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PJSK (1): Random Stuff

我手机在自动下载 好吧，我也不知道我为什么会开始玩这种任何方面都吸引不了我的游戏 （才不是因为某可爱粉毛） ，但它确实是很上头...随便说说好了，我水平非常烂，不必认真对待。 提示：这不是一篇完整的教程。 以下内容针对国服（全名懒得写了），别的服可能略有差别。 角色养成部分 卡牌带有分数，技能，属性，组合等属性，都是与生俱来的，但前两个可以升级，比如通过游戏或者使用道具升级或者观看角色剧情blabla，上下限取决于卡牌等级。一个队伍可以带有5张卡，但 必须不是同一个角色 。不同队伍可以带有相同的卡牌，也可以随时选择需要的队伍。 开局和观看剧情会送一些低等级卡(2*的对应组合的初音未来和4个1*的真人角色)，以及一些高等级(3*4*)的自选兑换券，可以选择一些喜欢的，最好是同一个组合的，因为虚拟世界除了角色加成道具还有组合道具可以给整个组合加成。现实世界的全局属性加成略，均衡发展即可。 这样是凑不齐一个4*队的，想要更多的卡牌需要抽卡。 抽卡部分 感觉概率有点问题，所以，如果想要什么具体的卡，最好等到有足够的水晶可以保底的时候再抽，毕竟水晶大体上不可再生。 最好只选择那个10连抽，因为各种东西都是以10为单位的，单抽没有好处。 一般4*概率都是3%，但可能是重复的。 由于每种卡牌至多只能持有一张，重复卡牌可以去休息室兑换成养成需要的紫色石头。 左上角有200抽保底的是常驻池，当期角色是其他时候随时可能随机出现的，抽卡可以兑换绿票，可以在同类卡池保底兑换的时候凑一部分，但这是有限的，上限都是10张。集齐30张也可以直接兑换此类4*卡牌，但似乎活动结束后需要一段时间才能这样做。 其他的是限定池，当期角色只会随机出现在(?)，抽卡可以兑换粉票，作用同上但不能直接兑换卡牌。标记了“回响”的是复刻从未存在的活动卡池，可以用20张粉票。 6%概率的是fes池，附加在某个限定池之上，有节日限定角色，建议抽。可以兑换20张粉票（因为它自己的兑换所和其依附的那个限定池是通用的），但兑换角色只能用10张。 生日池会把所有的4*替换成目标生日卡牌，3*2*也全部替换成该角色，保底100抽。 特殊的卡池略。 活动部分 活动会对特定的角色，属性和组合加成，按需组队即可，凑不齐可以用其他组合对应属性的来凑，还是凑不齐可以直接用最强的卡。 为此每次活动都需要培养一些卡，虽然标准放宽到3*也是可以的...
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Metal (1): Electrolysis

Well, since chemistry is nothing than electron transferring, electrolysis can act as both the strongest oxidant and reductant, but if controlled it can help us obtain high-quality crystals or something nearly impossible from normal chemical reactions. Since almost all of these were performed long long ago, we can only try our best to recall. Of course, having positive oxidation state, metals can only be deposited on the cathode. Na Being one of the most reactive metals, it reacts very violently with water and many weaker acids like NH3, so obtaining from aqueous solution is surely impossible. It's very soft so the shape does not matter. NaOH(l) - stainless steel anode The industrial way, but our equipment was very very terrible. However, when we do so the produced Na violently exploded. Too dangerous! NaCl - FeCl3 - Pt anode - propylene carbonate(PC) solvent FeCl3 reacts with NaCl to form soluble Na[FeCl4], while PC is a good polar aprotic solvent. Solution was nearly opaque and mu...
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Metal (2): From molten metal

Some metals have low melting point and melting them at home is possible. When this is frozen, crystals should form. Pb MP: 600.61 K ​(327.46 °C, ​621.43 °F) Pb is very similar to Bi in most properties so we used the same procedure as the one for Bi. Difference: higher melting point, higher toxicity, no radioactivity, higher chemical reactivity, higher density, solid density larger than liquid, lower hardness and strength, different crystal structure. We must remove asbestos to allow melting. We made some feather-like crystals on a hemisphere, but in almost all cases it freezes into hemisphere, not well-formed crystals. Maybe it freezes much faster than Bi due to the higher melting point. Also it quickly loses its silvery appearance and becomes dull upon exposure to air. Bi MP:  544.7 K ​(271.5 °C, ​520.7 °F) Bi can be molten easily with alcohol lamp in a stainless steel container. An asbestos mesh is put under it to reduce heating speed and make temperature uniform. When heate...
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Nitrite Chemistry (4): K2A[B(NO2)6] (A=Ba/Pb, B=Fe/Co)

