poj 2417 && poj3243(Baby-Step Giant-Step)

 

Discrete Logging

Time Limit: 5000MS		Memory Limit: 65536K
Total Submissions: 4624		Accepted: 2113

 

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

 

题意：

a^x = b(mod n) ，求解x(模板题)

 

#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <algorithm>
#include <cmath>
using namespace std;
typedef long long ll;
typedef long double ld;

using namespace std;
#define MOD 76543
int hs[MOD],head[MOD],Next[MOD],id[MOD],top;

void insert(int x,int y)
{
    int k = x % MOD;
    hs[top] = x,id[top] = y,Next[top] = head[k],head[k] = top++;
}

int find(int x)
{
    int k = x % MOD;
    for(int i = head[k];i != -1;i= Next[i])
    {
        if(hs[i] == x)
            return id[i];
    }
    return -1;
}

int BSGS(int a,int b,int n)
{
    memset(head,-1,sizeof(head));
    top = 1;
    if(b == 1)
        return 0;
    int m = sqrt(n*1.0),j;
    long long x = 1,p =1;
    for(int i = 0;i < m;i++,p = p*a%n)
    insert(p*b%n,i);
    for(ll i = m;;i+=m)
    {
        if((j = find(x = x*p % n)) != -1) return i-j;
        if(i > n) break;
    }
    return -1;
}

int main()
{
    int p,b,n;
    while(scanf("%d%d%d",&p,&b,&n) != EOF)
    {
        int ans = BSGS(b,n,p);
        if(ans == -1)
            printf("no solution\n");
        else
        printf("%d\n",ans);
    }
    return 0;
}