POJ 2417 Discrete Logging BSGS
Discrete Logging
| Time Limit: 5000MS | Memory Limit: 65536K | |
| Total Submissions: 4011 | Accepted: 1849 |
Description
Given a prime P, 2 <= P < 2
31, an integer B, 2 <= B < P, and an integer N, 1 <= N < P, compute the discrete logarithm of N, base B, modulo P. That is, find an integer L such that
BL == N (mod P)
Input
Read several lines of input, each containing P,B,N separated by a space.
Output
For each line print the logarithm on a separate line. If there are several, print the smallest; if there is none, print "no solution".
Sample Input
5 2 1 5 2 2 5 2 3 5 2 4 5 3 1 5 3 2 5 3 3 5 3 4 5 4 1 5 4 2 5 4 3 5 4 4 12345701 2 1111111 1111111121 65537 1111111111
Sample Output
0 1 3 2 0 3 1 2 0 no solution no solution 1 9584351 462803587
Hint
The solution to this problem requires a well known result in number theory that is probably expected of you for Putnam but not ACM competitions. It is Fermat's theorem that states
for any prime P and some other (fairly rare) numbers known as base-B pseudoprimes. A rarer subset of the base-B pseudoprimes, known as Carmichael numbers, are pseudoprimes for every base between 2 and P-1. A corollary to Fermat's theorem is that for any m
B(P-1) == 1 (mod P)
for any prime P and some other (fairly rare) numbers known as base-B pseudoprimes. A rarer subset of the base-B pseudoprimes, known as Carmichael numbers, are pseudoprimes for every base between 2 and P-1. A corollary to Fermat's theorem is that for any m
B(-m) == B(P-1-m) (mod P) .
Source
Waterloo Local 2002.01.26
/* ***********************************************
Author :CKboss
Created Time :2015年03月31日 星期二 19时39分34秒
File Name :POJ2417.cpp
************************************************ */
#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <string>
#include <cmath>
#include <cstdlib>
#include <vector>
#include <queue>
#include <set>
#include <map>
using namespace std;
const int MOD=76543;
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;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()
{
//freopen("in.txt","r",stdin);
//freopen("out.txt","w",stdout);
int b,n,p;
while(scanf("%d%d%d",&p,&b,&n)!=EOF)
{
int ans=BSGS(b,n,p);
if(ans==-1) puts("no solution");
else printf("%d\n",ans);
}
return 0;
}