A Simple Problem with Integers

A Simple Problem with Integers
Time Limit: 5000MS Memory Limit: 131072K
Total Submissions: 77964 Accepted: 24012
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
线段树的区间查询与区间更新,lazy优化,不然会超时

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <queue>
#include <algorithm>
#define LL long long
using namespace std;
const int MAX =  110000;
struct node
{
    LL lazy;
    LL sum;
} Tree[MAX*12];
LL a[MAX];
void Build(int L,int R,int site)
{
    if(L==R)
    {
        Tree[site].lazy=0;
        Tree[site].sum=a[L];
        return ;
    }
    Tree[site].lazy=0;
    int mid=(L+R)>>1;
    Build(L,mid,site<<1);
    Build(mid+1,R,site<<1|1);
    Tree[site].sum=Tree[site<<1].sum+Tree[site<<1|1].sum;
}
void update(int L,int R,int l,int r,int site,LL w)
{
    if(L==l&&r==R)
    {
        if(L==R)
            Tree[site].lazy=0;
        else
            Tree[site].lazy+=w;
        Tree[site].sum+=((R-L+1)*w);
        return ;
    }
    int mid=(L+R)>>1;
    if(Tree[site].lazy!=0)
    {
        update(L,mid,L,mid,site<<1,Tree[site].lazy);
        update(mid+1,R,mid+1,R,site<<1|1,Tree[site].lazy);
        Tree[site].lazy=0;
    }
    Tree[site].sum+=((r-l+1)*w);
    if(mid>=r)
    {
        update(L,mid,l,r,site<<1,w);
    }
    else if(l>mid)
    {
        update(mid+1,R,l,r,site<<1|1,w);
    }
    else
    {
        update(L,mid,l,mid,site<<1,w);
        update(mid+1,R,mid+1,r,site<<1|1,w);
    }
    Tree[site].sum=Tree[site<<1].sum+Tree[site<<1|1].sum;
}
LL Query(int L,int R,int l,int r,int site)
{
    if(L==l&&r==R)
    {
        return Tree[site].sum;
    }
    int mid=(L+R)>>1;
    if(Tree[site].lazy!=0)
    {
        update(L,mid,L,mid,site<<1,Tree[site].lazy);
        update(mid+1,R,mid+1,R,site<<1|1,Tree[site].lazy);
        Tree[site].lazy=0;
    }

    if(mid>=r)
    {
        return Query(L,mid,l,r,site<<1);
    }
    else if(mid<l)
    {
        return Query(mid+1,R,l,r,site<<1|1);
    }
    else
    {
        return Query(L,mid,l,mid,site<<1)+Query(mid+1,R,mid+1,r,site<<1|1);
    }
}
int main()
{
    int n;
    int Q;
    char s[5];
    int u,v;
    LL w;
    while(~scanf("%d %d",&n,&Q))
    {
        for(int i=1; i<=n; i++)
        {
            scanf("%I64d",&a[i]);
        }
        Build(1,n,1);
        while(Q--)
        {
            scanf("%s",s);
            if(s[0]=='Q')
            {
                scanf("%d %d",&u,&v);
                printf("%I64d
",Query(1,n,u,v,1));
            }
            else
            {
                scanf("%d %d %I64d",&u,&v,&w);
                update(1,n,u,v,1,w);
            }
        }
    }
    return 0;
}