K - K Crawling in process... Crawling failed Time Limit:1000MS Memory Limit:32768KB 64bit IO Format:%I64d & %I64u Submit Status Practice HDU 1875 Description 相信大家都听说一个“百岛湖”的
H - H Crawling in process... Crawling failed Time Limit:1000MS Memory Limit:65536KB 64bit IO Format:%I64d & %I64u Submit Status Practice POJ 3070 Description In the Fibonac
L - L Crawling in process... Crawling failed Time Limit:1000MS Memory Limit:32768KB 64bit IO Format:%I64d & %I64u Submit Status Practice HDU 1879 Description 省政府“畅通工程”的目标是使
生日猜猜猜 Time Limit: 3000/1000 MS (Java/Others) Memory Limit: 65535/32768 K (Java/Others) Total Submission(s): 442 Accepted Submission(s): 167 Problem Description 小明对生日十分看重,因为他可以得到祝福,可以和朋
Problem Description 有一棵 n 个节点的石榴树,点的下标为 1…n,下标为 m 的节点上有一颗石榴。石榴树有 2n−2 条有向边,每条有向边都连接两个不同的节点,不存在完全相同的两条有向边,且如果从节点 a 到节点 b 的有向边存在,反过来从节点 b 到节点 a 的有向边也必定存在。从任意一个节点都能(经过这些有向边中的若干条)到达任意一个其它节点。 现在有两只小蚂蚁去找石榴
题意: 一个小球在一个等边三角形内碰撞,碰撞速度不比,方向沿边镜像翻折。求发送 k k k次碰撞需要多少时间 思路: 一开始想着是模拟,然后估摸着最后会形成循环,经过起始点。但是太难模拟了。。 看了题解发现,真的特别巧妙。反射意味着穿过!那么就成了全是等边三角形铺成的平面,已知起点和速度。求经过 k k k条边的最短时间。 这个时间可以二分。 仅考虑平行x轴的边,那就直接用 a b s ⌊
Problem B: 题目描述 麻雀帕西和青蛙弗洛格是好玩伴,它们经常一起比赛唱歌。但冬天来了,青蛙弗洛格冬眠了,它的睡眠深度是D。麻雀帕西觉得好无聊,于是它想办法要唤醒弗洛格。麻雀帕西只会唱N首歌,第i首歌的音量是Si。每听完一首歌,青蛙弗洛格的睡眠深度就会减少,减少的值等于它听到的歌的音量。当青蛙弗洛格的睡眠深度大于0的时候,它会继续冬眠,当睡眠深度小于或者等于0时,它就会被唤醒了。麻雀帕西
题目描述 Eddy likes to play cards game since there are always lots of randomness in the game. For most of the cards game, the very first step in the game is shuffling the cards. And, mostly the randomnes
题目描述 Eddy was a contestant participating in ACM ICPC contests. ACM is short for Algorithm, Coding, Math. Since in the ACM contest, the most important knowledge is about algorithm, followed by coding(
链接:https://www.nowcoder.com/acm/contest/141/H 来源:牛客网 Eddy has solved lots of problem involving calculating the number of coprime pairs within some range. This problem can be solved with inclusion-e
链接:https://www.nowcoder.com/acm/contest/141/E 来源:牛客网 Eddy likes to play with string which is a sequence of characters. One day, Eddy has played with a string S for a long time and wonders how could
链接:https://www.nowcoder.com/acm/contest/141/A 来源:牛客网 Eddy was a contestant participating in ACM ICPC contests. ACM is short for Algorithm, Coding, Math. Since in the ACM contest, the most important