import java.util.Scanner;
public class namefinder {
public static void main(String[] args) {
System.out.println(" NAME PREDICTER ");
System.out.println(" Row 1 A B C D E F");
System.out.println(" Row 2 G H I J K L");
System.out.println(" Row 3 M N O P Q R");
System.out.println(" Row 4 S T U V W X");
System.out.println(" Row 5 Y Z . . . .");
System.out.println("enter the length of your first name in number!!! EX: vino , so length is 4");
Scanner scanner = new Scanner(System.in);
int y = scanner.nextInt();
int s[] = new int [20];
for(int i=1;i<=y;i++)
{
System.out.println("Enter whether the "+i+"letter of your name is in which row");
Scanner scanner1 = new Scanner(System.in);
s[i] = scanner1.nextInt();
}
char a[] = {'A','B','C','D','E','F'};
char b[] = {'G','H','I','J','K','L'};
char c[] = {'M','N','O','P','Q','R'};
char d[] = {'S','T','U','V','W','X'};
char e[] = {'Y','Z','.','.','.','.'};
for(int i=0;i<6;i++){
for(int j = 0; j <=y; j++) {
switch(s[j] ) { // Since counting starts from 0.
case 1: System.out.print("\t"+a[i]);break;
case 2: System.out.print("\t"+b[i]); break;
case 3: System.out.print("\t"+c[i]); break;
case 4: System.out.print("\t"+d[i]); break;
case 5: try {
System.out.print("\t"+e[i]);
} catch(ArrayIndexOutOfBoundsException ex) {
System.out.println("\t "); // Print an empty space with a \t character.
}
break; // Prone to ArrayIndexOutOfBoundsException
}
}
System.out.println(" "+"Row"+ (i+1)); // Print a newline character after printing every line
}
输出:
姓名预测员
第1行A B C D E F.
第2行G H I J K L
第3行M N O P Q R
第4行S T U V W X.
第5行Y Z. 。 。 。
输入您的名字的长度! EX:vino,长度为4
3
输入您姓名的1letter是否在哪一行
4
输入您姓名的2letter是否在哪一行
2
输入您姓名的3letter是否在哪一行
3
S G M Row1
T H N Row2
U I O Row3
V J P Row4
W K Q Row5
需要帮助: 现在我需要存储第1行的所有字母表以分隔数组。 即row1 [] = {S,G,M}和row2 [] = {T,H,N}等等............