A + B Problem II---hdu1002

 

A + B Problem II

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 260647    Accepted Submission(s): 50397

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

 

 

 

 1 #include<stdio.h>
 2 #include<string.h>
 3 #define as 1000
 4 int main()
 5 {
 6     int as1[as+20],as2[as+20],cot=0,j,t;
 7     char shuru1[as+20],shuru2[as+20];
 8     int n,i;
 9     scanf("%d",&n);
10     getchar();//吸收回车符
11     t=n;
12     while(n--)
13     {
14         cot++;
15         scanf("%s",shuru1);
16         scanf("%s",shuru2);
17         memset(as1,0,sizeof(as1));
18         memset(as2,0,sizeof(as2));
19         for(i=0,j=strlen(shuru1)-1;j>=0;j--,i++)
20         {
21             as1[i]=shuru1[j]-'0';
22             
23         }
24         for(i=0,j=strlen(shuru2)-1;j>=0;j--,i++)
25         {
26             as2[i]=shuru2[j]-'0';
27         }
28         for(i=0;i<as+20;i++)
29         {
30             as1[i]+=as2[i];
31             if(as1[i]>=10)//判断是否满十进一
32             {
33                 as1[i+1]++;
34                 as1[i]-=10;
35             }
36             
37         }
38         for(i=as+19;(i>=0)&&(as1[i]==0);i--);//去掉结果前面多余的0
39         printf("Case %d:\n",cot);
40         printf("%s + %s = ",shuru1,shuru2);
41          if(i>=0)
42          {
43              for(;i>=0;i--)
44              printf("%d",as1[i]);
45          }
46          else
47          printf("0");
48          printf("\n");
49          if(cot!=t)
50          printf("\n");//最后不要多换行
51     }
52     return 0;
53 }