中国剩余定理解同余方程组

ll exgcd(ll a,ll b,ll &x,ll &y)

{

    if(!b)

    {

        x = (ll)1,y = (ll)0;

        return a;

    }

    ll r = exgcd(b,a%b,x,y);

    ll t = x;

    x = y;

    y = t - a/b*y;

    return r;

}



//n个方程： x = a[i](mod a[i]) (0<=i<n)

ll china(int n,int *a,int *m)

{

    ll M = 1,d,y,x = 0;

    int i;

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

        M *= m[i];

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

    {

        ll w = M/m[i];

        d = exgcd(m[i],w,x,y)

        x = (x + y*w*a[i])%M;

    }

    return (x+M)%M;

}