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Commit a9b1719

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feat: add solutions to lc problem: No.2862 (#1645)
No.2862.Maximum Element-Sum of a Complete Subset of Indices
1 parent 17c2822 commit a9b1719

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7 files changed

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-18
lines changed

7 files changed

+226
-18
lines changed

‎solution/2800-2899/2862.Maximum Element-Sum of a Complete Subset of Indices/README.md‎

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@@ -56,14 +56,35 @@
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<!-- 这里可写通用的实现逻辑 -->
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**方法一:枚举**
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我们注意到,如果一个数字可以表示成 $k \times j^2$ 的形式,那么所有该形式的数字的 $k$ 是相同的。
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因此,我们可以在 $[1,..n]$ 范围内枚举 $k,ドル然后从 1ドル$ 开始枚举 $j,ドル每一次累加 $nums[k \times j^2 - 1]$ 的值到 $t$ 中,直到 $k \times j^2 > n$。此时更新答案为 $ans = \max(ans, t)$。
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最后返回答案 $ans$ 即可。
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时间复杂度 $O(n),ドル其中 $n$ 是数组的长度。空间复杂度 $O(1)$。
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<!-- tabs:start -->
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### **Python3**
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<!-- 这里可写当前语言的特殊实现逻辑 -->
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```python
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class Solution:
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def maximumSum(self, nums: List[int]) -> int:
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n = len(nums)
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ans = 0
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for k in range(1, n + 1):
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t = 0
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j = 1
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while k * j * j <= n:
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t += nums[k * j * j - 1]
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j += 1
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ans = max(ans, t)
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return ans
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```
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### **Java**
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return ans;
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}
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}
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```
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```java
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class Solution {
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public long maximumSum(List<Integer> nums) {
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long ans = 0;
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int n = nums.size();
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for (int k = 1; k <= n; ++k) {
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long t = 0;
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for (int j = 1; k * j * j <= n; ++j) {
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t += nums.get(k * j * j - 1);
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}
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ans = Math.max(ans, t);
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}
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return ans;
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}
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}
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```
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### **C++**
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```cpp
103-
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class Solution {
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public:
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long long maximumSum(vector<int>& nums) {
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long long ans = 0;
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int n = nums.size();
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for (int k = 1; k <= n; ++k) {
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long long t = 0;
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for (int j = 1; k * j * j <= n; ++j) {
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t += nums[k * j * j - 1];
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}
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ans = max(ans, t);
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}
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return ans;
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}
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};
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```
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### **Go**
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```go
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func maximumSum(nums []int) (ans int64) {
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n := len(nums)
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for k := 1; k <= n; k++ {
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var t int64
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for j := 1; k*j*j <= n; j++ {
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t += int64(nums[k*j*j-1])
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}
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ans = max(ans, t)
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}
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return
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}
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func max(a, b int64) int64 {
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if a > b {
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return a
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}
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return b
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}
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```
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### **TypeScript**
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```ts
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function maximumSum(nums: number[]): number {
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let ans = 0;
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const n = nums.length;
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for (let k = 1; k <= n; ++k) {
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let t = 0;
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for (let j = 1; k * j * j <= n; ++j) {
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t += nums[k * j * j - 1];
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}
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ans = Math.max(ans, t);
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}
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return ans;
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}
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```
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### **...**

‎solution/2800-2899/2862.Maximum Element-Sum of a Complete Subset of Indices/README_EN.md‎

