description stringlengths 171 4k | code stringlengths 94 3.98k | normalized_code stringlengths 57 4.99k |
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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 getMaxSubSum(arr):
maxSum = 0
partialSum = 0
for item in arr:
partialSum += item
maxSum = max(maxSum, partialSum)
if partialSum < 0:
partialSum = 0
return maxSum
t = int(input())
for _ in range(t):
n = int(input())
a = [int(x) for x in input().split()]
dif0 = list()
dif1 = list()
s = 0
for i in range(n // 2):
s += a[2 * i]
dif0.append(a[2 * i + 1] - a[2 * i])
if n % 2 == 1:
s += a[-1]
for i in range((n - 1) // 2):
dif1.append(a[2 * i + 1] - a[2 * i + 2])
print(s + max(getMaxSubSum(dif0), getMaxSubSum(dif1))) | FUNC_DEF ASSIGN VAR NUMBER ASSIGN VAR NUMBER FOR VAR VAR VAR VAR ASSIGN VAR FUNC_CALL VAR VAR VAR IF VAR NUMBER ASSIGN 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 ASSIGN VAR FUNC_CALL VAR VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL 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 BIN_OP BIN_OP NUMBER VAR NUMBER VAR BIN_OP NUMBER VAR IF BIN_OP VAR NUMBER NUMBER VAR VAR NUMBER FOR VAR FUNC_CALL VAR BIN_OP BIN_OP VAR NUMBER NUMBER EXPR FUNC_CALL VAR BIN_OP VAR BIN_OP BIN_OP NUMBER VAR NUMBER VAR BIN_OP BIN_OP 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 | def kadane(lis):
max_so_far = 0
size = len(lis)
max_curr = 0
for i in range(0, size):
max_curr = max(lis[i], lis[i] + max_curr)
if max_so_far < max_curr:
max_so_far = max_curr
return max_so_far
for _ in range(int(input())):
n = int(input())
l = list(map(int, input().split()))
s = sum(l[0:n:2])
even, odd = [], []
for i in range(n - 1):
if i % 2:
odd.append(l[i] - l[i + 1])
else:
even.append(l[i + 1] - l[i])
xtra = max(kadane(even), kadane(odd), 0)
print(s + xtra) | FUNC_DEF ASSIGN VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR NUMBER VAR ASSIGN VAR FUNC_CALL VAR VAR VAR BIN_OP VAR VAR VAR 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 FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR VAR NUMBER VAR NUMBER ASSIGN VAR VAR LIST LIST 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 ASSIGN VAR FUNC_CALL 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 | t = int(input())
for _ in range(t):
n = int(input())
a = list(map(int, input().split()))
even = [0]
odd = [0]
mino = 0
mine = 0
super = 0
i = 1
summer = a[0]
while i < n:
if i % 2 == 0:
summer += a[i]
olde = even[-1]
super = max(super, -(a[i] - a[i - 1]) - mine + olde)
mine = min(mine, -(a[i] - a[i - 1]) + olde)
even.append(-(a[i] - a[i - 1]) + olde)
else:
oldo = odd[-1]
super = max(super, a[i] - a[i - 1] + oldo - mino)
mino = min(mino, a[i] - a[i - 1] + oldo)
odd.append(a[i] - a[i - 1] + oldo)
i += 1
print(summer + super) | 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 NUMBER ASSIGN VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR VAR NUMBER WHILE VAR VAR IF BIN_OP VAR NUMBER NUMBER VAR VAR VAR ASSIGN VAR VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR BIN_OP BIN_OP BIN_OP VAR VAR VAR BIN_OP VAR NUMBER VAR VAR ASSIGN VAR FUNC_CALL VAR VAR BIN_OP BIN_OP VAR VAR VAR BIN_OP VAR NUMBER VAR EXPR FUNC_CALL VAR BIN_OP BIN_OP VAR VAR VAR BIN_OP VAR NUMBER VAR ASSIGN VAR VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR BIN_OP BIN_OP BIN_OP VAR VAR VAR BIN_OP VAR NUMBER VAR VAR ASSIGN VAR FUNC_CALL VAR VAR BIN_OP BIN_OP VAR VAR VAR BIN_OP VAR NUMBER VAR EXPR FUNC_CALL VAR BIN_OP BIN_OP VAR VAR VAR BIN_OP VAR NUMBER 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
input = sys.stdin.buffer.readline
T = int(input())
for _ in range(T):
n = int(input())
a = list(map(int, input().split()))
lr = []
rl = []
ans = 0
plus_lr = 0
last_lr = 0
plus_rl = 0
last_rl = 0
for i in range(0, n, 2):
ans += a[i]
if i + 1 < n:
e = a[i + 1] - a[i]
last_lr = max(e, last_lr + e)
if last_lr > plus_lr:
plus_lr = last_lr
if i - 1 >= 0:
e = a[i - 1] - a[i]
last_rl = max(e, last_rl + e)
if last_rl > plus_rl:
plus_rl = last_rl
print(ans + max(plus_rl, plus_lr)) | 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 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 VAR NUMBER VAR VAR VAR IF BIN_OP VAR NUMBER VAR ASSIGN VAR BIN_OP VAR BIN_OP VAR NUMBER VAR VAR ASSIGN VAR FUNC_CALL VAR VAR BIN_OP VAR VAR IF VAR VAR ASSIGN VAR VAR IF BIN_OP VAR NUMBER NUMBER ASSIGN VAR BIN_OP VAR BIN_OP VAR NUMBER VAR VAR ASSIGN VAR FUNC_CALL VAR VAR BIN_OP VAR VAR 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 | t = int(input())
def kadane(r):
sum_here, sum_max = 0, 0
for i in r:
sum_here = max(i, i + sum_here)
sum_max = max(sum_max, sum_here)
return sum_max
for _ in range(t):
n = int(input())
l = [int(i) for i in input().split()]
if n % 2:
l.append(0)
sum = 0
p, q = [], []
for i in range(0, n, 2):
sum += l[i]
p.append(l[i + 1] - l[i])
for i in range(1, n - 1, 2):
q.append(l[i] - l[i + 1])
print(max(kadane(p), kadane(q)) + sum) | ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR FUNC_DEF ASSIGN VAR VAR NUMBER NUMBER FOR 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 VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR VAR VAR FUNC_CALL FUNC_CALL VAR IF BIN_OP VAR NUMBER EXPR FUNC_CALL VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR VAR LIST LIST FOR VAR FUNC_CALL VAR NUMBER VAR NUMBER VAR VAR VAR 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 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 | t = int(input())
for u in range(t):
n = int(input())
a = list(map(int, input().split(" ")))
b = [(a[i] - a[i - 1]) for i in range(1, n, 2)]
b.append(0)
m1 = [b[0]]
for i in range(1, b.__len__()):
if m1[i - 1] >= 0:
m1.append(m1[i - 1] + b[i])
else:
m1.append(b[i])
ans = sum([a[i] for i in range(n) if i - 1 & 1]) + max(max(m1), 0)
c = [(a[i - 1] - a[i]) for i in range(2, n, 2)]
c.append(0)
m2 = [c[0]]
for i in range(1, c.__len__()):
if m2[i - 1] >= 0:
m2.append(m2[i - 1] + c[i])
else:
m2.append(c[i])
ans = max(ans, sum([a[i] for i in range(n) if i - 1 & 1]) + max(max(m2), 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 STRING ASSIGN VAR BIN_OP VAR VAR VAR BIN_OP VAR NUMBER VAR FUNC_CALL VAR NUMBER VAR NUMBER EXPR FUNC_CALL VAR NUMBER ASSIGN VAR LIST VAR NUMBER FOR VAR FUNC_CALL VAR NUMBER FUNC_CALL VAR IF VAR BIN_OP VAR NUMBER NUMBER EXPR FUNC_CALL VAR BIN_OP VAR BIN_OP VAR NUMBER VAR VAR EXPR FUNC_CALL VAR VAR VAR ASSIGN VAR BIN_OP FUNC_CALL VAR VAR VAR VAR FUNC_CALL VAR VAR BIN_OP BIN_OP VAR NUMBER NUMBER FUNC_CALL VAR FUNC_CALL VAR VAR NUMBER ASSIGN VAR BIN_OP VAR BIN_OP VAR NUMBER VAR VAR VAR FUNC_CALL VAR NUMBER VAR NUMBER EXPR FUNC_CALL VAR NUMBER ASSIGN VAR LIST VAR NUMBER FOR VAR FUNC_CALL VAR NUMBER FUNC_CALL VAR IF VAR BIN_OP VAR NUMBER NUMBER EXPR FUNC_CALL VAR BIN_OP VAR BIN_OP VAR NUMBER VAR VAR EXPR FUNC_CALL VAR VAR VAR ASSIGN VAR FUNC_CALL VAR VAR BIN_OP FUNC_CALL VAR VAR VAR VAR FUNC_CALL VAR VAR BIN_OP BIN_OP VAR NUMBER NUMBER FUNC_CALL 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 | def get(a):
dp = [0] * len(a)
if len(a) >= 1:
dp[0] = max(dp[0], a[0])
for i in range(1, len(a)):
dp[i] = max(0, dp[i - 1] + a[i])
return max(dp)
return 0
t = int(input())
for _ in range(t):
n = int(input())
arr = [int(i) for i in input().split()]
ans = 0
for i in range(n):
if i % 2 == 0:
ans += arr[i]
even = []
odd = []
for i in range(0, n - 1, 2):
even.append(arr[i + 1] - arr[i])
for i in range(1, n - 1, 2):
odd.append(arr[i] - arr[i + 1])
x = get(even)
y = get(odd)
ans += max(0, x, y)
print(ans) | FUNC_DEF ASSIGN VAR BIN_OP LIST NUMBER FUNC_CALL VAR VAR IF FUNC_CALL VAR VAR NUMBER ASSIGN VAR NUMBER FUNC_CALL VAR VAR NUMBER VAR NUMBER FOR VAR FUNC_CALL VAR NUMBER FUNC_CALL VAR VAR ASSIGN VAR VAR FUNC_CALL VAR NUMBER BIN_OP VAR BIN_OP VAR NUMBER VAR VAR RETURN FUNC_CALL VAR VAR RETURN NUMBER 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 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 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 | for _ in range(int(input())):
n = int(input())
arr = list(map(int, input().split()))
mn_e = 10**9
mn_o = 0
curr = 0
mx = 0
total = 0
for i in range(n):
num = arr[i]
if i % 2 == 0:
total += num
num = -num
curr += num
if i % 2:
mx = max(mx, curr - mn_o)
mn_o = min(mn_o, curr)
else:
mx = max(mx, curr - mn_e)
mn_e = min(mn_e, curr)
print(total + mx) | 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 NUMBER NUMBER ASSIGN VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR VAR ASSIGN VAR VAR VAR IF BIN_OP VAR NUMBER NUMBER VAR VAR ASSIGN VAR VAR VAR VAR IF BIN_OP VAR NUMBER ASSIGN VAR FUNC_CALL 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 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 max_subarray(a):
local_max = 0
global_max = float("-inf")
for i in range(len(a)):
local_max = max(a[i], a[i] + local_max)
if local_max > global_max:
global_max = local_max
return global_max
t = int(input())
for _ in range(t):
n = int(input())
a = list(map(int, input().split()))
first_arr = [(a[2 * i + 1] - a[2 * i]) for i in range(n // 2)]
first_max = max_subarray(first_arr)
second_arr = [(a[2 * i + 1] - a[2 * i + 2]) for i in range((n - 1) // 2)]
second_max = max_subarray(second_arr)
ans = sum(a[::2]) + max(first_max, second_max, 0)
print(ans) | FUNC_DEF ASSIGN VAR NUMBER ASSIGN VAR FUNC_CALL VAR STRING 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 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 BIN_OP VAR BIN_OP BIN_OP NUMBER VAR NUMBER VAR BIN_OP NUMBER VAR VAR FUNC_CALL VAR BIN_OP VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR BIN_OP VAR BIN_OP BIN_OP NUMBER VAR NUMBER VAR BIN_OP BIN_OP NUMBER VAR NUMBER VAR FUNC_CALL VAR BIN_OP BIN_OP VAR NUMBER NUMBER ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR BIN_OP FUNC_CALL VAR VAR NUMBER 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 i in range(t):
n = int(input())
a = [int(v) for v in input().split()]
v1 = []
v2 = []
m = 0
s = sum(a[::2])
for j in range(0, n - n % 2, 2):
v1.append(a[j + 1] - a[j])
for j in range(1, n - (1 - n % 2), 2):
v2.append(a[j] - a[j + 1])
dp = [0]
for j in range(len(v1)):
dp.append(max(v1[j] + dp[-1], v1[j]))
m = max(dp)
dp = [0]
for j in range(len(v2)):
dp.append(max(v2[j] + dp[-1], v2[j]))
m = max(max(dp), m, 0)
print(s + 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 LIST ASSIGN VAR LIST ASSIGN VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR NUMBER FOR VAR FUNC_CALL VAR NUMBER BIN_OP VAR 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 BIN_OP NUMBER BIN_OP VAR NUMBER NUMBER EXPR FUNC_CALL VAR BIN_OP VAR VAR VAR BIN_OP VAR NUMBER ASSIGN VAR LIST NUMBER FOR VAR FUNC_CALL VAR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR FUNC_CALL VAR BIN_OP VAR VAR VAR NUMBER VAR VAR ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR LIST NUMBER FOR VAR FUNC_CALL VAR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR FUNC_CALL VAR BIN_OP VAR VAR VAR NUMBER VAR VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL 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 | from sys import stdin
input = stdin.readline
INF = 10**9 + 7
MAX = 10**7 + 7
MOD = 10**9 + 7
for Ti in range(int(input().strip())):
n = int(input().strip())
a = [int(x) for x in input().strip().split()]
alen = len(a)
if alen % 2:
a.append(0)
alen += 1
os = 0
cs = 0
ts = 0
for i in range(0, alen, 2):
os += a[i]
for i in range(alen // 2):
cn = a[2 * i + 1] - a[2 * i]
cs = max(cs + cn, cn)
ts = max(cs, ts)
cs = 0
for i in range(1, alen - 1, 2):
cn = a[i] - a[i + 1]
cs = max(cs + cn, cn)
ts = max(ts, cs)
print(os + ts) | ASSIGN VAR VAR ASSIGN VAR BIN_OP BIN_OP NUMBER NUMBER NUMBER ASSIGN VAR BIN_OP BIN_OP NUMBER NUMBER NUMBER ASSIGN VAR BIN_OP BIN_OP NUMBER NUMBER NUMBER 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 VAR VAR FUNC_CALL FUNC_CALL FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR VAR IF BIN_OP VAR NUMBER EXPR FUNC_CALL VAR NUMBER VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR NUMBER VAR NUMBER VAR VAR VAR FOR VAR FUNC_CALL VAR BIN_OP VAR NUMBER ASSIGN VAR BIN_OP VAR BIN_OP BIN_OP NUMBER VAR NUMBER VAR BIN_OP NUMBER VAR ASSIGN VAR FUNC_CALL VAR BIN_OP VAR VAR VAR ASSIGN VAR FUNC_CALL VAR VAR VAR ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR NUMBER BIN_OP VAR NUMBER NUMBER ASSIGN VAR BIN_OP VAR VAR VAR BIN_OP VAR NUMBER ASSIGN VAR FUNC_CALL 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
from itertools import accumulate
def input():
return sys.stdin.readline().strip()
def list2d(a, b, c):
return [([c] * b) for i in range(a)]
def list3d(a, b, c, d):
return [[([d] * c) for j in range(b)] for i in range(a)]
def list4d(a, b, c, d, e):
return [[[([e] * d) for j in range(c)] for j in range(b)] for i in range(a)]
def ceil(x, y=1):
return int(-(-x // y))
def INT():
return int(input())
def MAP():
return map(int, input().split())
def LIST(N=None):
return list(MAP()) if N is None else [INT() for i in range(N)]
def Yes():
print("Yes")
def No():
print("No")
def YES():
print("YES")
def NO():
print("NO")
INF = 10**19
MOD = 10**9 + 7
class SegTree:
def __init__(self, n, func, intv, A=[]):
self.n = n
self.func = func
self.intv = intv
n2 = 1
while n2 < n:
n2 <<= 1
self.n2 = n2
self.tree = [self.intv] * (n2 << 1)
if A:
for i in range(n):
self.tree[n2 + i] = A[i]
for i in range(n2 - 1, -1, -1):
self.tree[i] = self.func(self.tree[i * 2], self.tree[i * 2 + 1])
def update(self, i, x):
i += self.n2
self.tree[i] = x
while i > 0:
i >>= 1
self.tree[i] = self.func(self.tree[i * 2], self.tree[i * 2 + 1])
def add(self, i, x):
self.update(i, self.get(i) + x)
def query(self, a, b):
l = a + self.n2
r = b + self.n2
s = self.intv
while l < r:
if r & 1:
r -= 1
s = self.func(s, self.tree[r])
if l & 1:
s = self.func(s, self.tree[l])
l += 1
l >>= 1
r >>= 1
return s
def get(self, i):
return self.tree[i + self.n2]
def all(self):
return self.tree[1]
def print(self):
for i in range(self.n):
print(self.get(i), end=" ")
print()
for _ in range(INT()):
N = INT()
A = LIST()
B1 = []
B2 = []
for i in range(N - 1):
if i % 2 == 0:
B1.append(A[i + 1] - A[i])
else:
B2.append(A[i] - A[i + 1])
acc1 = [0] + list(accumulate(B1))
acc2 = [0] + list(accumulate(B2))
N1 = len(acc1)
N2 = len(acc2)
st1 = SegTree(N1, max, -INF, acc1)
st2 = SegTree(N2, max, -INF, acc2)
base = sum([(a if i % 2 == 0 else 0) for i, a in enumerate(A)])
ans = base
for i in range(N1 - 1):
l = acc1[i]
r = st1.query(i + 1, N1)
ans = max(ans, base + r - l)
for i in range(N2 - 1):
l = acc2[i]
r = st2.query(i + 1, N2)
ans = max(ans, base + r - l)
