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); }
}