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
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".
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
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(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
1 #include<algorithm> 2 #include<map> 3 #include<cstdio> 4 #include<cstring> 5 #include<cmath> 6 #define ll long long 7 using namespace std; 8 ll p,b,n,s,x,y,m,k; 9 int exgcd(ll a,ll b){ 10 if (!b){ 11 x=1; y=0; 12 return a; 13 } 14 int d=exgcd(b,a%b); 15 ll t=x; x=y; y=t-(a/b)*y; 16 return d; 17 } 18 map<int,int> h; 19 int main(){ 20 while (~scanf("%lld%lld%lld",&p,&b,&n)){ 21 h.clear(); 22 ll t=(ll)sqrt(p); 23 s=1; h[1]=t; 24 for (int i=1;i<=t-1;i++){ 25 s=s*b%p; 26 if (!h[s]) h[s]=i; 27 } 28 s=s*b%p; 29 ll l=1e10,ans=n; 30 exgcd(s,p); 31 x=(x+p)%p; 32 for (int i=0;i<=t;i++){ 33 if (h[ans]){ 34 if (h[ans]==t) h[ans]=0; 35 l=i*t+h[ans]; 36 break; 37 } 38 ans=ans*x%p; 39 } 40 if (l!=1e10) printf("%lld ",l); 41 else puts("no solution"); 42 } 43 return 0; 44 }