http://poj.org/problem?id=3468ASimpleProblemwithIntegersTimeLimit: 5000MS MemoryLimit: 131072KTotalSubmissions: 83959 Accepted: 25989CaseTimeLimit:&n"/>

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poj3468asimpleproblemwithintegers（线段树+区间更新+区间求和）

clnchanpin clnchanpin     2022-09-06     767

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题目链接:http://poj.org/problem?

id=3468

A Simple Problem with Integers
Time Limit: 5000MS   Memory Limit: 131072K
Total Submissions: 83959   Accepted: 25989
Case Time Limit: 2000MS

Description

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

Hint

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

Source

POJ Monthly--2007.11.25, Yang Yi

题目大意：给定n个数，m个操作。Q a b表示输出【a。b】这段区间内的和，C a b c表示将【a。b】这段区间的全部值都加上c。经过一系列变化，依照要求进行输出。

解题思路：标准的线段树，建树、更新树以及查找树。

对于更新树是为了避免改动到最底下而导致超时问题。所以每次改动仅仅改动相相应的区间就可以。然后记录一个add。下次更新或者查询的时候，假设查到该节点，就把add直接加到子节点上去，在将add变为0，避免下次还会反复加。这样仅仅更新到查询的子区间，不须要再往下找了，所以时间复杂度为O（n），更新树和查询树都须要这样。

由于add不为0，该add从根一直加到了该节点，之前的都加过了，假设更新到时候不加到子节点。还要通过子节点更新当前节点，当前节点的sum值里面含有的add就会被“抹掉”，就不能保证正确性了。还须要注意的就是要用__int64。


详见代码。

#include <iostream>
#include <cstdio>

using namespace std;

#define LL __int64

struct node
{
    int l,r;
    LL sum;
    LL add;
    //int flag;//用来表示有几个加数
} s[100000*4];

void InitTree(int l,int r,int k)
{
    s[k].l=l;
    s[k].r=r;
    s[k].sum=0;
    s[k].add=0;
    if (l==r)
        return ;
    int mid=(l+r)/2;
    InitTree(l,mid,2*k);
    InitTree(mid+1,r,2*k+1);
}

void UpdataTree(int l,int r,LL add,int k)
{

    if (s[k].l==l&&s[k].r==r)
    {
        s[k].add+=add;
        s[k].sum+=add*(r-l+1);
        return ;
    }
    if (s[k].add!=0)//加数为0就不须要改变了
    {
        s[2*k].add+=s[k].add;
        s[2*k+1].add+=s[k].add;
        s[2*k].sum+=s[k].add*(s[2*k].r-s[2*k].l+1);
        s[2*k+1].sum+=s[k].add*(s[2*k+1].r-s[2*k+1].l+1);
        s[k].add=0;
    }
    int mid=(s[k].l+s[k].r)/2;
    if (l>mid)
        UpdataTree(l,r,add,2*k+1);
    else if (r<=mid)
        UpdataTree(l,r,add,2*k);
    else
    {
        UpdataTree(l,mid,add,2*k);
        UpdataTree(mid+1,r,add,2*k+1);
    }
    s[k].sum=s[2*k].sum+s[2*k+1].sum;
}

LL SearchTree(int l,int r,int k)
{
    if (s[k].l==l&&s[k].r==r)
        return s[k].sum;
    if (s[k].add!=0)
    {
        s[2*k].add+=s[k].add;
        s[2*k+1].add+=s[k].add;
        s[2*k].sum+=s[k].add*(s[2*k].r-s[2*k].l+1);
        s[2*k+1].sum+=s[k].add*(s[2*k+1].r-s[2*k+1].l+1);
        s[k].add=0;
    }
    int mid=(s[k].l+s[k].r)/2;
    if (l>mid)
        return SearchTree(l,r,2*k+1);
    else if (r<=mid)
        return SearchTree(l,r,2*k);
    else
        return SearchTree(l,mid,2*k)+SearchTree(mid+1,r,2*k+1);
}


int main()
{
    int n,q;
    LL w;
    while (~scanf("%d%d",&n,&q))
    {
        InitTree(1,n,1);
        for (int i=1; i<=n; i++)
        {
            scanf("%lld",&w);
            UpdataTree(i,i,w,1);
        }
        for (int i=1; i<=q; i++)
        {
            char ch;
            int a,b;
            LL c;
            getchar();
            scanf("%c%d%d",&ch,&a,&b);
            if (ch=='C')
            {
                scanf("%lld",&c);
                UpdataTree(a,b,c,1);
            }
            else if (ch=='Q')
            {
                LL ans=SearchTree(a,b,1);
                printf ("%lld
",ans);
            }
        }
    }
    return 0;
}




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