在控制台上输入a,b然后打印出一个b位数,每位都是a。。。。。。。。。。。。
String s = "";
for (int i = 0; i < b; i++)
{
s += a.toString();
}
System.out.println(s);
Scanner input = new Scanner(System.in);
int a = input.nextInt();
int b = input.nextInt();String s = "";
for (int i = 0; i < b; i++)
{
s += a.toString();
}
System.out.println(s);
把这些放在main里面就可以了。
public static void dowork(int a,int b)
{
for(int i=0;i<b;i++)
{
System.out.print(a);
}
}
注意,输出的不是一个数
首先,重复从控制台接受两个int类型的数据,而且限制a的大小为0-9;其次就是用字符串的累加b个a即可。
参考代码:
import java.util.Scanner;
public class PrintNumber {
public static void main(String[] args) {
Scanner input = new Scanner(System.in);
System.out.println("输入数字a(a<10):");
int a = input.nextInt();
while(a>9||a<0){
System.out.println("请重新输入数字a(a<10):");
a = input.nextInt();
}
System.out.println("输入任意数字b:");
int b = input.nextInt();
String result = generateNumber(a,b);
System.out.println("生成的数据为:"+result);
input.close();
}
public static String generateNumber(int a, int b) {
StringBuffer buffer = new StringBuffer();
for (int i = 0; i < b; i++) {
buffer.append(a);
}
return buffer.toString();
}
}
public class text {
public static void main(String[] args) {
System.out.println("请输入a");
Scanner a1 = new Scanner(System.in);
String a = a1.next();
System.out.println("请输入b");
Scanner b1 = new Scanner(System.in);
String b = b1.next();
int j = Integer.parseInt(b);
for(int i = 0;i<j;i++){
System.out.print("a");
}
}
}
import java.io.*;
public class test
{
public static void main(String args[])
{
try
{
BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
System.out.println("Enter your a");
String str1=br.readLine();
System.out.println("Enter your b");
String str2=br.readLine();
result(str1,str2);
}
catch(Exception e)
{System.out.println("io is error");}
}
public static void result(String str1,String str2)
{
int a=Integer.parseInt(str1);
int b=Integer.parseInt(str2);
int c=0;
int k=1;
for(int i=0;i<b;i++)
{
k=k*10;
System.out.println("K:"+k);
c=c+k*a;
}
System.out.println("so you result c is :"+c);
}
转成字符串叠加输出比较好