Timus Online Judge 1001. Reverse Root

Input

The input stream contains a set of integer numbers A_i (0 ≤ A_i ≤ 10¹⁸). The numbers are separated by any number of spaces and line breaks. A size of the input stream does not exceed 256 KB.

Output

For each number A_i from the last one till the first one you should output its square root. Each square root should be printed in a separate line with at least four digits after decimal point.

Sample

 1427  0   

   876652098643267843
5276538

2297.0716
936297014.1164
0.0000
37.7757

译文：

输入：

输入一个整数数组，这个数组中每个整数的取值范围为0<= Ai <=10的18次方。这些数字被一些空格和换行符分隔。输入的内存限制为256KB。

输出：

倒序输出数组的平方根。每个平方根结果保留最后四位小数。

==========================

第一次code:

 1 import java.math.BigDecimal;
 2 import java.math.RoundingMode;
 3 import java.util.Scanner;
 4
 5 public class Main
 6 {
 7     public static void main(String[] args)
 8     {
 9         Scanner input = new Scanner(System.in);
10         long[]a = new long[1000000000];
11         int i;
12         for (i=0; input.hasNext(); i++)
13         {
14              a[i] = input.nextLong();
15         }
16         for(i--;i>-1;i--)
17         {
18                         /**
19                         *   Math.sqrt()  :  对数据求平方
20                         *   BigDecimal适用于大型数据计算，精确度高
21                         */
22             BigDecimal   b   =   new   BigDecimal(Math.sqrt(a[i]));
23             System.out.println(b.setScale(4,RoundingMode.HALF_UP).toString());
24         }
25     }
26 }
27     

-------------------------------------我是万恶的分割线---------------------------------------------------------------------------------------------------------

/*

题目稍微修改一下，首先输入一个数，定义数组大小，然后倒序输出该数组的平方根。

*/

 1 import java.math.BigDecimal;
 2 import java.math.RoundingMode;
 3 import java.util.Scanner;
 4
 5 public class Main
 6 {
 7     public static void main(String[] args)
 8     {
 9         Scanner input = new Scanner(System.in);
10         int n =input.nextInt();
11         run();
12     }
13     public static void run()
14     {
15         Scanner input = new Scanner(System.in);
16         long[]a = new long[n];
17         int i;
18         for (i=0; i<n; i++)
19         {
20              a[i] = input.nextLong();
21         }
22         for(i--;i>-1;i--)
23         {
24             BigDecimal   b   =   new   BigDecimal(Math.sqrt(a[i]));
25             System.out.println(b.setScale(4,RoundingMode.HALF_UP).toString());
26         }
27     }
28 }