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feat: add solutions to lc problem: No.2507
No.2507.Smallest Value After Replacing With Sum of Prime Factors
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‎solution/2500-2599/2507.Smallest Value After Replacing With Sum of Prime Factors/README.md

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<!-- 这里可写通用的实现逻辑 -->
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**方法一：暴力模拟**
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根据题意，我们可以得到一个质因数分解的过程，即将一个数不断地分解为质因数，分解不能分解。过程中将每次分解的质因数相加，递归或者迭代地进行即可。
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时间复杂度 $O(\sqrt{n})$。
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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 smallestValue(self, n: int) -> int:
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while 1:
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t, s, i = n, 0, 2
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while i <= n // i:
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while n % i == 0:
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n //= i
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s += i
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i += 1
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if n > 1:
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s += n
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if s == t:
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return t
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n = s
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```
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### **Java**
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<!-- 这里可写当前语言的特殊实现逻辑 -->
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```java
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class Solution {
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public int smallestValue(int n) {
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while (true) {
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int t = n, s = 0;
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for (int i = 2; i <= n / i; ++i) {
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while (n % i == 0) {
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s += i;
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n /= i;
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}
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}
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if (n > 1) {
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s += n;
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}
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if (s == t) {
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return s;
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}
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n = s;
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}
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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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int smallestValue(int n) {
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while (1) {
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int t = n, s = 0;
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for (int i = 2; i <= n / i; ++i) {
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while (n % i == 0) {
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s += i;
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n /= i;
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}
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}
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if (n > 1) s += n;
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if (s == t) return s;
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n = s;
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}
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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 smallestValue(n int) int {
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for {
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t, s := n, 0
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for i := 2; i <= n/i; i++ {
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for n%i == 0 {
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s += i
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n /= i
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}
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}
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if n > 1 {
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s += n
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}
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if s == t {
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return s
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}
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n = s
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}
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}
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```
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### **...**

‎solution/2500-2599/2507.Smallest Value After Replacing With Sum of Prime Factors/README_EN.md

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### **Python3**
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```python
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class Solution:
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def smallestValue(self, n: int) -> int:
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while 1:
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t, s, i = n, 0, 2
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while i <= n // i:
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while n % i == 0:
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n //= i
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s += i
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i += 1
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if n > 1:
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s += n
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if s == t:
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return t
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n = s
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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 int smallestValue(int n) {
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while (true) {
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int t = n, s = 0;
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for (int i = 2; i <= n / i; ++i) {
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while (n % i == 0) {
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s += i;
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n /= i;
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}
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}
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if (n > 1) {
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s += n;
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}
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if (s == t) {
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return s;
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}
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n = s;
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}
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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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int smallestValue(int n) {
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while (1) {
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int t = n, s = 0;
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for (int i = 2; i <= n / i; ++i) {
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while (n % i == 0) {
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s += i;
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n /= i;
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}
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}
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if (n > 1) s += n;
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if (s == t) return s;
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n = s;
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}
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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 smallestValue(n int) int {
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for {
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t, s := n, 0
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for i := 2; i <= n/i; i++ {
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for n%i == 0 {
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s += i
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n /= i
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}
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}
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if n > 1 {
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s += n
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}
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if s == t {
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return s
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}
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n = s
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}
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}
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```
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### **...**

‎solution/2500-2599/2507.Smallest Value After Replacing With Sum of Prime Factors/Solution.cpp

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class Solution {
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public:
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int smallestValue(int n) {
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while (1) {
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int t = n, s = 0;
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for (int i = 2; i <= n / i; ++i) {
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while (n % i == 0) {
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s += i;
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n /= i;
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}
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}
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if (n > 1) s += n;
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if (s == t) return s;
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n = s;
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}
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}
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};

‎solution/2500-2599/2507.Smallest Value After Replacing With Sum of Prime Factors/Solution.go

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func smallestValue(n int) int {
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for {
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t, s := n, 0
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for i := 2; i <= n/i; i++ {
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for n%i == 0 {
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s += i
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n /= i
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}
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}
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if n > 1 {
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s += n
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}
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if s == t {
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return s
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}
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n = s
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}
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}

‎solution/2500-2599/2507.Smallest Value After Replacing With Sum of Prime Factors/Solution.java

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@@ -0,0 +1,20 @@
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class Solution {
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public int smallestValue(int n) {
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while (true) {
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int t = n, s = 0;
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for (int i = 2; i <= n / i; ++i) {
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while (n % i == 0) {
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s += i;
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n /= i;
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}
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}
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if (n > 1) {
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s += n;
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}
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if (s == t) {
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return s;
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}
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n = s;
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}
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}
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}

‎solution/2500-2599/2507.Smallest Value After Replacing With Sum of Prime Factors/Solution.py

Lines changed: 14 additions & 0 deletions
Original file line numberDiff line numberDiff line change
@@ -0,0 +1,14 @@
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class Solution:
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def smallestValue(self, n: int) -> int:
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while 1:
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t, s, i = n, 0, 2
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while i <= n // i:
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while n % i == 0:
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n //= i
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s += i
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i += 1
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if n > 1:
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s += n
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if s == t:
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return t
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n = s

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