Discrete Logging
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 Sample Output 0 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 |
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 ;
}
