日期:2014-05-20  浏览次数:21052 次

从键盘输入一个十进制数,转换成二进制数后输出
要求不用Integer,自己写,另输入一整数A,判断它是否质数。

------解决方案--------------------
Java code

public static void main(String args[]){
        Scanner input = new Scanner(System.in);
        int inputNumber = input.nextInt();
        erjinzhi(inputNumber);
    }
    public static String erjinzhi(int a){
        boolean flag = true;
        boolean isZhengshu = (a>=0?true:false);
        if(!isZhengshu){
            a = a&0xffff;
        }
        int temp = Math.abs(a);
        StringBuffer sb = new StringBuffer();
        while(flag){
            int yushu = temp%2;
            sb.append(yushu);
            temp = temp/2;
            if(temp == 0){
                flag = false;
            }
        }
        StringBuffer result = new StringBuffer();
        for(int i=0;i<sb.length();i++){
            result.append(sb.charAt(sb.length()-1-i));
        }
System.out.print(result);
        return result.toString();
    }

------解决方案--------------------
都是基础题,LZ应该好好思考
还是贴下代码吧。。。
Java code

import java.util.Scanner;

public class Test {
    public static void main(String[] args) {
        test1();
        test2();
    }
    
    //二进制
    static void test1(){
        System.out.print("二进制 输入一个数: ");
        Scanner scanner = new Scanner(System.in);
        try {
            int num = scanner.nextInt();
            int numTemp = num;
            //System.out.println(num + "的二进制数" + Integer.toBinaryString(num));
            StringBuffer sBuffer = new StringBuffer();
            while(numTemp != 0){
                int value = numTemp % 2;
                sBuffer.append(value);
                numTemp /= 2;
            }
            System.out.println(num + "的二进制数" + sBuffer.reverse().toString());
        } catch (Exception e) {
            System.out.println("输入数不合法");
        }

    }
    
    //质数
    static void test2(){
        System.out.print("质数 输入一个数: ");
        Scanner scanner = new Scanner(System.in);
        try {
            int num = scanner.nextInt();
            for(int i = 2;i < Math.sqrt(num);i++){
                if(num % i == 0){
                    System.out.println(num + "不是质数");
                    return;
                }
            }
            System.out.println(num + "是质数");
        } catch (Exception e) {
            System.out.println("输入数不合法");
        }

    }
}