abc374E Sensor Optimization Dilemma 2

生产某种产品有N道工序，对于工序i，有S[i]和T[i]两类机器可供选择，机器S[i]单价为P[i]，每台每天能处理A[i]件；机器T[i]单价为Q[i]，每台每天能处理B[i]件。在不超预算X的前提下，每天最多能生产多少件产品？
1<=N<=100; 1<=A[i],B[i]<=100; 1<=P[i],Q[i],X<=1E7

分析：最大产能为所有工序的最小值，可以二分答案，由于A[i],B[i]较小，可以按A[i],B[i]来枚举。

#include <bits/stdc++.h>
using i64 = long long;

void solve() {
	i64 N, X;
	std::cin >> N >> X;
	std::vector<i64> A(N + 1), B(N + 1), P(N + 1), Q(N + 1);
	for (int i = 1; i <= N; i++) {
		std::cin >> A[i] >> P[i] >> B[i] >> Q[i];
	}

	auto check = [&](i64 u) {
		i64 cost = 0;
		for (int i = 1; i <= N; i++) {
			i64 t = 1E9;
			for (int j = 0; j < A[i]; j++) {
				i64 k = std::max(0LL, u - j * B[i] + A[i] - 1) / A[i];
				t = std::min(t, j * Q[i] + k * P[i]);
			}
			for (int j = 0; j < B[i]; j++) {
				i64 k = std::max(0LL, u - j * A[i] + B[i] - 1) / B[i];
				t = std::min(t, j * P[i] + k * Q[i]);
			}
			cost += t;
			if (cost > X) {
				return false;
			}
		}
		return true;
	};

	i64 lo = 0, hi = 1E9, mid;
	while (lo < hi) {
		mid = lo + (hi - lo + 1) / 2;
		if (check(mid)) {
			lo = mid;
		} else {
			hi = mid - 1;
		}
	}
	std::cout << lo << "\n";
}

int main() {
	std::cin.tie(0)->sync_with_stdio(0);
	int t = 1;
	while (t--) solve();
	return 0;
}