HDU_oj_1002 A+B Problem Ⅱ

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

1 2

112233445566778899 998877665544332211

Sample Output

Case 1: 1 + 2 = 3

Case 2: 112233445566778899 + 998877665544332211 = 1111111111111111110

分析：

被加数与加数太大，因此无法用普通方式计算，这时应考虑用数组存储，模拟加法运算

注意点：

①数据长度不超过1000

②每个结果后面需多空出一行

改了n多次

 1 #include<iostream>
 2 #include<cstring>
 3 #define MAX 1000
 4 
 5 int main()
 6 {
 7     char a[MAX+1],b[MAX+1];
 8     int T;
 9     scanf("%d",&T);
10     for(int w=1;w<=T;++w)
11     {
12         memset(a,0,sizeof(a));
13         memset(b,0,sizeof(b));
14         
15         scanf("%s%s",a,b);
16         
17         printf("Case %d:
",w);
18         printf("%s + %s = ",a,b);
19         
20         int len1=strlen(a);
21         int len2=strlen(b);
22         
23         int fpla=MAX-(len1>len2 ? len1:len2);
24         
25         for(int i=len1-1,k=MAX;i>=0;--i,--k) 
26             a[k]=(a[i]-'0'),a[i]=0;
27              
28         for(int i=len2-1,k=MAX;i>=0;--i,--k) 
29             b[k]=(b[i]-'0'),b[i]=0; 
30             
31         for(int i=MAX;i>=fpla;--i)
32             a[i]+=b[i]; 
33             
34         for(int i=MAX;i>=fpla;--i)
35         {
36             int c=a[i]/10;
37             a[i-1]+=c;
38             a[i]%=10;
39         } 
40         for(int i=fpla+1;i<=MAX;++i)
41             printf("%d",a[i]);
42         puts("");
43         if(w<T)
44             puts("");
45     }
46     return 0;
47 }