from zero (i. e. the first element is $a_0$, the second one is $a_1$, and so on).
You can reverse at most one subarray (continuous subsegment) of this array. Recall that the subarray of $a$ with borders $l$ and $r$ is $a[l; r] = a_l, a_{l + 1}, \dots, a_{r}$.
Your task is to reverse such a subarray that the sum of elements on even positions of the resulting array is maximized (i. e. the sum of elements $a_0, a_2, \dots, a_{2k}$ for integer $k = \lfloor\frac{n-1}{2}\rfloor$ should be maximum possible).
You have to answer $t$ independent test cases.
-----Input-----
The first line of the input contains one integer $t$ ($1 \le t \le 2 \cdot 10^4$) — the number of test cases. Then $t$ test cases follow.
The first line of the test case contains one integer $n$ ($1 \le n \le 2 \cdot 10^5$) — the length of $a$. The second line of the test case contains $n$ integers $a_0, a_1, \dots, a_{n-1}$ ($1 \le a_i \le 10^9$), where $a_i$ is the $i$-th element of $a$.
It is guaranteed that the sum of $n$ does not exceed $2 \cdot 10^5$ ($\sum n \le 2 \cdot 10^5$).
-----Output-----
For each test case, print the answer on the separate line — the maximum possible sum of elements on even positions after reversing at most one subarray (continuous subsegment) of $a$.
-----Example-----
Input
4
8
1 7 3 4 7 6 2 9
5
1 2 1 2 1
10
7 8 4 5 7 6 8 9 7 3
4
3 1 2 1
Output
26
5
37
5 | def f(a):
total = 0
for i in range(0, len(a), 2):
total += a[i]
nb_pairs = len(a) // 2
even_offset_profit = 0
if nb_pairs >= 1:
d_0 = [0] * nb_pairs
for i in range(nb_pairs):
d_0[i] = -a[2 * i] + a[2 * i + 1]
s_0 = [0] * len(d_0)
s_0[0] = d_0[0]
for j in range(1, len(d_0)):
s_0[j] = max(s_0[j - 1] + d_0[j], d_0[j])
even_offset_profit = max(s_0)
nb_pairs = (len(a) - 1) // 2
odd_offset_profit = 0
if nb_pairs >= 1:
d_1 = [0] * nb_pairs
for i in range(nb_pairs):
d_1[i] = a[2 * i + 1] - a[2 * i + 2]
s_1 = [0] * len(d_1)
s_1[0] = d_1[0]
for j in range(1, len(d_1)):
s_1[j] = max(s_1[j - 1] + d_1[j], d_1[j])
odd_offset_profit = max(s_1)
return total + max(0, even_offset_profit, odd_offset_profit)
nb_tests = int(input())
for i in range(nb_tests):
_ = input()
a = [int(d) for d in input().split()]
print(f(a)) | FUNC_DEF ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR NUMBER FUNC_CALL VAR VAR NUMBER VAR VAR VAR ASSIGN VAR BIN_OP FUNC_CALL VAR VAR NUMBER ASSIGN VAR NUMBER IF VAR NUMBER ASSIGN VAR BIN_OP LIST NUMBER VAR FOR VAR FUNC_CALL VAR VAR ASSIGN VAR VAR BIN_OP VAR BIN_OP NUMBER VAR VAR BIN_OP BIN_OP NUMBER VAR NUMBER ASSIGN VAR BIN_OP LIST NUMBER FUNC_CALL VAR VAR ASSIGN VAR NUMBER VAR NUMBER FOR VAR FUNC_CALL VAR NUMBER FUNC_CALL VAR VAR ASSIGN VAR VAR FUNC_CALL VAR BIN_OP VAR BIN_OP VAR NUMBER VAR VAR VAR VAR ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR BIN_OP BIN_OP FUNC_CALL VAR VAR NUMBER NUMBER ASSIGN VAR NUMBER IF VAR NUMBER ASSIGN VAR BIN_OP LIST NUMBER VAR FOR VAR FUNC_CALL VAR VAR ASSIGN VAR VAR BIN_OP VAR BIN_OP BIN_OP NUMBER VAR NUMBER VAR BIN_OP BIN_OP NUMBER VAR NUMBER ASSIGN VAR BIN_OP LIST NUMBER FUNC_CALL VAR VAR ASSIGN VAR NUMBER VAR NUMBER FOR VAR FUNC_CALL VAR NUMBER FUNC_CALL VAR VAR ASSIGN VAR VAR FUNC_CALL VAR BIN_OP VAR BIN_OP VAR NUMBER VAR VAR VAR VAR ASSIGN VAR FUNC_CALL VAR VAR RETURN BIN_OP VAR FUNC_CALL VAR NUMBER VAR VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR FOR VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR VAR VAR FUNC_CALL FUNC_CALL VAR EXPR FUNC_CALL VAR FUNC_CALL VAR VAR |
You are given an array $a$ consisting of $n$ integers. Indices of the array start from zero (i. e. the first element is $a_0$, the second one is $a_1$, and so on).
You can reverse at most one subarray (continuous subsegment) of this array. Recall that the subarray of $a$ with borders $l$ and $r$ is $a[l; r] = a_l, a_{l + 1}, \dots, a_{r}$.
Your task is to reverse such a subarray that the sum of elements on even positions of the resulting array is maximized (i. e. the sum of elements $a_0, a_2, \dots, a_{2k}$ for integer $k = \lfloor\frac{n-1}{2}\rfloor$ should be maximum possible).
You have to answer $t$ independent test cases.
-----Input-----
The first line of the input contains one integer $t$ ($1 \le t \le 2 \cdot 10^4$) — the number of test cases. Then $t$ test cases follow.
The first line of the test case contains one integer $n$ ($1 \le n \le 2 \cdot 10^5$) — the length of $a$. The second line of the test case contains $n$ integers $a_0, a_1, \dots, a_{n-1}$ ($1 \le a_i \le 10^9$), where $a_i$ is the $i$-th element of $a$.
It is guaranteed that the sum of $n$ does not exceed $2 \cdot 10^5$ ($\sum n \le 2 \cdot 10^5$).
-----Output-----
For each test case, print the answer on the separate line — the maximum possible sum of elements on even positions after reversing at most one subarray (continuous subsegment) of $a$.
-----Example-----
Input
4
8
1 7 3 4 7 6 2 9
5
1 2 1 2 1
10
7 8 4 5 7 6 8 9 7 3
4
3 1 2 1
Output
26
5
37
5 | import sys
import time
from itertools import accumulate
buff_readline = sys.stdin.readline
readline = sys.stdin.readline
INF = 2**62 - 1
def read_int():
return int(buff_readline())
def read_int_n():
return list(map(int, buff_readline().split()))
def read_float():
return float(buff_readline())
def read_float_n():
return list(map(float, buff_readline().split()))
def read_str():
return readline().strip()
def read_str_n():
return readline().strip().split()
def error_print(*args):
print(*args, file=sys.stderr)
def mt(f):
import time
def wrap(*args, **kwargs):
s = time.time()
ret = f(*args, **kwargs)
e = time.time()
error_print(e - s, "sec")
return ret
return wrap
def max_subarray(numbers):
best_sum = 0
current_sum = 0
for x in numbers:
current_sum = max(0, current_sum + x)
best_sum = max(best_sum, current_sum)
return best_sum
def slv(N, A):
from itertools import accumulate
ans = sum(A[::2])
b = []
for i in range(0, N - 1, 2):
b.append(A[i + 1] - A[i])
c = []
for i in range(1, N - 1, 2):
c.append(A[i] - A[i + 1])
ans += max(max_subarray(b), max_subarray(c))
return ans
def main():
for _ in range(read_int()):
N = read_int()
A = read_int_n()
print(slv(N, A))
main() | IMPORT IMPORT ASSIGN VAR VAR ASSIGN VAR VAR ASSIGN VAR BIN_OP BIN_OP NUMBER NUMBER NUMBER 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 RETURN FUNC_CALL VAR FUNC_CALL VAR FUNC_DEF RETURN FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR FUNC_DEF RETURN FUNC_CALL FUNC_CALL VAR FUNC_DEF RETURN FUNC_CALL FUNC_CALL FUNC_CALL VAR FUNC_DEF EXPR FUNC_CALL VAR VAR VAR FUNC_DEF IMPORT FUNC_DEF ASSIGN VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR VAR VAR ASSIGN VAR FUNC_CALL VAR EXPR FUNC_CALL VAR BIN_OP VAR VAR STRING RETURN VAR RETURN VAR FUNC_DEF ASSIGN VAR NUMBER ASSIGN VAR NUMBER FOR VAR VAR ASSIGN VAR FUNC_CALL VAR NUMBER BIN_OP VAR VAR ASSIGN VAR FUNC_CALL VAR VAR VAR RETURN VAR FUNC_DEF ASSIGN VAR FUNC_CALL VAR VAR NUMBER ASSIGN VAR LIST FOR VAR FUNC_CALL VAR NUMBER BIN_OP VAR NUMBER NUMBER EXPR FUNC_CALL VAR BIN_OP VAR BIN_OP VAR NUMBER VAR VAR ASSIGN VAR LIST FOR VAR FUNC_CALL VAR NUMBER BIN_OP VAR NUMBER NUMBER EXPR FUNC_CALL VAR BIN_OP VAR VAR VAR BIN_OP VAR NUMBER VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL VAR VAR RETURN VAR FUNC_DEF FOR VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR EXPR FUNC_CALL VAR FUNC_CALL VAR VAR VAR EXPR FUNC_CALL VAR |
You are given an array $a$ consisting of $n$ integers. Indices of the array start from zero (i. e. the first element is $a_0$, the second one is $a_1$, and so on).