print(ans) | IMPORT FUNC_DEF RETURN FUNC_CALL FUNC_CALL VAR FUNC_DEF RETURN BIN_OP LIST VAR VAR VAR FUNC_CALL VAR VAR FUNC_DEF RETURN BIN_OP LIST VAR VAR VAR FUNC_CALL VAR VAR VAR FUNC_CALL VAR VAR FUNC_DEF RETURN BIN_OP LIST VAR VAR VAR FUNC_CALL VAR VAR VAR FUNC_CALL VAR VAR VAR FUNC_CALL VAR VAR FUNC_DEF NUMBER RETURN FUNC_CALL VAR BIN_OP VAR VAR FUNC_DEF RETURN FUNC_CALL VAR FUNC_CALL VAR FUNC_DEF RETURN FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR FUNC_DEF NONE RETURN VAR NONE FUNC_CALL VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL VAR VAR FUNC_DEF EXPR FUNC_CALL VAR STRING FUNC_DEF EXPR FUNC_CALL VAR STRING FUNC_DEF EXPR FUNC_CALL VAR STRING FUNC_DEF EXPR FUNC_CALL VAR STRING ASSIGN VAR BIN_OP NUMBER NUMBER ASSIGN VAR BIN_OP BIN_OP NUMBER NUMBER NUMBER CLASS_DEF FUNC_DEF LIST ASSIGN VAR VAR ASSIGN VAR VAR ASSIGN VAR VAR ASSIGN VAR NUMBER WHILE VAR VAR VAR NUMBER ASSIGN VAR VAR ASSIGN VAR BIN_OP LIST VAR BIN_OP VAR NUMBER IF VAR FOR VAR FUNC_CALL VAR VAR ASSIGN VAR BIN_OP VAR VAR VAR VAR FOR VAR FUNC_CALL VAR BIN_OP VAR NUMBER NUMBER NUMBER ASSIGN VAR VAR FUNC_CALL VAR VAR BIN_OP VAR NUMBER VAR BIN_OP BIN_OP VAR NUMBER NUMBER FUNC_DEF VAR VAR ASSIGN VAR VAR VAR WHILE VAR NUMBER VAR NUMBER ASSIGN VAR VAR FUNC_CALL VAR VAR BIN_OP VAR NUMBER VAR BIN_OP BIN_OP VAR NUMBER NUMBER FUNC_DEF EXPR FUNC_CALL VAR VAR BIN_OP FUNC_CALL VAR VAR VAR FUNC_DEF ASSIGN VAR BIN_OP VAR VAR ASSIGN VAR BIN_OP VAR VAR ASSIGN VAR VAR WHILE VAR VAR IF BIN_OP VAR NUMBER VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR VAR VAR IF BIN_OP VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR VAR VAR VAR NUMBER VAR NUMBER VAR NUMBER RETURN VAR FUNC_DEF RETURN VAR BIN_OP VAR VAR FUNC_DEF RETURN VAR NUMBER FUNC_DEF FOR VAR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR FUNC_CALL VAR VAR STRING EXPR FUNC_CALL VAR FOR VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR ASSIGN VAR LIST ASSIGN VAR LIST 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 EXPR FUNC_CALL VAR BIN_OP VAR VAR VAR BIN_OP VAR NUMBER ASSIGN VAR BIN_OP LIST NUMBER FUNC_CALL VAR FUNC_CALL VAR VAR ASSIGN VAR BIN_OP LIST NUMBER FUNC_CALL VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR VAR VAR VAR VAR ASSIGN VAR FUNC_CALL VAR VAR VAR VAR VAR ASSIGN VAR FUNC_CALL VAR BIN_OP VAR NUMBER NUMBER VAR NUMBER VAR VAR FUNC_CALL VAR VAR ASSIGN VAR VAR FOR VAR FUNC_CALL VAR BIN_OP VAR NUMBER ASSIGN VAR VAR VAR ASSIGN VAR FUNC_CALL VAR BIN_OP VAR NUMBER VAR ASSIGN VAR FUNC_CALL VAR VAR BIN_OP BIN_OP VAR VAR VAR FOR VAR FUNC_CALL VAR BIN_OP VAR NUMBER ASSIGN VAR VAR VAR ASSIGN VAR FUNC_CALL VAR BIN_OP VAR NUMBER VAR ASSIGN VAR FUNC_CALL VAR VAR BIN_OP BIN_OP 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 solve():
n = int(input())
a = list(map(int, input().split()))
n1 = []
n2 = []
for i in range(0, n - 1, 2):
n1.append(a[i + 1] - a[i])
for i in range(1, n - 1, 2):
n2.append(a[i] - a[i + 1])
ans = 0
dopans = 0
for i in range(n):
if i % 2:
pass
else:
dopans += a[i]
if len(n1):
dp1 = [(0) for i in range(len(n1))]
dp1[0] = max(n1[0], 0)
for i in range(1, len(n1)):
dp1[i] = max(0, dp1[i - 1] + n1[i])
else:
dp1 = [0]
if len(n2):
dp2 = [(0) for i in range(len(n2))]
dp2[0] = max(n2[0], 0)
for i in range(1, len(n2)):
dp2[i] = max(0, dp2[i - 1] + n2[i])
else:
dp2 = [0]
print(dopans + max(max(dp1), max(dp2)))
for i in range(int(input())):
solve() | 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 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 VAR IF BIN_OP VAR NUMBER VAR VAR VAR IF FUNC_CALL VAR VAR ASSIGN VAR NUMBER VAR FUNC_CALL VAR FUNC_CALL VAR VAR ASSIGN VAR NUMBER FUNC_CALL VAR VAR NUMBER NUMBER FOR VAR FUNC_CALL VAR NUMBER FUNC_CALL VAR VAR ASSIGN VAR VAR FUNC_CALL VAR NUMBER BIN_OP VAR BIN_OP VAR NUMBER VAR VAR ASSIGN VAR LIST NUMBER IF FUNC_CALL VAR VAR ASSIGN VAR NUMBER VAR FUNC_CALL VAR FUNC_CALL VAR VAR ASSIGN VAR NUMBER FUNC_CALL VAR VAR NUMBER NUMBER FOR VAR FUNC_CALL VAR NUMBER FUNC_CALL VAR VAR ASSIGN VAR VAR FUNC_CALL VAR NUMBER BIN_OP VAR BIN_OP VAR NUMBER VAR VAR ASSIGN VAR LIST NUMBER EXPR FUNC_CALL VAR BIN_OP VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL VAR VAR 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 | def solve(n, a):
n2 = n // 2
x = [None] * n2
x2 = [None] * n2
a.append(a[-1])
init = 0 if n % 2 == 0 else a[-1]
for i in range(n2):
init += a[i * 2]
x[i] = a[2 * i + 1] - a[2 * i]
x2[i] = a[2 * i + 1] - a[2 * i + 2]
return init + max(gs(x), gs(x2), 0)
def gs(a):
cur = 0
mm = 0
best = 0
for v in a:
cur += v
if best < cur - mm:
best = cur - mm
if cur < mm:
mm = cur
return best
t = int(input())
for i_t in range(t):
n = int(input())
a = list(map(int, input().split(" ")))
print(solve(n, a)) | FUNC_DEF ASSIGN VAR BIN_OP VAR NUMBER ASSIGN VAR BIN_OP LIST NONE VAR ASSIGN VAR BIN_OP LIST NONE VAR EXPR FUNC_CALL VAR VAR NUMBER ASSIGN VAR BIN_OP VAR NUMBER NUMBER NUMBER VAR NUMBER FOR VAR FUNC_CALL VAR VAR VAR VAR BIN_OP VAR NUMBER ASSIGN VAR VAR BIN_OP VAR BIN_OP BIN_OP NUMBER VAR NUMBER VAR BIN_OP NUMBER VAR ASSIGN VAR VAR BIN_OP VAR BIN_OP BIN_OP NUMBER VAR NUMBER VAR BIN_OP BIN_OP NUMBER VAR NUMBER RETURN BIN_OP VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL VAR VAR NUMBER FUNC_DEF ASSIGN VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR NUMBER FOR VAR VAR VAR VAR IF VAR BIN_OP VAR VAR ASSIGN VAR BIN_OP VAR VAR 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 FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR STRING 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.buffer.readline
t = int(input())
for _ in range(t):
n = int(input())
A = list(map(int, input().split()))
if n == 1:
print(sum(A))
continue
sumA = 0
for i in range(n):
if i % 2 == 0:
sumA += A[i]
E = []
for i in range(n):
if i % 2 == 1:
continue
if i + 1 < n:
E.append(A[i + 1] - A[i])
O = []
for i in range(1, n):
if i % 2 == 0:
continue
if i + 1 < n:
O.append(A[i] - A[i + 1])
m = len(E)
INF = 10**18
dpe = [-INF] * (m + 1)
for i in range(m):
dpe[i + 1] = max(E[i], dpe[i] + E[i])
e = max(dpe)
l = len(O)
dpo = [-INF] * (l + 1)
for i in range(l):
dpo[i + 1] = max(O[i], dpo[i] + O[i])
o = max(dpo)
ans = sumA + max([o, e, 0])
print(ans) | 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 IF VAR NUMBER EXPR FUNC_CALL VAR FUNC_CALL VAR VAR ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR VAR IF BIN_OP VAR NUMBER NUMBER VAR VAR VAR ASSIGN VAR LIST 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 ASSIGN VAR LIST FOR VAR FUNC_CALL VAR NUMBER VAR IF BIN_OP VAR NUMBER 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 BIN_OP NUMBER NUMBER ASSIGN VAR BIN_OP LIST VAR BIN_OP VAR NUMBER FOR VAR FUNC_CALL VAR VAR ASSIGN VAR BIN_OP VAR NUMBER FUNC_CALL VAR VAR VAR BIN_OP VAR VAR VAR VAR ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR BIN_OP LIST VAR BIN_OP VAR NUMBER FOR VAR FUNC_CALL VAR VAR ASSIGN VAR BIN_OP VAR NUMBER FUNC_CALL VAR VAR VAR BIN_OP VAR VAR VAR VAR ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR BIN_OP VAR FUNC_CALL VAR LIST 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 | def f(al):
ar1 = []
ar2 = []
i = 0
while i < len(al) - 1:
ar1.append(al[i + 1] - al[i])
i = i + 2
i = 0
while i < len(al) - 2:
ar2.append(al[i + 1] - al[i + 2])
i = i + 2
return ar1, ar2
def subarraymax(nums):
if len(nums) == 0:
return 0
sum = res = nums[0]
for i in range(1, len(nums)):
sum = max(nums[i], sum + nums[i])
res = max(res, sum)
return res
for tc in range(int(input())):
_ = input()
al = list(map(int, input().split()))
ini = 0
for i in range(0, len(al), 2):
ini = ini + al[i]
s1, s2 = f(al)
delta = max(subarraymax(s1), subarraymax(s2))
if delta <= 0:
print(ini)
else:
print(ini + delta) | FUNC_DEF ASSIGN VAR LIST ASSIGN VAR LIST ASSIGN VAR NUMBER WHILE VAR BIN_OP FUNC_CALL VAR VAR NUMBER EXPR FUNC_CALL VAR BIN_OP VAR BIN_OP VAR NUMBER VAR VAR ASSIGN VAR BIN_OP VAR NUMBER ASSIGN VAR NUMBER WHILE VAR BIN_OP FUNC_CALL VAR VAR NUMBER EXPR FUNC_CALL VAR BIN_OP VAR BIN_OP VAR NUMBER VAR BIN_OP VAR NUMBER ASSIGN VAR BIN_OP VAR NUMBER RETURN VAR VAR FUNC_DEF IF FUNC_CALL VAR VAR NUMBER RETURN NUMBER ASSIGN VAR VAR VAR NUMBER FOR VAR FUNC_CALL VAR NUMBER 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 ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR NUMBER FUNC_CALL VAR VAR NUMBER ASSIGN VAR BIN_OP VAR VAR VAR ASSIGN VAR VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL VAR VAR IF VAR NUMBER EXPR 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 | t = int(input())
for _ in range(t):
n = int(input())
a = list(map(int, input().split()))
tongchan, tongle = 0, 0
ans = 0
maxchan, maxle = 0, 0
minchan, minle = 0, 0
for i in range(len(a)):
if i % 2 == 0:
tongchan += a[i]
minchan = min(minchan, tongle - tongchan)
maxchan = max(maxchan, tongle - tongchan - minchan)
else:
tongle += a[i]
minle = min(minle, tongle - tongchan)
maxle = max(maxle, tongle - tongchan - minle)
print(tongchan + max(maxchan, maxle)) | 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 NUMBER ASSIGN VAR NUMBER ASSIGN VAR VAR NUMBER NUMBER ASSIGN VAR VAR NUMBER NUMBER FOR VAR FUNC_CALL VAR FUNC_CALL VAR VAR IF BIN_OP VAR NUMBER NUMBER VAR VAR VAR ASSIGN VAR FUNC_CALL VAR VAR BIN_OP VAR VAR ASSIGN VAR FUNC_CALL VAR VAR BIN_OP BIN_OP VAR VAR VAR VAR VAR VAR ASSIGN VAR FUNC_CALL VAR VAR BIN_OP VAR VAR ASSIGN VAR FUNC_CALL VAR VAR BIN_OP BIN_OP VAR 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 | from sys import stdin
def maxSubArraySum(a, size):
max_so_far = -100000000000
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 help():
n = int(stdin.readline())
arr = list(map(int, stdin.readline().split(" ")))
ans = 0
for i in range(0, n, 2):
ans += arr[i]
arr1 = []
arr2 = []
for i in range(1, n - 1, 2):
arr1.append(arr[i] - arr[i + 1])
for i in range(0, n - 1, 2):
arr2.append(arr[i + 1] - arr[i])
tt = max(maxSubArraySum(arr1, len(arr1)), maxSubArraySum(arr2, len(arr2)))
tt = max(0, tt)
print(ans + tt)
for _ in range(int(stdin.readline())):
help() | 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 FUNC_DEF ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR STRING ASSIGN VAR 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 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 FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL VAR VAR FUNC_CALL VAR VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR NUMBER VAR EXPR FUNC_CALL VAR BIN_OP VAR VAR 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
def getInt():
return int(input())
def getVars():
return map(int, input().split())
def getList():
return list(map(int, input().split()))
def getStr():
return input().strip()
t = getInt()
for _ in range(t):
n = getInt()
a = getList()
s = 0
maxZ = 0
z1 = 0
z2 = 0
for i in range(0, n, 2):
s += a[i]
if i < n - 1:
z1 += a[i + 1] - a[i]
if i > 0:
z2 += a[i - 1] - a[i]
z1 = max(z1, 0)
z2 = max(z2, 0)
maxZ = max(z1, z2, maxZ)
print(s + maxZ) | IMPORT ASSIGN VAR VAR FUNC_DEF RETURN FUNC_CALL VAR FUNC_CALL VAR FUNC_DEF RETURN FUNC_CALL VAR VAR FUNC_CALL 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 ASSIGN VAR FUNC_CALL VAR FOR VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR ASSIGN VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR NUMBER VAR NUMBER VAR VAR VAR IF VAR BIN_OP VAR NUMBER VAR BIN_OP VAR BIN_OP VAR NUMBER VAR VAR IF VAR NUMBER VAR BIN_OP VAR BIN_OP VAR NUMBER VAR VAR ASSIGN VAR FUNC_CALL VAR VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR NUMBER ASSIGN VAR FUNC_CALL 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()))
even = 0
for i in range(0, n, 2):
even += arr[i]
ans = even
left = []
for i in range(1, n, 2):
left.append(arr[i] - arr[i - 1])
if len(left) != 0:
s = left[0]
dp = [(0) for i in range(n + 1)]
dp[0] = left[0]
for i in range(1, len(left)):
dp[i] = max(dp[i - 1] + left[i], left[i])
temp = max(dp)
ans = max(ans, even + temp)
right = []
for i in range(1, n - 1, 2):
right.append(arr[i] - arr[i + 1])
if len(right) != 0:
s = right[0]
dp = [(0) for i in range(n + 1)]
dp[0] = right[0]
for i in range(1, len(right)):
dp[i] = max(dp[i - 1] + right[i], right[i])
temp = max(dp)
ans = max(ans, even + temp)
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 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 IF FUNC_CALL VAR VAR NUMBER ASSIGN VAR VAR NUMBER ASSIGN VAR NUMBER VAR FUNC_CALL VAR BIN_OP VAR NUMBER 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 FUNC_CALL VAR VAR BIN_OP 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 IF FUNC_CALL VAR VAR NUMBER ASSIGN VAR VAR NUMBER ASSIGN VAR NUMBER VAR FUNC_CALL VAR BIN_OP VAR NUMBER 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 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 | for test_i in range(int(input())):
n = int(input())
arr = list(map(int, input().split()))
max_dif, min_pref = 0, 0
cur = 0
for i in range(n // 2):
cur += arr[i * 2 + 1] - arr[i * 2]
if cur < min_pref:
min_pref = cur
elif cur - min_pref > max_dif:
max_dif = cur - min_pref
min_pref = 0
cur = 0
for i in range((n - 1) // 2):
cur += arr[i * 2 + 1] - arr[i * 2 + 2]
if cur < min_pref:
min_pref = cur
elif cur - min_pref > max_dif:
max_dif = cur - min_pref
sum_even = sum(arr[::2])
print(sum_even + max_dif) | 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 NUMBER ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR BIN_OP VAR NUMBER VAR BIN_OP VAR BIN_OP BIN_OP VAR NUMBER NUMBER VAR BIN_OP VAR NUMBER IF VAR VAR ASSIGN VAR VAR IF BIN_OP VAR VAR VAR ASSIGN VAR BIN_OP VAR VAR ASSIGN VAR NUMBER ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR BIN_OP BIN_OP VAR NUMBER NUMBER VAR BIN_OP VAR BIN_OP BIN_OP VAR NUMBER NUMBER VAR BIN_OP BIN_OP VAR NUMBER NUMBER IF VAR VAR ASSIGN VAR VAR IF BIN_OP VAR VAR VAR ASSIGN VAR BIN_OP 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 | def max_sum(A):
max_so_far = 0
max_ending = 0
for i in A:
max_ending += i
if max_ending < 0:
max_ending = 0
if max_ending > max_so_far:
max_so_far = max_ending
return max_so_far
for i in range(int(input())):
n = int(input())
arr = list(map(int, input().split()))
ans = 0
for i in range(n):
if i % 2 == 0:
ans += arr[i]
arr1 = []
arr2 = []
for i in range(n):
if i % 2 == 0:
if i - 1 >= 0:
arr1.append(arr[i - 1] - arr[i])
if i + 1 < n:
arr2.append(arr[i + 1] - arr[i])
print(ans + max(max_sum(arr1), max_sum(arr2))) | FUNC_DEF ASSIGN VAR NUMBER ASSIGN VAR NUMBER FOR VAR 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 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 VAR IF BIN_OP VAR NUMBER NUMBER IF BIN_OP VAR NUMBER NUMBER EXPR FUNC_CALL VAR BIN_OP VAR BIN_OP VAR NUMBER VAR VAR IF BIN_OP VAR NUMBER VAR EXPR FUNC_CALL VAR BIN_OP VAR BIN_OP VAR NUMBER VAR 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 exit, stderr, stdin
def rl():
return [int(w) for w in stdin.readline().split()]
def maxsum(b):
cur = r = 0
for x in b:
cur += x
if cur < 0:
cur = 0
elif cur > r:
r = cur
return r
(T,) = rl()
for _ in range(T):
(n,) = rl()
a = rl()
print(
sum(a[::2])
+ max(
maxsum(b)
for b in [
[(a[i] - a[i - 1]) for i in range(1, n, 2)],
[(a[i - 1] - a[i]) for i in range(2, n, 2)],
]
)
) | FUNC_DEF RETURN FUNC_CALL VAR VAR VAR FUNC_CALL FUNC_CALL VAR 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 ASSIGN VAR FUNC_CALL VAR FOR VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR EXPR FUNC_CALL VAR BIN_OP FUNC_CALL VAR VAR NUMBER FUNC_CALL VAR FUNC_CALL VAR VAR VAR LIST BIN_OP VAR VAR VAR BIN_OP VAR NUMBER VAR FUNC_CALL VAR NUMBER VAR NUMBER BIN_OP VAR BIN_OP VAR NUMBER VAR VAR VAR FUNC_CALL 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 | nb_cases = int(input())
for case in range(nb_cases):
n = int(input())
a = [int(x) for x in input().split()]
dp_pair = []
dp_impair = []
for x in range(n - 1):
if x % 2 == 0:
dp_pair.append(a[x + 1] - a[x])
else:
dp_impair.append(a[x] - a[x + 1])
p1 = 0
p2 = 0
somme = 0
p_min = -1
p_max = -1
somme_max = 0
while p2 < int(n / 2):
somme += dp_pair[p2]
if dp_pair[p2] > 0:
if somme > somme_max:
somme_max = somme
p_min = p1
p_max = p2
elif somme <= 0:
p1 = p2 + 1
somme = 0
p2 += 1
somme_max_pair = somme_max
p1 = 0
p2 = 0
somme = 0
p_min = -1
p_max = -1
somme_max = 0
while p2 < int((n - 1) / 2):
somme += dp_impair[p2]
if dp_impair[p2] > 0:
if somme > somme_max:
somme_max = somme
p_min = p1
p_max = p2
elif somme <= 0:
p1 = p2 + 1
somme = 0
p2 += 1
somme_max_impair = somme_max
somme_pair = sum([x for x in a[::2]])
print(somme_pair + max(somme_max_pair, somme_max_impair)) | 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 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 EXPR FUNC_CALL VAR BIN_OP VAR VAR VAR BIN_OP VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR NUMBER WHILE VAR FUNC_CALL VAR BIN_OP VAR NUMBER VAR VAR VAR IF VAR VAR NUMBER IF VAR VAR ASSIGN VAR VAR ASSIGN VAR VAR ASSIGN VAR VAR IF VAR NUMBER ASSIGN VAR BIN_OP VAR NUMBER ASSIGN VAR NUMBER VAR NUMBER ASSIGN VAR VAR ASSIGN VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR NUMBER WHILE VAR FUNC_CALL VAR BIN_OP BIN_OP VAR NUMBER NUMBER VAR VAR VAR IF VAR VAR NUMBER IF VAR VAR ASSIGN VAR VAR ASSIGN VAR VAR ASSIGN VAR VAR IF VAR NUMBER ASSIGN VAR BIN_OP VAR NUMBER ASSIGN VAR NUMBER VAR NUMBER ASSIGN 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 | debug = False
