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Commit 00e8b5d

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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

Lines changed: 89 additions & 5 deletions
Original file line numberDiff line numberDiff line change
@@ -47,12 +47,24 @@
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**方法一:枚举**
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50-
我们直接枚举 $[1,..n]$ 中的每一个数 $x,ドル如果 $x$ 能被 3ドル$、5ドル$、7ドル$ 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,ドル如果 $x$ 能被 3ドル$, 5ドル$, 7ドル$ 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),ドル其中 $n$ 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,ドル 2ドルx,ドル 3ドルx,ドル $\cdots,ドル $mx,ドル构成一个等差数列,首项为 $x,ドル末项为 $mx,ドル项数为 $m,ドル因此 $f(x) = \frac{(x + mx) \times m}{2}$。
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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),ドル空间复杂度 $O(1)$。
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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)
6678
```
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```python
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class Solution:
82+
def sumOfMultiples(self, n: int) -> int:
83+
def f(x: int) -> int:
84+
m = n // x
85+
return (x + m * x) * m // 2
86+
87+
return f(3) + f(5) + f(7) - f(3 * 5) - f(3 * 7) - f(5 * 7) + f(3 * 5 * 7)
88+
```
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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 {
83105
}
84106
```
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```java
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class Solution {
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private int n;
111+
112+
public int sumOfMultiples(int n) {
113+
this.n = n;
114+
return f(3) + f(5) + f(7) - f(3 * 5) - f(3 * 7) - f(5 * 7) + f(3 * 5 * 7);
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}
116+
117+
private int f(int x) {
118+
int m = n / x;
119+
return (x + m * x) * m / 2;
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}
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}
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```
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86124
### **C++**
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88126
```cpp
@@ -100,6 +138,19 @@ public:
100138
};
101139
```
102140
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```cpp
142+
class Solution {
143+
public:
144+
int sumOfMultiples(int n) {
145+
auto f = [&](int x) {
146+
int m = n / x;
147+
return (x + m * x) * m / 2;
148+
};
149+
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) {
113164
}
114165
```
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```go
168+
func sumOfMultiples(n int) int {
169+
f := func(x int) int {
170+
m := n / x
171+
return (x + m*x) * m / 2
172+
}
173+
return f(3) + f(5) + f(7) - f(3*5) - f(3*7) - f(5*7) + f(3*5*7)
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}
175+
```
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### **TypeScript**
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118179
```ts
@@ -127,16 +188,26 @@ function sumOfMultiples(n: number): number {
127188
}
128189
```
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```ts
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function sumOfMultiples(n: number): number {
193+
const f = (x: number): number => {
194+
const m = Math.floor(n / x);
195+
return ((x + m * x) * m) >> 1;
196+
};
197+
return f(3) + f(5) + f(7) - f(3 * 5) - f(3 * 7) - f(5 * 7) + f(3 * 5 * 7);
198+
}
199+
```
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### **Rust**
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```rust
133204
impl Solution {
134205
pub fn sum_of_multiples(n: i32) -> i32 {
135206
let mut ans = 0;
136207

137-
for i in 1..=n {
138-
if i % 3 == 0 || i % 5 == 0 || i % 7 == 0 {
139-
ans += i;
208+
for x in 1..=n {
209+
if x % 3 == 0 || x % 5 == 0 || x % 7 == 0 {
210+
ans += x;
140211
}
141212
}
142213

