逆元替換證明
證明 a/b%n=(a*c)%n,其中(b*c)%m=1
由於 (b*c)%m = 1，則 b*c = k*n+1，其中k = b*c/n
且a整除b
a/b%n = (a/b*1)%n = (a/b*(b*c-k*n))%n = (a*c-a/b*k*n)%n
令 t = (a*c-a/b*k*n)%n, (t < n) , k1 = (a*c-a/b*k*n)/n
a*c-a/b*k*n = k1*n+t
a*c = (a/b*k+k1)*n+t
t = (a*c)%n
因此，a/b%n=(a*c)%n,其中(b*c)%m=1