PAT 1001

来源：互联网发布：米多大数据引擎系统编辑：程序博客网时间：2024/05/22 14:06

原题目如下：

我的代码如下，题目很简单，考虑各种情况周全点就可以了

//输入任意两个整数，输出为标准格式如231000+322000=553,000 #include <iostream>#include <cmath>using namespace std;class add{private:int a,b;public:add(int,int);void out(void);};add::add(int temp1,int temp2){a=temp1;b=temp2;}void add::out(){int c=a+b;int sum=abs(c);int million=sum/1000000;int thousand=(sum-1000000*million)/1000;int hundred=sum-1000000*million-1000*thousand;if(c<0)cout<<'-';if(million){cout<<million<<',';if(thousand){if(thousand<100&&thousand>9)cout<<'0'<<thousand<<',';else if(thousand<10)cout<<"00"<<thousand<<',';elsecout<<thousand<<',';}elsecout<<"000,";if(hundred){if(hundred<100&&hundred>9)cout<<'0'<<hundred;else if(hundred<10)cout<<"00"<<hundred;elsecout<<hundred;}elsecout<<"000";}else{if(thousand){cout<<thousand<<',';if(hundred){if(hundred<100&&hundred>9)cout<<'0'<<hundred;else if(hundred<10)cout<<"00"<<hundred;elsecout<<hundred;}}elsecout<<hundred;}}int main(){int temp1,temp2;cin>>temp1>>temp2;add *p=new add(temp1,temp2);p->out();delete p;return 0;}

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