【BZOJ3239】Discrete Logging
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
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) .
Sample Input
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
1
3
2
0
3
1
2
0
no solution
no solution
1
9584351
题解：BSGS裸题
#include <cstdio>
#include <cstring>
#include <map>
#include <iostream>
#include <cmath>
using namespace std;
typedef long long ll;
map<ll,int> mp;
int main()
{
ll A,B,P,i,x,y,m;
while(scanf("%lld%lld%lld",&P,&A,&B)!=EOF)
{
mp.clear(),mp[B]=0,m=ceil(sqrt(P));
for(x=1,i=1;i<=m;i++) x=x*A%P,mp[x*B%P]=i;
for(y=1,i=1;i<=m;i++)
{
y=y*x%P;
if(mp.find(y)!=mp.end())
{
printf("%lld
",i*m-mp[y]);
break;
}
}
if(i==m+1) printf("no solution
");
}
return 0;
}