A - A Simple Problem with Integers POJ - 3468 线段树-优快云博客

本文链接：https://blog.youkuaiyun.com/jerans/article/details/72757921

本文介绍了一个涉及整数数组的操作问题，包括对指定区间内的整数进行批量加法操作及求和查询。通过使用线段树和懒惰传播优化算法实现高效处理大量操作。

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A Simple Problem with Integers

Time Limit: 5000MS		Memory Limit: 131072K
Total Submissions: 109423		Accepted: 34074
Case Time Limit: 2000MS

Description

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.

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, ... , 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.

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.

Source

POJ Monthly--2007.11.25, Yang Yi

线段树成段更新裸题lazy标记

#include<iostream>
#include<stdio.h>
#include<algorithm>

using namespace std;
#define lson l,m,node<<1
#define rson m+1,r,node<<1|1
#define maxn 100005
long long segtree[maxn<<2],lazy[maxn<<2];
void pushdown(int node)
{
    segtree[node]=segtree[node<<1]+segtree[node<<1|1];
    return ;
}
void pushback(int node,int l,int r)
{
    if(lazy[node]!=0)
    {
        int m=(l+r)>>1;
        segtree[node<<1|1]+=(r-m)*lazy[node];//这个地方要注意
        segtree[node<<1]+=(m-l+1)*lazy[node];
        lazy[node<<1|1]+=lazy[node];
        lazy[node<<1]+=lazy[node];
        lazy[node]=0;
    }
    return ;
}
void build(int l,int r,int node)
{
    lazy[node]=0;
    if(l==r)
    {
        scanf("%lld",&segtree[node]);
        return ;
    }
    int m=(l+r)>>1;
    build(lson);
    build(rson);
    pushdown(node);
    return ;
}
void update(int le,int ri,long long x,int l,int r,int node)
{
    if(le<=l&&ri>=r)
    {
        lazy[node]+=x;
        segtree[node]+=(r-l+1)*x;
        return ;
    }
    if(l==r)
    {
        segtree[node]+=x;
        return ;
    }
    pushback(node,l,r);
    int m=(l+r)>>1;
    if(m>=le)
    {
        update(le,ri,x,lson);
    }
    if(m<ri)
    {
        update(le,ri,x,rson);
    }
    pushdown(node);
    return ;
}
long long query(int le,int ri,int l,int r ,int node)
{
    if(l>=le&&r<=ri)
    {
        return segtree[node];
    }
    if(l==r)
    {
        return segtree[node];
    }
    pushback(node,l,r);
    int m=(l+r)>>1;
    long long sum=0;
    if(m>=le)
        sum+=query(le,ri,lson);
    if(m<ri)
    {
        sum+=query(le,ri,rson);
    }
    return sum;
}
int main()
{
    int n,i,q,l,r;
    long long x;
    char c;
    cin>>n>>q;
    build(1,n,1);
    for(int j=0;j<q;j++)
    {
        getchar();
        scanf("%c %d %d",&c,&l,&r);
        if(c=='Q')
        {
            printf("%lld\n",query(l,r,1,n,1));
        }
        else
        {
            scanf("%lld",&x);
            update(l,r,x,1,n,1);
        }
    }
    return 0;
}