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Discrete Logging
Description Given a prime P, 2 <= P < 231, 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 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) . Source |
裸的BSGS
b^L=b^k1*b^k2
已知b^k1 hash b^k2 复杂度O(sqrt(F)*hash(F)) hash(F)为哈希的复杂度
#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<algorithm>
#include<functional>
#include<iostream>
#include<cmath>
#include<cctype>
#include<ctime>
using namespace std;
#define For(i,n) for(int i=1;i<=n;i++)
#define Fork(i,k,n) for(int i=k;i<=n;i++)
#define Rep(i,n) for(int i=0;i<n;i++)
#define ForD(i,n) for(int i=n;i;i--)
#define RepD(i,n) for(int i=n;i>=0;i--)
#define Forp(x) for(int p=pre[x];p;p=next[p])
#define Forpiter(x) for(int &p=iter[x];p;p=next[p])
#define Lson (x<<1)
#define Rson ((x<<1)+1)
#define MEM(a) memset(a,0,sizeof(a));
#define MEMI(a) memset(a,127,sizeof(a));
#define MEMi(a) memset(a,128,sizeof(a));
#define INF (2139062143)
#define F (100000007)
#define MAXN (1000000)
typedef long long ll;
ll mul(ll a,ll b){return (a*b)%F;}
ll add(ll a,ll b){return (a+b)%F;}
ll sub(ll a,ll b){return (a-b+(a-b)/F*F+F)%F;}
void upd(ll &a,ll b){a=(a%F+b%F)%F;}
char s[]="no solution\n";
class Math
{
public:
ll gcd(ll a,ll b){if (!b) return a;return gcd(b,a%b);}
ll abs(ll x){if (x>=0) return x;return -x;}
ll exgcd(ll a,ll b,ll &x, ll &y)
{
if (!b) {x=1,y=0;return a;}
ll g=exgcd(b,a%b,x,y);
ll t=x;x=y;y=t-a/b*y;
return g;
}
ll pow2(ll a,int b,ll p)
{
if (b==0) return 1;
if (b==1) return a;
ll c=pow2(a,b/2,p);
c=c*c%p;
if (b&1) c=c*a%p;
return c;
}
ll Modp(ll a,ll b,ll p)
{
ll x,y;
ll g=exgcd(a,p,x,y),d;
if (b%g) {return -1;}
d=b/g;x*=d,y*=d;
x=(x+abs(x)/p*p+p)%p;
return x;
}
int h[MAXN];
ll hnum[MAXN];
int hash(ll x)
{
int i=x%MAXN;
while (h[i]&&hnum[i]!=x) i=(i+1)%MAXN;
hnum[i]=x;
return i;
}
ll babystep(ll a,ll b,int p)
{
MEM(h) MEM(hnum)
int m=sqrt(p);while (m*m<p) m++;
ll res=b,ans=-1;
ll uni=pow2(a,m,p);
if (!uni) if (!b) ans=1;else ans=-1; //特判
else
{
Rep(i,m+1)
{
int t=hash(res);
h[t]=i+1;
res=(res*a)%p;
}
res=uni;
For(i,m+1)
{
int t=hash(res);
if (h[t]) {ans=i*m-(h[t]-1);break;}else hnum[t]=0;
res=res*uni%p;
}
}
return ans;
}
}S;
int main()
{
// freopen("poj2471.in","r",stdin);
// freopen(".out","w",stdout);
ll p,b,n;
while(cin>>p>>b>>n)
{
ll ans=S.babystep(b,n,p);
if (ans==-1) cout<<s;
else cout<<ans<<endl;
}
return 0;
}
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