for _ in range(int(input())):
n = int(input())
arr = list(map(int, input().split()))
dp = [([0] * 3) for i in range(n + 1)]
for i in range(n):
dp[i + 1][0] = max(dp[i + 1][0], dp[i][0] + (0 if i & 1 else arr[i]))
if i + 2 <= n:
dp[i + 2][1] = max(
dp[i + 2][1],
max(dp[i][0], dp[i][1]) + (arr[i] if i & 1 else arr[i + 1]),
)
dp[i + 1][2] = max(
dp[i + 1][2], max(dp[i][0], dp[i][1], dp[i][2]) + (0 if i & 1 else arr[i])
)
print(max(dp[n])) | ASSIGN 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 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 FUNC_CALL VAR VAR BIN_OP VAR NUMBER NUMBER BIN_OP VAR VAR NUMBER BIN_OP VAR NUMBER NUMBER VAR VAR IF BIN_OP VAR NUMBER VAR ASSIGN VAR BIN_OP VAR NUMBER NUMBER FUNC_CALL VAR VAR BIN_OP VAR NUMBER NUMBER BIN_OP FUNC_CALL VAR VAR VAR NUMBER VAR VAR NUMBER BIN_OP VAR NUMBER VAR VAR VAR BIN_OP VAR NUMBER ASSIGN VAR BIN_OP VAR NUMBER NUMBER FUNC_CALL VAR VAR BIN_OP VAR NUMBER NUMBER BIN_OP FUNC_CALL VAR VAR VAR NUMBER VAR VAR NUMBER VAR VAR NUMBER BIN_OP VAR NUMBER NUMBER VAR 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 | import sys
input = sys.stdin.readline
t = int(input())
for _ in range(t):
n = int(input())
a = list(map(int, input().split()))
es = 0
for i in range(0, n, 2):
es += a[i]
e, o = 0, 0
me, mo = 10**18, 0
ans = -(10**18)
for i in range(n):
if i % 2 == 0:
e += a[i]
else:
o += a[i]
if i % 2 == 0:
me = min(me, o - e)
ans = max(ans, es + o - e - me)
else:
mo = min(mo, o - e)
ans = max(ans, es + o - e - mo)
print(ans) | 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 VAR NUMBER NUMBER ASSIGN VAR VAR BIN_OP NUMBER NUMBER NUMBER ASSIGN VAR BIN_OP NUMBER NUMBER FOR VAR FUNC_CALL VAR VAR IF BIN_OP VAR NUMBER NUMBER VAR VAR VAR VAR VAR VAR IF BIN_OP VAR NUMBER NUMBER ASSIGN VAR FUNC_CALL VAR VAR BIN_OP VAR VAR ASSIGN VAR FUNC_CALL VAR VAR BIN_OP BIN_OP BIN_OP VAR VAR VAR VAR ASSIGN VAR FUNC_CALL VAR VAR BIN_OP VAR VAR ASSIGN VAR FUNC_CALL VAR VAR BIN_OP BIN_OP 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 kadane(arr):
max_end = 0
max_far = -float("inf")
for i in arr:
max_end = max_end + i
max_far = max(max_far, max_end)
if max_end < 0:
max_end = 0
return max_far
t = int(input())
for _ in range(t):
n = int(input())
arr = list(map(int, input().split()))
pre = []
suf = []
su = 0
for i in range(0, n, 2):
su += arr[i]
if i + 1 < n:
pre.append(arr[i + 1] - arr[i])
if i + 2 < n:
suf.append(arr[i + 1] - arr[i + 2])
print(su + max(kadane(pre), kadane(suf), 0)) | FUNC_DEF ASSIGN VAR NUMBER ASSIGN VAR FUNC_CALL VAR STRING FOR VAR VAR ASSIGN VAR BIN_OP VAR VAR ASSIGN VAR FUNC_CALL VAR VAR VAR IF VAR NUMBER ASSIGN 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 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 IF BIN_OP VAR NUMBER VAR EXPR FUNC_CALL VAR BIN_OP VAR BIN_OP VAR NUMBER VAR BIN_OP VAR NUMBER EXPR FUNC_CALL VAR BIN_OP VAR FUNC_CALL VAR FUNC_CALL VAR 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 | def solve():
n = int(input())
a = list(map(int, input().split()))
if n == 1:
print(a[0])
return
bst = 0
bi1 = 0
bi2 = 0
end = -1
ei1 = 0
k = 0
api1 = 0
ap1 = []
ap2 = []
for i, x in enumerate(a):
if k:
ap1.append(x - api)
k = 0
else:
api = x
k = 1
k = 0
for i, x in enumerate(a):
if i == 0:
continue
if k:
ap2.append(api - x)
k = 0
else:
api = x
k = 1
bst = 0
end = 0
for p in ap1:
if p > 0:
end += p
elif end + p > 0:
bst = max(end, bst)
end += p
else:
end = 0
bst = max(bst, end)
end = 0
for p in ap2:
if p > 0:
end += p
elif end + p > 0:
bst = max(end, bst)
end += p
else:
end = 0
bst = max(bst, end)
s = 0
for i, x in enumerate(a):
if i % 2 == 0:
s += x
print(s + bst)
t = int(input())
for _ in range(t):
solve() | FUNC_DEF ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR IF VAR NUMBER EXPR FUNC_CALL VAR VAR NUMBER RETURN ASSIGN VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR LIST ASSIGN VAR LIST FOR VAR VAR FUNC_CALL VAR VAR IF VAR EXPR FUNC_CALL VAR BIN_OP VAR VAR ASSIGN VAR NUMBER ASSIGN VAR VAR ASSIGN VAR NUMBER ASSIGN VAR NUMBER FOR VAR VAR FUNC_CALL VAR VAR IF VAR NUMBER IF VAR EXPR FUNC_CALL VAR BIN_OP VAR VAR ASSIGN VAR NUMBER ASSIGN VAR VAR ASSIGN VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR NUMBER FOR VAR VAR IF VAR NUMBER VAR VAR IF BIN_OP VAR VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR VAR VAR VAR ASSIGN VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR VAR ASSIGN VAR NUMBER FOR VAR VAR IF VAR NUMBER VAR VAR IF BIN_OP VAR VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR VAR VAR VAR ASSIGN VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR VAR ASSIGN VAR NUMBER FOR VAR VAR FUNC_CALL VAR VAR IF BIN_OP VAR NUMBER NUMBER VAR VAR EXPR FUNC_CALL VAR BIN_OP 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 | n = int(input())
while n > 0:
arr = []
t = int(input())
arr = [int(tp) for tp in input().split()]
ans = 0
for i in range(t):
if not i % 2:
ans += arr[i]
dp, maxn = 0, 0
for i in range(0, t, 2):
if i + 1 < t:
dp = max(dp + arr[i + 1] - arr[i], 0)
maxn = max(dp, maxn)
dp = 0
for i in range(1, t, 2):
if i + 1 < t:
dp = max(dp + arr[i] - arr[i + 1], 0)
maxn = max(dp, maxn)
print(ans + maxn)
n -= 1 | ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR WHILE VAR NUMBER ASSIGN VAR LIST 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 VAR IF BIN_OP VAR NUMBER VAR VAR VAR ASSIGN VAR VAR NUMBER NUMBER FOR VAR FUNC_CALL VAR NUMBER VAR NUMBER IF BIN_OP VAR NUMBER VAR 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 VAR NUMBER IF BIN_OP VAR NUMBER VAR 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 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())
a = list(map(int, input().split()))
s = sum([a[i] for i in range(n) if i % 2 == 0])
dp = [0] * n
for i in range(2, n, 2):
dp[i] = max(dp[i - 2] + a[i - 1] - a[i], a[i - 1] - a[i])
for i in range(1, n, 2):
if i >= 2:
dp[i] = max(dp[i - 2] + a[i] - a[i - 1], a[i] - a[i - 1])
else:
dp[i] = a[i] - a[i - 1]
print(s + max(max(dp), 0)) | 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 VAR VAR FUNC_CALL VAR VAR BIN_OP VAR NUMBER NUMBER ASSIGN VAR BIN_OP LIST NUMBER VAR FOR VAR FUNC_CALL VAR NUMBER VAR NUMBER ASSIGN VAR VAR FUNC_CALL VAR BIN_OP BIN_OP VAR BIN_OP VAR NUMBER VAR BIN_OP VAR NUMBER VAR VAR BIN_OP VAR BIN_OP VAR NUMBER VAR VAR FOR VAR FUNC_CALL VAR NUMBER VAR NUMBER IF VAR NUMBER ASSIGN VAR VAR FUNC_CALL VAR BIN_OP BIN_OP VAR BIN_OP VAR NUMBER VAR VAR VAR BIN_OP VAR NUMBER BIN_OP VAR VAR VAR BIN_OP VAR NUMBER ASSIGN VAR VAR BIN_OP VAR VAR VAR BIN_OP VAR NUMBER EXPR FUNC_CALL VAR BIN_OP VAR 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())
nums = list(map(int, input().split()))
current_sum = 0
for i in range(0, n, 2):
current_sum += nums[i]
limite_par = n // 2 - 1
if n % 2 == 0:
limite_impar = limite_par - 1
else:
limite_impar = limite_par
ganancias_pares = []
ganancias_impares = []
max_eve_sum = max_odd_sum = 0
current_eve_sum = current_odd_sum = 0
for i in range(limite_par + 1):
voltear = nums[2 * i + 1] - nums[2 * i]
current_eve_sum += voltear
if current_eve_sum < voltear:
current_eve_sum = voltear
max_eve_sum = max(max_eve_sum, current_eve_sum)
for i in range(limite_impar + 1):
voltear = nums[2 * i + 1] - nums[2 * i + 2]
current_odd_sum += voltear
if current_odd_sum < voltear:
current_odd_sum = voltear
max_odd_sum = max(max_odd_sum, current_odd_sum)
print(current_sum + max(max_odd_sum, max_eve_sum)) | 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 BIN_OP BIN_OP VAR NUMBER NUMBER IF BIN_OP VAR NUMBER NUMBER ASSIGN VAR BIN_OP VAR NUMBER ASSIGN VAR VAR ASSIGN VAR LIST ASSIGN VAR LIST ASSIGN VAR VAR NUMBER ASSIGN VAR VAR NUMBER FOR VAR FUNC_CALL VAR BIN_OP VAR NUMBER ASSIGN VAR BIN_OP VAR BIN_OP BIN_OP NUMBER VAR NUMBER VAR BIN_OP NUMBER VAR VAR VAR IF VAR VAR ASSIGN VAR VAR ASSIGN VAR FUNC_CALL VAR VAR VAR FOR VAR FUNC_CALL VAR BIN_OP VAR NUMBER ASSIGN VAR BIN_OP VAR BIN_OP BIN_OP NUMBER VAR NUMBER VAR BIN_OP BIN_OP NUMBER VAR NUMBER VAR VAR IF VAR VAR ASSIGN VAR VAR ASSIGN VAR FUNC_CALL VAR 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()))
es = 0
ls = 0
rs = 0
s = 0
for i in range(0, n):
if i % 2 == 0:
es += a[i]
for i in range(1, n, 2):
s = max(0, s + a[i] - a[i - 1])
ls = max(ls, s)
s = 0
for i in range(1, n, 2):
if i + 1 >= n:
break
else:
s = max(0, s + a[i] - a[i + 1])
rs = max(rs, s)
print(max(es + ls, es + rs)) | 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 FOR VAR FUNC_CALL VAR NUMBER VAR IF BIN_OP VAR NUMBER NUMBER VAR VAR VAR FOR VAR FUNC_CALL VAR NUMBER VAR 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 ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR NUMBER VAR NUMBER IF BIN_OP VAR NUMBER VAR 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 FUNC_CALL VAR BIN_OP 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 | from sys import gettrace, stdin
if gettrace():
inputi = input
else:
def input():
return next(stdin)[:-1]
def inputi():
return stdin.buffer.readline()
def solve():
n = int(inputi())
aa = [int(a) for a in inputi().split()]
base = sum(aa[::2])
bb = []
for i in range(0, n - 1, 2):
bb.append(aa[i + 1] - aa[i])
me = 0
mx = 0
for b in bb:
me = max(me + b, 0)
mx = max(mx, me)
bb = []
for i in range(1, n - 1, 2):
bb.append(aa[i] - aa[i + 1])
me = 0
for b in bb:
me = max(me + b, 0)
mx = max(mx, me)
print(base + mx)
def main():
t = int(inputi())
for _ in range(t):
solve()
main() | IF FUNC_CALL VAR ASSIGN VAR VAR FUNC_DEF RETURN FUNC_CALL VAR VAR NUMBER FUNC_DEF RETURN FUNC_CALL 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 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 NUMBER ASSIGN VAR NUMBER FOR VAR VAR ASSIGN VAR FUNC_CALL VAR BIN_OP VAR VAR NUMBER ASSIGN VAR FUNC_CALL VAR 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 NUMBER FOR VAR VAR ASSIGN VAR FUNC_CALL VAR BIN_OP VAR VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR VAR EXPR FUNC_CALL VAR BIN_OP VAR VAR FUNC_DEF ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR FOR VAR FUNC_CALL VAR VAR EXPR 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 | t = int(input())
for _ in range(t):
n = int(input())
a = list(map(int, input().split()))
cur = a[0]
curodd = 0
cureven = 0
maxalt = 0
for i in range(1, n):
if i % 2 == 1:
curodd += a[i] - a[i - 1]
maxalt = max(maxalt, curodd)
if curodd < 0:
curodd = 0
else:
cureven -= a[i] - a[i - 1]
maxalt = max(maxalt, cureven)
cur += a[i]
if cureven < 0:
cureven = 0
print(cur + maxalt) | 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 NUMBER ASSIGN VAR NUMBER ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR NUMBER VAR IF 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 BIN_OP VAR VAR VAR BIN_OP VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR VAR 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 | def main():
t = int(input())
for _ in range(t):
n = int(input())
a = list(map(int, input().split()))
even_sum = 0
for i, v in enumerate(a):
if i % 2 == 0:
even_sum += v
temp = 0
best = 0
for i in range(1, n - 1, 2):
temp += a[i] - a[i + 1]
if temp < 0:
temp = 0
best = max(best, temp)
after_sum = even_sum + best
temp = best = 0
for i in range(0, n - 1, 2):
temp += a[i + 1] - a[i]
if temp < 0:
temp = 0
best = max(best, temp)
before_sum = even_sum + best
ans = max(even_sum, after_sum, before_sum)
print(ans)
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 FOR VAR VAR FUNC_CALL VAR VAR IF BIN_OP VAR NUMBER NUMBER VAR VAR 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 IF VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR VAR ASSIGN VAR BIN_OP VAR VAR 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 IF VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR VAR ASSIGN VAR BIN_OP VAR VAR ASSIGN VAR FUNC_CALL VAR 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 | t = int(input())
while t:
t = t - 1
n = int(input())
a = list(map(int, input().split()))
sum = 0
for i in range(0, n, 2):
sum += a[i]
m = 0
d = 0
for i in range(1, n, 2):
d += a[i] - a[i - 1]
if d < 0:
d = 0
m = max(m, d)
d = 0
for i in range(2, n, 2):
d += a[i - 1] - a[i]
if d < 0:
d = 0
m = max(m, d)
print(sum + m) | ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR WHILE VAR ASSIGN VAR BIN_OP 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 NUMBER ASSIGN 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 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
t = int(sys.stdin.readline())
for i in range(t):
num = 0
parr = 0
imparrr = 0
n = int(sys.stdin.readline())
a = list(map(int, input().split()))
suma = 0
for i in range(len(a)):
if i % 2 == 0:
suma += a[i]
for i in range(0, n - 1, 2):
num += a[i + 1] - a[i]
parr = max(parr, num)
if num < 0:
num = 0
num = 0
for i in range(1, n - 1, 2):
num += a[i] - a[i + 1]
parr = max(parr, num)
if num < 0:
num = 0
print(parr + suma) | IMPORT ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR FOR VAR FUNC_CALL VAR VAR ASSIGN VAR NUMBER ASSIGN VAR NUMBER ASSIGN 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 FUNC_CALL VAR VAR IF BIN_OP VAR NUMBER NUMBER VAR VAR VAR 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 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 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(a):
ans = res = 0
for i in a:
ans = max(ans + i, i)
res = max(res, ans)
return res
for _ in range(int(input())):
n = int(input())
a = [*map(int, input().split())]
b = [(a[i] - a[i - 1]) for i in range(1, n, 2)]
c = [(a[i - 1] - a[i]) for i in range(2, n, 2)]
print(sum(a[::2]) + max(kadane(b), kadane(c))) | FUNC_DEF ASSIGN VAR VAR NUMBER FOR VAR VAR ASSIGN VAR FUNC_CALL 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 LIST FUNC_CALL 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 EXPR FUNC_CALL VAR BIN_OP FUNC_CALL VAR VAR NUMBER 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 | def maxEvenLenSum(arr, n):
if n < 2:
return 0
dp = [(0) for i in range(n)]
dp[n - 1] = 0
dp[n - 2] = arr[n - 2] + arr[n - 1]
for i in range(n - 3, -1, -1):
dp[i] = arr[i] + arr[i + 1]
if dp[i + 2] > 0:
dp[i] += dp[i + 2]
maxSum = max(dp)
return maxSum
t = int(input())
for _ in range(t):
n = int(input())
A = [int(x) for x in input().split()]
B = []
for i in range(n):
if i % 2 == 0:
B.append(A[i])
for i in range(n):
if i % 2 == 0:
A[i] *= -1
max_so_far = maxEvenLenSum(A, n)
print(sum(B) + max_so_far) | FUNC_DEF IF VAR NUMBER RETURN NUMBER ASSIGN VAR NUMBER VAR FUNC_CALL VAR VAR ASSIGN VAR BIN_OP VAR NUMBER NUMBER ASSIGN VAR BIN_OP VAR NUMBER BIN_OP VAR BIN_OP VAR NUMBER VAR BIN_OP VAR NUMBER FOR VAR FUNC_CALL VAR BIN_OP VAR NUMBER NUMBER NUMBER ASSIGN VAR VAR BIN_OP VAR VAR VAR BIN_OP VAR NUMBER IF VAR BIN_OP VAR NUMBER NUMBER VAR VAR VAR BIN_OP VAR NUMBER ASSIGN VAR FUNC_CALL 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 LIST FOR VAR FUNC_CALL VAR VAR IF BIN_OP VAR NUMBER NUMBER EXPR FUNC_CALL VAR VAR VAR FOR VAR FUNC_CALL VAR VAR IF BIN_OP VAR NUMBER NUMBER VAR VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR VAR EXPR FUNC_CALL VAR BIN_OP 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 sum_even():
_sum = 0
for _i in range(0, len(array), 2):
_sum += array[_i]
return _sum
def max_sub_sum(_array):
dp = [(0) for _i in range(len(_array))]
dp[0] = _array[0]
for _i in range(1, len(_array)):
dp[_i] = max(_array[_i], dp[_i - 1] + _array[_i])
result = dp[0]
for _i in range(1, len(_array)):
if dp[_i] > result:
result = dp[_i]
return result
n_cases = int(input())
for n in range(0, n_cases):
length = int(input())
temp = input()
array = [int(n) for n in temp.split()]
if length == 1:
print(array[0])
continue
if length == 2:
print(max(array[0], array[1]))
continue
ori_arr = array.copy()
ori_sum = sum_even()
diff_left = []
diff_right = []
for i in range(0, length - 1, 2):
