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one method is to have:ZI 9 8 7 6 5 9 8 7 6 5 maps to
¥Ò¤A¤þ¤B¥³¤v©°¨¯¤Ð¬Ña and©½t¥Í³N¼Æ¬ã¨sªÀ -- ³N¼Æ¬ã¨s¡@¡@ &+( 9 8 7 6 5 4 9 8 7 6 5 4 maps toEx9{c ¤l¤¡±G¥f¨°¤x¤È¥¼ ¥Ó¨» ¦¦¥è
then e.g. forh ¥Ò¤l¤A¤¡, add them together, gets 9+8+9+8 = 34.=AH divided by 5, remainder is 4, and then, I forgot ... but it's documented in many books.C(RE@Q ©½t¥Í³N¼Æ¬ã¨sªÀ -- ³N¼Æ¬ã¨s¡@¡@ 2 the 2nd method is based on previous method, and using the "palm" to aid the calculation, and should be able to count the result within 2-5 seconds. It's quite hard for me to illustrate here though ... it's maths, anyway. not hard to derive oneself. This method is documented in 1 or 2 books only -- the books I have read so far.j.!U&r ©½t¥Í³N¼Æ¬ã¨sªÀ -- ³N¼Æ¬ã¨s¡@¡@ ;T But I have to admit that if you have good memory, ¤èªk¤@ is a good way.I ©½t¥Í³N¼Æ¬ã¨sªÀ -- ³N¼Æ¬ã¨s¡@¡@ ew|( And finally, my suggestion is, to prepare a cheat-sheet ... or, save it in PDA.../0U{.
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