POJ 3468——A Simple Problem with Integers——————【线段树区间更新, 区间查询】
Time Limit: 5000MS | Memory Limit: 131072K | |
Total Submissions: 86780 | Accepted: 26950 | |
Case Time Limit: 2000MS |
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
Source
#include<stdio.h>
#include<vector>
#include<queue>
#include<string.h>
#include<algorithm>
using namespace std;
typedef long long LL;
const int maxn = 1e6+20;
const int INF = 0x3f3f3f3f;
const int mod = 1e9+7;
#define mid (L+R)/2
#define lson rt*2,L,mid
#define rson rt*2+1,mid+1,R
struct SegTree{
LL val, lazy;
}segs[maxn*4];
void PushUp(int rt){
segs[rt].val = segs[rt*2+1].val + segs[rt*2].val;
}
void PushDown(int rt,int L,int R){
if(segs[rt].lazy){
segs[rt*2].val += segs[rt].lazy*(mid-L +1); //保证当前的正确性
segs[rt*2+1].val += segs[rt].lazy*(R-mid);
segs[rt*2].lazy += segs[rt].lazy; //累积
segs[rt*2+1].lazy += segs[rt].lazy;
segs[rt].lazy = 0;
}
}
void buildtree(int rt,int L,int R){
segs[rt].val = 0;
segs[rt].lazy = 0;
if(L == R){
scanf("%lld",&segs[rt].val);
return ;
}
buildtree(lson);
buildtree(rson);
PushUp(rt);
}
//void Update(int rt,int L,int R,int _idx,int _val){
// if(L == R && L == _idx){
// segs[rt].val += _val;
// return ;
// }
// if(_idx <= mid)
// Update(lson,_idx,_val);
// else
// Update(rson,_idx,_val);
// PushUp(rt);
//}
LL query(int rt,int L,int R,int l_ran,int r_ran){
if(l_ran <= L&&R <= r_ran){
return segs[rt].val;
}
PushDown(rt,L,R);
LL ret = 0;
if(l_ran <= mid){
ret += query(lson,l_ran,r_ran);
}
if(r_ran > mid){
ret += query(rson,l_ran,r_ran);
}
PushUp(rt);
return ret;
}
void Update(int rt,int L,int R,int l_ran,int r_ran,LL chg){
if(l_ran <= L&&R <= r_ran){
segs[rt].val += chg*(R-L+1);
segs[rt].lazy += chg;
return ;
}
PushDown(rt,L,R);
if(l_ran <= mid)
Update(lson,l_ran,r_ran,chg);
if(r_ran > mid)
Update(rson,l_ran,r_ran,chg);
PushUp(rt);
}
int main(){
int n,m;
while(scanf("%d%d",&n,&m)!=EOF){
buildtree(1,1,n);
int l,r;
LL c;
char s[12];
for(int i = 1; i <= m; i++){
scanf("%s",s);
if(s[0]=='Q'){
scanf("%d%d",&l,&r);
LL ans = query(1,1,n,l,r);
printf("%lld\n",ans);
}else{
scanf("%d%d%lld",&l,&r,&c);
Update(1,1,n,l,r,c);
}
}
}
return 0;
}
POJ 3468——A Simple Problem with Integers——————【线段树区间更新, 区间查询】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 ...
- A Simple Problem with Integers 线段树 区间更新 区间查询
Time Limit: 5000MS Memory Limit: 131072K Total Submissions: 115624 Accepted: 35897 Case Time Lim ...
- 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 ...
随机推荐
- ExposedObject的使用
ExposedObject可以将一个对象快速封装未一个dynamic using System; namespace ConsoleApp2 { class Program { static void ...
- sublime 快捷键 1到9的顺序
p{$$}*9
- ASP.NET Core 一个Json返回类
很多时候(如Ajax)我们需要从Web应用中得到标准的反馈以便进行数据分析. 为此,我将各类返回标准化封装,客户端将统一得到 JSON:[{"status":200,"m ...
- 【bzoj4836】二元运算 分治FFT
Description 定义二元运算 opt 满足 现在给定一个长为 n 的数列 a 和一个长为 m 的数列 b ,接下来有 q 次询问.每次询问给定一个数字 c 你需要求出有多少对 (i, j) 使 ...
- HBASE常用操作增删改查
http://javacrazyer.iteye.com/blog/1186881 http://www.cnblogs.com/invban/p/5667701.html
- nRF51822外设应用[2]:GPIOTE的应用-按键检测
版权声明:本文为博主原创文章,转载请注明作者和出处. 作者:强光手电[艾克姆科技-无线事业部] 1. nRF51822寄存器类型 nRF51822的寄存器和一般的单片机有所差别,nRF51822 ...
- 锐速破解版linux一键自动安装包
锐速破解版linux一键自动安装包(5月28日更新) 锐速破解版安装方法: wget -N --no-check-certificate https://github.com/91yun/server ...
- layer mobile开发layer.full
Layer For Mobile 之 layer.full() 背景介绍:layer mobile是专门针对手机页面开发的一套框架,具体介绍请看官方文档 http://layer.layui.com/ ...
- RNA-seq分析htseq-count的使用
HTSeq作为一款可以处理高通量数据的python包,由Simon Anders, Paul Theodor Pyl, Wolfgang Huber等人携手推出HTSeq — A Python fra ...
- C++_基础3-循环和关系表达式
这一部分内容节选自<C++ Primer Plus>的第五章 程序需要有执行重复的操作和进行决策的工具. ========================================= ...