poj 3468 A Simple Problem with Integers 线段树区间加，区间查询和(模板）

A Simple Problem with Integers

Time Limit: 1 Sec  Memory Limit: 256 MB

题目连接

http://poj.org/problem?id=3468

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

题意

区间加，区间查询和

题解：

入门模板题，学了这么久的线段树，竟然打了20分钟，而且还debug了，果然还是太菜了。

忘记建树导致re，忘记return ans。虽然是细节错误，但是说明自己还是不够熟练。什么时候才可以一遍打完，不用编译直接交，不需看结果，帅气切题啊。

代码：

 #include<cstdio>
 #include<iostream>
 #include<algorithm>
 using namespace std;
 #define N 200050
 #define ll long long
 int n,m; ll w[N];
 struct Tree{int l,r;ll j,sum;}tr[N<<];
 template<typename T>void read(T&x)
 {
   int k=;char c=getchar();
   x=;
   while(!isdigit(c)&&c!=EOF)k^=c=='-',c=getchar();
   if (c==EOF)exit();
   while(isdigit(c))x=x*+c-'',c=getchar();
   x=k?-x:x;
 }
 void read_char(char &c)
 {while(!isalpha(c=getchar())&&c!=EOF);}
 void push_up(int x)
 {
   ll len=tr[x].r-tr[x].l+;
   tr[x].sum=len*tr[x].j;
   if (len==)return;
   tr[x].sum+=tr[x<<].sum+tr[x<<|].sum;
 }
 void push_down(int x)
 {
   Tree &a=tr[x<<],&b=tr[x<<|];
   a.j+=tr[x].j;
   b.j+=tr[x].j;
   push_up(x<<);
   push_up(x<<|);
   tr[x].j=;
 }
 void bt(int x,int l,int r)
 {
   tr[x].l=l; tr[x].r=r; tr[x].j=;
   if (l==r)
     {
       tr[x].j=w[l];
       push_up(x);
       return;
     }
   int mid=(l+r)>>;
   bt(x<<,l,mid);
   bt(x<<|,mid+,r);
   push_up(x);
 }
 void update(int x,int l,int r,ll tt)
 {
   if (l<=tr[x].l&&tr[x].r<=r)
     {
       tr[x].j+=tt;
       push_up(x);
       return;
     }
   int mid=(tr[x].l+tr[x].r)>>;
   if(l<=mid)update(x<<,l,r,tt);
   if (mid<r)update(x<<|,l,r,tt);
   push_up(x);
 }
 ll query(int x,int l,int r)
 {
   if(l<=tr[x].l&&tr[x].r<=r)
     return tr[x].sum;
   ll ans=,mid=(tr[x].l+tr[x].r)>>;
   push_down(x);
   if (l<=mid)ans+=query(x<<,l,r);
   if (mid<r)ans+=query(x<<|,l,r);
   push_up(x);
   return ans;
 }
 int main()
 {
   #ifndef ONLINE_JUDGE
   freopen("aa.in","r",stdin);
   #endif
   read(n); read(m);
   for(int i=;i<=n;i++)read(w[i]);
   bt(,,n);
   for(int i=;i<=m;i++)
     {
       char id;
       int x,y; ll tt;
       read_char(id);read(x);read(y);
       if (id=='C')read(tt),update(,x,y,tt);
       if (id=='Q')printf("%lld\n",query(,x,y));
     }
 }