HDU 1452 Happy 2004 (Factor and integrable function)

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Test instructions: Give you an X that allows you to find the sum of all the factors of X 2004, and then take the remainder to 29.

Idea: Originally this is the integrable function, point here here, here is very detailed.

In the field of non-number theory, the integrable function refers to all functions that have the Properties F (AB) =f (a) f (b) for any and b .

The integrable function in number theory: for an arithmetic function f (n)of a positive integer n , if f ( 1) =1, and when a, b coprime f (AB) =f (a) f (b), In number theory, it is called the integrable function.

If for an integrable function f (n), even if A, b is not coprime, there is also F (AB) =f (a) f (b), it is said to be fully integrable.

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

S (=s) (4) *s (5) =7*6=42;

Look at s () = 1+2+5+10+25+50=93=3*31=s (2) *s (+), s (+) =1+5+25=31.

This is called the integrable function in number theory, when gcd (A, b) =1 s (a*b) =s (a) *s (b);

If P is prime: S (p^n) =1+p+p^2+...+p^n= (p^ (n+1)-1)/(P-1)-----is actually the geometric series summation formula (1)

Look at the subject again:

Calculation factor and s (2004^x) mod,

2004=2^2 *167

2004^x=4^x * 3^x *167^x

s (2004^x)) = (s (2^2x) )) * (S (3^x))) * (S (167^x))) and 167%29=22

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

a=s (2^2x) = (2^ (2x+1)-1)// according to (1)

b=s (3^x) = (3^ (x+1)-1)/2// according to (1)

c=s (22^x) = (22^ (x+1)-1)/21// according to (1)

% algorithm

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

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

s (2004^x) =(2^ (2x+1)-1)* (3^ (x+1)-1)/2 * (22^ (x+1)-1)/21

(a*b)//%m= a%m* b%m * INV (c)

C*INV (c) =1%M modulus is 1 of all the number inv(C) is the smallest can be divisible by C

INV (2) =15, INV (+) =18 2*15=1 mod, 18*21=1 mod 29

s (2004^x) =((2^ (2x+1)-1)* (3^ (x+1)-1)/2 * (22^ (x+1)-1)/21)mod = ((2^ ( 2x+1)-1) * (3^ (x+1)-1) *15 * (22^ (x+1)-1) *18) mod29

b^ ( -1) is the inverse element of b (%p), which is the inv above

The inverse element of 2 is the 2*15=30% 29=1%

The inverse of the element is, because 21*18=378% =1%

So

a=(powi (2,2*x+1,29)-1)%;

B= (Powi (3,x+1,29)-1) *15% 29;

C= (Powi (22,x+1,29)-1) *18% 29;

ans=(a*b)% 29*c%;

1 //14522#include <stdio.h>3#include <math.h>4#include <iostream>5 6 using namespacestd;7 8 intMultimod (intAintN//The exponentiation mode9 {Ten     intres =1 ; One      while(n) A     { -         if(N &1) -         { theRes *=A; -Res%= in ; -         } -A *=A; +A%= in ; -N >>=1 ; +     } A     returnRes; at } - intMain () - { -     intx; -      while(~SCANF ("%d",&x)) -     { in         if(x = =0) Break ; -         intA = (Multimod (2,2*x+1)-1) ; to         intb = (Multimod (3, x+1)-1)* the ; +         intc = (Multimod (167, x+1)-1)* - ; -printf"%d\n", (a*b*c)% in) ; the     } *     return 0 ; $}

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