一个函数的练习题，给大家分享分享，谁有好的想法说说 - 黑马程序员技术交流社区

/*
2.某班有5个学生，三门课。分别编写3个函数实现以下要求：
（1）求各门课的平均分；
（2）找出有两门以上不及格的学生，并输出其学号和不及格课程的成绩；
（3）找出三门课平均成绩在85-90分的学生，并输出其学号和姓名
*/
#include <stdio.h>
typedef struct {
char name[20];//课程名称
float score;//课程分数
}Score;
typedef struct {
int number;//学号
char name[20];//姓名
Score courses[3];//三门课程
}student;
//求三门课的平均成绩
void averageScore(student arr[] ,int arr_len ){
for(int j = 0; j<arr_len;j++){
float average = 0.0f;
for (int i = 0; i < 3; i++) {
average += arr[j].courses[i].score;
}
average /= 3;
if (average >= 85 && average <= 90) {
printf("学号：%d 姓名：%s 的平均分在 85-90之间\n",arr[j].number,arr[j].name);
}
}
}
//求每一门课总的平均成绩
void averageEvery( student arr[],int arr_len ){
for (int i = 0; i < 3; i++) {
float averageEveryScore;
float sum = 0.0f;
for (int j = 0; j < arr_len; j++) {
sum += arr[j].courses[i].score;
}
averageEveryScore = sum / arr_len;
printf("%s 的平均分数是 %.2f\n",arr[i].courses[i].name,averageEveryScore);
}
}
//至少两门低于60
void printPoorStudent(student arr[],int arr_len){
for (int i = 0; i < arr_len; i++) {
int count = 0;
for (int j = 0; j < 3; j++) {
if (arr[i].courses[j].score < 60) {
count++;
}
}
if (count > 1) {
printf("%s 的学号是: %d\n",arr[i].name,arr[i].number);
for (int k = 0; k < 3; k++) {
if (arr[i].courses[k].score < 60) {
printf("%s : %.2f\n",arr[i].courses[k].name,arr[i].courses[k].score);
}
}
}
}
}
int main(int argc, const char * argv[]) {
student person[5] = {
{1,"liujiaping","yuwen",99,"shuxue",98,"yingyu",59},
{2,"liuqiangsheng","yuwen",58,"shuxue",76,"yingyu",45},
{3,"malibo","yuwen",67,"shuxue",67,"yingyu",90},
{4,"duantaiyang","yuwen",65,"shuxue",56,"yingyu",45},
{5,"zhouyu","yuwen",45,"shuxue",36,"yingyu",45},
};
averageEvery( person,5 );
printf("------------------------\n");
printPoorStudent(person,5);
printf("------------------------\n");
averageScore(person ,5 );
return 0;
}

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