中国剩余定理及扩展中国剩余定理

TJOI2009 猜数字

HDU 1573 X问题

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中国剩余定理CRT

中国剩余定理是用来求线性同于方程组的。
\[ \begin{aligned} \left \{ \begin{matrix} x \equiv c_1 (mod \,\,m_1 )\\ x \equiv c_2 (mod \,\,m_2 )\\ ...\\ x \equiv c_n(mod \,\, m_n) \end{matrix} \right. \end{aligned} \]
中国剩余定理是这样来的。

我们先考虑如下几个方程组：
\[ \begin{aligned}\left \{ \begin{matrix} x_1 \equiv 1 (mod \,\,m_1 )\\ x_1 \equiv 0 (mod \,\,m_2 )\\...\\x_1 \equiv 0(mod \,\, m_n)\end{matrix} \right.,\left \{ \begin{matrix}x_2 \equiv 0 (mod \,\,m_1 )\\x_2 \equiv 1 (mod \,\,m_2 )\\...\\x_2 \equiv 0(mod \,\, m_n)\end{matrix} \right.,\left \{ \begin{matrix}x_n \equiv 0 (mod \,\,m_1 )\\x_n \equiv 0 (mod \,\,m_2 )\\...\\x_n \equiv 1(mod \,\, m_n)\end{matrix} \right.\end{aligned} \]
那么，在\(gcd(m1,m2,...,m_n)\)时，就显然有如下结论：

若
\[ \begin{aligned} \prod_{i \neq 1} m_i x_1' & \equiv 1(mod \,\, m_1)\\ \prod_{i \neq 2} m_i x_2' & \equiv 1(mod \,\, m_2)\\ \prod_{i \neq n} m_i x_n' & \equiv 1(mod \,\, m_n)\\ \end{aligned} \]
那么
\[ \begin{aligned} x_1 & =x_1'\prod_{i\neq 1} m_i \\ x_2 & =x_2'\prod_{i\neq 2} m_i \\ x_n & =x_n'\prod_{i\neq n} m_i \\ \end{aligned} \]
然后最终的结果就是
\[ x=\sum_{i=1}^nc_ix_i+k\prod_{i=1}^nm_i \]
那么程序就比较好写啦。

TJOI2009 猜数字

#include 
#define LL long long
using namespace std;

LL n;
LL A[ 20 ], B[ 20 ];
LL M, Ans, T[ 20 ];

LL QM( LL x, LL y ) {
    LL Ans = 0;
    for( ; y; y >>= 1, x = x * 2 % M ) 
        if( y & 1 ) Ans = ( Ans + x ) % M;
    return Ans;
}

void Expower( LL a, LL b, LL &x, LL &y ) {
    if( b == 0 ) {
        x = 1; y = 0; return;
    }
    Expower( b, a % b, y, x );
    y -= a / b * x;
    return;
}

LL INV( LL a, LL b ) {
    LL x, y;
    Expower( a, b, x, y );
    if( x < 0 ) x += b;
    return x;
}

int main() {
    scanf( "%lld", &n );
    for( LL i = 1; i <= n; ++i ) scanf( "%lld", &A[ i ] );
    for( LL i = 1; i <= n; ++i ) scanf( "%lld", &B[ i ] );
    for( LL i = 1; i <= n; ++i ) A[ i ] %= B[ i ];
    M = 1;
    for( LL i = 1; i <= n; ++i ) M *= B[ i ];
    for( LL i = 1; i <= n; ++i ) T[ i ] = QM( INV( M / B[ i ], B[ i ] ), ( M / B[ i ] ) );
    Ans = 0;
    for( LL i = 1; i <= n; ++i ) Ans = ( Ans + QM( A[ i ], T[ i ] ) ) % M;
    printf( "%lld\n", Ans );
    return 0;
}

