bzoj 3212 Pku3468 A Simple Problem with Integers 线段树基本操作

Pku3468 A Simple Problem with Integers

Time Limit: 1 Sec  Memory Limit: 128 MB
Submit: 2173  Solved: 951
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Description

You have N integers, A1, A2, ... , AN. 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, ... , AN. -1000000000 ≤ Ai ≤ 1000000000. 
Each of the next Q lines represents an operation. 
"C a b c" means adding c to each of Aa, Aa+1, ... , Ab. -10000 ≤ c ≤ 10000. 
"Q a b" means querying the sum of Aa, Aa+1, ... , Ab.

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

裸题

 #include<cstring>
 #include<algorithm>
 #include<iostream>
 #include<cmath>
 #include<cstdio>

 #define N 100007
 #define ll long long
 using namespace std;
 inline int read()
 {
     int x=,f=;char ch=getchar();
     while(!isdigit(ch)){if(ch=='-')f=-;ch=getchar();}
     while(isdigit(ch)){x=(x<<)+(x<<)+ch-'';ch=getchar();}
     return x*f;
 }

 int n,q;
 char ch[];

 #define ls p<<1
 #define rs p<<1|1
 struct seg
 {
     ll sum[N<<],add[N<<];
     void build(int p,int l,int r)
     {
         if (l==r)
         {
             sum[p]=read();
             return;
         }
         int mid=(l+r)>>;
         build(ls,l,mid),build(rs,mid+,r);
         sum[p]=sum[ls]+sum[rs];
     }
     void push_down(int p,int l,int r)
     {
         if (!add[p]) return;
         ll ad=add[p],mid=(l+r)>>;add[p]=;
         sum[ls]+=1ll*(mid-l+)*ad;
         sum[rs]+=1ll*(r-mid)*ad;
         add[ls]+=ad,add[rs]+=ad;
     }
     ll query(int p,int l,int r,int x,int y)
     {
         if (l==x&&y==r) return sum[p];
         push_down(p,l,r);
         int mid=(l+r)>>;
         if (y<=mid) return query(ls,l,mid,x,y);
         else if (x>mid) return query(rs,mid+,r,x,y);
         else return query(ls,l,mid,x,mid)+query(rs,mid+,r,mid+,y);
     }
     void modify(int p,int l,int r,int x,int y,int z)
     {
         sum[p]+=1ll*(y-x+)*z;
         if (l==x&&y==r)
         {
             add[p]+=z;
             return;
         }
         int mid=(l+r)>>;
         if (y<=mid) modify(ls,l,mid,x,y,z);
         else if (x>mid) modify(rs,mid+,r,x,y,z);
         else modify(ls,l,mid,x,mid,z),modify(rs,mid+,r,mid+,y,z);
     }
 }seg;
 #undef ls
 #undef rs
 int main()
 {
     n=read(),q=read();
     seg.build(,,n);
     for (int i=;i<=q;i++)
     {
         scanf("%s",ch);
         if (ch[]=='Q')
         {
             int x=read(),y=read();
             printf("%lld\n",seg.query(,,n,x,y));
         }
         else
         {
             int x=read(),y=read(),z=read();
             seg.modify(,,n,x,y,z);
         }
     }
 }