某c语言期末考试

题目

某班有5名同学，建立一个学生的数学课和英语课信息表，包括学号、姓名、数学课和英语课的成绩。
编写程序，计算每名同学的平均成绩和名次。
（注：一定一个结构体类型，可以用结构指针作为函数参数）
输入格式：
        %ld %s %f %f
输出格式：
        %-9ld%-10s%-10.1f%-5.1f\n
输出样式
        202201   Tom   91.8   85.5
        202202   Cat   96.5   87.6
输出样例
Number:   Name:     Score:         Average:   Rank:
202201     Tom     91.8     85.5     88.8         3
202202     Cat      96.5     87.6     92.1         1

代码实现

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

struct student {
	long long number;
	char name[64]; 
	float english_score; 
	float math_score; 
	}; 

int main() {
    struct student students[5] = {0};
	
	float array[5] = {0.0};
	int i = 0; 
	for (; i < 5; i++) {
		struct student *s = &(students[i]);
		int flag = scanf("%ld %s %f %f", &(s->number), s->name,
			&(s->math_score), &(s->english_score));
		if (flag <= 0) {
			break; 
		}
	}
    float scores[5] = {0.0};
    int ranks[5] = {1, 1, 1, 1, 1};
	int j = 0; 
    int k = 0;
    for (j=0; j<i; j++) {
        struct student s = students[j];
        float average = (s.english_score + s.math_score) / 2;
        scores[j] = average;
    }
    for(j=0; j<i; j++) {
        for(k=0; k<i; k++) {
            if (scores[k] > scores[j]) {
                ranks[j]++;
            }
        }
    }
    printf("Number: Name: Score: Average: Rank:\n")
    
    for(j=0; j<i; j++) {
		struct student s = students[j];
		printf("%-91d%-10s%-5.1f%-5.1f\n", s.number, s.name, s.math_score,
		s.english_score, scores[j], ranks[j]);
	}
	return 0; 
}

测试结果

wens-MBP:~ wenzhou$ ./a 
202201 Tom 91.8 85.5 
202202 Cat 96.5 87.6 
a
Number:  Name:     Score:    Average:  Rank:
202201   Tom       91.8 85.5 88.7      2 
202202   Cat       96.5 87.6 92.1      1 
wens-MBP:~ wenzhou$