3239: Discrete LoggingTime Limit:1 SecMemory Limit:128 MBSubmit:188Solved: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
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 15 2 25 2 35 2 45 3 15 3 25 3 35 3 45 4 15 4 25 4 35 4 412345701 2 11111111111111121 65537 1111111111
Sample Output
013203120no solutionno solution19584351462803587
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;}
人生并不在于获取，更在于放得下。放下一粒种子，收获一棵大树；
