LightOJ 1325 Distributing Chocolates

解高次同余方程：$a^n \equiv b \pmod{p}$
用baby-step giant-step算法，以给力的方法遍历$a^k$
复杂度$\log p$

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
const int p = 100000007;

ll fast_pow(ll x, int n)
{
    ll res = 1;
    while(n)
    {
        if(n & 1) res = (res * x) % p;
        x = (x * x) % p;
        n >>= 1;
    }
    return res;
}
int solve(int a, int b)
{
    int m = ceil(sqrt(p));
    map<int,int> M;
    ll mid = 1;
    for(int i = 0; i < m; i++)
        M[mid] = i, mid = (mid * a) % p;
    ll t = fast_pow(fast_pow(a, p - 2), m);
    for(int i = 0; i < m; i++)
    {
        if(M.find(b) != M.end()) return i * m + M[b];
        b = (b * t) % p;
    }
}

int main()
{
    int T;
    scanf("%d", &T);

    for(int ca = 1; ca <= T; ca++)
    {
        int k, res;
        scanf("%d%d", &k, &res);
        printf("Case %d: %d\n", ca, solve(k, res));
    }
    return 0;
}