POJ 题目3468 A Simple Problem with Integers（线段树成段更新，区间求和）

A Simple Problem with Integers

Time Limit: 5000MS		Memory Limit: 131072K
Total Submissions: 72251		Accepted: 22295
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.

Source

POJ Monthly--2007.11.25, Yang Yi

ac代码

#include
#include
struct s
{
	__int64 sum,val;
}node[410000];
void pushup(int tr)
{
	node[tr].sum=node[tr<<1].sum+node[tr<<1|1].sum;
}
void pushdown(int tr,int m)
{
	if(node[tr].val)
	{
		node[tr<<1].val+=node[tr].val;
		node[tr<<1|1].val+=node[tr].val;
		node[tr<<1].sum+=node[tr].val*(m-(m>>1));
		node[tr<<1|1].sum+=node[tr].val*(m>>1);
		node[tr].val=0;
	}
}
void build(int l,int r,int tr)
{
	node[tr].val=0;
	if(l==r)
	{
		scanf("%I64d",&node[tr].sum);
		return;
	}
	int m=(l+r)>>1;
	build(l,m,tr<<1);
	build(m+1,r,tr<<1|1);
	pushup(tr);
}
__int64 query(int L,int R,int l,int r,int tr)
{
	if(L<=l&&r<=R)
	{
		return node[tr].sum;
	}
	int m=(l+r)>>1;
	pushdown(tr,r-l+1);
	__int64 ans=0;
	if(L<=m)
		ans+=query(L,R,l,m,tr<<1);
	if(m>1;
	if(L<=m)
		update(L,R,add,l,m,tr<<1);
	if(m