diff_left.append(array[i + 1] - array[i])
for i in range(1, length - 1, 2):
diff_right.append(array[i] - array[i + 1])
sum_left = max_sub_sum(diff_left)
sum_right = max_sub_sum(diff_right)
if sum_left >= sum_right and sum_left > 0:
print(ori_sum + sum_left)
elif sum_right >= sum_left and sum_right > 0:
print(ori_sum + sum_right)
else:
print(ori_sum) | FUNC_DEF ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR NUMBER FUNC_CALL VAR VAR NUMBER VAR VAR VAR RETURN VAR FUNC_DEF ASSIGN VAR NUMBER VAR FUNC_CALL VAR 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 VAR VAR BIN_OP VAR BIN_OP VAR NUMBER VAR VAR ASSIGN VAR VAR NUMBER FOR VAR FUNC_CALL VAR NUMBER FUNC_CALL VAR VAR IF VAR VAR VAR ASSIGN 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 ASSIGN VAR FUNC_CALL VAR VAR VAR 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 FUNC_CALL VAR ASSIGN VAR 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 FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR VAR IF VAR VAR VAR NUMBER EXPR FUNC_CALL VAR BIN_OP VAR VAR IF VAR VAR 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 | tests = int(input())
for tst in range(tests):
n = int(input())
a = [int(i) for i in input().split()]
ans = sum([a[i] for i in range(0, n, 2)])
b1 = [(a[i] - a[i - 1]) for i in range(1, n, 2)]
m1, s1 = 0, 0
for i in range(len(b1)):
s1 += b1[i]
m1 = max(m1, s1)
if s1 < 0:
s1 = 0
b2 = [(a[i] - a[i + 1]) for i in range(1, n - 1, 2)]
m2, s2 = 0, 0
for i in range(len(b2)):
s2 += b2[i]
m2 = max(m2, s2)
if s2 < 0:
s2 = 0
print(max([ans, ans + m1, ans + 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 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 VAR NUMBER 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 BIN_OP VAR VAR VAR BIN_OP VAR NUMBER VAR FUNC_CALL VAR NUMBER BIN_OP VAR NUMBER NUMBER ASSIGN VAR VAR NUMBER 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 FUNC_CALL VAR LIST VAR BIN_OP 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())
while t > 0:
n, a = int(input()), list(map(int, input().split()))
mx_n, mx = 0, 0
for i in range(0, n - 1, 2):
mx_n += a[i + 1] - a[i]
mx = max(mx_n, mx)
mx_n = max(0, mx_n)
mx_n = 0
for i in range(1, n - 1, 2):
mx_n += a[i] - a[i + 1]
mx = max(mx_n, mx)
mx_n = max(0, mx_n)
even = sum(a[i] for i in range(n) if i % 2 == 0)
print(even + mx)
t -= 1 | ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR WHILE VAR NUMBER ASSIGN VAR VAR FUNC_CALL VAR FUNC_CALL VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR VAR NUMBER 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 ASSIGN VAR FUNC_CALL VAR NUMBER VAR 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 ASSIGN VAR FUNC_CALL VAR NUMBER VAR ASSIGN VAR FUNC_CALL VAR VAR VAR VAR FUNC_CALL VAR VAR BIN_OP VAR NUMBER NUMBER EXPR FUNC_CALL 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 | t = int(input())
def task1(list1):
ans = 0
temp_ans = 0
for i in range(1, n, 2):
temp1 = list1[i - 1]
temp2 = list1[i]
temp_ans += temp2 - temp1
if temp_ans <= 0:
temp_ans = 0
if temp_ans > ans:
ans = temp_ans
return ans
def task2(list1):
ans = 0
temp_ans = 0
for i in range(2, n, 2):
temp1 = list1[i - 1]
temp2 = list1[i]
temp_ans += temp1 - temp2
if temp_ans <= 0:
temp_ans = 0
if temp_ans > ans:
ans = temp_ans
return ans
for _ in range(t):
n = int(input())
list1 = list(map(int, input().strip().split()))
ans = 0
for i in range(0, n, 2):
ans += list1[i]
ans1 = task1(list1)
ans2 = task2(list1)
print(ans + max(ans1, ans2)) | ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR FUNC_DEF ASSIGN VAR NUMBER ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR NUMBER VAR NUMBER ASSIGN VAR VAR BIN_OP VAR NUMBER ASSIGN VAR VAR VAR VAR BIN_OP VAR VAR IF VAR NUMBER ASSIGN VAR NUMBER IF VAR VAR ASSIGN VAR VAR RETURN VAR FUNC_DEF ASSIGN VAR NUMBER ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR NUMBER VAR NUMBER ASSIGN VAR VAR BIN_OP VAR NUMBER ASSIGN VAR VAR VAR VAR BIN_OP VAR VAR IF VAR NUMBER ASSIGN VAR NUMBER IF VAR VAR ASSIGN VAR VAR RETURN 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 FUNC_CALL VAR ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR NUMBER VAR NUMBER VAR VAR VAR 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 | def maxSubArraySum(a, size):
max_so_far = -1000000000000000000
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
wqr = int(input())
for er in range(wqr):
a = int(input())
temp = list(map(int, input().split(" ")))
temp1 = []
ans = 0
i = 0
while i < a:
ans += temp[i]
i += 2
i = 0
while i < a - 1:
temp1.append(temp[i + 1] - temp[i])
i += 2
t = maxSubArraySum(temp1, len(temp1))
i = 1
temp2 = []
while i < a - 1:
temp2.append(temp[i] - temp[i + 1])
i += 2
t1 = maxSubArraySum(temp2, len(temp2))
m = max(t, t1)
if m > 0:
ans += m
print(ans) | 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 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 STRING ASSIGN VAR LIST ASSIGN VAR NUMBER ASSIGN VAR NUMBER WHILE VAR VAR VAR VAR VAR VAR NUMBER ASSIGN VAR NUMBER WHILE VAR BIN_OP VAR NUMBER 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 BIN_OP VAR NUMBER EXPR FUNC_CALL VAR BIN_OP VAR VAR VAR BIN_OP VAR NUMBER VAR NUMBER ASSIGN VAR FUNC_CALL VAR 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 f(a):
ans = -(10**20)
mx = -(10**20)
for i in a:
mx = max(mx, i)
ans = max(ans, mx - i)
return ans
T = int(input())
for _ in range(T):
n = int(input())
a = list(map(int, input().split()))
cum = []
cum.append(a[0])
sum = a[0]
for i in range(1, n):
if i % 2 == 0:
cum.append(cum[-1] + a[i])
sum = sum + a[i]
else:
cum.append(cum[-1] - a[i])
evn = []
for i in range(0, n, 2):
evn.append(cum[i])
od = [0]
for i in range(1, n, 2):
od.append(cum[i])
print(sum + max(f(evn), f(od))) | FUNC_DEF ASSIGN VAR BIN_OP NUMBER NUMBER ASSIGN VAR BIN_OP NUMBER NUMBER FOR VAR VAR ASSIGN VAR FUNC_CALL VAR VAR VAR ASSIGN VAR FUNC_CALL VAR VAR BIN_OP 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 FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR LIST EXPR FUNC_CALL VAR VAR NUMBER ASSIGN VAR VAR NUMBER FOR VAR FUNC_CALL VAR NUMBER VAR IF BIN_OP VAR NUMBER NUMBER EXPR FUNC_CALL VAR BIN_OP VAR NUMBER VAR VAR ASSIGN VAR BIN_OP VAR VAR VAR EXPR FUNC_CALL VAR BIN_OP VAR NUMBER VAR VAR ASSIGN VAR LIST FOR VAR FUNC_CALL VAR NUMBER VAR NUMBER EXPR FUNC_CALL VAR VAR VAR ASSIGN VAR LIST NUMBER FOR VAR FUNC_CALL VAR NUMBER VAR NUMBER EXPR FUNC_CALL VAR VAR 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 | import sys
input = sys.stdin.readline
t = int(input())
for tests in range(t):
n = int(input())
A = list(map(int, input().split()))
E = [0]
O = [0]
INIT = 0
for i in range(n):
if i % 2 == 0:
E.append(E[-1] + A[i])
O.append(O[-1])
INIT += A[i]
else:
E.append(E[-1])
O.append(O[-1] + A[i])
O_E = [(O[i] - E[i]) for i in range(n + 1)]
MAX0 = O_E[-1]
MAX1 = O_E[-2]
ANS = 0
for i in range(n, -1, -1):
if (n - i) % 2 == 0:
ANS = max(ANS, MAX0 - O_E[i])
MAX0 = max(MAX0, O_E[i])
else:
ANS = max(ANS, MAX1 - O_E[i])
MAX1 = max(MAX1, O_E[i])
print(ANS + INIT) | 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 LIST NUMBER ASSIGN VAR LIST NUMBER ASSIGN VAR NUMBER FOR 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 VAR VAR VAR EXPR FUNC_CALL VAR VAR NUMBER EXPR FUNC_CALL VAR BIN_OP VAR NUMBER VAR VAR ASSIGN VAR BIN_OP VAR VAR VAR VAR VAR FUNC_CALL VAR BIN_OP VAR NUMBER ASSIGN VAR VAR NUMBER ASSIGN VAR VAR NUMBER ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR VAR NUMBER NUMBER IF BIN_OP BIN_OP VAR VAR NUMBER NUMBER ASSIGN VAR FUNC_CALL VAR VAR BIN_OP VAR VAR VAR ASSIGN VAR FUNC_CALL VAR VAR VAR VAR ASSIGN VAR FUNC_CALL VAR VAR BIN_OP VAR VAR VAR ASSIGN VAR FUNC_CALL 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 i in range(int(input())):
am = int(input())
arr = list(map(int, input().split()))
nArr = []
for i in range(am):
if i & 1:
nArr.append(arr[i])
else:
nArr.append(-arr[i])
sums = 0
best = 0
for i in range(0, am - 1, 2):
sums = max(nArr[i] + nArr[i + 1], sums + nArr[i] + nArr[i + 1])
best = max(best, sums)
was = best
sums = 0
best = 0
for i in range(1, am - 1, 2):
sums = max(nArr[i] + nArr[i + 1], sums + nArr[i] + nArr[i + 1])
best = max(best, sums)
was = max(best, was)
s = 0
for i in range(0, am, 2):
s += arr[i]
print(s + was) | 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 VAR IF BIN_OP VAR NUMBER EXPR FUNC_CALL VAR VAR VAR EXPR FUNC_CALL VAR VAR VAR ASSIGN VAR NUMBER ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR NUMBER 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 VAR BIN_OP VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR VAR ASSIGN VAR VAR ASSIGN VAR NUMBER ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR NUMBER 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 VAR BIN_OP VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR VAR 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 | def read_list():
return list(map(int, input().strip().split(" ")))
def print_list(l):
print(" ".join(map(str, l)))
N = int(input())
for _ in range(N):
n = int(input())
a = read_list()
res = 0
tmp = 0
s = 0
for i in range(1, n, 2):
s += a[i] - a[i - 1]
res = max(res, s - tmp)
tmp = min(tmp, s)
tmp = 0
s = 0
for i in range(1, n - 1, 2):
s += a[i] - a[i + 1]
res = max(res, s - tmp)
tmp = min(tmp, s)
print(sum(a[::2]) + res) | FUNC_DEF RETURN FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL FUNC_CALL VAR STRING FUNC_DEF EXPR FUNC_CALL VAR FUNC_CALL STRING 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 ASSIGN VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR NUMBER VAR NUMBER VAR BIN_OP VAR VAR VAR BIN_OP VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR BIN_OP VAR VAR ASSIGN VAR FUNC_CALL VAR VAR VAR 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 BIN_OP VAR VAR ASSIGN VAR FUNC_CALL VAR 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 | for _ in range(int(input())):
n = int(input())
l = list(map(int, input().split()))
o = [0] * n
e = [0] * n
e[0] = l[0]
dp = [0] * n
dp[0] = o[0] - e[0]
ans = 0
emin = min(0, dp[0])
omin = 0
for i in range(1, n):
o[i] = o[i - 1]
e[i] = e[i - 1]
if i % 2 == 0:
e[i] += l[i]
else:
o[i] += l[i]
dp[i] = o[i] - e[i]
if i % 2 == 0:
temp = dp[i] - emin
ans = max(ans, temp)
emin = min(emin, dp[i])
else:
temp = dp[i] - omin
ans = max(ans, temp)
omin = min(omin, dp[i])
ans += e[-1]
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 BIN_OP LIST NUMBER VAR ASSIGN VAR BIN_OP LIST NUMBER VAR ASSIGN VAR NUMBER VAR NUMBER ASSIGN VAR BIN_OP LIST NUMBER VAR ASSIGN VAR NUMBER BIN_OP VAR NUMBER VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR FUNC_CALL VAR NUMBER VAR NUMBER ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR NUMBER VAR ASSIGN VAR VAR VAR BIN_OP VAR NUMBER ASSIGN VAR VAR VAR BIN_OP VAR NUMBER IF BIN_OP VAR NUMBER NUMBER VAR VAR VAR VAR VAR VAR VAR VAR ASSIGN VAR VAR BIN_OP VAR VAR VAR VAR IF BIN_OP VAR NUMBER NUMBER ASSIGN VAR BIN_OP VAR VAR VAR ASSIGN VAR FUNC_CALL VAR VAR VAR ASSIGN VAR FUNC_CALL VAR VAR VAR VAR ASSIGN VAR BIN_OP VAR VAR VAR ASSIGN VAR FUNC_CALL VAR VAR VAR ASSIGN VAR FUNC_CALL VAR VAR 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 | def max_sub(a):
total_max = 0
cur_max = 0
for v in a:
cur_max += v
cur_max = max(cur_max, 0)
total_max = max(total_max, cur_max)
return total_max
t = int(input())
for _ in range(t):
n = int(input())
a = list(map(int, input().split()))
a1 = list(a[::2])
a2 = list(a[1::2])
d1 = list(y - x for x, y in zip(a1, a2))
d2 = list(y - x for x, y in zip(a1[1:], a2))
print(sum(a1) + max(max_sub(d1), max_sub(d2))) | FUNC_DEF ASSIGN VAR NUMBER ASSIGN VAR NUMBER FOR VAR VAR VAR VAR ASSIGN VAR FUNC_CALL VAR VAR NUMBER 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 FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR NUMBER NUMBER ASSIGN VAR FUNC_CALL VAR BIN_OP VAR VAR VAR VAR FUNC_CALL VAR VAR VAR ASSIGN VAR FUNC_CALL VAR BIN_OP VAR VAR VAR VAR FUNC_CALL VAR VAR NUMBER VAR EXPR FUNC_CALL VAR BIN_OP FUNC_CALL VAR 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())
while t:
t -= 1
n = int(input())
a = list(map(int, input().split()))
sum = 0
for i in range(0, n, 2):
sum += a[i]
mmax = 0
cur = 0
for i in range(1, n, 2):
cur = max(0, a[i] - a[i - 1] + cur)
mmax = max(mmax, cur)
cur = 0
for i in range(2, n, 2):
cur = max(0, a[i - 1] - a[i] + cur)
mmax = max(mmax, cur)
sum += mmax
print(sum) | 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 NUMBER FOR VAR FUNC_CALL VAR NUMBER VAR NUMBER VAR VAR VAR ASSIGN VAR NUMBER ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR NUMBER VAR NUMBER ASSIGN VAR FUNC_CALL VAR NUMBER BIN_OP BIN_OP VAR VAR VAR BIN_OP VAR NUMBER VAR ASSIGN VAR FUNC_CALL VAR VAR VAR ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR NUMBER VAR NUMBER ASSIGN VAR FUNC_CALL VAR NUMBER BIN_OP BIN_OP VAR BIN_OP VAR NUMBER VAR VAR VAR ASSIGN VAR FUNC_CALL VAR 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 | __MULTITEST = True
def solve():
n = int(input())
a = list(map(int, input().split()))
l = 0
r = -1
da = []
db = []
for i in range(1, n, 2):
da.append(a[i] - a[i - 1])
glMax = 0
la = 0
ra = -1
curMax = 0
lc = 0
rc = -1
for i in range(len(da)):
curMax += da[i]
if curMax > 0:
rc = i
if curMax > glMax:
glMax = curMax
la, ra = lc, rc
else:
curMax = 0
lc = i + 1
db = []
for i in range(2, n, 2):
db.append(a[i - 1] - a[i])
glMaxB = 0
lB = 0
rB = -1
curMax = 0
lc = 0
rc = -1
for i in range(len(db)):
curMax += db[i]
if curMax > 0:
rc = i
if curMax > glMaxB:
glMaxB = curMax
lB, rB = lc, rc
else:
curMax = 0
lc = i + 1
if glMax < glMaxB:
l = lB * 2 + 1
r = rB * 2 + 2
else:
l = la * 2
r = ra * 2 + 1
a1 = a[:l]
a2 = a[l : r + 1]
a2.reverse()
a3 = a[r + 1 :]
a = a1 + a2 + a3
print(sum([a[i] for i in range(len(a)) if i % 2 == 0]))
t = int(input()) if __MULTITEST else 1
for tt in range(t):
solve() | ASSIGN VAR NUMBER 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 ASSIGN VAR NUMBER 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 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 VAR VAR VAR IF VAR NUMBER ASSIGN VAR VAR IF VAR VAR ASSIGN VAR VAR ASSIGN VAR VAR VAR VAR ASSIGN VAR NUMBER ASSIGN VAR BIN_OP VAR NUMBER ASSIGN VAR LIST FOR VAR FUNC_CALL VAR NUMBER VAR NUMBER EXPR FUNC_CALL VAR BIN_OP VAR BIN_OP VAR NUMBER VAR VAR ASSIGN 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 VAR VAR VAR IF VAR NUMBER ASSIGN VAR VAR IF VAR VAR ASSIGN VAR VAR ASSIGN VAR VAR VAR VAR ASSIGN VAR NUMBER ASSIGN VAR BIN_OP VAR NUMBER IF VAR VAR ASSIGN VAR BIN_OP BIN_OP VAR NUMBER 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 ASSIGN VAR VAR VAR ASSIGN VAR VAR VAR BIN_OP VAR NUMBER EXPR FUNC_CALL VAR ASSIGN VAR VAR BIN_OP VAR NUMBER ASSIGN VAR BIN_OP BIN_OP VAR VAR VAR EXPR FUNC_CALL VAR FUNC_CALL VAR VAR VAR VAR FUNC_CALL VAR FUNC_CALL VAR VAR BIN_OP VAR NUMBER NUMBER ASSIGN VAR VAR FUNC_CALL VAR FUNC_CALL VAR NUMBER 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 | for _ in range(int(input())):
n = int(input())
l = list(map(int, input().split()))
s = 0
for i in range(n):
if i % 2 == 0:
s += l[i]
x = []
for i in range(1, n, 2):
x.append(l[i] - l[i - 1])
prev = 0
ma = 0
for j in range(0, len(x)):
prev = max(prev + x[j], x[j])
ma = max(prev, ma)
y = []
for i in range(1, n, 2):
if i + 1 >= n:
break
y.append(l[i] - l[i + 1])
prev = 0
ma1 = 0
for j in range(0, len(y)):
prev = max(prev + y[j], y[j])
ma1 = max(prev, ma1)
ans = max(ma, ma1)
print(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 NUMBER FOR VAR FUNC_CALL VAR VAR IF BIN_OP VAR NUMBER 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 NUMBER FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR BIN_OP VAR VAR VAR VAR VAR ASSIGN VAR FUNC_CALL VAR VAR VAR 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 ASSIGN VAR NUMBER ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR NUMBER FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR BIN_OP VAR VAR VAR VAR VAR ASSIGN VAR FUNC_CALL 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 | def solve(n, a):