You can reverse at most one subarray (continuous subsegment) of this array. Recall that the subarray of $a$ with borders $l$ and $r$ is $a[l; r] = a_l, a_{l + 1}, \dots, a_{r}$.
Your task is to reverse such a subarray that the sum of elements on even positions of the resulting array is maximized (i. e. the sum of elements $a_0, a_2, \dots, a_{2k}$ for integer $k = \lfloor\frac{n-1}{2}\rfloor$ should be maximum possible).
You have to answer $t$ independent test cases.
-----Input-----
The first line of the input contains one integer $t$ ($1 \le t \le 2 \cdot 10^4$) — the number of test cases. Then $t$ test cases follow.
The first line of the test case contains one integer $n$ ($1 \le n \le 2 \cdot 10^5$) — the length of $a$. The second line of the test case contains $n$ integers $a_0, a_1, \dots, a_{n-1}$ ($1 \le a_i \le 10^9$), where $a_i$ is the $i$-th element of $a$.
It is guaranteed that the sum of $n$ does not exceed $2 \cdot 10^5$ ($\sum n \le 2 \cdot 10^5$).
-----Output-----
For each test case, print the answer on the separate line — the maximum possible sum of elements on even positions after reversing at most one subarray (continuous subsegment) of $a$.
-----Example-----
Input
4
8
1 7 3 4 7 6 2 9
5
1 2 1 2 1
10
7 8 4 5 7 6 8 9 7 3
4
3 1 2 1
Output
26
5
37
5 | def kadane(A):
ans = float("-inf")
cur = 0
for a in A:
cur = max(cur + a, a)
ans = max(ans, cur)
return ans
def solve(A, n):
S = sum(A[::2])
B1 = []
B2 = []
for i in range(0, n - 1, 2):
B1.append(A[i + 1] - A[i])
for i in range(1, n - 1, 2):
B2.append(-A[i + 1] + A[i])
a = kadane(B1)
b = kadane(B2)
return max(S + a, S + b, S)
T = int(input())
for _ in range(T):
n = int(input())
A = list(map(int, input().split()))
print(solve(A, n)) | FUNC_DEF ASSIGN VAR FUNC_CALL VAR STRING ASSIGN VAR NUMBER FOR VAR VAR ASSIGN VAR FUNC_CALL VAR BIN_OP VAR VAR VAR ASSIGN VAR FUNC_CALL VAR VAR VAR RETURN VAR FUNC_DEF ASSIGN VAR FUNC_CALL VAR VAR NUMBER ASSIGN VAR LIST ASSIGN VAR LIST FOR VAR FUNC_CALL VAR NUMBER BIN_OP VAR NUMBER NUMBER EXPR FUNC_CALL VAR BIN_OP VAR BIN_OP VAR NUMBER VAR VAR FOR VAR FUNC_CALL VAR NUMBER BIN_OP VAR NUMBER NUMBER EXPR FUNC_CALL VAR BIN_OP VAR BIN_OP VAR NUMBER VAR VAR ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR VAR RETURN FUNC_CALL VAR BIN_OP VAR VAR BIN_OP VAR VAR VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR FOR VAR FUNC_CALL 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 FUNC_CALL VAR VAR VAR |
You are given an array $a$ consisting of $n$ integers. Indices of the array start from zero (i. e. the first element is $a_0$, the second one is $a_1$, and so on).
You can reverse at most one subarray (continuous subsegment) of this array. Recall that the subarray of $a$ with borders $l$ and $r$ is $a[l; r] = a_l, a_{l + 1}, \dots, a_{r}$.
Your task is to reverse such a subarray that the sum of elements on even positions of the resulting array is maximized (i. e. the sum of elements $a_0, a_2, \dots, a_{2k}$ for integer $k = \lfloor\frac{n-1}{2}\rfloor$ should be maximum possible).
You have to answer $t$ independent test cases.
-----Input-----
The first line of the input contains one integer $t$ ($1 \le t \le 2 \cdot 10^4$) — the number of test cases. Then $t$ test cases follow.
The first line of the test case contains one integer $n$ ($1 \le n \le 2 \cdot 10^5$) — the length of $a$. The second line of the test case contains $n$ integers $a_0, a_1, \dots, a_{n-1}$ ($1 \le a_i \le 10^9$), where $a_i$ is the $i$-th element of $a$.
It is guaranteed that the sum of $n$ does not exceed $2 \cdot 10^5$ ($\sum n \le 2 \cdot 10^5$).
-----Output-----
For each test case, print the answer on the separate line — the maximum possible sum of elements on even positions after reversing at most one subarray (continuous subsegment) of $a$.
-----Example-----
Input
4
8
1 7 3 4 7 6 2 9
5
1 2 1 2 1
10
7 8 4 5 7 6 8 9 7 3
4
3 1 2 1
Output
26
5
37
5 | def main():
t = int(input())
for _ in range(t):
n = int(input())
a = tuple(map(int, input().split()))
m = 0
c = 0
start = None
curr_start = None
end = None
for i in range(0, n, 2):
if n - i < 2:
break
d = a[i] - a[i + 1]
if c + d <= 0:
if c >= 0:
curr_start = i
c += d
else:
c = 0
if c < m:
m = c
start = curr_start
end = i + 1
cc = 0
curr_start = None
for i in range(1, n, 2):
if n - i < 2:
break
d = a[i + 1] - a[i]
if cc + d <= 0:
if cc >= 0:
curr_start = i
cc += d
else:
cc = 0
if cc < m:
m = cc
start = curr_start
end = i + 1
result = 0
for i in range(n):
if start is None or end is None:
if i % 2 == 0:
result += a[i]
continue
if (i < start or i > end) and i % 2 == 1:
continue
if start <= i <= end and i % 2 == 0:
continue
result += a[i]
print(result)
main() | FUNC_DEF ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR FOR VAR FUNC_CALL 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 NUMBER ASSIGN VAR NUMBER ASSIGN VAR NONE ASSIGN VAR NONE ASSIGN VAR NONE FOR VAR FUNC_CALL VAR NUMBER VAR NUMBER IF BIN_OP VAR VAR NUMBER ASSIGN VAR BIN_OP VAR VAR VAR BIN_OP VAR NUMBER IF BIN_OP VAR VAR NUMBER IF VAR NUMBER ASSIGN VAR VAR VAR VAR ASSIGN VAR NUMBER IF VAR VAR ASSIGN VAR VAR ASSIGN VAR VAR ASSIGN VAR BIN_OP VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR NONE FOR VAR FUNC_CALL VAR NUMBER VAR NUMBER IF BIN_OP VAR VAR NUMBER ASSIGN VAR BIN_OP VAR BIN_OP VAR NUMBER VAR VAR IF BIN_OP VAR VAR NUMBER IF VAR NUMBER ASSIGN VAR VAR VAR VAR ASSIGN VAR NUMBER IF VAR VAR ASSIGN VAR VAR ASSIGN VAR VAR ASSIGN VAR BIN_OP VAR NUMBER ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR VAR IF VAR NONE VAR NONE IF BIN_OP VAR NUMBER NUMBER VAR VAR VAR IF VAR VAR VAR VAR BIN_OP VAR NUMBER NUMBER IF VAR VAR VAR BIN_OP VAR NUMBER NUMBER VAR VAR VAR EXPR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR |
You are given an array $a$ consisting of $n$ integers. Indices of the array start from zero (i. e. the first element is $a_0$, the second one is $a_1$, and so on).
You can reverse at most one subarray (continuous subsegment) of this array. Recall that the subarray of $a$ with borders $l$ and $r$ is $a[l; r] = a_l, a_{l + 1}, \dots, a_{r}$.
Your task is to reverse such a subarray that the sum of elements on even positions of the resulting array is maximized (i. e. the sum of elements $a_0, a_2, \dots, a_{2k}$ for integer $k = \lfloor\frac{n-1}{2}\rfloor$ should be maximum possible).
You have to answer $t$ independent test cases.
