[POJ2417]Discrete Logging（指数级同余方程）

Discrete Logging

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

HinThe 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：这道题目是Baby-step giant-step的裸题吧，嗯嗯，不要忘了开long long就可以了。

 #include<cstdio>
 #include<algorithm>
 #include<iostream>
 #include<cmath>
 #include<cstring>
 #include<map>
 #define    ll long long
 using namespace std;

 ll p,b,n;

 int ex_gcd(ll a,ll b,ll &x,ll &y)
 {
     if (!b){x=,y=;return a;}
     else
     {
         int fzy=ex_gcd(b,a%b,x,y);
         int t=x;x=y;
         y=t-a/b*y;
         return fzy;
     }
 }
 void solve()
 {
     ll m=(ll)sqrt(p);
     map<int,int>num;
     map<int,bool>app;
     app[]=,num[]=;
     ll zhi=;
     for (int i=;i<=m-;i++)
     {
         zhi=zhi*b%p;
         if (!app[zhi])
         {
             app[zhi]=;
             num[zhi]=i;
         }
     }
     zhi=zhi*b%p;
     ll x,y,nn=n;
     int fzy=ex_gcd(zhi,p,x,y);
     x=(x+p)%p;
     for (int i=;i<=m;i++)
     {
         if (app[nn])
         {
             printf("%lld\n",i*m+num[nn]);
             return;
         }
         else nn=nn*x%p;
     }
     printf("no solution\n");
 }
 int main()
 {
     while(~scanf("%d%d%d",&p,&b,&n))
         solve();
 }