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Jul 2, 2025
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[pull] master from wisdompeak:master #335

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4 changes: 2 additions & 2 deletions Binary_Search/3559.Number-of-Ways-to-Assign-Edge-Weights-II/Readme.md
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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
```
dp1[i] = dp2[i-1];
dp2[i] = dp1[i-1];
dp1[i] = dp2[i-1]+dp1[i-1];
dp2[i] = dp1[i-1]+dp2[i-1];
```
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。