HDU 4031 树状数组 区间更新及点询问

/*******************************************************************************
    去年成都赛区网络赛一道题，树状数组在区间更新中的应用。树状数组一般支持的是改点，查区间，但
是这道题要求的是改区间，查点，这就要变通一下，具体可以看代码，另外要考虑的一个问题是，如何统计？
这里的处理方法就是把总共的攻击次数-防御的次数，因为对单个点的询问可能有多次，所以为每个点都设
定一个游标来优化时间～
*******************************************************************************/
#include <iostream>
#include <functional>
#include <algorithm>
#include <complex>
#include <cstdlib>
#include <cstring>
#include <fstream>
#include <iomanip>
#include <sstream>
#include <utility>
#include <bitset>
#include <cctype>
#include <cstdio>
#include <limits>
#include <memory>
#include <string>
#include <vector>
#include <cmath>
#include <ctime>
#include <queue>
#include <stack>
#include <list>
#include <map>
#include <set>
using namespace std;

#define LOWBIT(x) ( (x) & ( (x) ^ ( (x) - 1 ) ) )
#define CLR(x, k) memset((x), (k), sizeof(x))
#define CPY(t, s) memcpy((t), (s), sizeof(s))
#define SC(t, s) static_cast<t>(s)
#define LEN(s) static_cast<int>( strlen((s)) )
#define SZ(s) static_cast<int>( (s).size() )

typedef double LF;
typedef __int64 LL;		//VC
typedef unsigned __int64 ULL;

typedef pair<int, int> PII;
typedef pair<LL, LL> PLL;
typedef pair<double, double> PDD;

typedef vector<int> VI;
typedef vector<char> VC;
typedef vector<double> VF;
typedef vector<string> VS;

template <typename T>
T sqa(const T & x) {return x * x;}
template <typename T>
T ll(const T & x) {return x << 1;}
template <typename T>
T rr(const T & x) {return x << 1 | 1;}
template <typename T>
T gcd(T a, T b) 
{ 
	if (!a || !b) 
	{
		return max(a, b);
	}
	T t;
	while (t = a % b)
	{
		a = b;
		b = t;
	}
	return b;
};

const int INF_INT = 0x3f3f3f3f;
const LL INF_LL = 0x7fffffffffffffffLL;		//15f
const double oo = 10e9;
const double eps = 10e-7;
const double PI = acos(-1.0);

#define  ONLINE_JUDGE

const int MAXN = 20004;

int test, n, q, t;
int tree[MAXN];
int crs[MAXN], def[MAXN];
vector<PII> vp;

void updateBIT(int x, int inc)
{
	for (int i = x; i > 0; i -= LOWBIT(i))
	{
		tree[i] += inc;
	}
	return ;
}
void updateSeg(int l, int r, int inc)
{
	updateBIT(l - 1, -inc);
	updateBIT(r, inc);
	return ;
}
int queryBIT(int x)
{
	int sum = 0;
	for (int i = x; i <= n; i += LOWBIT(i))
	{
		sum += tree[i];
	}
	return sum;
}
void ace()
{
	int cas = 1;
	char op[10];
	int l, r, p;
	for (scanf("%d", &test); test--; ++cas)
	{
		scanf("%d %d %d", &n, &q, &t);
		CLR(tree, 0);
		CLR(crs, 0);
		CLR(def, 0);
		vp.clear();
		int ind = 0;
		printf("Case %d:\n", cas);
		while (q--)
		{
			scanf("%s", op);
			if ('A' == op[0])
			{
				scanf("%d %d", &l, &r);
				if (l > r)
				{
					swap(l, r);
				}
				//l = max(l, 1);
				//r = min(r, n);
				updateSeg(l, r, 1);
				vp.push_back(PII(l, r));
				++ind;
			}
			else
			{
				scanf("%d", &p);
				if (1 == t)
				{
					puts("0");
					continue ;
				}
				int token = 0;
				while (crs[p] < ind)
				{
					if (vp[ crs[p] ].first <= p && p <= vp[ crs[p] ].second)
					{
						if (0 == token % t)
						{
							crs[p] += t - 1;
							++def[p];
							token = t - 1;
						}
						++token;
					}
					++crs[p];
				}
				printf("%d\n", queryBIT(p) - def[p]);
			}
		}
	}
	return ;
}
int main()
{
#ifndef ONLINE_JUDGE
	freopen("in.txt", "r", stdin);
	freopen("out.txt", "w", stdout);
#endif
	ace();
	return 0;
}
/*******************************************************************************
Test Data...
2
3 7 2
Attack 1 2
Query 2
Attack 2 3
Query 2
Attack 1 3
Query 1
Query 3
9 7 3
Attack 5 5
Attack 4 6
Attack 3 7
Attack 2 8
Attack 1 9
Query 5
Query 3
*******************************************************************************/