F. Lizard Era: Beginning Time Limit: 1 Sec Memory Limit: 256 MB 题目连接 http://codeforces.com/contest/586/problem/F Description In the game Lizard Era: Beginning the protagonist will travel with three companions: Lynn, Meliana and Worrigan. Overall the…

In the game Lizard Era: Beginning the protagonist will travel with three companions: Lynn, Meliana and Worrigan. Overall the game has nmandatory quests. To perform each of them, you need to take exactly two companions. The attitude of each of the com…

Lizard Era: Beginning 折半之后搜就完事了, 直接存string字符串卡空间, 随便卡卡空间吧. #include<bits/stdc++.h> #define LL long long #define fi first #define se second #define mk make_pair #define PLL pair<LL, LL> #define PLI pair<LL, int> #define PII pair<int, i…

D. Lizard Era: Beginning time limit per test 2 seconds memory limit per test 256 megabytes input standard input output standard output In the game Lizard Era: Beginning the protagonist will travel with three companions: Lynn, Meliana and Worrigan. Ov…

信息检索-科技报告是灰色文献获取渠道有限. 技术论文中因保密需要,会删除关键性技术. AD报告也产生较早,1951年开始出版.现由美国国防技术情报中心(DTIC:Defence Technical Information Center)负责收集整理和出版.报告内容以国防部各个合同户的研究报告为主,其报告号冠以AD,A原先为Armed.D则为Document之意,现在则可理解为入藏报告(Accession Document).AD报告是美国陆海空三军科研机构的报告,也包括公司企业及外国的科研机构和…

折半搜索,先搜索一半的数字,记录第一个人的值,第二个人.第三个人和第一个人的差值,开个map哈希存一下,然后另一半搜完直接根据差值查找前一半的答案. 代码 #include<cstdio> #include<map> #define ll long long #define N 100 using namespace std; map<long long,int> ma,Ma; ]; int n,i,a[N],b[N],c[N],Ans,A1,A2,ans[N]; vo…

嘟嘟嘟 题面我是不会咕的(没有真香):有\(n(n \leqslant 25)\)个任务和三个人,每次任务给出每个人能得到的值,每次任务选两个人,使\(n\)个任务结束后三个人得到的值是一样的,且尽量大.输出每次要派哪两个人,如果不行输出\(Impossible\). 暴力是\(O(3 ^ {25})\),必定过不去,但是如果一半\(O(3 ^ {13})\)就刚好可以过了,因此想到折半搜索. 令搜到的前一半的结果为\(a, b, c\),后一半为\(x, y, z\),那么我们需要的是\(a…

We spend a full five hours and 16 minutes a day in front of a screen, and that's without even turning on a television. So says a statistic from eMarketer, a research firm that focuses on digital media and marketing. It says that for the first time we…

新的博客,将会记录我在工作和学习中遇到的问题以及总结...…

一眼题...这个数据范围也太明显了吧... suma1==suma2 && sumb1==sumb2 && sumc1==sumc2 相当于suma1-sumb1==sumb2-suma2 && suma1-sumc1==sumc2-suma2 于是前一半O(3^(N/2))搜出所有情况的suma1-sumb1和suma1-sumc1,后一半搜出sumb2-suma2和sumc2-suma2,都丢到一个数组里作为两个关键字排序,在两个关键字都相同的一段里面找…