【BZOJ3239】Discrete Logging BSGS

【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

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".

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 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

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\n",i*m-mp[y]);
				break;
			}
		}
		if(i==m+1)	printf("no solution\n");
	}
	return 0;
}