q=%26%2340%2B%26%2341%E8%AF%BA%E6%9B%BC%C2%B7%E6%B1%89%E9%87%91%E6%96%AF%26%2340Norman%2BE.Hankins%26%2341&isFacet=true&view=standard&searchWay=author&rows=10&sortWay=score&sortOrder=desc&searchWay0=marc&logical0=AND
rows=10&searchWay0=marc&logical0=AND
(+)诺曼·汉金斯(Norman+E.Hankins)