斐波那契数列 面试算法(三)

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import java.math.BigDecimal; import java.util.Scanner; import java.util.function.BinaryOperator;   public class Fbnq {     /**      * 假设n为正整数，斐波那契数列定义为：      * f(n) = 1, n < 3;      * f(n) = f(n-1) + f(n-2), n>=3      *      * 现在请你来计算f(n)的值，但是不需要给出精确值，只要结果的后六位即可。      *      * 输入：一行，包含一个正整数n，且0<n<1000      * 输出：一行，f(n)的后6位（十进制，不足6位不补零）      * @param args      */     public static void main(String[] args) {         Scanner s = new Scanner(System.in);         int i = s.nextInt();         BigDecimal f2 = fb(i);         String v = String.valueOf(f2);         if (v.length() > 6) {             String substring = v.substring(v.length() - 6);             System.out.println(substring);         }else {             System.out.println(v);         }     }     static BigDecimal fb(int num){         BigDecimal m = new BigDecimal(1);         BigDecimal n = new BigDecimal(1);         if (num == 1 || num == 2){             return m;         }         BigDecimal t = new BigDecimal(0);         for (int i = 0; i <= num-3; i++) {             t = m.add(n);             m = n;             n = t;         }         return t;     }       /*public static int fb1(int num) {         if (num==1||num==2) {             return 1;         }else {             return fb1(num - 1) + fb1(num - 2);         }     }*/ }