凤凰之谜 1/4 潜行者

Part 1/4 – Rogue

Your deck needs to contain Bamboozle, Plagiarize, Yoink!, Coerce and Potion of Illusion.

Since they cost 12 mana in total, once you get at least two pieces in your hand, use Efficient Octo-bot + Prize Plunderer to discount them.

After you get everything and discount them, play them all in a single turn. Play only those five cards and nothing else! Just make sure that there’s a minion you can Coerce胁迫, because you can’t play it on an empty board and make sure you’ll have a hand space after Potion of Illusion. When you end your turn, you should get a card that starts a puzzle. Play it at the start of your next turn.

Here’s an example deck you can use, you can definitely make a stronger one, but I’ve tried to keep it cheap so everyone could use it. If you’re missing any cards OTHER than single copies of the necessary ones listed above, you can swap them as you wish:

AAEBAaIHAA/MuQPOuQO5vgOK0AOk0QPl0QPl0wPl3QPn3QPz3QOW6AOq6wOr6wOu6wOVnwQA

BT_042
偷天换日(Bamboozle)
Bamboozle
偷天换日
Secret: When one of your minions is attacked, transform it into a random one that costs 3 more.
奥秘：当你的随从受到攻击时，随机将其变形成为一个法力值消耗增加3点的随从。

 

SCH_706
抄袭(Plagiarize)
Plagiarize
抄袭
Secret: At the end of your opponent's turn, add copies of the cards they played to your hand.
奥秘：在你对手的回合结束时，将其使用的卡牌的复制置入你的手牌。

 

BAR_323
偷师学艺(Yoink!)
Yoink!
偷师学艺
Discover a Hero Power and set its Cost to 0. Swap back after 2 uses.
发现一个英雄技能，并使其法力值消耗变为0点。使用两次后换回。

 

SCH_521
胁迫(Coerce)
Coerce
胁迫
Destroy a damaged minion. Combo: Destroy any minion.
消灭一个受伤的随从。连击：消灭任意随从。

 

SCH_352
幻觉药水(Potion of Illusion)
Potion of Illusion
幻觉药水
Add 1/1 copies of your minions to your hand. They cost 1.
将你的所有随从的1/1的复制置入你的手牌，并使其法力值消耗变为1点。

 

BAR_320
高效八爪机器人(Efficient Octo-bot)
Efficient Octo-bot
高效八爪机器人
Frenzy: Reduce the cost of cards in your hand by 1.
暴怒：你的手牌法力值消耗减少1点。

 

DMF_519
奖品掠夺者(Prize Plunderer)
Prize Plunderer
奖品掠夺者
Combo: Deal 1 damage to a minion for each other card you've played this turn.
连击：在本回合中，你每使用一张其他牌，便对一个随从造成1点伤害。

 

Now, ignore your middle and right-most minion, use only the left one. The opponent has seven cubes on the board and you need to attack them exactly the right amount of times with your left minion.

  • 1st Cube – 1 time
  • 2nd Cube – 2 times
  • 3rd Cube – 3 times
  • 4th Cube – DON’T ATTACK
  • 5th Cube – DON’T ATTACK
  • 6th Cube – 3 times
  • 7th Cube – 2 times

After you do that, all Cubes will be set at 4 attack. Click your Hero Power to finish the puzzle.

 

这一关的要求是让上面7个方块变成相同攻击力即可过关。

解题方法是：用任意钥匙按照顺序攻击第一个方块1次、攻击第二个方块2次，攻击第三个方块3次，攻击第六个方块3次，攻击第七个方块2次。完成之后点技能即可过关。

研究己方第一个随从

对面

5,594 1,699 0,763 2,768 0,873 5,921 2,999

左边1,999攻击第一个
6,593 3,699 1,763 0,768 2,873 5,921 9,999
1,-1    2,0     1,0     -2,0    2,0    0,0      7,0【这里其实是12-3=9】

左边6,994第二次攻击第一个
7,587 5,699 2,763 8,768 4,873 5,921 6,999
1,-6 2,0 1,0 8,0 2,0 0,0 -3,0

左边第三次攻击
8,580 7,699 3,763 6,768 6,873 5,921 3,999
1,-7 2,0 1,0 -2,0 2,0 0,0 -3,0

左边第四次攻击
9,572 9,699 4,763, 4,768 8,873 5,921 0,999
1,-8 2,0 1,0 -2,0 2,0 0,0 -3,0

左边第五次攻击
0,563 1,699 5,763 2,768 0,873 5,921 7,999
9+1=10,563 9+2=11 -2,0 8+2=10 0,0 10-3=7

1,999-->第二次6,994-->第三次7,988-->第四次8,981-->第五次9，973

 

