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| 2134 | Codeforces Round 1045 (Div. 2) | FINISHED | False | 7200 | 20186723 | Aug. 26, 2025, 2:35 p.m. |
Solved |
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| ( 640 ) | F | Permutation Oddness | PROGRAMMING | combinatorics dp |
You are given four positive integers (c_0), (c_1), (c_2), and (c_3). Let (n = c_0 + c_1 + c_2 + c_3). Consider an array (a) of (n) integers with (x) ((0\le x\le 3)) appearing (c_x) times. For any distinct permutation (^{\text{∗}}) (b) of the array (a), define its oddness as(^{\text{†}})(^{\text{‡}}): ()\sum_{i = 1}^{n-1} \text{lowbit}(b_i \oplus b_{i+1})() Your task is to count, for each integer (k) from (0) to (2 \cdot (n-1)) (inclusive), the number of distinct permutations of (a) with an oddness equal to (k). Since the numbers might be too large, you are only required to find them modulo (10^9 + 7). (^{\text{∗}})A permutation of the array is an arrangement of its elements into any order. For example, (1,2,2) is a permutation of (2,2,1), but (1,1,2) is not. Two permutations are considered distinct if they differ in at least one position. (^{\text{†}})(\oplus) denotes the bitwise XOR operation . (^{\text{‡}})(\text{lowbit}(x)) is the value of the lowest binary bit of (x), e.g. (\text{lowbit}(12)=4), (\text{lowbit}(8)=8). Specifically, we define (\text{lowbit}(0) = 0). Each test contains multiple test cases. The first line contains the number of test cases (t) ((1 \le t \le 50)). The description of the test cases follows. The first and the only line of each test case contains four positive integers (c_0), (c_1), (c_2), and (c_3) ((1 \le c_0, c_1, c_2, c_3 < 800), (4 \le c_0 + c_1 + c_2 + c_3 \le 800)). Let (n = c_0 + c_1 + c_2 + c_3). It is guaranteed that the sum of (n) over all test cases does not exceed (800). For each test case, output (2 \cdot (n-1) + 1) integers in one line — the number of distinct permutations of (a) with an oddness equal to (0,1,\ldots, 2 \cdot (n-1)) respectively. Output the answers modulo (10^9+7). In the first test case, the ar |
| 145832 |
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