poj 2891 Strange Way to Express Integers 不互质的模同余方程组求解

给定同余方程组

　　X = r1 ( mod  a1 )

　　X = r2 ( mod a2 )

　　　　...

　　X = rn ( mod an )

求解正整数X， 若不存在输出 -1

 

关于 线性同余方程组求解， 看笔者前一篇博客关于求解方式，模板裸题

#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;

}



int main()

{

    int n;

    LL r[10], a[10];

    while( scanf("%d", &n) != EOF)

    {

        for(int i = 0; i < n; i++)

            scanf("%lld %lld",&a[i],&r[i] );

        printf("%lld\n", Modline( r, a, n ) );    

    }

    return 0;

}