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Easy Problem Once MoreTime Limit: 4000/2000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 109 Accepted Submission(s): 4 Problem Description Define matrix ![]() ![]() then matrix A can be called as partially negative matrix. Here matrix ![]() For example, matrix ![]() ![]() A symmetric matrix is a square matrix that equals to its transpose. Formally, matrix A is symmetric if A = AT. For example, ![]() Given two N-dimensional vector x and b, and we guarantee that there will be at least one 0 value in vector b. You task is to judge if there exists a symmetric partially negative matrix A, which fulfills Ax = b. Input There are several test cases. Proceed to the end of file. Each test case is described in three lines. The first line contains one integer N (2 ≤ N ≤ 100000) . The second line contains N integers xi (-1000000 < xi < 1000000, 1 ≤ i ≤ N), which is vector x. The third line contains N integers bi (-1000000 < bi < 1000000, 1 ≤ i ≤ N), which is vector b. There will be at least one bi which equals to zero. Output For each test case, output “Yes” if there exists such a matrix A, or “No” if there is no such matrix. Sample Input
Sample Output
Hint There exists a symmetric partially negative matrix ![]() ![]() Note Determinant of an n × n matrix A is defined as below: ![]() Here the sum is computed over all permutations σ of the set {1, 2, ..., n}. A permutation is a function that reorders this set of integers. The value in the ith position after the reordering σ is denoted σi. For example, for n = 3, the original sequence 1, 2, 3 might be reordered to σ = [2, 3, 1],with σ1 = 2, σ2 = 3, and σ3 = 1. The set of all such permutations (also known as the symmetric group on n elements) is denoted Sn. For each permutation σ, sgn(σ) denotes the signature of σ, a value that is +1 whenever the reordering given by σ can be achieved by successively interchanging two entries an even number of times, and −1 whenever it can be achieved by an odd number of such interchanges. Source | ||||||||||
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