可以发现规律是，对面随从主要是攻击力在变化。

如果攻击力增加超过10，就去掉十位数，只取个位数。比如9+1=10，9+2=11，分别取0和1

如果攻击力不够减，比如2-3=-1，那么就用12-3=9

1 2 3 4 5 6 7
+1 +2 +1 -2 +2 +0 -3

用己方中间那个随从攻击对面第一个随从，发现攻击力变化是一样的。

 

攻击对面第一个随从

5102052

攻击第一个一次

6310259

攻击第一个两次

7528456

 

攻击对面第二个随从

5102052-->3280330   -2,1,-2,-2,+3,-2,-2

推算第二次攻击第二个随从后的攻击力变化1368618 ，攻击后验证 推断成立

 

攻击对面第三个随从

5102052-->5913230   0,-2,1,1,2,-2,-2

推算第二次攻击第二个随从后的攻击力变化5724418 ，攻击后验证 推断成立

 

攻击对面第四个随从

5102052-->5113889   0,0,1,1,-2,3,-3

推算第二次攻击第二个随从后的攻击力变化5124616 ，攻击后验证 推断成立

 

攻击对面第五个随从

5102052-->2194135   -3,0,-1,2,1,-2,3

推算第二次攻击第二个随从后的攻击力变化9186218 ，攻击后验证 推断成立

 

攻击对面第六个随从

5102052-->7429063   2,3,2,-3,0,1,1

推算第二次攻击第二个随从后的攻击力变化9746074 ，攻击后验证 推断成立

 

攻击对面第七个随从

5102052-->3999083   -2,-2,-1,-3,0,3,1

推算第二次攻击第二个随从后的攻击力变化1786014 ，攻击后验证 推断成立

 

用代码推演解决方案

attack enemy 1 minion, 1 times
6,3,1,0,2,5,9
attack enemy 2 minion, 2 times
4,4,9,8,5,3,7
2,5,7,6,8,1,5
attack enemy 3 minion, 3 times
2,3,8,7,10,9,3
2,1,9,8,2,7,1
2,9,10,9,4,5,9
attack enemy 4 minion, 0 times
attack enemy 5 minion, 0 times
attack enemy 6 minion, 3 times
4,2,2,6,4,6,10
6,5,4,3,4,7,1
8,8,6,0,4,8,2
attack enemy 7 minion, 2 times
6,6,5,7,4,1,3
4,4,4,4,4,4,4
The final result is 4,4,4,4,4,4,4

 

   var attackArray = new int[] {5, 1, 0, 2, 0, 5, 2};
            int[,] attackChangeArray = new int[,]
            {
                {1, 2, 1, -2, 2, 0, -3}, //attack the first enemy minion
                {-2, 1, -2, -2, 3, -2, -2}, //attack the second enemy minion
                {0, -2, 1, 1, 2, -2, -2},
                {0, 0, 1, 1, -2, 3, -3},
                {-3, 0, -1, 2, 1, -2, 3},
                {2, 3, 2, -3, 0, 1, 1},
                {-2, -2, -1, -3, 0, 3, 1},
            };

这玩意看起来像是做矩阵的加法还是乘法

 

最少需要攻击11次，次少需要攻击21次

valid solution1, need attack 11 times: 1, 2, 3, 0, 0, 3, 2
valid solution2, need attack 21 times: 1, 2, 3, 0, 10, 3, 2  单个攻击10次太多了
valid solution3, need attack 21 times: 1, 2, 3, 5, 0, 3, 7   应该可以尝试这个
valid solution4, need attack 21 times: 1, 2, 3, 10, 0, 3, 2
valid solution5, need attack 25 times: 3, 8, 5, 4, 2, 1, 2
valid solution6, need attack 31 times: 1, 2, 3, 5, 10, 3, 7
valid solution7, need attack 31 times: 1, 2, 3, 10, 10, 3, 2
valid solution8, need attack 33 times: 7, 0, 9, 2, 6, 7, 2
valid solution9, need attack 35 times: 3, 8, 5, 9, 2, 1, 7
valid solution10, need attack 37 times: 9, 6, 1, 1, 8, 5, 7
valid solution11, need attack 37 times: 9, 6, 1, 6, 8, 5, 2
valid solution12, need attack 39 times: 5, 4, 7, 3, 4, 9, 7
valid solution13, need attack 39 times: 5, 4, 7, 8, 4, 9, 2
valid solution14, need attack 43 times: 7, 0, 9, 7, 6, 7, 7
valid solution15, need attack 43 times: 7, 10, 9, 2, 6, 7, 2
valid solution16, need attack 53 times: 7, 10, 9, 7, 6, 7, 7

 

posted @ 2021-05-09 15:28  ChuckLu  阅读(160)  评论(0)  编辑  收藏  举报