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Sep 12, 2024
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feat: add solutions to lc problem: No.1925 #3518

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91 changes: 57 additions & 34 deletions solution/1900-1999/1925.Count Square Sum Triples/README.md
View file
Open in desktop
Original file line number Diff line number Diff line change
Expand Up @@ -53,7 +53,13 @@ tags:

<!-- solution:start -->

### 方法一
### 方法一：枚举

我们在 $[1, n)$ 的范围内枚举 $a$ 和 $b$，然后计算 $c = \sqrt{a^2 + b^2}$，如果 $c$ 是整数且 $c \leq n$，那么就找到了一个平方和三元组，答案加一。

枚举结束后，返回答案即可。

时间复杂度 $O(n^2)$，其中 $n$ 是给定的整数。空间复杂度 $O(1)$。

<!-- tabs:start -->

Expand All @@ -62,32 +68,32 @@ tags:
```python
class Solution:
def countTriples(self, n: int) -> int:
res = 0
for a in range(1, n + 1):
for b in range(1, n + 1):
t = a**2 + b**2
c = int(sqrt(t))
if c <= n and c**2 == t:
res += 1
return res
ans = 0
for a in range(1, n):
for b in range(1, n):
x = a * a + b * b
c = int(sqrt(x))
if c <= n and c * c == x:
ans += 1
return ans
```

#### Java

```java
class Solution {
public int countTriples(int n) {
int res = 0;
for (int a = 1; a <= n; ++a) {
for (int b = 1; b <= n; ++b) {
int t = a * a + b * b;
int c = (int) Math.sqrt(t);
if (c <= n && c * c == t) {
++res;
int ans = 0;
for (int a = 1; a < n; a++) {
for (int b = 1; b < n; b++) {
int x = a * a + b * b;
int c = (int) Math.sqrt(x);
if (c <= n && c * c == x) {
ans++;
}
}
}
return res;
return ans;
}
}
```
Expand All @@ -98,36 +104,53 @@ class Solution {
class Solution {
public:
int countTriples(int n) {
int res = 0;
for (int a = 1; a <= n; ++a) {
for (int b = 1; b <= n; ++b) {
int t = a * a + b * b;
int c = (int) sqrt(t);
if (c <= n && c * c == t) {
++res;
int ans = 0;
for (int a = 1; a < n; ++a) {
for (int b = 1; b < n; ++b) {
int x = a * a + b * b;
int c = static_cast<int>(sqrt(x));
if (c <= n && c * c == x) {
++ans;
}
}
}
return res;
return ans;
}
};
```

#### Go

```go
func countTriples(n int) int {
res := 0
for a := 1; a <= n; a++ {
for b := 1; b <= n; b++ {
t := a*a + b*b
c := int(math.Sqrt(float64(t)))
if c <= n && c*c == t {
res++
func countTriples(n int) (ans int) {
for a := 1; a < n; a++ {
for b := 1; b < n; b++ {
x := a*a + b*b
c := int(math.Sqrt(float64(x)))
if c <= n && c*c == x {
ans++
}
}
}
return res
return
}
```

#### TypeScript

```ts
function countTriples(n: number): number {
let ans = 0;
for (let a = 1; a < n; a++) {
for (let b = 1; b < n; b++) {
const x = a * a + b * b;
const c = Math.floor(Math.sqrt(x));
if (c <= n && c * c === x) {
ans++;
}
}
}
return ans;
}
```

Expand Down
91 changes: 57 additions & 34 deletions solution/1900-1999/1925.Count Square Sum Triples/README_EN.md
View file
Open in desktop
Original file line number Diff line number Diff line change
Expand Up @@ -53,7 +53,13 @@ tags:

<!-- solution:start -->

### Solution 1
### Solution 1: Enumeration

We enumerate $a$ and $b$ in the range $[1, n)$, then calculate $c = \sqrt{a^2 + b^2}$. If $c$ is an integer and $c \leq n$, then we have found a Pythagorean triplet, and we increment the answer by one.

After the enumeration is complete, return the answer.

The time complexity is $O(n^2)$, where $n$ is the given integer. The space complexity is $O(1)$.

<!-- tabs:start -->

Expand All @@ -62,32 +68,32 @@ tags:
```python
class Solution:
def countTriples(self, n: int) -> int:
res = 0
for a in range(1, n + 1):
for b in range(1, n + 1):
t = a**2 + b**2
c = int(sqrt(t))
if c <= n and c**2 == t:
res += 1
return res
ans = 0
for a in range(1, n):
for b in range(1, n):
x = a * a + b * b
c = int(sqrt(x))
if c <= n and c * c == x:
ans += 1
return ans
```

#### Java

