// Solution for the sub‑task where T[i] = 1 for every coupon.
// The game reduces to buying as many coupons as possible while the total
// price does not exceed the initial number of tokens A.
// The optimal strategy is to purchase coupons in non‑decreasing order of price.
#include "festival.h"
#include <bits/stdc++.h>
using namespace std;
vector<int> max_coupons(int A, vector<int> P, vector<int> T) {
int N = (int)P.size();
// Pair each coupon with its index and price.
vector<pair<int,int>> coupons;
coupons.reserve(N);
for (int i = 0; i < N; ++i) {
coupons.emplace_back(P[i], i);
}
// Sort by price (ascending).
sort(coupons.begin(), coupons.end(),
[](const pair<int,int>& a, const pair<int,int>& b) {
return a.first < b.first;
});
long long tokens = A; // current amount of tokens (use 64-bit)
vector<int> answer; // indices of bought coupons in order
answer.reserve(N);
for (const auto& cp : coupons) {
long long price = cp.first;
if (tokens >= price) { // we can afford this coupon
tokens -= price; // after purchase (T[i] == 1)
answer.push_back(cp.second);
} else {
// All remaining coupons are at least as expensive,
// therefore we can stop.
break;
}
}
return answer; // maximal number of coupons bought
}
