实验2

实验任务一

源代码:

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

#define N 5

int main() {
    int number;
    int i;

    srand(time(0));    
    for (i = 0; i < N; ++i) {
        number = rand() % 100 + 1;
        printf("20240042%04d\n", number);
    }

    return 0;
}

运行截图:

回答问题:
1.生成一个介于1到100之间的随机整数,并存储在变量number里
2.输入数据为整型
3.生成随机202400420001~202400420100的编号

实验任务二

源代码:

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#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_s("%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_s("%d", &quantity);

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

        switch (choice) {
            case 1:
            case 2:
                total_price += 3 * quantity;
                break;
            case 3:
                total_price += 5 * quantity;
                break;
            case 4:
                total_price += 2 * quantity;
                break;
        }

        printf("请投入金额: ");
        scanf_s("%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:

回答问题:
1.方便下一次重新从零计数,如果删掉,下一次计数会累计计算
2.break之后是结束循环,continue是重新开始循环
3.没有必要,在输入switch语句前已经对数据进行了筛选与验证,不需要default语句再来处理

实验任务三

源代码:

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#include <stdio.h>

int main() {
    char color;
    char r, g, y;
    r = 'r';
    g = 'g';
    y = 'y';

    while (1 ) {
        color = getchar();
         

        scanf_s("%c", &color);

        if (color == r) {
            printf("stop!\n");
            continue;
        }

         else if (color == g) {
            printf("go go go\n");
            continue;
        }

         else if (color == y) {
            printf("wait a minute\n");
            continue;
        }

         else
        {
            printf("something must be wrong...");
        }

    }

    printf("感谢您的购买,欢迎下次光临!\n");
    return 0;
}

运行截图:

实验任务四

源代码:

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#include <stdio.h>

int main() {
    double expense, max = 0, min = 200000, total = 0;
    printf("输入今日开销,直到输入 - 1终止:\n");
    while (1) {
        scanf_s("%lf", &expense);
        if (expense == -1){
            break;
        }
        total += expense;
        if (expense > max){
            max = expense;
        }
        if (expense < min && expense > 0){
            min = expense;
        }   
    }
    printf("今日累计消费总额:%.lf\n", total);
    printf("今日最高一笔开销:%.lf\n", max);
    printf("今日最低一笔开销:%.lf\n", min);
    return 0;
}

运行截图:

试验任务五

源代码:

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

int main() {
    int day,guess;
    int chance = 3;
    srand((unsigned int)time(NULL));
    day = rand() % 30 +1;

    printf("猜猜2025年4月哪一天是你的lucky day\n");
    printf("开始喽,你有三次机会,猜吧(1-30):");

    while (chance>0) {
        scanf_s("%d",&guess);
        if (guess < day)
        {
            printf("你猜的日期早了,你的lucky day还没到呢\n");
            chance--;
        }
        else if (guess > day)
        {
            printf("你猜的日期晚了,你的lucky day在前面哦\n");
            chance--;  
        }
        else
        {
            printf("哇,猜中了:-)");
            return 0;
        }
        
        if (chance>0)
        {
            printf("再猜(1-30):");
        }
    }
    printf("次数用完啦。偷偷告诉你,4月你的lucky day是%d号\n",day);
    return 0;
}

运行截图:

实验任务六

源代码:

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

int main()
{
    int n;
    int i, k, m;
    i = 1;
    printf("input n: ");
    scanf("%d", &n);
    while(i <= n )
    {
        m = 0;
        while(m < i - 1)
        {
            printf("\t");
            m = m + 1;
        }
        k = 0;
        while(k < 2 * (n + 1 - i) - 1)
        {
            printf(" o \t");
            k = k + 1;
        }
        printf("\n");
        m = 0;
        while(m < i - 1)
        {
            printf("\t");
            m = m + 1;
        }
        k = 0;
        while(k < 2 * (n + 1 - i) - 1)
        {
            printf("<H>\t");
            k = k + 1;
        }
        printf("\n");
        m = 0;
        while(m < i - 1)
        {
            printf("\t");
            m = m + 1;
        }
        k = 0;
        while(k < 2 * (n + 1 - i) - 1)
        {
            printf("I I\t");
            k = k + 1;
        }
        printf("\n");
        i = i + 1;
    }
    system("pause");
    return 0;
}

运行截图:

posted @ 2025-03-23 12:41  黄卓尔  阅读(13)  评论(0)    收藏  举报