输入学生成绩,取平均,然后按成绩对学生进行排名

时间:2019-02-03 04:37:46

标签: java arrays loops

样本输入:

1年级学生:75
学生2年级:85
学生3年级:90
学生4年级:85
5年级学生:90

示例输出:

前1名:学生3
学生5
前2名:学生2
学生4
前3名:学生1

Scanner sc = new Scanner(System.in);
int[] grade = new int[5];
int[] gradeClone;

for(int i = 1; i <= 5; i++){
    System.out.print("Enter student " + i + " grade: ");
    grade[i-1] = sc.nextInt();
}

gradeClone = grade.clone();

int failed = 0;

for(int x = 1; x <= 5; x++){

    System.out.print("Student " + x + " Remark = ");

    if(grade[x-1] >= 60){
        System.out.println(" Passed");
    }else{
        System.out.println(" Failed");
        failed++;
    }

}

int total = 0;

for(int a = 0; a < 5; a++){
    total += grade[a];
}

int temp;

for(int j = 0; j < 5; j++){

    for(int k = 1; k < 5; k++){

        if(grade[k-1] < grade[k] ){
            temp = grade[k-1];
            grade[k-1] = grade[k];
            grade[k] = temp;
        }

    }

}
String ranking = failed <= 4 ? " then the ranking is: ": " all failed";
System.out.println("The average is " + total/5 + ranking ); 


int topCount = 5-failed;

for(int a = 1; a < 5; a++){
    if(gradeClone[a-1] == gradeClone[a]){
        --topCount;
    }
}

for(int i = 0; i < 5; i++){
    System.out.println(gradeClone[i]);
}

我无法提出一种对他们进行排名的方法。我应该怎么做?

1 个答案:

答案 0 :(得分:0)

我认为您需要这样的东西?从最高分到最低/最低到最高排名。顺便说一下,我对成绩进行了硬编码,您可以将其更改为用户输入。

int[] grade = {50,60,30,40,80};
int temp;
    for (int i = 0; i < grade.length; i++) 
    {
        for (int j = i + 1; j < grade.length; j++) 
        {
            if (grade[i] > grade[j]) 
            {
                temp = grade[i];
                grade[i] = grade[j];
                grade[j] = temp;
            }
        }
    }
    System.out.println("Rank from lowest to highest:");
    for (int i = 0; i < grade.length - 1; i++) 
    {
        System.out.print(grade[i] + "\n");
    }
    System.out.print(grade[grade.length - 1]);

    System.out.println("\n\nRank from highest to lowest:");
    for (int i = 4; i >= 0; i--) 
    {
        System.out.print(grade[i] + "\n");
    }