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
# include <iostream>
# include <cstring>
#include<algorithm>
using namespace std;
int main ()
{
int a;
cin>>a;
for(int b=0;b<a;b++)
{
char c [10000],d[10000],f[10000],maxArray[10000],g[10000],i[10000];
int c_long=0;
int d_long=0;
for(int e=0;e<10000;e++)
{
f[e]='0';
}
cin>>c>>d;
for(int e=0;e<strlen(c);e++)
{
g[e]=c[e];
c_long++;
}
for(int e=0;e<strlen(d);e++)
{
i[e]=d[e];
d_long++;
}
reverse(c,c+strlen(c));
reverse(d,d+strlen(d));
int min_long, max_long;
if(strlen(c)>strlen(d))
{
min_long=strlen(d);
max_long=strlen(c);
for(int e=0;e<strlen(c);e++)
{
maxArray[e]=c[e];
}
}
else
{
min_long=strlen(c);
max_long=strlen(d);
for(int e=0;e<strlen(d);e++)
{
maxArray[e]=d[e];
}
}
for(int e=strlen(c);e<10000;e++)
{
c[e]='0';
}
for(int e=strlen(d);e<10000;e++)
{
d[e]='0';
}
int temp=0;
int num=0;
for(int e=0;e<10000;e++)
{
temp = temp+(c[e]-'0')+(d[e]-'0');
f[e]=temp%10+'0';
temp=temp/10;
if(e>=max_long)
{
num++;
}
if(e>=max_long-1&&temp==0)
{
break;
}
}
cout<<"Case "<<b+1<<":"<<endl;
for(int e=0;e<c_long;e++)
{
cout<<g[e];
}
cout<<" + ";
for(int e=0;e<d_long;e++)
{
cout<<i[e];
}
cout<<" = ";
for(int e=max_long-1+num;e>=0;e--)
{
cout<<f[e];
}
cout<<endl;
if(b!=a-1)
cout<<endl;
}
}