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扩展中国剩余定理ExCRT

刚才提到，中国剩余定理适用于模数互质的时候。要是模数不互质，那么就需要用到扩展中国剩余定理。

ExCRT是这样工作的：

我们先观察两个线性同余方程组：
\[ \begin{aligned} x & \equiv c_1 (mod\,\,m_1)\\ x & \equiv c_2 (mod\,\,m_2) \end{aligned} \]
我们将它写成这种形式：
\[ \begin{aligned} x & =c_1+k_1m_1 \\ x & =c_2+k_2m_2 \end{aligned} \]
联立后得到：
\[ \begin{aligned} c_1+k_1m_1 &=c_2+k_2m_2\\ \Rightarrow k_1m_1-k_2m_2&= c_2-c_1 \end{aligned} \]
由裴蜀定理得，方程有解的充要条件是\(gcd(m_1,m_2)|(c_2-c_1)\)。

这样的话，我们又可以得到：
\[ \begin{aligned} &k_1\frac{m_1}{gcd(m_1,m_2)}-k_2\frac{m_2}{gcd(m_1,m_2)}=\frac{c_2-c_1}{gcd(m_1,m_2)}\\ \Rightarrow &k_1\frac{m_1}{gcd(m_1,m_2)}\equiv \frac{c_2-c_1}{gcd(m_1,m_2)}\,\,(mod\,\, \frac{m_2}{gcd(m_1,m_2)})\\ \Rightarrow & k_1\equiv\frac{c_2-c_1}{gcd(m_1,m_2)}\times(\frac{m_1}{gcd(m_1,m_2)})^{-1} \,\,(mod\,\,\frac{m_2}{gcd(m_1,m_2)}) \end{aligned} \]
然后将\(k_1\)代回\(x=c_1+k_1m_1\)中，得到
\[ x\equiv\frac{c_2-c_1}{gcd(m_1,m_2)}\times (\frac{m_1}{gcd(m_1,m_2)})^{-1}\times m_1+c_2\,\,(mod \,\, \frac{m_2}{gcd(m_1,m_2)}) \]
我们又得到了一个形如\(x\equiv c \,\,(mod\,\,m)\)的线性同余方程。所以迭代求解即可。

HDU 1573 X问题

#include 
#define LL long long
using namespace std;

const int Maxm = 20;

void Work();

int main() {
    int TestCases;
    scanf( "%d", &TestCases );
    for( ; TestCases; --TestCases ) Work();
    return 0;
}

int N, M, A[ Maxm ], B[ Maxm ];
struct equation {
    LL A, B;
};
equation T1, T2;

LL GCD( LL x, LL y ) {
    LL m = x % y;
    while( m ) {
        x = y; y = m; m = x % y;
    }
    return y;
}

void ExGCD( LL a, LL b, LL &x, LL &y ) {
    if( b == 0 ) {
        x = 1; y = 0; return;
    }
    ExGCD( b, a % b, y, x );
    y -= a / b * x;
    return;
}

LL Inv( LL a, LL b ) {
    LL x, y;
    ExGCD( a, b, x, y );
    if( x < 0 ) x += b;
    return x;
}

equation ExCRT( equation X, equation Y ) {
    LL Gcd = GCD( X.A, Y.A );
    if( ( Y.B - X.B ) % Gcd ) return ( equation ) { 0, 0 };
    LL A = X.A * Y.A / Gcd;
    LL B = Inv( X.A / Gcd, Y.A / Gcd ) * ( Y.B - X.B ) / Gcd % ( Y.A / Gcd ) * X.A + X.B;
    return ( equation ) { A, B };
}

void Work() {
    scanf( "%d%d", &N, &M );
    for( int i = 1; i <= M; ++i ) scanf( "%d", &A[ i ] );
    for( int i = 1; i <= M; ++i ) scanf( "%d", &B[ i ] );
    T1 = ( equation ) { A[ 1 ], B[ 1 ] };
    for( int i = 2; i <= M; ++i ) {
        T2 = ( equation ) { A[ i ], B[ i ] };
        T1 = ExCRT( T1, T2 );
        if( !T1.A ) {
            printf( "0\n" );
            return;
        }
    }
    if( T1.B < 0 ) T1.B += T1.A;
    if( T1.B > N ) {
        printf( "0\n" );
        return;
    }
    if( T1.B ) printf( "%d\n", ( int ) ( ( N - T1.B ) / T1.A + 1 ) ); 
    else printf( "%d\n", ( int ) ( N / T1.A ) );
    return;
}

转载于:https://www.cnblogs.com/chy-2003/p/11458559.html