leetcode-829

题目描述:

给定一个正整数 N，试求有多少组连续正整数满足所有数字之和为 N?

首先，刚开始读错题目了，以为是子集和就行，后来发现是连续子集；想到用子集树，但是超时了

class Solution:
    def consecutiveNumbersSum(self, N: int) -> int:
        self.res = 0
        nums = list(range(1,(N+1)//2+1))
        self.backtrack(nums,N,0,0,[])
        if N<=2:
            return 1
        return self.res+1
        
    def backtrack(self,nums,tar,index,pre,path):
        if tar==0:
            self.res += 1
            return 
        for i in range(index,len(nums)):
            if nums[i]<=tar and (len(path)==0 or nums[i] == path[-1]+1):
                self.backtrack(nums,tar-nums[i],i+1,i,path+[nums[i]])

参考答案：其实不难，开动脑筋想一想，就是一个等差数列问题，把公式写出来，就很好判断

class Solution:
    def consecutiveNumbersSum(self, N: int) -> int:
        
        if N<=2:
            return 1
        res = 0
        for i in range(1,N+1):
            m = N-i*(i-1)/2
            if m <= 0 :
                break
            if m%i==0:
                res += 1

        return res