思路: 题目链接http://poj.openjudge.cn/practice/C18A/ 先说一个结论,每一天要么7要么0,由此提供一种状态压缩dp的解法. 实现: #include <bits/stdc++.h> using namespace std; , INF = 0x3f3f3f3f; ][ << ]; int main() { int t, n; cin >> t; while (t--) { cin >> n; << ) - ;…

方法一:像哈利波特一样穿墙而出 无论是wife信号还是广播信号本质上都属于电磁波.x光穿透力强所以可以穿透人体给体内照相,但是wife信号作为电磁波虽然也可以穿透墙而过,但是他的穿透能力实在是太弱了.电磁波频率越高他的穿透力越强,我们常用的wife信号只有2.4赫兹.而同样是电磁波的可见光频率是wife的十几万倍,更不用说x光了.所以wife信号穿墙而过,会被墙里的钢筋和水泥吸收很多. 方法二:绕道走 wife作为一种波,在遇见障碍物会发生干涉和衍射现象.也就是绕道走.虽然绕道走距离远,但是相对…

How I explained OOD to my wife Learning Object Oriented Design principles through interesting conversations. Introduction My wife Farhana wants to resume her career as a software developer (she started her career as a software developer, but couldn't…

After reading The Time Traveller's Wife:      It's a tragedy,I think.But it's mixed with hope.Henry is dead.but he can time travel to the years after his death and see Clare again.So it might be a beautiful and imaginary tragedy.      The key of the…

第一季 The Good Wife Season 1 (2009)看点:在经受丈夫Peter的背叛以及因此而带来的公众羞辱后,Alicia Florrick选择重新继续自己原来的事业,一名辩护律师,以开始自己新的生活.她投靠了她以前的同学Will Gardner一起成为了芝加哥一家律师事务所的初级合伙人,而她的前同学则很想看看她在13年之后如何重新走上法庭. Alicia非常感谢事务所的首席律师Diane Lockhart给她提供了这个机会并让她明白她必须靠自己的努力来取得成功.Alicia的主…

I want a wife who will take care of my physical needs. I want a wife who will keep my house clean. A wife who will pick up after my children, a wife who will pick up after me. I want a wife who will keep my clothes clean, ironed, mended, replaced whe…

Introduction Me and my wife had some interesting conversations on Object Oriented Design principles. After publishing the conversation on CodeProject, I got some good responses from the community and that really inspired me. So, I am happy to share o…

Introduction My wife Farhana wants to resume her career as a software developer (she started her career as a software developer, but couldn't proceed much because of our first child's birth), and these days, I am trying to help her learn Object Orien…

题目:http://poj.openjudge.cn/campus2018/A 有正规的差分约束做法,用到矩阵转置等等. 但也有简单(?)的dp做法. 有一个结论(?):一定要么在一天一点也不选,要么在一天选了7个小时. 于是dp[ i ]表示第 i 天选了7个小时.之前合法 的方案数.可以从i-1到i-7转移.答案是n到n-6. #include<iostream> #include<cstdio> #include<cstring> #define ll long…

题目:http://poj.openjudge.cn/practice/C18A/ 据说正解是差分约束,转化的过程还要用到标准型.对偶型什么的知识,暂时还不太懂... 但也有贪心DP做法,有个结论:一个点要么没有,要么有7个小时(但不太会证明...): 然后就DP即可. 代码如下: #include<iostream> #include<cstdio> #include<cstring> using namespace std; typedef long long ll…