【bzoj 3239】【POJ 2417】Discrete Logging（BSGS）

3239: Discrete Logging

Time Limit: 1 Sec Memory Limit: 128 MB
Submit: 395 Solved: 252
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Description

Given a prime P, 2 <= P < 231, an integer B, 2 <= B < P, and an integer N, 2 <= 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”.

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

BP−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 == BP−1−m (mod P) .
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
Source

【题解】【BSGS模板题】
BSGS算法见：
[http://blog.csdn.net/reverie_mjp/article/details/51233630]

#include<map>
#include<cmath>
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
long long p,b,n;
long long ans;
map<long long,long long>mp;
inline long long poww(long long x,long long q)
{
   if (q==0) return 1;
   if (q==1) return x%p;
   if (q==2) return x*x%p;
   if (q%2==0)
       return poww(poww(x,q/2),2)%p;
    else
      return poww(poww(x,q/2),2)*x%p;
}
int main()
{
  long long i,j;
  while (scanf("%I64d%I64d%I64d",&p,&b,&n)==3)
    {
     if (b%p==0)
        {printf("no solution\n"); continue;}
     long long m,sum=0,k,x;
     bool t=false;
     mp.clear();
     m=ceil(sqrt((double)p));//sqrt在C++中是实数类型的函数，所以里面要进行计算的数据必须要是float或double类型的 
     n%=p;               
     sum=n; mp[sum]=0;
     for (j=1;j<=m;++j)                                          
        {
         sum=sum*b%p;
         mp[sum]=j;
        }
     sum=1; 
     x=poww(b,m);
     for (i=1;i<=m;++i)
      {
       sum=sum*x%p;
       if (mp[sum])
          {t=true; k=mp[sum]; break;}
      }
     ans=i*m-k;
     if (!t)
        printf("no solution\n");
      else
        printf("%I64d\n",(ans%p+p)%p);   
    }
  return 0;
}//BSGS模板题