大步小步攻击算法_完全版

先膜拜一下AekdyCoin大神....可以看看他的空间,有详细的讲解

不错的大步小步算法,可以秒掉poj_2417, poj_3243这种题

struct hash
{
    int a, b, next;
} Hash[MAXN << 1];
int flg[MAXN + 66];
int top, idx;
void ins(int a, int b)
{
    int k = b & MAXN;
    if (flg[k] != idx)
    {
        flg[k] = idx;
        Hash[k].next = -1;
        Hash[k].a = a;
        Hash[k].b = b;
        return;
    }
    while (Hash[k].next != -1)
    {
        if (Hash[k].b == b) return;
        k = Hash[k].next;
    }
    Hash[k].next = ++top;
    Hash[top].next = -1;
    Hash[top].a = a;
    Hash[top].b = b;
}
 
int find(int b)
{
    int k = b & MAXN;
    if (flg[k] != idx) return -1;
    while (k != -1)
    {
        if (Hash[k].b == b)
            return Hash[k].a;
        k = Hash[k].next;
    }
    return -1;
}
 
int ex_gcd(int a, int b, int& x, int& y)
{
    int t, ret;
    if (!b)
    {
        x = 1, y = 0;
        return a;
    }
    ret = ex_gcd(b, a % b, x, y);
    t = x, x = y, y = t - a / b * y;
    return ret;
}
 
int Inval(int a, int b, int n)
{
    int x, y, e;
    ex_gcd(a, n, x, y);
    e = LL(x) * b % n;
    return e < 0 ? e + n : e;
}
 
int pow_mod(LL a, int b, int c)
{
    LL ret = 1 % c;
    a %= c;
    while (b)
    {
        if (b & 1)
            ret = ret * a % c;
        a = a * a % c;
        b >>= 1;
    }
    return ret;
}
 
//A^x=B(mod C)
//使用前先B%=C
int BabyStep(int A, int B, int C)
{
    top = MAXN, ++idx;
    LL buf = 1 % C, D = buf, K;
    int i, tmp, d = 0;
    for (i = 0; i <= 100; buf = buf * A % C, ++i)
        if (buf == B)
            return i;
    while ((tmp = __gcd(A, C)) != 1)
    {
        if (B % tmp) return -1;
        ++d;
        C /= tmp, B /= tmp;
        D = D * A / tmp % C;
    }
    int M = (int)ceil(sqrt(C * 1.0));
    for (buf = 1 % C, i = 0; i <= M; buf = buf * A % C, ++i)
        ins(i, buf);
    for (i = 0, K = pow_mod(LL(A), M, C); i <= M; D = D * K % C, ++i)
    {
        tmp = Inval((int)D, B, C);
        int w;
        if (tmp >= 0 && (w = find(tmp)) != -1)
            return i * M + w + d;
    }
    return -1;
}