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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 , if for every m(1 ≒ m ≒ n), then matrix A can be called as partially negative matrix. Here matrix , and {i1,..,im} is a sub set of {1,..,n}.If you are not familiar with determinant of a matrix, please read the Note part of this problem. For example, matrix is a partially negative matrix because |-2|, |-6| andare negative. A symmetric matrix is a square matrix that equals to its transpose. Formally, matrix A is symmetric if A = AT. For example, is a symmetric matrix. 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 考 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 考 is denoted 考i. For example, for n = 3, the original sequence 1, 2, 3 might be reordered to 考 = [2, 3, 1],with 考1 = 2, 考2 = 3, and 考3 = 1. The set of all such permutations (also known as the symmetric group on n elements) is denoted Sn. For each permutation 考, sgn(考) denotes the signature of 考, a value that is +1 whenever the reordering given by 考 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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