BZOJ3239Discrete Logging

3239: Discrete Logging
Time Limit: 1 Sec Memory Limit: 128 MB
Submit: 372 Solved: 238
Description
Given a prime P, 2 <= P < 231, an integer B, 2 <= B < P, and an integer N, 2 <= 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”.
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) .
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
拓展BSGS。。
附上本蒟蒻的代碼：

#include<cstdio>
#include<map>
#include<cmath>
#include<iostream>
using namespace std;

long long gcd(long long x,long long y)
{
    return y==0?x:gcd(y,x%y);
}

long long exgcd(long long a,long long b,long long &x,long long &y)
{
    if (!b)
      {
        x=1,y=0;
        return a;
      }
    else
      {
        long long g=exgcd(b,a%b,x,y),t=x;
        x=y,y=t-a/b*y;
        return g;
      }
}

long long inv(long long a,long long p)
{
    long long x,y,d=exgcd(a,p,x,y);
    return d==1?(x+p)%p:-1;
}

long long pow(long long x,long long a,long long p)
{
    long long t;
    if (!a) return 1%p;
    if (a==1) return x%p;
    t=pow(x,a/2,p);
    if (a%2) return ((t*t)%p*x)%p;
    else return (t*t)%p;
}

int BSGS(long long a,long long b,long long p)
{
    long long m=0;
    for (;m*m<=p;m++);
    b%=p;
    map<long long,int>hash;
    hash[b]=0;
    long long e=b,v=inv(a,p),mul=pow(a,m,p);
    for (int i=1;i<m;i++)
      {
        e=e*v%p;
        if (!hash.count(e)) hash[e]=i;
        else break;
      }
    e=1;
    for (int i=0;i<=m;i++)
      {
        if (hash.count(e)) return hash[e]+i*m;
        e=e*mul%p;
      }
    return -1;
}

void solve(long long a,long long b,long long p)
{
    long long e=1;
    b%=p,a%=p;
    for (int i=0;i<100;i++)
      {
        if (e==b) 
          {
            printf("%d\n",i);
            return;
          }
        e=e*a%p;
      }
    int sum=0;
    while (gcd(a,p)!=1)
      {
        long long d=gcd(a,p);
        if (b%d) 
          {
            printf("no solution\n");
            return;
          }
        p/=d,sum++,b/=d;
        b=b*inv(a/d,p)%p;
      }
    int ans=BSGS(a,b,p);
    if (ans==-1) 
      {
        printf("no solution\n");
        return;
      }
    printf("%d\n",ans+sum);
}

int main()
{
    long long a,p,b;
    while (cin>>p>>a>>b)
      solve(a,b,p);
    return 0;
}