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You know, even the Second Hospital cannot contain me... We have gained much knowledge about these compounds so now we just need to deal with specific properties of these elements. Co2+ is very air-stable in weak field ligand environment, and [Co(H2O)6]3+ lies much higher than O2 so air would never oxidize it. However when the field strength increases(even to NH3) Co3+ quickly dominates since d6 is much more preferred than d7 in a low-spin configuration, the latter has an extra electron. Fe2+, on the other hand, is very air-sensitive unless in strongly acidic solution, or with strong field ligands like CN-. Although NO2- is also a strong field ligand, it cannot stop oxidation to hydrated Fe2O3 at all as shown below. Of course, besides oxygen, HNO2 can also oxidize these low-valent complexes, and is much more effective as shown in previous experiments. K2Ba[Co(NO2)6] Mixed acetate solution prepared as the Ni one mentioned before(in fact Co dissolves in H2SO4 even at RT so this is quite e...
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Nitrite Chemistry (3): K2A[B(NO2)6] (A=Ba/Pb, B=Ni/Cu)

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Not sure whether I can really achieve this as NaNO2 is the only source material I can use. Ref   ref2   ref3   ref4 The Mn complexes should be unstable so I won't prepare them here. Cd and Hg complexes might be stable but due to their extremely high toxicity we won't prepare them either. Zn and Pb complexes may only exist as solid solution. So, the actual range is quite small. Fe or Co complexes are extremely air-sensitive and would be separated into another page. If B is trivalent, then this is a rather different family and won't be discussed here. K2Ba[Cu(NO2)6] Attempt 1 12.50g(0.05mol) CuSO4.5H2O was dissolved and solution of 6.90g(0.05mol) K2CO3 was added very carefully(gas evolved!), then this was filtered and washed to give basic copper carbonate. Then this was added into a solution of 6.1g(>0.10mol) AcOH in ~80mL water. A very small excess is needed to prevent hydrolysis. After reaction we got a solution with different color than CuSO4 which indicates formation...
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Vanadium Chemistry (4): (NH4)2[VO(C2O4)2(H2O)].H2O

Well, as you know, due to some incidents, during the past months my technology has grown very much, so we can try again. In this compound, the O and H2O are cis, unlike the DMAP salt. ref   ref2 The data in the reference(the Manual also copied it) is completely a mess, but I can guess a little. (NH4)2C2O4 + 2NH4VO3 + 4H2C2O4 = 2(NH4)2[VO(C2O4)2(H2O)].H2O + 2CO2 Attempt 1 4.68g(0.04mol) NH4VO3, 2.48g(0.02mol) (NH4)2C2O4 and 10g(0.11mol) H2C2O4 was dissolved in minimal(~30mL) water and boiled until evolution of gas ceased. Solution became blue as demonstrated long long ago. Excess of H2C2O4 can maintain acidity and ensure full reduction, and it can be washed away with EtOH. After cooling to RT ~200mL EtOH was added and shaken. Solution immediately became opaque and black oil separated. The upper layer was poured away and the oil was shaken with EtOH to solidify(a chopstick or something can help you a lot). Then recrystallized with H2O/EtOH multiple times. However we found that while ...
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Nitrite Chemistry (1): General

Nitrite([NO2]-) is a common ion that has multiple modes of coordination. Mainly: -NO2 -ONO <O2N Among these, the -NO2 isomer(nitro) is of course the most interesting. As is known to all in organic chemistry, nitro is a strong electron-withdrawing or more specifically pi-accepting group. In coordination chemistry it is also so, for example, [Co(NO2)6]4- is a rather rare d7 low-spin while corresponding [Co(bipy)3]2+ is a normal d7 high-spin. However its properties are still different from other strong field ligands like CO or CN- whose complexes can hardly cross the 18e line. It is said to be a weaker sigma-donor and pi-acceptor than CN- at least for Fe2+. Its sodium salt, NaNO2, is hygroscopic and gradually oxidize into NaNO3 when exposed to air. It is rather hard to prepare, but can be bought. Although online shops has withdrawn it due to an all-known reason, it is in fact still possible to buy if you are careful enough. I bought a pack due to the same reason. It is commonly used as...
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Nitrite Chemistry (2): Na3[Co(NO2)6] & K2Na[Co(NO2)6].H2O(?)

