POJ 2417 Discrete Logging（离散对数 BSGS）-优快云博客

本文介绍了一种计算离散对数的有效方法，并提供了一个具体的C++实现案例。该方法利用了大数定理，特别是费马小定理，通过Baby-Step Giant-Step (BSGS) 算法高效地解决了离散对数问题。

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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) .

题目大意：计算离散对数，模板题，就是练练板子。。。

#include<cstdio>
#include<cstring>
#include<algorithm>
#include<cmath>
#define ll long long
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 != -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;
    ll x = 1,p = 1;
    for(int i = 0;i < m; ++i, p = p * a % n) insert(p * b % n,i);
    for(ll i = m;;i += m)
    {
        if((j = Find(x = x * p % n)) != -1) return i - j;
        if(i > n) break;
    }
    return -1;
}
int main()
{
    int p,b,n;
    while(scanf("%d %d %d",&p,&b,&n) != EOF)
    {
        int ans = BSGS(b,n,p);
        if(ans == -1) printf("no solution\n");
        else printf("%d\n",ans);
    }
    return 0;
}