시간 제한 | 메모리 제한 | 제출 | 정답 | 맞힌 사람 | 정답 비율 |
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1.5 초 | 512 MB | 81 | 11 | 9 | 14.516% |
Alice enjoys thinking about base-K numeral systems (don’t we all?). As you might know, in the standard base-K numeral system, an integer n can be represented as dm−1 dm−2 . . . d1 d0 where:
For example, in standard base-3, you would write 15 as 1 2 0, since (1) · 32 + (2) · 31 + (0) · 30 = 15.
But standard base-K systems are too easy for Alice. Instead, she’s thinking about weird-base-K systems.
A weird-base-K system is just like the standard base-K system, except that instead of using the digits {0, . . . , K − 1}, you use {a1, a2, . . . , aD} for some value D. For example, in a weird-base-3 system with a = {−1, 0, 1}, you could write 15 as 1 -1 -1 0
, since (1) · 33 + (−1) · 32 + (−1) · 31 + (0) · 30 = 15.
Alice is wondering how to write Q integers, n1 through nQ, in a weird-base-K system that uses the digits a1 through aD. Please help her out!
The first line contains four space-separated integers, K, Q, D, and M (2 ≤ K ≤ 1 000 000, 1 ≤ Q ≤ 5, 1 ≤ D ≤ 5001, 1 ≤ M ≤ 2500).
The second line contains D distinct integers, a1 through aD (−M ≤ ai ≤ M).
Finally, the i-th of the next Q lines contains ni (−1018 ≤ ni ≤ 1018).
Output Q lines, the i-th of which is a weird-base-K representation of ni. If multiple representations are possible, any will be accepted. The digits of the representation should be separated by spaces. Note that 0 must be represented by a non-empty set of digits.
If there is no possible representation, output IMPOSSIBLE
.
3 3 3 1 -1 0 1 15 8 -5
1 -1 -1 0 1 0 -1 -1 1 1
We have: (1) · 33 + (−1) · 32 + (−1) · 31 + (0) · 30 = 15, (1) · 32 + (0) · 31 + (−1) · 30 = 8, and (−1) · 32 + (1) · 31 + (1) · 30 = −5.
10 1 3 2 0 2 -2 17
IMPOSSIBLE