[Project Euler]加入歐拉 Problem 2

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上載者:User

Each new term in the Fibonacci sequence is generated by adding the previous two terms. By starting with 1 and 2, the first 10 terms will be:

1, 2, 3, 5, 8, 13, 21, 34, 55, 89, ...

By considering the terms in the Fibonacci sequence whose values do not exceed four million, find the sum of the even-valued terms.

求Fibonacci數列裡面偶數項的和

 

哥就暴力求解,無需技巧,電腦算的夠快,沒問題^_^

#include <stdio.h>
#include <stdlib.h>

double Fib(int n); //求第n個Fibonacci數列的值

int main(int argc, char *argv[])
{
    // 求出 < 4,000,000 對應的最大的Fib的n
    double fib = 0.0;
   
    int n = 1;
    while(fib < 4000000)
    {
        fib = Fib(n);
        n++;
    }
   
    // 暴力求解相加的和
    double sum = 0.0;
    int i = 2;
    while(i < n)
    {
        sum += Fib(i);
        i += 3;       
    }
   
    printf("%lf", sum);
 
  system("PAUSE");
  return 0;
}

double Fib(int n)
{
    if(n == 1)
    return 1;
    if(n == 2)
    return 2;
    else
    return(Fib(n-2) + Fib(n-1));
}

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