If AX=B, then let's multiply both sides by A inverse.

如果AX=B，在等号两边同时乘以A逆。

youdao

And in the even case, I'd square, divide by half, call this again: in the odd case, I go b minus 1 and then multiply by a.

就直接求a的平方的二分之b次方，如果是奇数的话，就去求a的b-1次方乘以a，以此递归下去。

youdao

Multiply a by a to the b minus 1.

这很不错对不对？

麻省理工公开课 - 计算机科学及编程导论课程节选

Let's multiply both sides by 2, I'll get 2q1* is equal to a - c over b - q1*.

等式两边同时乘以2,得到,2q1*=/b-q1

耶鲁公开课 - 博弈论课程节选

And in the even case, I'd square, divide by half, call this again: in the odd case, I go b minus 1 and then multiply by a.

就直接求a的平方的二分之b次方，如果是奇数的话，就去求a的b-1次方乘以a，以此递归下去。

麻省理工公开课 - 计算机科学及编程导论课程节选