-----Input-----
The first line of the input contains one integer $t$ ($1 \le t \le 2 \cdot 10^4$) — the number of test cases. Then $t$ test cases follow.
The first line of the test case contains one integer $n$ ($1 \le n \le 2 \cdot 10^5$) — the length of $a$. The second line of the test case contains $n$ integers $a_0, a_1, \dots, a_{n-1}$ ($1 \le a_i \le 10^9$), where $a_i$ is the $i$-th element of $a$.
It is guaranteed that the sum of $n$ does not exceed $2 \cdot 10^5$ ($\sum n \le 2 \cdot 10^5$).
-----Output-----
For each test case, print the answer on the separate line — the maximum possible sum of elements on even positions after reversing at most one subarray (continuous subsegment) of $a$.
-----Example-----
Input
4
8
1 7 3 4 7 6 2 9
5
1 2 1 2 1
10
7 8 4 5 7 6 8 9 7 3
4
3 1 2 1
Output
26
5
37
5 | for k in range(0, int(input())):
a = int(input())
l1 = list(map(int, input().split()))
l2 = []
l3 = []
A = 0
B = 0
C = 0
D = 0
s1 = 0
for u in range(0, len(l1), 2):
s1 += l1[u]
for i in range(0, len(l1) - 1, 2):
l2.append(l1[i + 1] - l1[i])
for j in range(1, len(l1) - 1, 2):
l3.append(l1[j] - l1[j + 1])
for l in l2:
A += l
if A < 0:
A = 0
if A > B:
B = A
for g in l3:
C += g
if C < 0:
C = 0
if C > D:
D = C
if B > D:
print(s1 + B)
else:
print(s1 + D) | FOR VAR FUNC_CALL VAR NUMBER FUNC_CALL VAR FUNC_CALL 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 LIST ASSIGN VAR LIST ASSIGN VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR NUMBER FUNC_CALL VAR VAR NUMBER VAR VAR VAR FOR VAR FUNC_CALL VAR NUMBER BIN_OP FUNC_CALL VAR VAR NUMBER NUMBER EXPR FUNC_CALL VAR BIN_OP VAR BIN_OP VAR NUMBER VAR VAR FOR VAR FUNC_CALL VAR NUMBER BIN_OP FUNC_CALL VAR VAR NUMBER NUMBER EXPR FUNC_CALL VAR BIN_OP VAR VAR VAR BIN_OP VAR NUMBER FOR VAR VAR VAR VAR IF VAR NUMBER ASSIGN VAR NUMBER IF VAR VAR ASSIGN VAR VAR FOR VAR VAR VAR VAR IF VAR NUMBER ASSIGN VAR NUMBER IF VAR VAR ASSIGN VAR VAR IF VAR VAR EXPR FUNC_CALL VAR BIN_OP VAR VAR EXPR FUNC_CALL VAR BIN_OP VAR VAR |
You are given an array $a$ consisting of $n$ integers. Indices of the array start from zero (i. e. the first element is $a_0$, the second one is $a_1$, and so on).
You can reverse at most one subarray (continuous subsegment) of this array. Recall that the subarray of $a$ with borders $l$ and $r$ is $a[l; r] = a_l, a_{l + 1}, \dots, a_{r}$.
Your task is to reverse such a subarray that the sum of elements on even positions of the resulting array is maximized (i. e. the sum of elements $a_0, a_2, \dots, a_{2k}$ for integer $k = \lfloor\frac{n-1}{2}\rfloor$ should be maximum possible).
You have to answer $t$ independent test cases.
-----Input-----
The first line of the input contains one integer $t$ ($1 \le t \le 2 \cdot 10^4$) — the number of test cases. Then $t$ test cases follow.
The first line of the test case contains one integer $n$ ($1 \le n \le 2 \cdot 10^5$) — the length of $a$. The second line of the test case contains $n$ integers $a_0, a_1, \dots, a_{n-1}$ ($1 \le a_i \le 10^9$), where $a_i$ is the $i$-th element of $a$.
It is guaranteed that the sum of $n$ does not exceed $2 \cdot 10^5$ ($\sum n \le 2 \cdot 10^5$).
-----Output-----
For each test case, print the answer on the separate line — the maximum possible sum of elements on even positions after reversing at most one subarray (continuous subsegment) of $a$.
-----Example-----
Input
4
8
1 7 3 4 7 6 2 9
5
1 2 1 2 1
10
7 8 4 5 7 6 8 9 7 3
4
3 1 2 1
Output
26
5
37
5 | t = int(input())
for i in range(t):
n = int(input())
a = list(map(int, input().split()))
b = [0]
psb = []
c = [0]
psc = []
tota = 0
for i in range(n):
if i % 2 == 0:
tota += a[i]
for i in range(n):
if i % 2 == 0 and i + 1 <= n - 1:
b.append(a[i + 1] - a[i])
for i in range(n):
if i % 2 == 1 and i < n - 1:
c.append(a[i] - a[i + 1])
psb.append(b[0])
psc.append(c[0])
for i in range(1, len(b)):
if psb[i - 1] + b[i] >= 0:
psb.append(psb[i - 1] + b[i])
else:
psb.append(0)
for i in range(1, len(c)):
if psc[i - 1] + c[i] >= 0:
psc.append(psc[i - 1] + c[i])
else:
psc.append(0)
x = max(psb)
y = max(psc)
z = max(x, y)
if z > 0:
ans = tota + z
else:
ans = tota
print(ans) | ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR FOR VAR FUNC_CALL 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 LIST NUMBER ASSIGN VAR LIST ASSIGN VAR LIST NUMBER ASSIGN VAR LIST ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR VAR IF BIN_OP VAR NUMBER NUMBER VAR VAR VAR FOR VAR FUNC_CALL VAR VAR IF BIN_OP VAR NUMBER NUMBER BIN_OP VAR NUMBER BIN_OP VAR NUMBER EXPR FUNC_CALL VAR BIN_OP VAR BIN_OP VAR NUMBER VAR VAR FOR VAR FUNC_CALL VAR VAR IF BIN_OP VAR NUMBER NUMBER VAR BIN_OP VAR NUMBER EXPR FUNC_CALL VAR BIN_OP VAR VAR VAR BIN_OP VAR NUMBER EXPR FUNC_CALL VAR VAR NUMBER EXPR FUNC_CALL VAR VAR NUMBER FOR VAR FUNC_CALL VAR NUMBER FUNC_CALL VAR VAR IF BIN_OP VAR BIN_OP VAR NUMBER VAR VAR NUMBER EXPR FUNC_CALL VAR BIN_OP VAR BIN_OP VAR NUMBER VAR VAR EXPR FUNC_CALL VAR NUMBER FOR VAR FUNC_CALL VAR NUMBER FUNC_CALL VAR VAR IF BIN_OP VAR BIN_OP VAR NUMBER VAR VAR NUMBER EXPR FUNC_CALL VAR BIN_OP VAR BIN_OP VAR NUMBER VAR VAR EXPR FUNC_CALL VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR VAR VAR IF VAR NUMBER ASSIGN VAR BIN_OP VAR VAR ASSIGN VAR VAR EXPR FUNC_CALL VAR VAR |
You are given an array $a$ consisting of $n$ integers. Indices of the array start from zero (i. e. the first element is $a_0$, the second one is $a_1$, and so on).
You can reverse at most one subarray (continuous subsegment) of this array. Recall that the subarray of $a$ with borders $l$ and $r$ is $a[l; r] = a_l, a_{l + 1}, \dots, a_{r}$.
Your task is to reverse such a subarray that the sum of elements on even positions of the resulting array is maximized (i. e. the sum of elements $a_0, a_2, \dots, a_{2k}$ for integer $k = \lfloor\frac{n-1}{2}\rfloor$ should be maximum possible).
You have to answer $t$ independent test cases.
-----Input-----
The first line of the input contains one integer $t$ ($1 \le t \le 2 \cdot 10^4$) — the number of test cases. Then $t$ test cases follow.
The first line of the test case contains one integer $n$ ($1 \le n \le 2 \cdot 10^5$) — the length of $a$. The second line of the test case contains $n$ integers $a_0, a_1, \dots, a_{n-1}$ ($1 \le a_i \le 10^9$), where $a_i$ is the $i$-th element of $a$.
It is guaranteed that the sum of $n$ does not exceed $2 \cdot 10^5$ ($\sum n \le 2 \cdot 10^5$).
-----Output-----
For each test case, print the answer on the separate line — the maximum possible sum of elements on even positions after reversing at most one subarray (continuous subsegment) of $a$.
