Number theory (a) rapid power

Issues into

Title Description

The input values a, n, b, find (A ^n- value)% b of. Wherein a, b, n * n is a long integer.

Input Format

Three integers a, n, b.

Output Format

Output "a ^ n mod b = s"

s is the result of the operation

Sample input and output

Entry

2 10 9

Export

2^10 mod 9=7

problem analysis

For this problem, we first need to have mathematical knowledge base

(a*b)%k=(a%k)*(b%k)%k

Obviously, we can break down the number of times n.

A ⁿ , if n is even, A ⁿ = A ^{n / 2} * A ^{n / 2} ; if n is odd, A ⁿ = A ^{n / 2} * A ^{n / 2} * A. We claim i.e. only the A ^{n-/ 2} , can be obtained by one multiplication A ^n- value. Continues to A ^{n / 2} decomposition eventually becomes n 1. This is the dichotomy of thinking, therefore, we need only logn times can become the b 1. This is what we call the power of fast, recursive and non-recursive writing.

 

[Recursive]

. 1  int quickPow ( int A, int n-, int B)
 2  {
 . 3      IF (n-== . 1 ) return A;
 . 4      
. 5      IF (! (& N- . 1 )) // even, the bit operation priority attention, to brackets 
6      {
 . 7          int T = quickPow (A, n-/ 2 , B);
 . 8          return T * T% B;
 . 9      }
 10      
. 11      the else 
12 is      {
 13 is          int T = quickPow (A, n-/ 2 , B);
 14         t=t*t%b;
15         t=t*a%b;
16         return t;
17     }
18 }

[Non-recursive]

 1 int quickPow(int a,int n,int b)
 2 {
 3     int  res=1;
 4     while(n)
 5     {
 6         if(n&1)res=res*a%b;
 7         a=a*a%b;
 8         n>>=1;
 9     }
10     return res;
11 }