First consider the case where all jump pads have a positive $v$. We may directly simulate Bessie's process. At a given power $p$, Bessie will jump at most $\frac n p$ times before she exits the number line or she hits a jump pad, in which her power increases. Therefore the number of bounces is bounded by $\sum_{i=1}^n \frac n i = O(n\lg(n))$.

When there are jump pads with with value $0$, then there's a possibility Bessie can get stuck in an infinite loop. This happens only when Bessie hits a jump pad of value $0$, switches direction, and after some bounces, the next jump pad she hits is also value $0$. It is possible to check for this infinite loop cycle, but it's easier to implement just simulating Bessie's bounces for enough iterations (about $O(n\lg(n))$) since Bessie won't break additional targets while stuck in a loop.

Sample implementation:

#include <bits/stdc++.h>
using namespace std;
 
int main() {
	ios::sync_with_stdio(0);
	cin.tie(0);
	int n, x;
	cin >> n >> x;
	vector<pair<int, int>> pad(n + 1);
	vector<bool> vis(n + 1);
	for (int i = 1; i <= n; ++i)
		cin >> pad[i].first >> pad[i].second;
	int dir = 1, power = 1, ans = 0;
	for (int reps = 0; reps < 5000000 && 1 <= x && x <= n; ++reps) {
		// Bessie breaks a target she hasn't broken before
		if (pad[x].first == 1 && power >= pad[x].second && !vis[x]) {
			vis[x] = true;
			++ans;
		}
		if (pad[x].first == 0) { // jump pad
			dir *= -1;
			power += pad[x].second;
		}
		x += dir * power;
	}
	cout << ans << "\n";
	return 0;
}