Can you answer these queries IV
Time Limit: 5000ms
Memory Limit: 262144KB
This problem will be judged on SPOJ. Original ID: GSS4
64-bit integer IO format: %lld Java class name: Main
You are given a sequence A of N(N <= 100,000) positive integers. There sum will be less than 1018. On this sequence you have to apply M (M <= 100,000) operations:
(A) For given x,y, for each elements between the x-th and the y-th ones (inclusively, counting from 1), modify it to its positive square root (rounded down to the nearest integer).
(B) For given x,y, query the sum of all the elements between the x-th and the y-th ones (inclusively, counting from 1) in the sequence.
Input
Multiple test cases, please proceed them one by one. Input terminates by EOF.
For each test case:
The first line contains an integer N. The following line contains N integers, representing the sequence A1..AN.
The third line contains an integer M. The next M lines contain the operations in the form "i x y".i=0 denotes the modify operation, i=1 denotes the query operation.
Output
For each test case:
Output the case number (counting from 1) in the first line of output. Then for each query, print an integer as the problem required.
Print an blank line after each test case.
See the sample output for more details.
Example
Input: 5 1 2 3 4 5 5 1 2 4 0 2 4 1 2 4 0 4 5 1 1 5 4 10 10 10 10 3 1 1 4 0 2 3 1 1 4 Output: Case #1: 9 4 6 Case #2: 40 26
Source
Own problem, used in ACM/ICPC Regional Contest, Shanghai 2011 preliminary
解题：线段树
1 #include <bits/stdc++.h>
2 using namespace std;
3 typedef long long LL;
4 const int maxn = 100010;
5 struct node{
6 int lt,rt;
7 LL maxv,minv,sum,lazy;
8 bool reset;
9 }tree[maxn<<2];
10 inline void pushup(int v){
11 tree[v].maxv = max(tree[v<<1].maxv,tree[v<<1|1].maxv);
12 tree[v].minv = min(tree[v<<1].minv,tree[v<<1|1].minv);
13 tree[v].sum = tree[v<<1].sum + tree[v<<1|1].sum;
14 }
15 inline void pushdown(int v){
16 if(tree[v].reset){
17 tree[v<<1].sum = tree[v].lazy*(tree[v<<1].rt - tree[v<<1].lt + 1);
18 tree[v<<1].maxv = tree[v<<1|1].minv = tree[v].lazy;
19 tree[v<<1|1].sum = tree[v].lazy*(tree[v<<1|1].rt - tree[v<<1|1].lt + 1);
20 tree[v<<1|1].maxv = tree[v<<1|1].minv = tree[v].lazy;
21 tree[v].reset = false;
22 }
23 }
24 void build(int lt,int rt,int v){
25 tree[v].lt = lt;
26 tree[v].rt = rt;
27 tree[v].reset = false;
28 if(lt == rt){
29 scanf("%lld",&tree[v].maxv);
30 tree[v].minv = tree[v].sum = tree[v].maxv;
31 return;
32 }
33 int mid = (lt + rt)>>1;
34 build(lt,mid,v<<1);
35 build(mid + 1,rt,v<<1|1);
36 pushup(v);
37 }
38 void update(int lt,int rt,int v){
39 LL a = sqrt(tree[v].maxv*1.0),b = sqrt(tree[v].minv*1.0);
40 if(lt <= tree[v].lt && rt >= tree[v].rt && a == b){
41 tree[v].reset = true;
42 tree[v].sum = a*(tree[v].rt - tree[v].lt + 1);
43 tree[v].maxv = tree[v].minv = a;
44 tree[v].lazy = a;
45 return;
46 }
47 pushdown(v);
48 if(lt <= tree[v<<1].rt) update(lt,rt,v<<1);
49 if(rt >= tree[v<<1|1].lt) update(lt,rt,v<<1|1);
50 pushup(v);
51 }
52 LL query(int lt,int rt,int v){
53 if(lt <= tree[v].lt && rt >= tree[v].rt) return tree[v].sum;
54 pushdown(v);
55 LL ret = 0;
56 if(lt <= tree[v<<1].rt) ret = query(lt,rt,v<<1);
57 if(rt >= tree[v<<1|1].lt) ret += query(lt,rt,v<<1|1);
58 pushup(v);
59 return ret;
60 }
61 int main(){
62 int n,m,op,x,y,cs = 1;
63 while(~scanf("%d",&n)){
64 build(1,n,1);
65 scanf("%d",&m);
66 printf("Case #%d:\n",cs++);
67 while(m--){
68 scanf("%d%d%d",&op,&x,&y);
69 if(op) printf("%lld\n",query(x,y,1));
70 else update(x,y,1);
71 }
72 }
73 return 0;
74 }