#include"stdio.h"
int main()
{
int i,n;
long int a,b;
scanf("%d",&n);
for(i=1;i<n;i++)
{
scanf("%ld%ld",&a,&b);
printf("Case %d:\n",i);
printf("%d+%d=%d",a,b,a+b);
printf("\n");
}
scanf("%ld%ld",&a,&b);
printf("Case%d:\n",n);
printf("%d+%d=%d",a,b,a+b);
return 0;
}





Sample Input
2
1 2
112233445566778899 998877665544332211



Sample Output
Case 1:
1 + 2 = 3

Case 2:
112233445566778899 + 998877665544332211 = 1111111111111111110


提交失败

[ 本帖最后由 yenga 于 2013-11-26 14:03 编辑 ]

http://acm.hdu.




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

输出的空格问题我也修改过了，但还是提交不了

楼主输入的数据太大了,long int类型在有些编译器下只分配4个字节,在有符号的情况下范围只有(十进制)-2147483648~2147483547

You may assume the length of each integer will not exceed 1000.

还“提交失败”！
你提交之前不先用给出的例子进行一下测试？

我运行没错啊

题目说了，数据可能是1000位的整数

#include"stdio.h"
int main()
{
int i,n;
long int a,b;
scanf("%d",&n);
for(i=1;i<n;i++)
{
scanf("%ld%ld",&a,&b);
printf("Case %d:\n",i);
printf("%d+%d=%d",a,b,a+b);    应该为printf("%ld+%ld=%ld",a,b,a+b);   
printf("\n");
}
scanf("%ld%ld",&a,&b);
printf("Case%d:\n",n);
printf("%d+%d=%d",a,b,a+b);   应该为printf("%ld+%ld=%ld",a,b,a+b);
return 0;
}

不把他们当做数字，当做字符串输入，然后自己设计算法手动加法