| Time Limit: 5000MS | Memory Limit: 65536K | |
| Total Submissions: 5865 | Accepted: 2618 |
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
BSGS模板题
1 #include<iostream>
2 #include<cstdio>
3 #include<cstring>
4 #include<cmath>
5 #include<map>
6 #define LL long long
7 using namespace std;
8 LL a,b,c;
9 map<LL,LL>mp;
10 LL fastpow(LL a,LL p,LL c)
11 {
12 LL base=a;LL ans=1;
13 while(p!=0)
14 {
15 if(p%2==1)ans=(ans*base)%c;
16 base=(base*base)%c;
17 p=p/2;
18 }
19 return ans;
20 }
21 int main()
22 {
23 // a^x = b (mod c)
24 while(scanf("%lld%lld%lld",&c,&a,&b)!=EOF)
25 {
26 LL m=ceil(sqrt(c));// 注意要向上取整
27 mp.clear();
28 if(a%c==0)
29 {
30 printf("no solution\n");
31 continue;
32 }
33 // 费马小定理的有解条件
34 LL ans;//储存每一次枚举的结果 b* a^j
35 for(LL j=0;j<=m;j++) // a^(i*m) = b * a^j
36 {
37 if(j==0)
38 {
39 ans=b%c;
40 mp[ans]=j;// 处理 a^0 = 1
41 continue;
42 }
43 ans=(ans*a)%c;// a^j
44 mp[ans]=j;// 储存每一次枚举的结果
45 }
46 LL t=fastpow(a,m,c);
47 ans=1;//a ^(i*m)
48 LL flag=0;
49 for(LL i=1;i<=m;i++)
50 {
51 ans=(ans*t)%c;
52 if(mp[ans])
53 {
54 LL out=i*m-mp[ans];// x= i*m-j
55 printf("%lld\n",(out%c+c)%c);
56 flag=1;
57 break;
58 }
59
60 }
61 if(!flag)
62 printf("no solution\n");
63 }
64
65 return 0;
66 }