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c++ pos機(jī)///小學(xué)期的作業(yè)

代碼片段和文件信息
#include?
#include
#include
using?namespace?std;
struct?Goods
{
int?num;
string?name;
double?price;
int?e_count;
int?count;
};
int?main()
{
Goods?g[20];
ifstream?file;
ofstream?FQ;
string?temp;
double?m;
int?t;
file.open(“清單.txt“);
for(int?i=0;i<20;i++)
if(!file.eof())
{
file>>t;
g[i].num=t;
file>>temp;
g[i].name=temp;
file>>m;
g[i].price=m;
g[i].count=0;
g[i].e_count=0;
}
file.close();
char?ch1ch2;
Goods?temp2;
double?sume_sumpay;
sum=e_sum=0;
ch1=ch2=‘y‘;
F.open(“明細(xì).txt“);
F<<“編號(hào)\t名稱\t單價(jià)\t數(shù)量\n“;
while(ch1==‘y‘)
{
while(ch2==‘y‘)
{
cout<<“請(qǐng)輸入商品的編號(hào)與數(shù)量:“< cin>>t>>m;
temp2.num=t;
temp2.e_count=m;
for(int?i=0;i<20;i++)
???????????????if(temp2.num==g[i].num)
???{
???temp2.name=g[i].name;
???g[i].count+=temp2.e_count;
???temp2.price=g[i].price;
???break;
???}
???e_sum+=temp2.price*temp2.e_count;
???cout<<“continue?y/n“< ???cin>>ch2; ???
???F< }
F<<“總計(jì):“< cout<<“總計(jì):“< sum+=e_sum;
cin>>pay;
cout<<“找零:“< cout<<“下一個(gè)顧客?y/n“< cin>>ch1;
e_sum=0;
ch2=‘y‘;
}
F.close();
Q.open(“每日.txt“);
Q<<“編號(hào)\t名稱\t單價(jià)\t數(shù)量\n“;
cout<<“名稱\t數(shù)量\t總計(jì)\n“;
for(int?j=0;j<20;j++)
{cout< ?????Q< }
Q<<“總營(yíng)業(yè)額為:“< Q.close();
return?0;
}
?屬性????????????大小?????日期????時(shí)間???名稱
-----------?---------??----------?-----??----
?????文件????????360??2011-07-03?17:44??每日.txt
?????文件????????127??2011-07-03?17:44??明細(xì).txt
?????文件???????1707??2011-07-03?17:43??text.cpp
?????文件????????380??2011-07-03?15:59??清單.txt
-----------?---------??----------?-----??----
?????????????????2574????????????????????4
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