POJ 3468 A Simple Problem with Integers(子区间更新维护)

A Simple Problem with Integers

Time Limit: 5000MS		Memory Limit: 131072K
Total Submissions: 67497		Accepted: 20813
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

4
55
9

15

有n个整数, v1,v2,v3...vn，可以进行两种操作，第一种是Q(x,y),输出区间x~y的和，第二种是C(x,y,z)，区间x~y内所有数+z，不多说了，直接上线段树。。。

#include <iostream>
#include <stdlib.h>
#include <cstring>
#include <cstdio>
#include <algorithm>
using namespace std;
const int N = 100010;
struct node
{
 __int64 sum,mark;             //区间和，懒惰标记
}q[N*4];
int v[N];                       //输入的序列
void build(int l,int r,int rt)     //建树
{
    q[rt].sum=0;
    q[rt].mark=0;
    if(l==r)
    {
        q[rt].sum+=v[r];
        return ;
    }
    int mid=(l+r)>>1;
    build(l,mid,rt<<1);
    build(mid+1,r,rt<<1|1);
    q[rt].sum=q[rt<<1].sum + q[rt<<1|1].sum;   //累计左右区间的和
}
void update(int l,int r,int rt)          //节点rt没标记，则推出，否则计算出左右区间的和，将标记传下去
{
    if(q[rt].mark)
    {
        int mid=(l+r)>>1;
        q[rt<<1].sum += q[rt].mark*(__int64)(mid-l+1);
        q[rt<<1|1].sum += q[rt].mark*(__int64)(r-mid);
        q[rt<<1].mark += q[rt].mark;
        q[rt<<1|1].mark += q[rt].mark;
        q[rt].mark=0;
    }
}
__int64 Query(int x,int y,int l,int r,int rt)
{
    if(x>r || y<l)
        return 0;
    if(x<=l && y>=r)
        return q[rt].sum;
    update(l,r,rt);
    int mid=(l+r)>>1;
    return Query(x,y,l,mid,rt<<1) + Query(x,y,mid+1,r,rt<<1|1);
}

void add(int x,int y,int z,int l,int r,int rt)
{
   if(x>r || y<l)
        return ;
    if(x<=l && y>=r)
    {
        q[rt].sum+=(__int64)(r-l+1)*z;
        q[rt].mark+=z;
        return ;
    }
    update(l,r,rt);
    int  mid=(l+r)>>1;
    if(x<=mid)
        add(x,y,z,l,mid,rt<<1);
     if(y>mid)
        add(x,y,z,mid+1,r,rt<<1|1);
     q[rt].sum=q[rt<<1].sum + q[rt<<1|1].sum;
}
int main()
{
    int n,m;
   while(~scanf("%d%d",&n,&m))
   {
       for(int i=1;i<=n;i++)
        scanf("%d",&v[i]);
        getchar();
       build(1,n,1);
       for(int i=0;i<m;i++)
       {
           char s;
           scanf("%c",&s);
         if(s=='Q')
         {
             int x,y;
             scanf("%d%d",&x,&y);
             __int64 ans=Query(x,y,1,n,1);
             printf("%I64d\n",ans);
         }
         else if(s=='C')
         {
             int x,y,z;
             scanf("%d%d%d",&x,&y,&z);
             add(x,y,z,1,n,1);
         }
         getchar();
       }
   }
    return 0;
}