模板題目：here

題意：給了一個數組，有兩種操作：0,x,y，把【x,y】之間的數都開平方；1,x,y，詢問【x,y】之間的所有數的和

AC_Code:

 1 #include <iostream>
 2 #include <cstdio>
 3 #include <algorithm>
 4 #include <cmath>
 5 #include <cstring>
 6 using namespace std;
 7 typedef long long ll;
 8 const int maxn = 1e5+10;
 9 
10 struct
 
 Tree{
11     ll l,r,val;
12 }tree[maxn<<2];
13 ll n,m;
14 
15 void pushup(ll rt){
16     tree[rt].val = tree[rt<<1].val + tree[rt<<1|1].val;
17     return ;
18 }
19 
20 void build(ll l,ll r,ll rt=1ll){
21     tree[rt].l=l; tree[rt].r=r;
22     tree[rt].val = 0;
23     if( l==r ){
24
 
         scanf("%lld",&tree[rt].val);
25         return ;
26     }
27     ll mid=(l+r)>>1;
28     build(l,mid,rt<<1);
29     build(mid+1,r,rt<<1|1);
30     pushup(rt);
31     return ;
32 }
33 
34 void modify(ll x, ll y, ll rt=1ll){
35     ll l=tree[rt].l, r=tree[rt].r;
36     if
 
( l==r ){
37         tree[rt].val = sqrt(tree[rt].val);
38         return ;
39     }
40     if( tree[rt].val==r-l+1 && x<=l && r<=y ) return ;//一個數最多開平方8次，所以加個剪枝，當區間和等於區間元素數時，不再遞迴下去
41     int mid = (l+r)>>1;
42     if( x<=mid )modify(x,y,rt<<1);
43     if( y>mid ) modify(x,y,rt<<1|1);
44     pushup(rt);
45 }
46 
47 ll query(ll x,ll y,ll rt=1ll){
48     ll l = tree[rt].l, r = tree[rt].r;
49     if( x<=l && r<=y ) return tree[rt].val;
50     ll mid=(l+r)>>1;
51     ll ans = 0;
52     if( x<=mid ) ans += query(x,y,rt<<1);
53     if( y>mid ) ans += query(x,y,rt<<1|1);
54     return ans;
55 }
56 
57 int main()
58 {
59     int cas=0;
60     while( ~scanf("%lld",&n) ){
61         build(1,n);
62         scanf("%lld",&m);
63         printf("Case #%d:\n",++cas);
64         while( m-- ){
65             int a,b,c; scanf("%d%d%d",&a,&b,&c);
66             int x=min(b,c);
67             int y=max(b,c);
68             switch( a ){
69             case 0:
70                 modify(x,y);
71                 break;
72             case 1:
73                 printf("%lld\n",query(x,y));
74                 break;
75             }
76         }
77         printf("\n");
78     }
79     return 0;
80 }