【POJ 2417】 Discrete Logging

【题目链接】

http://poj.org/problem?id=2417

【算法】

Baby-Step,Giant-Step算法

【代码】

#include <algorithm>
#include <bitset>
#include <cctype>
#include <cerrno>
#include <clocale>
#include <cmath>
#include <complex>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <ctime>
#include <deque>
#include <exception>
#include <fstream>
#include <functional>
#include <limits>
#include <list>
#include <map>
#include <iomanip>
#include <ios>
#include <iosfwd>
#include <iostream>
#include <istream>
#include <ostream>
#include <queue>
#include <set>
#include <sstream>
#include <stdexcept>
#include <streambuf>
#include <string>
#include <utility>
#include <vector>
#include <cwchar>
#include <cwctype>
#include <stack>
#include <limits.h>
using namespace std;
typedef long long ll;
const ll INF = 1e15;
const int MOD = 1e5 + ;

ll B,N,P,ans,tot;
int head[MOD],nxt[MOD],v[MOD];
ll num[MOD];

inline ll power(ll a,ll n)
{
        ll res = ,b = a;
        while (n)
        {
                if (n & ) res = res * b % P;
                b = b * b % P;
                n >>= ;
        }
        return res;
}
inline void clear()
{
        tot = ;
        memset(head,,sizeof(head));
}
inline void insert(ll val,int j)
{
        int h = val % MOD;
        tot++;
        num[tot] = val;
        v[tot] = j;
        nxt[tot] = head[h];
        head[h] = tot;
}
inline int query(ll val)
{
        int i,h = val % MOD;
        for (i = head[h]; i; i = nxt[i])
        {
                if (num[i] == val)
                        return v[i];
        }
        return -;
}
inline ll Baby_Step_Giant_Step(ll B,ll N,ll P)
{
        int i,j,t;
        ll val,ans = INF;
        clear();
        B %= P;
        t = (int)sqrt(P) + ;
        for (j = ; j <= t; j++)
        {
                val = N * power(B,j) % P;
                insert(val,j);
        }
        B = power(B,t);
        if (B == ) return N ==  ?  : -;
        for (i = ; i <= t; i++)
        {
                val = power(B,i);
                j = query(val);
                if (j >=  && i * t - j >= ) return i * t - j;
        }
        return -;
}

int main()
{

        while (scanf("%lld%lld%lld",&P,&B,&N) != EOF)
        {
                ans = Baby_Step_Giant_Step(B,N,P);
                if (ans == -) printf("no solution\n");
                else printf("%lld\n",ans);
        }

        return ;

}