只移动前后边界，二分搜索越界的机器人

#include<bits/stdc++.h>
using namespace std;
#define int long long
#define MAX_SIZE 333333

int a[MAX_SIZE];
int n, m, s;
int lower_(int x){
	int* t = lower_bound(a + 1, a + n + 1, x);
	return (t - a - 1);
}
int upper_(int x){
	int *t = upper_bound(a + 1, a + n + 1, x);
	return (t - a);
}

signed main() {
	ios::sync_with_stdio(false);
	cin.tie(0);
	cin >> n >> m >> s;
	for (int i = 1; i <= n; i++){
		cin >> a[i];
	}
	sort(a + 1, a + n + 1);
	int ss = -s;
	int q = 0;
	int p = n + 1;
	while(m--){
		int op, d, a, b;
		cin >> op;
		if (op == 1){
			cin >> d;
			ss -= d;
			s -= d;
			a = lower_(ss);
			b = upper_(s);
			q = max(q, a);
			p = min(p, b);
			//cout << "q = " << q << "; p = " << p << "\n";
		}else if (op == 2){
			cin >> d;
			ss += d;
			s += d;
			a = lower_(ss);
			b = upper_(s);
			q = max(q, a);
			p = min(p, b);
			//cout << "q = " << q << "; p = " << p << "\n";
		}else {
			cout << max(p - q - 1, 0LL) << "\n";
		}
	}
	return 0;
}

#include<bits/stdc++.h>
#define int long long
using namespace std;

const int N = 300000 + 10;
int n, m, k, op, x, tot, a[N];
//tot记录目前一共向正方向移动了多少个单位，可以是负数（负数代表向反方向移动） 
deque<int> q;//定义双向队列 

signed main(){
	cin >> n >> m >> k;
	for(int i=1;i<=n;i++)	cin >> a[i];//需要另开一个数组输入，因为要排序 
	sort(a + 1, a + 1 + n);//排序 
	for(int i=1;i<=n;i++)	q.push_back(a[i]);//进队 
	for(int i=1;i<=m;i++){
		cin >> op;
		if(op == 3)	cout << q.size() << endl;//输出队列长度，也就是元素个数 
		else if(op == 1){
			cin >> x;
			tot += x;//刷新tot 
			while(!q.empty()){
				int v = q.back();
				if(v + tot > k)	q.pop_back();//如果此埴轮移动完超过k，就被淘汰了 
				else	break;
			}
		}else if(op == 2){
			cin >> x;
			tot -= x;//刷新tot 
			while(!q.empty()){
				int v = q.front();
				if(v + tot < -k)	q.pop_front();//如果此埴轮移动完小于-k，也会被淘汰 （从队头弹出） 
				else	break;
			}
		}
	}
	return 0;
}