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java获取偶数或奇数

我有一个学校作业:我需要得到偶数或奇数,我做了一些工作来处理零以上的数,但我需要找出负数是偶数还是奇数

String[] evenodd = {" number is even", " number is odd"};
for(int i = 0; i < 100; i++)
{
    int x = i;
    while (x > 1) x = x - 2;
    System.out.println(i + evenodd[x]);
}

共 (4) 个答案

  1. # 1 楼答案

    尝试使用%运算符

    用法:x % y->;当y除以x时,这将得到余数。 在您的情况下,使用y=2。因此,如果余数=0,则该数字为偶数,否则该数字为奇数

    我不会给你确切的密码,因为这是你的任务。根据建议自己尝试一下

  2. # 2 楼答案

    偶数奇数数字:

    String[] evenodd = {" number is even", " number is odd"};
    for(int i = 0; i < 100; i++)
    {
        System.out.println(i + evenodd[i % 2]);
    }
    

    String[] evenodd = {" number is even", " number is odd"};
    for(int i = 0; i < 100; i++)
    {
        if(i % 2 == 0)
             System.out.println(i + evenodd[i % 2]);
        else
             System.out.println(i + evenodd[i % 2]);
    }
    

    对于负数是偶数还是奇数

    String[] evenodd = {" number is even", " number is odd"};
    for(int i = 0; i > -100; i--)
    {
        if(i % 2 == 0)
             System.out.println(i + evenodd[0]);
        else
             System.out.println(i + evenodd[1]);
    }
    
  3. # 3 楼答案

    找出奇数和偶数。遵循逻辑

    • divide your number by 2 if the reminder is 0 then it is even

    • else it is odd

    • this similar for both positive and negative numbers

    要查找提醒,请使用模除运算符(%)。如果您的\u编号%2=0,则您的\u编号为偶数,否则您的\u编号为奇数

    示例代码:

    for(int i=1;i<100;i++){
        if(i%2==0){//for positive number
            System.out.println(i+ " is Even Number");
        }else{
            System.out.println(i+ " is Odd Number");
        }
        if((-1)*i%2==0){//for negative number
            System.out.println((-1)*i+" is Even Number");
        }else{
            System.out.println((-1)*i+" is Odd Number");
        }
    }
    
  4. # 4 楼答案

    只需使用模块%运算符

    String[] evenodd = {" number is even", " number is odd"};
    for(int i = 0; i < 100; i++)
    {
    int x;
    
    if(i%2==0){
     x=0;/*pass 0 to print even number */
    }else{
     x=1;/*pass 1 to print odd number*/
    }
    System.out.println(i + evenodd[x]);
    }