实验2

task_1:
#include <stdio.h>
#include <stdlib.h>
#include <time.h>

#define N 100
#define CNT 5

int main() {
    int number;
  
    int i;

    srand(time(0));
  
    for(i = 0; i < CNT; ++i) {
        number = rand() % N + 1;
        printf("20490042%04d\n", number);
    }

    system("pause");
    return 0;
}

 

  运行结果:t1

问题1:程序随机生成 5 个 1~100 之间的整数,分别拼接到学号前缀 "20490042" 后面,输出 5 个学员编号

问题2:number = rand() % N + 1; —— rand()产生 0~RAND_MAX 的伪随机数,对 N (=100) 取余得到 0~99,再加 1 得到 1~100 之间的随机整数

问题3:%04d表示以 4 位宽度输出整数,不足 4 位时前面补 0。例如随机数是 5,输出就是0005,保证学号后四位始终占 4 个字符宽度

问题4:当前系统时间作为随机数种子初始化 rand ()。如果去掉这行,每次运行程序生成的随机数序列完全相同(因为默认种子固定为 1),多次运行结果不会变化。加上后,每次运行时间不同,随机序列也不同

 

task_2:

#include <stdio.h>

int main() {
    int choice, quantity;
    float total_price = 0, amount_paid, change;

    while (1) {
        printf("\n自动饮料售卖机菜单:\n");
        printf("1. 可乐 - 3 元/瓶\n");
        printf("2. 雪碧 - 3 元/瓶\n");
        printf("3. 橙汁 - 5 元/瓶\n");
        printf("4. 矿泉水 - 2 元/瓶\n");
        printf("0. 退出购买流程\n");
        printf("请输入饮料编号:");
        scanf("%d", &choice);

        if (choice == 0)
            break;

        if (choice < 1 || choice > 4) {
            printf("无效的饮料编号,请重新输入。\n");
            continue;
        }

        printf("请输入购买的数量:");
        scanf("%d", &quantity);

        if (quantity < 0) {
            printf("购买数量不能为负数,请重新输入。\n");
            continue;
        }

        if (choice == 1 || choice == 2)
            total_price += 3 * quantity;
        else if (choice == 3)
            total_price += 5 * quantity;
        else
            total_price += 2 * quantity;

        printf("请投入金额:");
        scanf("%f", &amount_paid);

        change = amount_paid - total_price;
        printf("本次购买总价:%.2f 元\n", total_price);
        printf("找零:%.2f 元\n", change);

        total_price = 0;
    }

    printf("感谢您的购买,欢迎下次光临!\n");

    return 0;
}

 

  运行截图:t2

问题1:total_price = 0;在每次完成一笔购买后重置总价。如果去掉,下一次购买时total_price会累加上次的金额,导致总价错误、找零错误

问题2:continue表示跳过本次循环剩余语句,直接进入下一次循环判断。两处 continue 分别在:①输入了无效饮料编号时,跳过后续购买流程,直接重新显示菜单;②输入了负数数量时,同样跳过后续步骤,重新让用户输入

 

task_3:

#include<stdio.h>

int main(){
    
    int i;
    
    char n;
    
    for(i=1;i<=4;++i){
        printf("请输入信号灯颜色:");
        scanf(" %c",&n);
    
    switch(n){
        case 'r':printf("stop!\n");break;
        
        case 'g':printf("go go go\n");break;
        
        case 'y':printf("wait a minute\n");break;
        
        default:printf("something must be wrong\n");
        break;
        
        
    }

 

    运行截图:t3

 

task_4:

#include <stdio.h>

int main() {
    double expense;
    double sum = 0, max = 0, min = 0;
    int first = 1;

    printf("输入今日开销,直到输入-1终止:\n");

    while (1) {
        scanf("%lf", &expense);
        
        if (expense == -1)
        
            break;

        sum += expense;

        if (first) {
            max = expense;
            min = expense;
            first = 0;
        } 
        else {
            if (expense > max)
                max = expense;
            if (expense < min)
                min = expense;
        }
    }

    printf("今日累计消费总额:%.1f\n", sum);
    
    printf("今日最高一笔开销:%.1f\n", max);
    
    printf("今日最低一笔开销:%.1f\n", min);

    return 0;
}

运行截图:

t4

 

task_5:

#include <stdio.h>

int main() {
    int a, b, c;
    int i;

    for (i = 0; i < 5; i++) {
        scanf("%d %d %d", &a, &b, &c);


        if (a + b <= c || a + c <= b || b + c <= a) {
            printf("不能构成三角形\n");
        } 
        else if (a == b && b == c) {
            printf("等边三角形\n");
        } 
        else if (a * a + b * b == c * c ||
                   a * a + c * c == b * b ||
                   b * b + c * c == a * a) {
            printf("直角三角形\n");
        } 
        else if (a == b || b == c || a == c) {
            printf("等腰三角形\n");
        } 
        else {
            printf("普通三角形\n");
        }
    }


    return 0;
}

运行截图:t5

 

task_6:

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

int main() {
    int lucky_day, guess;
    int i;

    srand(time(0));
    lucky_day = rand() % 30 + 1;
    
    

    printf("猜猜2026年11月哪一天是你的lucky day\n");

    for (i = 0; i < 3; i++) {
        if (i == 0)
            printf("开始喽,你有3次机会,猜吧(1~30):");
        else
            printf("再猜(1~30):");

        scanf("%d", &guess);

        if (guess == lucky_day) {
            printf("哇,猜中了:)\n");
            system("pause");
            return 0;
        } 
        else if (guess < lucky_day) {
            printf("你猜的日期早了,你的lucky day还没到呢\n");
        } 
        else {
            printf("你猜的日期晚了,你的lucky day在前面哦\n");
        }
    }

    printf("次数用光啦。11月你的lucky day是%d号\n", lucky_day);

    return 0;
}

运行截图:t6

 

posted @ 2026-11-03 20:37  仅予你的晴天  阅读(3)  评论(0)    收藏  举报