POJ2417 Discrete Logging

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本文作者：ljh2000

 

作者博客：http://www.cnblogs.com/ljh2000-jump/
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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 

    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) .

Source

Waterloo Local 2002.01.26

 

 

正解：BSGS算法

解题报告：

　　BSGS模板题。

　　BSGS又称大步小步算法(有人戏称之为拔山盖世算法)，其实应该算是一种优化暴力，是一种用空间换时间的办法。

　　首先我们想对于$a^{x} \equiv b$ ($mod p$)，$a、b、p$已知，求最小的正整数$x$。不妨设  $m= \sqrt{p}  $  取上整，令 $x=i*m+j$ ，那么我把原式化开之后就可以得到$a^{m*i}与b*a^{j}$关于p同余。对于右边值从$0$到$m$枚举$j$，把值插入哈希表，对于左边值从$1$到$m$枚举$i$，把值在哈希表中查询看是否存在，查询到的第一个答案即为所求。如果找不到的话，考虑因为我等于是枚举了$ a^{p} $以内的所有情况，但是还没有找到，根据费马小定理，指数大于$p$一定无解。

　　正确性的话应该是很好想通的，因为i枚举一开始就是$1$，乘上$m$之后显然一定比$b$大。

　　另外注意一点，因为插入哈希表时如果出现了相等的情况，显然$j$越大越好，所以j从小到大枚举时可以直接覆盖掉之前的结果。

 

//It is made by ljh2000
#include <iostream>
#include <cstdlib>
#include <cstring>
#include <cstdio>
#include <cmath>
#include <algorithm>
#include <ctime>
#include <vector>
#include <queue>
#include <map>
#include <set>
#include <string>
using namespace std;
typedef long long LL;
const int MOD = 300007;
const int MAXM = 100000;
LL p,b,ans,n,to[MAXM],next[MAXM];
int ecnt,first[MOD+12],block,w[MAXM];
inline LL gcd(LL x,LL y){ if(y==0) return x; return gcd(y,x%y);  }
inline LL fast_pow(LL x,LL y){ if(y==0) return 1; LL r=1; while(y>0) { if(y&1) r*=x,r%=p; x*=x; x%=p; y>>=1; } return r; }
inline int getint(){
    int w=0,q=0; char c=getchar(); while((c<'0'||c>'9') && c!='-') c=getchar();
    if(c=='-') q=1,c=getchar(); while (c>='0'&&c<='9') w=w*10+c-'0',c=getchar(); return q?-w:w;
}

inline void insert(LL x,int j){
	LL cc=x; x%=MOD; for(int i=first[x];i;i=next[i]) if(to[i]==cc) { w[i]=j; return ; }
	next[++ecnt]=first[x]; first[x]=ecnt; to[ecnt]=cc; w[ecnt]=j;
}

inline LL query(LL x){
	LL cc=x; x%=MOD; for(int i=first[x];i;i=next[i]) if(to[i]==cc) return w[i];
	return -1;
}

inline void work(){
	bool ok;
	while(scanf("%lld",&p)!=EOF) {
		b=getint(); n=getint();	ans=0;	if(n==1) { printf("0\n"); continue; }
		if(gcd(b,p)!=1) { printf("no solution\n"); continue; }
		memset(first,0,sizeof(first)); ecnt=0;
		block=sqrt(p); if(block*block<p) block++;
		for(int i=0;i<=block;i++) insert((n*fast_pow(b,i))%p,i);
		LL bm=fast_pow(b,block); ok=false;
		for(int i=1;i<=block;i++) {
			ans=query(fast_pow(bm,i)); 
			if(ans==-1) continue;
			ok=true; printf("%lld\n",(LL)i*block-ans);
			break;
		}
		if(!ok)	printf("no solution\n");
	}
}

int main()
{
    work();
    return 0;
}