-----Example-----
Input
4
8
1 7 3 4 7 6 2 9
5
1 2 1 2 1
10
7 8 4 5 7 6 8 9 7 3
4
3 1 2 1
Output
26
5
37
5 | t = int(input())
for hatt in range(t):
n = int(input())
lis = input().split()
arr = []
sum = 0
for i in range(n):
num = int(lis[i])
if i % 2 == 0:
sum += num
arr.append(-num)
else:
arr.append(num)
lis1 = []
lis2 = []
i = 1
while i < n:
lis1.append(arr[i] + arr[i - 1])
i += 2
i = 2
while i < n:
lis2.append(arr[i] + arr[i - 1])
i += 2
def maxsubsequence(lis):
allmax = 0
currmax = 0
for i in lis:
currmax += i
if currmax < 0:
currmax = 0
else:
allmax = max(currmax, allmax)
return allmax
print(sum + max(maxsubsequence(lis1), maxsubsequence(lis2))) | ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR FOR VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR LIST ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR VAR VAR IF BIN_OP VAR NUMBER NUMBER VAR VAR EXPR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR VAR ASSIGN VAR LIST ASSIGN VAR LIST ASSIGN VAR NUMBER WHILE VAR VAR EXPR FUNC_CALL VAR BIN_OP VAR VAR VAR BIN_OP VAR NUMBER VAR NUMBER ASSIGN VAR NUMBER WHILE VAR VAR EXPR FUNC_CALL VAR BIN_OP VAR VAR VAR BIN_OP VAR NUMBER VAR NUMBER FUNC_DEF ASSIGN VAR NUMBER ASSIGN VAR NUMBER FOR VAR VAR VAR VAR IF VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR VAR RETURN VAR EXPR FUNC_CALL VAR BIN_OP VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL VAR VAR |
You are given an array $a$ consisting of $n$ integers. Indices of the array start from zero (i. e. the first element is $a_0$, the second one is $a_1$, and so on).
You can reverse at most one subarray (continuous subsegment) of this array. Recall that the subarray of $a$ with borders $l$ and $r$ is $a[l; r] = a_l, a_{l + 1}, \dots, a_{r}$.
Your task is to reverse such a subarray that the sum of elements on even positions of the resulting array is maximized (i. e. the sum of elements $a_0, a_2, \dots, a_{2k}$ for integer $k = \lfloor\frac{n-1}{2}\rfloor$ should be maximum possible).
You have to answer $t$ independent test cases.
-----Input-----
The first line of the input contains one integer $t$ ($1 \le t \le 2 \cdot 10^4$) — the number of test cases. Then $t$ test cases follow.
The first line of the test case contains one integer $n$ ($1 \le n \le 2 \cdot 10^5$) — the length of $a$. The second line of the test case contains $n$ integers $a_0, a_1, \dots, a_{n-1}$ ($1 \le a_i \le 10^9$), where $a_i$ is the $i$-th element of $a$.
It is guaranteed that the sum of $n$ does not exceed $2 \cdot 10^5$ ($\sum n \le 2 \cdot 10^5$).
-----Output-----
For each test case, print the answer on the separate line — the maximum possible sum of elements on even positions after reversing at most one subarray (continuous subsegment) of $a$.
-----Example-----
Input
4
8
1 7 3 4 7 6 2 9
5
1 2 1 2 1
10
7 8 4 5 7 6 8 9 7 3
4
3 1 2 1
Output
26
5
37
5 | from sys import stdin
A = int(stdin.readline())
for u in range(0, A):
B = int(stdin.readline())
C = list(map(int, stdin.readline().split()))
even = 0
A = list()
D = list()
switch = 0
odd = 0
for y in range(0, len(C)):
if y % 2 == 0:
even += C[y]
if y % 2 == 1:
odd += C[y]
if y < len(C) - 1:
if switch == 0:
A.append(C[y + 1] - C[y])
switch = 1
else:
D.append(C[y] - C[y + 1])
switch = 0
MAX = 0
temp = 0
for r in A:
temp += r
MAX = max(MAX, temp, r)
if temp < 0:
temp = 0
temp = 0
for r in D:
temp += r
MAX = max(MAX, temp, r)
if temp < 0:
temp = 0
print(max(even + MAX, odd)) | ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR FOR VAR FUNC_CALL VAR NUMBER 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 NUMBER ASSIGN VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR ASSIGN VAR NUMBER ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR NUMBER FUNC_CALL VAR VAR IF BIN_OP VAR NUMBER NUMBER VAR VAR VAR IF BIN_OP VAR NUMBER NUMBER VAR VAR VAR IF VAR BIN_OP FUNC_CALL VAR VAR NUMBER IF VAR NUMBER EXPR FUNC_CALL VAR BIN_OP VAR BIN_OP VAR NUMBER VAR VAR ASSIGN VAR NUMBER EXPR FUNC_CALL VAR BIN_OP VAR VAR VAR BIN_OP VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR NUMBER FOR VAR VAR VAR VAR ASSIGN VAR FUNC_CALL VAR VAR VAR VAR IF VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR NUMBER FOR VAR VAR VAR VAR ASSIGN VAR FUNC_CALL VAR VAR VAR VAR IF VAR NUMBER ASSIGN VAR NUMBER EXPR FUNC_CALL VAR FUNC_CALL VAR BIN_OP VAR VAR VAR |
You are given an array $a$ consisting of $n$ integers. Indices of the array start from zero (i. e. the first element is $a_0$, the second one is $a_1$, and so on).
You can reverse at most one subarray (continuous subsegment) of this array. Recall that the subarray of $a$ with borders $l$ and $r$ is $a[l; r] = a_l, a_{l + 1}, \dots, a_{r}$.
Your task is to reverse such a subarray that the sum of elements on even positions of the resulting array is maximized (i. e. the sum of elements $a_0, a_2, \dots, a_{2k}$ for integer $k = \lfloor\frac{n-1}{2}\rfloor$ should be maximum possible).
You have to answer $t$ independent test cases.
-----Input-----
The first line of the input contains one integer $t$ ($1 \le t \le 2 \cdot 10^4$) — the number of test cases. Then $t$ test cases follow.
The first line of the test case contains one integer $n$ ($1 \le n \le 2 \cdot 10^5$) — the length of $a$. The second line of the test case contains $n$ integers $a_0, a_1, \dots, a_{n-1}$ ($1 \le a_i \le 10^9$), where $a_i$ is the $i$-th element of $a$.
It is guaranteed that the sum of $n$ does not exceed $2 \cdot 10^5$ ($\sum n \le 2 \cdot 10^5$).
-----Output-----
For each test case, print the answer on the separate line — the maximum possible sum of elements on even positions after reversing at most one subarray (continuous subsegment) of $a$.
-----Example-----
Input
4
8
1 7 3 4 7 6 2 9
5
1 2 1 2 1
10
7 8 4 5 7 6 8 9 7 3
4
3 1 2 1
Output
26
5
37
5 | def maxXsum(A, l, m, r):
x = 0
left = -1000000007
for i in range(m, l - 1, -1):
x = x + A[i]
if x > left:
left = x
x = 0
right = -1000000007
for i in range(m + 1, r + 1):
x = x + A[i]
if x > right:
right = x
return max(left + right, left, right)
def maxsubsum(A, l, r):
if l == r:
return A[l]
m = (l + r) // 2
return max(maxsubsum(A, l, m), maxsubsum(A, m + 1, r), maxXsum(A, l, m, r))
t = int(input())
for _ in range(t):
n = int(input())
A = list(map(int, input().split()))
ans = 0
for i in range(n):
if i % 2 == 0:
ans = ans + A[i]
L = [0]
R = [0]
for i in range(n):
if i % 2 == 1:
continue
if i < n - 1:
R.append(A[i + 1] - A[i])
if i > 0:
L.append(A[i - 1] - A[i])
print(
max(ans, ans + maxsubsum(L, 0, len(L) - 1), ans + maxsubsum(R, 0, len(R) - 1))
) | FUNC_DEF ASSIGN VAR NUMBER ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR VAR BIN_OP VAR NUMBER NUMBER ASSIGN VAR BIN_OP VAR VAR VAR IF VAR VAR ASSIGN VAR VAR ASSIGN VAR NUMBER ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR BIN_OP VAR NUMBER BIN_OP VAR NUMBER ASSIGN VAR BIN_OP VAR VAR VAR IF VAR VAR ASSIGN VAR VAR RETURN FUNC_CALL VAR BIN_OP VAR VAR VAR VAR FUNC_DEF IF VAR VAR RETURN VAR VAR ASSIGN VAR BIN_OP BIN_OP VAR VAR NUMBER RETURN FUNC_CALL VAR FUNC_CALL VAR VAR VAR VAR FUNC_CALL VAR VAR BIN_OP VAR NUMBER VAR FUNC_CALL VAR VAR VAR VAR VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR FOR VAR FUNC_CALL 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 NUMBER FOR VAR FUNC_CALL VAR VAR IF BIN_OP VAR NUMBER NUMBER ASSIGN VAR BIN_OP VAR VAR VAR ASSIGN VAR LIST NUMBER ASSIGN VAR LIST NUMBER FOR VAR FUNC_CALL VAR VAR IF BIN_OP VAR NUMBER NUMBER IF VAR BIN_OP VAR NUMBER EXPR FUNC_CALL VAR BIN_OP VAR BIN_OP VAR NUMBER VAR VAR IF VAR NUMBER EXPR FUNC_CALL VAR BIN_OP VAR BIN_OP VAR NUMBER VAR VAR EXPR FUNC_CALL VAR FUNC_CALL VAR VAR BIN_OP VAR FUNC_CALL VAR VAR NUMBER BIN_OP FUNC_CALL VAR VAR NUMBER BIN_OP VAR FUNC_CALL VAR VAR NUMBER BIN_OP FUNC_CALL VAR VAR NUMBER |
You are given an array $a$ consisting of $n$ integers. Indices of the array start from zero (i. e. the first element is $a_0$, the second one is $a_1$, and so on).
