Student Report

Lab C Advanced 30 points

Read students name m1 m2 m3. Print a report sorted by average descending (ties by name) as name avg grade (avg with 1 decimal; A ≥ 85, B ≥ 70, C ≥ 50, else F), then the class average.

Input

n, then n lines.

Output

n + 1 lines.

Example

Input

4
cy 90 80 100
ann 70 75 65
bob 40 55 50
dan 90 90 90

Output

cy 90.0 A
dan 90.0 A
ann 70.0 B
bob 48.3 F
class 74.6

Sample input available Sample output available

Editor

Hints

Sort the struct array with insertion sort and a two-level comparison.

Solution explanation

Structs, sorting, helper functions and formatted output together make a complete report.

Show solution

#include <stdio.h>
#include <string.h>

typedef struct {
    char name[30];
    double avg;
} Student;

char grade(double avg) {
    if (avg >= 85) return 'A';
    if (avg >= 70) return 'B';
    if (avg >= 50) return 'C';
    return 'F';
}

int main() {
    int n, a, b, c;
    Student s[50];
    double total = 0;
    scanf("%d", &n);
    for (int i = 0; i < n; i++) {
        scanf("%s %d %d %d", s[i].name, &a, &b, &c);
        s[i].avg = (a + b + c) / 3.0;
        total += s[i].avg;
    }
    for (int i = 1; i < n; i++) {
        Student key = s[i];
        int j = i - 1;
        while (j >= 0 && (s[j].avg < key.avg || (s[j].avg == key.avg && strcmp(s[j].name, key.name) > 0))) {
            s[j + 1] = s[j];
            j--;
        }
        s[j + 1] = key;
    }
    for (int i = 0; i < n; i++) printf("%s %.1f %c\n", s[i].name, s[i].avg, grade(s[i].avg));
    printf("class %.1f\n", total / n);
    return 0;
}

Output

Answer