Discrete Logging

Discrete Logging

Time Limit: 5000MS		Memory Limit: 65536K
Total Submissions: 5865		Accepted: 2618

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

Waterloo Local 2002.01.26

BSGS模板题

 

 #include<iostream>
 #include<cstdio>
 #include<cstring>
 #include<cmath>
 #include<map>
 #define LL long long
 using namespace std;
 LL a,b,c;
 map<LL,LL>mp;
 LL fastpow(LL a,LL p,LL c)
 {
     LL base=a;LL ans=;
     while(p!=)
     {
         if(p%==)ans=(ans*base)%c;
         base=(base*base)%c;
         p=p/;
     }
     return ans;
 }
 int main()
 {
     // a^x = b (mod c)
     while(scanf("%lld%lld%lld",&c,&a,&b)!=EOF)
     {
         LL m=ceil(sqrt(c));// 注意要向上取整
         mp.clear();
         if(a%c==)
         {
         printf("no solution\n");
         continue;
         }
         // 费马小定理的有解条件
         LL ans;//储存每一次枚举的结果 b* a^j
         for(LL j=;j<=m;j++)  // a^(i*m) = b * a^j
         {
             if(j==)
             {
                 ans=b%c;
                 mp[ans]=j;// 处理 a^0 = 1
                 continue;
             }
             ans=(ans*a)%c;// a^j
             mp[ans]=j;// 储存每一次枚举的结果
         }
         LL t=fastpow(a,m,c);
         ans=;//a ^(i*m)
         LL flag=;
         for(LL i=;i<=m;i++)
         {
             ans=(ans*t)%c;
             if(mp[ans])
             {
                 LL out=i*m-mp[ans];// x= i*m-j
                 printf("%lld\n",(out%c+c)%c);
                 flag=;
                 break;
             }

         }
         if(!flag)
         printf("no solution\n");
     }

     return ;
 }