[Prehistoric giant pitfall] Number Theory template integration and number theory template Integration

[Prehistoric giant pitfall] Number Theory template integration and number theory template Integration

This afternoon, Shen Yu told me a piece of number theory = I found that I couldn't play happily.
Sort out all the content templates.
This is a prehistoric giant pitfall, and the time is definitely not enough... Enter orz in the future

# Include <iostream> # include <cstdio> # include <cstring> # include <cmath> # include <algorithm> # define MAXN 1000000 using namespace std; bool not_prime [MAXN]; int prime_number [MAXN]; int nu; int factor [MAXN]; // record the minimum prime factor void prime () // line sieve {for (int I = 2; I <= MAXN; I ++) {if (! Not_prime [I]) {prime_number [++ nu] = I; factor [I] = I;} for (int j = 1; j <= nu & I * prime_number [j] <= MAXN; j ++) {not_prime [I * prime_number [j] = 1; f [I * prime_number [j] = prime_number [j]; if (I % prime_number [j] = 0) break ;}}} int mu [MAXN] = {0, 1}; void mu () // Mobius function {for (int I = 2; I <= MAXN; I ++) {if (! Not_prime [I]) mu [I] =-1; for (int j = 1; j <= nu & I * prime_number [j] <= MAXN; j ++) {not_prime [I * prime_number [j] = 1; if (I % prime_number [j] = 0) {mu [I * prime_number [j] = 0; break;} else {mu [I * prime_number [j] =-mu [I] ;}}} int phi [MAXN]; void phi () // Euler's function {for (int I = 2; I <= MAXN; I ++) {if (! Not_prime [I]) {phi [I] = I-1;} for (int j = 1; j <= nu & I * prime_number [j] <= MAXN; j ++) {not_prime [I * prime_number [j] = 1; if (I % prime_number [j] = 0) {phi [I * prime_number [j] = phi [I] * prime_number [j]; break ;} else {phi [I * prime_number [j] = phi [I] * (prime_number [j]-1) ;}}} int inv [MAXN]; // The following continuous reverse element void inv1 () // Fermat Theorem {for (int I = 1; I <= MAXN; I ++) inv [I] = pow (I, phi [p]-1) % p; // p is determined based on the subject requirements} void inv2 () // Method Based on line sieve {for (in T I = 2; I <= MAXN; I ++) {int a = p/I; int B = p % I; inv [I] = (p-) * inv [B] % p ;}} void inv3 () // factorial-Based Query {unsigned long a [MAXN] = {1}, rev [MAXN]; // This factorial is slightly larger than 0-0. I don't want to write high precision. unsigned, let's look at it. for (int I = 1; I <= MAXN; I ++) a [I] = a [I-1] * I; rev [MAXN] = pow (a [MAXN], phi [p]-1) % p; for (int I = MAXN-1; I> = 0; I --) rev [I] = rev [I + 1] * (I + 1) % p; for (int I = 1; I <= MAXN; I ++) inv [I] = a [I-1] * rev [I];} void inv4 () // The method based on a * I + B = p. This method is only used when p is a prime number {inv [1] = 1; for (int I = 2; I <= MAXN; I ++) {int a = p/I; int B = p % I; inv [I] = (p-) * inv [B] % p ;}// returns the Chinese Remainder theorem, which contains the original root, discrete logarithm, and Lucas theorem. the mobius Inversion is not complete =. = // Thank you for your teaching. He talked about the last three methods in the reverse element and the Mobius function... /* Int main (){}*/

It should be correct. QUQ. If there is any sand tea error, please tell me immediately that I am such a sand tea after all.