B. Light bulbs(2019 ICPC上海站)】的更多相关文章

There are NN light bulbs indexed from 00 to N-1N−1. Initially, all of them are off. A FLIP operation switches the state of a contiguous subset of bulbs. FLIP(L, R)FLIP(L,R)means to flip all bulbs xx such that L \leq x \leq RL≤x≤R. So for example, FLI…
题目链接:Light bulbs 比赛链接:The Preliminary Contest for ICPC Asia Shanghai 2019 题意 给定 \(N\) 个灯泡 (编号从 \(0\) 到 \(N - 1\)),初始都是关闭的. 给定 \(M\) 个操作,每个操作包含 \(L\) 和 \(R\),对 \([L, R]\) 内的所有灯泡改变状态. 求最后有几个灯泡是亮的. 思路 题目挺简单的,翻转奇数次的灯泡是亮的,所以要求每个灯泡翻转的次数. 容易想到可以用差分. 对所有操作的两…
Light Bulbs Time Limit: 2 Seconds      Memory Limit: 65536 KB Wildleopard had fallen in love with his girlfriend for 20 years. He wanted to end the long match for their love and get married this year. He bought a new house for his family and hired a…
MS    Memory Limit:65536KB    64bit IO Format:%lld & %llu SubmitStatusid=14946">Practiceid=14946">ZOJ 2976 Description Wildleopard had fallen in love with his girlfriend for 20 years. He wanted to end the long match for their love and…
2019 ICPC 南昌网络赛 比赛时间:2019.9.8 比赛链接:The 2019 Asia Nanchang First Round Online Programming Contest 总结 // 史上排名最高一次,开场不到两小时队友各A一题加水题共四题,排名瞬间升至三四十名 // 然后后三小时就自闭了,一题都没有突破...最后排名211 hhhh     B. Fire-Fighting Hero 题意 队友做的,待补.   AC代码 #include<cstdio> #includ…
B. Light bulbs There are NNN light bulbs indexed from 000 to N−1N-1N−1. Initially, all of them are off. A FLIP operation switches the state of a contiguous subset of bulbs. FLIP(L,R)FLIP(L, R)FLIP(L,R) means to flip all bulbs xxx such that L≤x≤RL \le…
2019 ICPC Asia Nanjing Regional A - Hard Problem 计蒜客 - 42395 若 n = 10,可以先取:6,7,8,9,10.然后随便从1,2,3,4,5里面选一个都肯定符合题意 若 n = 9,可以先取:5,6,7,8,9,然后随便从1,2,3,4里面选一个都肯定符合题意. 所以答案就是后半部分的数量+1 #include <cstdio> #include <iostream> #include <cmath> #inc…
题目:https://nanti.jisuanke.com/t/41399 思路:差分数组 区间内操作次数为奇数次则灯为打开状态 #include<bits/stdc++.h> using namespace std; map<int,int>mp; int main() { int T; scanf("%d",&T); int n,m; int l,r; ;i<=T;i++) { mp.clear(); scanf("%d%d"…
复杂度分析,询问一千次,区间长1e6,O(1e9)超时. 那么我们知道对于差分来说,没必要一个一个求,只需要知道区间长就可以了,所以我们定义结构体差分节点,一个头结点,一个尾节点. 这样tail.loc-head.loc就是整个区间长,该区间的实际的大小就是 Add标记的大小,Add标记就是从头加到尾. 排序2*m个差分节点,对于loc相同的节点,说明是同一个位置的差分,add合并就可以了,不需要进行统计,直接跳过本次循环. #include <bits/stdc++.h> using nam…
https://nanti.jisuanke.com/t/41399 题目大意: 有n个灯,m次操作,每次修改[l,r]内的灯,(off - on ,on - off),问最后有几盏灯亮着. 换种说法:n个点m个区间,每次区间内的数+1,最后n个点中计数为奇数的点的个数就是答案. 刚开始没注意,直接用线段树写,超内存了.... 这题因为外层有个T,并且n太大,还要卡内存,太过分了. 卡数据卡内存,每组样例一重循环都会超时.所以可以分块和对m处理来做. 对m处理的话,仔细想一想,只有区间次数被操作…