Rabbits Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 262144/262144 K (Java/Others)Total Submission(s): 1193    Accepted Submission(s): 628 Problem Description Here N (N ≥ 3) rabbits are playing by the river. They are playing on a number li…

整理代码... Little Boxes Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 262144/262144 K (Java/Others)Total Submission(s): 2304    Accepted Submission(s): 818 Problem Description Little boxes on the hillside.Little boxes made of ticky-tacky.Littl…

HDU 6222 Heron and His Triangle 链接:http://acm.hdu.edu.cn/showproblem.php?pid=6222 思路: 打表找规律+大数运算 首先我们可以打个表暴力跑出前几个满足题意的T 打表代码: #include<bits/stdc++.h> using namespace std; bool fun(double n) { if(abs(round(n) - n) < 0.000000000000001) ; ; } int ma…

Little Boxes Problem Description Little boxes on the hillside.Little boxes made of ticky-tacky.Little boxes.Little boxes.Little boxes all the same.There are a green boxes, and b pink boxes.And c blue boxes and d yellow boxes.And they are all made out…

今天做的沈阳站重现赛,自己还是太水,只做出两道签到题,另外两道看懂题意了,但是也没能做出来. 1. Thickest Burger Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 0    Accepted Submission(s): 0 Problem Description ACM ICPC is launching a thic…

C.Recursive sequence 求ans(x),ans(1)=a,ans(2)=b,ans(n)=ans(n-2)*2+ans(n-1)+n^4 如果直接就去解...很难,毕竟不是那种可以直接化成矩阵的格式,我们也因为这个被卡很长时间 事实上可以把这道式子化成几个基本元素的格式,然后就容易组合了,比如ans(n-2)*2+ans(n-1)+(n-1)^4+4*(n-1)^3+6*(n-1)^2+4*(n-1)^1+1 包含了所有的基本组成形式,化绝对为相对,并且除了一个n-2其他都是n…

废话: 这道题很是花了我一番功夫.首先,我不会kmp算法,还专门学了一下这个算法.其次,即使会用kmp,但是如果暴力枚举的话,还是毫无疑问会爆掉.因此在dfs的基础上加上两次剪枝解决了这道题. 题意: 我没有读题,只是队友给我解释了题意,然后我根据题意写的题. 大概意思是给n个字符串,从上到下依次标记为1——n,寻找一个标记最大的串,要求这个串满足:标记比它小的串中至少有一个不是它的子串. 输入: 第一行输入一个整型t,表示共有t组数据. 每组数据首行一个整型n,表示有n个串. 接下来n行,每行…

n pagodas were standing erect in Hong Jue Si between the Niushou Mountain and the Yuntai Mountain, labelled from 1 to n. However, only two of them (labelled aand b, where 1≤a≠b≤n) withstood the test of time. Two monks, Yuwgna and Iaka, decide to make…

题意:有\(n\)个点,\(m\)个集合,集合\(E_i\)中的点都与集合中的其它点有一条边权为\(t_i\)的边,现在问第\(1\)个点和第\(n\)个点到某个点的路径最短,输出最短路径和目标点,如果不满足条件则输出\(Evil John\). 题解:题目所给的边数关系太复杂了,我们可以让每个集合中的所有点都与一个虚拟节点连边,而这些点两两却不连,然后再去找\(1\)个和第\(n\)个点的最短路径,不难发现,最终得到的路径为\(dis[i]/2\),所以我们只要用虚拟节点建边然后跑两次dijk…

题意:给你\(n\)个字符串,\(s_1,s_2,...,s_n\),对于\(i(1\le i\le n)\),找到最大的\(i\),并且满足\(s_j(1\le j<i)\)不是\(s_i\)的子串. 题解:直接\(O(n^2)\)然后跑kmp匹配,这里注意要剪枝,不然会T,也就是说对于前\(i-1\)个串,如果它是后面某个串的子串,那么我们就不用对它跑kmp,因为它后面的某个串包含了它. 代码: int t; int n; string s[N]; int ne[N]; bool st[N]…

题意:有\(n\)个数,开始给你两个数\(a\)和\(b\),每次找一个没出现过的数\(i\),要求满足\(i=j+k\)或\(i=j-k\),当某个人没有数可以选的时候判他输,问谁赢. 题解:对于\(a\)和\(b\),我们能有他两得到的最小数一定是\(d=gcd(a,b)\),所以总共能选的数的个数为\(n/d\),判断奇偶即可. 代码: int t; int n,a,b; int main() { ios::sync_with_stdio(false);cin.tie(0);cout.ti…

1001题意:n个人,给m对敌对关系,X个好人,Y个坏人.现在问你是否每个人都是要么是好人,要么是坏人. 先看看与X,Y个人有联通的人是否有矛盾,没有矛盾的话咋就继续遍历那些不确定的人关系,随便取一个数3,与其相连的就是4,间隔就要相同,dfs搜过去就可以判断了 #include<bits/stdc++.h> using namespace std; #pragma comment(linker, "/STACK:102400000,102400000") #define…

题目链接:http://acm.hdu.edu.cn/search.php?field=problem&key=2016ACM%2FICPC%D1%C7%D6%DE%C7%F8%B4%F3%C1%AC%D5%BE-%D6%D8%CF%D6%C8%FC%A3%A8%B8%D0%D0%BB%B4%F3%C1%AC%BA%A3%CA%C2%B4%F3%D1%A7%A3%A9&source=1&searchmode=source A.染色乱搞. #include <bits/stdc…

