Discrete Logging ZOJ - 1898 (模板题大小步算法)

就是求A^x三B（mod C）当C为素数时

#include<cstdio>
#include<cstring>
#include<cmath>
#include<algorithm>
using namespace std;
typedef long long LL;
const int MAXINT = ((1 << 30) - 1) * 2 + 1;

int A, B, C;
struct Hashmap //哈希表代替map
{
    static const int Ha = 999917, maxe = 46340;
    int E, lnk[Ha], son[maxe + 5], nxt[maxe + 5], w[maxe + 5];
    int top, stk[maxe + 5];
    void clear() { E = 0; while (top) lnk[stk[top--]] = 0; }
    void Add(int x, int y) { son[++E] = y; nxt[E] = lnk[x]; w[E] = MAXINT; lnk[x] = E; }
    bool count(int y)
    {
        int x = y%Ha;
        for (int j = lnk[x]; j; j = nxt[j])
        if (y == son[j]) return true;
        return false;
    }
    int& operator [] (int y)
    {
        int x = y%Ha;
        for (int j = lnk[x]; j; j = nxt[j])
        if (y == son[j]) return w[j];
        Add(x, y); stk[++top] = x; return w[E];
    }
};
Hashmap f;

int exgcd(int a, int b, int &x, int &y)
{
    if (!b) { x = 1; y = 0; return a; }
    int r = exgcd(b, a%b, x, y), t = x; x = y; y = t - a / b*y;
    return r;
}
int BSGS(int A, int B, int C)
{
    if (C == 1) if (!B) return A != 1; else return -1;
    if (B == 1) if (A) return 0; else return -1;
    if (A%C == 0) if (!B) return 1; else return -1; //几种特判
    int m = ceil(sqrt(C)), D = 1, Base = 1; f.clear();
    for (int i = 0; i <= m - 1; i++) //先把A^j存进哈希表
    {
        f[Base] = min(f[Base], i);
        Base = ((LL)Base*A) % C;
    }
    for (int i = 0; i <= m - 1; i++)
    {
        int x, y, r = exgcd(D, C, x, y);
        x = ((LL)x*B%C + C) % C; //扩欧求A^j
        if (f.count(x)) return i*m + f[x]; //找到了
        D = ((LL)D*Base) % C;
    }
    return -1;
}
int main()
{
    while (~scanf("%d%d%d", &C, &A, &B))
    {
        int ans = BSGS(A, B, C);
        if (ans == -1) printf("no solution\n"); else
            printf("%d\n", ans);
    }
    return 0;
}