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#include <bits/stdc++.h>
using namespace std;
int gcd(int a, int b) {
while (b != 0) {
int temp = b;
b = a % b;
a = temp;
}
return a;
}
vector<int> gcdArrays(vector<int>& arr) {
int n = arr.size();
int max_gcd = 0;
int max_length = 0;
for (int i = 0; i < n - 1; ++i) {
int current_gcd = arr[i];
for (int j = i + 1; j < n; ++j) {
current_gcd = gcd(current_gcd, arr[j]);
if (current_gcd > max_gcd) {
max_gcd = current_gcd;
max_length = j - i + 1;
} else if (current_gcd == max_gcd && (j - i + 1) > max_length) {
max_length = j - i + 1;
}
}
}
return {max_gcd, max_length};
}
DE Shaw
Telegram:- @allcoding1
using namespace std;
int gcd(int a, int b) {
while (b != 0) {
int temp = b;
b = a % b;
a = temp;
}
return a;
}
vector<int> gcdArrays(vector<int>& arr) {
int n = arr.size();
int max_gcd = 0;
int max_length = 0;
for (int i = 0; i < n - 1; ++i) {
int current_gcd = arr[i];
for (int j = i + 1; j < n; ++j) {
current_gcd = gcd(current_gcd, arr[j]);
if (current_gcd > max_gcd) {
max_gcd = current_gcd;
max_length = j - i + 1;
} else if (current_gcd == max_gcd && (j - i + 1) > max_length) {
max_length = j - i + 1;
}
}
}
return {max_gcd, max_length};
}
DE Shaw
Telegram:- @allcoding1
👍6👎3
def max_benefit(n,A, K):
n = len(A)
dp = [0] * (n + 1)
prefix_sum = [0] * (n + 1)
for i in range(1, n + 1):
prefix_sum[i] = prefix_sum[i - 1] + A[i - 1]
for i in range(1, n + 1):
min_val = float('inf')
max_val = float('-inf')
for j in range(1, K + 1):
if i - j >= 0:
min_val = min(min_val, prefix_sum[i] - prefix_sum[i - j])
max_val = max(max_val, prefix_sum[i] - prefix_sum[i - j])
dp[i] = max(dp[i], dp[i - j] + max(max_val, -min_val))
return dp[n] % MOD
Alternate Segment
Infosys
Telegram:- @allcoding1
n = len(A)
dp = [0] * (n + 1)
prefix_sum = [0] * (n + 1)
for i in range(1, n + 1):
prefix_sum[i] = prefix_sum[i - 1] + A[i - 1]
for i in range(1, n + 1):
min_val = float('inf')
max_val = float('-inf')
for j in range(1, K + 1):
if i - j >= 0:
min_val = min(min_val, prefix_sum[i] - prefix_sum[i - j])
max_val = max(max_val, prefix_sum[i] - prefix_sum[i - j])
dp[i] = max(dp[i], dp[i - j] + max(max_val, -min_val))
return dp[n] % MOD
Alternate Segment
Infosys
Telegram:- @allcoding1
👍4