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, ... , A_N. 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, ... , A_N. -1000000000 ≤ A_i ≤ 1000000000.
Each of the next Q lines represents an operation.
"C a b c" means adding c to each of A_a, A_a+1, ... , A_b. -10000 ≤ c ≤ 10000.
"Q a b" means querying the sum of A_a, A_a+1, ... , A_b.

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[*N];

 void Build(int t,int x,int y)
 {
     a[t].x = x;
     a[t].y = y;
     a[t].sum = a[t].add = ;
     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)/;
     Build(t<<,x,mid);
     Build(t<<|,mid+,y);
     a[t].sum = a[t<<].sum + a[t<<|].sum;
 }

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

 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)/;
     if(y<=mid)
         return Query(t<<,x,y);
     if(x>mid)
         return Query(t<<|,x,y);
     else
         return Query(t<<,x,mid) + Query(t<<|,mid+,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+)*k;
         return ;
     }
     a[t].sum += (y-x+)*k;
     Push_down(t);
     int mid = (a[t].x+a[t].y)/;
     if(y<=mid)
         Add(t<<,x,y,k);
     else if(x>mid)
         Add(t<<|,x,y,k);
     else
     {
         Add(t<<,x,mid,k);
         Add(t<<|,mid+,y,k);
     }
 }

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