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Commit 534741b

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rain84yanglbme
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feat: add solutions to lc problem: No.0633 (doocs#3118)
* feat: add 2nd ts solution to lc problem: No.0633 * feat: add 3rd ts solution to lc problem: No.0633 * Update README.md * style: format code and docs with prettier * Update README_EN.md * Update Solution2.ts * Create Solution2.py * Create Solution2.java * Create Solution2.cpp * Create Solution2.go * style: format code and docs with prettier --------- Co-authored-by: Libin YANG <contact@yanglibin.info>
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‎solution/0600-0699/0633.Sum of Square Numbers/README.md‎

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<!-- solution:end -->
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<!-- solution:start -->
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### 方法二:数学
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这个问题实际上是关于一个数能否表示为两个平方数之和的条件。这个定理可以追溯到费马(Fermat)和欧拉(Euler),它在数论中是一个经典结果。
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具体来说,这个定理可以表述为:
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**一个正整数 \( n \) 能表示为两个平方数之和的充要条件是:\( n \) 的所有形如 \( 4k + 3 \) 的素数因子的幂次均为偶数。**
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这意味着,如果我们将 $n$ 分解成素数因子乘积的形式,即 $n = p_1^{e_1} p_2^{e_2} \cdots p_k^{e_k},ドル其中 $p_i$ 是素数且 $e_i$ 是它们对应的幂次,那么 $n$ 可以表示为两个平方数之和,当且仅当所有 4ドルk + 3$ 形式的素数因子 $p_i$ 的幂次 $e_i$ 都是偶数。
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更正式地,假设 $p_i$ 是形如 4ドルk + 3$ 的素数,则对于每一个这样的 $p_i,ドル要求 $e_i$ 是偶数。
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例如:
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- 数字 13ドル$ 是素数,且 13ドル \equiv 1 \pmod{4},ドル因此它可以表示为两个平方数之和,即 13ドル = 2^2 + 3^2$。
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- 数字 21ドル$ 分解为 3ドル \times 7,ドル其中 3ドル$ 和 7ドル$ 都是形如 4ドルk + 3$ 的素数因子,并且它们的幂次都是 1ドル$(奇数),因此 21ドル$ 不能表示为两个平方数之和。
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总结起来,这个定理在数论中非常重要,用于判断一个数是否可以表示为两个平方数之和。
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时间复杂度 $O(\sqrt{c}),ドル其中 $c$ 是给定的非负整数。空间复杂度 $O(1)$。
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<!-- tabs:start -->
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#### Python3
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```python
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class Solution:
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def judgeSquareSum(self, c: int) -> bool:
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for i in range(2, int(sqrt(c)) + 1):
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if c % i == 0:
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exp = 0
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while c % i == 0:
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c //= i
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exp += 1
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if i % 4 == 3 and exp % 2 != 0:
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return False
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return c % 4 != 3
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```
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#### Java
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```java
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class Solution {
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public boolean judgeSquareSum(int c) {
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int n = (int) Math.sqrt(c);
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for (int i = 2; i <= n; ++i) {
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if (c % i == 0) {
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int exp = 0;
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while (c % i == 0) {
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c /= i;
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++exp;
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}
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if (i % 4 == 3 && exp % 2 != 0) {
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return false;
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}
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}
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}
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return c % 4 != 3;
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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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bool judgeSquareSum(int c) {
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int n = sqrt(c);
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for (int i = 2; i <= n; ++i) {
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if (c % i == 0) {
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int exp = 0;
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while (c % i == 0) {
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c /= i;
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++exp;
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}
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if (i % 4 == 3 && exp % 2 != 0) {
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return false;
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}
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}
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}
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return c % 4 != 3;
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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 judgeSquareSum(c int) bool {
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n := int(math.Sqrt(float64(c)))
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for i := 2; i <= n; i++ {
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if c%i == 0 {
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exp := 0
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for c%i == 0 {
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c /= i
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exp++
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}
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if i%4 == 3 && exp%2 != 0 {
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return false
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}
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}
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}
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return c%4 != 3
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}
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```
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#### TypeScript
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```ts
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function judgeSquareSum(c: number): boolean {
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const n = Math.floor(Math.sqrt(c));
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for (let i = 2; i <= n; ++i) {
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if (c % i === 0) {
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let exp = 0;
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while (c % i === 0) {
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c /= i;
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++exp;
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}
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if (i % 4 === 3 && exp % 2 !== 0) {
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return false;
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}
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}
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}
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return c % 4 !== 3;
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}
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```
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<!-- tabs:end -->
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<!-- solution:end -->
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<!-- problem:end -->

