cf1354 D 树状数组+二分

题意：https://www.luogu.com.cn/problem/CF1354D

思路：首先我们要注意到每次加入的和原先在数组中的数都是处于1到n之间的，所以我们考虑树状数组标记每个值的状态，那么我们应该如何删掉第k个呢，我们每次删除操作的时候二分查找第k个数的下标即可。

/*keep on going and never give up*/
#include<cstdio>
#include<iostream>
#include<queue>
#include<algorithm>
using namespace std;
#define int long long
typedef pair<int, int> pii;
#define lowbit(x) x&(-x)
#define endl '\n'
#define wk is zqx ta die
int n, q;
int a[1000005];
int k[1000005];
int f[1000005];
void add(int x, int y) {
	while (x <= n) {
		f[x] += y;
		x += lowbit(x);
	}
}
int query(int x) {
	int res = 0;
	while (x) {
		res += f[x];
		x -= lowbit(x);
	}
	return res;
}
int pos(int x) {
	int l = 1;
	int r = n;
	int ans;
	while (l <= r) {
		int mid = (l + r) >> 1;
		if (query(mid) >= x) {
			ans = mid;
			r = mid - 1;
		} else {
			l = mid + 1;
		}
	}
	return ans;
}
signed main() {
	std::ios::sync_with_stdio(false);
	cin.tie(0);
	cout.tie(0);
	cin >> n >> q;
	for (int i = 1; i <= n; i++) {
		cin >> a[i];
		add(a[i], 1);
	}
	for (int i = 1; i <= q; i++) {
		int k;
		cin >> k;
		if (k > 0) {
			add(k, 1);
		} else {
			add(pos(-k), -1);
		}
	}
	if (query(n) == 0) {
		cout << "0" << endl;
	} else {
		cout << pos(1) << endl;
	}
	return 0;
}