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本文作者：ljh2000
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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) .
Source
正解：BSGS算法
解题报告：
BSGS模板题。
BSGS又称大步小步算法(有人戏称之为拔山盖世算法)，其实应该算是一种优化暴力，是一种用空间换时间的办法。
首先我们想对于$a^{x} \equiv b$ ($mod p$)，$a、b、p$已知，求最小的正整数$x$。不妨设 $m= \sqrt{p} $ 取上整，令 $x=i*m+j$ ，那么我把原式化开之后就可以得到$a^{m*i}与b*a^{j}$关于p同余。对于右边值从$0$到$m$枚举$j$，把值插入哈希表，对于左边值从$1$到$m$枚举$i$，把值在哈希表中查询看是否存在，查询到的第一个答案即为所求。如果找不到的话，考虑因为我等于是枚举了$ a^{p} $以内的所有情况，但是还没有找到，根据费马小定理，指数大于$p$一定无解。
正确性的话应该是很好想通的，因为i枚举一开始就是$1$，乘上$m$之后显然一定比$b$大。
另外注意一点，因为插入哈希表时如果出现了相等的情况，显然$j$越大越好，所以j从小到大枚举时可以直接覆盖掉之前的结果。
//It is made by ljh2000
#include <iostream>
#include <cstdlib>
#include <cstring>
#include <cstdio>
#include <cmath>
#include <algorithm>
#include <ctime>
#include <vector>
#include <queue>
#include <map>
#include <set>
#include <string>
using namespace std;
typedef long long LL;
const int MOD = 300007;
const int MAXM = 100000;
LL p,b,ans,n,to[MAXM],next[MAXM];
int ecnt,first[MOD+12],block,w[MAXM];
inline LL gcd(LL x,LL y){ if(y==0) return x; return gcd(y,x%y); }
inline LL fast_pow(LL x,LL y){ if(y==0) return 1; LL r=1; while(y>0) { if(y&1) r*=x,r%=p; x*=x; x%=p; y>>=1; } return r; }
inline int getint(){
int w=0,q=0; char c=getchar(); while((c<‘0‘||c>‘9‘) && c!=‘-‘) c=getchar();
if(c==‘-‘) q=1,c=getchar(); while (c>=‘0‘&&c<=‘9‘) w=w*10+c-‘0‘,c=getchar(); return q?-w:w;
}
inline void insert(LL x,int j){
LL cc=x; x%=MOD; for(int i=first[x];i;i=next[i]) if(to[i]==cc) { w[i]=j; return ; }
next[++ecnt]=first[x]; first[x]=ecnt; to[ecnt]=cc; w[ecnt]=j;
}
inline LL query(LL x){
LL cc=x; x%=MOD; for(int i=first[x];i;i=next[i]) if(to[i]==cc) return w[i];
return -1;
}
inline void work(){
bool ok;
while(scanf("%lld",&p)!=EOF) {
b=getint(); n=getint(); ans=0; if(n==1) { printf("0\n"); continue; }
if(gcd(b,p)!=1) { printf("no solution\n"); continue; }
memset(first,0,sizeof(first)); ecnt=0;
block=sqrt(p); if(block*block<p) block++;
for(int i=0;i<=block;i++) insert((n*fast_pow(b,i))%p,i);
LL bm=fast_pow(b,block); ok=false;
for(int i=1;i<=block;i++) {
ans=query(fast_pow(bm,i));
if(ans==-1) continue;
ok=true; printf("%lld\n",(LL)i*block-ans);
break;
}
if(!ok) printf("no solution\n");
}
}
int main()
{
work();
return 0;
}