hdu 1166 敌兵布阵 线段树 点更新

// hdu 1166 敌兵布阵 线段树 点更新
//
// 这道题裸的线段树的点更新，直接写就能够了
//
// 一直以来想要进线段树的坑，结果一直没有跳进去，今天算是跳进去吧，
// 尽管十分简单。十分的水，继续加油

#include <algorithm>
#include <bitset>
#include <cassert>
#include <cctype>
#include <cfloat>
#include <climits>
#include <cmath>
#include <complex>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <ctime>
#include <deque>
#include <functional>
#include <iostream>
#include <list>
#include <map>
#include <numeric>
#include <queue>
#include <set>
#include <stack>
#include <vector>
#define ceil(a,b) (((a)+(b)-1)/(b))
#define endl '\n'
#define gcd __gcd
#define highBit(x) (1ULL<<(63-__builtin_clzll(x)))
#define popCount __builtin_popcountll
typedef long long ll;
using namespace std;
const int MOD = 1000000007;
const long double PI = acos(-1.L);

template<class T> inline T lcm(const T& a, const T& b) { return a/gcd(a, b)*b; }
template<class T> inline T lowBit(const T& x) { return x&-x; }
template<class T> inline T maximize(T& a, const T& b) { return a=a<b?b:a; }
template<class T> inline T minimize(T& a, const T& b) { return a=a<b?a:b; }

const int maxn = 4 * 50008;
int n;
int sum[maxn];

void build(int root,int L,int R){
	if (L==R){
		scanf("%d",&sum[root]);
		return;
	}
	int M = L + (R - L) / 2;
	build(root * 2,L,M);
	build(root * 2 + 1, M+1,R);
	sum[root] = sum[root * 2] + sum[root * 2 + 1];
}

void init(){
	scanf("%d",&n);
	build(1,1,n);
}
int p,num;
void add_num(int root,int L,int R){
	if (L==R){
		sum[root] += num;
		return;
	}
	int M = L + (R - L) / 2;
	if (p<=M)add_num(root*2,L,M);
	else add_num(root*2+1,M+1,R);
	sum[root] = sum[root * 2] + sum[root * 2 + 1];
}

int query(int root,int ql,int qr,int L,int R){
	if (ql<=L&&R<=qr){
		return sum[root];
	}
	int M = L + (R - L) / 2;
	int ans = 0;
	if (ql<=M)	ans += query(root*2,ql,qr,L,M);
	if (M<qr)	ans += query(root*2+1,ql,qr,M+1,R);
	return ans;
}

void solve(){
	char s[10];
	int x,y;
	int i = 0;
	while(1){
		scanf("%s",s);
	//	cout << i << endl;
		if (s[0]=='Q'){
			scanf("%d%d",&x,&y);
	//	cout << i << endl;
			printf("%d\n",query(1,x,y,1,n));
		}else if (s[0]=='A'){
			scanf("%d%d",&x,&y);
			p = x;
			num = y;
			add_num(1,1,n);
		}else if (s[0]=='S'){
			scanf("%d%d",&x,&y);
			p = x;
			num = -y;
			add_num(1,1,n);
		}else {
			break;
		}
	}
}

int main() {
	int t;
	//freopen("G:\\Code\\1.txt","r",stdin);
	scanf("%d",&t);
	int kase=1;
	while(t--){
		init();
		printf("Case %d:\n",kase++);
		solve();
	}
	return 0;
}