‎solution/0600-0699/0633.Sum of Square Numbers/README_EN.md‎

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<!-- solution:end -->
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<!-- solution:start -->
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### Solution 2: Mathematics
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This problem is essentially about the conditions under which a number can be expressed as the sum of two squares. This theorem dates back to Fermat and Euler and is a classic result in number theory.
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Specifically, the theorem can be stated as follows:
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> A positive integer $n$ can be expressed as the sum of two squares if and only if all prime factors of $n$ of the form 4ドルk + 3$ have even powers.
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This means that if we decompose $n$ into the product of its prime factors, $n = p_1^{e_1} p_2^{e_2} \cdots p_k^{e_k},ドル where $p_i$ are primes and $e_i$ are their corresponding exponents, then $n$ can be expressed as the sum of two squares if and only if all prime factors $p_i$ of the form 4ドルk + 3$ have even exponents $e_i$.
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More formally, if $p_i$ is a prime of the form 4ドルk + 3,ドル then for each such $p_i,ドル $e_i$ must be even.
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For example:
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- The number 13ドル$ is a prime and 13ドル \equiv 1 \pmod{4},ドル so it can be expressed as the sum of two squares, i.e., 13ドル = 2^2 + 3^2$.
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- The number 21ドル$ can be decomposed into 3ドル \times 7,ドル where both 3ドル$ and 7ドル$ are prime factors of the form 4ドルk + 3$ and their exponents are 1ドル$ (odd), so 21ドル$ cannot be expressed as the sum of two squares.
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In summary, this theorem is very important in number theory for determining whether a number can be expressed as the sum of two squares.
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The time complexity is $O(\sqrt{c}),ドル where $c$ is the given non-negative integer. The space complexity is $O(1)$.
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<!-- tabs:start -->
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#### Python3
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```python
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class Solution:
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def judgeSquareSum(self, c: int) -> bool:
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for i in range(2, int(sqrt(c)) + 1):
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if c % i == 0:
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exp = 0
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while c % i == 0:
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c //= i
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exp += 1
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if i % 4 == 3 and exp % 2 != 0:
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return False
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return c % 4 != 3
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```
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#### Java
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```java
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class Solution {
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public boolean judgeSquareSum(int c) {
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int n = (int) Math.sqrt(c);
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for (int i = 2; i <= n; ++i) {
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if (c % i == 0) {
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int exp = 0;
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while (c % i == 0) {
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c /= i;
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++exp;
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}
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if (i % 4 == 3 && exp % 2 != 0) {
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return false;
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}
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}
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}
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return c % 4 != 3;
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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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bool judgeSquareSum(int c) {
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int n = sqrt(c);
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for (int i = 2; i <= n; ++i) {
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if (c % i == 0) {
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int exp = 0;
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while (c % i == 0) {
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c /= i;
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++exp;
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}
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if (i % 4 == 3 && exp % 2 != 0) {
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return false;
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}
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}
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}
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return c % 4 != 3;
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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 judgeSquareSum(c int) bool {
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n := int(math.Sqrt(float64(c)))
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for i := 2; i <= n; i++ {
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if c%i == 0 {
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exp := 0
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for c%i == 0 {
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c /= i
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exp++
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}
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if i%4 == 3 && exp%2 != 0 {
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return false
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}
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}
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}
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return c%4 != 3
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}
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```
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#### TypeScript
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```ts
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function judgeSquareSum(c: number): boolean {
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const n = Math.floor(Math.sqrt(c));
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for (let i = 2; i <= n; ++i) {
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if (c % i === 0) {
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let exp = 0;
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while (c % i === 0) {
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c /= i;
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++exp;
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}
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if (i % 4 === 3 && exp % 2 !== 0) {
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return false;
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}
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}
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}
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return c % 4 !== 3;
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}
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```
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<!-- tabs:end -->
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<!-- solution:end -->
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<!-- problem:end -->
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class Solution {
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public:
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bool judgeSquareSum(int c) {
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int n = sqrt(c);
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for (int i = 2; i <= n; ++i) {
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if (c % i == 0) {
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int exp = 0;
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while (c % i == 0) {
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c /= i;
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++exp;
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}
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if (i % 4 == 3 && exp % 2 != 0) {
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return false;
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}
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}
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}
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return c % 4 != 3;
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}
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};
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func judgeSquareSum(c int) bool {
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n := int(math.Sqrt(float64(c)))
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for i := 2; i <= n; i++ {
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if c%i == 0 {
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exp := 0
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for c%i == 0 {
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c /= i
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exp++
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}
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if i%4 == 3 && exp%2 != 0 {
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return false
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}
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}
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}
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return c%4 != 3
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}
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class Solution {
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public boolean judgeSquareSum(int c) {
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int n = (int) Math.sqrt(c);
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for (int i = 2; i <= n; ++i) {
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if (c % i == 0) {
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int exp = 0;
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while (c % i == 0) {
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c /= i;
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++exp;
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}
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if (i % 4 == 3 && exp % 2 != 0) {
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return false;
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}
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}
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}
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return c % 4 != 3;
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}
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}
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class Solution:
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def judgeSquareSum(self, c: int) -> bool:
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for i in range(2, int(sqrt(c)) + 1):
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if c % i == 0:
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exp = 0
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while c % i == 0:
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c //= i
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exp += 1
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if i % 4 == 3 and exp % 2 != 0:
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return False
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return c % 4 != 3
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function judgeSquareSum(c: number): boolean {
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const n = Math.floor(Math.sqrt(c));
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for (let i = 2; i <= n; ++i) {
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if (c % i === 0) {
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let exp = 0;
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while (c % i === 0) {
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c /= i;
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++exp;
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}
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if (i % 4 === 3 && exp % 2 !== 0) {
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return false;
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}
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}
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}
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return c % 4 !== 3;
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}

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