if n == 1:
return a[0]
elif n == 2:
return max(a)
even_sum = sum(a[i] for i in range(0, n, 2))
la, lb = [], []
for i in range(n - 1):
if i % 2 == 0:
la.append(-a[i] + a[i + 1])
else:
lb.append(a[i] - a[i + 1])
ta, tb = [la[0]], [lb[0]]
for e in la[1:]:
ta.append(max(e, ta[-1] + e))
for e in lb[1:]:
tb.append(max(e, tb[-1] + e))
return even_sum + max(max(ta), max(tb), 0)
for _ in range(int(input())):
n = int(input())
a = list(map(int, input().split()))
print(solve(n, a)) | FUNC_DEF IF VAR NUMBER RETURN VAR NUMBER IF VAR NUMBER RETURN FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR VAR VAR VAR FUNC_CALL VAR NUMBER VAR NUMBER ASSIGN VAR VAR LIST LIST 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 EXPR FUNC_CALL VAR BIN_OP VAR VAR VAR BIN_OP VAR NUMBER ASSIGN VAR VAR LIST VAR NUMBER LIST VAR NUMBER FOR VAR VAR NUMBER EXPR FUNC_CALL VAR FUNC_CALL VAR VAR BIN_OP VAR NUMBER VAR FOR VAR VAR NUMBER EXPR FUNC_CALL VAR FUNC_CALL VAR VAR BIN_OP VAR NUMBER VAR RETURN BIN_OP VAR FUNC_CALL VAR FUNC_CALL VAR VAR 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 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 | t = int(input())
for _ in range(t):
n = int(input())
l = list(map(int, input().split()))
o = 0
e = 0
c = 0
s = 0
m = 0
for i in range(n):
c += 1
if i % 2 == 0:
e += l[i]
s += l[i]
else:
o += l[i]
if c % 2 == 0:
if o - e > m:
m = o - e
if e > o:
e = 0
o = 0
c = 0
c = 0
e = 0
o = 0
for i in range(1, n):
c += 1
if i % 2 == 0:
e += l[i]
else:
o += l[i]
if c % 2 == 0:
if o - e > m:
m = o - e
if e > o:
e = 0
o = 0
c = 0
print(s + 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 NUMBER ASSIGN VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR VAR VAR NUMBER IF BIN_OP VAR NUMBER NUMBER VAR VAR VAR VAR VAR VAR VAR VAR VAR IF BIN_OP VAR NUMBER NUMBER IF BIN_OP VAR VAR VAR ASSIGN VAR BIN_OP VAR VAR IF VAR VAR ASSIGN VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR NUMBER VAR VAR NUMBER IF BIN_OP VAR NUMBER NUMBER VAR VAR VAR VAR VAR VAR IF BIN_OP VAR NUMBER NUMBER IF BIN_OP VAR VAR VAR ASSIGN VAR BIN_OP VAR VAR IF VAR VAR ASSIGN VAR NUMBER ASSIGN 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 | def ms(l):
m = 0
i = 0
s = 0
while i < len(l):
s += l[i]
m = max(s, m)
if s < 0:
s = 0
i += 1
return m
def solve():
n = int(input())
l = list(map(int, input().split()))
z = []
for i in range(0, n - 1, 2):
z.append(l[i + 1] - l[i])
o = []
for i in range(1, n - 1, 2):
o.append(l[i] - l[i + 1])
s = 0
for i in range(0, n, 2):
s += l[i]
print(max(ms(o), ms(z)) + s)
t = int(input())
for _ in range(t):
solve() | FUNC_DEF ASSIGN VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR NUMBER WHILE VAR FUNC_CALL VAR VAR VAR VAR VAR 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 FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL 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 NUMBER FOR VAR FUNC_CALL VAR NUMBER VAR NUMBER VAR VAR VAR EXPR FUNC_CALL VAR BIN_OP FUNC_CALL VAR FUNC_CALL VAR 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 | maxint = 10**10
t = int(input())
def kadane(arr):
if len(arr) == 0:
return 0
add = ans = arr[0]
for i in range(1, len(arr)):
add += arr[i]
if arr[i] >= add:
add = arr[i]
ans = max(add, ans)
return max(ans, 0)
for _ in range(t):
n = int(input())
arr = list(map(int, input().split()))
ans = 0
for i in range(0, n, 2):
ans += arr[i]
even, odd = list(), list()
for i in range(0, n - 1, 2):
even.append(arr[i + 1] - arr[i])
for i in range(1, n - 1, 2):
odd.append(arr[i] - arr[i + 1])
ans += max(kadane(even), kadane(odd))
print(ans) | ASSIGN VAR BIN_OP NUMBER NUMBER ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR FUNC_DEF IF FUNC_CALL VAR VAR NUMBER RETURN NUMBER ASSIGN VAR VAR VAR NUMBER FOR VAR FUNC_CALL VAR NUMBER FUNC_CALL VAR VAR VAR VAR VAR IF VAR VAR VAR ASSIGN VAR VAR VAR ASSIGN VAR FUNC_CALL VAR VAR VAR RETURN FUNC_CALL VAR VAR NUMBER 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 VAR FUNC_CALL VAR FUNC_CALL 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 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 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
sys.setrecursionlimit(10000)
t = int(input())
for _ in range(t):
n = int(input())
a = list(map(int, input().split()))
even_sum = 0
all_diffs_start = []
all_diffs_non_start = []
for i in range(n):
if i % 2 == 0:
even_sum += a[i]
else:
all_diffs_start.append(a[i] - a[i - 1])
try:
all_diffs_non_start.append(a[i] - a[i + 1])
except:
pass
max_sum = 0
curr_sum = 0
for adiff in all_diffs_start:
if curr_sum < 0:
curr_sum = 0
curr_sum += adiff
if curr_sum > max_sum:
max_sum = curr_sum
curr_sum = 0
for adiff in all_diffs_non_start:
if curr_sum < 0:
curr_sum = 0
curr_sum += adiff
if curr_sum > max_sum:
max_sum = curr_sum
print(even_sum + max_sum) | IMPORT EXPR FUNC_CALL VAR NUMBER 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 LIST ASSIGN VAR LIST FOR VAR FUNC_CALL VAR VAR IF BIN_OP VAR NUMBER NUMBER VAR VAR VAR EXPR FUNC_CALL VAR BIN_OP VAR VAR VAR BIN_OP VAR NUMBER EXPR FUNC_CALL VAR BIN_OP VAR VAR VAR BIN_OP VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR NUMBER FOR VAR VAR IF VAR NUMBER ASSIGN VAR NUMBER VAR VAR IF VAR VAR ASSIGN VAR VAR ASSIGN VAR NUMBER FOR VAR VAR IF VAR NUMBER ASSIGN VAR NUMBER 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 | for _ in range(int(input())):
n = int(input())
lis = list(map(int, input().split()))
if n == 1:
print(lis[0])
elif n == 2:
print(max(lis))
else:
dif1 = []
dif2 = []
tit = 0
for i in range(len(lis)):
if i % 2 == 0:
tit += lis[i]
i = 0
while i < len(lis) - 1:
dif1.append(lis[i + 1] - lis[i])
i += 2
j = 1
while j < len(lis) - 1:
dif2.append(-1 * (lis[j + 1] - lis[j]))
j += 2
def kadane(lis):
tolmax = -1 * float("inf")
locmax = 0
for i in range(len(lis)):
locmax = max(lis[i], locmax + lis[i])
if locmax > tolmax:
tolmax = locmax
return tolmax
a = max(kadane(dif1), kadane(dif2))
if a > 0:
print(tit + a)
else:
print(tit) | 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 EXPR FUNC_CALL VAR VAR NUMBER IF VAR NUMBER EXPR FUNC_CALL VAR FUNC_CALL VAR VAR ASSIGN VAR LIST ASSIGN VAR LIST ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR FUNC_CALL VAR VAR IF BIN_OP VAR NUMBER NUMBER VAR VAR VAR ASSIGN VAR NUMBER WHILE VAR BIN_OP FUNC_CALL VAR VAR NUMBER EXPR FUNC_CALL VAR BIN_OP VAR BIN_OP VAR NUMBER VAR VAR VAR NUMBER ASSIGN VAR NUMBER WHILE VAR BIN_OP FUNC_CALL VAR VAR NUMBER EXPR FUNC_CALL VAR BIN_OP NUMBER BIN_OP VAR BIN_OP VAR NUMBER VAR VAR VAR NUMBER FUNC_DEF ASSIGN VAR BIN_OP NUMBER FUNC_CALL VAR STRING 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 ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL VAR VAR 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 | def f(l):
res, end = 0, 0
for i in range(len(l)):
end += l[i]
if end < 0:
end = 0
elif res < end:
res = end
return res
for _ in range(int(input())):
n = int(input())
l = list(map(int, input().split()))
l1 = []
l2 = []
s = l[0]
for i in range(1, len(l)):
if i % 2 == 0:
s += l[i]
l2.append(l[i - 1] - l[i])
else:
l1.append(l[i] - l[i - 1])
print(s + max(f(l1), f(l2))) | FUNC_DEF ASSIGN VAR VAR NUMBER 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 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 LIST ASSIGN VAR LIST ASSIGN VAR VAR NUMBER FOR VAR FUNC_CALL VAR NUMBER FUNC_CALL VAR VAR 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 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, stdout
def maxium_sum_on_even_positions(n, a):
esum = 0
res1 = 0
dif1 = 0
res2 = 0
dif2 = 0
for i in range(0, n, 2):
esum += a[i]
if i + 1 < n:
dif1 = max(0, dif1 + a[i + 1] - a[i])
res1 = max(res1, dif1)
if i > 0:
dif2 = max(0, dif2 + a[i - 1] - a[i])
res2 = max(res2, dif2)
return max(esum, esum + max(res1, res2))
t = int(stdin.readline())
for i in range(t):
n = int(stdin.readline())
a = list(map(int, stdin.readline().split()))
stdout.write(str(maxium_sum_on_even_positions(n, a)) + "\n") | FUNC_DEF ASSIGN VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR NUMBER VAR NUMBER VAR VAR VAR IF BIN_OP VAR NUMBER VAR 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 IF VAR 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 RETURN FUNC_CALL VAR VAR 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 BIN_OP FUNC_CALL VAR FUNC_CALL VAR 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 solve():
n = int(input())
a = list(map(int, input().split()))
ans = 0
for i in range(0, n, 2):
ans += a[i]
mx = ans
cnt = 0
for i in range(1, n, 2):
cnt += a[i] - a[i - 1]
mx = max(mx, ans + cnt)
if cnt < 0:
cnt = 0
cnt = 0
for i in range(2, n, 2):
cnt += a[i - 1] - a[i]
mx = max(mx, ans + cnt)
if cnt < 0:
cnt = 0
print(mx)
t = int(input())
for _ in range(t):
solve() | 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 VAR ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR NUMBER VAR NUMBER VAR BIN_OP VAR VAR VAR BIN_OP VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR BIN_OP VAR VAR IF VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR NUMBER VAR NUMBER VAR BIN_OP VAR BIN_OP VAR NUMBER VAR VAR ASSIGN VAR FUNC_CALL VAR VAR BIN_OP VAR VAR IF VAR NUMBER ASSIGN VAR NUMBER EXPR FUNC_CALL 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 kadane(Arr, l, r):
sm = 0
x = 0
for i in range(len(Arr)):
x = x + Arr[i] if x + Arr[i] > 0 else 0
sm = max(sm, x)
return sm
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 + kadane(L, 0, len(L) - 1), ans + kadane(R, 0, len(R) - 1))) | FUNC_DEF ASSIGN VAR NUMBER ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR FUNC_CALL VAR VAR ASSIGN VAR BIN_OP VAR VAR VAR NUMBER BIN_OP VAR VAR VAR NUMBER 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 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 | from sys import maxsize, stdin, stdout
input = stdin.readline
def solve():
pass
test = 1
test = int(input().strip())
for t in range(0, test):
n = int(input().strip())
arr = [int(x) for x in input().strip().split()]
val = 0
for i in range(0, n, 2):
val += arr[i]
if n == 1:
print(arr[0])
continue
brr = [0] * n
brr[1] = max(0, arr[1] - arr[0])
for i in range(2, n):
if i % 2 == 0:
brr[i] = max(0, brr[i - 2] - arr[i] + arr[i - 1])
else:
brr[i] = max(0, brr[i - 2] + arr[i] - arr[i - 1])
print(max(val, val + max(brr)))
ans = solve() | ASSIGN VAR VAR FUNC_DEF ASSIGN VAR NUMBER ASSIGN VAR FUNC_CALL VAR FUNC_CALL FUNC_CALL VAR FOR VAR FUNC_CALL VAR NUMBER VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR VAR VAR FUNC_CALL FUNC_CALL FUNC_CALL VAR ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR NUMBER VAR NUMBER VAR VAR VAR IF VAR NUMBER EXPR FUNC_CALL VAR VAR NUMBER ASSIGN VAR BIN_OP LIST NUMBER VAR 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 VAR VAR BIN_OP VAR NUMBER ASSIGN VAR VAR FUNC_CALL VAR NUMBER BIN_OP BIN_OP VAR BIN_OP VAR NUMBER VAR VAR VAR BIN_OP VAR NUMBER EXPR FUNC_CALL VAR FUNC_CALL VAR VAR BIN_OP VAR FUNC_CALL VAR VAR ASSIGN 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 maxsub(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
t = int(input())
for _ in range(t):
n = int(input())
arr = list(map(int, input().split()))
pref1 = []
pref2 = []
for i in range(1, n, 2):
pref1.append(arr[i] - arr[i - 1])
for i in range(2, n, 2):
pref2.append(arr[i - 1] - arr[i])
a = maxsub(pref1, len(pref1))
b = maxsub(pref2, len(pref2))
ans = max(max(a, b), 0)
for i in range(0, n, 2):
ans += arr[i]
print(ans) | 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 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 ASSIGN VAR LIST 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 VAR NUMBER EXPR FUNC_CALL VAR BIN_OP VAR BIN_OP VAR NUMBER VAR VAR 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 FUNC_CALL VAR VAR VAR NUMBER FOR VAR FUNC_CALL VAR NUMBER VAR 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 | t = int(input())
for _ in range(t):
n = int(input())
a = list(map(int, input().split()))
m = (n + 1) // 2
x, y = [0] * m, [0] * (n // 2)
even = 0
for i in range(n):
if i % 2 == 0:
x[i // 2] = a[i]
even += a[i]
else:
y[i // 2] = a[i]
u, v = [0] * m, [0] * m
for i in range(m):
if not n % 2 == 1 or not i == n // 2:
u[i] = y[i] - x[i]
if not i == 0:
v[i] = y[i - 1] - x[i]
ans = 0
s1, s2 = 0, 0
for i in range(m):
s1 = max(s1 + u[i], u[i])
s2 = max(s2 + v[i], v[i])
ans = max(s1, s2, ans)
print(even + 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 BIN_OP BIN_OP VAR NUMBER NUMBER ASSIGN VAR VAR BIN_OP LIST NUMBER VAR BIN_OP LIST NUMBER BIN_OP VAR NUMBER ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR VAR IF BIN_OP VAR NUMBER NUMBER ASSIGN VAR BIN_OP VAR NUMBER VAR VAR VAR VAR VAR ASSIGN VAR BIN_OP VAR NUMBER VAR VAR ASSIGN VAR VAR BIN_OP LIST NUMBER VAR BIN_OP LIST NUMBER VAR FOR VAR FUNC_CALL VAR VAR IF BIN_OP VAR NUMBER NUMBER VAR BIN_OP VAR NUMBER ASSIGN VAR VAR BIN_OP VAR VAR VAR VAR IF VAR NUMBER ASSIGN VAR VAR BIN_OP VAR BIN_OP VAR NUMBER VAR VAR ASSIGN VAR NUMBER ASSIGN VAR VAR NUMBER NUMBER FOR VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR BIN_OP VAR VAR VAR VAR VAR ASSIGN VAR FUNC_CALL VAR BIN_OP VAR VAR VAR VAR VAR ASSIGN VAR FUNC_CALL 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 | import sys
input = sys.stdin.readline
def main():
t = int(input())
for _ in range(t):
N = int(input())
A = [int(x) for x in input().split()]
B = []
C = []
s = 0
for i in range(0, N, 2):
s += A[i]
if i != N - 1:
B.append(A[i + 1] - A[i])
if i != 0:
C.append(A[i - 1] - A[i])
ma = 0
r = 0
tmp = 0
for l in range(len(B)):
while r < len(B) and tmp >= 0:
tmp += B[r]
ma = max(ma, tmp)
r += 1
tmp -= B[l]
r = 0
tmp = 0
for l in range(len(C)):
while r < len(C) and tmp >= 0:
tmp += C[r]
ma = max(ma, tmp)
r += 1
tmp -= C[l]
print(s + ma)
main() | IMPORT ASSIGN VAR VAR 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 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 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 ASSIGN VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR FUNC_CALL VAR VAR WHILE VAR FUNC_CALL VAR VAR VAR NUMBER VAR VAR VAR ASSIGN VAR FUNC_CALL VAR VAR VAR VAR NUMBER VAR VAR VAR ASSIGN VAR NUMBER ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR FUNC_CALL VAR VAR WHILE VAR FUNC_CALL VAR VAR VAR NUMBER VAR VAR VAR ASSIGN VAR FUNC_CALL VAR VAR VAR VAR NUMBER VAR VAR VAR 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 | for _ in range(int(input())):
n = int(input())
lis = list(map(int, input().split()))
eve = []
odd = []
for i in range(1, n, 2):
eve.append(lis[i] - lis[i - 1])
for i in range(2, n, 2):
odd.append(lis[i] - lis[i - 1])
l = len(eve)
i = j = 0
l1 = len(odd)
while i < l and eve[i] <= 0:
i += 1
while j < l1 and odd[j] >= 0:
j += 1
ans1 = ans2 = tmp = 0
while i < l:
if eve[i] >= 0:
tmp += eve[i]
else:
ans1 = max(ans1, tmp)
if tmp + eve[i] < 0:
tmp = 0
else:
tmp += eve[i]
i += 1
ans1 = max(ans1, tmp)
tmp = 0
while j < l1:
if odd[j] <= 0:
tmp += odd[j]
else:
ans2 = min(ans2, tmp)
if tmp + odd[j] > 0:
tmp = 0
else:
tmp += odd[j]
j += 1
ans2 = abs(min(ans2, tmp))
ans = 0
for i in range(0, n, 2):
if i % 2 == 0:
ans += lis[i]
print(max(ans1, ans2) + 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 VAR NUMBER EXPR FUNC_CALL VAR BIN_OP 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 ASSIGN VAR VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR WHILE VAR VAR VAR VAR NUMBER VAR NUMBER WHILE VAR VAR VAR VAR NUMBER VAR NUMBER ASSIGN VAR VAR VAR NUMBER WHILE VAR VAR IF VAR VAR NUMBER VAR VAR VAR ASSIGN VAR FUNC_CALL VAR VAR VAR IF BIN_OP VAR VAR VAR NUMBER ASSIGN VAR NUMBER VAR VAR VAR VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR VAR ASSIGN VAR NUMBER WHILE VAR VAR IF VAR VAR NUMBER VAR VAR VAR ASSIGN VAR FUNC_CALL VAR VAR VAR IF BIN_OP VAR VAR VAR NUMBER ASSIGN VAR NUMBER VAR VAR VAR VAR NUMBER ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR VAR VAR ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR NUMBER VAR NUMBER IF BIN_OP VAR NUMBER NUMBER VAR VAR VAR EXPR FUNC_CALL VAR BIN_OP FUNC_CALL VAR 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.buffer.readline
