Question

Fibonacci序列中的每个新术语都是通过添加前两个术语生成的。从和开始，前10个术语将是：

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

在这里我们应该找到Fibonacci系列中的偶数并将它们加到总和

代码：

import java.util.*;

public class Abhi {


public static void main(String[] args) {

    Scanner in = new Scanner(System.in);
    int t = in.nextInt();
    int[] n = new int[t];

    int i,j;
    long sum;
    for(int a0 = 0; a0 < t; a0++){
        n[a0] = in.nextInt();            
    }
    //int an = n.length;
    int[] nn = new int[1000];
    nn[0]=1;
    nn[1]=2;
    for(i = 0 ; i<t;i++){
        sum = 2;
        for(j= 2;j<n[i];j++){                
            nn[j] = nn[j-2] + nn[j-1];
                if(nn[j]%2==0 && nn[j]<n[i])
                    {
                    sum += nn[j];
                 //System.out.println(sum);   
                 //the above line shows correct output 
                }
            }            
        System.out.println(sum);//this is printing different output
    }}}

示例输入：

1
100

示例输出：

Answer 1

这里的问题不在于外部的System.out.println（sum）;如你所说。这是因为int范围。

int的最大值是2 147 483 647，而Fibonacci系列2 971 215 073位于第47位，当它超出int范围时，它以意想不到的方式给出结果。

在代码数组中nn持有-1323752223而不是2971215073，这实际上导致了问题。

要解决此问题，请使用BigInteger，如下所示

    BigInteger sum;
    BigInteger[] nn = new BigInteger[1000];
    nn[0] = new BigInteger("1");
    nn[1] = new BigInteger("2");
    for (i = 0; i < t; i++) {
        sum = new BigInteger("2");
        for (j = 2; j < n[i]; j++) {
            nn[j] = nn[j - 2].add(nn[j - 1]);
            if (nn[j].mod(new BigInteger("2")).equals(new BigInteger("0")) && 
                nn[j].compareTo(new BigInteger(String.valueOf(n[i])))<0) {
                sum = sum.add(nn[j]);
                System.out.println(sum);
            }
        }
        System.out.println(sum);
    }

Answer 2

您还可以使用以下代码实现此目的：

import java.util.Scanner;

public class Fibo 
{
    public static void main(String[] args) 
    {
        Scanner sc = new Scanner(System.in);

        int No = sc.nextInt();

        int fNo1 = 1;
        int fNo2 = 1;
        int fNo3 = fNo1 + fNo2;
        int sumofEvenNo = 0;
        int i;
        while( fNo3 < No)
        {
            if(fNo3 % 2 == 0)
                sumofEvenNo += fNo3;
            fNo1 = fNo2;
            fNo2 = fNo3;
            fNo3 = fNo1 + fNo2;
        }

        System.out.println("Sum of all even nos = "+sumofEvenNo);
    }
}

Answer 3

我建议使用naming convention，您的代码将更加清晰，易于调试。我也使用static method来获得金额。

询问总和的计算次数。
获取索引并计算总和。
打印。

类：

import java.util.*;

public class Abhi {

    public static void main(String[] args) {

        System.out.println ( "This program will calculate the even sum of given fibonacci series index ( index starts at 1) " );
        System.out.println ( "Please enter how many times do you want to calculate the sum: " );
        Scanner scan = new Scanner(System.in);
        int iterationTimes = scan.nextInt() ;

        for (; iterationTimes > 0; iterationTimes-- )
        {
            System.out.println ( "Please, enter the index on fibonacci: " );
            System.out.println ( "Even sum: " + getEvenSum( scan.nextInt() ) );
        }

    }

    private static long getEvenSum( int index)
    {
        if ( index <= 2 )
        {
            return 0;
        }

        long n1=1, n2=1, n3, sum = 0;   

        for(int i = 2; i < index; i++) 
        {    
            n3=n1+n2;
            if ( n3 % 2 == 0)
            {
                sum += n3;
            }
            n1=n2;    
            n2=n3;    
        }  

        return sum;

    }


}

I / O示例：

This program will calculate the even sum of given fibbonacci series index ( index starts at 1 )
Please enter how many times do you want to calculate the sum: 
3
Please, enter the index on fibbonacci: 
3
Even sum: 2
Please, enter the index on fibbonacci: 
6
Even sum: 10
Please, enter the index on fibbonacci: 
12
Even sum: 188

注意： Fibbonaci：1,1,2,3 ......

两种不同的输出

3 个答案:

类：

I / O示例：