《逆水寒手遊》奇遇觸發指南 Part.1｜大吉級奇遇位置與條件詳解

正文摘要:

▶ 《逆水寒手遊》奇遇觸發指南 - Part 1：大吉級奇遇 👉 福 利 遊 戲 推 薦 ⬇️ 點 ⬇️⬇️ 免費玩各種動漫手遊 ▶ 《逆水寒手遊》大吉奇遇總覽 ▶ 《逆水寒手遊》大吉奇遇詳解 奇遇名稱座標位置觸發NPC觸發條件巴圖疑案杭州(1341,1019)宋文從NPC面前經過夜雨兩相別汴京(1202,887)青衣/烏衣直接對話觸發春日遊汴京(870,1170)錢六爺完成汴京主線後泡水韌筆如刀汴京(1020,1026)劉三對話選擇「不了不了」 觸發要點： 1. 巴圖疑案： - 位置精確度要求高 - 建議多次來回走動 - 無特殊前置條件 2. 夜雨兩相別： - 直接找到NPC對話 - 選項無特殊要求 - 可重複觸發機會 3. 春日遊： - 需要完成汴京主線 - 在指定位置泡水 - 注意時機把握 4. 韌筆如刀： - 對話選項很重要 - 必須選擇「不了不了」 - 位置要求精確 var custom_call_MIR = function (params) {if ( params === null || params.hasAd === false ) {var divRowSide;for (let index = 0; index < ONEADs.length; index++) {if (ONEADs[index].pub.player_mode === "mobile-inread") {divRowSide = ONEADs[index].pub.slotobj;}}var ins = document.createElement('ins');ins.className = 'clickforceads';ins.setAttribute('style', 'display:inline-block;');ins.setAttribute('data-ad-zone', '18264');divRowSide.appendChild(ins);var js = document.createElement('script');js.async = true;js.type = 'text/javascript';js.setAttribute('src', '//cdn.holmesmind.com/js/init.js');js.onload = function(){console.log('[ONEAD MIR] ClickForce Loaded');};divRowSide.appendChild(js);}};var _ONEAD = {};_ONEAD.pub = {};_ONEAD.pub.slotobj = document.getElementById("oneadMIRDFPTag");_ONEAD.pub.slots = ["div-onead-ad"];_ONEAD.pub.uid = "1000509";_ONEAD.pub.external_url = "https://onead.onevision.com.tw/";_ONEAD.pub.scopes = ["spotbuy", "speed"];_ONEAD.pub.player_mode_div = "div-onead-ad";_ONEAD.pub.player_mode = "mobile-inread";_ONEAD.pub.queryAdCallback = custom_call_MIR;var ONEAD_pubs = ONEAD_pubs || [];ONEAD_pubs.push(_ONEAD);var truvidScript = document.createElement('script'); truvidScript.async = true; truvidScript.setAttribute('data-cfasync','false'); truvidScript.type = 'text/javascript'; truvidScript.src = '//stg.truvidplayer.com/index.php?sub_user_id=597&widget_id=3309&playlist_id=2381&m=a&cb=' + (Math.random() * 10000000000000000); var currentScript = document.currentScript || document.scripts[document.scripts.length - 1]; currentScript.parentNode.insertBefore(truvidScript, currentScript.nextSibling);