挺简单的，正好能再复习一遍 $exgcd$~
按照题意一遍一遍模拟即可，注意一下 $pollard-rho$ 中的细节.
#include <ctime>
#include <cmath>
#include <cstdio>
#include <algorithm>
#define ll long long
#define ull unsigned long long
#define setIO(s) freopen(s".in","r",stdin), freopen(s".out","w",stdout)
using namespace std;
ll N,E,C,D,n,P,Q,R;
int array[20]={2,11,13,17,19};
ll exgcd(ll a,ll b,ll &x,ll &y)
{
if(!b)
{
x=1,y=0;
return a;
}
ll ans=exgcd(b,a%b,x,y),tmp=x;
x=y,y=tmp-a/b*y;
return ans;
}
ll mult(ll x,ll y,ll mod)
{
ll tmp=(long double)x/mod*y;
return ((ull)x*y-tmp*mod+mod)%mod;
}
ll qpow(ll base,ll k,ll mod)
{
ll tmp=1;
for(;k;k>>=1,base=mult(base,base,mod)) if(k&1) tmp=mult(tmp,base,mod);
return tmp;
}
int isprime(ll x)
{
if(x<=1) return 1;
int i,j,k;
ll pre,cur,a;
for(cur=x-1,k=0;cur%2==0;cur>>=1) ++k;
for(i=0;i<5;++i)
{
if(x==array[i]) return 1;
a=pre=qpow(array[i],cur,x);
for(j=1;j<=k;++j)
{
a=mult(pre,pre,x);
if(a==1&&pre!=1&&pre!=x-1) return 0;
pre=a;
}
if(a!=1) return 0;
}
return 1;
}
ll F(ll x,ll c,ll mod)
{
return (mult(x,x,mod)+c)%mod;
}
ll pollard_rho(ll x)
{
int step,k;
ll s=0,t=0,c=rand()%(x-1)+1,val=1,d;
for(k=1;;k<<=1,s=t,val=1)
{
for(step=1;step<=k;++step)
{
t=F(t,c,x);
val=mult(val,abs(t-s),x);
if(step%127==0)
{
d=__gcd(val,x);
if(d>1) return d;
}
}
d=__gcd(val,x);
if(d>1) return d;
}
}
void solve_d()
{
for(P=N;P>=N;)
P=pollard_rho(N);
Q=N/P;
R=(P-1)*(Q-1);
ll x,y,gcd;
gcd=exgcd(E,R,x,y);
x=(x+R)%R;
D=x;
}
int main()
{
int i,j;
// setIO("input");
srand((unsigned)time(NULL));
scanf("%lld%lld%lld",&E,&N,&C);
for(P=N;P>=N;)
P=pollard_rho(N);
Q=N/P;
R=(P-1)*(Q-1);
ll x,y,gcd;
gcd=exgcd(E,R,x,y);
x=(x+R)%R;
D=x;
n=qpow(C,D,N);
printf("%lld %lld
",D,n);
return 0;
}