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

  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 区间更新

    #include<iostream> #include<string> #include<algorithm> #include<cstdlib> #i ...

  9. 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 ...

  10. 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. QT 智能提示设置

    qt5.0的智能提示设置 qt默认的是Ctrl+空格 但这个是切换输入法,用着也不习惯 修改的地方是 工具->选项->环境 键盘选项把CompleteThis修改成自己习惯的快捷键

  2. web 前端常用组件【01】Pagination 分页

    分页组件几乎是一般网站都会涉及到的组件,网上有很多这样的插件,自己挑来跳去选择了这一款. 官方Demo网址:http://mricle.com/JqueryPagination 功能强大,可扩展性比较 ...

  3. Android中的Semaphore

    信号量,了解过操作系统的人都知道,信号量是用来做什么的··· 在Android中,已经提供了Semaphore来帮助我们使用~ 那么,在开发中这家伙有什么用呢? 用的地方不多,但是却真的是好用至极! ...

  4. ModernUI教程:MEF应用向导

    本文主要说明在Modern UI框架下使用MEF的必要步骤,关于MEF请自行脑补. MEF-INTO-MUI实例代码下载: MefMuiApp.zip 1:创建一个导出属性 ModernFrame用来 ...

  5. 通过Ajax实现增删改查

    项目链接:https://github.com/shuai7boy/Ajax_CRUD 简要截图:

  6. go println与printf区别

    Println 与Printf 都是fmt 包中的公共方法 Println :可以打印出字符串,和变量: Printf : 只可以打印出格式化的字符串,可以输出字符串类型的变量,不可以输出整形变量和整 ...

  7. ASP.NET 使用Ajax

    转载: http://www.cnblogs.com/dolphinX/p/3242408.html $.ajax向普通页面发送get请求 这是最简单的一种方式了,先简单了解jQuery ajax的语 ...

  8. 腾讯云CentOS 安装MediaWiki

    参考 : https://www.digitalocean.com/community/tutorials/how-to-install-mediawiki-on-centos-7 //安装好很多次终 ...

  9. RabbitMQ 工作队列

    创建一个工作队列用来在工作者(consumer)间分发耗时任务. 工作队列的主要任务是:避免立刻执行资源密集型任务,然后必须等待其完成.相反地,我们进行任务调度:我们把任务封装为消息发送给队列.工作进 ...

  10. RAP在centos上的部署

    在centos7上部署RAP(非官方) 作者批注:该部署文档为网友贡献,仅供参考.war请参考主页README.md下载最新版本哟~~~ 感谢热情网友的Wiki整理!万分感谢! 系统: centos7 ...