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Commit 00e8b5d

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yanglbmeyanglbme
authored
feat: add solutions to lc problem: No.2652 (doocs#1827)
No.2652.Sum Multiples
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‎solution/2600-2699/2652.Sum Multiples/README.md‎

Copy file name to clipboardExpand all lines: solution/2600-2699/2652.Sum Multiples/README.md
+89-5Lines changed: 89 additions & 5 deletions
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@@ -47,12 +47,24 @@
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**方法一：枚举**
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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。
50+
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。
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枚举结束后，返回答案即可。
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时间复杂度 $O(n)$，其中 $n$ 为题目给定的整数。空间复杂度 $O(1)$。
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**方法二：数学（容斥原理）**
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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 $f(x) = \frac{(x + mx) \times m}{2}$。
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根据容斥原理，我们可以得到答案为：
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$$
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f(3) + f(5) + f(7) - f(3 \times 5) - f(3 \times 7) - f(5 \times 7) + f(3 \times 5 \times 7)
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$$
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时间复杂度 $O(1)$，空间复杂度 $O(1)$。
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<!-- tabs:start -->
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### **Python3**
@@ -65,6 +77,16 @@ class Solution:
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return sum(x for x in range(1, n + 1) if x % 3 == 0 or x % 5 == 0 or x % 7 == 0)
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```
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```python
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class Solution:
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def sumOfMultiples(self, n: int) -> int:
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def f(x: int) -> int:
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m = n // x
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return (x + m * x) * m // 2
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return f(3) + f(5) + f(7) - f(3 * 5) - f(3 * 7) - f(5 * 7) + f(3 * 5 * 7)
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```
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6890
### **Java**
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<!-- 这里可写当前语言的特殊实现逻辑 -->
@@ -83,6 +105,22 @@ class Solution {
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}
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```
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```java
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class Solution {
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private int n;
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public int sumOfMultiples(int n) {
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this.n = n;
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return f(3) + f(5) + f(7) - f(3 * 5) - f(3 * 7) - f(5 * 7) + f(3 * 5 * 7);
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}
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private int f(int x) {
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int m = n / x;
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return (x + m * x) * m / 2;
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}
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}
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```
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### **C++**
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```cpp
@@ -100,6 +138,19 @@ public:
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};
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```
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```cpp
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class Solution {
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public:
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int sumOfMultiples(int n) {
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auto f = [&](int x) {
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int m = n / x;
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return (x + m * x) * m / 2;
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};
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return f(3) + f(5) + f(7) - f(3 * 5) - f(3 * 7) - f(5 * 7) + f(3 * 5 * 7);
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}
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};
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```
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103154
### **Go**
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```go
@@ -113,6 +164,16 @@ func sumOfMultiples(n int) (ans int) {
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}
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```
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```go
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func sumOfMultiples(n int) int {
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f := func(x int) int {
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m := n / x
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return (x + m*x) * m / 2
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}
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return f(3) + f(5) + f(7) - f(3*5) - f(3*7) - f(5*7) + f(3*5*7)
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}
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```
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### **TypeScript**
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```ts
@@ -127,16 +188,26 @@ function sumOfMultiples(n: number): number {
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}
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```
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```ts
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function sumOfMultiples(n: number): number {
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const f = (x: number): number => {
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const m = Math.floor(n / x);
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return ((x + m * x) * m) >> 1;
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};
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return f(3) + f(5) + f(7) - f(3 * 5) - f(3 * 7) - f(5 * 7) + f(3 * 5 * 7);
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}
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```
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### **Rust**
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```rust
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impl Solution {
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pub fn sum_of_multiples(n: i32) -> i32 {
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let mut ans = 0;
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137-
for i in 1..=n {
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if i % 3 == 0 || i % 5 == 0 || i % 7 == 0 {
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ans += i;
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for x in 1..=n {
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if x % 3 == 0 || x % 5 == 0 || x % 7 == 0 {
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ans += x;
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}
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}
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@@ -149,12 +220,25 @@ impl Solution {
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impl Solution {
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pub fn sum_of_multiples(n: i32) -> i32 {
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(1..=n)
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.filter(|&i| i % 3 == 0 || i % 5 == 0 || i % 7 == 0)
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.filter(|&x| x % 3 == 0 || x % 5 == 0 || x % 7 == 0)
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.sum()
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}
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}
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```
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```rust
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impl Solution {
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pub fn sum_of_multiples(n: i32) -> i32 {
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fn f(x: i32, n: i32) -> i32 {
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let m = n / x;
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(x + m * x) * m / 2
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}
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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)
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}
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}
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```
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### **...**
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```
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‎solution/2600-2699/2652.Sum Multiples/README_EN.md‎

