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 = (( << ) - ) *  + ;

int A, B, C;
struct Hashmap //哈希表代替map
{
    static const int Ha = , maxe = ;
    int E, lnk[Ha], son[maxe + ], nxt[maxe + ], w[maxe + ];
    int top, stk[maxe + ];
    void clear() { E = ; while (top) lnk[stk[top--]] = ; }
    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 = ; y = ; 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 == ) if (!B) return A != ; else return -;
    if (B == ) if (A) return ; else return -;
    if (A%C == ) if (!B) return ; else return -; //几种特判
    int m = ceil(sqrt(C)), D = , Base = ; f.clear();
    for (int i = ; i <= m - ; i++) //先把A^j存进哈希表
    {
        f[Base] = min(f[Base], i);
        Base = ((LL)Base*A) % C;
    }
    for (int i = ; i <= m - ; 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 -;
}
int main()
{
    while (~scanf("%d%d%d", &C, &A, &B))
    {
        int ans = BSGS(A, B, C);
        if (ans == -) printf("no solution\n"); else
            printf("%d\n", ans);
    }
    return ;
}