bzoj 3239: Discrete Logging （BSGS）

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3239: Discrete Logging

Time Limit: 1 Sec Memory Limit: 128 MB
Submit: 508 Solved: 324
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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

HINT

Source

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题解：BSGS

#include<iostream>#include<cstdio>#include<cstring>#include<cmath>#include<algorithm>#include<map>#define LL long long using namespace std;LL a,b,c;map<LL,int> mp;LL quickpow(int x){LL base=a%c; LL ans=1;while (x) {if (x&1) ans=ans*base%c;x>>=1;base=base*base%c;}return ans;}int main(){freopen("a.in","r",stdin);while (scanf("%I64d%I64d%I64d",&c,&a,&b)!=EOF) {mp.clear();if (a%c==0) {printf("no solution\n");continue;}int p=false;int m=ceil(sqrt(c));LL ans;for (int i=0;i<=m;i++) {if (i==0) {ans=b%c; mp[ans]=i; continue;}ans=(ans*a)%c; mp[ans]=i;}LL t=quickpow(m); ans=1; bool pd=false;for (int i=1;i<=m;i++) {ans=ans*t%c;if (mp[ans]) {int t=i*m-mp[ans];printf("%d\n",(t%c+c)%c);pd=true;break;}}if (!pd) printf("no solution\n");}}

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