小水题？？？但是时间限制异常鬼畜，跑了2min

\(P | (2^m)*(3^n)-1\)的意思就是\(2^m 3^n = 1 (\text{mod }P)\)

设f[i]表示3^k=i的最小的k

然后枚举2的次幂即可

#include<bits/stdc++.h>

#define il inline

#define vd void

typedef long long ll;

il int gi(){

	int x=0,f=1;

	char ch=getchar();

	while(!isdigit(ch)){

		if(ch=='-')f=-1;

		ch=getchar();

	}

	while(isdigit(ch))x=x*10+ch-'0',ch=getchar();

	return x*f;

}

int inv,P,ans,ans_m,f[10000010];

il int pow(int x,int y){

	int ret=1;

	while(y){

		if(y&1)ret=1ll*ret*x%P;

		x=1ll*x*x%P;y>>=1;

	}

	return ret;

}

int main(){

#ifndef ONLINE_JUDGE

	freopen("186b.in","r",stdin);

	freopen("186b.out","w",stdout);

#endif

	while(scanf("%d",&P)==1){

		memset(f,63,sizeof f);

		for(ll i=1,j=3;i<P;++i,j=j*3%P)f[j]=std::min(f[j],(int)i);

		ans=1e9;

		inv=pow(2,P-2);

		for(ll i=1,j=inv;i<P;++i,j=j*inv%P)if(i+f[j]<ans)ans=i+f[j],ans_m=i;

		printf("%d %d\n",ans_m,ans-ans_m);

	}

	return 0;

}