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This is the agent for K+ detection, as the K salt is insoluble while Na salt is extremely soluble. The whole reaction is rather confusing, and IDK whether HNO2 or O2 is the oxidant, all in all I decided to just follow the Manual, and just reduce the amount a little. Attempt 1 Our source material of Co is... Tutton's salt (NH4)2[Co(H2O)6](SO4)2 which contains NH4+, and as is known to all this reacts with nitrite, so we should remove it. Also Na2SO4 can be precipitated by ethanol. Converting to acetate as usual is a good choice. 19.75g(0.05mol) salt was dissolved in 150mL water. Containing large crystals, we heated to dissolve them, and then cooled to RT. 6.90g(0.05mol) K2CO3 was dissolved in 20mL water and added to the Co solution with vigorous stirring, filtered and washed carefully with water to give a purple solid of CoCO3. Solution is still slightly colored but large excess of K2CO3 is not allowed as evolved NH3 may attack CoCO3 and resulted [Co(NH3)6]2+ is very easily oxidized....
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Thiocyanate Chemistry (1): Basic Chemistry

Thiocyanate/rhodanide(SCN-) is a common ligand, which is ambidentate: both M-N=C=S and M-S-CN is possible. In general soft centers prefer S and hard centers prefer N. Such complexes generally can't be prepared in water due to the strong basicity of water. Ethanol is a good solvent. Although it seems to contain something, it normally does not release that. However, under acidic and especially acidic oxidizing conditions it definitely releases dangerous stuff so be careful. Due to its small size, homoleptic complexes with high coordination numbers are generally possible.
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Chromium Chemistry (1): Reaction of Metal

Bulk Cr, like Al but not Fe, forms a protective oxide layer of Cr2O3. It is extremely inert due to the combination of corundum(alpha-Al2O3) structure and exchange inert d3 Cr3+ centers. As a result, it resists further oxidation and attack from acid. Some sources claim that acid can destroy the layer, but at least for me, 30% H2SO4 is of no use even upon heating. Maybe, HCl is needed as Cl- is a good ligand. However it was found that reduction can effectively remove passivation. Passing electric current through it, using it as cathode and diluted H2SO4 as electrolyte, can remove passivation immediately, and now Cr normally reacts with acid. Cr dissolves in acid, forming a blue solution of Cr2+. Without exclusion of air it is quickly oxidized into dark green Cr3+. Paraffin oil cannot stop this especially when heated, and preparation of Tutton's salt (NH4)2[Cr(H2O)6](SO4)2 thus failed.
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Fibonacci数列的通项公式

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  很经典的问题，尽力讲的每个人都能懂，不需要专业知识。 题目：f(1)=f(2)=1, f(n+2)=f(n+1)+f(n), f(n)=? 首先让我们先不要管那个初始条件，只看差分方程。 它没有常数项，因此两边同乘以一个常数仍然是成立的。 f(n+3)=f(n+2)+f(n+1) kf(n+2)=kf(n+1)+kf(n) 如果我们假设每一项一一对应的话，f(n)就是一个等比数列，设为c*k^n。 代入差分方程，提取公因子，得到k^2=k+1，这个一元二次方程有两个根，k1=(1+sqrt(5))/2, k2=(1-sqrt(5))/2。 不过考虑到初始条件，貌似两个根都不对... 别急，再仔细看看差分方程。 显而易见如果f(n)和g(n)满足，那么它们的线性组合a*f(n)+b*g(n)也是满足的！ 假设f(n)=c1*k1^n+c2*k2^n，代入1和2处的值，就可以得到这么个方程组 （如果你懒得算那个平方，也可以改为代入0和1处的值，f(0)=0） c1*k1+c2*k2=1 c1*k1^2+c2*k2^2=1 二元二次方程组是有唯一解的，就是c1=1/sqrt(5), c2=-1/sqrt(5)。 组装一下就可以得到最终的公式了，可以随便代入几个值看看对不对。 是不是很简单？
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CF 1008 Div2

A 显然最终结果跟怎么操作无关，所以直接比较n*x是否等于sigma(a)即可。 B 猜测答案必然是1。 由于条件很弱，最终显然是引向n和n-1的二元环。 当k=1时，我们可以让所有的指向n，n指向n-1。这对于所有的奇数都是有效的。 k为偶数时可以颠倒一下，所有的指向n-1，n-1指向n。 C(DIV1A) 首先，排个序，方便处理。 你可能会认为a1或者其他奇数项是被删掉的那个数（记为x），但你会发现那样并不能保证x没有出现过。 但如果x是a2，那么x=a1+a3-a4+a5...，只要令偶数项为小的那些数，那么x>a1+a3>a1，构造成功。 D 显然我们要尽可能往乘法那边分配，并且要尽可能是大的乘法。 是否可以贪心按照最近的不平等位置进行决策呢？其实是的，因为只要遇到了乘法，你就可以把大于等于这条路线的人数全部放到对面去，即使遇到两次需要不同路线的情况，也不会影响到后面的决策。 所以只需要对于每一关，记录一下最近的不平等的位置和对应的正确选择即可。 E(DIV1B) 显然x和y具体是啥并不重要，重要的只是每一位是几个1，这个值可以是0-2。 这就等同于一个三进制的问题被强行挤压到了二进制之中，因此显然出现了损耗（进位）。 为了防止数据丢失，我们显然得问一个0。 接下来就很关键了，思考一下如果还有两次机会可以怎么做。 你会发现关键问题就是进位，因此必须构造一个不存在进位的情况，如果是4进制就不会出现进位了。 我们可以询问1010...和0101...，然后和0的答案相减，再取反，这样就消除了进位的问题。 好了，现在改成一次询问该怎么办？ 其实1010...是不必要的，因为当你问了0101...之后，得出了所有偶数位，就可以从0的答案之中减去这些位，从而将奇数位隔开。 得到所有位之后就不必多说了。 TODO:FG DIV1DFG
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Hub

