A + B Problem II

Problem Description

 I have a very simple problem for you. Given two integers A and B, your job is to calculate the Sum of A + B.  

Input

 The first line of the input contains an integer T(1<=T<=20) which means the number of test cases. Then T lines follow, each line consists of two positive integers, A and B. Notice that the integers are very large, that means you should not process them by using 32-bit integer. You may assume the length of each integer will not exceed 1000.  

Output

 For each test case, you should output two lines. The first line is "Case #:", # means the number of the test case. The second line is the an equation "A + B = Sum", Sum means the result of A + B. Note there are some spaces int the equation. Output a blank line between two test cases.  

Sample Input

2 1 2 112233445566778899 998877665544332211

 Sample Output

Case 1: 1 + 2 = 3

Case 2: 112233445566778899 + 998877665544332211 = 1111111111111111110

 Author

 Ignatius.L

代码：c++

#include<iostream>

#include<string> using namespace std;

int main()

{  void sum(string &a,string &b,string &c);

 int n;  

cin>>n;//输入

 for(int i=1;i<=n;i++)  

{   string a,b,c;  

 cin>>a>>b;  

 sum(a,b,c);  

 if(i>1)   

{   

cout<<" Case "<<i<<":"<<endl;  

 cout<<a<<" + "<<b<<" = "<<c<<endl;  

 }   

else   {   

 cout<<"Case "<<i<<":"<<endl;   

cout<<a<<" + "<<b<<" = "<<c<<endl;         }

     }

return 0;

}

void sum(string &a,string &b,string &c)//采用引用方便结果存储

{  int la,lb;

 int sum=0;

 char d;

 la=a.length()-1;

 lb=b.length()-1;

 for(;la>=0&&lb>=0;la--,lb--)  

{    sum=sum+a[la]-'0'+b[lb]-'0';//从尾部加起 即低位相加   

d=sum%10+'0';   

c.insert(0,d);//采用头插法，把底位的和记录  

 sum=sum/10;//记录进位  }

 while(la>=0)//判断加完没有

 {      sum=sum+a[la--]-'0';   d=sum%10+'0';   sum=sum/10;   c.insert(0,d);       }

 while(lb>=0)//判断加完没有

 {      sum=sum+b[lb--]-'0';   d=sum%10+'0';   sum=sum/10;   c.insert(0,d);       }

}