Discrete Logging POJ

Discrete Logging

题意：给P，B，N，求最小的L使得 B^L≡N (mod P)，其中P是素数。

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 1 #include <cstdio>
 2 #include <cstring>
 3 #include <iostream>
 4 #include <cmath>
 5 #define ll long long
 6 using namespace std;
 7 const int mod=99991;
 8 ll head[mod],nex[mod];
 9 ll cnt;
10 ll hs[mod],id[mod];
11 void init(){
12     memset(head,-1,sizeof(head));
13     cnt=0;
14 }
15 void insert_(ll x,ll i){
16     hs[cnt]=x;id[cnt]=i;
17     ll u=x%mod;
18     nex[cnt]=head[u];
19     head[u]=cnt++;
20 }
21 ll find_(ll x){
22     int u=x%mod;
23     for(int i=head[u];~i;i=nex[i]){
24         if(hs[i]==x) return id[i];
25     }
26     return -1;
27 }
28 //求解A^x≡B(mod C),返回x，若无解返回-1
29 ll bsgs(ll a,ll b,ll c){
30     if(b==1) return 0;
31     ll m=ceil(sqrt(c*1.0));
32     ll q=1;
33     for(ll i=0;i<=m;i++){
34         if(i==0) insert_(b%c,i);
35         else{
36             q=q*a%c;
37             insert_(q*b%c,i);
38         }
39     }
40     ll temp=1;
41     for(ll i=1;i<=m;i++){
42         temp=temp*q%c;
43         ll j=find_(temp);
44         if(j!=-1) return i*m-j;
45     }
46     return -1;
47 }
48 ll a,b,c;
49 int main(){
50     while(scanf("%lld%lld%lld",&c,&a,&b)!=EOF){
51         init();
52         ll ans=bsgs(a,b,c);
53         if(ans!=-1) printf("%lld
",ans);
54         else puts("no solution");
55     }
56 }

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