G. Lpl and Energy-saving Lamps 42.07% 1000ms 65536K   During tea-drinking, princess, amongst other things, asked why has such a good-natured and cute Dragon imprisoned Lpl in the Castle? Dragon smiled enigmatically and answered that it is a big secre…
题目链接:https://nanti.jisuanke.com/t/30996 中文题目: 在喝茶的过程中,公主,除其他外,问为什么这样一个善良可爱的龙在城堡里被监禁Lpl?龙神秘地笑了笑,回答说这是个大秘密.暂停后,龙补充道: - 我们有合同.租赁协议.他总是一整天都在工作.他喜欢沉默.除此之外,住在城堡还有更多的优势.比如说,很容易证明未接来电的合理性:电话铃声无法从手机离开的城堡的另一侧到达.因此,监禁只是一个故事.实际上,他思考一切.他很聪明.例如,他开始用整个城堡中的节能灯替换白炽灯.…
https://nanti.jisuanke.com/t/30996 题意 每天增加m个灯泡,n个房间,能一次性换就换,模拟换灯泡过程.询问第几天的状态 分析 离线做,按题意模拟.比赛时线段树写挫了..导致不断超时,我太弱了.每次询问符合要求的最左边的点. #include <iostream> #include <cstdio> #include <cstdlib> #include <cstring> #include <string> #i…
During tea-drinking, princess, amongst other things, asked why has such a good-natured and cute Dragon imprisoned Lpl in the Castle? Dragon smiled enigmatically and answered that it is a big secret. After a pause, Dragon added: — We have a contract.…
线段树节点维护区间最小值,查找时优先从左侧的区间寻找. 每一次循环都在树中不停寻找第一个小于等于当前持有数的值,然后抹去,直到找不到为止. #include<cstdio> #include<iostream> #include<algorithm> #include<cstring> #define lson rt<<1 #define rson rt<<1|1 #define Lson l,m,lson #define Rson…
题目链接:https://nanti.jisuanke.com/t/30996 During tea-drinking, princess, amongst other things, asked why has such a good-natured and cute Dragon imprisoned Lpl in the Castle? Dragon smiled enigmatically and answered that it is a big secret. After a pau…
J. Sum 26.87% 1000ms 512000K   A square-free integer is an integer which is indivisible by any square number except 11. For example, 6 = 2 \cdot 36=2⋅3 is square-free, but 12 = 2^2 \cdot 312=22⋅3 is not, because 2^222 is a square number. Some integer…
A. An Olympian Math Problem 54.28% 1000ms 65536K   Alice, a student of grade 66, is thinking about an Olympian Math problem, but she feels so despair that she cries. And her classmate, Bob, has no idea about the problem. Thus he wants you to help him…
M. Big brother said the calculation 通过线段树维护. 这个题和杭电的一道题几乎就是一样的题目.HDU5649.DZY Loves Sorting 题意就是一个n的排列,执行Q次操作,每次操作是对某个区间从小到大排序或者从大到小排序.最后只查询一次,输出第k个位置当前的数. 直接按HDU5649这个题写了. 因为只查询一次,而且这是n的全排列,所以直接二分答案,比mid小的赋值为0,大的赋值为1.区间查询判断的时候直接与0和1比较就可以了. 这个题写的简直要骂人…
题意:n个房间,每个房间有ai盏旧灯,每个月可以买m盏新灯,要求:按房间顺序换灯,如果剩下的新灯数目大于ai,那么进行更换,否则跳过该房间,判断下一个房间.如果所有房间都换完灯,那么久不会再买新灯. q次询问,每次询问该月已换的房间数以及剩余的新灯数量. 题解:线段树记录最小值,然后一遍模拟 #include<bits/stdc++.h> using namespace std ; #define lson l,mid,rt<<1 #define rson mid+1,r,rt&l…