我尝试制作一个执行Fibonacci序列的Java程序。
这是我的代码：
import java.io.*;
public class Fibonacci{
public static void main(String[]args){
BufferedReader Data=new BufferedReader (new InputStreamReader(System.in));
int ctr1=0;
int ctr2=0;
int num1=0;
int num2=0;
int num3=0;
try{
System.out.println("How many numbers would you want to see?");
ctr2=Integer.parseInt(Data.readLine());
for(int ans=0; ctr1==ctr2; ctr1++){
num1++;
System.out.println(num2 + "\n" + num1);
ans=num1+num2;
System.out.println(ans);
ans=num3;
}
}catch(IOException err){
System.out.println("Error!" + err);
}catch(NumberFormatException err){
System.out.println("Invald Input!");
}
}
}
显然，我是Java的初学者，我不知道如何正确使用for语句。有人会善待我的代码吗？或者可以使代码更短的代码。我是初学者，所以要冷静。谢谢:)）
答案 0 :(得分：1)
Java中的斐波那契数列实际上非常简单，只需一个for循环即可完成！！！
import java.io.*;
class fibonacci{
public static void main() throws NumberFormatException, IOException{
BufferedReader Data=new BufferedReader (new InputStreamReader(System.in));
int a,b,c,d;
System.out.println("Upto How many numbers do you want to see?");
d=Integer.parseInt(Data.readLine());
for (a=0,b=1,c=a;a<d;c=a,a+=b,b=c){
System.out.println(a);
}
}
}
这是使用缓冲读取器完成的.....如果据说只使用bufferedreader，则可以使用Scanner类，该类非常简单易用，因为您不必捕获或引发任何异常.....
扫描程序：-
import java.util.*;
class fibonacci{
public static void main(){
Scanner sc = new Scanner(System.in);
int a,b,c;
System.out.println("Upto How many numbers do you want to see?");
d=sc.nextInt();
for (a=0,b=1,c=a;a<d;c=a,a+=b,b=c){
System.out.println(a);
}
}
}
现在，正如我在一个循环中所说的，您可以执行此操作。...这是另一种方法，您可以在循环体内进行交换，而不是在循环的参数中进行交换... 对于初学者来说，这很容易理解，因为您不必在参数内传递多个变量，是的，它会更长一些
import java.util.*;
class fibonacci{
public static void main(){
Scanner sc = new Scanner(System.in);
int a = 0,b = 1,c,d;
System.out.println("Upto How many numbers do you want to see?");
d=sc.nextInt();
System.out.println(a +"\n" +b);//\n is used to go to next line....
for (c=0;c<d;c++){
c = a + b;//Doing and printing the fibonacci...
System.out.println(c);
a = b;
b = c;//Swapping the values...
}
}
}
所以在这里，我给了您三种方法，它们应该给出相同的输出（最有可能），选择对您方便的一种。
答案 1 :(得分：0)
请查看此代码段，这比您理解的要容易得多。解决方案提示很简单，你为前2个斐波那契数字保留2个指针并在循环中适当地更新它们。在下面的示例中，循环执行10次，您可以根据需要进行修改。
static void fibonacci() {
int ptr1 = 1, ptr2 = 1;
int temp = 0;
System.out.print(ptr1 + " " + ptr2 + " ");
for (int i = 0; i < 10; i++) {
System.out.print(ptr1 + ptr2 + " ");
temp = ptr1;
ptr1 = ptr2;
ptr2 = temp + ptr2;
}
}
<强>输出：
1 1 2 3 5 8 13 21 34 55 89 144
答案 2 :(得分：0)
扩展答案，如果你想看起来很酷，请使用递归。
public class Fibonacci {
public static long fib(int n) {
if (n <= 1) return n;
else return fib(n-1) + fib(n-2);
}
public static void main(String[] args) {
int N = 300; // how many numbers you want to generate
for (int i = 1; i <= N; i++)
System.out.println(i + ": " + fib(i));
}
}
以下是谷歌搜索它的内容，希望这些资源有所帮助：http://bit.ly/1cWxhUS
答案 3 :(得分：0)
我是java的初学者，但是我找到了一种使用数组创建斐波纳契数的简单方法。 Fibonacci数的基本原理是添加当前数字和之前的数字。 这是我的代码：
//Creation of array
int [ ] fib = new int[size];
//Assigning values to the first and second indexes of array named "fib"
fib [0] = 0;
fib [1] = 1;
//Creating variable "a" to use in for loop
int a = 1
//For loop which creates a Fibonacci number
for( int i = 2; i < size ; i++)
{
fib[i] = a;
a = fib[i] + fib[i-1];
}
答案 4 :(得分：0)
这是我发现的online的另一种算法，并从中简化了代码。
public static BigInteger fib(BigInteger x) {
if (x.intValue() < 0){return x.intValue() % 2 == 0 ?fib(x.multiply(BigInteger.valueOf(-1))).multiply(BigInteger.valueOf(-1)) : fib(x.multiply(BigInteger.valueOf(-1)));}
int n = Integer.valueOf(x.toString());
BigInteger a = BigInteger.ZERO,b = BigInteger.ONE;
for (int bit = Integer.highestOneBit(n); bit != 0; bit >>>= 1) {
BigInteger d = a.multiply(b.shiftLeft(1).subtract(a));
BigInteger e = a.multiply(a).add(b.multiply(b));
a = d;
b = e;
if ((n & bit) != 0) {
BigInteger c = a.add(b);
a = b;
b = c;
}
}
return a;
}
我知道您有可能不了解如何使用BigInteger，所以我给您这个link，只是想提供帮助。
答案 5 :(得分：0)
Here we get Fibonacci Series up to n.
