POJ 2417 Discrete Logging

Time Limit: 5000MS		Memory Limit: 65536K
Total Submissions: 6190		Accepted: 2746

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

 

BSGS模板 

屠龙宝刀点击就送

#include <cstdio>
#include <cmath>
#include <map>
using namespace std;
map<long long,int>q;
int P,B,N;
long long ans;
bool flag=false;
long long ksm(int m,int n,int ha)
{
    long long r=1,base=m;
    for(;n;n>>=1)
    {
        if(n&1)
            r=(r*base)%ha;
        base=(base*base)%ha;
    }
    return r;
}
void bsgs(long long y,long long z,long long p)
{
    q.clear();
    int m=(int)ceil(sqrt((double)p));//防止编译错误
    flag=false;
    q[1]=m+1;
    long long a=1;
    for(int i=1;i<m;i++)
    {
        a=a*y%p;
        if(!q[a]) q[a]=i;
    }
    long long inv=1;
    a=ksm(y,p-m-1,p);
    for(int k=0;k<m;k++)
    {
        long long an=z*inv%p;
        if(q[an])
        {
            int v=q[an];
            if(v==m+1) {ans=k*m;flag=true;return;}
            else {ans=k*m+v;flag=true;return ;}
        }
        inv=inv*a%p;
    }
}
int main()
{
    for(;scanf("%d%d%d",&P,&B,&N)!=EOF;)
    {
        if(B%P==0&&N==0) printf("1
");
        else if(B%P==0) printf("no solution
");
        else
        {
            bsgs(B,N,P);
            if(!flag) printf("no solution
");
            else printf("%lld
",ans);
        }
    }
    return 0;
}