IDE中的代码正确,但在CodeChef中给出错误

时间:2019-04-08 05:14:59

标签: java

我正在尝试解决codechef问题,我能够在IDE以及自定义输入中获得输出,当我尝试在其中运行输入时,它给了我错误

链接到问题: https://www.codechef.com/problems/HS08TEST

代码:

    /* package codechef; // don't place package name! */

import java.util.*;
import java.lang.*;
import java.io.*;
import java.text.DecimalFormat;
/* Name of the class has to be "Main" only if the class is public. */
class Codechef
{
    public static void main (String[] args) throws java.lang.Exception
    {
        Scanner input = new Scanner(System.in); 
        int numberOne = input.nextInt();
        float numberTwo = input.nextFloat();
        float reduction = 0;
        float result = 0;
        DecimalFormat df2 = new DecimalFormat(".00");
        if(numberOne > 0 && numberOne <= 2000 & numberTwo >= 0 && numberTwo <= 2000){
        if(numberOne % 5 == 0){
            reduction = (float)numberOne+(0.50f);
            if(reduction <= numberTwo){
                result = numberTwo-reduction;

                System.out.println(df2.format(result));
            }
            if(reduction > numberTwo){
                System.out.println(df2.format(numberTwo));
            }
        }
        else{
            System.out.println(df2.format(numberTwo));
        }
        }

    }

}

错误:

  

线程“ main”中的异常java.util.NoSuchElementException在   java.util.Scanner.throwFor(Scanner.java:862)在   java.util.Scanner.next(Scanner.java:1485)在   java.util.Scanner.nextInt(Scanner.java:2117)在   java.util.Scanner.nextInt(Scanner.java:2076)在   Codechef.main(Main.java:14)

3 个答案:

答案 0 :(得分:2)

“错误”是由于无法将输入解析为必需的类型(即Scanner不能将输入解析为intfloat

“ A”解决方案是获取输入并手动解析它。您可以使用nextLine并在其上运行另一个Scanner,或在公共定界符上拆分,或者可以简单地使用next,例如...

import java.text.DecimalFormat;
import java.util.Scanner;

class Codechef {

    public static void main(String[] args) throws java.lang.Exception {
        Scanner input = new Scanner(System.in);
        String element = input.next(); // Next value up to the next space or new line...
        int numberOne = Integer.parseInt(element);
        element = input.next(); // Next value up to the next space or new line...
        float numberTwo = Float.parseFloat(element);
        float reduction = 0;
        float result = 0;
        DecimalFormat df2 = new DecimalFormat(".00");
        if (numberOne > 0 && numberOne <= 2000 & numberTwo >= 0 && numberTwo <= 2000) {
            if (numberOne % 5 == 0) {
                reduction = (float) numberOne + (0.50f);
                if (reduction <= numberTwo) {
                    result = numberTwo - reduction;

                    System.out.println(df2.format(result));
                }
                if (reduction > numberTwo) {
                    System.out.println(df2.format(numberTwo));
                }
            } else {
                System.out.println(df2.format(numberTwo));
            }
        }

    }

}

这假定通常在一行上提供输入,但是此方法将允许您处理两个单独的输入。但是,如果不知道确切的输入是什么,很难提供更精确的解决方案

答案 1 :(得分:1)

您不会浪费输入值之间的空格。

只需使用nextLine读取第一行,然后相应地拆分并解析数字

答案 2 :(得分:0)

一件简单的事情对我有用.... 我只是试图抓住代码的周围。...

最终工作代码...

/* package codechef; // don't place package name! */

import java.util.*;
import java.lang.*;
import java.io.*;

/* Name of the class has to be "Main" only if the class is public. */
class Codechef
{
    public static void main (String[] args) throws java.lang.Exception
    {
        try{
        int n, sum = 0;
        Scanner s = new Scanner(System.in);
        n = s.nextInt();
        int a[] = new int[n];
        for(int i = 0; i < n; i++)
        {
            a[i] = s.nextInt();
        }

        int largest=0;
        int element=0;


        for(int i = 0; i < n; i++){
            for(int j=0;j<n;j++){
                element=a[i]%a[j];
                if(largest<element){
                    largest=element;
                }
            }
        }
        System.out.println(largest);
    }
            catch(Exception e){

    }
    }

}