实验2作业
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;
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++;
}
system("pause");
return 0;
}
问题1
会使第二次再购买饮料使的总价钱为第一次和第二次总价相加之和，使其购买总价计算错误
问题2
不再执行continue后面的语句，重新进行一次循环
任务3
#include <stdio.h>
#include<stdlib.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");
system("pause");
return 0;
}
任务4
#include#include#define _CRT_SECURE_NO_WARNINGSint main(){ float x,max=0,min=20000,ans=0; printf("输入今日开销，直到输入-1终止：\n"); scanf("%f",&x); while(x>0){ ans=ans+x; if(x>max) max=x; if(x
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