CF460C Present （二分 + 差分数列）

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CF460C Present （二分 + 差分数列）

Codeforces Round #262 (Div. 2) C C - Present C. Present time limit per test 2 seconds memory limit per test 256 megabytes input standard input output standard output Little beaver is a beginner programmer, so informatics is his favorite subject. Soon…

hdu4970 Killing Monsters （差分数列）

2014多校9 1011 http://acm.hdu.edu.cn/showproblem.php?pid=4970 Killing Monsters Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 131072/131072 K (Java/Others)Total Submission(s): 331 Accepted Submission(s): 198 Problem Description Kingdom Rush…

UVALive 4119 Always an integer （差分数列，模拟）

转载请注明出处: http://www.cnblogs.com/fraud/ ——by fraud Always an integer Time Limit:3000MS Memory Limit:0KB 64bit IO Format:%lld & %llu Combinatorics is a branch of mathematics chiefly concerned with counting discrete objects. For instanc…

UVA - 11478 - Halum（二分+差分约束系统）

Problem UVA - 11478 - Halum Time Limit: 3000 mSec Problem Description You are given a directed graph G(V,E) with a set of vertices and edges. Each edge (i,j) that connects some vertex i to vertex j has an integer cost associated with that edge. Deﬁn…

[CF 295A]Grag and Array[差分数列]

题意: 有数列a[ ]; 操作op[ ] = { l, r, d }; 询问q[ ] = { x, y }; 操作表示对a的[ l, r ] 区间上每个数增加d; 询问表示执行[ x, y ]之间的op. 打印最终数列. 思路: 用两次差分数列, 先处理出对于每个op调用了几次, 再对数列执行操作. #include <cstdio> #include <cstring> #include <algorithm> using namespace std; const i…

JZYZOJ1454 NOIP2015 D2T3_运输计划二分差分数组 lca tarjan 树链剖分

http://172.20.6.3/Problem_Show.asp?id=1454 从这道题我充分认识到我的脑子里好多水orz. 如果知道了这个要用二分和差分写,就没什么思考上的难点了(屁咧你写了一个下午). 下面讲述一下我写这道题的辛酸历程: 我最开始写了个树链剖分+线段树+二分+差分数组,tle了一个点,这完全搞不懂啊,什么鬼啊,为什么啊,不然你告诉我怎么写啊. 然后我去找了a了此题的Lcentury大神,然后他告诉我:“这个题要用tarjan求lca啊,都什么年代了还用树链剖分,tanj…