POJ 3468 A Simple Problem with Integers(线段树区间求和)
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
区间求和:
#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<vector>
typedef long long LL;
using namespace std;
#define REPF( i , a , b ) for ( int i = a ; i <= b ; ++ i )
#define REP( i , n ) for ( int i = 0 ; i < n ; ++ i )
#define CLEAR( a , x ) memset ( a , x , sizeof a )
const int maxn=100000;
int num[maxn];
LL sum[maxn<<2],add[maxn<<2];
int N,Q;
void pushup(int rs)
{
sum[rs]=sum[rs<<1]+sum[rs<<1|1];
}
void pushdown(int rs,int l)
{
if(add[rs])
{
add[rs<<1]+=add[rs];
add[rs<<1|1]+=add[rs];
sum[rs<<1]+=add[rs]*(l-(l>>1));
sum[rs<<1|1]+=add[rs]*(l>>1);
add[rs]=0;
}
}
void build(int rs,int l,int r)
{
if(l==r)
{
scanf("%I64d",&sum[rs]);
return ;
}
int mid=(l+r)>>1;
build(rs<<1,l,mid);
build(rs<<1|1,mid+1,r);
pushup(rs);
}
void update(int c,int x,int y,int l,int r,int rs)
{
if(l>=x&&r<=y)
{
add[rs]+=c;
sum[rs]+=(LL)c*(r-l+1);
return ;
}
pushdown(rs,r-l+1);
int mid=(l+r)>>1;
if(x<=mid) update(c,x,y,l,mid,rs<<1);
if(y>mid) update(c,x,y,mid+1,r,rs<<1|1);
pushup(rs);
}
LL query(int x,int y,int l,int r,int rs)
{
if(l>=x&&r<=y)
return sum[rs];
pushdown(rs,r-l+1);
int mid=(l+r)>>1;
LL ans=0;
if(x<=mid) ans+=query(x,y,l,mid,rs<<1);
if(y>mid) ans+=query(x,y,mid+1,r,rs<<1|1);
return ans;
}
int main()
{
int x,y,z;
std::ios::sync_with_stdio(false);
while(~scanf("%d%d",&N,&Q))
{
CLEAR(sum,0);
CLEAR(add,0);
build(1,1,N);
char str[2];
while(Q--)
{
scanf("%s",str);
if(str[0]=='C')
{
scanf("%d%d%d",&x,&y,&z);
update(z,x,y,1,N,1);
}
else
{
scanf("%d%d",&x,&y);
printf("%I64d\n",query(x,y,1,N,1));
}
}
}
return 0;
}
POJ 3468 A Simple Problem with Integers(线段树区间求和)的更多相关文章
- 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 ...
- [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 (线段树区间更新求和lazy思想)
A Simple Problem with Integers Time Limit: 5000MS Memory Limit: 131072K Total Submissions: 75541 ...
- 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 ...
- (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 线段树区间修改
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 ...
- 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<< ...
随机推荐
- iOS openURL方法实现打电话、发短信、发邮件、打开其他App
UIApplication有个功能十分强大的openURL:方法 - (BOOL)openURL:(NSURL*)url; 通过这个方法,我们可以实现: 先获取 UIApplication UIApp ...
- 怎么查看chrome网络日志
最近在分析一个页面访问慢的问题,在分析的除了wireshark抓包等手段外,还用到了chrome的日志辅助分析 使用 chrome://net-internals/#events 可以打开日志追踪窗口 ...
- UILocalNotification
- (BOOL)application:(UIApplication *)application didFinishLaunchingWithOptions:(NSDictionary *)launc ...
- 产生冠军(set,map,拓扑结构三种方法)
产生冠军 Time Limit: 1000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submis ...
- Tomcat可以在eclipse里启动,可是不能訪问首页的问题
今天在使用eclipse的时候发现一个问题.就是我在eclipse里面已经启动了tomcat.部署上去的项目也能够启动,可是就是没法訪问tomcat的首页.port等等都没有问题. 后来查到解决方式, ...
- C#获取本机IP方法,获取本机局域网IP地址方法
1. private void GetIP() { string hostName = Dns.GetHostName();//本机名 //System.Net.IPAddress[] address ...
- 使用FindControl("id")查找控件 返回值都是Null的问题
做了一个通过字符串ID查找页面控件并且给页面控件赋值的功能,过程中遇到了this.FindControl("id")返回值都是Null的问题,记录一下解决办法. 问题的原因是我所要 ...
- UINavigationController 和 UITabBarController
UINavigationController当设置根控制器的时候,意思就是把根控制器压入栈内,当我们push的时候,我们把下一个控制器压入栈内,当我们pop的时候把上面的控制器的内存释放 UITa ...
- 000-C#基础
C#中数据类型的继承关系如下 System.Object |-------------System.ValueType | |-------System.Boolean | |-------Syste ...
- java性能优化技巧
在JAVA程序中,性能问题的大部分原因并不在于JAVA语言,而是程序本身.养成良好的编码习惯非常重要,能够显著地提升程序性能. 1. 尽量使用final修饰符. 带有final修饰符的类是不可派生的. ...