@@ -149,12 +220,25 @@ impl Solution {
149220
impl Solution {
150221
pub fn sum_of_multiples(n: i32) -> i32 {
151222
(1..=n)
152-
.filter(|&i|i % 3 == 0 || i % 5 == 0 || i % 7 == 0)
223+
.filter(|&x|x % 3 == 0 || x % 5 == 0 || x % 7 == 0)
153224
.sum()
154225
}
155226
}
156227
```
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```rust
230+
impl Solution {
231+
pub fn sum_of_multiples(n: i32) -> i32 {
232+
fn f(x: i32, n: i32) -> i32 {
233+
let m = n / x;
234+
(x + m * x) * m / 2
235+
}
236+
237+
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)
238+
}
239+
}
240+
```
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158242
### **...**
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160244
```

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

Lines changed: 96 additions & 4 deletions
Original file line numberDiff line numberDiff line change
@@ -42,6 +42,26 @@
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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],ドル and if $x$ is divisible by 3ドル,ドル 5ドル,ドル and 7ドル,ドル we add $x$ to the answer.
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49+
After the enumeration, we return the answer.
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51+
The time complexity is $O(n),ドル 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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55+
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,ドル which are $x,ドル 2ドルx,ドル 3ドルx,ドル $\cdots,ドル $mx,ドル forming an arithmetic sequence with the first term $x,ドル the last term $mx,ドル and the number of terms $m$. Therefore, $f(x) = \frac{(x + mx) \times m}{2}$.
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57+
According to the inclusion-exclusion principle, we can obtain the answer as:
58+
59+
$$
60+
f(3) + f(5) + f(7) - f(3 \times 5) - f(3 \times 7) - f(5 \times 7) + f(3 \times 5 \times 7)
61+
$$
62+
63+
The time complexity is $O(1),ドル and the space complexity is $O(1)$.
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4565
<!-- tabs:start -->
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4767
### **Python3**
@@ -52,6 +72,16 @@ class Solution:
5272
return sum(x for x in range(1, n + 1) if x % 3 == 0 or x % 5 == 0 or x % 7 == 0)
5373
```
5474

75+
```python
76+
class Solution:
77+
def sumOfMultiples(self, n: int) -> int:
78+
def f(x: int) -> int:
79+
m = n // x
80+
return (x + m * x) * m // 2
81+
82+
return f(3) + f(5) + f(7) - f(3 * 5) - f(3 * 7) - f(5 * 7) + f(3 * 5 * 7)
83+
```
84+
5585
### **Java**
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5787
```java
@@ -68,6 +98,22 @@ class Solution {
6898
}
6999
```
70100

101+
```java
102+
class Solution {
103+
private int n;
104+
105+
public int sumOfMultiples(int n) {
106+
this.n = n;
107+
return f(3) + f(5) + f(7) - f(3 * 5) - f(3 * 7) - f(5 * 7) + f(3 * 5 * 7);
108+
}
109+
110+
private int f(int x) {
111+
int m = n / x;
112+
return (x + m * x) * m / 2;
113+
}
114+
}
115+
```
116+
71117
### **C++**
72118

73119
```cpp
@@ -85,6 +131,19 @@ public:
85131
};
86132
```
87133
134+
```cpp
135+
class Solution {
136+
public:
137+
int sumOfMultiples(int n) {
138+
auto f = [&](int x) {
139+
int m = n / x;
140+
return (x + m * x) * m / 2;
141+
};
142+
return f(3) + f(5) + f(7) - f(3 * 5) - f(3 * 7) - f(5 * 7) + f(3 * 5 * 7);
143+
}
144+
};
145+
```
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88147
### **Go**
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90149
```go
@@ -98,6 +157,16 @@ func sumOfMultiples(n int) (ans int) {
98157
}
99158
```
100159

160+
```go
161+
func sumOfMultiples(n int) int {
162+
f := func(x int) int {
163+
m := n / x
164+
return (x + m*x) * m / 2
165+
}
166+
return f(3) + f(5) + f(7) - f(3*5) - f(3*7) - f(5*7) + f(3*5*7)
167+
}
168+
```
169+
101170
### **TypeScript**
102171

103172
```ts
@@ -112,16 +181,26 @@ function sumOfMultiples(n: number): number {
112181
}
113182
```
114183

184+
```ts
185+
function sumOfMultiples(n: number): number {
186+
const f = (x: number): number => {
187+
const m = Math.floor(n / x);
188+
return ((x + m * x) * m) >> 1;
189+
};
190+
return f(3) + f(5) + f(7) - f(3 * 5) - f(3 * 7) - f(5 * 7) + f(3 * 5 * 7);
191+
}
192+
```
193+
115194
### **Rust**
116195

117196
```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
```
142221

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
### **...**
144236

145237
```

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

Lines changed: 3 additions & 3 deletions
Original file line numberDiff line numberDiff line change
@@ -2,9 +2,9 @@ impl Solution {
22
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
}
99
}
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