Shulou(Shulou.com)06/01 Report--

This article introduces the relevant knowledge of "the method of Java higher power module + product function + inverse". In the operation of actual cases, many people will encounter such a dilemma, so let the editor lead you to learn how to deal with these situations. I hope you can read it carefully and be able to achieve something!

Meaning of the title: sum of all positive factors of 2004 ^ x (S) to 29; output result

Original title link

Topic resolution: analyze the reference source: click on the open link

Factor sum

The factor of 6 is 1, 2, 3, 6, and the sum of 6 is s (6) = 1, 2, 3, 6, 12.

The factor of 20 is 1, 2, 4, 5, 10, 20, and the sum of 20 is s (20) = 1, 2, 4, 5, 10, 20, 42.

The factor of 2 is 1BI 2; the sum of the factors of 2 is s (2) = 1x 2m 3.

The factor of 3 is 1BI 3; the sum of the factors of 3 is s (3) = 1 million 3 4.

The sum of the factors of 4 is

S (4) = 1 / 2 / 4 / 7

The sum of the factors of 5 is

S (5) = 1 / 5 / 6

S (6) = s (2) * s (3) = 3 / 4 / 12

S (20) = s (4) * s (5) = 7 / 6 / 42

Is that a coincidence?

Then look at s (50) = 1 "2" 5 "10" 25 "50" 93 "3" 31 "s (2) * s (25), s (25) = 1" 5 "25" 31.

This is called a product function in number theory, when gcd (a) b) = 1 s (a) = s (a) * s (b)

If p is a prime

S (p ^ n) = 1 ^ p + p ^ 2 +... + p ^ n = (p ^ (naugh1)-1) / (pmur1) (1)

Example hdu1452 Happy2004

Calculation factor and s (2004 ^ X) mod 29

2004 = 2 ^ 2 * 3 * 167

S (2004 ^ X)) = (s (2 ^ 2X)) * (s (3 ^ X)) * (s (167 ^ X)

167) = 22

S (2004 ^ X)) = (s (2 ^ 2X)) * (s (3 ^ X) * (s (22 ^ X)

Astats (2 ^ 2X) = (2 ^ (2X+1)-1) / / according to (1)

Banners (3 ^ X) = (3 ^ (Xero1)-1) / 2max / according to (1)

Clocks (22 ^ X) = (22 ^ (Xero1)-1) / 21max / according to (1)

% algorithm

1. (aqb)% p = (a% p) * (b% p)

% algorithm

2. (a * b)% p = (a * b ^ (- 1)% p)

B ^ (- 1) Yes

Inverse element of b (% p)

The inverse element of 2 is 15 (), because 2 / 15 / 30 / 29 / 1 / 29

The inverse element of 21 is 18 (), because 21 / 18 / 378% 29 = 1% 29

therefore

A = (powi (2 and 2))-1)% 29

B = (powi (3) X-1 (29)-1) * 15% 29

C = (powi (22 ~ Xerox _ 1 ~ 29)-1) * 18% 29

Ans= (aqb) 293c 29

Data expansion: 1.

High-power fast module link

two。 A product function in number theory: an arithmetic function for a positive integer n

F (n), if f (1) = 1 and f (ab) = f (a) f (b) is called a product function in number theory. If

For a product function f (n), even if an and b are not prime, there is f (ab) = f (a) f (b), then it is called complete product. If n is expressed as a prime factorization

3. Find the inverse element:

When calculating (a _ Mod)% Mod, it is often necessary to first calculate the inverse p of b%Mod (the condition that b has an inverse is that gcd (br _ Mod) = = 1, obviously there must be an inverse of a prime), and then (aplomp)% Mod

Get the result c. this

The inverse p of b satisfies (bounded p)% Mod=1. Let's start with a simple proof:

(a Mod=c;)% Mod=c)% Mod=1; = = "(b) * (b)% Mod=c; = =" (ACPP)% Mod=c

We can see the correctness of the conclusion from the above, of course, here b needs to be a factor of A. Next we need to know how we calculate the inverse p based on b and Mod. The extended Euclidean algorithm, as we all know, is known to be a, b, and to find a set of solutions (x _ ray y) to make a*x+b*y=1. The x obtained here is the inverse of a% b and y is the inverse of b% a (think about why? Put b or an on both sides of the equation.

The following reasons are explained:

Modular m multiplicative inverse

Definition: if there is an integer b satisfying ab ≡ 1 (mod m), then b is the inverse of modular m multiplication of a.

Theorem: an exists multiplicative inverse of modulo m if and only if gcd (a) m = 1.

Adequacy:

Because

Gcd (aPerm) = 1

According to Euler's theorem, there are

A ^ φ (m) ≡ 1 (mod m)

therefore

A * a ^ (φ (m)-1) mod m = 1

So there is a modular m multiplicative inverse of a, that is, a ^ (φ (m)-1).

Necessity:

Suppose that the multiplicative inverse of a module m is b, then

Ab ≡ 1 (mod m)

So

Ab = km + 1

So

1 = ab-km

According to Euclid's theorem, there are

Gcd (aPerm) = 1

It is known by reason:

For ax + by = 1, we can see that x is the multiplicative inverse of a modular b and y is the multiplicative inverse of b modular a.

On the other hand, to calculate the multiplicative inverse of a module b is equivalent to finding the minimum positive integer solution of x of ax + by = 1, which is solved by linear indefinite equation.

Specific reference: http://blog.csdn.net/synapse7/article/details/9901195 calls ExtGcd (bmae Modjie x Magi y), and x is the inverse p of b%Mod.

There is another way to find the inverse p of b%Mod, that is, p = b ^ (Mod-2)% Mod, because b ^ (Mod-1)% Mod=1 (here Mod is a prime). Error analysis: 1:if (yellow1) ans*=x%29;// mistakenly misplaced ans=x*x%292. The data type should be _ _ int64

Code implementation:

# include#includeusing namespace std;typedef _ int64 ll;ll powmol (ll XLY) / / how to find the module x ^ ymod 29 {ll ans=1; x% 29; while (y) {if (y) ans*=x%29;//y is odd While (scanf ("% I64d", & x), x) {a = (powmol (2) powmol (2)-1)% 29; b = (powmol (3)) * 15% 29; c = (powmol (22))-1) * 18% 29; c = (powmol (22))-1) * 18% 29 Printf ("% I64d\ n", (aquib)% 29*c%29);} return 0;} "the method of Java higher power module + product function + inverse" is introduced here, thank you for reading. If you want to know more about the industry, you can follow the website, the editor will output more high-quality practical articles for you!

Welcome to subscribe "Shulou Technology Information " to get latest news, interesting things and hot topics in the IT industry, and controls the hottest and latest Internet news, technology news and IT industry trends.