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)


 

 1 #include <iostream>
 2 #include <cstdio>
 3 #include <cstdlib>
 4 #include <cstring>
 5 #include <algorithm>
 6 #include <cmath>
 7 using namespace std;
 8 typedef long long ll;
 9 typedef long double ld;
10 
11 using namespace std;
12 #define MOD 76543
13 int hs[MOD],head[MOD],Next[MOD],id[MOD],top;
14 
15 void insert(int x,int y)
16 {
17     int k = x % MOD;
18     hs[top] = x,id[top] = y,Next[top] = head[k],head[k] = top++;
19 }
20 
21 int find(int x)
22 {
23     int k = x % MOD;
24     for(int i = head[k];i != -1;i= Next[i])
25     {
26         if(hs[i] == x)
27             return id[i];
28     }
29     return -1;
30 }
31 
32 int BSGS(int a,int b,int n)
33 {
34     memset(head,-1,sizeof(head));
35     top = 1;
36     if(b == 1)
37         return 0;
38     int m = sqrt(n*1.0),j;
39     long long x = 1,p =1;
40     for(int i = 0;i < m;i++,p = p*a%n)
41     insert(p*b%n,i);
42     for(ll i = m;;i+=m)
43     {
44         if((j = find(x = x*p % n)) != -1) return i-j;
45         if(i > n) break;
46     }
47     return -1;
48 }
49 
50 int main()
51 {
52     int p,b,n;
53     while(scanf("%d%d%d",&p,&b,&n) != EOF)
54     {
55         int ans = BSGS(b,n,p);
56         if(ans == -1)
57             printf("no solution\n");
58         else
59         printf("%d\n",ans);
60     }
61     return 0;
62 }
View Code

 

posted @ 2016-01-14 22:51  Przz  阅读(155)  评论(0编辑  收藏  举报