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	`1`	`+### 3700.Number-of-ZigZag-Arrays-II`
	`2`	`+`
	`3`	`+此题是3699的进阶版,时间复杂度是o(NM)的动态规划是不可行的。考虑到n如此巨大,必然是用到了类似于快速幂的算法。`
	`4`	`+`
	`5`	`+此题的思路还是基于之前的DP思想:`
	`6`	+```
	`7`	`+up[i][x] = sum(down[i-1][y]) for y=x+1,...,m`
	`8`	`+down[i][x] = sum(up[i-1][y]) for y=1,2,...,x-1`
	`9`	+```
	`10`	+突破点在于第i轮的状态只依赖于第i-1轮,即{up[i]和down[i]}是{up[i-1],down[i-1]}的线性组合。因此我们令向量`vec[i]={up[1],up[2],...,up[m],down[1],down[2],...,down[m]}`,它与vec[i-1]之间必然存在转移矩阵T使得`vec[i]=Tvec[i-1]`. 这样的矩阵是一个2m2m的方阵。
	`11`	`+`
	`12`	`+我们不难构造出T的样子:`
	`13`	+```
	`14`	`+............... 0`
	`15`	`+ 1 0`
	`16`	`+ 1 1 0`
	`17`	`+ ...`
	`18`	`+ 1 1 1 .. 1 1 0`
	`19`	`+1 1 1 .. 1 1 0 ...........`
	`20`	`+...`
	`21`	`+1 1 0`
	`22`	`+1 0`
	`23`	`+0`
	`24`	+```
	`25`	`+右上部分体现了up[i]与down[i-1]的关系,左下部分体现了down[i]与up[i-1]的关系。`
	`26`	`+`
	`27`	+我们易知初始状态v[1] = {1,..0,0..,1},就可以得到 `v[n] = TTT...Tv[1]`,其中T要连乘n-1次幂,然后再与向量v[1]想乘,即得最后一轮的v。其向量的所有元素之和就是答案。

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