poj 3468 线段树区间更新/查询

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.

 

题意：裸线段树 区间更新/查询 

 

题解：线段树每次写都有不同程度的错误.too vegatable

1.延迟标记的add 都要初始化为0 

2.延迟的传递是累加而不是赋值

具体看代码。

 

 /******************************
 code by drizzle
 blog: www.cnblogs.com/hsd-/
 ^ ^    ^ ^
  O      O
 ******************************/
 //#include<bits/stdc++.h>
 #include<iostream>
 #include<cstring>
 #include<cstdio>
 #include<map>
 #include<algorithm>
 #include<cmath>
 #define ll long long
 #define PI acos(-1.0)
 #define mod 1000000007
 using namespace std;
 struct node
 {
     ll l,r;
     ll add;
     ll sum;
 } tree[];
 void  buildtree(ll root,ll left,ll right)
 {
     tree[root].l=left;
     tree[root].r=right;
     tree[root].add=;//wa点 所有的延迟标记都要初始化
     if(left==right)//不能放到下面的if中
     {

         ll exm;
         scanf("%lld",&exm);
         tree[root].sum=exm;
         return ;
     }
     ll mid=(left+right)>>;
     buildtree(root<<,left,mid);
     buildtree(root<<|,mid+,right);
     tree[root].sum=tree[root<<].sum+tree[root<<|].sum;
 }
 void pushdown(ll root)
 {
     if(tree[root].add==) return ;
     tree[root<<].add+=tree[root].add;//wa点 这里是增加而不是赋值
     tree[root<<|].add+=tree[root].add;
     tree[root<<].sum+=(tree[root<<].r-tree[root<<].l+)*tree[root].add;
     tree[root<<|].sum+=(tree[root<<|].r-tree[root<<|].l+)*tree[root].add;
     tree[root].add=;
 }
 void updata(ll root,ll left,ll right,ll c)
 {
     if(tree[root].l==left&&tree[root].r==right)
     {
         tree[root].add+=c;//wa点 这里是增加而不是赋值
         tree[root].sum+=(right-left+)*c;
         return ;
     }
      pushdown(root);
     ll mid=(tree[root].l+tree[root].r)>>;
     if(right<=mid)
         updata(root<<,left,right,c);
     else
     {
         if(left>mid)
             updata(root<<|,left,right,c);
         else
         {
             updata(root<<,left,mid,c);
             updata(root<<|,mid+,right,c);
         }
     }
     tree[root].sum=tree[root<<].sum+tree[root<<|].sum;
 }
 ll query(ll root,ll left,ll right)
 {
     if(tree[root].l==left&&tree[root].r==right)
     {
         return tree[root].sum;
     }
     pushdown(root);
     ll mid=(tree[root].l+tree[root].r)>>;
     if(right<=mid)
         return query(root<<,left,right);
     else
     {
         if(left>mid)
             return query(root<<|,left,right);
         else
             return query(root<<,left,mid)+query(root<<|,mid+,right);
     }
 }
 ll n,q;
 char what;
 ll l1,r1,ad;
 int main()
 {
     while(~scanf("%lld %lld",&n,&q))
     {
         buildtree(,,n);
         getchar();
         for(ll i=; i<=q; i++)
         {
             scanf("%c %lld %lld",&what,&l1,&r1);
             if(what=='Q')
                 printf("%lld\n",query(,l1,r1));
             else
             {
                 scanf("%lld",&ad);
                 updata(,l1,r1,ad);
             }
             getchar();
         }
     }
     return ;
 }
 /*
 10 10
 1 2 3 4 5 6 7 8 9 10
 C 1 10 1
 Q 2 3

 10 10
 1 2 3 4 5 6 7 8 9 10
 C 1 5 1
 Q 4 6
 */