def solution():
for _ in range(int(input())):
n = int(input())
l = list(map(int, input().split()))
ecnt = 0
for i in range(0, n, 2):
ecnt += l[i]
pre1 = [0] * n
pre2 = [0] * n
for i in range(1, n, 2):
pre1[i] = l[i] - l[i - 1]
if i < n - 1:
pre2[i] = l[i] - l[i + 1]
pre1x = pre2x = pre1y = pre2y = 0
for i in range(n):
pre1x += pre1[i]
pre2x += pre2[i]
if pre1x < 0:
pre1x = 0
if pre2x < 0:
pre2x = 0
if pre1x > pre1y:
pre1y = pre1x
if pre2x > pre2y:
pre2y = pre2x
print(ecnt + max(0, pre1y, pre2y))
solution() | IMPORT ASSIGN VAR VAR FUNC_DEF 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 BIN_OP LIST NUMBER VAR ASSIGN VAR BIN_OP LIST NUMBER VAR FOR VAR FUNC_CALL VAR NUMBER VAR NUMBER ASSIGN VAR VAR BIN_OP VAR VAR VAR BIN_OP VAR NUMBER IF VAR BIN_OP VAR NUMBER ASSIGN VAR VAR BIN_OP VAR VAR VAR BIN_OP VAR NUMBER ASSIGN VAR VAR VAR VAR NUMBER FOR VAR FUNC_CALL VAR VAR VAR VAR VAR VAR VAR VAR IF VAR NUMBER ASSIGN VAR NUMBER IF VAR NUMBER ASSIGN VAR NUMBER IF VAR VAR ASSIGN VAR VAR IF VAR VAR ASSIGN VAR VAR EXPR FUNC_CALL VAR BIN_OP VAR FUNC_CALL VAR NUMBER 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(A):
n = len(A)
if n == 0:
return 0
assert n & 1 == 0
B = []
for i in range(0, n, 2):
B.append(-A[i] + A[i + 1])
n = len(B)
S = [0]
for b in B:
S.append(S[-1] + b)
minn = 0
best = 0
for i in range(n):
minn = min(minn, S[i + 1])
best = max(best, S[i + 1] - minn)
return best
t = int(input())
for _ in range(t):
n = int(input())
A = [int(x) for x in input().split()]
if n & 1:
print(sum(A[::2]) + max(solve(A[1:][::-1]), solve(A[:-1])))
else:
print(sum(A[::2]) + max(solve(A), solve(A[1:-1][::-1]))) | FUNC_DEF ASSIGN VAR FUNC_CALL VAR VAR IF VAR NUMBER RETURN NUMBER BIN_OP VAR NUMBER NUMBER 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 ASSIGN VAR LIST NUMBER FOR VAR VAR EXPR FUNC_CALL VAR BIN_OP VAR NUMBER VAR ASSIGN VAR NUMBER ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR VAR VAR BIN_OP VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR BIN_OP VAR BIN_OP VAR NUMBER 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 BIN_OP VAR NUMBER EXPR FUNC_CALL VAR BIN_OP FUNC_CALL VAR VAR NUMBER FUNC_CALL VAR FUNC_CALL VAR VAR NUMBER NUMBER FUNC_CALL VAR VAR NUMBER EXPR FUNC_CALL VAR BIN_OP FUNC_CALL VAR VAR NUMBER FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL VAR VAR NUMBER 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
from itertools import accumulate
def input():
return sys.stdin.readline().strip()
def list2d(a, b, c):
return [([c] * b) for i in range(a)]
def list3d(a, b, c, d):
return [[([d] * c) for j in range(b)] for i in range(a)]
def list4d(a, b, c, d, e):
return [[[([e] * d) for j in range(c)] for j in range(b)] for i in range(a)]
def ceil(x, y=1):
return int(-(-x // y))
def INT():
return int(input())
def MAP():
return map(int, input().split())
def LIST(N=None):
return list(MAP()) if N is None else [INT() for i in range(N)]
def Yes():
print("Yes")
def No():
print("No")
def YES():
print("YES")
def NO():
print("NO")
INF = 10**19
MOD = 10**9 + 7
for _ in range(INT()):
N = INT()
A = LIST()
B1 = []
B2 = []
for i in range(N - 1):
if i % 2 == 0:
B1.append(A[i + 1] - A[i])
else:
B2.append(A[i] - A[i + 1])
acc1 = [0] + list(accumulate(B1))
acc2 = [0] + list(accumulate(B2))
accmn1 = list(accumulate(acc1, min))
accmx1 = list(accumulate(acc1[::-1], max))[::-1]
accmn2 = list(accumulate(acc2, min))
accmx2 = list(accumulate(acc2[::-1], max))[::-1]
N1 = len(acc1)
N2 = len(acc2)
base = sum([(a if i % 2 == 0 else 0) for i, a in enumerate(A)])
ans = base
for i in range(N1 - 1):
l = accmn1[i]
r = accmx1[i + 1]
ans = max(ans, base + r - l)
for i in range(N2 - 1):
l = accmn2[i]
r = accmx2[i + 1]
ans = max(ans, base + r - l)
print(ans) | IMPORT FUNC_DEF RETURN FUNC_CALL FUNC_CALL VAR FUNC_DEF RETURN BIN_OP LIST VAR VAR VAR FUNC_CALL VAR VAR FUNC_DEF RETURN BIN_OP LIST VAR VAR VAR FUNC_CALL VAR VAR VAR FUNC_CALL VAR VAR FUNC_DEF RETURN BIN_OP LIST VAR VAR VAR FUNC_CALL VAR VAR VAR FUNC_CALL VAR VAR VAR FUNC_CALL VAR VAR FUNC_DEF NUMBER RETURN FUNC_CALL VAR BIN_OP VAR VAR FUNC_DEF RETURN FUNC_CALL VAR FUNC_CALL VAR FUNC_DEF RETURN FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR FUNC_DEF NONE RETURN VAR NONE FUNC_CALL VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL VAR VAR FUNC_DEF EXPR FUNC_CALL VAR STRING FUNC_DEF EXPR FUNC_CALL VAR STRING FUNC_DEF EXPR FUNC_CALL VAR STRING FUNC_DEF EXPR FUNC_CALL VAR STRING ASSIGN VAR BIN_OP NUMBER NUMBER ASSIGN VAR BIN_OP BIN_OP NUMBER NUMBER NUMBER FOR VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR ASSIGN VAR LIST ASSIGN VAR LIST 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 EXPR FUNC_CALL VAR BIN_OP VAR VAR VAR BIN_OP VAR NUMBER ASSIGN VAR BIN_OP LIST NUMBER FUNC_CALL VAR FUNC_CALL VAR VAR ASSIGN VAR BIN_OP LIST NUMBER FUNC_CALL VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR VAR VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR VAR NUMBER VAR NUMBER ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR VAR VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR VAR NUMBER VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR BIN_OP VAR NUMBER NUMBER VAR NUMBER VAR VAR FUNC_CALL VAR VAR ASSIGN VAR VAR FOR VAR FUNC_CALL VAR BIN_OP VAR NUMBER ASSIGN VAR VAR VAR ASSIGN VAR VAR BIN_OP VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR BIN_OP BIN_OP VAR VAR VAR FOR VAR FUNC_CALL VAR BIN_OP VAR NUMBER ASSIGN VAR VAR VAR ASSIGN VAR VAR BIN_OP VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR BIN_OP BIN_OP 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 solve(s, e, a, mul):
maxi = 0
prev = 0
for i in range(s, e, 2):
prev = max(prev + mul * (a[i + 1] - a[i]), mul * (a[i + 1] - a[i]))
maxi = max(maxi, prev)
return maxi
for _ in range(int(input())):
n = int(input())
a = [int(x) for x in input().split()]
total = sum(a[i] for i in range(0, n, 2))
ans = total + solve(0, n - n % 2, a, 1)
if n % 2 != 0:
ans = max(ans, total + solve(1, n, a, -1))
else:
ans = max(ans, total + solve(1, n - 1, a, -1))
print(ans) | FUNC_DEF ASSIGN VAR NUMBER ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR VAR VAR NUMBER ASSIGN VAR FUNC_CALL VAR BIN_OP VAR BIN_OP VAR BIN_OP VAR BIN_OP VAR NUMBER VAR VAR BIN_OP VAR BIN_OP VAR BIN_OP VAR NUMBER 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 FUNC_CALL VAR VAR VAR VAR FUNC_CALL VAR NUMBER VAR NUMBER ASSIGN VAR BIN_OP VAR FUNC_CALL VAR NUMBER BIN_OP VAR BIN_OP VAR NUMBER VAR NUMBER IF BIN_OP VAR NUMBER NUMBER ASSIGN VAR FUNC_CALL VAR VAR BIN_OP VAR FUNC_CALL VAR NUMBER VAR VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR BIN_OP VAR FUNC_CALL VAR NUMBER BIN_OP VAR NUMBER 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()))
if n == 1:
print(a[0])
continue
if n == 2:
print(max(a))
continue
even, odd = [], []
for i in range(n):
if i % 2 == 0:
even.append(a[i])
else:
odd.append(a[i])
ans1, ans2 = 0, 0
dif1 = []
for i in range(len(odd)):
dif1.append(odd[i] - even[i])
tmp = 0
for i in range(len(dif1)):
tmp += dif1[i]
if ans1 < tmp:
ans1 = tmp
if tmp < 0:
tmp = 0
dif2 = []
for i in range(len(even) - 1):
dif2.append(odd[i] - even[i + 1])
tmp = 0
for i in range(len(dif2)):
tmp += dif2[i]
if ans2 < tmp:
ans2 = tmp
if tmp < 0:
tmp = 0
print(sum(even) + max(ans1, ans2)) | 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 IF VAR NUMBER EXPR FUNC_CALL VAR FUNC_CALL VAR VAR ASSIGN VAR VAR LIST LIST FOR 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 VAR NUMBER NUMBER ASSIGN VAR LIST FOR VAR FUNC_CALL VAR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR BIN_OP VAR VAR VAR VAR ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR FUNC_CALL VAR VAR VAR VAR VAR IF VAR VAR ASSIGN VAR VAR IF VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR LIST FOR VAR FUNC_CALL VAR BIN_OP FUNC_CALL VAR VAR NUMBER EXPR FUNC_CALL VAR BIN_OP VAR VAR VAR BIN_OP VAR NUMBER ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR FUNC_CALL VAR VAR VAR VAR VAR IF VAR VAR ASSIGN VAR VAR IF VAR NUMBER ASSIGN VAR NUMBER EXPR FUNC_CALL VAR BIN_OP 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 | import sys
def kadane(b, ch):
mx = b[0]
cur = 0
ms = 0
me = 0
start = 0
if ch == 1:
ms += 1
me += 1
for h in range(len(b)):
if cur + b[h] >= 0:
if cur < 0:
cur = b[h]
start = h
else:
cur += b[h]
else:
if cur > mx:
mx = cur
ms = 2 * start
me = 2 * h
if ch == 1:
ms += 1
me += 1
cur = b[h]
start = h
if cur > mx:
mx = cur
ms = 2 * start
me = 2 * h
if ch == 1:
ms += 1
me += 1
if cur > mx:
mx = cur
ms = 2 * start
me = 2 * (len(b) - 1)
if ch == 1:
ms += 1
me += 1
return mx, ms, me
def reverse(arr, i, j):
for p in range(i, j + 1, 2):
arr[p], arr[p + 1] = arr[p + 1], arr[p]
t = int(sys.stdin.readline())
ans_arr = []
for i in range(t):
n = int(sys.stdin.readline())
a = [int(x) for x in sys.stdin.readline().split()]
if n == 1:
ans_arr.append(str(a[0]))
elif n == 2:
ans_arr.append(str(max(a[0], a[1])))
else:
ans = 0
for e in range(n):
if e % 2 == 0:
ans += a[e]
b1 = []
b2 = []
for g in range(0, n, 2):
if g + 1 < n:
b1.append(a[g + 1] - a[g])
for h in range(1, n, 2):
if h + 1 < n:
b2.append(a[h] - a[h + 1])
mx1, ms1, me1 = kadane(b1, 0)
mx2, ms2, me2 = kadane(b2, 1)
ans1 = 0
ans2 = 0
if mx1 > mx2:
ms = ms1
me = me1
reverse(a, ms, me)
for p in range(len(a)):
if p % 2 == 0:
ans1 += a[p]
else:
ms = ms2
me = me2
reverse(a, ms, me)
for p in range(len(a)):
if p % 2 == 0:
ans2 += a[p]
ans = max(ans, ans1)
ans = max(ans, ans2)
ans_arr.append(str(ans))
print("\n".join(ans_arr)) | IMPORT FUNC_DEF ASSIGN VAR VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR NUMBER IF VAR NUMBER VAR NUMBER VAR NUMBER FOR VAR FUNC_CALL VAR FUNC_CALL VAR VAR IF BIN_OP VAR VAR VAR NUMBER IF VAR NUMBER ASSIGN VAR VAR VAR ASSIGN VAR VAR VAR VAR VAR IF VAR VAR ASSIGN VAR VAR ASSIGN VAR BIN_OP NUMBER VAR ASSIGN VAR BIN_OP NUMBER VAR IF VAR NUMBER VAR NUMBER VAR NUMBER ASSIGN VAR VAR VAR ASSIGN VAR VAR IF VAR VAR ASSIGN VAR VAR ASSIGN VAR BIN_OP NUMBER VAR ASSIGN VAR BIN_OP NUMBER VAR IF VAR NUMBER VAR NUMBER VAR NUMBER IF VAR VAR ASSIGN VAR VAR ASSIGN VAR BIN_OP NUMBER VAR ASSIGN VAR BIN_OP NUMBER BIN_OP FUNC_CALL VAR VAR NUMBER IF VAR NUMBER VAR NUMBER VAR NUMBER RETURN VAR VAR VAR FUNC_DEF FOR VAR FUNC_CALL VAR VAR BIN_OP VAR NUMBER NUMBER ASSIGN VAR VAR VAR BIN_OP VAR NUMBER VAR BIN_OP VAR NUMBER 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 VAR VAR FUNC_CALL FUNC_CALL VAR IF VAR NUMBER EXPR FUNC_CALL VAR FUNC_CALL VAR VAR NUMBER IF VAR NUMBER EXPR FUNC_CALL VAR FUNC_CALL VAR FUNC_CALL VAR VAR NUMBER VAR NUMBER 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 VAR NUMBER 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 VAR VAR FUNC_CALL VAR VAR NUMBER ASSIGN VAR VAR VAR FUNC_CALL VAR VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR NUMBER IF VAR VAR ASSIGN VAR VAR ASSIGN VAR VAR EXPR FUNC_CALL VAR VAR VAR VAR FOR VAR FUNC_CALL VAR FUNC_CALL VAR VAR IF BIN_OP VAR NUMBER NUMBER VAR VAR VAR ASSIGN VAR VAR ASSIGN VAR VAR EXPR FUNC_CALL VAR VAR VAR VAR FOR VAR FUNC_CALL VAR FUNC_CALL VAR VAR IF BIN_OP VAR NUMBER NUMBER VAR VAR VAR ASSIGN VAR FUNC_CALL VAR VAR VAR ASSIGN VAR FUNC_CALL VAR VAR VAR EXPR FUNC_CALL VAR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR FUNC_CALL STRING 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(0, t):
n = int(input())
l = list(map(int, input().split()))
d = 0
mi = 0
for j in range(0, n, 2):
mi = mi + l[j]
jj = mi
p = 0
mi = 0
for j in range(0, n - 1, 2):
if j + 1 < n:
if l[j] <= l[j + 1]:
p = p + l[j + 1] - l[j]
if p > mi:
mi = p
else:
p = p - l[j] + l[j + 1]
if p < 0:
p = 0
p = 0
for j in range(1, n, 2):
if j + 1 < n:
if l[j] > l[j + 1]:
p = p - l[j + 1] + l[j]
if p > mi:
mi = p
j = j + 1
else:
p = p + l[j] - l[j + 1]
if p < 0:
p = 0
j = j + 1
print(jj + mi) | 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 NUMBER FOR VAR FUNC_CALL VAR NUMBER VAR NUMBER ASSIGN VAR BIN_OP VAR VAR VAR ASSIGN VAR VAR ASSIGN VAR NUMBER ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR NUMBER BIN_OP VAR NUMBER NUMBER IF BIN_OP VAR NUMBER VAR IF VAR VAR VAR BIN_OP VAR NUMBER ASSIGN VAR BIN_OP BIN_OP VAR VAR BIN_OP VAR NUMBER VAR VAR IF VAR VAR ASSIGN VAR VAR ASSIGN VAR BIN_OP BIN_OP VAR VAR VAR VAR BIN_OP VAR NUMBER IF VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR NUMBER VAR NUMBER IF BIN_OP VAR NUMBER VAR IF VAR VAR VAR BIN_OP VAR NUMBER ASSIGN VAR BIN_OP BIN_OP VAR VAR BIN_OP VAR NUMBER VAR VAR IF VAR VAR ASSIGN VAR VAR ASSIGN VAR BIN_OP VAR NUMBER ASSIGN VAR BIN_OP BIN_OP VAR VAR VAR VAR BIN_OP VAR NUMBER IF VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR BIN_OP 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 | t = int(input())
def rec(mn, ind, l, b, dp):
if ind == 0:
return 0
else:
if dp[ind - 1] == 0:
if mn[ind - 1][0] < l and mn[ind - 1][1] < b:
dp[ind - 1] = max(
rec(mn, ind - 1, l, b, dp),
mn[ind - 1][2]
+ rec(mn, ind - 1, mn[ind - 1][0], mn[ind - 1][1], dp),
)
else:
dp[ind - 1] = rec(mn, ind - 1, l, b, dp)
return dp[ind - 1]
for _ in range(t):
n = int(input())
l = list(map(int, input().split()))
sm = 0
mx = 0
for i in range(0, len(l) - 1, 2):
sm += l[i + 1] - l[i]
mx = max(sm, mx)
if sm < 0:
sm = 0
sm1 = 0
mx1 = 0
for i in range(1, len(l) - 1, 2):
sm1 += l[i] - l[i + 1]
mx1 = max(sm1, mx1)
if sm1 < 0:
sm1 = 0
print(sum(l[::2]) + max(mx, mx1)) | ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR FUNC_DEF IF VAR NUMBER RETURN NUMBER IF VAR BIN_OP VAR NUMBER NUMBER IF VAR BIN_OP VAR NUMBER NUMBER VAR VAR BIN_OP VAR NUMBER NUMBER VAR ASSIGN VAR BIN_OP VAR NUMBER FUNC_CALL VAR FUNC_CALL VAR VAR BIN_OP VAR NUMBER VAR VAR VAR BIN_OP VAR BIN_OP VAR NUMBER NUMBER FUNC_CALL VAR VAR BIN_OP VAR NUMBER VAR BIN_OP VAR NUMBER NUMBER VAR BIN_OP VAR NUMBER NUMBER VAR ASSIGN VAR BIN_OP VAR NUMBER FUNC_CALL VAR VAR BIN_OP VAR NUMBER VAR VAR VAR RETURN VAR BIN_OP VAR NUMBER 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 FOR VAR FUNC_CALL VAR NUMBER BIN_OP FUNC_CALL VAR 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 FUNC_CALL VAR 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 EXPR FUNC_CALL VAR BIN_OP FUNC_CALL VAR VAR NUMBER 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())
oo = 10**18
for tc in range(t):
n = int(input())
a = list(map(int, input().split()))
sol = 0
best = ce = co = 0
loe = loo = 0
for i in range(n):
if i % 2 == 0:
sol += a[i]
ce += a[i]
co -= a[i]
best = max(best, -(ce - loe))
loe = max(loe, ce)
else:
ce -= a[i]
co += a[i]
best = max(best, co - loo)
loo = min(loo, co)
print(sol + best) | ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR BIN_OP NUMBER NUMBER 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 VAR VAR NUMBER ASSIGN VAR 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 VAR BIN_OP VAR VAR ASSIGN VAR FUNC_CALL VAR VAR VAR VAR VAR VAR VAR VAR VAR ASSIGN VAR FUNC_CALL VAR VAR BIN_OP 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
def rs():
return sys.stdin.readline().rstrip()
def ri():
return int(sys.stdin.readline())
def ria():
return list(map(int, sys.stdin.readline().split()))
def ws(s):
sys.stdout.write(s + "\n")
def wi(n):
sys.stdout.write(str(n) + "\n")
def wia(a):
sys.stdout.write(" ".join([str(x) for x in a]) + "\n")
def solve(n, a):
base = 0
for i in range(0, n, 2):
base += a[i]
ans = base
l = 0
best = 0
s = 0
for r in range(n // 2):
s += a[2 * r + 1] - a[2 * r]
while s < 0:
s -= a[2 * l + 1] - a[2 * l]
l += 1