public static void fibSequence(int n) {
int sum = 0;
for (int x = 0, y = 1; sum < n; x = y, y = sum, sum = x + y) {
System.out.print(sum + " ");
}
}
示例：
输入： n = 20
输出： 0 1 1 2 3 5 8 13
答案 6 :(得分：-1)
import java.util.*;
public class sequence1
{
public static void main(String[] args)
{
sequence1 fs=new sequence1();
fs.fibonacci();
}
public void fibonacci()
{
int numb1 = 1;
int numb2 = 1;
int temp = 0;
@SuppressWarnings("resource")
Scanner input=new Scanner(System.in);
System.out.println("How Many Terms? (Up To 45)");
int x=input.nextInt();
x=x-2;
System.out.println(numb1);
System.out.println(numb2);
for (int i = 0; i < x; i++)
{
System.out.println(numb1 + numb2 + " ");
temp = numb1;
numb1 = numb2;
numb2 = temp + numb2;
}
}
}
答案 7 :(得分：-1)
此函数返回斐波纳契数列
/**
* @param startElement
* @param secondElent
* @param length :length of fibonacci series
* @return fibonacciseries : contain the series of fibonacci series
*/
public int[] createFibonacciSeries(int startElement, int secondElent,
int length) {
int fibonacciSeries[] = new int[length];
fibonacciSeries[0] = startElement;
fibonacciSeries[1] = secondElent;
for (int i = 2; i < length; i++) {
fibonacciSeries[i] = fibonacciSeries[i - 1]
+ fibonacciSeries[i - 2];
}
return fibonacciSeries;
}
答案 8 :(得分：-1)
import java.util.*;
class MyFibonacci {
public static void main(String a[]){
int febCount = 15;
int[] feb = new int[febCount];
feb[0] = 0;
feb[1] = 1;
for(int i=2; i < febCount; i++){
feb[i] = feb[i-1] + feb[i-2];
}
for(int i=0; i< febCount; i++){
System.out.print(feb[i] + " ");
}
}
}
答案 9 :(得分：-1)
public class FibonacciExercitiu {
public static void main(String[] args) {
int result = fib(6); //here we test the code. Scanner can be implemented.
System.out.println(result);
}
public static int fib(int n) {
int x = 1;
int y = 1;
int z = 1; //this line is only for declaring z as a variable. the real assignment for z is in the for loop.
for (int i = 0; i < n - 2; i++) {
z = x + y;
x = y;
y = z;
}
return z;
}
/*
1. F(0) = 1 (x)
2. F(1) = 1.(y) =>Becomes x for point4
3.(z)F(2) = 2 (z) =>Becomes Y for point4 // becomes X for point 5
4.(z)F(3) = 3 // becomes y for point 5
5.(z)F(4) = 5 ..and so on
*/
}
答案 10 :(得分：-1)
public static int[] fibonachiSeq(int n)
{
if (n < 0)
return null;
int[] F = new int[n+1];
F[0] = 0;
if (n == 0)
return F;
F[1] = 1;
for (int i = 2; i <= n; i++)
{
F[i] = F[i-1] + F[i-2];
}
return F;
}
答案 11 :(得分：-1)
使用while循环
class Feb
{
static void Main(string[] args)
{
int fn = 0;
int sn = 1;
int tn = 1;
Console.WriteLine(fn);
Console.WriteLine(sn);
while (true)
{
tn = fn + sn;
if (tn >10)
{
break;
}
Console.WriteLine(tn);
fn = sn;
sn = tn;
}
Console.Read();
}
}
答案 12 :(得分：-1)
公共课Febonacci {
public static void main(String[] args) {
int first =0;
int secend =1;
System.out.print(first+","+secend);
for (int k=1;k<7;k++){
System.out.print(","+(first+secend ));
if(k%2!=0)
first+=secend;
else
secend+=first;
}
}
}
答案 13 :(得分：-1)
public class FibonacciSeries {
public static void main(String[] args) {
int a=0, c=0, b=1;
for(int i=0; i<10; i++) {
System.out.print(c+" ");
a = c + b;
c = b;
b = a;
}
}
}