POJ-2417-Discrete Logging（BSGS）

最新推荐文章于 2020-01-31 14:59:38 发布

转载最新推荐文章于 2020-01-31 14:59:38 发布 · 46 阅读

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原文链接：http://www.cnblogs.com/zhuchenrui/p/7632822.html

本文详细介绍了解决离散对数问题的BSGS算法，并通过一个具体的编程实例展示了该算法的具体实现过程。文章首先介绍了输入输出的要求，随后给出了一段C++代码实现，通过扩展欧几里得算法求解模逆元，再利用BSGS算法寻找离散对数。

Given a prime P, 2 <= P < 2 ³¹, 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

 == 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

^(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

^(-m)

 == B

^(P-1-m)

(mod P) .

题解

这道题是裸的BSGS，具体内容可以看hzw的博客—传送门

 1 #include<algorithm>
 2 #include<map>
 3 #include<cstdio>
 4 #include<cstring>
 5 #include<cmath>
 6 #define ll long long
 7 using namespace std;
 8 ll p,b,n,s,x,y,m,k;
 9 int exgcd(ll a,ll b){
10     if (!b){
11         x=1; y=0;
12         return a;
13     }
14     int d=exgcd(b,a%b);
15     ll t=x; x=y; y=t-(a/b)*y;
16     return d;
17 }
18 map<int,int> h;
19 int main(){
20     while (~scanf("%lld%lld%lld",&p,&b,&n)){
21         h.clear();
22         ll t=(ll)sqrt(p);
23         s=1; h[1]=t;
24         for (int i=1;i<=t-1;i++){
25             s=s*b%p;
26             if (!h[s]) h[s]=i;
27         }
28         s=s*b%p;
29         ll l=1e10,ans=n;
30         exgcd(s,p);
31         x=(x+p)%p;
32         for (int i=0;i<=t;i++){
33             if (h[ans]){
34                 if (h[ans]==t) h[ans]=0;
35                 l=i*t+h[ans];
36                 break;
37             }
38             ans=ans*x%p;
39         }
40         if (l!=1e10) printf("%lld\n",l);
41                 else puts("no solution");
42     }
43     return 0;
44 }

View Code

转载于:https://www.cnblogs.com/zhuchenrui/p/7632822.html