F - Detachment In a highly developed alien society, the habitats are almost infinite dimensional space. In the history of this planet,there is an old puzzle. You have a line segment with x units’ length representing one dimension.The line segment can…

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…

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…

A B C D E F G H I J K L M O O O O     $\varnothing$     $\varnothing$  $\varnothing$  $\varnothing$   $\varnothing$ [A. Relic Discovery] 签到 #include <bits/stdc++.h> int n; int main() { int T; scanf("%d", &T); while (T--) { scanf("…

A B C D E F G H I J K L M O O O $\varnothing$ $\varnothing$   $\varnothing$ $\varnothing$ $\varnothing$ $\varnothing$      $\varnothing$ [A. Thickest Burger] 签到. [B. Relative atomic mass] 签到 [C. Recursive sequence] $$f[i] = f[i - 1] + 2 * f[i - 2] +…

Detachment Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)Total Submission(s): 570    Accepted Submission(s): 192 Problem Description In a highly developed alien society, the habitats are almost infinite dimensiona…

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…

Convex Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)Total Submission(s): 294    Accepted Submission(s): 220 Problem Description We have a special convex that all points have the same distance to origin point.As y…

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…

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…