英文:
Fibonacci's Sequence using forloop
问题
我试图创建一个长度为 n 的数组（用户输入），我想我可以使用数组中关联的 i 值来计算斐波那契和。
到目前为止，这是我的代码，我无法弄清楚如何提取 i 值作为整数以便能够计算总和。
public class Fibonacci {
public static void main(String[] args){
System.out.println("请输入一个值作为 n：");
Scanner scanner = new Scanner(System.in);
int n = scanner.nextInt();
int[] newArray = new int[n];
int f1 = newArray[0];
int f2 = newArray[1];
int i;
for(i = 1; i <= n; ++i) {
System.out.print(f1 + " ");
int sum = f1 + f2;
f1 = f2;
f2 = sum;
}
}
}
如果有人对如何解决这个问题有任何建议，并能够解释一些理论，那将不胜感激。
英文:
I'm trying to create an array with length n (user input), and I thought that I could use the associated i values within the array to calculate my fibonacci sum.
Here is what I have so far, and I can't figure out how I should be extracting the i value as an int to be able to calculate the sum.
public class Fibonacci {
public static void main(String[] args){
System.out.println("Please enter a value for n: ");
Scanner scanner = new Scanner(System.in);
int n = scanner.nextInt();
int[] newArray = new int[n];
int f1 = newArray[0];
int f2 = newArray[1];
int i;
for(i = 1; i <= n; ++i) {
System.out.print(f1 + " ");
int sum = f1 + f2;
f1 = f2;
f2 = sum;
}
}
}
If anyone has any suggestions on how to approach this and can explain a bit of the theory that would be greatly appreciated.
答案1
得分: 0
如果您正在使用数组来存储斐波那契数列，您应该通过在该数组中使用索引来访问斐波那契数，并且不需要使用中间变量 f1、f2、sum。
int[] fiboNums = new int[n]; // 假设 n >= 2
fiboNums[0] = 1;
fiboNums[1] = 1;
System.out.printf("f(%2d)=%,13d ", i, fiboNums[i]);
for (int i = 2; i < n; i++) {
fiboNums[i] = fiboNums[i - 1] + fiboNums[i - 2];
System.out.printf("f(%2d)=%,13d ", i, fiboNums[i]);
if (i % 5 == 0) {
System.out.println();
}
}
然而，使用 int 来表示斐波那契数可能不是一个好主意，因为这个序列是指数增长的，当 i == 46 时会发生整数溢出。
f( 1)= 1 f( 2)= 2 f( 3)= 3 f( 4)= 5 f( 5)= 8
f( 6)= 13 f( 7)= 21 f( 8)= 34 f( 9)= 55 f(10)= 89
f(11)= 144 f(12)= 233 f(13)= 377 f(14)= 610 f(15)= 987
f(16)= 1,597 f(17)= 2,584 f(18)= 4,181 f(19)= 6,765 f(20)= 10,946
f(21)= 17,711 f(22)= 28,657 f(23)= 46,368 f(24)= 75,025 f(25)= 121,393
f(26)= 196,418 f(27)= 317,811 f(28)= 514,229 f(29)= 832,040 f(30)= 1,346,269
f(31)= 2,178,309 f(32)= 3,524,578 f(33)= 5,702,887 f(34)= 9,227,465 f(35)= 14,930,352
f(36)= 24,157,817 f(37)= 39,088,169 f(38)= 63,245,986 f(39)= 102,334,155 f(40)= 165,580,141
f(41)= 267,914,296 f(42)= 433,494,437 f(43)= 701,408,733 f(44)=1,134,903,170 f(45)=1,836,311,903
同样，使用 long 类型只能容纳前 91 个斐波那契数。
英文:
If you are using an array to store the Fibonacci sequence, you should access the Fibonacci numbers using index in this array and there is no need to use intermediate variables f1, f2, sum.
int[] fiboNums = new int[n]; // assuming n >= 2
fiboNums[0] = 1;
fiboNums[1] = 1;
System.out.printf("f(%2d)=%,13d ", i, fiboNums[i]);
for (int i = 2; i < n; i++) {
fiboNums[i] = fiboNums[i - 1] + fiboNums[i - 2];
System.out.printf("f(%2d)=%,13d ", i, fiboNums[i]);
if (i % 5 == 0) {
System.out.println();
}
}
However, using int to represent Fibonacci number may not be a good idea because this sequence grows exponentially and integer overflow occurs when i == 46.
f( 1)= 1 f( 2)= 2 f( 3)= 3 f( 4)= 5 f( 5)= 8
f( 6)= 13 f( 7)= 21 f( 8)= 34 f( 9)= 55 f(10)= 89
f(11)= 144 f(12)= 233 f(13)= 377 f(14)= 610 f(15)= 987
f(16)= 1,597 f(17)= 2,584 f(18)= 4,181 f(19)= 6,765 f(20)= 10,946
f(21)= 17,711 f(22)= 28,657 f(23)= 46,368 f(24)= 75,025 f(25)= 121,393
f(26)= 196,418 f(27)= 317,811 f(28)= 514,229 f(29)= 832,040 f(30)= 1,346,269
f(31)= 2,178,309 f(32)= 3,524,578 f(33)= 5,702,887 f(34)= 9,227,465 f(35)= 14,930,352
f(36)= 24,157,817 f(37)= 39,088,169 f(38)= 63,245,986 f(39)= 102,334,155 f(40)= 165,580,141
f(41)= 267,914,296 f(42)= 433,494,437 f(43)= 701,408,733 f(44)=1,134,903,170 f(45)=1,836,311,903
Similarly, using long would allow to fit only 91 Fibonacci numbers.
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