Number Theory Tour 4---the proof of Euler function and the implementation of code (I'll prove it's all a Lie ╮ (￣▽￣) ╭)

Euler function, denoted by φ (n)

The Euler function is the number of numbers in the number of n coprime less than n

Spicy, how to beg? ~ (~o￣▽￣) ~o

Can be found in 1 to n-1 with N not coprime number, and then cut them off

Like φ (12)

The decomposition of the 12 qualitative factor, 12=2*2*3, is actually obtained 2 and 32 of the quality factor

And then delete the multiples of 2 and the multiples of 3.

Multiples of 2:2,4,6,8,10,12

Multiples of 3:3,6,9,12

I was going to use 12-12/2 directly-12/3.

But 6 and 12 have been repeatedly reduced.

So we're going to add a number that is a multiple of 2 and a multiple of 3 (>﹏<)

So write 12-12/2-12/3 + 12/(2*3)

What is it called, this is called the repulsion, the law of tolerance.

such as φ (30), 30 = 2*3*5

So φ (30) = 30-30/2-30/3-30/5 + 30/(2*3) + 30/(2*5) + 30/(3*5)-30/(2*3*5)

But it's a hassle to write (￣.￣).

There's an easy way

φ (12) = 12* (1-1/2) * (1-1/3) = 12* (1-1/2-1/3 + 1/6)

φ (30) = 30* (1-1/2) * (1-1/3) * (1-1/5) = 30* (1-1/2-1/3-1/5 + 1/6 + 1/10 + 1/15-1/30)

You see (?∀?), and when you open it, you find that it helps you to do it automatically.

So φ (30) is calculated by first looking at the 30 quality factor.

2,3,5, respectively.

Then use 30* 1/2 * 2/3 * 4/5 to get it done.

The code is as follows:

1 //Euler functions2 intPhiintx) {3     intAns =x;4      for(inti =2; I*i <= x; i++)5         if(x% i = =0){6Ans = ans/i * (i-1);7              while(x% i = =0) x/=i;8         }9     if(X >1) ans = ans/x * (X-1);Ten     returnans; One}

(Phi is the pronunciation of φ)

Wit's code, WIT's Me (?? ' Ω´?

The complexity of this is O (√n), if you want to find the number of n Euler function, the complexity is O (n√n), which is too slow

There are faster ways to wire sieve Euler functions

Need to use the following properties

P is prime

1. Phi (p) =p-1 because the prime number p except 1 is only p, so the integer 1 to p is only p and p not coprime

2. If I mod p = 0, then phi (i * p) =phi (i) * p (I will not prove)

3. If I mod p≠0, then phi (i * p) =phi (i) * (p-1) (I will not prove)

(so I said I'd prove it was a lie. ╮ (￣▽￣) ╭)

The code is as follows:

1#include <cstdio>2 using namespacestd;3 Const intM = 1e6+Ten ;4 intPhi[m], prime[m];5 intTot//tot count, indicating how many prime numbers are in prime[m]6 intEuler () {7      for(inti =2; i < M; i + +) {8         if(!Phi[i]) {9Phi[i] = i1 ;Tenprime[++tot] =i; One         } A          for(intj =1; J <= tot && 1ll*i*prime[j] < M; J + +) { -             if(i% prime[j]) phi[i * Prime[j]] = phi[i] * (prime[j]-1) ; -             Else { thePhi[i * Prime[j]] = phi[i] *Prime[j]; -                  Break ; -             } -         } +     } - } +   A intMain () { at Euler (); -}

(Euler is Euler)

Wit's code, WIT's Me (?? ' Ω´?

Number Theory Tour 4---the proof of Euler function and the implementation of code (I'll prove it's all a Lie ╮ (￣▽￣) ╭)