学生成绩管理 14-3 程序设计

要求总分相同时，名次相同

#include <stdio.h>
#include <stdlib.h>
/*
 * 1 lili 88 99 77
2 sofa 77 88 99
3 lrtf 45 56 69
4 ipde 79 63 26
5 fghh 41 23 66
 */
struct Student{
    int num;
    char name[5];
    int score[3];
    int sum;
    int count;
};

void input(struct Student *stu){
    int n;
    printf("please input have how much student:\n");
    scanf("%d",&n);
    printf("num    name    math    yuwen    english :\n");
    for (int i = 0; i < n; ++i) {
        scanf("%d %s",&stu[i].num,stu[i].name);
        for (int j = 0; j < 3; ++j) {
            scanf("%d",&stu[i].score[j]);
        }
    }
}

void sum(struct Student *stu,int n){
    for (int i = 0; i < n; ++i) {
        stu[i].sum=0;
        for (int j = 0; j < 3; ++j) {
            stu[i].sum+=stu[i].score[j];
        }
    }
    for (int i = 0; i < n; ++i) {
        printf("%s de zongfen shi :%4d\n",stu[i].name,stu[i].sum);
    }
}

void sort(struct Student *stu,int n){
    struct Student max;
    for (int j = 0; j < n-1; ++j)
    {
        for (int i = 0; i < n-j-1; ++i) {
            if (stu[i].sum < stu[i + 1].sum) {
                max = stu[i];
                stu[i] = stu[i + 1];
                stu[i+1]= max;
            }
        }
    }

    for (int i = 0; i < 5; ++i) {
        if(i==0)
            stu[i].count=1;
        else if(stu[i].sum==stu[i-1].sum)
            stu[i].count=stu[i-1].count;
        else
            stu[i].count=stu[i-1].count+1;

        printf("\nNO .%d  : %4d%4s%4d%4d%4d%4d",stu[i].count,stu[i].num,stu[i].name,stu[i].score[0],stu[i].score[1],stu[i].score[2],stu[i].sum);

    }
}
void readdata(struct Student *stu){
    FILE *fp;
    fp= fopen("student.txt","w");
    if(NULL==fp){
        printf("can not open the file!");
        exit(-1);
    }
    for (int i = 0; i < 5; ++i) {
        fprintf(fp,"\nNO .%d : %4s%4d%4d%4d%4d",stu[i].count,stu[i].name,stu[i].score[0],stu[i].score[1],stu[i].score[2],stu[i].sum);
    }
}
int main() {
    struct Student stu[5];
    input(stu);
    sum(stu,5);
    sort(stu,5);
    readdata(stu);
    return 0;
}