```java
class Solution {
public int countTriples(int n) {
int res = 0;
for (int a = 1; a <= n; ++a) {
for (int b = 1; b <= n; ++b) {
int t = a * a + b * b;
int c = (int) Math.sqrt(t);
if (c <= n && c * c == t) {
++res;
int ans = 0;
for (int a = 1; a < n; a++) {
for (int b = 1; b < n; b++) {
int x = a * a + b * b;
int c = (int) Math.sqrt(x);
if (c <= n && c * c == x) {
ans++;
}
}
}
return res;
return ans;
}
}
```
Expand All @@ -98,36 +104,53 @@ class Solution {
class Solution {
public:
int countTriples(int n) {
int res = 0;
for (int a = 1; a <= n; ++a) {
for (int b = 1; b <= n; ++b) {
int t = a * a + b * b;
int c = (int) sqrt(t);
if (c <= n && c * c == t) {
++res;
int ans = 0;
for (int a = 1; a < n; ++a) {
for (int b = 1; b < n; ++b) {
int x = a * a + b * b;
int c = static_cast<int>(sqrt(x));
if (c <= n && c * c == x) {
++ans;
}
}
}
return res;
return ans;
}
};
```

#### Go

```go
func countTriples(n int) int {
res := 0
for a := 1; a <= n; a++ {
for b := 1; b <= n; b++ {
t := a*a + b*b
c := int(math.Sqrt(float64(t)))
if c <= n && c*c == t {
res++
func countTriples(n int) (ans int) {
for a := 1; a < n; a++ {
for b := 1; b < n; b++ {
x := a*a + b*b
c := int(math.Sqrt(float64(x)))
if c <= n && c*c == x {
ans++
}
}
}
return res
return
}
```

#### TypeScript

```ts
function countTriples(n: number): number {
let ans = 0;
for (let a = 1; a < n; a++) {
for (let b = 1; b < n; b++) {
const x = a * a + b * b;
const c = Math.floor(Math.sqrt(x));
if (c <= n && c * c === x) {
ans++;
}
}
}
return ans;
}
```

Expand Down
18 changes: 9 additions & 9 deletions solution/1900-1999/1925.Count Square Sum Triples/Solution.cpp
View file
Open in desktop
Original file line number Diff line number Diff line change
@@ -1,16 +1,16 @@
class Solution {
public:
int countTriples(int n) {
int res = 0;
for (int a = 1; a <= n; ++a) {
for (int b = 1; b <= n; ++b) {
int t = a * a + b * b;
int c = (int) sqrt(t);
if (c <= n && c * c == t) {
++res;
int ans = 0;
for (int a = 1; a < n; ++a) {
for (int b = 1; b < n; ++b) {
int x = a * a + b * b;
int c = static_cast<int>(sqrt(x));
if (c <= n && c * c == x) {
++ans;
}
}
}
return res;
return ans;
}
};
};
19 changes: 9 additions & 10 deletions solution/1900-1999/1925.Count Square Sum Triples/Solution.go
View file
Open in desktop
Original file line number Diff line number Diff line change
@@ -1,13 +1,12 @@
func countTriples(n int) int {
res := 0
for a := 1; a <= n; a++ {
for b := 1; b <= n; b++ {
t := a*a + b*b
c := int(math.Sqrt(float64(t)))
if c <= n && c*c == t {
res++
func countTriples(n int) (ans int) {
for a := 1; a < n; a++ {
for b := 1; b < n; b++ {
x := a*a + b*b
c := int(math.Sqrt(float64(x)))
if c <= n && c*c == x {
ans++
}
}
}
return res
}
return
}
18 changes: 9 additions & 9 deletions solution/1900-1999/1925.Count Square Sum Triples/Solution.java
View file
Open in desktop
Original file line number Diff line number Diff line change
@@ -1,15 +1,15 @@
class Solution {
public int countTriples(int n) {
int res = 0;
for (int a = 1; a <= n; ++a) {
for (int b = 1; b <= n; ++b) {
int t = a * a + b * b;
int c = (int) Math.sqrt(t);
if (c <= n && c * c == t) {
++res;
int ans = 0;
for (int a = 1; a < n; a++) {
for (int b = 1; b < n; b++) {
int x = a * a + b * b;
int c = (int) Math.sqrt(x);
if (c <= n && c * c == x) {
ans++;
}
}
}
return res;
return ans;
}
}
}
16 changes: 8 additions & 8 deletions solution/1900-1999/1925.Count Square Sum Triples/Solution.py
View file
Open in desktop
Original file line number Diff line number Diff line change
@@ -1,10 +1,10 @@
class Solution:
def countTriples(self, n: int) -> int:
res = 0
for a in range(1, n + 1):
for b in range(1, n + 1):
t = a**2 + b**2
c = int(sqrt(t))
if c <= n and c**2 == t:
res += 1
return res
ans = 0
for a in range(1, n):
for b in range(1, n):
x = a * a + b * b
c = int(sqrt(x))
if c <= n and c * c == x:
ans += 1
return ans
13 changes: 13 additions & 0 deletions solution/1900-1999/1925.Count Square Sum Triples/Solution.ts
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Original file line number Diff line number Diff line change
@@ -0,0 +1,13 @@
function countTriples(n: number): number {
let ans = 0;
for (let a = 1; a < n; a++) {
for (let b = 1; b < n; b++) {
const x = a * a + b * b;
const c = Math.floor(Math.sqrt(x));
if (c <= n && c * c === x) {
ans++;
}
}
}
return ans;
}
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