Commitaa0d535

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Original file line number	Diff line number	Diff line change
`@@ -1,5 +1,8 @@`
`1`	`1`	`package main`
`2`	`2`
	`3`	`+/**`
	`4`	`+双重遍历，可以排除10的倍数`
	`5`	`+*/`
`3`	`6`	`funcselfDividingNumbers(leftint,rightint) []int {`
`4`	`7`	`ret:= []int{}`
`5`	`8`	`fori:=left;i<=right;i++ {`

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Original file line number	Diff line number	Diff line change
`@@ -1,5 +1,20 @@`
`1`	`1`	`package main`
`2`	`2`
	`3`	`+/**`
	`4`	`+首先由于对称性，target是正是负影响不大。`
	`5`	`+当sum-target为偶数，1+...-（sum-target）/2+...+k=target，那么答案依然是k。#2`
	`6`	`+当sum-target为奇数，那么sum-target+1是一个偶数`
	`7`	`+类似#2的证明，1+...-(sum-target+1)/2+...k=target-1`
	`8`	`+此时再考虑k的奇偶性。`
	`9`	`+如果k是偶数并且k>sum-target+1`
	`10`	`+那么1+...-(sum-target+1)/2+....-(k/2)...+k+(k+1)=target`
	`11`	`+由#2相似可证，相当于在1+2....+k+(k+1)减去了sum-target+1和k。`
	`12`	`+等价于sum+（k+1）-sum+target-1-k====>target也就是答案是k+1.#3`
	`13`	`+如果k=sum-target+1，由#3可知依然是k+1.#4`
	`14`	`+如果k是奇数：`
	`15`	`+1+2+...-(sum-target+1)/2.....+k-(k+1)+(k+2)=sum-(sum-target+1)+1=target,`
	`16`	`+因此答案是k+2.#5`
	`17`	`+*/`
`3`	`18`	`funcreachNumber(targetint)int {`
`4`	`19`	`iftarget<0 {`
`5`	`20`	`target=-target`

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