POJ 3468 线段树

A Simple Problem with Integers

Time Limit: 5000MS		Memory Limit: 131072K
Total Submissions: 19890		Accepted: 5238
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

Hint

The sums may exceed the range of 32-bit integers.

线段数的经典题目，快速求区间总和，在每个线段节点加一个属性add，代表该段线段增加的总和，以后如果求该线段的子线段和

就可以用其子段的和加上 子险段长度*add即可。。

#include<iostream> using namespace std; struct LINE { int left,right; __int64 sum,add; }tree[500000]; void build_line_tree(int v,int left,int right) //建立线段树 { int mid=(left+right)/2; if(left>right) return ; tree[v].left=left; tree[v].right=right; tree[v].sum=0; tree[v].add=0; if(left==right) return ; build_line_tree((v+1)*2-1,left,mid); build_line_tree((v+1)*2,mid+1,right); } void add_line_tree(int v,int left,int right,int C) { int mid=(tree[v].left+tree[v].right)/2; if(left==tree[v].left&&right==tree[v].right) { if(left!=right) tree[v].add+=C; tree[v].sum+=(right-left+1)*C; return ; } tree[v].sum+=(right-left+1)*C; if(mid>=right) add_line_tree((v+1)*2-1,left,right,C); else if(mid<left) add_line_tree((v+1)*2,left,right,C); else { add_line_tree((v+1)*2-1,left,mid,C); add_line_tree((v+1)*2,mid+1,right,C); } } __int64 getsum_line_tree(int v,int left,int right) { __int64 mid=(tree[v].left+tree[v].right)/2,j; if(left==tree[v].left&&right==tree[v].right) { return tree[v].sum; } j=tree[v].add*(right-left+1); if(mid>=right) return j+getsum_line_tree((v+1)*2-1,left,right); else if(mid<left) return j+getsum_line_tree((v+1)*2,left,right); else { return j+getsum_line_tree((v+1)*2-1,left,mid)+getsum_line_tree((v+1)*2,mid+1,right); } } int main() { int n,i,c,t,a,b,d; char o; while(scanf("%d",&n)!=EOF) { build_line_tree(0,1,n); cin>>c; for(i=1;i<=n;i++) { scanf("%d",&t); add_line_tree(0,i,i,t); } while(c--) { cin>>o; if(o=='Q') { scanf("%d%d",&a,&b); printf("%I64d/n",getsum_line_tree(0,a,b)); } else { scanf("%d%d%d",&a,&b,&d); add_line_tree(0,a,b,d); } } } return 0; }