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

21 2112233445566778899 998877665544332211

Sample Output

Case 1:1 + 2 = 3Case 2:112233445566778899 + 998877665544332211 = 1111111111111111110我的答案是：#include<stdio.h>
int main()
{
 __int64 A,B;
 int T,i=1;
 while((scanf("%d",&T))!=EOF)
 {
  for(i=0;i<T;i++)
  {
   scanf("%I64d%I64d",&A,&B);
   printf("Case %d:\n",i+1);
   printf("%I64d + %I64d = %I64d\n\n",A,B,A+B);
  }
 }
 return 0;
} 还有就是：#include<stdio.h>
int main()
{
 int T,i,j;
 __int64 A[20],B[20];
 while((scanf("%d",&T))!=EOF)
 {
  for(i=0;i<T;i++)
   scanf("%I64d%I64d",&A[i],&B[i]);
  for(j=0;j<T;j++)
  {
   printf("Case %d:\n",j+1);
   printf("%I64d + %I64d = %I64d\n\n",A[j],B[j],A[j]+B[j]);
  }
 }
 return 0;
} 请大家多多指教，谢谢了

知道了
You may assume the length of each integer will not exceed 1000.的是意思是大数的位数要在1000位以内
题目理解错了   要用数组来做的

还是这个问题啊    要把每一位数字放入数组里的吧    应该怎么做呢？？
设A[1000]   如何使输入200位数字后判断输入结束呢？？？

哦  谢谢了

恩  知道了   应该用字符串才行的啊

为什么要写成str1[301],str2[301];这样呢？？？
为什么是301而不是1000？？？