if s > best:
best = s
ans = max(ans, base + best)
l = 0
best = 0
s = 0
for r in range((n - 1) // 2):
s += a[2 * r + 1] - a[2 * r + 2]
while s < 0:
s -= a[2 * l + 1] - a[2 * l + 2]
l += 1
if s > best:
best = s
ans = max(ans, base + best)
return ans
def main():
for _ in range(ri()):
n = ri()
a = ria()
wi(solve(n, a))
main() | IMPORT FUNC_DEF RETURN 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 EXPR FUNC_CALL VAR BIN_OP VAR STRING FUNC_DEF EXPR FUNC_CALL VAR BIN_OP FUNC_CALL VAR VAR STRING FUNC_DEF EXPR FUNC_CALL VAR BIN_OP FUNC_CALL STRING FUNC_CALL VAR VAR VAR VAR STRING FUNC_DEF ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR NUMBER VAR NUMBER VAR VAR VAR ASSIGN VAR VAR ASSIGN VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR BIN_OP VAR NUMBER VAR BIN_OP VAR BIN_OP BIN_OP NUMBER VAR NUMBER VAR BIN_OP NUMBER VAR WHILE VAR NUMBER VAR BIN_OP VAR BIN_OP BIN_OP NUMBER VAR NUMBER VAR BIN_OP NUMBER VAR VAR NUMBER IF VAR VAR ASSIGN VAR VAR ASSIGN VAR FUNC_CALL VAR VAR BIN_OP VAR VAR ASSIGN VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR BIN_OP BIN_OP VAR NUMBER NUMBER VAR BIN_OP VAR BIN_OP BIN_OP NUMBER VAR NUMBER VAR BIN_OP BIN_OP NUMBER VAR NUMBER WHILE VAR NUMBER VAR BIN_OP VAR BIN_OP BIN_OP NUMBER VAR NUMBER VAR BIN_OP BIN_OP NUMBER VAR NUMBER VAR NUMBER IF VAR VAR ASSIGN VAR VAR ASSIGN VAR FUNC_CALL VAR VAR BIN_OP 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 | import sys
input = sys.stdin.buffer.readline
def prog():
for _ in range(int(input())):
n = int(input())
a = list(map(int, input().split()))
even_s = 0
for i in range(0, n, 2):
even_s += a[i]
left = []
right = []
for i in range(n // 2):
left.append(a[i * 2 + 1] - a[i * 2])
for i in range(1, (n - 1) // 2 + 1):
right.append(a[i * 2 - 1] - a[i * 2])
mx1 = 0
curr = 0
for i in range(len(left)):
curr = max(curr + left[i], left[i])
mx1 = max(curr, mx1)
mx2 = 0
curr = 0
for i in range(len(right)):
curr = max(curr + right[i], right[i])
mx2 = max(curr, mx2)
even_s += max(mx2, mx1)
sys.stdout.write(str(even_s) + "\n")
prog() | IMPORT ASSIGN VAR VAR FUNC_DEF 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 ASSIGN VAR LIST FOR VAR FUNC_CALL VAR BIN_OP VAR NUMBER EXPR FUNC_CALL VAR BIN_OP VAR BIN_OP BIN_OP VAR NUMBER NUMBER VAR BIN_OP VAR NUMBER FOR VAR FUNC_CALL VAR NUMBER BIN_OP BIN_OP BIN_OP VAR NUMBER NUMBER NUMBER EXPR FUNC_CALL VAR BIN_OP VAR BIN_OP BIN_OP VAR NUMBER NUMBER VAR BIN_OP VAR NUMBER ASSIGN VAR NUMBER 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 ASSIGN VAR NUMBER 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 VAR FUNC_CALL VAR VAR VAR EXPR FUNC_CALL VAR BIN_OP FUNC_CALL VAR VAR STRING 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())
arr = list(map(int, input().split()))
s = 0
for i in range(n):
if i % 2 == 0:
s += arr[i]
a = []
b = []
for i in range(1, n):
if i % 2 == 1:
a.append(arr[i] - arr[i - 1])
for i in range(1, n - 1):
if i % 2 == 1:
b.append(arr[i] - arr[i + 1])
dp1 = [0]
dp2 = [0]
for i in range(len(a)):
dp1.append(max(a[i], dp1[-1] + a[i]))
for i in range(len(b)):
dp2.append(max(b[i], dp2[-1] + b[i]))
print(s + 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 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 VAR IF 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 IF BIN_OP VAR NUMBER NUMBER EXPR FUNC_CALL VAR BIN_OP VAR VAR VAR BIN_OP VAR NUMBER ASSIGN VAR LIST NUMBER ASSIGN VAR LIST NUMBER FOR VAR FUNC_CALL VAR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR FUNC_CALL VAR VAR VAR BIN_OP VAR NUMBER VAR VAR FOR VAR FUNC_CALL VAR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR FUNC_CALL VAR VAR VAR BIN_OP VAR NUMBER VAR 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 | def kadane_alg(a):
max_so_far = float("-inf")
max_ending_here = 0
for i in range(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 if max_so_far > 0 else 0
for _ in range(int(input())):
n = int(input())
arr = [int(c) for c in input().split()]
i = 0
newarr = [(0) for i in range(len(arr) // 2)]
j = 0
while i + 1 < len(arr):
newarr[j] = arr[i + 1] - arr[i]
i += 2
j += 1
i = 1
newarr2 = [(0) for i in range(len(arr) // 2)]
j = 0
while i + 1 < len(arr):
newarr2[j] = arr[i] - arr[i + 1]
i += 2
j += 1
evensum = 0
for i in range(0, len(arr), 2):
evensum += arr[i]
print(evensum + max(kadane_alg(newarr), kadane_alg(newarr2))) | FUNC_DEF ASSIGN VAR FUNC_CALL VAR STRING ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR 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 NUMBER 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 VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR NUMBER ASSIGN VAR NUMBER VAR FUNC_CALL VAR BIN_OP FUNC_CALL VAR VAR NUMBER ASSIGN VAR NUMBER WHILE BIN_OP VAR NUMBER FUNC_CALL VAR VAR ASSIGN VAR VAR BIN_OP VAR BIN_OP VAR NUMBER VAR VAR VAR NUMBER VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR NUMBER VAR FUNC_CALL VAR BIN_OP FUNC_CALL VAR VAR NUMBER ASSIGN VAR NUMBER WHILE BIN_OP VAR NUMBER FUNC_CALL VAR VAR ASSIGN VAR VAR BIN_OP VAR VAR VAR BIN_OP VAR NUMBER VAR NUMBER VAR NUMBER ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR NUMBER FUNC_CALL VAR VAR NUMBER VAR VAR 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 | for _ in range(int(input())):
n = int(input())
lis = list(map(int, input().split()))
if n % 2 == 0:
diff = []
su = 0
for i in range(0, n, 2):
su += lis[i]
diff.append(lis[i] - lis[i + 1])
mi = 0
fi = 0
for i in range(n // 2):
fi += diff[i]
if fi < 0:
if fi < mi:
mi = fi
else:
fi = 0
m1 = su - mi
dif2 = []
for i in range(1, n - 1, 2):
dif2.append(lis[i] - lis[i + 1])
ma = 0
fi = 0
for i in range(len(dif2)):
fi += dif2[i]
if fi > 0:
if fi > ma:
ma = fi
else:
fi = 0
m2 = su + ma
print(max(m1, m2))
else:
su = 0
su += lis[-1]
dif1 = []
for i in range(0, n - 1, 2):
su += lis[i]
dif1.append(lis[i] - lis[i + 1])
mi = 0
fi = 0
for i in range(n // 2):
fi += dif1[i]
if fi < 0:
if fi < mi:
mi = fi
else:
fi = 0
m1 = su - mi
dif2 = []
for i in range(1, n, 2):
dif2.append(lis[i] - lis[i + 1])
ma = 0
fi = 0
for i in range(len(dif2)):
fi += dif2[i]
if fi > 0:
if fi > ma:
ma = fi
else:
fi = 0
m2 = su + ma
print(max(m1, m2)) | 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 BIN_OP VAR NUMBER NUMBER ASSIGN VAR LIST ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR NUMBER VAR NUMBER VAR VAR VAR EXPR FUNC_CALL VAR BIN_OP VAR VAR VAR BIN_OP VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR BIN_OP VAR NUMBER VAR VAR VAR IF VAR NUMBER IF VAR VAR ASSIGN VAR VAR ASSIGN VAR NUMBER ASSIGN VAR BIN_OP 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 NUMBER ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR FUNC_CALL VAR VAR VAR VAR VAR IF VAR NUMBER IF VAR VAR ASSIGN VAR VAR ASSIGN VAR NUMBER ASSIGN VAR BIN_OP VAR VAR EXPR FUNC_CALL VAR FUNC_CALL VAR VAR VAR ASSIGN VAR NUMBER VAR VAR NUMBER ASSIGN VAR LIST FOR VAR FUNC_CALL VAR NUMBER BIN_OP VAR NUMBER NUMBER VAR VAR VAR EXPR FUNC_CALL VAR BIN_OP VAR VAR VAR BIN_OP VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR BIN_OP VAR NUMBER VAR VAR VAR IF VAR NUMBER IF VAR VAR ASSIGN VAR VAR ASSIGN VAR NUMBER ASSIGN VAR BIN_OP 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 IF VAR VAR ASSIGN VAR VAR ASSIGN VAR NUMBER ASSIGN VAR BIN_OP VAR 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 | from sys import stdin
t = int(stdin.readline())
for _ in range(t):
n = int(stdin.readline())
a = list(map(int, stdin.readline().rstrip().split(" ")))
prefixArr = [0]
out = 0
for i in range(1, n):
if i % 2 != 0:
diff = a[i] - a[i - 1]
else:
diff = a[i - 1] - a[i]
if diff > diff + prefixArr[max(0, i - 2)]:
prefixArr.append(diff)
else:
prefixArr.append(diff + prefixArr[max(0, i - 2)])
for i in range(n):
if i % 2 == 0:
out += a[i]
print(out + max(prefixArr)) | 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 FUNC_CALL VAR STRING ASSIGN VAR LIST 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 ASSIGN VAR BIN_OP VAR BIN_OP VAR NUMBER VAR VAR IF VAR BIN_OP VAR VAR FUNC_CALL VAR NUMBER BIN_OP VAR NUMBER EXPR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR BIN_OP VAR VAR FUNC_CALL VAR NUMBER BIN_OP VAR NUMBER FOR VAR FUNC_CALL VAR VAR IF BIN_OP VAR NUMBER NUMBER VAR VAR VAR EXPR FUNC_CALL 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
t = int(input())
for _ in range(t):
n = int(input())
arr = list(map(int, input().split()))
li = []
ans = 0
for i in range(n):
if i % 2 == 0:
ans += arr[i]
for i in range(1, n, 2):
li.append((i - 1, arr[i] - arr[i - 1]))
temp = 0
if len(li) != 0:
m = -(10**18)
min_so_far = 0
a = 0
b = 0
c = 0
p = 0
for i in range(len(li)):
p += li[i][1]
x = p - min_so_far
if x > m:
m = x
b = li[i][0] + 1
c = a
if p < min_so_far:
min_so_far = p
a = li[i][0] + 2
if m > 0:
for i in range(c, b + 1):
if i % 2 != 0:
temp += arr[i]
for i in range(c):
if i % 2 == 0:
temp += arr[i]
for i in range(b + 1, n):
if i % 2 == 0:
temp += arr[i]
ans = max(ans, temp)
li = []
for i in range(2, n, 2):
li.append((i - 1, arr[i - 1] - arr[i]))
temp = 0
if len(li) != 0:
m = -(10**18)
min_so_far = 0
a = 1
c = 0
b = 0
p = 0
for i in range(len(li)):
p += li[i][1]
x = p - min_so_far
if x > m:
m = x
b = li[i][0] + 1
c = a
if p < min_so_far:
min_so_far = p
a = li[i][0] + 2
if m > 0:
for i in range(c, b + 1):
if i % 2 != 0:
temp += arr[i]
for i in range(c):
if i % 2 == 0:
temp += arr[i]
for i in range(b + 1, n):
if i % 2 == 0:
temp += arr[i]
ans = max(ans, temp)
print(ans) | 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 LIST ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR VAR IF BIN_OP VAR NUMBER NUMBER VAR VAR VAR FOR VAR FUNC_CALL VAR NUMBER VAR NUMBER EXPR FUNC_CALL VAR BIN_OP VAR NUMBER BIN_OP VAR VAR VAR BIN_OP VAR NUMBER ASSIGN VAR NUMBER IF FUNC_CALL VAR VAR NUMBER ASSIGN VAR BIN_OP NUMBER 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 VAR VAR VAR NUMBER ASSIGN VAR BIN_OP VAR VAR IF VAR VAR ASSIGN VAR VAR ASSIGN VAR BIN_OP VAR VAR NUMBER NUMBER ASSIGN VAR VAR IF VAR VAR ASSIGN VAR VAR ASSIGN VAR BIN_OP VAR VAR NUMBER NUMBER IF VAR NUMBER FOR VAR FUNC_CALL VAR VAR BIN_OP VAR NUMBER IF BIN_OP VAR NUMBER NUMBER VAR VAR VAR FOR VAR FUNC_CALL VAR VAR IF BIN_OP VAR NUMBER NUMBER VAR VAR VAR FOR VAR FUNC_CALL VAR BIN_OP VAR NUMBER VAR IF BIN_OP VAR NUMBER NUMBER VAR VAR VAR ASSIGN VAR FUNC_CALL VAR VAR VAR ASSIGN VAR LIST FOR VAR FUNC_CALL VAR NUMBER VAR NUMBER EXPR FUNC_CALL VAR BIN_OP VAR NUMBER BIN_OP VAR BIN_OP VAR NUMBER VAR VAR ASSIGN VAR NUMBER IF FUNC_CALL VAR VAR NUMBER ASSIGN VAR BIN_OP NUMBER 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 VAR VAR VAR NUMBER ASSIGN VAR BIN_OP VAR VAR IF VAR VAR ASSIGN VAR VAR ASSIGN VAR BIN_OP VAR VAR NUMBER NUMBER ASSIGN VAR VAR IF VAR VAR ASSIGN VAR VAR ASSIGN VAR BIN_OP VAR VAR NUMBER NUMBER IF VAR NUMBER FOR VAR FUNC_CALL VAR VAR BIN_OP VAR NUMBER IF BIN_OP VAR NUMBER NUMBER VAR VAR VAR FOR VAR FUNC_CALL VAR VAR IF BIN_OP VAR NUMBER NUMBER VAR VAR VAR FOR VAR FUNC_CALL VAR BIN_OP VAR NUMBER VAR IF BIN_OP VAR NUMBER NUMBER 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 msas(a):
max_so_far = -float("inf") - 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 _ in range(t):
size = int(input())
arr = [int(i) for i in input().split()]
s = sum([arr[i] for i in range(0, len(arr), 2)])
ar = arr + [0] * (len(arr) % 2)
a = [(ar[i + 1] - ar[i]) for i in range(0, len(ar), 2)]
br = arr[1:-1] if len(arr) % 2 == 0 else arr[1:]
b = [(br[i] - br[i + 1]) for i in range(0, len(br), 2)]
print(max(s + msas(a), s, s + msas(b))) | FUNC_DEF ASSIGN VAR BIN_OP FUNC_CALL VAR STRING 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 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 FUNC_CALL VAR VAR NUMBER ASSIGN VAR BIN_OP VAR BIN_OP LIST NUMBER BIN_OP FUNC_CALL VAR VAR NUMBER ASSIGN VAR BIN_OP VAR BIN_OP VAR NUMBER VAR VAR VAR FUNC_CALL VAR NUMBER FUNC_CALL VAR VAR NUMBER ASSIGN VAR BIN_OP FUNC_CALL VAR VAR NUMBER NUMBER VAR NUMBER NUMBER VAR NUMBER ASSIGN VAR BIN_OP VAR VAR VAR BIN_OP VAR NUMBER VAR FUNC_CALL VAR NUMBER FUNC_CALL VAR VAR NUMBER EXPR FUNC_CALL VAR FUNC_CALL VAR BIN_OP VAR FUNC_CALL VAR 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 solve(n, ar):
ans = 0
sum = 0
for i in range(1, n, 2):
sum += ar[i] - ar[i - 1]
if ans < sum:
ans = sum
if sum <= 0:
sum = 0
sum = 0
for i in range(2, n, 2):
sum += -(ar[i] - ar[i - 1])
if ans < sum:
ans = sum
if sum <= 0:
sum = 0
for i in range(0, n, 2):
ans += ar[i]
print(ans)
t = int(input())
for _ in range(t):
n = int(input())
ar = list(map(int, input().split()))
solve(n, ar) | FUNC_DEF ASSIGN VAR NUMBER ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR NUMBER VAR NUMBER VAR BIN_OP VAR VAR VAR BIN_OP VAR NUMBER IF VAR VAR ASSIGN VAR VAR IF VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR NUMBER VAR NUMBER VAR BIN_OP VAR VAR VAR BIN_OP VAR NUMBER IF VAR VAR ASSIGN VAR VAR IF VAR NUMBER ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR NUMBER VAR NUMBER VAR VAR VAR EXPR FUNC_CALL 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 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()))
init = 0
o2 = []
e2 = []
for i in range(0, n, 2):
init += a[i]
if i < n - 1:
o2.append(a[i + 1] - a[i])
if i != 0:
e2.append(a[i - 1] - a[i])
me2 = 0
if len(e2) > 0:
me2 = e2[0]
cm = e2[0]
for i in range(1, len(e2)):
cm = max(e2[i], cm + e2[i])
me2 = max(me2, cm)
mo2 = 0
if len(o2) > 0:
mo2 = o2[0]
cm = o2[0]
for i in range(1, len(o2)):
cm = max(o2[i], cm + o2[i])
mo2 = max(mo2, cm)
print(max(init + max(me2, mo2), init)) | 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 LIST ASSIGN VAR LIST FOR VAR FUNC_CALL VAR NUMBER VAR NUMBER VAR VAR VAR 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 ASSIGN VAR NUMBER IF FUNC_CALL VAR VAR NUMBER ASSIGN VAR VAR NUMBER ASSIGN VAR VAR NUMBER FOR VAR FUNC_CALL VAR NUMBER 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 IF FUNC_CALL VAR VAR NUMBER ASSIGN VAR VAR NUMBER ASSIGN VAR VAR NUMBER FOR VAR FUNC_CALL VAR NUMBER 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 BIN_OP VAR FUNC_CALL VAR 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):
max_current = arr[0]
max_global = arr[0]
n = len(arr)
for i in range(1, n):
max_current = max(arr[i], max_current + arr[i])
if max_current > max_global:
max_global = max_current
return max_global
t = int(input())
for tc in range(t):
length = int(input())
if length == 1:
n = int(input())
print(n)
continue
if length == 2:
line = input().split()
print(max(int(line[0]), int(line[1])))
continue
line = input().split()
arr = []
diff1 = []
diff2 = []
for x in line:
arr.append(int(x))
if length % 2 == 0:
for x in range(0, length - 1, 2):
diff1.append(arr[x + 1] - arr[x])
for x in range(1, length - 2, 2):
diff2.append(arr[x] - arr[x + 1])
if length % 2 != 0:
for x in range(0, length - 1, 2):
diff1.append(arr[x + 1] - arr[x])
for x in range(1, length - 1, 2):
diff2.append(arr[x] - arr[x + 1])
add_this = max(kadane(diff1), kadane(diff2))
if add_this < 0:
add_this = 0
for x in range(0, length, 2):
add_this += arr[x]
print(add_this) | FUNC_DEF ASSIGN VAR VAR NUMBER ASSIGN VAR VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR FOR VAR FUNC_CALL VAR NUMBER VAR ASSIGN VAR FUNC_CALL VAR VAR VAR BIN_OP VAR VAR VAR 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 IF VAR NUMBER ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR EXPR FUNC_CALL VAR VAR IF VAR NUMBER ASSIGN VAR FUNC_CALL FUNC_CALL VAR EXPR FUNC_CALL VAR FUNC_CALL VAR FUNC_CALL VAR VAR NUMBER FUNC_CALL VAR VAR NUMBER ASSIGN VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR LIST ASSIGN VAR LIST ASSIGN VAR LIST FOR VAR VAR EXPR FUNC_CALL VAR FUNC_CALL VAR VAR IF BIN_OP VAR NUMBER 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 FOR VAR FUNC_CALL VAR NUMBER BIN_OP VAR NUMBER NUMBER EXPR FUNC_CALL VAR BIN_OP VAR VAR VAR BIN_OP VAR NUMBER IF BIN_OP VAR NUMBER 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 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 FUNC_CALL VAR VAR FUNC_CALL VAR VAR IF VAR NUMBER ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR NUMBER VAR 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 | t = int(input())