Twitter(1) Twitter(2) Twitter(3) Twitter(Club Official) GitHub GitHub Site(WIP) Novels Telegram(Personal) Telegram(Channel) PlanetMC Modrinth Codeforces Wikidot
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hospital-diary(Minecraft Server)

 title: hospital-diary author: VillagerVicky pages: #- Nov 26, 2023 晴 我生病了，被迫住进了医院。 我也说不清是什么问题，但我突然晕倒了，醒来后就在这里。 唉，来的第一天就觉得很无聊了。幸好我有手机，可以勉强解闷。 嘛，等出去了一定要好好逛一逛。这边风景很好，我还没怎么参观过。 #- Nov 27, 2023 晴 我看见了一群孩子，和我差不多年纪吧。 不知道为什么，我第一眼就注意到了她，似曾相识。 鼓起勇气，上去搭讪，他们很友好，我们成为了朋友。 这地方很小，没什么好逛的，我也不是来旅游的。 他们没什么玩的，却也很开心，真好啊。 医生说我的情况有些严重，看来一时半会走不了了。 #- Nov 28, 2023 雷雨 来到这里的第三天。 我无法再忍受了，且不提这里荒无人烟，而且晚上还有奇怪的怪物，周围除了电视塔什么都看不见...不是迫不得已谁想来医院呢？那些孩子们也没法缓解我的心情。 以及那黑暗的建筑是什么，窗户都没了... #- Dec 05, 2023 雨  果然，人在开心的时候是不会想写东西的。这么久过去了。 那些孩子很友好，但我还是和她玩的最多。虽然差不多大，但我们似乎更成熟一些，也更有话题。 有时看她治疗很痛苦，估计是很严重的病吧。我倒是还好，只是很虚弱，医生让我静养。我认为不过是没办法治疗吧。 #- 在对方最痛苦的时候，我们都互相陪伴着度过。这就是朋友的感觉吗？我很开心... 有些心疼，然而痛苦往往是不可避免的。 #- Dec 06, 2023 雪 下雪了，外面白茫茫的，要知道在这种地方雪并不常见。 她对我说，好想看看外面的世界是什么样子啊。 我很惊讶，问她，你难道没有出去过吗？或者说，你难道不是从外面来的吗？ 她说，自从自己记事以来，就是在医院里长大的，她听说自己是医生们捡来的孩子。 我不知道该如何安慰。 #- Dec 07, 2023 晴 我们又见面了，可是这次只有她一个人。 我问她，那些孩子们都到哪里去了，她说，都出去了，他们的病好了。 我说，好耶，等我们也好了，我们就一起出去玩，一起去找他们哦。 “好，好啊...可是，我出不去的，我活不了多久了。” “哈哈，那么我也活不了多久了，没关系的。” 希望你只是开玩笑。 #- Dec 10, 2023 阴 这几天，我们一起探...
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Vanadium Chemistry (3): The oxalato complexes

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Oxalate  V(V) forms a surprising complex: cis-[VO2(C2O4)2]3- whose ammonium salt (NH4)3[VO2(C2O4)2].2H2O is bright yellow and forms large, maybe even single crystals, but solution tends to be unsaturated till the end. Preparation of this is quite hard to describe: Mixing same molar NH4VO3, H2C2O4 and (NH4)2C2O4 together in solution is OK, but then it darkens soon even if you are very precise. You may use a few milliliter H2O2 to oxidize it back it turns red, and then in a few days it darkens and repeat H2O2 addition... finally it does not become dark but yellow, then go to vaporize at RT. Seems to be stable both as solid and solution, but wet reduced  dark blue parts of the solid turns into red insoluble substance in a few months. If you managed to get some good seeds and saturated solution, you may grow very large crystals as usual. Solubility is so high that visible growth is observed every few hours, and extremely large crystals can be obtained in only a few days. Crys...
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