feat: add solutions to lc problem: No.2652 (doocs#1827)

No.2652.Sum Multiples

Author: yanglbme

solution/2600-2699/2652.Sum Multiples/README.md

@@ -47,12 +47,24 @@
 
 **方法一：枚举**
 
-我们直接枚举 $[1,..n]$ 中的每一个数 $x$，如果 $x$ 能被 $3$、$5$、$7$ 整除，那么就将 $x$ 累加到答案中。
+我们直接枚举 $[1,..n]$ 中的每一个数 $x$，如果 $x$ 能被 $3$, $5$, $7$ 整除，那么就将 $x$ 累加到答案中。
 
 枚举结束后，返回答案即可。
 
 时间复杂度 $O(n)$，其中 $n$ 为题目给定的整数。空间复杂度 $O(1)$。
 
+**方法二：数学（容斥原理）**
+
+我们定义函数 $f(x)$ 表示 $[1,..n]$ 中能被 $x$ 整除的数之和，那么一共有 $m = \left\lfloor \frac{n}{x} \right\rfloor$ 个数能被 $x$ 整除，这些数字分别为 $x$, $2x$, $3x$, $\cdots$, $mx$，构成一个等差数列，首项为 $x$，末项为 $mx$，项数为 $m$，因此 $f(x) = \frac{(x + mx) \times m}{2}$。
+
+根据容斥原理，我们可以得到答案为：
+
+$$
+f(3) + f(5) + f(7) - f(3 \times 5) - f(3 \times 7) - f(5 \times 7) + f(3 \times 5 \times 7)
+$$
+
+时间复杂度 $O(1)$，空间复杂度 $O(1)$。
+
 <!-- tabs:start -->
 
 ### **Python3**
@@ -65,6 +77,16 @@ class Solution:
         return sum(x for x in range(1, n + 1) if x % 3 == 0 or x % 5 == 0 or x % 7 == 0)
 ```
 
+```python
+class Solution:
+    def sumOfMultiples(self, n: int) -> int:
+        def f(x: int) -> int:
+            m = n // x
+            return (x + m * x) * m // 2
+
+        return f(3) + f(5) + f(7) - f(3 * 5) - f(3 * 7) - f(5 * 7) + f(3 * 5 * 7)
+```
+
 ### **Java**
 
 <!-- 这里可写当前语言的特殊实现逻辑 -->
@@ -83,6 +105,22 @@ class Solution {
 }
 ```
 
+```java
+class Solution {
+    private int n;
+
+    public int sumOfMultiples(int n) {
+        this.n = n;
+        return f(3) + f(5) + f(7) - f(3 * 5) - f(3 * 7) - f(5 * 7) + f(3 * 5 * 7);
+    }
+
+    private int f(int x) {
+        int m = n / x;
+        return (x + m * x) * m / 2;
+    }
+}
+```
+
 ### **C++**
 
 ```cpp
@@ -100,6 +138,19 @@ public:
 };
 ```
 
+```cpp
+class Solution {
+public:
+    int sumOfMultiples(int n) {
+        auto f = [&](int x) {
+            int m = n / x;
+            return (x + m * x) * m / 2;
+        };
+        return f(3) + f(5) + f(7) - f(3 * 5) - f(3 * 7) - f(5 * 7) + f(3 * 5 * 7);
+    }
+};
+```
+
 ### **Go**
 
 ```go
@@ -113,6 +164,16 @@ func sumOfMultiples(n int) (ans int) {
 }
 ```
 
+```go
+func sumOfMultiples(n int) int {
+	f := func(x int) int {
+		m := n / x
+		return (x + m*x) * m / 2
+	}
+	return f(3) + f(5) + f(7) - f(3*5) - f(3*7) - f(5*7) + f(3*5*7)
+}
+```
+
 ### **TypeScript**
 
 ```ts
@@ -127,16 +188,26 @@ function sumOfMultiples(n: number): number {
 }
 ```
 
+```ts
+function sumOfMultiples(n: number): number {
+    const f = (x: number): number => {
+        const m = Math.floor(n / x);
+        return ((x + m * x) * m) >> 1;
+    };
+    return f(3) + f(5) + f(7) - f(3 * 5) - f(3 * 7) - f(5 * 7) + f(3 * 5 * 7);
+}
+```
+
 ### **Rust**
 
 ```rust
 impl Solution {
     pub fn sum_of_multiples(n: i32) -> i32 {
         let mut ans = 0;
 
-        for i in 1..=n {
-            if i % 3 == 0 || i % 5 == 0 || i % 7 == 0 {
-                ans += i;
+        for x in 1..=n {
+            if x % 3 == 0 || x % 5 == 0 || x % 7 == 0 {
+                ans += x;
             }
         }
 
@@ -149,12 +220,25 @@ impl Solution {
 impl Solution {
     pub fn sum_of_multiples(n: i32) -> i32 {
         (1..=n)
-            .filter(|&i| i % 3 == 0 || i % 5 == 0 || i % 7 == 0)
+            .filter(|&x| x % 3 == 0 || x % 5 == 0 || x % 7 == 0)
             .sum()
     }
 }
 ```
 
