Given a prime P, 2 <= P < 2 31, an integer B, 2 <= B < P, and an integer N, 1 <= 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".
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
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) .
题解
这道题是裸的BSGS，具体内容可以看hzw的博客—传送门
#include<algorithm>
#include<map>
#include<cstdio>
#include<cstring>
#include<cmath>
#define ll long long
using namespace std;
ll p,b,n,s,x,y,m,k;
int exgcd(ll a,ll b){
if (!b){
x=; y=;
return a;
}
int d=exgcd(b,a%b);
ll t=x; x=y; y=t-(a/b)*y;
return d;
}
map<int,int> h;
int main(){
while (~scanf("%lld%lld%lld",&p,&b,&n)){
h.clear();
ll t=(ll)sqrt(p);
s=; h[]=t;
for (int i=;i<=t-;i++){
s=s*b%p;
if (!h[s]) h[s]=i;
}
s=s*b%p;
ll l=1e10,ans=n;
exgcd(s,p);
x=(x+p)%p;
for (int i=;i<=t;i++){
if (h[ans]){
if (h[ans]==t) h[ans]=;
l=i*t+h[ans];
break;
}
ans=ans*x%p;
}
if (l!=1e10) printf("%lld\n",l);
else puts("no solution");
}
return ;
}