实验6

 

#include <stdio.h>
#include<string.h>
#define  N 100

typedef  struct { 
    char num[10];      
    int s1;           
    int s2;             
    double sum;        
    char level[10];     
} STU;


int fun(STU a[], int n, STU h[]);
int main() {
   STU s[N]={ {"GA05", 85, 76}, 
              {"GA03", 76, 90}, 
              {"GA02", 69, 90}, 
              {"GA04", 85, 56}, 
              {"GA01", 91, 95},
              {"GA07", 72, 80}, 
              {"GA08", 64, 45}, 
              {"GA06", 87, 98}, 
              {"GA015", 85, 86}, 
              {"GA013", 91, 97} };    
    STU h[N];                           
    int i, k, n = 10; 

    k = fun(s, n, h);

    printf("There are :\n");
    for(i = 0; i < k; i++)
        printf("%s %d %d %.2f %s\n", h[i].num, h[i].s1, h[i].s2, h[i].sum, h[i].level);
    
    return 0;
}

int fun (STU a[], int n, STU h[]) {
    int i,num=0,k=0;
    double sum1 = 0, sum2 = 0,sum;
    for (i = 0; i < n; i++)
    {
        sum1 = sum1 + a[i].s1;
        sum2 = sum2 + a[i].s2;
    }
    sum = sum1 / 10 * 0.7 + sum2 / 10 * 0.3;
    for (i = 0; i < n; i++)
    {
        a[i].sum = a[i].s1 * 0.7 + a[i].s2 * 0.3;
        if (a[i].sum >= sum)
        {
            strcpy_s(a[i].level, sizeof(a[i].level), "均分以上");
            h[k] = a[i];
            k++;
        }
    }
    return k;
}

 task5

#include <stdio.h>
#include <string.h>
#define N 5

typedef struct student {
    char name[10];
    int num;
    int maths;
    int computer;
    int english;
    int sum;
    char level[10];
} STU;

void fun(STU a[], int n);  // 函数声明

int main() {
    STU s[6*N]={ {"A001", 1, 34, 67, 80},
                 {"B003", 3, 78, 87, 90},
                 {"A002", 2, 90, 98, 99},
                 {"B002", 4, 56, 78, 98},
                 {"A005", 5, 35, 67, 79} };
    int i;

    fun(s, N);
    for(i = 0; i < N; i++)
      printf("%s %d %d %d %d %d %s\n", s[i].name, s[i].num, s[i].maths, s[i].computer, s[i].english, s[i].sum, s[i].level);
    
    return 0;
}

// 函数定义
// 功能：对包含n条学生成绩记录的数组a进行处理：
// 计算三门课程总分、总分最大值、总分最小值，并设置等级：
// 总分与总分最大值相等的同学的等级设置为优秀
// 总分与总分最小值相等的同学的等级设置为不及格
// 其余同学的等级设置为合格
void fun(STU a[], int n) {
    int i, max = 0, min = 300;
    
    
    for (i = 0;i < n; ++i)
    {
        a[i].sum = a[i].maths + a[i].computer + a[i].english;
        if(a[i].sum > max)
        {
        max = a[i].sum;
        printf("max = %d\n",max);
        }
        if(a[i].sum < min)
        {
        min = a[i].sum; 
        printf("min = %d\n",min);
        }
    }
    for (i = 0;i < n; ++i)
    {
        if(a[i].sum == max)
        strcpy(a[i].level,"优秀");
        else if (a[i].sum == min)
        strcpy(a[i].level,"不及格"); 
        else
        strcpy(a[i].level,"合格"); 
    }
}

 task6

#include <stdio.h>
#define N 5
// 定义结构体类型struct student, 并定义STU为其别名
typedef struct student {
long no;
char name[20];
int score;
} STU;
// 函数声明
void input(STU s[], int n);
int find_min_list(STU s[], STU t[], int n);
void output(STU s[], int n);
int main() {
STU stu[N], min_list[N];
int count;
printf("录入%d个学生信息\n", N);
input(stu, N);
printf("\n统计最低分人数和学生信息...\n");
count = find_min_list(stu, min_list, N);
printf("\n一共有%d个最低分，信息如下:\n", count);
output(min_list, count);
return 0;
}
// 输入n个学生信息，存放在结构体数组s中
void input(STU s[], int n) {
int i;
for(i = 0;i < n;++i)
{
    scanf("%ld %s %d",&s[i].no,s[i].name,&s[i].score);
}
}
// 输出结构体s中n个元素信息
void output(STU s[], int n) {
int i;
for(i = 0;i < n;++i)
{
    printf("%ld %s %d\n",s[i].no,s[i].name,s[i].score); 
}
}
// 在结构体数组s中，查找最低分学生的记录，将其存入结构体数组t中
// 形参n是结构体数组s中元素个数
// 函数返回最低分的学生人数
int find_min_list(STU s[], STU t[], int n) {
int min = s[0].score, i, k = 0;
for (i = 0;i < n;++i)
{
    if (s[i].score < min)
    min = s[i].score;
}
for(i = 0;i < n;++i)
{
    if (s[i].score == min)
    t[k++] = s[i];
}
return k;
}