+```rust
+impl Solution {
+    pub fn sum_of_multiples(n: i32) -> i32 {
+        fn f(x: i32, n: i32) -> i32 {
+            let m = n / x;
+            (x + m * x) * m / 2
+        }
+
+        f(3, n) + f(5, n) + f(7, n) - f(3 * 5, n) - f(3 * 7, n) - f(5 * 7, n) + f(3 * 5 * 7, n)
+    }
+}
+```
+
 ### **...**
 
 ```

solution/2600-2699/2652.Sum Multiples/README_EN.md

@@ -42,6 +42,26 @@
 
 ## Solutions
 
+**Solution 1: Enumeration**
+
+We directly enumerate every number $x$ in $[1,..n]$, and if $x$ is divisible by $3$, $5$, and $7$, we add $x$ to the answer.
+
+After the enumeration, we return the answer.
+
+The time complexity is $O(n)$, where $n$ is the given integer. The space complexity is $O(1)$.
+
+**Solution 2: Mathematics (Inclusion-Exclusion Principle)**
+
+We define a function $f(x)$ to represent the sum of numbers in $[1,..n]$ that are divisible by $x$. There are $m = \left\lfloor \frac{n}{x} \right\rfloor$ numbers that are divisible by $x$, which are $x$, $2x$, $3x$, $\cdots$, $mx$, forming an arithmetic sequence with the first term $x$, the last term $mx$, and the number of terms $m$. Therefore, $f(x) = \frac{(x + mx) \times m}{2}$.
+
+According to the inclusion-exclusion principle, we can obtain the answer as:
+
+$$
+f(3) + f(5) + f(7) - f(3 \times 5) - f(3 \times 7) - f(5 \times 7) + f(3 \times 5 \times 7)
+$$
+
+The time complexity is $O(1)$, and the space complexity is $O(1)$.
+
 <!-- tabs:start -->
 
 ### **Python3**
@@ -52,6 +72,16 @@ class Solution:
         return sum(x for x in range(1, n + 1) if x % 3 == 0 or x % 5 == 0 or x % 7 == 0)
 ```
 
+```python
+class Solution:
+    def sumOfMultiples(self, n: int) -> int:
+        def f(x: int) -> int:
+            m = n // x
+            return (x + m * x) * m // 2
+
+        return f(3) + f(5) + f(7) - f(3 * 5) - f(3 * 7) - f(5 * 7) + f(3 * 5 * 7)
+```
+
 ### **Java**
 
 ```java
@@ -68,6 +98,22 @@ class Solution {
 }
 ```
 
+```java
+class Solution {
+    private int n;
+
+    public int sumOfMultiples(int n) {
+        this.n = n;
+        return f(3) + f(5) + f(7) - f(3 * 5) - f(3 * 7) - f(5 * 7) + f(3 * 5 * 7);
+    }
+
+    private int f(int x) {
+        int m = n / x;
+        return (x + m * x) * m / 2;
+    }
+}
+```
+
 ### **C++**
 
 ```cpp
@@ -85,6 +131,19 @@ public:
 };
 ```
 
+```cpp
+class Solution {
+public:
+    int sumOfMultiples(int n) {
+        auto f = [&](int x) {
+            int m = n / x;
+            return (x + m * x) * m / 2;
+        };
+        return f(3) + f(5) + f(7) - f(3 * 5) - f(3 * 7) - f(5 * 7) + f(3 * 5 * 7);
+    }
+};
+```
+
 ### **Go**
 
 ```go
@@ -98,6 +157,16 @@ func sumOfMultiples(n int) (ans int) {
 }
 ```
 
+```go
+func sumOfMultiples(n int) int {
+	f := func(x int) int {
+		m := n / x
+		return (x + m*x) * m / 2
+	}
+	return f(3) + f(5) + f(7) - f(3*5) - f(3*7) - f(5*7) + f(3*5*7)
+}
+```
+
 ### **TypeScript**
 
 ```ts
@@ -112,16 +181,26 @@ function sumOfMultiples(n: number): number {
 }
 ```
 
+```ts
+function sumOfMultiples(n: number): number {
+    const f = (x: number): number => {
+        const m = Math.floor(n / x);
+        return ((x + m * x) * m) >> 1;
+    };
+    return f(3) + f(5) + f(7) - f(3 * 5) - f(3 * 7) - f(5 * 7) + f(3 * 5 * 7);
+}
+```
+
 ### **Rust**
 
 ```rust
 impl Solution {
     pub fn sum_of_multiples(n: i32) -> i32 {
         let mut ans = 0;
 
-        for i in 1..=n {
-            if i % 3 == 0 || i % 5 == 0 || i % 7 == 0 {
-                ans += i;
+        for x in 1..=n {
+            if x % 3 == 0 || x % 5 == 0 || x % 7 == 0 {
+                ans += x;
             }
         }
 
@@ -134,12 +213,25 @@ impl Solution {
 impl Solution {
     pub fn sum_of_multiples(n: i32) -> i32 {
         (1..=n)
-            .filter(|&i| i % 3 == 0 || i % 5 == 0 || i % 7 == 0)
+            .filter(|&x| x % 3 == 0 || x % 5 == 0 || x % 7 == 0)
             .sum()
     }
 }
 ```
 
+```rust
+impl Solution {
+    pub fn sum_of_multiples(n: i32) -> i32 {
+        fn f(x: i32, n: i32) -> i32 {
+            let m = n / x;
+            (x + m * x) * m / 2
+        }
+
+        f(3, n) + f(5, n) + f(7, n) - f(3 * 5, n) - f(3 * 7, n) - f(5 * 7, n) + f(3 * 5 * 7, n)
+    }
+}
+```
+
 ### **...**
 
 ```

solution/2600-2699/2652.Sum Multiples/Solution.rs

@@ -2,9 +2,9 @@ impl Solution {
     pub fn sum_of_multiples(n: i32) -> i32 {
         let mut ans = 0;
 
-        for i in 1..=n {
-            if i % 3 == 0 || i % 5 == 0 || i % 7 == 0 {
-                ans += i;
+        for x in 1..=n {
+            if x % 3 == 0 || x % 5 == 0 || x % 7 == 0 {
+                ans += x;
             }
         }