Java 第（N-2）次和第（N-1）次的第N次输出和

不管我怎么想，我都想不出这个问题，我可能只是想得太多了，我能得到一点帮助吗？我已经尝试了各种不同的for循环，我可以；我不知道如何得到这些值

for(int a = 1; a<1000; a++) 

---

for（int a=1；a您可以使用递归来执行此操作：

public int fibonacci(int n)  {
    if(n == 0) {
        return 0;
    } else if(n == 1) {
        return 1;
    } else {
        return fibonacci(n - 1) + fibonacci(n - 2);
    }
}

您可以使用递归来执行此操作：

public int fibonacci(int n)  {
    if(n == 0) {
        return 0;
    } else if(n == 1) {
        return 1;
    } else {
        return fibonacci(n - 1) + fibonacci(n - 2);
    }
}

如果要使用循环，可以执行以下操作（从answer中借用）：

public int fibonacci（int n）{
如果（n==0）
返回1；
如果（n==1）
返回3；
int祖父母=1；
int父代=3；
int curr=0；
对于（int i=2；i如果您想使用循环，可以执行以下操作（从答案中借用）：

public int fibonacci（int n）{
如果（n==0）
返回1；
如果（n==1）
返回3；
int祖父母=1；
int父代=3；
int curr=0；
对于（inti=2；i，这里是不使用递归的方法

    package fib;

    public class fib {

            public static void main(String[] args) {
                    int first  = 1;
                    int second = 1;
                    int thrid  = 2;
                    System.out.format("0, %d, %d, %d", first, second, thrid);
                    while ( thrid < 1000 ) {
                            first  = second;
                            second = thrid;
                            thrid  = first + second;
                            System.out.format(", %d", thrid);
                    }
            }
    }

下面是不使用递归的方法

    package fib;

    public class fib {

            public static void main(String[] args) {
                    int first  = 1;
                    int second = 1;
                    int thrid  = 2;
                    System.out.format("0, %d, %d, %d", first, second, thrid);
                    while ( thrid < 1000 ) {
                            first  = second;
                            second = thrid;
                            thrid  = first + second;
                            System.out.format(", %d", thrid);
                    }
            }
    }

尝试：

inta=1，b=1；
对于（b=1；b<10000；b+=a）{
系统输出打印（a+“”+b+“”）；
a+=b；
}
试试：

inta=1，b=1；
对于（b=1；b<10000；b+=a）{
系统输出打印（a+“”+b+“”）；
a+=b；
}
将while循环转换为for循环，将是：

for (int a=1, int b=1; a < 10000; b+=a+=b ){ 
   System.out.print(a+" "+b+" "); 
} 

for（inta=1，intb=1；a<10000；b+=a+=b）{
系统输出打印（a+“”+b+“”）；
} 
将while循环转换为for循环，将是：

for (int a=1, int b=1; a < 10000; b+=a+=b ){ 
   System.out.print(a+" "+b+" "); 
} 

for（inta=1，intb=1；a<10000；b+=a+=b）{
系统输出打印（a+“”+b+“”）；
} 
有关如何执行此操作的信息，请参阅，您可以使用Binet的公式（复制如下）在不递归的情况下执行此操作：

import java.lang.Math;
import java.math.BigInteger;
import java.math.BigDecimal;
import java.math.MathContext;
import java.math.RoundingMode;
class f3c {
// This program calculates the nth fibonacci number
// using alrogirhtm 3C: Binet's formula with rounding
//
// compiled: javac f3c.java
// executed: java f3c n
//

  // Method f3c.isqrt(n) finds the inverse root of n using the following
  // recurrent equation: y(n+1) = 0.5*y(n)*[3 - x*y(n)^2]
  // It is faster than typical Newton-Raphson square root finding because
  // it does not use division.
  // Funny how Java implemented exponentiation by repeated squaring
  // for BigDecimals, but did not implement any sort of square root.
 private static BigDecimal isqrt(BigDecimal x, MathContext mc) {
     BigDecimal guess = new BigDecimal(1/Math.sqrt(x.doubleValue()));
     BigDecimal three = new BigDecimal(3);
     BigDecimal half = new BigDecimal(0.5);
     BigDecimal oldguess;
     do{ oldguess = guess;
         guess = half.multiply(guess.multiply(
                   three.subtract(x.multiply(guess.pow(2)))), mc);
     } while (oldguess.compareTo(guess) != 0);

     return guess;
 }

 // method f3c.fib(n) calculates the nth fibonacci number
 // as floor( phi^n/sqrt(5) + 1/2), where phi = (1+sqrt(5.0))/2;
 private static BigInteger fib(int n) {
    MathContext mc = new MathContext(n/2, RoundingMode.HALF_DOWN);
    BigDecimal is5 = isqrt(new BigDecimal("5"), mc);
    BigDecimal one = BigDecimal.ONE;
    BigDecimal half = new BigDecimal(0.5);
    BigDecimal phi = half.multiply(one.add(one.divide(is5, mc)));
    return phi.pow(n, mc).multiply(is5).toBigInteger();
 }

 // Method f3c.f(n) handles the negative arguments: F(-n) = F(n)*(-1)^(n+1)
 private static BigInteger f(int n) {
    if(n<0)
       return (n%2==0) ? fib(-n).negate() : fib(-n);
    else
       return fib(n);
 }