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Original file line numberDiff line numberDiff line change
@@ -55,7 +55,18 @@ Hence, the maximum element-sum of a complete subset of indices is 19.
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### **Python3**
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```python
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class Solution:
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def maximumSum(self, nums: List[int]) -> int:
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n = len(nums)
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ans = 0
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for k in range(1, n + 1):
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t = 0
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j = 1
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while k * j * j <= n:
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t += nums[k * j * j - 1]
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j += 1
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ans = max(ans, t)
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return ans
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```
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### **Java**
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return ans;
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}
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}
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```
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```java
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class Solution {
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public long maximumSum(List<Integer> nums) {
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long ans = 0;
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int n = nums.size();
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for (int k = 1; k <= n; ++k) {
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long t = 0;
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for (int j = 1; k * j * j <= n; ++j) {
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t += nums.get(k * j * j - 1);
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}
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ans = Math.max(ans, t);
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}
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return ans;
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}
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}
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```
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### **C++**
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```cpp
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class Solution {
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public:
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long long maximumSum(vector<int>& nums) {
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long long ans = 0;
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int n = nums.size();
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for (int k = 1; k <= n; ++k) {
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long long t = 0;
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for (int j = 1; k * j * j <= n; ++j) {
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t += nums[k * j * j - 1];
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}
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ans = max(ans, t);
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}
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return ans;
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}
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};
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```
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### **Go**
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```go
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func maximumSum(nums []int) (ans int64) {
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n := len(nums)
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for k := 1; k <= n; k++ {
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var t int64
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for j := 1; k*j*j <= n; j++ {
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t += int64(nums[k*j*j-1])
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}
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ans = max(ans, t)
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}
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return
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}
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func max(a, b int64) int64 {
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if a > b {
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return a
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}
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return b
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}
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```
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### **TypeScript**
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```ts
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function maximumSum(nums: number[]): number {
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let ans = 0;
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const n = nums.length;
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for (let k = 1; k <= n; ++k) {
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let t = 0;
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for (let j = 1; k * j * j <= n; ++j) {
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t += nums[k * j * j - 1];
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}
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ans = Math.max(ans, t);
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}
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return ans;
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}
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```
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### **...**
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Original file line numberDiff line numberDiff line change
@@ -0,0 +1,15 @@
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class Solution {
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public:
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long long maximumSum(vector<int>& nums) {
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long long ans = 0;
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int n = nums.size();
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for (int k = 1; k <= n; ++k) {
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long long t = 0;
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for (int j = 1; k * j * j <= n; ++j) {
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t += nums[k * j * j - 1];
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}
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ans = max(ans, t);
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}
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return ans;
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}
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};
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Original file line numberDiff line numberDiff line change
@@ -0,0 +1,18 @@
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func maximumSum(nums []int) (ans int64) {
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n := len(nums)
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for k := 1; k <= n; k++ {
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var t int64
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for j := 1; k*j*j <= n; j++ {
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t += int64(nums[k*j*j-1])
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}
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ans = max(ans, t)
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}
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return
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}
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func max(a, b int64) int64 {
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if a > b {
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return a
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}
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return b
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}

‎solution/2800-2899/2862.Maximum Element-Sum of a Complete Subset of Indices/Solution.java‎

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Original file line numberDiff line numberDiff line change
@@ -2,21 +2,12 @@ class Solution {
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public long maximumSum(List<Integer> nums) {
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long ans = 0;
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int n = nums.size();
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boolean[] used = new boolean[n + 1];
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int bound = (int) Math.floor(Math.sqrt(n));
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int[] squares = new int[bound + 1];
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for (int i = 1; i <= bound + 1; i++) {
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squares[i - 1] = i * i;
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}
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for (int i = 1; i <= n; i++) {
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long res = 0;
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int idx = 0;
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int curr = i * squares[idx];
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while (curr <= n) {
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res += nums.get(curr - 1);
17-
curr = i * squares[++idx];
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for (int k = 1; k <= n; ++k) {
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long t = 0;
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for (int j = 1; k * j * j <= n; ++j) {
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t += nums.get(k * j * j - 1);
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}
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ans = Math.max(ans, res);
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ans = Math.max(ans, t);
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}
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return ans;
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}
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Original file line numberDiff line numberDiff line change
@@ -0,0 +1,12 @@
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class Solution:
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def maximumSum(self, nums: List[int]) -> int:
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n = len(nums)
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ans = 0
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for k in range(1, n + 1):
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t = 0
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j = 1
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while k * j * j <= n:
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t += nums[k * j * j - 1]
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j += 1
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ans = max(ans, t)
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return ans
Lines changed: 12 additions & 0 deletions
Original file line numberDiff line numberDiff line change
@@ -0,0 +1,12 @@
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function maximumSum(nums: number[]): number {
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let ans = 0;
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const n = nums.length;
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for (let k = 1; k <= n; ++k) {
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let t = 0;
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for (let j = 1; k * j * j <= n; ++j) {
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t += nums[k * j * j - 1];
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}
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ans = Math.max(ans, t);
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}
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return ans;
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}

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