Copy file name to clipboardExpand all lines: solution/2600-2699/2652.Sum Multiples/README_EN.md
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## Solutions
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**Solution 1: Enumeration**
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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.
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After the enumeration, we return the answer.
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The time complexity is $O(n)$, where $n$ is the given integer. The space complexity is $O(1)$.
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**Solution 2: Mathematics (Inclusion-Exclusion Principle)**
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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, $f(x) = \frac{(x + mx) \times m}{2}$.
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According to the inclusion-exclusion principle, we can obtain the answer as:
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$$
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f(3) + f(5) + f(7) - f(3 \times 5) - f(3 \times 7) - f(5 \times 7) + f(3 \times 5 \times 7)
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$$
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The time complexity is $O(1)$, and the space complexity is $O(1)$.
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<!-- tabs:start -->
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### **Python3**
@@ -52,6 +72,16 @@ class Solution:
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return sum(x for x in range(1, n + 1) if x % 3 == 0 or x % 5 == 0 or x % 7 == 0)
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```
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```python
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class Solution:
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def sumOfMultiples(self, n: int) -> int:
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def f(x: int) -> int:
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m = n // x
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return (x + m * x) * m // 2
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return f(3) + f(5) + f(7) - f(3 * 5) - f(3 * 7) - f(5 * 7) + f(3 * 5 * 7)
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```
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### **Java**
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```java
@@ -68,6 +98,22 @@ class Solution {
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}
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```
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```java
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class Solution {
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private int n;
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public int sumOfMultiples(int n) {
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this.n = n;
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return f(3) + f(5) + f(7) - f(3 * 5) - f(3 * 7) - f(5 * 7) + f(3 * 5 * 7);
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}
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private int f(int x) {
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int m = n / x;
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return (x + m * x) * m / 2;
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}
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}
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```
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### **C++**
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73119
```cpp
@@ -85,6 +131,19 @@ public:
85131
};
86132
```
87133
134+
```cpp
135+
class Solution {
136+
public:
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int sumOfMultiples(int n) {
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auto f = [&](int x) {
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int m = n / x;
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return (x + m * x) * m / 2;
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};
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return f(3) + f(5) + f(7) - f(3 * 5) - f(3 * 7) - f(5 * 7) + f(3 * 5 * 7);
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}
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};
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```
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88147
### **Go**
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90149
```go
@@ -98,6 +157,16 @@ func sumOfMultiples(n int) (ans int) {
98157
}
99158
```
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```go
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func sumOfMultiples(n int) int {
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f := func(x int) int {
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m := n / x
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return (x + m*x) * m / 2
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}
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return f(3) + f(5) + f(7) - f(3*5) - f(3*7) - f(5*7) + f(3*5*7)
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}
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```
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### **TypeScript**
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103172
```ts
@@ -112,16 +181,26 @@ function sumOfMultiples(n: number): number {
112181
}
113182
```
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```ts
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function sumOfMultiples(n: number): number {
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const f = (x: number): number => {
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const m = Math.floor(n / x);
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return ((x + m * x) * m) >> 1;
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};
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return f(3) + f(5) + f(7) - f(3 * 5) - f(3 * 7) - f(5 * 7) + f(3 * 5 * 7);
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}
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```
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### **Rust**
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```rust
118197
impl Solution {
119198
pub fn sum_of_multiples(n: i32) -> i32 {
120199
let mut ans = 0;
121200

122-
for i in 1..=n {
123-
if i % 3 == 0 || i % 5 == 0 || i % 7 == 0 {
124-
ans += i;
201+
for x in 1..=n {
202+
if x % 3 == 0 || x % 5 == 0 || x % 7 == 0 {
203+
ans += x;
125204
}
126205
}
127206

@@ -134,12 +213,25 @@ impl Solution {
134213
impl Solution {
135214
pub fn sum_of_multiples(n: i32) -> i32 {
136215
(1..=n)
137-
.filter(|&i| i % 3 == 0 || i % 5 == 0 || i % 7 == 0)
216+
.filter(|&x| x % 3 == 0 || x % 5 == 0 || x % 7 == 0)
138217
.sum()
139218
}
140219
}
141220
```
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222+
```rust
223+
impl Solution {
224+
pub fn sum_of_multiples(n: i32) -> i32 {
225+
fn f(x: i32, n: i32) -> i32 {
226+
let m = n / x;
227+
(x + m * x) * m / 2
228+
}
229+
230+
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)
231+
}
232+
}
233+
```
234+
143235
### **...**
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145237
```
Collapse file

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

Copy file name to clipboardExpand all lines: solution/2600-2699/2652.Sum Multiples/Solution.rs
+3-3Lines changed: 3 additions & 3 deletions
Original file line numberDiff line numberDiff line change
@@ -2,9 +2,9 @@ impl Solution {
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pub fn sum_of_multiples(n: i32) -> i32 {
33
let mut ans = 0;
44

5-
for i in 1..=n {
6-
if i % 3 == 0 || i % 5 == 0 || i % 7 == 0 {
7-
ans += i;
5+
for x in 1..=n {
6+
if x % 3 == 0 || x % 5 == 0 || x % 7 == 0 {
7+
ans += x;
88
}
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}
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