【BZOJ 3239】 Discrete Logging

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3239: Discrete Logging

Time Limit: 1 Sec Memory Limit: 128 MB
Submit: 188 Solved: 131
[Submit][Status]

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

BSGS算法模板题。

BSGS算法就是求满足a^x=b (mod c) （0<=x<c）的x的算法。【这里c为素数】

做法很简单：

设m=ceil(sqrt(c)) x=i*m+j 那么a^x=(a^m)^i*a^j

枚举i,可以在O(1)时间得出j。

如何在O(1)时间求出j?

设d=(a^m)^i 那么d*a^j=b (mod c)，求出d的乘法逆元，那么就可以求出a^j%c。

可以先把a^j用hash表预处理出来，然后O(1)就可以求出j。

整个算法的时间复杂度O(sqrt(c))

#include <iostream>#include <cstring>#include <cstdio>#include <algorithm>#include <cstdlib>#include <cmath>#include <map>#define LL long longusing namespace std;map<LL,LL> mp;LL pow(LL a,LL x,LL p){LL base=a,ans=1;while (x){if (x&1LL) ans=(ans*base)%p;base=(base*base)%p;x>>=1LL;}return ans;}int main(){LL p,a,b;        while (scanf("%lld%lld%lld",&p,&a,&b)!=EOF){mp.clear();a%=p;if (!a){if (!b) puts("1");else puts("no solution");continue;}LL m=(LL)ceil(sqrt(p));LL now=1;for (int i=1;i<=m;i++){now=(LL)(now*a)%p;if (!mp[now]) mp[now]=i;}LL ni=pow(a,p-m-1,p);bool f=false;mp[1]=0;for (int i=0;i<m;i++){if (mp.count(b)){printf("%lld\n",i*m+mp[b]);f=true;break;}b=(b*ni)%p;}if (!f)puts("no solution");}return 0;}

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