1 #include <cstdio>
 2 #include <cstring>
 3 const int N=50000*4+5;
 4 struct node{
 5     int l,r;
 6     int sum;
 7 }tree[N];
 8 int a[50005];
 9 int n;
10 void build(int bg,int ed,int i){//建树
11     if(i>4*n)    return ;//记得设返回条件（4*n个节点足以），否则会无限制建下去
12     tree[i].l=bg;
13     tree[i].r=ed;
14     if(bg==ed)  tree[i].sum=a[bg];//找到叶子节点
15     else{
16         build(bg, (bg+ed)/2, i*2);//建左节点
17         build((bg+ed)/2+1, ed, i*2+1);//建右节点
18         tree[i].sum=tree[2*i].sum+tree[2*i+1].sum;//区间总值
19     }
20 }
21 void update(int i,int x,int y){//更新
22     if(tree[i].l==tree[i].r)    tree[i].sum+=y;
23     else{
24         int mid=(tree[i].l+tree[i].r)/2;
25         if(x<=mid)  update(2*i, x, y);
26         else update(2*i+1, x, y);
27         tree[i].sum=tree[2*i].sum+tree[2*i+1].sum;//回溯更新父节点
28     }
29 }
30 int query(int bg,int ed,int i){
31     if(tree[i].l==bg && tree[i].r==ed) return tree[i].sum;//找到完全匹配的节点
32     else{
33         int mid=(tree[i].l+tree[i].r)/2;
34         if(ed<=mid) return query(bg, ed, 2*i);//查询区间在左子树中
35         else if(bg>mid)  return query(bg, ed, 2*i+1);//查询区间在右子树种
36         else if(bg<=mid && ed>mid)  return query(bg, mid, 2*i)+query(mid+1, ed, 2*i+1);//查询区间左右子树皆含有
37     }
38     return -1;
39 }
40 int main(){
41     int t,x,y;
42     char op[10];
43     scanf("%d",&t);
44     for (int i=1; i<=t; i++){
45         printf("Case %d:\n",i);
46         scanf("%d",&n);
47         for (int j=1; j<=n; j++) scanf("%d",&a[j]);
48         build(1, n, 1);
49         while(scanf("%s",op)!=EOF && strcmp(op,"End")!=0){
50             scanf("%d%d",&x,&y);
51             if(!strcmp(op, "Add"))   update(1,x,y);
52             else if(!strcmp(op, "Sub"))  update(1,x,-y);
53             else if(!strcmp(op, "Query"))   printf("%d\n",query(x,y,1));
54         }
55         
56     }
57     return 0;
58 }