[Euler's plan 2] Even Fibonacci numbers

Document directory

• Original ENGLISH
• Description
• My Solutions
• Direct Solution
• Another method
• Efficient Solution

This question comes from the second question of Project Euler, and calculates the sum of an even number less than 4 million. Continue with my Euler's plan today.

Problem description

Even Fibonacci numbers

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.

Description

In the Fibonacci series, each item is generated by the sum of the first two items. From 1 to 2, the top 10 items are:
1, 2, 3, 5, 8, 13, 21, 34, 55, 89 ,...
Consider the number of items in the Fibonacci series that cannot exceed 4 million, and obtain the sum of the items whose values are even numbers.

Solution my solution

This is a well-defined method. It uses the point loop queue idea, although it can be replaced by several variables. During the code writing process, I also made a silly mistake. Thanks to the help of the Forum friends.

# Include <stdio. h> # define bound 4000000int main (void) {int fib [3]; int I; long int sum; FIB [0] = 1; FIB [1] = 2; sum = 2; for (I = 2; FIB [(I-1) % 3] <bound; I ++) {fib [I % 3] = fib [(I-1) % 3] + fib [(I-2) % 3]; // queue idea if (FIB [I % 3] % 2 = 0) {sum + = fib [I % 3] ;}} printf ("% d \ n", FIB [(I-2) % 3]); printf ("% d", sum );}

The following are several solutions provided by question resolution.

Direct Solution

This method works the same way as my method.

# include <stdio.h>int main(void){int limit;int a, b, h;int sum;limit = 4000000;a = b = 1;sum = 0;while(b < limit) {if(b % 2 == 0) {sum += b;}h = a + b;a = b;b = h;}printf("%d", sum);}

Another method

After carefully observing the series, we can find that each of the three fibre nacci numbers has an even number. With this rule, conditional judgment can be omitted.

# include <stdio.h>int main(void){int limit;int a, b, c;int sum;limit = 4e6;a = 1;b = 1;c = 2;sum = 0;while(c < limit) {sum += c;a = b + c;b = a + c;c = a + b;}printf("%d", sum);}

Efficient Solution

If you take out the even number of items:
34,144 ......

They follow these rules:
E (n) = 4 * E (n-1) + E (n-2)

# Include <stdio. h> int main (void) {int limit; int A, B; int sum; Limit = 4e6; sum = 2; A = 2; B = 8; while (B <limit) {sum + = B; B = 4 * B + A; A = (B-a)/4; // if no third variable is used, the variable is moved back.} printf ("% d", sum );}

To prove that the above relation is true, you only need to prove:
F (n) = 4 * F (n-3) + f (n-6)

The verification process is as follows:
F (n) = f (n-1) + f (n-2)
= F (n-2) + f (n-3) + f (n-2) = 2 F (n-2) + f (n-3)
= 2 (f (n-3) + f (n-4) + f (n-3) = 3 F (n-3) + 2 F (n-4)
= 3 F (n-3) + f (n-4) + f (N-5) + f (n-6)
= 4 f (n-3) + f (n-6)

Final answer

4613732