实验2

task1.c
源代码:

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:生成一个1-100之间的随机整数,并赋值到number里面
问题2:将number输出4个字符宽度的十进制整数
问题3:生成并输出5个介于1-100之间的随机整数
运行结果截图:

task2.c
源代码:

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:去掉的话,total_price的值会保留上一次购买的总价,下一次购买时,新的总价会在上一次的基础上累加,导致结果不准确。
问题2:break是跳出当前循环结构,不再执行循环内的代码。continue是跳出本次循环中剩余代码,直接进入下一次循环。
问题3:没必要,在进入switch语句之前,已经通过if(choice < 1 || choice > 4)进行了处理,确保了choice的值只能是1、2、3、4、0
运行结果截图:

task3
源代码:

include <stdio.h>

int main() {
char input;
while (scanf("%c", &input) != EOF) {
switch (input) {
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;
}
// 清除输入缓冲区的换行符等多余字符
while ((getchar()) != '\n');
}
return 0;
}
运行结果截图:



task4
源代码:#include <stdio.h>

int main() {
double expense, total = 0, max = 0, min = 20000;
int flag = 0;

printf("输入今日开销,直到输入-1终止:\n");
while (1) {
    scanf("%lf", &expense);
    if (expense == -1) {
        break;
    }
    total += expense;
    if (!flag) {
        max = min = expense;
        flag = 1;
    } else {
        if (expense > max) {
            max = expense;
        }
        if (expense < min) {
            min = expense;
        }
    }
}

printf("今日累计消费总额: %.1f\n", total);
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;

}
运行结果截图:

task5
源代码:#include <stdio.h>

include <stdlib.h>

include <time.h>

int main() {
int luckyDay, guess, chance = 3;
srand(time(0));
luckyDay = rand() % 30 + 1;

printf("你有3次猜4月lucky day的机会,请开始猜测(1 - 30之间的数字):\n");
while (chance > 0) {
    scanf("%d", &guess);
    chance--;
    if (guess == luckyDay) {
        printf("你猜对了!\n");
        return 0;
    } else if (guess < luckyDay) {
        printf("猜早了,再试试。你还有 %d 次机会\n", chance);
    } else {
        printf("猜晚了,再试试。你还有 %d 次机会\n", chance);
    }
}
printf("你用光了三次机会,4月的lucky day是 %d\n", luckyDay);

return 0;

}
运行结果截图:

task6
源代码:#include <stdio.h>

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

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

return 0;

}

运行结果截图:
实验总结:
本次实验遇见的问题比较多,在函数的应用方面还需要改进,实验6中多用了if语句,造成结果不理想,对\t的使用还不是很熟练。总之需要提高的地方还有很多。

posted @ 2025-03-23 00:11  阿晏ing  阅读(33)  评论(0)    收藏  举报