GCD, extending Euclid, Chinese remainder theorem

1.GCD:

int gcd (int A,int  b) {    return b==0? a:gcd (b,a%b);}

2. Chinese remainder theorem:

Title: Student a gives n integers a[], student B gives n positive integers m[] and 22, the teacher asks a question: there is a positive integer ans, for each logarithm, there are: (Ans-a[i]) mod m[i]=0. Find out what the minimum is.

Input Sample:

1Ten231 2 32 3 581 2 3 4 5 6 7 8 the  the  the  A  -  -  -  the A1 2 3 4 5 6 7 8 9 Ten  One  A2 3 5 7  One  -  -  +  at  in  to Panax Notoginseng2-2 0999999999 10000000003-10000-20000-300009999 10000 100010

Implementation code:

1#include <fstream>2#include <iostream>3#include <algorithm>4#include <cstdio>5#include <cstring>6#include <cmath>7#include <cstdlib>8 9 using namespacestd;Ten  One #defineEPS 1e-6 A #definell Long Long - #defineINF 0x7fffffff -  the intN; -ll a[ *],m[ *]; -  -ll EXTENDGCD (ll a,ll b,ll &x,ll &y);//Expand Euclid +ll Crt (ll A[],ll m[],intn);//Chinese remainder theorem -  + intMain () A { at     //freopen ("d:\\input.in", "R", stdin); -     //freopen ("D:\\output.out", "w", stdout); -      while(SCANF ("%d",&N), N) { -          for(intI=0; i<n;i++) scanf ("%lld",&a[i]); -          for(intI=0; i<n;i++) scanf ("%lld",&m[i]); -printf"%lld\n", Crt (a,m,n)); in     } -     return 0; to } +ll EXTENDGCD (ll a,ll b,ll &x,ll &y) { -     if(!b) { thex=1, y=0; *         returnA; $}Else{Panax Notoginsengll R=EXTENDGCD (b,a%b,y,x); -Y-=x* (A/b); the         returnR; +     } A } thell Crt (ll A[],ll m[],intN) { +ll mm=1; -      for(intI=0; i<n;i++) mm*=M[i]; $ll ret=0; $      for(intI=0; i<n;i++){ - ll x, y; -ll tm=mm/M[i]; the EXTENDGCD (tm,m[i],x,y); -Ret= (Ret+tm*x*a[i])%mm;Wuyi     } the     return(ret+mm)%mm; -}

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Here's a brief extension of Euclid's derivation:

Basic algorithm: For a non-negative integer A,B,GCD (A, B) that is not exactly 0, the greatest common divisor of A/b is bound to have an integer pair of x, Y, which makes gcd (A, b) =ax+by.

Proof: Set A>b.

1, obviously when B=0,GCD (A, b) =a. At this time x=1,y=0;

When 2,ab!=0

Set AX1+BY1=GCD (A, b);

bx2+ (a mod b) y2=gcd (b,a mod b);

According to the naïve Euclid principle there is gcd (A, B) =gcd (b,a MoD);

Then: ax1+by1=bx2+ (a mod b) y2;

namely: ax1+by1=bx2+ (A-(A/b) *b) y2=ay2+bx2-(A/b) *by2;

According to the identity theorem: x1=y2; Y1=x2-(A/b) *y2;

This gives us a way to solve x1,y1: The value of X1,y1 is based on X2,y2.

The idea above is defined recursively, because the recursive solution of GCD will always have a time b=0, so recursion can end.

By the way, the Chinese remainder theorem:

GCD, extending Euclid, Chinese remainder theorem