?? cave.c
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// Room: /d/huashan/cave.c// Date: Look 99/03/25inherit ROOM;void create(){ set("short", "密洞"); set("long",@long你走進來,發現石壁后面,別有洞天,是個可容得千人之眾的大洞,洞中有七具骸骨,或坐,或臥,身旁均有兵刃。山洞四壁離地數丈突出一塊大石,似是一個平臺,大石之下刻著十六個大字(dazi)。你舉起火把更往石壁上照看時,只見一行字刻道:范松趙鶴破恒山劍法于此。這一行之旁是無數人形,粗略一計,少說也有五六百個。long); set("exits",([ "right" : __DIR__"rukou",])); set("item_desc",([ "dazi" : "\t五岳劍派,無恥下流,比武不勝,暗算害人。又見十六個,大字旁更刻了無數小字,都是些“卑鄙誣賴”、“可恥之極”、“低能”、“懦弱”等等詛咒的字眼。\n", ])); set("indoors", "華山" ); set("book_count", 1); setup();}void init(){ add_action("do_lingwu", "lingwu"); add_action("do_ketou", "ketou");}int do_lingwu(string arg){ object me, ob; me = this_player(); ob = this_object(); if(!living(me) ) return 0; if (me->is_busy() || me->is_fighting()) return notify_fail("你正忙著哪!\n"); if( present("sword book", me)) { if( me->query_skill("sword" , 1) < 100 && me->query_skill("sword", 1) > 30 ) { write("你對著石壁上的圖形領悟,覺得對你的基本劍法啟發不少。\n", me); me->improve_skill("sword", (int)me->query("int")*3/2); me->receive_damage("jing", 15 + random((int)me->query_skill("sword" , 1) / 4)); return 1; } } if( present("staff book", me)) { if( me->query_skill("staff" , 1) < 101 && me->query_skill("staff", 1) > 50) { write("你對著石壁上的圖形領悟,覺得對你的基本杖法啟發不少。\n", me); me->improve_skill("staff", (int)me->query("int")*3/2); me->receive_damage("jing", 15 + random((int)me->query_skill("staff" , 1) / 4)); return 1; } }}int do_ketou(string arg){ object me, ob; me = this_player(); ob = this_object(); if(!living(me) ) return 0; if (me->is_busy() || me->is_fighting()) return notify_fail("你正忙著哪!\n"); if(!arg || arg =="" || arg !="skeleton") return notify_fail("你要干什么?\n"); if(arg=="skeleton") { if( ob->query("book_count") < 1) { write("你跪在地上,對著骸骨虔誠的磕頭。\n", me); return 1; } if( me->query_temp("ketou/骸骨", 1) == 30 ) { ob=new(__DIR__"obj/staff_book"); if(!clonep(ob)) return notify_fail("你累的躺到在地,什么也沒有發生。\n"); ob->move(me); message_vision("突然從骸骨中掉出一本書出來。\n", me); me->delete_temp("ketou/骸骨"); this_object()->add("book_count", -1); return 1; } write("你跪在地上,對著骸骨虔誠的磕頭。\n", me); me->add_temp("ketou/骸骨", 1); return 1; } }
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