bsgs BSGS

Posted 2020-09-19 ioioioioioio

Question:
had know a,b,c;
ask min x;
made a^x=b(mod c);

think:
make m=ceil(sqrt(c));
make x=i*m-j;
so a^(i*m-j)=b(mod c)
so a^(i*m)/a^j=b(mod c)
so a^(i*m)=b*a^j(mod c)

easy to meijv i,j to full this deng shi

ANS:

answer=min(i*m-j);

 

 1 #include<iostream>
 2 #include<cstdio>
 3 #include<cmath>
 4 #include<algorithm>
 5 #include<map>
 6 #define ll longlong
 7 #define readln(n) scanf("%lld",&n);
 8 #define writeln(n) printf("%lld",n);
 9 
10 using namespace std;
11 
12 map<ll,int>mp;
13 
14 inline ll f(ll a,ll b,ll mod)
15 {
16     ll ans=1;
17     while(b)
18     {
19         if(b&1) ans=ans*a%mod;
20         a=a*a%mod;
21         b>>=1;
22     }
23     return ans;
24 }
25 
26 int main()
27 {
28     ll n;//a^x=b(mod)c,a^(i*m)=b*(a^j)(mod)
29     readln(n);
30     ll a,b,c;
31     readln(a);readln(b);readln(c);
32     if(!a%c){printf("!!!"); return 0; }
33     ll m=ceil(sqrt(c));
34     ll ans=b%c;
35     mp[ans]=0;
36     for(int i=1;i<=m;i++)
37     {
38         asn=(ans*a)%c;
39         mp[ans]=i;
40     }
41     ll ff=f(a,m,c);
42     ans=1;
43     bool flag=1;
44     for(int i=1;i<=m;i++)
45     {
46         ans=ans*f%c;
47         if(mp[ans])
48         {
49             ll answer=i*m-mp[ans];
50             writeln((answer%c+c)%c);
51             flag=0;
52         } 
53     }
54     if(!flag)
55     printf("!!!");
56 }