线段树区间修改（lazy数组）

先贴个old hand的博客
https://blog.youkuaiyun.com/u012469987/article/details/41357377
讲的很详细，主要看里面的lazy。

贴个题
A Simple Problem with Integers

You have N integers, A1, A2, … , 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.

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, … , 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.

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

题意：给你n个数， m个操作，Q操作是询问区间【ai，aj】的和， C是把区间【ai，aj】的值全都都加上 t；

#include<iostream>
#include<cstring>
#include<cstdio>
#include<string>
#include<cmath>
#define N  110000
using namespace std;
#define ll long long
struct node
{
    ll lazy, sum;
    int le;         这里的le表示节点的区间长度。
} e[N*5];

ll build(int l, int r, int num)   这个是建树很简单，初始lazy为0；
{
    int mid = (l+r)/2;
    e[num].lazy = 0;
    e[num].le = r-l+1;
    if(l == r) return e[num].sum = a[l];
    return e[num].sum = build(mid+1, r, num*2+1) + build(l, mid, num*2);
}

void pushdown(int num)  这个是lazy下放，也就是更新，把lazy下放给左右孩子，
{                        并且让左右孩子都加上lazy × le（孩子的区间长度）的值。 
    if(e[num].lazy)
    {
        e[num*2].lazy += e[num].lazy;
        e[num*2+1].lazy += e[num].lazy;
        e[num*2].sum += e[num].lazy * e[num*2].le;
        e[num*2+1].sum += e[num].lazy * e[num*2+1].le;
        e[num].lazy = 0;        本节点的lazy已经下放给孩子了，就没了，为0；
    }
}
下放的关键是每次查询到有lazy的节点就更新，不查询到不动他。


ll makesum(int i, int j, int l, int r, int num)  这个是求区间和。为求i，j在l，r区间的和。
{
    if(i <= l && j >= r)               所求的i，j的区间已经包括l，r直接就是l，r区间的sum。
         return e[num].sum;
    int mid = (r+l)/2;
    pushdown(num);
    ll ans = 0;
    if(mid >= i)
    ans+=makesum(i, j, l, mid, num*2);
    if(mid < j)
    ans+=makesum(i, j, mid+1, r, num*2+1);
    return ans;
}

void chang(int i, int j, int x, int l, int r, int num)   更新，让i，j在l，r区间的值都加r，num为这个节点编号。
{
    if(i<=l && j >= r)        所更新的i，j的区间已经包括l，r。
    {
        e[num].lazy += x;
        e[num].sum += x*e[num].le;
        return;
    }
    int mid = (l+r)/2;
    pushdown(num);
    if(mid >= i)
    chang(i, j, x, l, mid, num*2);
    if(mid < j)
    chang(i, j, x, mid+1, r, num*2+1);
    e[num].sum = e[num*2].sum + e[num*2+1].sum;
}

int main()
{
    int n, m;
    memset(a, 0, sizeof(a));
    memset(e, 0, sizeof(e));
    cin >> n >> m;
    for(int i = 1; i <= n; i++)
    {
        scanf("%lld",&a[i]);
    }
    build(1, n, 1);
    string s;
    while(m--)
    {
        cin >> s;
        int i, j, t;
        if(s == "C")
        {
            scanf("%d%d%d",&i, &j, &t);
            chang(i, j, t, 1, n, 1);
        }
        else if(s == "Q")
        {
            scanf("%d%d",&i, &j);
            printf("%lld\n",makesum(i, j, 1, n, 1));
        }
    }
    return 0;
}

完了-------------