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| 1 | +Let us sort A, B and C in non-decreasing order. |
| 2 | + |
| 3 | +A : A_1 >= A_2 >= ... >= A_x |
| 4 | +B : B_1 >= B_2 >= ... >= B_y |
| 5 | +C : C_1 >= C_2 >= ... >= C_z |
| 6 | + |
| 7 | +The optimum pairing will not have any (i, j, k) such that i*j*k > K. |
| 8 | + |
| 9 | +We need to search at most till i*j*k where i*j*k <= K. |
| 10 | + |
| 11 | +--- |
| 12 | + |
| 13 | +int main() |
| 14 | +{ |
| 15 | + int no_of_1_candles, no_of_2_candles, no_of_3_candles, no_of_chosen_cakes; |
| 16 | + cin >> no_of_1_candles >> no_of_2_candles >> no_of_3_candles >> no_of_chosen_cakes; |
| 17 | + |
| 18 | + vector <LL> cake_1(no_of_1_candles + 1, 0); |
| 19 | + read_and_sort(cake_1, no_of_1_candles); |
| 20 | + |
| 21 | + vector <LL> cake_2(no_of_2_candles + 1, 0); |
| 22 | + read_and_sort(cake_2, no_of_2_candles); |
| 23 | + |
| 24 | + vector <LL> cake_3(no_of_3_candles + 1, 0); |
| 25 | + read_and_sort(cake_3, no_of_3_candles); |
| 26 | + |
| 27 | + vector <LL> chosen_cakes; |
| 28 | + for(int i = 1; i <= no_of_1_candles; i++) |
| 29 | + { |
| 30 | + for(int j = 1; j <= no_of_2_candles; j++) |
| 31 | + { |
| 32 | + for(int k = 1; k <= no_of_3_candles; k++) |
| 33 | + { |
| 34 | + if(i*j*k > no_of_chosen_cakes) |
| 35 | + break; |
| 36 | + |
| 37 | + chosen_cakes.push_back(cake_1[i] + cake_2[j] + cake_3[k]); |
| 38 | + } |
| 39 | + } |
| 40 | + } |
| 41 | + |
| 42 | + sort(all(chosen_cakes), greater <LL>()); |
| 43 | + |
| 44 | + for(int i = 0; i < no_of_chosen_cakes; i++) |
| 45 | + cout << chosen_cakes[i] << "\n"; |
| 46 | + |
| 47 | + return 0; |
| 48 | +} |
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