实验2

实验任务1

#include <stdio.h>
#include <stdlib.h>
#include <time.h>
#define N 5
#define N1 80
#define N2 35

int main() {
int cnt;
int random_major, random_no;
srand(time(NULL));
cnt = 0;
while(cnt < N) {
random_major = rand() % 2;
if(random_major) {
random_no = rand() % N1 + 1;
printf("20256343%04d\n", random_no);
}
else {
random_no = rand() % N2 + 1;
printf("20256136%04d\n", random_no);
}
cnt++;
}
return 0;
}

实验结论:1.删去代码srand(time(NULL));会使最终运行的结果保持不变，而源代码中srand(time(NULL));可以让程序根据系统中的时间来改变运行结果

               2.根据系统中的时间来随机抽取学号，从而避免结果重复。

实验任务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;
}

实验结论2

1.若删去47行，则会使该程序在下一次循环中商品的总价钱加上前面全部商品的总价钱进行运算，因此第47行的作用是将本循环中商品的总价钱清零

2.终止当前循环的剩余执行过程，跳转至下一循环重新执行该循环。

实验任务3

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

    while(1)
    {
        char m;
        scanf("%c",&m);
        getchar();
        if(m == 'y')
        {printf("wait a minute\n");
        }
        else if(m == 'g')
        {printf("go go go\n");
        }
        else if(m == 'r'){printf("stop\n");
        }
        else{printf("something must be wrong\n");
        }
    }
    return 0;
}

实验任务4

#include <stdio.h>
int main()
{
    
        double expense;
        double total;
        double maxExpense = -1;
        double minExpense = 20000.1;
        printf("输入今日开销，直到输入-1终止:\n");
        while(1)
        {
            scanf("%lf",&expense);
            if(expense == -1)
            {break;
            }
            if(expense <= 0 || expense>20000){printf("每笔消费需满足0元<每笔消费<=2万元，请重新输入:\n");
            continue;
            }
            total += expense;
            if(expense > maxExpense){maxExpense = expense;
            }
            if(expense<minExpense){minExpense = expense;
            }
        }
        
    
 
 printf("今日累计消费总额:%.1f\n",total);
 printf("今日最高一笔消费:%.1f\n",maxExpense);
 printf("今日最低一笔消费:%.1f\n",minExpense);
 return 0;
}

运行结果

 

 

 

 

 

实验任务5

#include <stdio.h>
int main()
{
while(1)
{
int a,b,c;
scanf("%d%d%d",&a,&b,&c);
getchar();

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


else{printf("普通三角形\n");
}

}
return 0;
}

运行结果

实验任务6

#include <stdio.h>
#include <stdlib.h>
#include <time.h>
int main()
{
    srand((unsigned int)time(NULL));
    int lucky_day = rand() % 30 + 1;
    int guess, count = 0;
    printf("猜猜2025年11月哪一天是你的lucky day:\n");
    printf("开始喽，你有三次机会，猜吧(1—30):\n");
    while (count < 3)
    {
        scanf_s("%d", &guess);
        if (guess < lucky_day) {
            printf("你猜的日期早了，你的lucky_day还没到呢\n");
        }
        else if (guess > lucky_day) {
            printf("你猜的日期晚了，你的lucky_day在前面哦\n");
        }
        else {
            printf("哇，猜中了:)\n");
            break;
        }
        count += 1;

    }
    printf("次数用光了。偷偷告诉你，11月你的lucky_day是%d号\n", lucky_day);
    return 0;
}

运行结果