for you in range(t):
n = int(input())
l = input().split()
li = [int(i) for i in l]
if n == 1:
print(li[0])
continue
dp = [(0) for i in range(n)]
dp[0] = 0
dp[1] = li[1] - li[0]
for i in range(2, n):
if i % 2 == 0:
dp[i] = max(dp[i - 2] - li[i] + li[i - 1], -li[i] + li[i - 1])
else:
dp[i] = max(dp[i - 2] + li[i] - li[i - 1], li[i] - li[i - 1])
z = max(dp)
sumi = 0
for i in range(n):
if i % 2 == 0:
sumi += li[i]
print(sumi + max(0, 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 IF VAR NUMBER EXPR FUNC_CALL VAR VAR NUMBER ASSIGN VAR NUMBER VAR FUNC_CALL VAR VAR ASSIGN VAR NUMBER NUMBER ASSIGN 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 BIN_OP BIN_OP VAR BIN_OP VAR NUMBER VAR VAR VAR BIN_OP VAR NUMBER BIN_OP VAR VAR VAR BIN_OP VAR NUMBER ASSIGN VAR VAR FUNC_CALL VAR BIN_OP BIN_OP VAR BIN_OP VAR NUMBER VAR VAR VAR BIN_OP VAR NUMBER BIN_OP VAR VAR VAR BIN_OP VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR 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 | def kadane(lis):
if len(lis) == 0:
return 0
max_curr = lis[0]
max_global = lis[0]
for i in range(1, len(lis)):
max_curr = max(max_curr + lis[i], lis[i])
max_global = max(max_global, max_curr)
return max_global
t = int(input())
while t > 0:
t -= 1
n = int(input())
l = list(map(int, input().split()))
i = 0
even = []
odd = []
while i < n - 1:
even.append(l[i + 1] - l[i])
i += 2
i = 1
while i < n - 1:
odd.append(l[i] - l[i + 1])
i += 2
a = kadane(even)
b = kadane(odd)
sum_even = 0
for i in range(n):
if i % 2 == 0:
sum_even += l[i]
print(sum_even + max(0, a, b)) | FUNC_DEF IF FUNC_CALL VAR VAR NUMBER RETURN NUMBER ASSIGN VAR VAR NUMBER ASSIGN VAR VAR NUMBER FOR VAR FUNC_CALL VAR NUMBER 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 ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR WHILE VAR NUMBER 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 ASSIGN VAR LIST ASSIGN VAR LIST WHILE VAR BIN_OP VAR NUMBER EXPR FUNC_CALL VAR BIN_OP VAR BIN_OP VAR NUMBER VAR VAR VAR NUMBER ASSIGN VAR NUMBER WHILE VAR BIN_OP VAR NUMBER EXPR FUNC_CALL VAR BIN_OP VAR VAR VAR BIN_OP VAR NUMBER VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR VAR IF BIN_OP VAR NUMBER NUMBER VAR VAR VAR 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 | def f(L):
n = len(L)
if n == 1:
return L[0]
if n == 2:
return max(L[0], L[1])
S = []
P = []
bs = 0
for i in range(0, n, 2):
bs += L[i]
s = 0
for i in range(1, n, 2):
s += L[i] - L[i - 1]
S.append(s)
m = 0
for i in range(2, n, 2):
m += L[i - 1] - L[i]
P.append(m)
maxm1 = [S[0]]
maxm2 = [P[0]]
k = len(S)
m = len(P)
for i in range(1, k):
maxm1.append(max(S[i] - S[i - 1], S[i] - S[i - 1] + maxm1[-1]))
for i in range(1, m):
maxm2.append(max(P[i] - P[i - 1], P[i] - P[i - 1] + maxm2[-1]))
c = max(max(maxm1), max(maxm2))
if c >= 0:
return bs + c
return bs
n = int(input())
for i in range(n):
x = input()
L = list(map(int, input().split()))
print(f(L)) | FUNC_DEF ASSIGN VAR FUNC_CALL VAR VAR IF VAR NUMBER RETURN VAR NUMBER IF VAR NUMBER RETURN FUNC_CALL VAR VAR NUMBER VAR NUMBER ASSIGN VAR LIST ASSIGN VAR LIST ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR NUMBER VAR NUMBER VAR VAR VAR ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR NUMBER VAR NUMBER VAR BIN_OP VAR VAR VAR BIN_OP VAR NUMBER EXPR FUNC_CALL VAR VAR ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR NUMBER VAR NUMBER VAR BIN_OP VAR BIN_OP VAR NUMBER VAR VAR EXPR FUNC_CALL VAR VAR ASSIGN VAR LIST VAR NUMBER ASSIGN VAR LIST VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR VAR FOR VAR FUNC_CALL VAR NUMBER VAR EXPR FUNC_CALL 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 NUMBER FOR VAR FUNC_CALL VAR NUMBER VAR EXPR FUNC_CALL 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 NUMBER ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL VAR VAR IF VAR NUMBER RETURN BIN_OP VAR VAR RETURN 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 FUNC_CALL 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 | 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
return best_sum, best_start, best_end
for testcase in range(int(input())):
n = int(input())
nums = list(map(int, input().split()))
if len(nums) == 1:
print(nums[0])
continue
elif len(nums) == 2:
print(max(nums))
continue
s = sum(nums[::2])
diffs = [(b - a) for a, b in zip(nums[::2], nums[1::2])]
best_sum, best_start, best_end = max_subarray(diffs)
diffs2 = [(a - b) for a, b in zip(nums[1::2], nums[2::2])]
best_sum2, best_start, best_end = max_subarray(diffs2)
best_sum = max(best_sum, best_sum2)
print(best_sum + s) | 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 VAR RETURN 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 FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR IF FUNC_CALL VAR VAR NUMBER EXPR FUNC_CALL VAR VAR NUMBER IF FUNC_CALL VAR VAR NUMBER EXPR FUNC_CALL VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR VAR NUMBER ASSIGN VAR BIN_OP VAR VAR VAR VAR FUNC_CALL VAR VAR NUMBER VAR NUMBER NUMBER ASSIGN VAR VAR VAR FUNC_CALL VAR VAR ASSIGN VAR BIN_OP VAR VAR VAR VAR FUNC_CALL VAR VAR NUMBER NUMBER VAR NUMBER NUMBER ASSIGN VAR 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 | for _ in range(int(input())):
n = int(input())
ar = list(map(int, input().split()))
diff = []
s = sum(ar[::2])
for i in range(1, n):
if i % 2:
diff.append(ar[i] - ar[i - 1])
else:
diff.append(ar[i - 1] - ar[i])
a, b = diff[::2], diff[1::2]
s1, m1 = 0, 0
for i in a:
s1 = max(s1 + i, 0)
m1 = max(m1, s1)
s2, m2 = 0, 0
for i in b:
s2 = max(s2 + i, 0)
m2 = max(m2, s2)
su = max(m1, m2, 0)
print(su + s) | 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 FUNC_CALL VAR VAR NUMBER FOR VAR FUNC_CALL VAR NUMBER VAR 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 ASSIGN VAR VAR VAR NUMBER VAR NUMBER NUMBER ASSIGN VAR VAR NUMBER NUMBER FOR VAR VAR ASSIGN VAR FUNC_CALL VAR BIN_OP VAR VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR VAR ASSIGN VAR VAR NUMBER NUMBER FOR VAR VAR ASSIGN VAR FUNC_CALL VAR BIN_OP VAR VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR VAR ASSIGN VAR FUNC_CALL 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 | from sys import stdin, stdout
input = stdin.readline
print = stdout.write
for _ in range(int(input())):
n = int(input())
a = list(map(int, input().split()))
s, e = None, None
maxi = 0
curr = 0
si = None
for i in range(0, n - 1, 2):
if curr == 0:
si = i
curr += a[i + 1] - a[i]
if curr <= 0:
si = None
curr = 0
elif curr > maxi:
maxi = curr
s, e = si, i + 1
curr = 0
si = None
for i in range(1, n - 1, 2):
if curr == 0:
si = i
curr += a[i] - a[i + 1]
if curr <= 0:
si, ei = None, None
curr = 0
elif curr > maxi:
maxi = curr
s, e = si, i + 1
ans = 0
for i in range(n):
if i % 2 == 0:
ans += a[i]
if s != None and e != None:
a[s : e + 1] = a[s : e + 1][::-1]
ans1 = 0
for i in range(n):
if i % 2 == 0:
ans1 += a[i]
ans = max(ans, ans1)
print(str(ans) + "\n") | ASSIGN VAR VAR 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 VAR NONE NONE ASSIGN VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR NONE FOR VAR FUNC_CALL VAR NUMBER BIN_OP VAR NUMBER NUMBER IF VAR NUMBER ASSIGN VAR VAR VAR BIN_OP VAR BIN_OP VAR NUMBER VAR VAR IF VAR NUMBER ASSIGN VAR NONE ASSIGN VAR NUMBER IF VAR VAR ASSIGN VAR VAR ASSIGN VAR VAR VAR BIN_OP VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR NONE FOR VAR FUNC_CALL VAR NUMBER BIN_OP VAR NUMBER NUMBER IF VAR NUMBER ASSIGN VAR VAR VAR BIN_OP VAR VAR VAR BIN_OP VAR NUMBER IF VAR NUMBER ASSIGN VAR VAR NONE NONE ASSIGN VAR NUMBER IF VAR VAR ASSIGN VAR VAR ASSIGN VAR VAR VAR BIN_OP VAR NUMBER ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR VAR IF BIN_OP VAR NUMBER NUMBER VAR VAR VAR IF VAR NONE VAR NONE ASSIGN VAR VAR BIN_OP VAR NUMBER VAR VAR BIN_OP VAR NUMBER NUMBER ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR VAR IF BIN_OP VAR NUMBER NUMBER VAR VAR VAR ASSIGN VAR FUNC_CALL 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 | for _ in range(int(input())):
n = int(input())
s = list(map(int, input().split()))
e = s[::2]
o = s[1::2]
ans = cs = 0
for i in range(len(o)):
cs = max(0, cs + o[i] - e[i])
ans = max(cs, ans)
cs = 0
for i in range(len(e) - 1):
cs = max(0, cs + o[i] - e[i + 1])
ans = max(cs, ans)
print(sum(e) + 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 VAR NUMBER ASSIGN VAR VAR NUMBER NUMBER ASSIGN VAR VAR NUMBER FOR VAR FUNC_CALL VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR NUMBER BIN_OP BIN_OP VAR VAR VAR VAR VAR ASSIGN VAR FUNC_CALL VAR VAR VAR 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 VAR VAR BIN_OP VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR VAR EXPR FUNC_CALL VAR BIN_OP 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 maximum_sum_subarray(arr):
length = len(arr)
temp = [0] * length
if length == 0:
return 0
if length == 1:
return max(arr[0], 0)
temp[0] = max(arr[0], 0)
_max = temp[0]
for i in range(1, len(arr)):
temp[i] = max(0, temp[i - 1] + arr[i])
if temp[i] > _max:
_max = temp[i]
return _max
t = int(input())
for case in range(t):
n = int(input())
arr = list(map(int, input().split(" ")))
arr1 = []
arr2 = []
_sum = arr[0]
for i in range(1, len(arr)):
if i % 2 == 1:
arr1.append(arr[i] - arr[i - 1])
else:
arr2.append(arr[i - 1] - arr[i])
_sum += arr[i]
_max1 = maximum_sum_subarray(arr1)
_max2 = maximum_sum_subarray(arr2)
print(_sum + max(_max1, _max2)) | FUNC_DEF ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR BIN_OP LIST NUMBER VAR IF VAR NUMBER RETURN NUMBER IF VAR NUMBER RETURN FUNC_CALL VAR VAR NUMBER NUMBER ASSIGN VAR NUMBER FUNC_CALL VAR VAR NUMBER NUMBER ASSIGN VAR VAR NUMBER FOR VAR FUNC_CALL VAR NUMBER FUNC_CALL VAR VAR ASSIGN VAR VAR FUNC_CALL VAR NUMBER BIN_OP VAR BIN_OP VAR NUMBER VAR VAR IF VAR VAR VAR ASSIGN 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 FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR STRING ASSIGN VAR LIST ASSIGN VAR LIST ASSIGN VAR VAR NUMBER FOR VAR FUNC_CALL VAR NUMBER FUNC_CALL VAR 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 VAR VAR VAR 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 | from sys import stdin, stdout
for query in range(int(stdin.readline())):
n = int(stdin.readline())
a = [int(x) for x in stdin.readline().split()]
if n == 1:
stdout.write(str(a[0]) + "\n")
else:
prefix = [0] * n
count = 0
for x in range(n):
if x % 2 == 0:
count -= a[x]
else:
count += a[x]
prefix[x] = count
prefix.insert(0, 0)
minevens = [0] * (n + 1)
minimum = a[0]
for x in range(0, n + 1, 2):
minimum = min(minimum, prefix[x])
minevens[x] = minimum
minodds = [0] * (n + 1)
minimum = a[1]
for x in range(1, n + 1, 2):
minimum = min(minimum, prefix[x])
minodds[x] = minimum
maximum = 0
for x in range(1, n + 1, 2):
maximum = max(maximum, prefix[x] - minodds[x])
count = 0
for x in range(0, n + 1, 2):
maximum = max(maximum, prefix[x] - minevens[x])
if x <= n - 1:
count += a[x]
stdout.write(str(count + maximum) + "\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 VAR FUNC_CALL FUNC_CALL VAR IF VAR NUMBER EXPR FUNC_CALL VAR BIN_OP FUNC_CALL VAR VAR NUMBER STRING ASSIGN VAR BIN_OP LIST NUMBER VAR ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR VAR IF BIN_OP VAR NUMBER NUMBER VAR VAR VAR VAR VAR VAR ASSIGN VAR VAR VAR EXPR FUNC_CALL VAR NUMBER NUMBER ASSIGN VAR BIN_OP LIST NUMBER BIN_OP VAR NUMBER ASSIGN VAR VAR NUMBER FOR VAR FUNC_CALL VAR NUMBER BIN_OP VAR NUMBER NUMBER ASSIGN VAR FUNC_CALL VAR VAR VAR VAR ASSIGN VAR VAR VAR ASSIGN VAR BIN_OP LIST NUMBER BIN_OP VAR NUMBER ASSIGN VAR VAR NUMBER FOR VAR FUNC_CALL VAR NUMBER BIN_OP VAR NUMBER NUMBER ASSIGN VAR FUNC_CALL VAR VAR VAR VAR ASSIGN 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 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 VAR VAR VAR VAR IF VAR BIN_OP VAR NUMBER VAR VAR VAR EXPR FUNC_CALL VAR BIN_OP FUNC_CALL VAR BIN_OP 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 | for _ in range(int(input())):
n = int(input())
a = list(map(int, input().split()))
ev = 0
od = 0
b = []
c = []
for i in range(n):
if i % 2 == 0:
ev += a[i]
else:
od += a[i]
for i in range(0, n - 1, 2):
b.append(a[i + 1] - a[i])
for i in range(1, n - 1, 2):
c.append(a[i] - a[i + 1])
ma = 0
cur = 0
for i in range(len(b)):
cur += b[i]
if cur > ma:
ma = cur
if cur < 0:
cur = 0
ma2 = 0
cur = 0
for i in range(len(c)):
cur += c[i]
if cur > ma2:
ma2 = cur
if cur < 0:
cur = 0
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 NUMBER ASSIGN VAR LIST ASSIGN VAR LIST FOR VAR FUNC_CALL VAR VAR IF BIN_OP VAR NUMBER NUMBER VAR VAR VAR VAR 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 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 VAR ASSIGN VAR VAR IF 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 VAR ASSIGN VAR VAR IF VAR NUMBER ASSIGN VAR NUMBER 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())
while t > 0:
t -= 1
n = int(input())
li = list(map(int, input().split()))
if n > 2:
a = []
b = []
k = int(n / 2)
x = k
y = k
i = 0
j = 1
while x > 0:
z = li[i + 1] - li[i]
a.append(z)
i += 2
x -= 1
if n % 2 == 0:
while y > 1:
z = li[j] - li[j + 1]
b.append(z)
j += 2
y -= 1
else:
while y > 0:
z = li[j] - li[j + 1]
b.append(z)
j += 2
y -= 1
for i in range(1, len(a)):
a[i] = max(a[i], a[i - 1] + a[i])
for i in range(1, len(b)):
b[i] = max(b[i], b[i - 1] + b[i])
sum = 0
for i in range(len(li)):
if i % 2 == 0:
sum += li[i]
q = max(a)
r = max(b)
w = max(q, r)
if w > 0:
sum = sum + w
print(sum)
elif n == 1:
print(li[0])
elif n == 2:
print(max(li[0], li[1])) | ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR WHILE VAR NUMBER VAR NUMBER 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 ASSIGN VAR LIST ASSIGN VAR LIST ASSIGN VAR FUNC_CALL VAR BIN_OP VAR NUMBER ASSIGN VAR VAR ASSIGN VAR VAR ASSIGN VAR NUMBER ASSIGN VAR NUMBER WHILE VAR NUMBER ASSIGN VAR BIN_OP VAR BIN_OP VAR NUMBER VAR VAR EXPR FUNC_CALL VAR VAR VAR NUMBER VAR NUMBER IF BIN_OP VAR NUMBER NUMBER WHILE VAR NUMBER ASSIGN VAR BIN_OP VAR VAR VAR BIN_OP VAR NUMBER EXPR FUNC_CALL VAR VAR VAR NUMBER VAR NUMBER WHILE VAR NUMBER ASSIGN VAR BIN_OP VAR VAR VAR BIN_OP VAR NUMBER EXPR FUNC_CALL VAR VAR VAR NUMBER VAR NUMBER FOR VAR FUNC_CALL VAR NUMBER FUNC_CALL VAR VAR ASSIGN VAR VAR FUNC_CALL VAR VAR 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 VAR VAR BIN_OP VAR BIN_OP VAR NUMBER VAR VAR ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR FUNC_CALL VAR VAR IF BIN_OP VAR NUMBER NUMBER VAR VAR VAR 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 EXPR FUNC_CALL VAR VAR IF VAR NUMBER EXPR FUNC_CALL VAR VAR NUMBER IF VAR NUMBER EXPR FUNC_CALL VAR FUNC_CALL 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 kadane(arr):
mx, cur = 0, 0
for i in arr:
cur += i
if mx < cur:
mx = cur
if cur < 0:
cur = 0
return mx
for i in range(int(input())):
n = int(input())
arr = [int(x) for x in input().split()]
even = 0
for j in range(0, n, 2):
even += arr[j]
diff1, diff2 = [], []
for k in range(1, n, 2):
diff1.append(arr[k] - arr[k - 1])
if k + 1 < n:
diff2.append(arr[k] - arr[k + 1])
ans = even + max(0, kadane(diff1), kadane(diff2))
print(ans) | FUNC_DEF ASSIGN VAR VAR NUMBER NUMBER FOR VAR 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 VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR NUMBER VAR NUMBER VAR VAR VAR ASSIGN VAR VAR LIST LIST 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 BIN_OP VAR FUNC_CALL VAR NUMBER FUNC_CALL VAR VAR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR VAR |
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