贝祖定理
对于 a , b a,b a,b是任意两个正整数,则存在整数 s , t s,t s,t使得 s a + t b = ( a , b ) sa+tb=(a,b) sa+tb=(a,b)。其中 ( a , b ) (a,b) (a,b)是 a , b a,b a,b的最大公因数。
其证明也就是求最大公因数的逆过程。这里使用辗转相除法来证明。
不妨设 a > b a> b a>b,则有下列等式成立:
a = q 0 b + r 0 , ( 0 < r 0 < b ) b = q 1 r 0 + r 1 , ( 0 < r 1 < r 0 ) r 0 = q 2 r 1 + r 2 , ( 0 < r 2 < r 1 ) ⋮ r n − 2 = q n r n − 1 + r n , ( 0 < r n < r n − 1 ) r n − 1 = q n + 1 r n + r n + 1 , ( r n + 1 = 0 ) \begin{aligned} a&=q_0b+r_0,(0<r_0<b)\\ b&=q_1r_0+r_1,(0<r_1<r_0)\\ r_0&=q_2r_1+r_2,(0<r_2<r_1)\\ &\vdots \\ r_{n-2}&=q_nr_{n-1}+r_n,(0<r_n<r_{n-1})\\ r_{n-1}&=q_{n+1}r_n+r_{n+1},(r_{n+1}=0) \end{aligned} abr0rn−2rn−1=q0b+r0,(0<r0<b)=q1r0+r1,(0<r1<r0)=q2r1+r2,(0<r2<r1)⋮=qnrn−1+rn,(0<rn<rn−1)=qn+1rn+rn+1,(rn+1=0)
根据辗转相除法, r n = ( a , b ) r_n=(a,b) rn=(a,b)
现在,将上述等式倒过来,有
r n = − q n r n − 1 + r n − 2 , r n − 1 = − q n − 2 r n − 2 + r n − 3 , ⋮ r 2 = − q 2 r 1 + r 0 , r 1 = − q 1 r 0 + b , r 0 = − q 0 b + a . \begin{aligned} r_n&=-q_nr_{n-1}+r_{n-2},\\ r_{n-1}&=-q_{n-2}r_{n-2}+r_{n-3},\\ &\vdots\\ r_2&=-q_2r_1+r_0,\\ r_1&=-q_1r_0+b,\\ r_0&=-q_0b+a. \end{aligned} rnrn−1r2r1r0=−qnrn−1+rn−2,=−qn−2rn−2+rn−3,⋮=−q2r1+r0,=−q1r0+b,=−q0b+a.
然后逐步消去 r n − 1 , r n − 2 , r n − 3 , ⋯ , r 0 r_{n-1},r_{n-2},r_{n-3},\cdots,r_0 rn−1,rn−2,rn−3,⋯,r0,便找到了对应的 s , t s,t s,t。
如果令 s − 2 = 1 , s − 1 = 0 , s j = ( − q j ) s j − 1 + s j − 2 , t − 2 = 0 , t − 1 = 1 , s j = ( − q j ) t j − 1 + t j − 2 . s_{-2}=1,s_{-1}=0,s_j=(-q_j)s_{j-1}+s_{j-2},\\ t_{-2}=0,t_{-1}=1,s_j=(-q_j)t_{j-1}+t_{j-2}. s−2=1,s−1=0,sj=(−qj)sj−1+sj−2,t−2=0,t−1=1,sj=(−qj)tj−1+tj−2.
根据递推公式,就可以轻易算出 s , t s,t s,t了。而计算的过程就是扩展欧几里得算法。
c++示例代码
#include<iostream>
#include<vector>
using namespace std;
// ensure that a > b
vector<int> exgcd(int a,int b,int s1=1,int s2=0,int t1=0,int t2=1){
int r=a%b;
if(r==0){
return {
b,s2,t2};
}
int q=a/b;
int s3=-q*s2+s1;
int t3=-q*t2+t1;
return exgcd(b,r,s2,s3,t2,t3);
}
int main(){
int a,b;
cout<<"ensure that a>b."<<endl;
cout<<"please input a "<<endl;
cin>>a;
cout<<"please input b "<<endl;
cin>>b;
vector<int> res=exgcd(a,b);
cout<<"s is "<<res[1]<<endl;
cout<<"t is "<<res[2]<<endl;
cout<<"(a,b) is "<<res[0]<<endl;
cout<<"sa + tb = "<<res[1]*a+res[2]*b<<endl;
}
大整数扩展欧几里得算法
当 ( a , b ) = 1 (a,b)=1 (a,b)=1时,往往使用扩展欧几里得算法来求逆元,但是,通常的应用都是大整数,int类型显然太小了。在gmp库中实现了扩展欧几里得算法。
其函数原型为:
void mpz_gcdext (mpz_t g, mpz_t s, mpz_t t, const mpz_t a, const mpz_t b)
#include<iostream>
#include<gmpxx.h>
#include<gmp.h>
using namespace std;
int main(){
mpz_class a,b,s,t;
cout<<"Please input a and b (a<b)"<<endl;
cout<<"a:";
cin>>a;
cout<<"b:";
cin>>b;
mpz_class gcd;
mpz_gcdext(gcd.get_mpz_t(),s.get_mpz_t(),t.get_mpz_t(),a.get_mpz_t(),b.get_mpz_t());
cout<<"s is:"<<s<<endl;
cout<<"t is:"<<t<<endl;
cout<<"the gcd is:"<<gcd<<endl;
cout<<"sa + tb = "<<s*a+t*b<<endl;
return 0;
}