int exgcd(int a, int b, int &x, int &y)//求x,y使ax+by=gcd(a,b)
{
    if (!b)
    {
        x = 1, y = 0;
        return a;
    }
    int d = exgcd(b, a % b, y, x);
    y -= a / b * x;
    return d;
}

线性同余方程求解

int a, b, x, y, m;//求x使ax%m=b%m
cin >> a >> b >> m;
int d = exgcd(a, m, x, y);
if (b % d)
{
    cout << "impossible" << endl;
}
else
{
    x = b / d * x;
    int t = abs(m / d);
    cout << (x % t + t) % t << endl;       //求最小正整数解
    //cout << (ll)x * (b / d) % m << endl; //求出其中一个解
}