【玩法设计】材料副本：每日轮换表的供需平衡设计

local MATERIAL_ROTATION = {
    { day = 1, material = "玄铁矿石", count = 20, boss = "矿洞守卫" },
    { day = 2, material = "厚皮革",   count = 15, boss = "皮革商人" },
    { day = 3, material = "灵纹布",   count = 15, boss = "织法者" },
    { day = 4, material = "玄铁矿石", count = 20, boss = "矿洞守卫" },
    { day = 5, material = "龙筋",     count = 8,  boss = "龙脉守护" },
    { day = 6, material = "凤羽",     count = 8,  boss = "凤巢守卫" },
    { day = 0, material = "全能宝箱", count = 1,  boss = "混沌守卫" },
}
function MaterialDungeon.today()
    local wday = tonumber(os.date("%w"))
    for _, r in ipairs(MATERIAL_ROTATION) do
        if r.day == wday then return r end
    end
end

function MaterialDungeon.onClear(actor, material, count)
    actor:AddItemByName(material, count)
    QF_LogDungeon("material", actor:GetUserId(), material, count)
    sendmsg(actor, 1, "获得 " .. material .. " x" .. count)
end