(转)慢慢的,就没有了,就像从未存在过

GFW

几年以前,我曾经嘲笑过某科技界大佬。当时他说:也许90后、95后会慢慢不知道谷歌是什么网站。

  那一年,这对于我来说简直就是世界上最好笑的笑话。谷歌,全世界最卓越的互联网公司,活在互联网的一代中国人,会不知道他们的网站?

  今天,我收回这句嘲笑。因为这件不可能的事,它慢慢变成了现实。

  没有人再关注什么谷歌不谷歌。对他们来说,百度也蛮好用的,反正他们几乎没用过谷歌。没有谷歌又怎样?大家还是开心的刷微博,看微信,听歌,看娱乐节目。对于从来就不知道谷歌的人来说,少了谷歌又有什么影响?

慢慢的,就没有了,就像从未存在过

ZOJ month contest D.Determinant and Matrix

ACM

Time Limit: 2 Seconds Memory Limit: 65536 KB

##Description
Recently, LBH is learning the curse linear algebra. Thus he is very interested in matrix and determinant now. In order to practice his ability of solving the problem of linear algebra, he just invent some problems by himself. Once the problems was create, he would solve it immediately. However, he meet a problem that was so hard that he couldn’t work out even though racked his brains. The problem was described as follow:

To a integer martix Mnn(aij), we define two function add(Mnn(aij))=Mnn(aij + 1) and sub(Mnn(aij))=Mnn(aij - 1) which were exactly like this:

FFT求快速卷积的思考

ACM

离散型卷积的定义是:$$y(n)=\sum_{m=0}^{n} x(m)h(n-m)$$

注意,h函数是反转的。

在Chipher Messages一题中,b串需要反转再与a串匹配。

比如说:

a串: 110110110,则:

b`串:1011<——这里才是原来b串的头。但是向上对应到a串时,已经是m-1这个位置了。所以说,小于m-1的卷积是没有意义的。

于是,base=m。整体匹配。

baylor 6622 Absurdistan Roads( NWERC Contest)

ACM

原题pdf:click here

##Description
The people of Absurdistan discovered how to build roads only last year. After the discovery, every city
decided to build their own road connecting their city with another city. Each newly built road can be
used in both directions.

Absurdistan is full of surprising coincidences. It took all N cities precisely one year to build their
roads. And even more surprisingly, in the end it was possible to travel from every city to every other
city using the newly built roads.