计算两个数字之间的平方数,只能用小数字,为什么？

Question

问题是找到两个数字之间的平方数.

下面的代码可以处理小数字,但是数字很大.我怎么能纠正这个？

 import java.io.*;
 import java.util.*;
 import java.text.*;
 import java.math.*;
 import java.util.regex.*;

 public class NumOfSqrs {


public static void main(String[] args) {

    try{
    BufferedReader br = new BufferedReader(new InputStreamReader(System.in));

    String input;
    int line = 0;
    int testCases;
    int numOfSqrt = 0;      
    int j = 0;  

    while((input=br.readLine())!=null){

        if(line == 0){
          testCases = Integer.parseInt(input);
          line = line +1;
        }
        else{
          String[] splitter = input.toString().split(" ");

          //Here splitter gives two numbers, we need to find no of sqrs b/w these numbers for eg say 3 and 9

           for(int i = Integer.parseInt(splitter[0]); i<=Integer.parseInt(splitter[1]) ; i++){

            String value = ""+Math.sqrt(i);
            String[] isSqrt = value.toString().split("\\.");
            //System.out.println(""+isSqrt[0] + "" + isSqrt[1]);

            //Here lets say if  'i' is 4 (i.e 2.0) then isSqrt[0] = 2, isSqrt[1] = 0 and if isSqrt[1] != 1 then its obvious that its not a perfect square

            if(isSqrt[1].length() == 1){
              numOfSqrt++;
            }

          }
          System.out.println(""+numOfSqrt);
        }
     numOfSqrt = 0;
    }


    }catch(IOException io){
        io.printStackTrace();
    }   


  }
}

Answer 1

您确定方块的技术(转换为字符串并在点上拆分)是狡猾的.

这也是不必要的 - 你可以在一行中使用纯粹的数字方法:

int low, high; // fill from input
int numOfSqrt = (int) (Math.floor(Math.sqrt(high)) - Math.ceil(Math.sqrt(low)) + 1);