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 != -; i = next[i])
         if(hs[i] == x)
             return id[i];
     return -;
 }
 int BSGS(int a,int b,int n)
 {
     memset(head,-,sizeof(head));
     top = ;
     if(b == )return ;
     int m = sqrt(n*1.0), j;
     long long x = , p = ;
     for(int i = ; 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)) != - )return i-j;
         if(i > n)break;
     }
     return -;
 }
 int main()
 {
     //freopen("in.txt","r",stdin);
     //freopen("out.txt","w",stdout);
     int a,b,n;
     while(scanf("%d%d%d",&n,&a,&b) == )
     {
         int ans = BSGS(a,b,n);
         if(ans == -)printf("no solution\n");
         else printf("%d\n",ans);
     }
     return ;
 }