A Simple Problem with Integers(线段树区间更新复习，lazy数组的应用）-------------------蓝桥备战系列

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本文介绍了一种高效处理大规模数组中区间加法和查询求和的操作方法，通过使用线段树和懒惰传播技术，实现了对数级的时间复杂度，适用于大数据集的实时更新和查询。示例代码展示了如何构建线段树，进行区间更新和查询，以及如何通过懒惰传播避免不必要的节点更新。

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

Hint

The sums may exceed the range of 32-bit integers.

代码：

#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
#include<queue>
#include<stack>
#include<map>
#include<set>
#include<vector>
#include<cmath>

const int maxn=1e5+5;
typedef long long ll;
using namespace std;

struct node
{
	ll l,r,sum;
}tree[maxn<<2];
ll lazy[maxn<<2];
void pushup(int m)
{

	tree[m].sum=tree[m<<1].sum+tree[m<<1|1].sum;
}
void pushdown(int m,int l)
{
	if(lazy[m]!=0)
	{
		lazy[m<<1]+=lazy[m];
		lazy[m<<1|1]+=lazy[m];
		tree[m<<1].sum+=lazy[m]*(l-(l>>1));
		tree[m<<1|1].sum+=lazy[m]*(l>>1);
		lazy[m]=0;
	}
}
void build(int m,int l,int r)
{
	tree[m].l=l;
	tree[m].r=r;
	if(l==r)
	{
		scanf("%lld",&tree[m].sum);
		return;
	}
	int mid=(l+r)>>1;
	build(m<<1,l,mid);
	build(m<<1|1,mid+1,r);
	pushup(m);
}
void update(int m,int l,int r,int val)
{
	if(tree[m].l==l&&tree[m].r==r)
	{
		lazy[m]+=val;
		tree[m].sum+=(ll)val*(r-l+1);
		return;
	}
	if(tree[m].l==tree[m].r)
	return;
	int mid=(tree[m].l+tree[m].r)>>1;
	pushdown(m,tree[m].r-tree[m].l+1);
	if(r<=mid)
	{
		update(m<<1,l,r,val);
	}
	else if(l>mid)
	{
		update(m<<1|1,l,r,val);
	}
	else 
	{
		update(m<<1,l,mid,val);
		update(m<<1|1,mid+1,r,val);
	}
	pushup(m);
}
ll query(int m,int l,int r)
{
	if(tree[m].l==l&&tree[m].r==r)
	{
		return tree[m].sum;
	}
	pushdown(m,tree[m].r-tree[m].l+1);
	int mid=(tree[m].l+tree[m].r)>>1;
	ll res=0;
	if(r<=mid)
	{
		res+=query(m<<1,l,r);
	}
	else if(l>mid)
	{
		res+=query(m<<1|1,l,r);
	}
	else 
	{
		res+=(query(m<<1,l,mid)+query(m<<1|1,mid+1,r));
		
	}
	return res;
	
}
int main()
{
    std::ios::sync_with_stdio(false);
    std::cin.tie(0);
    int n,m;
    cin>>n>>m;
    build(1,1,n);

    char op[2];
    int l,r,val;
    for(int t=0;t<m;t++)
    {
    	scanf("%s",op);
    	if(op[0]=='Q')
    	{
    		scanf("%d%d",&l,&r);
    		printf("%lld\n",query(1,l,r));
		}
		else 
		{
			scanf("%d%d%d",&l,&r,&val);
			update(1,l,r,val);
		}
	}

	return 0;
}