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葛立恒数二吧 关注：762贴子：64,779
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因为希腊字母难以打出，我用p代替ψ，W代替Ω，w代替ω
(0) 1
(0)(0) 2
(0)(1) w
(0)(1)(0) w+1
(0)(1)(0)(1) w*2
(0)(1)(0)(1)(0)(1) w*3
(0)(1)(1) w^2
(0)(1)(1)(0)(1)(1) w^2*2
(0)(1)(1)(1) w^3
(0)(1)(1)(1)(1) w^4
(0)(1)(2) w^w
(0)(1)(2)(1) w^(w+1)
(0)(1)(2)(1)(2) w^(w*2)
(0)(1)(2)(2) w^w^2
(0)(1)(2)(2)(2) w^w^3
(0)(1)(2)(3) w^w^w
(0)(1)(2)(3)(4) w^w^w^w
(0)(1,1) p(W)
(0)(1,1)(0)(1,1) p(W)*2
(0)(1,1)(1) p(W+1)
(0)(1,1)(1)(2) p(W+w)
(0)(1,1)(1)(2)(3) p(W+w^w)
(0)(1,1)(1)(2,1) p(W+p(W))
(0)(1,1)(1)(2,1)(1)(2,1) p(W+p(W)*2)
(0)(1,1)(1)(2,1)(2) p(W+p(W+1))
(0)(1,1)(1)(2,1)(2)(3) p(W+p(W+w))
(0)(1,1)(1)(2,1)(2)(3,1) p(W+p(W+p(W)))
(0)(1,1)(1,1) p(W*2)
(0)(1,1)(1,1)(1)(2,1)(2,1) p(W*2+p(W*2))
(0)(1,1)(1,1)(1,1) p(W*3)
(0)(1,1)(2) p(W*w)
(0)(1,1)(2)(1,1)(2) p(W*w*2)
(0)(1,1)(2)(3) p(W*w^w)
(0)(1,1)(2)(3,1) p(W*p(W))
(0)(1,1)(2)(3,1)(3)(4,1) p(W*p(W*p(W)))
(0)(1,1)(2,1) p(W^2)
(0)(1,1)(2,1)(1,1)(2)(3,1)(4,1) p(W^2+W*p(W^2+W))
(0)(1,1)(2,1)(1,1)(2,1) p(W^2*2)
(0)(1,1)(2,1)(2) p(W^2*w)
(0)(1,1)(2,1)(2)(3,1)(4,1) p(W^2*p(W^2))
(0)(1,1)(2,1)(2,1) p(W^3)
(0)(1,1)(2,1)(2,1)(2,1) p(W^4)
(0)(1,1)(2,1)(3) p(W^w)
(0)(1,1)(2,1)(3)(4,1) p(W^p(W))
(0)(1,1)(2,1)(3,1) p(W^W)
(0)(1,1)(2,1)(3,1)(2) p(W^W*w)
(0)(1,1)(2,1)(3,1)(2)(3,1)(4,1)(5,1) p(W^W*p(W^W))
(0)(1,1)(2,1)(3,1)(2,1) p(W^(W+1))
(0)(1,1)(2,1)(3,1)(2,1)(2,1) p(W^(W+2))
(0)(1,1)(2,1)(3,1)(2,1)(3)(4,1)(5,1)(6,1) p(W^(W+p(W^W)))
(0)(1,1)(2,1)(3,1)(2,1)(3,1) p(W^(W*2))
(0)(1,1)(2,1)(3,1)(2,1)(3,1)(2,1)(3,1) p(W^(W*3))
(0)(1,1)(2,1)(3,1)(3) p(W^(W*w))
(0)(1,1)(2,1)(3,1)(3)(4,1)(5,1)(6,1) p(W^(W*p(W^W))
(0)(1,1)(2,1)(3,1)(3,1) p(W^W^2)
(0)(1,1)(2,1)(3,1)(3,1)(3,1) p(W^W^3)
(0)(1,1)(2,1)(3,1)(4) p(W^W^w)
(0)(1,1)(2,1)(3,1)(4)(5,1)(6,1)(7,1) p(W^W^p(W^W))
(0)(1,1)(2,1)(3,1)(4,1) p(W^W^W)
(0)(1,1)(2,1)(3,1)(4,1)(1,1) p(W^W^W+W)
(0)(1,1)(2,1)(3,1)(4,1)(2,1) p(W^(W^W+1))
(0)(1,1)(2,1)(3,1)(4,1)(3,1) p(W^W^(W+1))
(0)(1,1)(2,1)(3,1)(4,1)(4,1) p(W^W^W^2)
(0)(1,1)(2,1)(3,1)(4,1)(5,1) p(W^W^W^W)
(0)(1,1)(2,1)(3,1)(4,1)(5,1)(6,1) p(W^W^W^W^W)
(0)(1,1)(2,2) p(W_2)
IP属地:天津来自Android客户端1楼2023-12-14 17:05回复
    (0)(1,1)(2,2)(1,1)(2,1)(3,1) p(W_2+W^W)
    (0)(1,1)(2,2)(1,1)(2,2) p(W_2+p_1(W_2))
    (0)(1,1)(2,2)(1,1)(2,2)(1,1)(2,2) p(W_2+p_1(W_2)*2)
    (0)(1,1)(2,2)(2) p(W_2+p_1(W_2+1))
    (0)(1,1)(2,2)(2)(3,1)(4,2) p(W_2+p_1(W_2+p(W_2)))
    (0)(1,1)(2,2)(2,1) p(W_2+p_1(W_2+W))
    (0)(1,1)(2,2)(2,1)(3,1)(4,1) p(W_2+p_1(W_2+W^W))
    (0)(1,1)(2,2)(2,1)(3,2) p(W_2+p_1(W_2+p_1(W_2)))
    (0)(1,1)(2,2)(2,1)(3,2)(3,1)(4,2) p(W_2+p_1(W_2+p_1(W_2+p_1(W_2))))
    (0)(1,1)(2,2)(2,2) p(W_2*2)
    (0)(1,1)(2,2)(3) p(W_2*w)
    (0)(1,1)(2,2)(3,1) p(W_2*W)
    (0)(1,1)(2,2)(3,1)(4,1) p(W_2*W^W)
    (0)(1,1)(2,2)(3,1)(4,2) p(W_2*p_1(W_2))
