线段树（求区间和）

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#线段树

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本文介绍了一种使用线段树处理数组区间加法更新与区间求和查询的技术。通过递归分解问题并维护懒惰标记，可以高效地完成批量操作。适用于解决算法竞赛中的区间操作问题。

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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.

输入描述:

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.

输出描述:

You need to answer all Q commands in order. One answer in a line.

输入

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

输出

代码如下:

#include<iostream>
#include<cmath>
#include<cstdio>
#include<cstring>
#define ll long long
using namespace std;

const int maxn=1e5+100;
struct note
{
    int left,right;
    ll flag;
    ll sum;
}tree[maxn*4];
ll a[maxn];

void pushup(int root)
{
    tree[root].sum=tree[root<<1].sum+tree[root<<1|1].sum;
}
void build(int l,int r,int root)
{
    tree[root].left=l;
    tree[root].right=r;
    tree[root].sum=0;
    tree[root].flag=0;
    if (l==r) {
        tree[root].sum=a[l];
        return ;
    }
    int mid=(l+r)>>1;
    build(l,mid,root<<1);
    build(mid+1,r,root<<1|1);
    pushup(root);
}
void pushdown(int root)
{
    if (tree[root].flag!=0) {
        tree[root<<1].sum+=tree[root].flag*(tree[root<<1].right-tree[root<<1].left+1);
        tree[root<<1|1].sum+=tree[root].flag*(tree[root<<1|1].right-tree[root<<1|1].left+1);
        tree[root<<1].flag+=tree[root].flag;
        tree[root<<1|1].flag+=tree[root].flag;
        tree[root].flag=0;
    }
}
ll query(int l,int r,int root)
{
    int left=tree[root].left;
    int right=tree[root].right;
    if (left>=l&&right<=r) {
        return tree[root].sum;
    }
    pushdown(root);
    int mid=(left+right)>>1;
    ll sum=0;
    if (mid>=r) sum+=query(l,r,root<<1);
    else if (l>mid) sum+=query(l,r,root<<1|1);
    else {
        sum+=query(l,mid,root<<1);
        sum+=query(mid+1,r,root<<1|1);
    }
    return sum;
}

void update(int l,int r,int root,int val)
{
    int left=tree[root].left;
    int right=tree[root].right;
    if (left>=l&&right<=r) {
        tree[root].sum+=val*(right-left+1);
        tree[root].flag+=val;
        return ;
    }
    pushdown(root);
    int mid=(left+right)>>1;
    if (r<=mid) update(l,r,root<<1,val);
    else if (l>=mid+1) update(l,r,root<<1|1,val);
    else {
        update(l,mid,root<<1,val);
        update(mid+1,r,root<<1|1,val);
    }
    pushup(root);
}
int main()
{
    int n,q;
    cin>>n>>q;
    for (int i=1;i<=n;i++) {
        cin>>a[i];
    }
    build(1,n,1);
    while (q--) {
        string str;
        cin>>str;
        if (str=="Q") {
            int l,r;
            cin>>l>>r;
            cout<<query(l,r,1)<<endl;
        }
        else if (str=="C") {
            int l,r,x;
            cin>>l>>r>>x;
            update(l,r,1,x);
        }
    }
    return 0;
}