【学习】线段树的动态开点

就是用哪里开哪里，每次访问之前看看是否为空，为空就新建一个节点。
代码量小，非常好用。
###题目来源：BZOJ 1012
###代码：

#include <cstdio>
#include <iostream>
#define mid ((l+r)>>1)
typedef long long ll;
const int maxn = 200010;
struct Tree{
	ll mx;
	Tree* ch[2];
} T[maxn*20], *root;
ll cnt, D, m, Pre = 0, pp;
Tree* get(){return &T[++cnt];}
void insert(Tree *now, int l, int r, ll val, int pos){
	now->mx = std::max(now->mx, val);
	if(l == r) return;
	int wh = 1; if(pos <= mid) wh = 0;
	if(now->ch[wh] == NULL) now->ch[wh] = get();
	insert(now->ch[wh], wh == 0 ? l : mid+1, wh == 0 ? mid : r, val, pos);
}
ll que(Tree *now, int l, int r, int pos1, int pos2){
	if(pos1 == l && pos2 == r) return now->mx;
	if(pos2 <= mid) return que(now->ch[0], l, mid, pos1, pos2);
	if(pos1 >= mid+1) return que(now->ch[1], mid+1, r, pos1, pos2);
	return std::max(que(now->ch[0], l, mid, pos1, mid),que(now->ch[1], mid+1, r, mid+1, pos2));
}
int main(){
	scanf("%lld%lld", &m, &D);
	root = get();
	for(int i = 1; i <= m; i ++){
		char cc[2]; ll x;
		scanf("%s%lld", cc, &x);
		if(cc[0] == 'A'){
			x = (Pre+x)%D;
			insert(root, 1, m, x, ++pp);
		}else{
			Pre = que(root, 1, m, pp-x+1, pp);
			printf("%lld\n", Pre);
		}
	}
	return 0;
}