Euler's Function & Chinese Remainder Theorem in algorithm Summary

Euler's Function & Chinese Remainder Theorem in algorithm Summary

1. Euler's Function

　　Concept:In number theory, for a positive integer N, the Euler's function is the number of numbers that are less than or equal to the number of numbers that intersect with N.

　　General Formula: Phi (x) = x (1-1/P1) (1-1/P2) (1-1/P3) (1-1/P4 )..... (1-1/PN)

　　　　P1, P2 ...... All prime factors whose PN is x. x is an integer not 0.

　　Note:

1) PHI (1) = 1.

2) Each prime factor is only one. For example, 12 = 2*2*3, then PHI (12) = 12 * (1-1/2) * (1-1/3) = 4

(3) If n is the k power of prime p ), except for the multiples of P, all other numbers are in the same quality as N.

4) PHI (Mn) = PHI (m) PHI (N)

5) WHEN n is an odd number)

　　Code implementation:

1) calculate the number of ouarla directly:

1/* Euler's function value whose return value is N */2 int Euler (int n) 3 {4 int S = N, I, m; 5 m = SQRT (N ); 6 For (I = 2; I <= m; I ++) {7 if (N % I = 0) 8 S = s/I * (I-1 ); 9 While (N % I = 0) 10 N/= I; 11} 12 if (n> 1) 13 s = s/n * (n-1); 14 return S; 15}

2) create tables

1/* print the Euler function table of 1-maxn */2 int A [maxn] = {, 0}; 3 void Euler () 4 {5 Int I, J; 6 For (I = 2; I <= maxn; I ++) 7 if (! A [I]) 8 for (j = I; j <= maxn; j + = I) 9 {10 if (a [J] = 0) A [J] = J; 11 A [J] = A [J]/I * (I-1); 12} 13}

2. China Residue Theorem

Original article:
The question in Sun Tzu's computing Sutra: I don't know the number of things. I have three questions: One, two, five, one, three, and seven, and two. What is the total number of things?
Solution in Sun Tzu's computing Sutra: three or three numbers, take the number of 70, multiply by the number of two; five numbers, take the number of 21, multiply by the remainder three-phase; seven numbers, take the number 15 and multiply it by the remainder. Add the product and subtract the multiple of one hundred and five.

Conclusion:

　　　　Any fixed integer is M. When M/A is a, m/B is B, M/C is C, M/D is D ,..., When m/z is more than z, here a, B, c, d ,..., When Z is the divisor and the divisor is any natural number, the remainder a, B, c, d ,......, When Z is a natural integer.

1) when the proposition is correct, there is a solution within the least common multiples of the divisor, and there is a unique solution within each least common multiple.

2) When M is within the least common multiple of two or more divisor, the divisor and remainder of the two or more operators can locate the specific position of M within the least common multiple, that is, the size of M.

3) correct propositions: Divide them by a, B, c, d ,..., The number of different remainder combinations in Z = A, B, C, D ,..., Minimum Public multiples of Z = cycles of different remainder combinations.

　　Specific steps (take Sun Tzu's computing Sutra as an example ):

1) identify three numbers: Find the minimum number 15 of 7 Division remainder 1 from the public multiples of 3 and 5, find the minimum number 21 after division of 5 and 1 from the public multiples of 3 and 7, and find the minimum number 70 after division of 3 and 1 from the public multiples of 5 and 7.

2) Multiply by 15 by 2 (2 is the remainder of the final result divided by 7) and multiply by 21 by 3 (3 is the remainder of the final result divided by 5). Likewise, multiply 70 by 2 (2 is the remainder of the final result divided by 3), and then add the three products (15*2 + 21*3 + 70*2) to get and 233.

3) divide by 233 by the minimum public multiple of 3, 5, and 7 by 105, and obtain the remainder 23, that is, 233% 105 = 23. The remainder 23 is the minimum number that meets the condition.