模线性同余方程组求解

问题:模线性同余方程组:

  x = a1 ( mod n1 )

  x = a2 ( mod n2 )

    ....

  x = ak ( mod nk )

给定 A ( a1, a2 , ... , ak ) , N ( n1, n2, ..., nk ) 求 X 。

通常分为两种

  一, ( Ni, Nj ) 之间两两互质

  二, ( Ni, Nj ) 之间不都互质

一 ( Ni, Nj ) 之间两两互质

  定理( 见算法导论 P874 ):

  如果 n1, n2 , ... , nk 两两互质, n = n1*n2*..*nk ,则对任意整数 a1,a2,a3..,ak , 方程组

    x = ai ( mod ni )

  关于未知量 x 对 模n 有唯一解

从输入 ( a1, a2, ... , ak ) 计算出 a :

       

    定义   

    定义  

可以得到:

     

代码模板

  a = a[i] ( mod n[i] )   ( i = 0. 1. 2. .. k )       {  n[i] 之间两两互质 }

View Code
#include<stdio.h>
#include<string.h>
#include<stdlib.h>
#include<iostream>
#include<algorithm>
using namespace std;

typedef long long LL;
const int N =  1010;

LL a[N], n[N], nn, k;

LL Exgcd( LL a, LL b, LL &x, LL &y )
{
    if( b == 0 ){ x = 1; y = 0; return a; } 
    LL r = Exgcd( b, a%b, x, y );
    LL t = x;
    x = y; y = t - a/b*y;
    return r;
}
LL mul( LL a, LL b )
{
    LL res = 0;
    while( b )
    {
        if( b&1 ) if( (res+=a) >= nn ) res -= nn;
        a <<= 1; if( a >= nn ) a -= nn;
        b >>= 1;
    }
    return res;
}
LL inverse( LL A, LL B )
{//求逆元
    LL x, y;
    Exgcd( A, B, x, y );
    return (x%B+B)%B;
}
LL China(){
    scanf("%lld", &k);  // k个 模同余等式
    for(int i = 0; i < k; i++)
        scanf("%lld", &a[i] );
    for(int i = 0; i < k; i++)
        scanf("%lld", &n[i] );
    nn = 1;
    for(int i = 0; i < k; i++)
        nn *= n[i];
    
    LL A = 0;
    for(int i = 0; i < k; i++)
    {
        // ci = mi * (mi^-1  mod ni )
        // ai * ci 
        LL m = nn/n[i], m1 = inverse( m, n[i] );
        LL c = m*m1;
        A = ( A + mul( a[i], c ) ) % nn;
    }
    return A;
}

int main()
{
    printf("%lld\n",China() );    
    return 0;
}

二( Ni, Nj ) 之间不都互质  

  

    x = ai ( mod mi )  1 <= i <= k

    先考虑k==2的情况:

      x = a1 ( mod m1 )

      x = a2 ( mod m2 )

   方程组有解的充分必要条件是: d | (a1-a2) ,其中 d = (m1,m2)

证明如下:

    必要性:

        设 x 是上面同余方程组的解,从而存在整数q1,q2使得x=a1+m1*q1,x=a2+m2*q2,消去x即得a1-a2 = m2q2-m1q1。由于d= GCD(m1,m2),故d | (a1-a2)。

    充分性:

        若d=GCD(m1,m2) | (a1-a2)成立,则方程m1*x + m2*y = a1-a2有解。

        设解为x0,y0。那么m2*y0 = a1-a2 ( mod m1 )

        记x1 = a2+m2*y0,可以知道 x1=a2 ( mod m2 ),且x1 = a2+m2*y0 = a2 + ( a1-a2) = a1 ( mod m1 )

        所以 x1 = a2 ( mod m2 ) = a1 ( mod m1 ) 

        所以 x = x1 ( mod GCD[m1,m2] ) 

        另外,若x1与x2都是上面同余方程组的解,则 x1 = x2 ( mod m1 ), x1 = x2 ( mod m2 ),

        由同余的性质得 x1 = x2 ( mod [m1,m2] ),即对于模[m1,m2],同余方程组的解释唯一的。

    证毕。

C++代码模板:

  

#include<stdio.h>
typedef long long LL;
LL ExGcd( LL a, LL b, LL &x, LL &y )
{
    if( b == 0 ) { x=1;y=0; return a; }
    LL r = ExGcd( b, a%b, x, y );
    LL t = x; x = y; y = t - a/b*y;
    return r;
}
LL Modline( LL r[], LL a[], int n )
{
    //  X = r[i] ( mod a[i] ) 
    LL rr = r[0], aa = a[0];
    for(int i = 1; i < n; i++ )
    {
        // aa*x + a[i]*y = ( r[i] - rr );
        LL C = r[i] - rr, x, y;
        LL d = ExGcd( aa, a[i], x, y );
        if( (C%d) != 0 ) return -1;
        LL Mod = a[i]/d;    
        x = ( ( x*(C/d)% Mod ) + Mod ) % Mod;
        rr = rr + aa*x; // 余数累加
        aa = aa*a[i]/d; // n = n1*n2*...*nk
     }
    return rr;
}
// test 
int main()
{
    int n;
    LL r[10], a[10];
    while( scanf("%d", &n) != EOF)
    {
        for(int i = 0; i < n; i++)
            scanf("%lld", &r[i] );
        for(int i = 0; i < n; i++)
            scanf("%lld", &a[i] );
        printf("%lld\n", Modline( r, a, n ) );    
    }
    return 0;
}
原文地址:https://www.cnblogs.com/yefeng1627/p/2841058.html