LeetCode 829. Consecutive Numbers Sum--笔试题--C++解法_consecutive numbers题解-优快云博客

本文链接：https://blog.youkuaiyun.com/zhangpeterx/article/details/100735382

本文详细解析了LeetCode 829题连续整数求和的多种C++解法，包括暴力穷举、等差数列应用及优化算法，探讨了如何有效计算一个正整数可以表示为连续正整数之和的方式数量。

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LeetCode 829. Consecutive Numbers Sum–笔试题–C++解法

LeetCode题解专栏：LeetCode题解
LeetCode 所有题目总结：LeetCode 所有题目总结
大部分题目C++，Python，Java的解法都有。

题目地址：Consecutive Numbers Sum - LeetCode

Given a positive integer N, how many ways can we write it as a sum of consecutive positive integers?

Example 1:

Input: 5
Output: 2
Explanation: 5 = 5 = 2 + 3

Example 2:

Input: 9
Output: 3
Explanation: 9 = 9 = 4 + 5 = 2 + 3 + 4

Example 3:

Input: 15
Output: 4
Explanation: 15 = 15 = 8 + 7 = 4 + 5 + 6 = 1 + 2 + 3 + 4 + 5

Note: 1 <= N <= 10 ^ 9.

这道题目是2019-9-10的XX笔试原题，可惜我没有事先看到，听别人在牛客网上说才知道的。

这道题目最容易想到的解法就是暴力穷举，时间复杂度O(N^2)，肯定超时。
使用等差数列，可以快一些：
C++解法如下：

class Solution {
public:
    int consecutiveNumbersSum(int N) {
        int result = 1;
        int a = N;
        int temp;
        for (int i = 1; i < a / 2 + 1; i++) {
            for (int j = i + 1; j <= a - i; j++) {
                temp = (i + j) * (j - i + 1) / 2;
                if (temp == a) {
                    result += 1;
                    break;
                } else if (temp > a) {
                    break;
                }
            }
        }
        return result;
    }
};

但上面的做法会超时，换一个思路，首项是x，有m项，那么[x + (x + (m-1))]m / 2->x*m+m*(m-1)/2=n，只要遍历m从1到sqrt(n)即可。

C++解法如下：

class Solution {
public:
    int consecutiveNumbersSum(int N) {
        int ans = 0;
        for (int m = 1;; m++) {
            int m_x = N - m * (m - 1) / 2;
            if (m_x <= 0)
                break;
            if (m_x % m == 0)
                ans++;
        }
        return ans;
    }
};

更快的C++解法如下：

class Solution {
public:
    int consecutiveNumbersSum(int N) {
        int sum = 0, ans = 0;
        for(int i = 1; sum < N; i++) {
            sum += i;
            if (((N-sum) % i) == 0)
                ans++;
        }
        return ans;
    }
};

思路是：
let N = k + (k+1) + (k+2) + (k+3) + … + (k+i-1) = ik + (1+2+3+… + i-1)
let sum(i) = (1+2+3+…+i-1), then we have
N = sum(i) + ik ==>i*k = N - sum(i)
Which means: for each i, we can calculate N-sum(i). If N-sum(i) can be divided by i, there is an answer with length i.