线段树模板

好长啊，记不住。

链接：https://www.nowcoder.com/acm/contest/77/H
来源：牛客网

题目描述

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 A

₁

, A

₂

, ... , 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.

示例1

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

#include <bits/stdc++.h>
using namespace std;
typedef long long  ll;
const int INF = 0x3f3f3f3f;
const int maxn = 1e5+6;

ll n,a[maxn],m;

struct node
{
    int l,r;
    ll sum,lazy;
    void update(ll x)
    {
        sum += (r-l+1)*x;
        lazy += x;
    }
}tree[maxn*4];

void push_up(int x)
{
    tree[x].sum = tree[x<<1].sum+tree[x<<1|1].sum;
}
void push_down(int x)
{
    ll lazyval = tree[x].lazy;
    if(lazyval)
    {
        tree[x<<1].update(lazyval);
        tree[x<<1|1].update(lazyval);
        tree[x].lazy = 0;
    }
}
void build(int x,int l,int r) 
{
    tree[x].l = l,tree[x].r = r;
    tree[x].sum = tree[x].lazy = 0;
    if(l == r)
    {
        tree[x].sum = a[l];
    }
    else
    {
        int mid = (l+r)/2;
        build(x<<1,l,mid);
        build(x<<1|1,mid+1,r);
        push_up(x);
    }
}

void update(int x,int l,int r,ll val)        //x是根节点
{
    int L = tree[x].l;
    int R = tree[x].r;
    if(l <= L&&R <= r)
    {
        tree[x].update(val);
    }
    else
    {
        push_down(x);
        int mid = (L+R)/2;
        if(mid >= l) update(x<<1,l,r,val);
        if(r > mid) update(x<<1|1,l,r,val);   //注意此处的小于号，不是小于等于，写错会出错
        push_up(x);
    }
}

ll query(int x,int l,int r)
{
    int L = tree[x].l;
    int R = tree[x].r;
    if(l <= L&&R <= r)
        return tree[x].sum;
    else
    {
        push_down(x);
        ll ans = 0;
        int mid = (L+R)/2;
        if(mid >= l) ans += query(x<<1,l,r);
        if(r > mid) ans += query(x<<1|1,l,r);
        push_up(x);
        return ans;
    }
}

int main()
{
    scanf("%d%d",&n,&m);
    for(int i=1;i <= n;i++)
    {
        scanf("%d",&a[i]);
    }
    build(1,1,n);
    while(m--)
    {
        char s[2];
        int l,r,w;
        scanf("%s%d%d",s,&l,&r);
        if(s[0] == 'C')
        {
            scanf("%d",&w);
            update(1,l,r,w);
        }
        else
        {
            printf("%lld
",query(1,l,r));
        }
    }
    return 0;
}





/*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*/

区间加，查询和。