#include<iostream>
#include<string>
#include<algorithm>
#include<cstdlib>
#include<cstdio>
#include<set>
#include<map>
#include<vector>
#include<cstring>
#include<stack>
#include<cmath>
#include<queue>
//#include <bits/stdc++.h>
using namespace std;
#define LL long long
const int maxn=;
struct node
{
LL sum,val;
}tree[maxn*]; void pushup(int rt)
{
tree[rt].sum=tree[rt<<].sum+tree[rt<<|].sum;
}
void pushdown(int rt,int m)
{
if(tree[rt].val)
{
tree[rt<<].val+=tree[rt].val;
tree[rt<<|].val+=tree[rt].val;
tree[rt<<].sum+=(LL)(m-(m>>))*tree[rt].val;
tree[rt<<|].sum+=(LL)(m>>)*tree[rt].val;
tree[rt].val=;
}
} void build_tree(int l,int r,int rt)
{
tree[rt].val=;
if(l==r)
{
scanf("%I64d",&tree[rt].sum);
return ;
}
int m=(l+r)>>;
build_tree(l,m,rt<<);
build_tree(m+,r,rt<<|);
pushup(rt);
} LL query(int L,int R,int l,int r,int rt)
{
if(L<=l&&r<=R)
{
return tree[rt].sum;
}
int m=(l+r)>>;
pushdown(rt,r-l+);
LL ans=;
if(L<=m)
ans+=query(L,R,l,m,rt<<);
if(m<R)
ans+=query(L,R,m+,r,rt<<|);
pushup(rt);
return ans;
} void update(int L,int R,int add,int l,int r,int rt)
{
if(L<=l&&r<=R)
{
tree[rt].sum+=(LL)add*(r-l+);
tree[rt].val+=add;
return;
}
pushdown(rt,r-l+);
int m=(l+r)>>;
if(L<=m)
update(L,R,add,l,m,rt<<);
if(R>m)
update(L,R,add,m+,r,rt<<|);
pushup(rt);
} int main()
{
int n,a,b,q;
LL c;
while(~scanf("%d%d",&n,&q))
{
build_tree(,n,);
char s[];
while(q--)
{
scanf("%s",s);
if(s[]=='Q')
{
cin>>a>>b;
cout<<query(a,b,,n,)<<endl;
}
else if(s[]=='C')
{
cin>>a>>b>>c;
update(a,b,c,,n,);
}
}
}
return ;
}

POJ 3468 A Simple Problem with Integers 线段树 区间更新的更多相关文章

  1. 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   ...

  2. (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 ...

  3. [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 ...

  4. 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 ...

  5. 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   ...

  6. 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, ...

  7. 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<< ...

  8. 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 ...

  9. 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

  1. 详解MySQL中EXPLAIN解释命令(转)

    explain显示了mysql如何使用索引来处理select语句以及连接表.可以帮助选择更好的索引和写出更优化的查询语句. 使用方法,在select语句前加上explain就可以了: 如: expla ...

  2. PythonCrawl自学日志(4)

    2016年9月22日10:34:02一.Selector1.如何构建(1)text构建: body = '<html><body><span>good</sp ...

  3. ubuntu u盘安装

    参考博客:http://www.bjwilly.com/archives/325.html 1.下载光盘映像 目前可选12.04LTS(长期支持版本) http://www.ubuntu.org.cn ...

  4. hdu 2243 考研路茫茫——单词情结 ac自动机+矩阵快速幂

    链接:http://acm.hdu.edu.cn/showproblem.php?pid=2243 题意:给定N(1<= N < 6)个长度不超过5的词根,问长度不超过L(L <23 ...

  5. 洛谷1508 Likecloud-吃、吃、吃

    题目背景 问世间,青春期为何物? 答曰:“甲亢,甲亢,再甲亢:挨饿,挨饿,再挨饿!” 题目描述 正处在某一特定时期之中的李大水牛由于消化系统比较发达,最近一直处在饥饿的状态中.某日上课,正当他饿得头昏 ...

  6. The partner transaction manager has disabled its support for remote/network transactions.

    http://technet.microsoft.com/en-us/library/cc753510(WS.10).aspx

  7. JUnit测试工具在项目中的用法

    0:33 2013/6/26 三大框架整合时为什么要对项目进行junit测试: |__目的是测试配置文件对不对,能跑通就可以进行开发了 具体测试步骤: |__1.对hibernate进行测试 配置hi ...

  8. ios7 sdk 新特性

    iOS 7 is a major update with compelling features for developers to incorporate into their apps. The ...

  9. 使用struts2标签<s:action无法显示引用页面问题

    使用过程中参考:http://www.cnblogs.com/lihuiyy/archive/2012/03/23/2411601.html 个人使用: 1.引用页面 <s:action nam ...

  10. CSS3画三角形原理

    1.首先看一下画出一个下三角形完整的代码及效果图 #trangle1-up{ width:; height:; border-left:50px solid transparent; border-r ...