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POJ 2417 Discrete Logging BSGS

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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 < 2³¹, 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

   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時39分34秒
File Name     :POJ2417.cpp
************************************************ */

#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include

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;in) 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;
}