poj 3468 A Simple Problem with Integers 线段树第一次 + 讲解

A Simple Problem with Integers

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

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

报告
题目大意： 一串数，“C a b c" 表示在区间[a,b]中每个数 + c
              "Q a b"表示查询区间[a,b] 的和

线段树：   学习了线段树后的第一道题。这道题包含了线段树的Pushdown,Pushup,Query,Build,Update几个函数，很好地让初学者了解了线段树的用法和精髓--延迟更新，即每次更新时只是做一个标记，当要查找时才继续向下更新。还要注意的是，有些题不需要向下回溯（pushup），也不需要延迟更新（pushdown），甚至连更新都不用。但是需要根据题意改变要查找的内容和更新的内容，有的需要找最大值，有的需要找和，等等，需要随机应变。线段树能解决的问题，树状数组不一定能解决，但是树状数组能解决的，线段树一定能解决。

思路：  便不多说。上代码。

 #include<iostream>
 #include<cstring>
 #include<string>
 #include<cstdio>
 #define L(u) (u<<1)
 #define R(u) (u<<1|1)//没有分号
 using namespace std;
 int n,q;
 long long c;
 long long a[];
 struct node{
     int l,r;
     long long add,sum;
 };
 char s[];
 node pp[];/**数据范围**/
 void pushup(int u)
 {
     pp[u].sum=pp[L(u)].sum+pp[R(u)].sum;
     return ;
 }
 void pushdown(int u)
 {
     pp[L(u)].add+=pp[u].add;
     pp[L(u)].sum+=(pp[L(u)].r-pp[L(u)].l+)*pp[u].add;
     pp[R(u)].add+=pp[u].add;
     pp[R(u)].sum+=(pp[R(u)].r-pp[R(u)].l+)*pp[u].add;
     pp[u].add=;/*！！！*/
 }
 void update(int u,int left,int right,long long t)
 {
     if (left<=pp[u].l&&pp[u].r<=right)
     {
         pp[u].add+=t;
         pp[u].sum+=(pp[u].r-pp[u].l+)*t;
         return ;
     }
     pp[u].sum+=(right-left+)*t;
     if (pp[u].add) pushdown(u);
     int mid=(pp[u].l+pp[u].r)>>;
     if (right<=mid) update(L(u),left,right,t);
     else /***/if (left>mid) update(R(u),left,right,t);
     else
     {
         update(L(u),left,mid,t);
         update(R(u),mid+,right,t);
     }
     //pushup(u);
 }

 void build(int u,int left,int right)
 {
     pp[u].l=left;
     pp[u].r=right;
     pp[u].add=;
     if (pp[u].l==pp[u].r)
     {
         pp[u].sum=a[left];//*****
         return ;
     }
     int mid=(pp[u].l+pp[u].r)>>;
     build(L(u),left,mid);
     build(R(u),mid+,right);
     pushup(u);
 }
 long long query(int u,int left,int right)
 {
     if (left==pp[u].l&&pp[u].r==right)
         return pp[u].sum;
     if (pp[u].add) pushdown(u);
     int mid=(pp[u].l+pp[u].r)>>;
     if (right<=mid) return query(L(u),left,right);
     if (left>mid) return query(R(u),left,right);
     else
       return (query(L(u),left,mid)+query(R(u),mid+,right));
 }
 int main()
 {
     //freopen("tree.in","r",stdin);
     cin>>n>>q;
     for (int i=;i<=n;i++)
           scanf("%lld",&a[i]);//long long 读数
     build(,,n);
     for (int i=;i<=q;i++)
     {
         scanf("%s",s);
         if (s[]=='Q')
         {
             int k,b;
             scanf("%d%d",&k,&b);
             cout<<query(,k,b)<<endl;
         }
         else
         {
             int k,b;
             cin>>k>>b>>c;
             update(,k,b,c);
         }
     }

     return ;
 }

注意事项：1、数据范围：线段树一般定义4*n
            2、#difine 的使用  function() ( ** ) 注意括号没有分号；
　　　　　3、runtime error ：运行错误 long long a[i], %lld(linux) %I64d （windows)
　　　　　4、模板答题

顺便抱怨一下，poj的评测系统太慢了，waiting了好久.......