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
HINT
Source
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題解：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");	}}