C语言实现自动售货机
本文实例为大家分享了C语言实现自动售货机的具体代码,供大家参考,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 main(void) { 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 while((money!=-1)&&(money <= 10)) { sum = sum + money; scanf("%d",&money); } scanf("%d",&c[i]); while(c[i]!=-1)//将选的货物编号存储到数组c中并计算找零 { switch(c[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; } if(SUM>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; while(c[i]!=-1) { switch(c[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 if(flag==0) { printf("Total:%dyuan,change:%dyuan\n",sum, change); for(i=1; i<=10; i++) { if(b[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; while(1) { scanf("%d",&money); if(money==-1) { break; } sum=sum+money; } while(1) { scanf("%d",&num); if(num==-1) { break; } switch(num) { 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; } } } if(total>sum) { printf("Insufficient money"); } else { change=sum-total; printf("Total:%dyuan,change:%dyuan\n",sum,change); for(i=0;i<10;i++) { if(good[i].amount!=0) { printf("%s:%d;",good[i].name,good[i].amount); } } } }
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