feat: add solutions to lc problem: No.0829

No.0829.Consecutive Numbers Sum

Author: yanglbme

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

@@ -43,22 +43,88 @@
 
 <!-- 这里可写通用的实现逻辑 -->
 
+**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。
+
+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。
+
+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，我们枚举 $k$，累加满足条件的 $k$ 的个数即可。
+
+时间复杂度 $O(\sqrt{n})$。
+
 <!-- tabs:start -->
 
 ### **Python3**
 
 <!-- 这里可写当前语言的特殊实现逻辑 -->
 
 ```python
-
+class Solution:
+    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:
+                ans += 1
+            k += 1
+        return ans
 ```
 
 ### **Java**
 
 <!-- 这里可写当前语言的特殊实现逻辑 -->
 
 ```java
+class Solution {
+
+    public int consecutiveNumbersSum(int n) {
+        n <<= 1;
+        int ans = 0;
+        for (int k = 1; k * (k + 1) <= n; ++k) {
+            if (n % k == 0 && (n / k + 1 - k) % 2 == 0) {
+                ++ans;
+            }
+        }
+        return ans;
+    }
+}
+
+```
+
+### **C++**
+
+```cpp
+class Solution {
+public:
+    int consecutiveNumbersSum(int n) {
+        n <<= 1;
+        int ans = 0;
+        for (int k = 1; k * (k + 1) <= n; ++k) {
+            if (n % k == 0 && (n / k + 1 - k) % 2 == 0) {
+                ++ans;
+            }
+        }
+        return ans;
+    }
+};
+```
 
+### **Go**
+
+```go
+func consecutiveNumbersSum(n int) int {
+	n <<= 1
+	ans := 0
+	for k := 1; k*(k+1) <= n; k++ {
+		if n%k == 0 && (n/k+1-k)%2 == 0 {
+			ans++
+		}
+	}
+	return ans
+}
 ```
 
 ### **...**

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

@@ -45,13 +45,65 @@
 ### **Python3**
 
 ```python
-
+class Solution:
+    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:
+                ans += 1
+            k += 1
+        return ans
 ```
 
 ### **Java**
 
 ```java
+class Solution {
+    public int consecutiveNumbersSum(int n) {
+        n <<= 1;
+        int ans = 0;
+        for (int k = 1; k * (k + 1) <= n; ++k) {
+            if (n % k == 0 && (n / k + 1 - k) % 2 == 0) {
+                ++ans;
+            }
+        }
+        return ans;
+    }
+}
+```
+
+### **C++**
+
+```cpp
+class Solution {
+public:
+    int consecutiveNumbersSum(int n) {
+        n <<= 1;
+        int ans = 0;
+        for (int k = 1; k * (k + 1) <= n; ++k) {
+            if (n % k == 0 && (n / k + 1 - k) % 2 == 0) {
+                ++ans;
+            }
+        }
+        return ans;
+    }
+};
+```
 
+### **Go**
+
+```go
+func consecutiveNumbersSum(n int) int {
+	n <<= 1
+	ans := 0
+	for k := 1; k*(k+1) <= n; k++ {
+		if n%k == 0 && (n/k+1-k)%2 == 0 {
+			ans++
+		}
+	}
+	return ans
+}
 ```
 
 ### **...**

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

@@ -0,0 +1,13 @@
+class Solution {
+public:
+    int consecutiveNumbersSum(int n) {
+        n <<= 1;
+        int ans = 0;
+        for (int k = 1; k * (k + 1) <= n; ++k) {
+            if (n % k == 0 && (n / k + 1 - k) % 2 == 0) {
+                ++ans;
+            }
+        }
+        return ans;
+    }
+};

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

@@ -0,0 +1,10 @@
+func consecutiveNumbersSum(n int) int {
+	n <<= 1
+	ans := 0
+	for k := 1; k*(k+1) <= n; k++ {
+		if n%k == 0 && (n/k+1-k)%2 == 0 {
+			ans++
+		}
+	}
+	return ans
+}

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

@@ -1,11 +1,12 @@
 class Solution {
-    public int consecutiveNumbersSum(int N) {
-        int res = 0;
-        for (int i = 1; i * (i - 1) / 2 < N; ++i) {
-            if ((N - i * (i - 1) / 2) % i == 0) {
-                ++res;
+    public int consecutiveNumbersSum(int n) {
+        n <<= 1;
+        int ans = 0;
+        for (int k = 1; k * (k + 1) <= n; ++k) {
+            if (n % k == 0 && (n / k + 1 - k) % 2 == 0) {
+                ++ans;
             }
         }
-        return res;
+        return ans;
     }
-}
+}

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

@@ -0,0 +1,9 @@
+class Solution:
+    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:
+                ans += 1
+            k += 1
+        return ans