数据结构--线段树--lazy延迟操作

A Simple Problem with Integers

Time Limit: 5000MS		Memory Limit: 131072K
Total Submissions: 53749		Accepted: 16131
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

//更新某一段区域的时候，采用延迟标记~~
代码：

#include "cstdio" //poj 3468 lazy操作
#include "cstring"
#include "iostream"
using namespace std;

#define N 100005
#define LL long long

struct node{
int x,y;
LL sum;
LL add;   //记录以当前节点为根节点的树中需要增加的值
}a[3*N];

void Build(int t,int x,int y)
{
a[t].x = x;
a[t].y = y;
a[t].sum = a[t].add = 0;
if(a[t].x == a[t].y)  //到了叶子节点
{
scanf("%lld",&a[t].sum);
return ;
}
int mid = (a[t].x + a[t].y)/2;
Build(t<<1,x,mid);
Build(t<<1|1,mid+1,y);
a[t].sum = a[t<<1].sum + a[t<<1|1].sum;
}

void Push_down(int t)  //将add(增值)向下推一级
{
LL add = a[t].add;
a[t<<1].add += add;
a[t<<1|1].add += add;
a[t<<1].sum += add*(a[t<<1].y-a[t<<1].x+1);
a[t<<1|1].sum += add*(a[t<<1|1].y-a[t<<1|1].x+1);
a[t].add = 0;
}

LL Query(int t,int x,int y)
{
if(a[t].x==x &&a[t].y==y)
return a[t].sum;
Push_down(t);
int mid = (a[t].x + a[t].y)/2;
if(y<=mid)
return Query(t<<1,x,y);
if(x>mid)
return Query(t<<1|1,x,y);
else
return Query(t<<1,x,mid) + Query(t<<1|1,mid+1,y);
}

void Add(int t,int x,int y,int k)
{
if(a[t].x==x && a[t].y==y) //不在推到叶子节点，t下的子孙要增加的值存入a[t].add中
{
a[t].add += k;
a[t].sum += (a[t].y-a[t].x+1)*k;
return ;
}
a[t].sum += (y-x+1)*k;
Push_down(t);
int mid = (a[t].x+a[t].y)/2;
if(y<=mid)
Add(t<<1,x,y,k);
else if(x>mid)
Add(t<<1|1,x,y,k);
else
{
Add(t<<1,x,mid,k);
Add(t<<1|1,mid+1,y,k);
}
}

int main()
{
int n,m;
char ch;
int x,y,k;
scanf("%d %d",&n,&m);
Build(1,1,n);
while(m--)
{
getchar();
scanf("%c %d %d",&ch,&x,&y);
if(ch=='Q')
printf("%lld\n",Query(1,x,y));
else
{
scanf("%d",&k);
Add(1,x,y,k);
}
}
return 0;
}