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`@@ -4,7 +4,7 @@`
`4`	`4`
`5`	`5`	`我们可以用动态规划来推论一下。dp1[i]表示i条边组成的总长度为奇数的组合数,dp2[i]表示i条边组成的总长度为偶数的组合数。我们的转移方程是`
`6`	`6`	```
`7`		`-dp1[i] = dp2[i-1];`
`8`		`-dp2[i] = dp1[i-1];`
	`7`	`+dp1[i] = dp2[i-1]+dp1[i-1];`
	`8`	`+dp2[i] = dp1[i-1]+dp2[i-1];`
`9`	`9`	```
`10`	`10`	初始条件是`dp1[1]=dp2[1]=1`,显然会有对任意的i,都有`dp1[i]=dp2[i]`。故i条边组成的总长度为偶数和奇数的组合数一定相等。

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