poj 3468 A Simple Problem with Integers

该题是一道区间求和的问题，同时对区间值进行改变，这里我们用到lazy,以便我们不要更新到每个节点，直到我们要用到时我们才往下传递，这样就可以节省时间；

#include<cstdio>
#include<iostream>
#include<cstring>
using namespace std;
class Node
{
   public:
        int l,r,mid;
        long long lazy,sum;
};
Node tree[400024];
long long num[100024];
class Tree
{
 public:
        void Maketree( int l,int r,int cnt );
        void Change( int l,int r,int cnt,int m );
        long long  Qestion( int l,int r,int cnt );    
//        void Pushdown( int cnt );
        void Updata( int cnt );
};
void Tree::Maketree( int l,int r,int cnt )
{
   if( l == r )
   {
        tree[cnt].l = tree[cnt].r =tree[cnt].mid = l;
        tree[cnt].sum = num[l];
        tree[cnt].lazy = 0;
        return ;        
   }    
   else
   {
       tree[cnt].l = l;
       tree[cnt].r = r;
       tree[cnt].mid = ( l + r )>>1;
       Maketree( l , tree[cnt].mid , cnt*2  );
       Maketree( tree[cnt].mid + 1 , r , cnt*2 +1 );
       tree[cnt].sum = tree[cnt*2].sum + tree[cnt*2+1].sum;
       tree[cnt].lazy = 0;    
   }
}
void Tree::Change( int l, int r, int cnt,int m )
{
    if( l == tree[cnt].l && r == tree[cnt].r )
    {      
         tree[cnt].lazy+=m;   
//         Updata( cnt );
         tree[cnt].sum += ( tree[cnt].r - tree[cnt].l +1 )*m;
//         Pushdown( cnt );
         return;    
    }
       if( tree[cnt].lazy )
       Updata( cnt );
    if( r <= tree[cnt].mid )
    {
      Change( l ,r ,cnt*2 , m );
    }
    else
    {
      if( l > tree[cnt].mid )
        Change( l , r , cnt*2 +1 ,m );
      else
      {
         Change( l , tree[cnt].mid , cnt*2 , m );
         Change( tree[cnt].mid+1 , r ,cnt *2+1 ,m );        
      }    
    }
  tree[cnt].sum = tree[cnt*2].sum + tree[cnt*2+1].sum; 
}
void Tree::Updata( int cnt )
{
   tree[cnt*2].lazy += tree[cnt].lazy;
   tree[cnt*2+1].lazy += tree[cnt].lazy;
//   tree[cnt].sum=(tree[cnt*2].sum + tree[cnt*2+1].sum);    
//   tree[cnt].sum+=(tree[cnt].r - tree[cnt].l)*tree[cnt].lazy;    
  tree[cnt*2].sum+=( tree[cnt*2].r - tree[cnt*2].l +1 )*tree[cnt].lazy;
  tree[cnt*2+1].sum+=( tree[cnt*2+1].r - tree[cnt*2+1].l +1 )*tree[cnt].lazy;
  tree[cnt].lazy=0;
}
//void Tree::Pushdown( int cnt )
//{
   
//   tree[cnt].lazy = 0;    
//}
long long Tree::Qestion(int l,int r, int cnt )
{ 
   
   if( l == tree[cnt].l && r == tree[cnt].r )
   {
       return  tree[cnt].sum;
   }
   else
   {
          Updata( cnt );
      if( l>tree[cnt].mid )
        return Qestion( l , r ,cnt *2 +1  );
      else 
      if( r <= tree[cnt].mid )
        return Qestion( l , r, cnt *2 );
      else return Qestion( l , tree[cnt].mid , cnt*2 )+Qestion( tree[cnt].mid+1 , r ,cnt*2+1 );     
    }
}
int main( )
{
   int m,n,l,r;
   long long number;
   char c[5];
   while( scanf( "%d%d",&n,&m )==2 )
   {
       Tree e;
       memset( tree,0,sizeof( tree ) );
       for( int i=1; i<= n ; i++ )
       {
           scanf( "%lld",&num[i] );    
        }
        e.Maketree(1 , n , 1);
        while( m-- )
        {
           scanf( "%s",c );
           if( c[0]=='Q' )
           {
             scanf( "%d%d",&l,&r );
             printf( "%lld\n",e.Qestion( l , r, 1 ) );
           }
           else
           {
             scanf( "%d%d%lld",&l,&r,&number );
             e.Change( l ,r ,1, number );    
            }    
        }
   }
   return 0;    
}