I am searching for an algorithm that allow me to compute (2^n)%d with n and d 32 or 64 bits integers.
我正在寻找一种算法，允许我用n和d 32或64位整数计算（2 ^ n）％d。
The problem is that it's impossible to store 2^n in memory even with multiprecision libraries, but maybe there exist a trick to compute (2^n)%d only using 32 or 64 bits integers.
问题是即使使用多精度库也不可能将2 ^ n存储在内存中，但是可能存在仅使用32或64位整数来计算（2 ^ n）％d的技巧。
Thank you very much.
非常感谢你。
1 个解决方案
#1
24
Take a look at the Modular Exponentiation algorithm.
看一下Modular Exponentiation算法。
The idea is not to compute 2^n. Instead, you reduce modulus d multiple times while you are powering up. That keeps the number small.
这个想法不是计算2 ^ n。相反，您在通电时会多次减少模数d。这使数字变小。
Combine the method with Exponentiation by Squaring, and you can compute (2^n)%d in only O(log(n)) steps.
将该方法与Squaring的Exponentiation相结合，您只需在O（log（n））步骤中计算（2 ^ n）％d。
Here's a small example: 2^130 % 123 = 40
这是一个小例子：2 ^ 130％123 = 40
2^1 % 123 = 2
2^2 % 123 = 2^2 % 123 = 4
2^4 % 123 = 4^2 % 123 = 16
2^8 % 123 = 16^2 % 123 = 10
2^16 % 123 = 10^2 % 123 = 100
2^32 % 123 = 100^2 % 123 = 37
2^65 % 123 = 37^2 * 2 % 123 = 32
2^130 % 123 = 32^2 % 123 = 40
#1
24
Take a look at the Modular Exponentiation algorithm.
看一下Modular Exponentiation算法。
The idea is not to compute 2^n. Instead, you reduce modulus d multiple times while you are powering up. That keeps the number small.
这个想法不是计算2 ^ n。相反，您在通电时会多次减少模数d。这使数字变小。
Combine the method with Exponentiation by Squaring, and you can compute (2^n)%d in only O(log(n)) steps.
将该方法与Squaring的Exponentiation相结合，您只需在O（log（n））步骤中计算（2 ^ n）％d。
Here's a small example: 2^130 % 123 = 40
这是一个小例子：2 ^ 130％123 = 40
2^1 % 123 = 2
2^2 % 123 = 2^2 % 123 = 4
2^4 % 123 = 4^2 % 123 = 16
2^8 % 123 = 16^2 % 123 = 10
2^16 % 123 = 10^2 % 123 = 100
2^32 % 123 = 100^2 % 123 = 37
2^65 % 123 = 37^2 * 2 % 123 = 32
2^130 % 123 = 32^2 % 123 = 40