实验七

task 1

屏幕截图 2025-12-31 193652

task 2

屏幕截图 2025-12-31 194021

task 3

屏幕截图 2025-12-31 194252

task 4

#include <stdio.h>
#include <ctype.h>

int main() {
 
    FILE *file;
   
    file = fopen("data4.txt", "r");
    if (file == NULL) {
        printf("无法打开文件 data4.txt\n");
        return 1;
    }

    int line_count = 0;
    int char_count = 0;
    char ch;

    while ((ch = fgetc(file)) != EOF) {

        if (ch == '\n') {
            line_count++;
        }
        else if (!isspace(ch)) {
            char_count++;
        }
    }

    if (line_count > 0 || char_count > 0) {
        line_count++;
    }

    fclose(file);

    printf("data4.txt 统计结果:\n");
    printf("行数:%d\n", line_count);
    printf("字符数(不计空白符):%d\n", char_count);

    return 0;
}屏幕截图 2025-12-31 194949

 


task 5

#include <stdio.h>
#include <string.h>

#define N 10

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

// 函数声明
void read(STU st[], int n);
void write(STU st[], int n);
void output(STU st[], int n);
int process(STU st[], int n, STU st_pass[]);

int main() {
    STU stu[N], stu_pass[N];
    int cnt;
    double pass_rate;

    printf("从文件读入%d个考生信息...\n", N);
    read(stu, N);

    printf("\n对考生成绩进行统计...\n");
    cnt = process(stu, N, stu_pass);

    printf("\n通过考试的名单:\n");
    output(stu, N);  
    write(stu_pass, cnt);   

    pass_rate = 1.0 * cnt / N;
    printf("\n本次等级考试通过率: %.2f%%\n", pass_rate * 100);

    return 0;
}

 
void output(STU st[], int n) {
    int i;
    printf("准考证号\t姓名\t客观题得分\t操作题得分\t总分\t\t结果\n");
    for (i = 0; i < n; i++) {
        printf("%ld\t%s\t%.2f\t\t%.2f\t\t%.2f\t\t%s\n", st[i].id, st[i].name,
               st[i].objective, st[i].subjective, st[i].sum, st[i].result);
    }
}
 
void read(STU st[], int n) {
    int i;
    FILE *fin;

    fin = fopen("examinee.txt", "r");
    if (!fin) {
        printf("fail to open file\n");
        return;
    }

    for (i = 0; i < n; i++) {
        fscanf(fin, "%ld %s %f %f", &st[i].id, st[i].name, &st[i].objective,
               &st[i].subjective);
    }
    fclose(fin);
}

 
void write(STU s[], int n) {
    int i;
    FILE *fout;
    fout = fopen("list_pass.txt", "w");
    if (!fout) {
        printf("fail to open list_pass.txt\n");
        return;
    }
 
    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\t结果\n");
    for (i = 0; i < n; i++) {
        fprintf(fout, "%ld\t%s\t%.2f\t\t%.2f\t\t%.2f\t\t%s\n",
                s[i].id, s[i].name, s[i].objective, s[i].subjective,
                s[i].sum, s[i].result);
    }
    fclose(fout);
}
 
int process(STU st[], int n, STU st_pass[]) {
    int i, pass_cnt = 0;
   
    for (i = 0; i < n; i++) {
     
        st[i].sum = st[i].objective + st[i].subjective;
     
        if (st[i].sum >= 60) {
            strcpy(st[i].result, "通过");
            st_pass[pass_cnt++] = st[i];
        } else {
            strcpy(st[i].result, "不通过");
        }
    }
 
    return pass_cnt;
}

屏幕截图 2025-12-31 195339

task 6

#include <stdio.h>
#include<stdlib.h>
#include<time.h>
#include<string.h>
#define N 100
#define M 5
int main(){
    FILE * fp1,*fp2;
    char f[N];
    char name[N][N];
    char x[N][N];
    int n;
    int y[N] = { 0 };
    int random;
    fp1 = fopen("C:\\Users\\SYX\\Desktop\\实验7\\list.txt","r");
    if (!fp1) {
        perror("list.txt");
        return 1;
    }
    int i = 0;
    while (fgets(x[i], N, fp1) != NULL) ++i;
    n = i;
    srand(time(0));
    for (int i = 0; i < M; ) {
        random = rand() % n;
        if (y[random])
            continue;
        y[random] = 1;
        printf("%s", x[random]);
        strcpy(name[i], x[random]);
        ++i;
    }
    printf("请输入文件名:");
    gets_s(f);
    fp2 = fopen(f, "w");
    if (!fp2) {
        perror(f);
        return 1;
    }
    for (int i = 0; i <M ;++i) {
        fprintf(fp2, "%s", name[i]);
    }
    fclose(fp1);
    fclose(fp2);
    printf("保存成功");
    return 0;
}

 

posted @ 2025-12-31 19:56  菲洛特拉托  阅读(0)  评论(0)    收藏  举报