实验七

Posted 2020-11-26 saberzyk

Part1

结果正确

区别：

二进制文件需要特别的解码器，文本文件可以读写。

#include <stdio.h> 
#include <stdlib.h>

#define N 10
typedef struct student {
    int num;
    char name[20];
    int score;
}STU;


int main() {
    FILE *fin;
    STU st[N];
    int i;
     
    fin = fopen("file4.dat", "rb");
    if( !fin ) {  
        printf("fail to open file4.dat
");
        exit(0);
    }
    
    
    for(i=0; i<N; i++)
        fread(&st,sizeof(STU),N,fin);
    
    for(i=0; i<N; i++) 
        printf("%-6d%-10s%3d
", st[i].num, st[i].name, st[i].score);    
    return 0;
}

　　

Part2

#include <stdio.h>
#include <stdlib.h>
#include <string.h>
const int N = 10;


typedef struct student {
	long int id;
	char name[20];
	float objective;	
	float subjective;	
	float sum;
	char level[10];	
}STU; 


void input(STU s[], int n);
void output(STU s[], int n);
void process(STU s[], int n);

int main() {
	STU stu[N];
	
	printf("录入%个考生信息：准考证号，姓名，客观题得分（<=40),操作题得分（<=60)
", N); 
	input(stu, N);
	
	printf("
对考生信息进行处理：计算总分，确定等级
");
	process(stu, N);
	
	printf("
打印考生完整信息：准考证号，姓名，客观题得分，操作题得分，总分，等级
");
	output(stu, N); 
	
	system("pause");
	return 0;
} 

void input(STU s[], int n) {
	FILE *fin;
	int i; 
	fin=fopen("d:\\examinee.txt","r");
	if(!fin)
	{
		printf("fail to open examinee.txt
");
		exit(0);
	}
	for(i=0;i<n;i++)
	
	
		fscanf(fin,"%ld %s %f %f",&s[i].id,s[i].name,&s[i].objective,&s[i].subjective);
		fclose(fin);

}

void output(STU s[], int n) {
	
     FILE *fout;
     int i;
     fout=fopen("result.txt","w");
     if(!fout)
     {
     	printf("fail to open result,txt
");
     	exit(0);
	 }
	 fprintf(fout,"准考证号，姓名，客观题得分，操作题得分，总分，等级
"); 
	 for(i=0;i<n;i++)
	 {
	 	printf("%ld %s %f %f %f %s
",s[i].id,s[i].name,s[i].objective,s[i].subjective,s[i].sum,s[i].level);
	 	fprintf(fout,"%ld %s %f %f %f %s
",s[i].id,s[i].name,s[i].objective,s[i].subjective,s[i].sum,s[i].level);
	 	fclose(fout);
	 }
}


void process(STU s[], int n) 
{
	int i,j,m;
    STU t;
    
    for(i=0;i<n;i++)
    {
        s[i].sum = s[i].subjective + s[i].objective;
    }
    
    for(i=0;i<n-1;i++)
    {
        for(j=0;j<n-i-1;j++)
        {
            if(s[j].sum<s[j+1].sum)
            {
                t = s[j];
                s[j] = s[j+1];
                s[j+1] = t;
            }
        }
    }
    
    for(m=0;m<n; m++)
    {
        if(m<n*0.1)
        strcpy(s[m].level,"优秀");
        if(m>=n*0.1 && m<n*0.5)
        strcpy(s[m].level,"合格");
        if(m>=n*0.5)
        strcpy(s[m].level,"不合格");        
    }
}