Rummaged here all solution to a problem Luo Valley did not find to do with the Euler's theorem
Euler's theorem:
Euler's theorem there is a promotion:
That is, in this problem cycle length may be as \ (\ varphi (10 ^ k ) \)
Unfortunately, this cycle length is not the subject of the required minimum
However, it is clear that a minimum cycle length must be a legitimate any cycle length factor
For example \ (2 ^ i (\ bmod 7) \) is
2 4 1 2 4 1 2 4 1 ...
I find out a section of Euler's theorem cycle \ (\ varphi (. 7) = 6 \) , while the smallest section of the cycle \ (3 \) , \ (3 \) is \ (6 \) Factor
So we can see that by playing table
Then we can be enumerated by the number of prime factors 2 and 5 enumerate all \ (\ varphi (10 ^ k ) \) factor, and factor for each \ (n ^ {2 ^ i \ times 5 ^ j} \) for rapid detection of power is legitimate, if it is legitimate to update some answers
Calculate complexity of this time, the enumeration factor \ (K ^ 2 \) , do the fast power up \ (\ log_2 (10 ^ { 100}) = 332 \) multiplications set \ (N \) of \ (n- \) bits, each multiplier at most \ (N ^ 2 \) complexity
The total time complexity is \ (O (2 ^ K \ N ^ 2 Times \ Times 332) \) , if it is not OK alkylene sub
We found that the bottleneck in fast power above, we have to come up with a fast seek \ (n ^ {2 ^ i \ times 5 ^ j} \) algorithm
设 \(dp_{i,j}=n^{2^i\times 5^j}\)
Boundary conditions: \ (n-dp_ {0,0} = \)
Then we put \ (j = 0 \) is again determined recurrence
\(dp_{i,0}=dp_{i-1,0}^2\)
Factors为显natural, \ (n ^ {2 ^ i} = n ^ {2 ^ {i-1} + 2 ^ {i-1}} = n ^ {2 ^ {i-1}} \ times n ^ {2 ^ {i-1}} \ )
然后，\(dp_{i,j}=dp_{i,j-1}^5\)
因为，\(n^{2^i\times 5^j}=n^{5\times (2^i\times 5^{j-1})}=(n^{2^i\times 5^{j-1}})^5\)
We can get a very good \ (O (k ^ 2 \ times N ^ 2 \ times 5) \)
Still have to look TLE
The high-precision pressure 8 \ (N \) becomes \ (\ frac n 8 \)
So the total time complexity is \ (O (k ^ 2 \ times N ^ 2 \ times 5 + k ^ 2 \ times N ^ 2 \ times 3) \)
That is, ignoring constant \ (O (k ^ 2 \ times N ^ 2) \)
Constant is too great to be so open Ofast but still had to 450ms really dishes
// This code wrote by chtholly_micromaker(MicroMaker)
#pragma GCC optimize("Ofast")
#include <bits/stdc++.h>
#define reg register
#define int long long
using namespace std;
const int MaxN=150;
const int base=100000000;
const int pow10[]={1,10,100,1000,10000,100000,1000000,10000000,100000000};
int s[MaxN],k;
struct Bigint
{
int a[MaxN],n;
inline void clear()
{
memset(a,0,sizeof a);
n=1;
return;
}
inline void work()
{
while(n>1&&!a[n])
--n;
return;
}
inline Bigint read()
{
memset(s,0,sizeof s);
clear();
reg int cnt=0;
reg char c=getchar();
while(!isdigit(c))
c=getchar();
while(isdigit(c))
s[++cnt]=c^'0',c=getchar();
for(int i=1;i<=(cnt>>1);++i)
swap(s[i],s[cnt-i+1]);
n=0;
for(int i=1;i<=cnt;i+=8)
{
++n;
a[n]+=s[i+7]*10000000;
a[n]+=s[i+6]*1000000;
a[n]+=s[i+5]*100000;
a[n]+=s[i+4]*10000;
a[n]+=s[i+3]*1000;
a[n]+=s[i+2]*100;
a[n]+=s[i+1]*10;
a[n]+=s[i+0]*1;
}
work();
return *this;
}
inline void write() const
{
printf("%lld",a[n]);
for(int i=n-1;i;--i)
printf("%08lld",a[i]);
puts("");
return;
}
inline Bigint operator * (const Bigint &nt) const
{
Bigint res;res.clear();
res.n=min(105ll,n+nt.n);
for(int i=1;i<=n;++i)
for(int j=1;j<=nt.n;++j)
if(i+j-1<=k)
res.a[i+j-1]+=a[i]*nt.a[j];
for(int i=1;i<=res.n;++i)
res.a[i+1]+=res.a[i]/base,res.a[i]%=base;
res.work();
return res;
}
inline bool operator < (const Bigint &nt) const
{
if(n!=nt.n)
return n<nt.n;
for(int i=n;i;--i)
if(a[i]!=nt.a[i])
return a[i]<nt.a[i];
return false;
}
};
template <class t> inline void read(t &s)
{
s=0;
reg int f=1;
reg char c=getchar();
while(!isdigit(c))
{
if(c=='-')
f=-1;
c=getchar();
}
while(isdigit(c))
s=(s<<3)+(s<<1)+(c^48),c=getchar();
s*=f;
return;
}
inline bool issame(const Bigint &a,const Bigint &b)
{
int T=k/8;
for(int i=1;i<=T;++i)
if(a.a[i]!=b.a[i])
return false;
reg int lim=pow10[k-T*8];
return a.a[T+1]%lim==b.a[T+1]%lim;
}
inline Bigint fastpow(Bigint a,int b)
{
reg Bigint res;res.clear();
res.a[1]=1;
for(;b;b>>=1,a=a*a)
if(b&1)
res=res*a;
return res;
}
Bigint A,B,t,two,five;
Bigint ans;
Bigint fivepow[150],twopow[150];
Bigint dp[150][150];
signed main(void)
{
A.read();
read(k);
if(A.a[1]==1&&A.n==1)
{
puts("0");
return 0;
}
if(A.a[1]==0&&A.n==1)
{
puts("0");
return 0;
}
if(issame(A*A,A))
{
puts("1");
return 0;
}
two.clear();
five.clear();
ans.clear();
ans.a[1]=-1;
two.a[1]=2;
five.a[1]=5;
{
for(int i=0;i<k;++i)
fivepow[i].clear();
fivepow[0].a[1]=1;
for(int i=1;i<k;++i)
fivepow[i]=fivepow[i-1]*five;
}
{
for(int i=0;i<=k+1;++i)
twopow[i].clear();
twopow[0].a[1]=1;
for(int i=1;i<=k+1;++i)
twopow[i]=twopow[i-1]*two;
}
{
for(int i=0;i<=k+1;++i)
for(int j=0;j<k;++j)
dp[i][j].clear();
dp[0][0]=A;
for(int i=1;i<=k+1;++i)
dp[i][0]=dp[i-1][0]*dp[i-1][0];
for(int i=0;i<=k+1;++i)
for(int j=1;j<k;++j)
dp[i][j]=dp[i][j-1]*dp[i][j-1]*dp[i][j-1]*dp[i][j-1]*dp[i][j-1];
// dp[i][j]=fastpow(dp[i][j-1],5);
}
for(int i=0;i<=k+1;++i)
{
for(int j=0;j<k;++j)
{
if(!i&&!j)
continue;
B=twopow[i]*fivepow[j];
t=dp[i][j]*A;
if(issame(t,A))
{
if(ans.a[1]==-1||B<ans)
ans=B;
}
}
}
ans.write();
return 0;
}
// phi(10^k) = 4*10^(k-1)