A + B Problem II

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 98985 Accepted Submission(s): 18771

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 int main() 
 4 { 
 5     int n,i,l1,l2,max,a,j,t; 
 6     char s1[1001],s2[1001],s3[1001]; 
 7     scanf("%d",&n); 
 8     for(i = 0;i < n;i++) 
 9     { 
10         scanf("%s%s",s1,s2); 
11  
12         l1 = strlen(s1); 
13         l2 = strlen(s2); 
14         if(l1>l2) 
15             max = l1; 
16         else 
17             max = l2; 
18  
19         t=max; 
20         memset(s3,'\0',1001); 
21         memset(s3,'0',max+1); 
22         l1 = l1-1; 
23         l2 = l2-1; 
24         max = max; 
25         s3[0]='0'; 
26         while(l1>=0 && l2>=0) 
27         { 
28             a = s1[l1--]-'0'+s2[l2--]-'0'+s3[max]-'0';//问题可以直接简化为小学的算术题，从后往前算 
29             s3[max--] ='0'+a%10; 
30             if(a>=10) 
31                 s3[max]+=1;//大于十的时候记得向前加一，因为以上三个数永远不会大于等于20.。。。 
32              
33         } 
34         while(l1>=0) 
35         { 
36             a = s3[max]-'0'+s1[l1--]-'0'; 
37             s3[max--]='0'+a%10; 
38             if(a>=10) 
39                 s3[max]+=1; 
40         } 
41         while(l2>=0) 
42         { 
43             a = s3[max]-'0'+s2[l2--]-'0'; 
44             s3[max--]='0'+a%10; 
45             if(a>=10) 
46                 s3[max]+=1; 
47         } 
48  
49         printf("Case %d:\n",i+1); 
50         printf("%s + %s = ",s1,s2); 
51  
52         if(s3[0] > '0') 
53             printf("%c",s3[0]);//这一步很重要，假设是1和9想加，会上升一位 
54         for(j = 1;j < t; j++) 
55             printf( "%c", s3[j]); 
56         printf( "%c\n", s3[j]); 
57         if(i<n-1) 
58             printf("\n"); 
59     } 
60     return 0; 
61 }