实验2

task1

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

#define N 5

int main() {
    int number;
    int i;

    srand(time(0));     // 以当前系统时间作为随机种子
    for(i = 0; i < N; ++i) {
        number = rand() % 100 + 1;
        printf("20490042%04d\n", number);
    }

    return 0;
}

问题1：生成1~100随机整数

问题2：格式化number为至少4位宽度，不足四位时前面补0

问题3：随机生成5个学号，且学号只有后两位不同

task2

#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;
        }

        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("%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;
}

问题1：重置购买总价为0，如果去掉，上一次运行的购买总价会加入到下一次运行中

问题2：break会跳出整个循环且不再执行任何循环；continue则会跳过循环剩余部分执行下个循环

问题3：没必要，default子句是为了防止出现1~4以外的case，但是line17~30已经防止了这种情况的发生

 

task3

#include <stdio.h>
int main() {
    char ans;

    while (1) {
        ans = getchar();
        getchar();
        switch (ans) {
        case'g':printf("go go go\n");
            break;
        case'r':printf("stop\n");
            break;
        case'y':printf("wait a minute\n");
            break;
        default:
            printf("something must be wrong...\n");
            break;
        }
    }
    return 0;
}

task4

#include <stdio.h>

int main() {
    double ex, min = 20000, max = 0, totalEx = 0;
    int i = 1;

    printf("输入今日开销，直到输入 -1 终止：\n");

    while (1) {
        scanf_s("%lf", &ex);
        if (ex == -1) {
            break;
        }
        if (ex <= 0 || ex > 20000) {
            printf("输入的开销必须大于0且不超过20000元。\n");
            continue;
        }

        if (i) {
            max = min = ex;
            i = 0;
        }
        else {
            if (ex > max) {
                max = ex;
            }
            if (ex < min) {
                min = ex;
            }
        }
        totalEx += ex;
    }

    printf("今日累计消费总额：%.1f\n", totalEx);
    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;
}

 

task5

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

int main() {
    int luckyday, guess;
    int guesscount = 0;
    srand((unsigned int)time(NULL));
    luckyday = rand() % 30 + 1;
    printf("猜猜2025年4月哪一天是你的luckyday\n");
    printf("开始喽，你有三次机会，猜吧（1~30）：");
    while (guesscount < 3) {
        scanf_s("%d", &guess);
        guesscount++;
        if (guess == luckyday) {
            printf("哇，猜中了:-)\n");
            return 0;
        }
        else if (guess < luckyday)
            printf("你猜的日期早了，你的luckday还没到呢\n");
        else
            printf("你猜的日期晚了，你的luckday在前面哦\n");
        if (guesscount < 3)
            printf("再猜（1~30）：");
    }
    printf("次数用完啦。偷偷告诉你，4月你的luckyday是%d号\n", luckyday);
    return 0;
    }

task6

#include <stdio.h>

int main() {
    int n, i, j;
    printf("Input n: ");
    scanf_s("%d", &n);

    for (i = n; i >= 1; i--) {
        for (j = 0; j < (n - i); j++) {
            printf(" ");
        }
        for (j = 0; j < 2*i-1; j++) {
            printf(" 0 ");
            printf(" ");
        }
        printf("\n");
        for (j = 0; j < (n - i); j++) {
            printf(" ");
        }
        for (j = 0; j < 2*i-1; j++) {
            printf("<H>");
            printf(" ");
        }
        printf("\n");
        for (j = 0; j < (n - i); j++) {
            printf(" ");
        }
        for (j = 0; j < 2*i-1; j++) {
            printf("I I");
            printf(" ");
        }
        printf("\n");
    }
    return 0;
}