POJ - 3468 A Simple Problem with Integers

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.
 
用线段树表示一个区间上的和，注意区间更新时的lazy数组的处理。
 1 #include<iostream>
 2 #include<cstdio>
 3 #include<cstring>
 4 #define ls o<<1
 5 #define rs o<<1|1
 6 #define lson L,mid,ls
 7 #define rson mid+1,R,rs
 8 #define middnf int mid=(L+R)>>1;
 9 #define ll long long
10 
11 using namespace std;
12 
13 const int maxn=100005; 
14 ll mx[maxn<<2],lazy[maxn<<2];
15 int n,m,x,y,z;
16 
17 void pushup(int o)
18 {
19     mx[o]=mx[ls]+mx[rs];
20 }
21 
22 void pushdown(int o,int m)
23 {
24     if(!lazy[o]) return;
25     lazy[ls] += lazy[o];
26     lazy[rs] += lazy[o];
27     mx[ls]+=(m-(m>>1))*lazy[o];
28     mx[rs]+=(m>>1)*lazy[o];
29     lazy[o] = 0;
30 }
31 
32 void build(int L,int R,int o)
33 {
34     if(L==R)
35     {
36         scanf("%lld",&mx[o]);
37         return ;
38     }
39     middnf;
40     build(lson);
41     build(rson);
42     pushup(o);
43     //cout<<o<<" "<<mx[o]<<endl;
44 }
45 
46 void update(int l,int r,int L,int R,int o,int v){//区间更新
47     if(l<=L&&R<=r)
48     {
49         mx[o] += (ll)v*(R-L+1);
50         lazy[o] += v;
51         return;
52     }
53     pushdown(o,R-L+1);
54     middnf;
55     if(l<=mid) update(l,r,lson,v);
56     if(r>mid) update(l,r,rson,v);
57     pushup(o);
58 }
59 
60 ll query(int l,int r,int L,int R,int o)
61 {
62     if(l<=L&&R<=r) 
63         return mx[o];
64     pushdown(o,R-L+1);
65     middnf;
66     ll ret = 0;
67     if(l<=mid) 
68         ret = ret+query(l,r,lson);
69     if(r>mid) 
70         ret = ret+query(l,r,rson);
71     return ret;
72 }
73 
74 int main()
75 {
76     char c[5];
77     while(~scanf("%d %d",&n,&m))
78     {
79         build(1,n,1);   
80         for(int i=0;i<m;i++)
81         {
82             scanf("%s",c);
83             if(c[0]=='Q')
84             {
85                 scanf("%d%d",&x,&y);
86                 ll man=query(x,y,1,n,1);
87                 printf("%lld\n",man);
88             }
89             else
90             {
91                 scanf("%d%d%d",&x,&y,&z);
92                 update(x,y,1,n,1,z);
93             }
94         }
95     }    
96     
97     
98     return 0;
99 }

 

posted @ 2017-08-22 08:20  西北会法语  阅读(111)  评论(0)  编辑  收藏  举报