Count the Sheep 题意: 问题描述 开学翘课固然快乐,然而也有让呃喵抓狂的事,那当然就是考试了!这可急坏了既要翘课又想要打BC还要准备考试的呃喵. 呃喵为了准备考试没有时间刷题,想打BC又不想跌分,只得求助于BCround92的出题人snowy_smile,让他说点什么 ~~>_<~~. snowy_smile实在没有办法,但是又不好意思透题,只好告诉呃喵,当务之急是好好休息. "如果你按照下面这个办法睡着,那么第二天就绝对不会在BC的赛场上跌分—— 想象一片一望无际.…

Count the Sheep Time Limit: 3000/1500 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others) Total Submission(s): 686    Accepted Submission(s): 295 Problem Description Altough Skipping the class is happy, the new term still can drive luras an…

Count the Sheep Time Limit: 3000/1500 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others) Total Submission(s): 484    Accepted Submission(s): 212 Problem Description Altough Skipping the class is happy, the new term still can drive luras an…

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=6016 题意:给定男羊和女羊的朋友关系,即给定一个图,问从任意一只羊开始连续数四只不相同的羊的方法数. 这题挺简单的就是需要动一下脑子,我们只要找到关系A-B-C-D即可,且关系址建立在男女羊之间. 就是说我们只要找到 男羊-女羊-男羊-女羊,或者 女羊-男羊-女羊-男羊 即可. 于是那只要找到任意一种方式的总数然后乘2即可. 拿后一种关系为例. ans[i]表示i号女羊有几个男羊朋友vc[i]存储与…

Altough Skipping the class is happy, the new term still can drive luras anxious which is of course because of the tests! Luras became worried as she wanted to skip the class, as well as to attend the BestCoder and also to prepare for tests at the sam…

HDU 6187 Destroy Walls (思维,最大生成树) Destroy Walls *Time Limit: 8000/4000 MS (Java/Others) Memory Limit: 132768/132768 K (Java/Others) Total Submission(s): 1784 Accepted Submission(s): 692 * Problem Description Long times ago, there are beautiful histor…

题目 以下不是KMP算法—— 以下是kiki告诉我的方法,好厉害的思维—— 就是巧用标记,先标记第一个出现的所有位置,然后一遍遍从标记的位置往下找. #include<stdio.h> #include<string.h> #include<algorithm> using namespace std; int main() { ],shunxu; ]; scanf("%d",&t); while(t--) { memset(xiabiao,…

Count the string Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 8845    Accepted Submission(s): 4104 Problem Description It is well known that AekdyCoin is good at string problems as well as n…

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=6161 题意: 题目是给一棵完全二叉树,从上到下从左到右给每个节点标号,每个点有权值,初始权值为其标号,然后有两种操作: 1.把u点权值改为x 2.查询所有经过u点的路径中,路径上的点权和最大. 节点有n个,修改有m个,n<=1e8 ,m<= 1e5 解法:现场队友过的,orz,来自队友的思路. 我们首先对于一个点,如果没有访问我们不把它建出来,相反访问了就把它建出来,这个题的最小的子问题就是计算一…

Count primes 题目连接: http://acm.split.hdu.edu.cn/showproblem.php?pid=5901 Description Easy question! Calculate how many primes between [1...n]! Input Each line contain one integer n(1 <= n <= 1e11).Process to end of file. Output For each case, output…