POJ 2417 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 

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".
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
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

BSGS~

BSGS—Baby-Step-Giant-Step，用于解方程A^x≡B(mod C)：

设m=ceil(sqrt(C))（ceil即上取整），x=i*m-j，则A^(i*m)≡B*A^j(mod C)。

我们从0~m枚举j，将求得的B*A^j存入一个hash表中（用map替代），然后1~m枚举i，将对应的A^(i*m)与hash匹配，第一个出现再map中的即为答案。

（更详细的证明：http://blog.csdn.net/clover_hxy/article/details/50683832~）

（POJ上用C++交这题会CE，要用G++交~）

#include<cstdio>
#include<cstring>
#include<iostream>
#include<map>
#include<cmath>
using namespace std;
#define ll long long

ll a,b,c,ans,now;
int m;
bool tr;
map<ll,int> ma;

ll mi(ll u,ll v)
{
ll num=1;
while(v)
{
if(v&1) num=(num*u)%c;
v>>=1;u=u*u%c;
}
return num;
}

int main()
{
while(scanf("%d%lld%lld",&c,&a,&b)!=EOF)
{
ma.clear();
if(!(a%c))
{
printf("no solution\n");continue;
}
m=sqrt(c);tr=0;
if(c!=m*m) m++;
ans=b%c;ma[ans]=0;
for(int i=1;i<=m;i++)
{
ans=(ans*a)%c;ma[ans]=i;
}
ans=1;now=mi(a,m);
for(int i=1;i<=m;i++)
{
ans=(ans*now)%c;
if(ma[ans])
{
now=i*m-ma[ans];
printf("%lld\n",(now%c+c)%c);
tr=1;break;
}
}
if(!tr)
{
printf("no solution\n");continue;
}
}
return 0;
}