我们如何使用 println 打印 (n +nn +nnn) 这段代码?
how can we print (n +nn +nnn) using println for this code?
int n;
Scanner in = new Scanner(System.in);
System.out.print("Input number: ");
n = in .nextInt();
System.out.println(n " + " nn ); //this is not working
如何使用 println 进行打印
请将您的代码块格式化为代码。
如果你想像 if
那样打印两次相同的值
int n = 5;
System.out.println(String.valueOf(n) + "+" + String.valueOf(n) + String.valueOf(n);
这将打印 5 + 55
if you want to print the double of n then:
System.out.println(String.valueOf(n) + " + " + n*2;
这将打印 5 + 10
但我不确定您期望的结果是什么
这样的事情可能会有所帮助
n = in .nextInt();
for(int i=1;i<=n;i++){
for(int j=1;j<=i;j++){
System.out.print("n");
}
if(i!=n) {
System.out.print("+");
}
}
输出::
if input =3
Output => n+nn+nnn
if input=5
output => n+nn+nnn+nnnn+nnnnn
是这样的吗?
import java.util.Scanner;
public class Main {
public static void main(String[] args) {
Scanner in = new Scanner(System.in);
System.out.print("Input number: ");
String n = in.nextLine();
if (n.matches("[0-9]+")) {
System.out.println(n + " + " + n + n + " + " + n + n + n + " = "
+ (Integer.parseInt(n) + Integer.parseInt(n + n) + Integer.parseInt(n + n + n)));
} else {
System.out.println("Error: invalid input.");
}
}
}
样本运行-1:
Input number: 5
5 + 55 + 555 = 615
样本运行-2:
Input number: 1
1 + 11 + 111 = 123
样本运行-3:
Input number: a
Error: invalid input.
如果你想打印n+nn+nnn的模式,你可以考虑下面的代码
Scanner in = new Scanner(System.in);
System.out.print("Input number: ");
String n = in.nextLine();
String output = "";
for (int i = 1; i <= 3; i++) {
output += Stream.generate(() -> "" + n).limit(i).collect(Collectors.joining()) + "+";
}
System.out.println(output.substring(0, output.length() - 1));
输出:
Input number: 5
5+55+555
import java.util.Scanner;
public class Main {
public static void main(String[] args) {
System.out.println("Type your number : ");
Scanner input = new Scanner(System.in);
int n = input.nextInt();
String x=String.valueOf(n);
System.out.println(x+ " + "+ x+x +" + " +x+x+x+ " = "+ ((Integer.parseInt(x))+(Integer.parseInt(x+x))+ (Integer.parseInt(x+x+x))));
}
}
输出:(数字 = 5 的示例)
Type your number : 5
5 + 55 + 555 = 615
好吧,这是另一种略有不同的选择。
n
,要重复的数字
k
,重复次数
- 调用
successiveSum(n,k)
successiveSum(1,3);
successiveSum(5,3);
successiveSum(1,4);
版画
1 + 11 + 111 = 123
5 + 55 + 555 = 615
1 + 11 + 111 + 1111 = 1234
方法。
它的工作原理是将数字乘以 10,然后加上 n。
public static void successiveSum(int n, int k) {
int sum = 0;
int nextNumb = 0;
String s = "";
for (int i = 0; i < k; i++) {
nextNumb = nextNumb * 10 + n;
s = s + nextNumb + " + ";
sum += nextNumb;
}
// remove the last "+ " of the string.
System.out.println(s.substring(0,s.length()-2) + " = " + sum);
}
int n;
Scanner in = new Scanner(System.in);
System.out.print("Input number: ");
n = in .nextInt();
System.out.println(n " + " nn ); //this is not working
如何使用 println 进行打印
请将您的代码块格式化为代码。
如果你想像 if
那样打印两次相同的值int n = 5;
System.out.println(String.valueOf(n) + "+" + String.valueOf(n) + String.valueOf(n);
这将打印 5 + 55
if you want to print the double of n then:
System.out.println(String.valueOf(n) + " + " + n*2;
这将打印 5 + 10
但我不确定您期望的结果是什么
这样的事情可能会有所帮助
n = in .nextInt();
for(int i=1;i<=n;i++){
for(int j=1;j<=i;j++){
System.out.print("n");
}
if(i!=n) {
System.out.print("+");
}
}
输出::
if input =3
Output => n+nn+nnn
if input=5
output => n+nn+nnn+nnnn+nnnnn
是这样的吗?
import java.util.Scanner;
public class Main {
public static void main(String[] args) {
Scanner in = new Scanner(System.in);
System.out.print("Input number: ");
String n = in.nextLine();
if (n.matches("[0-9]+")) {
System.out.println(n + " + " + n + n + " + " + n + n + n + " = "
+ (Integer.parseInt(n) + Integer.parseInt(n + n) + Integer.parseInt(n + n + n)));
} else {
System.out.println("Error: invalid input.");
}
}
}
样本运行-1:
Input number: 5
5 + 55 + 555 = 615
样本运行-2:
Input number: 1
1 + 11 + 111 = 123
样本运行-3:
Input number: a
Error: invalid input.
如果你想打印n+nn+nnn的模式,你可以考虑下面的代码
Scanner in = new Scanner(System.in);
System.out.print("Input number: ");
String n = in.nextLine();
String output = "";
for (int i = 1; i <= 3; i++) {
output += Stream.generate(() -> "" + n).limit(i).collect(Collectors.joining()) + "+";
}
System.out.println(output.substring(0, output.length() - 1));
输出:
Input number: 5
5+55+555
import java.util.Scanner;
public class Main {
public static void main(String[] args) {
System.out.println("Type your number : ");
Scanner input = new Scanner(System.in);
int n = input.nextInt();
String x=String.valueOf(n);
System.out.println(x+ " + "+ x+x +" + " +x+x+x+ " = "+ ((Integer.parseInt(x))+(Integer.parseInt(x+x))+ (Integer.parseInt(x+x+x))));
}
}
输出:(数字 = 5 的示例)
Type your number : 5
5 + 55 + 555 = 615
好吧,这是另一种略有不同的选择。
n
,要重复的数字k
,重复次数- 调用
successiveSum(n,k)
successiveSum(1,3);
successiveSum(5,3);
successiveSum(1,4);
版画
1 + 11 + 111 = 123
5 + 55 + 555 = 615
1 + 11 + 111 + 1111 = 1234
方法。
它的工作原理是将数字乘以 10,然后加上 n。
public static void successiveSum(int n, int k) {
int sum = 0;
int nextNumb = 0;
String s = "";
for (int i = 0; i < k; i++) {
nextNumb = nextNumb * 10 + n;
s = s + nextNumb + " + ";
sum += nextNumb;
}
// remove the last "+ " of the string.
System.out.println(s.substring(0,s.length()-2) + " = " + sum);
}