feat: add solutions to lc problem: No.0829

No.0829.Consecutive Numbers Sum

Author: yanglbme

solution/0800-0899/0829.Consecutive Numbers Sum/README.md

@@ -58,15 +58,21 @@ tags:
 
 ### 方法一：数学推导
 
-连续正整数构成一个等差数列($d=1$)。我们假设等差数列的第一项为 $a$，项数为 $k$，则 $n=(a+a+k-1)*k/2$，即 $n*2=(a*2+k-1)*k$ ①。
+连续正整数构成一个公差 $d = 1$ 的等差数列。我们假设等差数列的第一项为 $a$，项数为 $k$，那么 $n = (a + a + k - 1) \times k / 2$，即 $n \times 2 = (a \times 2 + k - 1) \times k$。这里我们可以得出 $k$ 一定能整除 $n \times 2$，并且 $(n \times 2) / k - k + 1$ 一定是偶数。
 
-由于是连续正整数， $a>=1$，所以 ① 可以化为 $n*2>=(k+1)*k$，即 $k*(k+1)<=n*2$ ②。
+由于 $a \geq 1$，所以 $n \times 2 = (a \times 2 + k - 1) \times k \geq k \times (k + 1)$。
 
-因此，$k$ 的范围需要满足 $k>=1$ 并且 $k*(k+1)<=n*2$。另外，我们从 ① 式可以发现，$k$ 必须能整除 $n*2$。
+综上，我们可以得出：
 
-综上，我们枚举 $k$，累加满足条件的 $k$ 的个数即可。
+1. $k$ 一定能整除 $n \times 2$；
+2. $k \times (k + 1) \leq n \times 2$；
+3. $(n \times 2) / k - k + 1$ 一定是偶数。
 
-时间复杂度 $O(\sqrt{n})$。
+我们从 $k = 1$ 开始枚举，当 $k \times (k + 1) > n \times 2$ 时，我们可以结束枚举。在枚举的过程中，我们判断 $k$ 是否能整除 $n \times 2$，并且 $(n \times 2) / k - k + 1$ 是否是偶数，如果是则满足条件，答案加一。
+
+枚举结束后，返回答案即可。
+
+时间复杂度 $O(\sqrt{n})$，其中 $n$ 为给定的正整数。空间复杂度 $O(1)$。
 
 <!-- tabs:start -->
 
@@ -135,6 +141,21 @@ func consecutiveNumbersSum(n int) int {
 }
 ```
 
+#### TypeScript
+
+```ts
+function consecutiveNumbersSum(n: number): number {
+    let ans = 0;
+    n <<= 1;
+    for (let k = 1; k * (k + 1) <= n; ++k) {
+        if (n % k === 0 && (Math.floor(n / k) + 1 - k) % 2 === 0) {
+            ++ans;
+        }
+    }
+    return ans;
+}
+```
+
 <!-- tabs:end -->
 
 <!-- solution:end -->

solution/0800-0899/0829.Consecutive Numbers Sum/README_EN.md

@@ -57,7 +57,23 @@ tags:
 
 <!-- solution:start -->
 
-### Solution 1
+### Solution 1: Mathematical Derivation
+
+Consecutive positive integers form an arithmetic sequence with a common difference $d = 1$. Let's assume the first term of the sequence is $a$, and the number of terms is $k$. Then, $n = (a + a + k - 1) \times k / 2$, which simplifies to $n \times 2 = (a \times 2 + k - 1) \times k$. From this, we can deduce that $k$ must divide $n \times 2$ evenly, and $(n \times 2) / k - k + 1$ must be an even number.
+
+Given that $a \geq 1$, it follows that $n \times 2 = (a \times 2 + k - 1) \times k \geq k \times (k + 1)$.
+
+In summary, we can conclude:
+
+1. $k$ must divide $n \times 2$ evenly;
+2. $k \times (k + 1) \leq n \times 2$;
+3. $(n \times 2) / k - k + 1$ must be an even number.
+
+We start enumerating from $k = 1$, and we can stop when $k \times (k + 1) > n \times 2$. During the enumeration, we check if $k$ divides $n \times 2$ evenly, and if $(n \times 2) / k - k + 1$ is an even number. If both conditions are met, it satisfies the criteria, and we increment the answer by one.
+
+After finishing the enumeration, we return the answer.
+
+The time complexity is $O(\sqrt{n})$, where $n$ is the given positive integer. The space complexity is $O(1)$.
 
 <!-- tabs:start -->
 
@@ -69,7 +85,7 @@ class Solution:
         n <<= 1
         ans, k = 0, 1
         while k * (k + 1) <= n:
-            if n % k == 0 and (n // k + 1 - k) % 2 == 0:
+            if n % k == 0 and (n // k - k + 1) % 2 == 0:
                 ans += 1
             k += 1
         return ans
@@ -126,6 +142,21 @@ func consecutiveNumbersSum(n int) int {
 }
 ```
 
+#### TypeScript
+
+```ts
+function consecutiveNumbersSum(n: number): number {
+    let ans = 0;
+    n <<= 1;
+    for (let k = 1; k * (k + 1) <= n; ++k) {
+        if (n % k === 0 && (Math.floor(n / k) + 1 - k) % 2 === 0) {
+            ++ans;
+        }
+    }
+    return ans;
+}
+```
+
 <!-- tabs:end -->
 
 <!-- solution:end -->

solution/0800-0899/0829.Consecutive Numbers Sum/Solution.py

@@ -3,7 +3,7 @@ def consecutiveNumbersSum(self, n: int) -> int:
         n <<= 1
         ans, k = 0, 1
         while k * (k + 1) <= n:
-            if n % k == 0 and (n // k + 1 - k) % 2 == 0:
+            if n % k == 0 and (n // k - k + 1) % 2 == 0:
                 ans += 1
             k += 1
         return ans

solution/0800-0899/0829.Consecutive Numbers Sum/Solution.ts

@@ -0,0 +1,10 @@
+function consecutiveNumbersSum(n: number): number {
+    let ans = 0;
+    n <<= 1;
+    for (let k = 1; k * (k + 1) <= n; ++k) {
+        if (n % k === 0 && (Math.floor(n / k) + 1 - k) % 2 === 0) {
+            ++ans;
+        }
+    }
+    return ans;
+}