 // Method f3c.f_print(n) prints the nth Fibonacci number
 private static void fib_print(int n) {
  System.out.println(n + "th Fibonacci number is " + f(n));
 }
 // Method f3c.main is the program entry point
 // It makes sure the program is called with one commandline argument
 // converts it to integer and executse fib_print
 // If the conversion fails or if the number of arguments is wrong,
 // usage information is printed
 public static void main(String argv[]) {
  try {
      if(argv.length == 1) {
          fib_print(Integer.parseInt(argv[0]));
          System.exit(0);
      }
  } catch (NumberFormatException e) { }
  System.out.println("Usage: java f3c <n>");
  System.exit(1);
 }
}

…成为以下for循环：

for (int var = 0; i != n; i++) {
     ...
}

有关如何执行此操作的信息，请参阅，您可以使用Binet的公式（复制如下）在不递归的情况下执行此操作：

import java.lang.Math;
import java.math.BigInteger;
import java.math.BigDecimal;
import java.math.MathContext;
import java.math.RoundingMode;
class f3c {
// This program calculates the nth fibonacci number
// using alrogirhtm 3C: Binet's formula with rounding
//
// compiled: javac f3c.java
// executed: java f3c n
//

  // Method f3c.isqrt(n) finds the inverse root of n using the following
  // recurrent equation: y(n+1) = 0.5*y(n)*[3 - x*y(n)^2]
  // It is faster than typical Newton-Raphson square root finding because
  // it does not use division.
  // Funny how Java implemented exponentiation by repeated squaring
  // for BigDecimals, but did not implement any sort of square root.
 private static BigDecimal isqrt(BigDecimal x, MathContext mc) {
     BigDecimal guess = new BigDecimal(1/Math.sqrt(x.doubleValue()));
     BigDecimal three = new BigDecimal(3);
     BigDecimal half = new BigDecimal(0.5);
     BigDecimal oldguess;
     do{ oldguess = guess;
         guess = half.multiply(guess.multiply(
                   three.subtract(x.multiply(guess.pow(2)))), mc);
     } while (oldguess.compareTo(guess) != 0);

     return guess;
 }

 // method f3c.fib(n) calculates the nth fibonacci number
 // as floor( phi^n/sqrt(5) + 1/2), where phi = (1+sqrt(5.0))/2;
 private static BigInteger fib(int n) {
    MathContext mc = new MathContext(n/2, RoundingMode.HALF_DOWN);
    BigDecimal is5 = isqrt(new BigDecimal("5"), mc);
    BigDecimal one = BigDecimal.ONE;
    BigDecimal half = new BigDecimal(0.5);
    BigDecimal phi = half.multiply(one.add(one.divide(is5, mc)));
    return phi.pow(n, mc).multiply(is5).toBigInteger();
 }

 // Method f3c.f(n) handles the negative arguments: F(-n) = F(n)*(-1)^(n+1)
 private static BigInteger f(int n) {
    if(n<0)
       return (n%2==0) ? fib(-n).negate() : fib(-n);
    else
       return fib(n);
 }

 // Method f3c.f_print(n) prints the nth Fibonacci number
 private static void fib_print(int n) {
  System.out.println(n + "th Fibonacci number is " + f(n));
 }
 // Method f3c.main is the program entry point
 // It makes sure the program is called with one commandline argument
 // converts it to integer and executse fib_print
 // If the conversion fails or if the number of arguments is wrong,
 // usage information is printed
 public static void main(String argv[]) {
  try {
      if(argv.length == 1) {
          fib_print(Integer.parseInt(argv[0]));
          System.exit(0);
      }
  } catch (NumberFormatException e) { }
  System.out.println("Usage: java f3c <n>");
  System.exit(1);
 }
}

…成为以下for循环：

for (int var = 0; i != n; i++) {
     ...
}

我只是想说，你是如何在shell中实现这一点的：

red@cricket:~$ ./fib.sh 
0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144
red@cricket:~$ cat fib.sh 
#!/bin/sh

first=0
second=1
echo -n "$first, $second, "
thrid=`expr $first + $second`
echo -n $thrid 
while [ $thrid -lt 100 ]
do
        first=$second
        second=$thrid
        thrid=`expr $first + $second`
        echo -n ", $thrid"
done
echo ""

哈哈！有些人需要添加一个
python
answer！或者一个递归的
bash
解决方案。
在这里，你怎么能在shell中实现呢

red@cricket:~$ ./fib.sh 
0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144
red@cricket:~$ cat fib.sh 
#!/bin/sh

first=0
second=1
echo -n "$first, $second, "
thrid=`expr $first + $second`
echo -n $thrid 
while [ $thrid -lt 100 ]
do
        first=$second
        second=$thrid
        thrid=`expr $first + $second`
        echo -n ", $thrid"
done
echo ""

哈哈！有些人需要添加一个
python
answer！或者一个递归的
bash
解决方案。
问题是什么？问题是什么？唯一的问题是我应该用一个for循环来做这件事，这是我无法理解的。嗯，那会很慢。递归并不总是解决问题的最佳方法；-）唯一的问题是，我应该用for循环来做这件事，这是我无法理解的。嗯，那会非常慢。递归并不总是解决问题的最佳方法；-）也许删除TODO？也许删除TODO？这花了我很多时间才得到正确的答案：）。这花了我很多时间才得到正确的答案：）。