POJ 3468 A Simple Problem with Integers 区间更新

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本文介绍了一种使用线段树解决区间加法操作及区间求和查询的问题。通过维护懒惰标记，实现高效的区间更新和查询操作。适用于大量区间操作与查询的场景。

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

#include<cstdio>
#include<algorithm>
#include<iostream>
using namespace std;

const int maxn=111111;
long long lazy[100010<<2],seg[100010<<2];
void upGo(int rt)
{
    seg[rt]=seg[rt<<1]+seg[rt<<1|1];
}
void pushDown(int rt,int len)
{
    if(lazy[rt])
    {

        ///lazy[rt<<1]=lazy[rt<<1|1]=lazy[rt];///将延迟标志传给rt的左右子树。。。这样是错的
        lazy[rt<<1]+=lazy[rt];//必须要用累加
        lazy[rt<<1|1]+=lazy[rt];
        seg[rt<<1]+=lazy[rt]*(len-(len>>1));///直接计算更新后rt左右子树的值，而不是通过下向上累加来更新值
        seg[rt<<1|1]+=lazy[rt]*(len>>1);
        lazy[rt]=0;
    }
}
void build(int rt,int l,int r)
{
    lazy[rt]=0;
    if(l==r){
        scanf("%I64d",&seg[rt]);
        return;
    }
    int mid=(l+r)>>1;
    build(rt<<1,l,mid);
    build(rt<<1|1,mid+1,r);
    upGo(rt);
}
void upDate(int rt,int l,int r,int a,int b,int value)
{
    if(a<=l&&r<=b)
    {
        lazy[rt]+=value;///lazy[rt]=value
        seg[rt]+=(long long)value*(r-l+1);
        return;
    }
    pushDown(rt,r-l+1);
    int mid=(l+r)>>1;
    if(a<=mid)upDate(rt<<1,l,mid,a,b,value);
    if(b>mid) upDate(rt<<1|1,mid+1,r,a,b,value);
    upGo(rt);
}
long long query(int rt,int a,int b,int l,int r)
{
    if(a<=l&&r<=b)return seg[rt];
    pushDown(rt,r-l+1);///之气那没加
    int mid=(l+r)>>1;
    long long ans=0;
    if(a<=mid)ans+=query(rt<<1,a,b,l,mid);
    if(b>mid)ans+=query(rt<<1|1,a,b,mid+1,r);
    return ans;
}
int main()
{
    int n,m;
    while(scanf("%d%d",&n,&m)!=EOF)
    {
        build(1,1,n);
        char ch[2];
        int a,b,c;
        while(m--)
        {
            scanf("%s",ch);
            if(ch[0]=='Q'){
                scanf("%d%d",&a,&b);
                printf("%lld\n",query(1,a,b,1,n));
            }else{
                scanf("%d%d%d",&a,&b,&c);
                upDate(1,1,n,a,b,c);
            }
        }
    }
}