POJ 3468 A Simple Problem with Integers(线段树/区间更新)
题目链接: 传送门
A Simple Problem with Integers
Time Limit: 5000MS Memory Limit: 131072K
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 of numbers in a given interval.
Input
The first line contains two numbers N and Q. 1 ≤ N,Q ≤ 100000.
The second line contains N numbers, the initial values of A1, A2, ... , AN. -1000000000 ≤ Ai ≤ 1000000000.
Each of the next Q lines represents an operation.
"C a b c" means adding c to each of Aa, Aa+1, ... , Ab. -10000 ≤ c ≤ 10000.
"Q a b" means querying the sum of Aa, Aa+1, ... , Ab.
Output
You need to answer all Q commands in order. One answer in a line.
Sample Input
10 5
1 2 3 4 5 6 7 8 9 10
Q 4 4
Q 1 10
Q 2 4
C 3 6 3
Q 2 4
Sample Output
4
55
9
15
HINT
The sums may exceed the range of 32-bit integers.
#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
#define lson l ,m , rt << 1
#define rson m + 1 , r ,rt << 1 | 1
typedef __int64 LL;
const int maxn = 100005;
LL sum[maxn<<2],lazy[maxn<<2];
void PushUp(int rt)
{
sum[rt] = sum[rt<<1] + sum[rt<<1|1];
}
void PushDown(int rt,int m)
{
if (lazy[rt])
{
lazy[rt<<1] += lazy[rt];
lazy[rt<<1|1] += lazy[rt];
sum[rt<<1] += (m - (m>>1)) * lazy[rt];
sum[rt<<1|1] += (m>>1) * lazy[rt];
lazy[rt] = 0;
}
}
void build(int l,int r,int rt)
{
if (l == r)
{
scanf("%I64d",&sum[rt]);
lazy[rt] = 0;
return;
}
int m = (l + r) >> 1;
build(lson);
build(rson);
PushUp(rt);
}
void upd(int L,int R,int c,int l,int r,int rt)
{
if (L <= l && r <= R)
{
sum[rt] += (LL)(r - l + 1) * c;
lazy[rt] += c;
return;
}
PushDown(rt,r - l + 1);
int m = (l + r) >> 1;
if (L <= m) upd(L,R,c,lson);
if (R > m) upd(L,R,c,rson);
PushUp(rt);
}
LL qry(int L,int R,int l,int r,int rt)
{
if (L <= l && r <= R)
{
return sum[rt];
}
PushDown(rt,r - l + 1);
int m = (l + r) >> 1;
LL ret = 0;
if (L <= m) ret += qry(L,R,lson);
if (R > m) ret += qry(L,R,rson);
return ret;
}
int main()
{
int N,Q,a,b,c;
char opt;
while (~scanf("%d%d",&N,&Q))
{
build(1,N,1);
while (Q--)
{
getchar();
scanf("%c",&opt);
if (opt == 'C')
{
scanf("%d%d%d",&a,&b,&c);
upd(a,b,c,1,N,1);
}
else
{
scanf("%d%d",&a,&b);
printf("%I64d\n",qry(a,b,1,N,1));
}
}
}
return 0;
}
POJ 3468 A Simple Problem with Integers(线段树/区间更新)的更多相关文章
- poj 3468 A Simple Problem with Integers (线段树区间更新求和lazy思想)
A Simple Problem with Integers Time Limit: 5000MS Memory Limit: 131072K Total Submissions: 75541 ...
- (简单) POJ 3468 A Simple Problem with Integers , 线段树+区间更新。
Description You have N integers, A1, A2, ... , AN. You need to deal with two kinds of operations. On ...
- [POJ] 3468 A Simple Problem with Integers [线段树区间更新求和]
A Simple Problem with Integers Description You have N integers, A1, A2, ... , AN. You need to deal ...
- poj 3468 A Simple Problem with Integers 线段树区间更新
id=3468">点击打开链接题目链接 A Simple Problem with Integers Time Limit: 5000MS Memory Limit: 131072 ...
- POJ 3468 A Simple Problem with Integers(线段树,区间更新,区间求和)
A Simple Problem with Integers Time Limit: 5000MS Memory Limit: 131072K Total Submissions: 67511 ...
- POJ 3468 A Simple Problem with Integers(线段树区间更新)
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, ...
- POJ 3468 A Simple Problem with Integers(线段树区间更新,模板题,求区间和)
#include <iostream> #include <stdio.h> #include <string.h> #define lson rt<< ...
- POJ 3468 A Simple Problem with Integers 线段树 区间更新
#include<iostream> #include<string> #include<algorithm> #include<cstdlib> #i ...
- poj 3468 A Simple Problem with Integers 线段树区间加,区间查询和
A Simple Problem with Integers Time Limit: 1 Sec Memory Limit: 256 MB 题目连接 http://poj.org/problem?i ...
- poj 3468 A Simple Problem with Integers 线段树区间加,区间查询和(模板)
A Simple Problem with Integers Time Limit: 1 Sec Memory Limit: 256 MB 题目连接 http://poj.org/problem?i ...
随机推荐
- 翻译qmake文档 目录
利用空闲时间把qmke的文档翻译出来,翻译水平有限,有些地方翻译的不好,请谅解, 如果您能指出来,我会很感激并在第一时候做出修改. 翻译qmake文档(一) qmake指南和概述 翻译qmake文档( ...
- opencv6.3-imgproc图像处理模块之边缘检测
接opencv6.2-improc图像处理模块之图像尺寸上的操作 本文大部分都是来自于转http://www.opencv.org.cn/opencvdoc/2.3.2/html/doc/tutori ...
- C#异步编程二
上一异步编程的博客还是在9月份的,最近事情也比较多,烦恼事情一个接着一个,一个人的周末除了无聊就剩无聊了,也只有写写博客来打发下这无聊的时光.原本想着尽快把异步编程这块总结一下,之前把委托异步算是总结 ...
- ASP.NET Web API 安全验证之摘要(Digest)认证
在基本认证的方式中,主要的安全问题来自于用户信息的明文传输,而在摘要认证中,主要通过一些手段避免了此问题,大大增加了安全性. 1.客户端匿名的方式请求 (无认证) HTTP/ Unauthorized ...
- STM32 C语言,端口映射
static XX 有记忆的定义 typedef XX 可以多次定义一个 #ifedf XXX XXX(程序段1) #else XXX(程序段2)
- swfupload提示“错误302”的解决方法
1.关于图片上传控件,flash控件的显示效果要好一些,本人使用swfupload 2.swfupload上传控件使用方式详见文档 http://www.leeon.me/upload/other/s ...
- WinForm 问题集锦
[1]重用项目窗体解决方案: 1. 把FmMain.cs 和 FmMain.Designer.cs 和 FmMain .resx 三个文件复制到程序目录下: 2. 在vs里面添加现有项, 选择FmMa ...
- python学习笔记整理——列表
Python 文档学习笔记 数据结构--列表 列表的方法 添加 list.append(x) 添加元素 添加一个元素到列表的末尾:相当于a[len(a):] = [x] list.extend(L) ...
- alarm
AlarmManager的使用机制有的称呼为全局定时器,有的称呼为闹钟.通过对它的使用,它的作用和Timer有点相似.都有两种相似的用法:(1)在指定时长后执行某项操作 (2)周期性的执行某项操作 在 ...
- python环境搭建-pycharm2016软件注册码
pycharm软件下载地址 https://www.jetbrains.com/pycharm/ 方法一: pycharm 2016 注册码 43B4A73YYJ-eyJsaWNlbnNlSWQiOi ...