You can reverse at most one subarray (continuous subsegment) of this array. Recall that the subarray of $a$ with borders $l$ and $r$ is $a[l; r] = a_l, a_{l + 1}, \dots, a_{r}$.
Your task is to reverse such a subarray that the sum of elements on even positions of the resulting array is maximized (i. e. the sum of elements $a_0, a_2, \dots, a_{2k}$ for integer $k = \lfloor\frac{n-1}{2}\rfloor$ should be maximum possible).
You have to answer $t$ independent test cases.
-----Input-----
The first line of the input contains one integer $t$ ($1 \le t \le 2 \cdot 10^4$) — the number of test cases. Then $t$ test cases follow.
The first line of the test case contains one integer $n$ ($1 \le n \le 2 \cdot 10^5$) — the length of $a$. The second line of the test case contains $n$ integers $a_0, a_1, \dots, a_{n-1}$ ($1 \le a_i \le 10^9$), where $a_i$ is the $i$-th element of $a$.
It is guaranteed that the sum of $n$ does not exceed $2 \cdot 10^5$ ($\sum n \le 2 \cdot 10^5$).
-----Output-----
For each test case, print the answer on the separate line — the maximum possible sum of elements on even positions after reversing at most one subarray (continuous subsegment) of $a$.
-----Example-----
Input
4
8
1 7 3 4 7 6 2 9
5
1 2 1 2 1
10
7 8 4 5 7 6 8 9 7 3
4
3 1 2 1
Output
26
5
37
5 | def subs(lst):
if not lst:
return None
length = len(lst)
curr_max = lst[0]
global_max = lst[0]
for i in range(1, length):
curr_max = max(lst[i], curr_max + lst[i])
global_max = max(curr_max, global_max)
return global_max
def swap(lst):
if len(lst) == 1:
return lst[0]
evens = []
odds = []
for i in range(len(lst)):
if i % 2 == 0:
evens.append(lst[i])
else:
odds.append(lst[i])
diff = []
for i in range(len(odds)):
diff.append(odds[i] - evens[i])
diff2 = []
for j in range(len(odds)):
try:
diff2.append(odds[j] - evens[j + 1])
except IndexError:
pass
addition = subs(diff)
s = subs(diff2)
if not s:
s = addition - 1
net = max(addition, s)
if net < 0:
return sum(evens)
return sum(evens) + net
test = int(input())
matrix = []
for i in range(test):
n = int(input())
lst = list(map(lambda elem: int(elem), input().split(" ")))
matrix.append(lst)
for lst in matrix:
print(swap(lst)) | FUNC_DEF IF VAR RETURN NONE ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR VAR NUMBER ASSIGN VAR VAR NUMBER FOR VAR FUNC_CALL VAR NUMBER VAR ASSIGN VAR FUNC_CALL VAR VAR VAR BIN_OP VAR VAR VAR ASSIGN VAR FUNC_CALL VAR VAR VAR RETURN VAR FUNC_DEF IF FUNC_CALL VAR VAR NUMBER RETURN VAR NUMBER ASSIGN VAR LIST ASSIGN VAR LIST FOR VAR FUNC_CALL VAR FUNC_CALL VAR VAR IF BIN_OP VAR NUMBER NUMBER EXPR FUNC_CALL VAR VAR VAR EXPR FUNC_CALL VAR VAR VAR ASSIGN VAR LIST FOR VAR FUNC_CALL VAR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR BIN_OP VAR VAR VAR VAR ASSIGN VAR LIST FOR VAR FUNC_CALL VAR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR BIN_OP VAR VAR VAR BIN_OP VAR NUMBER VAR ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR VAR IF VAR ASSIGN VAR BIN_OP VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR VAR IF VAR NUMBER RETURN FUNC_CALL VAR VAR RETURN BIN_OP FUNC_CALL VAR VAR VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR LIST FOR VAR FUNC_CALL 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 STRING EXPR FUNC_CALL VAR VAR FOR VAR VAR EXPR FUNC_CALL VAR FUNC_CALL VAR VAR |
You are given an array $a$ consisting of $n$ integers. Indices of the array start from zero (i. e. the first element is $a_0$, the second one is $a_1$, and so on).
You can reverse at most one subarray (continuous subsegment) of this array. Recall that the subarray of $a$ with borders $l$ and $r$ is $a[l; r] = a_l, a_{l + 1}, \dots, a_{r}$.
Your task is to reverse such a subarray that the sum of elements on even positions of the resulting array is maximized (i. e. the sum of elements $a_0, a_2, \dots, a_{2k}$ for integer $k = \lfloor\frac{n-1}{2}\rfloor$ should be maximum possible).
You have to answer $t$ independent test cases.
-----Input-----
The first line of the input contains one integer $t$ ($1 \le t \le 2 \cdot 10^4$) — the number of test cases. Then $t$ test cases follow.
The first line of the test case contains one integer $n$ ($1 \le n \le 2 \cdot 10^5$) — the length of $a$. The second line of the test case contains $n$ integers $a_0, a_1, \dots, a_{n-1}$ ($1 \le a_i \le 10^9$), where $a_i$ is the $i$-th element of $a$.
It is guaranteed that the sum of $n$ does not exceed $2 \cdot 10^5$ ($\sum n \le 2 \cdot 10^5$).
-----Output-----
For each test case, print the answer on the separate line — the maximum possible sum of elements on even positions after reversing at most one subarray (continuous subsegment) of $a$.
-----Example-----
Input
4
8
1 7 3 4 7 6 2 9
5
1 2 1 2 1
10
7 8 4 5 7 6 8 9 7 3
4
3 1 2 1
Output
26
5
37
5 | for _ in range(int(input())):
n = int(input())
arr = list(map(int, input().split()))
odd = 0
even = 0
i = 0
j = 0
maxDif = 0
while j < n - 1:
even += arr[j]
odd += arr[j + 1]
if even > odd:
even = 0
odd = 0
i = j
maxDif = max(maxDif, odd - even)
j += 2
odd = 0
even = 0
i = 1
j = 1
while j < n - 1:
odd += arr[j]
even += arr[j + 1]
if even > odd:
even = 0
odd = 0
i = j
maxDif = max(maxDif, odd - even)
j += 2
print(sum([arr[i] for i in range(n) if i % 2 == 0]) + maxDif) | FOR VAR FUNC_CALL VAR FUNC_CALL VAR FUNC_CALL 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 NUMBER ASSIGN VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR NUMBER WHILE VAR BIN_OP VAR NUMBER VAR VAR VAR VAR VAR BIN_OP VAR NUMBER IF VAR VAR ASSIGN VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR VAR ASSIGN VAR FUNC_CALL VAR VAR BIN_OP VAR VAR VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR NUMBER WHILE VAR BIN_OP VAR NUMBER VAR VAR VAR VAR VAR BIN_OP VAR NUMBER IF VAR VAR ASSIGN VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR VAR ASSIGN VAR FUNC_CALL VAR VAR BIN_OP VAR VAR VAR NUMBER EXPR FUNC_CALL VAR BIN_OP FUNC_CALL VAR VAR VAR VAR FUNC_CALL VAR VAR BIN_OP VAR NUMBER NUMBER VAR |
You are given an array $a$ consisting of $n$ integers. Indices of the array start from zero (i. e. the first element is $a_0$, the second one is $a_1$, and so on).
You can reverse at most one subarray (continuous subsegment) of this array. Recall that the subarray of $a$ with borders $l$ and $r$ is $a[l; r] = a_l, a_{l + 1}, \dots, a_{r}$.
Your task is to reverse such a subarray that the sum of elements on even positions of the resulting array is maximized (i. e. the sum of elements $a_0, a_2, \dots, a_{2k}$ for integer $k = \lfloor\frac{n-1}{2}\rfloor$ should be maximum possible).
You have to answer $t$ independent test cases.
-----Input-----
The first line of the input contains one integer $t$ ($1 \le t \le 2 \cdot 10^4$) — the number of test cases. Then $t$ test cases follow.
