【題目鏈接】

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

【算法】

Baby-Step,Giant-Step算法

【代碼】

#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 + 7;

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 = 1,b = a;
        while (n)
        {
                if (n & 1) res = res * b % P;
                b = b * b % P;
                n >>= 1;
        }
        return res;
}
inline void clear()
{
        tot = 0;
        memset(head,0,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 -1;
}
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) + 1;
        for (j = 0; j <= t; j++)
        {
                val = N * power(B,j) % P;
                insert(val,j);
        }        
        B = power(B,t);
        if (B == 0) return N == 0 ? 1 : -1;
        for (i = 0; i <= t; i++)
        {
                val = power(B,i);
                j = query(val);
                if (j >= 0 && i * t - j >= 0) return i * t - j;
        }
        return -1;
}

int main() 
{
        
        while (scanf("%lld%lld%lld",&P,&B,&N) != EOF)
        {
                ans = Baby_Step_Giant_Step(B,N,P);
                if (ans == -1) printf("no solution\n");
                else printf("%lld\n",ans);        
        }
        
        return 0;
    
}

【POJ 2417】 Discrete Logging