【poj 3468】 A Simple Problem with Integers 线段树

A Simple Problem with Integers
Time Limit: 5000MS Memory Limit: 131072K
Total Submissions: 86938 Accepted: 26990
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

思路+题意：
线段数区间修改，求和

代码：

#include<iostream>
#include<stdio.h>
#include<string.h>
#define lson (id*2) 
#define rson (id*2+1)
using namespace std;
long long tr[1000000*4+5],lazy[1000000*4+5];

long long ans=0;


int n,m;
void push_up(int id)
{
  tr[id]=tr[lson]+tr[rson];
  return;
}
void push_down(int id,int l,int mid,int r)
{
  if(!lazy[id]) return ;
  long long t=lazy[id];
  //if(r==5) cout<<tr[rson]<<endl;
  lazy[id]=0;
  lazy[lson]+=t;
  lazy[rson]+=t;
  tr[lson]+=(mid-l+1)*t;
  tr[rson]+=(r-mid)*t;
  return;
}
void build(int id,int l,int r)
{
    lazy[id]=0;
    if(l==r)
    {
      scanf("%lld",&tr[id]);
      return;
    }
    int mid=(l+r)>>1;
    build(lson,l,mid);
    build(rson,mid+1,r);
    push_up(id);
}



void add(int id,int L,int R,int l,int r,long long  v)
{
  if(l>R||r<L||R<L) return ;
  if(L>=l&&R<=r)
  {
    tr[id]+=(R-L+1)*v;
    lazy[id]+=v;
    return ;
  }
     int mid=(L+R)>>1;
   push_down(id,L,mid,R);

  if(l<=mid) add(lson,L,mid,l,r,v);
  if(r>=mid+1) add(rson,mid+1,R,l,r,v);
  push_up(id);
}



void query(int id,int L,int R,int l,int r)
{
  if(l>R||r<L||R<L) return ;
  if(L>=l&&R<=r) 
  {
    ans+=tr[id];
    return ;
  }
     int mid=(L+R)>>1;
   push_down(id,L,mid,R);

  if(l<=mid) query(lson,L,mid,l,r);
  if(r>=mid+1) query(rson,mid+1,R,l,r);
}
int main()
{
    scanf("%d%d",&n,&m);
    build(1,1,n);
    char s[2];
    for(int i=1;i<=m;i++)
    {
        int  aa,bb;
        scanf("%s%d%d",s,&aa,&bb);
        if(s[0]=='Q')
        {
          ans=0;

          query(1,1,n,aa,bb);
          printf("%lld\n",ans);
        }
        else 
          {
              long long  vv;
              scanf("%lld",&vv);
              add(1,1,n,aa,bb,vv);
          }
    }
}