POJ-2417-Discrete Logging（BSGS）

Given a prime P, 2 <= P < 2 ³¹, 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) .

 

题解

这道题是裸的BSGS，具体内容可以看hzw的博客—传送门

 #include<algorithm>
 #include<map>
 #include<cstdio>
 #include<cstring>
 #include<cmath>
 #define ll long long
 using namespace std;
 ll p,b,n,s,x,y,m,k;
 int exgcd(ll a,ll b){
     if (!b){
         x=; y=;
         return a;
     }
     int d=exgcd(b,a%b);
     ll t=x; x=y; y=t-(a/b)*y;
     return d;
 }
 map<int,int> h;
 int main(){
     while (~scanf("%lld%lld%lld",&p,&b,&n)){
         h.clear();
         ll t=(ll)sqrt(p);
         s=; h[]=t;
         for (int i=;i<=t-;i++){
             s=s*b%p;
             if (!h[s]) h[s]=i;
         }
         s=s*b%p;
         ll l=1e10,ans=n;
         exgcd(s,p);
         x=(x+p)%p;
         for (int i=;i<=t;i++){
             if (h[ans]){
                 if (h[ans]==t) h[ans]=;
                 l=i*t+h[ans];
                 break;
             }
             ans=ans*x%p;
         }
         if (l!=1e10) printf("%lld\n",l);
                 else puts("no solution");
     }
     return ;
 }