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
.
题目理解:
将一个数字分解成一段连续自然数的和
解题思路:
连续自然数的和其实就是一个公差为1的等差数列的和,设这个数列是x,x+1,x+2,...,x+n,那么这个数列的和是(n+1)*x+n*(n+1)/2,这个数列的长度是n+1,第一个数字是x,然后就可以从数字长度为1开始逐渐尝试,看N能否分解成为长度为n的等差数列的和,如果可以,那么一定可以求得整数解x使得x=(N-n*(n+1))/(n+1)
代码如下:
class Solution {
public int consecutiveNumbersSum(int N) {
int res = 0;
int n = 0;
while(true) {
int top = N - (n * n + n) / 2;
if(top <= 0)
break;
if(top % (n + 1) == 0) {
res++;
//System.out.println(n);
}
n++;
}
return res;
}
}