您好我正在写一个code来计算的是P ^ Q其中

Hi I am writing a code to calculate P^Q where

P, Q are positive integers which can have number of digits upto 100000

我要的结果为

I want the result as

result = (P^Q)modulo(10^9+7)

例如：

P = 34534985349875439875439875349875 
Q = 93475349759384754395743975349573495
Answer = 735851262

我尝试使用的伎俩：

I tried using the trick:

 (P^Q)modulo(10^9+7) = (P*P*...(Q times))modulo(10^9+7)

 (P*P*...(Q times))modulo(10^9+7) = ((Pmodulo(10^9+7))*(Pmodulo(10^9+7))...(Q times))modulo(10^9+7)

由于两个P和Q是非常大的，我应该把它们存储在一个数组，并通过数字做模数字。

Since both P and Q are very large, I should store them in an array and do modulo digit by digit.

是否有这样做的任何有效的方式还是有些数论算法，我很想念？

Is there any efficient way of doing this or some number theory algorithm which I am missing?

在此先感谢

推荐答案

下面是一个比较有效的方法：

Here is a rather efficient way:

1）计算P1 = P模10 ^ 9 + 7

1)Compute p1 = P modulo 10^9 + 7

2）计算Q1 = Q模10 ^ 9 + 6

2)Compute q1 = Q modulo 10^9 + 6

3）则P ^ Q模10 ^ 9 + 7等于P1 ^ Q1模10 ^ 9 + 7。这种平等是因为费马小定理如此。需要注意的是P1和Q1足够小，适合在32位整数，这样你就可以实现二进制exponention与标准的整数类型（intermidiate计算，64位整数类型就足够了，因为初始值适合在32位）。

3)Then P^Q modulo 10^9 + 7 is equal to p1^q1 modulo 10^9 + 7. This equality is true because of Fermat's little theorem. Note that p1 and q1 are small enough to fit in 32-bit integer, so you can implement binary exponention with standard integer type(for intermidiate computations, 64-bit integer type is sufficient because initial values fit in 32-bits).