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 思路：字符串输入，将两个字符串倒叙输入两个新整型数组（一开始一直在纠结如何倒叙相加）。然后相加进位。注意最后要判断进位是否为0 . 注意！！！一定要将两个新的整型数组初始化为0（这长度不一的两个数倒叙相加时不会出错），另外最好用一个新的数组来接受相加结果。。
#include <iostream>
#include <string.h>
#include <stdio.h>
using namespace std;
int main()
{
int n , ans = 0;
cin >> n ;
while(n --)
{
ans++ ;
char a[10000] = {0}, b[10000] ={0} ;
int c[10000] ={0}, d[10000] = {0} , e[10000] ={0};
scanf("%s%s" , &a , &b);
int len1 = strlen(a) ;
int len2 = strlen(b) ;
int len = max(len1 , len2);
int j = 0 ;
for(int i = len1 - 1 ; i >= 0 ; i--)
{
c[j++] = a[i] - 48 ;
}
j = 0 ;
for(int i = len2 - 1 ; i >= 0 ; i--)
{
d[j++] = b[i] - 48 ;
}
int k = 0 ;
for(int i = 0 ; i < len ; i++)
{
e[i] = (c[i] + d[i] + k) % 10 ;
k = (c[i] + d[i] + k) / 10 ;
}
cout << "Case " << ans << ':' << endl ;
printf("%s + %s = " , a , b);
if(!k)
{
for(int i = len - 1 ; i >= 0 ; i--)
cout << e[i] ;
cout << endl ;
}
else
{
e[len] = k ;
for(int i = len ; i >= 0 ; i--)
cout << e[i] ;
cout << endl ;
}
if(n != 0)
printf("\n");
}
return 0;
}