耞11计
计abcdㄒ
abcd=1000a+100b+10c+d=(1001-1)a+(99+1)b+(11-1)c+d=
(1001a+99b+11c)+(-a+b-c+d)=
11(91a+9b+c)+(-a+b-c+d)
狦-a+b-c+d琌11计玥abcd琌11计
崩約n计狦(案计㎝)籔(计㎝)搭畉琌11计玥n计琌11计
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