Here is a template for the fast power of the matrix.
First of all, how do we remember the number of rows and columns of the transfer matrix, let's make a hand gesture, first horizontal, then vertical. It is an element of the answer matrix,
So the number of columns in the first matrix is equal to the horizontal number of the transfer matrix, and the number of columns in the transfer matrix is equal to the number of columns in the answer matrix ...
#include<bits/stdc++.h>
#define db double
#define ll long long
#define reg register
#define pb(x) push_back(x)
#define fup(i,x,y) for(reg int i=x;i<=y;++i)
#define fdw(i,x,y) for(reg int i=x;i>=y;--i)
using namespace std;
const int mod=10000;
int n;
struct wy
{
ll a[5][5];
wy(){memset(a,0,sizeof(a));}
inline void clear ()
{
memset(a,0,sizeof(a));
}
wy friend operator *(wy a,wy b)
{
wy c;
fup(i,1,2) fup(j,1,2) fup(k,1,2) c.a[i][j]=(c.a[i][j]+a.a[i][k]*b.a[k][j])%mod;
return c;
}
wy friend operator ^(wy a,ll y)
{
wy c;
fup(i,1,2) c.a[i][i]=1;
while(y)
{
if(y&1) c=c*a;
y>>=1;
a=a*a;
}
return c;
}
}A,B,C;
inline int read()
{
int x=0,ff=1;
char ch=getchar();
while(!isdigit(ch)) {if(ch=='-') ff=-1;ch=getchar();}
while(isdigit(ch)) {x=(x<<1)+(x<<3)+(ch^48);ch=getchar();}
return x*ff;
}
int main()
{
freopen("1.in","r",stdin);
B.a[1][1]=0;B.a[1][2]=1;
B.a[2][1]=1;B.a[2][2]=1;
while(1)
{
n=read();
if(n==-1) break;
if(n<=2)
{
if(n==0) puts("0");
else puts("1");
continue;
}
A.clear();
A.a[1][1]=1;A.a[1][2]=1;
C=B^(n-2);
A=A*C;
printf("%lld\n",A.a[1][2]);
}
return 0;
}