POJ 2417 Discrete Logging （Baby-Step Giant-Step）

Discrete Logging

Time Limit: 5000MS		Memory Limit: 65536K
Total Submissions: 2819		Accepted: 1386

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 

    B

^L

 == 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

 

 

 

 

 

 

 

模板题。

http://hi.baidu.com/aekdycoin/item/236937318413c680c2cf29d4

 

 

 

/* ***********************************************
Author        :kuangbin
Created Time  :2013/8/24 0:06:54
File Name     :F:\2013ACM练习\专题学习\数学\Baby_step_giant_step\POJ2417.cpp
************************************************ */

#include <stdio.h>
#include <string.h>
#include <iostream>
#include <algorithm>
#include <vector>
#include <queue>
#include <set>
#include <map>
#include <string>
#include <math.h>
#include <stdlib.h>
#include <time.h>
using namespace std;
//baby_step giant_step
// a^x = b (mod n) n为素数，a,b < n
// 求解上式 0<=x < n的解
#define 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 != -1; 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 a,b,n;
    while(scanf("%d%d%d",&n,&a,&b) == 3)
    {
        int ans = BSGS(a,b,n);
        if(ans == -1)printf("no solution\n");
        else printf("%d\n",ans);
    }
    return 0;
}