我試圖解決這個練習:如何獲得該數組的索引號[]
編寫一個叫做GradesAverage程序,該程序會提示用戶爲 學生數從鍵盤讀取它,並將其保存在一個int 變量名爲numStudents中。然後提示用戶輸入每個學生的成績 ,並將其保存在一個名爲成績的int數組中。 您的程序應檢查等級爲0之間和100樣本 會議情況如下:
Enter the number of students: 3
Enter the grade for student 1: 55
Enter the grade for student 2: 108
Invalid grade, try again...
Enter the grade for student 2: 56
Enter the grade for student 3: 57
The average is 56.0
但是,在我的文件,這是程序是如何工作(注意「學生「):
Enter the number of students: 3
Enter the grade for student10 55
Enter the grade for student20 56
Enter the grade for student30 57
你能看到10,20和30嗎?它不顯示student1
,student2
和student3
,它顯示student10
,student20
和student30
。
這裏是我的代碼:
import java.util.Scanner;
class GradesAverage {
public static void main (String[] args) {
Scanner miScanner = new Scanner(System.in);
System.out.println("Enter the number of students: ");
int numStudents = miScanner.nextInt();
int numberGrades[] = new int[numStudents];
int averageGrade = 0;
for (int i = 1; i <= numStudents; i++) {
System.out.println("Enter the grade for student" + i + numberGrades[numStudents - i]);
int grade = miScanner.nextInt();
averageGrade += grade;
if (grade < 0 || grade >100) {
System.out.println("Invalid grade, try again...");
break;
}
}
double average = averageGrade/numStudents;
System.out.println("The average is " + average);
}
}
更合適的是,它應該改爲:'System.out.print(「輸入學生成績」+ i +「:」);'匹配格式。 – CubeJockey