Wannafly挑战赛24 B 222333

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;
}