Fast power modulo
I.e., obtain fast (a ^ b) of the value of mod c. Because when the values of a, b very large directly find a ^ b may cause overflow and inefficient.
Thinking
The principle is based on \ (A * B \% C = ((A \% C) * (B \% C)) \% C \) , \ (A ^ B \% C = (A \% C) ^ B \% c \) formula.
Solving fast power:
Provided index represented in binary as b \ (b = (. 1-n-B_n B_ {...} b_2b_1b_0) _2 \) ,
\(b = b_0 + b_1*2^1 + b_2*2^2+...+b_{n-1}*2^{n-1} + b_n*2^n\)，
\ (a ^ b = a ^ {b_0 b_1 * 2 ^ 1 + b_2 * 2 ^ 2 + ... + b_ {n-1} * 2 ^ {n-1} + b_n * 2 ^ n} = a ^ {b0} * a ^ {b_1 * 2 ^ 1} * a ^ {b_2 * 2 ^ 2} * ... * a ^ {b_ {n-1} * 2 ^ {n-1}} * a ^ {b_n * 2 ^ n} \) ,
\(a^b \% c = a^{b0}*a^{b_1*2^1}*a^{b_2*2^2} *...*a^{b_{n-1}*2^{n-1}} * a^{b_n*2^n} \% c\)，
Set \ (K_n = (A ^ {B_n * 2 ^ n-}) \% C \) , seeking Kn words, when bn = 0 when Kn =. 1, BN =. 1 \ (Kn = (A ^ {2 ^ n- }) \% C \) , thus calculated and then consider \ ((A ^ ^ {n-2}) \% C \) .
\ ((a ^ {2 ^ n}) \% c = [(a ^ {2 ^ {n-1}} \% c) * (a ^ {2 ^ {n-1}} \% c)] \% c \) thus recursive.
Code
python
def quick_powmod(a, b, c):
a = a % c
ans = 1 # 存放结果
while b != 0:
if b & 1: # 二进制与
ans = (ans * a) % c
a = (a * a) % c # 取模是防止溢出
b >>= 1 # 二进制向右移动一位
return ansFor example, a = 2 b = 10 c = 3, b is the binary representation of 1010.
\ (^ 2 ^ {10} = {2 * 0+. 1. 1 + 0 * 2 ^ 2 ^ 2 + 2 ^ *. 3. 1} \) , the bits B taken from right to left, is 0, that multiplies a, is 1, the result is multiplied to take the tired ans years.
Rapid multiplication modulo
To transform the multiplication using binary adder.
Thinking
Fast modulo exponentiation and the like, converted to a binary multiplier calculation.
For example, \ (20 * 20 * 14 = (1110) 2 * 2 ^ 20 = 0 * 0 * 2 ^ 20 + 1 + 20 * 1 * 1 * 2 ^ 2 + 2 ^ 20 * 3 * 1 \)
Code
Python
def quick_mulmod(a, b, c):
ans = 0
a = a % c
while b != 0:
if b & 1 :
ans = (ans + a) % c
a = (2*a) % c
b >>= 1
return ansAt first glance the still very difficult to understand, for example derive again to understand.