HDU 6225 Little Boxes 题意 计算四个整数的和 解题思路 使用Java大整数 import java.math.BigInteger; import java.util.Scanner; /** * * @author reqaw */ public class Main { /** * @param args the command line arguments */ public static void main(String[] args) { Scanner in =…

任意门:http://acm.hdu.edu.cn/showproblem.php?pid=5979 按AC顺序: I - Convex Time limit    1000 ms Memory limit 65536 kB OS Windows We have a special convex that all points have the same distance to origin point. As you know we can get N segments after linki…

题面 题意:给你一堆点,求一个最大面积的空凸包,里面没有点. 题解:红书板子,照抄完事,因为题目给的都是整点,所以最后答案一定是.5或者.0结尾,不用对答案多做处理 #include<bits/stdc++.h> #define N 55 using namespace std; struct rec { double x,y; }; rec operator -(rec a,rec b) { rec c; c.x=a.x-b.x; c.y=a.y-b.y; return c; } doubl…

Recursive sequence Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)Total Submission(s): 249    Accepted Submission(s): 140 Problem Description Farmer John likes to play mathematics games with his N cows. Recently, t…

Counting Cliques Time Limit: 8000/4000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)Total Submission(s): 539    Accepted Submission(s): 204 Problem Description A clique is a complete graph, in which there is an edge between every pair…

Thickest Burger Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)Total Submission(s): 63    Accepted Submission(s): 60 Problem Description ACM ICPC is launching a thick burger. The thickness (or the height) of a piec…

Relative atomic mass Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others) Total Submission(s): 66    Accepted Submission(s): 59 Problem Description Relative atomic mass is a dimensionless physical quantity, the ratio of…

Pagodas Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)Total Submission(s): 1282    Accepted Submission(s): 902 Problem Description n pagodas were standing erect in Hong Jue Si between the Niushou Mountain and the…

Bazinga Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)Total Submission(s): 3509    Accepted Submission(s): 1122 Problem Description Ladies and gentlemen, please sit up straight.Don't tilt your head. I'm serious.Fo…

http://blog.csdn.net/queuelovestack/article/details/53055418 下午重现了一下大连赛区的比赛,感觉有点神奇,重现时居然改了现场赛的数据范围,原本过的人数比较多的题结果重现过的变少了,而原本现场赛全场过的人最少的题重现做出的人反而多了一堆,不过还是不影响最水的6题,然而残酷的现实是6题才有机会拿铜...(╥╯^╰╥) 链接→2016ACM/ICPC亚洲区大连站-重现赛  Problem 1001 Wrestling Match Accept…

题目链接 http://acm.hdu.edu.cn/showproblem.php?pid=4803 Problem Description Jenny is a warehouse keeper. He writes down the entry records everyday. The record is shown on a screen, as follow:There are only two buttons on the screen. Pressing the button i…

摘要 本文主要列举并求解了2016 ACM/ICPC亚洲区青岛站现场赛的部分真题,着重介绍了各个题目的解题思路,结合详细的AC代码,意在熟悉青岛赛区的出题策略,以备战2018青岛站现场赛. HDU 5984 Pocky 题意 给出一根棒子(可以吃的)的长度x和切割过程中不能小于的长度d,每次随机的选取一个位置切开,吃掉左边的一半,对右边的棒子同样操作,直至剩余的长度不大于d时停止.现在给出x和d,问切割次数的数学期望是多少. 解题思路 当看到第二个样例2 1时,结果是1.693147,联想到ln…

GPA http://acm.hdu.edu.cn/showproblem.php?pid=4802 签到题,输入两个表,注意细心点就行了. #include<cstdio> #include<cstring> ; char s[M],cp[M][M]={"A","A-","B+","B","B-","C+","C","C-"…

链接:传送门 A - Thickest Burger - [签到水题] ACM ICPC is launching a thick burger. The thickness (or the height) of a piece of club steak is A (1 ≤ A ≤ 100). The thickness (or the height) of a piece of chicken steak is B (1 ≤ B ≤ 100). The chef allows to add…

5510 Bazinga 题意:给出n个字符串,求满足条件的最大下标值或层数 条件:该字符串之前存在不是 它的子串 的字符串 求解si是不是sj的子串,可以用kmp算法之类的. strstr是黑科技,比手写的kmp快.if(strstr(s[i], s[j]) == NULL),则Si不是Sj的子串. 还有一个重要的剪枝:对于一个串,如果当前找到的串是它的母串,则下一次这个串不用遍历. #include <set> #include <queue> #include <cst…

为了这个题解第一次写东西..(我只是来膜拜爱看touhou的出题人的).. 首先以为对称性质..我们求出露琪诺的魔法值的期望就可以了..之后乘以3就是答案..(话说她那么笨..能算出来么..⑨⑨⑨⑨⑨⑨) 用dp表示方法数... 首先状态如此表示: 设dp(i,j,k)其中i代表节点的标号..j代表状态(就像官方题解一样..0表示这个颜色不选,1代表选而且和子节点形成的联通块的节点数是奇数,2代表偶数)...k代表x-y的值.. 这样的话..递推方程就能推咯..然后慢慢把子节点dfs出来后加到根…

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=4811 题目描述: Problem Description Jenny likes balls. He has some balls and he wants to arrange them in a row on the table. Each of those balls can be one of three possible colors: red, yellow, or blue. More…