POJ 3468 A Simple Problem with Integers 線段樹區間更新

POJ 3468 A Simple Problem with Integers 線段樹區間更新

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A Simple Problem with Integers Time Limit: 5000MS Memory Limit: 131072K Total Submissions: 67718 Accepted: 20897 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
給你n個數，Q（a,b)代表查詢區間[a,b]數字之和 C（a,b,c)代表在區間[a,b]增加值c 在點更新的基礎上加上一個mark標記。

//11412 KB	2704 ms
#include
#include
#include
#define ll __int64
#define M 100007
using namespace std;
struct node
{
    ll l,r,mid,val,mark;
}tree[M<<2];
ll s[M];
void build(ll left,ll right,ll i)//建樹
{
    tree[i].l=left;tree[i].r=right;
    tree[i].mid=(left+right)>>1;tree[i].mark=0;
    if(left==right){tree[i].val=s[left]; return;}
    build(left,tree[i].mid,i*2);
    build(tree[i].mid+1,right,i*2+1);
    tree[i].val=tree[i*2].val+tree[i*2+1].val;
}
void update(int left,int right,ll val,int i)//區間更新
{
    if(tree[i].l==left&&tree[i].r==right){tree[i].mark+=val;return;}
    tree[i].val+=val*(right-left+1);
    if(tree[i].mid=right)update(left,right,val,2*i);
    else
    {
        update(left,tree[i].mid,val,2*i);
        update(tree[i].mid+1,right,val,2*i+1);
    }
}
ll query(int left,int right,int i)//區間查詢
{
    if(tree[i].l==left&&tree[i].r==right)  return tree[i].val+tree[i].mark*(right-left+1);
    if(tree[i].mark!=0)
    {
        tree[i*2].mark+=tree[i].mark;tree[i*2+1].mark+=tree[i].mark;
        tree[i].val+=(tree[i].r-tree[i].l+1)*tree[i].mark;tree[i].mark=0;
    }
    if(tree[i].mid>=right){return query(left,right,i*2);}
    else if(tree[i].mid