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; }
来源:https://www.cnblogs.com/yefeng1627/archive/2012/12/31/2841058.html