The first line of the test case contains one integer $n$ ($1 \le n \le 2 \cdot 10^5$) — the length of $a$. The second line of the test case contains $n$ integers $a_0, a_1, \dots, a_{n-1}$ ($1 \le a_i \le 10^9$), where $a_i$ is the $i$-th element of $a$.
It is guaranteed that the sum of $n$ does not exceed $2 \cdot 10^5$ ($\sum n \le 2 \cdot 10^5$).
-----Output-----
For each test case, print the answer on the separate line — the maximum possible sum of elements on even positions after reversing at most one subarray (continuous subsegment) of $a$.
-----Example-----
Input
4
8
1 7 3 4 7 6 2 9
5
1 2 1 2 1
10
7 8 4 5 7 6 8 9 7 3
4
3 1 2 1
Output
26
5
37
5 | def helper(A):
MIN = []
MAX = []
temp_min = max(A) + 1
for i in range(len(A)):
temp_min = min(temp_min, A[i])
MIN += [temp_min]
temp_max = min(A) - 1
A = A[::-1]
for i in range(len(A)):
temp_max = max(temp_max, A[i])
MAX += [temp_max]
MAX = MAX[::-1]
ans = A[0] - A[-1]
for i in range(len(A) - 1):
ans = max(ans, MAX[i + 1] - MIN[i])
return ans
def find(A):
pre_even = []
pre_odd = []
temp_even = 0
temp_odd = 0
for i in range(len(A)):
if i % 2 == 0:
temp_even += A[i]
else:
temp_odd += A[i]
pre_even += [temp_even]
pre_odd += [temp_odd]
ans_odd = [0]
ans_even = []
for i in range(len(A)):
if i % 2 == 0:
ans_even += [pre_odd[i] - pre_even[i]]
else:
ans_odd += [pre_odd[i] - pre_even[i]]
ans1 = helper(ans_odd)
ans2 = helper(ans_even)
return max(pre_even[-1] + ans1, pre_even[-1] + ans2, pre_even[-1])
for _ in range(int(input())):
input()
A = list(map(int, input().strip().split()))
print(find(A)) | FUNC_DEF ASSIGN VAR LIST ASSIGN VAR LIST ASSIGN VAR BIN_OP FUNC_CALL VAR VAR NUMBER FOR VAR FUNC_CALL VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR VAR VAR VAR VAR LIST VAR ASSIGN VAR BIN_OP FUNC_CALL VAR VAR NUMBER ASSIGN VAR VAR NUMBER FOR VAR FUNC_CALL VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR VAR VAR VAR VAR LIST VAR ASSIGN VAR VAR NUMBER ASSIGN VAR BIN_OP VAR NUMBER VAR NUMBER FOR VAR FUNC_CALL VAR BIN_OP FUNC_CALL VAR VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR BIN_OP VAR BIN_OP VAR NUMBER VAR VAR RETURN VAR FUNC_DEF ASSIGN VAR LIST ASSIGN VAR LIST ASSIGN VAR NUMBER ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR FUNC_CALL VAR VAR IF BIN_OP VAR NUMBER NUMBER VAR VAR VAR VAR VAR VAR VAR LIST VAR VAR LIST VAR ASSIGN VAR LIST NUMBER ASSIGN VAR LIST FOR VAR FUNC_CALL VAR FUNC_CALL VAR VAR IF BIN_OP VAR NUMBER NUMBER VAR LIST BIN_OP VAR VAR VAR VAR VAR LIST BIN_OP VAR VAR VAR VAR ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR VAR RETURN FUNC_CALL VAR BIN_OP VAR NUMBER VAR BIN_OP VAR NUMBER VAR VAR NUMBER FOR VAR FUNC_CALL VAR FUNC_CALL VAR FUNC_CALL VAR EXPR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL FUNC_CALL VAR EXPR FUNC_CALL VAR FUNC_CALL VAR VAR |
You are given an array $a$ consisting of $n$ integers. Indices of the array start from zero (i. e. the first element is $a_0$, the second one is $a_1$, and so on).
You can reverse at most one subarray (continuous subsegment) of this array. Recall that the subarray of $a$ with borders $l$ and $r$ is $a[l; r] = a_l, a_{l + 1}, \dots, a_{r}$.
Your task is to reverse such a subarray that the sum of elements on even positions of the resulting array is maximized (i. e. the sum of elements $a_0, a_2, \dots, a_{2k}$ for integer $k = \lfloor\frac{n-1}{2}\rfloor$ should be maximum possible).
You have to answer $t$ independent test cases.
-----Input-----
The first line of the input contains one integer $t$ ($1 \le t \le 2 \cdot 10^4$) — the number of test cases. Then $t$ test cases follow.
The first line of the test case contains one integer $n$ ($1 \le n \le 2 \cdot 10^5$) — the length of $a$. The second line of the test case contains $n$ integers $a_0, a_1, \dots, a_{n-1}$ ($1 \le a_i \le 10^9$), where $a_i$ is the $i$-th element of $a$.
It is guaranteed that the sum of $n$ does not exceed $2 \cdot 10^5$ ($\sum n \le 2 \cdot 10^5$).
-----Output-----
For each test case, print the answer on the separate line — the maximum possible sum of elements on even positions after reversing at most one subarray (continuous subsegment) of $a$.
-----Example-----
Input
4
8
1 7 3 4 7 6 2 9
5
1 2 1 2 1
10
7 8 4 5 7 6 8 9 7 3
4
3 1 2 1
Output
26
5
37
5 | def maxSubArraySum(a, size):
max_so_far = -99999999999999999
max_ending_here = 0
for i in range(0, size):
max_ending_here = max_ending_here + a[i]
if max_so_far < max_ending_here:
max_so_far = max_ending_here
if max_ending_here < 0:
max_ending_here = 0
return max_so_far
t = int(input())
while t:
t += -1
n = int(input())
l = list(map(int, input().split()))
a1 = []
a2 = []
tp = 0
for i in range(0, n, 2):
tp += l[i]
if i + 1 < n:
a1.append(l[i + 1] - l[i])
for i in range(1, n, 2):
if i + 1 < n:
a2.append(l[i] - l[i + 1])
tmp = max(maxSubArraySum(a1, len(a1)), maxSubArraySum(a2, len(a2)))
print(max(tp, tp + tmp)) | FUNC_DEF ASSIGN VAR NUMBER ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR NUMBER VAR ASSIGN VAR BIN_OP VAR VAR VAR IF VAR VAR ASSIGN VAR VAR IF VAR NUMBER ASSIGN VAR NUMBER RETURN VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR WHILE VAR VAR NUMBER 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 LIST ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR NUMBER VAR NUMBER VAR VAR VAR IF BIN_OP VAR NUMBER VAR EXPR FUNC_CALL VAR BIN_OP VAR BIN_OP VAR NUMBER VAR VAR FOR VAR FUNC_CALL VAR NUMBER VAR NUMBER IF BIN_OP VAR NUMBER VAR EXPR FUNC_CALL VAR BIN_OP VAR VAR VAR BIN_OP VAR NUMBER ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL VAR VAR FUNC_CALL VAR VAR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR FUNC_CALL VAR VAR BIN_OP VAR VAR |
You are given an array $a$ consisting of $n$ integers. Indices of the array start from zero (i. e. the first element is $a_0$, the second one is $a_1$, and so on).
You can reverse at most one subarray (continuous subsegment) of this array. Recall that the subarray of $a$ with borders $l$ and $r$ is $a[l; r] = a_l, a_{l + 1}, \dots, a_{r}$.
Your task is to reverse such a subarray that the sum of elements on even positions of the resulting array is maximized (i. e. the sum of elements $a_0, a_2, \dots, a_{2k}$ for integer $k = \lfloor\frac{n-1}{2}\rfloor$ should be maximum possible).
You have to answer $t$ independent test cases.
-----Input-----
The first line of the input contains one integer $t$ ($1 \le t \le 2 \cdot 10^4$) — the number of test cases. Then $t$ test cases follow.
The first line of the test case contains one integer $n$ ($1 \le n \le 2 \cdot 10^5$) — the length of $a$. The second line of the test case contains $n$ integers $a_0, a_1, \dots, a_{n-1}$ ($1 \le a_i \le 10^9$), where $a_i$ is the $i$-th element of $a$.
It is guaranteed that the sum of $n$ does not exceed $2 \cdot 10^5$ ($\sum n \le 2 \cdot 10^5$).
-----Output-----
For each test case, print the answer on the separate line — the maximum possible sum of elements on even positions after reversing at most one subarray (continuous subsegment) of $a$.
