线段树（求区间和）

傷城~ 2022-05-29 09:38 284阅读 0赞

You have  *N*  integers,  *A* 1 ,  *A* 2 , ... ,  *AN* . 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, ... , AN. -1000000000 ≤ Ai ≤ 1000000000.
    Each of the next Q lines represents an operation.
    "C a b c" means adding c to each of Aa, Aa+1, ... , Ab. -10000 ≤ c ≤ 10000.
    "Q a b" means querying the sum of Aa, Aa+1, ... , Ab.

## 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 ##

4
    55
    9
    15

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;
    }