hash + bsgs模板

//POJ 2417
//baby_step giant_step
// a^x = b (mod n) n为素数，a,b < n
// 求解上式 0 <= x < n的解
#include <cmath>
#include <cstdio>
#include <cstring>
#define MOD 76543
using namespace std;
int hs[MOD], head[MOD], next[MOD], id[MOD], top;
void insert(int x, int y)
{
    int k = x % MOD;
    hs[top] = x;
    id[top] = y;
    next[top] = head[k];
    head[k] = top++;
}
int find(int x)
{
    int k = x % MOD;
    for (int i = head[k]; i != -1; i = next[i])
        if (hs[i] == x)
            return id[i];
    return -1;
}
int BSGS(int a, int b, int n)
{
    memset(head, -1, sizeof(head));
    top = 1;
    if (b == 1)
        return 0;
    int m = sqrt(n * 1.0), j;
    long long x = 1, p = 1;
    for (int i = 0; i < m; i++, p = p * a % n)
        insert(p * b % n, i);
    for (long long i = m; ; i += m)
    {
        if ((j = find(x = x * p % n)) != -1)
            return i - j;
        if (i > n)
            break;
    }
    return -1;
}
int main()
{
    int a, b, n;
    while (~scanf("%d%d%d", &n, &a, &b))
    {
        int ans = BSGS(a, b, n);
        if (ans == -1)
            printf("no solution\n");
        else
            printf("%d\n", ans);
    }
}