题意：每个顶点与其左边一个点和下面一个点与原点围成一个四边形，求最小删掉几个四边形，使得所有四边形互不相交。

思路：其实每个顶点可以看作占据了一个斜率区间，那么问题就变成的经典的区间问题，我们直接按照右端点排序然后贪心即可。

/*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
struct node {long double l, r;
} q[200005];
bool cmp(node a, node b) {return a.r < b.r;
}
signed main() {std::ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);int n;cin >> n;for (int i = 1; i <= n; i++) {long double x, y;cin >> x >> y;q[i].l = (x - 1) / y;q[i].r = x / (y - 1);}sort(q + 1, q + n + 1, cmp);long double last = -1;int ans = 0;for (int i = 1; i <= n; i++) {if (q[i].l >= last) {ans++;last = q[i].r;}}cout << ans << endl;return 0;
}