poj3468 A Simple Problem with Integers

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

Hint

The sums may exceed the range of 32-bit integers.
题目大意：给出一些数和一些操作，q表示询问，c表示修改。
思路：纯线段树
 1 /*
 2  * Author:  Joshua
 3  * Created Time:  2014年07月16日 星期三 15时19分46秒
 4  * File Name: poj3468.cpp
 5  */
 6 #include<cstdio>
 7 #include<algorithm>
 8 #define maxn 100005
 9 #define L(x) (x<<1)
10 #define R(x) (x<<1 |1)
11 typedef long long LL;
12 using namespace std;
13 struct node
14 {
15     int l,r;
16     LL sum,mark;
17 } e[maxn <<2];
18 int n,m;
19 int a[maxn];
20 LL ans;
21 void built(int t,int l,int r)
22 {
23     e[t].l=l;
24     e[t].r=r;
25     e[t].mark=0;
26     if (l==r)
27     {
28         e[t].sum=a[e[t].l];
29         return;
30     }
31     int mid=l+r >>1;
32     built(L(t),l,mid);
33     built(R(t),mid+1,r);
34     e[t].sum=e[L(t)].sum+e[R(t)].sum;
35 }
36 
37 void add(int t,int l,int r,int v)
38 {
39     if (l==e[t].l && r==e[t].r)
40     {
41         e[t].mark+=v;
42         return;
43     }
44     e[t].sum+=(r-l+1)*v;
45     int mid=e[t].l+e[t].r >>1;
46     if (r>mid) add(R(t),max(mid+1,l),r,v);
47     if (l<=mid) add(L(t),l,min(mid,r),v);
48 }
49 
50 void getSum(int t,int l,int r)
51 {
52     ans+=e[t].mark*(r-l+1);
53     if (e[t].l==l && e[t].r==r)
54     {
55         ans+=e[t].sum;
56         return;
57     }
58     int mid=e[t].l+e[t].r >>1;
59     if (r>mid) getSum(R(t),max(mid+1,l),r);
60     if (l<=mid) getSum(L(t),l,min(mid,r));
61 }
62 void solve()            
63 {
64     for (int i=1;i<=n;++i)
65        scanf("%d",&a[i]);
66     built(1,1,n);
67     char c;
68     int x,y,v;
69     for (int i=1;i<=m;++i)
70     {
71         scanf("%c",&c);
72         scanf("%c%d%d",&c,&x,&y);
73         if (c=='Q') 
74         {
75             ans=0;
76             getSum(1,x,y);
77             printf("%lld\n",ans);
78         }
79         else
80         {
81             scanf("%d",&v);
82             add(1,x,y,v);
83         }
84     }    
85 }
86 int main()
87 {
88     while (scanf("%d%d",&n,&m)==2)
89         solve();
90 
91     return 0;
92 }

 

posted @ 2014-07-18 14:42  一个大叔  阅读(140)  评论(0)  编辑  收藏  举报