-----Example-----
Input
4
8
1 7 3 4 7 6 2 9
5
1 2 1 2 1
10
7 8 4 5 7 6 8 9 7 3
4
3 1 2 1
Output
26
5
37
5 | import sys
input = sys.stdin.readline
f = lambda: list(map(int, input().strip("\n").split()))
res = []
for _ in range(int(input())):
n = int(input())
inp = f()
s = 0
for i in inp[0:n:2]:
s += i
ans = cur = 0
for i in range(1, n, 2):
cur += inp[i] - inp[i - 1]
if cur < 0:
cur = 0
ans = max(ans, cur)
cur = 0
for i in range(2, n, 2):
cur += inp[i - 1] - inp[i]
if cur < 0:
cur = 0
ans = max(ans, cur)
ans += s
res.append(ans)
print("\n".join(map(str, res))) | IMPORT ASSIGN VAR VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL FUNC_CALL VAR STRING ASSIGN VAR LIST FOR VAR FUNC_CALL VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR ASSIGN VAR NUMBER FOR VAR VAR NUMBER VAR NUMBER VAR VAR ASSIGN VAR VAR NUMBER FOR VAR FUNC_CALL VAR NUMBER VAR NUMBER VAR BIN_OP VAR VAR VAR BIN_OP VAR NUMBER IF VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR VAR ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR NUMBER VAR NUMBER VAR BIN_OP VAR BIN_OP VAR NUMBER VAR VAR IF VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR VAR VAR VAR EXPR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR FUNC_CALL STRING FUNC_CALL VAR VAR VAR |
You are given an array $a$ consisting of $n$ integers. Indices of the array start from zero (i. e. the first element is $a_0$, the second one is $a_1$, and so on).
You can reverse at most one subarray (continuous subsegment) of this array. Recall that the subarray of $a$ with borders $l$ and $r$ is $a[l; r] = a_l, a_{l + 1}, \dots, a_{r}$.
Your task is to reverse such a subarray that the sum of elements on even positions of the resulting array is maximized (i. e. the sum of elements $a_0, a_2, \dots, a_{2k}$ for integer $k = \lfloor\frac{n-1}{2}\rfloor$ should be maximum possible).
You have to answer $t$ independent test cases.
-----Input-----
The first line of the input contains one integer $t$ ($1 \le t \le 2 \cdot 10^4$) — the number of test cases. Then $t$ test cases follow.
The first line of the test case contains one integer $n$ ($1 \le n \le 2 \cdot 10^5$) — the length of $a$. The second line of the test case contains $n$ integers $a_0, a_1, \dots, a_{n-1}$ ($1 \le a_i \le 10^9$), where $a_i$ is the $i$-th element of $a$.
It is guaranteed that the sum of $n$ does not exceed $2 \cdot 10^5$ ($\sum n \le 2 \cdot 10^5$).
-----Output-----
For each test case, print the answer on the separate line — the maximum possible sum of elements on even positions after reversing at most one subarray (continuous subsegment) of $a$.
-----Example-----
Input
4
8
1 7 3 4 7 6 2 9
5
1 2 1 2 1
10
7 8 4 5 7 6 8 9 7 3
4
3 1 2 1
Output
26
5
37
5 | def maxSubArraySum(a):
max_so_far = -10000000000 - 1
max_ending_here = 0
for i in range(0, len(a)):
max_ending_here = max_ending_here + a[i]
if max_so_far < max_ending_here:
max_so_far = max_ending_here
if max_ending_here < 0:
max_ending_here = 0
return max_so_far
t = int(input())
for tt in range(0, t):
n = int(input())
ai = input().split(" ")
a = []
for i in range(0, n):
a.append(int(ai[i]))
b = []
c = []
i = 1
while i < n:
b.append(a[i] - a[i - 1])
i += 2
i = 1
while i + 1 < n:
c.append(a[i] - a[i + 1])
i += 2
max_so_far = max(maxSubArraySum(b), maxSubArraySum(c))
su = 0
for i in range(0, n):
if i % 2 == 0:
su += a[i]
print(su + max(0, max_so_far)) | FUNC_DEF ASSIGN VAR BIN_OP NUMBER NUMBER ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR NUMBER FUNC_CALL VAR VAR ASSIGN VAR BIN_OP VAR VAR VAR IF VAR VAR ASSIGN VAR VAR IF VAR NUMBER ASSIGN VAR NUMBER RETURN VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR FOR VAR FUNC_CALL VAR NUMBER VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL FUNC_CALL VAR STRING ASSIGN VAR LIST FOR VAR FUNC_CALL VAR NUMBER VAR EXPR FUNC_CALL VAR FUNC_CALL VAR VAR VAR ASSIGN VAR LIST ASSIGN VAR LIST ASSIGN VAR NUMBER WHILE VAR VAR EXPR FUNC_CALL VAR BIN_OP VAR VAR VAR BIN_OP VAR NUMBER VAR NUMBER ASSIGN VAR NUMBER WHILE BIN_OP VAR NUMBER VAR EXPR FUNC_CALL VAR BIN_OP VAR VAR VAR BIN_OP VAR NUMBER VAR NUMBER ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL VAR VAR ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR NUMBER VAR IF BIN_OP VAR NUMBER NUMBER VAR VAR VAR EXPR FUNC_CALL VAR BIN_OP VAR FUNC_CALL VAR NUMBER VAR |
You are given an array $a$ consisting of $n$ integers. Indices of the array start from zero (i. e. the first element is $a_0$, the second one is $a_1$, and so on).
You can reverse at most one subarray (continuous subsegment) of this array. Recall that the subarray of $a$ with borders $l$ and $r$ is $a[l; r] = a_l, a_{l + 1}, \dots, a_{r}$.
Your task is to reverse such a subarray that the sum of elements on even positions of the resulting array is maximized (i. e. the sum of elements $a_0, a_2, \dots, a_{2k}$ for integer $k = \lfloor\frac{n-1}{2}\rfloor$ should be maximum possible).
You have to answer $t$ independent test cases.
-----Input-----
The first line of the input contains one integer $t$ ($1 \le t \le 2 \cdot 10^4$) — the number of test cases. Then $t$ test cases follow.
The first line of the test case contains one integer $n$ ($1 \le n \le 2 \cdot 10^5$) — the length of $a$. The second line of the test case contains $n$ integers $a_0, a_1, \dots, a_{n-1}$ ($1 \le a_i \le 10^9$), where $a_i$ is the $i$-th element of $a$.
It is guaranteed that the sum of $n$ does not exceed $2 \cdot 10^5$ ($\sum n \le 2 \cdot 10^5$).
-----Output-----
For each test case, print the answer on the separate line — the maximum possible sum of elements on even positions after reversing at most one subarray (continuous subsegment) of $a$.
-----Example-----
Input
4
8
1 7 3 4 7 6 2 9
5
1 2 1 2 1
10
7 8 4 5 7 6 8 9 7 3
4
3 1 2 1
Output
26
5
37
5 | n = input()
for i in range(int(n)):
n_i = input()
numbers = list(map(int, input().split(" ")))
a = numbers[1::2]
b = numbers[:-1:2]
before = [(i - j) for i, j in zip(a, b)]
a = numbers[1:-1][::2]
b = numbers[2:][::2]
after = [(i - j) for i, j in zip(a, b)]
start = sum(numbers[::2])
max_ = 0
for diffs in [before, after]:
csum = 0
msum = 0
for i in range(len(diffs)):
csum = csum + diffs[i]
msum = min(msum, csum)
d = csum - msum
if d > max_:
max_ = d
print(start + max_) | ASSIGN VAR FUNC_CALL VAR FOR VAR FUNC_CALL VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR STRING ASSIGN VAR VAR NUMBER NUMBER ASSIGN VAR VAR NUMBER NUMBER ASSIGN VAR BIN_OP VAR VAR VAR VAR FUNC_CALL VAR VAR VAR ASSIGN VAR VAR NUMBER NUMBER NUMBER ASSIGN VAR VAR NUMBER NUMBER ASSIGN VAR BIN_OP VAR VAR VAR VAR FUNC_CALL VAR VAR VAR ASSIGN VAR FUNC_CALL VAR VAR NUMBER ASSIGN VAR NUMBER FOR VAR LIST VAR VAR ASSIGN VAR NUMBER ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR FUNC_CALL VAR VAR ASSIGN VAR BIN_OP VAR VAR VAR ASSIGN VAR FUNC_CALL VAR VAR VAR ASSIGN VAR BIN_OP VAR VAR IF VAR VAR ASSIGN VAR VAR EXPR FUNC_CALL VAR BIN_OP VAR VAR |
You are given an array $a$ consisting of $n$ integers. Indices of the array start from zero (i. e. the first element is $a_0$, the second one is $a_1$, and so on).
You can reverse at most one subarray (continuous subsegment) of this array. Recall that the subarray of $a$ with borders $l$ and $r$ is $a[l; r] = a_l, a_{l + 1}, \dots, a_{r}$.
