Fast power-take modulus algorithm

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Fast power modulus is a common algorithm, which is summarized here.
A^b%c (This is known as the RSA public Key encryption method), when a is large, directly solve the problem is not likely
Algorithm 1: Using the formula a*b%c= ((a%c) *b)%c, this processing is done at each step, which solves the problem that the a^b may be too large, but the time complexity of the algorithm is still not optimized
The code is as follows:

int Mod1 (int a,int b,int N)     {        int cnt = 1;    while (b--)    {        cnt = a * cnt% n;    }    return CNT;}  </span>

algorithm 2: Another algorithm uses two points of thought, can reach O (logn).
where P (i) (0<=i<=n) is 0 or 1

so a^b =  a^  (P (n) *2^n  +  p (n-1) *2^ (n-1)   +...+  P (1) *2  +  p (0))
                =  a^ (P (n) *2^n)   *  a^ (P (n-1) *2^ (n-1))   *...*  a^ (P (1) *)   *  a^p (0)
for the case of P (i) =0, a^ (P (i) * 2^ (i-1)  ) =  a^0  =  1, no processing
We have to consider only P (i) =1 the case (This is important!!) For details, see Qin Jiushao algorithm:Http://baike.baidu.com/view/1431260.htm)
With this, we can recursively calculate all the a^ (2^i)
Of course, by the conclusion of algorithm 1, we add modulo operations:
a^ (2^i)%c = ((a^ (2^ (i-1))%c) * a^ (2^ (i-1)))%c
Then the a^ (2^i)%c that satisfies P (i) =1 is multiplied by the algorithm 1 and%c is the result , that is, the binary scan has been scanned from the highest bit to the lowest bit .Instance code: Recursive

int Mod2 (int a,int b,int n) {    int t = 1;    if (b = = 0)        return 1;    if (b = = 1)    return a%n;    T=mod2 (A, b>>1, n);    t=t*t%n;    if (b&1)    {        t = t*a% n;    }    return t;}

Example code 2: Non-recursive optimizations:

int Mod3 (int a,int b,int y) {    int cnt=1;    while (b)    {        if (b&1) cnt=cnt*a%y;        a=a*a%y;        b>>=1;    }    return CNT;}

Fast power-take modulus algorithm