    (0)(1,1)(2,2)(3,1)(4,2)(5,1)(6,2) p(W_2*p_1(W_2*p_1(W_2)))
    (0)(1,1)(2,2)(3,2) p(W_2^2)
    (0)(1,1)(2,2)(3,2)(3) p((W_2^2)*w)
    (0)(1,1)(2,2)(3,2)(3,1) p((W_2^2)*W)
    (0)(1,1)(2,2)(3,2)(3,2) p(W_2^3)
    (0)(1,1)(2,2)(3,2)(4) p(W_2^w)
    (0)(1,1)(2,2)(3,2)(4,2) p(W_2^W_2)
    (0)(1,1)(2,2)(3,2)(4,2)(5,2) p(W_2^W_2^W_2)
    (0)(1,1)(2,2)(3,3) p(W_3)
    (0)(1,1)(2,2)(3,3)(1,1)(2,2)(3,3) p(W_3+p_1(W_3))
    (0)(1,1)(2,2)(3,3)(2,1)(3,2) p(W_3+p_1(W_2))
    (0)(1,1)(2,2)(3,3)(2,1)(3,2)(4,3) p(W_3+p_1(W_3))
    (0)(1,1)(2,2)(3,3)(2,2) p(W_3+W_2)
    (0)(1,1)(2,2)(3,3)(2,2)(3,3) p(W_3+p_2(W_3))
    (0)(1,1)(2,2)(3,3)(3,2)(4,3) p(W_3+p_2(W_3+p_2(W_3)))
    (0)(1,1)(2,2)(3,3)(3,3) p(W_3*2)
    (0)(1,1)(2,2)(3,3)(4,3) p(W_3^2)
    (0)(1,1)(2,2)(3,3)(4,3)(5,3) p(W_3^W_3)
    (0)(1,1)(2,2)(3,3)(4,4) p(W_4)
    (0)(1,1)(2,2)(3,3)(4,4)(5,5) p(W_5)
    (0)(1,1,1) p(W_w)
    IP属地:天津来自Android客户端2楼2023-12-14 17:20
    回复
      PSS？这是BMS数阵吧
      IP属地:河北来自Android客户端3楼2023-12-14 17:30
      收起回复
        顶
        IP属地:天津来自Android客户端4楼2024-05-03 12:08
        回复
          规律？看样子很像hybra。
          IP属地:浙江来自Android客户端5楼2024-05-03 12:44
          收起回复
            (0)(1,1)(1)(2,1)无法理解了，
            在这里出现了和prss规则的矛盾之处，
            IP属地:浙江来自Android客户端6楼2024-05-03 13:28
            回复
              @贴吧用户_aVbXJVN 这几张图介绍了BMS
              IP属地:天津来自Android客户端8楼2024-05-03 13:43
              回复
                (0)(1,1)(1)(2,1)是(0,0)(1,1)(1,0)(2,1)的简写，它展开成(0,0)(1,1)(1,0)(2,0)(3,0)(4,0)…
                IP属地:天津来自Android客户端9楼2024-05-03 13:45
                收起回复
                  我已经明白了一下是否正确
                  (0,0)(1,1)(2,2)
                  (0,0)(1,1)(2,1)(3,1)(4,1)……
                  (0,0)(1,0)(1,0)与单行一样
                  (0)(1)(0)(1)(0)(1)……
                  (0,0)(1,0)(1,1)
                  (0,0)(1,0)(2,0)(3,0)……
                  (0,0)(1,0)(2,1)(2,1)
                  (0,0)(1,0)(2,1)(2,0)(3,1)(3,0)(4,1)……
                  IP属地:浙江来自Android客户端10楼2024-05-03 23:06
                  收起回复
                    @贴吧用户_aVbXJVN BMS的一个特点是对于不标准的表达式也能正常展开。所以先不必理会表达式是否标准，先学会如何展开
                    IP属地:天津来自Android客户端11楼2024-05-04 07:49
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                      (0,0,0)(1,1,1)解如下
                      (0,0,0)(1,1,0)(1,1,0)(2,2,0)(2,2,0)(3,3,0)……
                      IP属地:浙江来自Android客户端12楼2024-05-04 11:41
                      收起回复
                        多行pss的顺序到底是什么
                        从小到大的顺序到底是什么
                        会展开了，但不知道分析哪个大。
                        IP属地:浙江来自Android客户端13楼2024-05-04 20:08
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                          @贴吧用户_aVbXJVN 如果你想分析BMS的话，本吧就有很多分析帖，可以参考前人的结果。另外，私发
                          IP属地:天津来自Android客户端14楼2024-05-05 18:06
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