Discrete Logging POJ - 2417（BSGS）

Posted 2020-10-03 yijiull

Discrete Logging

 POJ - 2417 

题意：给P，B，N，求最小的L使得 B^L≡N (mod P)

Baby Step Giant Step

 1 #include <cstdio>
 2 #include <cstring>
 3 #include <iostream>
 4 #include <cmath>
 5 #define ll long long
 6 using namespace std;
 7 const int maxn=76543;
 8 int head[maxn],nex[maxn],hs[maxn],id[maxn];
 9 int cnt;
10 void init(){
11     memset(head,-1,sizeof(head));
12     cnt=0;
13 }
14 void insert_(ll x,int i){
15     int u=x%maxn;
16     hs[cnt]=x;
17     id[cnt]=i;
18     nex[cnt]=head[u];
19     head[u]=cnt++;
20 }
21 int find_(ll x){
22     int u=x%maxn;
23     for(int i=head[u];~i;i=nex[i]){
24         if(hs[i]==x) return id[i];
25     }
26     return -1;
27 }
28 
29 ll a,b,c;
30 ll bsgs(ll a,ll b,ll c){
31     if(b==1) return 0;
32     ll m=ceil(sqrt(c*1.0));
33     ll x=1,p=1;
34     for(ll i=0;i<=m;i++){
35         if(i==0) insert_(b%c,i);
36         else{
37             p=p*a%c;
38             insert_(p*b%c,i);
39         }
40     }
41     ll ans=1,res=-1;
42     int u;
43     for(ll i=1;i<=m;i++){
44         ans=ans*p%c;
45         if((u=find_(ans))!=-1){
46             res=i*m-u;
47             break;
48         }
49     }
50     return res;
51 }
52 int main(){
53     while(scanf("%lld%lld%lld",&c,&a,&b)!=EOF){
54         init();
55         ll ans=bsgs(a,b,c);
56         if(ans!=-1) printf("%lld\n",ans);
57         else puts("no solution");
58     }
59 }

View Code