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 != -;i= Next[i])
     {
         if(hs[i] == x)
             return id[i];
     }
     return -;
 }

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

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