poj 3468 A Simple Problem with Integers(线段树)

H - A Simple Problem with Integers

Crawling in process...Crawling failedTime Limit:5000MS    Memory Limit:131072KB    64bit IO Format:%I64d & %I64u

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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

Hint

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

 

 

这道题是我最痛苦的一次，思路很快出来，敲的也快，可惜却因为敲少了个+号，少了个和，少了个long long ，却是绝大部分数据都OK，花了一个下午的狂调

 

#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cstring>
#define N 100300
using namespace std;
class node
{
  public:
  int l,r;
  long long sum,lazy;
}root[4*N];
long long shu[N];
void build(int t,int l,int r)
{
  root[t].l=l;root[t].r=r;root[t].lazy=0;
  if(root[t].l==root[t].r){cin>>root[t].sum;return;}
  build(t<<1,l,(l+r)/2);build(t<<1|1,(l+r)/2+1,r);
  root[t].sum=root[t<<1].sum+root[t<<1|1].sum;
}
void update(int t,int l,int r,long long date)
{
  if(root[t].l==l&&root[t].r==r)
  {root[t].sum+=(root[t].r-root[t].l+1)*date;root[t].lazy+=date;return;}//这里的+原下没加WA
  else
  {
    if(root[t].lazy!=0)
    {
      root[t<<1].lazy+=root[t].lazy;root[t<<1|1].lazy+=root[t].lazy;
      root[t<<1].sum+=(root[t<<1].r-root[t<<1].l+1)*root[t].lazy;
      root[t<<1|1].sum+=(root[t<<1|1].r-root[t<<1|1].l+1)*root[t].lazy;
      root[t].lazy=0;
    }
    if(root[t<<1].r>=r)update(t<<1,l,r,date);
    else if(root[t<<1|1].l<=l)update(t<<1|1,l,r,date);
    else {update(t<<1,l,root[t<<1].r,date);update(t<<1|1,root[t<<1|1].l,r,date);}
    root[t].sum=root[t<<1].sum+root[t<<1|1].sum;//这个很关键没加WA
  }
}
long long query(int t,int l,int r)
{
  long long s;
  if(root[t].l==l&&root[t].r==r)return root[t].sum;
  else
  {
    if(root[t].lazy!=0)
    {
      root[t<<1].lazy+=root[t].lazy;root[t<<1|1].lazy+=root[t].lazy;
      root[t<<1].sum+=(root[t<<1].r-root[t<<1].l+1)*root[t].lazy;
      root[t<<1|1].sum+=(root[t<<1|1].r-root[t<<1|1].l+1)*root[t].lazy;
      root[t].lazy=0;
    }
   if(root[t<<1].r>=r)s=query(t<<1,l,r);
   else if(root[t<<1|1].l<=l)
   s=query(t<<1|1,l,r);
   else s=query(t<<1,l,root[t<<1].r)+query(t<<1|1,root[t<<1|1].l,r);

  }
  return s;
}
int main()
{
  int n,m;

   scanf("%d%d",&n,&m);
  build(1,1,n);
  char f;
  long long a,b,c;
  for(int i=0;i<m;i++)
  {
    cin>>f>>a>>b;
    if(f=='C')
    {
      cin>>c;
      update(1,a,b,c);
    }
    else
    cout<<query(1,a,b)<<"\n";
  }

}