Lines of 残局文料/lua_lib/算机_程语_月_基类_能_传能_历_基类_表_排_取全.lua from check-in 4c19907139 that are changed by the sequence of edits moving toward check-in 1265aac793:

                         1: local s = {}
                         2: -- 返旧与返新与不返
                         3: -- 排数
                         4: --~ 排
                         5: --~ 办排中值的能
                         6: --~ 入参2
                         7: --~ 入参3
                         8: --~ 入参4
4c19907139 2024-02-13    9: function s.历排(u1_0, u1, u2, u3, o4)
                        10:   local k1_0
                        11:   --~ 集排
                        12:   k1_0 = {}
                        13:   for o5 = 1, #u1_0 do
4c19907139 2024-02-13   14:     k1_0[#k1_0 + 1] = u1(u1_0[o5], u2, u3, o4)
                        15:     end
                        16:   return k1_0
                        17:   end
                        18: --~ 排
                        19: --~ 排
                        20: --~ 办排中值的能
                        21: --~ 入参3
                        22: --~ 入参4
                        23: --~ 入参5
4c19907139 2024-02-13   24: function s.历二排(u1_0, u2_0, u1, u2, u3, o4)
                        25:   local k1_0
                        26:   --~ 集排
                        27:   k1_0 = {}
                        28:   for o5 = 1, #u1_0 do
4c19907139 2024-02-13   29:     k1_0[#k1_0 + 1] = u1(u1_0[o5], u2_0[o5], u2, u3, o4)
                        30:     end
                        31:   return k1_0
                        32:   end
                        33: --~ 排
                        34: --~ 排
                        35: --~ 排
                        36: --~ 办排中值的能
                        37: --~ 入参4
                        38: --~ 入参5
                        39: --~ 入参6
4c19907139 2024-02-13   40: function s.历三排(u1_0, u2_0, u3_0, u1, u2, u3, o4)
                        41:   local k1_0
                        42:   --~ 集排
                        43:   k1_0 = {}
                        44:   for o5 = 1, #u1_0 do
4c19907139 2024-02-13   45:     k1_0[#k1_0 + 1] = u1(u1_0[o5], u2_0[o5], u3_0[o5], u2, u3, o4)
                        46:     end
                        47:   return k1_0
                        48:   end
                        49: return s