实验三
正确输出了按分数由高→低排序的信息,同时,在当前路径下,生成了文本文件file3.dat。
实验四
子任务一:在屏幕上正确输出了按分数由高到底排序的学生信息。
同时,在当前路径下,生成了二进制文件file4.dat。
用记事本程序打开文件file4.dat,里面的数据信息不直观可读。
子任务二:
#include <stdio.h>
#include <stdlib.h>
#define N 10
// 定义一个结构体类型STU
typedef struct student {
int num;
char name[20];
int score;
}STU;
void sort(STU *pst, int n);
int main(){
FILE *fin;
STU st[N];
int i;
fin = fopen("file4.dat", "rb");
if(!fin){
printf("fail to open file4.dat\n");
exit(0);
}
fread(st,sizeof(STU),N,fin);
fclose(fin);
sort(st,N);
for(i=0; i<N; i++)
printf("%-6d%-10s%3d\n", st[i].num, st[i].name, st[i].score);
return 0;
}
void sort(STU *pst, int n) {
STU *pi, *pj, t;
for(pi = pst; pi < pst+n-1; pi++)
for(pj = pi+1; pj < pst+n; pj++)
if(pi->score < pj->score) {
t = *pi;
*pi = *pj;
*pj = t;
}
}
#include <stdio.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%d个考生信息: 准考证号,姓名,客观题得分(<=40),操作题得分(<=60)\n", N); input(stu, N); printf("\n对考生信息进行处理: 计算总分,确定等级\n"); process(stu, N); printf("\n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\n"); output(stu, N); return 0; } void input(STU s[], int n) { FILE* fin; int i; fin = fopen("examinee.txt", "r"); for (i = 0; i < n; i++) { fscanf(fin, "%ld %s %f %f",&s[i].id, &s[i].name, &s[i].objective, &s[i].subjective); } } void output(STU s[], int n) { int i; int a[N]; FILE* fout; printf("-------------------------------"); printf("\n"); printf("准考证号\t姓名\t客观题得分\t操作题得分\t总分\t等级\t"); printf("\n"); for (i = 0; i < n; i++) { printf("%ld\t%s\t%.2f\t%.2f\t%.2f\t%s", s[i].id, s[i].name, s[i].objective, s[i].subjective, s[i].sum, s[i].level); printf("\n"); } fout = fopen("result.txt", "w+"); fprintf(fout,"准考证号\t姓名\t客观题得分\t操作题得分\t总分\t等级\t\n"); printf("\n\n\n"); for (i = 1; i < n; i++) { fprintf(fout,"%ld\t%s\t%.2f\t%.2f\t%.2f\t%s\n", s[i].id, s[i].name, s[i].objective, s[i].subjective, s[i].sum, s[i].level); } void process(STU s[], int n){ int i,j,k; STU temp; for(i=0;i<n;i++){ s[i].sum=0.4*s[i].objective+0.6*s[i].subjective; } for(i=0;i<n-1;i++) for(j=0;j<n-1-i;j++) if(s[j].sum<s[j+1].sum) { temp = s[j]; s[j] = s[j+1]; s[j+1] = temp; } for(i=0;i<n;i++){ if(i<1) strcpy(s[i].level,"优秀"); else if(i<5&&i>0) strcpy(s[i].level,"合格"); else strcpy(s[i].level,"不合格"); } }