这个是我们的作业要求.....
1.实现数据输入,假设某班有N个学生,每个学生有M门课程,从文件中输入学生学号,姓名,M门课程成绩.
2.计算学生平均成绩,并按学生的平均成绩排序.
3.实现数据输出,输出学生原始成绩表(学号,姓名,成绩),输出学生平均成绩表(学号,姓名,平均成绩),输出按平均成绩排序后的成绩表(学号,姓名,平均成绩),输出到屏幕的同时输出到文件中.
4.按模块化程序设计要求将数据输入,数据处理,数据输出分别用不用的函数实现

我QQ:372712337......麻烦帮忙把文件和程序都发过来.....谢谢


急!

这个是第1次的作业`~~~~~~.应该会有用的


#include<stdio.h>
#define N 8
void main()
{
    int s[N],score[N][4];
    double average[N],t;
    char title[][10]={"学号","姓名","数学","物理","英语","计算机","平均成绩"};
    char xuehao[N][10],name[N][8];
    int k=0,a,b;

    for(;k<N;k++)
    {
        printf("\ninput xuehao %d:",k+1);
    scanf("%s",xuehao[k]);
    printf("\ninput name %d:",k+1);
    scanf("%s",name[k]);
    printf("\ninput score[%d][0] score[%d][1] scre[%d][2] score[%d][3]:",k,k,k,k);
    scanf("%d%d%d%d",&score[k][0],&score[k][1],&score[k][2],&score[k][3]);
    }
    printf("\n%-10s%-10s%-6s%-6s%-6s%-8s",
        title[0],title[1],title[2],title[3],title[4],title[5],title[6]);
    printf("\n-------------------------------------------------------------------");
    for(k=0;k<N;k++)
        printf("\n%-10s%-8s%4d %4d %4d %6d ",xuehao[k],name[k],score[k][0],score[k][1],score[k][2],score[k][3]);
for(k=0;k<N;k++)
    average[k]=(score[k][0]+score[k][1]+score[k][2]+score[k][3])/4.0;
   
        printf("\n%-10s%-10s%-6s%-6s%-6s%-8s",title[0],title[1],title[2],title[3],title[4],title[5],title[6]);
    printf("\n-------------------------------------------------------------------");
    for(k=0;k<N;k++)
            printf("\n%-10s%-8s%4d %4d %4d %6d %7.1f ",xuehao[k],name[k],score[k][0],score[k][1],score[k][2],score[k][3],average[k]);
for(k=0;k<N;k++)
        s[k]=k;
for(a=1;a<N;a++)
for(b=0;b<N-a;b++)
if(average[b]>average[b+1])
{
    t=average[b];
    average[b]=average[b+1];
    average[b+1]=t;
k=s[b];s[b]=s[b+1];s[b+1]=k;
}   
printf("\n%-10s%-10s%-6s%-6s%-6s%-8s%-8s",title[0],title[1],title[2],title[3],title[4],title[5],title[6]);
        printf("\n-------------------------------------------------------------------");
for(k=0;k<N;k++)
printf("\n%-10s%-8s%4d %4d %4d %6d %7.1f",xuehao[s[k]],name[s[k]],score[s[k]][0],score[s[k]][1],score[s[k]][2],score[s[k]][3],average[k]);
}