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 程式師世界 >> 編程語言 >> C語言 >> C++ >> 關於C++ >> POJ 2417/BZOJ 3239(Discrete Logging

POJ 2417/BZOJ 3239(Discrete Logging

編輯:關於C++

 

Discrete Logging Time Limit: 5000MS   Memory Limit: 65536K Total Submissions: 4236   Accepted: 1948

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

Waterloo Local 2002.01.26

 

裸的BSGS

 

b^L=b^k1*b^k2

已知b^k1 hash b^k2 復雜度O(sqrt(F)*hash(F)) hash(F)為哈希的復雜度

 

 

#include
#include
#include
#include
#include
#include
#include
#include
#include
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=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
;

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>>b>>n)
	{
		ll ans=S.babystep(b,n,p);
		if (ans==-1) cout<

 

 

 

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