瞎搞题啊.找出1 1 0 0这样的序列,然后存起来,这样的情况下最好的选择是1的个数除以这段的总和. 然后从前向后扫一遍.变扫边进行合并.每次合并.合并的是他的前驱.这样到最后从t-1找出的那条链就是最后满足条件的数的大小. Room and Moor Time Limit: 12000/6000 MS (Java/Others) Memory Limit: 262144/262144 K (Java/Others) Total Submission(s): 307 Accepted…
链接:https://www.nowcoder.com/acm/contest/104/B来源:牛客网 题意:A few days ago, WRD was playing a small game called Salty Fish Go. We can simplify the rules of the game as follows. 给你v,l,n,m代表有v个速度,下一行分别给出v个速度,l长的路,n个随机地点随机改变速度的加油站,m个随机地点宝藏.问拿到所有宝藏的时间期望. 题解:瞎…
Problem Description PM Room defines a sequence A = {A1, A2,..., AN}, each of which is either 0 or 1. In order to beat him, programmer Moor has to construct another sequence B = {B1, B2,... , BN} of the same length, which satisfies that: Input The i…
传送门 Room and Moor Time Limit: 12000/6000 MS (Java/Others) Memory Limit: 262144/262144 K (Java/Others)Total Submission(s): 1288 Accepted Submission(s): 416 Problem Description PM Room defines a sequence A = {A1, A2,..., AN}, each of which is eit…