Poj-3070-fibonacci

Topic links

http://poj.org/problem?id=3070

Just learned the matrix, I wrote a Fibonacci, the problem looked at the input and output, directly affixed to the code, on AC. Matrices are a great way to do this.

Code

#include <stdio.h>
#include <string.h>
#define MOD 10000

__int64 A[5];
__int64 B[3][3];

__int64 Quick_mod (__int64 N)
{
__int64 i,j,k;
A[1]=1; A[2]=1;
b[1][1]=0; B[1][2]=1;
B[2][1]=1; B[2][2]=1;
while (n)
{
if (n&1)
{
__int64 S[3];
Memset (s,0,sizeof (s));
for (i=1;i<=2;i++)
{
for (j=1;j<=2;j++)
{
S[i]= (S[i]+a[j]*b[j][i])%mod;
}
}
for (i=1;i<=2;i++)
A[i]=s[i];
n--;
}
N=N/2;
__int64 Kk[5][5];
memset (kk,0,sizeof (KK));
for (i=1;i<=2;i++)
{
for (j=1;j<=2;j++)
{
for (k=1;k<=2;k++)
{
Kk[i][j]= (Kk[i][j]+b[i][k]*b[k][j])%mod;
}
}
}
for (i=1;i<=2;i++)
{
for (j=1;j<=2;j++)
{
B[I][J]=KK[I][J];
}
}
}
return a[1];
}

int main (void)
{
__int64 N;
while (scanf ("%i64d", &n) ==1)
{
if (n==-1)
return 0;
if (n==0)
{
printf ("0\n");
Continue
}
printf ("%i64d\n", Quick_mod (n-1));
}
return 0;
}

The method of using the problem can also

Code

#include <stdio.h>
#include <string.h>
#define MOD 10000

__int64 A[3][3],b[3][3];

__int64 Quick_mod (__int64 N)
{
__int64 i,j,k;
__int64 kk[3][3];
A[1][1]=1; a[1][2]=1;
A[2][1]=1; a [2] [2]=0;
B[1][1]=1; b[1][2]=1;
B[2][1]=1; b[2][2]=0;
while (n)
{
if (n&1)
{
memset (kk,0,sizeof (KK)),
for (i=1;i<=2;i++)
{
for (j=1;j <=2;j++)
{
for (k=1;k<=2;k++)
{
kk[i][j]= (kk[i][j]+a[i][k]*b[k][j])%mod;
}
}
}
for (i=1;i<=2;i++)
{
for (j=1;j<=2;j++)
{
A[i][j]=kk[i][j];
}
}
n--;
}
N=n/2;
Memset (kk,0,sizeof (KK));
for (i=1;i<=2;i++)
{
for (j=1;j<=2;j++)
{
for (k=1;k<=2;k++)
{
kk[i][j]= (kk[i][j]+ B[I][K]*B[K][J])%mod;
}
}
}
for (i=1;i<=2;i++)
{
for (j=1;j<=2;j++)
{
B[i][j]=kk[i][j];
}
}
}
return a[1][2];
}

int main (void)
{
__int64 I,j,k,k1,k2,n;
while (scanf ("%i64d", &n) ==1)
{
if (N==-1) return 0;
if (n==0)
{
printf ("0\n");
Continue
}
printf ("%i64d\n", Quick_mod (n-1));
}
return 0;
}

Poj-3070-fibonacci