[POJ2417]Discrete Logging（指数级同余方程）

Discrete Logging

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

HinThe 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

B^-m== B^(P-1-m)

(mod P) .

题解：这道题目是Baby-step giant-step的裸题吧，嗯嗯，不要忘了开long long就可以了。

 1 #include<cstdio>
 2 #include<algorithm>
 3 #include<iostream>
 4 #include<cmath>
 5 #include<cstring>
 6 #include<map>
 7 #define    ll long long
 8 using namespace std;
 9 
10 ll p,b,n;
11 
12 int ex_gcd(ll a,ll b,ll &x,ll &y)
13 {
14     if (!b){x=1,y=0;return a;}
15     else
16     {
17         int fzy=ex_gcd(b,a%b,x,y);
18         int t=x;x=y;
19         y=t-a/b*y;
20         return fzy;
21     }
22 }
23 void solve()
24 {
25     ll m=(ll)sqrt(p);
26     map<int,int>num;
27     map<int,bool>app;
28     app[1]=1,num[1]=0;
29     ll zhi=1;
30     for (int i=1;i<=m-1;i++)
31     {
32         zhi=zhi*b%p;
33         if (!app[zhi])
34         {
35             app[zhi]=1;
36             num[zhi]=i;
37         }
38     }
39     zhi=zhi*b%p;
40     ll x,y,nn=n;
41     int fzy=ex_gcd(zhi,p,x,y);
42     x=(x+p)%p;
43     for (int i=0;i<=m;i++)
44     {
45         if (app[nn])
46         {
47             printf("%lld
",i*m+num[nn]);
48             return;
49         }
50         else nn=nn*x%p;
51     }
52     printf("no solution
");
53 }
54 int main()
55 {
56     while(~scanf("%d%d%d",&p,&b,&n))
57         solve();
58 }