2.用二维数组存储若干名学生3门课程的考试成绩，计算每名学生的平均成绩（按右下角运行图实现编程）

#include<iostream>
using namespace std;

int main(){

    int n,m=0,i,j;
    printf("请输入学生的人数：");
    scanf("%d",&n);
    float score[n][3],sum,ave;
    for (i = 0; i < n; i++)
    {
        printf("请输入第%d名同学的三门功课成绩：",i+1);
        scanf("%f%f%f",&score[i][0],&score[i][1],&score[i][2]);
    }
    printf("-------------------------------------------------\n");
    printf("序号\t成绩1\t成绩2\t成绩3\t平均成绩\n");
    printf("-------------------------------------------------\n");
    for (i = 0; i < n; i++)
    {
        sum = 0;
        for (j = 0; j < 3; j++)
        {
            sum += score[i][j];
        }
        ave = sum / 3;
        printf("%d\t%5.1f\t%5.1f\t%5.1f\t%7.1f\n",i+1,score[i][0],score[i][1],score[i][2],ave);
    }
    return 0;
}

#include "stdio.h"

#define N 2
struct student{
    int id;
    char name[20];
    int kaoqun;
    int biaoxian;
    int zuoye;
    int biji;
    int sum;
}stud[N],t;

int main()
{
    int i,j;
    for(i=0;i<N;i++){
        printf("请输入第%d个学生信息\n",i+1);
        scanf("%d %s",&stud[i].id,&stud[i].name);
        fflush(stdin);
        scanf("%d %d %d %d",&stud[i].kaoqun,&stud[i].biaoxian,&stud[i].zuoye,&stud[i].biji);
        stud[i].sum = stud[i].kaoqun*0.3+stud[i].biaoxian*0.3+stud[i].zuoye*0.3+stud[i].biji*0.1;
    }
    
    //排序
    
    for (i = 0; i < N - 1; i++){
        for (j = 0; j < N - 1 - i; j++){ //按成绩对学生信息进行排序
            if (stud[j].sum > stud[j + 1].sum){ //整型数字的比较
                t = stud[j];
                stud[j] = stud[j + 1];
                stud[j + 1] = t;
            }
        }
    }
    //打印
    printf("学号\t姓名\t考勤\t表现\t作业\t笔记\t总分\t\n");
    for (i = 0; i < N; i++){
        printf("%d\t", stud[i].id);
        printf("%s\t", stud[i].name);
        printf("%d\t", stud[i].kaoqun);
        printf("%d\t", stud[i].biaoxian);
        printf("%d\t", stud[i].zuoye);
        printf("%d\t", stud[i].biji);
        printf("%d \n", stud[i].sum);
    }
    
    return 1;
}