Define a function f(n)=(f(n-1)+1)/f(n-2). You already got f(1) and f(2). Now, give you a number m, please find the value of f(m).
There are several test cases. Each case contains three integers indicating f(1), f(2) and m ( 1 <= f(1), f(2), m <= 1000,000,000).
For each case, please output the value of f(m), rounded to 6 decimal places.
#include <iostream> #include <stdio.h> using namespace std; int main() { //freopen("fun.in", "r", stdin); //freopen("fun.out", "w", stdout); int m; double a[100]; while(~scanf("%lf %lf %d", &a[0], &a[1], &m)){ for (int i=2; i<10; i++){ a[i]=(a[i-1]+1.0)/a[i-2]; } //for(int i=0; i<10; i++) //printf("%.6lf", a[i]); printf("%.6lf\n", a[(m-1)%5]); //printf("%.6lf\n", a[m]); } return 0; }