从键盘读入个100以内的正整数,进行乘法计算并输出。
- 输入
-
多组测试数据，每组测试数据包括两个整数m,n仅一行，以0 0 结尾。两个数的第一位和最后一位都不是0.
- 输出
-
输出格式: 两个乘数的末位数对齐 乘号后面紧跟着第二个乘数,线的长度以最长的数的位数为准,每两组测试数据中间，输出一个空行。
- 样例输入
-
89 13 2 3 0 0
- 样例输出
-
89 x13 ---- 267 89 ---- 1157 2 x3 -- 6
#include <iostream>
using namespace std;
int main()
{
int m, n;
cin>>m;
cin>>n;
while(m!=0 || n!=0)
{
if(m<10 && n< 10)
{
int k = m*n;
if(k>=10)
{
cout<<" "<<m<<endl;
cout<<"x"<<n<<endl;
cout<<"--"<<endl;
cout<<k<<endl;
cout<<endl;
}
else
{
cout<<" "<<m<<endl;
cout<<"x"<<n<<endl;
cout<<"--"<<endl;
cout<<" "<<k<<endl;
cout<<endl;
}
}
else if(m<10 && n>=10)
{
int k = m*n;
if(k<100)
{
cout<<" "<<m<<endl;
cout<<"x"<<n<<endl;
cout<<"---"<<endl;
int temp = n%10*m;
if(temp<10)
cout<<" "<<temp<<endl;
else
cout<<" "<<temp<<endl;
cout<<" "<<n/10*m<<endl;
cout<<"---"<<endl;
cout<<" "<<k<<endl;
cout<<endl;
}
else
{
cout<<" "<<m<<endl;
cout<<"x"<<n<<endl;
cout<<"---"<<endl;
int temp = n%10*m;
if(temp<10)
cout<<" "<<temp<<endl;
else
cout<<" "<<temp<<endl;
temp = n/10*m;
if(temp<10)
cout<<" "<<temp<<endl;
else
cout<<temp<<endl;
cout<<"---"<<endl;
cout<<k<<endl;
cout<<endl;
}
}
else if(m>=10 && n<10)
{
int k = m*n;
if(k<100)
{
cout<<m<<endl;
cout<<"x"<<n<<endl;
cout<<"--"<<endl;
cout<<k<<endl;
cout<<endl;
}
else
{
cout<<" "<<m<<endl;
cout<<" x"<<n<<endl;
cout<<"---"<<endl;
cout<<k<<endl;
cout<<endl;
}
}
else if(m>=10 && n>=10)
{
int k = m*n;
if(k<1000)
{
cout<<" "<<m<<endl;
cout<<"x"<<n<<endl;
cout<<"---"<<endl;
int temp = n%10*m;
if(temp<10)
cout<<" "<<temp<<endl;
else if(temp <100)
cout<<" "<<temp<<endl;
else
{
cout<<temp<<endl;
}
temp = n/10*m;
if(temp<10)
cout<<" "<<temp<<endl;
else
cout<<temp<<endl;
cout<<"---"<<endl;
cout<<k<<endl;
cout<<endl;
}
else
{
cout<<" "<<m<<endl;
cout<<" x"<<n<<endl;
cout<<"----"<<endl;
int temp = n%10*m;
if(temp<10)
cout<<" "<<temp<<endl;
else if(temp<100)
cout<<" "<<temp<<endl;
else
{
cout<<" "<<temp<<endl;
}
temp = n/10*m;
if(temp<10)
cout<<" "<<temp<<endl;
else if(temp<100)
cout<<" "<<temp<<endl;
else
cout<<temp<<endl;
cout<<"----"<<endl;
cout<<k<<endl;
cout<<endl;
}
}
cin>>m;
cin>>n;
}
return 0;
}