题目链接：
http://acm.hdu.edu.cn/showproblem.php?pid=1002
A + B Problem II
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 409136 Accepted Submission(s): 79277
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
1 2
112233445566778899 998877665544332211
Sample Output
Case 1:
1 + 2 = 3
1 + 2 = 3
Case 2:
112233445566778899 + 998877665544332211 = 1111111111111111110
思路：
采用字符数组储存两个加数，模拟小学的加法竖式计算
注意点：
1.俩个加数长度不等的时候，长度短的加数前面加0，0的个数为二者长度相减的绝对值
2.输出格式问题，只有输出最后一组数据的结果的时候，一个回车，其余都是两个回车
代码如下：
#include<bits/stdc++.h>
int main()
{
int n;
scanf("%d",&n);
int y=;
while(y<=n)
{
char a[]= {''},b[]= {''},C[],A[],B[];
getchar();
scanf("%s %s",a,b);
int l1=strlen(a);
int l2=strlen(b);
if(l1>l2)
{
int k=l1-l2;
char d[k];
for(int i=; i<k; i++)
d[i]='';
d[k]='\0';
B[]='\0';
strcat(B,d);
strcat(B,b);
A[]='\0';
strcat(A,a);
}
else if(l1<l2)
{
int k=l2-l1;
char d[k];
for(int i=; i<k; i++)
d[i]='';
d[k]='\0';
A[]='\0';
strcat(A,d);
strcat(A,a);
B[]='\0';
strcat(B,b);
}
else if(l2==l1)
{
A[]='\0';
B[]='\0';
strcat(A,a);
strcat(B,b);
}
l1=strlen(A);
l2=strlen(B);
int k=,cc=;
for(int i=l1-,j=l2-; i>=&&j>=; i--,j--)
{
int t=A[i]-''+B[j]-''+cc;
if(t>=)
{
cc=;
t=t-;
}
else
{
cc=;
}
C[k]=t+'';
k++;
}
if(cc==)
{
C[k]='';
C[k+]='\0';
}
else
{
C[k]='\0';
}
int l3=strlen(C);
printf("Case %d:\n",y);
printf("%s + %s = ",a,b);
for(int i=l3-; i>=; i--)
{
printf("%c",C[i]);
}
printf("\n");
if(y!=n)
printf("\n");
y++;
}
return ;
}