8楼所说方法是否这样:
[java5=300,330]
[param=scrptLanguage]PirzerENG[/param]
[param=scrpt]MRR2 U2 MLL U2 MRR' U2 MRR U2 F2 MRR F2 MLL' MRR2 \n MRR2 B F CF'(MRR2 U2 MLL U2 MRR' U2 MRR U2 F2 MRR F2 MLL' MRR2 )CF F' B' MRR2 \n F U' R CU (MRR2 U2 MLL U2 MRR' U2 MRR U2 F2 MRR F2 MLL' MRR2 )CU' R' U F'[/param]
[param=stickersFront]0,5,0,5,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0[/param]
[param=stickersUp]5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,0,5,0,5[/param]
[/java5]
其实,如果要用这个对棱交换公式,也只要做一次即可,不必三次的:
[java5=300,300]
[param=scrptLanguage]PirzerENG[/param]
[param=scrpt]MRR2 B F CF'(MRR2 U2 MLL U2 MRR' U2 MRR U2 F2 MRR F2 MLL' MRR2 )CF F' B' MRR2[/param]
[param=stickersFront]0,5,0,5,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0[/param]
[param=stickersUp]5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,0,5,0,5[/param]
[/java5]
这个公式的作用演示如下;
[java5=300,300]
[param=scrptLanguage]PirzerENG[/param]
[param=scrpt]MRR2 U2 MLL U2 MRR' U2 MRR U2 F2 MRR F2 MLL' MRR2[/param]
[/java5]
或者可以先四棱轮换,再两两对棱换:
[java5=300,300]
[param=scrptLanguage]PirzerENG[/param]
[param=scrpt](MRR2 B2 MRR' U2 MRR' U2 CR' U2 MRR' U2 MRR U2 MRR' U2 MRR2 U2 CR) \n (TR2 F2 U2 MRR2 U2 F2 TR2 ) [/param]
[param=stickersFront]0,5,0,5,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0[/param]
[param=stickersUp]5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,0,5,0,5[/param]
[/java5]
你问的“不过还遇到一种情况,就是把LZ这种情况的一个愣移到对面...”一种情况就是9楼的第三图演示的公式,另一种情况如下,先要预调动几步,才可以执行公式:
[java5=300,300]
[param=scrptLanguage]PirzerENG[/param]
[param=scrpt]F U' R CU'(TR2 U2 MLL U2 MRR' U2 MRR U2 F2 MRR F2 MLL' TR2 )CU R' U F'[/param]
[param=stickersFront]0,3,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0[/param]
[param=stickersBack]3,0,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3[/param]
[/java5]作者: zaybxc 时间: 2009-8-28 21:45:58