【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