Can you answer these queries?
Problem Description
A lot of battleships of evil are arranged in a line before the battle. Our commander decides to use our secret weapon to eliminate the battleships. Each of the battleships can be marked a value of endurance. For every attack of our secret weapon, it could decrease the endurance of a consecutive part of battleships by make their endurance to the square root of it original value of endurance. During the series of attack of our secret weapon, the commander wants to evaluate the effect of the weapon, so he asks you for help.
You are asked to answer the queries that the sum of the endurance of a consecutive part of the battleship line.
Notice that the square root operation should be rounded down to integer.
Input
The input contains several test cases, terminated by EOF.
For each test case, the first line contains a single integer N, denoting there are N battleships of evil in a line. (1 <= N <= 100000)
The second line contains N integers Ei, indicating the endurance value of each battleship from the beginning of the line to the end. You can assume that the sum of all endurance value is less than 263.
The next line contains an integer M, denoting the number of actions and queries. (1 <= M <= 100000)
For the following M lines, each line contains three integers T, X and Y. The T=0 denoting the action of the secret weapon, which will decrease the endurance value of the battleships between the X-th and Y-th battleship, inclusive. The T=1 denoting the query of the commander which ask for the sum of the endurance value of the battleship between X-th and Y-th, inclusive.
Output
For each test case, print the case number at the first line. Then print one line for each query. And remember follow a blank line after each test case.
Sample Input
10
1 2 3 4 5 6 7 8 9 10
5
0 1 10
1 1 10
1 1 5
0 5 8
1 4 8
Sample Output
Case #1:
19 7 6
题意：给出一个n的数组，支持2种操作：
1.0 a b :把a到b之间的数都开平方
2.1 a b :查询区间a~b的数之和
注意：这道题求出的平方根要下取整。
然后，每个样例之后有空行。
有个技巧，写在了注释。
1 #include<cstdio>
2 #include<cstring>
3 #include<algorithm>
4 #include<cmath>
5
6 using namespace std;
7
8 #define lson l,m,rt<<1
9 #define rson m+1,r,rt<<1|1
10 #define LL long long
11 const int maxn=100000+5;
12
13 LL sum[maxn<<2];
14
15 //因为2^63总和最多2^63
16 //如果只有1个数，也就最多开7次根号
17 //即开着开着就恒为1
18 //所以，当该区间的和等于区间长度（r-l）+1的时候，
19 //该区间的值都不会再更新
20
21 //1表示这个区间不需要更新了
22
23 void pushup(int rt)
24 {
25 sum[rt]=sum[rt<<1]+sum[rt<<1|1];
26
27 }
28
29 void build(int l,int r,int rt)
30 {
31 if(l==r)
32 {
33 scanf("%lld",&sum[rt]);
34 return ;
35 }
36
37 int m=(l+r)>>1;
38
39 build(lson);
40 build(rson);
41 pushup(rt);
42 }
43
44 void update(int L,int R,int l,int r,int rt)
45 {
46 if(sum[rt]==r-l+1)
47 return ;
48
49 if(l==r)
50 {
51 sum[rt]=(LL)(sqrt((long double)sum[rt]));
52 //printf("%lld ",sum[rt]);
53 return ;
54 }
55
56 int m=(l+r)>>1;
57
58 if(L<=m)
59 update(L,R,lson);
60 if(R>m)
61 update(L,R,rson);
62
63 pushup(rt);
64 }
65
66 LL query(int L,int R,int l,int r,int rt)
67 {
68 if(L<=l&&R>=r)
69 {
70 return sum[rt];
71 }
72
73 int m=(l+r)>>1;
74
75 LL ret=0;
76
77 if(L<=m)
78 ret+=query(L,R,lson);
79 if(R>m)
80 ret+=query(L,R,rson);
81
82 return ret;
83 }
84
85 int main()
86 {
87 int cas=1;
88
89 int n;
90
91 while(scanf("%d",&n)!=EOF)
92 {
93 //memset(cal,0,sizeof(cal));
94
95 build(1,n,1);
96
97 int m;
98 scanf("%d",&m);
99
100 int T,x,y;
101
102 printf("Case #%d:\n",cas++);
103
104 for(int i=1;i<=m;i++)
105 {
106 scanf("%d %d %d",&T,&x,&y);
107
108 if(x>y)
109 swap(x,y);
110
111 if(T==0)
112 {
113 update(x,y,1,n,1);
114 }
115 else
116 {
117 printf("%lld\n",query(x,y,1,n,1));
118 }
119 }
120
121 printf("\n");
122 }
123 return 0;
124 }