【算法乱讲】BSGS

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

显然，这是一道bsgs的裸题
那么bsgs是什么玩意呢，
我们先玩一玩式子
令 m=ceil(sqrt(p))设a^(m*i+j)=b(mod p)

显然 a^j*a^(m*i)=b(mod p)　
　    a^j=b*a^(-m*i) (mod p)

因此，我们预处理所有可能的a^j丢进哈希表中然后我们枚举i，
看看有没有可能对应的j所以我们的算法时间复杂度为O(n^0.5)

 #include<stdio.h>
 #include<stdlib.h>
 #include<iostream>
 #include<string>
 #include<string.h>
 #include<algorithm>
 #include<math.h>
 #include<queue>
 #include<map>
 #include<vector>
 #include<set>
 #define il inline
 #define re register
 using namespace std;
 typedef long long ll;
 struct hash_set{
     ll v[];
     int next[],g[],w[],tot;
     il void clear(){
         memset(g,false,sizeof(g));tot=;
     }
     il void insert(ll h,int f){
         v[++tot]=h;
         w[tot]=f;
         next[tot]=g[h%];
         g[h%]=tot;
     }
     il int find(ll h){
         for(re int i=g[h%];i;i=next[i])
             if(h==v[i]) return w[i];
         return -;
     }
 } p;
 ll A,B,P,m,t,s;
 il ll ksm(re ll base,re ll pow){
     if(pow<){
         cout<<"-1";exit();
     }
     ll ans=;
     for(;pow;pow>>=){
         if(pow&) ans=ans*base%P;
         base=base*base%P;
     }
     return ans;
 }
 il ll rev(re ll a){
     return ksm(a,P-);
 }
 il void init(){
     p.clear();
     m=ceil(sqrt(P));t=;
     for(int i=;i<m;i++){
         if(p.find(t)<) p.insert(t,i);
         t=t*A%P;
     }
     //cout<<endl;
     for(int i=,l;i<=P/m;i++){
         t=rev(ksm(A,m*i));
     //    cout<<t<<" "<<m*i<<" ";
         s=t*B%P;
     //    cout<<s<<endl;
         l=p.find(s);
         if(l>=){
             printf("%lld\n",m*i+l);
             return;
         }
     }
     printf("no solution\n");
 }
 int main(){
     while(scanf("%lld%lld%lld",&P,&A,&B)!=EOF){
         init();
     }
     return ;
 }