POJ3468 线段树（区间更新，区间求和，延迟标记）

A Simple Problem with Integers

Time Limit: 5000MS		Memory Limit: 131072K
Total Submissions: 97196		Accepted: 30348
Case Time Limit: 2000MS

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：

一串数，有两种操作，在区间a,b内每一个数加c,求区间a,b数的和。

线段树模板题。

代码：

 #include<iostream>
 #include<string>
 #include<cstdio>
 #include<cstring>
 #include<algorithm>
 using namespace std;
 int n,q;
 long long sum[];//数组多乘4倍。
 long long add[];
 void pushup(int rt)
 {
     sum[rt]=sum[rt<<]+sum[rt<<|];
 }
 void pushdown(int rt,int lne)
 {
     if(add[rt])
     {
         add[rt<<]+=add[rt];
         add[rt<<|]+=add[rt];
         sum[rt<<]+=add[rt]*(lne-(lne>>));
         sum[rt<<|]+=add[rt]*(lne>>);
         add[rt]=;
     }
 }
 void build(int l,int r,int rt)
 {
     add[rt]=;
     if(l==r)
     {
         scanf("%lld",&sum[rt]);
         return;
     }
     int mid=(l+r)>>;
     build(l,mid,rt<<);
     build(mid+,r,rt<<|);
     pushup(rt);
 }
 void update(int L,int R,int c,int l,int r,int rt)
 {
     if(L<=l&&R>=r)
     {
         add[rt]+=c;
         sum[rt]+=(long long)c*(r-l+);
         return;
     }
     pushdown(rt,r-l+);
     int mid=(l+r)>>;
     if(L<=mid)
     update(L,R,c,l,mid,rt<<);
     if(R>mid)
     update(L,R,c,mid+,r,rt<<|);
     pushup(rt);
 }
 long long query(int L,int R,int l,int r,int rt)
 {
     if(L<=l&&R>=r)
     {
         return sum[rt];
     }
     pushdown(rt,r-l+);
     int mid=(l+r)>>;
     long long ans=;
     if(L<=mid)
     ans+=query(L,R,l,mid,rt<<);
     if(R>mid)
     ans+=query(L,R,mid+,r,rt<<|);
     return ans;
 }
 int main()
 {
     int a,b,c;
     scanf("%d%d",&n,&q);
     build(,n,);
     while(q--){
     char ch[];
     scanf("%s",ch);  //不能用scanf("%c".....),字符后有空格，可以scanf一个字符数组/cin。
     if(ch[]=='Q')
     {
         scanf("%d%d",&a,&b);
         printf("%lld\n",query(a,b,,n,));
     }
     else if(ch[]=='C')
     {
         scanf("%d%d%d",&a,&b,&c);
         update(a,b,c,,n,);
     }
     }
     return ;
 }