Nefu number of bits of fibs 461 (matrix)

fibs number of digits

problem:461 Time limit:1000ms Memory limit:65536k

Description

The generalized Fibonacci sequence is defined as follows: R0=a; R1= b; rn= uRn-1 + vRn-2 (n >= 3) Here a,b,u,v are real numbers, called RN for the generalized Fibonacci sequence. Now let you calculate the number of bits of the generalized Fibonacci sequence.

input

There are several groups of input data, each group of 1 rows Total 5, respectively, representing N,a,b,u,v,n is a positive integer, A,b,u,v is a real number, and u*u+4v>=0,200<=n<=100000000.

Output

The number of digits of the output RN.

Sample_input

1 2 1 1 250 1 2 1 1

Sample_output

53

hint

Source #include <stdio.h> #include <string.h> #include <math.h> #include <algorithm > #define LL long long using namespace std; int main () { ll a,b,u,v,n; while (scanf ("%lld%lld%lld%lld%lld", &n,&a,&b,&u,&v)!=eof) { double c=sqrt (u*u+4*v) ; Double x= (u+c)/2.0; Double y= (u-c)/2.0; Double ans=n*log10 (x) +log10 (b-y*a)-log10 (c); printf ("%lld\n", (ll) ans+1); } return 0;