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

 1 #include<cstring>
 2 #include<algorithm>
 3 #include<iostream>
 4 #include<cmath>
 5 #include<cstdio>
 6 
 7 #define N 100007
 8 #define ll long long
 9 using namespace std;
10 inline int read()
11 {
12     int x=0,f=1;char ch=getchar();
13     while(!isdigit(ch)){if(ch=='-')f=-1;ch=getchar();}
14     while(isdigit(ch)){x=(x<<1)+(x<<3)+ch-'0';ch=getchar();}
15     return x*f;
16 }
17 
18 int n,q;
19 char ch[4];
20 
21 #define ls p<<1
22 #define rs p<<1|1
23 struct seg
24 {
25     ll sum[N<<2],add[N<<2];
26     void build(int p,int l,int r)
27     {
28         if (l==r)
29         {
30             sum[p]=read();
31             return;
32         }
33         int mid=(l+r)>>1;
34         build(ls,l,mid),build(rs,mid+1,r);
35         sum[p]=sum[ls]+sum[rs];
36     }
37     void push_down(int p,int l,int r)
38     {
39         if (!add[p]) return;
40         ll ad=add[p],mid=(l+r)>>1;add[p]=0;
41         sum[ls]+=1ll*(mid-l+1)*ad;
42         sum[rs]+=1ll*(r-mid)*ad;
43         add[ls]+=ad,add[rs]+=ad;
44     }
45     ll query(int p,int l,int r,int x,int y)
46     {
47         if (l==x&&y==r) return sum[p];
48         push_down(p,l,r);
49         int mid=(l+r)>>1;
50         if (y<=mid) return query(ls,l,mid,x,y);
51         else if (x>mid) return query(rs,mid+1,r,x,y);
52         else return query(ls,l,mid,x,mid)+query(rs,mid+1,r,mid+1,y);
53     }
54     void modify(int p,int l,int r,int x,int y,int z)
55     {
56         sum[p]+=1ll*(y-x+1)*z;
57         if (l==x&&y==r)
58         {
59             add[p]+=z;
60             return;
61         }
62         int mid=(l+r)>>1;
63         if (y<=mid) modify(ls,l,mid,x,y,z);
64         else if (x>mid) modify(rs,mid+1,r,x,y,z);
65         else modify(ls,l,mid,x,mid,z),modify(rs,mid+1,r,mid+1,y,z);
66     }
67 }seg;
68 #undef ls
69 #undef rs
70 int main()
71 {
72     n=read(),q=read();
73     seg.build(1,1,n);
74     for (int i=1;i<=q;i++)
75     {
76         scanf("%s",ch);
77         if (ch[0]=='Q')
78         {
79             int x=read(),y=read();
80             printf("%lld\n",seg.query(1,1,n,x,y));
81         }
82         else
83         {
84             int x=read(),y=read(),z=read();
85             seg.modify(1,1,n,x,y,z);
86         }
87     }
88 }

 

posted @ 2018-04-15 15:39  Kaiser-  阅读(129)  评论(0)  编辑  收藏  举报