POJ 2417 Discrete Logging ( Baby step giant step )

Discrete Logging

Time Limit: 5000MS		Memory Limit: 65536K
Total Submissions: 3696		Accepted: 1727

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 )

 

 

 

题意要求解一个   k^D = n ( mod p )  的一个最小D

一开始也不会解 。 

在网上看了一下，原来是用到一个 Baby step giant step 的算法 。

先要把 D 分解 ，  D = i * m + j  (  m = ceil( sqrt (p - 1  ) ) )

原式   ：  k^D = n ( mod p ) 

       ->   k^i^m * k^j   = n (mod p )

       ->   k^j = n * ( k ^(-m)^ i ) ( mod p )

根据题目给的 Hint ( 费马小定理 )可以求出 k^m 的逆元 .

然后枚举 i  ， 查找是否存在 k^j 与 n * ( k ^(-m)^i ) 相等

所以预处理 k^j (  mod p  ) 排序以后 ， 就可以进行二分查找了 （复杂度降为log(m)）。 

加上枚举 ， 那么总复杂度就是 m*log(m) .

 

 

#include <iostream>
#include <cstdio>
#include <string>
#include <cstring>
#include <algorithm>
#include <cmath>
using namespace std;
typedef long long LL ;
const int N = 100010;

struct B
{
    LL num , id ;
    bool operator < ( const B &a ) const{
        if( num != a.num ) return num < a.num;
        else return id < a.id ;
    }
}baby[N];

LL n , k , p ;
int tot ;

LL quick_mod( LL a , LL b ,LL mod )
{
    LL res =1 ;
    while( b )
    {
        if( b &1 ) res = res * a % mod ;
        a = a * a % mod ;
        b >>= 1 ;
    }
    return res ;
}
int find( LL n )
{
    int l = 0 , r = tot - 1 ;
    while( l <= r ){
        int m = (l + r) >> 1;
        if( baby[m].num == n){
            return baby[m].id;
        }
        else if( baby[m].num < n )
            l = m + 1;
        else
            r = m - 1 ;
    }
    return -1;
}
void run()
{
    int m = (int)ceil(sqrt((double)(p-1)));
    baby[0].num = 1 , baby[0].id = 0;
    for( int i = 1 ; i < m ; ++i ){
        baby[i].id = i ;
        baby[i].num = baby[i-1].num * k % p ; //   k^j
    }
    sort( baby , baby + m );
    tot = 1 ;
    for( int i = 1 ; i < m ; ++i ){
        if( baby[tot-1].num != baby[i].num ) baby[tot++] = baby[i];
    }

    LL bm = quick_mod( k , p - 1 - m , p ) ; //  k^(-m) ;
    LL temp = n ;

    for( int j = 0 ; j < m ; ++j ){
        // k^(-m)^j
        int pos = find( temp );
        if( pos != -1 ){
            printf("%d\n" , ( m * j + pos ) );
            return ;
        }
        temp = temp * bm % p ;
    }
    puts("no solution");
}
int main()
{
    #ifdef LOCAL
        freopen("in.txt","r",stdin);
    #endif // LOCAL
    while( ~scanf("%lld%lld%lld",&p,&k,&n) ) run();
}