这篇文章将为大家详细讲解有关如何使用C语言实现自动售货机,小编觉得挺实用的,因此分享给大家做个参考,希望大家阅读完这篇文章后可以有所收获。
具体内容如下
如图所示的简易自动售货机,物品架1、2上共有10样商品,按顺序进行编号分别为1-10,标有价格与名称,一个编号对应一个可操作按钮,供选择商品使用。如果物架上的商品被用户买走,储物柜中会自动取出商品送到物架上,保证物品架上一定会有商品。用户可以一次投入较多钱币,并可以选择多样商品,售货机可以一次性将商品输出并找零钱。
用户购买商品的操作方法:
(1)从“钱币入口”放入钱币,依次放入多个硬币或纸币。钱币可支持1元(纸币、硬币)、2元(纸币)、5元(纸币)、10元(纸币),放入钱币时,控制器会先对钱币进行检验识别出币值,并统计币值总额,显示在控制器显示屏中,提示用户确认钱币放入完毕;
(2)用户确认钱币放入完毕,便可选择商品,只要用手指按对应商品外面的编号按钮即可。每选中一样商品,售货机控制器会判断钱币是否足够购买,如果钱币足够,自动根据编号将物品进行计数和计算所需钱币值,并提示余额。如果钱币不足,控制器则提示“Insufficient money”。用户可以取消购买,将会把所有放入钱币退回给用户。
输入格式:
先输入钱币值序列,以-1作为结束,再依次输入多个购买商品编号,以-1结束。
输出格式:
输出钱币总额与找回零钱,以及所购买商品名称及数量。
输入样例:
1 1 2 2 5 5 10 10 -1
1 2 3 5 1 6 9 10 -1
输出样例:
Total:36yuan,change:19yuan
Table-water:2;Table-water:1;Table-water:1;Milk:1;Beer:1;Oolong-Tea:1;Green-Tea:1;
解法一:
#include<stdio.h> int mainvoid) { char a[10][20] = {"Table-water","Table-water","Table-water","Coca-Cola","Milk","Beer","Orange-Juice","Sprite","Oolong-Tea","Green-Tea"}; int b[11] = {0,0,0,0,0,0,0,0,0,0,0}; int c[50]; int i=1, k, sum = 0, money, SUM = 0, change, flag=0; scanf"%d",&money);//输入币值并计算总币值 whilemoney!=-1)&&money <= 10)) { sum = sum + money; scanf"%d",&money); } scanf"%d",&c[i]); whilec[i]!=-1)//将选的货物编号存储到数组c中并计算找零 { switchc[i]) { case 1: case 2: case 3: SUM = SUM + 1;break; case 4: case 5: SUM = SUM + 2;break; case 6: case 7: case 8: SUM = SUM + 3;break; case 9: case 10: SUM = SUM + 4;break; default:break; } ifSUM>sum) { printf"Insufficient money"); flag = 1; break; } i++; scanf"%d",&c[i]); } change = sum-SUM; //用数组b统计各种商品数量 i = 1; whilec[i]!=-1) { switchc[i]) { case 1: b[1]++;break; case 2: b[2]++;break; case 3: b[3]++;break; case 4: b[4]++;break; case 5: b[5]++;break; case 6: b[6]++;break; case 7: b[7]++;break; case 8: b[8]++;break; case 9: b[9]++;break; case 10: b[10]++;break; default:break; } i++; } //输出结果 ifflag==0) { printf"Total:%dyuan,change:%dyuan\n",sum, change); fori=1; i<=10; i++) { ifb[i]==0) continue; else { printf"%s:%d;",a[i-1],b[i]); } } } return 0; }
解法二:
#include"stdio.h" struct goods { int num; char name[20]; int price; int amount; }; int main) { struct goods good[10]= { {1,"Table-water",1,0}, {2,"Table-water",1,0}, {3,"Table-water",1,0}, {4,"Coca-Cola",2,0}, {5,"Milk",2,0}, {6,"Beer",3,0}, {7,"Orange-Juice",3,0}, {8,"Sprite",3,0}, {9,"Oolong-Tea",4,0}, {10,"Green-Tea",4,0} }; int sum=0,num,change,total=0,money=0,i=0; while1) { scanf"%d",&money); ifmoney==-1) { break; } sum=sum+money; } while1) { scanf"%d",&num); ifnum==-1) { break; } switchnum) { case 1: { total=total+good[0].price; good[0].amount=good[0].amount+1; break; } case 2: { total=total+good[1].price; good[1].amount=good[1].amount+1; break; } case 3: { total=total+good[2].price; good[2].amount=good[2].amount+1; break; } case 4: { total=total+good[3].price; good[3].amount=good[3].amount+1; break; } case 5: { total=total+good[4].price; good[4].amount=good[4].amount+1; break; } case 6: { total=total+good[5].price; good[5].amount=good[5].amount+1; break; } case 7: { total=total+good[6].price; good[6].amount=good[6].amount+1; break; } case 8: { total=total+good[7].price; good[7].amount=good[7].amount+1; break; } case 9: { total=total+good[8].price; good[8].amount=good[8].amount+1; break; } case 10: { total=total+good[9].price; good[9].amount=good[9].amount+1; break; } } } iftotal>sum) { printf"Insufficient money"); } else { change=sum-total; printf"Total:%dyuan,change:%dyuan\n",sum,change); fori=0;i<10;i++) { ifgood[i].amount!=0) { printf"%s:%d;",good[i].name,good[i].amount); } } } }