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| 1 | +Let us consider a monster (D, H) and a unit (d, h). |
| 2 | + |
| 3 | +Time taken by monster to kill 1 unit = h/D |
| 4 | +Time taken by unit to kill monster = H/d |
| 5 | + |
| 6 | +We need to ensure that the time taken by the monster is strictly more than the time we take to kill the monster |
| 7 | + |
| 8 | +h/D > H/d |
| 9 | +h*d > H*D |
| 10 | + |
| 11 | +We can treat all the monsters as products (D x H) |
| 12 | + |
| 13 | +For a given unit, we need to find the minimum integer c, such that |
| 14 | + |
| 15 | +h x (c x d) > H x D |
| 16 | + |
| 17 | +----- |
| 18 | + |
| 19 | +Another condition is that c <= C/C[i] |
| 20 | + |
| 21 | +----- |
| 22 | + |
| 23 | +Now, instead of checking the amount of damage a monster can take, let us try to calculate the maximum |
| 24 | +product we can reach with C[i] coins |
| 25 | + |
| 26 | +Maximum_product[C[i]] = max(H[i] x D[i]) initially |
| 27 | + |
| 28 | +The key insight to propagate this DP is that we can buy multiple sets of the same unit |
| 29 | + |
| 30 | +If we buy C coin sets of the i-th unit |
| 31 | + |
| 32 | +Maximum_product[C[i] x C] = max(C x H[i] X D[i]) |
| 33 | + |
| 34 | +----- |
| 35 | + |
| 36 | +We can then binary search the answer for each query |
| 37 | + |
| 38 | +----- |
| 39 | + |
| 40 | +int main() |
| 41 | +{ |
| 42 | + int no_of_elements, max_cost; |
| 43 | + cin >> no_of_elements >> max_cost; |
| 44 | + |
| 45 | + vector <long long> cost(no_of_elements + 1), damage(no_of_elements + 1), health(no_of_elements + 1); |
| 46 | + vector <long long> max_product(max_cost + 1, 0); |
| 47 | + for(int i = 1; i <= no_of_elements; i++) |
| 48 | + { |
| 49 | + cin >> cost[i] >> damage[i] >> health[i]; |
| 50 | + |
| 51 | + max_product[cost[i]] = max(max_product[cost[i]], damage[i]*health[i]); |
| 52 | + } |
| 53 | + |
| 54 | + |
| 55 | + for(int c = 1; c <= max_cost; c++) |
| 56 | + { |
| 57 | + for(int coin_sets = 1; c*1LL*coin_sets <= max_cost; coin_sets++) |
| 58 | + { |
| 59 | + max_product[c*coin_sets] = max(max_product[c*coin_sets], max_product[c]*coin_sets); |
| 60 | + } |
| 61 | + } |
| 62 | + |
| 63 | + vector <long long> max_product_till(max_cost + 1); |
| 64 | + for(int i = 1; i <= max_cost; i++) |
| 65 | + { |
| 66 | + max_product_till[i] = max(max_product[i], max_product_till[i - 1]); |
| 67 | + } |
| 68 | + |
| 69 | + int no_of_monsters; |
| 70 | + cin >> no_of_monsters; |
| 71 | + |
| 72 | + for(int i = 1; i <= no_of_monsters; i++) |
| 73 | + { |
| 74 | + long long damage_here, health_here; |
| 75 | + cin >> damage_here >> health_here; |
| 76 | + |
| 77 | + long long product = damage_here*health_here; |
| 78 | + |
| 79 | + int minimum_coins = upper_bound(all(max_product_till), product) - max_product_till.begin(); |
| 80 | + |
| 81 | + cout << (minimum_coins <= max_cost ? minimum_coins : -1) << "\n"; |
| 82 | + } |
| 83 | + |
| 84 | + return 0; |
| 85 | +} |
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