Your task is to reverse such a subarray that the sum of elements on even positions of the resulting array is maximized (i. e. the sum of elements $a_0, a_2, \dots, a_{2k}$ for integer $k = \lfloor\frac{n-1}{2}\rfloor$ should be maximum possible).
You have to answer $t$ independent test cases.
-----Input-----
The first line of the input contains one integer $t$ ($1 \le t \le 2 \cdot 10^4$) — the number of test cases. Then $t$ test cases follow.
The first line of the test case contains one integer $n$ ($1 \le n \le 2 \cdot 10^5$) — the length of $a$. The second line of the test case contains $n$ integers $a_0, a_1, \dots, a_{n-1}$ ($1 \le a_i \le 10^9$), where $a_i$ is the $i$-th element of $a$.
It is guaranteed that the sum of $n$ does not exceed $2 \cdot 10^5$ ($\sum n \le 2 \cdot 10^5$).
-----Output-----
For each test case, print the answer on the separate line — the maximum possible sum of elements on even positions after reversing at most one subarray (continuous subsegment) of $a$.
-----Example-----
Input
4
8
1 7 3 4 7 6 2 9
5
1 2 1 2 1
10
7 8 4 5 7 6 8 9 7 3
4
3 1 2 1
Output
26
5
37
5 | import sys
input = lambda: sys.stdin.readline().rstrip()
for _ in range(int(input())):
n = int(input())
a = list(map(int, input().split()))
ans = extra = 0
s = [0, 0]
mn = [10000000000.0, 10000000000.0]
t = 0
for i in range(n):
s[t] += a[i]
if t == 0:
ans += a[i]
diff = s[1] - s[0]
extra = max(extra, diff)
mn[t] = min(mn[t], diff)
extra = max(extra, diff - mn[t])
t ^= 1
print(ans + extra) | IMPORT ASSIGN VAR FUNC_CALL FUNC_CALL VAR FOR VAR FUNC_CALL VAR FUNC_CALL VAR FUNC_CALL 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 VAR NUMBER ASSIGN VAR LIST NUMBER NUMBER ASSIGN VAR LIST NUMBER NUMBER ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR VAR VAR VAR VAR VAR IF VAR NUMBER VAR VAR VAR ASSIGN VAR BIN_OP VAR NUMBER VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR VAR ASSIGN VAR VAR FUNC_CALL VAR VAR VAR VAR ASSIGN VAR FUNC_CALL VAR VAR BIN_OP VAR VAR VAR VAR NUMBER EXPR FUNC_CALL VAR BIN_OP VAR VAR |
You are given an array $a$ consisting of $n$ integers. Indices of the array start from zero (i. e. the first element is $a_0$, the second one is $a_1$, and so on).
You can reverse at most one subarray (continuous subsegment) of this array. Recall that the subarray of $a$ with borders $l$ and $r$ is $a[l; r] = a_l, a_{l + 1}, \dots, a_{r}$.
Your task is to reverse such a subarray that the sum of elements on even positions of the resulting array is maximized (i. e. the sum of elements $a_0, a_2, \dots, a_{2k}$ for integer $k = \lfloor\frac{n-1}{2}\rfloor$ should be maximum possible).
You have to answer $t$ independent test cases.
-----Input-----
The first line of the input contains one integer $t$ ($1 \le t \le 2 \cdot 10^4$) — the number of test cases. Then $t$ test cases follow.
The first line of the test case contains one integer $n$ ($1 \le n \le 2 \cdot 10^5$) — the length of $a$. The second line of the test case contains $n$ integers $a_0, a_1, \dots, a_{n-1}$ ($1 \le a_i \le 10^9$), where $a_i$ is the $i$-th element of $a$.
It is guaranteed that the sum of $n$ does not exceed $2 \cdot 10^5$ ($\sum n \le 2 \cdot 10^5$).
-----Output-----
For each test case, print the answer on the separate line — the maximum possible sum of elements on even positions after reversing at most one subarray (continuous subsegment) of $a$.
-----Example-----
Input
4
8
1 7 3 4 7 6 2 9
5
1 2 1 2 1
10
7 8 4 5 7 6 8 9 7 3
4
3 1 2 1
Output
26
5
37
5 | def KadanesAlgorithm(a):
res = -float("inf")
s = 0
for i in range(len(a)):
s = max(s + a[i], a[i])
res = max(s, res)
return res
for t in range(int(input())):
n = int(input())
(*a,) = map(int, input().split())
ans = 0 if (n - 1) % 2 == 1 else a[n - 1]
evens, odds = [], []
for i in range(n - 1):
if i % 2 == 0:
ans += a[i]
evens.append(a[i + 1] - a[i])
else:
odds.append(a[i] - a[i + 1])
d = max(KadanesAlgorithm(evens), KadanesAlgorithm(odds))
ans = max(ans, ans + d)
print(ans) | FUNC_DEF ASSIGN VAR FUNC_CALL VAR STRING ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR BIN_OP VAR VAR VAR VAR VAR ASSIGN VAR FUNC_CALL VAR VAR VAR RETURN VAR FOR VAR FUNC_CALL VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR BIN_OP BIN_OP VAR NUMBER NUMBER NUMBER NUMBER VAR BIN_OP VAR NUMBER ASSIGN VAR VAR LIST LIST FOR VAR FUNC_CALL VAR BIN_OP VAR NUMBER IF BIN_OP VAR NUMBER NUMBER VAR VAR VAR EXPR FUNC_CALL VAR BIN_OP VAR BIN_OP VAR NUMBER VAR VAR EXPR FUNC_CALL VAR BIN_OP VAR VAR VAR BIN_OP VAR NUMBER ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR VAR BIN_OP VAR VAR EXPR FUNC_CALL VAR VAR |
You are given an array $a$ consisting of $n$ integers. Indices of the array start from zero (i. e. the first element is $a_0$, the second one is $a_1$, and so on).
You can reverse at most one subarray (continuous subsegment) of this array. Recall that the subarray of $a$ with borders $l$ and $r$ is $a[l; r] = a_l, a_{l + 1}, \dots, a_{r}$.
Your task is to reverse such a subarray that the sum of elements on even positions of the resulting array is maximized (i. e. the sum of elements $a_0, a_2, \dots, a_{2k}$ for integer $k = \lfloor\frac{n-1}{2}\rfloor$ should be maximum possible).
You have to answer $t$ independent test cases.
-----Input-----
The first line of the input contains one integer $t$ ($1 \le t \le 2 \cdot 10^4$) — the number of test cases. Then $t$ test cases follow.
The first line of the test case contains one integer $n$ ($1 \le n \le 2 \cdot 10^5$) — the length of $a$. The second line of the test case contains $n$ integers $a_0, a_1, \dots, a_{n-1}$ ($1 \le a_i \le 10^9$), where $a_i$ is the $i$-th element of $a$.
It is guaranteed that the sum of $n$ does not exceed $2 \cdot 10^5$ ($\sum n \le 2 \cdot 10^5$).
-----Output-----
For each test case, print the answer on the separate line — the maximum possible sum of elements on even positions after reversing at most one subarray (continuous subsegment) of $a$.
-----Example-----
Input
4
8
1 7 3 4 7 6 2 9
5
1 2 1 2 1
10
7 8 4 5 7 6 8 9 7 3
4
3 1 2 1
Output
26
5
37
5 | from sys import stdin
for _ in range(int(stdin.readline())):
n = int(stdin.readline())
a = list(map(int, stdin.readline().split()))
res = 0
max_so_far_before = 0
max_ending_before = 0
max_so_far_after = 0
max_ending_after = 0
for i, v in enumerate(a):
if i % 2 == 0:
res += v
else:
max_ending_before += v - a[i - 1]
if max_ending_before < 0:
max_ending_before = 0
if max_so_far_before < max_ending_before:
max_so_far_before = max_ending_before
if i < n - 1:
max_ending_after += v - a[i + 1]
if max_ending_after < 0:
max_ending_after = 0
if max_so_far_after < max_ending_after:
max_so_far_after = max_ending_after
res += max(max_so_far_after, max_so_far_before)
print(res) | FOR VAR FUNC_CALL VAR FUNC_CALL VAR FUNC_CALL 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 NUMBER ASSIGN VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR NUMBER FOR VAR VAR FUNC_CALL VAR VAR IF BIN_OP VAR NUMBER NUMBER VAR VAR VAR BIN_OP VAR VAR BIN_OP VAR NUMBER IF VAR NUMBER ASSIGN VAR NUMBER IF VAR VAR ASSIGN VAR VAR IF VAR BIN_OP VAR NUMBER VAR BIN_OP VAR VAR BIN_OP VAR NUMBER IF VAR NUMBER ASSIGN VAR NUMBER IF VAR VAR ASSIGN VAR VAR VAR FUNC_CALL VAR VAR VAR EXPR FUNC_CALL VAR VAR |
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