摘要:
/* Consider the following algorithm to generate a sequence of numbers. Start with an integer n. If n is even, divide by 2. If n is odd, multiply by 3 阅读全文
摘要:
/* calculate a num y which is base x to exp n 2^n=2*2^n-1 (n is an odd) 2^n=2^(n/2)*2^(n/2) (n is an even) judge method: if:b&1 b%2==1 (b is an odd) else:!(b&1) b%2==0 (b is... 阅读全文