Problem Description The inversion number of a given number sequence a1, a2, ..., an is the number of pairs (ai, aj) that satisfy i < j and ai > aj. For a given sequence of numbers a1, a2, ..., an, if we move the first m >= 0 numbers to the end of
Balanced Lineup Time Limit: 5000MS Memory Limit: 65536K Total Submissions: 36813 Accepted: 17237 Case Time Limit: 2000MS Description For the daily milking, Farmer John's N cows (1 ≤ N ≤ 50,000) always line up in the same order. One day Farmer John de
Group Time Limit: 4000/2000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 1323 Accepted Submission(s): 703 Problem Description There are n men ,every man has an ID(1..n).their ID is unique. Whose ID is i and i-
原文地址:http://www.cnblogs.com/GXZlegend/p/6809743.html 题目描述 设计数据结构支持:1 x 若x不存在,插入x2 x 若x存在,删除x3 输出当前最小值,若不存在输出-14 输出当前最大值,若不存在输出-15 x 输出x的前驱,若不存在输出-16 x 输出x的后继,若不存在输出-17 x 若x存在,输出1,否则输出-1 输入 第一行给出n,m 表示出现数的范围和操作个数接下来m行给出操作n<=10^6,m<=2*10^6,
此题题意很好懂: 给你N个数,Q个操作,操作有两种,‘Q a b ’是询问a~b这段数的和,‘C a b c’是把a~b这段数都加上c. 需要用到线段树的,update:成段增减,query:区间求和 介绍Lazy思想:lazy-tag思想,记录每一个线段树节点的变化值,当这部分线段的一致性被破坏我们就将这个变化值传递给子区间,大大增加了线段树的效率. 在此通俗的解释我理解的Lazy意思,比如现在需要对[a,b]区间值进行加c操作,那么就从根节点[1,n]开始调用update函数进行操作,如果