无论我尝试什么，我都无法弄清楚这个问题，我可能只是想多了，我能得到一点帮助吗？我已经尝试了各种不同的 for 循环，但我不知道如何获取这些值
for(int a = 1; a<1000; a++)
我想让它显示为 a+b 但我不认为这是可能的
以下代码以斐波那契而闻名，其中第 N 个输出是 (N- 2)th 和 (N-1)th 值。编写 for 表达式来完成相同的任务。
int a=1, b=1;
while (a < 10000 ){
System.out.print(a+" "+b+" ");
b+=a+=b;
}
【问题讨论】：
标签： java for-loop while-loop
将while循环转换为for循环，就是：
for (int a=1, int b=1; a < 10000; b+=a+=b ){
System.out.print(a+" "+b+" ");
}
【讨论】：
你可以使用递归来做到这一点：
public int fibonacci(int n) {
if(n == 0) {
return 0;
} else if(n == 1) {
return 1;
} else {
return fibonacci(n - 1) + fibonacci(n - 2);
}
}
【讨论】：
如果要使用循环，可以执行以下操作（借用this 答案）：
public int fibonacci(int n) {
if (n == 0)
return 1;
if (n == 1)
return 3;
int grandparent = 1;
int parent = 3;
int curr = 0;
for(int i=2; i <= n; i++){
curr = 3 * parent - grandparent;
grandparent = parent;
parent = curr;
}
return curr;
}
【讨论】：
试试：
int a=1, b=1;
for (b=1; b < 10000; b+=a ){
System.out.print(a+" "+b+" ");
a+=b;
}
【讨论】：
以下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);
}
}
}
输出：
0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, 233, 377, 610, 987, 1597
【讨论】：
请参阅algorithm 3 了解如何执行此操作，您可以使用 Binet 的公式在不递归的情况下执行此操作，复制如下：
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);
}
}
但是，如果您只想将 while 循环转换为 for 循环，则一般原则是，如下代码：
int var = 0;
while (var != n) {
...
var++;
}
... 变成下面的 for 循环：
for (int var = 0; i != n; i++) {
...
}
【讨论】：
在这里很傻，你怎么能在 shell 中做到这一点：
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！有些需要添加python答案！或递归bash 解决方案。
【讨论】：