q=%26%2340%E7%BE%8E%26%2341%E7%BD%97%E6%9D%B0%C2%B7%E6%A2%85%E9%87%8C%E5%B0%94%26%2340A.Roger%2BMerrill%26%2341&searchType=standard&isFacet=true&view=standard&searchWay=author&rows=10&sortWay=score&sortOrder=desc&searchWay0=marc&logical0=AND
rows=10&searchWay0=marc&logical0=AND
(美)罗杰·梅里尔(A.Roger+Merrill)