A Simple Problem with Integers Time Limit: 5000MS   Memory Limit: 131072K Total Submissions: 47944   Accepted: 14122 Case Time Limit: 2000MS Description You have N integers, A1, A2, ... , AN. You need to deal with two kinds of operations. One type of…

A Simple Problem with Integers Time Limit: 1 Sec  Memory Limit: 256 MB 题目连接 http://poj.org/problem?id=3468 Description You have N integers, A1, A2, ... , AN. You need to deal with two kinds of operations. One type of operation is to add some given nu…

A Simple Problem with Integers Time Limit: 1 Sec  Memory Limit: 256 MB 题目连接 http://poj.org/problem?id=3468 Description You have N integers, A1, A2, ... , AN. You need to deal with two kinds of operations. One type of operation is to add some given nu…

A Simple Problem with Integers   Description You have N integers, A1, A2, ... , AN. You need to deal with two kinds of operations. One type of operation is to add some given number to each number in a given interval. The other is to ask for the sum o…

id=3468">点击打开链接题目链接 A Simple Problem with Integers Time Limit: 5000MS   Memory Limit: 131072K Total Submissions: 63565   Accepted: 19546 Case Time Limit: 2000MS Description You have N integers, A1, A2, ... , AN. You need to deal with two kinds of…

A Simple Problem with Integers Time Limit: 5000MS   Memory Limit: 131072K Total Submissions: 75541   Accepted: 23286 Case Time Limit: 2000MS Description You have N integers, A1, A2, ... , AN. You need to deal with two kinds of operations. One type of…

A Simple Problem with Integers Time Limit: 5000MS   Memory Limit: 131072K Total Submissions: 67511   Accepted: 20818 Case Time Limit: 2000MS Description You have N integers, A1, A2, ... , AN. You need to deal with two kinds of operations. One type of…

题目地址:POJ 3468 打了个篮球回来果然神经有点冲动. . 无脑的狂交了8次WA..竟然是更新的时候把r-l写成了l-r... 这题就是区间更新裸题. 区间更新就是加一个lazy标记,延迟标记,仅仅有向下查询的时候才将lazy标记向下更新.其它的均按线段树的来即可. 代码例如以下: #include <iostream> #include <cstdio> #include <cstring> #include <math.h> #include &l…

题目链接:http://poj.org/problem?id=3468 以前用线段树做过,现在用Splay Tree A了,向HH.kuangbin.cxlove大牛学习了各种Splay各种操作,,,Orz.. Splay Tree的区间操作和线段树的操作差不多,也是保存子树的值,然后懒惰操作,在Rotate()最后维护节点信息的时候,只要Push_Up(y)的,因为x还需要网上旋转到根节点,最后更新下就可以了,并且在下一次Rotate()的时候,还会Push_Down(x)的信息,因此不能Pu…

http://poj.org/problem?id=3468 题目大意: 给你N个数还有Q组操作(1 ≤ N,Q ≤ 100000) 操作分为两种,Q A B 表示输出[A,B]的和   C A B X表示把[A,B]的所有数加上X 思路: 线段树的区间修改..... 昨天晚上改了老半天. 然后关机准备睡觉毕竟今天有实验..去洗个头..突然有灵感..急急忙忙的开电脑改了就对了~哈哈哈 PS:POJ AC 100了~ 因为混迹各个OJ,SO才100 用位运算优化*2  1600+MS,不用2200…