x^A=B(mod C)的解 (离散对数与原根)
#include <iostream>
#include <string.h>
#include <algorithm>
#include <stdio.h>
#include <math.h>
#include <bitset>
#include <map>
using namespace std;
typedef long long LL;
const int N = 1000005;
/**以下为求原根部分 */
bitset<N> prime;
int p[N],pri[N];
int k,cnt;
void isprime()
{
prime.set();
for(int i=2; i<N; i++)
{
if(prime[i])
{
p[k++] = i;
for(int j=i+i; j<N; j+=i)
prime[j] = false;
}
}
}
void Divide(LL n)
{
cnt = 0;
LL t = (LL)sqrt(1.0*n);
for(int i=0; p[i]<=t; i++)
{
if(n%p[i]==0)
{
pri[cnt++] = p[i];
while(n%p[i]==0) n /= p[i];
}
}
if(n > 1)
pri[cnt++] = n;
}
LL multi(LL a,LL b,LL m)
{
LL ans = 0;
a %= m;
while(b)
{
if(b & 1)
{
ans = (ans + a) % m;
b--;
}
b >>= 1;
a = (a + a) % m;
}
return ans;
}
LL quick_mod(LL a,LL b,LL m)
{
LL ans = 1;
a %= m;
while(b)
{
if(b&1)
{
ans = multi(ans,a,m);
b--;
}
b >>= 1;
a = multi(a,a,m);
}
return ans;
}
LL broot(LL p)
{
Divide(p-1);
for(int g=2 ;; g++)
{
bool flag = true;
for(int i=0; i<cnt; i++)
{
LL t = (p - 1) / pri[i];
if(quick_mod(g,t,p) == 1)
{
flag = false;
break;
}
}
if(flag) return g;
}
}
/**以上为求原根部分 */
/** 以下为Baby_Step */
LL gcd(LL a,LL b)
{
return b ? gcd(b,a%b):a;
}
void extend_Euclid(LL a,LL b,LL &x,LL &y)
{
if(b == 0)
{
x = 1;
y = 0;
return;
}
extend_Euclid(b,a%b,x,y);
LL tmp = x;
x = y;
y = tmp - (a / b) * y;
}
LL Inv(LL a,LL p)
{
return quick_mod(a,p-2,p);
}
LL Baby_Step(LL A,LL B,LL C)
{
map<LL,int> mp;
LL M = ceil(sqrt(1.0*C));
LL t = Inv(quick_mod(A,M,C),C);
LL ans = 1;
for(int i=0; i<M; i++)
{
if(!mp.count(ans))
mp[ans] = i;
ans = multi(ans,A,C);
}
for(int i=0; i<M; i++)
{
if(mp.count(B))
return i * M + mp[B];
B = multi(B,t,C);
}
return -1;
}
/** 以上为Baby_Step */
LL ans[1005];
void Work(LL A,LL B,LL C)
{
LL root = broot(C);
LL t1 = Baby_Step(root,B,C);
LL t2 = C - 1;
LL d = gcd(A,t2);
if(t1 % d)
{
puts("-1");
return;
}
LL x,y;
extend_Euclid(A,t2,x,y);
t2 /= d;
t1 /= d;
ans[0] = (x * t1 % t2 + t2) % t2;
for(int i=1; i<d; i++)
ans[i] = ans[i-1] + t2;
for(int i=0; i<d; i++)
ans[i] = quick_mod(root,ans[i],C);
sort(ans,ans+d);
for(int i=0; i<d; i++)
cout<<ans[i]<<endl;
}
int main()
{
int T = 1;
LL A,B,C;
isprime();
while(cin>>A>>C>>B)
{
printf("case%d:\n",T++);
Work(A,B,C);
}
return 0;
}