我有大量的二进制值，我想（更多）紧凑地存储为字符串，所以我使用二进制 - ＆gt;使用
进行base36转换 parseInt('0101010101010 etc', 2).toString(36)
但是，尝试使用
反转此计算时 parseInt('base36 value', 36).toString(2)
由于浮点数限制，我会假设因为一些数字后的所有数字都被切断了，所以准确度会有所下降。
完成此任务的正确/更好的方法是什么？
Ex：我的基地36是dkmgiancyx44so4ck8gwgkoc8gk4ks8ks48kook0co8w04sk4scokkgk800kk4w08s4s0oso0g8w0g8kkggskosw48s00wgo04wgw0g4okg04kg0okswc8ogg4s4skkscccg44o8so084ckocosoo4sokkgco8gg4ggocscg08goo4c0cw0kcscso
但是运行parseInt（text，36）.toString（2）会产生
10000010100101100000000000000000001100011000001101111000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000